AITS-2425-PT-II-JEEM-Sol

FIITJEE
ALL INDIA TEST SERIES

PART TEST – II
JEE (Main)-2025
TEST DATE: 01-12-2024
ANSWERS, HINTS & SOLUTIONS

Physics PART – A
SECTION – A
1. B
Sol. Use principle of superposition, 
 2K    
E2    sin  
 R  2
2K 1 ˆ K ˆ /2 /2
  i i R
R 2 R
E2
2. C
Sol. dq  (2xdx)
dq 2xdx  
di    xdx
dt 2
0 2(x3 dx)
dB  
4 (y2  x 2 )3/2
0   r 2  2y 2 
B  2y 
2  r 2  y2 
 
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AITS-PT-II-PCM(Sol.)-JEE(Main)/2025 2
3. D
Sol. Equivalent circuit is 2RT RT
V
P (90C)
RT R (30C)
4. C
Sol. by solving
x T
dT Q
Q  KA
dx

A 
0

dx  K 0 (1  T)dT
0
T
Q  T2 
 x  K0  T  
A  2 
 0
By solving
Q  T2 
x  K0  T  
A  2 

Q  (300)2

So, x0  K 0  300  
A  2
 
So, at x = 2x0 temperature T  425 K
5. B
Sol. P = VI
6. B
Sol. Let x be the temperature of block. In steady state
x  10 x  5 x  3
   0  x = 6C
R R R
7. C
n1(4) 1
Sol. Now,   n1 = 2n2
n2 (32) 4
3  5 
n1  R   n2  R 
2   2   11R
Now, Cv  
n1  n2 6
17
CP  CV  R  R
11
CP 17 6
    1
CV 11 11
8. D
d R2
Sol.  B
dt 2
R2
Where, is area swept in unit time perpendicular to the magnetic field.
2
3 AITS-PT-II-PCM(Sol.)-JEE(Main)/2025
9. D
Sol. The voltage across the resistor R is equal to the voltage across the coil
UR = UL
Voltage across the resistor
UR= IRR
Voltage across the inductor coil:
dI
UL  L L
dt
Current through a resistor
dq
IR  R
dt
Then
dqR dL
R  L L  RdqR  LdIL
dt dt
According to Kirchhoff's second law
  INCT  IR  R
Then the current through the resistor at the moment of opening


(2R)r 
IR  
3r 6r
Then the current through the coil from Kirchhoff's first law:
  
IL  INCT  IR   
2r 6r 3r
We sum (integrate) (1)
qR
IL

R dqR  L
0
0
dIL  RqR  LIL
Taking into account (2)

L  L
qR   2
R 3r 9r
10. D
2
Sol. i
10
 2  12
VBD  6     1.2 V
 10  10
11. C
X 40
Sol.   R = 6
R0 60
78R1
and 6   Rt= 6.5 
R t  78
R t  R0
  8.3  104 K 1
R0 t
12. B
Sol. R  37  102  5%
=  (3700  185)
13. D
Sol. Potential difference A and C = 4V
6
Current in above circuit =  1A
51
So, resistance of AD = 4
Hence length = 80 cm
14. C
1 q q q  1  5q 
Sol. VC   R  2R  3R   4  
40   0  6R 
15. C
1 1 1 1 1
Sol.    
R 20 20 30 30
16. A
  
Sol. Eq  E8q  0
Kqx K(8q)x
 
(R2  x2 )3/2 (16R2  x2 )3/2
 x = 2R
1 Kq  q K(8q)  q
 mv 2   
2 2
(R  x )2
(16R 2  x 2 )
v = 20 m/s
17. A
3 4
Sol. Ex  , Ey 
0 0
5
 Enet 
0
1 4 
 U 0E2  R3   0.45 J
2 3 
18. A
0 id sin(90   ) d
dr
Sol. dB  
4 r2 rd
 i 
dB  0 2 d cos 
4r r

