JEE MAINS MODEL PRACTICE TEST

GRAND TEST - 2
(KEY SHEET)

PHYSICS

01) 1 02) 2 03) 3 04) 1 05) 2

06) 1 07) 1 08) 1 09) 4 10) 2

11) 4 12) 1 13) 3 14) 2 15) 1

16) 1 17) 2 18) 3 19) 1 20) 2

21) 1 22) 4 23) 1 24) 1 25) 1

CHEMISTRY

26) 4 27) 2 28) 4 29) 2 30) 3

31) 2 32) 4 33) 4 34) 3 35) 2

36) 4 37) 2 38) 3 39) 2 40) 1

41) 1 42) 3 43) 1 44) 4 45) 3

46) 3 47) 4 48) 4 49) 3 50) 2

MATHEMATICS

51) 3 52) 2 53) 1 54) 3 55) 1

56) 1 57) 3 58) 3 59) 1 60) 4

61) 3 62) 1 63) 1 64) 1 65) 4

66) 4 67) 1 68) 4 69) 3 70) 3

71) 2 72) 3 73) 2 74) 4 75) 1

 PHYSICS HINTS & SOLUTIONS 3R/11
.
01. Sol: Since [Fv]  M1L2T 3
04. Sol. In minimum incidence case the angles will
 be as shown in figure.
So  2  should also be [M1L2T 3 ]
x 

[]  
2
 M1L2T 3  []  M1L4 T3 and  Fv  x 2  A
[x ] imin A-ic
will also have dimension [M1L2T 3 ] ic

[]
So 2
 [M1L2T 3 ]  []  M1L2T 1 .
[t ]

2V2 V
1
02. Sol. P  aV  W   aV dV  aV 3
2 2
3 V1
V 1 Applying snell’s law:
(P V )
1 2 2
1 7
 PV  (P2 V2  P1V1 )  1 R(T2  T1 ) . 1  sinimin  sin(A  ic )
3 ( P1V1 ) 3 3 3
03. Sol: Here we are drawing step by step 7
simplified circuit diagrams  (sin A cosic  cos A sinic )
3

7 3 3 1
  sin60 1   cos60 
In parallel 3 7 7 2

 imin  30
  max  imin  90  A  30  90  60  60 .
A B
In parallel /2 /2
A B
05. Sol: T1 C k1 k2 T2 C

C T0 C
R
R Let temperature of junction be T0 C
Temperature difference across slab
A R O R B
 A  (T1  T0 )C

R/2 Temperature difference across slab

B  (T0  T2 )C
Since both slabs are connected in series, heat
R current through both will be equal

R k1A(T1  T0 ) k 2 A(T0  T2 )

/2  /2
A,C R O B

 k2 k
R2 As k1   2 (T1  T0 )  (T0  T2 )k 2
2 2

R T1  T0  2T0  2T2

T1  2T2  3T0 ––––––– (1)

R/2 R As T1  T2  12C  T2  T1  12 –––––– (2)
A B From (1) and (2), we get

R/3
 T1  2(T1  12)  3T0 As the maximum current is through last bulb
and equal to 2A so for maximum power
3T1  24  3T0 dissipation with proper functioning
T1  T0  8C . I  2A
06. Sol: Maximum distance upto which a bullet can So, total power dissipated in circuit is
be fired is its maximum range, therefore I2 I2
P  2  R  3  R  I2 R
v2 2 v 4 4 9
Rmax  . Maximum area  (R max )  .
g g2 12 12
P   12  22 W .
07. Sol.   W / Q  (2000  100) / 10  103  20% . 2 3

F
11. Sol: Acceleration of two mass system is a 
2m
leftward
B
N
r

08. Sol.
60°

J 30°

Magnetic field intensity at any inside point of FBD of block A

  
current carrying is given by B  0 [J  r] . The mF
N3F .
2 N cos60  F  ma 
2m
above expression can be easily found by using
ampere’s law. Now using superposition 12. Sol. As elevator accelerates up at 2 ms–2, tension
principle we can find, the magnetic field at in the string is
any point inside the cavity. T  m(g  a)  4(10  2)  48 N
P Linear mass density of string is
  19.2  103 kg / m
O'
Speed of transverse waves on string is
O
 
T 48
   50 ms1 .
 19.2  103
Using principle of superposition
Magnetic field at any point inside the cavity
         dF B
is, B  0 [J  OP]  0 [J  OP]  0 [J  OO']
2 2 2
Which is uniform i.e., constant at all points in
the cavity and non-zero. 13. Sol. A
09. Sol. Conceptual/Basic Question. x dx
10. Sol: The current distribution in various branches
of the circuit are shown in figure.

