1. A uniform rod AB of length l and mass m is free to rotate about point A. The rod is released from rest in horizontal position. Given that the moment of inertia of the rod about A is (ml^2)/3 the initial angular acceleration of the rod will be

a)2g/3lb)mg l/2c)3/2 gld)3g/2l

2. Two boys of masses 10 kg and 8 kg are moving along a vertical rope, the former climbing up with acceleration of 2 ms^(-1). 2 while later coming down with uniform velocity of 2 ms^(-1). Then tension in rope at fixed support will be(Take g=10ms^(-2))

a)200 Nb)120 Nc)180 Nd)160 N

3. Four spheres of diameter 2 a and mass M are placed with their centres on the four corners of a square of side b. Then the moment of inertia of the system about an axis along one of the sides of the square is

a)4/5 M a^2+2M b^2b)8/5 M a^2+2M b^2c)8/5 M a^2d)4/5 M a^2+M b^2

4. From a circular ring of mass M and R, an arc corresponding to a 90^0 sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the ring and perpendicular to the plane of ring is  k times  〖MR〗^2. Then the value of k is

a)3/4b)7/8c)1/4d)1

5. The ratio of rotational and translatory kinetic energies of a sphere is

a)2/9b)2/7c)2/5d)7/2

6. Angular momentum is conserved

a)Always b)Never

c)When external force is absentd)When external torque is absent

7. When a ceiling fan is switched off, its angular velocity falls to half while it makes 36 rotations. How many rotations will it make before coming to rest?

a)24b)36c)18d)12

8. A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness t/4. Then the relation between the moment of inertia I_(X )and I_Y is

a)I_Y= 〖32 I〗_Xb)I_Y= 〖16 I〗_Xc)I_Y= I_Xd)I_Y= 〖64 I〗_X

9. A circular disc rolls down an inclined plane. The ration of rotational kinetic energy to total kinetic energy is

a)1/2b)1/3c)2/3d)3/4

10. The angular momentum of a particle describing uniform circular motion in L. If its kinetic energy is halved and angular velocity doubled, its new angular momentum is

a)4Lb)L/4c)L/2d)2L

11. A motor is rotating at a constant angular velocity of 600 rpm. The angular displacement per second is

a)3/50π  radb)3π/50  radc)25π/3  radd)50π/3  rad

12. A wheel has angular acceleration of 3.0 rad/s〖ec〗^2 and an initial angular speed of 2.00 rad/sec. In a time of 2 sec it has rotated through an angle (In radian) of

a)6b)10c)12d)4

13. A disc is rolling on the inclined plane, what is the ration of its rotational KE to the total KE?

a)1:3b)3:1c)1:2d)2:1

14. The moment of inertia about an axis of a body which is rotating with angular velocity 1〖 rads〗^(-1) is numerically equal to

a)One-fourth of its rotational kinetic energyb)Half of the rotational kinetic energy

c)Rotational kinetic energyd)Twice the rotational kinetic energy

15. What is the moment of inertia of solid sphere of density ρ and radius R about its diameter?

a)105/176 R^5 ρb)105/176 R^2 ρc)176/105 R^5 ρd)176/105 R^2 ρ

16. If ‘I’ is the moment of inertia of a body and ‘ω’ is its angular velocity, then its rotational kinetic energy is

a)1/2×Iωb)1/2×I^2 ωc)1/2×Iω^2d)1/2×I^2 ω^2

17. A rigid body of mass m rotates with the angular velocity ω about an axis at a distance ‘a’ from the centre of mass G. The radius of gyration about G is K. Then kinetic energy of rotation of the body about new parallel axis is

a)1/2 mK^2 ω^2b)1/2 ma^2 ω^2c)1/2 m(a^2+K^2)ω^2d)1/2 m(a+K^2)ω^2

18. Three rods each of length L and mass M are placed along X,Y and Z axes in such a way that one end of each rod is at the origin. The moment of inertia of the system about Z-axis is

a)(ML^2)/3b)(2ML^2)/3c)(3ML^2)/2d)(2ML^2)/12

19. Two masses of 200 g and 300 g are attached to the 20 cm and 70 cm marks of a light metre rod respectively. The moment of inertia of the system about an axis passing through 50 cm mark is

a)0.15 kg m^2b)0.03 kg m^2c)0.3 kg m^2d)Zero

20. Three identical rods, each of length x, are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the triangle is

a)x/√3 b)x/2 c)√(3/2) x d)x/√2

21. A couple produces

a)No motion b)Linear and rotational motion

c)Purely rotational motiond)Purely linear motion

22. When a mass is rotating in a plane about a fixed point, its angular momentum is directed along

a)A line perpendicular to the plane of rotationb)The line making an angle of 45° to the plane of rotation

c)The radiusd)The tangent to the orbit

23. Centre of mass is a point

a)Which is geometric centre of a bodyb)From which distance of particles are same

c)Where the whole mass of the body is supposed to concentratedd)Which is the origin of reference frame

24. A system of three particles having masses m_1=1 kg and m_3=4 kg respectively is connected by two light springs. The acceleration of the three particles at any instant are 1ms^(-2),2ms^(-2) and 0.5 ms^(-2) respectively directed as shown in the figure. The net external force acting on the system is

a)1 Nb)7 Nc)3 Nd)None of these

25. The total kinetic energy of a body of mass 10 kg and radius 0.5 m moving with a velocity of 2 m/s without slipping is 32.8 joule. The radius of gyration of the body is

a)0.25 mb)0.2 mc)0.5 md)0.4 m

26. A solid sphere of radius R has moment of inertia I about its geometrical axis. If it is melted into a disc of radius r and thickness t. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to

a)2/√15  Rb)2/√5  Rc)3/√15  Rd)√3/√15  R

27. A disc of mass 2 kg and radius 0.2 m is rotating with angular velocity 30 rads^(-1)  . What is angular velocity, if a mass of 0.25 kg is put on periphery of the disc?

a)〖24 rads〗^(-1) b)〖36 rads〗^(-1) c)〖15 rads〗^(-1) d)〖26 rads〗^(-1)

28. A solid sphere rolls down without slipping on an inclined plane at angle 60° over a distance of 10 m. the acceleration (in ms^(-2)) is

a)4b)5c)6d)7

29. The curve between log_e⁡L and log_e⁡P is (L is the angular momentum and P is the linear momentum)

a)b)c)d)

30. The centre of mass of three particles of masses 1 kg, 2kg and 3kg is at (2, 2, 2). The position of the fourth mass of 4 kg to be places in the system as that the new centre of mass is at (0, 0, 0) is

a)(-3,-3,-3)b)(-3,3,-3)c)(2,3,-3)d)(2,-2,3)

31. If there is change of angular momentum from J to 5 J in 5 s, then the torque is

a)3J/5b)4J/5c)5J/4d)None of these

32. A tennis ball bounces down flight of stairs striking each step in turn and rebounding to the height of the step above. The coefficient of restitution has a value

a)1/2b)1c)1/√2d)1/2√2

33. Two solid discs of radii r and 2r roll from the top of an inclined plane without slipping. Then

a)The bigger disc will reach the horizontal level first

b)The smaller disc will reach the horizontal level first

c)The time difference of reaching of the discs at the horizontal level will depend on the inclination of the plane

d)Both the discs will reach at the same time

34. A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity ω. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring the ring. The ring now rotates with an angular velocity  ω’ is equal to

a)ω(m+2M)/m

b)ω(m-2M)/((m+2M) )

c)ωm/((m+M) )

d)ωm/((m+2M) )

35. Three bricks each of length L and mass M are arranged as shown from the wall. The distance of the centre of mass of the system from the wall is

a)L/4b)L/2c)(3/2)Ld)(11/12)L

36. A 2 kg body and a 3 kg body are moving along the x-axis. At a particular instant the 2 kg body has a velocity of 3 ms^(-1)and the 3 kg body has the velocity of 2 ms^(-1). The velocity of the centre of mass at that instant is

a)5 ms^(-1)b)1 ms^(-1)c)0d)None of these

37. A thin rod of length L and mass M is bent at the middle point O at an angle of 60°. The moment of inertia of the rod about an axis passing through O and perpendicular to the plane of the rod will be

a)(ML^2)/6b)(ML^2)/12c)(ML^2)/24d)(ML^2)/3

38. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity ω_0. When the tortoise moves along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platform ω(t) will vary with time t as

a)b)c)d)

39. Moment of inertia of a disc about an axis which is tangent and parallel to its plane is I. Then the moment of inertia of disc about a tangent, but perpendicular to its plane will be

a)3I/4b)5I/6c)3I/2d)6I/5

40. Four particles of mass 1 kg, 2 kg, 3 kg and 4 kg are placed at the corners A,B,C and D respectively of a square ABCD of edge X-axis and edge AD is taken along Y-axis, the co-ordinates of centre of mass in SI is

a)(1,1)b)(5,7)c)(0.5,0.7)d)None of these

41. A body of moment of inertia of 3 kg-m^2 rotating with an angular velocity of 2 rad/sec has the same kinetic energy as a mass of 12 kg moving a velocity of

a)8 m/sb)0.5 m/sc)2 m/sd)1 m/s

42. Two bodies A and B of definite shape (dimensions of bodies are not ignored). A is moving with speed of 10 ms^(-1) and B is in rest, collide elastically. The

a)Body A comes to rest and B moves with speed of 10ms^(-1)

b)They may move perpendicular to each other

c)A and B may come to rest

d)They must move perpendicular to each other

43. Moment of inertia of a ring of mass M and radius R about an axis passing through the centre and perpendicular to the plane is I. What is the moment of inertia about its diameter

a)Ib)I/2c)I/√2d)I+MR^2

44. A particle is projected with 200 ms^(-1), at an angle of 60°. At the highest point it explodes into three particles of equal masses. One goes vertically upward with velocity 100 ms^(-1), the second particle goes vertically downward with the same velocity as the first. Then what is the velocity of the third particle?

a)120 ms^(-1) with 60° angleb)200 ms^(-1) with 30° angle

c)50ms^(-1) vertically upwardsd)300 ms^(-1) horizontally

45. Two bodies of identical mass m are moving with constant velocity v but in the opposite directions and stick to each other, the velocity of the compound body after collision is

a)vb)2vc)Zerod)v/2

46. What remains constant in the field of central force

a)Potential energyb)Kinetic energyc)Angular momentumd)Linear momentum

47. A bullet of mass m hits a target of mass M hanging by a string and gets embedded in it. If the block rises to a height h as a result of this collision, the velocity of the bullet before collision is

a) v=√2ghb)v=√2gh (1+m/M)c)v=(1+M/m) √2ghd)v=√2gh (1-m/M)

48. The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axes is

a)√2 ∶1

b) √2 ∶√3

c) √3 ∶√2

d) 1∶√2

49. (1) Centre of gravity (C. G.) of a body is the point at which the weight of the body acts

(2) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius

(3) To evaluate the gravitational field intensity due to any body at an external point, the centre mass of the body can be considered to be concentrated at its C.G.

(4) The radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the C.G. of the body to the axis

Which one of the following pairs of statements is correct

a)(4) and (1)b)(1) and (2)c)(2) and (3)d)(3) and (4)

50. A binary star consists of two stars A (mass 2.2 M_s) and B (mass 11 M_s), where M_s is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is

a)7b)6c)9d)10

51. A T shaped object with dimension shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.

a)2/3 lb)3/2 lc)4/3 ld)l

52. Three identical blocks A,B and C are placed on horizontal frictionless surface. The blocks A and C are at rest. But A is approaching towards B with a speed 10 ms^(-1). The coefficient of restitution for all collisions is 0.5. The speed of the block C just after collision is

a)5.6 ms^(-1)b)6 ms^(-1)c)8 ms^(-1)d)10 ms^(-1)

53. A wheel of mass 8 kg and radius 40 cm is rolling on a horizontal road with angular velocity of 15 rad s^(-1). The moment of inertia of the wheel about its axis is 0.64 kg m^(-2). Total KE of wheel is

a)288 Jb)216 Jc)72 Jd)144 J

54. The two bodies of mass m_1 and m_2 (m_1>m_2) respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is

a)[(m_1-m_2)/(m_1+m_2 )]^2 gb)[(m_1-m_2)/(m_1+m_2 )]^2c)gd)zero

55. Moment of inertia of big drop in I. If 8 droplets are formed from big drop, then moment of inertia of small droplet is

a)I/32 b)I/16 c)I/8 d)I/4

56. A thin wire of mass M and length L is bent to form a circular ring. The moment of inertia about its axis is

a)1/(4π^2 ) 〖ML〗^2   b)1/12 〖ML〗^2 c)1/〖3π〗^2  〖ML〗^2 d)1/π^2  〖ML〗^2

1 (d)
Weight of the rod will produce the torque
τ=Iα⇒mg×l/2=(ml^2)/3×α

Angular acceleration
α=3g/2l
2 (b)
Since, m_2 moves with constant velocity

∴ f_2=m_2 g
f_2=8×10=80N
Since, boy of mass m_1 moves with acceleration a=2ms^(-2) in upward direction

∴f_1=m_1 g=m_1 a
∴f_1=m_1 g=m_1 a
=10×10+10× =120 N
3 (b)
We calculate moment of inertia of the system about AD

Moment of inertia of each of the sphere A and D about
AD=2/5 Ma^2
Moment of inertia of each of the sphere B and C about AD
=(2/5 Ma^2+Mb^2 )
Using theorem of parallel axes
∴ Total moment of inertia
I=(2/5 Ma^2 )×2+(2/5 Ma^2+Mb^2 )×2
=8/5 Ma^2+2Mb^2
4 (a)
The moment of inertia of ring =MR^2
The moment of inertia of removed sector =1/4 MR^2
The moment of inertia of remaining part =MR^2-1/4 MR^2
=3/4 MR^2
According to question, the moment of inertia of the remaining part =kMR^2
then, k=3/4
5 (c)
(Rotational kinetic energy)/(Translatory kinetic energy)=(1/2 mv^2 K^2/R^2 )/(1/2 mv^2 )=K^2/R^2 =2/5
6 (d)
According to law of conservation of angular momentum, if there is no torque on the system, then the angular momentum remains constant.
7 (d)
From third equation of angular motion,
ω^2=ω_0^2=2α (Here,ω=ω_0/2,θ=36×2π)
∴ (ω_0/2)^2=ω_0^2-2α×36×2π
or 4×36πα=ω_0^2-(ω_0^2)/4
or 4×36πα=(3ω_0^2)/4
or α=(ω_0^2)/(16×12π) …(i)
According to question again applying the third equation of angular motion
ω^2=ω_0^2-2αθ (Here,ω=0)
∴ 0=(ω_0/2)^2-2×(ω_0^2.θ)/(16×12π)
or θ=24π or θ=12×2π
But 2π=1 cycle
So, θ=12 cycle
8 (d)
Mass of disc (X),m_X=πR^2 tρ
Where, ρ= density of material of disc
∴ I_X=1/2 m_X R^2=1/2 πR^2 tρR^2
I_X=1/2 πρtR^4 …(i)
Mass of disc (Y)
m_Y=π(4R)^2 t/4 ρ=4πR^2 tρ
and I_Y=1/2 m_Y (4R)^2=1/2 4πR^2 ρt.16R^2
⇒ I_Y=32πtρR^4 …(ii)
∴ I_Y/I_X =(32πtρR^4)/(1/2 πρtR^4 )=64
∴ I_Y=64 I_X
9 (b)
The ratio of rotational kinetic energy to total kinetic energy is
(KE_R)/(KE_T )=(1/2 Iω^2)/(1/2 Iω^2+1/2 Mv^2 )
The moment of inertia of a ring
= 1/2 MR^2
∴ (KE_R)/(KE_T )=(1/2 (1/2 MR^2 ) [v/R]^2)/(1/2×1/2 MR^2 [v/R]^2+1/2 Mv^2 )

=(1/2 (1/2 Mv^2 ))/(1/2 (1/2 Mv^2 )+1/2 Mv^2 )
(KE_R)/(KE_T )=1/3
10 (b)
We know
L=Iω …(i)
L^2=2KI
From Eq. (i)
L^2=2K L/ω
L=2K/ω
L^’=2(K/2)/2ω=L/4
11 (d)
θ=ωt=(500×2π)/60=50π/3 rad
12 (b)
θ=ω_0 t+1/2 αt^2=2×2+1/2×3×(2)^2=10rad
13 (a)
The rotational kinetic energy of the disc is
K_rot=1/2 Iω^2=1/2 (1/2 MR^2 ) ω^2=1/4 MR^2 ω^2
The translational kinetic energy is
K_trans=1/2 Mv_CM^2
where v_CM is the linear velocity of its centre of mass.
Now, v_CM=Rω
Therefore, K_trans=1/2 MR^2 ω^2
Thus, K_total=1/4 MR^2 ω^2+1/2 MR^2 ω^2=3/4 MR^2 ω^2
∴ K_rot/K_total =(1/4 MR^2 ω^2)/(3/4 MR^2 ω^2 )=1/3
14 (d)
Rotational kinetic energy,
KE=1/2 Iω^2
Here, ω=1 rads^(-1)
∴ KE=1/2 I×I
or I=2 KE
ie, moment of inertia about an axis of a body is twice the rotational kinetic energy.
15 (c)
I=2/5 MR^2=2/5 (4/3 πR^3 ρ) R^2=8/15×22/7 R^5 ρ
I=176/105 R^5 ρ
17 (c)
M.I. of body about centre of mass =I_cm=mK^2
M.I. of a body about new parallel axis

I_new=I_cm+ma^2=mK^2+ma^2
I_new=m(K^2+a^2)
K_R=1/2 I_new ω^2=1/2 m(K^2+a^2 ) ω^2
18 (b)
Moment of inertia of a rod about one end =(ML^2)/3
As, I=I_1+I_2+I_3
∴ I=0+(ML^2)/3+(ML^2)/3=(2ML^2)/3
19 (b)
I=m_1 r_1^2+m_2 r_2^2
=200/1000 (30/100)^2+300/1000 (20/100)^2=0.03 kg m^2
20 (a)
The radius of gyration is the distance from the
axis of rotation at which if whole mass of the
body is supposed to be concentrated.
Here, the whole mass of the equilateral triangle
acts at point O. So the distance OA is the radius of gyration of
this system. Now from triangle ADB

x^2=BD^2+(x/2)^2
or BD^2=x^2-x^2/4
or BD^2=(3x^2)/4
or BD=√3 x/2
Hence, the distance, OB=(√3 x)/2×2/3
⇒ OB=x/√3
But, the distances OA,OB and OC are the same.
So, OA=x/√3
Hence, the radius of gyration of this system is x/√3.
21 (c)
A couple consists of two equal and opposite forces acting at a separation, so that net force becomes zero. When a couple acts on a body it rotates the body but does not produce any translatory motion. Hence, only rotational motion is produced.
22 (a)
It’s always in axial direction
24 (c)
∴ a ⃗_CM=F ⃗_eq/((m_1+m_2+m_3))=(m_1 a ⃗_1+m_2 a ⃗_2+m_3 a ⃗_3)/((m_1+m_2+m_3))
∴F ⃗_eq=m_1 a ⃗_1+m_2 a ⃗_2+m_3 a ⃗_3
=1×1+2×2+4×(-0.5)=1+4-2=3 N
25 (d)
Total kinetic energy =1/2 mv^2 (1+K^2/R^2 )=32.8 J
⇒1/2×10×(2)^2 (1+K^2/(0.5)^2 )=32.8⇒K=0.4 m
26 (a)
2/5 MR^2=1/2 Mr^2+Mr^2
or 2/5 MR^2=3/2 Mr^2
∴ r=2/√15 R
27 (a)
If no external torque acts on a system of particle then angular momentum of the system remains constant, that is,
τ=0
⇒ dL/dt=0
⇒ L=Iω=constant
⇒ I_1 ω_1=I_2 ω_2
∴ 1/2 Mr^2 ω_1=1/2 (M+2m) r^2 ω_2 …(i)
Here, M=2kg, m=0.25 kg, r=0.2m.
ω_1=30 rads^(-1)
Hence, we get after putting the given values in Eq. (i)
1/2×2×(0.2)^2×30=1/2×(2+2×0.25) (0.2)^2×ω_2
⇒ 60=2.5ω_2
∴ ω_2=24 rads^(-1)
28 (c)
Here, θ=60°,l=10 m,a=?
a=(g sin⁡θ)/(1+K^2/R^2 )
For solid sphere, K^2=2/5 R^2
a=(9.8 sin⁡〖60°〗)/(1+2/5)=5/7×9.8×√3/2=6.06 ms^(-2)
29 (a)
L=r P⇒log_e⁡〖L=log_e⁡〖P+log_e⁡r 〗 〗
If graph is drawn between log_e⁡L and log_e⁡P then it will be straight line which will not pass through the origin
30 (a)
m_1=1 kg,m_2=2 kg,m_3=3 kg
Position of centre of mass (2, 2, 2,)
m_4=4 kg
New position of centre of mass (0, 0, 0).
For initial position
X_CM=(m_1 x_1+m_2 x_2+m_3 x_3)/(m_1+m_2+m_3 )
2=(m_1 x_1+m_2 x_2+m_3 x_3)/(1+2+3)
m_1 x_1+m_2 x_2+m_3 x_3=12
Similarly, m_1 y_1+m_2 y_2+m_3 y_3=12
and m_1 z_1+m_2 z_2+m_3 z_3=12
For new position,
X_CM^’=(m_1 x_1+m_2 x_2+m_3 x_3+m_4 x_4)/(m_1+m_2+m_3+m_4 )
0=(12+4×x_4)/(1+2+3+4)
4x_4=-12
x_4=-3
Similarly, y_4=-3
z_4=-3
∴ Position of fourth mass (-3, -3, -3)
31 (b)
We know that rate of change of angular momentum (J) of a body is equal to the external torque (τ) acting upon the body.
ie. dJ/dt=τ
Given, J_1=J,J_2=5J
∴ ∆J=J_2-J_1=5J-J=4J
Hence, τ=4/5 J
32 (c)
As shown in adjoining figure ball is falling from height 2h and rebounding to a height h only. It means that velocity of ball jus before collision
u=√((2(2h))/g)=√(4h/g) and velocity just after collision

v=-√(2h/g)
∴e=(-v√(2h/g))/(u √(4h/g ))=1/√2
34 (d)
As no external torque is acting about the axis, angular momentum of system remains conserved.