 i  id
dB  0 2 rd  0
4r 4r
0id
dB 
 c 
4  b   
  
B  /2
0I0 d I  c 
  dB   = 0 0 n  1 
 c  4c  2b 
0 0 4  b   
  
19. C

2
Sol. d  B sin  xdx 
0
x
I22 

  d = (B sin )
2 dx
1 1
=  4  1   1
2 2 p
20. A
     y
Sol. B  B1  B2  B3  B 4
   i
B2  B3  0 (cos 1  cos 2 )jˆ
4a
  0 i ˆ 4
B2  B3  j
4 2 a
  i
B4  0 kˆ O
4 a O 1 x
(a, 0, 0)
  i
 B  0 ( 2ˆj  k) ˆ
v2
3
4a (0, 0, a) v1
2
1
SECTION – B
21. 2
Q
Sol. Let us consider a cube of double side length of same density. Also, V  and V becomes 4
r
1
times on doubling the side length. Let the potential at center due to of this cube is V1. This
8
point lies at corner of each of eight cubes of original size.
22. 8
Sol. Potential of plate 4 is zero 2Q
2Q
Q
 (V3  V4) = V3
 2Q   Qd 
(V3  V4 )    2d  4  
 A 0   A 0 
V3 = 8 volt
2Q Q 2Q
23. 4
Sol. H  i2Rt
2
200 = 2  R 1
 200 = 4R
1
H2 = 1  R  8 = 400 Joule
24. 3
 0 J 0 J
Sol. B   0 J
2 2
B2
U  6L3  30 J2L3
20
25. 9
Sol. Hint: According to stefan's law, the power radiated by a black body at absolute temperature T is
given by
 = AT4 …(i)
According to wein's displacement law
b
m T  b  T 
m
From (1) and (2)
4
 b  Ab4
  A    2
 m  m
For a sphere of radius r, A = 4r2
b4 4r 2 r2
Hence  = 2
 K 2
m m
Where K = 4b4 is a constant.
42
Hence 1  K 1
m2
  1
r22
2  K
 
m4
2
2 2 4 2
1  r1   3   500  5
2  r2 
4
    m   2
    4
  m
 5   300 
  1
 
3
Chemistry PART – B
SECTION – A
26. B
Sol. Possible products
OH
OH
O O O O
H3C H H3C H
Plane of symmetry Two chiral centres

(Optically inactive) (4-isomers)
27. C
Sol. OH
CH It gives yellow ppt. of CHI3 with I2  NaOH.
H3C CH3
Q
2 o
alcohol 
28. C
Sol. O
O O O O
Et3 N MeO Cl
Ph C OH   Ph C O  Ph C O C OMe
 Cl
O NH2
MeO C OH
Ph C NH
29. A
Sol. Factual
30. D
Sol. Me Me Me Me
Br H H Br Br H H Br
Br H H Br H Br Br H
Me Me Me Me
(i) (ii) (iii) (iv)

(iii) and (iv) are non-superimposable mirror image.
31. A
Sol. Factual
32. B
Sol. O OD
1. LiAlD4
H3C C H 
2. D O
 H3C C H
3
D
O OH
1. LiAlD4
H3C C OCH3 
2. H O
 H3C C D
3
D
O OD
1. LiAlH4
H3C C H 
2. D O
 H3C C H
3
H
33. C
Sol. O OH O O
H H
N N N N
N O N O N OH N OH
H H H
34. D
Sol. CF3 CF3 CF3
NH
2

  
Cl NH2
CH3 CH3 CH3
NH
2

  
Cl NH2
35. B
Sol. O
S S HO Me S HO Me Me
S
Me Me
H H
(A) (B) (C) (D)
O
HO Ph
O HO Ph
(E) (F)
Symmetrical (3-isomers)
36. C
COOH
Hydrolysis
Sol. N C C N  
Cyanogen COOH
Oxalic acid
37. C
Sol.
HO OH H2 SO4
 
 HO OH2 
38. D
Sol. 1 CH
2
3 4
H3C CH2 C CH CH3 2-ethyl-3-methyl but-1-ene
2
CH3
39. C
Sol. Order of electron withdrawing nature
NO2  OCH3  H
40. A
Sol. Enolate in option ‘A’ will form a stable six membered compound.
41. C
Sol. O
I2 + NaOH
 CHI3 
Ph
O
I2 + NaOH
Ph Ph  No ppt.
42. B
Sol. Ph
O can't be synthesized
43. C
Sol.
O
O

Cl
Cl
Cl
Cl
44. C
Sol. (i)
H3C CH CH CH CH CH3
Cl Cl
(ii)
Ph CH2 CH CH CH3
(iii) H2C CH CH2 Ph
(iv)
Cl CH2 CH CH CH CH3
Cl
45. A
Sol. O O O O
HO OH
O O
H H H OH
(A) (B) (C) (D)
SECTION – B
46. 3
Sol. Cl Cl

H3C C C H  2Zn   CH3  C  C  H + 2ZnCl2
0.02 mol
Cl Cl
0.02 mol
CH3  C  C  H  AgNO3  NH4 OH 
 CH3  C  CAg + NH4NO3  H2 O
0.02 mol 0.02 mol
Moles of CH3  C  CAg  0.02
Mass of CH3  C  CAg  0.02  147  2.94  3 g
47. 4
Sol. O
C CH3
O O
NaOH
H3C C (CH2)4 C H 

