A body of mass `m` slides down an incline and reaches the bottom with a velocity `v`. If the same mass were in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been

Options

A. v,v and v
B. `sqrt2v`
C. `(1)/(sqrt2)v`
D. `sqrt((2)/(5))v`

Answer

C

Explanation

`v= sqrt(2gh)`
`v.= sqrt((2gh)/(1+ (K^(2))/(R^(2))))= sqrt(gh)= (v)/(sqrt2) [ because "for ring" (K^(2))/(R^(2))=1]`