SRI CHAITANYA IIT ACADEMY, INDIA 03‐01‐23_ Sr.

S60_NUCLEUS&STERLING_BT _ Jee‐Main_GTM‐01_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS&STERLING_BT JEE-MAIN Date: 03-01-2023
Time: 09.00Am to 12.00Pm GTM-01 Max. Marks: 300
KEY SHEET
PHYSICS
1) 3 2) 2 3) 4 4) 1 5) 2
6) 2 7) 2 8) 2 9) 4 10) 2
11) 3 12) 1 13) 3 14) 1 15) 2
16) 1 17) 1 18) 2 19) 3 20) 4
21) 50 22) 45 23) 2 24) 15 25) 175
26) 9 27) 22 28) 8 29) 78 30) 30

CHEMISTRY
31) 2 32) 3 33) 4 34) 4 35) 2
36) 3 37) 1 38) 4 39) 1 40) 2
41) 2 42) 4 43) 1 44) 1 45) 4
46) 1 47) 2 48) 4 49) 2 50) 3
51) 6 52) 2 53) 3 54) 20 55) 3
56) 4 57) 356 58) 4 59) 2 60) 5

MATHEMATICS
61) 2 62) 3 63) 3 64) 1 65) 3
66) 1 67) 3 68) 2 69) 4 70) 1
71) 1 72) 4 73) 4 74) 2 75) 2
76) 1 77) 1 78) 4 79) 2 80) 3
81) 1 82) 167 83) 84 84) 5 85) 7
86) 2 87) 17 88) 7 89) 27 90) 7

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 SRI CHAITANYA IIT ACADEMY, INDIA 03‐01‐23_ Sr.S60_NUCLEUS&STERLING_BT _ Jee‐Main_GTM‐01_KEY &SOL’S

SOLUTIONS
PHYSICS
1.   60 . Solid angle subtended by BCD is

  2 1  cos    
Solid angle subtended by ABDE is
 ABCDE    BCD  2    
q  q
Hence, flux through ABDE is    
0 4 4 0
2. 4T
The pressure due to surface tension 
R
2
The pressure due to electrostatic forces 
2 0
Just before the bubble bursts.
4T  2 8T
 or R  20
R 2 0 
3. hc
K  0 i

4hc
and K '   0  ii 
3
4hc hc
From Eq/s  i  &  ii  , we get  K ' K  
3 
hc
K ' K 
3
hc K 
But from Eq.  i   K  0  K ' K   0
 3 3
4K 0 4K
 K ' K   Or K ' 
3 3 3
4.  D 
y9  position of 9th bright fringe  9  
 d 
 1  D 3 D
y 2  position of 2nd dark fringe =  2   
 2 d 2 d

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 SRI CHAITANYA IIT ACADEMY, INDIA 03‐01‐23_ Sr.S60_NUCLEUS&STERLING_BT _ Jee‐Main_GTM‐01_KEY &SOL’S

D  3 3
y9  y 2  7.5 mm   9    7.5  10
d  2
 2   0.5  10 
3
    7.5  103     2 
 15   100  10 
 2
  75     5  1044   50  108 m
o
 5000A
 15 
5. Since there is no parallax, it means that both images (By plane mirror and convex
mirror) coinciding each other.

According to property of plane mirror it will form image at a distance of 30 cm

behind it. Hence for convex mirror u  50cm, v  10cm
1 1 1 1 1 1 4
By using      
f v u f 10 50 50
25
 f  cm  R  2f  25 cm
2
6. t/T
N  1  1/ 2
Here T1/2  20 minutes; we know  
N0  2 
t /20
N 80  1  1
For 20% decay    i
N 0 100  2 
t /20
N 20  1  2
For 80% decay    .... ii 
N 0 100  2 
Dividing  ii  by  i 
 t 2  t1 
1 1 20
  ; On solving we get t 2  t1  40 min.
4 2
7. 1 1
Energy, R  K  2  2  (K=constant)
 n1 n 2 
1 1 5
n1  2 and n 2  3. So E  K  2  2   K  
2 3   36 
For removing an electron n1  1 to n 2  
36
Energy E1  K 1  E  7.2E
5
 Ionization energy  7.2E
8. Clearly, i c  60
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 SRI CHAITANYA IIT ACADEMY, INDIA 03‐01‐23_ Sr.S60_NUCLEUS&STERLING_BT _ Jee‐Main_GTM‐01_KEY &SOL’S
So, maximum possible value of ic is 60.

1 g 1
Now, 1  g   Or 1   g sin ic
sin ic 1 sin i c
3
 1.5 sin60  1.5   1.5  0.8666  1.299  1.3
2
9.   XL  XC  ;
R 
2 2
Z
R  10, X L  L  2000  5  103  10
1 1
XC    10 i.e Z  10
C 2000  50  106
V 20
Maximum current i 0  0   2A
Z 10
2
Hence i rms   1.4A And Vrms  4  1.41  5.64V
2
10. To free electron from the metal surface minimum energy required is equal to the work
function of that metal. So Assertion A, is correct
h  W0   K.E max If h  W0  K.E max  0
Hence reason R, is correct, But R is not the correct explanation of A.
11. Rate of decay of current between t  5 ms to 6 ms
di  5 
    slope of the line BC     3 
 5  103 A / s
dt  1  10 
di
Hence induced emf    L  4.6   5  103   23  103 V
dt
12. b b
I  Iv  b 
Induced emf  Bvdx   0 vdx  induced emf, E  0 ln  
0 a
2x 2  a 
E2
 power dissipated 
R
2
E2 0 Iv  b   1
Also, power  F.V  F  F ln   
vR  2   a   vR
13. X  X C  X L and average power in ac circuit can be zero.
14.

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 SRI CHAITANYA IIT ACADEMY, INDIA 03‐01‐23_ Sr.S60_NUCLEUS&STERLING_BT _ Jee‐Main_GTM‐01_KEY &SOL’S
0 I 
Bx  2
 0 3  102  3  105 T
2 2  10 4
 I'
By  0 2
 4  105 T
2 2  10
Bnet  B2x  B2y  25  1010  5  105 T
15. In case of internal resistance measurement by potentiometer,
V1 1  ER1 /  R 1  r   R 1  R 2  r 
  
V2  2  ER 2 /  R 1  r   R 2  R1  r 
Here  1  2m,  2  3m, R 1  5, and R 2  10. So
2 5 10  r 
 or r  10
3 10  5  r 
16. Circuit can be redrawn as follows:

 5  25   10  2  35 35
Now, Vx  Vy , Ceq      F
 5  25   10  2  6 6
17. When energy on both is same, means energy on capacitor is half of its maximum
q 2 1 Q2 Q Q 1
energy.  q  Q cos t   cos t 
2C 2 2C 2 2 2
  
 t   t   LC
4 4 4
18. r1  A1 
1/3
3  27 
1/3
27 27
r A       
1/3
  A  125
r2  A 2  5  A 125 A
Number of nuclei in atom X  A  52  125  52  73 .
19. For lens u  30cm,f  20cm, hence by using
1 1 1 1 1 1
      v  60cm
f v u 20 v 30
The final image will coincide the object, if light ray falls normally on convex mirror
as shown. From figure, it could be seen clearly that separation between lens and
mirror is 60 – 10 = 50 cm.

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 SRI CHAITANYA IIT ACADEMY, INDIA 03‐01‐23_ Sr.S60_NUCLEUS&STERLING_BT _ Jee‐Main_GTM‐01_KEY &SOL’S
20. 5 20
R eq   i   5A
2 5
 1.5
2
Potential difference between X and P,

5
Vx  VP     3  7.5V i 
2
5
Vx  VQ   2  5V  ii 
2
On solving  i  &  ii  VP  VQ  2.5volt; VQ  VP .
 
21. Using dV   E.dr

V
 V  E. r cos   E
r cos 
  20  10  10 102
 E    200V / m
10  102 cos120 10  10   sin 30  1 / 2
22. 2
Number of atoms in 2 kg fuel  6.02  1026  5.12  1024
235
Fission rate = number of atoms fissioned in one second
5.12  1024
  1.975  1018 s 1
30  24  60  60
Each fission gives 185 MeV. Hence, energy obtained in one second,
E  185  1.975  1018 MeV s 1  142  1.975  1018  1.6  1019  106 J s 1  44.87MW
23.

 2  4  1012 4
Fnet  2 F31 cos   2     0.46N
40  0.5
2
5
24. Total range is doubled, i.e., 4Ig
Now shunt required is

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Ig 1 1 1
S  G  10   or x  15
4Ig  Ig 30 x 10
25. For wire if
M  mass,   density, A  Area of cross sec tion
V  volume,l  length, l  change in length
M Al
Then mass per unit length m    A
l l
T/A YlA
And Young’s modulus of elasticity Y   T .
l / l l
Hence lowest frequency of vibration
 l 
Y A
1 T 1  l   1 Yl  1 9  1010  4.9  104
n  n  35 Hz
2l m 2l A 2l l 2 1 1  9  103
26. The intensity of the wave is proportional to the area of the slit. Thus we can use the
intensities I1 & I 2 from the slits on screen are in ratio
I1 b1 b1 1
  
I 2 b 2 b 2 4
I1 a12 1 a 1
If a1 &a 2 are the amplitudes of the waves,  2  1 
I2 a 2 4 a2 2
The ratio of maximum to minimum intensity is given as
I max  a1  a 2  1  2
2 2
9
  2 
Imin  a1  a 2  1  2 1
27. n1  Frequency of the police car horn observer heard by motorcyclist
n 2  Frequency of the siren heard by the motorcyclist.
V = speed of motorcyclist.
330  V 330  V
n1   176 n 2   165  n1  n 2  0  V  22 m / s
330  22 330
28. V2
Acceleration a 
r
Z n2 Z3
Where V  & r   a  4  both are in ground state i.e n = 1
n Z n
3 3
a
3 He   ZHe   2 
So, a  Z    8
a H  ZH   1 
29. Probable frequency of A is 390 Hz and 378 Hz and after loading the beats are
decreasing from 6 to 4 so the original frequency of A will 390 Hz
30.  0i rB   4  10 2  300  106
B i   30A
r 0 4  107
30 A in opposite direction

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CHEMISTRY
31. Conceptual
32.  t1/2 1   a 2  n 1 ; 120   4  102  n 1 ;n  2
 t   a  240  8  102 
1/2 2 1

33. Assertion is false but Reason is true. The enthalpy of chemisorption is of the order
of 40  400kJmol1 while for physical adsorption it is of the order of
20  40 kJmol1 .
34. 3 2d 2  4.52 0
For bcc, d  a or a    5.219 A  522 pm
2 3 1.732
zM 2  39
   0.91g / cm3  910kg m 3
a  N A  10 30
 522    6.023  10   10
3 3 23 30

35. Cell reaction is, Zn  Cu 2  Zn 2  Cu

RT  Zn 
2

E cell  E 0
cell  n
nF  Cu 2 

 Zn 2
Greater the factor 
   , less is the EMF
 Cu
2
  
Hence E1  E 2
36. As colloidal particles move towards anode so these particles are negatively charged
and coagulated by cations of electrolyte.
According to Hardy Schulze rule,
Coagulation power  charge of ion
 Order of coagulation power is Al3  Ba 2  Na 
37. Oxide in which central atom has higher charge and more electronegativity is more
acidic, i.e.
N 2O5  N 2O 4  P2O5  As 2O3
38. 2Ag 2O  s  4Ag  s  O2  g 

2Pb3O4  s  6PbO  s  O2  g 

2PbO 2  s  2PbO  s  O 2  g 
39. Metal halides with higher oxidation state are more covalent than the one in lower
oxidation state.
40. XeF4 is planar
41. Conceptual
42. Conceptual
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43. Conceptual
44. Alkyl iodides are often prepared by the reaction of alkyl chlorides/bromides with
NaI in dry acetone. This reaction is known as Finkelstein reaction.
R  X  NaI  R  I  NaX
X  Cl,Br
NaCl or NaBr thus formed is precipitated in dry acetone.
It facilitates the forward reaction according to lechatelier’s principle. The synthesis
of alkyl fluorides is best accomplished by heating an alkyl chloride/ bromide in the
presence of a metallic fluoride such as AgF, Hg 2 F2 ,CoF2 orSbF3 . The reaction is
termed as Swarts reaction.
H 3C  Br  AgF  H 3C  F  AgBr
45. Conceptual
46. OH

3 2 1
H3C COOH
6 4
5 CH3
47. i) Gives Propanoic acid
ii) Methyl isocyanide on hydrolysis give methyl amine
48. For the preparation of Me3CNH 2 , the required alkylhalide is Me3CX which will
react with potassium phthalmide, a strong base, to form alkene rather than
substituted product. For preparing C6 H 5 NH 2 ,C6 H 5Cl will be starting halide in
which Cl is non-reactive.
49. Nylon-66 is an example of first synthetic fibres produced from the simple
molecules. It is prepared by condensation polymerization of adipic acid and
hexamethylene diamine.
50. Conceptual
51. Convulsion is caused by deficiency of vitamin B6
52. i) 5Fe 2  MnO 4  8H   Mn 2  4H 2O  5Fe3

iv) 5NO 2  2MnO 4  6H   2Mn 2  5NO3  3H 2O

53. Cell constant   / a
Unit  m / m2  m1
54. Presence of catalyst does not affect enthalpy change of reaction
H R  E f  E b  180  200  20kJ / mol

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55. 1 1
No.of M atoms   4   8 11 2
4 8
1 1
No.of X atoms   6   8  3 1 4
2 8
So, formula  M 2 X 4  MX 2 , a = 1 ; b = 2 ; a + b = 3.
56. Cu  NH3 3 Cl   Pt  NH3  Cl3 

Cu  NH3  Cl3   Pt  NH3 3 Cl 

CuCl4  Pt  NH3 4 

& Cu  NH3 4   PtCl4 

The isomers Cu  NH3 2 Cl2   Pt  NH3 2 Cl2  does not exist due to both parts being
neutral
57. From Raoult’s law
P  P w M 121.8  120.2 15 78
   
P m W 121.8 m 250
15  78 21.8
Or m    356.2
250 1.6
58. 2, 3, 5, 8
59. COCH3

C6H6
CH3COOH + PCl5 CH3COCl
Anh.AlCl3
(A)
Friedle Craft
reaction
OH OMgBr
H + MgBrC2H5
C2H5 C CH3 C2H5 C CH3
Ether
hydrolysis

(C)
60. Conceptual

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MATHEMATICS
61. x x x x

f  x   x   e f  x  t  dt  x   e
2 t 2  x  t 
f  t  dt  x   e .e f  t  dt  x  e
2 x t 2 x
 e f  t  dt
t

0 0 0 0
x

f 1  x   2x  e  x .e x f  x   e  x  e t f  t  dt  2x  f  x   f  x   x 2
0
3
x x3
f  x  x   c f  0  0 f  x   x 
2 2

3 3
f  3  9  9  18
62.
 tan  sin x  1 
1 2

c
2
tan 1  sin x  1 cos x
Let, I   dx
3  2sin x  1  sin 2 x  
Substituting, 1  sin x  t  cos dx  dt
tan 1  t   
2
tan 1  t  dt
tan 1  t  dt
I   c
2   t  1  2  t  1 1  t2
2
2

 tan  sin x  1 
1 2

I c
2
63.  
2x 3 ydy  1  y2 x 2 y2  y2  1 dx  0 
2y dy y2 1 1
  
1  y  2 2 dx 1  y 2 x x 3
y2
Put  u.
1  y2
2y dy du
Then 
 
2
1  y 2 dx dx
1
du u 1  dx
   3  Integrating factor  e x  eln x  x
dx x x
1
Solution is u.x   2 dx  c  x 2 y 2   cx  1 1  y 2
x
 
64. 2n  3 1
tn  .
n  n  1 3n
Therefore,
2n  3 1  2  n  1  1 1  2 1  1
tn  . n  .  
n  n  1 3  n  n  1  3  n n  n  1  3n
n

 2  n  1  n  1  2 1 1  1
      
n n  n  1  3  n n n  1 3n
n

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3 1  1 1 1 1 1
tn     . n  . n 1  .
 n n  1 3 n 3 n  1 3n
1 1
Sn   t n  1  . n
n 1 n  1 3
65. The equation of a straight line which is at a unit distance from the origin is
x cos   ysin   1  1
Differentiating w.r.t. x
dy
cos   sin   0   2
dx
On eliminating  from (1) and (2), we get
 dy 
cos ec  y  x   1
 dx 
 dy 
  y  x   cos ec   3
 dx 
dy
Also, slope    cot  {using (2)}
dx
2
 dy 
cos ec  1  cot   1      4 
2

 dx 
 xdy 
2
  dy  2 
From (3) and (4), we have  y    1    
 dx   dx 
66. Since x 2  y 2  4y  5  0
Centre is C1  0, 2 and Radius r1  4  5  3
Let C2  h,k  be the centre of the smaller circle and its radius r2 .
Since, C1C 2  4  h 2   k  2  3  r2  4  r2  1
2

But k  r2  1
Hence, using equation (1), we have
4  h 2  1  2  16  h 2  9
2
h   7
Since, h > 0
 
2
  y  1  2
2
Hence, required circle is x  7
67. a 2 x 2  ax  1 is positive for all real values of x.
1

 Area    a 2 x 2  ax  1 dx
0
2
a a 1

3 2

  1  2a 2  3a  6
6

1  2 3 a 18 
 2 a  a    6  
6  2 16  16 

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1  39 
2
3 3
  2 a     , which is minimum for a  
6  4 8 4
ba
68. 16
b

 2  2 a x  adx  4  b b  xdx
3 a ba

8 16  a  b 
 ab  a  b     ab  1
3 3  2 
69.
 
1 x2  1 
2
e x 2x  x 3 1
 e   2  2x
  2  3  x 2  
2
3  x2 3  x 2 

  dx  1
2
e x 2x  x 3
2
ex
Hence,  3  x  2

2 3  x2
c

70. The point (–2k, k + 1) is the interior point of the circle and parabola

So  2k    k  1  4  0
2 2

 4k 2  k 2  2k  1  4  0  5k 2  2k  3  0
 k  1  k 
3  3
  0  k   1,   1
5 5
Now,  k  1  4  2k   0
2

 k 2  2k  1  8k  0  k 2  10k  1  0

k  5  2 6, 5  2 6   2 
So from (1) & (2) k  1, 5  2 6  
71. Total formed numbers that begin with a odd digit  5C1. 8 P4  5  8 7  6 5
Total formed numbers that end with a odd digit =  5C1. 8 P4  5  8 7 6 5
Total formed number that begin with an odd digit and also end with an odd digit
 5C 2 .2!. 7 P3  5. 4 7  6 5
Thus total formed numbers that begin with an odd digit or end with an odd digits is
equal to
5.7.6.60
Total formed numbers  9 P5  9.8.76.5
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5
Thus, required probability 
6
72. If z1 be the new complex number then z1  z  2  2 2
z1 z1 i3 /2  3 3 
Also  e  z1  z.2  cos  isin   2 1  i 0  i  2i  2  2 1  i
z z  2 2
73. See figure, the given equation is written as
 3
 4 , when x  1
arg  z  1  i   
  , when x  1
 4

74. Distance CP = CQ = OC = 5 units

3
Slope of OC 
4
4 4
Slope of PQ   tan   
3 3
Q

O C

P
Co-ordinates of point P and Q are  4  5cos ,3  5sin  and  4  5cos ,3  5sin 
= (4 – 3, 3 + 4) and (4 + 3, 3 – 4)
(1, 7) and (7, –1)
75. Let, t  211x


2 11x 3

 
2
 211x.22  211x .2  1
22
t3
 4t  2t 2  1
4
 t 3  8t 2  16t  4  0
Cubic it t has roots t1 , t 2 , t 3
i.e. t1t 2 t 3  4  211x1.211x 2 .211x3  4
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11 x1  x 2  x 3 
2  22
2
 11 x1  x 2  x 3   2  x1  x 2  x 3 
11
76. Let, a = number of red balls.
b = number of blue balls,
p1 = probability of drawing two red balls
a
C a  a  1
 ab 2 
C2  a  b a  b  1
C2 b  b  1 b
p 2  probability of drawing two blue balls  
ab
C2  a  b a  b  1
p3  probability of drawing one red and one blue ball
a
C1. b C1 2ab
 ab 
C2  a  b a  b  1
Given that p1  5p 2 & p3  6p 2
 a  a  1  5b  b  1 and 2ab = 6b (b – 1)
 a  6,b  3  Total number of balls = 9
77. 
P c
A  P A   A  B

 c
  P A  B c

 A B c 
 P  A  B
c
   1  P  A  B
4 3
P  A   P  A  B 7  28 13
   
1  P  A  B 25 25
28
78. Let n consecutive odd integers are
2m + 1, 2m + 3, 2m + 5, ……, 2m + 2n – 1
Given that,
452  212   2m  1   2m  3   2m  5  .......   2m  2n  1
 2mn  1  3  5.....   2n  1   2mn  n 2  m 2  2mn  n 2  m 2
 452  212   m  n   m 2
2

 m  n  45&m  21
 n  24&m  21
Hence, the numbers are
43, 45, ……, 89
79. We have
 1  1 1 1  1 1  1  1  1 1 1 1 1  1  1  1 1 1  1 1
S , , , , , , , , ,
 1 1  1 1  1  1 1  1 1 1 1  1  1  1  1 1 1  1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 
, , , , , 
1 1 1 1 1 1 1 1 1 1 1 1 
 n  S  16
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Now, number of favorable cases are given by (   0 , that is either   2or  2 )
 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
E , , , , , , , 
 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
n  E 8 1
Hence probability that the system of equations has a unique solution   
n  S 16 2
80. Number of ways  9C5 - no.of ways in which she selects 5 from only of two
categories.
 9C5  3. 6C5  126  18  108
Alternatively: Number of ways 1 1 3 or 2 2 1 i.e.,
3. 3C1. 3C1. 3C3  3. 3C2 . 3C2 . 3C1  27  81  108 .
81. Let x  u 6 ,dx  6u 5du
dx 6u 5du u3
 x  3 x  u 3  u 2  u  1 du
  6

 1 
 6  u 2  u  1   du
 u  1
 2u 3  3u 2  6u  6ln  u  1  e
 2 x 3  x   6 x   6ln 
3 6 6

x 1  e
 a  2,b  3,c  6,d  6  a  b  c  d  1  a  b  c  d  1
82. x 1
x16 x 6
 f  t  dt   t .f  t  dt    a  1
2

0 x
8 3
1
1 1 11
For x = 1,  f  t  dt  0    a  a
0
8 3 24
Diff. both sides of (1) w. r. t. x we get;
f  x   0  x 2f  x   2x15  2x 5
1 1
x15  x 5 11 11
 2
1  x 2
dx 
24

 a  2 x13  x11  x 9  x 7  x 5 dx 
24
a 
0 0

 1 1 1 1 1  11 167
 2       aa
 14 12 10 8 6  24 840
83. Any point on hyperbola xy = 1 (t, 1/t)
Now image of (t, 1/t) by the mirror y = 2x is
x  t y 1/ t  2t  1 / t 
  2  
2 1  5 
4 3
 5x   3t and 5y  4t 
t t
Now eliminating t from above two equations we get,
12x 2  12y 2  7xy  25  0
 bc   7 12  84
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84.  
r  4n
 
1 .1
lim  
n      n
2

 r/n 3 r 4 
r 1

  n  

4
1
 dx
 
2
0 x 3 x 4
Put, 3 x  4  t
3 1 2
dx  dt  dx  dt
2 x x 3
When x = 0 then t = 4
When x = 4 then t = 10
10 10
2 1 2  1 2  1 1 1
  2 dt          
34t 3 t4 3  10 4  10
85. 3
4 p
Req.Area    x 2  4x  3 dx     Pq7
3 q
1
86. 200  C0  C1  ......  n Cn  400
n n

 200  2n  400
n 8

   
8 r r
Tr 1  8C r 4
x 3 a 4 x5
 Tr 1  8Cr a r x 2r 6
For this term to be independent of x,
2r  6  0  r  3
T4  8C3a 3  448  56a 3  a 3  8
a2
87. 1  x  n   r0 Cr x r
n

 x 1  x    r 0 Cr x r 1
n n

Differentiating w.r.t x we get

xn 1  x   1  x    r 0  r  1 Cr x r
n 1 n n

Again multiplying both sides by x

1  x  n 1  nx  1  x  x   r 0  r  1 Cr x r 1
n

Again differentiating w.r.t x we get,

d
dx
 
1  x  n 1 nx 2  x 2  x   r0  r  1 2 Cr x r
n
  
 
 r0  r  1 Cr x r  1  x   2nx  2x  1  nx 2  x 2  x  n  11  x 
n 2 n 1 n2
 
Putting x = 1 on both sides, we get,
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  r  1 C  2
 2n  3   n  2 . n  1 2n 2  2n 2  n 2  5n  4
n 2 n 1
r 0 r

Now, f  x   x 2  5x  4   x  1 x  4    1,   4

Hence,  2   2  17
88. Given the sum S  1!   2!   3!  ........   2023!
2 2 2 2

Since the digit at units place is zero in n! for n  5

Hence, S  1   2   6   24  numbers having zero at units place = 617 + all
2 2 2 2

other numbers having zero at units place.

89. 2nd win in 4th test  In the first 3 matches there must be exactly 1 win so, required
1 2
 1  3  1
probability  C1      
3
 4  4  4
27
P
256
90. x 2 y2
Given hyperbola is 3x 2  2y 2  6 or  1
2 3
Slope from of tangent is y  mx  a 2 m 2  b 2 or  y  mx   a 2 m 2  b 2
2

Tangent from the point  ,  is given by,

  m  2  2m 2  3
 
 m2  2  2  2m  2  3  0
 3 2
m1m 2   2  tan  tan 
2  2
 2  3  2 2  4  7  2 2   2

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS & ALL_BT JEE-MAIN Date: 05-01-2023
Time: 09.00Am to 12.00Pm GTM-02 Max. Marks: 300
KEY SHEET
PHYSICS
1) 2 2) 2 3) 1 4) 1 5) 2
6) 4 7) 2 8) 3 9) 3 10) 1
11) 1 12) 1 13) 1 14) 3 15) 4
16) 4 17) 1 18) 3 19) 1 20) 3
21) 6 22) 4 23) 2 24) 4 25) 2
26) 5 27) 4 28) 9 29) 1 30) 1

CHEMISTRY
31) 4 32) 1 33) 2 34) 1 35) 2
36) 3 37) 4 38) 1 39) 1 40) 3
41) 2 42) 3 43) 4 44) 2 45) 2
46) 3 47) 3 48) 3 49) 1 50) 1
51) 5 52) 1 53) 2 54) 8 55) 1
56) 5 57) 9 58) 3 59) 7 60) 5

MATHEMATICS
61) 1 62) 4 63) 2 64) 1 65) 4
66) 1 67) 2 68) 1 69) 3 70) 2
71) 2 72) 2 73) 2 74) 3 75) 4
76) 2 77) 2 78) 3 79) 3 80) 2
81) 3 82) 5 83) 15 84) 2 85) 5
86) 1 87) 13 88) 7 89) 7 90) 31

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SOLUTIONS
PHYSICS
1. During Collision Kinetic Energy Is Not Conserved.
2. F 2  Agx
a  net    2 x
m m
2
T

3. Work done by the friction on block is   216J
Therefore the work done by the motor is  2  216 J  432 J
4. Conceptual
5. In cae (i) work done by friction is zero, while in case (ii) it is non-zero as it will
forward slip (may be for some time) on BC.
6. du 
F 
dr r2
3 b2
Eccenricity e  1  
a2 5
so, SO = 3m, SA= 2m, SC = 8m

UA    20 J ,U c  5 J
2
rAmV A  rc mVc 2V A  8Vc
1 2 1
mv A  20  mvc2  5
2 2
Solving we get V A  4m / s
7.  x2

W   2  e kT
 x2
As exponents are dimensionless, so, should be dimensionless.
kT
 kT  ML2T 2
     2   2  MT 2
x  L
From the dimensional homogeneity,  2  should have dimension of work.
2 2
2 2 2  ML T
 2 
    ML T        M 0 LT 0
    MT  2
8. vu 1
Using equation, a  and S  ut  at 2
t 2
1  45 675
Distance travelled by car in 15 sec  152  m
2 15 2
Distance travelled by scooter in 15 seconds  30  15  450
 distance  speed  time 
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 SRI CHAITANYA IIT ACADEMY, INDIA 05‐01‐23_ Sr.Super60_NUCLEUS & ALL_BT_ Jee‐Main_GTM‐02_KEY &SOL’S
Difference between distance travelled by car and scooter in 15 sec, 450-337.5 = 112.5m
Let car catches scooter in time t;

675
 45  t  15   30t  337.5  45t  675  30t  15t  337.5
2
 t  22.5sec
9.  a a
From figure, r G  iˆ  kˆ
2 2
   a a  a a  a ˆ ˆ
a
2
a
r H  ˆj  kˆ  r H  r G   ˆj  kˆ    iˆ  kˆ  
2 2 2  2 2  2
j i  
10.
Net acceleration
dv
dt

 a   g   v2 
Let time t required to rise to its zenith (v = 0) so,
0 t
dv
 g   v2   dt for H max,v  0 
v0 0

1   
t  tan  
 g 0 
v
g  
11.

dy
y  4Cx 2   tan   8Cx
dx
At P, tan   8Ca
For steady circular motion
g tan 
m 2a cos  mg sin    
a
g  8aC
   2 2 gC
a
12. Ax
Given : Gravitational field, EG  ,V  0
2 2 3/2
(x  a )
Vx x   x
Ax
 dV    E G .d x  Vx  V   
2 2 3/2
dx
V   (x  a )
A A
Vx  0
( x 2  a 2 )1/2 ( x 2  a 2 )1/2

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 SRI CHAITANYA IIT ACADEMY, INDIA 05‐01‐23_ Sr.Super60_NUCLEUS & ALL_BT_ Jee‐Main_GTM‐02_KEY &SOL’S
13. Let V f is the final speed of the body. From questions,
1 1 V
mV f2  mV02  V f  0  5m / s
2 8 2
 dV 
 (102 )
dV
F  m    kV
2
 kV 2
 dt  dt
5 10
1 1
or , K  104 kgm 1
dV
 V 2  100 K  dt   100 K 10 
5 10
10 0
14.  du d  A B
At equilibrium, F  0  0    0
10
dr dr  r r5 
A 5B 1 10 A 5B 
  10  0    0
11 6 6 5 1
r r r  r 
1
10 A 5 10 A  2 A 5
 5
5 B  r   r  
r 5B  B 
15. Using Bernoulli’s equation
1 1
P1  12   gh1  P2  22   gh2
2 2
For horizontal pipe, h1  0 and h2  0 and taking
P 1 P 1
P1  P, P2  ,we get  P   2   V 2
2 2 2 2

p 1 2 1 P
    V 2  V   2 
2 2 2 
16. The thermal resistance is given by
x 4x x 2 x 3x
   
KA 2 KA KA KA KA
Amount of heat flow per second,
dQ T (T2  T1) KA 1  A(T2  T1) K  1
     f 
dt 3x 3x 3 x  3
KA
17. Equation of the BC is given by two point straight line equation
P  P0 3P0  P0 2P
   P  P0  0 (V  2V0 )
V  2V0 V0  2V0 V0
using PV = nRT
2 p0V 2
P0V   4 P0V
V0
Temperature, T 
1 R
( n  1mole given)

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 SRI CHAITANYA IIT ACADEMY, INDIA 05‐01‐23_ Sr.Super60_NUCLEUS & ALL_BT_ Jee‐Main_GTM‐02_KEY &SOL’S

P0  2V   dT 4V 5
T  5V  , 05  0  V  V0
R  V0  dV V0 4
P  5V 2 25  25 P0V0
T  0 5  0   V02  
R 4 V0 16  8 R
18. From, ideal gas equation PV  nRT
 25  2
 400  103  500  106  n   (300)  n 
 3  25
Let n1 & n2 be no. of moles of hydrogen & oxygen respectively
Also, n1  n2  n
m m1 m2 2
Using n   
M 2 32 25
m 16
and m1  m2  0.76 gm 2 
m1 3
19. Total energy,
2
1 E '  2a 
E  m 2a 2 i.e., E  a 2 ,   E '  4 E.
2 E  a 
20. Equate torques about centre of mass
21. Surface area of bubble of radius r  4 r 2 .Surface are of bubble of radius
2
2r  4  2r   16 r 2 . Therefore, increases in surface area  16 r 2  4 r 2  12 r 2
.since a bubble has two surface, the total increase in surface are  24 r 2
Energy spent = work done  24 r 2
22. 2mv2  3mv,2  6   3v, v  4m / s
23. MI of plat form  200kg m 2
MI of man  mR 2  200kgm 2
For system (plat form + man) by using COAM
50 1 2 1
Ip0  mvR 0   rad / s
200 2
V
Angular velocity of man w.r.t platform   0
R
1 1 2 rad
  1rad / sec   2 s
2 2 1rad / sec
Time taken by the man to complete one revolution
24. m1  1kg , m2  2kg , u1  3ms 1, u2  2ms 1
Initial momentum  (m 1u1)2  (m2u2 )2  9  16  5kg ms 1
5
If combined velocity is v(m1  m2 )v  5, v  ms 1
3

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1 1 1 13
Heat liberated = loss in kinetic energy  m1 u12 m2 u22  (m1  m2 )v 2  J
2 2 2 3
25. Diameter  2.4  2  LC  (0.4  2  LC )  2cm
26.  2
L final  0, so mR 20  mv0 R
5
27. Let the velocity acquired by A and B be V, then
v 1 1 1 1
mv  mV  mV  V  , Also mv 2  mV 2  mV 2  kx 2
2 2 2 2 2
Where x is the maximum compression of the spring. On solving the above equations,
1/2
m
we get x  v  
 2k 
At maximum compression, kinetic energy of the
2
1 2 1 2 2 mv
A  B system  mV  mV  mV 
2 2 4
28. Taking point A as origin and the axes as shown in for path A to B, we have R=200m.
H=540 – 50=490 m
2H
For horizontal projection, we have R  u
g

g 9.8
uR  200   20m / s N  9
2H 2  490
29. Let a =edge of the block  3cm x  height of the block above the water surface in
floating condition.
From F.B.D., we have
 a3 g   wa 2 (a  x ) g ...(1)
 a3 g  W  kx   wa3 g ...(2)
  
From eqn. (1), we get x  a 1    3 1  0.8   0.6 cm
  w 
From eqn. (2). we get
W  kx    w    a 3 g  50  0.006  (1000  800)  (3  102 )3  10
 0.354  (20) Newtons
30. Fl
l  =1
Ay

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CHEMISTRY
31. Eq. H 2O2  eq. KMnO4
50  N  10  0.2
0.2 1 N 1
N  ;M  
5 25 2 50
S 1
M  S   34  0.68 g / L
M .wt. 50
32. Pb PV
At high P,Z Z  1  
RT RT
PV vs P, slope =b, intercept =RT
Vander waals Constant “a” values are negligible & neglected (for H 2 & He ) in
comparison with a values of other gases.
33. Kp  Kp  1
log  log RT  0     ( RT )
Kc  KC 
  ng  1
34. Before mixing,
 Ba 2   4  103 M   SO 2   6  104 M
   4 
after mixing,
4  103  100  M 2  400
 Br 2   103 M
 
6  104  300  M 2  400
 SO 2   4.5  104 M
 4 
IP   Ba 2   SO42   4.5  107 ; Ksp  I .P.
  
35. E1 Z12 n22
 
E2 Z 22 n12
36. More the magnitude of negative charge, more the size
37. With non-metals like Cl2 , Br2 , I 2 , P4 , S8 NaOH gives disproportionated products.
38. BeSO4 to BaSO4 : Solubility decreases
BeCO3 to BaCO3 : Thermal stability increases
BeF2  MgF2  CaF2  SrF2 : Solubility in water.
39. ( Be; A)( Li; Mg )( B, Si) Diagonally related, in properties. Pairs
40. Conceptual

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41.

42. 2-ethyl-3-methylpentanal
43. a) lone pair of electrons is not involved in resonance
b) +I  basic nature
c) Cyclic compounds are more basic then acylic camp
44. 
R  C  O  octet filled structure. where as R  C  O has “C” with 6e .
45. CH3

Ph  C  CH 2 D
OH
46.
It is E2 elimination and antiperiplanarity is required

Hence X is

47. NaNH 2 is a stronger base than KOH . Intermediate vinyl Bromide needs stronger
Base NaNH 2 for better yield.
48.

49. Phenol (OH) group is strongly activating and electron releasing group, so it prefers
elctrophilic substitution.

50. In graph  Point B is

Non aromatic,  Complex, Arenonium ion intermediate.
Nitroso is an o, p director.
51. no.of equivalents of = no. of equivalents of
acid base
2 1 100
  n5
600 6 1000
n
52. Al  OH 3 acts as acid and accepts 1 NaOH
 n  f 1

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53. (uav ) A TA M B 300 16
    2
(uav ) B M A TB 600 2
54. AB  A  B
at eqm1-1/3 1/3 1/3; total moles =4/3
2/3 1 1/ 3 1
PAB  P  P; PA  PB  P P
4/3 2 4/3 4
P / 4 P / 4 P
KP  
P/2 8
55. [S ]
pH  5  pKa  log
[ A]
pKa  pKb  14  pKa  14  10  4
[S ] [ A ]
5  4  log   10
[ A] [ H A ]
56. A  O2  H1  10 Kcal
B  O2  H 2  15Kcal
3 A  xB  105
 310   x 15   105  x  5
57. The given reaction is
6 Na  Al2O3  3Na2O  2 Al
So G of this reaction
3(377)  (1582)  451kJ
We know that G  2.303 RT log k
log k  79.04  k  9  1080
58. Configuration ( Z  33)  1s 2 2 s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p3
Each subshell has one orbital with m = 0
 No.of electrons  2  7  1  15
59. XeF2 , SCl4 , PCl5 , SF4 , ClF3 , ICI 2 , XeOF2
60. H 2 , O2 , O2 , C2 , N 2

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MATHEMATICS
61. 1 x
Replacing x by in the equation
1 x

 f  x   f  11  xx   x3 ....(i)
2

 
2 3
  1 x    1 x 
We have  f   f  x    ...(ii)
  1 x    1 x 
Solving (i) and (ii), we have
2
 x3   f  x    1  x 3
3 3
 1  x  2  1 x 
  f  x        f  x  x  
  f  x  
2
 1 x  x 6
 1 x   1 x 
 
 1  4
f  2   4    
 3  3

 f  2    43   43    43  2  23
 
62. f  x    x  b   2c 2  b 2  min f  x   2c 2  b 2
2

Also g  x    x 2  2cx  b 2  b 2  c 2   x  c   max g  x   b2  c 2

2

As min f  x   max g  x  we get 2c 2  b 2  b 2  c 2

 c 2  2b 2  c  2 b
63. Taking eiA , common from R , e from R2 and eiC from R3 , we get
iB

e 
i A B  C 
1
where
e 
i A C 
e 
i A  B 
eiA
1  e 
i B  C 
e 
i A  B 
eiB
e 
i B  C 
e 
i A C 
eiC

But A+ B+ C =  , so that e    ei

i A B  C

 cos   i sin   1. Also,

A+C =   B  ei ( A C )  e  i eiB  eiB
Thus,
eiA eiB eiC 1 1 1
iA iB iC i  A B  C 
1  e e e e 1 1 1
eiA eiB eiC 1 1 1

Using C1  C1  C2 , we get
0 1 1
1   1 0 1 1   1 2  2   4
2 1 1
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Therefore ,    1 1  4
64.
 
1/3
f ( x  T )  1  1  3 f ( x)  3( f ( x)) 2  ( f ( x))3


 1  1  f ( x)  
3 1/3

f  x  T    f ( x)  2 …(i)
 f ( x  T )  f ( x)  2 …(ii)
 f ( x  2T )  f ( x  T )  2
 from (ii)-(i) we get
f  x  2T   f ( x)  0
 f ( x  2T )  f ( x)
 Period of f ( x) is 2T
Assertion (A) & Reason (R) both are true & Reason (R) is correct explanation of
Assertion (A).
65.  1 1 2 1 1
Let   1  1    2  1 C1  C1  C2  C3 
1 1  2 1 
1 1 1 1 0 0
    2 1  1  1  1 0
1 1  1 0  1
2
 using C2  C2  C1and C3  C2  C1      2    1
If   0, then   2or   1
But when  =1, the system of equation becomes x1  x2  x3  1 which has infinite
number of solutions. When   2 , by adding three equations, we obtain 0  3 and
thus, the system of equations is inconsistent.
66. 6 A  A  cA  dI
1 2

 6 I  A  cA  dA 3 2

1 0 0  1 0 0  1 0 0
A  0 1 1  0
2
1 1   0 1 5 
 
0 2 4 0 2 4 0 10 14 
and A3  A2 A
1 0 0  1 0  1 0
0 0
 0 1 5  0
 1   0 11 19 
1  

0 10 14  0 2 4  0 38 46 

Now,
6I  A3  cA2  dA
6  1  c  d , 0  19  5c  d
6  11  c  d
6  46  14c  4 d , 0  38  10c  2d
 d  11, c  6

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67.    1
1 n  n  1 1 n 1 
  sin 1 
n
  sin  
1 n(n  1) 1
n 1
 n n 1 n 
  1 1 1 1
  sin 1  1  1 
1  n n 1 n 1 n
n

1 1 1 
 Lt  sin  sin 1
n  1 n n  1 
 1  
 Lt   sin 1  (by telescopic)
n  2 n  1  2
68. We have sin 2 x  sin 4 x  sin 6 x  ...
sin 2
  tan 2 x
1  sin x
2


Therefore,  exp  sin 2 x  sin 4 x  sin 6 x  ...upto  ln 2 
 
2
 exp tan 2 x ln 2  exp ln 2tan  2 tan x
2
x

As  satisfies the equation x  17 x  16  0 we get

2

2
Since 0  x   / 2, tan 2 x  0    2 tan x  1. Therefore,
2
2tan x  16  24  tan 2 x  4  tan x  2  tan x  0
2cos x 2 2 1
Thus,   
sin x  2cos x tan x  2 2  2 2

69. 1
    x  is discontinuous at x  1 ,
x 1
1 1
y  f    is discontinuous at 
1
  2    2    1
1
 2,   1. If   2 then 2 
x 1
1
 x  1 / 2. If   1 then 1   x  2. Hence the composite function is
x 1
discontinuous only at x  1,1 / 2, 2.
70. x
Since x  sin x, for x  0 and lim 1
x 0  sin x
11x 11x
so  11 as x  0  but  11. Thus
sin x sin x
 11x 
 sin x   11 for value x  0  . Similarly
21sin x 21sin x  21sin x 
 21 as x  0  but 20   21   20. Hence
x x  x 
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  11x   21sin x  
lim      x    31
x 0    sin x   
 11x 
Similarly x  sin x for x  0 , so   11
 sin x 
 sin x 
as x  0  and  21  20 as x  0  . Thus
 x 
lim
  11x   21sin x  
x 0       31
  sin x   x  
71. 1 1
y  tan 1  tan 1  ...  upto n terms
2 2
1 x  x x  3x  3
 tan 1
 x  1  x  tan 1  x  2    x  1  ...n terms
1  x  x  1 1   x  1 x  2 
 tan 1  x  1  tan 1 x  tan 1  x  2   tan 1  x  1  ... 
tan 1  x  n   tan 1  x   n  1 
 tan 1  x  n   tan 1 x.
1 1
y'  x   
2
1   x  n 1  x2
1 n2
 y 0 
'
1   .
1  n2 1  n2
72. Conceptual
73. 1 1 dy dy y
 0   .
2 x 2 y dx dx x
y
Equation of tangent at any point  x, y  of the curve is Y  y    X  x  .So
x
intercepts of X  axis and Y  axis are x  xy and y  xy .
Therefore, the sum of intercepts  x  y  2 xy  ( x  y ) 2  ( a ) 2  a
74. f  a   f 1 c
 f ' c 
a 1 a2  c2
a2  a2  a2  1 c

a 1 a2  c2


a 1 c a 1 c2
   
a 1 2
a c 2 a  1 a2  c2

 
  a  1 a 2  c 2  c 2  a  1

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 c 2   a  1  1  a     a  1 a 2

 c 2  2a    a  1 a 2  c2 
 a  1 a
2

Since c  1, a  so c 
 a  1 a
2
75. y  2x  1
Distance of any point  x, y  from y  2 x  1 is: . If  x, y  is on
5
x 4  3x 2  1
y  x 4  3 x 2  2 x then this distance is S 
5
dS 4 x3  6 x ds
   0  x  0.
dx 5 dx
Als , S '  x   0 for x  0 and S '  x   0 for x  0. Thus S is minimum when x  0, and
1
min. S is .
5
76. For Rolle’s theorem & L.M.V theorem, the function f ( x) must be continuous &
differentiable in the interval (a, b) if Rolle’s theorem is applicable for f ( x) then
L.M.V. theorem is applicable for f ( x)
   
 f ( x )  sin x in  , . is non differentiable at x  0
 3 3 
 Rolle’s theorem & L.M.V theorem can not be applicable
   
for f ( x)  sin x x   ,  .
 3 3
77. x y X Y
Equation of the line L in the two coordinate system is   1,   1
a b p q
Where  X , Y  are the new coordinates of a point  x, y  when the axes are rotated
through a fixed angle, keeping the origin fixed. As the length of the perpendicular
from the origin has not changed
1 1 1 1 1 1
    
1 1 1 1 2 2 2 2
  a b p q
a 2 b2 p2 q2
1 1 1 1
or     0.
2 2 2 2
a p b q
78. We are given l 2  m2  n 2  0 and l  m  n  0 also we have l 2  m2  n 2  1
1
So that 2n 2  1  n   and l  m  n
2
2
  l  m   n2  l 2  m2  2lm  0

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1
 Either l  0 or m  0 if l  0, m  n  0  m   n  
2
1 1
So the direction cosines of one of the lines are, 0,  , and if
2 2
1
m  0, l  n  0  l  n   and the direction cosines of the other line are
2
1 1
 ,0,  .
2 2

Hence the required angle is
3
79. Volume of the parallelepiped is
1 a 1
V  a   0 1 a  a3  a  1
a 0 1
1
V '  a   3a 2  1  0 if a  
3
1
V "  a   6 a  0 if a 
3
1
Thus V  a  is minimum when a  .
3
80.
 
2
A  sin 2   1  sin 2 

 
 1  sin 2  sin 2   1

 1  sin 2  cos 2   1
1 1 3 3
Also A  1    sin 2 2  1       . Hence  A  1
4 4 4 4
81.   x2 x4    x 2 x 4   x 2 x3  
  1    ...   1 1    ...   1  x   ..  
 2! 4!    2! 4!   2! 3!  
      
2
 1 x 2  2 x
3 
 x    ...   x  x  ... 
 2 4!  3! 
 
 1 x 2  x2 
 x3    ...  1  x  ... 
 2 4!  3! 
 
 1 
(cos x  1)(cos x  e x )   terms containig x 
  2
x3  
 and higher powers of x 
82. Any point on the first line is  2r1  1,3r1  1,4r1  1 and on the second line is

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 r2  3,2r2  k , r2 
The lines will intersect when
2 r1  1  r2  3,3 r1  1  2 r2  k , 4 r1  1  r2
 2r1  r2  2 , 4r1  r2  1
3
 r1   , r2  5 .
2
9 9
and k  3 r1  2 r2  1    10  1  .
2 2
83. The given expression is equal to
  
1  tan 2 tan 1 2  1  cot 2 cot 1 3 
 1  4  1  9  15.
84. The given equation can be written as
sin 4 x  cos 4 y  2  4 sin x cos y  0

     2sin 2 x  2cos2 y  4sin x cos y  0

2 2
 sin 2 x  1  cos 2 y  1

  sin 2 x  1   cos 2 y  1  2  sin x  cos y   0

2 2 2

which is true if sin 2 x  1,cos 2 y  1and sin x  cos y,

so sin x  cos y  2 as 0  x, y   / 2 .
85. The given equation can be written as 1  2sin 2 x  a sin x  2a  7
 2sin 2 x  a sin x  2a  8  0
a  a 2  8  2a  8 
a   a  8
 sin x  
4 4
a4 a4
 sin x  which is possible if 1   1 or 2  a  6.
2 2
so the required values of a are 2,3,4,5,6 and hence the required number is 5.
86.
 
Given equation is possible if cos  x  4  1 and cos  x  1  
Since x  4  0  x  4 and x  0
So x  4 is the only solution.
87. Any point on the given line is  3r  2,4r  1,12r  2  which lies on the given plane is
3r  2   4r  1  12r  2  5  r  0
So the point of intersection of the line and the plane is (2, -1, 2) and the required
distance is
 2  12   1  52   2  10 2  13
88. Let r be the length of the line segment which makes angles  ,  ,  respectively with
x, y and z  axes, then r cos   2, r cos   3, r cos   6

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 
 r 2 cos 2   cos 2   cos 2   2 2  32  6 2  49

 r  7 as cos 2   cos 2   cos 2   1

89. Squaring and adding the given equations of the lines we get x 2  y 2  a 2  b 2 as the
locus of the point of intersection of these lines.
Since (3,4) lies on the locus, we get
9  16  a 2  b 2 i.e a 2  b 2  25 (i)
Also, (a, b) lies on 3x  4 y  0
so 3a  4b  0  b  (3 / 4)a (ii)
From (i) , a 2  (9 / 16)a 2  25  a 2  16
2
So that a  b  (7 / 4)2 a 2  49 a  b  7
90. We have
4  2  3  3  26 3 2  4  3  p

42  32 42  32
  25  6  12  p
  25  6  12  p
 p   25  6
 p  31 or  19.

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 SRI CHAITANYA IIT ACADEMY, INDIA 07‐01‐23_ Sr.S60_NUCLEUS&ALL_BT _ Jee‐Main_GTM‐03_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 07-01-2023
Time: 09.00Am to 12.00Pm GTM-03 Max. Marks: 300
KEY SHEET
PHYSICS
1) 2 2) 4 3) 3 4) 3 5) 3
6) 3 7) 1 8) 4 9) 2 10) 2
11) 1 12) 2 13) 4 14) 1 15) 3
16) 2 17) 1 18) 2 19) 1 20) 1
21) 815 22) 50 23) 25 24) 46 25) 3
26) 8 27) 2 28) 3 29) 12 30) 8

CHEMISTRY
31) 2 32) 1 33) 4 34) 1 35) 3
36) 1 37) 1 38) 1 39) 1 40) 2
41) 2 42) 1 43) 2 44) 1 45) 3
46) 4 47) 1 48) 1 49) 1 50) 4
51) 4 52) 222 53) 7 54) 3 55) 56
56) 18 57) 1 58) 7 59) 8 60) 3

MATHEMATICS
61) 4 62) 4 63) 1 64) 1 65) 1
66) 2 67) 3 68) 3 69) 1 70) 2
71) 1 72) 3 73) 3 74) 3 75) 4
76) 3 77) 3 78) 1 79) 3 80) 2
81) 11 82) 15 83) 2 84) 43 85) 14
86) 112 87) 35 88) 5 89) 38 90) 15

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SOLUTIONS
PHYSICS
1.  Pr 4  8Vl
We know that coefficient of viscosity,    P
8Vl  r4
Let P1, P2 be the pressures at the ends of the two tubes and V1 be the rate of volume
 8V1l
flow through them, then pressure difference across the first tube, P1 
 r4
 8V1l
Pressure difference across the second tube, P2 
  r / 2
4

The two tubes are connected in series. So, the total pressure, P  P1  P2
 8Vl  8V1l  8V1l Q
 4
 4
 4
 16  Q  V1  16V1  17V1 
r r r 17
So, the correct answer is (B).
2. V  V1  V2
1
 Pr 4  Pr14  Pr24
  or r 4  r14  r24 or r   r14  r24  4
8l 8l 8 l
3. For Si  0.7V , Ge  0.3V (barrier potential)
net emf E  VSi  VGe 12  0.7  0.3 11
i   3
  103  2.2mA
R R 5  10 5
4. T2 40 3 5 5
 1   1    T1  T2  T1   300  500 K
T1 100 5 3 3
New efficiency   60%
T2 60 2
 1   1  
T1 100 5
5
T1   300  750 K T  750  500  250 K
2
5. T T
1  1  2  1 
T1 800
300
2  1  1   2  T  489.9 K
T
6. y   A  B  .B  A.B  B.B  A.B
7. Given, T2  40 C  277 K
T1  300 C  303K
Q2  600 cal per second
Coefficient of performance,
T 277 277
 2  
T1  T2 303  277 26

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Q2
Also,  
W
 Work to be done per second = power required
Q 26
W 2  600 cal per second
 277
26
  600  4.2 J per second  236.5W
277
8. ‘A’ is false and ‘R’ is false
9. Q
Given work, W 
4
Where, Q  input heat.
W Q/4 1
Efficiency,     
Q Q 4
T
Also,   1  2
T1
Where, T2  temperature of sink
And T1  temperature of source.
T 1
1  2   1
T1 4
 When temperature of sink is reduced by 50 K,
100
efficiency becomes %
3
100 1 1
  
3 100 3
T  50 1
So, 1  2 
T1 3
T 50 1
1 2     2
T1 T1 3
Subtracting Eq. (2) from Eq. (1), we get
50 1 1 1
  
T1 3 4 12
T1  600 K
3
T2   600  450 K
4
10. Comparing the given equation with
  x t   
B y  B0 sin  2     
   T   
2
We get,   m  1.26cm
0.5  103

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11. b  b 2  4ac
2
a  b  c  0    ; For single frequency b 2  4 ac
2a
12. 2
For critical damping b  4 mK  b  4mK  2 Km
13. K
  5 rad s 1
m
2 2
1  b 
2   12  1
      5    4 rad s
 2m   2 2 
 Frequency decreases by 20%.
14. 1 1 1 1 1 1 1 1 1
  ,   ;  
u0 v0 f 0 ue ve f e 2 v0  2.1
1 1 10
   v0  42 cm
v0 2 21
1 1 1 1 1 1
  , ve  25cm;  
f e ve ue 5 25 ue
1 1 1 25
  ; ue  cm
5 25 ue 6
L  v0  ue  42  4.16; L  46.16 cm
15. The total energy falling on the surface is
   
U  18W / cm2  20cm2   30  60   6.48  105 J
Therefore, the total momentum delivered (for complete absorption) is
U 6.48  105 J
p  8
 2.16  103 kgm / s
c 3  10 m / s
The average force exerted on the surface is
p 2.16  103
F  4
 1.2  106 N
t 0.18  10
16. In case of diffraction at single slit, the position of minima is given by d sin   n .
Where d is the aperture size and for small  : sin      y / D 
 y D
 d    n , i.e, y   n 
D d
D D
So that, y3  y1   3      2  and hence,
d d

d

0.50  2  6  10 7

 2  104 m  0.2 mm
3
3  10
17. Both ‘A’ and ‘R’ are true and ‘R’ is the correct explanation of ‘A’
18. I I I
I  0 cos 2   0 cos 2 450  0
2 2 4

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I 1 I
   0.25; %  25%
I0 4 I0
19. m 80
V   10cc
 8
M 10
I  6
 106 A / m
V 10  10
20. A
We know that A f 
1  A
Here, A f  10 and A  100
100
 10  or 10  1000  100
1  100
100  10 90 9
    0.09
1000 1000 100
21. Let the speed of the flow be v and the diameter of the tap be d  1.25 cm . The
volume of the water flowing out per second is
Q  v  d 2 / 4
v  4Q / d 2
We then estimate the Reynolds number to be
Re  4  Q /  d

 4  103 kg m3  Q / 3.14  1.25  102 m  103 Pas 
 1.019  108 m 3sQ
Since initially (a)
Q  0.48 L / min  8 cm3 / s  8  106 m3s 1, e obtain,
Re  815
22. T 300
  1  2  100  1   100
T1 600
600  300 300
  100   100
600 600
1
  500    50%
2
23. (a) Signal Amplitude, B  15
B 15
Carrier amplitude, A  60 m    0.25
A 60
 Percentage modulation  0.25  100  25%
24. d m  2  64  105  32  2  64  105  50m
 64  102  10  8  103  10m
 144  102  10m  45.5 km

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25.  bt
For damped oscillator A  A0e .
A0
 A0e6b 6b
Here 27 That is e  1 / 27 .
1/3
1/3  1 
A2  A0e 2b  A0 e 6b   A0    A0 / 3
 27  .
26.  f0
m  ,
 fe
f0  
m  4, m  i ,  4  0i ;i  8
fe 0 2
27. E 60
B0  0  8
 2  107 T
C 3  10
28. Intensity of emerging light is I  3W / m
2

I0
cos 2  cos 2  90     3
2
32 2
cos  sin 2   3
2 ,
 
4 4cos 2  sin 2   3  sin 2 2 
3
4
3
 sin 2   sin 600 ;  300
2
29. In a common – base arrangement, the current amplification factor.
 I  I
  C   C;
 I E VCB I E
Given   0.88, I E  1mA .
 Collector current I C   I E  0.88  1  0.88mA
Now since I E  I B  I C
 base current I B  I E  I C  1  0.88  0.12 mA .
30. 0 Ni H  4  103  12
B  0 H  ;i    8A
 N 60  100

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CHEMISTRY
31. Electrolysis of molten NaH
NaH  Na  H 
1
At Anode H   H 2  e
2

32. By warm aqueous Barium hydroxide by using Ni electrode.

33. Potassium ion present within the cell fluid.
34. Suspension of slaked lime in water is called milk of lime
35. Co  Hb is 300 times more stable then O2  Hb

36. Liquid CO2 solvent for drycleaning due to less harm to ground water
37. Photo chemical smog is called oxidizing smog
38. Conceptual
39. Both statement I and statement II are true
40. Bakelite – electrices switches
Glyptol – paints & Lacquers
Urea – formdehyde – Lamineted sheets
HDPE – Buckets, Dustbins
41. Sucralose, aspartame are artificial sweetners
SO2  is antioxidant

42. Cetyl trimethyl ammonium bromide is cationic detergent

43. In sucrose - C1 of   D  glucose C2 of   D  fructose
Maltose - C1 of   D glucose C4 of   D glucose
Lactose - C1 of   D  galctose C4 of   D  glucose
Amylose - C1 , C4 of glycosidic linkage
44. Vitamin C – scury
Vitamin D – Rickets
Vitamin K – increases blood clotting time
45. A and R is true R is the correct explanation of A, Antiseptic property is due to free
iodine
46. They are chemically Inert

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47. A and R is true R is the correct explanation of A
48. Colour of ions due to f-electrons.
49. log
x 1 1
 .log p  log k : slope  intercept  log k
m n n
50. CFSE of octahedral complexes is greater for 5d series metal when compared to 3d & 4d
series.
51. 8.256 103  4 significant figures.
52. E  W  KE
hc
 W  KE


6.63 1034  3 108

9
 4.411019  KE
300 10

KE  2.22 1019  222 1021

53. PN 2  K H  X N 2

0.987
X N2  3
 0.0129 103  1.29 105
76.48 10
nN 2 nN 2
X N2   1.29 105   nH 2  71.595 104
nH 2O 55.5

54. Pepsin – proteins in to peptides

ZSM=5 convert alcohol to gasoline by dehydration
55. 1.4  N  V 1.4  2 10
% of N 2    56
wt of 0.c 0.5
56. 12  wt of CO2 100 12  0.20  100 240
% of C     18
44  wt of O.C 44  0.30 13.2
57. CH 3

CH 3 OH
CH 3

58. 4
 M nO  OH   NH 3
At cathode M nO2  NH 4  e 
3

x  4 y  3

59. Gd 2   Xe 4 f 7 5d 1 6 s 0

60. Buna – N, Buna – S,- Bakelite

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MATHEMATICS
61. Given expression is    p  q    q
   p  q   q ~ p  q  q ~ p  t
Make truth table
p q p  q ~  p  q ~  p  q  q pq p   p  q
T T T F T T T
T F F T T T T
F T F T T T T
F F F T T F T
p   p  q  is also a tautology.
62. 50 n
 X i   Yi  T ;  n  X i   10, n Yi   5
i 1 i 1
50 n 500 5n
So,  X i  500,  Yi  5n    n  30
i 1 i 1 20 6
63. Given a set 1, 2,.....50
Possible choices of P are
2,3,5,7,11,13,17,23,29,31,37,41,43 and 47.
So, we can calculate no. of elements in R1 as
 2,2  ,  2,2 .. 2,2 
0 1 5

3,3  ,......3,3 
0 3

5,5  ,......5,5 
0 2

 7,7  ,....... 7,7 

0 2

11,11  ,........11,11 
0 1

Every number of P n should lie in the given set

1,2,.....50 .
Total number of relations are
n  R1   6  4  3  3   2  10   36
Similarly for relation R2 ,
 2,2  ,  2,2 ...... 47,47  ,  47,47 
0 1 0 1

Total number of relations are

n  R2   2  14  28
Required difference
n  R1   n  R2   36  28  8

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64.  n2 1 
 
Req  2n  1  2 2  for n  5
 
 
 
65. Let probability of success is p and q  1  p .
Here, n  33
3P  x  0   P  x  1
3.33 C0  q  33 C1 pq32
33

1 11 q
p  , q  ,  11 ……………… (i)
12 12 p
P  x  15  P  x  16 

P  x  18  P  x  17 
33 3
C15 p15q18 33
C16 p16 q17  q   q 
33
    
C18 p18q15 33
C17 p17 q16  p   p 
 11  11
3
{From (i)}  1320
66. Let p  Mean & q  Variance  p  q 
So, p  q  24, pq  128  p  16 & q  8
1 1
 np1  16; np1q1  8  q1   p1  , n  32
2 2
1

p  x  n  3  n n Cn  2  n Cn 1  n Cn
2

32  31
 k 32 C30 32 C31 32 C32   32  1  496  33  529
2
67. 
Given mean   , variance  and
3
4
P  x  1 
243
X  nP  
  1 2
 2  npq    q   q   P  1  q 
3 3 3 3
4
Here, P  X  1 
243
1 n 1
 2 1 4
n
C1     
 3 3 243
 2  1  4
n   n 1   n6
 3  3  243
4 2 5 1
 2 1 2 1
So, P  X  4 or X  5   C4      6 C5    
6
 3 3  3 3

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4
 2 1 16
    9 
 3 3 27
68. Given
Mean x  np
Variance  x2  npq
np  npq  24 ………… (i)
np.npq  128 …… (ii)
Solving (i) and (ii)
1 1
We get p  , q  , n  32
2 2
1 1
Now, get p  , q  , n  32
2 2
33
Now, P  X  1  P  X  2  32 C1 pq31 32 C2 p 2 q30 
228
69. Take conditional probability
 1  x   P 1  x  4  x  2  P 1  x  2  P  x  2 
P   
 x2  P  x  2 P  x  2 P  x  2
4k 4
From distribution table  
k  2k  4k 7
70. Since X has a binomial distribution, B  n, p 
n2
 P  X  2  n C2  p  1  p 
2

n 3
And P  X  3 n C3  p  1  p 
3

Given P  X  2   P  X  3
n2 n 3
C2 p 2 1  p  n C3  p  1  p 
3
 n

p 2 1  p  p 3 1  p 
n n
n! n!
 .  .
2! n  2 ! 1  p 2 3! n  3! 1  p 3
1 1 p
  .  31  p   p  n  2 
n  2 3 1 p
 3  3 p  np  2 p  np  3  p
71. Definition of reflexive relation.
72. B

C 2h
h
 2
H P x A
d  7h

h 2h
tan 2  and tan 2 
x x  7h

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2
tan  
1  tan   
2
7
2 tan 
On solving we get  tan   7  2
73. Definition of mathematical induction.

74. 150
P
150
h

600
A B C Q

 PB bisects APC, therefore

AB : BC  PA : PC
h
Also in APQ, sin 300   PA  2h
PA
h 2h
And in CPQ,sin 600   PC 
PC 3
2h
 AB : BC  2h :  3 :1
3
75. T
y
0 0
30 60 R
P 40m Q x
In right QTR
y
tan 600   y  3 x ……………… (i)
x
In right PTR
y x  40
tan 300  y …………….. (ii)
x  40 3
x  40
From (i) and (ii), 3x 
3  x  20m
76. Given  p  r    p   ~ q     ~ p  taking r  q
p q ~p ~q pq p ~ q  p  r    p   ~ q 
T T F F T F F
T F F T F T F
F T T F F F T
F F T T F F T
 p  r    p   ~ q    ~ p 
77. pqq
 ~  p  q  q
  ~ p ~ q   q
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 ~ p  q  ~ q  q
  ~ p  q   t ~ p  q
From options we can say option c is correct
 p  q  ~ p  q ~ p ~ q ~ p  q  t
Hence C  
78. Given conditional statement is
 
 p  q    ~ p  r   ~ p   r    p  q  
 Here,  is equal to  ~ A  B  .
From given statement,
  ~ p ~ q    ~ p  r    p  q 
 ~ p   r ~ q   p    ~ r  q 
If negation of p and only p is present with union, then it represents tautology.
79. Given boolen expression are
~ p  q  p  q and ~ q  p ~  p  q 
Negation of ~  p  q    ~  p  q   i.e., ~  p  q 
80. Negation of given statement ~  p   ~ p  q  
~ p  ~  ~ p  q  ~ p   p  ~ q 
  ~ p  q    ~ p ~ q 
 F   ~ p  ~ p  ~ pK 7 ~ p
81. 2k  12
Median k 6
2
 Mean deviation about median
k  3  k  1  k  1  6  k  k  6  k  9  k  15  k  18
 6
8
60  2k  7 
  6   k  6 
8  2 
 k  6, Median  6  6  12
82. x 24
x  i 34   xi  24
5 5
2
xi2  24  194
  2
    xi2  154
5  5  25
x1  x2  x3  x4  14  x5  10
2 x12  x22  x32  x42 49
   a
4 4
x12  x22  x32  x42  4a  49
x52  154  4a  49

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5
 100  105  4a  4a  5  a 
4
4a  x5  10  15
83. Let two remaining observations are x1, x2 .
2  4  10  12  14  x1  x2
So, x   8 (given)
7
 x1  x2  14 ………….. (i)
2
2 xi2  xi 
Now,      16 (given)
N  N 
4  16  100  144  196  x12  x22
  64  16
7
 460  x12  x22  16  64   7
 x12  x22  100 ……….. (ii)
  x  y 2  x2  y 2  2 xy  xy  48 …………. (iii)
  x  y 2   x  y 2  4 xy  196  192  4
 x y 2 x y 2
84. Number of observations are = 50
Mean  x   15
Standard deviation    2
Let incorrect observation is x1 & correct observation is  y1 
Given x1  y1  70
x  x  ....  x50
 x 1 2  15 (given)
50
 x1  x2  .......x50  750 ………… (i)
Now
Mean of correct observation is 16
y1  x2  ......  x50
 16
50
y1  x2  x3  .....x50  16  50 …………….. (ii)
Eq. (i) – eq. (i)
 y1  x1  16  50  15  50
y1  x1  50 & x1  y1  70
y1  60
x1  10
x12  x22  .......  x50
2
 4  152 ………. (iii)
50
y 2  x22  .....x50
2
 2  1  162 …….. (iv)
50
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From (iii)

 4
10 
2

x22 32 ....  x50
2
 225
50 50

 4  2  225 
x 2
2  x32  .....  x50
2

50

 227 
 x22  x32 2
 .......  x50 
50
From (iv)
 60   x22  x32  ......  x50
2 2 
  16 
2 2
   
50  50 
60  60
2   227  256
50
 2  72  227  256
 2  43

85. Since 0  y  x  2 y
x x
 y  x y
2 2
 x  y  y  x  2x  y
yx
Hence median   10
2
 x  y  20
And range   2 x  y    x  y   x  2 y …………………. (i)
But range = 28
 x  2 y  28 …………………. (ii)
From equations (i) and (ii),
x  12, y  8

 Mean 
 x  y   y  x  2x  y   4x  y
4 4
y 8
 x   12   14
4 4
86. Given sets are A  1,2,3,4,5,6,7 and B  3,6,7,9
Total subset of A  27  128
Here, C  B   when set C contains the element 1,2,4,5 .
Therefore, S  C  A; C  B  
Total number of elements = Total   C  B   
 128  2 4  112

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87. x1  x2  .......  x25
Let;  x  40
25
 x1  x2  .......  x25  1000
 x2  x2  .........  x25  60  A  39  25
 1000  60  A  975  A  975  940  35
88. 1
P E 
2
8 r r 8
8
8 1 1 1 8
8 1 1
  Cr        Cr   
r n 2 2 2 r n 2 2
8

1 8
 8 8
  Cn  Cn 1  .......  Cn 
2
1
2

8 8 8
 Cn  Cn 1  ......  C8  128
 256   8

C0 8 C1  ...... 8 Cn 1  128
8
 C0 8 C1  ..... 8 Cn 1  128
 n 1  4  n  5
89. Given, n  C   73, n T   6n  C  T   x
 65  n  C  T   65  73  100
 65  x  38  x  36
90. Number of equivalence equals to numbers of partitions of set A.

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 SRI CHAITANYA IIT ACADEMY, INDIA 09‐01‐23_ Sr.S60_NUCLEUS & STERLING_ Jee‐Main_GTM‐04_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS&STERLING_BT JEE-MAIN Date: 09-01-2023
Time: 09.00Am to 12.00Pm GTM-04 Max. Marks: 300
KEY SHEET
PHYSICS
1) 1 2) 2 3) 2 4) 1 5) 2
6) 3 7) 1 8) 4 9) 4 10) 1
11) 4 12) 3 13) 2 14) 4 15) 2
16) 4 17) 1 18) 1 19) 4 20) 3
21) 3 22) 2 23) 6 24) 28 25) 3
26) 30 27) 72 28) 1 29) 9 30) 6

CHEMISTRY
31) 3 32) 3 33) 2 34) 1 35) 1
36) 3 37) 4 38) 2 39) 4 40) 4
41) 2 42) 1 43) 4 44) 1 45) 3
46) 1 47) 4 48) 2 49) 1 50) 3
51) 5 52) 2 53) 1 54) 10 55) 23
56) 2 57) 1 58) 500 59) 2 60) 10

MATHEMATICS
61) 3 62) 2 63) 4 64) 3 65) 3
66) 2 67) 3 68) 3 69) 1 70) 1
71) 3 72) 4 73) 3 74) 1 75) 2
76) 2 77) 4 78) 2 79) 3 80) 2
81) 6 82) 8 83) 2 84) 5 85) 9
86) 4 87) 1 88) 2 89) 3 90) 7

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 SRI CHAITANYA IIT ACADEMY, INDIA 09‐01‐23_ Sr.S60_NUCLEUS & STERLING_ Jee‐Main_GTM‐04_KEY &SOL’S

SOLUTIONS
PHYSICS
1. As  increases saturation current also increases
2. Let x be the depth of point P from surface
x
App. depth of point P from surface 

x  2h
App. depth of image of P from surface 

x  2h x 2h
So, separation between two   
  
3.
Velocity of point ‘A’ VA  V 2  2R 2  v 2 Normal acceleration of point A,

2R V2
a A  n   2R cos 45  R cos 45  a cos 45 , a A n  

2 2R
 radius of curvature of trajectory of point ‘A’ relative to the ground is

 
2
2
r
 VA 

V 2
 2 2R
a A n  V2
2R
4. : From ohm’s law electric field  current density
5. A  b  2  T
6. Conceptual
7. Conceptual
8.
SOL:

V 2  2 a 2  x 2 
 V12  2  a 2  y  ......................1
2
 1

 V22  2  a 2  y  ..................... 2 
2
 2
From 1 and  2  , weget
2 2
y1  y 2
T  2
2 2
v 2  v1
9. Conceptual
10. Conceptual
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 SRI CHAITANYA IIT ACADEMY, INDIA 09‐01‐23_ Sr.S60_NUCLEUS & STERLING_ Jee‐Main_GTM‐04_KEY &SOL’S
11. Conceptual
12. tan 
tan 1 
cos 
tan  tan 
and tan 2  

cos 900    sin 

tan 
 cos   .................1
tan 1
tan 
and sin   ................. 2 
tan 2
Dividing  2  by 1 , we have
tan 1
tan  
tan 2
13. Conceptual
14. Ic  100  0.04mA  4mA
8
Vc  20  12  8V  R c   2000
4  103
15. Conceptual
16. Conceptual
17. The density of lead is 1.13  104 kg / m3 , so we should expect our calculated value to
be close to this value. The density of water is 1.00  103 kg / m3 , so we see that lead is
about 11 times denser than water, which agrees with our experience that lead sinks.
Density is defined as   m / V . We must convert to SI units in the calculation.
3
 23.94g   1kg  100cm 
   
 2.10cm3  1000g  1m 
 23.94g   1kg   1000 000cm3 
       1.14  104 kg / m3
 2.10cm   1000g  
3
1m 3 
   
18. I. Induced emf in the rod   BIv
:
  t / RC BIv  t / RC
Current in the circuit I  e  e
R R
Since the net force on the rod should be zero, the external force will be equal in
magnitude but opposite to the magnetic force.
B2 I 2 V  t / RC
 F  IIB  e
R
19. Conceptual
20. SOL: (i), (ii) are true but (iii) is false, as we know that viscosity in gaseous is about
100 times less than viscosity in liquids.
21. V2 V3
R .V  ;RV3
gv0 gV0
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 SRI CHAITANYA IIT ACADEMY, INDIA 09‐01‐23_ Sr.S60_NUCLEUS & STERLING_ Jee‐Main_GTM‐04_KEY &SOL’S
22. C1 V2 2
Capacitors are in series therefore  
C2 V1 3
23. Conceptual
24. 2
0  I0  mR  f  2mg
5
&f  ma  maR
10 10 10 2
f  mg  mg  N  mg    5mg  ,    
7 7 7 7

25. SOL: Required kinetic energy =  m  N   m  n   m  H   m  C    931MeV

 2.99MeV
26. SOL: When the extension is maximum, their velocities are equal. From the law of
conservation of Momentum,
Pf  pi   6     3   6  2   3  1
  1ms 1
This energy is also conserved
1 2 1 2 1 1 2 1 2
E f  Ei   6 1   31  Kx 2m   6  2    31
2 2 2 2 2
1
3  1.5   200  x 2m  12  1.5
2
100x 2m  9  x 2m  0.09  x m  0.3m  30m
27. A  2 c
28. E1 31.25  10
E2  
R1  R AB 50
29. If initial velocities are: u1  2gh    ,u 2  2gh   
Then final velocities: v1  3 2gh , v 2 2gh
By using conservation of momentum and equation of e & m <<M.
30. I I
  10log 3  log  I  109 W / m2
I0  12
10
2 2
Now, I 
 P0  V  BAK  V B2A 2 BA 2 42f 2
  
2B 2B 2V 2V
IV I 0
A   5.55 A
2 2
B2 f v22f 2

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CHEMISTRY
31. Refer NCERT- P -block Group – 15, pg-179,180 and solutions, pg- 49
32. Refer f - block
33. Conceptual
34. Conceptual
35. NCERT-Hydrogen
36. C5H10

(trans) (cis)
, +

, ,
37. A is CH 2  CHCl , B is HC  CH
HC  CH has active hydrogen so CH 4 is liberated
38. Cl Cl
NO 2
Cl

NO 2 NO 2
A= B= C=
OH
OH
NO 2 NO 2
NO 2

NO 2 NO 2
D= E=
39. Solution :
N 2 Cl

 CH3CH 2OH  CH3CHO  HCl  N 2

40. Carboxylate ion is more stable than phenoxide ion

41. Ethanolic NaCN2 H / Ni
CH3  Cl   CH3CN  CH3  CH 2  NH 2
compound ‘F’ is CH3  NH 2
42. Refer NCERT page no – 413,414,415 12 th class part II
43. NCERT, Histamine is used for secretion of pepsin and HCl in stomach
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44. ion/molecule NO.of e in No of e in Bond order
BMO ABMO
O 2 10 5 2.5
O2 10 6 2
O 2 10 7 1.5
O 22 10 8 1
45. textbook pg no . 285,290
46. 1. Li    76 pm  , Mg 2   72 pm   NCERT pg  304 s  block 
2. Cu 2  Zn 2 3. Na   F  Isoelectronic 
4. Ce3  Pr 3  Lathanoide contraction 
47. Hybridisation of Al is sp3d 2 , shape-Octahedral
b)Boran is unable to from BF63
c)preparation of diborane(ncert)
SiF6 2 is know(ncert)
48. ncert
49. NCERT
50. Weight of AgCl  35.5  100
% of chlorine 
Weight of substance 143.5
0.7175  35.5  100 5 1000
    100 =45.45
0.3905  143.5 1000 11
51.  Radial nodes are obtained for
 13r r2   r  r 
 1    0  1  1  0
 36a 0 36a 2   4a 0  9a 0 
 0
 r  4a 0 and r  9a 0 , Distance between nodes r  5a 0
52. ng 47.9
sol: 2 K p  K c  RT   Kc  2
 0.083  2881
53. Po  Ps  n solute  650  640 0.25  78
 i   1  M  solute   32gm
Ps  n solvent  640 m  39
0.25  1000
Now, Tf  K f xm  5.12  1
32  39
54. Sol: x  10 , Ecell  Eocathode  Eanode
 Eanode R  EIn 3  /In 2 , EIn 3 / In 2  2EIn 3 / In 1  EIn 2 / In 1
E oIn  3 / In  2  2  0.42    0.4   0.44
 E cell  0.15   0.44   0.59

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nf
So, RT ln keq  nfEcell  log keq  Ecell
2  303RT
0.59
 log keq   keq  1010
0.059
55. Sol: Y  23
Unit of k represent first order kinetics 2N 2O5  2N 2O 4  O 2
t=o 1 0 0
t=t 1-p p p/2
 1  p  p  p / 2  1.45  P  0.9
2.303 1 2.303  1 
t  log t log  
2K 1 P 2  5  104  0.1 
 t  2.303  103 sec  t  23  102 sec
1 d  N 2O5 
Note: Here ‘2K’ should be considered instead of K because  K.
2 dt
56. Sol: 2 Pv  nRT
3.12  0.0821  300 2.4
 V   2.4L  Volume adsorbed per gram  2
32  1 1.2
57. Sol: 1
2Mno4  5C2O24  16H  2Mn 2  10CO2  8H2O   5  16    2  10  8   1
58. Sol: 500
1 3 
So  30    40   20  30   50   20JK 1
2 2 
 
 At equilibrium, TSo  Ho  T  20JK 1  10  1000J  T  500K
59. Sol: K  2
OV
x
a/2
a/2

2 2
a a a
x      
2 2 2
60. Sol: 10
H 2CO3  NaOH  NaHCO3  H 2O
Millimole 10 10 - -
At end 0 0 10+10=20.
 Final mixture has 20 milli moles NaHCO3 and 10 millimoles Na 2CO3
salt
P H  Pka 2  log
Acid
 Buffer : Na 2CO3  NaHCO3 
 10 
 PH  10.31  log    10
 20 
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MATHEMATICS
61. Let the total population of town be x
25x 15x 65x 105x
   1500  x  x  1500  x  30000
100 100 100 100
62. Given system has infinitely many solutions
k  26 11k  18
 ,  and verify
19 19
63.
r  5x 2  8x  14
r  3 x  
64. h a h b
:  , 
y xy x xy
a b
h ;h 
x y
1 1
y x
A Q

x a
1 a
T
y h b
x a ah h
 1 
y h h B x M y P
y h

x ah
b ab
h h
h ab
1
ab
65. A

C
R

E r B D

Let AB=h
BC=h-R
1
r 2  4hR  h 2 and v  r 2 h
3
1

 v   4hR  h 2 h
3

dv
0r 4
dh

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 SRI CHAITANYA IIT ACADEMY, INDIA 09‐01‐23_ Sr.S60_NUCLEUS & STERLING_ Jee‐Main_GTM‐04_KEY &SOL’S
66. Point of intersection of the given curves is (6, 12)
 Equation of the normal at P(6, 12) to y2  24x is x  y  18  0
67. Conceptual
68. 9y 2  x 3 ...........1
x2
'
y   1  Given slope of normal  1
6y
 6y  x 2 ......... 2 
 8
Solving 1 and  2  weget P  4, 
 3
 Equation of tan gent at Pis3x  3y  4  0
69. Take x  t
1
x    
1 3 1  T2  3  T2
 1  x  3  x dx   1  T 2 3  T 2 dt
0 0   
70. dy dx

 dx dy  y 2 x 2
   0
 x y   1 1 2
  
y x
 
dx dy  1 
  d 0
x y 11
y x
 
1
log x  log y   c  int igrated 
1 1

y x
x xy
log  c
y xy

71. 2 4
 x 2   x2 
A  2   7  x    1    dx  2   7  x     1  dx  32
 4    4 
0    2   

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 0,1

 2,0  2,0   4,0   7,0 

72.   20
sin ce  sin x i  and  sin 1 x i  10
1
2 2 i 1

 sin 1 x i  1  i  20
2
 x i 1, 1 i  20
20
  x i  20
i 1
20
 xi
 i 1  2
10
73. : Centre of S1   2,4 
Radius of S1  radius of S2  4
Centre of circle S2   4,2 
2 2
S2   x  4    y  2   16
 x 2  y 2  8x  4y  4  0.........1
Equation of the circle touching y=x at (1, 1) can be taken as
 x  12   y  12    x  y   0...... 2 
 2    2 
2   4   2  242 1 and 2 orthogonal 
 2   2 
2
2  2  6
74.  1
 x sin   , x  0
F x    x
 0 , x0

1
lim F  x   lim x sin    0
x 0 x 0 x
Also F (0) = 0
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 SRI CHAITANYA IIT ACADEMY, INDIA 09‐01‐23_ Sr.S60_NUCLEUS & STERLING_ Jee‐Main_GTM‐04_KEY &SOL’S
 lim F(x)  F(0)
x 0
 F  x  is continuous at x=0
 F  x  is continuous for all real numbers
Statement-1 is true
f1  x   x
 it is continuous on R
 1
sin ,x0
f 2  x     x 
 0 , x0

1
lim sin does not exist
x 0 x
 it is not continuous at x=0
 f 2  x  is discontinuous on R
Thus statement-2 is false.
75.
sin

  m
 
  
 m  1  
6    
 4  4 
   m  1  m  
4
m 1 sin   sin( 
 
 4 4 
6 
 
  cot   
 m  1    cot    m    4
  
 
m 1 
4   4  


 cot   cot   3
2 4
 cot   tan   4
 tan   2  3
    , 5 n , 
12 12 2  

   
12 12 3

76.
   
Ans- 1  3c1 x92  2 3c2  9  3  1  300

77. b3  4b 2  3b1
b1h 2  4b1r  3b1

 r2  4r  3  0
r  1 or r  3
78. CONCEPTUAL
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79. USE EXPANSIONS
80.
 
Ans - f 3   2
2
1 3
gl  2   
fl 3   
2
7

81. 99 99
P  x  2   1  P  x  0   P  x  1 
100 100
1 3n  1
  n6
100 4n
82.

