A body of mass `m` slides down an smooth incline and reaches the bottom with a velocity, Now smooth incline surface is made rough and the same mass was in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been:

Options

A. `vsqrt(2)`
B. `v/sqrt(2)`
C. `2sqrt(2)v`
D. `v/sqrt(2+1)`

Answer

B

Explanation

`mgh =1/2 mv^(2) =1/2 mv_(2)^(2) +1/2(mR^(2))(v_(2)/R)^(2) rArr v^(2) = 2v_(2)^(2) rArr v_(2) = v//sqrt(2)`