Number of – CH2 – unit ‘4’.
48. 20
Sol. Number of H-bond between A and T are 2.
Number of H-bond between G and C are 3.
The complimentary strand is “TATACGCG”
Total H-bond = 4 × 2 + 4 × 3 = 20
49. 4
Sol. Copolymers are Bakelite, Buna-S, Melamine, Terylene.
50. 8
Sol. Br Cl Br Br Br Br
Cl
Cl
Cl
Br Br Br Br Br Cl
(1) (2) (2) (2) (1)
Mathematics PART – C
SECTION – A
51. D
2
Sol. x12   x 2  1  0
x1  0, x 2  1
 y1  12   y 2  12  0
y1  1, y 2  1
52. A
2
 41  i 
Sol. Since BiCi is parallel to B0C0 triangles ABiCi are similar to AB0C0. So area of ABiCi is  
 41 
2
1 1  41  i  41  i
of the area of the area of ABiCi. So the area of BiCi Ci+1 is    .
41  i 41  i  41  412
41
i 41 42 21
The sum of all triangles BiCi Ci+1 is then  41
i1
2

2

41
. The height of AB0C0 is
412
1
412  9 2  40 , so its area is
 40  18  360 .
2
21 7560
Hence total area  360  .
41 41
53. D
Sol.
x2 y2
 1 
P 2 cos , 3 sin  
4 3
 3h = 2cos, 3k  3 sin 
cos(h, k)
2
x y2
  1.
4 1/ 3
S(–1, 0) S (1, 0)
9
54. C
Sol. yt = x + t2 2
P(t , 2t)
 pQ   SN  1  t2  10 (0, k)
N 90º
 t = 3, – 3. 90º Q
90º
S (1, 0)
55. B
Sol. y2 = 8x  y3 = – 8  y = – 2, x = ½
equation of tangent is y + 2 = –2(x – ½)
y intercept = – 1
y = cos(x + y)(1 + y)
– 2 = cos(x + y)(– 1)
cos(x + y) = 2 not possible.
56. C
Sol. x 2  y 2  25  y  0

0 c
2 2
  25
1 2 4
 2 – 2c + 150 – 2c2 = 0

1 and 2 are the roots of equation
  
2  0  1 2   50, 1 2  100
 4 
 2c2 = 250  c2 = 125  c = 5 5   c   11.
57. D
Sol. x 2 + y2 = 8
x  3 cos    y  3 sin    8 … (i)
2 2 (h, k)
Also, hx + ky = h + k … (ii)
3 cos  3 sin  8 (3 cos, 3sin)
  2
h k h  k2
8h 8k
 cos   , sin  
2
3 h k 2
 3 h  k2
2
 
2
8
Locus is S : x 2  y 2   
3
The given line mult pass through centre of circle
 c=2
hx + ky = 1 touches the ellipse
1 h2
  8
k 2 k2
The locus is x2 + 8y2 = 1
Eccentricity of conjugates hyperbola 3.
58. A
xx1 yy1
Sol. The equation of tangent at (x 1, y1) is  2 1
a2 b
It passes through (0, – b), so
y1
0  1  y1= b
b
a2 x b2 y
Normal at (x1, y1) is   a 2e 2
x1 y1
It passes through 2 2a,0  so

2 2a
x1 
e2
Now x1, y1 lies one hyperbola
8a2 b2
  1
e4 a2 b2
 e4 = 4,  e2 = 2.
59. C
Sol. Quadrilateral PACB is cyclic and PC will be the A
diameter of any circle passing through any of given 4
points.
 diameter will be PC 2(3, 4) C P 1(0, 0)
Locus of C is (x – 3)2 + (y – 4)2 = 1
Minimum distance O1O2 – r1 – r2 = 3. B
60. B
Sol. By using cosine formula we get, A
a 17  3 21
a2  3 3a  5  0  2  3 2
a1 10
30º 
B C
61. D
Sol. Let the variable line be lx + my + n = 0
3al1 3am 1
 n
P1 = a  b  c a  b  c
l2  m2
3bl 2 3bm 2
 n
P2 = a  b  c a bc
l2  m2
3cl 3 3cm 3
 n
P3 = a  b  c a bc
l2  m2
P1 + P2 + P3 = 0
3l  a1  b 2  c3  3m  a1  b2  c3 
   3n = 0
abc abc
62. B
Sol. Let 768  32cos 
16 3  32cos 
3 
cos =  
2 6