A  4,3 C  0,3
B  4,3

AB  8
0 2 5

83. 12  14
Mean   13
2
122  142
2   132  1  var iance 
2
12  14    
mean   13
4
    26................1
122  142   2  2
2   132  1
4
  2  2  340..... 2 
by 1 and  2 
  2
84. 1 x2 4
1 0 3  0
3 3 2
85. Solution: 2m  56  2n
 2m  2n  56
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 2n  (2m  n  1)  56
m  6,n  3

 
86.  1 
Ans- 2f x 2  3f    x 2  1
 x2 
 1 
x 
2
2f  2   3f x  2 
1  x2
x
Bysoluvingweget

1  x 2  3  2x 2 
 
f x 2

5x 2
1
Take x 
52
87. 1  z  z2 22
Ans: 1 R
1  z  z2 1  z  z2
z
 R
1  z  z2
1  z  z2
 R
z
1
  z  1 R
z
1 1
 z 2
z 2
1 1
zz 
z z
zz
zz
zz
zz  1  z  1 z  z

88.
Ans- Put x 2  11    x 2  2  11
t 2    11  2    11  4  1
   
Clearly t 2  t  11  t 2  t  11  2t  2

2

 2t  11  2t  11


1 2    11  2    11 2
T
 2  t  11  2    11   3
2

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2
2  
1  2      11 2  
 4
   4  x 2  1  16
x2  5
x 5
89.
Ans -Let y  3log3 91x  21
 y  31x  21
7
 G.Eis  y  z 
7
T6  C Y 2 Z5
5
567  21. 3  2 x 2
1.3 x 2
9 
3 3 2 2 x 2
27  43  93
x 2
 43  92  27  0 where   3
  3 satisfies
1
z  7 5 log 7  4.3  9
x 2
 

 
11
x 2
z  4.3 9 5
x 2
3 3
x  2 1
=3,1

90. Ans –
 x  1, 1  x  0
 0, x0

 x, 0  x 1
f x  
 2x  1, 1  x  2
x 1 2x3

5 x 3
 a  3,b  4

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS&STERLING_BT JEE-MAIN Date: 11-01-2023
Time: 09.00Am to 12.00Pm GTM-05 Max. Marks: 300
KEY SHEET
PHYSICS
1) 4 2) 3 3) 2 4) 2 5) 1
6) 2 7) 2 8) 1 9) 2 10) 3
11) 4 12) 1 13) 1 14) 4 15) 1
16) 2 17) 4 18) 2 19) 1 20) 1
21) 10 22) 100 23) 2 24) 7 25) 2
26) 6 27) 1 28) 5 29) 7 30) 45

CHEMISTRY
31) 1 32) 3 33) 3 34) 2 35) 3
36) 3 37) 2 38) 1 39) 4 40) 1
41) 2 42) 4 43) 2 44) 4 45) 3
46) 2 47) 1 48) 4 49) 2 50) 2
51) 57 52) 1 53) 3 54) 3 55) 6
56) 8 57) 3 58) 3 59) 3 60) 4

MATHEMATICS
61) 3 62) 1 63) 1 64) 2 65) 3
66) 2 67) 4 68) 2 69) 1 70) 2
71) 4 72) 1 73) 3 74) 1 75) 1
76) 3 77) 1 78) 2 79) 2 80) 4
81) 216 82) 8 83) 2 84) 1 85) 7
86) 2 87) 4 88) 64 89) 5 90) 4

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SOLUTIONS
PHYSICS
1. 53  52  55  54  51
tmean  ttrue   53sec
5
0 1 2 1 2 6
Mean error   1.2
5 5
Least count is 1 sec, means round off 1.2 to 1 sec.
 t  53  1sec
2. The rolling sphere has rotational as well as translational kinetic energy.
1 1 1 1 2 
KE  mu 2  I  2  mu 2   mR 2   2
2 2 2 2 5 
1 1 7
 mu 2  mu 2  mu 2   u  r 
2 5 10
7 7u 2
From energy conservation i.e., mgh  mu 2 or h 
10 10
3. V  VCE  I C RC
 15  7  I C  2 103  iC  4mA
i 4
   C  iB   0.04mA
iB 100
4. Changing polarity is termed as AC.
5. Velocity of ball when it reaches to surface of liquid
V 1000  g

V  500  g

1000 gV  500 gV
a where V is the volume of the ball
500V
a  10m sec2 ()
Apply v  u  at  0  2 gh  10t
0  2 gh  10t 2 gh  10   2 
2 10  h  400  h  20m
6. V  BE sin 
H  BE cos 
H

V
BE

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7.  Kq
For a metal sphere Ein  0 and E out  2 r
r
E

Ein  0

r
8. 0.61 0.61 0.5 m
Numerical aperture  d   0.24  m
d 1.25
9. Fd  dm 2 R
Fd    ARd   2 R
dm 2 R

F d F
R
O 

 stress, F A   2 R 2
F l R
Now,  Y Y
A l R
 l  2 R, l  2R
R  2 R3
 2 R 2  Y R 
R Y
10.

2
m 1
FT  V 2   2  g 
L 2
mg 3mg
FT  mg FNet  mg 
2 2
11. GMm 1 1
Te i  Te   m  3ve   m  v 
2 2

R 2 2
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1 1  1
 mve2  9  mve2   m V  V  2 2Ve
2

2 2  2
12. workdone 200  600  300 5
  
heat suplied 200  600 8
13. U  x    x 2  3x  J
du
For a conservative field, Force F  
dx
F  
d 2
dx
 x  3x     2 x  3  2 x  3
3
At equilibrium position, F  0  2 x  3  0  x  m  1.5m
2
14. Eddy current effect is not used in electric heater
15. Let 1  4eV , then 2  2eV
 E    represent kinetic energy of most energetic electron.
E  2  2  E  1   E  6eV
16. Let a  b  c, so
V 2bc V 2 ac V 2 ab
P1  ,P  ,P 
a 2 b 3 c
m m m
Volume of cuboid  abc   4 2c 3  c  3
d d 4 2d
17. 5 4L 5
L  k 
4 5 2L
2 4 L 5
 k
k 5 2L
L

O
Equation of wave from open end:
P 1 5 x  2L L
At t  0 P  0   sin x  
2 2 2L 6 5 15
18. Range  x  2hT R
 2  150  6400  103
Area   x 2
50 105
Population density 
  2 150  6400 103
1 106
 km   828.6km2
2
 m 2 
6  64 6  64
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    
19. 
F  q v B 
F  v
 
Work done = F .S
Work done = 0
20. C p is always greater than Cv in gases.
Work done at constant pressure is more than at constant volume.
21. Mass of rope  10  0.5  5kg
Given force  25n
Acceleration  F m  25 5  5 m s 2
Length of remaining rope  4m
Hence mass of remaining rope  4 2  2kg
Hence tension on the rope at a point 6m away  ma  2  5  10
22. V
Under resonance iA    5 A Voltmeter reading V  500V
R
23. h hc 2mc 2
 and 0   n2
2meV eV h
24. For maximum intensity on the screen d sin   n
n n  2000  n
 sin   sin   
d 7000 3.5
Since sin   1 n  0,1,2,3 only
Thus only seven maximas can be obtained on both sides of the screen
25. The two bodies will collide at the highest point if both cover the same vertical height
v12 sin 2 30 v22
in the same time. So, 
2g 2g
v2 1
Or  sin 30   0.5
v1 2
26. Energy  FAT x 3
 T 
x3
M 1L2T 2   M 1LT
1 2
  LT
1 2

By equating power of time, 2  4  x 3  x  6.00

27. dN B N  N
A 1
 B 2
C ,  1 N A  2 N B  0 , A  2  1 A  1
dt N B 1 2 N B
28. Energy of diatomic gas due to its thermal motion is
 5 5  5 5 m
U  nRT  PV   PV  P    5  10 J
4

 2 2  2 2 
29. 

T  75 7
 constant (or) P  T  
 1
 1
C    .
P  1 2 5 2
30. I I
A and B are parallel so cos 2   cos 2      45
2 8

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CHEMISTRY
31. C x H 2 y O y  2 xO2  xCO2  yH 2O  xO2
Number of moles after cooling =2x
Volume after cooling=2.24 litres
Number moles of CO2 =0.05
 EF= CH 2O
P 0  17.5mm
P 0  P  0.104mm
M=151.4
MF= C5 H10O5
32. For the given question only C & D compounds are possible. In that H of C is less
acidic.
33. O
Cl Cl

N2H4 OH,

34. Both are true statements.

35. Basic strength order 3>4>2>1>5.
36. XeF6  H2O  XeOF4  HF
XeF4 : sp3d
37. HNO3
Cu2 S2 CuS Cu2 S
NH3 deepblue solution
38. NaCl  H2 SO4  HCl  NaHSO4
MnO2  HCl  MnCl2  Cl2  H2O
39. A is false but R is true
40. 3 2
[Cr ( H 2O)6 ]3 - d 2 sp3 ,3 , [Co(CN ) 4 ]2 - dsp2 ,1 , [ Ni  NH 3 6 ]2 - sp d , 2 , [ MnF6 ]4 - sp3d 2 ,5
41. Assertion and reason both are correct statements and reason explains the assertion.
42. A is correct but R is incorrect
43. For formation of NH 3 and PCl5 change in entropy in negative
44. NH 2   OH   NH 3 is correct order of basic strength.
45. AB2 g   A s   2 AB g 
Initial 0.7 0
0.7-x 2x
Final y (0.4 - y)
 2x
2
1 1
Kp   
0.7  x 4 0.45
Ptot  0.95
 0.7  x  x  0.25

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 0.4  y 
2
5
 y  0.13
y 9
0.13
At second equilibrium VAB   100  32.5%
2
0.4
46. m  Zit
1
q  Area of figure  [100  10  10 3 ]  2[  10  103 ]  2
2
m
m  Z  2 Z 
2
47. d  a .No
3
Z a3  volume  10 A0 10 A0  sin 600 15 A0
M
2 3  750 3 1024  6 1023
 Z 6
450
48. A is Prop –1–en–2–ol. On tautomerization, it changes to ketone
49. Cl Cl OH OCH3 OCH3
SOH
3
Nitration NaOH /1500C CH3 I H2SO4
O O O O O

NO2 NO2 NO2 NO2

50. ph
PhMgBr I2 / KOH
CHCOCH
3 3 CH3 CCH3 -ve

A) OH
MeMgBr I2 / KOH
CHCH
3 2 CHO
CH3 CH2 CH CH3 ve

B) OH

PhMgBr PhMgBr I2 / KOH

HCOOEt PhCHO PhCHPh ve

C) OH

Ph

PhMgBr I2 / KOH
CHCOOE
3 t CH3 CPh ve

D. OH

51. 21.4 103

 conc. of salt solution
M 2 (molecular weight=1070)
1
P H  7  [log c  log K b ]
2
 3 
1 21.4  10
5  7  log  9
2  2M 
52. Only (iv) is wrong.
53. 1000  K f  w
T f 
m w
For the solution in benzene using the data given
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1000  5.12  w
1.28  ....1
mw 100
For the solution in water in which solute dissociates
1000 1.86  w
1.40  ....2
mexp 100
Dividing eq. (ii) by (i)
mN 1.40 5.12
i    3.01  3.0
mexp 1.28 1.86
Now, suppose that formula of solute is
Ax By  xA  yB 
1 0 0
1    x y
i  1    x  y
i  3 and   1  Given that   1
No of ions given (x+y) = 3
54. 1 micelle  2.4 1013 molecules
3
1.2  103 M  1mm3  103 cm3  1.2  10  106 moles
= 1.2  109 moles
1.2 109 moles  1.2  6 1014 molecules
1.2  6  1014  2.4  1013 x
1.2  6  10
x  30
2.4

55. With no 2s – 2p mixing the order of energy of MO’s would be

 1s   1*s   2 s 2*s   2 pz   2 P   2 P   2* p   2* p   2 p
x y y x 2

H 2  D  , He 
2  P  , Li2  D  , Be2  D  , C2  P  , C  D  , N  P  , N 2O2  P  , O2  P  , S2  P  , F2  D  .
2
2

2

56.
and
both can show G.I.
 8 alkenes of x
57. 2.303 R 2.303 1.24 102 2.303
K log 1  log  log 6.2
t2  t1 R2 60 0.2 102 60
= 0.0304  3  102
[ ref : NCERT solved example 4.5]
58. H2O2 in basic medium reduces Fe3 to Fe 2 .
59. Mg3 N2 with D2O gives ND3 having M .wt 20
Al4C3 and Be2C gives CD4 having M .wt 20
60. Group 1 bicarbonates exists in solid state except lithium.

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MATHEMATICS
61. f  x and f 1
 x can only intersect on the line yx
 y  x must be tangent
Solving 3 x 2  7 x  c  x  3x 2  8 x  c  0
The above equation has real and equal roots
16
 64  12c  0 c 
3
62. Let g( x)  f ( x  T / 2)  f ( x)
then g(k)  f (k  T / 2)  f (k) …... (1)
and g(k  T / 2)  f (k  T)  f (k  T / 2)
 f (k)  f (k  T / 2)  g(k)
Hence by intermediate value property there exist an x0 [k,k  T / 2] for which g(x) = 0
63.
(0,2)

(1,1)

(2,0)
(1,0)

64. f  4  f  0 
By LMVT, a   0,4   f 1  a   f  4  f  0   4f 1  a 
40
f  4  f  0 
 lies between f  0  and f  4 , by Intermediate value theorem
2
f  4  f  0 
b   0,4   f  b hence,  f  4    f  0   8
2
f1 a  f b
2

2
65. x A
We know that if di  i then  x  h  d .
h
x 3/ 2
In this case 2 xi  3  i .
1 / 2
1
So. h 
2
1
This.  d   x  2  3.5  7
h
66.  2008 8  224  2518 has 25  9  225 positive divisors, including

 20084   20088 . There is a one to one correspondence between the positive

4 4
divisors less than  2008  and those larger than  2008  . It follows that there are
1
 225  1  112 positive divisors less than  20084 .
2

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67.
C
P
24 43
20
19 29

n    50 M

Since, n  M  P  C   50
n  M   37, n  P   24, n  C   43
n  M  P   19, n  M  C   29
And n  P  C   20
Since, we know that
n  P  C  M   n  P   n C   n  M 
n  P  C   n  C  M   n  M  P   n  P  C  M 
 50  37  24  43  n  M  P   n  P  C  n  M  C   n  M  P  C 
 n  M  P  C   N  M  P   n  M  C   M  P  C   54
19  29  20  54  14
68. Equation of the chord AB having (a, b)
as M .P S1  S11  ax  by  a 2  b 2  0  
Chord length  2 r 2  a 2  b 2
 ar 2 br 2 
c , 
 a 2  b2 a 2  b2 
 
r 2  a 2  b2
h
a 2  b2
1
Area  bh
2
69.
 
16 sin 5 x  cos5 x  11 sin x  cos x   0

   
  sin x  cos x  16 sin 4 x  sin 3 x cos x  sin 2 x cos 2 x  sin x cos3 x  cos4 x  11  0

 
  sin x  cos x  16 1  sin 2 x cos 2 x  sin x cos x  11  0  
  sin x  cos x  4sin x cos x  1 4sin x cos x  5   0
As 4sin x cos x  5  0,We have
The required values are
 / 12,5 / 12,9 / 12,13 / 12,17 / 12, 21 / 12
They are 6 solutions on  0, 2 

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70. z  2i 2z i  1 z  i  1
 i   ,   z 2
z  2i 4i i  1 2i i  1
71. Tangent on ellipse having slope 2 will be
y  2x  4a 2  b 2  It is normal to circle
 (–2, 0) is on it.  (–2, 0) is on it.
 0  4  4a 2  b2  4a 2  b2  16
2 2
 A.M.  G.M.  4a  b  2ab  8  2ab  ab  4
2
 Maximum value of ab  4 Ans.
72. 2
Equation of tangent y  mx  , Passes through  2,3
m
y

(-2,3)

 y3  1
 2m2  3m  2  0   2m  1 m  2   0 , m    
 x  2  max . 2
73. S1 :    p   q      p   r    p   q  r 
S 2 : p   q  r   p   q  r  By Conditional Law
S1  S 2
74. 5 5
 
4
sin x  cos x 4
 cos x  sin x  e x  4 dx
I=
3
  
 x  
dx ; I= 
3
 
 x 
 4  4
4 1 e 4 1 e
5
5
4

4
  1 1   1 1 
  cos x  sin x  dx   sin x  cos x  
 2
   
2  2
 
2
3 
4  3
4

2I  0  I  0 . Ans.
75. dy
 f  x y  0
dx
dy
 f  x  dx
y
In y   f  x  dx

y1  x   e    Then for given equation e  

f x dx f x dx
I.F =

1
y. y1  x    r  x  y1  x  dx , y  r  x  y1  x  dx
y1x 
Hence Solution

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76. 1  x  y  1
x  y 1 x  y 1

2
Required area   1  

 
77. 1001
1  x  x 2  x3  x 4 1  x 1002

 
1001
 1  x  1  x5

so all the powers of x will be of the 5m or 5m  1 m 1

So coff. of x 2009 will be 0
78. Coordinates of any point Q on the given line are
 2r  1, 3r  1,8r  10  for some r  R
So, the direction ratios of PQ are 2r , 3r  1, 8r  10
Now PQ is perpendicular to the given line
if 2  2r   3  3r  1  8  8r  10   0
 77r  77  0  r  1
and the coordinates of Q, the foot of the perpendicular from P on the line
are (3, 4, 2).
Let R (a, b, c) be the reflection of P in the given lines when Q is the mid-point of PR
a 1 b c
  3,  4,  2
2 2 2
 a  5, b  8, c  4
and the coordinates of the required point are  5, 8, 4  .
79. Since u , m, r are mutually perpendicular vectors of same magnitude, we can resolve
e along the directions u , m, r let u  m  r  K
Let e  au  bm  cr
u   e  m   u    u .u  e  m    u . e  m   u

 K 2 (e  m)   u .e  u  u .m  m.r  m.u  0 
 K 2  e  m    u .e  u

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Similar m  (e  r )  m   m.m  e  r   m. e  r  m

 K 2  e  r    m.e  m
r   e  u   r    r .r  e  u    r . e  u   .r

 K 2  e  u    r .e  r
1
Substitute above, in the given question e  u  m  r 
2
80.   
Pqr a3  b3  c3  abc p3  q3  r 3 
  
 pqr a3  b3  c3  3abc  abc p3  q3  r 3  3 pqr 
 pqr  a3  b3  c3  3abc   abc  p  q  r   p 2  q 2  r 2  pq  qr  rp 

 pqr  a3  b3  c3  3abc 

81. Toys in group 112 

4!
 3!  36
1! 1! 2! 2!
Marbles O O O O = 4 C2  6
 Total ways = 36  6  216
82. H 1 : Three numbers drawn are 1, 2 and 3 in any order.
H 2 : Three numbers drawn are 1, 2 and 2.
1 1 1 2 1 1 1 1
P  H1   6      think! P  H 2     
3 3 3 9 3 3 3 27
P  H2  1 27 1 a
 P  H 2 H1  H 2      
P  H1   P  H 2  27 7 7 b
  a  b   8.
83. The equation of any plane containing the given line is
 x  y  2z  3    2x  3y  4z  4   0
 1  2  x  1  3  y   2  4  z   3  4   0 .... 1
If the plane is parallel to z-axis whose direction cosines are 0, 0, 1; then the normal
to the plane will be perpendicular to z-axis
1
 1  2  0   1  3  0    2  4 1  0  
2
Put in eq. (1), the required plane is
1
 x  y  2z  3   2x  3y  4z  4   0  y  2  0 ...  2 
2
2
 S.D. = distance of any point say (0, 0, 0) on z-axis from plane (2)  2
1
2

84. adj B adj  adjA  A

 312
A
3

  
C 5A 53 A 125
adj B
Now |A| = 5  1 Ans.
C

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85. The equation of the tangent at  5cos , 2sin   is

x y
cos   sin   1
5 2
1 10 3
If it is a tangent to the circle then  4  cos   ,sin  
cos 2  sin 2  4 7 2 7

25 4
Let A and B be the points where the tangent meets the coordinate axis then
 5   2  25 4 14
A , 0  , B  0, , L  2 
 cos    sin   cos  sin 
2
3
86.
D C(1,1)
0
(0.1) 45
Q(1,y)

A P(x,0) B(1,0)
(0,0)

tan 1  x and tan  2  1  y

tan 1  tan 2
Since, 1  2  450  1
1  tan 1 tan 2
x  1  y  2x
 1  y  .... i 
1  x 1  y  1 x

Now, Perimeter = 1  x  y  1  x 2  y 2
By using (i), we get
Perimeter=2
87. c
Normal to xy = c 2 is y - = t 2 ( x - ct )
t
Solving with xy = - c 2 we get
æc 2 ö 2 2 2 æc 3ö 2
ç + t ( x - ct )÷÷ + c = 0 t x + ç
xç ç - ct ÷÷ x + c = 0
è t ø è t ø
2
æ1 ö
For equal roots çç - t 3 ÷÷ - 4t 2 = 0 Þ 4 values are possible
èt ø
88. 2
b
b bx bx 2tan  a
 tan ,  tan2 or  
a a a 1  tan2  b2
1 2
a
bx 2ab 2a 2b
 2  x  b
a a  b2 a 2  b2

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C
Flag
x

b
Tower


P a A

or x

2
a 2 b  b3 b a  b

2

a 2  b2 a 2  b2
(0,8)
89. 

x
(-1/2,0) O

2

 z2  2  
arg  2 
 z   2
 1  3
z 2 lies on a circle whose center is  , 0  and radius is equal to units.
 2  2
z  2 1  i  z  2 1  i   z 2  4  2i   z 2  8i
Minimum value of z 2  8i is equal to
1 3 257  3 a  b a  b 257  3
 64      5
4 2 2 2 2 52
90. R : A  B under given condition a < b is given by
 1,3 1,5   2,3    3,1  5,1  3, 2  
, R  
1
R 
 2,5   3,5   4,5    5, 2   5,3  5, 4  
RoR - 1 : for composing RoR - 1 we will pick up an element of R - 1 first and then of R
 3,1  R 1 , 1,3  R,   3,3  R o R 1
 R o R 1  { 3,3 ,  3,5  ,  5,3 ,  5,5 } only
Elements are not repeated in a set

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 SRI CHAITANYA IIT ACADEMY, INDIA 13‐01‐23_ Sr.S60_NUCLEUS & STERLING_BT_ Jee‐Main_GTM‐06_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS&STERLING_BT JEE-MAIN Date: 13-01-2023
Time: 09.00Am to 12.00Pm GTM-06 Max. Marks: 300
KEY SHEET
PHYSICS
1) 3 2) 1 3) 3 4) 3 5) 1
6) 4 7) 2 8) 4 9) 4 10) 2
11) 1 12) 1 13) 3 14) 4 15) 3
16) 1 17) 3 18) 2 19) 3 20) 1
21) 20 22) 2 23) 60 24) 175 25) 144
26) 177 27) 31 28) 4 29) 10 30) 6

CHEMISTRY
31) 2 32) 2 33) 1 34) 1 35) 4
36) 4 37) 3 38) 2 39) 4 40) 3
41) 3 42) 3 43) 1 44) 1 45) 2
46) 4 47) 1 48) 3 49) 2 50) 1
51) 2 52) 1 53) 6 54) 1 55) 2
56) 6 57) 5 58) 5 59) 2 60) 3

MATHEMATICS
61) 1 62) 1 63) 1 64) 3 65) 4
66) 1 67) 1 68) 3 69) 1 70) 2
71) 1 72) 4 73) 2 74) 2 75) 2
76) 3 77) 3 78) 2 79) 1 80) 4
81) 1 82) 4 83) 2 84) 6 85) 113
86) 3 87) 16 88) 7 89) 9 90) 5

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SOLUTIONS
PHYSICS
1. Conceptual
   
2. r f = r i + DS1 + S 2 + ...........

( ) ( ) (
r f = 2i + 3j + 5i + 8j + -2i + 4j + -6j )

r f = 5i + 9i

Distance from thrower = 52 + 92 = 106

3. Dimension of z is dimension of time and only option C has dimension of time.
coefficent of t 6 ´108
4. speed = 3´108 = k=2
coefficent of x k
I
5. Intensity after passing through 1st polaride = 0
2
Let 2 polaride is at angle q from first polaride & 3rd polaride is at angle 90- q from 2nd
nd
I
one then intensity after passing through 2nd polaride = 0 cos 2 q
2
I
Intensity after passing through 3rd polaride = 0 cos 2 q cos 2 (90 - q )
2
I I I
I final = 0 cos 2 q sin 2 q = 0 sin 2 2q = 0 sin 2 2wt
2 8 8
6. For first dark fringe on either side, d sin q = l
dy lD
Or =l \ y=
D d
2l D
Therefore, distance between two dark fringes on either side = 2y =
d
Substituting the values, we have

Distance = =
( )(
2 600 ´10-6 mm 2 ´103 mm ) = 2.4mm
(1.0mm)
m NI
7. D= 0
2p R
meffective
8. T = 2p
k
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 SRI CHAITANYA IIT ACADEMY, INDIA 13‐01‐23_ Sr.S60_NUCLEUS & STERLING_BT_ Jee‐Main_GTM‐06_KEY &SOL’S
If mass increases time period increases. If collision take place of extreme position then no
energy loss take place hence amplitude remains same.
If collision take place at mean position then due to inelastic collision energy loss take
place & Amplitude decreases.
9. W = area enclose in the cycle
1
W = 4 ´10-4 ´ 2 ´105 = 40 J
2
10. Zero error is 3 division
0.5
LC = = 0.01 mm
50
zero error = 0.03 mm
Reading = 5.0 + 29 ´0.01
d = 5.29-0.03 = 5.26 mm
11. For initial condition :
1 é1 1ù 1 é3 ù é 2 ù
= [m -1] ê - ú  = ê -1ú ê ú
f ê R1 R2 ú 12 êë 2 úû êë R úû
ë û
R = 12 cm
When liquid is poured
u = -12, V = -24, f eq = ?
1 1 1
+ = [Mirror Formula]  f eq = -8
v u f
Peq = 2 PL1 + PL2 + PM
1 æ1ö 1 é1ù
= 2çç ÷÷÷ + 2 PL2 + 0  PL2 = = [m -1] ê ú
8 çè12 ø 48 ëê R ûú
1 é1ù 5
PL2 = = [m - 1] ê ú  m =
48 êë R úû 4
12. For isothermal process

PV = C
-P
PdV + Vdp = 0 dp = .dV
V
-p æ Pa 2 ö÷
dp = ( Ax)  Fnet = (dP) A = -ççç ÷÷ x  F = - kx
V çè V ø÷÷
1 g RT
13. f=
4 M
df 1 dT 1 1
Since DT << T  =  df = ´ ´10,000  df = 16.67
f 2 T 2 300

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 SRI CHAITANYA IIT ACADEMY, INDIA 13‐01‐23_ Sr.S60_NUCLEUS & STERLING_BT_ Jee‐Main_GTM‐06_KEY &SOL’S
14. As temperature decrease
E = eAsT 4  Intensity reduce
and lmaxT = constant
lmax : increase
15. Let resistance per unit length of wire ' AB ' be k
Case (i) :
x = 0  balance length :  1
Rh = 2W
40
i=  i = 5A
4+2+2
\ At balance length
e = 5k  1 ………(1)
Case (ii) :
x = 2  Balance length  2
Rh = 6W
40
i=  i = 10 / 3 A
4+2+6
\ At balance length,
e = (10 / 3) k  2 ……..(2)
 2
From (1) and (2), 5 x 1 = (10 / 3) k  2  1=
2 3
æ s ÷ö
16. Fy = mg -j ( ) Fx = çç ()
÷ q i
çè 2e ø÷÷
0
æ s q ö÷
()
a y = -g i a x = çç ÷i
çè m2e ÷÷ø0
As S y = H max ,V y = 0
Let the time taken to reach max. height be
u sin q
t1 . v y = u y + a y t1  0 = u sin q - gt1  t1 =
g
æ sq ö÷ u sin q
At t = t1, v x = u x + a xt1  vx = u cos q + çç ÷
çè 2me0 ÷÷ø g
é sq tan q ù
= u sin q ê1 + ú
ê 2 m e g ú
ë 0 û

17.
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 SRI CHAITANYA IIT ACADEMY, INDIA 13‐01‐23_ Sr.S60_NUCLEUS & STERLING_BT_ Jee‐Main_GTM‐06_KEY &SOL’S
Applying bernouli’s theorem between point on surface of water and point at orifice taking
ground as reference,
1 1
Patm + rV12 + r gH = Patm + rV22  V22 - V12 = 2 gH
2 2
æA ö
 V22 - çç 2 ÷÷÷V22 = 2 gH [ A1V1 = A2V2 ]
çè A1 ÷ø
2 gH
 V22 =
æ A ö÷2
1 - çç 2 ÷÷
çè A1 ÷ø
A2 1
Substituting = , H = 0.3m  V2 = 2 2
A1 2
If q = 300
3
V22 sin 2(90 - q ) 8´ 2 2 3
Range = = =
g 10 5

18.
Let ' M ' be total mass of earth.
Consider a shell of thickness ' dr ' and mass
' dm ' at a distance ' r ' from centre inside earth,
 dm = r 4pr 2dr
R
3 4pkR 4
M = ò dm = ò 4pkr dr = = pkR 4
4
0
Let field due to earth’s gravity at a distance '2 R ' from centre be
T , I ´ A = 4pGminside .
2
 I ´ 4p (2 R ) = 4pG pkR 4 ( )
pkR 4G
 I=
4R2
For a satellite of mass ' m ' moving in orbit or '2 R ' radius.
mv 2
mI =
(2 R)
V2
I=
2R

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 SRI CHAITANYA IIT ACADEMY, INDIA 13‐01‐23_ Sr.S60_NUCLEUS & STERLING_BT_ Jee‐Main_GTM‐06_KEY &SOL’S

pkR 2G V 2
 =
4 2R
pkR3G
V=
2
19. Conceptual
20. Conceptual
¶v
21. Ex = - = -6i
¶x
¶v
E y = - = -8j
¶y
¶v
E z = - = -8 zk
¶z

E net at origin = 62 + 82 = 10
 
 F = qE = 20 N
22. Induced field in rod, E = vB
KQ
electric field on surface of sphere =
R2
KQ kQ 9 ´109 ´ 30
= vB  R 2 = =
R2 vB 9 ´1
R = 3 ´105 = 1.73´105
23. For all collision to take place electron has to excite to n = 3.
4m
m Perfectly inelastic collision
EnergyE m+4m
V
mu + 0 = 5m V
1 2 1 æ 4 ö÷2 4 E
Loss = mu - 5m ´çç ÷÷ =
2 2 çè 5 ø 5
4E 2
= 12.09 ´(2) eV
5
 E = 60.45eV
f max
24. = Quality factor
Df half of max power
XC
= Quality factor
R
æ IA ö
25. Impulse on block = çç ÷÷÷ cos 2 53 ´(Dt )
çè C ø

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(20)(10 ´10-4 ) 2
= ´(0.6) ´ 6 ´10-3
3´108
72
= ´10h -14kg m / s
5
Now we have
Impulse = mv
72
´10-14 = 1´10-9 v
5
72
v = ´10-5 m / s
5
Now we have
1 2 1 2
kx = mv
2 2
10 x = 10-9 ´ 24 ´10-5 ´ 24´10-5
-5 2
72
x = ´10-7 m
5
7.2
N= = 1.44
5
26. mv min = L  2mneutron K neutron = 4 ´10-34
  min = 1.25´10-14 m = 125´10-16 m
2 3
27.
0.2
3
DV 2Dr Dl Dl DV Dr
28. = +  = -2 = 0.204%
V r l l V r
0.204
\ Stress = 2 ´1011 ´ = 4.08´108 N / m2
100
29. The diode is forwards biased, so the equivalent resistance of the circuit is 15KW .
w1 l1
30. =
w2 l2

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CHEMISTRY
31. Polarizing power of the cation
32. Steam volatile and water insoluble
33. Aromatic aldehydes and ketones do not give positive Fehling’s test
34. E1 CB mechanism
35. Presence of unpaired electrons
36. SP mixing
37. Chromyl chloride test
38. X  B2 H 6 ;Y  B3 N 3 H 6 ; Z  H 2
39. Calgon treatment
40. Half filled f orbital configuration
41. SRP vaues
42. Ozone is component of photo chemical smog
43. LiF has high lattice energy
44. Ethene is produced (gas)
45. Adsorbate is concentrated on surface of adsorbent
46. Addition of copper rod does not change direction of current flow.
47. NCERT XI part-1 Page No 181.
48. State-I to state- II is spontaneous, NCERT XI, Equilibrium
49. NCERT lab manual 12 class.
50. One motif for unit cell (NCERT 12 Page No. 6)
51. Two P-H bonds are present
52. Observe chiral carbons in the product
53. There are 6 atoms present in straight line
54. H 2 S 2O8
55. Nucelic acids
56. XeO3
57.  = h/mv
NCERT (text question)
58. 0.005 moles of Barium chloride in 2L
59.   CRT   gh
0.2 / M 1.013  1000  0.2463
Or,  0.0821  300 
100 / 1000 1.013  106
 M  2  105
-4
éH +ù 5´10
60. êë úû = ´1000 = 10-3 M  p H = 3.0
final 500

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MATHEMATICS
 z  2i   
61. arg     arg  z  2i   arg  z  2i  
 z  2i  6 6
 y  2  y  2  xy  2 x  xy  2 x 1
 tan 1    tan 1     
 x   x  6 x2  y2  4 3
 x 2  y 2  4   4 3x  x2  y 2  4 3x  4  0


 x2 3 2  y 2  12  4  42  
centre 2 3,0 , radius = 4

P  x, y 

 

1 2

O  0,0 
62.

1     2 1  2 
2
 
1  2  2  1   2 2  1 
2 2
dy
2 tan  2  1
2
dx   x


1  tan  2  
2 tan 
1  dy 
1  
2 y
 dx 
2 2
dy  dy   dy  dy
2y  x   x  x   2y  x
dx  dx   dx  dx
63. Three b’s and four a’s can arrange abababa
babababa
abbababa

Fourth ‘b’ can take 5 places, ababbaba
abababba

 abababab
Take 8 place from 12 places and arrange letters in same order, cc dd can take remaining
4!
4 place 12C8  5 
2!2!
6
C1 6 C1 6 1 1 175
64. Required probability = 1  5  5  C2 . 2C1 5 . 4 
2 2 2 2 216

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 SRI CHAITANYA IIT ACADEMY, INDIA 13‐01‐23_ Sr.S60_NUCLEUS & STERLING_BT_ Jee‐Main_GTM‐06_KEY &SOL’S
tan 3x - tan 2 x
65.   tan (3x - 2 x) = 1
1 + tan 3x tan 2 x
Or tan x = 1
p
Or x = np + , n Î I
4
sin x
But at this value, tan 2x is undefined, hence there is no solution and tan x =
cos x
\ tan x is not defined when cos x = 0
p
Or x = np + , n Î I
2
66. There are two possibilities : either the curves y  x 2  u and x  y 2  u intersect in
exactly one point, or they intersect in two points but one of the points occurs on the
branch y   x  u .
Case-1 : The two curves are symmetric about y = x, so they must touch that line at
exactly one point and not cross it. Therefore, x  x 2  u , so x 2  x  u  0 . This has
1
exactly one solution if the discriminant,  1  4 1 u   1  4u , equals 0,so u  .
2
4
Case-2 : y  x 2  u intersects the x–axis at  u , while y  x  u starts x = u and
goes up from there. In order for these to intersect in exactly one point, we must have
 u  u , or u  u 2 (note that u must be positive in order for any intersection
points of y  x 2  u and x  y 2  u to occur outside the first quadrant). Hence we have
u  u  1  0 , or u  1,0 .
1
67. Equation of diagonal AC is y  2   x  1  x  2 y  5
2
 5 7
On solving with y = x we get A , 
 3 3
 1 7
On solving with y = 7x we get C  , 
 2 2
2 4
1 1 20 10
Clearly B = (24) required area = 4 2    sq. units
2 2 3 3
3 3

68. As lines are perpendicular to each other ‘C’ moves on a circle with AB as diameter.
Now P, mid–point of AB (which is fixed) when joined with C is median.

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1
 Centroid is moving at a constant distance  PA from P.
3
 Locus is a circle
 6 1
A is point of intersection of x + 4y + 2 = 0 and x – y + 1 = 0 i.e.,   ,  
 5 5
 9 6 
B is point of intersection of 4x – y + 6 = 0 and x + y + 3 = 0 i.e.,  ,  
 5 5
2 2
  3 7   3  7 17
 P   ,  therefore locus is
 2 10   x     y   
2 10 50
69. y  cos   sin   sin 2   sin  
 
y
 sin   sin 2   sin 2 
cos 
y sec  sin   sin 2   sin 2  4 y 2 1  y 2  sin 2    0
 
Squaring on both sides  1  y 2  sin 2   0
y 2 sec 2   sin 2   2 y tan   sin 2   sin 2  y 2  1  sin 2   0
y 2  y 2 tan 2   2 y tan   sin 2   0  
y 2  1  sin 2   0
tan   R    0

 
4 y 2  4. y 2 . y 2  sin 2   0 y    1  sin 2  , 1  sin 2  
 
70. Since, f(x) is differentiable and hence continuous x  R
   
 f 0  f 0  P  0  0 and    
 f 0   f / 0   P  0  0
Similarly, continuity at x  1  P 1  1 and differentiability at x  1  P 1  0 .
Since, P(x) is a polynomial of least degree and P /  x vanishes x = 1 and x = 0.
Hence, P(x) must be cubic.
 x3 x 2 
 P  x  kx  x  1  P  x   k     C
/
 3 2
Since, P  0  0  C  0 and P 1  1  k  6

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 0 if x0

Hence, P  x   2 x3  3 x 2 .  f  x  2 x3  3 x 2 if 0  x 1
 1 if x 1


2 4 6 2  n  1 
71. S   n  1 cos   n  2 cos   n  3 cos  .....  cos
n n n n
2 4 2  n  1 
S  1cos  2cos  ......   n  1 cos
n n n
 2 4 2  n  1  
2 S  n  cos  cos  .......  cos 
 n n n
 2 2  n  1  
sin  n  1 
2S  n n cos  n n 
  2   n
sin  
n  
 
72. a  tan A ; b  tan B ; c  tan C ;0  A, B, C  0 AC 
2 2
P  2cos A  3cos C  2cos  A  C 
2 2 2

3 3
=1  cos 2 A   cos 2  cos  2 A  2C 
2 2
3 3
=1  cos 2 A   cos 2C  cos 2 A cos 2C  sin 2 A sin 2C
2 2
3 3
=  cos 2C  1  cos 2C  cos 2 A  sin 2C .sin 2 A
2 2
3 3
  cos 2C  1  cos 2C    sin 2C 
2 2
2 2
3 3 5 3
=  cos 2C  2  2cos 2C   cos 2C  2sin C
2 2 2 2
3 3

 1  2sin 2 C  2sin C
2 2

3 3
  3sin 2 C  2sin C
2 2
2
3 3  2 1 1 1  1
  3 sin 2 C sin C       3 sec 
2 2  3 9 9 3  3
10
=
3
73. Using A.M  G.M
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1/3
x x  x2 y 
x  y    y  3 
2 2  4 
x
Equality holds if and only if  y  x  2y .
2
Also, 2 x 2  2 xy  3 y 2
2 x2 2 x 2 2 zy 2 xy
 .....    .....   y2  y2  y2
8 8 4 4
1 2
 2 8 4 
3 15  x y 5
  
2
 2x  2 xy  2
 15     y  15  
 8  4  4 

2 x 2 2 xy
Equality holds if and only if   y 2 or x = 2y. Thus
8 4
1/3
 x2 y 

k  x  y  2 x 2  2 xy  y  
2 1/2
 3  15 
 4 

74. Let the numbers be 1, 2, 3, 4, ……, n and the erased number be x then 1  x  n
n  n  1
x 7
Now, 2  35
n 1 17
n  n  1 n  n  1
n 7 1
 2  35  2
n 1 17 n 1
n 7 n2 n 7 n2
  35    35  
2 17 2 2 17 2
14
 n  70   n  2  n  69  or  70
17
69  70
2 7
at n = 69 ;  35  x  7
68 17
at n = 70 ; x I
10
75.   xi  x yi  y  80
i 1
10 10 10 10 10
 xi yi  y  xi  x yi   xy  80 which implies  xi yi  10 yx  80
i 1 i 1 i 1 i 1 i 1
10
  xi  yi 
2

 2  i 1
10

 yx 2  9
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76. Statement -1

x y x y x y x y ~  x  y x y ~  x  y
T T T T T F F T
T F F T F T T F
F T F T F T T F
F F T F F T F T
Statement -2 :  ( p  q ) « ( p   q )

 ( p  q ) «  ( p  q )

( p  q ) « ( p  q ) º p (Neither a tautology nor a contradiction)

1 x x2 1 1 1
2
77. 1 y y 2x 2 y 2z
1 z z 2 x2 y2 z2

78. f  0  lt
1 f  0  h   f  0
= lt 

  h 3e 1/ h  4
 
 1
 0    2
h n 0   1/ h  h 
h 0 2e
 
f  0  h   f  0
Rf 1  0  lt
h 0 h

= lt 

 h 3e1/ h  4

0

  1  = lt 
 3  4e1/ h 
  3
h 0  2  e1/ h   h h 0  2e 1/ h  1 
 
Since Lf 1  0  Rf 1  0
f(x) is differentiable at x = 0. But f(x) is continuous at x = 0
dy x12
79. 
dx y12
Tangent equation is x12 x  y12 y  x13  y13
 x12 x  y12 y  a 3
Since, it passes through  x2 , y2 
 x12 x2  y12 y2  a3 (1)
and x13  y13  a3 (2)
x23  y23  a3 (3)
By solving (1), (2), (3) we get result
80. Let g ( x) = x n f ( x)(n > 0)
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g (0) = 0, g (3) = 0
By using rolls theorem, there exist some a Î (0,3) such that
g1 (a ) = 0  a f 1 (a ) + n f (a ) = 0, n > 0 .
2æ 1 ö÷ 0 3
çç 4
81. ( 2
) ÷
÷
ò ççç x - 2 x + 1 -1÷÷÷dx + ò ( x + 1)2 dx = 1
3 +1

1 çè ø -1
1 3
2 0
ò( )
4 2 2
 3
x - 2 x + 1 dx + ò ( x + 1) + 1dx = 1 + 2 -1  I = 2
1 -1
82. Let P and Q be two points t1 and t2 respectively whose abscissas are in the ration m : 1.
at 2 m
\ 1 = or t1 = t2 (m)
at22 1
If (h, k) be the point of intersection of tangents at P and Q, then h = at1t2 k = a (t1 + t2 )
Or h = a (m)t22 k = a(1 + m )t2
2
Eliminating t2 we get the required locus as y 2 = ax m1/4 + m-1/4( )
83. We first use partial fraction decomposition on this function. Doing so gives us

5 x 2 - 2 xy + y 2
=
5 x 2 - 2 xy + y 2
=
(
3x 2 - 2 xy + 3 y 2 + 2 x 2 - y 2 )
x2 - y2 ( x + y )( x - y ) ( x + y )( x - y )

=
(
x 2 + 2 xy + y 2 + 2 x 2 - 2 xy - y 2 ) + 2 = ( x + y)2 + 2( x - y)2 + 2
( x + y )( x - y ) ( x + y )( x - y )
x+ y 2( x - y )
= + + 2.
x- y x+ y
We can then apply AM-GM to the first two terms to get
x + y 2( x - y )
+ ³2 2.
x- y x+ y
Thus, the minimum is 2 + 2 2 , which is achieved when x, y satisfy the equation
( x + y )2 = 2( x - y )2 , which has solutions y = (3  2 2 ) x . When y = 3(3 - 2 2 ) x
and x > 0 , then condition x > y > 0 is satisfied.
84. Since MN is tangent to C1 at M, NMQ = MPQ . Since MN = PN, DMNP is isosceles
so MPN = PMN . It follows that NPQ = PMQ . But MN is tangent to C2 at N, so
NPQ = MNQ . Hence, MNQ = PMQ . Combining this with the fact that

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PQ QM
NMQ = MPQ , we see that DPMQ  DMNQ . Then = , so
QM QN
QM 2 = PQ.QN = 3.2 = 6 .

{ }
85. Since { x} + x 2 = 1 , the value x must satisfy x + x 2 = n for some integer n. The
quadratic equation the gives us
-1  1 + 4 n
x=
2
-1 + 1 + 4 n
If we consider when 0 £ x £ 8 , then we must have £ 8 . Solving the
2
inequality, we find that x satisfies the equation when 0 £ n £ 72 , giving us 73
-1 - 1 + 4n
possibilities. Likewise, when -8 £ x < 0 , we must have ³-8 , which has
2
solutions when 0 £ n £ 56 , for a total of 57 possibilities.
{ }
Since { x} + x 2 < 2 , we must also eliminate the cases when { x} + x 2 = 0 , which { }
happens only when -8 £ x £ 8 is an integer, for a total of 17 possibilities.
Therefore, the total number of solutions is 73 + 57 – 17 = 113 .
2 2 2 2
86. We know that, 
a +
b +
c =a +
b +c +2 
a .
b +
b .
c +
c .
a ( )
ïìï  2  2  2 ïüï
í2 a + b + c ý - 2 
ïïî ïïþ
a .
b + (
b .
c+
c .
a )
2 2 2 2
=
a -
b +b -
c +c -
a  9 = 3´ 3 - 
a +
b +
c

 
a + b + c=0  
b + c = - a
2
a + 5 b + 5 c = 2 ( a +5 ) b + c = 2 ( ) a + 5 - a -3
a =3
     
87. é c d a ù b - é c d bù a + é d b a ù c - é d b c ù a + éb c a ù d - éb c d ù a + k a = 0
ëê ûú ëê ûú  ëê  ûú ëê ûú ëê ûú ëê ûú
Taking dot with b ´ c ,
            
-3 éêb c d ùú éê a b c ùú + k éê a b cùú + éêb c a ùú éêb c d ùú = 0
ë ûë û ë û ë ûë û
 
é ù é
 -48 ê a b c ú + k ê a b c ú = 0 ù  k = 48
ë û ë û
88. Let N be (3l + 6,2l + 7, -2l + 7) such PN is perpendicular to the line.
Then l = -1 \ N = (3,5,9)
\ PN = 7
89. n( F ) = 38, n( B) = 15, n (C ) = 20
n( F È B È C ) = 58, n ( F Ç B Ç C ) = 3
n ( F È B È C ) = n ( F + n ( B ) + n (C ) - n ( F Ç B ) - n ( F Ç C ) - n ( B Ç C ) + n ( F Ç B Ç C ))
 n ( F Ç B) + n ( F Ç C ) + n ( B Ç C ) = 18
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 SRI CHAITANYA IIT ACADEMY, INDIA 13‐01‐23_ Sr.S60_NUCLEUS & STERLING_BT_ Jee‐Main_GTM‐06_KEY &SOL’S
a, denote number of men who got medals in foot ball and basket ball.
b, denote number of men who got medals foot ball and cricket.
c, denote number of men who got medals basket ball and cricket.
d, denote number of men who got medals in all the three sports.
 a  d  b  d  c  d  18 (d = 3) or a  b  c  9 .
90. 175 = 52.7.245 = 5.72.875 = 537.1715 = 5.73
Let a = log5, b = log 7
log175 2a + b log875 3a + b
a= = b= =
log 245 a + 2b log1715 a + 3b
1 - ab (a + 2b )(a + 3b ) - (2a + b )(3a + b )
=
a - b (2a + b )(a + 3b ) - (a + 2b )(3a + b )

=
(
5 b 2 - a2 )=5
b 2 - a2

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 SRI CHAITANYA IIT ACADEMY, INDIA 17‐01‐23_ Sr.S60_NUCLEUS&ALL_BT _ Jee‐Main_GTM‐07_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 17-01-2023
Time: 09.00Am to 12.00Pm GTM-07 Max. Marks: 300
KEY SHEET
PHYSICS
1) 1 2) 4 3) 4 4) 3 5) 3
6) 2 7) 1 8) 2 9) 1 10) 2
11) 3 12) 1 13) 3 14) 1 15) 2
16) 2 17) 4 18) 2 19) 2 20) 1
21) 18 22) 1 23) 16 24) 48 25) 136
26) 1 27) 2 28) 3 29) 2400 30) 412

CHEMISTRY
31) 1 32) 3 33) 1 34) 3 35) 1
36) 2 37) 2 38) 1 39) 2 40) 3
41) 4 42) 4 43) 1 44) 3 45) 2
46) 4 47) 1 48) 3 49) 2 50) 1
51) 42 52) 30 53) 1 54) 1 55) 3
56) 0 57) 4 58) 2 59) 24 60) 2

MATHEMATICS
61) 4 62) 3 63) 4 64) 3 65) 3
66) 3 67) 2 68) 3 69) 4 70) 4
71) 3 72) 1 73) 2 74) 3 75) 1
76) 2 77) 1 78) 4 79) 4 80) 4
81) 75 82) 0 83) 1062 84) 5376 85) 1
86) 2 87) 12 88) 1552 89) 12 90) 2

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SOLUTIONS
PHYSICS
1.
0 0 0
1 1 1
0 1 0
1 0 0
From the truth table
Y = A.B
So it is “AND” gate
2
h1  d 1 
2. d  2 Rh d  h     h1  900 m
h d 
h
3. Given,  
mv0
Velocity of an electron after time ‘t’
E  e  Ee
V  V0  t  V0  t  Wave length
m m
h h 0
  
mv  Ee 
1
eE0
m  v0  t t
 m  mv0
4. A  A0 e  t
A 1 t 30
Then  4 T   7.5
A0  2  t /T
T 4

Y V

E q
X


B
5. Z
E0
C
B0
FE  E0 q
FB  q B0
FE C
  10
FB V
2 tan 
6. R.P  
1.22
R.P medium  m 2
 
R.P air  air 1
1
7. tan  1   tan 
cos 
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1
tan 60   tan 
cos 450
 3
Actual dip   tan 1  
 2
 R   4   3  16  3 J
2
8. H  DC   I DC
2


H  AC    I rms   R1  4 / 2  2  16 J 
2 2

 ratio of heat produces is 3:1

 r1 E r2

I V1  0

9. R
  2
I  ____(1)
R  r1  r2 R  r1  r2
But V1    Ir1 _____(2)
Given V0  0 _____(3)
From (1),(2) and (3), R  r1  r2
I
10. T  2
MBH
2
M 1 I1  T2  8
   
M 2 I 2  T1  3

3RT
11. A. RMS velocity 
M
T is same & M is same so, RMS velocities will be same
So, A is correct
B. n1 : n2  1: 4
n1 RT1 n2 RT2
p1 : p2  :
V1 V2
n1 : n2  1: 4
T1  T2 ,V1  V2 
So, B is correct
C. P1 : P2  1: 4 and not 1 : 1
So, C is wrong
D. rms velocities are equal so D is wrong
ax
Q ax q0 Q
12.
KQq0 KQq0
Fnet   towards right
a  x a  x
2 2

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 SRI CHAITANYA IIT ACADEMY, INDIA 17‐01‐23_ Sr.S60_NUCLEUS&ALL_BT _ Jee‐Main_GTM‐07_KEY &SOL’S
When displacement is x towards left
KQq0 KQq0
Sq Fnet    x
a  x a  x
2 2

 a  x  2   a  x  2 
  KQq0     KQq 4ax
 a2  x2 
0
2
a4
KQq0 4ax Qq0
  x
4 0 a 3  0 a 3
F Qq0
a  net   x
m  0 ma 3
2  0 ma 3 4 3 0 ma 3
T  2 
 Qq0 Qq0
L L L K A 1
13. R1 : R2  1 : 2  1 . 2 . 2  2.9.  9
K1 A1 K 2 A2 L2 K1 A1 2
Let junction temperature be T
T  450 T  0
Then ,  0
R1 R2
R1 R2  450 0 
T    
R1  R2  R1 R2 
1  450 R2  9 450
T       450 C
1  R2 / R2  R1  10 9
14. A. Small temperature difference allows use of newton’s law of cooling
dQ
 kA   0 
dt
dQ
 0  is doubled  doubled
dt
dQ dQ
B. :  TA4 : TB4  2834 : 2934
dt P dtQ
4 4
 293   10 
 1:    1:  1  
 283   283 
40
 1:1  = 1 : 1.15
283
[considering same emissivities]
15.

150
100 125

Initial
Effectively , 25 cm column of water from top of right vessel entered the left G  mgh (h is
height reduced of the COM)
 16 25103 g  25 102  1 J

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1
16. Total mechanical energy   (potential energy)
2
[ for circular orbits under central forces]
SQ T .M .E A : T .M .EB
GMm1 GMm2
 :
2r1 2r2
 m1r1 : M 2 r2   4m 4r  :  3m 3r  = 16 : 9
dm
17. Force needed     0.5 5  2.5
dt
Power needed  F  2.5  5  12.5W
N

f f

18. mg

N – mg = 0, f  N
f = ma  a   g  4 m / s2
22  02
u = 0,   2   u  2as  s 
2 2
 0.5m
2 4
19.    ML2T 2
 M L 2T
 2   % error = 5% + 2(5%) – 2 (-5%) = 25%
 M L T
20. Consider downward –ve
Then, u = - 100 a = -10
  100  10t and after collision , the velocity becomes zero from – 200 almost suddenly
so, option A is correct
21. F  12t  3t 2 ;  1.5 12t  3t 2 
1.5 12t  3t 2  d t3
  4t  t 2   4t  t 2     2t 2 
4.5 dt 3
To change the direction of motion, pulley need to come to rest momentarily,
t3
2t 2   0  t  6sec
3
d t3 2t 3 t 4 18
 2t 2        t 6sec  36 rad  rev
dt 3 3 12 
2u 2u sin 
22. Given, T1  T2  1  2
g g
H1 u12 2g
u1  u2 sin    2 2 1
H 2 2 g u2 sin 
K
23. Initial maximum velocity at mean position, v1  A11 , 1 
m1
By LCLM, m1v1   m1  m2  v2
m1v1 K
V2   A  , 
 m1  m2  2 2 2  m1  m2 
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v1 A11 A v  m  m2  m1 A m1  m2 1.024 10.24

  1  1 1   1   
v2 A22 A2 m1v1 m1  m2 A2 m1 0.9 9
A1 3.2 
   3.2  3.2  3    16
A2 3  1

F x 18 104  60 102 18 103

24.  ,   25 109   x   48 106 m  48 m
A l 15  60 104  x 15  25
i i 1.7 108 1
25. E  J  ; F  Ee  F  e  6
1.6 1019  136 1023 N
A A 2 10
  q  x l
2
26.  E.ds  0  E.2 xl  0
x 2 0
E ; Given, x   E  1V / m
2 0 

27. I A  I  4 I  2 4 I 2 cos  5I
3

I B  I  4 I  2 4 I 2  cos  7I  I B  I A  7 I  5I  2 I
3
V2 100 100 50 1
28. P R  200 i 2 R  50  i 2  i  A
R 50 200 2
R C

200V1 50 Hz

1 1 1 V 200
xC   i  
C 100 C 2 Z X C2  R 2
 400 
2
 X C2  200 2  X C  100 12
1 100 50
 100 12  C  F  F
100 C  12  3

S l S 50
29.     S  2400
R 100  l 5600 700
30. 1 MSD = 1 mm,
10VSD = 9 MSD  1VSD = 0.9 mm
LC = 1MSD – 1VSD = 0.1 mm = 0.01 cm
Zero error = +4 divisions
MSR = 4.1 cm, VC = 6
Diameter = MSR + (VC – zero error) LC
 4.1   6  4  0.01  4.12cm  412 102 cm

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CHEMISTRY
80 0.36
31. % Br   100  34.04
188 0.45
32.  PhSO2Cl  Hinsberg reagent
Ph  C  NH 2 Ph  N  C  O
O
Hoffmann bromamide reaction
 R  NH 2  carbyl amine reaction
i amine

 Saytzeff product x hoff man product

33. Aldoses give Seliwanoff test slowly and furfural has 5-carbon atoms
So aldopentose
34. Morphine is narcotic drug. Saccharin is 550 times sweeter than sucrose chloroxylenol is
antiseptic phenelizine in antidepressant
35.
CH 2OH CH 2  Cl CH 2  I


 HCl   
NaI

OH OH OH
A B

36. Novolac  PhOH + HCHO

Glyptal  Glycol + phthalic Acid
Buna –s  Butadiene + Styrene
Dacron (or) terylene  Terphthalic acid + ethylene glycol
37. Excess sulphates in water have laxative effect
NO3  methemoglobinemia
Pb2  Kidneydamage
F   Brown mottling of teeth
38. Cis –(10)- annulene is not aromatic due to lack of planarity
39. In neutral solutions I  is oxidized to IO3 by KMnO4
So statement –I is false
In MnO42 P  d  bonding is present
O O
O
Mn Mn
O
O
O
40. O

41. BeCl2 & AlCl3 acts as lewis acids and Be  OH 2 , Al  OH 3 are amphoteric
42. Pyrophosphoric acid H 4 P2O7
43. Calcination and leaching are used in concentration of ore not in the purification of metal
44. H 2O2 is used as OA & RA in both medium and dH 2O2 : 1.44 gm / cc , d D O  1.1059 gm / cc 2

45. Same B.pt means same concentration

2 1000 8 1000
  
M A 100 M B 100

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MB
4 MB  4 MA
MA
46. IE1 of Zn is more than Ga due to stable E.C
47. With increase in nuclear charge orbitals come closer to nucleus and their energy
decreases
48. Gases with greater inter molecular attractions are easily liquifiable have higher TC and are
readily adsorbed
250 10.8
49. amount of ‘C’ in solutions   2.5 10.8  27 gm of ‘C’
100
180 gm of glucose has 72 gm of ‘C’
27 180
Amount of glucose with 27 gm of ‘C’   67.5
72
67.5 1000 67.5 1000
m    2.0548 gm
180  250  67.5 180 182.5
50. O2 , Cu 2 , Fe 3 are paramagnetic NaCl, H 2O are diamagnetic
So both a statements are correct
12.6 126
51. % purity  100   42
30 3
60 50
52. Overall yield   100 = 30%
100 100
53. 1:1 electrolyte means
CO  NH 3 4 Cl2  Cl
Primary valency is 3  CO 
3

54. C2O42 is oxidized to CO2

O.NO changes from 3 to 4
55.
Species CN  NO  O2 O2 O22
B.O 3 3 2 2.5 3
3
2.34 10 1000
56. M   3 104  solubility
78 100
 
3
K sp of Ca F2  4 S 3  4  3 104
 4  27 1012 m3  108 1012 m3  0.0108 108 M
57. KO2 , NO2 , ClO2 , NO are paramagnetic
58. U  nCv T 5000  n (20.785-8.314) (500-300)
25
n  2.0046  2
12.471
59. K 2Cr2O7  6 Fe 2  6 Fe 3  2Cr 3
MV M V
 K2Cr2O7  1 1  2 2 Fe2
1 6
 
20  0.02 M 2 10
 M 2  0.024 M  24 102 M
1 6
60. From 2 to 1 when initial pressure of ‘NO’ is doubled by keeping PH const, initial rate 2

increases by four times so order w.r.t ‘NO’ is 2.

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MATHEMATICS
61. Circle is x  y  2 gx  6 y  19c  0
2 2

It passes through (6,1) 36  1  12 g  6  19c  0

12 g  19c  43(1)
Line x  2cy  8 passes through Centre S  g , 3
g  6c  8
By (1) & (2) g = 2, c = 1
 Circle is x 2  y 2  4 x  6 y  19  0
 x-intercept  2 a 2  19c  2 4  19  2 23
62. Since f ( x) is continues at x  4
 f (4)  f (4)
4 3 4
16  4b    5 | t  3 | dt   (2  t ) dt   (8  t )dt
0 0 3
2 3 4
 t   t2  9  9
  2t     8t    6   0  (32  8)   24  
 2 0  2 3 2  2
1
 16  4b  15  b 
4
 x

   5  t  3  dt , x  4

f  x   0
 2 x
 x  4 , x4

5  x  3 , x  4 8  x , x4
 
f ' x    1   1
 2 x  , x  4 2 x  4 , x  4
 4
 1
f  x  is decreasing in  ,    8,  
 8
63. 2 x  ky  5 z  1, 3kx  ky  z  5 are two perpendicular planes
6k  k 2  5  0
k  1,5
But k  3  k  1
 2 x  y  5 z  1, 3x  y  z  5
P   2 x  y  5 z  1    3 x  y  z  5   0
 x  3  2   y 1     z    5    5  1
1
x- intercept = 1  5  1  3  2   
2
5  1
y  intercept = 7
1 
64. A 1,1 B  4,3 C  2, 5 
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Area  ABC   18
Let P  ,   lies on BC
1 1 1
1 1
Area APB    1  2  5  7
2 2
4 3 1
Area APB 4 144
Given   2  5  7 
Area ABC 7 7
144
 2  5  7   ------(1)
7
Equation of AC is 2 x  y  1  0 ----- (2)
1 
It cuts x-axis at M  ,0 
2 
Equation of BC 4 x  y  13  0 --- (3)
Solving (1) & (3) we get
 36 53   20 11 
P ,  or  , 
 7 7   7 7 
Since x  coordinates of B,C are 4 and 2 respectively
 20 11 
 P ,   4  x  intercept of P  2
 7 7 
1
Equation AP is 2 x  3 y  1  0 y  0x 
2
 1 
Let N  ,0 
 2 

1 1 1
1 1 1
 Area of  NAM   0 1
2 2 2
1
0 1
2

65.
h
in ABCtan2 
x
2h
in  ADE tan  
x  7h

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2h 2 tan 2
 tan   
h
 7h 1  7 tan 2
tan 2
4t 4t
tan   t t   ,t  0
1  t2 1  t  2 7t
2

2t
1 7
1  t2
 1  t 2  2 7t  4  t 2  2 7t  3  0
2 7  28  12 2 74
t   72 t  7  2
2 2
66. truth table
P q ~p ~q p^~q
T T F F F
T F F T T
F T T F F
F F T T F
 p ^ r    p^ ~ q   ~p  p ^ q
T F F T T T
F T F T F F
F F T F T/F F
F F T F T/F F

ris equivalent to q.
    
67. a  2i  j  5k b   i  j  2k
 iˆ ˆj kˆ 
   
 
a  b   2 1 5   iˆ  2  5   ˆj  4  5   kˆ  2   
  2
 
i j k
 
 
a  b  i  2  5 5  4   2  (  2) j  (4  5)k
1 0 0
  ˆ ˆ  ˆ ˆ
 ˆ
[( a  b )  i ]  k   (  2) j  (4  5 ) k   k
 
23 3
 4  5 
2 2
ˆ ˆ ˆ
i j k
 
b  2 j    2  4i  2k
0 2 0
 ˆ
| b  2 j | (4)2  (2)2  16  4 2

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2
 3 
 16  4    16  9  25  5.
 2 
68. Terms divisible by 7 between 10000 and 99,999 are
{10,003;10,010;99,995}
No. of terms  12857
Terms divisible by both 7 and 5 is which are divisible by LCM(7,5)  35
 these are {10,010;10045;9995}
No. of terms  2572
12857  2572 1.0285
p 
90000 9
9 p  1.0285
69. y 2  8x 4A = 8, A = 2
A 2
P(a, b) bca Point or the Parabola y  mx   y  mx 
m m
S  x 2  y 2  10 x  14 y  65  0
2
C (5,7) lies on y  mx 
m
2 2
7  5m   5m 2  7 m  2  0 m  1,
m 5
y  8 x diff. both sides w.r.t ‘x’
2

dy dy 4
2y  8  m
dx dx y
dy 4 2
at P (a, b)   1,  or 
dx b 5
From (1) b  4 or 10
25
b2  8a  a  2,
2
 25
A  2    25
 2 
B  4  10  40 A  B  65

70. a  i  j   k
 ˆ ˆ ˆ
b  3i  5 j  4 k
 
a  b  i (4  5)  j (4  3)  k (5  3)
Given
  ˆ
a  b  i  9 j  12 k
By comparison
4  5  1 5  3  12
5  5 5  15

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 SRI CHAITANYA IIT ACADEMY, INDIA 17‐01‐23_ Sr.S60_NUCLEUS&ALL_BT _ Jee‐Main_GTM‐07_KEY &SOL’S

  1   3

a  3i  j  k

b  3i  5 j  4 k
 
b  2a   3  6  i  7 j  6 k
 9i  7 j  6k
 
b  a  0i  4 j  3k
Projection of  b  2a  on  b  a 
   
(b  a )  (b  2a )
  
(b  a )
28  18 46
 
5 5
71. y 2  8 x  4 ; x 2  y 2  4 3x  4  0
Points of intersection   0, 2
2

-2

x2  4 3x  y 2  4 x 2  22 3 x  y 2  12  4  12

 x  2 3  y  4  
2 2
2 2
y  42  x  2 3

 x  2 3   16  y
2 2

  y2  4  
 
2
1
A    16  y 2  2 3     dy  4  12 3  8
2   8  3
dy
72. yx
dx
I.F1  e   e x
 dx

Solution y.I .F   x.e  x dx

y  e  x  x  e  x dx  (1  x)e  x  c
y  1  x  ce x ------ (1)
y1 (0)  0  0  1  0  c  c  1
 y  1  x  e x ------ (2)
y2 (0)  1  1  1  0  c  c  2
 y  1  x  2e x ----- (3)
If (2) and (3) are intersect then
  1  x  e x  1  x  2e x
e x  0 Not possible
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 SRI CHAITANYA IIT ACADEMY, INDIA 17‐01‐23_ Sr.S60_NUCLEUS&ALL_BT _ Jee‐Main_GTM‐07_KEY &SOL’S
 No point of intersection.
  
73. R.H.L = lim f ( x)asin    2  a  2
x 1  2 
L.H.L = lim f ( x)  0  3  3
x 1
lim f ( x) exist a  2  3  a  1
x1

 
 f ( x)   sin  [ x]   2  [  x]
2 
4 1 2 3 4
Now  0
f ( x ) dx   f ( x) dx   f ( x) dx   f ( x) dx   f ( x) dx
0 1 2 3
1 2 0 4
  (0  2  1) dx   ( 1  2  2) dx   (0  2  3) dx   (1  2  4) dx
0 1 2 3
 (1  0)  1(2  1)  (3  2)  (4  3)
 1  1  1  1  2
 /3
 8sin x  sin 2 x 
74. Given I 
/4

  x
 dx

8sin x  sin 2 x
f  x 
x
21 3
2r

4 4 2 1 


 
4 3

5
min area 
12
f  x  is an increasing function
 /3  /3

 f  / 4   I   f  / 3 
 /4  /4

4 2 1 4 2 1 4 2 1
f  / 4   
 /4  /4 1
 4 2  1      /3  /3

      f  / 4   I   f  r / 3
 1   12   / 4  /4

   5
 4 2 1   
 12  12
____(1)
5
I
12
 2023  2    2023  1
2022 2021
75.
  7k1  2    7k1  1  7 N  22022  1
2022 2021

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 SRI CHAITANYA IIT ACADEMY, INDIA 17‐01‐23_ Sr.S60_NUCLEUS&ALL_BT _ Jee‐Main_GTM‐07_KEY &SOL’S

Remainder  2  1 2022

  8   1   7  1  1
674 674

Remainder = 1-1 = 0
5
76. S5  [2a  4d ]  5a  10d
2
9
S9  [2a  8d ]  9a  36d
2
S 5 5a  10d 5
Now  5   
s9 17 9a  36d 17
 17a  34d  9a  36d
 8a  2 d  d  4 a ---- (1)
Now a15  a  14d  57a
given 110  a15  120  110  57a  120
 1.929  a  2.105
10
Now S10  2a  9d   190a from (1)
2
= 380 for a = 2.
V  x  y 2   x  3  y 2  x 2  ( y  6) 2
2 2
77.
 3( x  1) 2  3( y  2) 2  30
V is min at Z 0  1  2i v0  30


2 2
2Z 02  z03  3  v02  2( 3  4i )  ( 11  12i )  3  900
| 8  6i |2 900  100  900  1000.
1 2
78. A  ,  R
 2 5 
Trace  1  5  4,| A | 5  4  1.
A2  4 A  I  0
2 A2  8 A  2 I  0
2 A2  8 A  2 I .
A2  A  2I      2  8  10.
79. i) aR1b  ab  0
reflexive; let a  R
aR1 a  a 20
 R1 is reflexive.
Symmetric; let a, b  R
aR1b  ab0
 ba0
 bR1a.
 R1 is symmetric

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 SRI CHAITANYA IIT ACADEMY, INDIA 17‐01‐23_ Sr.S60_NUCLEUS&ALL_BT _ Jee‐Main_GTM‐07_KEY &SOL’S
Transitive let a,b,c  R
aR1b  ab0 and bR1c  bc0
 ab0 and bc0  ac0
 ac0  aR1c
 R1 is transitive  R1 is equivalence
(ii) given aR2b  ab
reflexive; let a R .
aR2 a  a  a
 R2 is reflexive.
Symmetric; let a, b  R .
aR2b  a  b
 b  a is need not be true
 R2 is not symmetric
 R2 is not equivalence.
80. Given f (a )  
where a  N  {1} and  is the Maximum of the powers of those primes p such that
for a  2 and p  2 then
p  is not divisible by a for any  N
 f is Not 1  1  f  g is Not1  1.
f ( a )   , such that p  divides a.
g (a)  a  1 for all a  N  1
  f  g  a     a  1
now, g (a)  a  1 g : N  1  N
in co-domain is N.
but is 1  a   , 2  a  1  
Range of g is  2,   range is not equal to co-domain hence g is not onto.
 f  g is not onto.
81. Ellipse  x  12  4  y  12    5,   5
 5
a 2    5, b 2 
4
Given latus rectum = 4
2b 2
  4  b 4  4a 2
a
    5   64    5 
2

   5  64    5
   59 a  642

l  length of major axis = 2a = 16

   l  75

82. Z 2  Z  0   x  y   x  iy  0
2

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x  y  x  i  2 xy  y   0
2 2

y  2 x  1  0 x2  y2  x  0
and
y = 0 (or) 2x-1 = 0
of y = 0, x = 0,-1
1 3
of x  , y  
2 2
  Re Z  Im Z 
Z S
1 3 1 3
  0  0    1  0           0
2 2  2 2 
83. f  x   2 x2  x 1
f  x   800  800  2 n 2  n  1  800
 19  n  20
20

 f  x     2n
ns n 19
2
 n  1 , replacing with n – 20
40
   2n 2  81n  819 
n 1

2  40  41 81 81 40  41

   819  40  10620
6 2
 a1 a2 a3 
84. A   a4 a5 a6  , ai  1 or 1
 a7 a8 a8 
Sum of elements of principal diagonal in AT A  6
 a12  a22  a32  a42  a52  a62  a72  a82  a92  6
Number of matrices  9c  26 ( 6 places, each in 2 ways)
6

= 5376
85. Put x 2  t , then M + m = 2
dy y
Given equation 
dt sin t cos t log tan t
sin t  cos t

sin t cos t.log tan t
dt
 sin t cos t log tan t
I.F  e
 log tan t
sin t  cos t
Solutions is y log tan t   dt    sec t  cos ect  dt
sin t cos t
 sec t  tan t 
 log  C
 cos ect  cot t 
   
 ,1 is on this  C   log 3 2  3 
6 
 

When t  , y  1
3

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  
 y    1
 3
86. y 5  9 xy  2 x  0 -----(1)
dy dy dy 9y  2
4 y4  9x  9 y  2  0   4
dx dx dx 5 y  9 x
Tangent is parallel to x-axis
dy 2
  0  y  , not satisfying (1)
dx 9
Tangent is parallel to y-axis, if 5 y 4  9 x  0
10
From (1), y 5  5 y 5  y 4  0
9
5
 y  0,
18
 M  0, N  2, M  m  2

87. Equation y plane passing through the line

4ax  y  5 z  7a  0  2 x  5 y  z  3 is
4ax  y  5 z  7a  (2 x  5 y  z  3)  0 ----- (1)
x  4 y 1 z
The line   lies in this plane
1 2 1
 Substituting (4, 1,0) in ---- (1)
10d  9a  1  0 ----- (2)
D  R ' s of normal of (1) are (2  4a, 5  1,5   )
1(2  4a)  2(5  1)  1(5  )  0
 11  4a  7 ------- (3)
Solving (2), (3)   1, a  1
The equation of plate is 2 x  4 y  6 z  4  0
x  2 y  3 z  2  0  4( from (1))
x3 y 2 z 3
thy first on the line   is P(7t  3,2  t ,3  4t ) , lies on (4), then t  2
7 1 4
P(, , )  (17,0, 5)        12
88. vertices of hyperbola  (0, 8)
For ellipse, B '  (0, 8), B  (0,8)
 2b  16, b  8
49  64
e1 is eccentricity of hyperbola 
8
e2 is eccentricity of ellipse
1
Given e1e2 
2
4
 e2 
113
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 SRI CHAITANYA IIT ACADEMY, INDIA 17‐01‐23_ Sr.S60_NUCLEUS&ALL_BT _ Jee‐Main_GTM‐07_KEY &SOL’S

 16 
For ellipse a 2  b 2 1  e22   64 1  
 113 
2a 2 2  64  97 64  97
d  
b 113  8 113
113d  1552
89.    
cos sin 1 x cot tan 1 cos  sin 1 x     k ,0 | x | 1
2
1  2 x2
 k
1  x2
1  2 x2
  k 2  1  2x2  k 2  k 2 x2
1 x 2

  k 2  2 x2  k 2  1
k 2 1 
x 2  sum of the roots     0   1 ------- (1)
k  2 
x  bx  5  0
2

1 1 
Sum of the roots = 2  2   b ---(2)
  
2
 2 1  b (from(1))

2
2  -----(3)
b 1
Product of roots  5
 1 1  2
  2  2    5   2  1  5  b  4 (from(3))
 
2  k2 1  2 1 b
   2
2
 3k  1  k 
2
 12
5 k 2 3 k2
90. x  15, M  15

x  150 15   xi2
 225
10
xi2  2400
x1   xn2  15  25 xi  10
x new    14
10 10
2400  (25) 2  (15) 2
M new   (14) 2
10
2000
M new   196  4  S  D  2
10

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 SRI CHAITANYA IIT ACADEMY, INDIA 18‐01‐23_ Sr.Super60_(NUCLEUS,STERLING)&LIIT __BT_ Jee‐Main_GTM‐08_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_(NUCLEUS,STERLING)&LIIT _BT JEE-MAIN Date: 18-01-2023
Time: 09.00Am to 12.00Pm GTM-08 Max. Marks: 300
KEY SHEET
PHYSICS
1) 4 2) 4 3) 4 4) 1 5) 4
6) 2 7) 3 8) 1 9) 1 10) 4
11) 2 12) 2 13) 4 14) 4 15) 3
16) 4 17) 2 18) 1 19) 1 20) 2
21) 150 22) 100 23) 40 24) 48 25) 3
26) 4 27) 25 28) 2 29) 30 30) 500

CHEMISTRY
31) 3 32) 1 33) 2 34) 1 35) 1
36) 1 37) 4 38) 3 39) 1 40) 3
41) 3 42) 4 43) 4 44) 2 45) 1
46) 1 47) 3 48) 3 49) 4 50) 3
51) 16 52) 32 53) 6 54) 536 55) 6
56) 55 57) 9 58) 6 59) 7 60) 7

MATHEMATICS
61) 4 62) 2 63) 1 64) 1 65) 3
66) 2 67) 2 68) 2 69) 3 70) 2
71) 4 72) 3 73) 3 74) 2 75) 2
76) 4 77) 3 78) 3 79) 4 80) 4
81) 17 82) 9 83) 6 84) 21 85) 2
86) 8 87) 7 88) 191 89) 6 90) 8

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 SRI CHAITANYA IIT ACADEMY, INDIA 18‐01‐23_ Sr.Super60_(NUCLEUS,STERLING)&LIIT __BT_ Jee‐Main_GTM‐08_KEY &SOL’S

SOLUTIONS
PHYSICS
1. Independent of R.
Intensity of radiation at sphere
The radiation pressure acting on sphere =
speed of light
Po
= where P0 is the power radiated by sun
4  x 2c
Po
Force on sphere due to radiation pressure =  R2
4 x c
2

GMm
Gravitational pull on sphere =
x2
The gravitational force and the force due to light pressure both decrease with the square
of the distance from the Sun. If a ball of radius R floats, it must be in neutral equilibrium
and will float at any height above the sun. This is quite independent of the radius of the
sphere.
2. The plot shows a linear relationship between x and log y, which means that y is an
exponential in x. Since y decreases as x increases, the answer must be (4).
3. Gravity exerts a force mg downward on the block, which means that the wedge must
exert a force mg upward on the block. Thus, the block exerts a force mg downward on the
wedge, and gravity also exerts a force mg downward on the wedge. Since these forces
have no horizontal components, no friction with the ground is necessary to keep the
wedge static.

 1 
4. X  L – . Hence the graph of X vs. f is . Therefore the plot of
 C 

magnitude of X, that is |X| vs. f is

5. The gravitational force vanishes at the midway point between the planets, so the rocket
only needs to have enough energy to get there. The initial and final gravitational
potential energies are
GMm GMm 4GMm 2GMm GMm
Ui     and U f   
R 3R 3R 2R R

Thus, the initial kinetic energy needed is

1 2 GMm
mv  U f  Ui 
2 3R
2GM
which implies v 
3R
6. Momnetum of system is zero.
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tan 
7.  cos 
tan  '
tan 
tan  ' 
cos 
since cos   1
 tan  '  tan 
'
8. Sky wave signals are used for long distance communication. The sky wave signals are
less stable than ground wave signals because of the variation of state of ionosphere.
9. The potential difference across the inductor is e = E– iR.
Hence the plot of e versus i is a straight line with negative slope.
10. Since both charged particles move along same straight line, the magnetic field due to one
particle at location of other is zero. Hence there is no magnetic interaction amongst the
charged particles.
11. At any time t, the charge on right capacitor be q. Applying Kirchoff's law
Q0  q q Q0  2q dq
iR  
C C 2CR dt
integrating and evaluating the constant we get
i
R
Q0-q q
C C

2t
Q0  dq Q0  22RCt
Hence q  (1  e 2 RC
) or i  e
2 dt RC
12. From symmetry of induced charges on sphere, the net electrostatic force Fe on induced
charges on sphere due to point charge is along line joining centre of sphere and point
charge as shown.

13.
Given NED  30
 BED  120
BCDE is cyclic quadrilateral
 BCD  60
The line CE will be angle bisector of BCD
a BE a 3 2
 BE = a tan30º =  now tan i   
3 AB a / 2 3
2 2 1 4
 sin i  now by snell’s law 1  sin i  n sin r  n  n
7 7 2 7

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14. Net heat absorbed by one mole of diatomic gas in going from A  B (isochoric process)
and B  C (isobaric process) is
5 7
Q  CV T  CP T  RT0  RT0 Q  6 RT0
2 2
2T
 T   2  1  gh
r
15. P0  1 gh  2 gh   P0
r 2
16. From thermodynamics second law, the relation between COP of a Carnot refrigerator and
heat pump is given as
 COP heat pump   COP REF  1
17. The
 angular momentum of disc about point A is
  
LA   cm  m r  vcm

 1
vcm = velocity of centre of mass of disc = 0.  L  I cm  MR 2
2
18. From archimedes principle statement-2 is correct explanation of statement-1.
19. mT  const. Fu; r  ln  m  ln T  C
d  m dT d m dT
 0  
m T m T
\ =
d m 1
Now  1%   (–ve sign indicates decrease)
m 100
dT = 1 (given)  T = 100 K.
20. The magnitude of phase difference between the points separated by distance 10 metres
= k×10 = [10 × 0.] ×10= 
1
21. mVm2  15  103 Vm  0.150 m / s
2
g
A  0.150 m / s Lqm   0.150 m / s
L
0.150 0.150
gL  3
  1.5 m
100  10 0.1
my A4 A4
22. We have K   .Q  K   .Q  48  .50  A  100
m y  m A A

23. Strain    

 
  T  10 –5  200  2  10 –3
Stress = Y (strain)
Stress = 1011 × 2 × 10–3 = 2 × 108 N/m2
 Required force = stress × Area = (2 × 108) (2 × 10–6) = 4 × 102 = 400 N

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400
 Mass to be attached = = 40 kg
g
24. Let the initial velocity given to the mass at A be u.

Then the velocity of mass at B is u/2

As the system moves from initial the final position
Increase in potential energy is = 4 mgl + 2mgl
2
1 2 1  u 5
Decrease in kinetic energy = mu  m    mu 2
2 2  2 8
From conservation of energy
5 2 48
mu = 6 mgl or u g
8 5
25. n  t    1
18  106
30 × 600 × 10 = 36 × 10 × ( -1) =
-9 -6
 .5
36  106
   1  .5    1.5
8 1F 10V 8 1F 10V
4 2 4 1

4V 6V 6V 6V

26. (Moderate) 
8 6 4
 
E 4 2 4  6V ; Q  4V  1 F  4C
1 1 1
 
4 2 4

27. e  (v  B). 
e  [ iˆ  (3iˆ  4 ˆj  5kˆ)] .5 ˆj
 e  25 volt
c 0 E02 3  108  8.85  1012  362
28. I av  
2 2
 1.72W/m 2

29. Vin  V  VZ
band width
30. n
signal band width

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CHEMISTRY
31. Conceptual
32. Conceptual
33. Conceptual
34. Negative pole of water molecules will attract the positive charged rod and positive pole of
water can attract negative charge of rod. Hexane is non polar.
35. Water sample is reported to be highly polluted if BOD value of sample is more than 17
ppm
36. Due to steric hindrance BrCl3 cannot be formed.
37. Cyclopentadienyl anion carry a negative charge. Since the complex contains three
negative groups(2 cyclopentadienyl anion ligands and one BF4 counter ion) the oxidation
state of iron is +3
38. I) Incorrect. As the reaction is exothermic, increase in temperature decreases Kc.
II) Correct. Addition of an inert gas at constant pressure drives the reaction in the
direction of increasing number of moles.
III) Correct. Increase in pressure moves the reaction in the direction of decreasing
number of moles.
IV) Incorrect. Value of Kc does not provide any information on rate of the reaction and
hence regarding catalyst.
39. Refer NCERT
40. Conceptual
41. The colloidal particle formed is AgI | I , thus Pb2 will be the most effective in
coagulating it.
42. Refer NCERT
43. Diamnine and dicarboxylic acid must give a polyamide. The connectivity must be
-NH-ring-NH-CO-ring-CO- or the reverse at meta positions of both the rings. Only 4
satisfies this criterion
44. Proline does not have primary amino group while histidine has. With NaNO2/HCl,
histidine gives effervescence of nitrogen.
Serine has an alcohol group and hence gives red colouration with ceric ammonium
nitrate.
Tyrosine has phenolic group and hence would give characteristic colouration with FeCl3.
As both Lysine and glutamic acids both have acidic groups, they both would evolve CO2
with NaHCO3 and hence can’t be differentiated.
45. E2 elimination. Zaitsev product
46. Equanil is a tranquilizer
47. Wolf-Kishner reduction selectively reduces carbonyl group.
48. The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than carbon
atom of the carbonyl group present in propanal. The polarity of the carbonyl group is
reduced in benzaldehyde due to resonance as shown below and hence it is less reactive
than propanal.
49. Refer NCERT
50. The gausche form of N2H4 is stable due decrease in setric repulsion of lone pairs and
nitrogen atoms
51. Conceptual
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s
52. cos  
s 1
s
0.47 
s  0.47
53. Except I2Cl6 the remaining are angular with oxygen bridge.
Initial Mass 1.43  1.5
54. As the gas behaves non-ideally, Final density =   536 g / L
Final Volume 0.004
55.
At the first nearest position there are 8 atoms while at the second nearest position there
are 6 atoms.
56. MnO2 + C2O24- + 4H+ ¾¾ 2Mn2+ + 2CO2 + 2H2O
2MnO-4 + 5C2O24- + 16H+ ¾¾ 2Mn2+ + 10CO2 + 8H2O
0.1´ 30.0
2 ´ nC O2- (unused) = 5 ´ nMnO-  2 ´ x = 5 ´  x = 7.5 ´10-3 mol
2 4 4 1000
1.89
nC O2- used) = - 7.5 ´10-3 = 7.5 ´10-3 = nMn (in pyrolusite)
2 4 126
7.5 ´10-3 ´ 55
 % of Mn= ´100 = 55%
0.75

57.
CH4  10 OH  CO32  7 H2O  8 e 
1 80  3600  1
No.of Faradays required  
8 96500
1 80  3600  1
VCH4    24 L  9L
8 96500

58.