∴ I_1 ω=I_2 ω’
⇒ mR^2 ω=(mR^2+2MR^2)ω’
⇒ ω^’=(m/(m+2M)))ω
35 (d)

From figure, x_1=L/2,x_2=L/2+L/2=L
x_3=L/2+L/4+L/2=5L/4
∴X_CM=(m_1 x_1+m_2 x_2+m_3 x_3)/(m_1+m_2+m_3 )
=(M×L/2+M×L+M×5L/4)/(M+M+M)=(11/4 ML)/3M=11L/12
36 (d)
v ⃗_cm=(m_1 v ⃗_1+m_2 v ⃗_2)/(m_1+m_2 )=(2×3+3×2)/(2+3)=12/5=2.4m/s
37 (b)
Moment of inertia of a uniform rod about one end=(ml^2)/3
∴ Moment of inertia of the system
=2×(M/2) (L/2)^2/3=(ML^2)/12
38 (b)
Since there is no external torque, angular momentum will remain conserved. The moment of inertia will first decrease till the tortoise moves from A to C and then increase as it moves from C to D. Therefore ω will initially increase and then decrease

Let R be the radius of platform m the mass of tortoise and M is the mass of platform
Moment of inertia when the tortoise is at A
I_1=mR^2+(MR^2)/2
and moment of inertia when the tortoise is at B
I_2=mr^2+(MR^2)/2
Here r^2=a^2+[√(R^2-a^2 )-vt]^2
From conservation of angular momentum ω_0 I_1=ω_((t)) I_2
Substituting the values we can found that the variation of ω_((t)) is non linear
39 (d)
The moment of inertia of the disc about an axis parallel to its plane is

I_t=I_d+MR^2
⇒ I=1/4 MR^2+MR^2
=5/4 MR^2
or MR^2=4I/5
Now, moment of inertia about a tangent perpendicular to its plane is
I^’=3/2 MR^2=3/2×4/5 I=6/5 I
40 (c)
x_CM=(m_A x_A+m_B x_B+m_C x_C+m_D x_D)/(m_A+m_B+m_C+m_D )