4  8  32  32cos
6

= 4  8  8 cos
12

= 4  4 sin
24
11
= 4  4 cos
24
11 b
= 2 2 cos   24
48 a
63. D
Sol. The tangent 3x + 4y  25 = 0 is tangent at vertex and axis is 4x  3y = 0
so PS = a = 5
L.R = 20
64. D
Sol. m3 + (2p + 5) m2  6m  2p = 0
m1 + m2 + m3 = (2p + 5)
m1m2 = 6
m1m2m3 = 2p
For A
3
P   mi  1
i 1
 P  2P  5 = 1  P = 4
 m1m2m3 = 8
 m1 = 1, m2 = 2, m3 = 4
For B
 P  2P  5 = 5  P = 0
 m1m2m3 = 0
 m1 = 1, m2 = 0, m3 = 6
For D
P + 2P = 32 not possible.
65. A
2
 3x  6y 
Sol. 2x2 + 2xy + 3y2    0
 P 
 2P2  9 + 3P2  36  P2 = 9
66. B
x2 y 2
Sol.  1
a2 b2
Tangent at P(asec, btan)
x sec  y tan 
 1
a b
b
y=  x
a
M = [a(sec  tan), b(sec  tan)]
N = [a(sec + tan), b(sec + tan)]
 ON = a2  b2 (sec + tan) = ae(sec + tan) and OM = ae(sec  tan)
 OM + ON = 2ae sec
 a  a
SP + SP = e  a sec     e  a sec    = 2ae sec
 e  e
67. C
Sol. Orthocentre lies on the rectangular hyperbola and
2 1
H(, ) G(h, k) O(3x1, 3y1)
2  3x1   2  3y1  
 h ,k 
3 3
 = 3h  6x1 ,  = 3k  6y1
2 2
9(h  2x1)  9(k  2y1) = 36   = 4
68. C
Sol. l1l2 = 2 and t1 + t2 + t3 = 0 and a = 2
Let the circumcentre be (h, k)
at t  at 32
h= 12  h  2  t 23
2
a  t1  t 2   2at 3
k=  k = t3
2
 h = 2 + k2
y2 = x  2
69. D
Sol. Let A be the vertex
AR = a2 t 42  4a2 t 22  at 2 t 22  4
a=1
2
|t2| = t1   2 2
t1
AR  4 6
70. B
Sol. ATC is isosceles, BHFC is cyclic
BFH = BHC. Then TBF ~ THC
Since TBF is isosceles, so THC
1
Area =  10  63  15 7
2
SECTION – B
71. 0

Sol. Let  =
28
cos2 cos6 cos18
 
sin3 sin9 sin27
1  2cos2  sin  2cos6  sin3 2cos18  sin9 
=   
2  sin   sin  sin9  sin3 sin27  sin9 
1  sin3  sin  sin9  sin3 sin 27  sin9 
=   
2  sin   sin  sin9  sin3 sin27  sin9 
1
= [cosec  cosec3 + cosec3  cosec9 + cosec9  cosec27]
2
1
= [cosec  cosec27] = 0
2
72. 2
sin x 1 sin x  sin y 3
Sol.   
sin y 2 sin x  sin y 1
x y xy
2 sin   cos  
  2   2  = 3
xy xy
2cos   sin  
 2   2 
cos x 3

cos y 2
cos x  cos y 3  2
 
cos x  cos y 3  2
xy xy
2cos   cos  
  2   2  = 5
xy xy
2sin   sin  
 2   2 
xy 3
 tan2  
 2  5
k=2
73. 2
Sol. m2sin2  2mtan + tan2 + cos2 = 0
2 tan 
m1 + m2 =
sin2 
tan2   cos2 
m1m2 =
sin2 
m1  m2 = m1  m2 2  4m1m2
4 tan2  4 tan2   4 cos2 
= 
sin4  sin2 
2
= tan2    tan2   cos2   sin2  = 2
sin2 
74. 1
1  sin  46  45  sin  48  47  sin  50  49  sin 134  133  
Sol.     ..... 
sin1  sin 45 sin 46 sin 49 sin 48 sin 49 sin50 sin133 sin134 
1
= cot 45  cot 46  cot 47  cot 48  cot 49  cot 50  .....  cot133  cot134  1
sin1 sin1
75. 0
Sol. Perpendicular distance = 2
Now sec2 + 2cosec2 = 2
No value of  is possible.

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ALL INDIA TEST SERIES

ANSWERS, HINTS & SOLUTIONS

Taking into account (2)

Plane of symmetry Two chiral centres

(i) (ii) (iii) (iv)

(A) (B) (C) (D)

(A) (B) (C) (D)

 2 – 2c + 150 – 2c2 = 0

It passes through 2 2a,0  so

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