Q 11-06-156JEE MainJEE Main 2021 (20 Jul, Shift 2)Easy
A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is:
Answer: (B) solid cylinder
$\dfrac{K_{rot}}{K_{trans}} = \dfrac{\frac12I\omega^2}{\frac12mv^2} = \dfrac{k^2}{R^2} = \dfrac12$, which is the solid cylinder ($I = \frac12mR^2$).
Solution by Sreeraj P, M.Sc Physics