A uniform sphere of mass m and radius R is placed on a rough...

The sphere will roll without slipping when (\omega=\frac{v}{r}), where $v$ is linear velocity and (\omega) is angular velocity of the sphere. Now, angular momentum of sphere, about centre of mass [We are applying conservation of angular momentum just before and after struck] $$\begin{aligned} m v(h-R) & =I \omega=\left(\frac{2}{5} m R^2\right)\left(\frac{v}{R}\right) \\ m v(h-R) & =\frac{2}{5} m v R \\ h-R & =\frac{2}{5} R \Rightarrow h=\frac{7}{5} R \end{aligned}$$) Therefore,$the$sphere will roll without slipping with a constant velocity and hence, no loss of energy, so (\quad D \rightarrow p) Torque due to applied force, $F$ about centre of mass (\tau=F(h-R)) (clockwise) For (\tau=0, h=R), sphere will have only translational motion. It would lose energy by friction. Hence, (\mathrm{B} \rightarrow \mathrm{~s}) The sphere will spin clockwise when (\tau>0 \Rightarrow h>R) Therefore, (C \rightarrow q) The sphere will spin anti-clockwise when (\tau < 0 \Rightarrow h < R, \Lambda \rightarrow \mathrm{r})$