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A circular ring starts rolling down on a...

A circular ring starts rolling down on an inclined plane from its top. Let `v` be velocity of its centre of mass on reaching the bottom of inclined plane. If a block starts sliding down on an identical inclined plane but smooth, from its top, then the velocity of block on reaching the bottom of inclined plane is

A

`v/2`

B

`2v`

C

`v/(sqrt(2))`

D

`sqrt(2)v`

Text Solution

Verified by Experts

The correct Answer is:
D
for ring `v_(1)=sqrt((2gh)/(1+(k^(2))/(R^(2))))=v` for block `v_(2)=sqrt(2gh)`
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Knowledge Check

  • A solid cylinder starts rolling down on an inclined plane from its top and V is velocity of its centre of mass on reaching the bottom of inclined plane. If a block starts sliding down on an identical inclined plane but smooth, from its top, then the velocity of block on reaching the bottom of inlined plane is

    A
    `v/(sqrt(2))`
    B
    `sqrt(2)v`
    C
    `sqrt(3/2)v`
    D
    `sqrt(2/3)v`

  • A disc of mass 3 kg rolls down an inclined plane of height 5 m. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is

    A
    50 J
    B
    100 J
    C
    150 J
    D
    175 J

  • A solid cylinder rolls down from an inclined plane of height h. What is the velocity of the cylinder when it reaches at the bottom of the plane ?

    A
    `sqrt((2gh)/3)`
    B
    `sqrt(2gh)`
    C
    `sqrt((4gh)/3)`
    D
    `sqrt((3gh)/2)`

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