AIITS 2224 OT JEEM Sol

FIITJEE
ALL INDIA INTEGRATED TEST SERIES

OPEN TEST
JEE (Main)-2024
TEST DATE: 25-03-2023
ANSWERS, HINTS & SOLUTIONS

Physics PART – A
SECTION – A
1. A
displacement
Sol. time t 
component of velocity along the direction of displacement

 2  
v cos 45 v
2. C
Sol. Acceleration a  2 x
On solving by integration v    2  a2
3. A
Ftangential F
Sol. Shear stress =  = s.
A 2rh
4. B
Sol. R to be maximum.
3H Total height
Y=  .
2 2
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AIITS-OT-PCM(Sol.)-JEE(Main)/2024 2
5. C
Sol. For breaking off the hemisphere
mv 2
mgcos  
R
2
or, v  Rgcos  v
Conserving mechanical Energy 
1
Or, 0  mgR  mv 2  mgR cos 
2
1
Or, mgR  1  cos    mv 2
2
mRgcos 
Or mgR  1  cos   
2
2
Or cos  
3
2
h  R cos   R
3
6. A
Sol. mg = Vg
= 3 × 2 × (0.01 m) × (103 kg/m3) × 10
Mg = 600
 m = 60 kg
7. D
Sol. Due to Mg sin , Vo will increase. Net torque about centre of mass is zero. So, o
remains constant.
8. B
Sol. Antinodes are formed at a distance 0.6r from the end of the tube, where r is radius of the
tube.
47 + 2  .6r = /2
 = v/n = 100 cm
47 + .6D = 100/2
D = 5 cm
9. D
1
mv 2 gR2m gR 2  gR2  2
Sol.  2
 V2  V 
 R  x  R  x  R  x  R  x 
10. A
IA a A 2 fA 2
Sol.  1
IB aB 2 fB 2
11. D
Sol. U = T(A)
3 AIITS-OT-PCM(Sol.)-JEE(Main)/2024
2
 R 
U = T  1000  4    4 R2 
  10  
 
= T × 4(9R2)
U = 36R2T
12. D
Sol. Each block completes half oscillation.
13. C
Sol. When centre of cube is directly above the edge of cube.
  = 45°.
14. A
dT  1
Sol. 4R2  6l2 and ms    AT 4  Rate of cooling (R) 
 dt  m
3 3 3/2
R  l 3  l   3   4  
 1    R    4   6  
R2 4  4  6
  R3      
 3 
15. C
1
Sol. E mv 2
2
  m   E1v 2 T 0
16. D
Sol.   I2
Ky L ML2 2 y 15K
KyL     
2 2 3 L 4M
17. B
Sol. 54  S W  90  30   mS  536  mS  SW   100  90 
54  1  60  mS  536  1  10 
54  60
mS   5.93
546
mass of mixture = 60 gm.
18. A
Sol. Upward force exerted by liquid at the bottom of conical flask
= ghA = 1200 × 10 × 10–1 × 0.1 = 120N
Weight of flask = 100 N
Since flask is in equilibrium, net force acting on it should be equal to zero.
So force exerted by liquid on the curved surface of the conical flask
= 20 N in downward direction.
19. D
Sol Net force is zero
3F2L 2FL 8FL
Total elongation  =
Y3A 3YA 3YA
20. D
Sol. (Fmin  80)  60 ; Fmin = 200 N
SECTION – B
21. 7
7
Sol. As air comprises of O2, N2 & H2 all have  =
5
7
  of air can be taken to be
5
For pv = constant
p v

p v
7
=  5% = 7%.
5
22. 5
1
Sol. EPE per unit volume =  stress  strain
2
23. 0
Sol. Force is conservative and displacement zero so work done is zero.
24. 6
61  59  61  59 
Sol. k  30 
4  2 
51  49  51  49 
k  30  ; t = 6 min.
t  2 
25. 3
Sol. New resistance becomes (2R)
H H
 Then time taken =  = 12 min
Power ( T)2
2R
Resistance is parallel becomes (R/2)
H
 Time taken = = 3min.
( T)2
R/2
26. 1
v
Sol. fopen 
2
v
fclosed  =f ; K=1

