Q 11-06-170JEE MainJEE Main 2020 (8 Jan, Shift 2)Easy
A uniform sphere of mass $500$ g rolls without slipping on a plane horizontal surface with its centre moving at a speed of $5.00$ cm s$^{-1}$. Its kinetic energy is:
Answer: (A) $8.75\times10^{-4}$ J
For rolling without slipping, $\omega = v/R$ and $I = \tfrac{2}{5}mR^2$ for a solid sphere:
$$K = \frac{1}{2}mv^2 + \frac{1}{2}\cdot\frac{2}{5}mR^2\cdot\frac{v^2}{R^2} = \frac{7}{10}mv^2$$
$$K = 0.7\times0.5\times(0.05)^2 = 8.75\times10^{-4}\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics