SRI GURU JUNIOR COLLEGE

JEE ADVANCE MODEL PAPER-1 Date: 24-04-2025

GRAND TEST - 01

ANSWERS, HINTS & SOLUTIONS

Physics PART – I

Section – A

1. AD
Sol. So, velocity of first particle
= 3 cos 37° î + 3 sin 37° ĵ
12 ˆ 9 ˆ
= i j
5 5
Velocity of second particle
= 4 cos 53o ˆi  4 sin 53o ˆj
12 ˆ 16 ˆ
= i j
5 5
So, relative horizontal velocity is zero. So their relative velocity is in vertical direction
only. Since, both particles are moving under gravity, so their relative acceleration is zero.
16 9 7
Their relative velocity =   = 1.4 m/s
5 5 5

2. BC
Sol. Just after BP is cut.
For block A no force has changed.
 acceleration of m1 = 0
for m2 downward force is being reduced
 m2 will move upwards.
3. AC 

Sol. dw  f  ds
Since, body is hauled slowly, so
f = mg sin  + mg cos
W =  (mg sin   mg cos )ds

  mgds sin    mgds cos 

  mgdy   mgdx

= mgh + mgL

4. AC
Sol. 5 – F1 = 1 × 2  F1 = 3 N
Taking torque about CM:
5x = 3( + x)  2x = 3 × 20  x = 30 cm

Length of rod = 2 × ( + x) = 100 cm = 1 m

5. AC
Sol. AC = 5 m
kq 9  109  1 10 6
V 
AC 5
3
= 1.8 × 10 = 1.8 kV
VB = (VB)due to q + (VB)i
(VB)i = –0.45 kV
So, (A) and (C) are correct.

6. BC
V2
Sol. For a bulb R 
W
 RB  RA
when switch is open IA = IB
2
PA  R AIA
2
PB  RBIB
 PB < PA and VA > VB
 VA > 12V and VB < 12V
After closing the switch
VA = VB = 12 V
7. A
Sol.
(I)

Pi  Pf  60(1  V)  100 V
3 3
 V opposite to velocity of Ram i.e. m/s towards right.
8 8
(II) 80 V = 80 (1 – V)
1
 V = m / sec left.
2
(III) 80(1 + V) + 60(–1 + V) + 20V = 0
1
V   m/s
8
(IV) After jump of Ram 3/8
3
Now (80  20)  80(1  V)  20 V
8
17
V m /s
40

8. B
3
Sol. (I) Velocity of fish in air = 4  = 3
4
Velocity of fish w.r.t. bird = 3 + 6 = 9
3
(II) Velocity of image of fish after reflection from mirror in air = 4  = 3
4
w.r.t. bird = –3 + 6 = 3
4
(III) Velocity of bird in water = 6  = 8
3
w.r.t. fish = 8 + 4 = 12
(IV) Velocity of bird in water after reflection from mirror = 8
w.r.t. fish = 8 – 4 = 4

9. D
r
2br 3
Sol. I   2rdrbr 
0
3
Use ampere law for B.

10. C
 dB 
Sol.  B  d  A  dt 
 Section – B

11. 7.50
Sol. The centre of the wheel is moving with constant speed on a circular path of radius 6R.
V2
Hence, it has a centripetal acceleration of ac  directed towards the centre of
6R
curvature of the convex surface.
V2
With respect to the centre of the wheel, the contact point has acceleration equal to
R
directed towards the centre of the wheel.
 Acceleration of the contact point in reference frame of ground is
V 2 V 2 5V 2
aP    .
R 6R 6R
12. 2.56
Sol. For a jump of h0 = 1 m on the earth, speed required is given by
1
mV 2  mgh0  V  20 m /s
2
Escape speed on the surface of a planet is
2GM 8 GR2
Vesc  
R 3
Vesc planet Rplanet
 
Vesc  earth Rearth
planet
We want Vesc = 20 m /s
planet
And it is given that Vesc = 11.2 km/s
20  (6400 km)
Rplanet  = 2.56 km
11200
13. 400.00
Sol. If force exerted on piston of area A2 is F2 then, the force acting on the other piston will be
F
= 2  A1 [Pascal’s law]
A2
A 
= 5F2   1  5
 A2 
To raise the load 5F2 = 20000  F2 = 4000 N
Since lever bar is light, net torque on it (about the hinge) must be zero.
 F2a = F(a + b)
Fa 4000  4
 F 2  = 400 N
ab 4  36
14. 283.40
Sol. The product nucleus 198Hg is in excited state and possesses extra 1.088 MeV energy. If
198
Hg would had been in ground state, the kinetic energy available to electron and
antineutrino must have
Q = (mAu – mHg) 931 MeV
 = (197.968233 – 197.966760) 931 MeV
= 1.3714 MeV
Since 198Hg is in excited state, actual kinetic energy available to electron and
antineutrino is
K = (1.3714 – 1.088) MeV
= 0.2834 MeV
As -ray and antineutrino has continuous spectrum starting from zero value, therefore,
this is also the maximum kinetic energy of the electron emitted.
15. 7.00
Sol. x = (1 – 1)t1 – (2 – 1)t2
= (2 – 1)200 – (1.5 – 1)200
= 200 – 100 = 100 nm
2 2 
  x   100 
 600 3
 P  P1  P2  2 P1P2 cos 