4 
2
27. 4
1
Sol. T and P × T4

28. 1
K
Sol. Maximum Acc = 2A =   A = 4m/s2
m
24
A  4  A = 1m.
6
29. 6
Sol. W = Wg
 M   L  MgL
=   g   72
 6   12 
30. 2
 m  em2   m2  em2 
Sol. V1   1  u1    u2
 m1  m2   m1  m2  V1 V2
m 6 6 m m m
 1 e   1 e 
 2   6   2   6 
   
2 4
  6    6   2 m / s
6 6
Chemistry PART – B
SECTION – A
31. B
Sol. Due to existence of p – p back bonding in BF3.
F
B
F F
32. B
H2 O
Sol. KO2   KOH  H2O2  O2
H2 O
K 2S   KOH  H2S
33. D
Sol. CH3 CH3 CH3
H CH3 HO H
H Br H OH
P =: C C Q= R or S = or
H Br HO H H OH
H3C H
CH3 CH3 CH3
34. B
Sol. For a reversible reaction
S(system) = -S(surrounding)
 S(total) = zero
35. B
Sol. Force of repulsion develops at very high pressure. So the attractive intermolecular
force(a) is assumed to be zero.
36. B
Sol.  NH4  Cl
NH4Cl 
NH4  H2O   NH4OH  H
H+ shift the ionization reaction toward backward direction due to common ion effect.
37. A
Sol. The solubility of the sulphates of Gr-2 ions decrease down the group.
38. C
Sol. CH3 CH3
HBr
CH2  CH  C  CH2CH2Br  BrCH2 CH2  C  CH2CH2Br
R2O 2
C2H5 C2H5
Optically inactive
39. B
Sol. KP = KC[RT]n
For KP = KC, n  0
It is possible for (B)
40. B
Sol. P3– is not the conjugate base of acid H3PO4.
41. A
Sol. Rate of bromination ter: pri = 1600 :1
Rate of clorinaction ter : pri = 5 : 1
42. C
Sol. As from list of decreasing order of –I effect
43. D
 22  32  4 2  5 2  54 54
Sol. RMS    cm/sec
4 4 2
44. D
h
Sol. Orbital angular momentum = l  l  1
2
for 11th electron, l  0
45. A
Sol. In IF7 , 72 0  5 ; 900  10 ; 1800  1
46. C
Sol. R R
n
H O
HO  Si  OH 
2
O  Si  O
n
R R
47. D
Sol. Abnormal behavior of Be
48. B
3RT
Sol. The highest velocity is Crms which is and the velocity possessed by the highest
M
2RT
fraction of molecule is Cmp which is
M
49. B
Sol. H+ (aq) + OH-(aq) → H2O(l)
∆H = -13.7 = [-68] – [0 + ∆Hf OH- ]
∆Hf OH-+ = -54.3 Kcal mol1
50. D
H2O 
Sol. Na2B 4 O7   B  OH 3  Na B  OH 4 
SECTION – B
51. 10
Kw
Sol. KH 
Ka
K w 10 14
or, K a   2.6  10 11.4
K H 10
pK a   logK a   log10 11.4  11.4
x + (1.4) = 11.4
or x = 11.4 – 1.4 = 10
52. 400
3RT 2RT
Sol. 
30 20
3R400 2RT
or  , T  400K
30 20
53. 19
Sol. The products
, and
Have respectively 3, 10 and 6 allylic hydrogen atom(s).
54. 5
2 2
Sol. KC 
NO2 

