Question

A spherical ball of radius $r$ and relative density $0.5$ is floating in equilibrium in water with half of its immersed in water. The work done in pushing the ball down so that whole of it is just immersed in water is $1\rho$ is the density of water-

• A. $0.25\rho rg$
• B. $0.5\rho rg$
• C. $\frac{4}{3}\pi r^3\rho g$
• D. $\frac{5}{12}\pi r^4\rho g$

Answer 1

The correct option is D $\frac{5}{12}\pi r^4\rho g$

When the ball is pushed down the water, it gains potential energy.

The gained potential energy of water$=\left(V\rho \right)rg-(\frac{V}{2}\times \rho )(\frac{3}{8}\times r)g$

When the half of the spherical ball is immersed, rise of c.g. of displaced water$=\frac{3}{8}\times r$

$=V\rho rg\left(\frac{1-3}{16}\right)=\frac{4}{3}\pi r^3\rho rg\times \frac{13}{16}=\frac{13}{12}\pi r^4\rho g$

Lost potential energy$=V\rho rg=\frac{4}{3}\pi r^4\rho g$

Work done$=\frac{13}{12}\pi r^4\rho g-\frac{4}{3}\pi r^4\rho g=\frac{5}{12}\pi r^4\rho g$