I/3
I/2
Consider an element of wire AB as shown.
I I/3 I Force experienced by this element is,

 I 
dF  I  0  
dx 5 I2 dx
I/2  0 
 2x  cos37 8 x
I/3
 Force experienced by all the elements of the  2  100    10 rad / s
wire would be in same direction so integration  10 5
can be done directly. f   .
2 2 
13a
5 I2 dx 5 0I2  13 
F   dF  0  5
 n   . BvL BvL
8 a x 8  5 18. Sol: VC  VD  and VC  VA 
2 2
14. Sol. E1 (for 4th to 3rd excited state) So VA  VD  0 .
1 1 dN
 13.6  32  2  2   2.75 eV  N
4 5  19. Sol.Activity 
dt
E2 (for 3rd to 2nd excited state) These activities are same at t  0 , so

1 1 N1 : N2 : N 3  1 : 2 : 4 where N1 , N2 , N3 are the

 13.6  3 2  2  2   5.95 eV
3 4  number of initial nuclei of  ,  and deuteron
emission respectively.
For shorter wavelength i.e., for E 2 ,
Probability of  emission at t  1600 s is
V01  3.95 V
From eV01  hv    3.95  5.95      2eV 1N1
 16 1 1
For longer wavelength, 1N1  2N2  3N3  1  4  8  13 .
 
eV02  2.75  2  0.75 eV 16 4 2

So V02  0.75 V . 20. Sol. From the i  t graph, area from t  0 to

1
 6.28  104 t  2 sec   2  10  10 Amp. sec
15. Sol.      500 2
0 4   107
10
  1    Average Current   5 Amp .
2
      1  500  1  499 . L L
21. Sol.At equilibrium mg  A2g  Ag
16.Sol. At the time of Collision of spring is 2 2
m2 g 3
elongated by, x  mg  LAg
k 2
From momentum conser vation, If the block is depressed by the distance x
m1u  (m1  m1 )v where v is the speed of
 L  L  
combined block just after collision. At this F    x  A2g    x  Ag   mg
instant total energy of system is,  2  2  

(m1  m2 )v 2 kx 2 3
E   LAg  Agx  mg  F  Agx
2 2 2

m12u2 m 2g2  A g  m
  2 a  x  T  2 .
2(m1  m2 ) 2k  m  Ag
This is the oscillation energy.
17. .Sol. Given that potential lenergy is U  mV
22. Sol: V(O1 )  Kq  Kq
2 2
 U  (50x  100)10 R R 2  d2

dU Kq Kq
F  (100x )102 V(O2 )  
dx R d2 2 R
2 2
 m x  (100  10 )x
1 1 
3 2 2 V  V(O1 )  V(O2 )  2Kq   
10  10  x  100  10 x R R 2  d2 
 q q species]
Now, K.E.   T.E.
R R
 T.E.  4x and I.E.  T.E.  ( 4x)  4x .
O1 O2 30. Sol. Conceptual/Basic Question.
31. Sol. Polytetrafluoroethene (teflon) is a chain
d
growth polymer formed by polymerisation of
tetrafluoroethene in presence of persulphate
catalyst.
q 1 1 
   . 32. Sol.
20 R
2 2
R d 
OH
23. Sol. The electrostatic energy density is equal
CHO CH COOH
to magnetostatic energy density.
HCN
CN H O
24. Sol. For common base circuit current gain is, (CHOH)4  (CHOH)4  2
 NH3
 CHOH

