A small object of uniform density rolls up a curved surface with an initial velocity 'v' . It reaches upto a maximum height of `(3v^(2))/(4g)` with respect to the initial position . The object is

Options

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

Answer

B

Explanation

The kinetic energy of the rolling object is converted into potential energy at height `h (= (3v^2)/(4g))`
So by the law of conservation of mechanical energy , we have
`(1)/(2) mv^(2) + (1)/(2) I omega^(2) = Mgh`

`(1)/(2) Mv^(2) + (1)/(2) I ((v)/(R))^(2) = Mg ((3v^2)/(4g)) (because omega = (v)/(R))`
`(1)/(2) I (v^(2))/(R^(2)) = (3)/(4) Mv^(2) - (1)/(2) Mv^(2) , or I = (1)/(2) MR^(2)`
Hence the object is disc.