59. A dodecapeptide is made up of 12 amino acid units. On hydrolysis they take up 12 moles
of water.
89  x
 100  52.9  x  7
980  11  18
60. Primary amines give carbylamine test.

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MATHEMATICS
n 2  n 1 1
x  x 
g n  e  d  x   n 2  n  1  ex / 2  dx
x /2
61.  0 2  0 2 
1
x 
  n  n  1  ex / 2  dx
2
 n 2  n  1  4  2e1/2 
0 2 
So, minimum value is 12  6 e
2
et t 5 e
62. f 't    f ' 1 
t 
2
4
 2t 2  2 25

dv 1 ev v 2
Let x 2  v   2x   dv
dx 2 v 2  2v  2 2  
  2
1 v 1 2v  2 1 ev 1 ex
 e  2    
 v  2v  2 v  2v  2    2 v  2v  2 2 x4  2 x2  2
2 2
2 2

t
 1 1
2 2
1 ex et e 1
 f t    
 t 4  2t 2  2 2  2  f 1  
2 x4  2 x2  2   10 4
0

e e 1 7e 1
 f 1  f ' 1 
   
10 25 4 50 4
2
dt 1 t sin 2022 tdt
63. Let x  t   2 x  I   2022
2

dx 2 0 cos t  sin 2022 t

1
2
 2  t  sin 2022 t 2
 sin 2022 t
I
2  cos
0
2022
t  sin 2022 t
 2I   cos
0
2022
t  sin 2022 t
2 2
 sin 2022 t  cos 2022 t
 2I  4   2I  4 
0
cos 2022 t  sin 2022 t 0
cos 2022 t  sin 2022 t
2
2
 4 I  4   dt  I  .
0
2
dy  x  2   y  2
64. 
dx  x  2   y  2
Put x  2  h, y  2  k
dk h  k dv 1  v
 put k  vh  v  h 
dh h  k dh 1  v
1 v 1

1 v 2
dv  
dh
h
 tan 1 v  ln 1  v 2  ln h  c
2
 
 y  2  1   y  2 
2

 tan 1   ln 1   ln  x  2  c
 x  2  2   x  2 2 
 it passes through (3, 2)
1
 tan 1 0  ln1  ln1  c  c  0
2
 it also passes through (p + 2, 3)
 1 1  1  1
tan 1    ln  1  2   ln p
 p 2  p 
 2 tan 1    ln 1  p 2
 p
 
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65. From the diagram, AOB  BOC  COD  60

 YOD 
6

 Let X be the centre of right hand circle, OX sin 30  a  OX  2a

r
Now r  OY  2a  a a 
3
1 1
66. a 2  , b 2  let P be  x1 , y1 
4 9
1x 1 y 5
Equation normal  
4 x1 9 y1 36
5 y1 5 1
It cuts y-axis at  0,  1  
5y 1
  y1   x1 
 4  4 8 3 2 3 4
5
Thus  
36
1 5 5 25
Area of OAB    
2 36 8 3 576 3
67. All the coefficient of odd powered terms is an odd number of odd degree terms multiplied
together and all coefficient of even powered terms has an even number of odd degree
terms multiplied together. Since every odd degree term is negative and every even degree
243
is positive so we sum is just Q  1  P  1 
5
.
32
68. A3  4
G2  5
5
H1 
4
So equation is  x  A3   x  G2  x  H 1   0
69. Conceptual
70. Sum  51C1  52C2  53C3  ....  99C49
 coefficient of x in 1  x  x 1 1  x  x 2 1  x  ......  x 48 1  x 
51 52 53 99
 
 1 x 49

1  x 51    1
 x 
 coefficient of x in
 1 x 
  1
x 
 coefficient of x in  x 48 1  x   x 1  x 
100 51
 100C9  1 100 C51  1
71. f   x   f  x for the condition given in option (1)
f   x   f  x for the condition given in option (2)

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f  x  T   f  x for the condition given in option (3)
Hence (4) is the correct option.
72. Combined mean,
n1 x1  n2 x2 200  25  300  10
x   16
n1  n2 500
Let d1  x1  x  25  16  9 and d 2  x2  x  10  16  6

2 
  
n1  12  d12  n2  22  d 22  
200  9  81  300 16  38

33600
 67.2
n1  n2 500 500
1
 m 1
m  1 
m m
1
 1 1
73. lim  am     1   
m
  
m    m  m   e 1
h
74. In ECD, tan 3 
CD
 CD  h cot 3 ____________ 1
h
In EBD, tan 2 
BD
 BD  h cot 2 ____________  2
h
In EAD, tan  
AD
 AD  h cot  ____________  3

From equation (2) and (3), we have AD  BD  h cot   h cot 2

 AB  h  cot   cot 2  ____________  4
From equation (1) and (4), we have BD  CD  h cot 2  h cot 3
 BC  h  cot 2  cot 3  ____________  5
cos  cos 2 sin  2   

From equation (4) and (5), we have
AB

h  cot   cot 2   sin  sin 2  sin  sin 2
BC h  cot 2  cot 3  cos 2  cos 3 sin  3  2 
sin 2 sin 3 sin 2 sin 3
sin 3
  3  4sin   3  2 1  cos 2   1  2 cos 2
2

sin 
75. Define A as the number of the elements in S, we have
 2006 2005  2006 2003  2006
A  9   9  .....   9
 1   3   2005
2006
 2006 2006 k 2006
 2006
On the other hand,  9  1 2006     9 and  9  1 2006
     1 k 92006 k
k 0  k  k 0  k 

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 2006  2006  2006 1
So, A   9 2005   92003  .....   9  102006  82006
 1   3   2005 2
76. Since each has equally 9 different possible results for A and B to draw a ball from the
packet independently, the total number of possible events is 92  81. From a  2b  10  0
we get 2b  a  10. We find that when b  1, 2,3, 4,5 a can take any value in 1, 2,3,.....,9 to
make the inequality hold. Then we have 9  5  45 admissible events
When b = 6, a can 3, 4, ….., 9 and there are 7 admissible events
When b = 7, a can 5, 6, 7, 8, 9 and there are 5 admissible events
When b = 8, a can 7, 8, 9 and there are 3 admissible events
When b = 9, a can 9 and there are 1 admissible events
45  7  5  3  1 61
So, the required probability is 
81 81
77. ( a, a )  R
If (a, b)  R  (b, a)  R
If (a, b)  R,(b, c)  R  (a, c)  R
78. Differential equation can be written as,  p  x p  2sin x 2 p  cos x  0 which has solution
as  2 y  x 2  c   y  2 cos x  c 2 y  sin x  c  0
 A  1 P  A  B 1 x
79. P    2  1  x2  x 1  0
 
B P  
B x

5 1 1  5 5 1
 x or x   x is positive x
2 2 2

80. Equation of AG
4 1 3
y4  x  0 ; y  4    x
05 5
3
5 y  20  3x; y   x  4
5

4
Equation AC y   x4
5
PQ 10
Equation EF, y  2 x  7  
EF 91
81. Conceptual
 2  2cos x cos 2 x 
 x  3
  x  3
82. 
 lim 3  2 cos x cos 2 x
x 0
  x 2 
is of the form1
lim 
 it is equal to e x0 x2 

2  2 cos x cos 2 x
Now, lim
x0 x2

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 SRI CHAITANYA IIT ACADEMY, INDIA 18‐01‐23_ Sr.Super60_(NUCLEUS,STERLING)&LIIT __BT_ Jee‐Main_GTM‐08_KEY &SOL’S
2 cos x
  sin 2 x  2  cos 2 x   sin x
 lim 2 cos 2 x
x0 2x
2 cos x sin 2 x
 2sin x cos 2 x  2  2cos x cos 2 x 
  x  3
cos 2 x lim  x2 
 lim =2+1=3  e x0  e3 3  e 9  a  9
x 0 2x
a 2  b2
83.  C1 : x 2  y 2  a 2  b 2  C2 : x 2  y 2 
3
a 2  b2
Director circle of C1 is x 2  y 2  2  a 2  b2   2  a 2  b2  
3
 6a 2  6b2  a 2  b2  5a 2  7b 2
2 12
  7
 
5a 2  7a 2 1  e12  e12  and 5a 2  7a 2 e22  1  e12 
7
84. Asymptotes 3(x – 1) – 2(y + 3) = 0 and 3(x – 1) + 2(y + 3) = 0
3x – 2y – 9 = 0 and 3x + 2y + 3 = 0
It touches x 2  y 2  2 x  k /13  0
 5 12   31 12 
 A   ,   and B  ,  
 13 13   13 13 
 k  23
3
85. On each roll, the probability that Kiran decides to go to Russia is times the probability
2
3
she decides to go to Ukraine, so the total probability that she goes to Russia is times
2
the total probability that she goes to Ukraine. Since the total probability is 1(they are the
2
only two eventual outcomes) Kiran goes to Ukraine with probability and Russia with
5
3
probability .
5
86. Let Z  m  ni
Z  m  ni
i.e., x3  ax  b has roots – 20, Z , Z and x3  cx 2  b has roots – 21, Z , Z
  20  Z  Z  0
m  10
21Z  21Z  ZZ  0
21 Z  Z   ZZ  0  n  320
87. Equation reduces to  cos x  cos 4 x   cos 2 x  cos 3x  0
5x  x 5x x
 2 cos  2 cos x cos   0  cos  0, or cos x  0 or cos  0
2  2 2 2
 2n  1  or x   2n  1   0 or x  2n  1  , n  z
 x  
5 5
  3 7 9  3 
 x   , , , , , ,  . Hence, number of solution 7
5 5 5 5 2 2
10 10
Ck 10 k C10
88.   2
k 0
k

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 SRI CHAITANYA IIT ACADEMY, INDIA 18‐01‐23_ Sr.Super60_(NUCLEUS,STERLING)&LIIT __BT_ Jee‐Main_GTM‐08_KEY &SOL’S
 10
Coefficient of x in  C0 1  x  C1
10 10 10 1  x
11

 C2
10 1  x
12

 .....  C10
10 1  x 
20


 2 22 210 
 10 2
 1  x  10  1  x  10  1  x 
3
1 x 
10

Coefficient of x in 1  x 
10 10
 C0  C1 
10
 C2   C3   .....  C10 
10

  2   2   2   2  
10
  1 x 
Coefficient of x in 1  x
10
1   2  
10

Coefficient of x10 in
1  x 
2 10
C5 252

10

63
10 10
2 2 1024 128
i
   1 i 3
89. f  e 3    a  c   b  c;   
  2
 1 i 3
  a  c     b  c  4038 3i  4032
 2 2 
 c  a
 a  c  8076;   b  4032  a  c  4038and b  4032
 2 
90. A   3, 4,5,...., ; B  ...., 7, 6

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 SRI CHAITANYA IIT ACADEMY, INDIA 19‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐09_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 19-01-2023
Time: 09.00Am to 12.00Pm GTM-09 Max. Marks: 300
KEY SHEET
PHYSICS
1) 2 2) 3 3) 4 4) 4 5) 4
6) 4 7) 4 8) 3 9) 2 10) 1
11) 2 12) 3 13) 2 14) 2 15) 3
16) 2 17) 4 18) 4 19) 2 20) 1
21) 234 22) 12 23) 1234 24) 134 25) 134
26) 23 27) 123 28) 13 29) 2 30) 3

CHEMISTRY
31) 3 32) 2 33) 3 34) 4 35) 3
36) 4 37) 1 38) 4 39) 3 40) 3
41) 1 42) 2 43) 1 44) 3 45) 4
46) 4 47) 2 48) 3 49) 2 50) 4
51) 4 52) 6 53) 3 54) 2 55) 2
56) 5 57) 3 58) 3 59) 6 60) 6

MATHEMATICS
61) 2 62) 4 63) 3 64) 3 65) 2
66) 1 67) 4 68) 1 69) 1 70) 3
71) 1 72) 3 73) 2 74) 4 75) 3
76) 4 77) 3 78) 4 79) 4 80) 1
81) 2 82) 7 83) 4 84) 3 85) 2
86) 24 87) 2 88) 6 89) 4 90) 1

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SOLUTIONS
PHYSICS
1. Conceptual
2. Conceptual
3. Conceptual
4. Conceptual
5. Conceptual
6. Conceptual
7. Conceptual
8. Conceptual
9. Conceptual
10. Conceptual
11. Conceptual
12. Conceptual
13. Conceptual
14. Conceptual
15. Conceptual
16. Conceptual
17. Conceptual
18. Conceptual
19. Conceptual
20. Conceptual
21. Conceptual
22. Conceptual
23. The correct answers are
b) continuous if there is no charge at that point
d) discontinuous if there is a charge at that point
24. Conceptual
25. Conceptual
26. Conceptual
27. (a) a larger angle to be subtended by the object at the eye and hence viewed in
greater detail.
(b) the formation of a virtual erect image
28. Conceptual
29. Conceptual
30. Conceptual

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 SRI CHAITANYA IIT ACADEMY, INDIA 19‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐09_KEY &SOL’S

CHEMISTRY
31. Greater the magnitude of charge on the electrolyte added more will be the
coagulating power towards oppositely charged colloidal solutions.
32. Extent of physical adsorption is directly proportional to critical temperature of a
gas.
33. In an atom In an atom the Z eff experienced by the electron decrease with increase in
azimuthal quantum number.
34. In pentaacetate form of glucose as it cannot revert back in to open chain form with
free aldehyde group
35. For alkali metals melting and boiling points decreases in the order
F  Cl  Br  I
36. Magnetic moment is directly proportional to number of unpaired electrons available
37.
OCH 3 OH OH

SOCl2 CH 3OH

CH 2OH CH 2Cl CH 2OCH 3

38. NCERT
39. P is a tertiary alcohol, which is not oxidized by chromic an hydride.
40. Addition of inert gas at constant temperature and pressure will favour the reaction
where more number of moles of reaction where more number of moles of gases are
prepared
41. O OH

CH 3  CH 2  C  CH 3 
 CH 3  CH 2  C  CH 3
HCN

CN

42. NCERT
43. Greater the splitting energy lower will be the wavelength of radiation absorbed
44. Conceptual
45. LiAIH 4 reduces both  C  O as well as COOH to alcohols, while NaBH 4 reduces only
 C  O without affecting COOH .
46. Chromic acid will oxidise primary alcohol in to carboxylic acid.
47. Conceptual
48. LiH , CaH 2 , AlH 3 , SiH 4
49. For an irreversible spontaneous process  dS   0 and  dG   0 .
50. Localized lone pair on nitrogen atom and no inversion will make it a better base.
51. Chloramphenicol, Ofloxacin, Ampicillin, Amoxycillin
 3 PtCl6   4 NO  8H 2O
52. 2
3Pt  16 H   4 NO3  18Cl  


53. I, iv, vii

54. CCl4 , SF6
55. K and M = 0; L=2: N=4
56. Bronze Cu  Sn
57. CH 2  CH  CH 2Cl , p  NO2C6 H 5CH 2Cl , CH 3Cl
58. Statements-I,II,III are correct
59. Conceptual
60. CCl3COOH , C6 H 5COOH , HCOOH ,  COOH 2 , CH 3COCOOH ,  2, 4, 6  trinitro phenol 

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MATHEMATICS
61. Conceptual
62. Conceptual
63. Conceptual
64. Conceptual
65. Conceptual
66. Conceptual
67. Conceptual
68. Conceptual
69. Conceptual
70. Conceptual
71. Conceptual
72. Conceptual
73. Conceptual
74. Conceptual
75. Conceptual
76. Conceptual
77. Conceptual
78. Conceptual
79. Conceptual
80. Conceptual
81. Conceptual
82. Conceptual
83. Conceptual
84. Conceptual
85. Conceptual
86. Conceptual
87. Conceptual
88. Conceptual
89. Conceptual
90. Conceptual

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 SRI CHAITANYA IIT ACADEMY, INDIA 20‐01‐23_ Sr.Super60_(NUCLEUS,STERLING) & LIIT _BT _ Jee‐Main_GTM‐10_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_(NUCLEUS,STERLING) & LIIT _BT JEE-MAIN Date: 20-01-2023
Time: 02.00Pm to 05.00Pm GTM-10 Max. Marks: 300
KEY SHEET
PHYSICS
1) 2 2) 3 3) 4 4) 2 5) 1
6) 4 7) 4 8) 3 9) 1 10) 3
11) 2 12) 3 13) 2 14) 3 15) 4
16) 2 17) 2 18) 4 19) 2 20) 2
21) 5 22) 2 23) 8 24) 3 25) 9
26) 2 27) 5 28) 2 29) 5 30) 5

CHEMISTRY
31) 4 32) 3 33) 3 34) 1 35) 1
36) 3 37) 4 38) 3 39) 3 40) 2
41) 2 42) 3 43) 3 44) 4 45) 3
46) 1 47) 3 48) 3 49) 2 50) 2
51) 3 52) 16 53) 2 54) 4 55) 8
56) 500 57) 1 58) 7 59) 3 60) 600

MATHEMATICS
61) 2 62) 1 63) 2 64) 3 65) 1
66) 1 67) 1 68) 2 69) 1 70) 3
71) 2 72) 4 73) 1 74) 1 75) 3
76) 1 77) 4 78) 3 79) 3 80) 3
81) 1 82) 1 83) 1 84) 0 85) 78
86) 2 87) 0 88) 4 89) 2 90) 8

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SOLUTIONS
PHYSICS
1. v cos(900   )  u cos  or v sin   u cos  ; v  u cot 
vT2 u 2 cot 2 
 ac ; R
R g sin 
2. mg sin 
a) when t  , f will be upwards & f= mg sin   kt
k
mg sin 
b) At t  , f 0
k
mg sin 
C) At t  till the body starts to moves f will be down wards and f= kt  mg sin 
k
d) Once it start moving f = constant
3. Conceptual
4. a)  mvo sin   5R  mv1R....(1)
1 2 GMm 1 2 GMm
b) 2 mv0  5 R  2 mv1  R ....(2)

1 8GM 
Solving   sin 1  1 
5 5v02 R 
5. 2 gyL2  2 g (4 y)   R 2
L
R
2
6. 1 1 
2T      gh
 r1 r2 
 h  11.36 mm
7. Resolving power is directly proportional to diameter of objective

8. Electromagnetic waves interact with matter via their electric and magnetic field which
in oscillation of charges present in all matter. The detailed interaction and so the
mechanism of absorption, scattering, etc. depend of the wavelength of the
electromagnetic wave, and the nature of the atoms and molecules in the medium.
9. Conceptual
10. Potential drop across C1 is maximum
Hence energy stored in C1 is maximum as Energy  (potential drop)
11. r
4R
9

By symmetric method

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 SRI CHAITANYA IIT ACADEMY, INDIA 20‐01‐23_ Sr.Super60_(NUCLEUS,STERLING) & LIIT _BT _ Jee‐Main_GTM‐10_KEY &SOL’S
The internal resistance must be equal to external resistance for maximum power transfer
4R
The Req for circuit=
9
4R 4R
Thus, Req  .Thus r 
9 9

12. Current density J  j0 rkˆ

d d d3
Current within a distance d I  r 0 j.ds  0 J 0 r.2 rdr  2 J 0
3
 3
From ampere’s Law  B.dl   0 2 J 0 d
C
3
Here loop is a circle of radius, r, B is magnetic field at r
d3
2 r.B  0 2 J
3
 J d2
 Br (  d )  0 0
3

13. Use Fleming’s left hand rule, we find that a force is acting in the radially outward
direction through the circumference of the conducting loop.

14. i1rms 
Erms

130
 10 A
xc2  R12 13
Erms
i2 rms   13 A
xL2  R22
i1 R1

R2
i2

Power dissipated
i12rms R1  i22rms R2  102  5  132  6 = 1514 W = power delivered by battery

15.   (450  300 )  (1800  600 )  (450  300 )

= 1500 clockwise
16. D
fringe width=
d
12  600
121  k 2 . Hence k   18
400

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 SRI CHAITANYA IIT ACADEMY, INDIA 20‐01‐23_ Sr.Super60_(NUCLEUS,STERLING) & LIIT _BT _ Jee‐Main_GTM‐10_KEY &SOL’S
17. N 1
 t /t1/2
N0 2
18. Ferro-magnetic substances become paramagnetic above Curie temp.

19. CONCEPTUAL
20. Conceptual
21. Given Q  x 2/5 y 1t 1/2 z 3
Q 2 x y 1 t z
100  100  100   100  3  100
Q 5 x y 2 t z
2 1
  2.5  2  3  0.5   1  5%
5 2
22. lose in gravitational P.E= gain in spring P.e
1
mgh  k ( h cot   h) 2
2
2mg
or (cot   1) 
kh
2mg
cot   1 
kh
23.

Sol: mg

(mg  f )r  (3mr 2  mr 2 ) ……(1)

f =4ma --- (2)
g
Solving  
8r

24. restoring torque is given by kx    I

l
2

x  (l / 2)

2
 l  l  ml 3k 3k
kx            
 2  2  2 m m
25. 2
Er  Ar   v2  v1 
2

     1/ 9
Ei  Ai   v1  v2 
E
Therefore, r  8 / 9
Ei

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 SRI CHAITANYA IIT ACADEMY, INDIA 20‐01‐23_ Sr.Super60_(NUCLEUS,STERLING) & LIIT _BT _ Jee‐Main_GTM‐10_KEY &SOL’S
26. 330  v 330  v
f   176; f 2   165
330  22 330
f1  f 2  0  v  22m / s
27. K  1 27
a)  PAV ......(1)
9 L
L
b) PA  nR  300.......(2)
2
K  3 2nR  300V K
Hence  V 
L L 100 R

1,2 
28. v
2
dv   
 0,0 
E.d r d r  dxiˆ  dyjˆ

E .d r  (2 xy  y ) dx  ( x 2  x ) dy  d ( x 2 y  xy )
v (1,2)
2
dv   
(0,0)
( x 2 y  xy )

V  2   12  2  1 2   0 
V  2  4, V  2volts
29. 20  
10 t
 3
I1   1  e 510    1.5 A
3

10   2
4 5

20V 20V

L  5mH I1 I2 5

6 C  0.1mF
t
20 1103
I1  e  1.0 A
10
From superposition I  I1  I 2  2.5 A

30. for first line of balmar series

1 1 1 36
 R    R 
 4 9 5
Wave length of the first line  of the Lyman series is given by
L

1  1  36 3 27 5
 R 1       L 
L  4  5 4 5 27

CHEMISTRY
31. Greater the polarity of solvent more will be its interaction with substance which will
effect R f . TLC is an example of adsorption chromatography.
32. Isomeric ethers have relatively low boiling point than alcohols. Cresol is soluble in
aqueous NaOH .
33. BF3 , CO32 , NO3  Planartriangular
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2 2
SO , CrO , CF4  tetrahedral
4 4

NH 3  Pyramidal
SF4  sea  saw
34. CH 3

CH 3  CH  CHBr i ) NaNH 2


ii ) red .hot . Fe

CH 3
CH 3
For electrophilic substitution on the ring
all positions are similar
35. Due to low hydration enthalpies of both ions in CsI it is less soluble in water
36. Deacon’s process------ CuCl2 , Ostwald’s process------ Pt  Rh wire guage
Contact process----------- V2O5 , Haber’s process----------- Fe
37. Aldehydes will give silver mirror with Tollens reagent
38. Ring expansion and more stable tertiary carbocation formation is involved
39. Mg (OH ) 2 is less soluble in water than Ca(OH )2
40. NH2 NaNO2  2 HCl

N2

 H 2O

OH
+


H 2O


41. PbS  black ppt

Pb( NO3 ) 2  soluble in water
PbI 2 -yellow ppt
42. Diborane is non-planar molecule
43. More number of resonance structures formed, greater will be its stability
44.
( CH 2  CH )n

( CH 2  CH )n 
H 3O

OCOCH 3 OH
45. Benzaldehyde is produced
46. XeF5  has square pyramidal shape with different Xe  F bond lengths due to lone
pair-bond pair repulsions.
47. electron withdrawing groups will increase reactivity towards SN 2 mechanism
48. Cu  8 HNO3 (dilute)  3Cu ( NO3 )  2 NO  4 H 2O
49. In amylose   D  glucose units were joined via C1  C4 glycosidic link
50. calamine--- ZnCO3 Cryolite---- Na 3AlF6
Siderite --- FeCO3 Magnetite--- Fe3O4
51. it gives two geometrical isomers out of which one is optically active
52. 0.06  [ Ag  ] saturated 
0.42  log  
1   0.1 
[ Ag  ]  1108
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 SRI CHAITANYA IIT ACADEMY, INDIA 20‐01‐23_ Sr.Super60_(NUCLEUS,STERLING) & LIIT _BT _ Jee‐Main_GTM‐10_KEY &SOL’S
53. 1
T f  (1.86)    0.93K
2
Tb  (0.52)  2   1.04K
Tb  T f  1.97  2
54. 15 moles of NaOH will neutrilize 5 moles of KHC2O4 .H 2C2O4 .2 H 2O
55. 3P  6 electrons
4 S  2 electrons
56. Kp 
80
 4 (2)(T )  n(4)   1400  5.6T
20
2.8T  1400  5.6T
2.8T  1400
1400
T K  500K
2.8
57. K
1  2 
n  
0.2  1.6 
58. Cubic  3
Tetragonal  2
Hexagonal  1
R hom bohedral  1
59. weight of 1 lit of solution=1100g
Weight of solvent in 1 lit solution
=1100-360
=740g
1000
Molality  2 
740
= 2.7mole.Kg 1

60. A2  B2  2 AB H  300 KJ
H r   B.E A2    B.E B2     2( B.E AB ) 

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MATHEMATICS
61.  p ~ q    ~ p  q    ~ q  p    ~ p  q
 ~ q   p ~ p   q  ~ q  F  q   ~ q  q   F F F F
Statement-1 is true  p  q    ~ q ~ p 
~ q  ~ p  ~ (~ q) ~ p  q  ~ p  ~ p  q
Statement-2 is true
Thus, both the statements are true and
statement-2 is not the correct explanation for statement-1
62. The given equation is dx – x (ydx+ xdy )= x5y4(ydx + xdy)
1
dx 1 5 5 xy  x 5 y 5
  (1  x 4 y 4 )d( xy)  lnx = xy + x y  ln c  x= ce 5 .
x 5
63. The roots of first equation are –1 and a2 –1. Now the roots of second equation are 1,
a2  4a.
According to given condition a2 – 1 < 1 and a2 – 1 < a2  4a
a    2, 2  and a >  a    , 2 
 1 1
4 4  
64. 1 
Since 2
x  3x  2  0, 0  tan 2
x  3x  2 
2

And 4 x  x 2  3  0  0  cos 1 4 x  x 2  3  .
2
Adding, we have 0  L.H.S.  
 The given equation has no solution.
65. 2 4 6 2(n  1)
S  (n  1) cos  (n  2) cos  (n  3) cos  ......  cos
n n n n
2 4 2(n  1)
S  1cos  2 cos  ..........  (n  1) cos
n n n
 2 4 2(n  1) 
2 S  n  cos  cos  .......  cos 
 n n n 
  2 2(n  1) 
sin(n  1) 
2S  n n cos  n n 
    n
sin  2 
n  
66. Let be the required angle, then

67. dy x2
  12
dx y1

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Tangent equation is x x  y y  x13  y13
2
1
2
1  x12 x  y12 y  a 3
Since, it passes through  x2 , y2 
 x12 x2  y12 y2  a3 1
and x13  y13  a3  2
x23  y23  a3  3
By solving (1), (2), (3) we get result
68. There will exist two common tangents when both the circles are intersecting.
2 2
13  13   13 
Solving the equation 4 – 4x + 9 = 0  x   y2    4 or y =  4-   .
4  4   4 
2

It should have two real and distinct values so 4     0

13
 4 
69. 2 6( x  2)  2(2 x  1) 10 x  10
f '( x)  ( x  2) 2/3 2  (2 x  1) ( x  2) 1/3  
3 3( x  2)1/3 3( x  2)1/3
x  1 is a point of local maximum and x  2 is a point of local minimum
 No.of extremum points is 2
70. We have to find the number of integral solutions if x1  x2  x3  x4  x5  6 and that
equals 561 C51 10 C4
Thus Statement-1 is false.
Number of different ways of arranging 6A’s and 4B’s in a row
10 10
  C4 = Number of different way the child can buy the six ice-creams.
6 4
Statement-2 is true
So, Statement-1 is false, Statement-2 is true.
71. 1
1 2
x2 1 1
 2  1  2 1 dx   x
2
dx put x   t
x
x x  x  2  1   1 
 x x x  x  2
 x  x
dt 2 sec  tan  d 1
 t  2 sec     C
t t 2
2
2 sec  2 tan  2
72. 5x+3y-2=0, 3x-y-4=0
(x, y)=(1, -1)
x-y+1=0, 2x-y-2=0
(x, y)=(3, 4)
Required line passing through (1, -1) and (3, 4)
73. ln f  x   ln  x  1  ln  x  2   .....  ln  x  100  Differentiating
f '  x 1 1 1
   ....... 
f  x x 1 x  2 100
f  x f "  x  f '  x
2
1 1 1
Again differentiating   ...... 0
f  x  x  1  x  2   x  100 
2 2 2 2

 f  x  f "  x   f '  x  0  g  x  0

2
 g  x   0 has no solution

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74.  1  1
For f to be defined we must have log1/ 2 1  4   1  1  4   21   2 which is
1

 x x
1
possible if and only if 4
 1 i.e. 0<x<1. Hence the domain of the given function is
x
 x : 0  x  1
75. The chord of contact y yo  2  x  xo  of the point P  xo , yo  w.r.t the parabola is tangent
to the hyperbola x 2  y 2  1 iff 2 x02  y02  4 . Locus of P is the ellipse 2 x 2  y 2  4
76.  1 1
x 1 2   x2 x 
 
C=AB=  1  2 x 2 
3x 1 
 4 x 2  1  3 x 
 
 
1 x 1 1
 ( x )   cij = c11  c12  c22 = +2x+6+1+2+2x+3-2+ 2 = +4x+10+
x 4x 1 x 1
1i , j  2
4x 
x
1
Let 4x   t  t  4
x
1 57
 ( x ) min  4  10  
4 4

77. Given xRy  sin 2 x  cos 2 y  1

Now sin 2 x  cos 2 x  1, So R is Reflexive. i.e, xRx
Let xRy  sin 2 x  cos 2 y  1
 1  cos 2 x  1  sin 2 y  1
 sin 2 y  cos 2 x  1
So xRy  yRx  R is symmetrix
Now Let xRy and yRz holds
i.e, sin 2 x  cos 2 y  1 and sin 2 y  cos 2 z  1
So sin 2 x  cos 2 z  1  from above '2' equations 
So xRy & yRz  xRz
 R is Transitive
Hence R is an equivalence relation
78. Using L.M.V.T for x   7, 1 , we have
f  1  f  7  f  1  3
2  2  f  3  3  f  1  9
1  7 6
Also using L.M.V.T for x   7,0 , we have
f  0   f  7  f  0  3
2  2  f  0   11  f  0   f  1  20
07 7
79. n is even so n = 2m
m ! m!  2
. C0  2C12  3C22 ......   1  2m  1 C22m   1
2m
E  2.
 2m !  

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C2 m  C0 , C2m-1  C1
Write the terms in reverse order
2m ! m !
E  2m  1 C02  2mC12   2m  1 C22  .....C22m    2 
 2m  !
2 m !m !
2E   2m  2   C02  C12  C22  .....C22m 
 2m  !
2m !m ! m  1
 1 2mCm C02  C12  C22 ....  Cn2    1 .nCn / 2
n/2

m

 2m !
n 
 2  1  m  1  2  1   1 n  2
n/2 n/2
  1
m

2 
80. Let the distance between each of the pole be x
h
 tan 
a  9x
h cos   a sin 
x=
9sin 

81. eia + eib = (cos a + i sin a) + (cos b + i sin b ) = eig

1 1 1
Let a = eia , b = eib , c = eig then a + b = c . Also + = .
a b c
 ab = c 2  e (
i a +b )
=e(
i 2g)
 sin (a + b ) = sin 2g .

82. x  y   z   x  ' y   ' z  '

S.D. between the lines   and '   is given by
l m n l m' n'
  '  '   '
   mn'  m' n 
2
S.D. = l m n
l m' n'
1 2 2
1
  mn  m n    1   2   1   6
2 2 2 2
' '
 S .D.  2 3 4  6 
  6
3 4 5
83. Required area
1 1 1
 2 x3   x2 x2 
=   2x  2x  x log x  dx =  x2 
2
  log x  
0  3 0  2 4 0
1 1 7  
   =0.583  xlog x 2 log x  0
3 4 12 0 

84. a = P(A getting 6), b = P(B getting 7)
  (1  a )(1  b)

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5
a 30
P ( A)  = 36  =0.4918
1   1  31 . 5 61
36 6
85. We have x  170, x 2  2830
Increase in x  10
 x1  170  10  180
Increase in x2  900  400  500
 x2  2830  500  3330
2
1 1 
    3330     180   222  12   78
2 2

15  15 
  x 2  x  
2

     
2

 n  n  
86. g 2  x  g 2  x
& g  2  x  sin x is an odd function  I1  0
Now g  2   2  x    g  2  2  x   g  x   g  4  x   g1  x    g1  4  x 
4 4 4
1 dx dx
So I 2   g1  x 
dx  1  g1  4  x 
  g1  x 
0 1 e 0 1 e 0 1 e
4 g1 x  4
e
 g1 x 
dx   2  1   2  : 2I 2   1dx  I 2  2
0 1 e 0

87.     
 3 i  k , i   3 j and i   2  sin   j   k
 

3 0 1
1  3
0 0
1 2  sin  

 2  sin    3     6   0   7   3  2  sin   0    0
88. Max values of sin x + cos x and 1 + sin 2x are 2 and 2 respectively.
 2
2
Also 2
 the equation can hold only when sinx + cos x = 2 and 1+sin 2x=2
Now sin x + cos x = 2
  
 cos  x    1  x  2n  
 4 4
n 
1  sin 2x  2  sin 2x  1  x    1 n
2 4

The value in [-, ] satisfying both the equations is . [when n = 0 ]
4
89. 1 1

4  ex
sin x 4e x  1 sin x
lim
x 0
f ( x ) = lim
x 0 
4  2 = lim
x 0 
1 3 2 =0+2=2
1 e x x e e
x x x

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1

4e x
sin x
lim f ( x ) = lim 4 2 =4-2=2
x 0 x 0 
1 e x x
90. 1
We have sin x 8cos 2 x  1  sinx | cosx| =
2 2
Case – 1 when cosx > 0
1 1  3 9 13
In this case sinx cosx =  sin 2 x   2x  , , ,
2 2 2 4 4 4 4
 3 9 13
 x , , ,
8 8 8 8
 3
As x lies between 0 and 2 and cos x  0 , x  ,
8 8
Case – 2 when cosx < 0
1 1 1
In this case  sinx cosx =  or sin2x  
2 2 2 2 2
5 7  13 15 5 7 
 x , , ,  x , as cosx < 0
8 8 8 8 8 8
Thus the value of x satisfying the given equation which lie between 0 and 2 are
 3 5 7  
, , , These are in A.P. with common difference .
8 8 8 8 4

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 21-01-2023
Time: 09.00Am to 12.00Pm GTM-11 Max. Marks: 300
KEY SHEET
PHYSICS
1) 4 2) 4 3) 1 4) 1 5) 3
6) 2 7) 1 8) 3 9) 2 10) 3
11) 2 12) 3 13) 2 14) 3 15) 1
16) 1 17) 3 18) 1 19) 1 20) 4
21) 10 22) 600 23) 500 24) 60 25) 1400
26) 41 27) 10 28) 3 29) 8 30) 100

CHEMISTRY
31) 4 32) 2 33) 3 34) 4 35) 3
36) 2 37) 3 38) 2 39) 4 40) 4
41) 1 42) 1 43) 2 44) 1 45) 2
46) 2 47) 3 48) 3 49) 4 50) 1
51) 80 52) 50 53) 0 54) 40 55) 200
56) 18 57) 3 58) 69 59) 3 60) 5

MATHEMATICS
61) 2 62) 1 63) 2 64) 3 65) 1
66) 1 67) 3 68) 1 69) 3 70) 1
71) 1 72) 2 73) 2 74) 3 75) 3
76) 2 77) 1 78) 1 79) 4 80) 1
81) 8 82) 1 83) 9 84) 1 85) 9
86) 3 87) 23 88) 166 89) 9 90) 24

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SOLUTIONS
PHYSICS
1. No of divisions on main scale  N
No of divisions on vernier scale  N  1
Size of main scale division  a
Let size of vernier scale division be b then we have
aN
aN  b  N  1  b 
N 1
aN
Least count is a  b  a 
N 1
 N 1 N  a
 a   N  1
 N 1
2. 2u sin 300 2 10 1 / 2  2
t   sec
g cos30 0 10  3/2  3
1
R  10cos300 t  g sin300 t 2
2
10 3  2  1 14 10 20
    10     10   m
2  3 2 23 3 3
3. Limiting friction between block and slab  smAg  0.6 10  9.8  58.8N
But applied force on block A is 100 N. So the block will slip over a slab.
Now kinetic friction works between block and slab
Fk  k mAg  0.4 10  9.8  39.2N
This kinetic friction helps to move the slab
39.2 39.2
 Acceleration of slab    0.98 m / s 2
mB 40
4. 1
By conservation of energy mg  3h   mg  2h   mv 2 ( v  velocity at B)
2
1
mgh  mv2 ; v  2 gh
2
From free body diagram of block at B
mv 2
B h

N mg

mv2
N  mg   2mg; N  mg
h
5. By Maxwell’s law, time varying electric field produce time – varying magnetic field
C
and vice-versa. So statement I is correct and, V  . So statement II is
r r
incorrect.
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 SRI CHAITANYA IIT ACADEMY, INDIA 21‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐11_KEY &SOL’S
6. GMm 1 1 
P.E  RR dr  GMm   
0
r2  R R0 
1 2
The K.E. acquired by the body at the surface  mv
2
1 2 1 1 
 mv  GMm   
2  R R0 
 1 1
v  2GM   
 R0 R 
7. The volume flow rate (Q) of an incompressible fluid in steady flow remains constant
From equation of continuity,
av  constant
1
 Q  a  v  constant or, a  v
Where a  area of cross-section and v  velocity
 If v decreases a increases and vice – versa.
When stream of water moves up, its speed   decreases and therefore ' a ' increases
v
i.e. the water spreads out as a fountain. When stream of water from hose pipe moves
down, its speed increases and therefore area of cross-section decreases.
8. Vertical distance covered by water before striking ground   H  h  . Time taken is,
t  2H  h / g :
Horizontal velocity of water coming out of hole at P,
u  2 gh
 Horizontal range  ut  2 gh  2  H  h  / g
 2 h H  h
9. The linear momenta given by,
p  2 mE  2 mqV  E  qV 
p m q 4m p .2q p 4 2 2 2
    
pp m pq p m pq p 1 1 1
10. The wavelength of spectral line of the third member of Lyman series is given by
1 1 1  16
 R     1  …………. (i)
1 1 16  15R
The wavelength of spectral line of the first member of paschen series is given by
1 1 1  144
 R     2  …………. (ii)
2  9 16  7R
Dividing (ii) by (i) we get
1 16 7R 7
  
 2 15 R 144 135
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 SRI CHAITANYA IIT ACADEMY, INDIA 21‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐11_KEY &SOL’S
11. For A 1/ 2  20 min, t  80 min, number of half lifes n  4
t
N
 Nuclei remaining g  0 . Therefore nuclei decayed
4
2
N
 N0  0
24
For B 1/ 2  40 min, t  80 min, number of half lifes n  2
t
N0
 nuclei remaining  . Therefore nuclei decayed
22
N
 N0  0
22
N 0  40 1  1
N
 Required ratio  2  16  15  4  5
N 1 16 3 4
N 0  20 1 
2 4
12. As we know, current density,
vd
j   E  nevd    ne  ne 
E
1 1
 
 ne ee Resistivity
1

or   0.4  m
10  1.6  10  19  19  1.6
19

13. Time in which left arm reach at x  10cm

0.1 1
t   0.01sec r1  10  0.01  0.1m
10 100
  i i 
e  lV  B1  B2   0.1  10  0  0 
 2 r1 2 r2 

r1  
B1 B2

r2

 1 1 
 0.1  10  2  10  7  1  
 0.1 2  0.1 
 2 1 
 2  10  7  6
  10 V
 2  0.1 
B1lV B2lV

14. Real depth  5 cm 1cm  6 cm

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Water
  1.33
5 cm

  1.5 1cm
Glass

d1 d2 5 1
Apparent depth      3.759  0.666
1 2 1.33 1.5
 4.425  Shift  6 cm 4.425 cm 1.575 cm
So most appropriate option will be (c).
15. The two springs are in parallel.
 Effective spring constant, k  k1  k2
Initial frequency of oscillation is given by
1 k1  k2
f  ………. (i)
2 m
When both k1 and k2 are made four times their original values, the new frequency is
given by
1 4k1  4k2 1 4  k1  k 2   1 k1  k 2 
f  f '  2   fv
2 m 2 m  2 m 
16. We know that velocity in string is given by
T
v ……………. (i)

Where   m  mass of string
1 length of string
m
The tension T   x g …………..(ii)

From (i) and (ii)
 
 gx  x 1/2dx  gdt   x 1/2 dx  g  dt 
dx
2 1
dt 0 0
 20
 g t  t2 2 2 2
g 10

l T

17. Given, Electric field, E  3  104

Mass of the drop, m  9.9  10 15 kg
At equilibrium coulomb force on drop balances weight of drop.
qE  mg

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9.9  1015  10
 3.3  1018 C
mg
 q  q
4
E 3  10
18. We have given two metallic hollow spheres of radii R and 4R having charges Q1 and
Q2 respectively.
kQ kQ
Potential on the surface of inner sphere (at A) VA  1  2
R 4R
Potential on the surface of outer sphere (at B)
kQ kQ  1 
VB  1  2  Here, k  
4R 4R  4 0 
Potential difference,  V  V A  V B  3 . kQ1  3 Q
. 1
4 R 16 0 R
Q2
Q1

R A
4R B

19. Equivalent circuit

Capacitor blocks the DC current. So current will flow in lower loop only.
5
So, VPQ   4  4V
4 1
Charge on 4  F  V ' C   4  4   4  2  4  8  C
44
20. Resistance of a metal conductor at temperature t0C is given by Rt  R0 1   t  ,
R0 is the resistance of the wire at 00 C and  is the temperature coefficient of
resistance.
Resistance at 50 0 C , R50  R 0 1  50  ……….. (i)
Resistance at 100 0 C , R100  R0 1  100  ……….. (ii)
From (i), R50  R0  50 R0 …………. (iii)
From (ii), R100  R0 100 R0 ………… (iv)
Dividing (iii) by (iv), we get
R50  R0 1

R100  R0 2
Here, R50  5 and R100  6

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 SRI CHAITANYA IIT ACADEMY, INDIA 21‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐11_KEY &SOL’S
5  R0 1 6  R0  10  2R0 or, R0  4
  or,
6  R0 2

   R11  R12 

21. 1
By lens maker formula,  g a  1
f
For Lens 1:
1 1 1 
 1.5  1   
f1  R R 
1 2
  0.5 
f1 R
 f1  R  R 15 cm
For Lens 2:
1  1 1 
 1.25  1   
f2  R  R 
2 0.5 1
 0.25    
R R 2R
 f2  2R
For Lens 3:
Similarly like lens 1, f3  R
1 1 1 1 1 1 1
So,      
f net f1 f 2 f 3 R  2 R R
3 2 1
    f net  10 cm
2 R 2  15 10

22. D

d
D
  
d
d 3  105  103
   0.6  106  600  109 m
d 5  102
 600 nm

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 SRI CHAITANYA IIT ACADEMY, INDIA 21‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐11_KEY &SOL’S
23.   
E  50 sin   t  . x   E0  50 NC  1
 c 
1
Energy density  0 E02
2
Energy = energy density volume
1
 0 E02 .V  5.5  1012
2
1
8.8  1012   50   V  5.5  1012
2

2
5.5  2
 V  500 cm3 .
2500  8.8
24. Volume, V  Ibh
V l b h
   
V l b h
V  b h
    
V T T bT hT
   5  10  5  5  10  6  5  10  6
 60  106 / 0 C
 Value of C  60.00
25. Work done  P  V
 400  PV
 400  nRT [ PV  nRT at constant pressure]
Now, Q  nCPT
R  1.4
n  T  400   400   1400 J
 1  1 0.4
26. 3 RT
Root mean square speed is given by vrms 
M
Here, M = Molar mass of gas molecule
T = temperature of the gas molecule
We have given v N 2  v H 2
3RTN2 3RTH 2
 
M N2 M H2
TH 2 573
   TH 2  41K
2 28
27. P S
As per Wheatstone bridge balance condition 
Q R
Let resistance R’ is connected in parallel with resistance S of 10

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15 10 R '
 
12 10  R '
4
10 R '
 5
10  R '
 50  5R' 10R'
50
 R'   10
5

28. Magnetic field at centre of ring/coil is given as,

N 0 I
Bcentre 
2R
100  4  107  I
 37.68  104   I  3A
2  5  102
29. Given: Magnetic moment, M  9.85  10  2 A / m 2 , moment of inertia,
I  5  10  6 kgm 2
10 oscillations in 5 seconds
5
 Time period   0.5s
10
2
I  T  I
T  2   
MB  2  MB
I  4 2 5  106  4  9.85  102
 B 
M T 2 9.85  102   0.5
2

 B  80  10  4  8 mT
30. For minimum impedance, we have
1
X L  Xc  L 
c
1 1
L   0.01  101  103 H  100 mH
2 2
 c  2  500 

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CHEMISTRY
31. Cu2S .Fe2S3  or  CuFeS2
32.  2 Pb  NO3 2  2H 2O  O2 
2 PbO2  4 HNO3 
33. Si contains vacant d-orbtials so it accepts lone pair of H2O and gets hydrolysed. On
the other hand C in CCl4 can’t do so due to absence of vacant d-ortbials.
34. It is Ionic compound. Remaining are only covalent.
35. 2XeF2  2H2O  2Xe  4HF  O2
XeF6  H2O  XeOF4  2HF
XeF6  2H2O  XeO2F2  4HF
XeF6  3H2O  XeO3  6HF
36. 3Hg  2HNO3  6HCl 3HgCl2  2NO  4H2O
37. 6 0
d 6 4s 0 4 p0
Co3  3d 7 4s2 t2 g eg 3  
d 2 sp3
Fe3  W .L : t2 g 3eg 2
5
d
38. Mabcd  3 isomers (2 cis + 1 trans)
a )  Pt C l  N O 2   N O 3   SN        3
b )  Pt cl  NO 2   N O3   N CS         3
c )  Pt C l  O N o   N O 3  N C S       3
d )  Pt C l  O N o   N O 3  SC N        3
39. Intramolecular H-bond  ortho effect are more acidic nature
40. Alternate double bonds with lone pair present in the compound.
41. Order of  I  effect.
42. K a1 K a2
I P P
P  6
2
43. Cl


SoCl2
 A (Major product)

HO CH 2  OH

OH (O lone pairs present conjugation)

SoCl2 / P4

Cl CH 2Cl

Only non-phenolic – OH will be replaced according to darzens reaction.

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44. CH 3
O

Con.H SO
2 4



Aldol reaction H 3C CH 3
H in acidic medium
CH 3  H 2O

H 2CH  C  O H 2CH  C  CH 3

O
O  C  CH H2
CH 3

45. O O

NH 2 NH NH
NH 2

 CH3CO 2 O / Py
 3

KBrO / HBr
 3  H O


Br Br

NaNO2  2 HCl
 
N 2 Cl 0  50 C
Br

Cu / HBr
 

Br Br

46.  Ph  NH 2  A
Ph  Co  NH 2 
Br2  KOH

CHCl3  KOH

3  Ph  N C H O
HCOOH  Ph  NH 2 
 B
(C)
47. Dinitrogen and dioxygen combine to form nitric oxide when the mixture is heated to
2273 – 3273 K in an electric arc.
48. Br Br Br

Br

Br2 / Fe


1,2-dibromobenzene
Br
1,4-dibromobenzene

49. p-hydroxy benzoic acid has higher b. pt. than o-hydroxybenzoic acid due to
intermoleucalr H-bonding.
50. Schotky in the pressure leads to increased collision frequency, missing from their
respective site.
51. For adiabatic process q 0 U     Pext V2  V1   6  25  40   90L  bar
Now H  U    PV   90  150  160   90  10 H  80 L  bar
52.
For w.A & S.B OH    K .C
Ka  
53. For zero order reaction, the rate law is written as
x 0.25 x 0.06
KI    5 K II   5
t 0.05 t 0.12
54. 2  20  60 24
Eq. of OH  produced  
96500 965

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24
Eq. of NaOH produced   Eq. of HNO3needed
965
24 V
 0.62   V  40ml
965 1000
55. T f  K f m 0.2325  1.86m m  0.125
 0.125 mole Glucose present with 1000 grm water
 18 grms of Glucose present with _______?
1000
  18  800 grm of water
0.125  180
 weight of ice  initial wt of H2O  final 1000  800  200 grms
56. Let x atoms are in  2 state x  2  1.90  x 1  2  0 x  0.1
nCu  2 0.1 1 n 1
   Cu  18
nCu  1 1.90  0.1 18 nCu  2
57. Millimoles of surfactant molecules  100  0.01  1m.mol
 Mole of surfactant molecules  103 moles
3 23 20
 No. of surfactant molecule  10  6 10  6 10 molecules
 Area occupied by a surfactant molecule
0.54 cm 2
a 2
 9  1022 a 2  9  1022 cm2 a  3  1011cm
6  1020
58.
OH
OH

CHO
CHCl3 
NaOH

 P  Salicylaldehyde

Molecular formula of product P  C7H6O2

12  7
So, mass % of C in P   100  68.85  69%
84  6  32
59. O O

O  H 2 N  NH  C  NH 2 
H O N  NH  C  NH 2
2

60. N 2O3 
N N
O O O
O
P
P
P4O6 O O O O
O P O P
P4O10

O P O P O
O
N O N O O O
O O
N 2O5 
O P
P
O O

S
OH OH
O H 2 S2O3 HO S OH
P P
H 4 P2O5

h O O
O O OH

O O O S O
H 5 P3O10 HO P O P O P OH
H 2 S 2O5 HO S S OH

OH OH OH O
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MATHEMATICS
61. Given that f  x   2  cos x which is continuous and differentiable every where
Also, f '  x    sin x,  f '  x   0  x  n
 There exists c t , t    for t  R such that f '  x   0
 Statement – 1 is true.
Also f  x  being periodic 2 , statement – 2 is true, but statement – 2 is not a
correct explanation of statement – 1.
62. x2   x  B  0  D  0   2  4  0   2  4
 ,    1,11,2.......1,6 ,  2,2 ,  2,3 ,  2,4 ,  2,5 ,  26
 3,3 ,  3,4 ,  3,5 ,  3,6 ,  4,5 ,  4,6
63. A, D , I , K , M , N , N
M
A
N
K
I
N
D
 6!  3    2  3!  1 2!  1! 0!
4 4!
Rank 
2 2
 1492
64. Mean + variance  24  np  npq  24 …………. (i) np.npq =128 ……(ii)
33
 p  x  1  p  x  2   32C1 pq31  32C2 p 2q30 
228
65. Let PR  x
15
tan 600  ………… (1)
AQ
x  15
tan 750 
AQ

On dividing (i) by (ii) 

15 3
 x  15 
15  2  3
 x  10 3
 
2  3 x  15 3
PQ  15  10 3  5 3  2 3  
P

150 15 m
600
Q

66. cos x  sin x, x   0, 2 

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For interval  0, 2  there are 2 solutions

(1)
(2)

  2
2

67. Given: a1, a2 , a3 , a4 are in G.P.

Then b1, b2 , b3 , b4 are the numbers a1, a1  a2 , a1  a2  a3 , a1  a2  a3  a4 or
a, a  ar , a  ar  ar 2 , a  ar  ar 2  ar 3
Since, above numbers are neither in A.P. nor in G.P.
Therefore, statement 1 is true.
1 1 1 1
Also, , , , are not in A.P.
a a  ar a  ar  ar 2 a  ar  ar 2  ar 3
 b1, b2 , b3 , b4 are not in H.P.
 Statement 2 is false.
68. Let point P  h, k  & Q  p, q 
Equation of circle by using diameter form.
 x  h  x  p    y  k  y  q   0
(where h, p are the roots of x 2  4 x  6  0 and k , q are the roots of y 2  2 y  7  0 )
x 2  y 2  4 x  2 y  13  0
Now,
Compare it with the given equation, we get
a  2, b  1, c  13, Now, a  b  c  12
69. Given series is
x  1  a  ......... and it forms G.P.
a
Apply S  for a1  1 & r  a .
1 r
1 1 1 1
x  a  1  ; Similarly, y   b 1
1 a x 1 b y
1 1
z  c 1
1 c z
Given that a, b, c are in A.P.
1 1 1
 , , are in A.P.
x y z
70. According to question,
 x  12   y  2 2   x  2 2   y  12  14
 x2  y 2  x  3 y  2  0 …………. (i)
Put x  0 . In eq. (i)

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 y2  3y  2  0
3  17
Apply quadratic formula,  y 
2
2
Put y  0, in eq. (i).  x  x  2  0
x  2,1 .
 3  17   3  17 
Then, point A  2,0  , B 1,0  C  0,  & D  0, 
 2   2 

71.
lim tan 2 x  2sin 2 x  sin x  4   sin 2 x  6sin x  2
x
  
2
Rationalize the functions apply the limit in the denominator.
tan 2 x sin 2 x  3sin x  2  tan 2 x  sin x  1 sin x  2 
 lim    lim
 9  9  6
x x
2 2
1 1 sin x 1  sin x 
2
1
 lim tan 2 x 1  sin x   lim 
6 x 6 x   1  sin x 1  sin x  12
2 2
72. Given statement is
 p  q    p  r    ~ p  q     p  r  ~  p   q  r  
 p   q  r    c  is true.
Now,  A  p  ~ r   q
~  p ~ r   q    p  r   q
~ p   r  q   p   q  r 
From option (D).
 p ~ q   r  p   q  r 
73. For getting infinite solutions   0, 1   2    0 then check all the three equations
2 1 1
Let   1 3 2 0    3
1 4 
7 1 1
And 1  1 3 2 0 K 6   K 3
K 3 4
74. x6
We have y 
 x  2  x  3
At y  axis, x  0  y  1
On differentiating, we get
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dy

 
x 2 5 x  6 1   x  6  2 x  5 

 
dx 2
x2  5x  6
dy
 1 at point  0,1  Slope of normal  1
dx
Now equation of normal is y  1  1 x  0 
 y 1  x  x  y  1
1 1
  ,  satisfy it.
2 2
75.  dx
ecos x sin x
Let I   ……………… (i)
 2
0 1  cos x  ecos x  e cos x 
b b
Applying identity  f  x  dx   f  a  b  x  dx
a a
  cos x
e sin x
I ……………. (ii)
  
dx
2  cos x
0 1  cos x e  ecos x
Add equations (i) and (ii), we get
 sin x
2I   dx
2
0 1  cos x
On putting cos x  t , we get

 0  4
1 dt 1
2I  2   tan 1 t
2
01  t
76. 3 2
13 11 13 11  5
3x  2x 3x  2x 3
x x
I  dx   dx I   dx
 2x 
4 4 4
4
 3x  1 2  3 1   3 1 
x16  2  2  4  2  2  4 
 x x   x x 
3 1  3 2 
Let 2    t , 2    dx  dt
x2 x4  x3 x5 
dt
 1  t 41  1 1 1
Then, I   2     C, I   C
t4 2  4  1  2  3  3 1 
3
2  2
 4
 x x 
1 x12
I C
 2x 
6 4 2 3
 3x  1

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77. A  0 /2   sin x  cos x   cos x  sin x  dx
A  0 /4   sin x  cos x    cos x  sin x   dx
  /4
/2
  sin x  cos x    sin x  cos x   dx
A  20 /4 sin xdx  2 /4
/2
cos x dx
 1   1 
A  2   1  2 1  
 2   2
A  4  2 2  2 2 2 1  
78. dy  1 
Given,  y2
dx  x log x 
1
 dx
e 
log log log x 
I .F .  e x x  log x
y.log x   2log xdx  c
y log x  2  x log x  x   c
Put x  1, y.0  2  c
c2
Put x  e
y log e  2e  log e  1  c
y e  c  2
 
79. Given a   iˆ  ˆj  kˆ & b  2iˆ  ˆj   kˆ .
 
 
a  b  1    iˆ   2  2 ˆj    2  kˆ
 
Projection of a  b on iˆ  2 ˆj  2 kˆ
 

 
a  b . iˆ  2 ˆj  2kˆ  30
3
13
 2 2    91  0    7, .
 2

80. Here, P, Q, R are collinear  PR   PQ
2iˆ   y  3  ˆj   z  4  kˆ    6iˆ  3 ˆj  6kˆ 
1
 6  2, y  3  3 , z  4  6    , y  2, z  6
3
 Point R  4, 2,6 
2 2 2
Now, OR   4    2    6   56  2 14
81. There are 5 students in class 10, 6 students in class 11 and 8 students in class 12.
Number of ways, in which 10 students can be selected, according to given conditions
10th class 11th class 12th class

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2 2 6  5C2  6C2  8C6
2 3 5  5C2  6C 3  8C 5
3 2 5  5C3  6C 2  8C 5
 total number of ways
 5C2  6C2  8C6  5C2  6C3  8C5  5C3  6 C  8C2  23800
2
But total number of ways = 100k [given]
 100k  23800  k  238
82. On comparing with  x 2  2 hxy  by 2  0 we have a  11,2h  100, b  5 . If m1 and
m2 be slopes of lines represented by 11x 2  100 xy  5 y 2  0 then
2h a 11
s  m1  m2    20; p  m1m2  
b b 5
 11 
 5  p  20 s   5   400   2011
5 
83. 3x 2 3x  12
We have,  y
4 2
For point of intersection, we have 3 x 2  6 x  24
 x 2  2 x  8  0   x  4  x  2   0
 x  4 and x  2

4  3 x  12 3 x 2 
 Required area      dx
 2 4 
2  
4
 3x2 x3 
  6x  
 4 4 
2
 12  24  16    3  12  2   20   7   27
84. For circle, x 2  y 2  10 x  10 y  41  0
Centre C1   5,5  and radius, r1  3
For circle, x 2  y 2  24 x  10 y  160  0
Centre C2  12,5  and radius, r2  3
C1C2  7  r1  r2  Circles are separated
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 Required minimum possible distance, p1 p2
 7   3  3  1
85. If the line cuts the circle, then
CP  r. ……….. (i)
Since, equation of circle is x 2  y 2  2 x  4 y  4  0
 Centre  1,2  and radius  1  4  4  1
Also, the equation of line is 3x  4 y  k .
38k
 Using (i), 1
9  16
11  k
  1  11  k  5
5
 11  k  5 and  11  k   5
 k  6 and 11  k  5  k  16  k   6,16 
Integral values of k  7,8,9,10,11,12,13,14,15
Thus, number of integral values of k is 9.
86. Equation of straight line passes through P  3,2, 1 and parallel to given line is
x  3 y  2 z 1
   ……………. (i)
2 2 1
 Q  2  3, 2  2,   1 be any point in line (i) and it is also lies on given plane.
 3  2  3   2  2   4    1  1  0
  4  4  0    1
87. 100  a
Here d 
n 1
Now, A1  a  d and, An  100  d
A 1 ad 1
So,  1     7 a  8d  100
An 7 100  d 7
 100  a 
 7a  8    100 ……….. (i)
 n 1 
 a  n  33 ……….. (ii)
Now, by Eq. (i) and (ii)
7 n 2  132n  667  0
29
n  23 and n  reject.
7
88. 10 k m
Given expression is  
4 2
k 1 k  k  1 n

 
2
  2
1 10 k  k  1  k  k  1 
 
2 k 1 k 2  k  1 k 2  k  1

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  
1  10  1 1    55  m
  
    
2  k 1 k 2  k  1 k  k  1  
 2

111 n

 m  n  166
89. x2 y 2
Given ellipse is  1
16 9
So equation of tangent to the ellipse is,
y  mx  a 2m2  b2
y  mx  16m2  9 ……………. (i)
Now, given circle is
x 2  y 2  12
So equation of tangent to the circle is
ymx  12 1  m2 ……………… (ii)
For common tangent on equating equations (i) and (ii)
 16m 2  9  12 1  m 2  
16m2  12m2  3 4m 2  3 12m 2  9
90. Given  ,  roots of equation 3 x 2   x  1  0
 1
So,     , 
3 3
According to given relation of roots,
1 1     2  2
   15  2  9    3
2 2 2 2
   
2
2  3  2 5
Now,          2
2 2 2 2
     
 3  3 3

    
2
Take, 6  3   3  6      2   2  

   3   5 1 2  2
 6        6 1  2   6  4  24 .
 3  3 3  
 

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_(NUCLEUS,STERLING) & LIIT _BT JEE-MAIN Date: 22-01-2023
Time: 09.00Am to 12.00Pm GTM-12 Max. Marks: 300
KEY SHEET
PHYSICS
1) 2 2) 4 3) 2 4) 1 5) 1
6) 2 7) 4 8) 4 9) 2 10) 3
11) 2 12) 2 13) 3 14) 2 15) 1
16) 1 17) 1 18) 2 19) 2 20) 2
21) 3 22) 3 23) 2 24) 40 25) 5
26) 0 27) 4 28) 40 29) 51 30) 9

CHEMISTRY
31) 3 32) 1 33) 4 34) 3 35) 3
36) 4 37) 1 38) 4 39) 3 40) 1
41) 4 42) 2 43) 4 44) 4 45) 2
46) 4 47) 2 48) 1 49) 4 50) 1
51) 2 52) 0 53) 5 54) 6 55) 5
56) 2 57) 5 58) 8 59) 2 60) 9

MATHEMATICS
61) 3 62) 1 63) 3 64) 1 65) 2
66) 4 67) 1 68) 3 69) 1 70) 3
71) 3 72) 1 73) 2 74) 2 75) 3
76) 1 77) 1 78) 1 79) 3 80) 2
81) 2 82) 25 83) 7 84) 1 85) 8
86) 2018 87) 2 88) 26 89) 385 90) 100

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SOLUTIONS
PHYSICS
1. vy u y  gt 4 1
tan    ux  m/s 3  u y 1  10 12 uy  8 m / s
vx ux 1 2
8  10 1
 tan   
4 2
2. ke  2mg  mg
3.  P dt  KE
4. Density is given as p  x   a 1  bx 2  where a and b are constants and 0  x  1
Let b  0, in this cases p  x   a  constant Hence, centre of mass will be at x  0.5m.
(middle of the rod)
 
5. 
L  I cm  m r  v cm 
1 5

2
  
MR 2 kˆ  m 4 Riˆ  3Rjˆ   Riˆ   MR 2 kˆ
2

6. F  F12  F22  2 F1 F2 cos 
3GMm a
2
 mr 2 Here, r 
a 3
7. Motion of a simple pendulum is simple harmonic for small angular displacement only,
not for all angular displacement.
8.
l 
Fl
;  BC 
 70 103  1
 3.5 107 m
Ay 1 2 10 11

9. 1
P0  Pg  2h    2 p  gh  P0   2 P  v 2
2
Liquid of density 2p is coming out
10. V  Vs 1
n11  n
V  Vs

11. P1  d1 
 
P2  d 2 
12. Q T
  2  2 ;P 
W
W T1  T2 t
13. Conceptual
14. 1 1 1  1 1 1
    1    ;  
f  R1 R2  f v u
15. 1 qx qx d 2x 1 qex
E  , m 2  
4   a 2  x 
2 3/2 4 0 a 3
dt 4 0 a 3

1 qe
So motion is S.H.M.  2 
4 0 ma 3

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16. L T g 2
L 2T  T T 
T  2 L 2  or   2   20 0 
g 4 L T  T0 
T20  T0 1 1
Or       20C  10C  
T0 2 2
Here T20 is the correct time period . Since the length L of the rod decreases when
cooled, the time period decreases. So the clock runs fast at 0C . The time gained in a
day   24hrs  24  60  60 s  is
10  24 hrs  10   24  60  60 s  Thus
12 s
12s  10C  24  60 s     14 106 per C
10C   24  60  60 s 
17. R  2100  X  X

p R
Applying 
Q S
2 X R X  20
We have  i  
R 10  X 2 80  x
Solving equations (i) and (ii) we get R= 3
 Correct option is (a)
18. B1 n1  i1 
  
B2 n2  i2 
19. For given circuit current is lagging the voltage by  / 2 so circuit is purely inductive
and there is no power consumption in the circuit. The work done by battery is stored as
magnetic energy in the inductor.
20. 1

1

1
 3  ?
3 1 2
21. Dimensional formula of Y  ML1T 2
Dimension of L,b,t,e=L
22. h h p ma 4
      2 :1
mv 2mE a m p 1
23. F applied  f K 0t   mg K 0 3 mg
a    g
m m k0 m
 3 g   g  2 g  a  2 g

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24. x  n11  n2  2  n11  n2 2

As, n11 60  4000
n2    40
2 6000
25. ax 
qE dv q
 x  5  2x 
m dt m
At time t 
v E   5  2 x   106
dv q
v  5  2x 
dx m
vx2 q 2q
   5 x  x 2   vx2  5x  x2 
2 m m
2q

dx
dt

m
5x  x2 
For x to be maximum
dx
 0  5 x  x 2  0  x  5m
dt

26. K.C.L
27. B 2l 2V
F
R
28.   1  2 or 
1 1 1
 , where t is T1/ 2 T1/ 2  20 years
t t1 t2
29. 12  8 8
Current in zener diode    6.4mA
500 5 103
Power lost in zener diode  VI  8  6.4mw  51.2 mw

30. m
Vm
Ve
3
Vm   m  Ve  12  9V
4

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CHEMISTRY
31. O.P = (6.35/12.7)100 = 50%
32. All others are used for this purpose and they form cyclic acetal or cyclic thioacetals

at

33. Phenol forms 2,4,6-tribromophenol on treatment with Br2 in water.

In H2O, phenoxide ion is formed which highly activates it towards electrophilic
substitution reaction.
34. 10 aromatic amine on coupling with Benzene diazonium chloride gives azo dye test.
35. The monomer of orlon is acrylonitrile.
36. SN2 order of reactivity is methyl halide>primary>secondary>tertiary
37.

38. 1s2 1p6 2s2 2p63s2 2d8

39. Assertion is true but Reason is false.

40. Assertion and Reason are true and reason is the correct explanation of assertion.
41. nHc 1 400
   2 3,Also nH c  nN 2  8, No of moles in first vessel and second
nN 2 3 200
16 24 1 p
vessel respectively are  ,  ,P  1.066atm
5 5 300  2 200  16
5
42. FACT
43. Conceptual
44. Conceptual
45. Due to large size Cs + is least hydrated among alkali metal cation due high hydration
Li(+aq ) mobility is less. I.P. of Li is more than sodium due to small size.

46. Conceptual
47. strong heating
(A) 2H3PO2  H3PO4  PH3

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(B) PCl3  3H2O   H3PO3  3HCl

Hydrolysis

(C) 2NO2  H2O  HNO2  HNO3

(D) 4HNO3  P4O10  4HPO3  2N2O5
PH3 and HNO2 act as reducing agents.
48. V  CO 6 can be reduced easily as the effective number of V is 35
49. Cl2 is minor product
50. 5- membered stable ring drives reaction forward
51. NO2 & CN groups increase the electrophilicity of diazonium ion
52. Oxygens are in resonance with the rings. So, no cleavage of bond takes place due to
double bond character
53. (ii),(iii),(vi) do not have  hydrogens.
(v),(viii) under go oxidation at side chain = and 
54. n f of Ax B y is xy and xy=6, x=3 and y=2
55. TlI 3  aq  Tl  (aq)  I 3 (aq)
x 2  2.5 109
56. Formula of the compound = X 4Y4 Z8
Final formula = X 3.5Y3Z 4
57. I law of thermodynamics
58. With HCl + H 2S only group-II cations are precipitated but with NH 4 Cl  NH 4 OH  K 2S
along with group –IV, group –II cations are also precipitated if they are not
precipitated earlier
59. O O O
+
O Br Br Br O
+6 O O O

+4
60. VSEPR theory , back bond application.

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MATHEMATICS
61. Clearly n  
 x2 1 2
l t  x  (cos ) x 
x  e x 

1
l t x 2 (cos 1)
 0  e x x

 e1/2
62. dy  3 1  1 1 
   x  ,P   , 
dx  p 4 8  8 16 
63. f '(x)  0  f (x) is one one
64. Tangent at (0,1) is 3x  y  1  0
Area = 500/81
65. Point of intersection is (a  b)
Length of the  ar
AB  AC

AB
66. The plane  contains
L1 and  ar to the plane containing L1 & L 2
67. f (x) is real  0  [x]  2  0  x  3
g(x) is real
 x 
68. 
r  z1 cos
8
R  z1

ratio  cos 2
8
69. pq   p  (p  q)
t
(p  q)  q   p(p  q)  q
t
70. ab  0
and a  3 b
(or) b  3 a
71. A lies inside the director circle and outside the ellipse
72. A  B  C  D  2   cot A   cot A cot B cot C
  cot A  cot A cot B cot C cot D tan A


 tan A    tan A 
 cot A
73. abc  total  (numbers difer by  10)  100c 2  90c2  945

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74. 5
1 1 1
6

  .     .. 1
2 2 2 
3 4
1 1 1 1 4
      ....
2
  2 2
  2
75. b r 1  b r 2 1  br br 2
br   
b r 1  b r  2 1  b r b r 1
n 1  1  b 
G.E.  b n  2  r
  b0
 1  br 
76. Conceptual
77. If ,  are the
roots, then
a 2  b 2  ( 2  1)(2  1)
78. (A  B)  c1  15(2n  1) / n  , n  50
79. PQ  distance between the lines
80. (2, 2)  R
(2, 0)(0, 2)  R, but(2, 2)  R
81. Sketch the graph
82. 2x  y1/5  y 1/5
diff w.r.t. x
10y   y1/5  y 1/5  y1
 100y 2  4(x 2  1)(y1 ) 2  (x 2  1)(y1 ) 2  25y 2
diff w.r.t x
(x 2  1)  xy1  25y
83. n 5
5
1  55
A    3  5
2  2
B2 5

84.  1 1   1 1 
(x, y)   ,  (or)  , 
2 2   2 2
85. A = (2,4),B = (-4,-2)
86. Apply by parts twice
87.   0  a = -1 only

Sum  n  1 n n  n  1
88. 2 2

 2
 2n 
2 2
89. a 5  8  no of numbers  7C4
a 5  8  no of numbers  8C4  5C4
90. y1 (x 1 )  0
 x 12  2cx 1  1  0
Now x 1 .y(x 1 )
x 12  cx 1 1  cx 1 1
  
1  x 12 2  2cx 1 2

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_(NUCLEUS,STERLING) & LIIT _BT JEE-MAIN Date: 23-01-2023
Time: 09.00Am to 12.00Pm GTM-13 Max. Marks: 300
KEY SHEET
PHYSICS
1) 2 2) 3 3) 1 4) 2 5) 2
6) 2 7) 4 8) 3 9) 3 10) 1
11) 4 12) 1 13) 3 14) 3 15) 4
16) 4 17) 2 18) 2 19) 1 20) 4
21) 1 22) 6 23) 10 24) 8 25) 9
26) 18 27) 9 28) 2 29) 1 30) 1

CHEMISTRY
31) 2 32) 2 33) 4 34) 1 35) 1
36) 2 37) 3 38) 3 39) 1 40) 1
41) 3 42) 4 43) 2 44) 4 45) 4
46) 3 47) 1 48) 1 49) 4 50) 1
51) 6 52) 1 53) 2 54) 90 55) 1
56) 34 57) 5 58) 8 59) 3 60) 1

MATHEMATICS
61) 3 62) 2 63) 4 64) 4 65) 1
66) 4 67) 3 68) 4 69) 1 70) 3
71) 2 72) 1 73) 3 74) 2 75) 3
76) 3 77) 3 78) 4 79) 4 80) 2
81) 1 82) 9 83) 10 84) 6 85) 1
86) 21 87) 81 88) 19 89) 120 90) 2

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SOLUTIONS
PHYSICS
1. 1 M .S .D 1
L.C    0.1mm
no.of V .S .D 10
Side of cube L  10mm  1  0.1mm  1.01cm
m 2.736
   2.66 gcm3
v 1.01 3

2.
u

 H

O R
2

 u 2 sin 2 
2 
 2 g  sin 2  1
  
H
Tan  Tan  tan 
 R / 2  u sin 2 sin 2
2 2
g
3. f

mgsin37 f1
f

3mgsin37
For equilibrium of plank 3mg sin37  f  f1
Where f1  umg cos 37 = 0.5 (4 mg cos37) = 2 mg cos37°
9 8 mg
mg  mg  f  f   2m
5 5 5
For equillibrium of man mg sin37 + f = ma
6m + 2m = ma
a  8ms 2
4. V = Vmaxwhen a = 0
Let dispt. of block is x0 when its speed is max
mg sin    mg cos  k x0
mg sin    mg cos   k x0 ____(1)
1
By W.E.T, K .E f  K .Ei  mgx0 sin   kx02   mgx0 cos
2
1 2 1
mVmax  0  mgx0 sin   kx02   mgx0 cos
2 2
1 1
2
mVmax   mgx0 sin    mg cos  x0  kx02 ____(2)
2 2

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1 2 1 1
(1) or (2)  mVmax  kx02  kx02  kx02
2 2 2
K
Vmax  x0 substituting eq (1), we get
m
K  mg sin    mg cos 
Vmax 
m  K 
m
Vmax   sin    cos  g
k
5. m x  m2 x2 100  25  50  55
X cm  1 1   35 cm
m1  m2 100  50
Distance from centre of rod = 35 – 25 = 10 cm to the right
6. 1 2 1 2 1 2 1 2
mv  I   mV 1  I  2  mv1  mgh
2 2 2 2
2
v
v1  v / 2  h 
4g
7. I R  I  I P  IQ but I P   I Q  I R  0
8. Conceptual
9. L
W  LA g OT 
2sin 
 L 
FB    A w g
 2sin  
OT L
FB acts at P, where OP  
2 4sin 
FB

Q T

P W
O 

R
Balancing torque about O,
L  L  L 
W  OQ cos   FB  OP cos  LA g     A w g  
 2   2sin   4sin  
 1
sin 2   w  sin 2      450
4 2
10. kA 1   2  t
=  mlice
l
11. Conceptual
12. Conceptual
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13. Y  ( A  B )ꞏ( AꞏB )

 ( A  B)ꞏ( A  B )  AA  AB  BꞏA  BB  0  AꞏB  BꞏA  0  AꞏB  BꞏA
14. 1
N  N 0et  0  0  10  2   
N N
e2 e10 5
ln 2
T1/2   5ln 2

15. At t   circuit will be in steady state. Then circuit becomes as below
2A
4A
12V 6
3

I=6A

I=6A

At t  0 current through indicator is

A B
I
12V
3 6

1A

VL L=1H
9

D C

Applying K V L to loop ABCDA

di di
6I  L  12 6 I  12    L  1H 
dt dt
di i dI t 1
 dt      dt  [ln(i  2)]i6  [t  0]
6i  12 6 6i  12 0 6
i 2 i  2 6t
ln    6t   e  i  2  4e6t
 4  4
16. Wave is traveling along x -axis.
B is directed along z -axis
E should be directed along kˆ  iˆ  ˆj along y-axis
E0  B0C  2  107  3  108  60V / m
17. l  AC  2 R (cos 45iˆ  sin 45 ˆj )  2(iˆ  ˆj )
F  i ( i  B )  1( 2(iˆ  ˆj )  (3iˆ  4 ˆj  kˆ)] F  2(iˆ  ˆj  kˆ)
18. when x is connected to y balancing length l is proportional to p.d on
100  V11  i (100)  l1  (1)
When x is connected to z , l is proportional to p.d on (100+R)
100  R 588
 V2  i (100  R )  l2  (2) ,   R  47
100 400
19.  v  v0 
f  f0  
V  Vs  
 340  20 
320  f0    0  1210 Hz.
 340  10 
20. W= area of cycle
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  4  3 4  2    2  1 3  2.5  2.5 5

W   watmL
2 2 2 4
5
As cyclic process is in anticlockwise direction w   atmL
4
21. V O / O  V O  V O  10cos 60 i  10sin 60 j   10cos 600 i  10sin 600 j 

0  0 
    
V O / O  10i
O VO / O A

O'
(assumed to be rest)
Separation will be minimum when particle O is at A, i.e releative dispt OA = 10 m
OA 10
t   1sec
VO / O 10
22. dV dV dV y
V 2  Vx2  V y2 2V  2Vx x  2V y
dt dt dt
 15 
V a  V a  3  5   4  
dV

x x y y
  4
 6ms 1
dt 2 2 2 2
V Vx y 3 4
23. FB

6 rv

v
First case : mg

FB  6 rv  mg ____(1)

F 2 mg FB 
FB
6rvt

vt

For second case : mg

2mg  2 FB  FB  mg  6 rvt

mg  FB  6 rvt ____(2)

From (1) & (2) V  Vt  10cm / s

24. Velocity of a man Vm  V sin  iˆ  V cos ˆj
Velocity of image VI  v sin  iˆ  v cos ˆj
Velocity of a man with respect to image
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xv
VMI  VM  VI  2v sin  iˆ  sin  (given)
4

V
vcos vcos
y
 

man vsin vsin

-x +x
x
 2 x 8
Image
-y

4
25. V  Vz Vz
I z  in 
R RL
90  30 30
Vz    15  103  6  103  9  103 A  9mA
3 3
4  10 5  10
26.  1 1 3
E  13.6     13.6   10.2eV
 4 1 4
1 2
KEman  E  W0 mvmax  10.2  1.2  9eV
2
2  9  1.6  1019
Vmax   2  1.6  1012  3.2  106  1.78  106
9.1  1031
 17.8  105  N  105 N  18
27. Band width 90
No. of stations   9
2  highest band width 25
c
28. c c 2c
A c
=
A B C
c

C  2C 2 2 A 2  2  3 / 2 2 0 x 0
C AC   C  0  C AC   0   x2
3C 3 3 d 3 d d d
29. Force on Q2 will be due to induced charge on outer surface only which will be same in
both cases along with its distribution
B
30. 5V 2 A 3
5V
I1 I2
C

3 3

5 5 5
i1   1A , i2   A. v A  0  3i1  3  1  3 V
23 33 6
5
vB  0  3i2  3   2.5 V
6
1
P.d on capacitor 3 V  2.5 V  0.5 V , q  CV  2   1C
2

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CHEMISTRY
31. 2k2CrO4  H 2 SO4  k2Cr2O7  Na2 SO4  H 2O
Oxidation state of Cr in k2CrO4 and k2Cr2O7 is +6.i.e no change in o.s so, the
explanation is wrong.
32. Compound can be chiral even in the absence of chiral atoms
33. Due to its higher oxidation state, B3 ion does not exists.
34. Activation enthalpy to form C is 5 kJ more than that to form D
35. The PH of NaOH is more than 1 and during the titration it decreases so graph(1) is
correct
36. For the extraction of iron, haematite ore in used
Haematite= Fe2O3
37. RT
P
Vm  b
PVm pb pb
PVm  pb  RT 1  Z 1
RT RT RT
b
Slope of Z vs P curve (straight line)=
RT
Higher the value of b, more steep will be the curve and b=size of gas molecules
38. O
CH3 CCH3 CH3 CH2 CHO
Generally, aldehydes are more reactive than ketone in nucleophillic addition
reactions. Rate if reaction with alcohol to form acetal and ketal is
O
CH3 CH2 CHOCH3 CCH3

39. BeCl2 in vapour phase exist as dimer(below 1200 K temperature)

BeCl2 in solid state has chain structure.
40. HF has highest boiling point among the hydrogen halides due to strong H-bonding
between HF molecules
41. S N 1 reaction proceeds via formation of carbocation.
42. Cumenes gives phenols
43. Arsenious sulphide sol is negatively charged, so according to Hardy-schulze rule the
cation which is in more oxidation state will be most effective.
44. N 2 Diamagnetic  N 2 (Paramagnetic)
O2 paramagnetic  O2 (Paramagnetic)
O2 Paramagnetic  O22 (Diamagnetics but bond order decreases from 2 to 1)
NO(Paramagnetic bond order 2.5)  NO (Diamagnetic, bond order bond order 3)
45. Second electron gain enthalpy of oxygen is positive
46. PH of rain water is around 5.6
47. It is Neurotransmitter

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48. Z M0 
Density 
N A  a3
Z=4 (FCC) M 0  63.5 g N A  6  1023
a  x  108 cm
4  63.5 422 g
d .
6  1023  x3  1024 x3 cm3
49. U  n.cvm T  5  28  100 =14 kJ
  PV   nR T2  T1   5  8  100  4kJ
50. NH2 NC

C 
HCl2.KOH

CH2

O CN O
CN
H2 / Pd C
NH CH3

OH
H2N

51. 0.27 gm in 100 ml of hexane

In 10 ml of aqueous solution only 0.027 gm acid is present.
0.027
Volume of 0.027 g acid=  ml
0.9
0.027
 r 2h  (Given r=10cm,   3 )
0.9
h  104 cm  106 m
52. ICl5 is sp3d 2 hybridized (5 bond pairs, 1 lone pair)

Cl Cl

square pyramidal

Cl Cl
Cl

ICl4 is sp3d 2 hybridized (4 bond pairs,2 lone pairs)

-
Cl Cl

square planar

Cl Cl

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53. eq n1 S  O2  SO2 K1  1052

eq n 2 2S  3O2  2SO3 K 2  10129
eq n 3 2SO2  O2  2SO3
10129
eq 3  eq 2  2 eq 1 
n n
  n  1025
1052 
2

54. Volume occupied by atoms in solid 2

4 4 3
  r 3    2r   12 r 3
3 3
Relationship between edge length (a) and edge of atom(r)
6r
 6r  3a  a  a  2 3r
3
4 4 3
P.E   r 3    2r 
3 3
12 r 3
Packing efficiency=   100  90%
3
 6r 
 
 3
55. CF3 CH CH2

H
+ +
CF3 CH CH3 CF3 CH2 CH2
(More stable)

(Less stable due to -1 effect Cr

of CF3)
CF3 CH2 CH2 Cl
(Anti markovnikov)
56. 11.2 V of H 2O2
1
H 2O2  H 2O  O2
2
11.2 L of O2 at STP =0.5 mol
It means 1 L of given H 2O2 solution consist 1 mole of H 2O2 (i.e 34g)
34
Strength=  100  3.4%
100
57. N
NH2
N
NH2

H3I
base
  C 
xN2


N N N N
(-)
H(-)
NH3
N
N

N N
CH3

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In the given compound, H-atom attached to secondary N-atom is more acidic. The
base removes the more acidic H-atom and the conjugate base of the given compound
attackes at CH 3 group to give the final product shown above.
58. O

Cr
H

1BuCl

O
Conc
H
 
2SO4

H
+


59. Ag  aq   Fe2 aq   Fe3 aq   Ag aq 

0
Ecell E   E0
Ag / Ag Fe2  / Fe2 
To calculate E 0 3 2 
Fe / Fe
y

Fe3 Fe2 Fe

E0  z
EFe2 /Fe2 3z 2y

E0 x
Ag  / Ag
0
ECell  x  3z  2 y
60. According to Henrys law

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MATHEMATICS
61. Hint: PQ  3( AB)  7 2
C (2,1)r  5  C
3 3 1
M  ,  CM 
2 2 2
1
7 2  2 5  c   C  20  r  5
2
62. A2  A  A
 ( AB )( AB )  A( BA) B  ( AB) B  AB  A
Similarly B2  B  A  A2  A3 
B  B 2  B3  

 A2021  B2021 
2022
 ( A  B) 2022

( A  B)2  ( A  B)( A  B)  A2  AB  BA  B2
 A2  A  B  B2  2( A  B)
( A  B)3  22 ( A  B)  ( A  B)2022  22021( A  B)
63. x c2 x c3 x 2
y  
x  c1  x  c1  x  c2   x  c1  x  c2   x  c3 
x2 c3 x 2
 
( x  4c1)  x  c2   x  c1  x  c2   x  c3 
x3

 x  c1  x  c2   x  c3 
 ln y  ln x3  ln  x  c1   ln  x  c2   ln  x  c3 
 ln y  3ln x  ln  x  c1   ln  x  c2   ln  x  c3 
y' 3 1 1 1
   
y x x  c1 x  c2 x  c3
y x x x  y  c1 c c 
y'  3       2  3 
x x  c1 x  c2 x  c3  x  c1  x c2  x c3  x 
64. 1
T p  a  ( p  1)d  ------- (1)
q
1
Tq  a  (q  1)d  --------(2)
p
pq 1 1
(1)-(2)  ( p  q)d  d  a  TPq  1
pq pq pq
Clearly 1 is one of the roots of given equation
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65. Statement 2 is correct as when 3 ,3 ,3 are in G.P., we have

a b c

 3    3  3   2b  a  c  a, b, c are in A.P. Thus, selecting three numbers in G.P.