 0.2 
5
NO2 O2   0.2 2  0.2
Equilibrium constant will not change by adding or reducing the quantity of reacting
species.
55. 5
Sol. Sodium perborate is NaBO3.
56. 6
rCH4 nCH4 MSO2 80 / 16 64
Sol.    5 :1  x : y
rSO2 nSO2 MCH4 128 / 64 16
x+y=6
57. 8
Sol. x = 6, y = 2, z = 0
58. 281
Sol. CH2Cl CH3
Cl
P= Q=
CH3 CH3
59. 3
Sol. Rate = K[X]2[Y2]1
 overall order = 2 + 1 = 3
60. 4
Sol. Due to double bond, the geometrical isomers are:
CH3
CH3CH  CH  CH  CH  CH  CH3
Cis Cis
Trans Trans
Cis Trans
Trans Cis Identical
When the double bonds are Cis and trans, the molecule becomes optically active.
 For trans Cis (two optical isomers are possible)
 Total no. of stereoisomers are:
Geometrical (2) + Optical (2) = 4
Mathematics PART – C
SECTION – A
61. D
41
Sol. C0  41C1  42C2  .........  60C20
42
C1  42C2  .........  60C20
61 60 m
. C20  . 60C20
41 n
62. B
Sol. Equation of tan to hyp : y  mx  a2m 2  b2  mx  16m2  9
 m2 (h2  16)  2mhK  K 2  9  0 .....(i)
K2  9
m1m2  1   x2  y 2  25
h2  16
63. A
Sol.  x  22   y  12    x  2y   0
C : x 2  y 2  x    4   y  2  2   5  0
C1 : x 2  y 2  2y  5  0
S1  S2  0
(Equation of PQ)
   4  x   2  4  y  10  0
Press through (0, –1)
   7
C : x 2  y 2  11x  12y  5  0
245

4
Diameter  7 5
64. B
Sol. Let length of common chord  2x
5 12
25  x 2  144  x 2  13
12  5 x
After solving x 
13
x
120
2x  5 12
13
65. C
2 2at1 2at 2
Sol. t2 = t1 – also  2  1
t1 at12 at 2
 t1t2 =  4
4 2
    t1 
t1 t1
 t12 + 2 = 4 and t1 =  2
so, point can be (2a,  2 2 a).
66. A
x sec  y tan 
Sol. Let the tangent  1
a b
A  (a cos, 0), B  (0, b cot)
 P  (a cos, b cot) = (h, k)
a b
sec = , tan =
h k
2 2
a b
 2  2 1
h k
a2 b2
 Locus is 2  2  1.
x y
67. B
Sol. Given parabolas are symmetric about the line y = x so they have a common normal with
slope -1
 1 13   13 1
it meets the parabolas at  ,  ,  ,  hence the req circles is
 2 4  4 2
11 11 13
x2+y2  x y 0
4 4 4
68. C
Sol.  pi  1  6k2  5k  1
6k 2  5k  1  0
6k 2  6k  k  1  0
1
 6k  1k  1  0  k  1 (rejected); k 
6
P  x  2   k  2k  5k 2
1 2 5 6  12  5 23
    
6 6 36 36 36
69. A
Sol. S  x2 + y2 – 2x – 2y + 1 = 0
S  Considering tangent for boundary value
y  2x  ( 1  5) will be tangent for S
y  2x  15  2 5
a  (15  2 5,  1  5 )
70. A
xb
Sol. Equation of AB is  ky  1
4
A
Centroid equation of as OB = 4
2
2 2  xb xy 
x  4y  4    0 ………..(i)
 4 1
x 2  4y 2  xy  0 ………..(ii)
2 2
h  4 k  4 br
  B
16 4 2h
 2
 
1
h  4  4 k 1 
Locus is x 2  4y 2  0
71. C
 h2  k 2 
Sol. P  0,
 k 
 h2  k 2 
Q ,0
 h 
PQ = 2a
72. A
3  
Sol. sin2 2  cos4 2  ,    0, 
4  2
3
 1  cos2 2  cos4 2 
4
4 2
 4 cos 2  4 cos 2  1  0
2

 2 cos2 2  1  0
1 
 cos2 2   cos2
2 4

 2  n  , n  I
4
n 
 
2 8
  
 , 
8 2 8
73. C
Sol. P(,   3)
y(  3)  4(  2)
4x  3y  (4  y)  0
y  4, x  3
74. B

1 1
Sol. Sn   n2  n  2 ;   n  2n  1
n 3
1 1 1 
   
3 n 1 n  2
1  1 1   1 1   1 1   1 1   1 1  
                     ...
3  2 5   3 6   4 7   5 8   6 9  
1  1 1 1  13
   
