JR IIT *CO SC (MODEL-A &B) WTM-16 DATE: 06-11-2023

TIME: 3Hrs Max Marks:300

KEY
PHYSICS

1 B 2 B 3 A 4 A 5 A
6 D 7 B 8 C 9 A 10 D
11 B 12 C 13 C 14 B 15 D
16 B 17 A 18 A 19 B 20 B
21 3 22 20 23 3 24 5 25 9
26 4 27 48 28 27 29 75 30 45

CHEMISTRY
31 B 32 A 33 A 34 B 35 C
36 B 37 C 38 B 39 B 40 C
41 C 42 C 43 B 44 A 45 D
46 A 47 D 48 D 49 D 50 C
51 15 52 0 53 5 54 3 55 2
56 4 57 5 58 5 59 4 60 5

MATHMATICS
61 B 62 A 63 A 64 C 65 C
66 D 67 B 68 A 69 A 70 B
71 D 72 C 73 B 74 A 75 B
76 B 77 B 78 C 79 C 80 C
81 5 82 25 83 0 84 2 85 7
86 1 87 4 88 720 89 1 90 24
PHYSICS

I
1. T  2
mgL

ml 2 ml 2 2mgl 2
Here I   
3 3 3
L
From figure : sin 45o 
l/2

l
L
2 2

2ml 2 2 2l
 T  2  2
l 3g
3 mg
2 2
2. Energy  ( Amplitude)2
E  KA2 , K ( Constant)
dE dA 2  5
2   10%
E A 100

l
3. T  2
mgd

5 2 mR 2 
3 mR   mR 2 
4
 2  
 R 
2
 2m  g   R 2  
 16 

 35R 
  2 
 6 17 g 
 
4. Time period of 1st pendulum be T1

Time period of 2nd pendulum T2

Now T1  |1

T2  |2

|1  length of 1st pendulum

2|Page
 | 2  length of 2nd pendulum

Again T= 9 T1  7T2

T  7
 1 
 T2  9
|1 7
 
|2 9
|1 49
 
|2 81
2c k
5. 2

mR m

2c
R
k
3mR 2 d 2  3R 
6. 2
 kR 2  3kR  
2 dt  2 
3mR 2 d 2 11kR 2

2 dt 2 2
d 2  11K 
2
  
dt  3m 

3m
 Time period, T  2
11k

4x
7. When block is further pushed by x in liquid buoyant force increases due to height of liquid
3
displaced as  R 2 X   4 R 2   R 2  y

x
 y
3

Hence extra buoyant force = + .2

8.

3|Page
 L
x 
2
L
Restoring torque    2kx  .
2
kL  L / 2   kL2 / 2   6k 
    2      
I  ML /12  M 

1  1 6k
f  
2  2 M

9. Let plank is slight displacement a distance x.

32
F1  kx
3
32k
a x
3m
32k

3m
10. Let, T1 and T2 time period of 1st and 2nd pendulum.
We know, T  L

T1 1.44 144 12
  
T2 1 100 10

 5T1  6T2
r 1.5
11. We know b    30
2m 2  25  10 3
k 25
&    1000
m 25 10 3
time period
2 2
T   0.628sec
2 2
 b 1000  900

12. F  Kv
F
K
V
kgms 2
Unit of K  K  1
 kgs 1
ms

13. Tmin  mg  cos  0 …….(i)

4|Page
 mv 2
Tmax  mg  …….(ii)
R
1 2
mv  mg  R  R cos  )  …….(iii)
2
Tmax
4
Tmin
1
 0  cos1  
 2

1 mgl
14. f0
2 I
Where l is the distance between point of suspension and centre of mass of the body.
Thus, for the stick of length L and mass m,
g  L
m
1 4  1 12 g  2 f
f0 2 0
2 L 2 L
m  2 
2 12

15. x  A sin t
 2   2    A
x  t  1S   A sin  t   A sin    A sin   
 T   8   4 2
 2   2   
x  t  25   A sin  t   A sin   2   A sin    A
 T   8  2
And the required ratio is:
A
x  t  1S  2  1  2 1

x  t  2 S   x  t  1S  A  A 2 1
2
I
16. T  2 Support
mglcm
2 3 2
I Support  m  l sin 60o   ml
2
3l
lcm  l sin 60o 
2
3 2
 4 ml  3L
 T  2    2
 3l  2g
mg  
 2 
17. x  A cos wt

5|Page
 1 1
K .E .  k  A2  x 2   kA2 sin 2 wt
2 2
2
1  1  cos 2 wt  kA
 kA2    1  cos 2wt 
2  2  4
Frequency of K. E. is double of acceleration

18. For small angular displacement  

Net torque on body = I 
 ML2 
=  k1a sin   a   k2b sin   b   mL2  
 3 
k1a 2  k 2b 2 1 k1a 2  k 2b 2
For small   a    frequency 
ML 2
2 L2  m  M / 3
mL2 
3

19. The point x  a and b are stable equilibrium point.

20. Friction force is F, having mass m and acceleration of its SHM f is given by:

k1a 2  k 2b 2
 a 
2 ML2
mL 
3
F  ma
F  mf
F  m 2 A
k k
  
mm 2m
k
F m A
2m
k
F  A
2

21.

fv  3 fh

22.