2
b a c

from 3 ,3 ,3 ,,3 ,3  is equivalent to selecting 3 numbers from {1, 2,3,,101}

1 2 3 100 101

which are in A.P. Now, a, b, c are in A.P. if either a and c are odd or a and c
are even.
Number of selection ways of 2 odd numbers is 51 C2 . Number of selection
ways of 2 even numbers is 50 C2 . Hence, total number of ways is
51
C2  50 C2  1275  1225  2500 .

66. 7 9
1 x 2  1 x 2
dx
I2       
0 5  x   5  x  (5  x)2
x 5 dx 1
Put t dx  dt   dt
5 x 2 2 5
(5  x) (5  x)
1 7 9
dt

 I 2  6 (t ) 2 (1  6t ) 2
0 (5)11/2
; Now Put 6t  

1
and simplify we get I 2  I we Conclude a  30
9/2 7/2 1
5 6
67. m n
 sin 2  ;  cos 2   s  mcosec 2 t  n sec2 
s t
s  t  mcosec2  n sec2   m  n  m cot 2   n tan 2 
3 m cot 2  n
  n  n tan 2

Use Am Gm
 3  2 mn  mn  2 mn3
m  1 n  2 ( m  n) Point is (1,2)
use S1  S11
68.
 (2n  1)  n 2  2n  n 2  1 
 
 sin 1
n 1 
n(n  1)  n 2  2n  n 2  1 
  
 
  2 2 
1  n  2 n  n  1 
 sin  n(n  1) 
n 1  
  2 2 
1  ( n  1)  1  n  1 
 sin  n(n  1) 
n 1  

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 
1 1 1 1 1 
 sin  1 


2 n 1
1 
2

n 1 
 n ( n 1) n 

 1 1 
  sin 1 n  sin 1 n  1 
n 1
 1  1 1  1 1
 Sn   sin 11  sin 1    sin 1  sin 1    sin 1  sin 1 
 2  2 3  3 4
 1 1 
   sin 1  sin 1 
 n n 1
 1 
Sn   sin 1 lim Sn  .
2 n  1 n  2
69. The direction cosines of segment OA are
2 1 3
, and . OA  14n
14 14 14
This means OA will be normal to the plane and the equation of the plane is
2 x  y  3 z  14
70. 2
 2    1 (  1)     1
 

 3 1 

 2    1 (  1)  2    1 (  1)  2    1   
2 2  6 2 (  3) 3  
  
3
  2  2  1 2 4 2
Similarly remaining

 (3   )   (3   )   (3   )


3(     )   2   2   2  7
2 2
71.  4r 2 
n n
1  4r 2 
lim  ln  n   lim  ln  
n  r 1  n 2  n  r 1n  n 2 
 
1 1
 ln 4 x   dx  (2ln 2  2ln x)dx
2

0 0
 2 x ln 2  2 x ln x  2 x]10  2ln 2  2
72. 1
1  2 x   2x  1
3 cos    sin 1 
2 3
2
1 x  1 x    1
I  ex  1

2 1 e  1
x
dx
1
3 3

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1
3
 ex  1 
 2I       dx  I 
2  x x  2 3
1  e 1 e 1

3
73. 1
x  f ' ( x)  f ( x)  1  f ( x)  f ' ( x) 
x
f ( x)  ln x  k
But f (1)  1 k  1  f ( x )  ln x  1
10 10 10
 f e k
  k1ln e k
1   (k  1)  65
k 1 k 1
74. P is true; q is true, r is false verify the options
75. r 2 r
 2  i
9
ar   cis  or e r  1, 2,3,
 9 
 a1, a2 , ......are in GP

a1 a2 a3 a1 a12 a13 1 a1 a12

a4 a5 a6  a14 a15 a16  a1  a14  a17 1 a1 a12  0
a7 a8 a9 a 7 a18 a19 1 a1 a12
1

Now a1a9  a3a7  a110  a110  0

76.  (3 x  2)2
 ( x  3)( x  1) e x  (3, )
 2
f ( x)  ( x  3)( x  1) e(3 x  2) x  [1,3]
 2
( x  3)( x  1) e(3 x  2) x  ( , 1)

Clearly non differentiable of x  1&3
77. | 2a  3b || 3a  b |
S.O.B.S
4a 2  9b 2 ∣2a  b  9a 2  b 2  6a  b
5a 2  6a  b  8 | b |2
1
5a 2  6  8 | b | cos 60  8 | b |2 | a | 1
8
40  3 | b || b |2
| b |2 3 | b | 40  0
(| b | 8)(| b | 5)  0
| b | 5

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78. 6
 6 10 15 
  P  xi   3     
i 4  36 36 36 
xu 2 3 4 5 6 7 8 9 10 11 12
P( x  x) 1 2 3 4 5 6 5 4 3 2 1
36 36 36 36 36 36 36 36 36 36 36
31 77
3 
36 36
79. n1  10, n2  10
Average m1  60 m2  40
1  4; 2  to
2
n  2  n2 22 m1n2  m1  m2 
Standard deviation of Combined series   1 1 
n1  n2  n  n 2 1 2
 126
80. Let height of pole  10l
3l l 10l 2 tan  10l
tan    , tan 2  , 
18 6 2
18 1  tan  18
l 72
use tan    l 
6 5

height of pole  10l  12 10

81.
 
2

h   2 k  2 2   2  1 
1 1 2
2
h1  2  1; k    1
2
 h 1
eliminating  , k    1 4 K  h 2  2h  3
 2 
Locus is ( x  1)2  4( y  1)
82. : |   1  i | 2 circle with c(1, 1)r  2
|  |max  2  2
|   4  5i | 3 circle with c(4, 5) with r  3
(  )max  41  3

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 k  164 
 2 |  |max 3 |  |max  164  18   1    9
 K 2   18 
83. 1 2 1  32 
  8  8   32 Δ   32 [Δ]     10
2 3 3 3

84. f ( x)  ( x  1)( x  2)( x  3)( x  4)( x  5)  x  f (6)  5! 6

85. dx 1 d 2 x d  dx  d  1  dt 1
        
dy t dy 2 dy  dy  dt  t  dy 12t 5
 d 2x 
 2 4
 dy 
   const  n  5 , Sum  5  1
 dx 
n 1
1
 dy  n
 
86.  2 ,  2 are Roots of 4 x2  21x  1  0
Let c   2 , d   2  sn  c n  d n
  
 4 S2021  S2019  4 c 2021  d 2021  c 2019  d 2019 
  
 c 2019 4c 2  1  d 2019 4d 2  1 
 c 2019 (21c)  d 2019 (21d )  21S 2020
87. 5!
  5!
a
coefficient of x5  x 2  ( x)b  (2)c  (2)c (1)b  x 2a b
a !b!c ! a !b!c !
a b c
Now a  b  c  5 (i ) 0 5 0

2a  b  5  (ii ) 1 3 1
(iii ) 2 1 2
5! 5! 5!
Required coefficient  (2)0 (1)5  (2)1(1)3  (2) 2 ( 1)1
0!5!0! 1!3!1! 2!1!2!
 1  40  120  81
88. The no. of Common roots are the no of Common vertices
G.D of 2973&1982 is 991
89. A function f : A  B such that f (0)  f (1)  f (2) in the following Cases
case 1: f (0)  f (1)  f (2) 8 C3
case2 : f (0)  f (1)  f (3) 8 C1

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case3: f (0)  f (1)  f (2) 8 C 2

Case4 : f (0) f (1)  f (2) 8 C2
8C3  8C2   8C2  8 C1 9 C3  CC2 10 C3
90. log18 2log 3  log 2
a  a
log12 2log 2  log 3
log54 3log3  log 2
b  b Let log 3 be y ad log 2 be x
log 24 2log 2  log 3

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 24-01-2023
Time: 09.00Am to 12.00Pm GTM-14 Max. Marks: 300
KEY SHEET
PHYSICS
1) 1 2) 4 3) 1 4) 3 5) 2
6) 4 7) 2 8) 2 9) 2 10) 1
11) 1 12) 3 13) 2 14) 1 15) 4
16) 1 17) 1 18) 2 19) 4 20) 4
21) 8 22) 280 23) 3 24) 11 25) 3
26) 625 27) 45 28) 3 29) 251 30) 7

CHEMISTRY
31) 3 32) 4 33) 1 34) 2 35) 2
36) 1 37) 1 38) 3 39) 4 40) 3
41) 1 42) 2 43) 1 44) 3 45) 2
46) 2 47) 3 48) 2 49) 2 50) 1
51) 6 52) 6 53) 4 54) 10 55) 36
56) 7 57) 300 58) 2 59) 8 60) 2000

MATHEMATICS
61) 3 62) 4 63) 2 64) 4 65) 1
66) 1 67) 4 68) 3 69) 3 70) 2
71) 1 72) 3 73) 1 74) 2 75) 3
76) 4 77) 2 78) 2 79) 4 80) 3
81) 9 82) 5 83) 6 84) 9 85) 9
86) 2 87) 5 88) 3 89) 1 90) 8

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SOLUTIONS
PHYSICS
1. When final image is formed at infinity.
Length of the tube  v0  f e  15  v0  3  v0  12cm
1 1 1 1 1 1
For objective lens       u0  2.4cm
f0 v0 u0  2   12  u0
2.

Given
Direct current I dc  a
Alternative current I ac  b sin t
Total current

 i 2dt T

 2
I  I dc  Iac  I  a  b sin t  I rms  
1
 a  b sin t 2 dt


T

dt
0

 T T T  T

   
 
2 1 
 I rms   a 2 dt  b2 sin 2 t  2ba sin t   sin tdt  0
T 
 
 0 0 0
 0
T


1/2
T 1 T   b2 
 sin 2 tdt  2
 I eff   a 2T  b 2  0   I rms   a 2  
2 T 2   2 

3. conceptual
4. Given
Radius of hollow sphere = R
Specific gravity  
Now mass of spherical shell
M1 Solid volume of the sphere × density of material M1  4 R 2t 
Where M1 =Mass of the hollow sphere
t = Thickness of the hollow sphere
While the mass of water having same volume
M 2 =Whole volume of the sphere × density of the water
4
M 2   R3  w , For the floatation of sphere M1  M 2
3
4
3
R
 4 R 2t    R3 w  tp    w  1g / cm3
3
 
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5.

1
Before 1 K opened the energy stored over capacity C1  CV 2
2
1
  3  106  102  150 J
2
 
2. Now this energy will be dissipated E  i 2 Rt in the two resistors in the ratio of
R1 : R2 as they are in series (current flowing through the resistors is same) when switch
K 2 is closed.
150 J is lost at the radio 2:4=1:2
1 2
In 2 rd of 150  Jand in 4 rd of 150  Jwill be lost
3 3
Therefore, the correct answer is  2 
6.

The induced electro motive force  v  across ends of rotating rod about one end in
B
magnetic field  12  , 2.  can be calculated according to law of conservation
2
energy. Increase in rotational kinetic energy = decrease in gravitational potential energy
1
3. I  2  mgh
2
1 ml 2 2 1 3g
   mg. sin     sin 
2 3 2 1
1 3g
NowV  Bl 2 sin  ......1 V  sin 
2 1
3
Also from 1 , V 12
Therefore, the correct answer is  2 
7. E
tan   45   v
Er
8. Given
Weight of a body at the equator of planet is half of that at the poles

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g g g
i.e.., g    g  R 2  R 2  g 
2 2 2
 g  2 R 2
We know v0  R The escape velocity ve  2 gR

 
 ve  2 R 2 R 2  ve  4 R 2 2  ve  2 R  2V0
9. W0  mg  46 gm
At 1  27 C
W1  30 gm  w0  B1
B1  46  30
B1  16  V11g
 2  42 C  W2  30.5 gm  w0  B2
B 1.2
B2  15.5  V2  2 g 2 
B1 1.24
 15.5 1.24   1
 s      1 
 16 1.2   45
1
 s  2.31 105 I 0C 
43200
10. Velocity of ball when it reaches to surface of liquid

100 gV  500 gV
a where V is the volune of the ball
500V
a=10m/sec2
Apply v=u+at  0= 2 gh  10t
2 gh  10   2 
2  10  h  400  h  20m

11. Given
Density of water  
Cross-sectional area = A
Constant speed = v
Along x and y directions along in momentum of water is

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3
Px mv sin 60  mv
2

mv 3
Px   mv  mv
2 2
9 3
 Pnet  Px2  Py2     mv
4 4
 Pnet  3mv
Force on bend is rate of change of momentum
 dm  2  dm 
 Pnet  3   .v  3 Av    Av 
 dt   dt 
12.

L 2u sin  gL
So   u2  u
2u cos g 2sin  2  min 
gL
2
13. H  H
F  mg  mg  F  mg 1  
h  h
14.

15. 1 2 2u g
h  ut1  gt1  t1  t2 , So h= t1t2
2 g 2
16. Let n moles of diatomic gas dissociate into monoatomic. Then mono 2n moles and dia
5 Pfinal 160
1 – n moles n  
11 Pinitial 33
17. emf in PQ=0
emf in QR= B a 2
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18. Mr 2 3
w0  MR 2 w
2 2
w 2 g Rw0
 w  0 ,  ,t 
3 R 3mg
19. v  vc
Wavelength of sound reaching the hill
f
 v  v0   v  v0  2vc  v  v0 
f1    f , f2    f , f  f1  f 2 
 v  vc   v  vc  v 2  vc2
20. Dimensional analysis
21. 1
fc   6.37kHz
2 RC
1  x2 4
Modulation factor on= 
n 3
 as n=0.6, fmax  m  fc
22. bt

A  A0e 2m
23. k
cut-off potential  max
e
24. X 53
  X  10.60
10 50
25. i  2 2A
26.
Optical source frequency f 
c

 
 3  108 / 800  109  3.8  1014 Hz

Bandwidth of channel (1% of above)= 3.8  1012 Hz

Number of channels = (Total bandwidth of channel)/ (Bandwidth needed per channel)
(a) Number of channels for audio signal)
  
 3.8  1012 / 8  103  4.8  108
27. Given
Electric field E  3.2  105V / m
Magnetic induction  2.0  103T
Mass of the electron  9  1031 kg
Initially we have
E 3.2  105
eE=Evb V    1.6  108 m / s
B 2  103
When electric field is removed, electron will only be moving in the influence of
magnetic field and it follows a circular motion with radius given as

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mV 9  1031  1.6  108 9  101

r  r  0.45m
eB 1.6  1019  2  103 2
28. Pradiation  3% of 100W
P 1
=3W  I= andI   0 E02C
4 r 2 2
P 1 2P
   0 E02C  E02 
4 r 2 2 4 0r 2C

2  3  9  109
23 6
   2.678 . 2.68Vm 1
8 5 2.24
25  3  10
29. Instrument has negative error
e=-5×0.01=-0.05 cm
Measured reading =2.4+0.04  2.46 cm
True reading =2.46+0.05=2.51 cm
30. For maximum intensity on the screen
d sin   n
n
 sin  
d
n  2000  n
sin   
7000 3.5
Since sin >  1
n=0,1,2,3 only
Thus only seven maximas can be obtained on both sides of the screen

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CHEMISTRY
31. (A) The process of precipitating colloidal solution is called coagulation. Hence false.
(B) For colloidal solutions concentration is very small due to very large molar mass
and hence their colligative properties are very small as compared to true
solutions T f is lesser for colloidal solution. Hence false.
(C) At CMC surfactant from micelles. Hence true
(D) Micelles and macromolecular colloids are two different types of colloids. Hence
false.
32. ( A)2 Na  Al  OH 4   CO2  Na2CO3  H 2O  2 Al  OH 3   
 aq 
(B) Function of Na 3AlF6 is to lower the melting point of electrolyte.
(C) During electrolysis of Al2O3 the reactions at anode are:
 2 Al 3     3O 2     
At anode
O2  gas   2e  
 
C  graphite   O2  CO     CO2   
(D) The steel vessel with a lining of carbon acts as cathode.
33.
PV

P According to Boyle’s law

1
P PV=constant
V
So PV vs P is constant
34. A. Natural rubber is polyisoprene containing cis alkene units
 HN   CH 2   C 
 5 n

B. Nylon-6 has amide likage O

C. Cellulose has only   D glucose units

 CF2  CF2 n
Per sulphate
F2C  CF2 
D.
35. Probability will be maximum at a and c.
36. Chemical formula
Sphalerite ZnS
Calamine ZnCO3
Galena PbS
Siderite FeCO3
37. (A) Energy of 2s orbital H atom> 2s orbital of Lithium.
(R) Energy of orbital in same sub-shell decreases with increase in atomic number
38.
O O

is a conjugated diketone

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39. Penicillin G is a narrow spectrum antibiotic
40. Li  has high polarizing power so lithium halides are covalent in nature.
41. Sodium hydrogen carbonate (Baking soda), NaHCO3 is used in the fire
extinguishers.
42. The melting point of carboxylic acid has no trend.
Water solubility of carboxylic acid decreases as we increase the molar mass, due to
increase in hydrophobic part of the chain.
43. Ion Colour of the flame
(A) Cu 2 green flame with blue centre
2
(B) Sr Crimson Red
2
(C) Ba Apple green
44. Buna-N is an addition copolymer of 1,3-butadiene and acrylonitrile.
45. With an increase of temperature, the ionization of water increases implies the
hydrogen and hydroxyl ion concentrations increase. Hence, P H and P OH decrease.
H 2O  H   OH  Is an endothermic reaction.
46. Sodium fusion extract is boiled with concentrated HNO3 to remove sodium cyanide
and sodium sulphide
47. Enamine formation is an example of nucleophilic addition elimination reaction since
O
CH3 CH3
C
C C CH3
H3C

in ketone CH CH carbonyl group is highly sterically hindered hence attack

3 3

of nucleophile will not be possible.

48. Dettol is mixture of
CH3

OH

and

H3C CH3
H3C C OH

CH3
CH3

Chloroxylenol Terpineol
(Compound A) (Compound B)
It has 6 e It has 2 e

Hence compound ‘B’ is Terpineol. O

49. O C CH3
HC CHO
O C CH3

O
H 2O / H 
CrO3 / Acctic an hydride
2CH 3COOH
Hydrolysis

50.

is most stable as it is aromatic

51. I   Pt  NH3  Cl2  Br2  G.I.  2
 4 
 II   Pt  NH3 4 Br2  Cl 2,
   G.I.  2

 III   Pt  NH3 4 BrCl  Br.Cl  G.I.  2

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I, II, III are ionisation isomers of each other, each having 2 geometrical
isomers. Total possible isomers will be 6
52. KMnO 4  KI  MnO 2  I 2
Eq of KMnO4  Eq of I2
43  n  2
n=6
53. K 2CrO 4 +H 2O 2 
 CrO5
Amyl alcohol
 In acidic medium  X 
 Blue liquid 
Here the structure of is:-
O
O O
Cr
O O
Here, single bonded O-atoms with Cr is=04

54. O
*
O O O

* * *
O O
1  2 1  2  3 1
 R+S  R+S R,R
R,S
S,S

55.   C.R.T
7.47=C  0.083  450
C=0.2M=0.2  108gL1
21.6gL1
56. X=1 Y=6
OC OC CO
OC CO CO CO
OC OC CO
57. G=H-TS=0 at equilibrium
 -165  103  T   505   0  T  300 K
The answer is 300
58. Tb  iKb m
4 K 1.5 K
T f  iK f m,  b , b  2
4 Kf 3 Kf
59. c and e are the graphs which represent first order i.e, x=2 and a,b and d are graphs
which represent zero order i.e, y=3. Then 23  8 .
60. 200 times

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MATHEMATICS
61.  ,  are roots of the equation
x 2  px  2  0
62. 8x+y+4z=-2
X+y+z=0 ….(1)
 x  3y   .....(2)
From (1)-(2), we get ….(3)
7x+3z=-2 ….(4)
From 3(2)+(3), we get
 3    x  3z  
For infinitely many solutions, we have
   4,   2
 1  1
  ,  ,     4, 2,  
 2  2
63. z 5  4

 x  52  y 2  16
So, points (x,y) lie inside or on the circle whose center is (-5,0) and radius is4.
Comparing thes with    2, we get
  32,   16
    48
64. S1 :   p  q    q  p   q   p  p   q  t  t
S2 :  p   q     p  q    p   q    p   q   f
So, both S1 and S 2 are true.
65.

In the figure, circle S1, S2 and S3 have b, c and a, respectively.

AC1  b, BC2  c and CC3  a
From the figure,
DC1  b  a, EC2  c  a, FC2  c  b
2 2 1 1 1
  b  c    b  c   bc  ab  ac   
a c b
66. f  x
L  lim  finite value   f 1  0 and L= lim f ' 1
x 1 x  1 x 1
Now f  x   f '  x   f "  x   x5  64 ….(1)
So, clearly f(x) is polynomial of degree 5.
Differentiating f '  x   f "  x   f "'  x   5 x 4

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 f "  x   f "'  x   f iv  x   20 x3 ....(2)

 f "'  x   f iv  x   f v  x   60 x 2 ...  3
From (2)-(3), we get
f v  x   f "  x   60 x 2 -20x3 ...  3
Also, from (3), differentiating two more times we get
f v  x   120
Now, putting x=1 in (4), we get  f ' 1  15
67. Normals to these two curves are
y  m  x  c   2bm  bm3 ,
y  mx  4am  2m3
If the curves have a common normal, then
  4a  c  2b  m   b  2a  m3
For normal other than x-axis, m  0.
  4a  c  2b  m   b  2a  m2
c c
 m2  20 2
2a  b 2a  b
So, option (4) is correct.
68.  1  x 2 x  1

A  x 1 x 
 
 2 x  1  x 1 
 A  4 x3  4 x 2  4 x  f  x  let 
Let f '  x   0
 x  1, 1 / 3
Sum of minimum and maximum values is
20 88
f 1  f  1 / 3  4  
27 27
69.
e

2 x 2  bx  c   1  2  x2  bx  c 
L  lim
x  x   2
e

 2 x 2  bx  c  
 1  2 x 2  bx  c 

x  
2
 
L  lim 
x  x   2  x   2

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2
 2  

 2       4   2 b 2  4c

2
  
70. M (3, 5, 7) satisfies the line L1 .
3 a 52 7 b
 
l 3 4
b=3 and a+1=3
M is foot of perpendicular from P (4, 3, 8) on line L1 .

Also,  2iˆ  3 ˆj  4kˆ and v2  3iˆ  4 ˆj  5kˆ
 
 v1  v2  iˆ  2 ˆj  kˆ
  
AB.  v1  v2  1
Shortest distance    
v1  v2 6
71. Given differential equation
dy 2e4 x
  8  4 cot 2 x  y   2sin 2 x  cos 2 x 
dx sin 2 2 x

IF . 

e
Solution is
8  4 cot 2 x  dy  e8 x  2loge  sin 2 x   e8 x .sin 2 2 x


y e8 x .sin 2 2 x 

e4 x

 2e4 x  2sin 2 x  cos 2 x  dx  C

 y  x 
sin 2 x

4 2
  e 6 2  3
 y    e
 6  sin  2.   3
 
 6
      
72.
 
a  b  c  a  b c sin
6
 
Now,
  
a  b  2iˆ  ˆj  2kˆ  iˆ  ˆj 
 2iˆ  ˆj  2kˆ
 
From c  a  2 2 , we get

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 c 2  2c  1  0

 c 1 c
Thus, from (1),
  
 
a  b  c  3  1 1 / 2  3 / 2
73. Period of both sin 2 x and  x /   is  .
100 

So, I 
  0
e 
sin 2 x
x /
dy  100

0
sin 2 x
e x /
dx

   

 
 
 
 50  e  x / dx  e  x /  cos 2 xdx 
 
 
 0 0 




  1  e 1  2 
0
e  x /  sin 2 xdx

  1  e1   

 I  50  1  e1 

 1  4 2
 
 



200 1  e1  3 
1  4 2
74. Let C,S,B and T be the events of Ramesh using car, scooter, bus or train.
Let L be the event of Ramesh reaching offices late. By hypothesis
1 3 2 1
P  C   , P  S   , P  B   and P T  
7 7 7 7
L 4 5 L 1 8
P    1  , P   1 
C  9 9 T  9 9
By baye’s theorem,

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L
P C  P  
C  C 
P  
L L L L L
P  C  P    P  S  P    P  B  P    P T  P  
C  S B T 
75. 1
A1. A3. A5 . A7 
1296
4 1 1
  A4    A4 
1296 6
7 1
A2  A4   A2 
36 36
 A6  1, A8  6 and A10  36  A6  A8  A10  43
76. Let xi  5  yi
9 9
2
  xi  5  9 and   xi  5  45
i 1 i 1
So, required standard deviation is
 9 
9
  yi  2
2  i 1  45  9 
   yi    9 9  2
i 1 
9  

 
 
77. In the figure PQR is triangular park with PQ=PR=200 TV tower of height ‘h’ stands at
midpoint M of QR.

In triangle TMR, MR-h cos

In triangle TMR, PM=TM=h
In triangle PMR, PR 2  PM 2  MR 2
 2002  h 2  3h 2
 h  100m
78.

From the figure,

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 MD 2   MC 2  h2  64   H  10 2  121
 2h 2  20h  64  100  121
2
 2  h  5   235
79. Conceptual
80. x2 y 2
Equation of hyperbola is  1
4 2
a 2 e 2  a 2b 2  4  2  6
So, focus F   6, 0 
Equation of tangent at p 4, 6 is  
4 x  2 6 y  4 or 2x- 6 y  2
Solving it with latus rectum line, x  6 we get
 
R   6,

2
6
 6 1 


Also, Q  1, 0 

So, area of QFR 

1
2
6 1 . 
2
6

6 1 
7
6
2  
81. General term in the expansion
10!  
 a  2b  .  4ab 
 !  ! !
For term containing a 7b8 , we have
   7
  8
      10
Solving we get and   5,   2 and   3
10! 3 10
So, coefficient= 2 .2
2!3!5!
 315  216  K  315
82. 1 2



1
 5  x 2  1  x  dx 


 


1
 5  x 2  1  x  dx


 


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 5 1   5 2 5 1 
 2 1  sin 1   2  1  sin 1  1  sin 1 
 2 5  2 5 2 5
5 1  1
  tan 1  tan 1 2  
2 2  2
5 1
 
4 2
83.

  max 82sin 3x.44cos3x
xR

 
 max 26sin 3 x 8cos3 x  210

 
  max 26sin 3 x 8cos3 x  210

 1/5  22 ,  1/5  22

 b  34 and c  8
So, c  b  42
84. Circle x 2  y 2  2 x  4 y  4  0
Center of the circle is C(1,2) and radius, r=1.
Line 3x+4y-k=0 intersects the circle at two distinct points. So, distance of the center of
the circle from the line must be less than ‘r’.
3 1  4  2  k
 1
2 2
3 4
 k  11  5
 k  {7,8,9,.......,15} since k  Z
85. 36  22  32
So, the digits of 5-digit numbers can be
(1,1,1,9,4),(1,1,1,6,6),(1,1,2,2,9),(1,1,3,3,4),(1,1,6,2,3),(1,2,2,3,3)
So, total number of numbers
5! 5! 5! 5!
   3   180
3! 3! 2! 2! 2! 2!
86. 9  x2 16 dy 16
Let y  5 x  1
5 x x  5 dx  x  5 2
In [0, 2], derivative vanishes at x=1.

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5
So   2 and  =
3
3


 9  x2 
I  max  ,x
 5 x 
 

1
9/5 3

I 

1

5  x 


1  18 and  2  16
16 
 dx  I 
x 5

9/5
xdx

87. We must have 0  P  Ei   1for i=1,2,3

2  3p 2 4
0 1   p  ....(1)
6 3 3
2 p
0  1  6  p  2 ....(2)
8
1 p
0  1  1  p  1 ....(3)
2
From (1), (2) and (3), we get
2 / 3  p  1 ....  4 
2
 p ....  5
3
From (4) and (5), we get
2 / 3  p  1 p1  1 and p2  2 / 3 , So, p1  p2  5 / 3
88. From the given information, p '  x   a  x  1 x  2 
 x3 x2 
 p  x  a   3  2x   C
 3 2 
 
5
Now, p 1  8  a  C  8 .......(1)
6
2
p  2  4  a  C  4 .......(2)
3
Solving (1) and (2), we get a=24, C=-12P(0)=C=-12
89. For r=1, equilateral triangle is formed.
Let 0<r<1.
For this, obviously 5  5r  5r 2 and 5+5r 2  5r. So, we must have
5+5r 2  5 or r 2 +r-1>0
 1  5  1  5  1  5 1  5
 r   r  0r  or r 
 2  2  2 2
 5 1 
 r  ,1 .........1
 2 
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Let r>1
Cleary, for this case 5  5r  5r 2 and 5+5r 2  5r. So, we must have
5  5r  5r 2 or r 2  r  1  0
 1 5  1 5 
  r    r    0
 2  2 
 1 5 
 r  1,  .....................  2 
 2 
Form (1) and (2) along with r=1, we have
 5 1 5 1
r   , 
 2 2 


90.
dx
I
2
 1
2
3
 x    
 2 4

1 3
Put x   tan 
2 2

I


2
3

4 3

 9
16
sec2 

sin 2 
4
sec 

c
d


9  2 
4 3  2 x  1  1   2 x  1 
 tan 1    c
9  3  3  x2  x  1 
9  
3a  b  15

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 SRI CHAITANYA IIT ACADEMY, INDIA 25‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐15_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 25-01-2023
Time: 09.00Am to 12.00Pm GTM-15 Max. Marks: 300
KEY SHEET
PHYSICS
1) 3 2) 2 3) 1 4) 3 5) 3
6) 1 7) 1 8) 3 9) 2 10) 1
11) 3 12) 1 13) 3 14) 3 15) 1
16) 4 17) 2 18) 4 19) 2 20) 1
21) 60 22) 5 23) 25 24) 6 25) 400
26) 210 27) 5 28) 1 29) 0 30) 363

CHEMISTRY
31) 4 32) 4 33) 3 34) 2 35) 3
36) 1 37) 1 38) 3 39) 2 40) 2
41) 3 42) 1 43) 1 44) 2 45) 2
46) 4 47) 3 48) 2 49) 4 50) 2
51) 875 52) 143 53) 173 54) 167 55) 108
56) 144 57) 47 58) 19 59) 7 60) 7

MATHEMATICS
61) 1 62) 4 63) 3 64) 1 65) 3
66) 4 67) 3 68) 3 69) 2 70) 3
71) 1 72) 4 73) 1 74) 3 75) 2
76) 3 77) 2 78) 4 79) 3 80) 2
81) 98 82) 11 83) 100 84) 216 85) 5
86) 0 87) 4 88) 2 89) 7 90) 12

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 SRI CHAITANYA IIT ACADEMY, INDIA 25‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐15_KEY &SOL’S

SOLUTIONS
PHYSICS
1.   kt 
P log e  
  x
kt kt ML2T 2  1 
1     E  kt 
x x L  2 
As P is dimensionless         MLT 2 
 
2.

mg  N  ma  N  m  g  a 
Person experiences weightloss, whenacceleration of lift is downward.
3. At maximum height, V = 0
Momentum of object is zero.
4. V  2 gR sin 
mv 2
N  ma sin    2mg sin 
R

N 1 3
 1 
2mg sin  2 2

 A  constant
5. Applying conservation of angular momentum

 
MR 2  MR 2  2mR 2  '  ' 
2m
M  2m
6.  GMr
 3 , r  R
g R
 GM , r  R
 R 2

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7.  T 
  1  L   100%
 Tn 
TL  00 C  273K , Tn  373K   26.809%
8. 
T 2
geff


a)When a  0,T  2
g
g '  6
b) when a  , T  2 T '  T
6 g
g 7
6
9. CP 2
 1   1.4  F  5
CV F
By conservation of energy
F 1 mv 2 mv 2
nRT   nm  v 2 T  
2 2 FR 5R
10. Charge on capacitor C2
C  Qtotal C2 C1V  C1C2V
 2  
Qtotal C1  C2 C1  C2
11. S1 : In nonpolar molecules, centre of +ve chargecoincides with centre of –ve charge,
hence netdipole moment is comes to zero.
S2 : When non polar material is placed in externalfield, centre of charges does not
coincide, hencegive non zero moment in field
12. d
  5t 3  4t 2  2t  5 e   15t 2  8t  2
dt
e 78
At t  2, e  15  22  8  2  2  e  78V  I    15.60
R 5
13.   R  A
R  
A R  A
A  k

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 SRI CHAITANYA IIT ACADEMY, INDIA 25‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐15_KEY &SOL’S
 A R 2
 0 
 A R 
R
 2  0.4  0.8%
R
14. R M  q p R 4 1
    2
RP M P q RP 1 2
15. Conceptual
16. a 1

 100
For reflection size of obstacle must be much larger than wavelength, for diffraction size
should be order of wavelength.

Since the object is of size , much smaller than wavelength, so scattering will occur.
100
17. h h
e   photon   Pphoton  mv
mv Pphoton
1 2
Ee mv 1 mv v
 2  v 
E ph hc 2 p phC 2C

18. 238 4 0 206
92 U  82 He  61e 82 Pb
19. V R1 v1 i2 0.1 50
R    5
i R2 v2 ii 0.2 5
20. Conceptual
21. Both should have same horizontal component of velocity
200  400cos   600
22. v 2  u 2  2as 100  0  2 10  s S  5m
Height from ground  10  5  5m
23. stress
y  2.0  1010
strain
1
Energy density  stress  strain
2

 
1 1 2 kJ
  strain  y  5  104  20  1010  25  102  10  25
2
2 2 m3
24.  earth g 104
  g  earth 
 planet g planet 6  105
g planet  6 ms 2
25.
     dt    Ldi / dt  dt  L  di    Li
 dt  dt  dt i

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V 20
i0    2 A, if i  0 A
R 10
20  103   2  0  2  103
T  100  s, L  20 mH    
100  106 5
   400V
26.  total  3600  2  3600  2  750
 total  2100

1  450

  1200  900  2100

27. 20MSD=1cm
1
1MSD= cm
20
10VSD  9 MSD
9 9 1
1VS  MSD   cm
10 10 20
9
1VSD  cm
200
1 9 1
VC  1 MSD  1VSD  cm  cm   10mm
20 200 200
VC  5  102 mm  5
28.

V  IRnet
10  I  10
I  1A
29. At resonance I L  I C

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2
1  1 1 
Alternatively,    
Z  X L XC 
At resoancne, X L  X C and Z  
 Z total circuit   i.e, I  0
30. From continuity equation
a
av1  v2 v2  2v1
2
From Bernoulli's theorem,
1 1
P1   gh1   v12  P1   gh2   v22
2 2
 v 2  v 2  
P1  P2    2 1   g  h2  h1  
 2  
 4v 2  v 2  
4100  800  1 1   10   0  1 
 2 
 
41 3v12 121 2
 10    v12
8 2 8 3
I 21 3
v1  
43 3
363
v1  m/s
6
X=363

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CHEMISTRY
31. At the same temperature, lighter gases (with lower molar mass, M) will have higher
values of vmp . But on increasing temperature, peak shift forward showing that vmp
2 RT T
increases. vmp  ; vmp 
M M
300 400 300
For N 2 , O2 , H 2  
28 32 2
vmp of N 2  300 K   vmp of O2  400 K   vmp of H 2  300 K 
32. Number of radial nodes  n  l  1
For 3s-orbital n  3, l  0 Number of radial nodes  3  0  1  2
33. Cupric metaborate is formed by heating boric anhydride with CuSO4 in an oxidizing
(non-luminous) flame.
B2O3  CuSO4  Oxidising
flame
 Cu  BO2 2  SO3
(Cupric metaborate)
(Blue-green)
Blue cupric metaborate is reduced to colourless cuprous metaborate in reducing
(luminous) flame.
2Cu  BO2 2  2 NaBO2  C 
Reducing
flame
 2CuBO2  Na2 B4O7  CO
(Blue) (Colourless)
34. Sucrose is disaccharide and a non-reducing sugar. Hence, assertion is true. In sucrose,
linkage is in between C1 of  - glucose and C2 of  - fructose. Hence, reason is false.
35. NaH acts as reducing agent.
The lone pair on nitrogen in pyridine does not get delocalized and is available easily
for donation, making it basic.
NaH Na   H 

(Pyridine)

36. O
CH 2  Br COOH C
 O 
 
 O

CH 3 C
COOH
 A  B C  O
37. Basicity is inversely proportional to electronegativity.
38. Biochemical oxygen demand (BOD) is the amount of dissolved oxygen needed by
aerobic biological organism to break down organic material present in a certain
volume of given water sample at certain temperature.
39.
 NiCl4 2 : Oxidation state of Ni in  NiCl4 2  2

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Cl  is a weak field ligand and cannot take part in pairing of electrons.

Hence, the complex is tetrahedral and paramagnetic with two unpaired electrons.
 Ni  CO   : Oxidation sate of Ni in  Ni  CO   is zero. CO is a strong field ligand.
 4  4

Hence, the complex is tetrahedral and diamagnetic.

40. 2.675
No.of moles of CoCl3.6 NH 3   0.01
267.5
4.78
No.of moles of AgCl   0.03
143.5
Since 0.01 moles of the complex CoCl3.6 NH 3 gives 0.03 moles of AgCl on treatment
with AgNO3 , it implies that 3 chloride ions are ionisable, in the complex. Thus, the
formula of the complex is Co  NH 3 6  Cl3 .
41. In 3d series elements, only Cu shows positive value for electrode potential of
M 2 / M .
42. Hydrochloric Acid(HCl) is usually not used in the process of titration because it reacts
with the indicator potassium permanganate  KMnO4  used in the process. It reacts
with KMnO4 solution and gets oxidized resulting in the liberation of chlorine gas.
43. The gases which is more easily liquefiable such as NH 3 , H 2 S and SO2 are more
soluble in water and are more easily adsorbed. Easily liquefiable gas has higher critical
temperature and thus strong van der Waals’ forces of attraction.
44. Spontaneous process  GT , P  0
Process with P  0, T  0  Isothermal and isobaric process
 reaction  [Bond energies of molecules in reactants]- [Bond energies of product
molecules]
Exothermic process  H  0
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45.

It has phenolic functional group, alcoholic functional group and a terminal alkyne.
Hence, it can react with Br2 / water due to presence of unsaturation, with
ZnCl2 / HCl due to presence of alcoholic group and with FeCl3 due to presence of
phenolic group.
46. Both I.E. and electron gain enthalpy increase on going from left to right in a period.
47. In the given Ellingham diagram, below 1200K the C  CO curve lies below the
M  MO curve hence, carbon can reduce MO.
48. Conceptual
49. Due to the inert pair effect (the reluctance of ns 2 electrons of outermost shell to
participate in bonding) the stability of M 2  ions (of group 14 elements) increases as
we go down the group.
50. Conceptual
51. 2SO2( g )  O2( g )  2SO3( g )
Initial pressure: 250 m bar 750 mbar 0bar
At completion:250-250=0 m bar 750  125  625 m bar
250 m bar

Here, SO2 is a limiting reagent

Total pressure  PSO2  P O2  OSO3  0  625  250  875 mbar
52. ZM
 3
a  NA
Z  2.5  102 1 pm  1012 m 
2.7  103 
 405 10   6.023 10
12 3 23  

or, Z  4 , Z  4 ,indicates that it is a fcc unit cell.

405  1012
m  143.2  1012 m  143  1012 m
a
a  2 2r , r  
2 2 2 2
53. B2 :  1s 2 *1s 2 2 s 2 * 2 s 2 2 p1x
  n  n  2   11  2   1.73BM  173  102 BM
54. On mixing two solutions,
 V  V 0.1  1  0.2  2
 final  1 1 2 2   0.1667 or 167  103
V1  V2 3
Thus, x  167
55. A3 B2 3 A2  2 B 3
3x 2x 3 2 5
M  3x   2 x   x 
M K sp       108  
M  M  M 
56. For reaction,
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2Cu(aq ) Cu( s )  Cu(2aq )

o RT
Ecell  ln KC
nF
o
Ecell  Eo   Eo  0.52  0.16  0.36V
Cu / Cu Cu 2  / Cu 
0.025
0.36V   ln KC
1
0.36
ln KC   14.4  144  101
0.025
57. According to Arrhenius equation
k  Ae Ea / RT
Ea
log k  log A  _____(i )
2.303RT
2.47  103
log k  20.35  _____(ii )
T
Comparing (i) and (ii)
Ea
 2.47  103
2.303R
Ea  2.47  103  R  2.303  2.47  8.314  103  2.303
 47  103 J / mol or 47 kJ / mol
58. Reaction(1);
2 Fe2  H 2O2 
 2 Fe3  2OH 
Reaction(2);
2MnO4  6 H   5H 2O2   2Mn2  5O2  8H 2O
Thus, x  2, y  2, x '  2, y '  5, z '  8
Sum  2  2  2  5  8  19
59. According to Dumas method:
 y y z  y
C x H y N z   2 x   CuO   xCO2  H 2O  N 2   2 x   Cu
 2 2 2  2
For C2 H 7 N
x  2, y  7, z  1
60.

Number of Co-Co bond  X   1

Number of terminal CO ligands  Y   6
X Y  7

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MATHEMATICS
61. We have, z0   or  2

(where  is a non real cube root of unity)

Now, z  3  6iz081  3iz093
81 93
 3  6i    3i    3  6i  3i  3  3i
3 
 arg  z   tan 1    tan 1 1 
3 4
62. 2 2 3 x  a 
A   3 1 5  , X   y  and B  b 
     
1 3 2   z  c 
Here, A  0
Since, the system of linear equations has more than one solution, so  adj A  B   0
13 13 13   a  0
  1 1 1 b   0  b  a  c
    
 8 8 8 c  0
63. 10 10 10 10  1 8
S1   j  j  1 10C j   j  j  1 . . C j  2  90  28
j 1 j 1
j  j  1
10 10
S2   j. 10
C j  10  8 C j 1  10  29
j 1 j 1
10 10
S3   j 2 .10C j    j  j  1  j .10 C j
j 1 j 1
10 10
  j  j  1 10C j   j.10 C j
j 1 j 1
 90.28  10.29   45  10  29  55.29.
64. 
10
x 1 x 1 
 2/3 1/3  
 x  x  1 x  x1/2 

   
10
 x1/3  1 x 2/3  x1/3  1
 

x 1 x 1 
 

 x 2/3
 
x1/3
 1    x x 1 


  x1/3  1  1  1 / x1/2    x1/3  1 / x1/2 

10 10

General term in given expansion

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r 10 r r
 
 
10 1/3 10  r  1  10 3 3 2  1r
 Cr x  1/2   C r x
x 
10 r r
Now for x 5 ,    5
3 3 2
10  1 1  25 5 25 6
  5  r     r  r    10
3 3 2 3 6 3 5
Coeff.of x 5  10C10 1 1  1
10

65. 100  199   100   100  1100  1  100  2 100  2   .....  100  99 100  99 
2

2
 2 2
 
 100   100  1  100  2 2 2
  .....  100 2 2
 99 
 100   99 100   12  22....  992 
2 2

2  99  100  199  
 100  .100   100 3  199 1650 
6
   3 and   1650
1650  0
So, required slope, m   550
30
66. We have
 p  q     p  q   q  simplifying as
 p  q    p  q   q 
 p  q     p  q     q  q 
 p  q    p  q  which is a tautology
67. 
Given, tan 1 a  tan 1 b  0  a, b  1
4
ab
  1  a  b  1  ab   a  1 b  1  2
1  ab
 a 2 a3  b 2 b3 
Now,  a    ..... b    .....
 2 3   2 3 
 loge 1  a   loge 1  b   loge 1  a 1  b    loge 2
68. Let height of the pole AB be h.
Then BC  h cot 600  h / 3
BD  h cot 450  h
As BD  BC  CD

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h
h
3
7h  
3 1  7 3

7 3 3 1 7 3  
h
7 3
3 1

2

2
3 1 m  
69. Let number of trials be n, probability of success be p and that of failure be q.
Given, mean  np   8 , variance  npq   4
4 1 1 1
 q    p  1    p  q  1
8 2 2 2
So, n  16
16
C0  16C1  16C2
Now, P  X  2  
216
1  16  120 137 k
  
 k  137
216 216 216
70. Total number of observations are 9 which is odd and it means median is 5th item.
Now we are increasing 2 in the last four items which does not effect its value. So,
new median remains unchanged.
71.  2   4 
Let a cos  b cos      c cos      k ( say )
 3   3 
k k k
a ,b  ,c 
cos cos   2 / 3 cos   4 / 3
 
Now, ai  b j  ck bi  c j  ak 
 ab  bc  ac
 cos  cos   2 / 3  cos   4 / 3 
 k2  
 cos .cos   2 / 3 .cos   4 / 3 
 cos  2cos      cos  / 3 
 k2  
 cos .cos   2 / 3 .cos   4 / 3 
 cos  cos  
 k2  0
 cos  .cos   2 / 3  .cos   4 / 3  

Angle between ai  b j  ck and bi  c j  ak is
2
72. x  2 y 1 z
The given line is     ( say )
3 0 4
So, the coordinates of any point D on the given line are  3  2,1,4  .
Let AD is perpendicular to the given line.

 AD. 3i  4k  0 
  3  2  1 3  1  1 0    4  2  4  0
 9  9  16  8  0
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17 24 18
 25  17     i  2 j  k
25 25 25
2 2
 24  2  18 
So, AD      2   
 25   25 
576 324 900  2500 3400 34 2
 4     10  34
625 625 625 625 25 5
1
Now, area of ABC   BC  AD
2
1 2
  5 34  34 sq.units
2 5
73. 5
Let a tangent to the parabola be y  mx   m  0
m
As it is a tangent to the circle x 2  y 2  5 / 2 , we have
 5
 
 m 
5
2

1  m2  1  m2 m2  2 

Which gives m4  m2  2  0  m2  1 m2  2  0  
As m  R, m2  1 m  1
Also m  1 does satisfy m4  3m2  2  0
Hence common tangents are y  x  5 and y   x  5
74. x
 
Given line are y   x  0  and y  3x  x  0  using a, a 2 in these lines
2
a
a 2   0.....(i) and a 2  3a  0.......(ii )
2
1
Solving (i) and (ii), we get  a  3
2
75. dy y  2  x 
The given differential equation can be written as, 
dx x 2  x  1


dy

 2  x  dx  log y   1  2  1 dx
y x 2  x  1
  x x 2 x  1 
2
 log y   log x   log  x  1  c......(i )
x
Now, y  2   e
 log e   log 2  1  0  c  c  log 2
 i  becomes,
2
log y   log x   log  x  1  log 2
x
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 xy  2
 log  
 2  x  1  x
xy 2  x  1 2/ x
  e2/ x  y  .e
2  x  1 x
2.3.e1/2 3
 y  4   e
4 2
76. 2 4d sin   2
We have , det  A  1 sin   2 d
5 2sin   d  sin   2  2d
Applying R1  R1  R3  2 R2 , we get
1 0 0
det  A  1 sin   2 d
5 2sin   d 2  2d  sin 
  2  sin   2  2d  sin    d  2sin   d 
 4  4d  2sin   2sin   2d sin   sin 2   2d sin   d 2
2
 d 2  4d  4  sin 2    d  2   sin 2 
2
For a given d, minimum value of det  A   d  2   1  8  d  1or  5
77. Given y  x  sin y
dy dy
  1  cos y
dx dx
dy dy 1
 1  cos y   1   ____(i )
dx dx 1  cos y
 3 1 
Since tangent is parallel to line joining  0,  and  ,2
 2 2 
3 1
2
 
dy 2  2  1____(ii )
  
 dx  ( a,b) 1  0 1
2 2
1
From (i) and (ii), 1
1  cos b

 1  cos b  1  cos b  0  b   2n  1 , n  Z .
2
Since point  a, b  lies on curve y  x  sin y .
 b  a  sin b
 b  a  sin b  b  a  1

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78.  /3
Let I   
tan 3 x.sin 2 3x 2sec2 x.sin 2 3 x  3tan x.sin 6 x dx 
 /6
 /3
 tan 3 x.sin 4 3 x.2sec2 x  3tan 4 x sin 2 3 x.sin 6 x dx
   
 /6
d
 /3
  
4
 4
 dx tan x .sin 3 x  d
 tan 4 x. dx
sin 4 3 x   dx

2 2
 /6  
 
 /3
 
1 d 1  /3
  tan 4 x.sin 4 3x dx   tan 4 x.sin 4 3x 
2 dx 2  /6
 /6
1
4 
   1  1  1
4 1
  3 0   1      
2
  3  2  9  18
79. The given differential equation is,
 x  x3  dy   y  yx2  3x4  dx, x  2
  xdy  ydx   3x 4dx  yx 2dx  x3dy
 xdy  ydx  2
   3 x dx  ydx  xdy
2
 x 
  d  y / x    3 x 2dx   d  xy 


y
x
 x3  xy  c  x 4  y x 2  1  cx  
Now, y  3  3 (given)
57
 81  24  3c  c   19
3

Solution is x 4  y x 2  1  19 x 
Putting x  4 , we get
 4 4  y  42  1  19  4
256  76
 y  4   12
15
80. 1
d 2x d  dx  d  dy  
      
dy 2 dy  dy  dy  dx  
 
1  2 1 3 2
d  dy   dx  dy  d 2 y  dy   dy  d y
   .     .    
dx  dx 
 
dy  dx  dx 2  dx   dx  dx 2

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81. Let z  x  iy .

 z2   x  iy  2  
Since, arg     arg   4
 z 2 4  x  iy  2 
  x  2   iy  x  2   iy  
 arg   
  x  2   iy  x  2   iy  4

 arg 

 x 2  4  y 2  4iy 

    tan 1 

4y  

 2   x 2  4  y 2  4
  x  2  y2  4  
 
 4 y  x 2  4  y 2  tan  / 4   1

 
2 2 2
 x2  y 2  4 y  4  8   x  0    y  2   2 2 ,
Which represents a circle with centre, A  0, 2  and radius, r  2 2
 
Let P 9 2, 2 be point which lies outside the circle.

 
2 2
Then, minimum value of z  9 2  2i  AP  r
2
 
9   
2 2 2
 2 0   2  2  2 2   7 2  98
 
 
82. Let f  x    c  5 x 2  2cx  c  4
 f  0  c  4
f  2    c  5  4  4c  c  4  c  24
f  3   c  5  9  6c  c  4  4c  49
Now, f  0  f  2   0 and f  2  f  3  0
  c  4  c  24   0 and  c  24  4c  49   0
49
 4  c  24 ___(i ) and  c  24 ____(ii )
4
 49 
From (i) and (ii), c   , 24 
 4 
 S  13,14,15,....,23
So, number of elements in S is 11.
83. For number to be divisible by 55, it should be divisible by both 5 and 11.
So, the number should be of the form 5abba5.
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Total possible values that a can take = 10
Total possible values that b can take = 10
 Total number of possible 6 digit palindromes divisible by 55  10  10  100
84. x  y  z  12
Now, A.M  G.M .
 x  y  z
3   4    5    3 4 5 1/12
3 4 5      
              
x y z
12  3   4   5  
x3 y 4 z 5
  1  x3 y 4 z 5  33.44.55
334455
3
But,given x3 y 4 z 5   0.1 600 
 all the numbers are equal.
x y z
    k ( say )  x  3k , y  4k , z  5k
3 4 5
But, x  y  z  12  3k  4k  5k  12  k  1
 x  3; y  4; z  5 So, x3  y3  z 3  216
85. We have AD  8m , BC  11m, AB  10m .
Let AM  x m , then BM  10  x  m
In DAM , by Pythagoras theorem, we have

MD 2  AD 2  AM 2  82  x 2
 MD 2  64  x 2 ____(i )
In CBM , by Pythagoras theorem, we have
2
MC 2  BC 2  BM 2  121  10  x 
 MC 2  121  100  x 2  20 x
 MC 2  221  x 2  20 x ____(ii )
From(i) & (ii), we have MD 2  MC 2
 64  x 2  221  x 2  20 x  285  20 x  2 x 2
dS
Let S  285  20 x  2 x 2   20  4 x .
dx
dS
Now  0  4 x  20  0  x  5
dx

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d 2S
Also,  4  0 . So, S is minimum for x  5 .
dx 2
Thus, distance of M from the point A is 5m.
86. 4/ n2
Let L  lim U n 
n
 4  
 log L  lim   log U n 
n  n 2  
 4    1   22   32   n 2  
 lim   log 1   2log  1    3log  1    ....  log 1  2  
2
n
n   n 2      n2   n2   n 
 n       
 4  n   r 2 
 lim     r log 1  
n   n 2  r 1   n2 
  
 n 1 r  r 2 
  4  lim   . log 1  
n   r 1 n n  n2  
  
1

 log L   4   x log 1  x 2 dx 
0
2
Put 1  x  t  2 xdx  dt
When x  0, t  1 & when x  1, t  2
2
4 2
 log L   log dt  2t log t  t 1  2 2log 2  1
2
1
1
log
Le 4 log 2  2
e 16 .e 2  1 e 2
16
87. x2 y2 z 5 x 1 y  4 z  4
Let L1 :   and L2 :  
3 5 7 1 4 7
Equation of plane containing L1 and L2 is
x 1 y  4 z  4
3 5 7 0
1 4 7
  x  1 35  28    y  4  21  7    z  4 12  5   0
 7 x  14 y  7 z  77  0  x  2 y  z  11  0
11 11
Perpendicular distance from the origin to the plane is  units
1 4 1 6
88. 1
Given, probability of any machine to be out of service 
4

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1 3
So, probability of no machine is out of service  1 
4 4
Required probability = when no machine is out of service + when only one machine
is out of service + when only two machines are out of service.
5 4 2 3
3  1  3  1 3
 C0    5C1     5C2    
5
4  4  4  4 4
3
3  9  3  1 
     5    10   
 4  16  16   16  
3 3 3 3
 3   9  15  10   3   34   3   17   3 
           k
4  16   4   16   4   8   4 
17
k 
8
89. cos x  cos 2 x  cos3 x  cos 4 x  0
Using sum-product formula, we have
 cos x  cos3x    cos 2 x  cos 4 x   0
 2cos x cos 2 x  2cos x cos3x  0
 2cos x  cos 2 x  cos3 x   0
5x x
 2cos x.2cos cos  0
2 2
If x   0,2  we have the solution as
 3  3 7 9
x  , , , , , ,
2 2 5 5 5 5
Thus we have 7 solutions.
90.
 
Let t ,  3t be the point of intersection.

So, t 2  3t 2  4b  t 2  b
t 2 3t 2 b 3b
Also,  1  1
16 b 2 16 b2
 b2  16b  48  0  b  4,12
But b  4  b  12

Sec: Sr.Super60_NUCLEUS & ALL_BT Page 20

 SRI CHAITANYA IIT ACADEMY, INDIA 26‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐16_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 26-01-2023
Time: 09.00Am to 12.00Pm GTM-16 Max. Marks: 300
KEY SHEET
PHYSICS
1) 1 2) 1 3) 3 4) 2 5) 2
6) 4 7) 1 8) 1 9) 4 10) 3
11) 2 12) 4 13) 2 14) 1 15) 1
16) 4 17) 4 18) 2 19) 1 20) 4
21) 3 22) 10 23) 10 24) 200 25) 1
26) 8 27) 16 28) 25 29) 1055 30) 9

CHEMISTRY
31) 2 32) 1 33) 4 34) 4 35) 2
36) 1 37) 4 38) 1 39) 1 40) 1
41) 2 42) 1 43) 1 44) 2 45) 3
46) 4 47) 1 48) 1 49) 3 50) 4
51) 3 52) 6 53) 2 54) 16 55) 8
56) 3 57) 3 58) 3 59) 5 60) 1

MATHEMATICS
61) 4 62) 3 63) 1 64) 1 65) 2
66) 1 67) 3 68) 3 69) 4 70) 2
71) 2 72) 1 73) 2 74) 2 75) 3
76) 4 77) 4 78) 2 79) 1 80) 3
81) 2 82) 8 83) 5 84) 0 85) 20
86) 1 87) 5 88) 1 89) 4 90) 3

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 SRI CHAITANYA IIT ACADEMY, INDIA 26‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐16_KEY &SOL’S

SOLUTIONS
PHYSICS
1. Q  AT 
2

Charge Q=CV  C   2 2
 M 1 L2T 4 A2
V ML T
F MLT 2
Electric field , E    LMT 3 A1
q AT
q1q2
Force ,= F  0  M 1 L3T 4 A2
4 0 r 2
1 1 1
Speed of light , C   0  
0 0 0 C 2
 M L T A2   LT 1 
1 3 4 2

  M 1 L1T 2 A2 
2. Conceptual
3. For Statement
The maximum speed by which cyclist can take a turn on a circular path
 v   rg  0.2  2  9.8  v  3.92
5
Speed of cyclist, 7 kmh-1 = 7  18  1.94m / s
The maximum safe speed on a banked frictional road
   tan     98  0.2  tan 450  2  9.8  1.2
vallowable  rg v 
1   tan  1  0.2  tan 45 0
0.8
 5.42 m / s
Speed of cyclist, 18.5kmh-1=5.13 m/s
So, both the statements are true.
4. Here M 0  200kg , m  80kg Using conservation of angular momentum Li  L f
 M 0 R2 
I11  I 22 I1   I M  I m     mR 2 
 2 
1  M R2  5
I2  M 0 R 2 and 1  5rpm  w2   0  mR 2   0 2
2  2  M R
2
5 R 2  80  100 
   9rpm.
R2 100

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5.

For point O to be the centre of mass of the system, moment about O should be zero
d
 2mx  m  d  x   3mx  md  x 
3
For equilibrium, F  gravitational  Fcentripetal
G  2m  m d 
F  2
  2m   2  
d 3
Gm d 3Gm 3Gm
  2  2  3   
d 2
3 d d3
2 d3
 Period of revolution, T   2
 3Gm
6. Stress 
Normal force N
 
N
Area A  2 a  b
V   a 2b
V  2 aba
Stress  B  strain
N 2 aa  b  V 
B  Strain 
 2 a  b a b 
2
V 
 2 a 
2
ab 2
NB
 a 2b
Force need to push the cork.
  B  4 2 a 2b 2 a
f  N   4  Bb  a
 a 2b
7. Let R be the radius of curvature of common surface
4T 4T
P1  P0  and P2  P0 
a b
4T  4T   4T  4T 1 1 1 ab
And P1  P2   P0     P0      R 
R  a   b  R a b R b  a 

8. As the rods are identical, so they have same length (l) and area of cross-section (A).
They are connected in series. So, heat current will be same for all rods.
Q Q  Q 
Heat current   
 
 
  
 t  AB  t  BC  t CD


100  70  K1 A   70  20  K 2 A   20  0  K3 A
l l l
 K1 100  70   K 2  70  20   K3  20  0 

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K1 K 2 K 3
 K1  30   K 2  50   K3  20    
10 6 15
 K1 : K2 : K3  10 : 6 :15  K1 : K3  2 : 3.
9. Cp  2
The ratio of specific heats at constant pressure  Cv     1  
Cv  f 
Cp  2 7
Where f is degree of freedom  1   
Cv  5  5
10. Net torque due to spring force:
Suppose the rod is tilted by angle d then,
1 1 Kl 2
Fres  Kx  K d  res  Fres  arm of couple  K . d  l  d
2 2 2
Kl 2 Kl 2 ml 2 2 6K
As,  res  I  2 d  I  2 d   
2 2 12 m

11. Frequency of unknown fork=known frequency  Beat frequency =288+4 cps or 288-
4cps i.e. 292 cps or 284 cps. When a little wax is placed on the unknown fork, it
produce 2 Beats/sec. When a little wax is placed on the unknown fork, its frequency
decreases and Simultaneously the beat frequency decreases confirming that the
frequency of the unknown Fork is 292 cps.
Note: Had the frequency of unknown fork been 284 cps, then on [lacing wax its
frequency Would have decreased thereby increasing the gap between its frequency and
the frequency of Known fork. This would produce high beat frequency.
12. Time period of the pendulum (T) is given by
 mg    qE 
2 2 2
L  gE  L
T  2  geff   g eff  g 
2
  T  2
geff m  m   qE 
2

g 
2

 m 
13. Electric potential is constant inside a charged spherical shell and outside it vary in
inversely with distance.
14. 0 I I x 
B1  sin 90  sin 1   0 1  2 2  ....  i 
4 y 4 y  x y 

0 I  I y 
B2   sin 90  sin  2   0 1   .....  ii 
4 x 4 y  x 2  y 2 
10 I x 1 y 
BNet  B1  B2 B      
4
 y y x 2  y 2 x x x 2  y 2 

0 I  x  y x 2  y 2  0 I  x  y x2  y 2 
B      
4  xy xy x 2  y 2  4  xy xy 
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0 I  2
B x  y 2   x  y 
4 xy  

15. 1 L 1 80  103 1 200

Quality factor, Q   6
 40  103  2
R C 100 2 10 100 100
16. Temperature coefficient of resistance is negative for pure semiconductor. And no. of
charge carriers in conduction band increases with increases in temperature.
17. It will be concentric circles because locus of all the point having same path difference
Lies on concentric circle
18. The liner momenta given by
p m q 4m p .2q p 4 2 2 2
p  2mE  2mqV  E  qV      
pp mp q p mp q p 1 1 1
19. In the given circuit, the zener diode is used as a voltage regulating device. Hence, the
voltage across 2000 is 30 V.

100  50
Current flowing through 1000, I   50mA
1000
50
Current flowing through 2000, I '   25mA
2000
Using Kirchhoff’s current law,
I  I 2  I '  I Z  I  I  50  25  25mA
20. As , d  2Rh
d2 h2 3d h2
So, d  h     h2  9h1  h2  9 100  900m
d1 h2 d h1
21. Along horizontal
2 1 1
F1  1cos 450  2sin 450 F1   
2 2 2
Along vertical
3
F2  1sin 45  2 sin 45 F2  3sin 45 
2
F1 1
so, ,  .so, x  3
F2 3
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 SRI CHAITANYA IIT ACADEMY, INDIA 26‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐16_KEY &SOL’S
22. From the condition of equilibrium
f  F cos 600
N  mg  F sin 600
1
For no movement of the block
M
F cos 60  F  F cos 60    mg  F sin 600 
0 0

 mg
F  Fcritical  10 N
cos 60   sin 600
0

23. From momentum conservation in perpendicular direction of initial motion

mu1 sin 1  10mv1 sin  2 ......  i 
It is given that energy of m reduced half . If u1 be velocity of m after collision, then
1 21 1 2 u
 mu   mu1  u1 
2 2 2 2
If v1 be the velocity of mass 10m after collision, then
1 1 u2 u
10m  v12  m  v1 
2 2 2 20
For equation (i), we have
sin 1  10 sin  2
24. Tsink 300
Initially,   0.25  1   0.5  0.5 
Tsource Tsource
T T
1  sink  0.25  0.75  sink
Tsource Tsource
T 300
Tsource  sin k   400k
0.75 0.75
Finally,   0.25  100% of 0.25
  0.5  Tsource
'
 600 K
T   600  400  K  200 K  2000 C
25. Current does not pass through 5 as potential difference across it is zero. Then circuit
becomes

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 SRI CHAITANYA IIT ACADEMY, INDIA 26‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐16_KEY &SOL’S
26. Given: Magnetic moment, M  9.85 10 A / m 2 , 2

Moment of inertial, I  5 106 kgm 2

5
10 oscillations in 5 seconds  Time period   0.5s
10
2
1  T  I 1 4 2 5 106  4  9.85 102
T  2     B  
 2  M T 2 9.85  102   0.5 
2
MB MB
 B  80  10 4  8 mT
27. The moving rod will cut the vertical component of magnetic field perpendicularly.
So, e  Bv lV  BH  lV  Dip  450 K  Bv  BH 
 4  10 3  0.2  20  16  10 3 V
28. In case of simple microscope ,
D D 25
Magnification, m  1  or , 6  1   5   f 0  5cm
f0 f0 f0
As total magnification double using an eyepiece along with the given lens i.e, case of
.D
compound microscope, Magnification, m m 
f0 . fe
60  25
or, 12   f e  25cm
5. f e
29. Form Bohr’s formula for hydrogen atom,
1  1 1 
 R  2  2  , R  1.097 107 m 1
  n1 n2 
For Lyman series:
1
 R 1  R  n2   and n1  1
min
1  1  3R 4 1 1
 R 1     n1  2, n1  1  max  min.     304  given 
max  4 4 3R R 3R
For paschen series:
1 1 1  7R
 'min  R   and  'max  R    
 
9  9 16  16  9
16  9 9 81
 'min   'max   
7R R 7R
81 81 81 3  1 
or ,  'max   'min     304   304 A 
7 R 7  3R 7  3R 
 For Pachen series,  'max   'min  10553.14
30. Radioactive material NA have decay constant, 1  25
Radioactive material NB have decay constant 2  16
N B e  2t
N  N 0 e t     2t  e   2t .e  1t
NA e
1 1 1
e1  e 1 2 t   1  1  t  1  t   
1  2 25  16 9
1
Compare with , so a  9
a

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CHEMISTRY
31. Differ at C-4 configuration
32. Conceptual
33. Conceptual
34. The strength of nucleophile depends upon the nature of alkyl group R on which
nucleophile has to attack and also in the nature of solvent. The order of strength of
nucleophiles flows the order: Br   CN   ter  butoxide  methoxide
35. Mass number of element  81 i.e., p  n  81
31.7
Let the number of protons be x. Number of neutrons  x   x  1.317 x
100
81
 x  1.317 x  81or 2.317 x  81Or x   35
2.317
Symbol of the element 81 35 Br
36. Radiation coming from sun ore outer space have high energy or short wavelength,
which are allowed to enter by green house gases. However, radiation emitted by
earth is in infrared region, having, long wavelength, are reflected back by the
envelope of green house gases.
37. 3
A) Cr2 O3  2 Al  Al2 O3  2Cr
1 
B) Au  2CN   H 2O  O2   Au  CN 2    2OH 
2
 2
2  Au  CN 2   Zn   Zn  CN 4    2 Au
C) Statement is true
D) Ag 2 S  4 NaCN O2
 2 Na  Ag  CN 2   Na2 S
2 Na  Ag  CN 2   Zn  Na2  An  CN 4   2 Ag
38. For mono electronic species energy depends upon principal quantum number only
39. Zinc rod dipped in blue copper sulphate solutions is oxidized to Zn 2 and Cu 2 are
reduced to Cu and get deposited on zinc rod.
40. OH

OH
41. Fluorine has high reduction potential due to low bond enthalpy high hydration
energy
42. 2CuSO4  2 Na2CO3  H 2O  CuCO3 .Cu  OH 2  2 Na2 SO4  CO2
43. Initially adsorption increases with increases in pressure at a particular temperature,
then get slow. After attaining equilibrium adsorption it becomes independent of
pressure.
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44. Conceptual
45. Conceptual
46. For strong electrolytes, the plot between  m and C 1/2 is a straight line.
47. Conceptual
48. Conceptual
49. Conceptual
50. I) 1,1,1- trichloro II) 1,1,2- trichloro III) 1,2,2- trichloro IV) 1,2,3- trichloro
V) 1,1,3- trichloro
51. M 2 Ox 
Re duction
M
Wt.of M 2Ox Wt. of Metal
Eq. of M 2Ox  eq.of metal 
Eq.wt.of M 2Ox Eq.wt.of Metal
4 2.8
 x = n-factor of metal
2  56  x  16 56
2x x
4 2.8 1 1
On solving we get,      2x  6  x  3
56  8 x 56 14  2 x 20
Hence, the oxide is M 2O3 .
52. I,II,III,IV,V,VII
53. 20  102 200
  1000   2m.mol ' s / l
20  80 100
54. 16σ Bonds are present
55. Functional groups are present carboxylic, alcohol, alkene, aldehyde, amine,
ketone,ether. Sec-amine
56. 13.6 Z eff2 13.6 Z 2 13.6 Z 2 
Eionisation  E  En  eV   2
 
n2  n 2 n12 
13.6  12 13.6  12
E  hv   ; hv  13.6  0.85
1  4
2 2

13.6  0.85
 h  6.625  1034 v  34
 1.6  1019  3.08  1015 s 1
6.625  10
57. II,III,IV
58. III,IV,V
59. Total volume =100Ml
1.0 0.5
 acid   10mL   0.1  salt   20mL   0.1
100 100
pH of acidic buffer =  pK a  log
 salt   4.76  log 0.1  4.76
acid 0.1
60. γ  form fcc Z  4 a  386 pm
ργ 4M / N A  386  2  290 
3 3

β  form bcc Z  2 a  29 pm    0.8481

ρβ 2M / N A  290 3  386 
3

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MATHEMATICS
61. Since ,  are the roots of the equation
x 2  px  q  0      p and a  q.
Now, ( 1/ 4
  )  [(  1/ 4  1/ 4 ) 2 ]2
1/ 4 4

[  1/ 2  1/ 2  2(  )1/ 4 ]2
2 2
      2   2()1/4    p  2 q  2(q)1/4 
   
1/4
 p  6 q  4q1/4 p  2 q  1/4  1/4  p  6 q  4q1/4 p  2 q 
 
62. The total numbers of factors of xnyz…. is equal to the number of ways of
selecting one or more out of n identical quantities of one type and
remaining m distinct quantities. Hence, the required numbers of factors =
(n + 1)2m
63. F(x) is defined for
[x]2  [x]  6  0  ([x]  3)([x ]  2)  0  [x]  2 or [x]  3
But [x]  2  [x]  3, 4, 5,......
 x  2  Domain of f  ( , 2)  [4, )
64. Let the roots of the given equation be 1 + ip and 1 – ip, where p  R
   product of roots
 (1  ip)(1  ip)  1  p 2  1,  p  R    (1,  )
65. Given condition A 2  B 2  (A  B)(A  B)
 A 2  B 2  A 2  AB  BA  B 2  AB  BA .
66. Since a 1 b 1 c1 , a 2 b 2 c 2 and a 3 b 3 c 3 are divisible by k, therefore
100a1  10b1  c1  n1k
100a 2  10b 2  c 2  n 2 k
100a 3  10b3  c3  n 3 k
where n1, n2, n3 are integers.
a1 b1 c1
Now   a 2 b2 c2
a3 b3 c3
a1 b1 100a1  10b1  c1
 a2 b 2 100a 2  10b 2  c 2
a3 b3 100a 3  10b3  c3
(Applying C 3  C 3  10C 2  100C 1 )
a1 b1 n1k a1 b1 n1
 a2 b2 n 2k = k a 2 b2 n 2  k1
a3 b3 n 3k a3 b3 n3
  is divisible by k
(Since elements of  1 are integers,   1 is an integer)
67. 4!
 6 ways and the
The four digits 3, 3, 5, 5 can be arranged at four even places in
2!2!

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5!
remaining digits viz., 2, 2, 8, 8, 8 can be arranged at the five odd places in  10
2!3!
ways. Thus, the number of possible arrangements is (6) (10) = 60.
68. 10
C5 5
term independent of x will be 10 C5 (sin )5 (cos )5  sin (2)
25
10
C5
So maximum value will be 
25
69. 1 1 1 
1 1 1 1
Given: .... 4 
1 2 34
 4
   i 1 (2i  1)
4
 4  4  4  .....
1 3 5
1 1 1 1 1 1 
 4  4  4  4  ...   4  4  ... 
1 2 3 4 2 4 
1 1 1 1  1 15
   4  4  4  4  ...     
2 1 2 3  16 16
70. A.M. > G.M.
1/3
x y z x y z x y z
   /3 . .     3
y z x y z x y z x
71. x  y is an integer
x  x  0 is an integer  A is Reflexive
x  y is an integer  y  x is an integer  A is symmetric x  y, y  z are
integers
As sum of two integers is an integer.
 (x  y)  (y  z)  x  z is an integer
 A is transitive. Hence statement 1 is true.
x
Also  1is a rational number  B is reflexive
x
x y
  is rational  need not be rational
x x
0 1
i.e., is rational  is not rational
1 0
Hence B is not symmetric
 B is not an equivalence relation.
72. Let Ei denote the event that Ai dies in a year, then P(E i )  p and
P(Ei' )  1  p for I = 1, 2, …..n P (none of A1, A2, …. A3 dies in a year)
 P(E1'  E '2  E 'n )  P(E1' )P(E'2 )....P(E'n )  (1  p)n
Because E1, E2,……En are independent.
Let E denote the event that at least one of A 1 , A 2 ,.......A n dies in a year, then
P(E)  1  P(E1'  E'2  ....  E'n )  1  (1  p)n
1
Let F denote the event that A1 is the first to die, then P(F / E)  .Also, P(F) =
n
1
P(E).P(F/E) = 1  (1  p)n 
n
73. 1
Since, P(A  B) 
3
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1 1
 P(A  B)   1  P(A  B) 
3 3
2 2 1 2
 P(A  B)   P(A)  P(B)  P(A  B)   p  2p  
3 3 2 3
74. x 3 z 2  (z  x) 2 x 3 2zx  x 2
lim  lim
   
x 0 4 x 0 4
3
8xz  4x 2  3 8xz
3
x 3 8z  4x  3 8z 3 x

x 4/3 3 2z  x 3
2z 1
 lim 4
 4
 23/3
x 0
x 4/3
 3 8z  4x  3 8z   2 3 8z  2 .z
   
75. In the neighbourhood of x = 0, f(x) = log 2 – sin x
 g(x) = f(f(x)) = log2 – sin (f(x))
log 2 – sin(log 2 – sin x)
Since g(x) is differentiable at x = 0,
 g ' (x) = – cos(log2 – sinx) (–cosx)
 g '(0)  cos(log 2)
76. f (x)  x 3  x 2 f '(1)  xf "(2)  f "'(3)
 f '(x)  3x 2  2xf '(1)  f "(2)
 f "(x)  6x  2f '(1)
 f "'(x)  6
Now, f '(1)  3  2f '(1)  f "(2)
 f "(2)  f '(1)  3  0 (1)
Again f "(2)  12  2f '(1)
 f "(2)  2f '(1)  12  0 (2)
Again f "'(3)  6 (3)
From (1) and (2)
2f "(2)  f "(2)  6  0  f "(2)  2 (4)
 (1) gives f '(1)  2  3  0
 f '(1)   5
 f (x )  x 3  5x 2  2x  6 (5)
 f (0)  6
f (1)  1  5  2  6  4
f (2)  8  20  4  6   2
f (3)  27  45  6  6   6
f (0)  f (2)  6  2  f (1)  (a) is false
f (0)  f (3)  6  6  0
f (1)  f (3)  (1  5  2  6)  6  2  f (2)
f (1)  f (3)   2  f (0) [ f (0)  6]
 (4) is false
77. Let, f (x)  x 3  3x  a
 f '(x)  3x 2  3  3(x  1)(x  1)
Now, f (1)  a  2, f ( 1)  a  2
The roots would be real and distinct if,
f (1)f (  1)  0  (a  2) (a  2)  0 or   2  a  2

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Thus the given equation would have real and distinct roots if a  ( 2, 2)
78. x n 1 xn x n 1 x2
Let f (x)  a 0  a1  a2  ...  a n 1  a n x
n 1 n n 1 2
Then f(x) is continuous and differentiable in [0, 1], as it is a polynomial
function of x.
Also, f(0) = 0
a0 a a a
and f (1)   1  2  ...  n 1  a n  0 . (Given)
n 1 n n 1 2
Hence, by Rolle’s theorem, there exists atleast one real number
c  (0,1) such that f '(c)  0 i.e., c is a root of the equation
a 0 x n  a1x n 1  ...  a n 1x  a n  0 .
79.  1 
1  2  dx
I  x 
 1 2  1  1
 (x  )  1 tan  x  
 x   x
1
Put x   t
x
80. Conceptual
81. 1
A ,B
1
8 4 2
82. Use LHopital
83. x2  y 2   x  0
84. n
n 1
2

85.       4 
r   b   c, c    xi  yj  so c   x  i  j 
 3 
6
We can prove that 1  5  x   . Four vectors possible.
5
Which are r    2i  j  ,   i  j 
2 11
 5 5 
86. y  xt  2at  at 3

f 1  l  ; 9t 2  2;  0  Only one normal

87. 13 2 x  5 y  1
 x  2    4  3
2 2

5 13
88. The given curve represent a point   , 2 
1
 2 
89. Keep the least element on diagonal line and greatest element on the other positions
90. f  x  k x

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 27-01-2023
Time: 09.00Am to 12.00Pm GTM-17 Max. Marks: 300
KEY SHEET
PHYSICS
1) 3 2) 3 3) 2 4) 1 5) 3
6) 2 7) 4 8) 1 9) 1 10) 1
11) 4 12) 1 13) 2 14) 3 15) 4
16) 3 17) 4 18) 3 19) 3 20) 1
21) 8 22) 2 23) 4 24) 8 25) 4
26) 0 27) 4 28) 2 29) 3 30) 3

CHEMISTRY
31) 1 32) 4 33) 2 34) 3 35) 4
36) 1 37) 1 38) 1 39) 1 40) 3
41) 2 42) 2 43) 1 44) 3 45) 3
46) 1 47) 1 48) 2 49) 4 50) 1
51) 298 52) 100 53) 14 54) 1 55) 3
56) 1 57) 2 58) 4 59) 10 60) 1

MATHEMATICS
61) 4 62) 2 63) 1 64) 2 65) 1
66) 3 67) 2 68) 3 69) 2 70) 2
71) 1 72) 4 73) 1 74) 2 75) 3
76) 4 77) 2 78) 2 79) 3 80) 2
81) 2 82) 8 83) 2 84) 1 85) 2
86) 2 87) 20 88) 2 89) 15 90) 1

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SOLUTIONS
PHYSICS
1. 3b 3b 2
cos    , Which gives 3b 2  2ab cos   0
2a 2ab
 2  2  2
Given, P  a 2  4b 2  4ab cos  , And P  Q
Equation (1) can be achieved if
 2
P  a 2  4b 2  4ab cos   a 2  b 2  2ab cos 
 2   

So, Q  a 2  b 2  2ab cos  Q  a  b 
2.  A   0 A   0 A    MLT 4   given 
     
 0    0 0  1 / speed of light 
2

 ML2T 2 
  0 A   ML T  
1  2
  0 A is the energy per unit volume.
 L3 
 
3. u and at are perpendicular vectors. Acceleration is constant. Equations of motion can
be applied.
 
At t  2sec, u  2a, at  a 2  2   2 2a

 2  2 1/2 
 
2 1/2

    
Velocity at t  2sec  v   u  at    2a   2 2a   a 12
2


  tan 1 
 
 at 
   tan 

1 2 2a

  tan
1
 2
 
 u 

 2 a 
4. gx 2
Compare with y  x tan   2
2u cos 2 
3 4 3
tan   ,cos  ,sin  
4 5 5
5. T  mg
f

mg
For upper block to move along with block A ,

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0 mg  mg , 0  1.0
For block A to move along with other blocks,
f 1  f But f 1    4m  m  g
 1.0
So, 5 mg  0 mg   0   0.20
5 5
f  0 mg

A
f1

 4m  g
6. Sphere just reaches the top of the wedge, It means relative velocity of sphere w.r.t
wedge is zero mu   M  m V
mu
So, V=horizontal velocity of sphere and wedge 
 M  m
From conservation of mechanical energy,
1 2 1
mu   M  m V 2  mgR
2 2
1

u 
2  M  m   m 2u 2 
   2 
 2gR  M  m   2
  M  m 2 
gR u  
m    M 
7. M  L
L  2 R
R  L / 2
MR 2 3 3  L2  3 L3
I x  x1  MR 2   MR 2    L   2  
 4  8
2
2 2 2
8.  
 GM m GM m  2G  M  M  2GM
E   1
 2
 1 2
  M1  M 2 
  d   d   d d
  2    
  
2
1 2Gm G  M1  M 2 
m e 2   M1  M 2  ,  e  2
2 d d
9. Equivalent spring constant of a wire is given by
YA Y  2 A  Y1 A Y2 A
K Keq  K1  K 2  
L L L L
Y Y
Or Y  1 2
2
The system can be replaced as if suspended from a single wire of Young’s modulus Y.

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1  stress 
2
1
U   stress  strain  volume   AL
2 2 Y
2
1  Mg  1 M 2g 2L
    AL 
2  A  Y A Y1  Y2 
10. V V 
f1  , f2  n   , f 2  f1 and n is odd :
L  4L 
5V  5
For n  5 , it is possible f 2     f1.
4 L  4
11. TB  TA
The heat will flow C to B to maintain steady state. Applying condition formula to the
 T   T  T  TC TC  T 2
sides AC andCB .    a  
 a 2  AC  CB a 2 a
T 3
On solving, C 
T 2 1 

12. From figure r1  r2  900

n2
 r1 max  900   r2 min  900  C for TIR at AD sin C 
n1
n1 sin  max sin  max sin  max
Snell’s law at A, B   
n2 sin  r1 max sin  90  C  cosC
0

n  n  n 
sin  max   1   cosC  1 .cos  sin 1 2 
 n2  n2  n1 
n  n 
 max  sin 1  1 cossin 1  2  
 n2  n1  

13. At distance ' a ' apart, kinetic energy of the particles is assumed zero and at infinity
their kinetic energy will be maximum.
q1q2 1 q1q2
 mVmax 2  2,Vmax 
4 0 ma 2 4 0 ma

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14. dL
  0, L  cons tan t
dt
No torque will act on planet under any condition otherwise L  constant and planet
will become unstable
15. For satellite revolving in a circular orbit.
mv 2 GMm
 2
r r
or mv r  GMm  cons tan t
2

 mvr   Lv  cons tan t

Air friction will provide a retarding torque to satellite and thus velocity of satellite will
not change
L1  mv1r1 and L2  mv2 r2
If L2  L1 and r2  r1 then v2  v1
16. No current through capacitor in steady state.