= I0 r12  I0 r22  2 I0 r12  I0 r22 cos
3
= 7 W
16. 4.01
Sol. As here volume of gas remains constant,
(Q)v = CVT, Here Cv = 5 cal/mol K
And T = (400 – 300) = 100 K
And so for ideal gas PV = RT,
(10)5  (0.2)
 = 8.0224
8.31 300
(Q)v =  × 5 × 100 = 4.01 kcal.
17. 3.14
F L mg
Sol. Y   Y()
A L A
AY( ) r 2 Y( )
m 
g g
(10 3 )2  1011  10 5  10
= =
10
18. 30.00
Sol. f  T for strings.
On increasing the tension by 1%
f   1.01 T
1
f 1.01 T 1
  (1  0.01) 2  1 
f T 200
 f 
Beat frequency, f   f  f   1 = 1
f 
Number of beats in 30 seconds = 1 × 30 = 30.
Chemistry PART – II

Section – A

19. A
Sol. O O
O
O

O O O O O O

O O

O O
O O O O O

20. BC
Sol. 2C  3H2  C2 H6
s g  g
0 0
H  2  H  3  B.E H  H   B.E.  C  C   6  B.E  C  H
f sub

 85   2  718    3  436     x  6y 

 x  6y  2829 …..1
Similarly for C3H8  g 
2x  8y  4002 …..2
Solving (1) & (2), x = 345
y = 414

21. AB
E a
Sol. k  Ae  Ea /RT or nk  nA
RT

22. ABC
Sol. The outermost electronic configuration of Yb is 4f 146s2
 Yb2+ has full-filled 4f14 configuration radius of Yb3+ <Yb2+

23. AD
H2O
Sol. Be 2C   Be  OH2  CH4
H2O
Al4 C3   Al  OH 3  CH4

24. ABCD
H 40,600
Sol. S    108.84 j / K for one mole
T 373
9 108.84
for mole, S   54.42 j / K
18 2
25. A
Sol. I – can undergo Nu– substitution, elimination
II – can undergo Nu– substitution, esterification, dehydrogenation & oxidation.
III – can undergo Nu– addition, esterification and oxidation
IV – can undergo Nu– substitution

26. A
Sol. Fact based

27. C
Sol. Elements having filled d-orbitals are not true transition elements.

28. B
Sol. SO32  dil.H2SO 4 
 H2O  SO 2  SO24
CO32  dil.H2SO 4 
 H2O  CO 2  SO 42

NH4  NaOH  Na   H2O  NH3
S 2  H2SO 4 
 H2S  SO 42

Section – B

29. 162.00
2

Sol. K f  of camphor  
 
R Tfo  M2

2  152   450 
2

 40.5
1000  Hof 1000  1.52  1000
Tf  i  K f  m
0.04  1000
450  430  1 40.5 
Msolute  0.5
0.04  1000  40.5
Msolute   162
0.5  20

30. 81.81
Sol. NaOH  HCl  NaCl  H2O
Millimoles 0.1 .001 x 0 0
-0.001 x -0.001x
_________________________
t=t 0.1 - .001 /x x
10–3 M NaOH have pH = 11
New pH after adding acid be 10
Hence, 10–4 M (100 + x) = 0.1 - .001 x
0.1  0.001x
100 + x =
10 4
 11x = 1000 – 100 = 900
900
x=  81.81 mL
11
31. 390.55

Sol.  0mCH3COOH 
 0
m Ca  CH3 COO 2  2 m0 HCl   m0 CaCl2 
2
200.8  2  425.95   271.6
  390.55
2

32. 1.50
Sol. Vander waals equation for 1 mole of real gas, when b = 0
 a 
  P  2   V   RT
 V 
1
 PV  a   RT
V
y  mx  c
Slope = tan(–) = –a
21.6  20.1
So, tan  = a =  1.5
32

33. 11.00
Sol. The product contains three phenyl group and two multiple bonds
 Total number of pi-bonds = (3  3) + 2 = 11

