A spherical ball of radius r and relative density 0.5 is floating in equilibrium in water with half of it immersed in water. The work done in pushing the ball down so that whole of it is Just immersed in water is

 When the ball is pushed down, the water gains P.E whereas the ball losses P.E. Hence gain in P.E. of water 
$$\begin{gathered} =\mathrm{V\rho rg}-\frac{\mathrm{V}}{2}\mathrm{\rho }\left(\frac{3}{8}\mathrm{r}\right)\mathrm{g} \\ =\frac{13}{12}{\mathrm{\pi r}}^4\mathrm{\rho g}\left(\because \mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3\right) \\ \text{ Loss in P.E. of ball }={\mathrm{V\rho }}^{\mathrm{\prime }}\mathrm{rg} \\ =\frac{4}{3}{\mathrm{\pi r}}^4{\mathrm{\rho }}^{\mathrm{\prime }}\mathrm{g} \\ \text{ Work done }=\frac{13}{12}{\mathrm{\pi r}}^4\mathrm{\rho g}-\frac{4\mathrm{\pi }}{3}{\mathrm{r}}^4{\mathrm{\rho }}^{\mathrm{\prime }}\mathrm{g} \\ ={\mathrm{\pi r}}^4\mathrm{\rho g}\left[\frac{13}{12}-\frac{4}{3}\frac{{\mathrm{\rho }}^{\mathrm{\prime }}}{\mathrm{\rho }}\right] \\ ={\mathrm{\pi r}}^4\mathrm{\rho g}\left[\frac{13}{12}-\frac{4}{3}\times 0\cdot 5\right] \\ =\frac{5}{12}{\mathrm{\pi r}}^4\mathrm{\rho g} \end{gathered}$$