A solid sphere of mass M and radius 2 R rolls down an inclin

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 Multiple Choice QuestionsMultiple Choice Questions

151.

Match the following
Angular momentum 1. [M-1 L2 T-2 ]
B. Torque 2 [M1 L2 T-2
C. Gravitational constant 3.[M1 L2 T-2]
D. Tension 4.[M1 L2 T-1]

  • C- 2, D - 1

  • A - 4, B - 3

  • A - 3, C -2

  • B-2, A - 1


152.

A tangential force acting on the top of sphere of mass m kept on a rough horizontal place as shown in figure

                  

If the sphere rolls without slipping, then the acceleration with which the centre of sphere moves, is

  • 10 F7 m

  • F2 m

  • 3 F7 m

  • 7 F2 m


153.

A solid sphere is set into motion on a rough horizontal surface with a linear speed v in the forward direction and an angular speed vR in the anticlockwise direction as shown in figure. Find the linear speed of the sphere when it stops rotating and ω = vR

  

  • 3v5

  • 2 v5

  • 4 v3

  • 7 v3


154.

Two blocks of masses m1 and m2 are connected by a spring of spring constant k. The block of mass m2 is given a sharp empulse so that it acquires a velocity vtowards right. Find the maximum elongation that the spring will suffer.

           

  • m1 m2m1 + m212 v0

  • m1 + m2m1 -  m2 v0

  • m1 + m2m1 - m212 v0 

  • 2m1 + m2m1 m212 v0


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155.

Particles of masses m, 2m, 3m, ... , nm are placed on the same line at distances L, 2L, 3L, ... , nL from O. The distance of centre of mass from O is


156.

A ball of radius R rolls without slipping. Find the fraction of total energy associated with its rotational energy, if the radius of the gyration of the ball about an axis passing through its centre of mass is K.

  • K2K2 + R2

  • R2K2 + R2

  • K2 + R2R2

  • K2R2


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157.

A solid sphere of mass M and radius 2 R rolls down an inclined plane of height h without slipping. The speed of its centre of mass when it reaches the bottom is

  • 67 gh

  • 3 gh

  • 107 gh

  • 43 gh


C.

107 gh

When solid sphere rolls down on an inclined plane, then it has both rotational and translational kinetic energy

       K = Krot + Ktrans

       K = 12 2 + 12 M v2

where,

I= moment of inertia of solid sphere = 25 MR2

∴     K = 12 25 M  2R 2 ω2 + 12 Mv2                      [R =2R ]

           = 45MR2 v2R + 12 Mv2                           [ v = Rω and R = 2R ]

        K  = 15 Mv2 + 12 Mv2

        K = 710 Mv2

Now, gain in KE= loss in PE

          710 Mv2 = Mgh

 ⇒               v = 107 gh


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158.

A wheel starts rotating from rest at time t = 0 with a angular acceleration of 50 radians/s2. The angular acceleration (α) decreases to zero value after 5 seconds. During this interval, a varies according to the equation

      

The angular velocity at t = 5 s will be

  • 10 rad/s

  • 250 rad/s

  • 125 rad/s

  • 100 rad/s


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159.

Assertion:  A solid sphere is rolling on a rough horizontal surface. Acceleration of contact point is zero. 

Reason:  A solid sphere can roll on the smooth surface.

  • If both assertion and reason are true and reason is the correct explanation of assertion.

  • If both assertion and reason are true but reason is not the correct explanation of assertion.

  • If assertion is true but reason is false.

  • If both assertion and reason are false.


160.

A sphere of mass 10 kg and radius 0.5 m rotates about a tangent. The moment of inertia of the sphere is

  • 5 kg m2

  • 2.7 kg m2

  • 3.5 kg m2

  • 4.5 kg m2


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