2V  V V
I 
2 R  R 3R
VP  V  V  IR  VQ
VQ  VP  IR
V  V
 R 
 3R  3
17. 2 r  4a
4a
r
2
2 2
 2a  4a I
M  IA  I .   
  
18. At a distance r , an elements of thickness dr is considered. Number of turns in the
N N
element  .dr  .dr
b  a  b  a 
0  dN  I 0 I  N  dr
dB  
 .
2r 2 ba r
b
 NI b
B   dB  0 in  
a
2b  a   a 
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19.  di 
VA  1  5  15   5  103       VB
 dt 
VA  VB ( given)
di
 4  103 A / sec.
dt
20. Let R1  R2
Current flows in large coil of radius R2
i
Magnetic field at centre  0
2 R2
 0i
Flux linked with smaller coil   R 21B   R 21
2 R2
  Mi
 0
2
R 1
M  / i   
 2  R2
M R 21 / R2
Hence, R1  R2 otherwise, R1  R2
21. qt 4 2

dq di
i  2t and 2
dt dt
At t  3sec, q  5c and i  6 A
q di
VA   L iR  VB
c dt
q di
VA  VB   L  iR
c dt
5
   0.5  2    6 1  8V
3
22. di
VL  L
dt
Ldi  VL dt
L  i f  i0   Area under VL  t  graph
1
4  i f  i0    8 2 
2
if  2A
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 SRI CHAITANYA IIT ACADEMY, INDIA 27‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐17_KEY &SOL’S
23.  1 
E photon  13.6 1   eV  13.0eV
 25 
E / c  mv (momentum conserved )
E 13 1.6  1019 
v  4 m / sec
mc 1.67  1027   3 108 
24. a  v2 / r
Z2
so, a
1 / 2 
Thus aZ 3
3 3
a1  Z1   2 
    8
a2  Z 2   1 
25. q  CV
dq  dv  8
i  C     2   4 A
dt  dt  4
26. W  U f  Ui
1
Ui  
4 0 a
 2q  2q    2q  q  2q    0

Also U f  0
So, W  0
27. 1   2   
   1 2  0 
t  2 
75  65  75  65 
   30 
2  2 
For second case,
55  45  55  45 
   30 
2  2 
1
Divide eqn 1 by eqn. 2  ,  2
2
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 SRI CHAITANYA IIT ACADEMY, INDIA 27‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐17_KEY &SOL’S
t  4min ute
28. 1
Given Vsound  Vrms
2
1 1
  RT  2  3RT  2
   
 M   2M 
3
So,  
2
C n1C p1  n2C p 2
But   p 
Cv n1Cv1  n2Cv 2
3  3 5R / 2    n  7 R / 2 

2  2  3R / 2   n  5 R / 2 
n2
29. PV since graph is a straight line
PV 1  constant
For the processes PV x  Constant , molar heat capacity is
R R
Given by C  
  1 1  x 
But here,   1.4 and x  1
R R
C   3R  IR( given)
1.4  1 1  1
So, I  3
30. No change in temperature
So,  n1  n2 i   n1  n2  f
PV PV PV PV
1 1
 2 2 1 2
RT RT RT RT
PV  PV
P 1 1 2 2
RT

 4 P0   3V   1.5P0   2V 
5V
 3P0  IP  given 
So, I  3

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CHEMISTRY
31. At critical temperature slope of pressure volume curve is maximum and it is zero
32. H 2 A  2 NaOH  Na2 A  H 2O
4
 Na2 A  n NaOH   2n H 2 A
 32  20 
4
  0.25  V  2  0.2  20 
52
 0.0769  V  32 ml
n Na2 A  n H 2 A  20  0.2  4
33. 0.693 0.693
K1300 k   k2320 k  
20 5
k Ea  1 1  2.303RT2T1 k
log 2      Ea  log 2
k1 2.303R  x1 x2  T2  T1  k1
2.303  8.314  300  320 20
Ea   log  55.14
20  10 3
5
34. On dilution number of ions decreases in unit volume hence specific conductance
decreases. But separation between ions also increases hence equivalent conductance
increases
35. i  1    1.25,T f  K f .m.i  i  0.633:

i 1     0.4650 C
2
36. a
Distance b/w two nearest Cl  
2
a
Distance b/w two nearest Na  
2
37. IF7  P.B.P.,   0 , SO2  Bent ,   0
CS 2  linear ,   0 , SF4  See  saw,   0
38. Due to electronegativity difference between F and H bond length is smaller.
39. IIA Cations have high hydrate formation nature due to their high charges.
40.

41. H
CH 3 CH CH C* CH 3

OH
n2
2n  22  4
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It exhibits both geometrical and optical isomerism
42. Cl CH CH NO2
In , double bond character in carbon chlorine bond is
maximum due to resonance and so the bond length is shortest
43.  
O O

A dipolar resonance structure has aromatic character In the ring and would be
expected to make a major contribution to the overall structure
44. Lewis acids are the most common compounds used for initiation of cationic
polymerization. Some other lewis acids are SnCl4 , AlCl3 , BF3 , TiCl4
45. Weak bases are good leaving Groups
46. RNA controls the synthesis of proteins
47. Formaldeyde is more reactive for nucleophilic addition reaction
48. CH 3 CH 3
CH 3
ONa OCH 3
OH NaOH CH 3 I
 NaI

49. OH OH

SOCl2

OH CH 2 OH Cl Cl
CH 2
In the case of OH group present on the aromatic ring, carbon - oxygen bond has
partial double bond character, hence, it is a strong bond and difficult to cleave.
50. O

 CH3  C  
Acetone contains methyl ketonic group and undergoes Haloform reaction
51. S  206.5  114.6  180.7  88.8
0

G 0  H 0  T S 0
 H 0  271.8  298  88.8  1013  298.3
52. Coagulation value
Number of millimoles of electrolyte required

volume of colloidal solution (in litre)
10  0.5
  1000  100
50
53. nh
The angular momentum 
2

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34
n  6.626  10
.1652  1034  ,n  3
2
 1 1   1 1  5R
 v  R.Z 2 . 2  2   v  R.22. 2  2  
 n1 n2  2 3  9
X=5,y=9,x+y=14
54. Mole of pure CuSO4  0.12  mole of Na2 s2O3 react with I 2
Na2 s2O3  120  103  M  0.12 , M  1.0
55. 2 NOCl 2 NO g   Cl 2
g
At eqm p-2x
p  x  1; p  2 x  0.64, x  0.12
 0.24   0.12   16.875  103
2

Kp  atm
 0.64 
2

56. Aspirin is Acetyl salicylic acid

OCOCH 3
COOH

3
57. The number of optical isomers for Cr  C2O4 3  is two
58. PbS , CuS , As2 S3 CdS are soluble in 50% HNO3 . Where as HgS , Sb2 S3 are
insoluble in 50% HNO3
59. Ho  67   4 f 11 6s 2
Ho 3  4 f 10
60. One hydrogen bonded H 2O molecule.

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MATHEMATICS
61.  a  b   b  b.a  b  b.b  a  2(2i  j  k )  6(i  j  k )  2i  8 j  8k
62. x  1  a  x2
Right hand side is non negative x  R such that a  x 2  0
So left hand side x  1  0  x  1
 x  1  a  x2
1  2a  1
 x  1
2
 a  x2 x 
2
1  2a  1
the root x  does not satisfy x  1
2
1  2a  1
So it is not a root and x  satisfies the equation for a  1
2
63. AM  GM
 a1  a3  a5 ......   a2  a4  a6 ........
2
  a1  a3  a5 ..... a2  a4  a5 ......
n 1
20
or
2
  a  a   k for k  0
i 1
i i 1

n 1 n 1
100   ai  ai 1   k   ai  ai 1 
i 1 i 1
Hence S = 100
64. At point of maxima f '  x   0 and f " x   0
 f " x   x 2  f 2  x   0
Since the curve is x 2  y 2  a 2 and x 2  f 2  x   0
2 2
x1  y1  a 2
Point lies outside hyperbola
Hence 2 tangents
65.  
 
 1 1 2 x 
I   e x  tan 1 x     dx
  1  x 
2 2 2
1 x 1 x 2
  
 
 f  x
 f ' x  
 1 
I  e x  tan 1 x  c
 1  x2 
Using  e x  f  x   f '  x   dx  e x f  x   c
66. n  S   1003
x1  x2  x3  75
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xi  1
74
C72
PE 
1003
67. hx
 tan 30
y
y  h  x  3
h
 tan 600  h  3 y
y
1 2h
y h x
3 3

68. Nidhi intelligent = p

Nidhi pass = q
Not pass  q
p  q
69. 1  2 2 2 4 2 6 
 2sin cos  2sin cos  2sin cos 
2  7 7 7 7 7 7 
2sin
7
1  4 6 2 sin8 sin 4 
  sin  sin  sin   
2  7 7 7 7 7 
2sin
7
1

2
70. P1 1,1,1 PM
1 1  2
M 1 1,0,0 
1 1 1 2
P2  , ,  P2 M 2 
3 3 3 3
1  2
M 2  ,0,0  P3 M 3 
3  9
2 3
Hence PM 1  P2 M 2 .....  
1
1 2
1
3
71. Matrices are

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1 0 0 0
0 ,
 0  0 1 
1 0  1 1 
0 ,
 1  0 0 
0 1 1 1 1 0
0 , , ,
 1 0 1 1 0 
0 0  1 0  0 1
1 , , ,
 1  1 1  1 0 
1 1  0 1 1 1
1 0  , 1 1 , 1 1
     
72. Substitute y in equation of circle and put   0
x 2   ax  b   2 x  2  ax  b   2  0
2

x 2 1  a 2   x  2ab  2  2a    b 2  2b  2   0
 a  1  ab   1  a 2  b2  2b  2 
2

Hence  a  b   2  a  b  a  b  1  0
2

73. 2 4 6 2(n  1)

S  (n  1) cos  (n  2) cos  (n  3) cos  ......  cos
n n n n
2 4 2(n  1)
S  1cos  2cos  ..........  ( n  1) cos
n n n
 2 4 2( n  1) 
2 S  n  cos  cos  .......  cos 
 n n n 
  2 2(n  1) 
sin(n  1)  n  
2S  n n n

cos    n
sin  2 
n  
74. 8
Let x 2  4sin  , y 2  cos   x 2  y 2  4sin   cos 
8
3 3
64 4 13
 maximum value  16  
9 3
75.

(2,3)

Equation of tangent at origin is

2( x  0)  3( y  0)  0
 2x  3 y  0

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7
tan θ =
4
2
m
 3 7
2m 4
1
3
3m  2 7
 
3  2m 4
 12m  8  21  14m
1
 26m  13  m 
2
1
 y  x x  2y  0
2
76. n( s )  8C3  56
Choosing a vertex 3 triangles can be formed from the diagonals of the faces not
8 3
passing through the vertex. But each triangle is repeated 3 times ....n( E )  8
3
8 1
 p( E )  
56 7
77. A is symmetric  A '  A now X'AY=   X'AY  '   (Y ' A ' X )
 (Y ' AX ) ( X ' AY is 11 matrix)
1  0 
Let E1    , E2   
0  1 
If X  E1 and Y  E2 then
E 11 AE2  E 12  E1  a12  a21
 A is symmetric.
78. 1
1 2
x2 1 1
 2  1  2 1 dx  
x dx put x   t
2 x
x x  x  2  1   1 
 x x x  x  2
 x  x
dt
 t  2 sec 
t t2  2
2 sec  tan  d 1
   C
2 sec  2 tan  2
79. Let xi  8  
18 18

   9,  
i 1 i 1
2
 45

x   xi
2
2

 S .D  i
  
n  n 
9 3
 
4 2

80. Normal’s  3, 3 2  

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 SRI CHAITANYA IIT ACADEMY, INDIA 27‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐17_KEY &SOL’S
Centre of given circle =(2,3/2)
Radius =5/2
Radius of required circle = 5 + 5/2
= 15/2
Diameter =15.
81. x

  tan x 
1 2
dx

lim 0
form
x 
x 12 
2
1  x2   
lim  tan x 
1 2
 
x  x 2
   2 
lim     2
x 
 2  
82. let x1  2 cos
y1  2 sin 
x2  2 sin 
y2  2 cos
P  2  x1  y1  x2  y2  x1 y1  x2 y2
 2  2  cos  sin    2  cos  sin    2sin  cos  2sin  cos
 
 2  4cos      2sin 2
 4
2428
83. x2 y2
The product of perpendicular from the foci on any tangent of 2  2  1 is a 2  a  b 
a b
2 2
x y
Here  1
4 9
ac  4
4

 x   x dx
0
4 4
  xdx    x dx
0 0
86
2
84.  2 x   x  1  2 x …………………(1)
  x  1  2 x
0  2 x  1
  x  1  0
 x  1  0
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 SRI CHAITANYA IIT ACADEMY, INDIA 27‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐17_KEY &SOL’S
1  x  0
2  2 x  0
 2 x   2, 1
from equation (1)
2 x  0  2 x
2 x  2, 1
1
x  1, 
2
85. y   x cos x 
x

ln  h cosh  h sinh  cosh

ln y  lim  0
h 0 1/ h  h cosh   1/ h2 
 y 1
Similarly lim  x sin x   1
x

x 0
86. 2 3 4 1
1 1 1 6  4  3 13
I    x    x    x   x  x 2  x3 
2 3
   
1 2 3 0
2 3 4 12 12
87. Number of reflexive relations
2
 2n n
 2255
 220
88. 2 6( x  2)  2(2 x  1) 10 x  10
f '( x)  ( x  2)2/3 2  (2 x  1) ( x  2)1/3  
3 3( x  2)1/3 3( x  2)1/3
x  1 is a point of maximum and x  2 is a point of minimum
 No.of extremum points is 2
89. x 1 y  2 z  3
  
1 2 2
po int on the 
    1, 2  2, 2  3
plane 
   6  11   5
 6, 8,13
Hence
Dis tan ce  25  100  100
 15
90. If x  x  y then y 2  5y  6  0
2

y 2

 6  y  1  0 x 2  x  6  0, x 2  x  1  0
x  3, 2, x  ,  2
Sum of non-real roots =    2 =-1

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 SRI CHAITANYA IIT ACADEMY, INDIA 28‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐18_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 28-01-2023
Time: 09.00Am to 12.00Pm GTM-18 Max. Marks: 300
KEY SHEET
PHYSICS
1) 1 2) 1 3) 3 4) 1 5) 3
6) 1 7) 4 8) 3 9) 2 10) 2
11) 3 12) 3 13) 3 14) 1 15) 4
16) 3 17) 3 18) 2 19) 2 20) 3
21) 9 22) 2 23) 3 24) 4 25) 2
26) 4 27) 6 28) 4 29) 0 30) 55

CHEMISTRY
31) 2 32) 2 33) 1 34) 3 35) 3
36) 2 37) 3 38) 4 39) 1 40) 2
41) 3 42) 3 43) 3 44) 2 45) 4
46) 4 47) 2 48) 1 49) 4 50) 1
51) 144 52) 17 53) 16 54) 2 55) 4
56) 6 57) 28 58) 1 59) 6 60) 5

MATHEMATICS
61) 4 62) 4 63) 1 64) 1 65) 3
66) 2 67) 2 68) 1 69) 4 70) 3
71) 2 72) 2 73) 3 74) 3 75) 4
76) 4 77) 3 78) 3 79) 1 80) 4
81) 7 82) 101 83) 36 84) 6 85) 9
86) 9 87) 5 88) 21 89) 9 90) 28

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 SRI CHAITANYA IIT ACADEMY, INDIA 28‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐18_KEY &SOL’S

SOLUTIONS
PHYSICS
1. V=xyz
Where x,y and z are the three dimensions then
 V   x y z 
    
 V   x y z 
 V   x y z 
or   X 100      X 100
 V   x y z 
 0.1 0.1 0.1 
    X 100
 5 10 5 
 5%
2. dx dy d 2x d 2 y
Given that  c   0
dt dt 2 2
dt dt
3 2
Further z  ax  by
dz dx dy
  3ax 2  2by
dt dt dt
 dx dy 
 3acx 2  2bcy   c  
 dt dt 
d 2z  dx   dy 
  6acx    2bc    6ac 2 x  2bc 2
dt 2  dt   dt 
Now, acceleration of particleis
 d 2x d 2 y d 2z 
a i 
j k
dt 2 dt 2 dt 2
 (6ac 2 X  2bc 2 )k
3. From work energy thermo
KE Wnet
or K1  K1   Pdt
2
1 3 
or mv 2    t 2  dt
2 2 
0
2
 t3 
or v2    or v  2m / s
2
  0
4. 4 j Component is perpendicular to velocity. So, it is radial (or normal) acceleration.

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 SRI CHAITANYA IIT ACADEMY, INDIA 28‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐18_KEY &SOL’S

v2
 an 
R
v 2 (2) 2
or R   1m
an 4
5. Mass of the remaining portion is
 2 R2 
  R  p
 2 
 
Here p= mass per unit area

 m1x1  m2 x2
 R2  R2 R
   R2   px  . p.
 2  2 2
 
R
or x
2(2  1)
6. I remaining  I whole I removed

1  1  R 2  2R  
2
or I  (9m)( R) 2   m    m   
2  2  3   3  
2
9M R
Here, m  x    M
2 3
R
Substituting In Eq.(1), we have
I  4 MR 2

7.

From conservation of mechanical energy,

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 SRI CHAITANYA IIT ACADEMY, INDIA 28‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐18_KEY &SOL’S

 1 2   Gm Gm 2 
2
3  mv   3   
2   2 R d 

1 2 1 2
v 2  Gm     v  Gm   
R d  R d 
8. Using PV1 1  PV
1 1 (for the air bubble)
we have
4T   4  r  
3
 4T  4 3  
 P1    r    P2      
 r  3   r 2 3  2  
 
Solving this, we get
24T
P2  8 P1 
r
9. V
n1 
2 /1
V
 /1 
2n1
V
n2 
2 /2
/2  V
2 n2
V
Now, n
2(/1  / 2 )
Substituting the values we get
n1n2
n
n1  n2
10. 5 3
1  means gasis monoatomic or Cv1  R
3 2
7 5
 2  means gasis diatomic or Cv2  R
5 2
Cv (of the mixture)
3  5 
(1)  R   (1)  R 
n Cv  n Cv
 1 1 2 2 
2   2   2R
n1  n2 11
C p (of the mixture)  Cv  R  3R
Cp 3R
  mixture    1.5
Cv 2R

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 SRI CHAITANYA IIT ACADEMY, INDIA 28‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐18_KEY &SOL’S
11. 5
QAB  nC p T  n R (2T0  T0 )
2
QBC  WBC  nR.2T0ln 2
QAB 5
 
WBC 4ln 2
12. sin i 2sin(i / 2)cos(i / 2)
n 
sin i / 2 sin(i / 2)
 i  2cos 1(n / 2)

i/2

13. 
Intensity at the centre will be zero it path difference =
2
 
or (   1)t  or t 
2 2(   1)
14. In equilibrium electrostatic attraction between the plates = spring force
q2
  kx
2 0 A
(CE )2
  k (d  0.8d )
2 0 A
2
 0 A  2
  E
  0.8d   0.2dk
2 0 A
 0 AE 2 4 0 AE 2
 k 
0.256d 3 d3
15. VMN  4 Volt on the basis of this currents in different resistors can be obtained.

4
A
3

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 SRI CHAITANYA IIT ACADEMY, INDIA 28‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐18_KEY &SOL’S
2
VP  1X  4  VQ
3
Now
2 10
VP VQ   4   V
3 3
16. Resistance across capacitor
4 X 8 32
R  
4  8 12
 C  CR  16  s
t t
c  160 / 6   c
i  ioe  e
 32 / 12 
t
 10e c
10
After t  c,i 
A
e
Thisis distributed ininverse ratio of resis tan ce.
2  10  20
 i1    A  A
3 e  3e
17. Concept
18. Concept.
19. E
c 0
B0
E 18
 B0  0 
c 3 x108
 6  108T
20. By the emission of one  particle, atomic number decreases by 2 and mass number by
4. By the emission of one   particle atomic number increases by 1 and mass number
remain same.
21. When force is applied on A
FA  mA g

mA  mB mB
When force is applied on B
FB  mA g

mA  mB mA
Dividing these two equations. We get

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 SRI CHAITANYA IIT ACADEMY, INDIA 28‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐18_KEY &SOL’S
FB mB

FA m A
m 8
 FB  B .FA  X 112  24 N
mA 4

22.

a /a R
TR  / , TR  / or T  T
R R2
Mg  T  Ma T

 /  2kg M
or Mg  Ma  T   M  a
 R2 
Mg 2 x10
or a   2m / s 2
 1  0.32
 M  2  2  0.04
 R 
23.  N  m 2 A
N (0.4)m.g
 Amax  
2
m m 2
 0.4  (10)  2 L 
   
(6 / 1.5)  M m
 1m
1
Maximum KE  KA2  3 J
2
24. Initially the rods are in parallel
Q (1   2 )

t R
 Q  A
   
 t 1 

 q1L 
100  0     (1)
R/2
Finally when rods areinseries
 Q mL (100  0)
     q2 L     (2)
 t 2 t 2R
From Eqs.(1) and (2)
q1 4

q2 1

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 SRI CHAITANYA IIT ACADEMY, INDIA 28‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐18_KEY &SOL’S
25. W  U  U f  U i
 (3U )  U  2U
26. M q

L 2m
qL
M
2m
q( I )
M
2m
1 
q  mr 2  
M 
2   1 q r 2
2m 4
27. 2
i 2rms  
i dt

 dt
4

4
(4t )dt
4
4 t dt t2 
 2 2
4

2
2
 
2 dt  2 2
 12 A2
irms  12  2 X 6
28. hc h c W
ev   W V   . 
 e  e
h c W
V1    
 e  1 e
h c W
V2    
 e  2 e
Solving thesetwoequation, we get ,
h 12 (V1  V2 ) (0.6 X 0.4 X 1012 )(1.0)
 
e c(2  1) (3 X 108 )(0.2 X 108 )
 4 X 1015V  s
29. The diode is in reverse biasing, so no current flows through it
30. d max  2 Rh

 2  6.4 X 106  240

 554  103 m
 55.4 km

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CHEMISTRY
31. a n  3 : l  0 : m  0  3s  3  0  3
b n  4 : l  0 : m  0  4s  4  0  4
c n  3: l  1: m  0  3 p  3  1  4
d n  3: l  2 : m  1  3d  3  2  5
If two orbitals having same  n  l   value, Priority will be more for lesser ‘n’-value.
3s  3 p  4s  3d  a  c  b  d
32.  RT  1
PM  dRT  P  d    PT and d So, Straight line is formed, with
M  T
T3  T2  T1
33. Bi2S3  K SP  108S 5  1.08 1073  S 5  1075  S  1015 M
34. 2C  3H 2  C2 H 6 H  ? C2 H 6  7 / 2 O2  2CO2  3H 2O
2  394  3  286 
H   1646  1560  86 KJ / mole
788
858
1646 KJ / mole
35. PCl5  PCl3  Cl2 PV  nRT  N 
PV
RT
x x x 2.46  200 2.46
n 
5 x x x 0.082  600 0.246
Total no.of moles  5  x  2 moles of N 2 Total no.of moles  10 moles
7 x
 7  x  10  x  3 K p  K c .  RT 
n

2 9
nPCl5  2moles  CPCl5   Kp   0.082  600
200 400
3
nPCl3  nCl2  3moles   PCl3   Cl2   54
200   0.082
3 3 4

 KC 
 PCl3  Cl2   200 200  9  1.1 atm
 PCl5  2 400
200
36. N2  3H 2  2 NH3
56Lt 
10Lt 20Lt
46Lt of N 2 is left un reacted

 4  So4 , H 2O
37. 
CuSo4 .5 H 2 O  Cu H 2 O


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H2O OH 2 H O O
Cu O S
H2O OH 2
H O O

38. I 2  HNO3  HIO3  NO2  H 2O : In HIO3  O.s.of I  5

39. Cast iron  Pig iron +scrap iron + coke (wrong statement-I)
% Pig iron: 4% Cast iron : 3% (Statement-II also wrong)
40. 2
 Ni  CO    SP3 :  Ni  CN    dsp 2
 4  4
3 3
CO  CN  
 6
 d 2 sp3; CO F6   sp3d 2
41. Conceptual
42. O O
 
  CH 3  COO 2 Ca 


CH 3  CN 
H 2O / H
 CH 3  C  OH 
Ca
 H2
 CH 3  C  CH 3
 A  B C 

Zn  Hg
Hcl
Cl
Cl2 / h
CH 3  CH  CH 3  CH 3CH 2CH 2  Cl   CH 3  CH 2  CH 3
 Major %  Pr opane

43. O



CH 3  CH 2  C  CH 2  C  H  NaBH
 4
 OH
PBr3

Br  H 3 PO3
MgBr 
M g / d .E


 Nu  add  n CO2 / H 3O 
O

C OH
44. R  CH 2  OH 
 R  CH 2  Cl 
SOCl2 NaCN
 NaCl
 R  CH 2  CN 
H 2 / Ni

 R  CH 2  CH 2  NH 2

i 0  alchol i 0  a min e
45. Compound “A” exhibits plane of symmetry but compound “B” do not exhibits any
symmetry. Hence A  optically inactive
B  optically active
46. Conceptual
47. Penicillin is a bactericidal antibiotic.
48. OH OH CH 2OH
CHO OH
 
1) CHCl  KOH aq

3
  
PCC

2) H

 A  B C 
49. Conceptual

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50. i ) CH COONa  CH  Na CO
NaOH  CaO
3 4 2 3
 
X
Cl2 / h
ii) CH 4   CH 3  Cl
X  Y 
iii ) 2CH 3  Cl Na
dryether
CH 3  CH 3  2 NaCl
Y  Z 
Kolbe ' s Electrolysis : 
 2CH 3COOK  2 H 2O  CH 3  CH 3  2CO2  2 KOH  H 2   
 At anode 

51. E 0cell 
RT
ln Kc
nF
E 0 cell  E 0 cu  / cu  E 0 cu 2 / cu   0.52  0.16  0.36V
0.36
 0.36V  0.025  ln Kc  ln Kc   14.4
0.025
 144 101
52. Fe0.93O  Fe93O100
x  2    93  x  3  200  x  79
79 Fe 2 93 79 Fe2
14 Fe3 100 ? Ans: 17
Ans: 85% Fe 2 5x  5 17  85
53. 2.303  100  10
67%   t 1 log     t 1  log 3
0.693  100  67  3
2 2
10
  t 1  0.48  1.6  t 1  x  101  16  101  1.6 Ans: 16
3
2 2
54. 20 1 1
200 ml  0.1M  20 m.moles   moles   60 gm CH 3COOH
1000 50 50
 1.2 gm acetic acid  0.6 gm wood charcoal
?  1gm
 2gm
55. NO, N2O, CO  Neutral Oxides
B2O3 , N2O5 , SO3 , P4O10  AcidicOxides
56. II, III, IV,VII, IX and X Compounds are aromatic.
57. Major product is ethene
58. (No of amino acids  4 and No of peptide bonds 3)
59. (Compounds 1,2,4,5,6 & 8-gives +ve Iodo form test)
60. ( i,ii,iii,iv,v  are exothermic)

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MATHEMATICS
61. z2  iz1  z1  z2  z2 , iz1,0 are collinear
arg  iz1   arg z2

 arg z2  arg z1 
2
z  iz1
 z3  2  z3  z2  i  z3  z1 
1 i
z z  
 Arg  3 2   and z3  z2  z3  z1
 z3  z1  2
 BC  AC and AB 2  AC 2  BC 2
 AB 2  25  2 AC 2  25
1 25
Re quired area  AC  BC   Sq.units
2 4
62. 1  sin 2 x Cos 2 x Sin 2 x 1  Sin 2 x Cos 2 x Sin 2 x
Sin 2 x 1  Cos 2 x Sin 2 x  1 1 0
Sin 2 x Cos 2 x 1  Sin 2 x 0 1 1

 
 1  Sin 2 x 1  0   Cos 2 x  1  0   Sin 2 x 1  0 
 2  Sin 2 x
   2  1  3,   2  1  1
A triangle can be constructed having its sides as   3,   1 and     2 is false
63. 3x  y  4 z  3  1
x  2 y  3 z  2  2
6 x  5 y  kz  3  3
1 2  2  7 x  5z  4
1  5  3  21x   20  k  z  12
21 20  k 12
    20  k  15  k  5
7 5 4
64. 12C2  10C3  23  63360
65.  x 2 if o  x  1

f  x   x x  0 if x  0
 2
 x if  1  x  0
x1, x2  A, f  x1   f  x2   x1  x2  f is one  one
f  x   y, x  A  y  A  B  f is on to
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66. A

A  4,7,8  , B  2,3, 4  , C  2,5,7 

D Divides BC in AB : AC  6 : 6  2 :1
 2  2   1 2  2  5   1 3 2  7   1 4   B D
C
D , , 
 2 1 2 1 2 1 
1

 OD  6i  13 j  18k
3

67. 1
Tangent to parabola y 2  4 x is y  mx 
m
Tangent to parabola x2  4 y is y  mx  m2
1
 m2  m  1
m
Equation of common tangent is x  y  1  0 is tangent to x2  y 2  c2
1
C 
2
68. 2 2
The circle is  x  4    y  4   42  x 2  y 2  8 x  8 y  16  0
Image of centre  4, 4  w.r.t.o y  0 is  4, 4 
2 2
Required circle is  x  4    y  4   16  x 2  y 2  8 x  8 y  16  0
69. x2  y 2  1, x2  2 y 2  4
R  x1, y1  is point of intersection of tangents drawn at P, Q to ellipse PQ chord
equation is xx1  2 yy1  4  0 touches x 2  y 2  1
16  4 3
 12  x12   2 y1    16  x 2  4 y 2  16 is ellipse with e 
2

  16 2
2  4
Latus rectum length  2
4
70. 5 
Centre C  ,20  , foci S 10,24 
2 
2
5  225 17
 16  , another foci s1  5,16 
2
ae  cs    10    20  24  
2  4 2
289 89
Hyperbola touches y-axis  1.5 10   b2  a 2e2  a 2  50   a2  a2 
4 4
 2a  89
71. The equation of plane containing the given line is
 x  y  2 z  3    2 x  3 y  4 z  4   0
 1  2  x  1  3  y  (2  4 ) z  3  4  0 parallel to z  axis
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1
 1 2  4   0  x  y  2 z  3   2x  3 y  4z  4  0
2
 y  2  0  y20
2
S.D = Distance from  0,0,1 on z  axis from y  2  0 is 2
2
1
72. x 1 y  2 z  3
Line passing through P 1, 2,3 is   r
3 m 1
Foot of perpendicular Q  3r  1, mr  2, r  3
x  2 y  3 z  10  0  3r  1  2  mr  2   3  r  3  10  0
Q lies on
 mr  1  W  3r  1, 1, r  3
7 2 7 5 1
PQ 2    3r   1  r 2   10r 2   r 
2 2 2 2
1
 mr  1  m    1  m  2
2
73.  x3 , x  1
f  x   f  x  is continuous x  R, but not differentiable at x  1
1 , x  1
y

y  x3 y  x
2

y 1
1
x1 x
1

y1
74. f 11  x   g11  x   0  f 1  x   g1  x   c
 f 1 1  g1 1  c  2  4  c
 f 1  x   g1  x   2  f  x   g  x   2 x  d
 f  2   g  2   4  d
 3  9  4  d  d  2  f  x   g  x   2 x  2
 f  4   g  4   10

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75. f  x  x  1
1   5   10   5  n  1  
lim 1
   1 
   1   ........    1 
n  n   n   n   n 
1  1  2  3  .....  n  1   5  n  1 5 7
 lim  n  5     lim 1  1 
n  n   n   n 2n 2 2
76. 
 2
I  2  x   Sinx  dx    x    Sinx dx 
2 
2
   
2 2 2 2
I   x    x dx    Sinx   Sinx dx    1 dx    1 dx
0 0 0 0
   
I  2   x 0 2  2    
 2 
77. ~  ~ p  q    p  q    ~  ~ p    ~ q     p  q 
  p ~ q    p  q 
 p  ~ q  q
 pF  p
78. Conceptual
79.  
Cos 1x  2Sin 1x  Cos 1 2 x   3Sin 1x   Sin 1 2 x  3Sin1x  Sin1 2 x
2 2
 Sin 1
3x  4 x   Sin
3 1
2 x  3 x  4 x  2 x  4 x3  x  0
3

 Sum of x values  0
80. dy dy x2
  x  1 y  1     x  1 dx  log e  y  1 
y 1 
xc
dx 2
1 x2
x  0, y  0  log e  0  0  c  log e  y  1  x
2
x2
x
 y 1  e 2
1 1
1
 y 1  e 2  1  e 2  1
81. a  3  0,   o
 122  4  a  3 a  6   0 
 36  a 2  6a  3a  18  0 
 a 2  3a  54  0   a  9  a  6   0
 a  9 or a  6 but a  3  least value of a is 7
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82. The coefficient of x 4 in
404 303 202 101
4co 1  x   4c1 1  x   4c2 1  x   4c3 1  x   4c4 in
4 4
1  x 101  1  101c x  101c x 2  .....  101c x101  is 1014
   1 2 101 

83. E1 , E2 , E3 be the events that two headed coin, biased coin, unbiased coin
1
 p  E1   p  E2   p  E3  
3
E be the event that head shows
1
E 1 1 4 4
p   3    p
 E1  1 1  1  3   1  1  1  3  1 4  3  2 9
    4 2
3 3 4  3 2 
 81 p  36
84. 2
v2  v1  4   p 1  3p2 1  2 p2  2 p  4  0  p  2 p 1  0  p  2 p  o
2 3 p   p  1 4 3 3 4 4
Cos     9
 4  p  12 3 p 2  1  13 13 43 2 9
    
 32
   4 
2 2
 Tan   Sec 2  Tan 2  1  169   3  2 33
4 3 3
 169   2  3  4  4 3  48  9  24 3
 165  3 2  57,   6
85. 1  17  9
Slope of BC  , D  2,  , Equation of SD is 2 x  y  
2  2 2
9  9   
x  0  y    0,    0,    9
2  2  2 
A

B D C

86. y  ax 2  bx  c 
dy
 2ax  b  2a  b  3

dx
 1
1,0   a  b  c  0   4a  2  a   , b  4
 2
 3,4   9a  3b  c  0 2a  3  b  1 

7
c    2a  b  4c  1  4  14  9
2
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87. n
1 n
r  1 1  n  n  1 n
Sn   1  2  3....  r  
    n   n  3
r 1
r r 1
2 2 2 4
S20  15 115  15
 5
20 20
88. 1 14
BE  2  49  81  64   7 m
2 2
h
Tan   h  21m
BE
A Tan  3
4
9 h
 E
4

B C
89. 2
  x 
 1   2 x  2   dx
2


1
2
x3
 
2
Area  x  2 x  3 dx   x 2  3 x]21
3
1
8  1 
  4  6    1  3  9
3  3 
y1
y  x2 1

y  2x 2

1  0,1 x
x o
x1 1

x 2

y1

90. 2m  112  2n  2m  2n  112

 m  7, n  4  mn  28

Sec: Sr.Super60_NUCLEUS & ALL_BT Page 17

 SRI CHAITANYA IIT ACADEMY, INDIA 29‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐19_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 29-01-2023
Time: 09.00Am to 12.00Pm GTM-19 Max. Marks: 300
KEY SHEET
PHYSICS
1) 3 2) 1 3) 2 4) 1 5) 1
6) 3 7) 2 8) 2 9) 4 10) 4
11) 4 12) 2 13) 3 14) 1 15) 1
16) 3 17) 2 18) 1 19) 1 20) 2
21) 3 22) 5 23) 9 24) 4 25) 14
26) 1 27) 3 28) 11 29) 5 30) 94

CHEMISTRY
31) 2 32) 2 33) 1 34) 4 35) 2
36) 4 37) 4 38) 2 39) 1 40) 2
41) 4 42) 3 43) 2 44) 2 45) 2
46) 4 47) 3 48) 2 49) 4 50) 3
51) 2 52) 3 53) 25 54) 20 55) 3
56) 9 57) 2 58) 4 59) 4 60) 7

MATHEMATICS
61) 4 62) 3 63) 4 64) 2 65) 4
66) 4 67) 2 68) 1 69) 2 70) 4
71) 2 72) 3 73) 3 74) 4 75) 4
76) 2 77) 2 78) 4 79) 1 80) 2
81) 18 82) 41 83) 4 84) 30 85) 6
86) 96 87) 3 88) 5 89) 80 90) 248

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 SRI CHAITANYA IIT ACADEMY, INDIA 29‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐19_KEY &SOL’S

SOLUTIONS
PHYSICS
1.

1 m.6m 7
WD  K , mgl  1  2 2  01  2  gl
2 m  6m 3
2.   2eV
hc
 8eV
1
T2  2T1
If 1 is the wavelength corresponding to maximum intensity at T1 & 2atT2 :
Then

2  1 (by wein ' s displacement Law)
2
hc hc hc
  16eV ,  2eV  K .Emax     14eV .
2 1 1
3. 4 4 1 4/3
I'  I  ( given) whereG  16, I ''  I
4G 4G 5 4
 16
3
1 1 5 i 4
I ''  I , or I ''  (5I ')  I ' , G  4 X i  G  16W
13 13 3 5 5
4
In sec ond case(i  i ') X 16  i '
3
2X 4 4 i' 13i 12
Req   , 4i  4i '   4i   i'  i
24 3 3 3 13
1 1
 i  i '  i  X 0.65mA  0.05mA
13 13

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 SRI CHAITANYA IIT ACADEMY, INDIA 29‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐19_KEY &SOL’S
4. 1 2 1 2
Li  Cv
2 2
9 X 1012 12 X 3
108
C
iv  12 
3 5
L 2.5 X 10
i  7.2 X 104 A .
5. MgL
y
 r 2e
y  L r e 
X 100   2   X 100
y  L r e 
 0.1 0.001 0.001 
  2X   X 100  4.89%
 110 0.05 0.125 
6. V V
V  L dI  dI  0 sin t  I   0 cos 
dt dt L L
Instantaneous power,
V2 V2
P  VI   sin t cos t   0 sin 2t
0
L 2 L

7. The total translational kinetic energy of all the molecules of a gas is given as
3 3
E  nRT  PV
2 2
Thus statement-1 is correct .In a gas ,molecule are in Brownian motion and travel
randomly in all directions and at every collision direction of motion changes so
velocity changes .Thus statement -1 and 2 are true but statement-2 is not explanation
for statement-1.
8.

Electric field due to lower part at B is= electric field due to full sphere –electric field
KQ 1  (4 / 3) R3 R
due to upper part E E  E.
 2 R 2 4 0 4R 2 12 0

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 SRI CHAITANYA IIT ACADEMY, INDIA 29‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐19_KEY &SOL’S
9. PV 
m
RT
M
P mT .
10 28 330
 
5 m 300
28  330  5 231 77
m  gm  gm
10  300 15 5
77 63
Gas leaked  28   gm
5 5
10. Number of beats n  n1  n2
For no beats observed , n1  n2  0
 V  V0   V  V0 
  324    320  0
 V   V 
Where , V0 =speed of observer and V=344 ms 1
On solving V0 =2.1 ms 1 .
11. 1 2 1
By conservation of energy , we have mvc  I c 2  mgh
2 2

Moment of inertia of solid and hollow cylinders are given as

MR 2
I solid  , I hollow  MR 2
2
For pure rolling , we have
v
  c , As I solid  I hollow ,  vsolid  vhollow
R
Hence solid cylinder will reach the bottom first.
12.
 
N x  N 0e  t  , N y  N 0 1  e  t  .
13. mr 2 2 r2
l   ( r t )
2 2
4
I r t
4 4 3
I A  rA  t A  1  4  1 
  .  . 
I B  rB  tB  4  1  4 
I
Hence, I A  B or I A  I B
43
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 SRI CHAITANYA IIT ACADEMY, INDIA 29‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐19_KEY &SOL’S
14. I
Time period 
mgd
Mgd = constant for both situation.
Moment of inertia about x-axis is less than the
Moment of inertia of about z-axis.
So Ix  Iz, so T1  T2.
15. f '' image
M .P  0  .
f e '' object
16. Conceptual.
17. 1 n 2 B cos1
T ; n B and 1  1 .
B n22 B2 cos 2
18. Conceptual
19. Lower NOT gate inverts input to zero. NOT gate from NAND gate inverts this output
to 1 upper NAND gate converts this input 1 and input 0 to 1. Thus A=1 and B=1
become inputs of NAND gate giving final output as zero.

20. In pulling case when force is applied at some angle to horizontal on a body then its
vertical upward component reduces the normal reaction so limiting friction will be less
in the case compared to the case of pushing because in case of pushing the vertical
component of force increases the normal reaction On body and the limiting friction
increases. So if is easier to pull a heavy object than to push so, statement-1is right and
of friction force also depends on the nature of surface. Therefore statement-1 &
statement-2 both are right & Statement-2 is not the correct explanation of statement-1.
21.
  
210   Ae 5004  3004 ,700   A 5004  3004  e  .  3
10
22. Path of satellite after reducing velocity is as shown.
GMm GMm 1 2 GMm 2
E   mv  v    .
5R 4R 2 10 R 5

23. If intensity of light incident on slits S1 and S2 is I0 .

The intensity through S4 will be I1  4I0

 
The intensity through S3 Will be I 2  4 I 0 cos 2   I 0
2
22
I max  4 I 0  I 0  3 9
K      .
I min  4 I 0  I 0  1 1
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 SRI CHAITANYA IIT ACADEMY, INDIA 29‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐19_KEY &SOL’S
24.
1 2 1 2 2 mgx  kx 2 4
mv  kx   mgx  v  , v   N  4.
2 2 m 10
25. For horizontal equilibrium of the block,
T1 sin 600  T2 sin 600  m 2r  m 2 (1sin 600 )
 T1  T2  m 21 ........(i)
For vertical equilibrium of the block
T1 cos600  T2 cos600  mg
 T1  T2  mg  2mg ..............(ii)
cos 60 0
Dividing (i) by (ii)

TA
2 1
T1  T2  1 T2  21
 or 
T1  T2 2 g TA
1 2 g
T2
2
i.e., 4  1   1   2  10 g  10  9.8 32  196    14rads 1
4 1 2 g 31 3 50 10
26. a a
y  tan  x  x 2 , compare y  px  qx 2 .q  .
2u 2 cos 2  2u 2 cos 2 
27. 1.02
Vs   2.04
0.5
3
iX 1  2.04  i  1.53
4
28. Heat lost =heat gain.
29. q q q
  .
R 2 R 3R
30. 2
m 
2
Psb  Pc  a   Pc
 0.5  0.0625P
c
 2  4
 m2    0.5 2 
Also P  Pc 1  a   Pc 1    1.125Pc
 2   2 
   
 % saving 
1.125 pc  0.0625 pc  X 100  94.4%.
1.125 pc

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 SRI CHAITANYA IIT ACADEMY, INDIA 29‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐19_KEY &SOL’S

CHEMISTRY
31. Fact
32. The slope change in the Ellingham diagram is due to phase change of metal. And,
the higher values of G 0 generally show its inability to form oxide.
33. Amount of charge used =10×60×60=36000C=0.373F
Amount of cu 2 available =0.5×0.5=0.25 mole
Amount of charge required to deposit all cu 2 =0.5F
 only cu 2 will be deposited(partially), zn 2 will not discharge
34. From
1  1 1 
 Rz 2   
  n2 n2 
 1 2

1  1 1  1 
2 1 1  3  27 R 
so  R(2)2     R and  R (3)     9R    
1  (2) 2 ()2  2  (1)2 (2)2  4  4 
   
 2  R  4  4   4 
      2   1 
 1  27 R  27   27 

35.  a 
 p  2  (V  b)  RT
 V 
a
 0(neglected )
V2
 p(v  b)  RT
pv  pb  RT
 pv  RT  pb
R 
v   T  b
p 
isobar is made at cons tan t pressurebetweenV ( y  axis)and T ( x  axis)

slope   tan   0.1642 dm3 k 1

R
p
R 0.0821dm3 mol 1 k 1
p 
tan  0.1642 dm3 k 1
 0.5atm mol 1

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 SRI CHAITANYA IIT ACADEMY, INDIA 29‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐19_KEY &SOL’S
36.  H   CN  ka  4.5  1010
HCN 

 H   CH 3COO  , k1a  1.5  105

CH 3COOH 
1
CN   H  
 HCN ka2 
ka

CH 3COOH  CN     HCN  CH 3COO  , keq  ka' ka"

1
 1.5  105   3.33  104
10
4.5  10
37. d [c ]
let rateof reaction   k[ A]x [ B] y
t
now fromthe given data
1.2  103  k[0.1]x [0.1] y ...(i )
1.2  103  k[0.1]x [0.2] y .....(ii )
2.4  103  k[0.1]x [0.2] y .......(iii )
Dividing equation (i ) and (ii )
1.2  103 k[0.1]x [0.1] y
 
1.2  103 k[0.1]x [0.1] y
we find , y  0
nowdividing equatin (i ) by (iii )
1.2  103 k[0.1]x [0.1] y
 
1.2  103 k[0.2]x [0.1] y
we find x  1
d [C ]
Hence  k[ A]1[ B ]0
dt
38. Conceptual.
39. Conceptual.
40. According to Hardy Schulze Rule statement 1 is correct. (Generally, the greater the
valence of the flocculating ion added, the greater is its power to cause
precipitation)
According to Hardy Schulze Rule statement 2 is incorrect.
41. 4 6
3 7
5
2 8
1
OH 9
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42. B is

43.

and .
44. More the number of fused benzene ring more is the resonance energy. Angularly
fused phenanthrene has five canonical forms whereas linearly fused anthracene
has only four forms.
Anthrarene :
8 9 1
7
2

3
6
10 4
5

Resonance energy 83 k.cal/mole.

Phenanthrene:
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3
2
4
5
1 6

10 7
9 8

Resonance energy 91 k.cal/mol.

45. Presence of hydroxyl group at the anomeric carbon is needed for the reaction with
hydroxyl amin
46.
47. Square planar complexes with asymmetric ligands are optically active.
48. Bridging ligands as hydroxo, OH  and Imido NH 2
49. 1)
Co  en   Cr  ox   ,
 3  3
Co  en   ox   Cr  ox   en  
 2  2 
Cr  ox  en   Co  ox  en 
 2 2 
Cr  en   Co  ox  
 3  3
Cu  NH 3   CuCl4 
 4
2)
Cu  NH 3  Cl  Cu  NH 3  Cl3 
 3 

3)
Cr  en   Co  NO2  
 3  6
Cr  en   NO2   Co(en)  NO2  
 2 2 4
Co(en)2  NO2   Cr  en  NO2  
 2 4
Co  NO2   Cr  en  
 6 3

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50.

51. CrO42  2 H   2 H 2O  CrO5  3H 2O

O O
+6
Cr
O
O
O
52. The reaction given is
Cr2O72  Fe2  C2O42  Cr 3  Fe3  CO2
Cr2O72  2Cr 3
On balancing
14 H   Cr2O72  6e  2Cr 3  7 H 2O ..............(i )
Fe 2  Fe3  e  ................(ii )
C2O42  2CO2  2e .................(iii )
On adding all the three equations.
Cr2O72  Fe2  C2O42  14 H   3e  2Cr 3  Fe3  2CO2  7 H 2O
Hence the total no.of electrons involved in this reaction =3.

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53. H 0  nFE 0
G 0   nFE 0  H 0  T S 0 
T


 825.2 X 
103   2 X 96487 X 4.315 

825.2 X 103  832.682 X 10
3

298 298
7.483 X 103
  25.1JK 1mol 1.
298
54. weight / M w 6.3 / 126
 H 2C2O4.2 H 2O    x X 102   20.
V ( L) 250 / 1000
55. The number of halogen/(s) forming halic (V) acid is
HClO3
HBrO3
HIO3.
56. O

C
O
O
C
C
Cr

C C
C O

O
O
57. Conceptual
58. 5d series have higher density than 4d series and d‐block have higher density than
s‐block elements
Ag, Au ii) Zn, Hg iii) Na, Cu iv) Ca, Co.
59. OH OH OH

OH + HCOONa
A is (C)
CHO

OH (D)
D
D
60. 7(carboxylic acid, anhydride, lactone, ether, ketone, amide and amine).

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MATHEMATICS
61. R1  {ab  0, a, b  R}.
Obviously R1 is reflexive symmetric but not Transitive
If a  2, b  0, c  3, ab  0; bc  0 but ac  0.
for R2 : a  b does not imply b  a.
62. sin x  0,cos x  1  x  0,2 ,4
7 7
sin x  cos x  1 is possible when  5
or sin x  1,cos x  0  x  ,
2 2
63. 2 4 2
e(cos x  cos x .....) loge 2 2cot x  1,8
2
 2cot x  cot 2 x  0,3
t 2  9t  8  t  1,9
2sin x 2 2 1
  
sin x  3 cos x 1  3 cot x 4 2
64. I : m | a  ib |2 ; n | c  id |2
mn  (| a  ib || c  id |)2 | (ac  bd )  i(ad  bc) |2
= e2  f 2
Where e=ac- bd & f=ad +bc  I
T1 :| z1  z2 |2  (| z1 |  | z2 |)2
 Re( z1 z 2 ) | z1 z 2 | z1 z 2  0
65.
sin 2 x  2sin x  5  2sin 2 y  (sin x  1)2  4  2sin 2 y
thisistrueif sin x  1and | sin y | 1 ,
sin x | sin y |
66. Total matches of boys=7×4=28ways
Total matches of girls=n×6=6nways
28+6n=52 n=4
67. 2403  2400.23  8(24.100 )  8(1  15)100  8  15
8
Fractional part=
15
68. Statement 2 is a property of an A.P so it is true.
a1  a4  a7    a16  147,
 (a1  a16 )  (a4  a13)  (a7  a10 )  147
 3(a1  a16 )  147
 a1  a16  49.
 a1  a6  a11  a16  (a1  a16 )  (a6  a11)
 2(a1  a16 )  98
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 SRI CHAITANYA IIT ACADEMY, INDIA 29‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐19_KEY &SOL’S
69. x y
  1 passes through(2,3)
h k

2 3
 1
h k
2 3
Locus:   1
x y
70. c  (3,5) r  34  p  34  p  3  p  25.
s11  0  p  29
71. Tangent to y  x 2 : tx  y  at 2 is tanget to y  ( x  2)2
t2 2 2 t2
 tx   ( x  2)  x  (t  4) x   4  0
4 4
D  0  t  0 or t  4
y  4( x  1)
72. x2 y2
  1  Q(2cos , 2 sin  )
4 2

2cos  4 2 sin   3
R  ( x, y )  ( , )
2 2
( x  2)2 ( y  3 / 2) 2
Eliminate ' ' :  1
1 (1 / 2)
1
e  11 / 2 
2
73. lim sin 2 ( (1  sin 2 x) 2 )
x0
x4

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 SRI CHAITANYA IIT ACADEMY, INDIA 29‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐19_KEY &SOL’S
lim sin 2 ( (2sin 2 x  sin 4 x))
 x0
x4
lim sin 2 ( (2sin 2 x  sin 4 x  2 (2sin 2 x  sin 4 x) 2
 x  0( )
 (2sin 2 x  sin 4 x x4
 4 2
74. p q p q  pq
T T F F T
T F F T F
F T T F T
F F T T T

options :
1 2 3 4
T T T T
T F T T
F T T T
F T F T
75. ( x  1)2  (2 x  5)2  ( x 2  3x)  x  20  x  3
32  3  21
Average marks   2.8
20
76.    2a
a  (sin 1 x)2  (cos 1 x)2  (  2cos 1 x)  2cos 1 x  
2 2 2 
 2a  2a
 cos 1(2 x 2  1)    2 x 2  1  cos(  )
2  2 
77.   a  ab  ab 
2
  1 0 0 
A A  I   ab  a  ab   0 1 0 
T  2
   
 ab  0 0 1
  ab  a 2  
 
  a 2  1;  ab  0
a3  b3  c3  3abc  (a  b  c)(a 2  b2  c 2  ab  bc  ca)
 a  b  c  (a  b  c)2   a 2  2 ab
 1
2  3abc  1  abc  1 / 3
or 2  3abc  1  abc  1
78. For infinite solution, D  0, D1  D2  D3  0

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 SRI CHAITANYA IIT ACADEMY, INDIA 29‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐19_KEY &SOL’S
79. Differentiable  continuous  k2  1
f 1( x)  2k1( x   ), x   LHD  RHD
 k2 sin x, x    k1  1/ 2

80. dy e0
( )  (1  log x) x c  1  log c 
dx e 1
 e  e1/ e1
81. f ( x)
lim 4t 3 0
lim g ( x)  x  2
x 2
 ( x  2)
dt ( )
0
6
4( f ( x)) f x 4  63
lim 3 1
 x2  18
1 12
82. 2 0 1
2 41
A  ( x  6)dx   x dx   xdx 
6
3 2 0
83. 1
put y  vx  sin ( y / x)  log x  c
e
put x  1, y  0, c  0. y  x sin(log x) A   x.sin(log x)dx put log x  t
1

1 1
  e2t .sin t dt   .e2    e2 
5 5
0
84. b.c  10 | c | 4
a.(b  c)  a  b  c

| a  (b  c) || a || b  c |sin .1
3
 30
85. (a  c).(b  d )
 9   6
|bd |
86. x 1 y 1 z  3
eq.of plane : 6 7 8  0  x  2y  z  0
3 5 7
7  4  13
dis tan ce   4 6  PQ
6
87. p  1/ 10; q  9 / 10
n
9 1
p (atleast 1hit )  1      (0.9) n  0.75
 10  4
 nmin  3
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 SRI CHAITANYA IIT ACADEMY, INDIA 29‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐19_KEY &SOL’S
88. let AM  x
( MD)2  ( MC )2  64  x 2  121  (10  x)2  f ( x)( say)
f 1( x )  0  x  5
f 11( x)  0
89. h
tan 450   BC  h
BC

BD  EF  h  x
y
sin 300   y  40
80
x
cos300   x  40 3
80
h y h  40
tan 750  2 3   h  80m
hx h  40 3
90. f ( x  y)  2 x. f ( y)  4 y f ( x)
put y  2 2 x f (2)  42 f ( x)  f ( x  2)
f 1( x  2)  16 f 1( x)  3.2 x.log 2
f 1(4)  16 f 1(2)  12ln 2.....................(1)
f ( y  2)  4 f ( y )  3.4 y
f 1(4)  4 f 1(2)  96ln 2........................(2)
solving : f 1(2)  7ln 2 f 1(4)  24.31.ln 2
f 1(4) 14.24.31.ln 2 14  124
14    248 .
f 1(2) 7 ln 2 7

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 SRI CHAITANYA IIT ACADEMY, INDIA 30‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐20_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 30-01-2023
Time: 09.00Am to 12.00Pm GTM-20 Max. Marks: 300
KEY SHEET
PHYSICS
1 1 2 1 3 4 4 2 5 2
6 4 7 3 8 4 9 2 10 3
11 1 12 4 13 1 14 4 15 1
16 4 17 2 18 4 19 1 20 1
21 1 22 3 23 16 24 20 25 4
26 1 27 2400 28 100 29 27 30 3

CHEMISTRY
31 4 32 3 33 3 34 1 35 1
36 2 37 1 38 4 39 3 40 4
41 3 42 2 43 2 44 1 45 4
46 4 47 4 48 4 49 1 50 4
51 6 52 3 53 9 54 7 55 108
56 5 57 3 58 9 59 9 60 3

MATHEMATICS
61 4 62 4 63 3 64 1 65 3
66 1 67 3 68 3 69 4 70 2
71 2 72 3 73 4 74 3 75 2
76 4 77 1 78 3 79 2 80 2
81 10 82 21 83 77 84 766 85 2
86 5 87 96 88 124 89 8 90 6

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 SRI CHAITANYA IIT ACADEMY, INDIA 30‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐20_KEY &SOL’S

SOLUTIONS
PHYSICS
1. For minimum speed, observer is symmetric between the 2 positions

A x B

300 300 h  3400 m

O tan 300 
 x / 2   2h tan 300 V 
680
m / s So, option (1) can’t be
min
h 3
2. From momentum conservation equation, we have
j
u02  2 gH


 u0 cos 
I
Pi  Pf
 m  u0 cos   i  m  
u02  2 gH j   2m  v ....... i 

u02 sin 2 
H ........ ii 
2g
 u cos  u cos 
From Equations (i) and (ii) v 0 i  0 j
2 2
Since both components of v are equal. Therefore, it is making 450 with horizontal.
3. V1 T
 1
V2 T2
4. Velocity of light in vacuum CV 
1
0 0
1
Velocity of light in the medium Cm 

CV  3
Refractive index of the medium is n    r r  2   3
Cm 0 0 2
5. Conceptual
6. 
h
 p  h Graph is rectangular hyperbola.
p
7. R
Since 
l1
R
Sl1

100  2.9cm   3
S 100  l1  100  l1  100  2.9  cm
The accuracy of measuring R can be improved if S and R are of the same order, i.e. S
should be changed to 3 .
8. Suppose straight length of the rod is l . The distance of the end B from O
OB 2  OA2   2 R   l 2   2 R 
2 2

The potential difference between A and B can be calculated as:

VA  VB  VO  VB   VO  VA  ...... i 

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 SRI CHAITANYA IIT ACADEMY, INDIA 30‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐20_KEY &SOL’S

B  OB 
2
B 2 B Bl 2
 OA 
2
Here V0  VB   l  4 R 2  And V0  VB 
2 2 2 2
Substituting these values in equation (i), we get
B 2 3
VA  VB   l  4 R 2   l 2    4 R 2  2 B R 2 .
2   2
  
9. v M  4i  5 j  8k , v 0  3i  4 j  5k  Plane of the mirror is xy   v I  3i  4 j
    
  
and for kv  2v  v  2  8  5  11  v I  3i  4 j  11k
IZ MZ OZ

q
10. E E
q

E
q q

Kq Kq Kq Kq
E  0,V     0
x x x x
It is not necessary that if potential is zero then electric field is also zero.
11. X Q Q2  A2T 2 
X  5YZ 2  Y ......... 
i X  capacitance   
Z2 V W  ML2T 2 
 M 1L2T 4 A2 
Z   MT 2 A1  Y  
F
X   M 1L2T 4 A2  Z  B  2
IL  MT 2 A1 
Y   M 3 L2T 8 A4 
12. V  V1  V2
4 3 4 3 4 3
 R   R1   R2 3
R13  R23
3 2 3
13. 1 4 2
Maximum KE=TE- PEmin Or kA2  9  5  k   8 104 N / m
2 0.012
m 2 
Time period T  2  2  s
k 8 10 4
100
14. When two capacitors with capacitance C1 and C2 at potential V1 and V2 connected to
each other by wire, charge begin to flow from higher to lower potential till they
acquire common potential. Here, some loss of energy takes place which is given by
C1C2
V1  V3 
2
Heat loss, H 
2  C1  C2 
C
In the equation, put V2  0, V1  V0 C1  C , C2 
2
C
C 1
Loss of heat  2 V  0 2  C V 2 H  CV02
 C 0 6
0
6
2 C  
 2
15. Currents in same direction attract
16. Mass of object remains same
W earth  9 g earth 
Weight of object  acceleration due to gravity  
W planet  4 g planet 
2 2
9 GM  earth  R  planet  M  earth  R  planet 
    
4 GM  planet  R 2 earth  M  planet  R 2  earth 

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2 2
R  planet  R  earth  R
 9 2
 R 2 planet   
R  earth  2 2
17. The induced emf across the element of length dx is
de  Bvx dx  B  x  dx.
O P Q
  
x
a dx
b
b
B  b 2  a 2 
The emf across the whole rod e  B  xdx 
a
2
18. k1 
hc
 k2  3k1 
2hc
 
3hc
 3
  
hc hc
So 2  so 
 2
19. h h h2 hc
  k  But, 0 
p 2mk 2m 2
k
2mc 2
From equation (i) and equation (ii) 0 
h
20.  1  T2 T2 
  1   1  
 6  T1 T1 
T2 5 6T
  T 1  2 ......... 1
T1 6 5
1  T2  62 
 1   .............. 2 
3  T1 
From eqn (1) and (2)
T2 310K and T1  372K
21. T1 cos37 0  T2 cos 530
4
 T2
3
T1 4T1  3T2 
T1 3

5 5 T2 4
22.
 a 2   0     a 2  a3
rcm 
   a 2    a 2 
23. 110 Mg 63 10
P1  P0  105 Pa  4
 2 105 Pa ; P2  P0   105  4
 64 105 Pa
10 A 10
5 5
T T
For adiabatic process 1 1  21
 400  3 
T23
2 2
P1 P2
 2 10  5 3
 64 10 5 3

2
T2  400  32  5  400  22  1600 K
24. Ex  
dV
   6  8 y2  Ey  
dV
  16 xy  8  6 z  Ez  
dV
   6 y  8z 
dx dy dy
At origin x  y  z  0 so Ex  6, E y  8 and Ez  0  E  E x2  E y2  10 N / C
Hence force F  QE  2  10  20 N
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 SRI CHAITANYA IIT ACADEMY, INDIA 30‐01‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐20_KEY &SOL’S
25.  t  t
N  No.e  1  8.e
 t l n2
e  8  t  3 l n 2  .t  3l n 2
t1/2
 t  3t1/2  3  1.4 109  4.2  109 Years
26. i
Using vd we get ;
ude ie nh 7 5 5
    
n udh ih ne 4 7 4
27. 
V
K
28. 5 V
Potential gradient   p
1000 1200
V p  6V
Vp 6
And R p    100
I 60 103
VP

L  1200 cm

1000cm

I  60 mA
G

5V 20 

29.  p
Electric field due to dipole on equatorial plane, E  k 3
r

 p
At point P, F p  k 3 Q ...............  i 
y

  p 
At point P , F p    k
'
Q  ...............  ii 
  y / 3 3

 
From

equation

(i) and (ii),

F p'  27 F p'  27 F
30. Both V and I are in the same phase. So, let us calculate the time taken by the voltage
to change from peak value to rms value. Now 220  220sin100 t1
 1 220
 or  100  t1   or  t1 
s again,  220 sin 100 t2
2 200 2
1 3 3
 or   sin 100 t2  or  100 t2   or  t2  s
2 4 400
1
Required time t2  t1  s  2.5 103 s
400

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CHEMISTRY
31. 
Na2 B4O7   B2O3  NaBO2 3B2O3  Cr2O3  2Cr BO2   3
X Y
32. 
1
; V
1
K .E n
33. Apply MO theory. For CO32 : BO :1+
No.of π-bonds 1
 1   1.33
Total No σ bonds 3
34. S sys
V 10
 2.303 nR log 2  2.303  1 8.314  log  19.14 J / k
V1 1
Ssurr  0; Stotal  19.14 J / K
35. Carbonions are stabilized by e  with drawing groups
36. At high pressure ‘a’ is neglected due to repulsion between gas molecules.
 P V  b   RT or PV  RT  Pb
37. Lattice energy  Melting point
( L.E.) ( M.P.)
L.E . : LiF  LiCl
L.E. : MgO  NaCl
Lattice energy depends on the size of the ion. Smaller the size of anion or cation,
z z
higher the value of L.E. LE 
d
38. The products are due to coupling of different alkyl groups. CH 3CH 2CH 2Cl is RCl and
CH 3 C HCH 3 is R ' Cl , the products will be RR,R’R’ and RR’.
|
Cl
39. In cyclopentadienyl anion, each carbon is sp2 hybridized
H H

H
: H

40. Cl O  Na 

623 K 
 NaOH 
300 atm
Dow’s process
41. Xenates(+6), Perxenates(+8)
XeO3  NaOH  Na   HXeO4  2  HXeO4   2OH    XeO6   Xe  O2  2 H 2O
  4

42. Factual
43. KOBr - Hoffmann’s bromamide degradation with stepping down of series.
LiAlH 4 Reduces amides to 10 amines with same no. of carbon atoms
44. Flocculating value 
1
Flocculating power
According to Hardy-Schulze rule: Flocculating power : Sn 4  Al 3  Ba 2  Na  for
negatively charged sol
45. T f  KCl  T f  X  1
mKCl  mx   i X   For trimerisation of 75% extent
iKCl i x  2

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1  3 1
i  1    1  i 
6  4 2
46. Conceptual
47. 5% of sulphur is used for manufacturing of tyre rubber
48. Due to weak Bi – H bond.
49. Conceptual
50. Statement I is correct as +3 oxidation state of lanthanoids is more stable. Thus
E 0 4 3 is positive.
Ce /Ce
Statement II is incorrect as lanthanoids are more stable in +3 oxidation state.
51. P4  3NaOH  3H 2O  PH3  3NaH 2 PO2
X y6
52. The intially formed carbocation will undergo rearrangement. The 20 -carbocation will
form an optical isomer when attached with benzene.
53. p y  pz 3 3
Kp    3 Pequilibrium  px  p y  pz  3  3  3  9atm
px 3
54. 32 Ge 1s 2 2s 2 2 p 6 3s 2 3 p 6 4s 2 3d 10 4 p 2
ml  0, 1, 1 ml  1,0, 1  2, 1,0,1,2
ml  0  s  orbital and one p-orbital out of three in subshell, one d-orbital.
55. For any sparingly soluble salt
Ax By  x A y   y B x
In question its A3B2 K sp  108 S 5
So a = 108
56. 1 100
K 2  ln
1 100
........ i  K1  ln
K
........ ii  2 
ln 25
 4.65
t 4 t 50 K1 ln 2
57. M
 3   M  3  e n-factor of MgCl2 .KCl.H 2O  3.
e
58. O
||
O
||
O
||
OH  C*  CH 2  CH
|
 C*  NH  CH  C*  OCH 3
|
NH 2 NH 2

*
* *
* *
All marked carbon atoms of aspartame are sp
* 2

hybridized. Aspartame is methyl ester of dipeptide formed from aspartic acid and
phenylalanine.
59.

Nucleotide with uracile base present only in RNA.

60. The diene is CH 3CH  CH  CH 2  CH  CHCH 3

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MATHEMATICS
61. n
5 3  4
n n

5.       
5.5  3  4 9 9 9 000
 lim   n   n   
n n n
lim n 0
n 5  2 n  27.9 n n
5 2 0  0  27
      27
9 9
2
62. 3
f  x    x  2 f 1  x    x  2 5 
3/5 3
Non – differentiable at x = 2
5 x  2
2/5
5
63. 750

750
400

tan 750 
h
40

h  40 tan 750 h  40 2  3 mts 
64. 1  5 7 11
2sin 2   3sin   2  0  sin   2  2sin   1  0 sin  
 , , , Sum  2
2 6 6 6 6
65. Clearly origin is equidistant to the vertices  circumcentre   0,0,0   S , centroid
 3 1 5 4  0  5 5  7 
 , ,    3,3, 4  Centroid divides HS in the ratio 2:1
 3 3 3 

Let H   ,  ,   OG 
2OS  1 OH    3,3, 4    0   ,  ,  
 
2 1  3 3 3
  9,   9,   12 HS  92  92  122  306
66. 2
x2
 x 1 x 1
2
f  x     , x  0 log e f  x   x 2  log 2  log x  f 1  x   0  log e      e 2  x 


 x 2 2 2 e
4
 e
 2  4

 
2
Local max.value     e e
 ee
 2 
 
 e
67. We know that if f  x   f 1  x  then f  x   x   x  1  1  x  x 2  x  0  x  x  1  0
2

 x  0, x  1
68. x y h  1 k  4 2  1  12  3
Line   1  x  3y  3  0  
3 1 1 3 1  9 
h  1 16 k  4 16 11 48 28
Image   , 
11 28
 ,  h ,k  4  k
1 5 3 5 5 5 5 5 5 
69.  2 2
For no solution 2 3 5  0   18  5   2 12  20   2  2 2  12   0
4  6
 2  6  16  0    8   2   0   2
70. x  y  z 1 (1) Required plane x  y  z  1    x  2 y  3z   0
1    x  1  2  y  1  3  z  1  0 (2) (1) and (2) are perpendicular

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3
 (1) 1     11  2   11  3   0  2  3   
2
3
 Required plane is x  y  z  1  x  2 y  3z   0 2 x  2 y  2 z  2  3 x  6 y  9 z  0
2
 x  8 y  7 z  2 x  8 y  7 z  2
71.  a  ib     i Taking complex conjugate  a  ib     i
5 5


 i a  ib     i   i   b  ai     i  b  ai   
5 5 5 5
2
   i  
i
72.

 2,3

Equation of tangent at origin is

2
7 m
2  x  0   3  y  0   0  2 x  3 y  0 tan    3 7
4 2m 4
1
3
3m  2 7 1 1
   12 m  8  21  14 m  26m  13  m   y  x  x  2 y  0
3  2m 4 2 2
73. Equation of normal to the ellipse is
5x

4y
9
cos sin 
9
 r distance from centre  1
25sec   16cos ec 2
2

But radius of circle  3  No. of normals is 0.

74. 1
1
dt 1
1
e1 1
Put x 3  t  I 2    
3 0 e  2  t  3 0 e 1  t 
t
dt  I1
3e
I1
  3e
I2
75. Let the truth values of P and Q be T,F q  p is T  p   q  p  is T
p  q is T  P   p  q  is T
76. Let f  x   x3  3x  1 Then f 1  x   3 x 2  1 Let f 1  x   0  x  1 and f 1 f  1  0
2
77.
f  x  is symmetric about y  x Required area  2x  f  x   x dx
0
 2
 
 2   sin x dx     sin x  dx   2  2  2   2 K  3
3

0 

78.  5  1   5  1 1  5  1  5
87 87 87 87
487  687 2  87 C1.5 87 C3.53  .....87 C87 .587 
10.87  2  87 C3.53  ......87 C87 .587  =870+an expression divisible by 25
870 20
  34   Reminder is 20
25 25
79. Given hyperbola are conjugate and quadrilateral formed by their foci is a square.

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y

 0,be2 

x
 ae1,0 0  ae2 ,0

 0, be2 

a 2 2 a  b 
2 2 2
2
x2 y 2 x2 y 2 b2 2
Now 2  2  1 and 2  2  1 e1  1  2 ; e2  1  2 ; e1 e2 
2
:
a b a b a b a 2b 2
a 2  b2
e1e2 
ab
 2ae1  2be2   2ab  a 2  b2   2 a 2  b 2
A
2 ab
 
80. R   2,3 ,  3,3 ,  2,3 ,  3, 2  ,  3, 4  ,  4,3 ,  4, 4 
81. Given M  25, N  100, F  45, C  10, l  20
N
F
M l 2  c  f  10
f
82. n
2r  1 n 
1 1  1 440
Sn   2     2 
 1   n  1  21
r 1 r  r  1
2

r 1 r
2
 r  1  441 441

83. The required numbers contain
7!
1,1,1,1,1,2,3   42
5!
7!
1,1,1,1,2,2,2,   35
4!3!
Total no of ways  77
84. a b g  a 2  b 2  c 2 ad  be  cf ag  bh  ci 
c a d
 
A   d e f   A   b

h  AAT   da  eb  fc d 2  e2  f 2 dg  eh  fi 
T
e
 g h f i   ga  hb  ic gd  he  if g 2  h 2  i 2 
i   c
 
Given, a  b  c  d  e  f  g  h  i  9 a, b, c, d , e, f , g , h, i  0,1, 2,3
2 2 2 2 2 2 2 2 2

(i) One of them is 3, remaining are 0s 9 ways (ii) 2 of them are 2s, one of them is 1
and remaining 0s 9C2 .7C1  252 ays
iii) One of them is 2’s five of them are 1’s and remaining 0’s 9C .8C  504 ways 1 5

(iv) All are 1s1 waysTotal  9  252  504  1  767 ways

85. y
dy
 sin  x  y   sin  x  y   2sin x cos y sec2 ydy  2sin xdx
dx
  
tan y  2cos x  2  y  0   0  x   tan y  2  sec 2 y  y1  2sin x  5 y1    2
2 2
86. 2
y  eabx passes through (1, 1)  1  e ab
(1) Given slope  2  2bx  .e abx   2 2b.e a b  2 2b  2  b  1
2
 ab  0
  1,1
2a  3b  2 1  3  1  5

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87. x 1 y 1 z  3
Eqn.of the plane containing the given lines is 6 7 8 0
3 5 7
7  2  2   13
24
 4 6  PQ   96
2
 x  2 y  z  0 PQ  
1  2 1
2
6
2 2

88. f  x  y   2 f  y   4 f  x  put
x y
y  2 f  x  2   2 x. 3  16 f  x 
f '  x  2   3.2 x log 2  16 f '  x 
put x  2 f  2  y   4 f  y   4 y f  2  f 1  2  y   4 f 1  y   3.4 y log 4
f 1  4   4 f 1  2   48log 22 f 1  4   4 f 1  2   96log 2      (2) From (1) and (2)
f 1  4  7 124 log 2 
f 1  2   7 log 2 f 1  4   124 log 2 2   124
f 1  2 7 log 2
89. h 2  ab h 2   9  4  h  6 af 2  bg 2  9  f 2   4  9  f 2  4 f  2
h f  62 8
90. Applying R1  R1  R2 , R2  R2  R3
2 2 0
f  x  2 0 1  4  2 cos 2x Max.value =4+2
sin x cos x 1  cos 2 x
2 2

6

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 SRI CHAITANYA IIT ACADEMY, INDIA 31‐03‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐31_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 31-03-2023
Time: 09.00Am to 12.00Pm GTM-31 Max. Marks: 300
KEY SHEET
PHYSICS
1) 3 2) 4 3) 1 4) 4 5) 3
6) 1 7) 2 8) 4 9) 3 10) 2
11) 1 12) 3 13) 2 14) 3 15) 3
16) 2 17) 1 18) 2 19) 2 20) 2
21) 4 22) 12 23) 140 24) 10 25) 16
26) 8 27) 6 28) 8 29) 5 30) 2

CHEMISTRY
31 3 32 2 33 1 34 4 35 4
36 3 37 4 38 4 39 3 40 3
41 1 42 3 43 4 44 3 45 4
46 4 47 2 48 4 49 3 50 3
51 5 52 4 53 4 54 6 55 6
56 5 57 3 58 6 59 8 60 3

MATHEMATICS
61 3 62 3 63 1 64 4 65 2
66 3 67 3 68 1 69 1 70 1
71 1 72 2 73 1 74 4 75 3
76 4 77 1 78 2 79 4 80 2
81 3 82 15 83 8046 84 17 85 0
86 1 87 4 88 22 89 1 90 176

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SOLUTIONS
PHYSICS
1. Acceleration = 0

2.

3.

Time for maximum height from ground

T 4 7
t   t '  1   sec
2 3 3

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 3 5
4. Conservation of energy mg  2mg  m 2 2
2 2 2

1 2Gmm0 2Gmm0
5. m0v 2  
2  
  3
2 2
8Gm 8Gm 8Gm  1 
    1 
 3   3

6. Buoyant force
11
FB   R 3  g
24

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7. y  A  A cos t

 A1  cos t 

8.

9.

 
2 Rt 
E 
10. Current through inductor, I  1  e 3L 
2R  
 

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2
 Bv  2 B 2 2 v 2 2  4  1  1
11. Heat     R2  
 R  R 2
12.

A 5
13. Modulation index   m   0.5
Ac 10
2  2000
Band width  2 f m   2000 Hz
2
14.

15.

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t 0.693 t0
16.  0  
2  2

 t2  t1   20
17. Radius of gyration, RG a hbc c

Comparing powers, a = 1/2, b = 1/2, c = -3/2

18.

19. Least count = 1 M.S.D - 1 V.S.D = 0.05 - 0.047 = 0.003

Reading = 6.10 + 24  0.003 = 6.172
100 R
20. 100    60
 5R  3
 
 6 
(where R is thermal resistance of each rod)

21.

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22. Let x be the amount of water and y be the amount of ice at the end of second minute
yLice   x  y  Cwater T  30  60
400 y  ( x  y ) 4  2  1800 ……………….(1)
 x  y   4  6  30  60 …………(2)
Solving (i) and (ii)
1800
400 y   2  1800 , 400 y  1200  y  3g
6
30  120
Initial amount of ice   3  12 g
400
P  P 
23. W  nR  400  n  B   nR  300  n  D 
 PC   PA 
24.

4
I 0  3  100
25.     15.82%  16%
2 4 I0
26.

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8
Work done = q0 VB  VA    KPq0   8 Joule
125
Power 1
27. Intensity,   0 E02C
Area 2
2.4 100 1
   0 E02C
100 4  2 2 2
 E  6V / m
d
28. B R 2 I  Mg
2
Mgd
I
2 R 2 B
V 50 5
29. cos   R  
V 130 13
2
VR2  VL2  100 
2
130  VR2  VL  VC 
16900  VR2  1400
VR  2500  50V

30. Apply Bernoulli’s between B and C

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CHEMISTRY
31.

32. P(V  b)  RT
PV  bP  RT
Divide by RT
PV bP
 1
RT RT
 b 
Z 1  P
 RT 
Slope  b  size of atom
33. Na ( g )  e  Na  ( g )
Na ( g )  e  Na  ( g )
Rest B, C and D are endothermic
34. A, B and C are correct options
35. n2
rn  r
Z
r

 R  rn2  rn1   n22  n12
Z

He , R1    (32  22 )
r
2
Li 2 , R2    (42  32 )
r
3
 R1 3 5 15
  
 R2 2 7 14
36.

37. Biuret test detects presence of peptide bonds.

Barfoed’s test based on reduction of copper (II) acetate to copper (I) oxide which
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forms brick red precipitate.
Molisch’s test is for presence of carbohydrate.
38. Glyptal is used to manufacture of paints and lacquers
39. Due to deficiency of Vitamin E
40. B.H.T. and BHA are most familiar anti oxidants
41. CdS, ZnO, Graphite are the examples of hexagonal
42. AB  A  B 
K sp   A   B    5  109  10  109
  
 50  1018  5  1017 M 2
43.

44. Reverbertory furnace is used in sulphides of copper.

A, B and D are correct statements.
45. A – B has highest bond enthalpy so stiffest bond.
A – C has the largest bond length.
46. Let us assume reactivity of C – D is a and C – H is b.
2a 50

2a  6b 100
a
 0.5  a  0.5a  1.5b
a  3b
a
0.5a  1.5b   3
b 1
47. Wave length range is 200 – 315 nm (UV region)
48. A, B and C are correct statements. Repeated used of DDT in insects is not good. So,
Aldrin and Dieldrin were introduced in place of DDT in the market
49. In option A, B and D entropy is increasing where as in C option when pressure is
increased O2 molecules come closer to each other hence entropy decreases ∆S < 0
50. G   2.303RT log10 Keq.
413.15  2.303  2  298log10 K eq.
 0.3010  log10 K eq
K eq  2
Hence, graph (C) matches correctly
51. AB2 ( g )  A ( g )  B  ( g )  B( g )
t =0
t 240 hr ab b/3
b 3b 3a
a b   a  b 
3 4 4
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a
 a to time given is 240 hours which is equal to 2t1/2 for A.
4
Hence, 2t1/2  240  t1/2  120hrs  5 days
52. Degree of unsaturation of C4 H 7Cl is 1 hence alkenes and cyclo compounds are possible.
Alkene isomers = 10

x – y = 10 – 6 = 4
53. Only CH3CHO will react with Fehling solution.

143 gm of Cu2O is formed by = 44 gm CH3CHO

44
0.975 gm of Cu2O is formed by=   0.975
143
0.3 gm of CH3CHO
Mass of acetone = 0.2 gm
0.2
% of acetone    100  40%
0.5
x = 40
x
So,  4
10
54.

55. NH 2CONH 2  2 NaOH  2 NH 3  Na2CO3

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100  2
Moles of NH 2CONH 3  moles of HCL   0.2 moles
1000
1 1
Moles of NH 2CONH 2  moles of NH 3   0.2  0.1
2 2
Mass of urea =0.1  60=6 gm
56. P  P n nA 9
   P  P
P N nB 10
9 
P  P
10 W W
 A B
P WB WA
1 a 1 b
   5
10 b 2 a

57. AB2 ( g )  A ( g )  B  ( g )  B( g )
58. 
2 NaClo3   2 NaCl  3O2
NaCl  AgNO3  AgCl  NaNO3
2 2
Moles of O2  moles of NaCl  moles of AgCl
3 3
2 2 2
Moles of AgCl  moles of O2  
3 3 32
2 2
Moles of AgCl    143.5 gm
3 32
 5.97
6
59. Seven chiral carbon atoms have hydrogen

60.

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MATHEMATICS
61.
For one to 6, 3 black and 3 white and 7th ball is white
62.

63.

64. It is differentiable at x = 0 also differentiable at x = π

65. Circum centre is mid point of AB

66.

67.

68.

69.

70. See the figure ∠AOB = 1200

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71.

72. The mid point lie on x + y = 0

Find the equation of the chord by T = S1 and by taking mid point (-t, t)
73.

74.

75.

76.

77.

78.

79.

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80.

81.
82. Here focus chord is normal. Hence the focal chord is the major axis.
83. 6 3 9 4 12
(
Write general terms of 1 + x 2 ) (1 + x ) (1 + x )
84.

85.

86.

87.

88.

89.

90. Apply vector triple product

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 01-04-2023
Time: 09.00Am to 12.00Pm GTM-32 Max. Marks: 300
KEY SHEET
PHYSICS
1 2 2 1 3 2 4 1 5 3
6 1 7 1 8 1 9 4 10 3
11 2 12 1 13 4 14 4 15 1
16 3 17 4 18 2 19 2 20 3
21 34 22 13 23 13 24 13 25 14
26 24 27 34 28 14 29 24 30 123

CHEMISTRY
31 3 32 1 33 3 34 3 35 3
36 2 37 4 38 1 39 1 40 2
41 4 42 3 43 1 44 1 45 1
46 2 47 3 48 2 49 2 50 3
51 7 52 5 53 4 54 7 55 1
56 6 57 1 58 4 59 6 60 3

MATHEMATICS
61 1 62 4 63 1 64 2 65 2
66 4 67 3 68 4 69 4 70 1
71 1 72 1 73 2 74 4 75 3
76 1 77 3 78 2 79 3 80 4
81 66 82 4 83 2 84 5 85 4
86 8 87 3 88 2 89 5 90 8

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SOLUTIONS
PHYSICS
1. Angular momentum of the system is conserved during standing about the axis of swinging.
2. Conceptual
3. Conceptual
4. Thermal expansion depends on shape and dimensions of configuration of given material.
5. Conceptual
6. Increase in speed of air will reduce pressure above B, Hence level will raise.
7. In initial state inductor keeps its path open.
8. Conceptual
9. For beats to be heard frequencies must be different
10. Conceptual
11. The current in the inductor decays gradually hence bulbs will glow even after battery is
disconnected.
12. modulus oƒ elasticity Y 
V= VX  VY 
density P 
In general for solids Y    rider fly along X-axis
13. h

mv
14. Reading = kx which is same in both.
15. Conceptual
16. Conceptual
17. Low specific heat ensures quick/faster growth in temperature and high conductivity
leads to good transfer of heat.
18. Speed of light is more than speed of sound
19. Conceptual
20. Conceptual
21. Conceptual
22. Electrostatic field lines, gravitational field lines are not closed loops.
23. Fresnel distance = Z F  a  Z F2  1
2

 a
24. Introducing plate decreases effective width between plates, hence capacitance increases
and energy decreases.
25. For maximum tension spring should either have maximum compression at the rate of
maximum elongation.
26. 20
  s  max   N  20N Fnet  152   20   25N a  25  20  2.5 m s 2 Tan    53
2

2 15
27. PV   constant
He, Ne are monoatomic, oxygen is diatomic. Hence different.
28. Conceptual
29. Pressure depends on the no. of collisions of molecules with walls.
30. For   1 LC , circuit is inductive hence, current lags behind voltage.