=(1×0+2×1+3×1+4×0)/(1+2+3+4)
=(2+3)/10=1/2=0.5 m
Similarly, y_CM=(m_A y_A+m_B y_B+m_C y_C+m_D y_D)/(m_A+m_B+m_C+m_D )
=(1×0+2×0+3×1+4×1)/(1+2+3+4)
=7/10=0.7 m
41 (d)
1/2 Iω^2=1/2 mv^2⇒1/2×3×(2)^2=1/2×12×v^2
⇒v=1m/s
42 (b)
(a) This is only possible when collision is head on elastic.
(b) When collision is oblique elastic, then in this case, both bodies move perpendicular to each other after collision
(c) Since, in elastic collision, kinetic energy of system remains constant so, this is not possible.
(d) The same reason as (b).
43 (b)
Moment of inertia of a ring of mass M and radius R about an axis passing through the centre and perpendicular to the plane
I=MR^2 …(i)
Moment of inertia of a ring about its diameter
I_diameter=(MR^2)/2=I/2 [Using (i)]
44 (d)
At the highest point momentum of particle before explosion p ⃗\=mv cos⁡〖60°〗
=m×200 1/2=100m horizontally
Now as three is no external force during explosion, hence
p ⃗=p ⃗_1+p ⃗_2+p ⃗_3
However, since velocities of two fragments, of masses m/3 each, are 100 ms^(-1) downward and 100ms^(-1) upward.
hence, p ⃗_1=-p ⃗_2 or p ⃗_1+p ⃗_2
p ⃗_3=m/3.v_3=p ⃗_ =100 m horizontally
v_3=300ms^(-1) horizontally
45 (c)
mv-mv=(m+m)v
v=0
46 (c)
In the field central force, Torque = 0 ∴ angular momentum remains constant
47 (c)
If initial velocity of bullet be v then after collision combined velocity of bullet and target is
v^’=mv/((M+v) ) and h=v^’2/2g or v^’=√2gh
∴ mv/((M+m))=√2gh
⇒ v=((M+m)/m)∙√2gh=(1+M/m) √2gh
48 (d)
Radius of gyration of circular disc k_disc=R/√2
Radius of gyration of circular ring k_ring=R
Ratio =k_disc/k_ring =1/√2
50 (b)
L_Total/L_B =((I_A+I_B)ω)/(I_B.ω) (as ω will be same in both cases)
=I_A/I_B +1=(m_A r_A^2)/(m_B r_B^2 )+1
=r_A/r_B +1 (as m_A r_A=m_B r_B)
=11/2.2+1 (as r∝1/m)
=6
∴ The correct answer is 6.
51 (c)
For translator motion the force should be applied on the centre of mass of the body so we have to calculate the location of centre of mass of T shaped object.
Let mass of rod AB is m so the mass of rod CD will be 2m.
Let y_1 is the centre of mass of rod AB and y_2 is the centre of mass of rod CD. We can consider that whole mass of the rod is placed at their respective centre of mass ie, mass m is placed at y_1 and mass 2m is placed at y_2.
Taking point c at the origin position vector of points y_1 and y_2 can be written as
r_1=2lj,r_2=lj
and m_1=m and m_2=2m
Position vector of centre of mass of the system
r_CM=(m_1 r_1+m_2 r_2)/(m+m_2 )=(m2lj ̂+2mlj ̂)/(m+2m)=(4mlj ̂)/3m=(4lj ̂)/3
52 (a)
For collision between blocks A and B,
e=(v_B-v_A)/(u_A-u_B )=(v_B-v_A)/(10-0)=(v_B-v_A)/10
∴v_B-v_A=10e=10×0.5=5 ….(i)
from principle of momentum conservation,
m_A u_A+m_B u_B=m_A v_A+m_B v_B
Or m×10+0=mv_A+mv_B
∴ v_A+v_B=10 ….(ii)
Adding Eqs. (i) and (ii), we get
v_B=7.5 ms^(-1) …(iii)
Similarly for collision between B and C
v_C-v_B=7.5e=7.5×0.5=3.75
∴ v_C-v_B=3.75ms^(-1) …(iv)
Adding Eqs. (iii) and (iv) we get
2v_C=11.25
∴ v_C=11.25/2=5.6ms^(-1)
53 (b)
Here, m=8 kg,r=40 cm=2/5 m,
ω=12 rad s^(-1),I=0.64 kg m^2
Total KE=1/2 Iω^2+1/2 mv^2
=1/2 Iω^2+1/2 mr^2 ω^2
=1/2×0.64×15^2+1/2×8×(2/5)^2×15^2=216 J
54 (a)
In the pulley arrangement |a ⃗_1 |=|a ⃗_2 |=a=((m_1-m_2)/(m_1+m_2 ))g
But a ⃗_1 is in downward direction and in the upward direction ie, a ⃗_2=-a ⃗_1
∴ Acceleration of centre of mass
a ⃗_CM=(m_1 a ⃗_1+m_2 a ⃗_2)/(m_1+m_2 )=(m_1 [(m_1-m_2)/(m_1+m_2 )]g-m_2 [(m_1-m_2)/(m_1+m_2 )]g)/((m_1+m_2))
=[(m_1-m_2)/(m_1+m_2 )]^2 g
55 (a)
Moment of inertia of big drop is I=2/5 MR^2. When small droplets are formed from big drop volume of liquid remain same
n 4/3 πr^3=4/3 πR^3
⇒ n^(1⁄3) r=R
as n=8
⇒ r=R/2
Mass of each small droplet =M/8
∴ Moment of inertia of each small droplet
=2/5 [M/8] [R/2]^2
=1/32 [2/5 MR^2 ]=I/32
56 (a)
Here a thin wire of length L is bent to form
a circular ring.

Then, 2πr=L (r is the radius of ring)
⇒ r=L/2π
Hence, the moment of inertia of the ring about its axis
I=Mr^2⇒ I=M(L/2π)^2⇒ I=(ML^2)/(4π^2 )