A mass `m` of radius `r` is rolling horizontally without any slip with a linear speed `v`. It then rolls up to a height given by `(3)/(4)(v^(2))/(g)`

Options

A. solid sphere
B. hollow sphere
C. disc
D. ring

Answer

C

Explanation

Let I bt the moment of inertia of the body. Then total `KE=1/2mv^(2)+1/2I omega^(2)`
or `KE=1/2mv^(2)+1/2 I (v^(2))/(R^(2))" "(omega=v/R)`
According to energy conservation loss in KE =gain in PE.
or `1/2(m+1/(R^(2)))v^(2)=mgh=mg((3v^(2))/(4g))`
Solving this we get `I=1/2mR^(2)`
i.e. the solid body is a disc.