3  2 3 4  36
75. A
1
Sol. n  n  1 3m  n  1
6
1
18 17  60  18  1
6
1
  18  17
6
= 2091
76. B
1  sin 1  0  sin  2  1 sin  45  44  
Sol.    ...  
sin1  cos 0 cos1 cos 2 cos 2 cos 44 cos 45 
1
=
sin1
 tan1  tan 0  tan 2  tan 2  tan 3  tan 2  ...  tan 45  tan 44
1
= . 1  0 
sin 1
1
=
sin1
1
= .
x
77. D
Sol. ~  ~ s   ~ s  
s  r  ~ s 
 s  r    s ~ s 
s  r   
s  r 
78. A
4 tan x  4 tan3 x
Sol. tan 4x  1
1  6 tan2 x  tan 4 x
 tan4 x  4 tan3 x  6 tan2 x  4 tan x  1  0

5
Harmonic mean  1
1 1 1 1
  
tan  tan  tan  tan 
79. D
1
Sol. R
sin30o
1
r
tan30o
R 2
Now  sec 30o 
r 3
80. C
Sol. AB  30m  NP M
MN
In ANM tan 45o  1
AN
 MN  AN
PM  MN  30 30o
P B
 AN  30
PM AN  30
In BPM tan 30 o  
PB AN
1 AN  30 45o

3 AN
N A
AN  3AN  30 3
30 3 30 3  3 1   15 3  3
AN 
3 1

2
 
SECTION – B
81. 17
Sol. Let  be a double root of f  x   0
f '  x   4  x  1 x  2  x  3 
If x  1 is double root, then f 1  0  p  9
If x = 2 is double root, then f  2   0  p  8

If x = 3 is double root, then f  3   0  p  9
 Sum of all possible values of p = 17
82. 6
Sol. y  ax 2  bx  c
y is max. at x = 2
b
  2  b  4a
2a
D
max value 
4a
b 2
 4ac 1
4a 2
4ac  b2  2a
4ac  16a 2  2a
2c  8a  1
 2c  8a  5  6
83. 1320
Sol. Coefficient of x
n 2
in  x  1 x  2  .......  x  n  
 
n n2  1  3n  2 
24
10  99  32 10  33  32
Here, n  10,     40  33
24 8
84. 84
10 9
Sol. 1  x 1  x  1  x  x 2 
3 9
1  x 1  x 
2
9
C6  84
85. 3

Sol. B  , by sine Rule
3
1
sin A 
2
 A  30o ,a  2, b  2 3, c  4
1
   2 3  2  2 3 sq.cm
2
86. 4
2 2
Sol. x 2   y  1  x 2   y  2 
2y  1  4y  4
1
6y  3  y  
2
25 24
x2  y2   x2  6
4 4
i
 z 6
2
25 49
z  3i  6  
4 4
7
z  3i 
2
87. 463
Sol. PA = x, PB = y, PC = z
Applying consine laws:
cosine
 x 2  z2  b2  2bz cos 

  y 2  x 2  c 2  2cx cos 
 2 2 2
z  y  a  2ay cos 
2 2 2
Adding 2  cx  ay  bz  cos   a  b  c
1
Also for ABC   cx  ay  bz  sin 
2
4 168
tan   2   m  n  463
a  b2  c 2 295
88. 132
Sol. Let number of balls A gets be 2x
2x  y  z  21 where x   0, 20 
y  z  21  2x y, z  0
for non negative integer solution
21 2x  21
C21  222x C1
10
  22  2x   132
x 0
89. 10
Sol. y  mx  a2m2  b2
A
4 16
y     18   32
3 9
4
y    8
3
Distance between tangents 0 B
16 16  3 48
  
16 5 5
1
9
90. 25
25 25 25
Sol. z  z   24  z 
z z z
25 2
 24  z   24 or 24 z  z  25  24 z
z
2 2
 z  24 z  25  0 and z  24 z  25  0
  z  25  z  1  0 and  z  25  z  1  0
 z  1  0 and z  25  0
Hence 1  z  25
or 1  z  0  25

AIITS 2224 OT JEEM Sol

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ALL INDIA INTEGRATED TEST SERIES

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