IC CH2OH (CHOH)4

IE  IC  0.96 IE
CH2OH
COOH
From KCK, IC  IB  IE  IB  0.041 IE
HI, heat
(CH2 )5 

60 A
 IE   1.5 mA CH3
0.04
Heptanoic acid
IC  1.5 mA  60A  1.44 mA .
33. Sol. Detergents containing branched hydrocarbon
2
mv hc chains are non-biodegradable.
25. Sol. Let m is the mass of particle, 
2 photon 34. Sol. Conceptual/Basic Question.
where symbols have their usual meaning. 35. Sol. Conceptual/Basic Question.
36. Sol.
p2 hc h
  and p N 2 : ( 1s2 )(  *1s2 )( 2s2 )(  * 2s2 )( 2p2x  2p2y ) ( 2p1z )
2m  photon  particle
1
photon 2mc B.O.   (9  4)  2.5
    particle  2mc  h 2
 particle h h mv
1
2c N2 : B.O.   (10  4)  3
  40 . 2
0.05c
O2 : ( 1s2 )(  *1s2 )( 2s2 )(  * 2s2 )( 2p2z )
CHEMISTRY HINTS & SOLUTIONS
26. Sol. Conceptual/Basic Question. ( 2p2x  2p2y ) (  * 2p1x )
27. Sol. Conceptual/Basic Question.
1
B.O.   (10  5)  2.5
CH3 CH3 2
28. Sol.
 CH3  C  CH2  Br 
CH 3  C  CH2  Br  1
O2 : B.O.   (10  6)  2
H H 2
2°-Carbocation
Since N2 has lower bond order than N2 , bond
CH3 CH3 1,2-hydride
shift
CH3O
CH3  C  CH3 

CH 3  C CH3 length of N  N in N2 increases. In O2 , bond
CH3OH 

OCH3 3°-Carbocation order increases from 2 to 2.5 hence, bond

length decreases.
37. Sol. Conceptual/Basic Question.
K.E.1 Z12  n22 x 22  3 2
29. Sol. K.E.  n2  Z2 or K.E.  22  62 38. Sol. [Co(py)2 (H2O)2 C 2 ]C
2 1 2 2
and [Co(py )2 (H2O) C 3 ] H2O show hydrate
5
 K.E.2  4x [Z for M  6 as it is H-like isomerism.
 
39. Sol. 4 LinO3   2 Li2O  4NO2  O2 ;
COCH 3 COOH
 43. Sol.
2 NaNO3   2 NaNO2  O2 .
I2 /NaOH
40. Sol. 1) (F); As the size of halogen atom increases  H
 CHI3 
crowding on Si atom will increase, hence, (Y)
tendency of attack of Lewis base decreases. (X) (Z)
COOH
2) (T); M.P. of NH 3 is highest due to
intermolecular H-bonding in it. Next lower KMnO4 / H 
M.P. will be of SbH3 followed by AsH3 due to
high mol. wt. of SbH3 . (Z)

3) (F); M.P. and B.P. of increase from PH3 to 44. Sol. For Buffer solution, we should have two
SbH3 via AsH3 due to increase in mol. wt. NH3 constituents i.e., a weak & it’s conjugate base .
does not follow this trend due to inter alc.KOH
molecular H-bonding. 45. Sol. H2C  CH2 
 KBr
 CH2  CHBr
 H2O
4) (T); Value of bond moment decreases. Br Br (X)
41. Sol. Work done, w  Pext .dV NaNH2

