a
(a) When the ball is pushed down the water, it gains potential energy.
The gained potential energy of water $=(V \rho) r g-\left(\frac{V}{2} \times \rho\right)\left(\frac{3}{8} \times r\right) g$
When the half of the spherical ball is immersed, rise of $\mathrm{c}$. $\mathrm{g}$. of displaced water $=\frac{3}{8} \times r$
$=V \rho r g\left(\frac{1-3}{16}\right)=\frac{4}{3} \pi r^{3} \rho r g \times \frac{13}{16}=\frac{13}{12} \pi r^{4} \rho g$
Lost potential energy $=V \rho r g=\frac{4}{3} \pi r^{4} \rho g$
Work done $=\frac{13}{12} \pi r^{4} \rho g-\frac{4}{3} \pi r^{4} \rho g=\frac{5}{12} \pi r^{4} \rho g$