A solid sphere of radius R made of material of bulk modulus K is surrounded by a liquid of cylindrical container. A massless piston of area A floats on the surface of the liquid when a mass M is placed on the piston to compress the liquid, the fractional change in the radius of the sphere is

See fig.

When the mass M is placed on the piston, then a pressure (or stress) is applied which is given by $∆$P =(Mg/ A), This pressure is applied equally in all sides of the liquid (Pascal's law). So, the volume of the sphere is reduced ,i.e., its radius is reduced. Iet the new radius become R -$∆$r. The new volume of sphere
$\begin{array}{l}{\mathrm{V}}^{\mathrm{\prime }}=\frac{4}{3}\mathrm{\pi }(\mathrm{R}-\mathrm{\Delta R}{)}^3=\frac{4}{3}{\mathrm{\pi R}}^3(1-\mathrm{\Delta R}/\mathrm{R}{)}^3\\ =\frac{4}{3}{\mathrm{\pi R}}^3\left(1-\frac{3\mathrm{\Delta R}}{\mathrm{R}}\right)=\mathrm{V}\left(1-\frac{3\mathrm{\Delta R}}{\mathrm{R}}\right)\end{array}$
where V $=\frac{4}{3}{\mathrm{\pi R}}^3$, initial volume.
So, the change in volume (V'- V) is given by
$\mathrm{\Delta V}=\left[\mathrm{V}\left(1-\frac{3\mathrm{\Delta R}}{\mathrm{R}}\right)\right]-\mathrm{V}=-3\mathrm{V}\left(\frac{\mathrm{\Delta R}}{\mathrm{R}}\right)$
Now, $\mathrm{K}=\frac{\mathrm{\Delta P}}{-(\mathrm{\Delta V}/\mathrm{V})}=\frac{\mathrm{Mg}/\mathrm{A}}{3(\mathrm{\Delta R}/\mathrm{R})}$
or $\frac{\mathrm{\Delta R}}{\mathrm{R}}=\frac{\mathrm{Mg}}{3\mathrm{KA}}$