A small object of uniform density rolls up a curved surface with an initial velocity ‘v’. It reaches up to a maximum heights of \(\frac{3v^2}{4g}\) with respect to the initial position. The object is : 

Disc

From conservation of mechanical energy

\(\frac{I \omega^2}{2}+ \frac{1}{2}mv^2 = mgh \)

We know that v = ωR

\(\frac{1}{2}\)\(\frac{v^2}{R^2}\) + \(\frac{1}{2}\)mv² = mgh

∴ h = \(\frac{3v^2}{4g}\)

\(\frac{1}{2}\)\(\frac{v^2}{R^2}\) + \(\frac{1}{2}\)mv² = mg\(\frac{3v^2}{4g}\)

On calculating we get

I = \(\frac{mR^2}{2}\) which is for disc