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

Answer

D

Explanation

For a sliding body of mass `m`
`v_("body")=sqrt(2gh)=v`………..i
For a rolling of same mass `m`
`v_("ring")=sqrt((2gh)/g)`
`=sqrt((2gh)/(1+I/(mR^(2))))=sqrt((2gh)/(1+(mR^(2))/(mR^(2))))=sqrt((2gh)/2)=sqrt(gh)=v/(sqrt(2))`