 NaBr
CH  CH
 NH3
W  2  2.5  5 L atm  506.3 J (Y)
Because this work is used in raising the
temperature of water, so work done is equal 46. Sol. pH   log [H ]  14  [H ]  1014
to the heat supplied i.e., w  q  m  c sT Hence, [OH ]  1 M [H ][OH ]  104
Given that, m  18g (  1 mole) ,
Cu(OH)2  Cu2  2OH
cs  4.184 J g 1K 1 ,
K sp  [Cu2 ][OH  ]2
q  506.3 J , T  ? (Heat is given to water)
For the reaction, Cu2  2e   Cu , n  2
q 506.3 0.0591 1
T    6.72 o
E cell  Ecell  log
cs  m 4.184  18 n [Cu2 ]
 Final temperature, Tf  Ti  T  293  6.72
0.0591 1
299.72 K  300 K . Ecell  0.34  log  0.22 V .
2 1 1019
42. Sol: Initially, number of moles of gas in each 47. Sol. For bcc, Z  2
pi V pi V
bulb is n1  RT and n2  RT d
ZM d  NA  a3
1 1 3 M
NA  a Z
After the temperature of second bulb is
raised to T2 then the number of moles of 4  6  1023  (500  1010 )3
 M
pf V 2
gas in both the bulbs are n'1  and
RT1 4  6  1023  125  1024
  150 g mol1 .
pf V 2
n'2 
RT2 Now, the total number of moles of 48. Sol. Molarity of Na 2CO3 solution 2.65 1000
   0.1 M
106 250
gas in both the bulbs remains same in both
10 mL of this solution is diluted to 500 mL
the cases. n1  n2  n'1  n'2
M1V1  M2 V2
2pi V pf V pf V 2p V p V  T  T1 
   i  f  2  0.1  10  M 2  500  M2  0.002 M .
RT1 RT1 RT2 RT1 R  T1T2 
3.6
2piT2 49. Sol. Mass of O 2 per gram of adsorbent  3
pf  1.2
T1  T2 .
Number of moles of O 2 per gram of adsorbent
 3 n 
    n  ( 1)   
32  6
Volume of O2 per gram of adsorbent 7
n
or n  ( 1) .
3 0.0821  273 6
   2.10 .
32 1 60. Sol: Let A  (h, k)
50. Sol. Conceptual/Basic Question.
k 4 4
MATHEMATICS HINTS & SOLUTIONS m AH  , mOB 
51. Sol: Conceptual/Basic Question. h 1 3
52. Sol: Conceptual/Basic Question. (k  4) 4
  1  4k  3h  19 ––––– (1)
53. Sol: Conceptual/Basic Question. (h  1) 3
54. Sol: Conceptual/Basic Question.
k 44
55. Sol: 2x  3y  1  0 mOA  , mBH  0
h 3 1
3x  y  2  0  x  1 , y  1 Since OA and BH are mutually perpendicular,
If the equations are consistent, then it implies that h  0 .
a(1)  2( 1)  b  0  a  b  2 . 19
Putting h  0 in eq. (1), we get k 
1
56. Sol: sin (sin5)  x  4x 2 4

 sin1[sin(5  2 )]  x 2  4x  19 
Thus, orthocentre is  0,  .
2
 4
 x 2  4x  5  2  x  4x  (2  5)  0
61. Sol: Given, (x  y )  ( y  x )  a –––– (1)
 (x  2)2  9  2  x  2  9  2 .
2x
 y2 y2 
 x  y  y x  ––––– (2)
57. Sol: f  2x 2  , 2x 2    xy a
 8 8  Adding eqns. (1) and (2), then

2 2 2x
 y 2   2 y2  2 xy a
  2x 2     2x   a
 8   8 
2
Squaring, 4x  4y  a 2  4x  4x
 f(60, 48)  f(80, 48)  f(13, 5) a2
 (60) 2  ( 48 )2  ( 80) 2  ( 48 )2  (13) 2  ( 5) 2 dy 8x
 44 0 2 4
 36  64  12  112 . dx a
2
A B 5  04d y  8
58. Sol: tan  tan  ,
2 2 6 dx 2 a 2

A B 1  A B d2 y 2
tan  tan   tan   1   .
2 2 6  2  dx 2 a 2

 62. Sol: We have, z  (1  icot8)3

 A  B  2 .
4  cosec3 8(sin8  icos8)3
59. Sol: Given 2sin2   3sin   2  0 3
3    
1  cosec 8 cos   8   isin   8  
 sin   2 ,   2  2 
2
  3   3 
1   7  cosec3 8 cos   24   isin   24  
 sin     sin    or sin   2   2 
2  6 6
3 i( 24  3  /2)
 cosec3 8  ei( 3 /2 24 )  cosec 8  e .
63. Sol: Let the ‘n’ consecutive integers be P
2k  1, 2k  3, 2k  5,...., 2k  2n  1

n O (0, 0)
their sum  [2k  1  2k  2n  1]
2 q
Q R
 n(2k  n)  n2  2kn  (n  k)2  k 2 (3, 4) ( 4, 3)

Given, (n  k)2  k 2  252  112 3 4

Slope of OR  , slope OQ 
 k  11 , n  k  25  n  14 . 4 3
 OR and QO are perpendicular..
64. Sol: Value  ( 40 C9  40C10 )  41 C11  ...  50C20

