c
Only the translational kinetic energy of disc changes into gravitational potential energy. And rotational \(\mathrm{KE}\) remains unchanged as there is no friction.
\(\frac{1}{2} \mathrm{mv}^2=\mathrm{mgh}\)
\(\mathrm{h}=\frac{\mathrm{v}^2}{2 \mathrm{~g}}\)