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
B. `sqrt(2) v`
C. `v//sqrt(2)`
D. `sqrt((2//5))v`

Answer

C

Explanation

( c) `v = sqrt(2 gh)`…(1)
For second case : `(1)/(2) mv'^2 + (1)/(2) I omega^2 = mgh`
`(omega = v'//r, I = mr^2)`
`rArr v' = sqrt(gh)`…(2)
from (1)/(2) `v' = v //sqrt(2)`.