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CHEMISTRY
31. For ideal binary solution Hmin  0 , G  ve , S  ve

When pure A & B will be there then all changes will be 0

32. Na2Cr2O7 is hygroscopic and that is why, it is not used as a primary standard in
volumetric analysis.
33. If above TB gas shows positive deviation.
34. Assertion is True, Reason is False.
35. 5 Br
Br
4

due to sterric crowding have high energy

36. Assertion is true Reason is true: Reason is not the correct explanation for Assertion.
37. Ferrimagnetism is in 
38. Both Statement-I and Statement-II are correct.
39. The order of bond angles is HCH  HCF  FCF
40. Stability of carbocation: IV < I < III < II due to SIR
41. Composition changes with external pressures
42. Pb2  Na2 S2O3  PbS2O3  2Na

PbS2O3  H 2O  PbS  H 2 SO4 -NCERT text book information

43. Assertion is True, Reason is true: Reason is the correct explanation for Assertion.
44. Dilute H2SO4 solution with Cu electrodes.
45. All(1,2,3,4)
2
46. Cl  is weak ligand so no pairing of electrons take place. so  NiCl4  is tetrahedral.
pph3 group is bulkier one so it favours tetrahedral geometry, through pph3 is strong
field ligand.
47. Two boron atoms have four B  O bonds while other two have three B  O bonds
Each boron atom has one OH group
48. Pb2 and Hg22
49. H3PO2 reduces diazonium salt into benzene derivative.
50. k k
m   1000;  x  y    1000
M M
1000  k
M 
 x  y

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
Solubility in gL 
1  k 1000 188
x y
51. 7 compounds
52. I to V factors influence the electronegativity of the atom of an element
53. In PF3 , OCl2 , N  SiH 3 3 and  SiH 3  2 O , the lone pairs delocalize in to the vacant d-
orbital of the surrounding atom in bond. Due to this reason the bond will get double
bond character and the effect of double bond character and the effect of lone pair at
central atom decreases. So the bond angles increases than expected
54. 7 statements are correct.
55. All are incorrect
56. n- function = 6
57. Equivalents of oxidizing agent must be same 5x=6y
58. SOLUTION: Conceptual, Modified question from NCERT.

H3 C CH3 CH3 CH3

OH

CH3
59. 2 XeF6  SiO2  2 XeOF4  SiF4
60.


Remaining reactions are conceptual

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MATHEMATICS
61. Conceptual
62. px  qy qx  py a a
  .is of the form X  Y 
p2  q 2 p2  q 2 p2  q 2 p  q2
2

The area bounded by the curve x  y  a  a  0  is 2a 2

63. 1
x n1  x n  x 2 n1  1 n 1 
0 1  x n dx  0  1  x n  1  x  dx  0
1 1 1
1
  1  x n1  dx  
dx dx
  1 ;  n  N.
0
1 x n
0 0
1 x n
n
64. Conceptual
65. Assume g  x     x   x Rolls theorem
2 2

66. RoS is symmetric if (RoS)-1 = RoS

But (RoS)-1 = S-1oR-1 = SoR
\ RoS is symmetric iff RoS = SoR.
Similarly SoR is symmetric iff SoR = RoS
67. Conceptual
68. Conceptual
69. Conceptual
70. Write equation of chord joining t1 & t2 where t1t2  4 It will come in form of
family of lines.
71. Conceptual
72. Conceptual
73. Conceptual
74. f ' x =3ax2  2bx  c  0
f  x =ax3  bx2  cx+d Rolls theorem
75. k  1,   1 k    2
76. 1     2   3   4  0 , a  b  1 and ab  1
77. Conceptual
78. Conceptual
79. x2
 2  x 2   i
n
80. Conceptual
81. x1+ x2 + x3 = 15 0  x1  5, 0  x2  10, 0  x3  15
n = co-efficient of x15 in (1 – x6) (1 – x11) (1 – x16) (1 – x)–3 n = 66
82. n
n  k 1 3
n
Ck 
2
n
 n  k 1
2 n

 k  n  1  k 
n
Ck
 
2
  k n  
3
k   
 
n
Ck 1 k k 1  Ck 1  k 1 k k 1

 k  n  1 
n n
 2  n  1 k  k 2   k  n  1  2  n  1 k 2  k 3 
2 2

k 1 k 1


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2     
n n n
  n  1   k   2  n  1   k 2     k 3 
 k 1   k 1   k 1 
 n  n  1  n  n  1 2n  1  n  n  1  n  n  1 
2 2
2 n
  n  1    2  n  1  n  1   2n  1  
2
  
 2  6  2  2  3 2
n  n  1 6  n  1  4  2n  1  3n n  n  1  n  2 
2 2

(i)   
2 6 12
83. Use vector products
84. 13 12 x  5 y  1 13
 x  2    y  3
2 2
Equation can be rewritten as  So, e  .
5 13 5

85. 

 
2
I     x   sin 2  sin x    cos 2  cos x  dx  2 I  2   sin 2  sin x   cos 2  cos x   dx
0 0
 
2 2
 I     sin 2  sin x    cos 2  cos x  dx     sin 2  cos x   cos 2  sin x   dx
0 0

2
2
 2 I    2dx  I 
0
2
86. Let   120 L.H.S 
1
sin120 sin 480 sin 720 sin 540
0
sin 72
1 sin 3 12  sin 54
0 0
sin 360 sin 540 cos 360 1
   
4 sin 720 8sin 360 cos360 8cos 360 8
87. Equation of normal is y = mx – 2am – am3 to the curve y2 = 4ax.
1 m m3  m2 1 
a=  y  mx   passes through (c, 0) then m    c   0
4 2 4  4 2 
 Remaining normals are perpendicular.
m2 1
 Product of the roots of equation   c  0 will be 1 .
4 2
1
c
2 1 1 3
  1  c    so 4c = 3
1 2 4 4
4
88. If x  (0, 1)
Then – 1  x + y < 0
And if x  [1, 2)
0x+y<1

Required area  4  .1. 2 sin   2 sq. units
1
2 4

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1
1 2 3
0
–1
–1

x+y=1
x+y=0
x + y = –1

89. Roots are 2 w,  2  3w   2  3w2  2  w  w2  ,  2  3w  and 2  3w2 are conjugate each other
2w is complex root, then other root must be 2w2 (as conjugate root occur in
conjugate pair)
2  w  w2  2   1  3 which is real.
Hence least degree of the polynomial : 5.
90. on putting log3 x  t, we get 2t  t  a  0 …(i)
2

If t  0, then 2t  t  a  0 …(ii)
2

If t  0, then 2t  t  a  0 (iii)
2

If Eq. (i) has four roots then Eq. (ii) must have both roots positive and Eq. (iii) has
both roots negative. Now, Eq. (ii) has both roots positive, if D  0
a/20
 1  8a  0, a  0
 1
 a   0,  on taking intersection.
 8
Again, Eq. (iii) has both roots negative, if D  0, a / 2  0.
  1
We again get a   0, 8   K  8
 

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 SRI CHAITANYA IIT ACADEMY, INDIA 02‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐32_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 02-04-2023
Time: 09.00Am to 12.00Pm GTM-33 Max. Marks: 300
KEY SHEET
PHYSICS
1) 1 2) 4 3) 4 4) 3 5) 1
6) 3 7) 1 8) 4 9) 1 10) 2
11) 1 12) 1 13) 3 14) 4 15) 4
16) 3 17) 4 18) 3 19) 4 20) 3
21) 4 22) 5 23) 1 24) 4 25) 2
26) 7 27) 4 28) 3 29) 5 30) 5

CHEMISTRY
31) 1 32) 1 33) 4 34) 4 35) 1
36) 2 37) 1 38) 1 39) 3 40) 4
41) 3 42) 1 43) 1 44) 3 45) 1
46) 1 47) 2 48) 2 49) 2 50) 3
51) 4 52) 4 53) 36 54) 9 55) 9
56) 500 57) 25 58) 2 59) 6 60) 22

MATHEMATICS
61) 2 62) 3 63) 2 64) 1 65) 1
66) 1 67) 3 68) 1 69) 1 70) 1
71) 1 72) 4 73) 1 74) 2 75) 3
76) 4 77) 1 78) 1 79) 3 80) 1
81) 71 82) 5 83) 9 84) 0 85) 12
86) 2000 87) 24 88) 25 89) 128 90) 1

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 SRI CHAITANYA IIT ACADEMY, INDIA 02‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐32_KEY &SOL’S

SOLUTIONS
PHYSICS
1 1 1 2 L
1.     Lp 
Lp L L L 2
1.8  104
Where L is inductance of each part,   0.9  104 H
2
L 0.9  104
 Lp    0.45  104 H
2 2
6 1 1 1 2
Resistance of each part, r   3 Now ,   
2 rp 3 3 3
L 0.45  104
Time constant of circuit,  p   3  105 s
rp 1.5
2. When the spherical conductors are connected by a conducting wire, charge is
redistributed and the spheres attain a common potential V.
1 QA 1  C AV  4 0 RA  V
 Intensity E A  or E A   
4 0 RA
2
4 0 RA2 4 0 RA2 RA
V E R 2R 2
Similarly EB   A  B  
RB EB RA R 1
3. If final temp is T
Cv  4T0  T   0 0 Cv T  T0 
PV0 0 PV
4 RT0 RT0
4T0  T 8T0
 T  T0 4T0  T  4T  4T0 8T0  5T T 
4 5
8
T0
Tf 2
Final pressure In left Pf  Pi  5 P0  P0
To 4To 5
8
T0
Force   0  0  A  0
8 8P 2 P 6P A
In right Pf  5 P0  P0 ,
To 5  5 5  5
A sin   B cos 
4. Z ; Dimensions of A and B would be same as two quantities can be added
A B
only when they have same dimensions. So from given equation, we can say that Z is
dimensionless.
5. The acceleration of ball during upward motion is g  a , where a is the acceleration due to
u
resistive force. The time of accent, t1  1
ga
v
The acceleration during descent g-a and time taken is t2  ......  2
ga
The time taken can also be written as
u2
2h ga u2
t2    t22  .......  3
ga ga  g  a g  a

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 SRI CHAITANYA IIT ACADEMY, INDIA 02‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐32_KEY &SOL’S
t2 u
Dividing equation (3) by equation (1) and (2) 
t1 v
 v 
6. Let velocity of car  vn '  n   ...... i 
 v  vs 
 v  vs   v  vs 
n ''  n '   ......  ii  from (i) & (ii) n ''  n  
 v   v  vs 
n ''  v  vs  v
Now  2   2  v  vs  2v  2vs  3vs  v  vs 
n  v  vs  3
7.  A  B  vA  vB , wa  w  Vg   wa  A   wa  B
8.

0
B  2  2M  cos 60
4 d 3
 A  B   0.5 0.01 
9. Error in AB=     AB      0.25  0.075  0.08
 A B   2.5 0.10 
 AB   0.25  0.08 m
10.

mv
d 2qB 1 T m
sin        45, t  
r mv 2 8 4qB
q2
11.

b bx
tan   y  mx  c y  b
f f
12. For potential to be made zero both capacitor must have equal charge and positive plate of
one must be connected to –Ve plate of other
120C1  200C2
3C1  5C2
P  100 
13. ig  3   P  Vi ig  3  ig  1.36 A
V   200 
14. For process (2) U  0 (isothermal zero) so 1 & 2 options are wrong so 4th option correct
V
15. i  neAU d  neAe E  neAe
l
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 SRI CHAITANYA IIT ACADEMY, INDIA 02‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐32_KEY &SOL’S
1.5  10  1.6  10  1 cm  0.14  2
16 19 2
  6.72  107 A
10 cm
16. Applying loop law to the outermost loop,
VBE  I B RB  Vcc  0
 VBE  5.5  10 5  5  105
 0.5V
bt ln 2 20
 A0
17. A  A0 e 2 m A  A0 e10 2 2
 A0 e  ln 2

2
18. f  f 2  f1  220 Hz. These many number of beats are formed but cannot be heard.
19. Applied emf e  e0 sin  t
1
L 
Produced current I  I0 sin  t    where tan   C
R

If    res then tan   0  0    current lags behind e. But in given graph, ‘I’ leads
2
ahead of e. So ‘1’ is wrong statement
20. WNC   TE 
If WNC  0, then TE  PE  KE is conserved. Work done by non conservative force can be +
ve or – ve or zero.
2
 3R 
21. I Ab  I disc  I plate  M  
 2
2
1 MR 2  3R  31
(From Parallel Axis Theorem) MR 2   M    MR 2
4 12  2 12
22.

V  20 V
  V  5  0  V  20  V  2 V  5  0
2 2
30 15 15  10 5
 4V  20  10  0  V   V,I   A  2I  5 A
4 2 2 2
23. 64   T  2 rl  r  10 m  1mm
4 3

power
24. F ('c'is speed of light)
c
60 F
F  2  107.Pressure =Force/area =
3  10 8
4 R 2
25. 20VSD  16MSD  5VSD  4MSD
1MSD 1mm
LC    2  101 mm
no.of divisions on vernier scale 5
2
 30 
10   
gT 2  20  9
26. l   m
4 2
4  10 16

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 SRI CHAITANYA IIT ACADEMY, INDIA 02‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐32_KEY &SOL’S
l l l
g  4 2 2
 4 2 2
 4 2 N 2 2
T  t t
 
N
dg dl 2dt
 
g l t
 
 dg   10 3
1  2 1.6 20 
 g  100    2   10    
 9  30   9 3
  16  
61.6
 7
9
 9  108
27. V   1.5  108 m / s
k 6
C 3  108
Refractive index    K 
V 1.5  108
 K 4
1 hc
28. mu12    0 ______ 1
2 1
1 hc
mu22    0 ______  2
2 2
1240
1  2  496   0
 2 1240
 0
620
1240 1240  1 1 
  2 0    0   0  1240  
310 496  310 496 
x
 0  1.5ev   x  3ev
2
29. H  B / 0
ni  2  104
40  100i  2  104
20
i A  5A
4
dy
30.    1 t0 
D
D
y     1 t0  velocity of central maxima,
d
dy 2.4  104  2
v   2t  12 
dt 2  103
At t = 4 sec
v  4.8 m / s

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 SRI CHAITANYA IIT ACADEMY, INDIA 02‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐32_KEY &SOL’S

CHEMISTRY
31. Alkali metal hydrides are ionic hydrides due to more difference in electronegativity. So
their fused hydrides can conduct electricity due to movement if ions in liquid state.
32. The intense blue colour is due to charge transfer between Fe(II) and Fe(III) in out but side
of the coordination sphere.
33. Oxidation states of Co in (A) is + 3, (B) is – 1 (C) is + 2 and (D) is 0.
34. Internal energy of CD molecule is less than AB since C – D bond distance is smaller tan
A–B
35. Quartz contain three dimensional diamond like structure in which all the atoms form
strong covalent with all the neighbouring atoms.
36.
O

HOS OOH

O
Contain four S – O and one O – O bonds
37. The coordination isomers of same complex compound have same coordination numbers
and have same empirial formula
38. Primary pollutants include ammonia, sulfur dioxide, nitrogen dioxide and carbon
monoxide. Secondary pollutants include ground-level ozone, acid rain and nutrient
enrichment compounds. So, the correct option is O3 .

39.
Nitriles are selectively reduced by DIBAL–H to imines followed by hydrolysis to
aldehydes similarly, esters are also reduced to aldehyde with DIBAL–H.
h
40. 
mv
According to Einstein’s theory of photoelectric effect :
1 2h  v  v0 
hv  hv0  KE hv  hv0  mv 2 2h  v  v0   mv 2  v2
2 m
1
h 1
v   v  v0  2  1
 1
 v  v0  2  v  v0  2
41.

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 SRI CHAITANYA IIT ACADEMY, INDIA 02‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐32_KEY &SOL’S

42. STotal  C p ln
T1  T2   C ln
T1  T2 
p
2T1 2T2
43. Co  Vitamin B12
Zn  Carbonic anhydrase
Rh  Wilkinson catalyst
Mg  Chlorophyll
44.

45.

46. Formula of the compounds

No. of octahedral voids are equal to the number of atoms forming lattice
A occupy octahedral void i.e. 2/3 of them
B forms crystal lattice
A2/3 B  A2 B3
47. Electron withdrawing group enhances the rate of hydrolysis.
48.

49.   COCH 3 is present it will show both 2, 4-DNP & iodoform test. Due to steric inhibition
of resonance. I.P of ‘N’ is not involved in delocalization so coupling reaction will not
take place.
50. A2  g   B2  g  
  2 AB  g  ______ 1

K1

6 AB  g  
  3 A2  g   3B2  g  ______  2

K2

3
 1
Reaction(2) = - 3 × reaction (1)  K 2     K 2  K13
 K1 
51. H 2SO 4  H 2 NCONH 2  H 2 NSO 3 H
Sulphamic acid NH 2SO3H can react with nitrite ion liberating N 2 gas, so used to remove
from the mixture of NO 2 and NO 3 before testing for NO 3 by brown ring test.
H 2 NSO 2 OH  HNO 2  N 2  H 2SO 4  H 2 O

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 SRI CHAITANYA IIT ACADEMY, INDIA 02‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐32_KEY &SOL’S
H2 NSO4OH  H2O  NH4 HSO4
can give ppt with BaCl2 .
NH4 HSO4
52. Antimony pentafluoride has also been used in the first discovered chemical reaction that
produced fluorine gas from fluoride compounds
4 SbF5  2 K 2 MnF6  4 KSbF6  2 MnF3  F2
Fe2O3  3CO  2 Fe  3CO2
53.
36 118
To get 118 gm of Fe we required 36 gm of CO
To get 118 tons of Fe we required 36 tons.
54.

55. H 2O  s   H 2O  l   H 2O  l   H 2O  g   H 2O  g 
1kg 1kg 1kg
at 273 k at 273 k at 373 k at 273 k at 383 k

S  S1  S 2  S3  S 4
334 373 2491 383
  4.2 ln   2 ln  9.267 kJ Kg 1 K 1
273 273 373 373
 2n  3
56. n1T1  n2T2 , n  300   n   T2 , 300  T2  T2  500 K
 5 5
57. Apply law of equivalence 25  N  30  0.1  2
30  0.2 6 1.2
N HCl    0.2 
25 5 5
1.2
For the 2nd titration  VHCl  30  0.2
5
6  5 30
VHCl    25 ml
1.2 1.2
58. 2A + B products 
Rate = K[A]x[B]y
r = K[A]x[B]y ______(i)
0.3 = K[A]x[B]y - - - - (1)
2.4 = K[2A]x[2B]y - - - - (2)
0.6 = K [2A]x[B]y - - - - (3)
From (1), (2) & (3)
x = 1, y = 2
Overall order = 2 + 1 = 3
Order w.r.t A = 1
Order w.r.t B = 2
59.

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60.

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 SRI CHAITANYA IIT ACADEMY, INDIA 02‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐32_KEY &SOL’S

MATHEMATICS
ln 2 ln 2

f "  x  dx f '  x  0  2  e2 x f '  x dx

ln 2
61. e   e
2 x 2 x

0 0

1  ln 2
  f '  ln 2  3  2 e 2 x f  x   2 e 2 x f  x  dx   13
4  0

62. Conceptual
63. Conceptual
64. The line can be written as y  mx and curve as x 2  y 2  4
Let C(h, k) be a point on the circles and A  3,1 be given point, then
h2 3

3

 h  3  2 3
k2
 m  k  3m  2
3
Now, this point (h, k) lies on the circle
 
2
 3  2 3   3m  2  4
2

9 2  12  12 3  9m 2 2  4  12m  4

 9 1  m 2  2  12   3  m  12  0

3 1  m 2  2  4   3m 40 
   
2
16 3  m  4  3 1  m 2  4  0

   
2
3  m  3 1  m2  0
2 3m  2m 2  0
2m 2  2 3m  0
m  0, 3  
65. Conceptual
2
abcd a 2  b2  c2  d 2
66.   
4 4
 8  e 2  16  e2 64  e2  16e  64  4e2
16 4
16
5e2  16e  0 0e
5
1 1 2 3 
67.  2e     .....   2eS
x  3! 5! 7! 

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
r 1   2  r  1  1 1   1 1 
S      
r 1  2  r  1 ! 2 r 1  2  r  1 ! 2 r 1   2  r  !  2  r  1 !
1 1 1 1 1  1 1 1 2 e
      .......   e1    2eS  1
2 2 ! 3 ! 4 ! 5 !  2 2 e x 2 e
ex sin 2 x 2 x sec 2 x 2
ex 2 cos 2 x 2 x sec 2 x 2 ex sin 2 x tan x 2
1
68.  '  x  ln 1  x cos x sin x   sin x cos x  ln 1  x cos x sin x
1  x
cos x 2
e 1
x
sin x 2
2 x sin x 2
e x
2 x cos x 2
cos x 2 ex  1 sin x 2
 B  2Cx  .....
1 2 0 1 0 0 1 0 0
Put x  0, B  0 1 0  1 0 1  0 1 0  0
1 0 0 1 0 0 0 1 0
69. Perfect square   100   1  9  excluding one
Perfect cubes  1001/3   1  3
Perfect 4th powers  1001/4   1  2
Perfect 5th powers  1001/5   1  1
Perfect 6th powers  1001/6   1  1
Now, perfect 4th powers have already been counted in perfect squares and perfect 6th
powers have been counted with perfect squares as well as with perfect cubes
Hence the total ways = 9 + 3 + 1 – 1 = 12
dV  t  K T  t 
2

70.   K T  t  V t   C
dt 2
KT 2
At t  0,V  I  C  I 
2
KT 2
 Scrap value V T   V  t  T   C  I 
2
  
71. OAB  APO   AP  BP  r cot and AB  2r cos .
2 2 2
  
 Perimeter of APB, L  2r  cot  cos 
 2 2

dL   
 r   cos ec 2  sin   0
dr  2 2
  2 
for all    ,  .
3 3 
 
 L is maximum when   and then OP  r cos ec  2r.
3 2
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 SRI CHAITANYA IIT ACADEMY, INDIA 02‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐32_KEY &SOL’S
So, the co-ordinates of P are  2r cos 45, 2r sin 45 .
72. q  2and p  3

ln x  ln x
73. x
0
2
t 2
dt 
2x
74. Let origin be centre of larger circle and O ' be the centre of smaller circle.

Equation of (1) : x 2  y 2  2
5 7
Equation of (2) :  x  2  y 2  1
2
 A , 
4 4 
x A  xC y  yC
xB  , yB  A  xC2  yC2  2
2 2
2
  7
2
 5
xC   yC 

  4  2  4  1  AC 
7
  
2 2 2
   
 
75. Ortho-centres of triangles formed by three tangents and corresponding normals to a
parabola are equidistant from axis of parabola.
76.  PQT  QT PT  QP  PQ
PQ is symmetric  PQP 1  QPP 1  Q
 P 1 PQP 1  P 1Q  QP 1  P 1Q
 P Q  
1 T T
 QT P 1  P 1Q
77. Conceptual
78. Consider statement 2.
p  n  n 7  n is divisible by 7.  n N
 p 1  1  1  0 is divisible by 7.
7

 p 1  1is true.

Suppose p  k  : k 7  k is divisible by 7 is true. Let k 7  k  7 m, m  N 1
 7 6  7 5
 k  1 7   k  1  k 7   k    k  .....  7 k  1  k  1
 1  2
 7  7  7
  1

 k 7  k    k 6    k 5  .....    k
 2  6
 7
k 7  k is divisible by 7 and   is divisible by 7. For1  r  6.
 r
 By 1
 p  k  1 is true.
 p  n  is true,n  N.
 Statement (2) is true.

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 SRI CHAITANYA IIT ACADEMY, INDIA 02‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐32_KEY &SOL’S
Now consider statement 1.
From statement 2, n2  n is divisible by 7.  n N
  n  1   n  1 is divisible by 7.
7

  n  1   n  1 =7m, m  N and n 7  n  7 p.
7
p N
  n  1  n7  n7  n  1  7m
7

  n  1  n7  1  n7  n  7m

7
  
  n  1  n7  1  7  m  P
7

  n  1  n7  1 is divisible by 7. n  N
7

 Statement 1 is also true.

 Statement 1 and statement 2 both are true and statement 2 is the correct explanation for
statement 1.
79. l = 20, N/2 = 20, c.f. = 13, f = 14, h = 10
use formula
80. ~  p  q    ~ p  q 
  ~ p  ~ q   ~ p  q
~ p   ~ q  q
~ p
81. Probability that calculator of brand r is selected and is defective
6
 7  r k 6 8k
   kr      7r  r 2 
 21  21 r 1 3

_____ 1 
1

 Let Er denote the event that calculator of brand r is selected P  Er   kr

Since Er   r  1, 2,...., 6 are mutually exclusive and exhaustive events we must have
6 6
1
 P  Er   1   kr  1  k 
r 1 r 1 21
8k 8 p
 required probability   
3 63 q
  p  q   71

82. lim
t 2
2  1
 1 1
 
x 
1 1
t  2   2  1  2 
2
2 1
 t t t t 
Now x  1  6  k has four distinct solutions if k   0, 6
 Number of integral values of k is 5
83. Conceptual
84. Second curve is parabola with focus  3, 3 and directrix x  y  2  0 equation of axis is
x  y  0 and foot of directrix 1, 1
Vertex =  2, 2
 Shortest distance = 0
85. Let first term of G.P. be a then
t1  t2  ......  t109  t1  t2  ......  t100  12
t101  t102  ......  t109  12

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 SRI CHAITANYA IIT ACADEMY, INDIA 02‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐32_KEY &SOL’S
aq 100
 aq 101
 .....  aq 108
 12 _______  i 

a  aq  aq 2  .....  aq108 
q100
  12
a 1  q  q 2  .....  q8   100
_____  ii  100
   12
q q q100
86. PQ is focal chord  t1t2  1

 1  200 1 10
a  t1     t1  
 t1  9 t1 3
3 3
a2 1  10  2000
Area of PQR  t1   2  
2 t1  3  27
87. x  75600  24335271
Let divisor is 2a1.3b1.5c1.7 d1 and 2a2 .3b2 .5c2 .7 d2
Product of divisors = 26.33.51.71
 a1  a2  6  3 ways  b1  b2  3  4 ways
 c1  c2  1  2 ways  d1  d 2  1  2 ways
3 4 2 2
Total way N   24
2
88. k
Cr  k Cr 1  k 1 Cr
n
1 1
89. 2 lim L  2 lim 
r 1 n    2r   3r 
n n  r
 1   1    1  
n n n
1 1
1 1/ 2 4 9/2
 2 dx  2   dx
0 
1  x1  2 x1  3 x 0 
1  x  1  2 x 1  3 x
1
1 3  128 lim 2 L 128
 2  ln 1  x   2 ln 1  2 x   ln 1  3 x   ln 2  4 ln 3  3ln 4  ln  en   1.58
2 2 0 81 81
90. x2  2x  4  2  x x 2  3x  2  0
1

x  1, 2 Required area =   2  x   x 2  2 x  4 dx

2

1 1
3x 2 x3
  
2  3 x  x dx  2 x  
2

2
2 3 2
1
 3x x  2
3 1 3
8   5  2  1
   2x        2      4  6             
 2 3  2  2 3  3   6  3  6

Sec: Sr.Super60_NUCLEUS & ALL_BT Page 14

 SRI CHAITANYA IIT ACADEMY, INDIA 03‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐34_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 03-04-2023
Time: 09.00Am to 12.00Pm GTM-34 Max. Marks: 300
KEY SHEET
PHYSICS
1 3 2 3 3 2 4 1 5 2
6 3 7 4 8 4 9 2 10 1
11 1 12 2 13 3 14 2 15 1
16 3 17 1 18 2 19 3 20 3
21 24 22 6 23 4 24 5 25 5
26 2 27 13 28 5 29 6 30 2

CHEMISTRY
31 1 32 3 33 1 34 1 35 3
36 2 37 1 38 2 39 3 40 1
41 4 42 4 43 2 44 1 45 1
46 3 47 2 48 1 49 3 50 2
51 19 52 5 53 222 54 6 55 2
56 4 57 5 58 10 59 8 60 3

MATHEMATICS
61 2 62 2 63 3 64 3 65 1
66 1 67 3 68 2 69 1 70 4
71 1 72 3 73 3 74 4 75 4
76 3 77 4 78 1 79 3 80 3
81 3 82 8 83 3 84 4 85 8
86 2 87 7 88 1 89 7 90 2

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 SRI CHAITANYA IIT ACADEMY, INDIA 03‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐34_KEY &SOL’S

SOLUTIONS
PHYSICS
1. Condition for light to transmit is sin C  1 /  and  V  g   r
2. The electric field due to P, S, Q and T = 0 (charge). So, charges due to U and R will
only be added up. i.e. -,+,+,-,+,-
3. According to Lenz’s law, induced emf are in a direction such as to attempt to
maintain the original magnetic flux when a change occurs. When the switch is
opened, the sudden drop in the magnetic field in the circuit induces an emf in a
direction that attempts to keep the original current flowing. This can cause a spark
as the current bridges the air gap between the poles of the switch. (The spark is more
likely in circuits with large inductance).
4.  dQ  5(T1  T2 ) KA  dQ  2 (T1  T2 ) KA  dQ  (T1  T2 ) KA
   i1        
 dt  AB 14l  dt  BE 7 l  dt  BC 14l
 dQ  (T1  T2 ) KA
  
 dt CH 7l
5. BC, CD and BA are the known resistances. Thus the unknown resistance has to be
connected between A and D.
6. z
should be dimensionless, hence   MLT 2
k
 /   ML1T 2  P , Hence   M 0L2T 0
7. Gm A m B m A rA 42 m BrB 42
   m A rA  m B rB  TA  TB
 rA  rB 2 TA2 TB2

rA rB
mA mB
C

8. L D
M 1  
f 0  e 
9. d  3 5
l  1  cos 2   2d cos   or or
cos  2 2 2
10. Conceptual
11. 1 T
Fundamental frequency f 
2l 
1 T 1 stress 1 2.2  1011  102
    178.2Hz
2l AP 2l  2  1.5 7.7  103
12. The temperature goes on decreasing with time (non-linearly). The rate of decrease
will be more initially which is depicted in the second graph.
13. Loss of energy is maximum when collision is inelastic as in an inelastic collision
there will be maximum deformation.
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 SRI CHAITANYA IIT ACADEMY, INDIA 03‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐34_KEY &SOL’S

1  Mm  2 1  Mm  2
KE in COM frame is   Vrel KEi   V KEf  0  Vrel  0 
2 M  m  2 M  m 
1  Mm  2 M
Hence loss in energy is  V  f 
2 M  m Mm
14. A  A0e kt  0.9A 0  A 0e5k and A0  A0e15k solving    0.729
15. dy 1 x2 1 1
mg sin    mg cos  tan     tan      , x  1 y  m
dx 2 2 2 6
16. L
aL2 bL3
2
F  ax  bx dw  Fdx W   ax  bx  2
 dx W 
2

3
0
17.   P, i.e It is an isothermal process (T = constant), because  
PM
RT
P  1 PM
WAB  RTA ln  A   RTA   =  0 ln(2)
 PB  2 0
18.   f  15  10 
For M1 ; v=   30cm
 f 5  10
For M 2   =
10  10   5cm m  v    5   1  h i  1.5 cm
 

2
10  10  10  2 h 0
19.  vA 2
2ˆ
E  30x i dV   E.dx dV   30x 2dx VA  V0  80 Volt
  
v0 0
20. Item No. Power
40 W bulb15600 Watt
100 W bulb 5500 Watt
80 W fan 5400 Watt
1000 W heater 11000 Watt
Total Wattage = 2500 Watt
So current capacity
P 2500 125
i    11.36  12 Amp
V 220 11
21. A1  N1  N0e t1 , A 2  N 2  N0et 2
A 0  N 0  24000dps.
22. Object is at centre of curvature, hence image will also be at centre of curvature.
23. m  m 2r l 0.003 2  0.005 0.06 4
         4%
2  m r 0.3 0.5 6 100
lr l
24. 1 T1 1 T2
 T2  T1 / 4
2   
For rotational equilibrium,
T1x  T2  L  x   x  L / 5

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 SRI CHAITANYA IIT ACADEMY, INDIA 03‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐34_KEY &SOL’S
A C

T1 T2

B O D

25. de  B  x  .dx

3L
2L L 
e  B  xdx
2L
x 5BL2

2
26. 2 3 2R
MR 2  Mr 2 r 
5 2 15
27. Heat is extracted from the source in path DA and AB is
3  P V  5  2P V  13
Q  R  0 0   R  0 0   P0V0
2  R  2  R  2
28. 120C1  200C 2 6C1  10C 2 3C1  5C 2
29. E 0  CB0  3  108  20  109 = 6 V/m
30. Net force on any one particle
GM 2 GM3 GM 2  1 1 
0 GM 2 0
  cos 45  cos 45  
2  4 2 
 2R 2 R 2 2    
2
R 2 R
M
u
u
450
M M
450

u u
M
This force will be equal to centripetal force so
Mu 2 GM 2 1  2 2 
R

2
R 

4
 u
GM 
4R 1  2 2  
1 GM
2 R

2 2 1 


CHEMISTRY
31. Acidic strength order:

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 SRI CHAITANYA IIT ACADEMY, INDIA 03‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐34_KEY &SOL’S
O

R  C  OH  R  OH  R  C  CH
32. When we are moving from left to right in a periodic table acidic character of oxides
increases (as well as atomic number of atom increases)
 X  Y  Z (acidic character)
X  Y  Z ( atomic number)
33. Conceptual
34. Conceptual
35. In fact products of reaction D are  CH3 3 C  I  C2H 2OH it follows SN1
mechanism
36. SBC  0; q  0; Scyclic  SAB  SBC  SCA  0 (state function)
 SAB  SCA  SAC
37. Formic acid can reduce Tollens reagent but not acetic acid. All carboxylic acids
liberates CO2 from NaHCO3
38. Terylene (Dacron) is a polyester polymer of ethylene glycol and terephathalic acid.
39. P  CH 3CHCl CH 3 and S  CH 3CO CH 3
40. H 3 PO4 is a tribasic acid and H 3 PO3 is a dibasic acid
41. H 3 BO3 is a weak monobasic Lewis acid .

4 LiNO3  2 Li2O  4 NO2  O2
 SiCl6 2 is not known, but  SiF6 2 is known.
42. German silver: Cu  50% Zn  30% Ni  20%
43. Conceptual
44. 1 1
Resistivity  K  ohm1cm1
K 18.5
K  1000 1 1000  1  49  1000
c     49  c   0.507
N 18.5 15  0 18.5  15  348
 %of ionization  50.7%
45. Conceptual
46. 120
t93.75%  4  t50%  120 min  t50%   30min
4
0.693 0.693
K    2.31  102 min 1 Amount left after 120 minits is 0.125
t1 30
2
mole/liter. Rate  K  R   2.88  103 M min 1

   6.023  1023  4
47. 3
3
  a  NA 2.75  660  1010
Z 
M 119
 Number of formula units per unit cell =4No. of unit cells

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 SRI CHAITANYA IIT ACADEMY, INDIA 03‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐34_KEY &SOL’S
23.8
  6.023  1023  0.05  6.023  1023  3.011  1022
119  4
48. 1
PbBr2  Pb  2Br 2 
2  104  3
K sp   0.85  2  0.85   S   
0.85 2  0.85  1.6  1.6 
49. 1
protective power 
gold number
50. Lactose is Disaccharide which is formed.

51. K 2Cr2O7 2  1  2 x  7  2   0 x  6

In K 2Cr2O7 , Transition metal  Cr  present in +6 oxidation state.
KMnO4  1  y  4  2   0 x  7 In KMnO4 , transition metal (Mn) present in +7
oxidation state K 2 FeO4 2  1  z  4  2   0

x  6 In K 2FeO 4 , transition metal (Fe) present in+6 oxidation.

x  6
y  7
z  6
So, x  y  z  19

52. C : H  4 :1
Mass ratioC : H : O =12 : 3 : 16Mole ratio
C : O  3: 4
C : H : O = 1: 3 : 1Empirical formula = CH 3O Molecular formula = C2 H 6O2 (
5
saturated acyclic organic compound) C2 H 6O2  O2  2CO2  3H 2O
2
2 mole 5 mol Mole of O2 required = 5 moles
53.
 4.41  1019
hc
E  W  K .Emax K .Emax  E  W 

6.63  1034  3  108
  4.41  1019  2.22  1019 J  222  1021 J
9
300  10
54. S2O32 is oxidized to SO42 ionHence answer is 6

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55. OH OH

* *
O Cl
NH CH
N
C Cl
O
O
56. Acetylene, benzene, cyclobutadiene, cyclooctatetraene forms only glyoxal by
ozonolysis
NH2 NH 2
57. NO 2
Br Br
NH4HS Br2/H2 O

Br NO 2
NO 2 NO 2
Br Br

Br Br NaNO 2+ HC l

NH 2

Br Br
H2 O N 2Cl
Br2
Br Br Br Br Br
Br
Sn CuBr
HCl HBr

NH2 NO 2 NO 2

58. CH 2  CHO

CH 3

CHO
CH 3 CH 3

 O / m / p  O / m / p
59.

60. T f  K f  Molality  i
i 3

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MATHEMATICS
61. 2
Det value   tan x  2x  tan x  x   0  tan x  2 x and tan x  x
Draw the graphs In the given intervals 2 solutions possible
62. M 4k  I k  N
63. f  x  is continuous in x  R  b  0,a  1
g '  x     x  1 x  1  g  x   x 3  3x  1
64. f f  x   x
65. 45r r
Tr 1  45Cr  4  5  7 10
For rational terms r = 0, 10, 20, 30, 40 No. of irrational terms = 46 – 5 = 41
66.   p  q    p  q 
67. dy xy 1
 
dx x  1 x  1
I .F  e x  x  1
G.S is y.e x  x  1  e x  c
1
 0, 1  C  0  y 
x 1
68. 1  n1n2 
2  2 2
 n11  n2 2   x1  x2 2 
n1  n2  n1  n2 
69.  a  b sin  
f    log   is odd function
 a  b sin  
70. n  s   216
5 5 5 125
p  x  0  . . 
6 6 6 216
1 5 5 3! 75
p  x  1  . . . 
6 6 6 2! 216
1 1 5 3! 15
p  x  2  . . . 
6 6 6 2! 216
1 1 1 1
p  x  3  . . 
6 6 6 216
1  75  2  15  3  1 108 1
Mean  x1P  X  x1    
216 216 2
71. Conceptual
72. m n
 sin 2 ,  cos 2 , mn  2,m  n  3
s t

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73. 10 10 10
S1   j j  1 10C j  9  10 8C j 2  90  2 S2  10  9C j1  10  29
8
j1 j1 j1
S3  S1  S2
74. 1 9
q    2
 2
75.
1  sin 360  sin180  cos180
76. Let the required line be L1   L2  0
77.
 
2 2 2 2
a  b  a b  a.b  116   4  12
78. dy 3 x 2
y 2  x3 and 
dx 2 y
79. LHS  2, RHS  2  cot 2 x  0and sin x  1
80.
 
f  0   f 0
81.  x  2 2  y 2  9  x  2 2   y  32  16 wherez = x + iy
Equation of radical axis is S  S  0  3 y  1  0
82. L  1 ,M  2 , N  2
6 3
83. g '  2   3 , g" 2   0
7
84. 3 1
p  ;q 
4 4
p  x  1  0.99  1  P  X  0   0.99
1
1  nC0 p 0q n  0.99   0.01
4n
 4n  100  n  4,5,6,....., 
85. dy y  y
  cos2  
dx x x
dx
y  vx   sec2 v.dv   
v
 y
 tan     log x  c
x
 
1,   c  1
 4
 y e
 tan     log x  1  log  
x  x
  e 
y  x tan 1  log   
  x 
x  e  y  e tan 1  0   0
86. x 2  2px  3p 2  5   x 2  2px  2p  3q D  0  p 2  2p  3q  5  0 p  R

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D0q2
87.  3 7  11
cos 2  0 and sin 2  1   , , ,
2 4 4 12 12
88.  AB AC AD   0
 
89. 725
16ET  18MT  16E  18M EM  20I  E 2  M 2  I  a  b  743
18
90.  1 1 
sin x  t, J n  2   
 n 1 n  2 

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 04-04-2023
Time: 09.00Am to 12.00Pm GTM-35 Max. Marks: 300
KEY SHEET
PHYSICS
1 2 2 1 3 2 4 4 5 4
6 1 7 2 8 3 9 3 10 3
11 4 12 2 13 1 14 3 15 1
16 2 17 4 18 1 19 4 20 1
21 9 22 6 23 20 24 3 25 2
26 2 27 1 28 80 29 6 30 50

CHEMISTRY
31 4 32 4 33 1 34 2 35 2
36 4 37 2 38 4 39 1 40 1
41 3 42 2 43 4 44 4 45 1
46 1 47 2 48 1 49 4 50 1
51 2 52 1 53 2 54 2 55 3
56 3 57 2 58 2 59 1 60 3

MATHEMATICS
61 3 62 3 63 3 64 3 65 3
66 2 67 3 68 2 69 1 70 3
71 1 72 2 73 3 74 3 75 2
76 1 77 3 78 2 79 4 80 1
81 8 82 5 83 1 84 12 85 4
86 200 87 192 88 2 89 1 90 11

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SOLUTIONS
PHYSICS
1. Conceptual
2. f
m 0
fe
3. 3/2 5 3
3/2  PV 
= constant P   =constant  P 2V 2 = constant
 nR 
PT
3
R
 PV 5 = constant C  Cv  for a process PV x = constant
1 x
5R R 5R R
C      5R
2 1 3 2 2/5
5
4. For non-uniform circular motion the total acceleration will be the resultant of radial
and tangential. So it need not be directed towards the centre but for the radial
acceleration we need a variable force as the acceleration is always directed towards
the center.
5. The nuclear force favors parallel spin. So the nuclear force between the protons will
be stronger but the protons are also going to repel each other. So in order to find the
net force we need to have numerical values to compare them.
6. F 0i1i2
 , Using the concept, parallel currents attract and opposite connects repel,
 2 d
Force acting on wire i2 will be highest followed by i3 and then i1
7. V AL dh 1 dL
Vin water g  V wax g ,Vout  Ah  ,   2cm / hr
2 2 dt 2 dt
8.

4M
 M 2  M
B1  0 3  B0 B2  0 . 3 3 3  0 3  B0
4 R 4  2 R  4 R
 
3
60 3
Bnet  2 B0 cos  2 B0 .  B0 3
2 2
9. hc 1242eV  nm 1242
Using, E    eV  59 KeV
 0.021nm 21  103
10. Conceptual
11. Fact Based
12.
 
2
An   r 2   r0 n 2   r02n4

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A
 
n n  n n4  4 n n .
A1
13.

at t = 1 second, slope of curve is some as slope of line AB

10  0
 a  10m / s 2  F  0.1  10  1N
1 0
Similarly at t = 3 second, slope of curve is same as slope of line BC.
15  15
a  0m / s 2  F  0.1  0  0 N at t = 5 second, slope of curve is same
42
0  10
as that of CD.  a   10m / s 2 F  0.1   10  1N
65
14. So block ‘Q’ is moving due to force while block ‘P’ due to friction.
Friction direction on both +Q blocks as shown.

First block ‘Q’ will move and P will move with ‘Q’ so by FBD taking ‘P’ and
‘Q’ as system F  9  0  F  9N
When applied force is 4 N then FBD

4kg block is moving due to friction and maximum friction force is 8

8
N.So acceleration =  2m / s 2  amax .
4
Slipping will start at when Q has +ve acceleration equal to maximum acceleration of
P i.e. 2m / s 2
F  17  5  2  F  27 N .
15. R
In AOB :  cos600   OB  2 R
OB
Here gravitational force will provide the required centripetal force.
GMm
Hence  m  OB  2
 OB 2

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Gm GM GM
   
 OB 3
 
2 R
3
8 R3
16. 1 R
T1  2  2  
1 1 g
g  
  R
2 GmMx R R
T2  F2   F2   mw22 x  T2  2  T3  2
2 R 3 g g
17.

E||
tan   tan   2cot    tan 1 2
E
18. AB  isothermal PAV A  PBVB ……..(i)BC  Adiabatic
PBVB  PCVC ……..(ii) CD  Isothermal PCVC  PDVD ………(iii)
DA  Adiabatic PDVD  PAV A ………..(iv) From (i), (ii), (iii) and (iv)
VB VA

VC VD
19. 1 q
By the principal of superposition, we have V   .
4 0 r
2 q 2 q 2q  1 
V  A  .  . V  A   1 
4 0 L 4 0 5L 4 0 L  5
20.

2 1 sin i
In ABC ; sin (i) = In  xyz;sin  r     2 .
d d sin r

21.

 a2  æ j ö  3

Here M = -ia 2i - i j - ia 2  k ÷÷÷
k M = -ia 2 çççi + +   M = ia 2
2 çè 2 ÷ø 2
3  9
= ´ 3´1 M = amphere - meter 2
2 2

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22. æt t ö
Apparent depth 1 + 2 Shift = (t1 + t2 ) - çç 1 + 2 ÷÷÷
t t
m1 m2 çè m1 m2 ø÷
æt 10 ´3ö÷ t1
3.0 = (t1 + 10) - çç 1 + ÷ 3.0 = + 2.5 t1 = 1.5cm
çè1.5 4 ø÷ 3
23. é A1 ù
æI ö æ A + A ö÷ ê + 1ú
êA ú
SL1 - SL2 = 10log10 çç max ÷÷÷ = 10log10 çç 1 2÷
÷  SL1 - SL2 = 20log10 ê 2 ú
çè I ç
min ø÷ è A1 - A2 ø÷ ê A1 - 1 ú
ê ú
ëê A2 ûú
= 20log10 10 = 20dB
24. 1 æ -1ö -1 1 +1 1 2 - 3
Reflection through M1 + ççç ÷÷÷ = = - =
v è 15 ø 10 v 15 10 30
1 1 1 1 -2
v = -30cm Reflection through M 2 + = - =  v = -5
v 10 10 v 10
-v 5 1 h 1 hi 3
M= = = m= i , =  hi = hi = 1.5cm \ Distance of
u 10 2 ho 2 ho 2
image from AB = 3 – 1.5 = 1.5 cm
25. 12
i
6
26. v1 = 3sin w t ; v2 = 5sin ( w t + f1); v3 = 5sin ( w t - f2 )
æ 5 3 ö÷
vmax ççç ÷÷ + (1.5)2 = 21
çè 2 ÷ø

I max Vmax 21 21´ 3 3

= = = = 3 A I rms = = 2.14
R R 7 7 1.4
3
27. a 2 dB æ a 2 dB ö÷
ç
E= t = çç l 2p R ÷÷ R J =t dt = I w
2 R dt çè 2 R dt ÷÷
ø
28. 2
df
2 
2
df = 4t y dy f = 4t  = 4t 3 = 80mV
= 2t 2 3 e =
2 dt
29. N 50 H 3´103
L = 10 cm, = 0.1 m N = 50 h= = = 500 i = = = 6A
L 0.1 n 500
30. t = M .B = 12.5´ 4 = 50.00

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CHEMISTRY
31. Graphite is polymer
32. Hydrogen is lighter than air
33. Be is anomalous due to its small size
34. Compare bond order
35. Water acts as Lewis base towards H 2 S .
36. P4 reacts with air, lithium also reacts with water
37. Addition polymerization
38. S is not present
39. More substituted
40. Al does not show +2 oxidation state
41. In aqueous solution Cu  in unstable.
42. No elimination
43. Electron deficient hydride donor reagent
44. Steric hindrance
45. NCERT XII part-I page no 9
46. NCERT XII Part-I page no 134
47. NCERT XI part-I
48. water is formed during the reaction and concentration of electrolyte decreases

and boiling point decreases

49. NCERT 12 part I page no 136
50. For strong electrolytes, Debye-Huckel -onsager euation is followed
51. Three GI
52. 17 electrons are present in the valency shell
53. Two lone pair of electrons are present
54. 3 F’s are equatorial and two Cl’s are axial
55. Cannizzaro reaction
56. three moles are necessary
57. 2F produces one mole of PbO2
58. phOH(solution II)  phOH(solution I)

æ -0.02 ö æ -0.03 ö
DH = ççç ´ 94÷÷÷ - çç ´ 94÷÷÷ = -2 kcal / mol
è 0.47 ø èç 1.410 ø
59. Total number of nodes = n-1
60. 1-n
Half life is proportional to (a0 )

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MATHEMATICS
61.
lim
(
f x2 + x + 1 ) æ0 ö
çç form÷÷ As. f (x) is differentiable then apply
x0 f ( x 4 - x 2 + 2 x + 4) - f ( 4) çè 0 ÷ø

(2 x + 1) f '( x 2 + x + 1) f '(1)
L ' Hospital rule lim = =4
x0 (4 x3 - 2 x + 2) f '( x 4 - x 2 + 2 x + 4) 2 f '(4)

62. 2
Let h = 2t12 &12 = 4t2 We know t2 = -t1 -  t1 = -1 or -2
t1
 h = 2 or 8
63. Focus is foot of perpendicular drawn from vertex on latus rectum i.e. (4, 3)
and equation of the directrix is x + y + c = 0
1+ c
i.e. =3 2  c = 5 or -7
2
\ equation of the directrix is x + y + 5 = 0 \ equation of the required parabola is
2 2æ x + y + 5 ö÷2
( x - 4) + ( y - 3) = ççç ÷  x 2 + y 2 - 2 xy - 26 x - 22 y + 25 = 0
è 2 ÷ø
Compare with x 2 + y 2 - 2 xy + px + qy + r = 0
 p = -26, q = -22, r = 25 \ p + q + r = -23
64. dy
y 2 = 2cx + 2c3/2 ...............(1) 2 y = 2c
dx
dy
y =c ................(2) from (1) and (2)
dx
dy æ ydy ö÷3/2 æ ö æ ö3/2
2
y = 2 y ( x) + 2ççç ÷ çç y - 2 x dy ÷÷ = 2çç dy ÷÷ y1/2
dx è dx ÷ø èç dx ÷ø èç dx ÷ø
æ dy ö÷ æ dy ö÷3/2 1/2
çç y - 2 x ÷ = 2çç ÷ y
èç dx ø÷ èç dx ø÷
on squaring degree = 3order = 1
65. 1/2 1024-r
( ) ((7) ) will be
r
1024 1/8
Statement – 1 : General term = Cr 5
integer. if r is multiple of 8
 r = 0,8,16,24,............,1024
 Number of integral terms = 129.
10! a 2 b 3 T 6
Statement – 2 : General term = 2 3 5 for rational terms
a!b !g !
a = 0,2,4,6,8,10 b = 0,3,6 g = 0,6
Hence possible sets = (4,6,0),(4,0,6);(10,0,0)
Hence, there are s rotational terms.
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10! 2 10! 5
\ required rdom = 2 5 2 = 12632 .
4!6! 10!
Statement – 3 : tr +1 , the (r + 1) in the expansion of
10 100-r 1/8 r
(51/6 + 21/8 ) ( )
is tr +1 =100 Cr 51/6 ( )
2
100 - r r
As 5 and 2 are relatively prime, tr +1 will be rational if and are
6 8
both integers. i.e if 100 – r is a multiple of 6 and r is a multiple of 8. As
0 £ r £ 100 , multiple of 8 upto 100 and corresponding value of 100- r
r = 0, 8, 16, 24, ……….., 88, 96
100 – r = 100, 92, 84, 76,…….., 12, 4
Out of 100 – r, multiple of 6 are 84, 60, 36, 12
\ there are just four rational terms
 number of irrational terms is 101-4=97
66. 3
1 1 æç 2 1 ö÷ æ 1ö
g ( x) = x + 2 f ( g ( x )) = x + 6 = ç x + 2 ÷÷ - 3çç x 2 + 2 ÷÷÷
2 6
x x çè x ø èç x ø
3
f ( g ( x)) = ( g ( x)) - 3g ( x)  f ( x) = x3 - 3x f '( x) = 3x 2 - 3
67. f ( x) = cos x + cos-1 (sgn ) x + nx f ( x) = cos x + cos-1 (1) + nx " Î (0,2p )
p 3p
= cos x + nx "x Î (0,2p ) \ f ( x) is not differentiable at x = 1, ,
2 2
68. xh yk h 2 k 2
Let middle point of chord be (h, k) T = S1  + = +
3 4 3 4
2h h 2 k 2
Chord passing through (2, 0)  = +
3 3 4
(h -1)2 k2
2
( x -1) y2
 + =1 + =1
1 4 1 4
3 3
69. 6! 5!
A, I 
 P, P, L, C , T , N ways = .7 C5.
, A, I , O &  = (45)7!
2! 2!2!
5 6
70. 2
2 2 å di2 æç å di ÷ö 125 æç 5 ö÷
2
25 1 49
Let xi - 5 = di x s =s = -ç ÷ ÷÷ = - ç ÷÷ = - =
d
n ç
çè n ø 10 çè10 ø 2 4 4
71. \ VA = 0 (Null matrix)  (log a ) x + (log b) y + (log c) z = 0
(log b) x + (log c) y + (log a ) z = 0 (log c) x + (log a ) y + (log b) z = 0
which is homogenesis system of linear equations in x, y and z.  x, y, z ¹ 0
log a log b log c
 system possesses non trivial solutions.  D = log b log c log a = 0
log c log a log b
Sec: Sr.Super60_NUCLEUS & ALL_BT Page 8
 SRI CHAITANYA IIT ACADEMY, INDIA 04‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐35_KEY &SOL’S
 log a + log b + log c = 0 OR log a + log b + log c
 a, b, c, are distinct
 log a + log b + log c = 0  a b c = 1  b3 = 1 ( b2 = ac)
1
 b =1  ac = 1  c=  System become(log a) x – (log a )
a
z = 0 - (log a) y + (log a) z = 0 -(log a) x – (log a) y = 0  x - z = 0
a b g
y-2= 0 x- y = 0  x= y=2  a : b : g = 1:1:1 + + = 3
b g a
72. é x 1 1ù
ê ú
Here A = 4 where A = ê 2 x 3ú , then
ê ú
ê x -1 2 ú
ë û
73. z1 z1  4, z2 z 2  9, z3 z 3  16
16 z1z2  9 z1z3  4 z2 z3  z3 z 3 z1z2  z2 z 2 z1z3  z1 z1 z2 z3  96
 z1 z2 z3 z1  z 2  z 3  96  z1  z2  z3  4
74. The plane containing the given line is
(2 x + 3 y + 5 z + 1) + l (3x + 4 y + 6 z + 2) = 0
 (2 + 3l ) x + (3 + 4l ) y + (5 + 6l ) z + (1 + 2l ) = 0
3
This plane is parallel to y – axis  3 + 4l = 0  l =-
4
 A point on y – axis is the origin.  perpendicular distance of the origin from
the
2
plane x - 2 z + 2 = 0 is
5
75. 1 1 1 1
S = + + + + .............
15 35 55 75
æ1 1 1 1 ö æ1 1 1 ö
= çç + + + + ........÷÷÷ - çç + + + .......÷÷÷
çè15 25 35 45 ø çè 25 45 65 ø
1 æ1 1 1 ö k 31k
= (k ) - çç + + + ...........÷÷÷ = k - =
32 çè15 25 35 ø 32 32
76.

ïì x -1 + x; x > 1 ìïï2 x -1; x > 1

y = ïí =í
ïïî-x + 1 + x; x £ 1 ïïî 1; x £1
Required area
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 SRI CHAITANYA IIT ACADEMY, INDIA 04‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐35_KEY &SOL’S
x= 2 y=3
é y + 1ù 7+4 2
= ò
é
( 2 ù
)
ê1 - x -1 ú dx +
ë û ò êë
ê y + 1 - ú dy =
2 úû 3
x=- 2 y=1
77. 2 2
S1 : ( x -1) + ( y -1) + l ( x + y - 2) = 0 S2 : x 2 + y 2 + 4 x + 5 y - 4 = 0
chord PQ : S1 - S 2 = 0 (-6 x - 7 y + 6) + l ( x - y - z ) = 0  x = 8, y = -6
78.

x2 + 1
=4  x2 = 7 x2 + y2 = 5  7 + y2 = 5  y2 = - 2
2
\ ( x2 , y2 ) º (7, -2)
x + 1 y3 + 2
x3 = 4 3 + =5  4 + y3 = 10 - 3 = 7
2 2
 y3 = 3 \ ( x3 , y3 ) º (4,3)
79. ( 2
n-1)
3
n-1) =38
Statement-1 : adj (adjA) = A adj (adj (adj ( A))) = A (
2
( ) (
Statement-2 : A-1.adj B-1 .adj 2 A-1 = ) 1 1
.
A B2
. 2 A-1

1 1 64 64 64 8
= . = = =
A B 2 A2 3 2
A B 8´ 9 9

Statement-3 : 3 ABA-1 + 3 A = 2 A-1BA + 2 A

3 ABA-1 + 3 AAA-1 = 2 A-1BA + 2 A-1 AA
3 A( B + A) A-1 = 2 A-1 ( B + A) A

3n A + B = 2n A + B  A + B = 0
80. 2
n (nx - ny ) = e x y (1 - nx ) ...........(1)
diff. w. r. to x.
1 æ 1 1 dy ö÷
.çç - ÷
(nx - ny ) çè x y dx ÷ø
2 æ 1ö 2 æ dy ö
= e x y çç- ÷÷÷ + (1 -  n x)e x y çç x 2 + y.2 x÷÷÷ ..........(2)
èç x ø çè dx ø
Put x = e in (1), (1)  n (1 - ny ) = 0  y = 1 Now put x = e, y = 1 in (2)
2 2
æ1 ö ee 1 + ee
then (2)  çç - y '(e)÷÷÷ = -  y '(e) =
çè e ø e e

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 SRI CHAITANYA IIT ACADEMY, INDIA 04‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐35_KEY &SOL’S
81. sin 3 xdx
I =ò  let cos x = t  sin xdx = -dt
2
1 + sin x

 I =ò
(1 - t 2 ) dt
=ò
( 2 - t 2 ) -1
dt
2
t -2 (t - 2)
2

1 t- 2 1 cos x - 2
= -t - n = -cos x - n
2 2 t+ 2 2 2 cos x + 2
  
82. Let c = l a + mb

take dot by b
   2
( ) ()
0 = l a .b + b = -l + 5m
 l - 5m = 0 ..........(1)
  2  
again a . c = 7  l a + m a .b = 7 ( )
 3l - m = 7 .........(2)
5 1
solving (1) and (2) l = , m =
2 2
 5 1   3i 5  
2
(
  
) (
\ c = -i + j + k + 2i + k = - + j + 3k
2 2 2
)
2  2 1
 c = ´ 35 = 5
7 7
83. Equation of chord is
x -5 y -3
= =r
cos a sin a
Point (5 + r cos a,3 + r sin a ) of chord is one of its end point
2 2
2(5 + r cos a ) - 3(3 + r sin a ) = 6
(5, 3) is midpoint of chord iff sum of roots = 0
20cos a -18sin a = 0
cos a sin a 10
=  tan a = = m
9 10 9
\ p - q =1
84. 9cos12 x + cos 2 2 x + 1 = 6cos6 x cos 2 x + 6cos6 x
2
( )
-2cos2 x = 0 3cos6 x -1 - cos 2 x = 0 3cos6 x = 1 + cos 2 x = 2cos 2 x
2 2
 cos x = 0 or cos 2 x = cos x = 0,cos x = 
3 3
2
cosx = 0 has 4 soltuions. cos x =  has 8 soltuions.
3
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 SRI CHAITANYA IIT ACADEMY, INDIA 04‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐35_KEY &SOL’S
Hence total 12 solutions
85.

ABC = 900 BC = 2 5 AB = 9 + 1 - 5 = 5
 AC = 5 Power of point  AB 2 = AD. AC  AD = 1  CD = 4
86.

h 3h 2 1
tan q = & tan 2q =  tan 2q = 3tan q =3  tan q = or
d d 1 - tan 2 q 3
p p
q=  h = 50 3.tan = 50 CD = 4h = 200m
6 6
87. 1 1æ 1 1 ö÷
Tn = Tn = ççç - ÷÷
n (n + 2)(n + 4) 4 èç n (n + 2) (n + 2)(n + 4)ø÷
1æ 1 1 ö 1æ 1 1 ö÷ 1 æç 1 1 ö÷ 22
T1 = çç - ÷÷÷ T2 = çç - ÷ T3 = ç - ÷ = 192
4 çè1.3 3.5 ø 4 çè 2.4 4.6 ø÷ 4 çè 3.5 5.7 ÷ø S
1 æ 1 1 ö 11
S¥ = çç + ÷÷÷ =
4 çè 3 8 ø 96
88. a x 0 x a b 1 -1 1
b y 0 ´ y b c + cos q cos q sin q = 2
c z 0 z c a -sin q -sin q cos q
89. 1 1+ y 1 1
y=  x=  equation whose roots are , are
2 x -1 2y 2a -1 2b -1
æ1 + x ö÷2 æ1 + x ö÷
a çç ÷ + ç ÷ + c = 0 absolute term of a above equation is ' a '
èçç 2 x ø÷
b
çè 2 x ÷ø
C
 C = a  =1
a
90. y = sin-1 (sin 6) - tan-1 ( tan8) + cos-1 (cos6) = 6 - 2p - (8 - 3p ) + 2p - 6
= 3p - 8 a = 3, b = –8

Sec: Sr.Super60_NUCLEUS & ALL_BT Page 12

 SRI CHAITANYA IIT ACADEMY, INDIA 05‐04‐23_ Sr.Super60_NUCLEUS & ALL_BT_ Jee‐Main_GTM‐36_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS & ALL_BT JEE-MAIN Date: 05-04-2023
Time: 09.00Am to 12.00Pm GTM-36 Max. Marks: 300
KEY SHEET
PHYSICS
1) 4 2) 4 3) 2 4) 3 5) 1
6) 3 7) 4 8) 1 9) 1 10) 3
11) 2 12) 3 13) 4 14) 3 15) 4
16) 1 17) 3 18) 1 19) 4 20) 4
21) 31 22) 16 23) 0 24) 3 25) 5
26) 2 27) 5 28) 3 29) 2 30) 4

CHEMISTRY
31) 3 32) 1 33) 1 34) 2 35) 3
36) 1 37) 1 38) 3 39) 2 40) 3
41) 2 42) 3 43) 4 44) 4 45) 3
46) 2 47) 1 48) 2 49) 2 50) 2
51) 125 52) 6 53) 6 54) 2 55) 2
56) 5 57) 8 58) 4 59) 5 60) 2

MATHEMATICS
61) 4 62) 2 63) 1 64) 2 65) 2
66) 4 67) 2 68) 1 69) 2 70) 3
71) 4 72) 2 73) 2 74) 2 75) 3
76) 1 77) 3 78) 4 79) 3 80) 1
81) 150 82) 6 83) 2 84) 5 85) 1
86) 4 87) 1 88) 16 89) 7 90) 90

Sec: Sr.Super60_NUCLEUS & ALL_BT Page 1

 SRI CHAITANYA IIT ACADEMY, INDIA 05‐04‐23_ Sr.Super60_NUCLEUS & ALL_BT_ Jee‐Main_GTM‐36_KEY &SOL’S

SOLUTIONS
PHYSICS
1. For a projectile, horizontal component velocity remains constant.
2. In magnetic field speed remains same, where as in electric field speed may increase or
decrease depending on the direction of field.
3. it’s periodic but not SHM as acceleration is not directly proportional to displacement.
4. In case of capillary tube of insufficient length,
hR = constant.
5. In helical path, velocity is never parallel to axis.
6. F  weight of the body.
B
7. i1  i0 1  e  t /T1  i2  i0 e  t /T2
if T1  T2 the i1  i2  2i0
8. ib 
20
 10 A
2
p  vi  200W
9. 1 1
 
1 dR dR1 dR2
 
R R1 R2 R2 R12 R22
10. Error is –ve, so correction is +ve.
d  2.4cm   6  5 0.01 cm = 2.51 cm
11. eV0  h   , slope 
h
c
12. Difference in PE is same. So PE increases by 27.2 eV.
13. N  N 0  0.9   in 5 days
N  N 0  0.9   in 20 days
4

14.
t
2 A
g A0
 H1  H 2 
15. I  xH and x 
C I 2 H 2 T1

T I1 H1 T2
16. In the case, zener diode is not operated in break down region, so current through it is
zero.
17. According to law of conservation of linear momentum S-1 is correct.
Kinetic energy can be increased by internal force also.
Eg: Explosion.
18. A
Ao
3n
19.   tan i & 
c

20. If temperature increased, volume need not increase
21. i  io 
Rsh
Rg  Rsh
22. 2 KP  KP
E1  3 & E2 
 2r 
3
r

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 SRI CHAITANYA IIT ACADEMY, INDIA 05‐04‐23_ Sr.Super60_NUCLEUS & ALL_BT_ Jee‐Main_GTM‐36_KEY &SOL’S
E1  16 E2
23. z2
BE  Eo  2
n
2
n
r  ro 
z
1
 BE   r  Eo  ro 
z
24. TA  0  mg 
2l
 T l
3
2
T  mg
3
T  RA  mg
mg
RA 
3
25. dQ
 F 
dt
 r 2
 r5
26. With reference to life
u  10m / s, a  10  2  12
2u 20 5
T     1.67
a 12 3
27. 1 1
mgh  m 2  I  2
2 2
1 2 5
KEr   mgh KET  mgh  mgh  mgh
3 3 3
28.
C1

C2
O2
O

O1C1  O2C2
29. Electro
 
static filed is conservative. So work done by filed is independent of path. So
 C. dr should be proper integral which does not change path but only depends on limits.
i.e  xdy  ydx
30.  
at , x 
4 4

 
2
I  I o  cos 2   4   I2 o

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 SRI CHAITANYA IIT ACADEMY, INDIA 05‐04‐23_ Sr.Super60_NUCLEUS & ALL_BT_ Jee‐Main_GTM‐36_KEY &SOL’S

CHEMISTRY
31. Eit
W
96500
32. Mw
i). MnO4  Mn 2  acidic  Ew 
5
Mw
ii). MnO4  MnO2  neutral  Ew 
3
Mw
iii ). MnO4  MnO4 2  basic  Ew 
1
1 1 1
Ratio is : :  3: 5 :15
5 3 1
33. Graham’s law of diffusion.
34. 1 100 1 100
 ln  x  ln
k 1 k 10
35. 1  1
 RZ 2 1   for lyman line.
  4
36. CO2  H 2O  HCO3  H 
When conc. of CO2 decreases then position of equilibrium moves back ward, then H 
conc. decreases pH increases
37.

38. Consequences of lanthanide contraction.

39. Storage of H 2O2
40. Ionic mobilities of hydrated alkali metal ions are as:
Cs   Rb  K   Na   Li 
HO
41. - Na


B O
OH
O B
O
B O
HO .8H 2O
O B - Na 
OH
42. Poling
43. S8 and S6 structures in NCERT.
44. Axial and Equatorial bond lengths are different.
45. Priority   
Lowest set of locants rule and alphabetical order of name of substitutents.
46. Conjugate base of I,II, and III:

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HC  C  (II) (III)
(I)  Negative charge 
 
 on sp C atom  Aromatic Anti-caromatric
acidic character : II >I >III
47. Group stabilise the carbanion character on   carbon

48. Methoxy group stabilises the cationic intermediate by +Meffect

49. 2 molecules of RMgX are consumed by acid chloride group as it undergoes addition
elimination first and then addition. The other aldehydic & ketonic group consume one
RMgX molecule each.
50. OH
OH
A) B)

OH OH

51. 2  rcation  ranion   a

52. x
log
1
 log k  log p
m n
x 1
 plot of log versus log p is linear with slope  and intercept = log k
m n
1
Thus  tan   tan 45  1or n  1
n
Log k= 0.301 or k = antilog 0.301 = 2
At p = 3 atm
x
 kp1/ n  2   3  6
1

m
53. CrO2Cl2
54. U  3 1.5R 100  2  2.5R 100  1900cal
 1.9K cal
55. Cu 2 & NO2
56. Paracetamol, Aspirin, Morphine, Heroin, Codeine
57. CONCEPTUAL
58. Et

Me * H
Me * Br

Et
Number of isomers  2n  22  4
(n is the number of chiral C-atoms)
59. A, B, C , E and F gives positive response.
60. Bond breaks between oxygen and carbon of alkyl group
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 SRI CHAITANYA IIT ACADEMY, INDIA 05‐04‐23_ Sr.Super60_NUCLEUS & ALL_BT_ Jee‐Main_GTM‐36_KEY &SOL’S

MATHEMATICS
61.  2 2 
n

 cos 5  sin 5  1 0 
We have,    
 sin 2 cos 2   0 1 
 5 5 
 2 n 2 n 
 cos 5  sin
5  1 0 
     n5
 sin 2n cos 2n   0 1 
 5 5 
62. Centre  2, 6 . Substituting in L 2  k  7   6  k  1  4  k  5
  2k  6k  4k   14  6  20  0
Hence, every member of L passes through the centre of the circle
 Cuts at 900 .
Hence, S-1 is true and S-2 is false.
63. Conceptual
64. Required area = Area of rectangle OABC – Area of curve OBCO

 / 4 
  tan y dy    log cos y 0
/ 4

4 0 4
65. Given equation is
dy  x y  x y
 sin    sin  
dx  2   2 
dy  x y  x y
  sin    sin  
dx  2   2 
dy  y  x
  2sin   cos  
dx 2 2
 y x
 cosec   dy  2 cos   dx etc
2 2
66. Let u  x  1, v  y  1, w  z  1 and p  t  1
 u, v, w, p  0 and u  v  w  p  24 .
So, required number of solutions is
24 41
C41 27 C3 .
67. The contra positive of “If x  A  B, then x  A and x  B” is
“If x  A or x  B, then x  A  B” .
68. Put x = y find point and tangent equation
69. s  1 2  4  .  tn1  tn ------1
s  1 2  4  .  tn1  tn ------2
1-2
0  1  1  2  ..  tn
Find tn , sn  tn
Sec: Sr.Super60_NUCLEUS & ALL_BT Page 6
 SRI CHAITANYA IIT ACADEMY, INDIA 05‐04‐23_ Sr.Super60_NUCLEUS & ALL_BT_ Jee‐Main_GTM‐36_KEY &SOL’S
70. x  2 y 1 z  3
  
2 2 1
x  2  2 y  2 1 z    3
Put in the given plane
Find Q, P  2, 1,3 find PQ
71.  1 
e
x  x 1 1
 x 1  2  e x  x dx
 x 
 f  x   x f   x  dx
72. 3 5
 log 2 x   log 2 x   log x 2
2

4 4
3 2 5 1
 a a   where a  log 2 x 
4 4 2a
 3a3  4a 2  5a  2  0
 a  1,  2,  1/ 3  x  2,1/ 4, 21/3
73. By Lagrange’s Mean Value Theorem,  c  1, 6  such that
f  6   f 1 f  6   2
f c   2
6 1 5
74. V  ax   a 2 V  x 
75. f  x  f  x 
1
 log x
x
Solve Linear D.E
76. Find centre r 2  154 find r
77. Let x1 , x2 , x3  R be the roots of f  x   0
 f  x    x  x1  x  x2  x  x3 
f  i    i  x1  i  x2  i  x3 
f  i   x1  i x2  i x3  i  1

 x12  1 x22  1 x32  1  1

This is possible only if x1  x2  x3  0
 f  x   x3
a0bc
 abc  0
(Sum of coefficients zero can not imply that all zero roots).
78. A  3, 0  B  0,1
x y
Solve   1 and curve find M
3 1
79.

x
tan 45  find x + y
10
80. As the slope of incident ray is 
1
so the slope of reflected ray has to be
1
.
3 3

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 SRI CHAITANYA IIT ACADEMY, INDIA 05‐04‐23_ Sr.Super60_NUCLEUS & ALL_BT_ Jee‐Main_GTM‐36_KEY &SOL’S

The point of incidence is  

3, 0 . Hence the equation of reflected ray is y 
1
3

x 3 . 
 3 y  x   3.  x  3 y  3  0
81. Let Three types be A, B, C
A A A B C
 5!
B B B C A3
3!
C C C A B 
A A B B C
 5!
A A C C B3
2!2!
B B C C A 
82. 1
sin x 
By solving 3
Total 6 values
83. Determinant = 0
84. r
nP
Pq
1
P q  2 n  10
2 find Tr 1
85. f 1  f  1
1
f    0
2 Solve
86. Probability of getting at least one head is at least 0.9.
n
1
 1     0.9
2 [ Probability of getting at least one head = 1 – Probability of
getting no head]
87. Re  z   0
1
 cos  
2


4
88. PO  PA2  PB 2  PC 2  18 find circle and radius
2

89. a  r 7Cr
b    7  r  Cr
7

nCr  2n
r nCr  n2n1
90. b c  5 3
a .b  0 find , 

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 SRI CHAITANYA IIT ACADEMY, INDIA 06‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐37_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 06-04-2023
Time: 09.00Am to 12.00Pm GTM-37 Max. Marks: 300
KEY SHEET
PHYSICS
1) 2 2) 2 3) 2 4) 3 5) 1
6) 4 7) 4 8) 1 9) 1 10) 3
11) 2 12) 2 13) 4 14) 2 15) 2
16) 1 17) 1 18) 1 19) 4 20) 1
21) 2 22) 20 23) 5 24) 25 25) 8
26) 3 27) 4 28) 3 29) 200 30) 8

CHEMISTRY
31) 1 32) 1 33) 2 34) 1 35) 1
36) 1 37) 4 38) 2 39) 2 40) 1
41) 4 42) 3 43) 4 44) 2 45) 2
46) 2 47) 3 48) 4 49) 3 50) 4
51) 5 52) 78 53) 4 54) 24 55) 336
56) 1 57) 2 58) 18 59) 45 60) 2

MATHEMATICS
61) 4 62) 3 63) 3 64) 1 65) 3
66) 4 67) 1 68) 1 69) 4 70) 4
71) 2 72) 2 73) 1 74) 4 75) 4
76) 3 77) 1 78) 3 79) 3 80) 3
81) 1 82) 6 83) 2 84) 9 85) 0
86) 17 87) 16 88) 105 89) 7 90) 10

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SOLUTIONS
PHYSICS
1. Conceptual
2. Conceptual
3. Conceptual
4.  
 v   340 
f ' f   , f '  640  100 
 v  vs cos   340  cos 
 3 
 
 340 
f '  640   f '  680 Hz
100 3 
 340   
 3 5
5. Conceptual
6. Conceptual
7. The bar is at equilibrium. The net force from night or left of a section of BC is 70
KN.

 
F


70  103  1
AY 1  2  1011
  3.5  107 m
8. 1
F   Area 
C

  C1
Effective area   R 2 , F   R 2

 
22 2 1 1
F  21  102   . F  42  1013 N
7 110 3  108
9. 1 1 1
 
v 200 40
v = 50 cm
d = 50+4=54cm

10. As t is increasing  is decreasing so refraction is taking place from denser

to rarer medium as light is proceeding in the medium. Using continuous varying
refractive index, let the ray reflect back after traveling t  t0 along the normal, then

at this location, angle of refraction is .
2
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 
 
So, 1  sin   0  kt02 sin    t0 
2
0  sin 
k
11. Compare the given equation with standard equation
E y  E0 cos t  kx    2  10
2 v  2  10  v  106 Hz
2
k    102 or    102 or   200 m

12. 
For first diffraction minimum a sin     a 
sin 
For first secondary maximum, we use
3 3 1 3 sin  3 3 3
a sin  '  sin  '       sin 30o   '  sin 1  
2 2 a 2  2 4 4
13. u = velocity of water through the hole = 2 gh
R = rate of mass flow =  Au [A = area of the hole]
F = reaction force on the wall  R  u  0    Au 2
For cylinder to just move
 d 2 2 w
 
2 2
 w   Au  2 gh d 
4  hg
14. General equation of traveling wave is
y  f  ax  bt   f  x  vt 

y e
  ax 2 bt 2  2 ab xt   e  ax  bt 
2

b
So, the wave will be traveling along –x axis with speed
a
15. By conservation of angular momentum of the body about O

 
mv sin 30o R  mv '  R  h 
vR v
  R  h  h  R
2 4

16. T  m2 g sin 30o .... i 

 Av
T  m1g sin 30o  .... ii 

From (i) and (ii)

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
 m2  m1  g  sin 30o
 3.46  103 Pa sec
vA

17. As all the points on the periphery of either ring are at the same distance from point
P, the potential at point P due to the whole ring can be calculated as
GM
V  , where x is the axial distance from the centre of the ring. This
2 2
R x
expression is independent of the fact whether the distribution of mass is uniform or
non-uniform
GM G 2 M GM  1 2 
So, at P,V      
R 2 R 5 R  2 5 
18.  v

 c
v
1
c
v=c
19. Acceleration when block B is in the liquid
 3m 
m A g   3mg  g 
m g   mB g  upthrust   2 
a1  A 
m A  3m m A  3m
Acceleration when block B is outside of the liquid.
3mg  m A g
a2 
m A  3m
3 9
Given a1  a2  m A g  mg  3mg  m A g m A  m
2 4
20. Output of the gates are marked as y, y1 and y2 .

A NAND gate gives low output when both the inputs are high, otherwise it gives a
high output. Now the truth table.

21. 0.25
Least count  cm  5  104 cm
5  100
Sec: Sr.Super60_NUCLEUS & ALL_BT Page 4
 SRI CHAITANYA IIT ACADEMY, INDIA 06‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐37_KEY &SOL’S

Reading  4  0.05cm  30  5  104 cm = 2015 mm

22. T = 20 sec
at t = 2, v =5 cm/s v  A cos t
 2   2   2  A
v  A   2 f  cos  t  5  A    cos  t   cos36o  5
 T   20   20  10
23. p 100  103
The current through circuit i    0.2 A
v 0.5
1
Voltage drop across resistance = 1.5 – 0.5 =1V  R   5
0.2
24. Here T1/2  270 days
After 540 days, it means two half lives have been completed n = 2
n n
N 1 m 1 m 1
       m  25 mg
N 0  2  m0  2  100 4
25. If I is the intensity of the incident unpolarised light, the intensity transmitted by the
first is I/2. This is the intensity of incident light on the second polaroid. Intensity
1
transmitted by the second polaroid is cos 2  , where is  the angle between the
2
3 4
axes. Here sin   therefore cos  
5 5
2
1 I 4 8 8
cos 2      I  is the required ratio
2 2 5 25 25
26. from first reaction
2 1 2  1 n 1 n  1
   
v u R v  R
n n 1 nR
 AI '  v  For 2nd refraction
v R n 1
..

u  BI '  AI ' 2 R 
nR
 2R 
2  n R
n 1  n  1
and v = R For second surface
2 1 2  1 1 n 1  n 1 n  n  1 n  1
     
v u R v u R R  2  n R R
(2 – n) – n (n – 1) = (n – 1) (n – 2) N = 4/3
27. Let ground state energy (in eV) be E,
E
Then from given condition, E2 n  E1  204 eV 1  E1  204 eV
4n 2
 1 
E1   1  204 eV .... i 
 4n 2 
And E2 n  En  40.8 eV
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 SRI CHAITANYA IIT ACADEMY, INDIA 06‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐37_KEY &SOL’S
E1E 3
 1  40.8 eV  E1    40.8 eV
4n 2 n 2 4n 2
1
1 2
From (i) and (ii) 4n  5 n = 2
3
4n 2
Now from equation (ii)
4
E1   n 2  40.8  Or E1  217.6 eV . And E1  13.6Z 2
3
E 217.6
Therefore Z 2  1   16 Z = 4
13.6 13.6
28. 1 1 1 3 1 1
     u  50m
v u f 25 u 10
7 1 1
and for final position of the object    u  25m
50 u 10
Therefore, the shift in object  50  25  25 m
25
Speed  m / s  3 km / hr
30
29. v
Fundamental frequency of open pipe f0 
2
v
Third harmonic of the closed pipe fc  3 
4
 v  v v
Given 3     100 Or  200 Hz
 4  2 2
30.  
Shift X     1 t . Shift due to one plate  X1   1  1
 

Shift due to another plate  X 2    2  1 t , Net shift X   X 2  X1 


 X   2  1  t ...... i 

And also X  5 .... ii 
 5
Hence 5   2  1  t  t   8  106 m t  8  m
  2  1 
p

S1

S2

D
screen

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CHEMISTRY
31. Ga has similar Ionisation enthalpy as Al because of inert pair effect ( or completely
filled d-orbital in Ga).
32. Butylated hydroxy anisole (BHA) is an antioxidant. It is added to butter to increase its
shelf life from months to years. BHA reacts with O2 present in air in preference to butter.
So both the assertion reason are correct.
33.  
Strength of ligand F  NCS  NH 3

As given graph : A < B < C  max 

 Correct matching A-(iii), B-(i), C-(ii)
34. From rate law
1 d  A d  B 
r   K  A  B 
x y

2 dt dt
6 10  K  0.1  0.1
3 x y

2.4  102  K  0.1  0.2 

x y

1.2 102  K  0.2   0.1

x y

 3  1  x  1
 2    3  x  2
So, order with respect to A = 1
Order with respect to B = 2
(4)  (3)
2
 x   0.2  7.2  102
    2
 0.2   0.1  1.2  10
6  0.2
x
4
x = 0.3M
5   4 
2
 y  2.88  10 1
  
 0.2  7.2  102
y 2  4  0.22
Y = 0.4M
35. Conceptual
36. 2 N 2O5  g   2 N 2O4  g   O2  g 

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Pt  0.5  2 x  2 x  x  0.5  x  x  Pt  0.5
 PN 2O5  0.5  2 x  0.5  2  Pt  0.5   0.5  2 Pt  1  1.5  2 Pt
at 100 sec, Pt  0.6 atm
 PN 2O5 at 100 sec  1.5  2  0.6   0.3 atm
2.303 0.5
k  log  k  5.1 103 S 1
100 0.3
37. 8a a R.TC T
TC  and PC  2
b and C is minimum for He.
27 Rb 27b 8.PC PC
38. mixture of CaF2 and Na3 AlF6 decreases the melting point of Al2O3
39. Penicillin is bactericidal and bacteriostatic.
40. In halide anion analysis AgNO3 added in carius method to detect halide anion.

41. High leaving tendency corresponds to high reactivity towards hydrolysis. Hence order is.

42. Due to the larger acid dissociation constant (Ka) sulphuric acid acts as an acid and HNO3
acts as a base.

43.

