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. `v/2`
B. `2v`
C. `v/(sqrt(2))`
D. `sqrt(2)v`

Answer

D

Explanation

for ring , `v_(1) = sqrt((2gh)/(1+(k^(2))/(R^(2))) ) = v ` for block `v_(2) = sqrt(2gh ) `