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

Answer

C

Explanation

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