34. 2920.40
Sol. Meq of MnO 4 = Meq of Fe2+
w
VMn=  1000
E
w
or, 104.3  1000  0.1  5 = 1000
56
1
On solving w = 2920.4 g
35. 8.33
200
Sol. The average O.S of iron in Fe0.96O is
96
Let the % of Fe(II) be x
100  200
x  (+3) + (100 - x)  (+2) =
96
3x + 200 – 2x = 208.33
 x = 8.33
36. 1.41
ZM
Sol. d
Na  a3
45 z  27
 3
16 6.02  1023  4  10 8
 
 Z = 4 (so fcc)
2a o
r  1.41 A
4
Mathematics PART – III

Section – A

37. AC
  
Sol.  
V  a. b  c  a12  a22  a32 b12  b 22  b32 c12  c 22  c 23 …….(i)
1/3
 a1  a 2  a3  b1  b2  b3  c1  c 2  c 3   (Using A.M.  G.M.)
L3   a1  a2  a3 b1  b 2  b3  c1  c 2  c 3 
2
now  a1  a 2  a3   a12  a 22  a32  2  a1a2  a2a3  a3a1 
2
a1  a2  a3   a12  a22  a32
 a1  a2  a3   a12  a22  a32
similarly b1  b2  b3   b1  b2  b3
2 2 2

c1  c 2  c 3  c12  c 22  c 32
1/2
  
L3   a12  a 22  a32 b12  b22  b23 c12  c 22  c 23  
L3  V
38. BC
3 3 2
Sol. We supposed to find m and n such that lim 3 8x  mx  nx  1 or
x 
3
lim 8x 3  mx  nx  1.
x 
We compute

3 3
8x  mx  nx  2 8  n  x
3 3
 mx 2
.
2
3
8x 3
 mx   nx
2 3 3
8x  mx  n x 2 2 2

8  n3 must be equal to 0
n2
m
Now f  x   .
2
 m m
 8  x   23 8  x  4
3
 
m
We see that lim f  x   . For this to be equal to 1, m must be equal to 12. Hence the
x  12
answer to the problem is (m, n) = (12, 2).
39. ABCD
x4  1 x 4  x2  1 x2 1 1 x '
3
 
Sol.  x 6  1 dx   x 6  1 dx  x 6  1 dx   x 2  1 dx   3 x3 2  1 dx
 
1
 arc tan x  arc tan x 3 .
3
To write the answer in the required form we should have
P x
3 arctan x  arc tan x 3  arc tan
Q x
Applying the tangent function to both sides, we deduce
3x  x 3
2
 x3  P x 
1  3x  tan  arctan  .
3x  x 3 3  Q  x 
1 .x  
1  3x 2
P x 3x  3x 5
From here arctan  arc tan , and hence
Q x 1  3x 2  3x 4  x 6
P  x   3x  3x 5 ,Q  x   1  3x 2  3x 4  x 6 . The final answer is
1 3x  3x 5
arctan  C.
3 1  3x 2  3x 4  x 6
40. AC
ab
Sol. Denote the value of the integral by I. With the substitution t  we have
x
b t t b
t a a t
b e  e ab be e
I . 2 dt    dt  I .
a ab t a t
t
Hence, I = 0.

41. AB

Sol. 3 tan3x 

3 3 tan x  tan3 x   3 tan 3
x  9 tan x
2
1  3 tan x 3 tan2 x  1
8 tan x
 .
3 tan2 x  1
Hence
1 tan x 1 
 2
  3 tan 3x  tan x  for all x  k , k   .
cot x  3 tan x 1  3 tan x 8 2
It follows that the left – hand side telescopes as
1
8

3 tan 27o  tan9o  9 tan81o  3 tan 27o  27 tan243o  9 tan81o  81 tan729o  27 tan 243o 
 1

8
 
81tan9o  tan9o  10 tan9 o .

42. BC
Sol.

A
O
(4, 8)

B
P
(–2, 0)

2 5 1
sin   
10 5
8 4
Slopes of PA and PB are tan      where tan   
6 3
4 1 4 1
 
 3 2 , 3 2
4 1 4 1
1  . 1 .
3 2 3 2
11 5
 ,
2 10
  11   2 
 A,B   4  2 5   ,8  2 5   ,
  5 5   5 5 
  1   2    2 44 
4  2 5  , 8  2 5     5 , 5 
  5   5   
  6, 4 

43. B
y
Sol. (I) A, B, C are the 3 critical points of y  f  x  . A

f "  x   0 for x  2 and fails to exists at x = 0.

c
B
x
0 1 2
 1
(II) x  and 2. Make a quadratic in log2 x and
4 3x + 4y = c
interpret the result.
x1, y1

dy 3 1
(III)  1  2x13    x1 
dx 4 2
1 1 15 3
   y1 or y1   c
32 2 32 8
3
(IV) f '  x   2x  3x  1 this is always positive in (1, 2).
 Increasing in [1, 2]
 f  2  will be the greatest value.