 ( 41 C10  41C11 )  42C12  ...  50C20  
2
 ...  50C19  50C20  51C20 .  
Hence, QPR   .
65. Sol: Let the required plane be ax  by  cz  0 2 4
We have 3a  5b  2c  0 , a  2b  3c  0 70. Sol: Shortest distance will take place along the
common normal.
a b c
  
15  4 2  9 6  5 y
 a : b : c  11 : 11 : 11 A
Thus plane is x  y  z  0 .
x
66. Sol: Total number of even numbers in the set is O (6, 0)

18 and total number of odd numbers is 16. C

A : Sum of selected numbers is even.
B : Selected numbers are odd. y'

18
C2  16 C2 16
C2 Equation of normal for y 2  4x at (at2 , 2t) is
P(A)  34
, P(A  B)  34
C2 C2
y  tx  t  t3
16
 B  P(A  B) C2 40
P    16 18
 . If it passes through (6, 0), then t3  5t  0
A P(A) C2  C2 91
 t  0, t  5
2 2
67. Sol: I   (ln x ) dx  x  (ln x )  2  ln x dx
 A  (5, 2 5 ) , C  (5,  2 5 )
2
 x(ln x )  2(x  ln x  x)  C
Now, PA  PC  1  20  21 , OP  6
2
 x[(ln x )  2 ln x  2]  C .
 Minimum distance  ( 21  5 ) units .
2 2
68. Sol: (2y  6xy )dx  (3x  8x y)dy  0 NUMERICAL VALUE QUESTIONS:
 2y dx  3x dy  6xy 2 dx  8x 2 y dy  0 1
71. Sol: Let sin x   , then x  sin 
 2xy dx  3x 2 dy  6x 2 y 2 dx  8x 3 y dy  0  
Since,   sin1 x 
3 2 2 2
 2xy dx  3x y dy  6x y dx  8x y dy  0 4 3 3 2 2

 d(y 3  x 2 )  2d(x 3 y 4 )  0  
  
2 2
 x 2 y 3  2x 3 y 4  c .

69. Sol: Let O(0, 0) be the centre of the given circle. Also as x  1 ,   h
2
1  
Clearly, QPR  QOR  . 2
2 2  Required limit  lim

  cos 
2
 h (m, n)  {(0, 0), (1, 0), (0, 1), (1, 1), (2, 0), (3, 0)} .
 lim
h 0
   Number of terms free from radical sign is 6.
cos   h 
 2  Second Method :
General term in the given expansion
  
Putting   2  h, h  0 
 
10!  3 7
 1  3  7 , where       10
 !!  !
h h  Possible value of  are : 0, 3, 6, 9
 lim  lim 1.
h 0 sinh h  0 sin h
Possible values of  are : 0, 7
72. Sol: y  f(x )  x  1  x  2  x  3 Possible value of  are : 0, 1, 2, ..., 10
or y  3x  6 ,   x  1 Since,       10

 x  4 , 1  x  2  Possible triplets ( , ,  ) will be (0, 0, 10),

 3x  6 , 2  x   (0, 7, 3), (3, 0, 7), (3, 7, 0), (6, 0, 4), (9, 0, 1).
75. Sol: After shifting the origin at the point
dy
  3 ,   x  1  1, 1  x  2 (2,  1) , the equation of curve becomes
dx
x  y  1 . This curve will represent a square
e
 3, 2  x  
as shown in the figure below.
dy Y
Sign scheme for is
dx
B
dec. min. inc y x 1 x  y 1
 
 ve 1  ve 2  ve C A
X
O
Thus, y has least value at x  2 . x  y  1 x y 1
Least value of f(x)  f(2)  2 .
D
b c
73. Sol: x1  x 2   , x1 x 2  Clearly, area of this square  4 times the area
a a
of triangle OAB  2 sq. units.
1 1 x12  x 22
Now,   2 2
x12 x 22 x1 x 2
***THE END***
(x  x 2 )2  2x 1x 2
 1
(x 1x 2 )2

2
 x  x2  2 b2 2a
 1    2 
 x1x 2  x1x 2 c c

b2 2a b
We have, 2
 
c c a

b2 2ac bc b2 bc
    2   2
c c a ac a

b2 bc
   2.
ac a 2
74. Sol: The term free from radical sign must be of the
form (constant) (31/3 )3m  (71/7 )7n where
3m  7n  10 and m, n are non-negative integers.
This inequality has 6 solutions