6|Page
 I
T  2
mg 
TA 3 9 B
  
TB 4 4 A
3 3 B

4 2 A
B 1

A 4
4
 A   25  20 cm
5

23.
a: acceleration ring w.r.t. ground
b: acceleration of particles w.r.t. ring

Hence w.r.t ‘C’, for the particle

N cos   mg .......  i 
N sin   ma  mb ........  ii 
And for the ring
 N sin   f  Ma .....  iii 
f .R  MR 2 .a / R
 f  Ma ......  iv 
 N sin   2 Ma
N  mg
mg.  2 ma
mg
 a 
2M

From (ii)
N sin    ma  mb
 2Ma  ma  mb  a  2M  m   mb
mg
  2 M  m   m.b
2M
x
Put   [x: displacement of particle w.r.t. ring]
R
g x g m 
  2M  m   b  b  1   .x
2M R R  2M 

= 1+

7|Page
 3a
24. OP  a sin 60o 
2
2 a
OC  1   OP  
2 3
 ma 2  2 ma 2
I  I0  2   m  OP  
 3  12
2 3 1
 ma 2  ma 2 
3 4 12ma 2
3
 ma 2
3
I
T  2
 3m  gl

 2
 3 / 2ma  2

 a 
 3m  g  
 3
3a
 2
2 g
1
Pitting a  m and g  10m / s 2
3

We get, T  sec
5
25. This is a case damped vibration as the amplitude of vibration is decreasing with time. Amplitude of
vibrations at any instant t is given by a  a0 e bt , where a0 is the initial amplitude of vibrations and
b is the damping constant.
Now, when t  100T , a  a0 / 3 T is time period  Let the amplitude be a ' at t  200 T . i.e. after
completing 200 oscillations.

 a  a0 / 3  a0e 100Tb .....  i 

and a '  a0e 200Tb . .....  ii 
1
From  i  ,  e 100Tb .  e 200Tb  1/ 9.
3
1 a
From  ii  , a '  a0   0 .
9 9
 The amplitude will be reduced to 1/9 of initial value.

26. When the block A moves with velocity V and collides with the block B, it transfers all
energy to the block B (because it is an elastic collision). The block A will move a distance x against
the spring, again the block b will return to the original point and completes half of the oscillation. So,

The time period of = =

The block B collides with the block A and comes to rest at that point. The block A again moves a
further distance L to return to its original position.
L L 2L
The time taken   
V V V

8|Page
 So that periods of the periodic motion = +
27. For first case : T1  mg  0.3  
For second case: T2  mg  0.5  
Let the time period of the pendulum be T, by using the parallel axis theorem, Moment of inertia is
I1  I  2  0.32
 I 2  2  0.52
Because time period of blue points is equal,
I 1  I  2  0.32
 I 2  I  2  0.52
I1 I
 2  2 2
T1 T2
I1 I1
 
T1 T2
Putting out of the value of the moment of inertia,
I  2  0.32 0.3mg

I  2  0.52 0.5mg
I  2  0.32 3
 
I  2  0.52 5
 I  0.30kgm 2
So, the value of moment of inertia comes out to be 0.30kgm2
Moment of inertia about an axis passing through first point can be find a:
I1  0.3  2  0.32
 0.48kgm2
28.  p  k  2R  2R
lp  4kR 2
7
mR 2  4kR 2
5
 20 k 
   
 7 m
7m
T  2
20k

b
 b 
29.  2  02   
 2m 
2
k  b 
  
m  2m 

9|Page
 100 100
 
1 4
-1
= 75 s

30. Half of the oscillation is completed with length l and rest half with l / 4 .
T T
 Time period  1  2
2 2
1 l l/4
  2  2 
2 g g 
3 l  3
   2   T = 45s
4 g 4

CHEMISTRY

31 . To 60 Conceptual

MATHS
 5x  3 y  5 f  x   3 f  y 
61. Given  f   , which satisfies section formula for abscissa on LHS and for
 53  53
ordinate of RHS. Hence f  x  must be the linear function (as only straight line satisfies such section
formula) hence, f  x   ax  b . But f  0   3  b  3, f '  0   2  a  2
62. Put x  1  t
1  t3 t5 
f  x   t   ......
4 3 5 
1
f '  x   1  t 2  t 4 ......
4
1 1 
 
4 1  t 2 
1 1 
=   
4 1   x  1 2 
 x, x2

63. g  x   x  2  2   4  x, 2  x  4
 x  4, 4 x

g '  1  g ' 1  g '  3  g '  5   1  1  1  1  0
64. Let P  t , t 4  be a point on the curve equation of tangent at P, y  t 4  4t 3  x  t  it is drawn from  2, 0 
8
, then t  0,
3
4096 2048  8
So, equation of required tangent is y   x 
81 27  3
65. Tangent to y  x 2  bx  b at 1,1 is
b 1
x - intercept = and y – intercept =   b  1
b2

10 | P a g e
 1  b 1 
ATQ Ar     2       b  1   2  b  3
2 b 2 
66. For y 2  4ax , y – axis is tangent at  0, 0  , while for x 2  4ay , x – axis is tangent at  0, 0  . Thus the
two curves cut each other at right angles.
3x cos 
67. Equations of tangent to the ellipse  y.sin   1
9
 Sum of intercepts =
dS
For min. value of S , 0
d
1  
 tan    tan   
3 6 6
dy  x12
68.  2
dx y1
x12
Tangent y  y1    x  x1  _______ 1
y12
x23  y23  a3 _________  2
x13  y13  a3 _________  3

Putting y  y2 and x  x2 and solving with (2) and (3) we get result.