44. D
1
tan
 n2  n o
Sol. Let L  lim 
n n  1
   
 
1  n2  2ln  ln  n  1   
In L  lim tan ln    lim  
n n  n  1  n  1  
cot  
 n
2
Put n  t  lim
4ln t  ln t 2  1  
2 t2  2  
t   1  1  1 
 
cot    lim t t 2  1 cos ec 2    2 
 t  t  t  t 
2
 1
2 t  2  sin t 
 2

 lim  1  0

t  t t 2  1
  
 t 
L  1   Q
3
(III) sin 2x  
2
2x   0, 4 
Hence, 8 solutions.
(IV) Since g  x  is differentiable  x  R
 f  x  must have the factor x  x  1 x  2  atleast once.
 minimum 3 roots of f  x   0  R 
45. A
Sol. x bbb x c x c x c x
4! 5
Number of ways   C4  20
3!
(II) 2b, 1b; 2c, 1c or 2b,1b; 1c,1c,1c or 2b, 1b; 3c
(same way starting with c)
cbbcbc, cbcbbc
bbcccb, bcccbb
Number of ways  12  6C4  180
7
(III) bcbcbc  C4
bccbcb or bcbccb  2  6C3
bbccbc or bccbbc or bbcbcc or bcbbcc  4  5C2
bcccbb or bbcccb  2  4C1
bbbccc  4C4
Total ways  2  35  40  40  8  1  248
(IV) bcbcbc, cbcbcb
bccbcb, cbbcbc
bcbccb, cbcbbc
bcbccb, cbcbbc
7
number of ways  C1  2  4  18

46. D
1 11C5 1
Sol. (I) P  12 
C6 2
2
C1  10C5 6
(II) P  12

C6 11
10
C4 5
(III) P  
12
C5 22
10
(IV) P 
11
 Section – B

47. 252.00
2018 2018
Sol. x  3 1  and y   3 1 
2018 2018
 x   x  y   3 1    3 1 
1009 1009 
 x   1  21009  2  3  
 2 3  

 21009  2  1009 C0 21009  1009C2 2100731  .....  1009C1006 23 3503  1009C1008 2  3504 

 x   1  21011  1009 C0 21008  .....  1009C1006 22 3503  1009C 1008 3504 



odd
[ {x} + y  (0, 2)  {x} + y = 1]
 N is divisible by 21011, hence divisible by (16)252

48. 13.00
n n 1
7
Sol. Coefficient of x 2  7  1  6  2  5  3  4   1  2  4   13

49. 18.00
2
Sol. We have          2  2   2  2       
 4  6  2      
       1 .
Also, 3  3   3  3 

        2  2   2       
50. –1.59
Sol. The critical point of f are solutions to the system of equations
f 9
 x, y   4x3  12xy2   0,
x 4
f 7
 x, y   12x 2 y  4y 3   0
y 4
divide the two equations by 4 and then add, respectively, subtract them, we obtain
1
x 3  3x 2 y  3xy 2  y 3  1  0 and x 3  3x 2  3xy 3  y 3 
. We write these as
8
3 3 1 1
 x  y   1 and  x  y   , from which we obtain x  y  1 and x  y  . We
8 2
3 1
find a unique critical point x  , y  . The minimum of f is attained at this point, and
4 4
 3 1 51
it is equal to f  ,    .
4 4 32
51. 126.00
  5  2 
Sol. We have, lim gn  3   lim n  f  3    f 3  
n n
  n  n 
f  3  h   f 3  f 3  f 3  h
 5 lim  2 lim
h 0 h h 0 h
 5f '  3   2f '  3   5  18  2  18  90  36  126 .

52. 1.01
Sol. y  x2 y  2  x2 y2

2 2
Required area    
 x2  2  x2  
  2   x
2

 2  dx
1 2

20
  4 2  1.01
3
53. 18.00
Sol. The points  x1, x2  and  y1, y2  lie on the circle of radius 2 2  c centered at the
origin. We can write  x1, x2    c cos ,c sin   and  y1, y2    a cos ,c sin   .
Then
s  2  c  cos   sin   cos   sin    c 2  cos  cos   sin  sin  
     
 2  c 2   sin      sin  0     c 2 cos      to 1 by choosing
  4  4 
5 2

4
2
. The maximum of S is 2  2c 2  c  c  2 .  
54. 11.00
 1  1 
Sol.  x  2   1 or  x  2  1
   
1 1
x   1,2  x   1, 2 
2 2
3 5  1 3 
x   ,  ….(1) x   ,  ……..(2)
2 2  2 2 
1   2 
 1 3  3 5 
x ,  , 
2 2  2 2 
1 9 9 25
x12  x 22  x 32  x 24      11
4 4 4 4