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

Options

A. `sqrt3 v`
B. `(v)/(sqrt3)`
C. `(v)/(sqrt2)`
D. `sqrt2 v`

Answer

D

Explanation

Given, `v = sqrt((2gh)/(1+(K^(2))/(R^(2))))`
`= sqrt((2gh)/(2))=sqrt(gh)`
for the block, `v' = sqrt(2gh)=sqrt(2)(sqrt(gh))`
`sqrt(2)v`.