44. Conceptual
45. 1)
F

Cl F

F (Planar distorted-T shape and polar)

2)
F F

Xe
F (Pentagonal planar and non-polar)
F F

3)

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 SRI CHAITANYA IIT ACADEMY, INDIA 06‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐37_KEY &SOL’S
Cl Cl

P
(Non - planar and non - polar)
Cl Cl
Cl

46. IMF are responsible for deviation from Raoult’s law.

OH
47. OH

OH HIO4 NaBH4

48. When red phosphorus is heated under high pressure, a series of phases of black
phosphorus is formed.
49. Due to high screening effect
50. P  K H . X  1 bar  K H 
0.2
 K H  277.78 bar  2.78  10 7 Pa
1000
0.2 
18
51. C:H=4:1
C:O=3:4
Mass ratio
C : H : O = 12 : 3 : 16
Mole ratio
C:H:O=1:3:1
Empirical formula  CH 3O
Molecular formula  C2 H 6O2
(saturated acyclic organic compound)
5
C2 H 6O2  O2  2CO2  3H 2O
2
2 mole 5 mol
Moles of O2 required = 5 moles
52. (Given atomic masses C = 12, K = 1.0 , O = 16.0, Br = 80.04 )
6.1
Moles of benzoic acid  = moles of m-bromo benzoic acid.
122
6.1
So, weight of m-bromo benzoic acid =   201g  10.05 g
122
actual weight
% yield  100
Theretical weight
7.8
 100  77.61%  78%
10.05
53. I 2  H 2O  2OH   2 I   2 H 2O  O2
2 MnO4  6 H   5 H 2O2  2 Mn 2  8 H 2O  5O2
2 MnO4  3H 2O2  2 MnO2  3O2  2 H 2O  2OH 
HOCl  H 2O2  H 3O   Cl   O2
PbS s   4 H 2O2 aq   PbSO4 s   4 H 2O s 
54. Let M be the molar mass of the element
Volume of the unit cell  a 3   290 1010 cm   24.4 1024
3

In a body centred cubic (bcc) structure n = 2

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n M
Density 
Av.no  a 3
2 M 6.8  6.023 1023  24.4 1024
6.8  M   50 g
6.023  10 23  24.4  10 24 2
 50g of the element contain 6.023 1023 atoms
6.023 1023  200
200 g of the element would contain  24.09 1023 atoms.
50
55. H  G
From H  G  T S  S 
T
 51.4   49.4   1 1 1 1
S     1000 JK mol  336 JK mol
 300 
56. Total pressure of mixture  PTotal  X A  PAo  X B  PBo  90  0.6  15  0.4  60mm
X B'  PBo
Mole fraction of B in vapour phase,  X B'  
PTotal
0.4  15
 X B'   0.1  1 101
60
57. NHCOCHCl2
*
O2 N C H  C H  CH 2OH
*

OH
In amides, nitrogen lone pair is involved in conjugation
58. OH
OH

Phenolpthalein
59. Resonance energy  5  28.8   99  45 Kj / mol : Expected enthalpy – actual enthalpy
60. 1
Effective number of atoms of X  4  4   3.5
8
Effective number of atoms of Y  4  1  3
Effective number of atoms of Z  8  4  4
3 4
2.
7
2

MATHEMATICS
61.  3  2i sin    1  2i sin  
Let z    
 1  2i sin    1  2i sin  

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3  4sin 2   8i sin 
(Rationalising the denominator) 
1  4sin 2 
 a 2  b 2   a  b  a  b  and i 2  1
 3  4 sin 2    8sin  
  i
 1  4 sin    1  4 sin  
2 2

As z is purely imaginary, so real part of z = 0

3  4sin 2  3 3
  0  3  4sin 2   0  sin 2    sin  
1  4sin 2  4 2

   2 
    , , 
 3 3 3 
2
Sum of values of   .
3
62. Here, xH 2  2 x sec   1  0 has roots 1and 1.
2sec   4sec2   4
 1 , 1 
2 1
2sec   2 tan 

2
   
Since,    ,  ,
 6 12 
2sec   2ta
i.e.   IV quadrant 
2
 1  sec   tan  and 1  sec   tan   as 1  1 
and x 2  2 x tan   1  0 has roots  2 and  2 .
2 tan   4 tan 2   4
i.e.  2 , 2 
2
  2   tan   sec 
and  2   tan   sec   as  2   2 
Thus, 1   2  2 tan 
63. Key idea use n th term of APi.e., an  a   n  1 d , if a, A, b are in
Ap , then 2 A  a  b and n th term of G.P.i.e., an  ar n 1.
It is given that, the terms a,b, care in GP with common
1
Ratio r, were   0 and 0  r  .
2
So, let, b = ar and c = ar 2
Now, the terms 3a, 7b and 15c are the first three terms
Of an AP, then
2 (7b) = 3a + 15c
 14ar  3a  15ar 2  as b  ar , c  ar 2 
 14r  3  15r 2  as a  0
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 15r  14r  3  0
2

 15r 2  5r  9r  3  0
 5r (3r  1)  3(3r  1)  0
 (3r  1) (5r  3)  0
1 3
 r  or
3 5
 1  1
as, r   0,  , so r 
 2 3
Now, the common difference of AP = 7b-3a
= 7ar - 3a = a   3   
7 2a
3  3
2a 
So, 4th term of AP  3a  3  a
 3 
64. Given Boolean expression is
~  p   ~ q 
~   ~ p  v  ~ q    p  q ~ p  q 
 pq
65.   1  1 
6


Given binomial is x  1 log10 x 
x 12 
 
 
Since, the fourth term in the given expansion is 200.
3
2
 1 2  1 
 C3  1 log10 x   12   200
6

x  x 
 3 1
  
 21 log10 x  4 
 20  x  200
3 1

21 log10 x  4
 x  10
 3 1
    log10 x  1
 2 1  log10 x  4 
 applying log10 both sides 
 6  1  log10 x   log10 x  4 1  log10 x 
  7  log10 x  log10 x  4  4 log10 x
 t 2  7t  4  4t let log10 x  t 
 t  3t  4  0
2

 t  1, 4  log10 x
 x  10, 104
Since, x>1 x = 10
66. 1' P, 2' A, 1R, 3' L, 1E

4 diff : 5c4  5

3 alike of 1 kind & 1 diff = 1c1.4c1  4

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2 alike of 1 kind & 2 diff = 2c1.4c2  2.6  12

2 alike of 1 kind & 2 diff of 2nd kind = 2c2  1

Total = 22
67. Since, the system of equations has infinitely many
Solution, therefor D  D1  D2  D3  0
Here,
1 1 1
D  1 2 3  1 2  9   1  3  1 3  2 
1 3 
1 1 1
And D3  1 2 3  1 2  27   1   9   1 3  2 
1 3 
   13
Now , D=0
  5  0    5
And D3  0    13  0
   13
     13  5  8
68. We have, f  x   ax 2  bx  c
Now, f  x  y   f  x   f  y   xy
Put y  0  f  x   f  x   f 0  0
 f  0  0
 c0
Again, put y = -x
 f  0  f  x   f   x   x2
 0  ax 2  bx  ax 2  bx  x 2
 2ax 2  x 2  0
1
 a
2
Also, a + b + c = 3
1 5
 b0 3 b 
2 2
x  5x
2
 f  x 
2
n  5n 1 2 5
2
Now, f  n    n  n
2 2 2
10
1 10 2 5 10
  f  n   n   n
n 1 2 n 1 2 n 1
1 10  11 21 5 10  11
 .  
2 6 2 2

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385 275 660
    330
2 2 2
69. Given, f (0) = 2 = g (1), g (0) = 0 and f (1) = 6
f and g are differentiable in (0,1).
Let h(x) = f (x) – 2g(x) …(i)
h (0) = f (0) -2g (0) = 2 - 0 = 2
and h (1) = f (1) -2g (1) = 6 – 2(2) = 2
h (0) = h (1) = 2
Hence, using Rolle’s theorem,
h’(c) = 0, such that
Differentiating Eq.(i) at c, we get
 f 'c  2g 'c  0
 f 'c  2g 'c
70.

71.

72.

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73.

74. When differential equation

dx
  tan x  y  sec 2 x
dy
  sec2 x  y  sec 2 tan x
dx

dy
which is linear differential equation if the
dx
form  py  Q
dy
where p  sec2 x and Q  sec2 x tan x
2
I . f  esec x dx
 e tan x
So, solution at given differential equation is
y   I . f     Q  I . f  dx  c
 y.e tan x   e tan x  sec 2 x  tan x dx
 y.e tan x   e tan x  sec 2 x  tan x dx
 y.e tan x  e tan x  tan x  1  1

Now at x  
4
y e  e  1  1  1
1 1

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1 1
y e  2e  1  y  e  2
75.

76.

77. Conceptual
78. Given equation is

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79.

80. Key idea standard deviation is remain unchanged, if observations are added
or subtracted by a fixed number
we have,

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81.

 abc  1
82.

83.

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84.

85.

86. Apply Leibnitz rule

87. Conceptual
88. Multinomial theorem
89. Conceptual
90. Conceptual

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 SRI CHAITANYA IIT ACADEMY, INDIA 07‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐38_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 07-04-2023
Time: 09.00Am to 12.00Pm GTM-38 Max. Marks: 300
KEY SHEET
PHYSICS
1 1 2 2 3 3 4 4 5 4
6 3 7 4 8 3 9 1 10 4
11 2 12 1 13 1 14 4 15 3
16 3 17 4 18 2 19 2 20 1
21 0 22 4 23 2 24 4 25 2
26 4 27 8 28 7 29 5 30 1

CHEMISTRY
31 2 32 1 33 1 34 2 35 4
36 3 37 3 38 1 39 1 40 4
41 4 42 4 43 1 44 4 45 3
46 3 47 2 48 4 49 3 50 3
51 6 52 9 53 6 54 4 55 4
56 95 57 10 58 68 59 5 60 16

MATHEMATICS
61 2 62 4 63 1 64 2 65 3
66 1 67 1 68 2 69 4 70 1
71 2 72 3 73 3 74 3 75 1
76 1 77 4 78 4 79 1 80 2
81 97 82 2 83 0 84 16 85 2
86 24 87 0 88 602 89 12 90 4

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SOLUTIONS
PHYSICS
1.  Z   0 0 0   K   
 K    M L T  ,     Z  ,  P     
     
2 2 
    K   ML T
       1 2    M 0 L2T 0 
   
 P   ZP   LML T 
  
2. P, Q and R forms the sides of a right angled triangle in which


P  A 2
tan 300   P 
Q 3
3. The component of velocity perpendicular to inclined plane at the point of projection
and the point of collision with the plane remains same, i.e, v sin 
v sin   10
Given 10 10
v   20m / sec
sin  sin 300

4. T T

Ina position as shown 2T cos  W For AB to be horizontal   900 T  

5. 1 1 v 
2
8 1 1 8 2
mvB 2  m  B   mgr , vB 2  gr , v A  vB  gr  gr
2 2  2  3 2 2 3 3
mv 2 A m 2
0 2
TA  mg cos90   . gr  mg
r r 3 3
2
13
 mg 2   mg   mg
2
net force 
3  3

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6. dv v 2 K 2 r 2t 2
P  mK 2r 2t  mat .v at v  K 2r 2t , v  K 2r 2t ,  or v  Krt
dt 2 2
dv v2
at   Kr so, ac   K 2rt 2
dt r
 
7.  M a1  ma2   Mg sin 600  mg sin 300
a com  , But a1  a2 Accelerating of system=
m  M  m  M 
 
M a1  ma2
  0 , M  0, m  0
m  M 
  m
It is possible if a1  a2  0 Or Mg sin 600  mg sin 300 M 
3
8. Conservation of angular momentum,
I1
I11  I 22 2  .1 

ML2 / 12  0
I2
  
 ML2 / 12  2m  L / 2 2 

 ML2   M 
   0   0
 ML2  6mL2   M  6 m 
 
9. Total height attained by block after detaching from spring above the mean position
h  y  v2 / 2 g
Where v is the velocity of block at a distance y above the mean position

v   A2  y2  dh
0
 2  A2  y 2 
For h to be maximum, dy
h y or y0  g /  2
2g
10. k a
ax2  bv2  k bv2  k  ax2 v 2   x 2 Compare with v2  A2 2   2 x2
b x
k
2 2
a k A k
 2  and A2 2  A   b 
b b 2 a a
b
11. GMm
Before collision, PE.  mv   After collision, velocity will be zero. The
r
wreckage will come to rest. The energy will be only the potential energy.
GM  2m  2GMm 1
P.E    Ratio 
r r 2
12. Speed is more near the mean position and it is less near the extreme positions in SHM .
Time taken by the particle to go from 0 to A / 2 will be thus less then the time taken
to go from A / 2 to A. SoT1  T2 .
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13. V   Vin If A is cross-section of the block, L is its thickness and out of L, h is inside

water, then V  AL, Vin  Ah  AL  Ah h    L  0.90  5  4.5 m 4.5 m of ice
 
will be submerged in water. Level of water in hole  5  4.5  0.5 m below top of ice.
Length of rope required is 0.50 m
14. Due to Doppler effect, apparent frequency changes. Intensity depends on frequency on
and amplitude. So intensity also changes
15. Intensity   amplitude  it is non linear physical quantity. The intensities cannot added
2

directly
16. I  I 0  I 0  2 I 0 .I 0 cos   2 I 0  given 
n  2 2 
So, cos   0     n  1 for S2 Pto be min imum      x    x 
2 2     4  2
 x  1m  x  S2 P  S1P  1 S2 P  S1P  1   x  1 metre
17. K KCP / Cv KCP
Q   
  1  CP / Cv   1 CP  Cv
K  nCP T  K Q K Q K
   Or 1  K W
 nCP T  nCv T  Q U  W W
18. R
CP  , dQ  nCP dt , dU  nCv dT
 1
Q
dW  dQ  dU  n  CP  Cv  dT  nRdT , dQ  Q, ndT 
CP
RQ  1 Q  1 1
dW  nRdT   RQ    given      1.5
CP R 3  3
19.

One face AB, i  0 and r  0, PQ will travel undeviated . At face AC , angle of

incidence = angle of prism  i For one colour to be separated, TIR should take plcace.
1 1
i  c sin i  sin C Or sin  Or   For i  300 ,   2 No colour will be
 sin i
separated. For i  450 ,   2  1.414 The colour for which refractive index of prism
is 1.40 will be separated. This colour will be transmitted, while others will be reflected.
20. r  r '  900  1800  By geometry  , r '  900  r

 
 D sin i   R sin r '   R sin 900  r   R cos r
R  
 tan r , c  sin 1  R   sin 1  tan r 
D  D 

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B  
21. WA B  q VB  VA  ,  dV VB  VA    E.d r
B

A A

 4,1    4,1

 2, 2 
 yi  x j  . dxi  dy j  dzk    
 2, 2 
 ydx  xdy 
 4,1 4,1
 d  xy     xy 2,2  0
 2, 2 
WAB 0
22. VE  0 and VD  0 VB  2V and VC  10  2  12V
VA  12  6  6V and VAVB   6  2  4V VC  VD   12  0  12V
12
Current through 3 resistar   4 A
3
23. Resistance in upper branch is 10 and in lower branch 5 . Thus, if current in lower
2 H 1
branch is i, current in upper branch is  i / 2  Now H 4   i / 2   4  H 5  i 2  5  4 
H5 5
H 10
Or H 4  5   2cal / sec
5 5
24. mv 2 1 m 2ev0
Bqv   1 ev0  mv 2  2 R 
R 2 Bq m
25. In critical condition, torque on loop is equal to gravitational torque about an axis
 
  M  B  MB sin 900   r 2iBx
tangent to the loop,  mgr Bx is contributing a torque,
mg  7  11  10 
i     2A
 rBx  22  3.5  5 
26. di
V AB  iR  L 10  2R  L And 6  2R  L on solving, R  4
dt
27. 8
Motional emf e  BvL   4  2 1  8V Current through connector i   2A
22

Magnetic force on connector Fm  iLB   2 1 4   8 N . To keep the connector with

constant velocity, a force of 8 N will be applied towards right.
28. x and y number of   decays and  -decays respectively
92  2 x  y  85 Or, 2 x  y  7 ........1 Similarly,
238  4 x  210 ...... 2  x  7
29. When electron jumps form nth state to ground state, number of possible emission lines
n n 1
 . Here, number of possible emission lines   n  1 n  2   10 On solving, n  6
2 2
30. P 1 P 1  pe  1
r , r  re ,    Or, P  pe  . So,   e . Or, n  1
q 2 2 2 1  p
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CHEMISTRY
31.  1 1   n 2  22 
n1  2 , n2  6 v  R 1     R 
  2 2  n 2   4n 
2

32. Since dq=0 the change in entropy equals zero S  dqrev / T

Volume increases in expansion so entropy increases in order to maintain
equilibrium the entropy is decreased by fall in temperature
33.

pKb  NH 4OH   14  pK a NH 4  4.74 , So, buffer range, pKb  1 
34. Raoult’s law is applicable for ideal solution
35. Na   126.5  151.5  150  78  50
36. The concentration of the salt content in the cell decreases.
37. The sols obtained in the two cases will be oppositely charged so coagulate each other.
38. 1.67  1024 gm  1 amu  mass of 1 molecule of glu cos e  180 amu
 
1.67  1024  N A  1gm  mass of 1 mole glu cos e  180 gm
39. Smaller the orbital, stronger will be the strength of   bond . 3d  as compared to 3 p 
orbital’s.
40. Aqua regia  Mix of 1 part HNO3 conc.  3 part HCl conc.
HNO3  3HCl  2 H 2O  NOCl  2Cl  , Au  3Cl   AuCl3 
 H  AuCl4 
HCl

2 HCl
Pt  4Cl   PtCl4  H 2  PtCl6 
41. Acidic strength order, acidic strength  stability of conjugate base
   
O COO COO COO

NO2 CH 3

Equivalent resonating structure II  III  IV  I

42. 
Cl
DI
Cl

DCl  CH 3  C  CHD2
CH 3  C  CH CH 3  C  CHD Cl CH 3  C  CHD
M .R , I

43. (Be; Al )(Li;Mg)(B,Si)Diagonally related, in properties.

44. Greater the polarity of solvent more will be its interaction with substance
which will effect R f . TLC is an example of adsorption chromatography.
45. CH 3 CH 2 CH 3
Mo2O3
CH 3 CH 2 CHO
46. Isocyanide on reduction gives secondary amines
47. HC  CH
HgSO4 / H 2 SO4
CH 2  CH
H 2O
CH 3  CHO
H 2O
OH
Acetaldehyde
Vinyl alcohol

48. Phenol forms 2,4,6-tribromophenol on treatment with Br2 in water. In H 2O ,

phenoxide ion is formed which highly activates it towards electrophilic substitution
reaction.
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49. The reason for double helical structure of DNA is Hydrogen bonding
50. 2NaOH  Cl2  NaCl  NaOCl  H 2O
51. Structure of Aspartame
O O O

H O C CH 2 CH C 2 NH CH C 2 OCH 3
Sp 2 Sp Sp
CH 3 CH 2
Sp 2
Sp 2 Sp 2

Sp 2 Sp 2
Sp 2

52. 4
53. The reaction that takes place is NaCl  AgNO3  AgCl   NaNO3
143.5g of AgCl is produced from 58.5g NaCl 14g of AgCl will be produced from
58.5  14
 5.70g NaCl This is the amount of NaCl in common salt;
143.5
5.70
% purity   100  95%
6
54.  PCl3  Cl2
Given : M Mix  62.M H 2  124 For PCl5 
 M  P C l 5   M  M ix 2 0 8 .5  1 2 4  0.681  68%
  
 2  1  .  M  M ix 124

55. It is cyclic silicate.

56. The reaction that takes place is NaCl  AgNO3  AgCl   NaNO3
143.5g of AgCl is produced from 58.5g NaCl 14g of AgCl will be produced from
58.5  14
 5.70g NaCl This is the amount of NaCl in common salt;
143.5
5.70
% purity   100  95%
6
57. 4 4
b  4    r 3  N A ; 4    104  1000  4     r 3  6  1023
3 3
r  5  109 cm Distance of closest approach =2r 108 cm or 1010 m
58.  PCl3  Cl2
Given : M Mix  62.M H  124 For PCl5 
2
 M  PCl5   M Mix 208.5  124
   0.681  68%
 2  1. M Mix 124
59. B  CO2 : D  C2 H 2 Difference  4   1  5
60. It is cyclic silicate.

MATHEMATICS
61. If the lines a1 x  b1 y  c1  0 and a2 x  b2 y  c2  0 meets the axes is concyclic
points then a1a2  b1b2
62. Sol : A  x1, y1 be any point on parabola y 2  4 x and B (h,k) is its image

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h  x1 k  y1
w.r.t x-y+13 = 0 then   13  0 and
2 2
 k  y1  2
  (1)  1  x1  h  13, y1   k  13   h  13  4  k  13
 h  x1 
63.  ,  ,  are roots of  x  13  8
   1,   1,   1 are equal to 2,  2 ,  2 2 in any order then
 1  1  1 3k 1 3k  2
   3 2 or 3  3   or 3  where k  Z
 1  1  1
 p  3, q  3k  1 or 3k  2
11 10
p  q  15  k  or which is not possible
3 3
64. Formula application
65. Taking the coordinates of vertices O,P,Q,R as (0,0), (a,0), (a,a), (0,a)
 a a 
respectively get the coordinates of M as  a,  and of N as  , a 
 2 2 
66. Shortest distance is 0.
67. 1 
foot of the perpendicular A (1,0,2) to the line is B  ,1,  3 / 2 
2 
68.  1 1
  
 0  f  0  Continuous at x = 0 Let
x x
Clearly Lt
x0 xe 
 1 1
  
f  x   f  0  Lt
x x
xe  0
Lt
x0  x0 1
x0 x0
 1 1
  
f  x   f  0  Lt
x x
xe  0 2/ x
Lt  x0 = Lt e 0
x0 x0 x0 x0
Not differentiable at x = 0
69. 1 1
, are the roots of cx 2  bx  a  0
 
 1  1
cx 2  bx  a  c  x   x  
   
70. T : x –x = 0 is integers reflexives y – x is integer symmetric (x-y) + (y-z) = x
-z is integers transitive  T is equivalence and S : y – x = 1  x – x  1
not reflexive.
71.     /2
cos 2 x cos   x  
2
cos 2 x 2 2 
 1  a x dx    1  a x  1  a  x  dx   cos x dx  2  cos x dx  2
 0   0

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72. 1
We have  e x cos xdx  e x  sin x  cos x 
2
1 x 1
Therefore, I  x. e  sin x  cos x    e x  sin x  cos x dx   e x sin xdx
2 2
x 1
 e x  sin x  cos x    e x  cos x  sin x  dx
2 2
1 1
 e x  sin x  cos x   e x cos x   c  e x  x sin x  x cos x  cos x   c
x
2 2 2
73. from truth table p   q  p   p   p  q 
74. total no of ways of answering is 15. Out of 15 combinations only one is
1
correct. The probability of ticking the answer at the first .. is , that of 2nd is
15
14 1 1 14 13 1 1
.  and that of third is . .   probability that the student
15 14 15 15 14 13 15
1 1
will get marks for the question if he is allowed into 3 chances  3    .
 15  5
75. Let f  x   x 2  2 ax  a 2  a  3  0
Then (i)   0 (ii) sum of root <6 (iii) f  3  0
76. cot   tan   m  sec 2  .  m tan   1

 
1/3
sec  cos  n  tan 2  .  n sec   2  tan 2   n nm2

 
1/3
from (1) & (2)  similarly sec2   m n2m eliminate 

X i i 
77. f x 12  22  .....  52 11

i f 15 3

M .D from X 
 fi  xi  X    18   2 5   3 2   41   5 4   16
 
 fi  3  15  15
78. log e 2
 
log e 2 1
A1   e x  e  x dx  e x  e x  
 0 2
0
e
A2  2   log e x  dx  2  x logx  x 1  2  A2  4 A1
e

1
79. f  x   f  x  3  f  x  6   f  x  9  Replacing x with x + 3
f  x  3  f  x  6   f  x  9   f  x  12  Adding we get f  x   f  x  12 
x 12
Now g  x    f  t  dt  g |  x   f  x  12   f  x   0 x  g  x  is a constant
x
function
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80. for no solution or infinitely many solutions
 1 1
2
1  1  0    1   2   0    1,1,  2
1 1 
but for  = 1, there are infinite solutions when  = -2, we have
2 x  y  z  3, x  2 y  z  3 x  2 y  2 z  3 adding 0 = - 9,
which is not true  no solution.
81. Tr 1  100cr (51/6 )100 r (21/8 ) r As 5 and 2 are relatively prime, Tr+1will be rational
100  r r
if and are both integers  100-r is a multiple of 6 and r is a multiple of
6 8
8, 100-r is multiple of 6 if r = 4, 12, 20… 100 of which 12, 36, 60, 84 are divisible
by 6. Hence, there are just four rational terms.  no. of irrational terms is 101 – 4 =
97.
82. r 2 i r2
i i 2
Z = re Then (e  e )  1 
12 3
r 2 cos 2  3  r 2 3
   r2   3 max value of |z| = r = 3
3 3 1  cos 2 
83. 100!  50   50 
We have 100 c50  The exponent of 7 in 50! is       7  1  8
50!50!  7   72 
100  100 
And the exponent of 7 in 100!      14  2  16 ,Thus, exponent of 7 in
 7   72 
100
C50 is 16  2  8  0
84. x

x 2
i
400 80
i 1
   n  16 hence, least possible value of n = 16.
n n n
85. sin 1  sin 3    3 sin 1  sin 4     4 sin 1  sin 5   5  2 sum  2
86. Conceptual
87. Differentiating x 2  2 y 2  y  c, 2 x  4 yy1  y1  0  2 x    4 y  1 y1
1 2 xdy 4dy 2dx
Replacing y by  ,  4 y 1  
y1 dx 4 y 1 x
1
ln  4 y  1  lnx 2  ln c1 4 y  1  c1x 2 y  cx 2  ,4c  c1.
4
88. 3 3 3
By observation Sn  n  S11  S9  11  9  1331  729  602
89. 1 1 1
 
By R3  R3  100 R1, det A1  4 6 8  2
0 0 1

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Adj  2 A   4 Adj; A

 
det A1 
1
det  A 
 det  A  
1
2
 43. A
2

3 1
 4 .  42  16
4
and 2 A  2 A  4  Adj  2 A  2 A  12
3

90.
a 2  b2
e  ae  16  9  7
2
a
 
 Foci  7, 0 lies on a circle with centre  0,3
r  7  9  4

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 08-04-2023
Time: 09.00Am to 12.00Pm GTM-39 Max. Marks: 300
KEY SHEET
PHYSICS
1 4 2 2 3 4 4 4 5 1
6 3 7 3 8 4 9 4 10 3
11 2 12 2 13 1 14 2 15 3
16 3 17 1 18 3 19 4 20 3
21 1 22 1 23 2000 24 5 25 3
26 10 27 2 28 18 29 580 30 20

CHEMISTRY
31 2 32 3 33 2 34 3 35 1
36 4 37 1 38 4 39 3 40 3
41 1 42 1 43 3 44 2 45 4
46 4 47 4 48 4 49 4 50 2
51 1001 52 16 53 2 54 2 55 5
56 58 57 25 58 3 59 3 60 1

MATHEMATICS
61) 2 62) 3 63) 2 64) 2 65) 1
66) 1 67) 4 68) 2 69) 4 70) 2
71) 1 72) 1 73) 1 74) 2 75) 2
76) 4 77) 2 78) 3 79) 4 80) 2
81) 65 82) 28 83) 64 84) 1 85) 6
86) 166 87) 4 88) 481 89) 51 90) 2

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SOLUTIONS
PHYSICS
1. The least count of screw guage is smallest length which can measured accurately
1
with it as L.C is 0.001cm  hence measured value should places i.e 5.320cm
1000
2. GM  G  2M   3GM
Vm  Vshall    
R  R  R
3. a
F
 mass of rod 
M T  M 2 
 2   g  a 
480   20  12  8 10
a  12  4 12 
20  12  8
192 N
 2m / sec 2
4. mgl  1 1  mgl  1 1  mgl  25  4  21mgl
W  2  2      
2  n1 n2  2  4 25  2  100  200
5. Conceptual
6. Volume disc = volume of sphere
4 2
 r12  t   r23 I sp  mr2 2
3 5
r 4 2 r 
2
2 mr12
 r12  1   r23  m 1  
6 3 5 2 5 4
3
 r2  1 1 mr12
   
 r1  8 5 2
r I
r2  1 
2 5
7. The adiabatic relation between  and v for a perfect gas is
V r    1
RT
Again from standard gas equation V r  nRT  V 

Putting in e.q.(1) we got

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
R
 k

k
or  1   another cons tan t
RT
i.e.  1  cons tan t
11   21
1
Here,  2    1
R
T1  27 0 C  273  27  300 K
5
T2  ?,  
3
 5
T      T 3
 2    1  or  2    8   8
1 5/3 2/3


 T1   2   300 
 T2  130.6 K
T2  1420 C

8. VD D 1 2
R r  R
 2  
15 103 m3  5 min  5  60sec
15 103 1 1
Rate of flow   A V , 103    104 v
5  60 20 
V  0.5
2
103  0.5   102
R   0.55  10 4  5500
3
10
9. 1 1 1 1 1 1.5  1  1  1 2  1.5  1
      
f f1 f 2 f3 60 12 12 12
1 1 16 8
  1  2 1  2, 2 1  3   , 1   1.60
5 5 5 5
10. 4
r1 r1
Qr E  1 q  1  Qr1
4

q   r 3  , dq  4 r 2 dr. , q   4 r 2 dr. , q   4 r 2 dr 
3 0 0
 R4 4 0 r12 4 0 R 4 r12
4Q r 4 Qr 4 1 Qr14
q  4  1  14 
R 4 R 4 0 R 4
11. Conceptual
12. 0 A  1 1 1 
Cef  C1  C2  C3     
3  d d  b d  2b 
0 A

3d  d  b  d  2b 
  d  b  d  2b   d  d  2b   d  d  b  
0 A

3d  d  b  d  2b 
 3d 2
 6bd  2b 2 

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 SRI CHAITANYA IIT ACADEMY, INDIA 08‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐39_KEY &SOL’S
13. x

l1
y 100  l1
x 39.5

1.5 100  39.5
x  8.16
y l2

x 100  l2
12.5 l2

8.16 100  l2
1.53 100  l2   l2
153
l2   60.5cm
2.53
14. K1 A 100  80  K 2 A  80  0 

16 8
K1  20
 K 2  80
2
K1  8 K
15. Final momentum,
E P  t 30  103  100  109
P  
c c 3  108
 1.0  1017 kg ms 1
16. 
1 1 1
1  1 1   9
Balmer series : n1  2; n2  3, 4.......    2  2   l arg est  2   4 
2

  n1 n2  shortest 1 1
 2
5 5
2
2 3 36
17. 180
1e
A 
0

 A1 
180

 A2   A3
176 172

He4 He4
72 73 2 71 2 69

18. Conceptual
19. i  i1  i2  a  b sin wt
T
T I rm2    a  b sin wt  dt
2

0
T
   a 2  b 2 sin wt  2ab sin wt  dt
0
T T T
T I rmg 2   a 2 dt   b 2 sin 2 wt   2ab sin wtdt
0 0 0

1  cos 200t
T T
T I rmg 2  a 2T  b 2  dt   2ab sin wtdt
0
2 0

 T 1  sin wt  T   cos wt  
T

T I rmg 2  a 2T  b 2      2 ab  w  
 2 2  w  0 0 
2 T 
 a T  b   0  0
2

2 
b2
I rmg 2  a 2 
2
1/2
 2 b2 
I rmg 2 a  2 
 
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20. Ma 
Am 20 2
   0.67
Ac 30 3
21. V 2
i    1A
R 2
22. I1 B1 T2 0.8  20
  
I 2 T1 B2 0.4  5
0.8 0.8  20

I2 0.4  5
4 I 2  0.4
I 2  0.1
23. W L 2 fL
Q 0 
R R
2  3.14 107  2 104

6.28
4  3.14 103

6.28
 2 103
 2000
24. d sin   n
d  2
2d  sin   n
n2
no.of max imus  2n  1
 2 2 1
 4 1
5
25. x 2  at 2  2bt  c, diff w. r. t ' t ', 2 x
dx
 2at  2b  0
dt
dx 2at  2b at  b
v   , Again diff w. r. t ' t '
dt 2x x
d dx
x  at    at  b 
a  dt dt  a 1  a  x 3
2
x x3
a  xn
n3
26. Conceptual
27. 10  5  2 1  10
di di
 3  0,  10  20,
di
 2 A / C
dt dt dt

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 SRI CHAITANYA IIT ACADEMY, INDIA 08‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐39_KEY &SOL’S
28. F  12t  3t 2

  F r
 12t  3t 2  1.5   18t  4.5t 2
 18t  4.5t 2
 
I 4.5
  4t  t 2

w t t
, dw   dt ,  dw    dt    ut  t 2 dt
dw

dt 0 0 0

t3 t3
w  2t 2  , w  0; 2t 2   t  b
3 3
d  t3 
6 6
36
w   d   w dt ,     2t 2   dt  radi 18 rad
dt 0  0
3 2
29. For pariticle ' A ' For particle ' B '
X A  3t 2  8t  10 YB  5  8t 3
 
VA   8  6t  i VB  24t 2 j
 
a A  6i a  48ti
B

At t  1 sec
 
VA   8  6t  i  2i and VB  24 j
  
 VA  VA  VB  2i  24 j
Speed of B w. r. t A, V  22  242
 4  576  580
V  580  m / s 
30. V  iR, I mole 
Vmole
, PZ  VZ I Z I Z 
2
 0.2
R 10
V 14  10 4
I Z  I mole , R  max  ,R   20
I max 0.2 0.2

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CHEMISTRY
31. 1
H 2C2O4  l   O2  g   H 2O  l   2CO2  g 
2
1 3
n  2   In Bomb Calorimeter 'V ' is constant
2 2
 Z    M 8.75  0.312  90
E    122.85KJmol 1  Enthalpy of combustion
m 2
3 8.3
 H   E  n.RT  122.85    300   126.6 KJ / mol
2 1000
32. T f  i  k f  m 0.558  i  1.86  0.1  i  3  1     n. If   1 i  n ; n  3
no.of ions n  3  formulae  Cocl3 .5 NH 3  or  CO  NH 3 5 Cl  Cl2 x5
33. AgNO3  KI   AgI  I  KI  AgNO3   AgI  Ag 
 excess   excess 
The preferential adsorption of common ion on AgI particles provides ve and  ve
charge.
34.  5 
I 2 0  H I O3


MW MW
EW of I 2    n factor of I 2  10  x
n  factor 10
o
S8  H 2 SO4


MW
EW of S8   n  factor of S8  48  y  x  y 10  48  58
48
35. Conceptual
36. 3.a
Diamond cubic  DC  Radius of atom, r  ; Z eff  8
8
3

4 4  3.a 
8  r3 8 3   8 
PF  3     3  0.34 hcp ; radius of atom r  a ; z  6
3 3 eff
a a 16 2
4 4
height  4r 
2
; Base area  6 3.r 2 6   r3 6   r3
3 3 
3 P.F  3
 3
  0.74
a 24 2r 3 2
volume of unit cell  24 2r 3

37. (1)
N aO H  C H 3 C O O H  C H 3 C O O N a  H 2 O
50 m l , 0.1 M 50 m l , 0.1 M 0
5m Eq 5m Eq 0
0 0 5m Eq
5
C H 3 C O O N a    Salt  
 0.05 M
100
1
for salt hydrolysis P H   Pkw  Pka  log C 
2
1 1
for univalent salt  14  4.74  log  0.05    14  4.74  2  0.7   8.72
2 2
2.5
(2)  NaOH   OH   
 2.5 102 M
100
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For base,
P OH   log OH     log 2.5  102   log 2.5  2  4  2  1.60
(3) For buffer solution
P H  Pka  log
S   4.74  log
 2.5  4.74
 A  2.5
38. Conceptual
39. Conceptual
40. CH3

CH3 C CH2 CH3

H
1 6 1 5 3.8  2 1 3
2 Chloro 3-Methyl butane is major
41.
HO H

H H

C2H5

S as the hydrogen on horizontal

S-R it is R
42. For Assertion : Acetal and ketals are basically ethers hence they must be stable in
basic medium but should break down in acidic medium.
Hence assertion is correct
For reason : Alkoxide ion  RO   is not considered a good leaving group hence reason
must be false.
43. Statement – I is true pure aniline is colourless liquid
Statement – II is false Aniline becomes dark brown due to action of air and light
oxidation
44. H CH3 H CH3

CH3 C CCH2 CH2 

 CH3 urc
peroxide
C CCH2CH3

Br H
2-Chiral carbon so it has 4 optical Isometal
45. Pinacol, pinacolone rearrangement
46. Cannizzaro reaction is self oxidation and self reduction reaction
47. Carbyl – Amine reaction
48. Alitame is 2000 times more sweeter than sucrose
49. When carbondioxide passes in to insoluble carbonates they are convert in to soluble
bicarbonates
50. Statement II : The corssed arrow symbolizes the direction of the shift of electron
density in the molecule.
51. Equalent conductance  0 
1 0 1 0
.   .
n n
Where n and n are charge on each ion

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1 0 1 1 1
 0 Al  so  

 Al 3    0 so 2   0 Al 3   0 so 2
2
n4 3
n 4
3 2 4

1 1
  189   160  143 S cm 2 eq 1  x
3 2
Molar conductance at infinite dilution 0    . 0    . 0
Where   and   are the number of  ve and ve ions
 0 Al2  so4     . 0 Al 3    . 0 so 2  2  189  3  160
3 4

 858S cm 2 mol 1  y, x  y 143  858  1001

52. Using Einstein’s photo electric equation
KE2 h 2  0  2  0 KE2
  But given that 3
KE1 h 1  0  1  0 KE1
2  0 2  1016
3  3 1  0   2  0  31  2  30  0  20 0   11016 s 1
1  0 2
53. meq. of KMnO4  meq. of Fe2  C2O4 3
Fe3 not oxidised , where as C2O42 oxidises to CO2
M
3C2O4 2  6CO2  6e  , E 
6
n  factor  6
N1V1  N 2V2
24
x  5  100  0.2  6  200  x  M  0.48M
50
Normality of KMno4  n  factor  M  5  0.48  2.4  2.0 N
54. F  32 C

9 5
98.4  32 C
  C  36.880 c  309.88 K
9 5
W 168 W
PV  RT ,  6   0.0821 309.88. W  1.67 g  2.0
M 760 32
55. Conceptual
56. 1.25 104 K
log K  s 1   14 
Ea
Like log K  log A 
T 2.303RT
14 1
log A  14  A  10 s
Ea 1.25  104 K

2.303RT T
 Ea  1.25  10  2.303  2  103 kcal  57.6k cal
4

57.
% N  1.4 
 N1V1  N 2V2 
w
58. 13  8
BO   2.5
2
59. K 2 SO4 , Al2  SO4 3 24 H 2O, x  2, y  2, z  3, p  24 
24
3
2  2  3 1
60. 25
 1.25
20
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MATHEMATICS
61.
L.H .L Lt f  x   Lt 10  x 2  1
x 3 x 3
R.H .L Lt f  x   Lt 2  e x 3  2  e 0  1
x 3 x 3
f  3  1  f is continuous at x  3
f  3  h   f  3
 
Lf 1 3  Lth0
h
2
10   3  h   1 1  6h  h 2  1
 Lth0  Lth0
h h
6h  h 2 h6  h
 Lth0  Lth0
h  1  6h  h 2  1 h  1  6h  h 2  1
   
6
  3
2
f  3  h   f  3
 
Rf 1 3  Lth 0
h

 Lth0
 2  eh   1  Lt 
 eh  1   1
h 0
h h
     f is not differentiable at x  3
f 1 3  f 1 3

4  5w334  3w365  4  5  w3  w  3  w3  w2
62. 111 121

 4  5w  3w2


 1  3 1  w  w2  2w 
 1  3 0   2w


 1  2w  1  1  i 3  i 3 
63. y  3x  2 touches ellipse ; m  3, c  2
c 2  a 2 m 2  b 2  4  9a 2  b 2
9a 2  b 2
AM  GM   9a 2b 2  9a 2  b 2  6ab
2
 4  6ab
 2  3ab
2 2
  ab    ab
3 3

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 SRI CHAITANYA IIT ACADEMY, INDIA 08‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐39_KEY &SOL’S
64.  13  2
Mid point of QR   ,1 PS  
2  9
2
Equation of the line is y  1   x  1  2x  9 y  7  0
9
65. p q p  q  p   p  q  p   p  q  q
T T T F T
T F T F T
F T T T T
F F F F T
66. I   esin x  x cos x  tan x sec x  dx

  xe 
sin x
 cos x dx   esin x sec x tan xdx

  xesin x   1esin x dx    esin x sec x   esin x cos x sec xdx 

   
 xesin x   esin x dx  esin x sec x   esin x dx

 xesin x  esin x sec x  c

67. f  x   ax 2  bx  c, g  x   px  q
f  g  x    8 x 2  2 x, g  f  x    4 x 2  6 x  1
2
 a  px  q   b  px  q   c  8 x 2  2 x
ap 2  8, 2apq  bp  2, aq 2  bq  c  0

 
and p ax 2  bx  c  q  4 x 2  6 x  1  ap  4, bp  6, cp  q  1

ap 2  8 and ap  4  p  2, a  2 b  3
c  1, q  1 f  x   2 x 2  3 x  1, g  x   2 x  1
f  2   g  2   15  3  18
68.     k  2,   k 2  3k  5
 2   2     2  2   k  2 2  2k 2  6k  10
  k 2  10k  6
Let f  k   k 2  10k  6
f 1  k   0   2k  10  0  k  5
f 11  k   2  0

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 SRI CHAITANYA IIT ACADEMY, INDIA 08‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐39_KEY &SOL’S

f  5   25  50  6
Maximum value
19
69.
 
100 100 100
3400  34   81  1  80 

 100c0  100c180  100c2 802      100c100 80100

 1  100  Integer 
 Last two digits = 01
70. 2n  1  25  1  31
71.
   
a . c b  a . b c  
1  
2
bc 
1   1
a , c  , a. b  
2 2
 1 

a . b  
2  1 3
cos     
a b 11 2 4
72. y 2  4  x  1 , y 2  8  x  2 
Equation of the tangents are
L1 : yt1   x  1  t12
L2 : yt2   x  2   2t22
t1t2 y   x  1 t2  t2t12
t1t2 y   x  2  t1  2t22t1
x  3  x  3  0
73. 1  sin t  u  cos tdt  du
0 1
an   u n 2 1  u  du   2  u n 1  u  du
1 0
1
 
 2 u n  u n 1 du  2 
0
 1

1 
 n  1 n  2 
an  1 1 
 2  
n  n  n  1 n  n  2  
n
an  1 1 1   1 1 1 
 n
 2
1.2

2.3

n  n  1
  2
1.3

2.4

n  n  2 

n 1    
 1   1 1  1 1   2  1   1 1   1 1  
 2  1                   1                 
 2   2 3   n n  1   2  3   2 4   3 9  

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 SRI CHAITANYA IIT ACADEMY, INDIA 08‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐39_KEY &SOL’S

lim n ak 3 1

n   k 1 k
2 
2 2
74. 
sin 1 x  sin 1 y 
2
sin 1 x  cos 1 y  sin 1 1  y 2
x 2  y 2  1

   2
2
1 x  y4 4 1  x2  y 2  2 x2 y2 2 1  x2 y 2
 
x2  x2 y 2  y 2 1  x2 y 2 1  x2 y 2
75. a11
  1 b 1  abc   a  b  c   2
11 c
  0  abc  2  a  b  c
1  abc 1
abc  2  3  abc  3    abc  3 
 3 
1
Let  abc  3  x
x3  2  3 x  x3  3 x  2  0
2
  x  1  x  2  0  x  2  0
1
 x  2   abc  3  2  abc  8
76.
cos2
3
2 1
 sin x  2 2
4
1  2sin x 
1
4
1

 sin 2 x  1  4sin 4 x  4sin 2 x
4

1 1
 sin 2 x   sin 4 x  sin 2 x
4 4
sin 4 x  0  sin x  0
 3 ,  2 ,   , 0,  , 2 , 3 ,
 Number of solutions  7
77. dx x  2 y 3 dx 1
   x  2 y2
dy y dy y
1
  y dy 1
I .F  e  G.S is x  I .F    Q. I .F  dy  c
y

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 SRI CHAITANYA IIT ACADEMY, INDIA 08‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐39_KEY &SOL’S
1 1
G.S is x   2 y 2 . dy  y 2  c
y y
78. sin x
lim  sin x  x sin x
   1 form
x  0 x 
lim sin x  sin x  x 
x 0 x sin x  x  1
e 
e
79. 
If f  x  has an extremum at x 
3
 
Then f 1    0
3
Now f 1  x   a cos x  cos3x
  a
f 1   0  1  0 a  2
3 2
80. Let E1, E2 , E3 be respectively events that P, Q, R ride the horse A.
A= event that horse A win the race
1 1 1
P  E1   , P  E2   , P  E3  
2 3 6
1 2 3
P  A / E1   , P  A / E2   , P  A / E3  
8 8 8
1 1
.
2 8 3
P  E1 / A   
1 1 1 2 1 3 10
.  .  .
2 8 3 8 6 8
81. Number of boys  70
Average marks of boys  75
Total marks of boys  75  70  5250
Total marks of the class  72  100  7200
Total marks of girls  1950
 1950
Average marks of the girls  65
30
82. a a 5
, a, ar  GP  .a.ar   a  8
r r 2
8
 4, 8  4, 8r  AP
r
8 1
24   4  8r  r  2,
r 2
 terms 4, 8,16 sum  28
83. 2 2
2n  n  23 3  26  64

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 SRI CHAITANYA IIT ACADEMY, INDIA 08‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐39_KEY &SOL’S
84. 1 0
2  x  2  dx  1 x
2
dx
1 1 5
  
2 3 6
85. x3 y 4 z 5
   t , P 1, 1, 9 
1 2 2
Q  t  3, 2t  4, 2t  5  lies on the plane x  y  z  17 then
t  3  2t  4  2t  5  17  5t  5  t  1
Q  4, 6, 7  : PQ  9  25  4  38  6
86. 10
k
 k4  k2 1  2  2
2
2
1 10 k  k  1  k  k  1   
k 1
2

k 1 k  k  1 k  k  1  
1 1  55
 1  
2  111  111
m  n  166
87. bc3  gc2  168
b  b  1 b  2  g  g  1
  168
3 2 2 1
b  b  1 b  2  g  g  1  2  6  168
 2  6  8  21
 2  6  8 3 7
  8  7  6  3  2 
 b  8, g  3
b  39 8  9
  4.12
4 4
88.

f  x   sin 2 tan 1 2 x 

f 1  x   cos 2 tan 1 2 x .  2
2 x ln 2
 
1 2 x 2


f 1 1  cos 2 tan 1 2  1 2 4  2  ln 2
 2
1 1  2 4
 coscos   ln 2
 2 5
1 2 
12
 ln 2
25
a 2  b 2  625  144  481
Sec: Sr.Super60_NUCLEUS & ALL_BT Page 15
 SRI CHAITANYA IIT ACADEMY, INDIA 08‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐39_KEY &SOL’S
89. Intersection of two lines B 1, 2 
Let C  h, 4  2h 
AB  AC  AB 2  AC 2
2 2
 25  1   6  h   1  4  2h 
 26  36  h 2  12h  9  4h 2  12h
 5h 2  24h  19  0
19
h ,1
5
 19 18 
C  , 
5 5 
 18 
 6  1  19 / 5 1  2  5 
 ,     , 
 3 3 
 
 54 3 
 , 
 15 15 
 51 
15      15    51
 15 
90. Let the lines be y  m1 x and y  m2 x
2c 1
m1  m2   m1m2  
7 7
Given that m1  m2  4m1m2
2c  4
  c2
7 7

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 SRI CHAITANYA IIT ACADEMY, INDIA 09‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐40_KEY &SOL’S

Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.Super60_NUCLEUS &ALL_BT JEE-MAIN Date: 09-04-2023
Time: 09.00Am to 12.00Pm GTM-40 Max. Marks: 300
KEY SHEET
PHYSICS
1) 3 2) 4 3) 3 4) 3 5) 3
6) 2 7) 3 8) 3 9) 3 10) 3
11) 3 12) 4 13) 1 14) 4 15) 3
16) 2 17) 3 18) 4 19) 3 20) 2
21) 48 22) 4 23) 9 24) 3 25) 1
26) 4 27) 6 28) 5 29) 8 30) 6

CHEMISTRY
31) 3 32) 4 33) 4 34) 2 35) 3
36) 1 37) 2 38) 2 39) 1 40) 4
41) 3 42) 4 43) 1 44) 4 45) 3
46) 3 47) 2 48) 3 49) 2 50) 4
51) 9 52) 3 53) 15 54) 1440 55) 481
56) 1 57) 18 58) 60 59) 195 60) 8

MATHEMATICS
61) 2 62) 2 63) 1 64) 3 65) 3
66) 3 67) 1 68) 3 69) 2 70) 3
71) 4 72) 3 73) 1 74) 2 75) 4
76) 2 77) 1 78) 1 79) 1 80) 4
81) 96 82) 15 83) 87 84) 8 85) 269
86) 2 87) 9 88) 1 89) 8 90) 1

Sec: Sr.Super60_NUCLEUS & ALL_BT Page 1

 SRI CHAITANYA IIT ACADEMY, INDIA 09‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐40_KEY &SOL’S

SOLUTIONS
PHYSICS
1. Consider the expression for the energy of a photon E  hv to get the dimensions of h as
those of ( energy time ). Again the expression for the force 𝐹 acting on a charge 𝑞
moving with velocity 𝑣 in a magnetic field is qvB sin  . This gives the dimensions of
qB as those of (force/velocity). Use these to determine the dimensions of the quantity
under consideration, noting that n is a dimensionless quantity.
2.

1
Applying Kirchhoff's second law for closed loop AP2 P1DA , we get
10 I1  2 I  5  0 2 I  10 I1  5 ......(i)
Applying Kirchhoff's second law for closed loop P2 BCP1P2 ,we get
2  1( I  I1)  10 I1  0
I  11 I1  2 .........(ii )
or 2 I  22 I1  4 .........(iii )
1
Subtracting eq. (iii) from eq. (i), we get 32 I1  1 or I1 
A  0.03 A from P2toP1
32
3. Here, 2 F and 3 F capacitors are connected in series.
1 1 1 6
Their equivalent capacitance is   or CS   F
Cs 2 3 5
Net voltage,V  16V  6V  10V
The equivalent circuit diagram as shown in figure below.

Charge on each capacitor,

6
q  CsV   10  12  C
5
The potential difference
12  C
between A and B   6V
2 F
4. Nuclear stability depends upon the ratio of neutron to proton. If the n / p ratio is more
than the critical value, then a neutron gets converted into a proton forming a  
particle in the process. n  p  e

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 SRI CHAITANYA IIT ACADEMY, INDIA 09‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐40_KEY &SOL’S

The   particle (e ) is emitted from the nucleus in some radioactive

transformation. So electrons do not exist in the nucleus but they result in
some nuclear transformation.
5. Total time of flight is T  4 s and if  its initial speed and 𝜃 the angle of projection.
u sin 
Then T  4
g
or u sin   2 g …(i)
After 1 s velocity vector makes an angle of 450 with horizontal i.e., vx  v y
u cos  u sin   gt
u cos  u sin   g ( t  1s )
u cos  2 g  g (U sin g (i ))
or u cos  g .....(ii )
Squaring and adding (i) and (ii), we get
u 2 sin 2   u 2 cos 2   (2 g )2  ( g ) 2
u 2  5 g 2  5(10)2 m2 s 2
 u  22.36ms 1
Dividing (i) by (ii), we get,
u sin  2 g
 2
u cos g
tan   2 or   tan 1(2)
6.

As field due to current loop 1 at an axial point

0 I 2 R 2
 B1 
2( d 2  R 2 )3/2
Flux linked with smaller loop 2 due to B1 is
0 I1R 2
2  B1 A2  2 2 3/2
 r2
2(d  R )
The coefficient of mutual inductance between the loops is
1 0 R 2 r 2
M 
2 2(d 2  R 2 )3/2
Flux linked with bigger loop 1 is
0 R 2 r 2 I 2
1  MI 2 
2(d 2  R 2 )3/2

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 SRI CHAITANYA IIT ACADEMY, INDIA 09‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐40_KEY &SOL’S
Substituting the given values, we get
4  107  (20  102 )    (0.3  102 )  2
1 
2[(15  102 )2  (20  102 )2 ]3/2
1  9.1  1011 weber
7. 1 T
Fundamental frequency of vibration of wire is v 
2L 
where 𝐿 is the length of the wire, 𝑇 is the tension in the wire and 𝜇 is the mass per
length of the wire As 𝜇 𝜌𝐴 where 𝜌 is the density of the material of the wire and 𝐴
is the area of cross-section of the wire.
1 T
v 
2L  A
Here tension is due to elasticity of wire
 L   Stress TL 
T  YA    As Y  
 L  Strain AL 
1 Y L
Hence, v 
2L  L
L
Here, T  2.2  1011 Nm2 ,   7.7  103 kgm3  0.01, L  1.5m
L
Substituting the given values, we get
1 2.2  1011  0.01 103 2
v  Hz  178.2 Hz
2  1.5 3 3 7
7.7  10
8. Depth of 𝑃 below the free surface of water in the vessel  ( H  h)
Since the liquid exerts equal pressure in all directions at one level, hence the pressure at
P  ( H  h)  g .
9. For image formed by the surface on the right side
R
μ1  1,μ 2  1.5,u  ,R1  R
2
μ 2 μ1 μ 2  μ1
 
v u R
1.5 2 1.5  1 3R
  v
v R R 5
 3R 
The image is at a distance  R    0.4R from the centre P towards the
 5 
righthand side. For the surface on the left hand side
3R 1.5 2 1.5  1 9R
u ,R 2  R,μ1  1,μ 2  1.5 So,   or v 
2 v 3R R 7
 9R  2R
The image is at a distance of  R from the centre P towards the right
 7  7
hand side.
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 SRI CHAITANYA IIT ACADEMY, INDIA 09‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐40_KEY &SOL’S
∴ The distance between the two images is
2R
 0.4R   0.114R
7
Therefore option (c) is correct.

10. Using Gauss's theorem for spherical surface of radius r outside the sphere with the
r
1 r 1 r 

0
ε0 0 R
α
charge q  Eds ρ(r)dV E4πr 2   q   4πr 2 dr 
ε 0  r 

 
E4πr 2 
 q  2παR 2  4παr 2

ε0 2ε 0
The intensity E does not depend on R if
q  2παR 2
 0 or q  2παR 2
ε0
11. Efficiency of a perfect engine working between 300 C and 27 C
T 270K
(i.e., T2  270 K and T1  300 K ) ηengine  1  2  1   0.1
T1 300K
Since efficiency of the refrigerator (ηref . ) is 50% of ηengine
 ηref .  0.5ηengine  0.05
(Using (i))
If Q is the heat transferred per second at higher temperature by doing work W, then
W W 1kJ
ηref.  or Q1    20kJ
Q1 ηrec. 0.05
( as W  1kW  1s  1kJ)
Since ηref . is 0.05, heat removed from the refrigerator per second, i.e.,
Q2 
 Q1  ηref . Q1  Q1 1  ηref . 
 20kJ(1  0.05)  19kJ
12. 3/2
 2 nI  2 nIr 2 Bh  h2 
As Bc  0 and Bh  0 so  1  
4 r 4
 2
r h 2 3/2
 Bc  r 2 

Fractional decrease in the magnetic field will be

B  Bh  B 
 c  1  h 
Bc  Bc 
  2 3/2   3h 2  3h 2

 1  1 
h
   1  1  
  r2    2r 2  2 r 2
     

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 SRI CHAITANYA IIT ACADEMY, INDIA 09‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐40_KEY &SOL’S
13. Bulk modulus of elasticity measures how good the body is to regain its original volume
on being compressed. Therefore, it represents incompressibility of the material.
 PV
K where P is increase in pressure, V is change in volume.
V
14. Let V  V0 sin t
(as V  0 at t = 0 )
Then VR  V0 sin t
and VC  V0 sin(t   / 2)
V and VR are in same phase. While VC lags V (or VR ) by 90 . Now VR is in same
phase with initial potential difference across the capacitor for the first time when,
 3
t    2 
2 2
3
t 
2
15. 13.6 Z 2
Using, En   eV
n2
13.6(3) 2
( For Li  ) Z  3 E1   eV
2
(1)
 122.4eV
13.6(3) 2
and E3   13.6eV
(3)2
E  E3  E1  13.6  122.4  108.8eV
16.

As, I  e1000V /T  1 mA  ….(i)
Here, I  5mA at T  300K
dV  0.01V

  
 5  e1000V /T  1  e 1000V /T  6mA 
Differentiating eqn. (i), we get
 1000  (1000V /T ) 1000
dI   e dV  (6)(0.01)  0.2mA
 T  300
17. 2GM k
Centripetal force is provided by gravitational force F  m; F    mk is some
Lr r
constant.

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mv 2 km
So, 
r r
 v  cons tan t
2 r
T T  r
v
18. Terminal velocity  v viscous force upwards = weight of sphere downwards or
4 
6 rv    r 3  (    ) g
3 
For gold and silver spheres falling in viscous liquid,
vg  g   19.5 18 2 v g 0.2
     or vs    0.1 ms 1
vs  s   10.5  1.5 9 1 2 2
19. BE of deuteron (1 H 2 )  2  1.1MeV  2.2 MeV BE of helium atom
(2 He4 )  4  7 MeV  28 MeV 1H 2 1 H 2 2 He4  Energy released
Energy released  BE of helium  2  BE of deuteron
 28MeV  2  2.2MeV  28MeV  4.4 MeV
 23.6MeV
20.
Here, m  100 g  0.1kg , v  900ms 1 de Broglie wavelength,  
h
mv
6.63  1034 Js
  7.4  1036 m
1
0.1kg  900ms
21. If 𝐹 is the upward force applied, 𝑎 is the net upward acceleraton and 𝑚 is the mass of
each link. Then, F  5mg  5mg or F  5m( g  a)....(i ) . If N is force of interaction
between the top link and link immediately below it, then
ma  F  mg  N
N  F  mg  ma  5( g  a )  m( g  a )(U sin g (i ))
N  4m( g  a )  4  1  (10  2)  48 N
22. Here, m  0.1 kg ; L  1m; A  4.9  107 m 2 ;  140 rads 1
Let l be the extension in wire when mass is suspended from one end of wire. Then
mg L
Y 
A l
When mass attached at the end of wire executes SHM, its angular frequency,
g g
 or l  . Substituting this value of l in eq. (i), we get
l 2
mg L mL 2
Y  
A (g / 2) A

0.1  1  (140)2
  4  109 Nm 2
4.9  107
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23. According to Wien's displacement law,
mT  cons tan t
 (m ) A TA  (m ) B (TB )
T ( ) 1500nm TA
or A  m B  or 3
TB (m ) A 500nm TB
According to Stefan's Boltzmann law, rate of energy, radiated by a black body
E   AT 4   4 R 2T 4[ Here, A  4 R 2 ]
2 42
E A  RA   TA   6cm  4
        (3)  9
EB  RB   TB   18cm 
24. As P  V diagram is a straight line passing through origin, therefore, P  V or PV 1 =
constant
In the process PV x  constant, molar heat capacity is given by
R R
C 
 1 1 x
where x  1 here and   1,4 for diatomic gas.
R R 5 R
C    R   C  3R
1.4  1 1  (1) 2 2
25. For the equilibrium of a small part of semicircular arc subtending an angle of d at the
centre,

 d 
or ,2T sin    BIr0d (dl  r0d )
 2 
T 1.5
B   1T
Ir0 (10)(0.15)
 1
26. For an adiabatic process, TV  1  cons tan t TV
i i  T f V f 1
Substituting the given values, we get
 1
 1 v 
TiVi  aTi  i   a  32 1
 32 
7
For diatomic gas,  
5
7
1
a  (32) 5  (32)2/5  (2) 2  4

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27. 1 2 1 2
u   0 Erms  Brms
2 2 0
1 2 1  Erms
2 
1 2 1 2
  0 Erms   2    0 Erms  Erms 0 0
2 2 0  c  2 2 0
1 2 1 2 2
  0 Erms   0 Erms   0 Erms
2 2
1 1
u ε E B
2 2μ
 (8.85  1012 )  (720)2  4.58  106 Jm 3
28. w
P
t
W  Pt  (0.5W )(5s )  2.5 J
According to work-energy theorem
W  K f  Ki
1
W  mv 2 ( u  0)
2
1 2  2.5 2  2.5
 mv 2  2.5 or v 2  
2 m 0.2
or v  5ms 1
29. A
Modulation index,   m
Ac
Here,
  80%  0.8, Ac  10V
 Am  (0.8)(10V )  8V
30. 30 Here,
Mass of the disc, M  1 kg
Radius of the disc, R  2 m
Moment of inertia of the circular disc about XY is
MR 2
I XY   2kg m 2 (Given)
2
According to theorem of parallel axes, the moment of inertia of the circular disc
about X 1Y1 is
2MR 2
I XY1  I XY  MR   MR 2
2
3 3
 MR 2   (1kg )  (2m) 2  6kg m 2
2 2

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CHEMISTRY
31. For the unpaired electrons of carbon n  2, l  1, m  0, 1, s  ½
32. In PCl2 F3 , F atoms are present in axial as well as equatorial positions. So it is
unsymmetrical and the dipole vectors are not cancelled. Therefore it's dipole moment is
non zero. Other molecules have zero dipole moments.
33. 3RT
The highest velocity is Crms which is and the velocity possessed by the highest
M
2 RT
fraction of molecule is Cmp which is
M
34. 4 KO2  2CO2  2 K 2CO3  3O2
35. PO2  0.5 atm
1 1
KP   2
PO2 0.5
36. Volume  Area  thickness
Mass  Volume  density
80  0.005
 Mass of Ag to be deposited   10.5  0.42 g
10
Ewt  i  t
Amount deposited 
96500
108  3  t
 0.42   t  125.1sec onds
98500
37. Rate  K [ X ]2[Y2 ]1  overall order  2  1  3
38. Conceptual
39. Conceptual
40.

41. 1000
Moles of water  55.5
18
55.5
Mole fraction of water   0.982
55.5  1
PA  0.982  92.5; PA  90.8torr
42. The electrophiles that can attack a benzene ring are

C H 2 CH 2Cl , PhCH 2CH 2 , PhCH 
43. Conceptual
44. Conceptual
45.

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46.

47. CF2Cl2 and NO can form free radical in presence of radiation.

48. Rate= k[Concentration]
Or Rate  [Concentration]
49. Maltose is compound of two   D - glucose units in which Cl of one glucose is linked
to C4 of another glucose unit.

50. Conceptual
51. 1 K 1
Since it is an amphiprotic salt pH   P a1  P a2    (7  10)  8.5
K
2   2
52. Due to identical terminal groups, the number of isomer is three.
53. Number of s-electrons with s   1  4
2
Number of p-electrons with s   1  6
2
Number of d-electrons with s   1 5
2
54. 3 3
EK  nRT  (1)(2)(480)  1440cal
2 2
55. M eq of MnO4  M eq of Fe2 or ,V  N  mole  1000
961.2
or ,V  n  M  0.9612  1000 or ,V  5  0.4  961.2 or ,V   480.6 mL
2
56. Moles of
240  0.2 360  0.2  2
H    0.192  Moles of NaOH required  0.192
1000 1000
V  0.16
 0.192  or ,V  1200mL  1.2 L
1000
57. n 3
1 1 142
Pt    P0    P0   17.75
2 2 8
58.

59. Formula of beryllium pyrosilicate is Be3Si2O7

60. nRT ( R )400
P   100 R  100(0.0821)  8.21
V 4

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MATHEMATICS
61. f ( x)  f ( y)  ( x  y)  xy
x2  y 2 x2  y2
f ( x)  f ( y )   f ( x  y )  xy 
2 2
2 2 2
x y ( x  y)
f ( x)   f ( y)   f ( x  y) 
2 2 2
2 2
x y
Let g ( x)  f ( x)  , g ( y )  f ( y ) 
2 2
g ( x)  g ( y )  g ( x  y )
Let y  o g ( x)  g (o)  g ( x)
g (o)  o put y  x g ( x)  g ( x)  o
x2
g ( x)  odd function  g ( x)  kx  kx  f ( x) 
2
x2
f ( x)   kx  f ( x) is quadratic hence it’s into and many-one
2
62.  a1 a2 
Let A   
b1 b2 
Aadj ( A)  A  adj ( A)  I  4
a1  b2 0
Aadj ( A)  A I
0 a1  b2
(a1  b2 )2  4
63. np  2 npq  1
1 1
q , p n4
2 2
p( x  1)  1  p( x  0)  p( x  1)
64. P( E  F  G c )  P ( E  F )  P ( E  F  G )

P( E  G  F c )  P( E  G )  P( E  F  G )
65. 4 f 3 f 1  xf 3

4 f 1( x)  x   xdx
x2 x2
4 f ( x)  c c0 f ( x) 
2 8
66. 2 2
2 3 1 2 3 1 dy
x  y 0
3 3 dx
dy dx
Now replace by and then solve the differential equation to get orthogonal curve
dx dy

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67. Point of intersection is 0, 1 now find and hence slope of tangents.
At this point and then apply angle between two line formula.
68. use expansion
 2 x2  2 x2
sin( x  (  x  ) (   )  
2  2 1
2 2
9x 9x
1  3x  x 1
2 2
   ,   3,  3
69. 
In(1  b sin x)
f (b)   2 dx
 sin x
2

1
f '(b)   2 dx f (b)   sin 1(b)
 1  b sin x
2
70. 9
Put x 2  t
71.
(4n  4n 2  1)( 2n  1  2n  1)
Tn 
( 2n  1  2n  1)( 2n  1  2n  1)
1
Tn  ((2n  1) 2n  1  (2n  1) 2n  1)
2
72.  13   13  13( 12    12 ) 5S11
 5
 11   11 S11
73. 1 2 3
D  1 3 4  0, D y  0 for k  3
1 1 2
74. 2
2  1 3x  4 y  7 
2
( x  2)  ( y  3)   
2 5 
1
Focus (2, 3) directrix 3 x  4 y  7  0, e 
2
75. (1  x)((1  x)(1  x  x  x ))  (1  x)(1  x 4 )4
2 3 4

76.  10 
0
4cos36  5  1,cot  7   6  4  3  2
 2 
 
77. h( x)  ( f (a )  f ( x))( g (b)  g ( x))e x
h(a)  h(b)  0 c  (a, b), h '(c)  0
78. Either m  1, m '  1or m  1, m '  1
79. Coefficient of x674in(1  x)674 x674 ( x  2)674  2274

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80. DR of normal vector of P3 should be 1, 1,1 so a  b  3
81. r 3r
Line y  3 x can be written as x  , y 
2 2
in parametric form now for point of intersection put these values in the equation of curve
82. Slope of normal
dx
mn    2 x  1
dy
1 1
x  , y  & mn  1
2 16
Equation of normal is y  2  ( x  1)
83. Now
2 2 2
z1  z2  z2  z3  z3  z1
2 2 2
 2( z1  z2  z3 )( z1 z 2  z1z2  z2 z 3  z3 z1  z3 z1)
 58  ( z1 z 2  z1z2  z2 z 3  z 2 z3  z3 z1  z 3 z1) ....(1)
2
z1  z2  z3  0
2 2 2
 z1  z2  z3  z1 z 2  z1 z 2  z2 z 3  z 3 z1  z 3 z1  0
 ( z1 z 2  z1z2  z2 z 3  z 2 z3  z 3 z1  z 3 z1)  29 .....(2)
From (1) and (2)
Maximum value  58  29  87
84. a1e1  a2e2
PF1  PF2  2a1
PF1  PF2  2a2
( PF1  PF2 ) 2  4(a12  a22 )  4a12e12  2(a12  a22 )
PF12  PF22  4a12e12
 2(2 PF12  PF22 )  4(a12  a22 )  4a12e12  2(a12  a22 )
2 2
 a2  e  1 1
 2e1  1     2e12  1   1
2
  2  2 2
 a1   e2  e1 e2
1 1
Let  2 cos ;  2 sin 
e1 e2
9 1
E  9e12  e22  sec2   cos ec 2
2 2
1 1
E  (10  9 tan 2   cot 2  ); Emin  (10  6)  8
2 2

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 SRI CHAITANYA IIT ACADEMY, INDIA 09‐04‐23_ Sr.S60_NUCLEUS & ALL_BT _ Jee‐Main_GTM‐40_KEY &SOL’S
85. p 90 90
 
q 180  1 179
86. Let tan x  t Equation reduces to
4t 2 2
 t2   5  (t 2  1)(t 4  t 2  2)  0
2 2
1 t t
 3
 t  1, 2 x  ,
4 4
87.   ˆ
v  (i  7 ˆj  2kˆ)
9
 3 6 
v  6    3  v.a  9
9
88. Here
x
 cos t  sin t
4
y
and  cos t  sin t
7
Square and adding equation (1) and (2)
x2 y 2
  2(sin 2 t  cos 2 t )
16 7
x2 y 2
   1 ,which is an ellipse
32 14
b2 14 3
eccentricity e  1   1 
a2 32 4
89. 1 a 1
3 2 4 6 66 3 2 4
 (4 x f ( x)  f ( x)  4 x  4 x )dx  7 ;  4 x dx   ( f ( x)  2 x ) dx  7
0 0 0
1 2
3 2 15
 ( f ( x)  2 x ) dx  0  f ( x)  2 x3;  f ( x)dx 
2
0 0
90. Unit place is decided by unit place only
Unit place of (17)1995 and 71995 is same.
Unit place of (11)1995 is 1 since it’s unit place is 1

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 SRI CHAITANYA IIT ACADEMY, INDIA 25-03-22_Sr SUPER 60,ELITE,TARGET&LIIT_Jee-Main-GTM-02_KEY &SOL’S

Sri Chaitanya IIT Academy., India.

A.P, TELANGANA, KARNATAKA, TAMILNADU, MAHARASHTRA, DELHI, RANCHI
A right Choice for the Real Aspirant
ICON Central Office – Madhapur – Hyderabad
SEC: Sr SUPER 60,ELITE,TARGET&LIIT JEE-MAIN Date: 25-03-2022
Time: 09.00Am to 12.00Pm GTM-02 Max. Marks: 300
KEY SHEET
PHYSICS
1 4 2 2 3 1 4 4 5 2
6 4 7 1 8 3 9 3 10 3
11 2 12 1 13 1 14 2 15 2
16 2 17 1 18 3 19 2 20 1
21 90 22 3 23 5 24 1 25 50
26 200 27 2 28 1200 29 1 30 600

CHEMISTRY
31 2 32 2 33 1 34 1 35 4
36 2 37 3 38 2 39 4 40 4
41 3 42 2 43 3 44 4 45 4
46 1 47 3 48 2 49 1 50 1
51 2 52 10 53 5 54 74 55 148
56 75 57 5 58 8 59 5 60 10

MATHEMATICS
61 2 62 1 63 1 64 1 65 4
66 3 67 3 68 3 69 2 70 2
71 2 72 3 73 2 74 4 75 4
76 2 77 2 78 2 79 2 80 1
81 8 82 2 83 6 84 1 85 10
86 8 87 1440 88 5 89 5 90 21

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 SRI CHAITANYA IIT ACADEMY, INDIA 25-03-22_Sr SUPER 60,ELITE,TARGET&LIIT_Jee-Main-GTM-02_KEY &SOL’S

SOLUTIONS
PHYSICS
1. bt
From A  A0e 2 m
b120

6  12e 21  6  12e b60
1 60b ln 2
 e  ln 2  60b  b  .
2 60
0.693
Or, b   1.15  102 kg s 1
60
2. Amplitude of vibration at time t  0 is given by
A  A0e0.10  1  A0  A0
A 1
Also at t  t , if A  0   e0.1t
2 2
t  10 ln 2  7 s
3.  TL
Coefficient of performance  2 
W TH  TL
d 2
263 263 d 2 263 dW 263
 dt        35
dW 298  263 35 dt 35 dt 35
dt
d2
  263 watt
dt
4. H  mL  5  336  103  Qsin k
Qsin k T
 sin k
Qsource Tsource
T
 Qsource  source  Qsin k
Tsin k
Energy consumed by freezer
T 
 woutput  Qsource  Qsin k  Qsin k  source  1
 Tsin k 
Given : Tsource  27 0 C  273  300 K ,
Tsin k  00 C  273  273k
 300 
Woutput  5  336  103   1  1.67  105 J
 273 
5. Heat is extracted from the source in path DA and AB is
3  P V  5  2P V  3 5  13 
Q  R  0 0   R  0 0   P0V0  2 P0V0    P0V0
2  R  2  R  2 2 2
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 SRI CHAITANYA IIT ACADEMY, INDIA 25-03-22_Sr SUPER 60,ELITE,TARGET&LIIT_Jee-Main-GTM-02_KEY &SOL’S
6. Efficiency of engine
T 1
1  1  2 
T1 6
T2 5
  …………. (i)
T1 6
T  62
When T2 is lowered by 62K, then . Again, 2  1  2
T1
T 62 1
1 2   ……….. (ii)
T1 T1 3
Solving (i) and (ii), we get,
5
T1  372 K and T2   372  310 K
6
7. When final image is formed at infinity.
Length of the tube  v0  f e
 15  v0  3  v0  12cm
1 1 1
For objective lines  
f0 v0 u0
1 1 1
    u0  2.4cm
 2   12  u0
8. 1 1.22
Minimum angular separation   
R.P. d
1.22  5000  1010
 0.3  106 rad
2
9. The direction in which the first minima occurs is  (say). Then e sin    or e  

or,      sin  when  small 
e


Width of the central maximum  2b  e  2b. e
e
10. Position of first minima = position of third maxima i.e.,
1 1D  2  3  1 2 D
  1  3.52
d 2 d
11. 1 I 1
I '  cos 2   or cos     550
2 6 3
12. No light is emitted from the second Polaroid, so P1 and P2 are perpendicular to each
other

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 SRI CHAITANYA IIT ACADEMY, INDIA 25-03-22_Sr SUPER 60,ELITE,TARGET&LIIT_Jee-Main-GTM-02_KEY &SOL’S
P3

 900  
P1 P2

Let the initial intensity of light is I 0 . So intensity of light after transmission from
I I
first Polaroid  0 . Intensity of light emitted from P3 I1  0 cos 2 
2 2
Intensity of light transmitted from last Polaroid i.e. from
I I 2 I
 2

P2  I1 cos 2 900    0 cos 2  .sin 2   0  2sin  cos   0 sin 2 2
8 8
13. The situation is as shown in the figure. As the point O lies on broad, side position
with respect to both the magnets. Therefore,

The net magnetic field at point O is

Bnet  B1  B2  BH
 M m M 
Bnet  0 1  0 2  BH  0  M1  M 2   BH
4 r 3 4 r 3 4 r 3
Substituting the given values, we get
4  107
Bnet  1.2  1  3.6  105
3

4  10  102 
107
  2.2  3.6  105  2.2  104  0.36  105
3
10
 2.2  104  0.36  104  2.56  104Wb / m2
14. V  V  I R  15  7  I  2  103  i  4mA
CE C L C c
i 4
   C  iB   0.04mA
iB 100
15. Y  AB. A Y  AB  A  0  0  0
16. du 1 1
 0 E 2 1.02  108   8.85  1012 E 2
dv 2 2
E 48
E  48V B    16  108  160  109  160 nT
C 3  10 8
17. V  5 1  0.6cos 6280t sin 211  104 t
m    
Vm  55  3cos6820t  sin 211 104 t  
Vmax  5  3  8 Vmin  5  3  2
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 SRI CHAITANYA IIT ACADEMY, INDIA 25-03-22_Sr SUPER 60,ELITE,TARGET&LIIT_Jee-Main-GTM-02_KEY &SOL’S
18. 
E  c  3.1eV
e
19. Retentivity  1.0T
Co – ercivity = 50 A/m
20.

A B C
0 0 0 0
1 0 0 0
0 1 0 0
0 0 1 0
1 1 0 0
1 0 1 0
0 1 1 0
1 1 1 1
21. Efficiency of Carnot engine
T 1 T T2 1 9 T1 10 T 
n  1  2 i.e.,  1  2  1     w  Q2  1  1
T1 10 T1 T1 10 10 T2 9  T2 
 10  1
i.e., 10  Q2   110  Q2    Q2  90 J
9  9
22. B
Magnetic field in solenoid B  0 ni   ni
0
B Ni
(Where n  number of turns per unit length)  
0 L
100i
 3  103   i  3A
10  102
23. Erms  755
1 1 2
The average total energy density  0 E02  0  2 Erms  0 Erms
2
2 2
2
 8.85  1012   755  5.027  106 J / m3
24. 96
No. of electrons reaching the collector, nC   1010  0.96  1010
100
n e
Emitter current, I E  E
t
n e
Collector current, IC  C  Current transfer ratio,
t
I n 0.96  1010
 C  C  10
 0.96
I E nE 10

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 SRI CHAITANYA IIT ACADEMY, INDIA 25-03-22_Sr SUPER 60,ELITE,TARGET&LIIT_Jee-Main-GTM-02_KEY &SOL’S
25. 24 8
Here Vmax   12 mV and Vmin   4 mV
2 2
V  Vmin 12  4 8 1
Now, m  max     0.5  50%
Vmax  Vmin 12  4 16 2
26. Condition for minima
d
d sin   n n

n 6 105
sin   
dt 6  107
n
1  100
dt
Total number of minima on one side = 100
Total number of minima =200
27. 12
With battery polarity, D2 is off. So, I   2A
42
28. R V d
Vc  R c
d 
29.  Pr 4
Q
8 
r1  a P1  n  g
r2  a / 2 P2  n  g
l1  
2   Q1  16
Q r 4
4
Q1  r1 
 
Q2  r2 
Q2  1CC
30. 900

Q1  Q  W

W=1200 J
Q2  Q
300

Q1 T1
For Carnot engine 
Q2 T2
Q  1200 900

Q 300
Q  1200  3Q
Q  600 J .

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 SRI CHAITANYA IIT ACADEMY, INDIA 25-03-22_Sr SUPER 60,ELITE,TARGET&LIIT_Jee-Main-GTM-02_KEY &SOL’S

CHEMISTRY
CH 2OH CH 2OH
31. 5
H H H H H H
4 1
HO OH H 0 OH H OH

3 2
H OH H OH
  D  glu cos e   D  glu cos e

Maltose.
6 6
CH 2OH CH 2OH

5 5
OH H H H OH
4 1 4 1
0
H OH H H OH H H
3 2 3 2
H OH H OH
  D  Galactose   D  glu cos e

Lactose
32. h1  h0  K .E1
h 1  0  K .E1
h2  h0  K .E2 
h 2  0  K .E2
1  0 1  K  2
 2  0  K 1  0   0  1
2  0 K K 1
33. 1 poise  101 kg m 1s 1
1 poise 101 pa sec
Viscosity coefficient  1.2  103 pascal sec
1 millipoise  103 poise 1.2  103  10 poise
103 millilpoise = 1 poise 1.2  103  10  103  12 milli poise
34. ice  water
 more
volume  less
water 
On increasing pressure, equilibrium shifts forward
35. 673K
CO g   H 2O g  CO2 g   H 2 g 
catalyst
This reaction is called water gas shift reaction.
36. Global warming and acid rain
37. I – Freon (12) CCl2 F2 is one of the most common ions in industrial use.
Cl
Cl C
Cl
CH

II) Cl Cl

III) When CCl4 is released into the air it rises to the atmosphere and depletes the
ozone layer
IV) CH 2Cl2 is used as a solvent.
38. Vitamin K deficiency increases blood clotting time.
39. H 2O D2O
Dielectric constant 78.39 78.06

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Boiling point 373 374.4
40. Nylon 6,6 is a condensed polymer
41. 1.4 NV 1.4  0.5 2.5  102
%N  NV  milli equivalent of NH 3    25
Worg 28  103 103
42. x 1
log  log k  log p
m n
43. 2d sin   n
44. CrO2 is ferro magnetic compound
45. N 2O and NO illustrate law of multiple proportion
46. Lanathanide contraction
47. Low spin d 5 t2g 5eg 0 C.F .S .E   0.4  5   o  2o
48. I is  Pt  NH 3 5 Cl  Cl3 it gives 4 ions in water.
49. Ofloxacin is Bactericidal antibiotic
50. He  O2 mixture is used by deep sea divers since Helium is much less soluble in
blood than nitrogen.
51. NH COCH Cl2
O2 N CH  CH  CH 2  OH

OH

Chloramphenicol
52. Dissolved oxygen can reach 10 ppm
53. Conceptual
54. y  3 z  5 x  0.2
a  0.2
5 3
 5  0.2
0.2
15
1
0.2
75  1  74
55. x  143 m.mol L1
y  5 m. mol L1
56. 12 WCO 2
%c   100
44 Worg
12  0.55
 100  75
44  0.2 
57. Bakelite, nylon6,6 are not addition polymers
58. It gives octa acetyl derivative
59. NCERT
60. HOOC   CH 2 4  COOH

H 2 N  CH 2 6  NH 2

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MATHEMATICS
61. Use the property
~  p  q   p ~ q
Hence (2) is correct option
62. Statement : p  q
Contra positive of statement  Nq  Np
63. .
p  Nq q Np  q  p  Nq   q 
 Np  q 
F T F T
T F F T
F T F T
F F F F

a) p  q pq p  Nq Np  q
T T T F
T F T F
T F F F
F F T F
 Clearly ‘a’ is ans
64. ~  p  q  ~ p ~ q (By De’Morgan’s law)
 A is a tautology. Also R is true. R is correct explanation for A.
65. Write down the truth table.
66. n 1
Mean x 
2
n 1 n 1 n 1 n 1
  2  3  .....  n 
2 2 2 2
 MD 
n
n 1 n 1 n 1 n 1
  2  ....  
2 2 2 2

n
n 1 n 1 n  3 n 1 2n  2 n  1 n 1
    ....    n
2 2 2 2 2 2 2

n
 n  1 n  1  
 1 
2 n  3 n  1  n  1  2  2  2 
 1  2  .....      terms  
n n 2  2  n 2 
 

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n  n  1
[ sum of first n natural numbers  ]
2
2 1  n  1  n  1   1  n 2  1  n 2  1
 .       .
n 2  2  2   n  4  4n
67. R is reflexive as  3,3 ,  6,6  ,  9,9  , 12,12   R
R is not symmetric as  6,12   R but 12,6   R
R is transitive as the only pair which needs verification is  3,6  and
 6,12  R   3,12   R .
68. 30
Since  AL  S
L 1
Let n  s   x
10 x  150
x  15
n
 BL  S
L 1
3n  9 x
9x
n  3x  45
3
69. 2
Given, n  18,   x  5  3 and   x  5   43
  x  5
 Mean  A 
18
3
5  5  0.1666  5.1666  5.17
18
2 2
  x  5    x  5 
And SD   
n  n 
2
43  3 
  
18  18 
2
 2.3944   0.166   2.3944  0.2755  1.59
70. Median will go up by 2 and S.D. will remain same.
71. ACB         
In ABC , applying sine rule
AB CA

sin          sin 
a sin 
Clearly CDA   , DCB   ,  CA 
sin      

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Now, in CAD, applying sine rule
CA CD CA.sin  a sin  sin 
  CD  
sin  sin  sin  sin  sin      
C D
 


 

A a B

72. Let h be the height of a tower.

Since, AOB  600
h C
O 900
a 60 0 0
a 30
B a A

So, OAB is an equilateral.