2
69. xy   a  x  y  xy '  2  a  x 

Now y '  1 or y  x  2  a  x 

2 2
a  x  x  2  a  x  or  x  2  a  x  
a  x
x x

or  x 2   a  x  x  a  or  x 2  x 2  a 2

a
2 x 2  a 2 or x  
2

dx a2  y2  dy  y1
70.    
dy y  dx  x1 , y1  a 2  y12

y1 y12
LT = 1  m2 ,  a 2  y12 . 1  a
m a 2  y12

1
71. Equation of normal is y  y1   x  x1 
m

y1 y
72. L.S.T = , L.T  1 1  m 2
m m

73. S  t n  s  k  t n  , differentiate

11 | P a g e
 a y dy dy ab
74.   ax  ay  hy   h  a   ab  
h x y dt dt  h  a 

75. Given x and y are the sides of two squares such that y  x  x 2

2
 Areas of the first square, A1  x 2 and area of the second square, A2  y 2   x  x 2 

dA2 dx
dA2  2 x  2 x 2  1  2 x 
 dt  dt
dA1 dA1 2x
dx
dt dt

1  2 x  2 x 1  x  
1  2 x 1  x   2 x2  3 x  1
2x

76. OM  r cos 

dv
77.  2c.c / sec; V  288
dt

4 3
r  288,  r  6
3

d 4 3 4 2 dr dr 1
  r   .3r  
dt  3  3 dt dt 72

dv
78.  50cm3 / min ,
dt

d 4 3 dr 50
 r   50  
dt  3  dt 4r 2

dr 50 1
   cm / min
dt 4  225  18

H h 10 h h
79.    r
R r 5 r 2

1 2 h3
Use V  r h 
3 12

dv
Then at h  4 mts ?
dt

1 2
80. r  h;V  r 2 h  r 3 ;V  r 3  h3
3 3

dv dh
 3h 2
dt dt

12 | P a g e
 1 1
81. g ' y    1 x 
 dy   
  2e  2   x 2  x  1
 dx 

7  7 1
 y    x  1; g '     , k  5
6  6 5

1
82. f  x  x 
x  f  x

 f 2  x   x2  1

 2 f  x . f ' x   2x  0  f  x  . f '  x   x

 f  50 . f '  50   50

f  50  . f '  50 
  25
2

83. Since f  x  is an even function f '  0   0

84. y  xn

dy
 nx n 1  na n 1
dx

1
Slope of normal =  equation of normal
na n 1

1
 y  an   x  1
na n 1

Put x  0 to get y – intercept

1 1
y  an  b  an 
na n  2 na n 2

 0 if n  2
1

lim b   if n  2
a 0
 2
 if n  2

85. x  t2; y  t3

dy dy dy 3t
 2t ;  3t 2  
dt dt dx 2

3t
y  t3 
2
 x t2 

13 | P a g e
 2k  2t 3  3th  3t 3  t 3  3th  2 k  0

Product of roots, t1t2t3  2k

Take t1t2  1, t3  2k

Now, t3 must satisfy equation 1 , Therefore,

3
 2k   3  2k  h  2k  0

i.e., 4 y 2  3x  1  0 or 4 y 2  3 x  1

or a  b  7

86. At x  0, y  1

dy
Evaluate
dx at x 0& y 1

Find equation of tangent at x  0 and y  1 .

87. We have f  0   2

Now, y  f  a   f '  a   x  a 

For x intercept, y  0 , so

f a f a
xa  a  2 or 2
f 'a  f 'a

f 'a 1
or 
f a 2

 on integrating both sides with respect to a, we get

a
ln f  a   C
2

f  a   Cea /2  f  x   Ce x /2

f  0   C or C  2  f  x   2e x /2

1 k
Hence, k  2, p  or  4
2 p
88. Let the body is at a height h1 at a time ‘t’ and is at a height “h” at a time  t  4  from above.
h1  h  400
1 1 2
 gt 2  g  t  4   400
2 2
2 2
 t   t  4   80  t  12sec

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 1 2
h  g  t  4   320m
2

Hence, total distance

= 320 + 400 = 720 m

89.   t 2    kt 2 (k constant)

2 
k k
64 32

d   
 k .2t   2  8  R / sec K
dt 32 2 2

ds 3 3
90. V  12t  t 2 , a  12  3t  0, t  4 v  12.4  16   24
dt 2 2

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