From a circular ring of mass M and radius R, an arc corresponding

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 Multiple Choice QuestionsMultiple Choice Questions

41.

A body of mass 1.5 kg rotating about an axis with angular velocity of 0.3 rad s-1 has the angular momentum of 1.8 kg m2 s-2. The radius of gyration of the body about an axis is

  • 2 m

  • 1.2 m

  • 0.2 m

  • 1.6 m


42.

In a two-particle system with particle masses m1 and m2 , the first particle is pushed towards the centre of mass through a distance d, the distance through which second particle must be moved to keep the centre of mass at the same position is

  • m2dm1

  • d

  • m1dm2

  • m1dm1 + m2


43.

The principle involved in the performance of a spinning-chair circus acrobat is

  • conservation of angular momentum

  • conservation of linear momentum

  • conservation of energy

  • principle cf moment


44.

A billiard ball of mass m and radius r, when hit in a horizontal direction by a cue at a height h above its centre, acquired a linear velocity v0. The angular velocity ω0 acquired by the ball is

  • 5v0r22h

  • 2v0r25h

  • 2v0h5r2

  • 5v0h2r2


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45.

The radius of gyration of a solid cylinder of mass M and radius R about its own axis is

  • R2

  • R2

  • R3

  • R3


46.

Two masses m1 = 1 kg and m2 = 2 kg are connected by a light inextensible string and suspended by means of a weightless pulley as shown in the figure.

                

Assuming that both the masses start from rest, the distance travelled by the centre of mass in 2s is (Take g = 10 ms-2)

  • 209 m

  • 409 m

  • 23 m

  • 13 m


47.

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

  • L/4

  • L/2

  • (3/2)L

  • (11/12)L


48.

A fly wheel of moment of inertia 3 x 102 kg mis rotating with uniform angular speed of 4.6 rads-1. If a torque of 6.9 x 102 Nm retards the wheel, then the time in which the  wheel comes to rest is

  • 1.5 s

  • 2 s

  • 0.5 s

  • 1 s


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49.

Moment of inertia of a ring of mass M and radius R about a tangent to the circle of the ring is

  • 52 MR2

  • 32 MR2

  • 12 MR2

  • MR2


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50.

From a circular ring of mass M and radius R, an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is k times MR2. Then the value of k is

  • 34

  • 78

  • 14

  • 1


A.

34

The moment of inertia of circular ring

             = MR2

The moment of inertia of removed sector

             = 14 MR2

The moment of inertia of remaining part

             = MR2 - 14 MR2= 34 MR2

According to question, the moment of inertia of the remaining part

              = kMR2

then     k = 34


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