 OA  OB  AB  a
In OAC ,
h a
tan 300   h 
a 3
73. In APC ,
 r 
sin   AC  r cos ec
2 AC 2

r
C
 /2
H
 P

A B

BC 
In ABC ,sin    H  r sin  cos ec
AC 2
74. Let l be the length of the ladder, then AD  BE  l
E
y

wall
 
C B x A
CE
In BCE ,sin    CE  l sin 
BE
BC
And cos   BC  l cos
BE
CD
In ACD,sin    CD  l sin 
AD
AC
And cos    AC  l cos 
AD

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 x  AC  BC  l cos   l cos 
 x  l  cos   cos   …………………… (i)
And y  CD  CD  y  l sin   l sin 
 y  l  sin   sin   …………….. (ii)
From Eqs. (i) and (ii), we get
       
2sin   sin  
x cos   cos   2   2 
 
y sin   sin         
2cos   sin  
 2   2 
x    
  tan   …………………. (iii)
y  2 
   
 x  y tan  
 2 
 Option (1) is correct.
   
1  tan 2  
 2 
 cos     
   
1  tan 2  
 2 
x2
1 2
y
 cos      2
[from Eq. (iii)]
x
1 2
y
y 2  x2
 cos     
x2  y 2
  
y 2  x 2  x 2  y 2 cos    
 Option (2) is correct.
   
2 tan  
 2 
 sin     
   
1  tan 2  
 2 
x
2.
y
 sin      [From Eq. (iii)]
x2
1 2
y
2 xy
 sin     
x2  y 2

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  x2  y2  sin      2 xy
Option (3) is correct.
75. P

Q
 
 

B c A
Let P.Q be objects. A.B observation point. When PQ subtend maximum angle  at
A. AB is tangent to circle passing through P, Q, A.
Let APB  
          1800
180    
 
2
From PQA & BQA
PQ AQ AB AQ
 & 
sin  sin  sin     sin 
c sin  sin 
 PQ 
sin  sin    
c sin  sin  2c sin  sin 
 
            cos   cos 
sin    sin   
2 2  2 2 
76.
20 m 20 m
450
300
x  20 m p
20 1
tan 300 
20  x

3
 20 3  20  x  x  20 3  1  
77. EF 28.5
From BFE ,cot 600  , EF 
28.5 3
BE
From BDE ,cot 300  , BE  28. 3
28.5
BF  BE  EF
28.5  1   2  57
 28.5 3   28.5  3    28.5    19 3
3  3  3 3
D

28.5

1300 600
B E
1.5 1.5

A C

78. A  1,2,3,.....,100
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And B  C  3k  1: k  event
 B  C  7,13,19,......,97
So, A  B  C  7,13,19,.......97
16
Sum of all elements   7  97   832
2
79. 0  x  y  1
 is symmetric
0  y  x  1
Let 0  x  y  1 ……….. (i)
1,2   R and  2,3  R satisfying the equation (i)
But 1,3  R not satisfied
Hence 0  x  y  1 is symmetric but not transitive relation
80.  p  q    r  q   p 
~  p  q    r  q   p 
~  p  q     r  q    p  q  
  ~  p  q    p  q    ~  p  q    r  p  
 t     p  q    r  p  
~  p  q    r  p 
  p  q   r  p 
81.  2, 3
 4, 1
82. A1  1,1 2,2  3,31, 2  2,1
And A2  1,1 2,2  3,31,2  2,1 2,3 3,1 3,2 1,3
83. Given, n1  20,1  5, x1  17 and
n2  20, 2  5, x2  22
We know that,
2
n112  n2 22 n1n2  x1  x2 
 
n1  n2  n1  n2 
2 2
20   5  20   5  20  20 17  22 
 
20  20  20  20 2
1000 400  25 25 125
   25    31.25  5.59
40 1600 4 4
84. Sum of frequencies
2
x  2  x  x 2   x  1  2 x  x  1  60

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 2 x  2  x 2  x 2  1  2 x  2 x  x  1  60
 2 x 2  7 x  60
 2 x 2  7 x  60  0
 2 x 2  15 x  8 x  60  0
 x  2 x15  4  25  15   0
  2 x  15  x  4   0
15
 x,4
2
15
 x [inadmissible]  x  I 
2
xi fi di  xi  3 fi di fi di2
0 2 3 6 18
1 4 -2 -8 16
2 16 -1 -16 16
A=3 25 0 0 0
4 8 1 8 8
5 5 2 10 20
Total fi  60 fi di  12 f i di2  78
fi di  12 
Mean  A  3   2.8
f i  60 
2 2
fi di2  fi di  78  12 
       13  0.04  1.26  1.12
fi   f i  60  60 
85. xi
Given, n  100, x  40,  10 and x  40   40
n
xi
  40  xi  4000
100
Now, Corrected xi  4000  30  70  3  27  4030  100  3930
3930
Corrected mean   39.3
100
xi2 2 xi2
Now,  2    40   100   1600
n 100
 xi2  170000
Now, corrected
2 2 2
xi2  170000   30    70   32   27   164939
164939 2
 Corrected     39.3  1649.39  39.3  39.3
100

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 1649.39  1544.49  104.9  10.24

86. Since, the first 10 positive integers are 1,2,3,4,5,6,7,8,9 and 10
On multiplying each number, we get
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
On adding 1 in each number, we get
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
9  10
 xi  0  1  2  3  4  5  6  7  8  9   45 and
2
2 2 2 9  10  19
xi2  02   1   2   .....   9    285
6
2
285  45  285 2025 2850  2025
 SD      .  8.25
10  10  10 100 100
2
Now, variance   SD    8.25   8.25
2
 
87. 3
For OA, A, OA1   1 km
0
tan 60
 II nd Position  I st Position 
B A

3km

30 0 600
B1 A1 O

3
For OB1, B, OB1   3km
tan 300
As a distance of 3  1  2 km is covered in 5 seconds.
Therefore the speed of the plane is
2  3600
 1440 km / hr
5
88. R   2,3 ,  3,3  ,  2,3 ,  3, 2  ,  3,4  ,  4,3 ,  4,4 
89. 28  2 x  22  40
Eng
11-x x 12-x Mat

2
3 12

5
Eco
2 x  10 x5
A
90.

9
90
4 8E
Tan 13
B 7 C

cos c  2 / 7 (from cosine rule in triangle ABC)

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BE=7 (From cosine rule in triangle BCE) h  21 m

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.SUPER60,ELITE,TARGET&LIIT JEE-MAIN Date: 20-05-2022
Time: 09.00Am to 12.00Pm GTM-13 Max. Marks: 300
KEY SHEET
PHYSICS
1) 1 2) 3 3) 2 4) 1 5) 4
6) 3 7) 4 8) 4 9) 3 10) 3
11) 3 12) 2 13) 2 14) 2 15) 2
16) 4 17) 1 18) 1 19) 1 20) 2
21) 35 22) 8 23) 4 34) 0 25) 9
26) 3 27) 6 28) 20 29) 198 30) 6

CHEMISTRY
31) 1 32) 1 33) 1 34) 1 35) 1
36) 3 37) 2 38) 1 39) 3 40) 3
41) 4 42) 3 43) 1 44) 4 45) 2
46) 3 47) 2 48) 1 49) 3 50) 2
51) 68 52) 2 53) 434 54) 65 55) 0
56) 2 57) 6 58) 28 59) 5 60) 3

MATHEMATICS
61) 1 62) 3 63) 4 64) 2 65) 1
66) 2 67) 3 68) 3 69) 2 70) 2
71) 1 72) 2 73) 1 74) 2 75) 1
76) 1 77) 4 78) 1 79) 3 80) 1
81) 41 82) 0 83) 18 84) 576 85) 132
86) 7 87) 9 88) 2 89) 3 90) 54

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SOLUTIONS
PHYSICS
1. Particle velocity v p is related o the displacement of the particle from the mean
position as
 v
v p  2v A 2  y2 v p  2   A 2  y 2
 

2 2 2 3
  0.1  0.1   0.05  jm / s
0.5 50
Since the wave is sinisodial moving in positive x-axis the point will move parallel to
y-axis therefore options (c) and (d) are ruled out. As the wave moves forward in
positive X-direction, the point should move upwards i.e. in the positive Y-direction.
2. Since r .p  0

E must be antiparallel to p  Ê is parallel to ˆi  3jˆ  2kˆ
3. Potential energy of a dipole is given by
U   P.E  PEcos 
[Where   angle between dipole and perpendicular to the field]
  
  1029 103 cos 450  0.707  10 26 J  7  10 27 J
4.  
Equivalent capacitances in series combination C1 is given by
1 1 1 CC
1
   C1  1 2
C C1 C2 C1  C 2
For parallel combination equivalent capacitance
C11  C1  C 2
For parallel combination
q  10  C1  C2 
q1  500 C
500  10  C1  C2 
C1  C2  50 F ………. (i)
For series combination
C1C2
q 2  10
 C1  C2 
CC
80  10 1 2 from equation ……… (ii)
50
C1C 2  400 ……… (iii)

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From equation (i) and (ii)
C1  10F C2  40F
5. Number 2 is associated wit the red colour. This colour is replaced by green.
 Colour code figure for green is 5 New resistance  500 
6. Given resistance of each resistor ‘R’
5 4 3
R PQ  R,R QR  R &R PR  R  R PQ is maximum
11 11 11
7.

Let R be the resistance of the whole wire potential gradient of the potentiometer
wire
dV I  R  60  R 
AB      mv / m
d    AB 
 dV  60  R
VAP     AP   1000mV  VAP  50R mV
 d AB  1200
Also, VAP  5V (for balance point at P)
VAP 5
R    100
50  103 50  10 3
8. v

+q

Length of the circular path,   2r

q qv
Current, i  
T 2r
qv
Magnetic moment M = Current × Area  i  r 2   r 2
2 r
1
M  q.v.r
2
mv
Radius of circular path in magnetic field, r 
qB
1 mv mv 2
KtM  qv  M
2 qB 2B
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Direction of M is opposite of B therefore

 mv 2 B
M
2B2
(By multiplying both numerator and denominator by B).
9. V  i g  G  R   4  103  50  5000  20V
10.  0  i1i2 i1i 2   0i1i 2
F     a 
2 a 2a 4
0
11. Here,   30 ,   0.018N  m,B  0.06T
Torque on a bar magnet:
  MBsin 
0.018  M  0.06  sin 300
1
 0.018  M  0.06   M  0.6A  m 2
2
Position of stable equilibrium   00 
Position of unstable equilibrium   1800 
Minimum work required to rotate bar magnet from stable to unstable equilibrium
U  U f  U i   MBcos1800   MBcos00  
W  2MB  2  0.6  0.06
 W  7.2  10 2 J
12. B  N
Corecivity, H  & B   0 ni  n  
0  
N 100
or, H  i   5.2  2600A / m
 0.2
13. d
According to faraday’s law of electromagnetic induction, c 
dt
di 15 5
L  25  L   25or L  H
dt 1 3
Change in the energy of the inductance,
1 1 5 5
   
E  L i12  i 22    252  10 2   525  437.5J
2 2 3 6
14.

Let angle between the two mirrors be  .

Ray PQ || mirror M1 and Rs || mirror M 2
 M1Rs  ORQ  M1OM 2  

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Similarly, M 2QP  OQR  M 2OM1  
1800
 In ORQ,3  1800     600
3
15. From lens maker’s formula,
1  1 1 1 1 1  1
    1    ;   1  1   
f  R1 R 2  f1    R  2f 2
Similarly, for plano-concave lens
1  1 1
   2  1   
f2  R  
1 1
Dividing by we get,
f1 f2
 1  1    2  1
or 21   2  1
R 2R
16. 1  1 1  3R
 R 2  2 
1  2 4  16
1  1 1  5R
And  R 2  2 
1  2 3  36
 5R / 36 20
Now 1  
 2 3R / 16 27
17. Let N1 & N 2 be the number of radioactive nuclei of substance at anytime t.
N1  at t   N 0e5t   i
N 2  at t   N0 e t   ii
Dividing equation (i) ny (ii), we get
N1 1
 2  e4 t  4t  2
N2 e
2  1
t  
4  2 
18. P  P  P R 
Power gain  60  10log  0   6  log  0   0  106  2  out 
P
i P P i iR  in

 10000 
 106   2  [as R out  10,000  R in  100  ]    100
 100 
19. Given: modulation index m = 80% = 0.8
E c  14V,E m  ?
E
Using, m  m  E m  m  E c  0.8  14  11.2V
Ec
20. As we know, average power Pavg  Vrms I rms cos 

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 V  I   100   20 
  0   0  cos   
 2  2  
 2   2  cos 450    450  
1000
Pavg  watt
2
Wattless current I  I rms sin 
I 20
 0 sin    sin 450  10A
2 2
21. Densit of wire,   9  103 kg cm 3
Young’s modulus of wire, Y  9  1010 Nm 2
Strain  4.9  104
Stress T / A T
Y    Y  Strain  9  109  4.9  104
Strain Strain A
Also, mass of wire, m  A
m
Mass per unit length,    A

Fundamental frequency in the string
1 T 1 T 1 9  1010  4.9  104
f  
2  2 A 2  1 9  103
1 1
 49  101053   70  35Hz
2 2
22. dV
E  8X  8
dX
23. Time Constant, T= RC
2
1
Impedance Z= R    , Given Z= R 1.25
2
 C 
2
 1  2 2 2
 R 1.25  R     RC    1000 ms  RC  4ms
 C   500
24.  A  m   60  m 
sin   sin 
 2   2 
 , 2
sin A / 2 sin 60 / 2
60  m
A  60 o

for an equilateral triangle 
2
 45o  m  30o

The condition is for minimum deviation. In this case the ray inside the prism
becomes parallel to base. Therefore the angle made by the ray inside the prism with
the base of the prism is 0o
25. In Youngs double slit experiment, intensity at a point is given by

I  I 0 cos 2 ..... 1
2
Where ,   phase difference,
2
Using phase difference,    path difference


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For path difference , phase difference 1  2
 
For path difference , , phase difference 2 
6 3
Using equation (i),
   2 
cos2  1  cos2  
I1  2  2 K 1 4 3K 9K
      I2    n  9.
I2 
2 2 2  I2 3 3 4 12
cos   cos  
 2  6
26. h 1
We know that,   or,  
2mqV qm
p m q  4 2
     83
 mp q p 1 1
27. From question,
B0  20 nT  20  10 9 T ( Velocity of light in Vaccum C = 3  108 ms 1 )
     
E0  B0  C E0  B . C  20  10 9  3  108  6 V / m.
28. 20 R

We know, 200V
V2 V 2 100  100
Power P =  R bulb    20
R P 500
This would be possible only when R = 20 is in series with the bulb resistance
 R bulb  20 because in that case both resistances will share equal p.d of 100 V each.
29. For obtaining secondary minima at a point path difference should be integral
n
multiple of wavelength d sin   n  sin  
d
d 6  105
For n to be maximum sin   1 , n    100
 6  10 7
Total number of minima on one side = 99
Total number of minima = 198.
30. The observer and source are moving towards each other. The image of the source
serves as source of reflected sound .
The frequency of sound reflected from the wall

Vs 5ms1 Vs  5ms1

(The image of the bus

formed by the wall
behaves as source)

 v  v0   342  5 
v'  v   v '  200   206 Hz.
 v  vs   342  5 
 Frequency of beats = v '  v  206  200  6 Hz.

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CHEMISTRY
31. In fifty minutes, the concentration of H 2O 2 decreases from 0.5 to 0.125 M or in one
half-time, concentration of H 2O 2 decreases from 0.5 to 0.25 M. In two half-lives,
concentration of H 2O 2 decreases from 0.5 to 0.125 M or 2t1/ 2  50 min, t1/ 2  25 min
 0.693 
 k min 1
 25 
d O2  1 d  H 2O 2  k  H 2O 2 
Or    6.93  10 4 mol min 1
dt 2 dt 2
32. According to Kohlrausch’s law,
 eq Na 2SO4   eq  NaCl   eq K 2SO4   eq

KCl
 123.7  152.1  147.0  1cm 2eq 1  128.8  1cm 2 eq 1
33. According to Dalton’s law, p  x A p 0A  x Bp o B Or  2  3
550  p 0 A    p o B  
 5  5
or 2p 0 A  3p o B  2750 ......  i   When 1 mole of B is added to it,
 2  4
560  po A    p o B  
 6  6
or 2p o A  4p o B  3360
From Eqs (i) and (ii), we get
po A  460 mm
34. Given, angle of diffraction ( 2 )= 90o ,   45o
Distance between two planes,
0
d  2.28 A, n  2
Bragg’s equation is n  2d sin 
2    2  2.28  sin 45o
  1.612
35. MZ
Mass of one unit-cell (m)= Volume  density = a 3  d  a3 
N 0a 3
Mz 58.5  4
m g
N0 6.02  1023
1 6.02  10 23
 Number of unit cell in 1 g    2.57  1021
m 584  4
36. As coagulating power is inversely proportional to coagulation value, the ratio of
their coagulating powers will be
Coagulating power of AlCl3 Coagulation Value of NaCl 52
= =  559 :1
Coagulating power of NaCl Coagulation Value of AlCl3 0.093
37. Conceptual
38. XeF4 oxidises potassium iodide.
XeF4  4l   2l2  4F  Xe
XeF6 oxidises hydrogen like other xenon fluorides.
39. As the size of the halogen atom increases from F to l, the bond length of H – X
increases. This increase in bond length decreases the bond strength and thus the
thermal stability. Therefore, the order is
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HF  HCl  HBr  Hl
40. N
P
 Group 15
As Valence
 =5
Sb Electrons
MCl3 has sp3 hybridized M-element with one lone pair.
Lone pair and bond pair repulsion decreases bond angle. However, the bond pair of
electrons are much farther away from the central atom in SbCl3 than they are in NCl3
. Thus, lone pair causes even greater distortion of PCl3 , AsCl3 and SbCl3 . Hence, bond
angle decreases from NCl3 (Maximum) to SbCl3 (minimum)
41. II is most reactive as it produces an aromatic carbocation while IV is less reactive as
it produces a non-resonance stabilized carbocation.III is least reactive than I as
former involve an anti-aromatic carbocation.
42.

Electron releasing group decreases while electron withdrawing group increases

acidic strength by destabilising and stabilizing the phenoxide ion formed
respectively
43.

The very first reaction in the above road map looks like Kolbe’s reaction which
results to salicylic acid as :

The salicylic acid with acetic anhydride[  CH3CO 2 O ] in the presence of catalytic
amount of conc. H 2SO 4 undergoes acylation to produce aspirin as:

Aspirin is a non-narcotic analgesic (pain killer).

44. The reaction takes places as follows :

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45.

46. Teflon is Thus, it is fully fluorinated polymer.

47. Glucose and fructose give positive Tollen’s test as they reducing sugar because of
the presence of free (—CHO) group, expl van Ekenstein rearrangement
48. 0.2 per cent solution of phenol acts as an antiseptic and its 1% solution is a
disinfectant.
49.  o  0.4  n t  0.6  n e [where, n = number of electrons]
2g g

a) For high spin d ,  o  CFSE   0.4  4  0.6  2  0.4

6

b) For low spin d 4 ,  o  CFSE   0.4  4  0.6  0  1.6

c) For low spin d 5 ,  o  CFSE   0.4  5  0.6  0  2.0
d) For high spin d 7 ,  o  CFSE   0.4  5  0.6  2  0.8
The calculations reveal that maximum CFSE is for low spin d5 configuration.
50. Co  NH3 5 Cl3 is an octahedral complex ionising in aqueous solutions as
2
Co  NH3  Cl  Cl2  Co  NH3  Cl   2Cl
 5   5 
0.1
Moles of Co  NH 3 5 Cl  Cl2   100  0.01 mol
1000
Moles of AgCl formed  2  moles of Cl  in complex  0.01  2  0.02 mol
51. For isotonic solutions, 1   2
n1RT n 2 RT n n
  1 2
V1 V2 V1 V2
w1 w 2 w1 w
Thus,  ,  1, 2  5
V1m 1 V2 m 2 V1 V2
1 5
m1  ?, m 2  342, 
m1 342
342
m1   68.4  x
5
i.e. molar mass of solute x is 68.4  68 g mol-1
52. i.t
Equivalents of tin =
96500
22.2 2  5  60  60
Or 
eq.wt 96500
Eq. Wt = 59.5
 Valency of tin
at. wt 118.69
=   2 (an integer)
eq. wt 59.5
Therefore, Sn 2  2e  Sn
Thus, oxidation state of tin (Sn) in salt is +2.
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53. k1  k 2
 2000   1000 
15  T  14  T 
10 e  10 e
2000 1000
Or 15   14 
T  2.303 T  2.303
1000
Or 1
T  2.303
 T  434.22 K
According to Arrhenius equation,
ln k  ln A  E a / RT
Ea
Or log k  log A 
2.303RT
Differentiating both sides, we get
d log k Ea

dt 2.303RT 2
Both statements(a) and (b) are correct.
54. Formula of sulphide is MS because oxidation number of metal ion being +2.
3
MS  O 2  MO  SO 2
2
 M  32  M  16
1.164 0.972
M  32 M  16

1.164 0.972
Or M = 65 g
55. Conceptual
56. Complex  Co  NO 2 3  NH 3 3  is of type MA 3B3 which have 2 geometrical isomers that
is fac and mer isomers.
57. The number of glycine units present in the decapeptide is 6. Given, molecular mass
of decapeptide = 796. Since, it gives glycine, alanine, phenylalanine on hydrolysis.
Thus, it needs 9 molecule of water for hydrolysis.
The mass of product  796  18  9  958
47
So, the contribution by glycine   958
100
450
 Number of glycine unit  6
75
58. 64  103
Moles of CaC 2   1  103 mole
64
 From the balanced chemical equation,
Moles of C 2 H 4  moles of C 2 H 2  moles of CaC 2  1  103
1
 Moles of polythene   1  103
n
1
 Weight of polythene   1  28 n kg  28 kg
n
59.

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Since, each COCH3 group displace one H atom in the reaction of one mole of

with one  NH 2 group , the molecular mass increases with 42 unit.

Since, the mass increases by (390 -180)=210 hence the number of  NH 2 group is
210
 5.
42
60. The given reaction is an example of repeated aldol condensation followed by
Cannizzaro reaction.

In the last step, formaldehyde is oxidised and the other aldehyde is reduced giving
the desired product.

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MATHEMATICS
i 2 r
61. 20
ar  e
a1 a3 a5 e e 3 e5 e e 3 e5
a r  e r  a 7 a9 a11  e 7 e9  e11  e6 e e 3 e5  0
a13 a15 a17 e13 e15 e17  e13 e15 e17 
62. Given equation can be rewritten as
2
2 2  5x  12y  17 
 x  1   y  3   
13
Here, focus is (1,3), directrix is 5x – 12y + 17 = 0
5  36  17 14
 the distance of the focus from the directrix  
25  144 13
14 28
 Latusrectum  2  
13 13
63. n
2
  x  k
k 1
0

64.  5
As 1  2x 2  3  2 x  2, 
 2
5/2

   2x 2  3 dx  0 x  0, 2
2

 g  x   0 should have at least one real root in (0, 2)  g  x   0

1

65.  2  sin x  dy dy  cos x

We have,     cosx, y  0  1    dx
 y  1  dx y 1 2  sin x
4  4 4 4 1   1
  y  1  at x   y  1    y  1  y  
sin x  2 2   2 3
sin  2 3 3 3
2
66. Distance CP = CQ = OC = 5 units
3
Slope of OC 
4
4 4
Slope of PQ   tan   
3 3

Co-ordinates of point P and Q are  4  5cos ,3  5sin  and  4  5cos ,3  5sin 
  4  3,3  4 &  4  3,3  4  1,7  &  7, 1
69
67. C3r 1  69C3r  69C r2 1  69Cr 2

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70
C3r  Cr 2 ........  C r  Cr 1  n 1Cr 
70 n n

r 2  3r or r 2  3r  70
r = 0, 3, 7, -10
Valid values of r = 3 or 7.
68.

3
69.

No of ways of getting sum 10 is = coefficient of x10 is x  x 2  ......x 6   27
27 1
Probability  
216 8
70. xdy  ydx
x 2 
 xe x  e x  1 dx 
y
 xe x  x  c x 1 y  e 1 c  0
x

y  x 2 ex  1 
71. e2x e 2x e 2x 1
2 2x 2
 x e dx  x . 2  2x. 4  2 8  c  4 2x  2x  1 e  c 
2 2x

1

f  x   2x 2  2x  1
4

1
Min value of f(x) is
8
72. A tangent to ellipse is
x y
cos   sin   1
3 2
It intersects x = 3 and x = -3 at points
 2 1  cos    2 1  cos  
P  3,  &Q  3,
 sin    sin  
The circle with PQ as diameter is
 2 1  cos    2 1  cos  
 x  3 x  3   y    y 0
 sin    sin  
 4 
 x 2  y2  5   y0
 sin  
It is a family of circles passing through the intersection of the circle x 2  y 2  5  0
and the line y = 0, which is  5,0  
73. Since E1  E 2  E1  E 2 &  E1  E 2   E1  E 2   
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1

P  E1  E 2    E1  E 2   P    0   4
74. 1 1
P  A  B  P  A  .P  B  3K 2   K 
3 3
P(exactly one of A, B occurs)  P  A   P  B  2P  A  B
75. 30 30
S   r  C r  30 r 29C r 1
2 30

r 1 r 1
30 30
 
S  30   r  1 C r 1   29 Cr 1 
29

 r 1 r 1 
 30 
S  30   29  28Cr 2  229   30  29  228  2 29 
 r 1 
76. Since,   z  2z  3z  18z  12  0   n  2 n 1  3 n 2  18  12  0
2 n 1 n2

  
  n  1  2  n 1  1  3  n 3  1  18    1   
  n  1   n 1  1   n 3  1
  2  3    1   18
   1     1 

  
 1     2  .......   n 1  2 1     2  ......   n 2 

3 1     2  ......   n 3  18 

 1      ......  
2 n 1
  2 1      .....    2 n 2


3 1    
2
 .....     18
n 3

As n   &   1
1 2 3 1
    18  3
1  1  1  1 
1 2
1      &  
3 3
77.
 1 x2  1  
2
e x 2x  x 3 1
 e    2x
2 2 2 2 2
 3 x 2 
 3  x 3 x   
  dx  1
2
e x 2x  x 3
2
ex
Hence,  3  x  2

2 2 3  x2
x

n
78. Let E   a  b  c   a   b  c 
n

 2 4  n  1 
 E  2  n C0 a n  n C2 a n  2  b  c  n C4 a n  4  b  c  .......   terms 
  2  
 n  1
The number of terms in the expansion will be  terms
 2 
 n is odd
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 n  1
Hence, the number of Distinct terms are: 1   2  1   4  1  ....... terms i.e.
 2 
2
 n  1
 
2 
79. The equation of a straight line which is at a unit distance from the origin is
x cos   ysin   1  1
Differentiating w.r.t. x
dy
cos   sin   0   2
dx
On eliminating  from (1) and (2), we get
 dy 
sin   y  x   1
 dx 
 dy 
  y  x   cos ec   3
 dx 
dy
Also, Slope    cot  {using(2}
dx
2
 dy 
cos ec  1  cot   1      4 2
 dx 
2 2
 xdy    dy  
From (3) and (4), we have  y    1    
 dx    dx  
80. The value of the number xyz is 100x + 10y + z, where as x and z can be any two of
the digits 1 to 9. But y can only be any one of the digits 1, 4, 9.
 Total value of the 3 digit numbers so formed is
V  100 1  2  3  ....  9 .9.3  10 1  4  9 .9.9  1  2  ....   .9.3
 V  121500  11340  1215  134055
10 r r
81. 10
.
2 35
Tr 1  Cr .2
r  0,10
Sum  10C0 .25.30  10C10 .20.32  32  9  41
 
82. cos2 x
2
I  e cos  2n  1 xdx   ecos   x cos3  2n  1   x  dx
3

0 0
 
cos2 x 2
 e cos  2n      2n  1 x  dx    e cos x cos3  2n  1 xdx   I
3

0 0
Hence 2I  0  I  0
x x x x
83.  x  t 
f  x   x   e f  x  t  dt  x   e
2 t 2
f  t  dt  x   e .e f  t  dt  x  e
2 x t 2 x
 e f  t  dt
t

0 0 0 0
x

f 1  x   2x  e x .e x f  x   e  x  e t f  t  dt  2x  f  x   f  x   x 2
0

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x3
f  x   x   c f  0  0
2

3
x3
f  x  x 
2
f  3  9  9  18
3
f  3  9  9  18
84. The number of ways to fill the three even places by 4 consonants  4P3
After filling the even places, remaining places can be filled in 4 P4 ways.
So, the required number of ways  4 P3  4 P4  576
Hence, 576 is the correct answer.
85. Required ways  12  6  1. 6  7 6
7 6 1
 Required probability  
12 132
86. dy
 y  3cot x   sin 2x
dx
1
Integrating factor  e 
3cot dx
 e 3log sin x 
sin 3 x
1 sin 2x
y. 3   3 dx   2cot x cos ecx dx  2cosecx  c
sin x sin x
y  f  x   c.sin 3 x  2sin 2 x
 
f  3 c2c 5
 2
 
f     5  1  2 1  7
 2
87.


Area   2  x  x 2 dx
2
88. cos 4x  1
Let I   dx
cot x  tan x
2cos 2 2x 1
I  4
sin x cos xdx  I  cos 2xsin 2xdx  sin 4x dx
cos 2 x  sin 2 x
1 1
 I  cos 4x  c  A   &B  R ; SoA  2
8 8
89. Any point on circle x  y  4 is  2cos  ,2sin  
2 2

Equation of directrix is x  cos    y  sin    2  0

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Let focus be  x1 , y1  . Then as A 1,0 ,B  1,0 lie on parabola we must have

 x1  1 2  y12   cos   2 2 
2 2
 x1  2cos , y1   3 sin 
 1  1 
x  1  y 2
 cos   2  
Length of semi latus-rectum of parabola = perpendicular distance from focus to
directrix 2  3 sin 2 
Hence, maximum possible length  2  3
2 2 2 2
90.  1  1  1  1 
Let S   z     z 2  2    z 3  3   ....   z 27  27 
 z  z   z   z 
1 1 1 
S  z 2  z 4  z 6  .....  z54   2  4  ....  54   54
z z z 
1 1 
1  54 
S
2

z 1 z 
54


2 
z  z 
 54
1  z2 1 2
1
z
z   (where  is an imaginary cube root of unity)
S = 54

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.S60, ELITE ,TARGET & LIIT JEE-MAIN Date: 25-05-2022
Time: 09.00Am to 12.00Pm GTM-15 Max. Marks: 300
KEY SHEET
PHYSICS
1) 2 2) 2 3) 1 4) 3 5) 2
6) 2 7) 4 8) 1 9) 3 10) 3
11) 4 12) 2 13) 3 14) 4 15) 2
16) 1 17) 1 18) 3 19) 4 20) 1
21) 2 22) 20 23) 4 24) 35 25) 3
26) 4 27) 19 28) 1 29) -113 30) 40

CHEMISTRY
31) 2 32) 2 33) 3 34) 1 35) 1
36) 2 37) 4 38) 1 39) 2 40) 4
41) 3 42) 2 43) 4 44) 4 45) 4
46) 2 47) 4 48) 3 49) 3 50) 1
51) 3 52) 8 53) 462 54) 6 55) 4
56) 3 57) 30 58) 2 59) 3 60) 8

MATHEMATICS
61) 3 62) 3 63) 3 64) 1 65) 4
66) 3 67) 2 68) 1 69) 1 70) 1
71) 2 72) 4 73) 3 74) 1 75) 3
76) 3 77) 3 78) 3 79) 4 80) 4
81) 1 82) 5 83) 3 84) 6 85) 1
86) 0 87) 13 88) 1 89) 11 90) 10

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SOLUTIONS
PHYSICS
1. Conceptual
2. Conceptual
3. L L 1 1 1/2  1  1 1 dL
t  dt  L dL   dL  dt  dL   0.01
5 300 52  300  2 5 20 300
 1 1   15 
dL     0.01 dL    0.01 dL  3
 20 300   300  16
dL 3 1 15
 100   100   1%
L 16 20 16
2
4. Given, g  10 m / s
2
Equation of trajectory of the projectile. y  2 x  9 x ...(i )
In projectile motion, equation of trajectory is given by
gx 2
y  x tan  0  ...(ii )
2v02 cos2  0
By comparison of Eqs. (i) and (ii), we get
tan 0  2 ...(iii)
g g
2
9
2
v02  ...(iv)
and 2 v cos  0 0 or 9  2 cos 2  0
From Eq. (iii), we can get value of cos  and sin 
1 2
cos 0  sin 0  (v)
5 and 5

5
2

0
1

Using value of cos0 from Eq. (v) to Eq. (iv), we get

2

v02 
10   5
10  5
2 25  5
2  1  9 2  9  v0  9 or v0  3 m / s ...(vi )
2

 1 
From Eq. (v), we get  0  cos 1  
 5
5. Components of velocity at an instant of times t of a body projected at an angle  is
vx  u cos   g xt and v y  u sin   g y t
Here, components of velocity at t  1s , is
1
vx  u cos 60 0  0  as g x  0   10   5 m / s and v y  u sin 600   10   (1)
2
 5 3  10  Vy  10  5 3 m / s
Now, angle made by the velocity vector at time of t  1s

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Vy 10  5 3
tan     tan   2  3 or   150
Vx 5
2
 Radius of curvature of the trajectory of the projected body, R  v / g cos 
2


 5
2

 10  5 3   v 2
 vx2  v y2 and cos 150  0.97   R  2.77 m  2.8 m
10  0.97
6. The given situation is shown below.

h


O M B

u 2 sin 2
Here, range, OB  2h tan  
g
u 2 sin 2
u 2 sin 2  2h tan  g
and h    2 2  tan   2cot 
2g h u sin 
2g
2 cos  u 2 sin 2 
 sin    h
1  cot 2  2g
u2 h h tan 2 
or  
2 g sin 2  4 cot 2 
2
1  4 cot    2
1 
u 2 tan 2  u 2 tan 2 
  2   2
gh 2 gh 2
7. Conceptual
8. Conceptual
9. Let acceleration of mass m relative to wedge down the plane is ar . Its absolute
acceleration in horizontal direction is ar cos 600  a (towards right.) Hence, let N
be the normal reaction between the mass and the wedge. Then
N sin   Ma  m( ar cos 60 0  a )
 M  m a  2  M  m  a
or ar 
m cos 600 m
10. T1   mg
2T2 T2 2T1 T1
F m m m
T2

T2  2T1   mg
or T2  2 mg   mg (Putting T1   mg )
or T2  3 mg
Now, F  2 T2   mg
or F  6 mg   mg (Putting T2  3 mg )
or F  7 mg
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11. Conceptual
12. Let, N1 = normal reaction at A and N2 = normal reaction at B
3N1 mg N 2 3
   ma 3 N1 3N 2 mg N1  N 2 mg 3
2 2 2   ma   
2 2 2 2 2
3 N1 3 N 2 3mg
  
2 2 2
From Eqs. (i) and (ii) , we have 3N1  2mg  ma , 3 N 2  mg  ma
Since cylinder does not loose contact at B, so
N2  0  a  g  amax  g
13. Conceptual
14. F  s 1/3
i.e. acceleration a  s 1/3
dv
or v  Ks 1/3 or v 2  s 2/3
ds
or v  s1/3
Now, P  F .v
or P  s 1/3 , s1/3 or P  s 0
i.e. power is independent of s.
15. Conceptual
16. Fnet mg sin    0 x  (mg cos  )
a   g sin     0 g cos   x
m m
dv
 v.  g sin     0 g cos   x
dx
0 X max

  vdv    g sin    0 g cos   x  dx

0 0

Solving this integration, we get

2 tan 
xmax 
0
17. From work energy theorem,
Wg  Wpsudo force  K f  Ki
N

ma

mg v

1 2
mv  v  2 gR  sin   cos   1
mgR cos   maR 1  sin   
2
dv
v is maximum when  0    450
d
18. Apply work energy theorem
19. Apply work energy theorem
20. In equilibrium, (with respect to cylinder)

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2mg
(Pseudo force)


Net
mg
2mg
tan  
mg
or   tan 1 (2)
 Maximum angular displacement  2  2 tan 1 (2)
21. Conceptual
22. Conceptual
23.  4   4 3 4 3 
G   .  R3  G  2     2 R     R  
EQ  
3    3 3 
2 2
2R  2R 
 4 
G    R3 
3
EP   2  EQ / E P  15 / 4  3.75
R
24. kg
Density of wire, d  9  103 3
 9  103 kg / m3
cm
Strain in the wire,   4.9 104
N
Young’s modulus of wire is Y  9  1010 2
m
Lowest frequency of vibration in wire will be f  1 T
2L  M / L 
T T T T
Now,   
M / L V  / L  LA.  A
 
 L 
T T Y 
But =stress Y  strain  Y    
A M /L 
So, from Eq (i), frequency will be
1 T 1 Y  1 9  1010  49  10 4
f      35 Hz
2L  M / L  2L  2 1 9  103
25. Conceptual
26. Conceptual
27. Conceptual
28. Conceptual
29. Conceptual
30. Conceptual

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CHEMISTRY
31. Formation of NaCl is an exothermic reaction.
32. 1 (mv )2

2 ( mv)1
33. [ClF2O ]  sp 3 ,[ClF4 O ]  sp 3 d 2
34. C2 H5OH  2Cr2O72  2CO2  4Cr3
Eq. of Cr2 O72  eq of C 2 H 5 OH
8.0  .05  6 10 8.0  .05  46 0.0092
 W  6  .0092 %   100  0.092%
1000 46 /12 1000  12 10
35. As per Le Chatelier’s principle, equilibrium moves in forward direction.
36. 1
C ( graphite)  O2 ( g )  CO( g ); H  ve Is not a combustion reaction
2
37. V2
S  2.303nR log
V1
38. De-excitation of electron from 5th orbit to 4th orbit gives first line in Brackett
series.
39. HI n  H   I n
Acidic basic
Red Blue
 H   I n 


Ka 
 HI n 
75
Red from 75% blue from 25%   H    3  10 5   9  105
25
When 75% blue and 25% red, then
25
 H    3 105   1 105  change is   H    9  105  110 5 = 8  105
75
40. Localised electron pair on nitrogen.
41. 2 2
1 * 4 4 * 1
3 -2S and 3 -3R
42. Most stable alkene is the major product.
43. BeF2 Soluble in water due to grater hydration energy of Be+2 , as compared to L.E
44. A3 ion further can not lose electrons so it cannot act as reducing agent.
45. H3C H

(X)

optically active.
H3C H
(Y)
optically inactive.
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 SRI CHAITANYA IIT ACADEMY, INDIA 25-05-22_ Sr.S60, ELITE ,TARGET & LIIT _ Jee-Main_GTM-15_KEY &SOL’S
46. For lead, (+2) is more stable due to inert pair effect
47. As BOD values increases, pollution increase.
48. As the electronegativity of central atom increases, bond angle increases.
49. Friedel Crafts alkylation is generally not preferred for the preparation of alkyl
benzenes because
electrophile undergoes rearrangement giving mixture of products and poly
alkylation takes place
as alkyl group is activating
50. More hyper conjugations.
51. I, III and IV
52. 2S4 N4  4N2 ( g )  S8 ( g )
53. H
G  O at equilibrium hence H  T S and T 
S
54. Borax- Na 2 B4 O 7 ·10H 2 O and Kernite- Na 2 B 4 O 7 ·4H 2 O
55. a, b, e, g
56.

, ,
57. 1 1  P2V2
PV
58. q  ms T
59. i, ii , iii.
60. KCN HCN
Volume = ‘x’ ml volume = 10ml
Molarity = 5 Molarity = 2m
 P H   log Ka  log
 salt 
 Acid 
5 x /(10  x )
9.6020   log 5  1010  log
20 /(10  x)
x
9.6020  9.3010  log
4
x
log  0.3010  log 2
4
x  8ml

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MATHEMATICS
61. Lines are y  1  tan   x  1 and  y  1  cot   x  1 multiply and compare
2
sin 2 
a2
n
62. 
lim  tan 1  r 2  r   tan 1  r 2  r  
n
r 1 2
63.
Let g  x   e x f  x  As g ''  x   0 so g '  x  is increasing
So, for x  1 / 4, g '  x   g ' 1 / 4   0
  f  x   f  x   e  x  0  f  x   f  x  in  0,1/ 4 
64. ST 2  AS 2  AT 2  2 AS . AT . cos  and AS . AT 
1
bc
2
65. Use half angle formulae and Heron’s formula
66. Y

 6  3 
, 
 6  3   2  yx
 0, 
 2 

 0,   3x  2 y  6
 ,  

6  3  6  5  dA 3
Area of rectangle    
     0  
 2   2  d 5
3  3 9
AMax      Sq units
5  2  10
67. Since f ''  x   0
 f '  x  is always increasing.
g '  x   2 f '  2 x 3  3x 2    6 x 2  6 x   f '  6 x 2  4 x 3  312 x  12 x 2 


 12  x 2  x  . f '  2 x 3  3 x 2   f '  6 x 2  4 x 3  3 
 12 x  x  1 . f '  2 x 3  3 x 2   f '  6 x 2  4 x 3  3
For increasing g '  x   0
 2 x 3  3x 2  6 x 2  4 x 3  3 f '  x  is incresing
2 1 1  1 
  x  1  x    0  x    x    ,0   1,  
 2 2  2 
Case-II If 0  x  1
f '  2 x 3  3x 2   f '  6 x 2  4 x 3  3
2  1 1
 x  1 x 0 x   , so there is no solution
 2 2
 1 
 Hence the values are x    ,0   1,  
 2 

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 SRI CHAITANYA IIT ACADEMY, INDIA 25-05-22_ Sr.S60, ELITE ,TARGET & LIIT _ Jee-Main_GTM-15_KEY &SOL’S
68. 5  0 0
0 3 2  0    1, 4,5
0 1 2 
69. m21  m11  m22 2012  32012  32012  32012
   2010
32012 32012
70. 1 2x 3x 2
Given y  1   
x  1 ( x  1)( x  2) ( x  1)( x  2)( x  3)
 1  2x 3x 2
 1    
=  x  1  ( x  1)( x  2) ( x  1)( x  2)( x  3)
x3
y=
( x  1)( x  2)( x  3)
x   x   x 
log y=log    log    log  
 x 1  x2  x3
Diff. both sides w.r. t x, we get
dy y  1 2 3 
     9y(4) = 104
dx x 1  x 2  x 3  x 
71. m sin 2  m sin           cos2      m cos     sin    
m sin 2   m sin           cos2      m cos     sin    
1 1
Now, 
1  m sin 2 1  m sin 2 
1 1 2
 2 
sin      m cos     sin     sin      m cos     sin     1  m2
2

72. 
Continuous at x  1 p  q  2 
4
1 1 1  13
differentiable at x  1 3p   p  ;q  
11 2 6 4 6
2
73. 1  y  cos 2 x  4cos 2 x  3 y  1  0
Since cos 2x is real, 16  4  3 y  11  y   0 Or 3 y 2  2 y  5  0

But y  1  cos 2 x  1 i.e. x  which is not permissible.
2
74. Total disks in An is S n  3n  n  1 and length of side of hexagon is
2 rn 1
1  2 rn  n  1   rn 
3 2
2n  2 
3
2 2
75.   
d 1 .d 2  0  a  2b . 2 a  b  0  2 a  3a.b  2 b  0
2 2
d .d  0   2 a  b  .  a  2b   0  2 a  3a.b  2 b  0
3 4
2 2
 a.b  0 Now  a b a  2b    a  b  .  a  2b   2 a b
76. et et cos t e t sin t 
 
A  et e t (sin t  cos t ) e t (cos t  sin t ) 
et 2e t sin t 2e t cos t 

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1 cos t sin t
A  et .e 2t 1  sin t  cos t cos t  sin t
1 2sin t 2cos t
1 cos t sin t
A  e  t 0  sin t  2 cos t cos t  2sin t
0 2sin t  cos t 2 cos t  sin t
A  e  t  sin 2 t  4 cos 2 t  4 sin t cos t  4 cos t  4 sin 2 t  cos 2 t 
A  e t  5sin 2 t  5.cos 2 t   5.et  ot  R
77. 0 1 0 1   0 1   3 0
A    A2  A. A   
3 0 3 0   3 0  0 3
3 0 3 0  32 0 
 A 4  A2 . A 2     
 0 3  0 3  0 32 
834 0 6 4 2 32 0  3 0  33 0
A   and A  A . A   2   
0 34   0 3  0 3  0 33 
x
Let V   
 y
A8  A6  A4  A2  I
81 0   27 0  9 0   3 0   1 0  121 0 
 0 81   0 27   0 9    0 3   0 1    0 121
           
0
 A8  A6  A4  A2  I V  11
 
121 0   x   0  121x   0 
           
 0 121  y  11 121y  11
78.
Let y 
cos3  sin 3  1

 cos  sin  1  sin  cos    1
2 2
 sin   cos   1  sin   cos  1
 x2  1 
x 1  1
2 
2
 2  3x  x 3  2  x  x  1 x
Let cos  sin   x y  2
 2
 2
 1
 x  1 2  x  1 2  x  1 2
cos   sin  2
 y 1 1
2 2
79. x y z 1 x y z 1
The lines are   ,  
1 1 0 1 1 0
Let C   ,  ,  
 
    c  a   b
The distance of C from r  a  sb is 
b
2

Distance of C from the first line is   1

2

    .... i 
2

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2

The distance of C from the second line is   1

2

    ....  ii 
2
2
2 2     4
i  ,  ii     1    1   2 . Locus of C is xy  2 z.
2
80. f '  x   2sin  x    cos  x     2sin  x    cos  x     2cos
    cos  x    sin  x     2 cos     sin  x    cos  x   
 sin  2 x  2   sin  2 x  2   2cos     sin  2 x     
 2sin  2 x      cos      2 cos     sin  2 x       0
 f  x  is a constant function.
81. Let equation of plane be x  2 y  3z  7   x  0
Since  1,0, 2  lies on it    2
Equation of plane is  x  2 y  3z  7  0
6 p
cos  
7 q
82. 3
Plane is  x  z  1  d 
2
83. f  x  y  f  x f  y
2 3
f  2    f 1  , f  3   f 1  and so on
n

 f  a  r   f  a  f 1 1  f 1   f 1  

2 n 1
 .....   f 1 
r 1

3.  3n  1
 f a
2
84. If f    0  sin   0 or sin 2  0 or cos 3  0 or cos 4  0 or cos 5  0
85.  t  1 x  8 y  4t  0
tx   t  3 y   3t  1  0
t 1 8 4t
 
t t  3   3t  1
t3
86.   
5
  
5  4 2   2 cos  x    
4 2   cos x  sin x   10 cos x 1  sin 2 x    4  
lim  lim   lim10cos x
x 1  sin 2 x x  1  sin 2 x  x
4
4   4

 
 5  
 1  cos  x  4   10 4 2 1  cos5 h 
 lim 4 2      lim 5 2
x
4
 1  sin 2 x  2 h 0 1  cos 2h
 
 
4 2 1  cosh  cos2 h  cos3 h  cos4 h 
 lim 5 2  5 2 5 2  0
x 0 2 1  cosh 
87. Taking dot product with a, b , c, we get 5  a b c   a . a, 3  a b c   a . c and a . b  0

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 2a  c  3d  .a  13 as d . a 
So.
a b c
 b   a  b .a  0
1 2
 
88. Check the continuity and differentiability at integral points. At x = 1, it is not
differentiable f  x   x sin   x  1   x sin  x x  1  x sin  x
It is continuous but not differentiable
89. Let P and Q be points on the lines, then
P  2  2 s, 6  3s,34  10 s  , Q  6  4t ,7  3t ,7  2t 
The d .r ' s of PQ are  4  4t  2 s,13  3t  3s, 27  2t  10s 
PQ is perpendicular to the two lines with d .r ' s2, 3, 10 and 4, 3, 2.
 2  4  4t  2 s   3 13  3t  3s   10  27  2t  10s   0
4  4  4t  2 s   3 13  3t  3s   2  27  2t  10 s   0
 113s  19t  301, 29  19 s  1
Solving s  3, t  2, P   4,3, 4    a, b, c   a  b  c  11
90. R Q

T
, ,

S P
 iˆ ˆj kˆ 
1   1  
Area of base (PQRS)  PR  SQ  3 1 2 
2 2
4 
1 3
1
 10iˆ  10 ˆj  10kˆ  5 iˆ  ˆj  kˆ  5 3
2
1 2  3 2
Height = proj. of PT on iˆ  ˆj  kˆ  
3 3
 2 
 
volume  5 3    10 cu. Units
 3

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.S60, ELITE, TARGET&LIIT JEE-MAIN Date: 27-05-2022
Time: 09.00Am to 12.00Pm GTM-16 Max. Marks: 300
KEY SHEET
PHYSICS
1) 1 2) 1 3) 3 4) 1 5) 2
6) 3 7) 2 8) 1 9) 2 10) 1
11) 4 12) 3 13) 4 14) 1 15) 3
16) 1 17) 4 18) 2 19) 3 20) 1
21) 20 22) 2 23) 6 24) 5 25) 2
26) 35 27) 75 28) 25 29) 100 30) 50

CHEMISTRY
31) 3 32) 2 33) 2 34) 2 35) 3
36) 2 37) 4 38) 1 39) 2 40) 3
41) 2 42) 2 43) 4 44) 2 45) 3
46) 3 47) 3 48) 2 49) 3 50) 4
51) 3 52) 1 53) 36 54) 9 55) 1
56) 14 57) 5 58) 3 59) 144 60) 4

MATHEMATICS
61) 2 62) 4 63) 3 64) 1 65) 1
66) 4 67) 4 68) 3 69) 4 70) 1
71) 4 72) 2 73) 4 74) 1 75) 1
76) 1 77) 4 78) 1 79) 1 80) 4
81) 50 82) 40 83) 100 84) 75 85) 4
86) 0 87) 2 88) 1 89) 30 90) 11

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SOLUTIONS
PHYSICS
1. f nimage
M .P  0 
fe nObject
2. 1.22

a
3. Conceptual
4. For 2nd minima d sin   2
3  3
sin   (given)   ……………. (i)
2 d 4
 3
So for 1st minima is d sin   2 In    (from equation (i))
d 4
  25.650 (form the sin table)   250
5. M
I
V
6. K .E  W  MB 1  cos 
7. 1 n12 B1 cos1
T ; n B and 2 
B n2 B2 cos 2
8. V V2  V1  V1  V2  0.2  0.1  10
4
We know that B   , ,
r r r 0.1sin 300
9. V  VCE  I C RL  15  7  I C  2  103  iC  4mA
i 4
 S  C  iB   0.04mA
iB 100
10. Lower NOT gate inverts input to zero. NOT gate from NAND gate inverts this
output to 1 upper NAND gate converts this input 1 and input 0 to 1. Thus A = 1 and
B = 1 become inputs of NAND gate giving final output as zero.

11. The P-N junctionwill conduct only when it is forward biased i.e., when -5V is fed to
it, so it will conduct only for 3rd quarter part of signal shown and when it conducts
potential drop 5 volts will be across both the resistors, so output voltage across R2
is 2.5 V  V0  2.5V
12. 2 2
m 
Psb  Pc  a   Pc
 0.5  0.0625P
c
 2  4
2  2
Also P  P  1  ma   P 1   0.5    1.125 P
c  c c
 2   2 
 
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 % solving 
1.125Pc  0.0625 Pc   100  94.4%
1.125Pc
13. Given, Power of laser beam  P   27mW  27  103W
Area of cross-section  A  10mm2  10  106 m2
Permittivity of free space V0   9  1012 SI unit
1
Speed of light  c   3  108 m / s I  ncV0 E 2
2
Where, n is refractive index, for air n  1
1
 I  C.V0 E 2 …………… (i)
2
P
Also, I  ………….. (ii)
A
From Eqs. (i) and (ii), we get
1 2 P 2 2P 2  27  103
cV0 E   or  E   or  E 
2 A AcV0 10  108  3  108  9  1012
 1.4  103V / m  1.4kV / m
14.
  
The volume of the needle, V  3  102  1  103  0.5  103   1.5  108 m3
The mass of the needle  .......V
8
 7900  1.5  10  1.183  104 kg
1.185  104 
The number of atoms in the needle     6.02  10
23
 1.27  1021
3
 56  10 
1
The needle’s dipole moment M 
10
  
1.27  1021  2.1 10 23  2.7  103 J / T
15. B BV
tan 450  V tan d  2
BH BH cos 60
16. The conductivity of a semiconductor is given by t  nee ~ e  nhe ~ h
Since in the problem semiconductor is pure germanium (intrinsic semiconductor)
 ne  nh  nl  t  nl e  ~ e  ~ h 
As per given data, n1  1.6  106 per cubic meter
~ e  0.4m 2V 1s 1 ~ h  0.2m 2V 1s 1
 t  1.6  106  1.6  1019  0.4  0.2   1.6  1.6  0.6  1013
Current produced in germanium plate,
V   5 
I  JA  tEA  t   A  1.6  1.6  0.6  1013     2  104
d  1.2  1013 
 I  1.28  1013 ampere

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17.
1
12  8 A
4
 103 A
 200  200  400
Power loss in each diode = VI
18. Magnetic field in solenoid B ~ 0 ni
B
  ni (where n = number of turns per unit length)
~0
B Ni 3 104 i
   3  10   i  0.24 A
~0 L 10  102
19. Here, H E  0.16G  0.16  104 T , dip angle  u   600
Then, magnitude of earth’s field.
H E 0.16  104 0.16  10 4
BE   T BE  T  0.32  104T  0.32G
cos u cos600 1/ 2
20. As final image is at 25 cm in front of eye-piece
1 1 1 25
  , i.e., ue  
25 ~ e 5 6
Now for objective,  Lue  20   25 / 6    95 / 6 
So if object is at a distance u from the objective,
6 1 1 95
  i.e., u   cm
95 ~ 0.95 94
i.e., object is at a distance  95 / 94  cm inform of field lens.
21. }
We have a }  or "  (a = width of each slit)
a
} } d 1
10  2  a    0.2 mm
d a 5 5
22. The transmitted intensity of unpolarised light will be constant for all the orientations
of the polarized sheet whereas intensity of polarized light will be expressed by the
law of Malus, I P  I P .cos 2  .
Let us consider the intensity of transmitted, polarized and unpolarised components
to be I P and I 0 respectively , For   f / 2, I P  O and for   O 0 , I P  I P while
I
for all orientations I 0  0 . From given condition
2
I I
I max  I P  0 when   O0 and I min  0 when   f / 2
2 2
I I 3I I 3
 I P  0  4. 0 or I P  0 i.e., P 
2 2 2 I0 2
I 3I 1 I 5I
For   450 , I  I P cos 2 450  0  0  0 0
2 2 2 2 4

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23. W 2.4  105
B  W / A and B ~ H  ~   7.5  104 H / m
 4
AH 0.2  10  1600
~ 7.5  104
 x 1   1  600
~0 7
4 f  10
24.  CW  MBH sin 
Where W is the twist of the wire and  is the deflection of magnet.
 
For 1st magnet, C 1800  300  M1BH sin 300
M1 150 5
 
For IInd magnet, C 2700  300  M 2 BH sin 300   
M 2 240 8
25. The current density J is given by
I 4.8  103 4.8  103
J    4800 A / m 2
A
 
4  103 25  105 10 6
J 4800
The drift velocity vd is given by vd 
  3m / s
ne 1022  1.6  1019
l 6  102
The time taken t is given by t   0.02 sec
vd 3
26. bt
 A bt
A  A0e 2 m  0 ;  ln 2  0.693
2 2m
2m 500
 t  0.693  t  2   0.693
b 20
 50  0.693  34.6 sec
27. 3
I  I 0 cos 2   I 0 cos 2 300  I0    75
4
28. D D 25
For simple microscope m  1   6 1  5  f 0  5cm
f0 f0 f0
1.D 60  25
For compound microscope m   12   f e  25cm
f0 . fe 5. f e
29. 6
Power gain  I 2 R , Power gain   2 .R  106   2  100 ,  2  10  104    100
2
10
30. Here hr  20 m
In case (i) is hk  0 and in case (ii) hR  5m
Range in case (i) is d1  2RhT and in case (ii) is d 2  2 RhT  2 RhR
Percentage increase in the range is
d 2 d1 2 RhR 5
 100   100   100  50%
d1 2 RhT 20

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CHEMISTRY
31. VO,VO2 ,VO3 & TiO3 show metallic (or) insulators depending upon temperature.
32. Conceptual
33. Argon is used mainly to provide an inert atmosphere in high temperature
metallurgical process
34. BeH 2 and MgH 2 are polymeric in structure.
35. Zn  2 HCl 
 ZnCl2  H 2
36. Cermic – honey comb coated with precious metals Pt , Pd and Rh .
37. Unbranched chain can be biodegraded more easily and hence pollution is prevented.
38. Vitamin B6 is also called Pyridoxine
39. Nylon 6 is used in manufacture of tyre cords, fabrics and ropes
40. Conceptual
41. Conceptual
42.
33.3  3.11  0.313  36.723  36.7
After decimal point one significant figure.

43. Hydrogen and oxygen is used to generate electricity in fuel cell:

44. Flourine
45. Silica gel, Alumina and Starch are used as an adsorbent in adsorption
chromatography
46. n  2d sin 
2 1  2  d  sin 600
2 2
d   1.15 A0
3 1.732
47. pH 8 pH  4
Cr2O72 
 CrO42 
 Cr2O72 .
OH  H
48. Ni 2  DMG  Rose pink coloured observed.
49. Pu and Np show highest of oxidation states.
50. Methylene chloride is used metal cleaning and finishing agent solvent
51. CH 3 CH 3
O O CH 3 C
HO  C  CH 2  CH  C  NH  CH  C  NH  CH S
NH 3 O C
CH 3 CH 3

52. 1.014 
 4 significant figures
0.07 
1 significant figures
5.11 
 3 significant figures
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53. PV
1 1  P2V2
T1 T2
700  100 760  V2 7  273
  V2 
250 273 19
28 7  273 1
%N     100
22400 19 0.35
 35.71%  36%
54. 1
  450  1
n
log k  log3  k  3
1
x 1
 k . p n   3 3  9
m
55. hc 6.625  10 34  3  108
E  eV  3.45 eV
 3600  1010  1.6  1019
E    kemax
kEmax  E    3.45  2  1.45 eV
56. Enthalpy of ionization of HCO3 = Enthalpy of neutralization of WA & SB  -
Enthalpy of neutralization of (SA&SB)
 42   56   14 KJ / mole .
57. potassium ion participate in the oxidation of glucose to produce ATP.
58. O
CH 2  CH 2  C

CH 2

CH 2  CH 2  NH

COCH 3
59. COONa

I  NaOH
2
 CHI3 

60. b) High concentration of SO2 leads to stiffness of flower buds.

e) Rate of production of NO2 is faster, when nitric oxide reacts with ozone on the
stratosphere
f) Carbon monoxide combine Haemoglobin to form carboxyhaemoglobin, which is
about 300 times more stable than oxygen-Haemoglobin
g) The concentration of dissolved oxygen of water is below 6 ppm, the growth of
fish gets inhibited.

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MATHEMATICS
61. n  A ' B '  n  A  B  '  n U   n  A  B 
 n U    n  A   n  B   n  A  B  
 700   200  300  100  300
62. Let F, B and C denote the set of men who received medals in football, basketball
and cricket, respectively.
Then n  F   38, n  B   15, n  C   20
n  F  B  C   58 and n  F  B  C   3
Therefore, n  F  B  C   n  F   n  B   n  C 
n  F  B   n  F  C   n  B  C   n  F  B  C  , gives
n  F  B   n  F  C   n  B  C   18
Consider the Venn diagram as given in figure.
U
F B
a
d
b c
C

Here, a denotes the number of men who get medals in football and basketball only,
b denotes the number of men who get medals in football and cricket only, c
denotes the number of men who got medals in basket ball and cricket only and d
denotes the number of men got medal in all the three.
Thus, d  n  F  B  C   3 and a  d  b  d  c  d  18
Therefore a  b  c  9 , which is the number of people who got medals in exactly
two of the three sports.
63. n  P   25%, n  C   15% ,

 
n P c  C c  65%, n  P  C   2000


Since n P c  C c  65% 
c
 n  P  C   65%  n  P  C   35%
Now n  P  C   n  P   n  C   n  P  C 
 35  25  15  n  P  C   n  P  C   40  35  5
Then n  P  C   5%
But n  P  C   2000
 5% of the total = 2000
2000  100
 Total number of families   40000
5
Therefore, statements 2 and 3 are correct.
64. As R is reflexive relation on A,  a, a   R for all a  A . The minimum number of
ordered pairs in R is n . Hence, m  n

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65. As 1,1 ;  2, 2  ;  3,3  R, R is reflexive
As 1, 2   R but  2,1  R, R is not symmetric
It can be easily seen that R is transitive
66. Z Z
Z1RZ1  1 1  0  R is Reflexive because 0 is real.
Z1  Z1
Z  Z2  Z  Z2 
Z1RZ 2  1 is real   1  is real
Z1  Z 2  Z1  Z 2 
 R is symmetric
Z1  a1  ib1 : Z 2  a2  ib2
Z  Z2
Z1RZ 2  1 is real
Z1  Z 2
  a1  a2  b1  b2    a1  a2  b1  b2   0
a1 a2
 2a2b1  2b2a1  0  
b1 b2
a a
 Z 2 RZ3  2  3  Z1RZ3  R is transitive.
b2 b3
67. OB  100 cot 450
OA  100 cot 300
AB  OA2  OB2   200 m

B 450 100 m
O
300

68. Let BC be the declivity and BA be the tower. Using sine rule in ABC, we have
BC AB 80sin 300
  AB 
sin 750 sin 300 sin 750
40  2 2

3 1
 40 6  2  
A
750
300 C
750
150
B

69. S

R
c
bQ
a 
 
P x

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a b c
We have tan   , tan   and tan  
x x x
       1800 , so
tan   tan   tan   tan  tan  tan 
a b c a b c
Or    . .
x x x x x x
abc
Or x 2 
a bc
70. Marks obtained from three subjects out of 300 is 75  80  85  240
If the marks of another subject is added, then the marks will be  240 out of 400.
240
 Minimum average marks   60% [When marks in the fourth subject =0]
4
71. th
 9 1
Since n  9, median term     5th term. Now, the last four observations are
 2 
increased by 2. Since the median is the 5th observation, which remains unchanged,
there will be no change in median.
72. Given, observations are 3, 10, 14, 4, 7, 10 and 5.
3  10  10  4  7  10  5 49
 x  7
7 7
xi di  xi  x
3 4
10 3
10 3
4 3
7 0
10 3
5 2
Total  di  18

 di 18
Now, M .D.    2.57
N 7
73.  xi
2 2
x 
Variance   i 
n  n 
18000  960 2
    300  256  44
60  60 
74. If p : A number is a prime
q : It is odd
We have p  q
The inverse of q is ~ p  ~ q
i.e., if a number is not a prime then it is not odd.
75. If p  q is false only when p is true and q is false
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p   q  r  is false when p is true and q  r is false
And q  r is false when both q and r are false
Hence, truth values of p, q, r are respectively T, F, F.
76. If  p  ~ q    ~ p  q    p  ~ p    ~ q  q    f  f   f
 f  false 
(By using associative laws and commutative laws)
  p ~ q    ~ p  q  is a contradiction.
77. It is clear from table that ' ~ '  ~ p  q  is equivalent to ~ p  ~ q .
p q ~ p ~ q p  q p ~ q ~ p  q ~ p ~ q ~ p  q ~  ~ p  q 
T T F F T F F F T F
T F F T F T F F T F
F T T F F F T F T F
F F T T F F F T F T
78. Consider the following statements
p : Two triangles are identical.
q : Two triangles are similar.
Clearly, the given statement in symbolic form is p  q .
Therefore, its contrapositive is given by ~ q  ~ p
Now, ~ p : two triangles are not identical.
~ q : two triangles are not similar.
 ~ q  ~ p : If two triangles are not similar, then these are not identical.
79. Let p : I become a teacher
q : I will open a school
Negation of p  q is ~  p  q   p  ~ q
i.e. I become a teacher and I will not open a school.
80. The event follows binomial distribution with n  5, p  3 / 6  1 / 2, q  1  p  1 / 2
Therefore, variance is npq  5 / 4
81. tan   tan    
20  h h

1 70 70
 
6
1
 20  h  h
 70 2
  70 2  20h  h2   6  70  20 
 h 2  20h  70  70  120   0
 h 2  20h   50  70   0
20  400   4  50  70 
 h  50m
2

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
20



 h

 
70 m

82. 3
h
4


 h/4
  
40
   
   
tan   tan 
 tan  
1  tan  tan 
h h

3
  40 160
5 1 h  h
40 160
 h2  200h  6400  0
 h  40 or 160 m
Hence, the possible height is 40 m.
83. In the figure PQR is triangular park with PQ  PR  200 TV tower of height ‘h’
stands at midpoint M of QR.

In triangle TMR, MR  h cot 300

In triangle TMP, PM  TM  h
In triangle PMR, PR 2  PM 2  MR 2
 2002  h2  3h2
 h  100 m
84.
82 
 27  x    31  x   89  x   107  x   156  x 
5
 82  5  410  5 x  410  410  5 x  x  0
Therefore, the required mean is
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130  x  126  x  68  x  50  x  1  x
x
5
375  5 x 375  0 375
    75
5 5 5
85. 25  26
From the given data, median  a  25.5a
2
Required mean deviation about median
2 0.5  1.5  2.5  ....  24.5
 a  50
50
 a 4
86. Here,  CV 1  50,  CV 2  60, x1  30 and x2  25

  CV 1  1  100
x1

 50  1  100
30
30  50
 1   15
100

And CV2  2  100
x2

 60  2  100
25
60  25
 2   15
100
Now, 1   2  15  15  0
87. Let xi  5  yi
9 9
2
  yi  9 and  yi  45
i 1 i 1
So, required standard deviation is
2
9 9
2  
 yi   yi  2
i 1 45  9 
   i 1     2
 9 
9 9 9
 
 
88. Let P be the summit of the mountain and Q be the foot. Let A be the first position
and B the second position of observation. BN and BM are perpendiculars from B to
PQ and AQ, respectively,
Then, AB  1000 m  1 km ,
MAB  300 ,
MAP  450 , NBP  600

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Now, BAP  MAP  MAB  450  300  150

APB  APN  BPN  450  300  150
P

150 450

600
B N
150
450
A Q
M

Therefore, ABP is isosceles and AB  BP .

But AB  1 km, so BP  1 km .
Now, PQ  PN  NQ
 PN  BM
 BP sin 600  AB sin300
3 1 3 1
 1  1  m
2 2 2
89.  60  h  cot 600  h cot 300
 h  30 m
T
60 m

600
300
P O
90. According to the question, we have
2m  2n  112
 
 n 2m  n  1  2 4  7

 2n  24 and 2m n  1  7
 n  4 and 2m n  8
 n  4 and 2m n  23
 n  4 and m  n  3
 n  4 and m  4  3
 n  4 and m  7

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Sri Chaitanya IIT Academy.,India.

 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Sr.S60, ELITE,TARGET & LIIT JEE-MAIN Date: 27-07-2022
Time: 09.00Am to 12.00Pm GTM-42 Max. Marks: 300
KEY SHEET
PHYSICS
1) 2 2) 1 3) 4 4) 1 5) 2
6) 4 7) 2 8) 3 9) 3 10) 2
11) 3 12) 3 13) 1 14) 2 15) 3
16) 3 17) 4 18) 3 19) 2 20) 2
21) 3 22) 70 23) 1 24) 20 25) 2
26) 8 27) 4 28) 2 29) 80 30) 5

CHEMISTRY
31) 1 32) 3 33) 2 34) 2 35) 4
36) 3 37) 1 38) 1 39) 4 40) 2
41) 1 42) 2 43) 1 44) 4 45) 4
46) 2 47) 3 48) 2 49) 3 50) 1
51) 1 52) 3 53) 4 54) 4 55) 1
56) 0 57) 3 58) 3 59) 102 60) 2

MATHEMATICS
61) 2 62) 1 63) 4 64) 2 65) 2
66) 2 67) 4 68) 4 69) 2 70) 4
71) 1 72) 3 73) 4 74) 3 75) 2
76) 1 77) 2 78) 2 79) 1 80) 2
81) 0 82) 1 83) 5 84) 3 85) 16
86) 420 87) 2 88) 0 89) 5 90) 7

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SOLUTIONS
PHYSICS
1. As magnets are perpendicular to each other, the resultant magnetic moment
2I
M '  M 2  M 2  2M T1  2
 2M H 
I T2 I 4
In the second case, T2  2 ,  1/4
T2  1/4
 3.36s
MH T1  2  2
2. 1 1 1 2 L
    Lp 
Lp L L L 2
1.8  104
Where L is inductance of each part,   0.9 104 H
2
4
L 0.9 10
 Lp    0.45  104 H
2 2
6 1 1 1 2
Resistance of each part, r   3 Now,   
2 rp 3 3 3
Lp 0.45 104
Time constant of circuit,    3 105 s
rp 1.5
3. When the spherical conductors are connected by a conducting wire, charge is
1 QA
redistributed and the spheres attain a common potential V.  Intensity E A 
4 0 RA2
1 C AV  4 0 RA  V
or E A   
4 0 RA2 4 0 RA2 RA
V E R 2
Similarly EB   A 0 
RB EB RA 1
4.  to theplane. Area vector is at 60 0 angle with x-axis and 30 0
Areal vectors will be 
angle with – z-axis. M  NI A

M  10 102 Am2 M  0.1 Am 2

M  M x i  M z k  M cos 60i  M cos 300 k

M  0.05 i  3 k
 
5. The block will just start sliding when the angle  made by the inclined plane with the
horizontal will be equal to the angle of repose    respose  tan 1     tan    ,
3

4
6. If final temp is T

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PV PV
0 0
Cv  4T0  T   0 0 Cv T  T0 
4 RT0 RT0
4T0  T 8T
 T  T0 4T0  T  4T  4T0 8T0  5T T  0
4 5
8
Tf T0
5 2
Final pressure In left Pf  Pi  P0  P0
T0 4T0 5
8
T0
5 8
In right Pf  P0  P0
T0 5
 8P 2 P  6P A
Force   0  0  A  0
 5 5  5
7. p  h g  3 p
h  g  2 p ____(1)
4
p '  p  h  g ____(2)
5
From (1) and (2)
4 13 p
p'  p  2p  p' 
5 5
8. F
Stress 
A
400 400
Elastic limit is 379 MPa  379  106  
A d2 / 4
400  4
d 6
 1.16  10 3 m  1.16mm
  379  10
9. 1 2 hc
mv    .....(i)
2 
1 2 4 hc
mv    .....(ii)
2 3 
hc 1 2
From Eqn. (i)  mv   On putting this Eqn. (ii)
 2
1 2 4 1 2  4 2 4
mv   mv      v 2  v 2   v'  v
2 3 2  3 3 m 3
10.  K  U  mgH & K  2U
mgH mgH
 2U  U  mgH  U   mgh  h H /3
3 3
1 2mgH gH
 mv 2  2U  v2
2 3 3
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11. A sin   B cos 
Z ; Dimensions of A and B would be same as two quantites can be
A B
added only when they have same dimensions. So from given equation, we can say that Z
is dimensionless.
12.

2h
The time, the particle takes to go through h is t  and in one complete oscillation,
g
the particle travels thorugh h four times.
For the motion of the particle through the tunnel, the force acting on the be like an SHM.
GMm GM R3 R
F   3 x a   3 x TSHM  2  2
R R GM g
2h R 2h
So the time period of the particle is T  TSHM  4 T  2 4
g g g
13.

dy
For maxima x  n  n
D
dy 0.032 102 1.06  102
For n=4     530 nm
4D 4  1.6
14. In steady state, the capacitors are fully charged and act as open circuit, so the equivalent
circuit in steady state would be as shown:

Steady state current is

12
I  2 A Potential difference across AB is
24
V  2  4  8V Sum of potential difference across 2  F and 4  F capacitors is 8V.
As these are in series, so charges on them would be same. Let q be the charge on them,
q q 32
then, From KVL,   8  q  C
2 4 3
15. dN
Probability for a particular nucleus to decay in any time interval dt is   dt
N
0.693
  4  12.4  10 4
T1/2
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16. 2
V  VR2  VL  VC   VR VL  VC  , VR  V  200 volt
17.  v   340   340 
f1    f   f   f
 v  vs   340  34   306 
 340   340  f1 19
f2    f    f 
 340  17   323  f 2 18
18. Fraction of supplied energy which increases the internal energy is given by
u CV T 1
f   
 Q  P CP T 
[Here according to definition of sp. Heat at constant pressure  Q  p c  nc p T
7 5
For diatomic gas    f 
5 7
19. 

Given E  E0 sin 1.57 107   x  ct  .....(i)  E  E
c
 x  ct  ......(ii )
0 sin
 
On comparing Equations (i) and (ii), we get  1.57  107   107
c 2
hc
  4  10 7 m  400mn As, 0   Ek

1242
1.9   Ek Ek   3.1  1.9  eV  1.2eV
400
20.

Velocity of image of the point in the first mirror with respect to the image of the point in
 
 
the Second mirror is V I1  V0 3 i  V0 3 j   V I1  V0 6
I2 I2

21. 4
Wavelength of K line is given by   2
3R  Z  1
For  , between 250 and 179 pm three elements are possible (Z = 24, 25 and 26)
22. Q nC p T Cp 7
  Q  .W   20  70 J .
W nRT R 2
23.

Conservation of linear momentum:

3u   3u / 2   mv  mv   3u / 2  ......(1)

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Newton’s experimental law: u  e  u / 2  v 

3u
Putting e = 1 for elastic collision we obtain, v  .....(2)
2
Using (1) and (2), we obtain m = 1 kg.
24. k 2k
Using the concept of reduced mass, we obtain :-   
mreduced m
 mm m  2  200
 mreduced      20rad s 1
 mm 2  1
L
25.
M
dm   0

x  dx For centre of mass X cm 
 dmx  2L
0
L
 L 
 dm 3
0
26. 4
x AD   100  80cm  8  10cm
5
27. 2
 3R 
I AB  I disc  I plate  M  
 2 
2
1 2 MR 2  3R  31 2
(From Parallel Axis Theorem) MR  M   MR
4 12  2  12
28. V V V V
Vopen  , Vclosed   Vopen  Vclosed  4   4
2L 4L 2L 4L
 V  16 L
V V
Now in 2 nd case Vopen '
 '
Vclosed 
4L 8L
' ' V V V 16 L
Vopen  Vclosed     2
8L 4 L 8L 8L
29.

F  10  g  20  g  10a.......(i)
10  g  10a..........(ii)
From (i) and (ii)
F  40  g
F  80 N
30.

V  20 V
  V  5  0 V  20  V  2 V  5  0
2 2
30 15 15  10 5
 4V  20  10  0 V   V, I   A  2I  5 A
4 2 2 2
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CHEMISTRY
31. 1 1 d  A 1 d  B 
A   2B  
2 1/ 2 dt 2 dt
2 d  A 1 d  B   d  A 1 d  B 
or    
dt 2 dt dt 4 dt
32. The substances which have lower reduction potentials are strong reducing agents while
those which have higher reduction potentials are stronger oxidising agents.
 EM0 n  M for V,Fe and Hg are lower than that of NO3 , so NO3 will oxidize V,Fe and
Hg
33. A)d 6  High spin  ; t24g eg2
CFSE   0.4  4  0.6  2   0  0.4 0
B ) d 5  Low spin  ; t25g
CFSE   0.4  5  0.6  0   0  2.0 0
C ) d 4  Low spin  ; t24g
CFSE   0.4  4  0.6  0   0  1.6 0
D ) d 7  High spin  ; t 25g eg2
CFSE   0.4  5  0.6  2   0  0.8 0
Magnitude of CFSE is maximum for d 5 (low spin) complex
34.

35. The more nucleophile compound among the given will undergo more rapid electrophile
substitution reaction.  NHCOCH 3 group is more electron releasing group than
CH  CH 3 2 grope, exhibit more rapid electrophilic substituation reaction. Hence, the
correct order is
NHCOCH3
NO2

> > >

(iii) (iv) (i)

(ii)
36. The value of van’t Hoff factor for the given solutions will be the same, i.e., iA  iB  iC
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due to complete dissociation of NaCl (strong electrolyte) in dilute solutions. On
complete dissociation value of i for NaCl is 2.
37. -2 +4 0
2 H 2 S  SO2 
 2 N 2O  3s
38. Benzene can be prepared by cyclisation of long chain alkanes on heating at
500  5500 C under high pressure in presence of catalyst CrO3 supported on alumina or
Pt  Al2O3 (i.e., catalytic reforming).
CrO3 / Al2O3
C6 H14 
773 K
 C6 H 6  4H 2

n-hexane at high pressure (A)

Br MgBr

Br2 / Fe Mg / ether
    H 2O
 

(A) (B) (C) (D)

39. Conceptual
40. 
2NaHCO3   Na2CO3  CO2  H 2O
(X)
(Y) white residue,soluble water

Na2CO3  2 HCl 
 2 NaCl  CO2  H 2O
(Y)
(dil.)
41. The change given is occurring at the boiling point of the liquid, where, at given pressure
and temperature, the liquid-vapour system virtually remains at equilibrium and hence,
G  0 . Also due to absorption of heat as latent heat of vaporisation, or due to change
from liquid to gaseous sate where randomness has also increased, S  0 .
42. Physical adsorption results into multimolecular layers on adsorbent surface under high
pressure. While chemical adsorption results into unimolecular layer.
43. 2 HCl
CH 3 NH 2  COCl2   CH3  N  C  O
Methyl amine Phosgene Methyl isocyanate

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44. Cl

Toluene
 SbCl5 
Ph Me

SbCl6 
Ph Me

 form

stable benzyl carbocation

45. (+)-Lactose is a reducing sugar and shows mutarotation.

46. Conceptual
47.

48. Conceptual
49. Conceptual
50. Among the following options, only K 2Cr2O7 exhibits triclinic crystal system in which
      900 and a  b  c .
51. 1 xA xB moles of A 1
  XA    0.5  1
PT PoA PoB Total moles 2
52. I, II and III only
Ma4b2 , M (aa)2 b2 , M (ab)2 - show GI
M ( aa)3 - does not show GI.
53. ClF3 , ICl3 , HNO3 , SO2
54. HNO3  Sn  H 2 SnO3  4 NO2  H 2O
55. 1 18 18 dThe 41 50
dThe   Z    1.219  1.22
(0.0821)500 41 d Exp 36 41
100
56. Moles of CH3 MgBr  5 103
V = 0.112 L
57. Acetal form is non reducing while hemiacetals are reducing in nature.
58. H  E  ng RT  2.1  2  0.002  300  3.3kcal
G  H  T S  3.3  300  (0.02)  2.7 kcal
59. d 8 is Ni 2 and d 7 is Ni 3
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Total charge of Nickel is (0.96  2  0.04  3)  2.04
So number of oxide ion is 1.02.
Therefore formula should be Ni1.00O1.02
60. 8.4
Normality =  1.50
5.6

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MATHEMATICS
3
61. 16  3  3  1/2 x2  45
A  total area   48 , A1    2 3 x   dx 
3 0
12  4
45
4 15
Ratio  
45 49
48 
4
62. dy
 1  2 x  dx Integrate both sides
y2
63. 
1 1 1 
I   2  2  dx
0
2  x  1 x  3 

1 1  x   
I   tan 1 x  tan 1   I  
2 2 3  3  0 4 4 3
64. dx 1
I  , Put t  1 
8 1 
4/7
x7
x 1  7 
 x 
65. 7
2  a  b  ____(1)
2
1
2 a  b   ____(2)
2
Solve (1) and (2)
66. 50
200   2a  49d  .......(1)
2
100
2900   2a  99d  .......(2)
2
Solve (1) and (2)
67.  1   1   1  4      
cos  sin  tan  cos  sin  tan       
     3      
  
    4 
cos  sin 1  tan  cos 1     
    5 
  3  7
cos  sin 1    
  4  4
68. Conceptual
69. Conceptual

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70.  a, b 

 0,0 

1,5   2,4 
 b0  5 4
   1_____(1)
 a  0  1 2
 b  4   5  0  1_____(2)
 a  2 1 0
Solve (1) and (2)
71.

a 0 1
16
2 3 1  0 a 
5
6 7 1
72. 2
Equation of circle  x  3   y 2  36
2
 x  3  12 x  36 x  3, 9 x  3  y  6
73. Use R3  R2 and R2  R1
a2 b2 c2
  2a   2 2b   2 2c   2
4a 4b 4c
a2 b2 c2 a2 b2 c2
  4 3 1 1 1     4 2  a b c k  4 2
a b c 1 1 1
74. Tangent at point p  2,4  to the parabola y 2  8 x
y.4  4  x  2  y  x  2
Solve y  x  2 & y 2  8 x  5 and find mid point.
75. Conceptual
76. Conceptual
77. 2 2
 x 1   y  2  1, b2  a 2 1  e 2  a 2e2  16  9 ae  7 distance  2 7
2 2
4 3
78. 1 1

lim 2 cos1 x 1  x 2  lim  1

1

1
x 1 1 x1
cos1 x 1  x  2 2
2 x 1
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79. 1  11
x   a , x  7  4 , x  11  0 ,  r , r  12
2 2
1 0
80. 2
x  x 1 2
x  1 y  3 y  , x  y  1   y  1 x   y  1  0
x2  x  1
1 1 
 y  3  y   ,3 
3 3 
81. Conceptual
82. Conceptual
2 2
83.
2 f ' x f ' x  2 f  x  f ' y  ,
 f ' x  
 f  x  c
2 2
5 2 2
Put x  0  c 
2
 f '  x     f  x  5
dy
dx 
 y 2  5  n y  y 2  5  x  c 
Use f  0   2 y 2  5  y  5e x
19
Put x  n2 y   4.75
4
84. c
3a  b 
3   abc 1/3 abc  1000 log abc  3
10
3
85. c  c2 d1  d 2
Area of parallelogram  1
m1  m2
86.  8C2  6C2  28  15  420
87.  a 2  a  1 2a  1 
L , 
 2 2 

a 2  a  1 2a  1
Since L lines on line AB 3   6a
2 2
1
 3a 2  7 a  2  0  a  2,
3
1
Also, PQ  AB  3 2  1  a 2  a  2  0  a  2, 1
a  a 1
Common value of a  2
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88. Conceptual
89. Let h  x   f  x   g  x 
h ''  x   f ''  x   g ''  x   0
 h ''  x   f '  x   g '  x   k (constant)
h ' 1  f ' 1  g ' 1  k  2  4  k  k  2
h  x   f  x   g  x   kx  c
h  2   f  2   g  2   2k  c  3  9  2  2   c  6  4  c
c  2
h  x   2 x  2
3 3
h    2   2  5
2 2
90. cos6 x  tan 2 x  cos6 x tan 2 x  1  0
 
cos 6 x 1  tan 2 x  1  tan 2 x
1  tan 2 x
cos 6 x   cos 2 x
1  tan 2 x
cos 6 x  cos 2 x  0
2sin  4 x  sin  2 x   0
sin 4 x  0(or )sin 2 x  0
n n
x x (rejected )
4 2
Sol are 0,  / 4,3 / 4,  ,5 / 4,7 / 4,2

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