Explanation

(c) In volume of sphere in liquid,
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V = $\frac{4}{3}\pi r^{3}$ …….(i)
When mass m is placed on the piston, the increased pressure $p = \frac{mg}{a}.$ since this increased pressure is equally applicable to all directions on the sphere, so there will be decrease in volume of sphere, due to decrease in its radius. From Eq.(i), change in volume is
$\mathrm{\Delta}V = \frac{4}{3}\pi\ \times 3r^{2}\ \mathrm{\Delta}r = 4\pi\ \mathrm{\Delta}r$
$\therefore\ \ \ \frac{\mathrm{\Delta}V}{V} = \ \frac{4\pi r^{2}\mathrm{\Delta}r}{(4/3)\pi r^{3}}\ \frac{3\mathrm{\Delta}r}{r}$
Now, $K = \ \frac{p}{dV/V} = \frac{mg}{a} \times \frac{r}{3\mathrm{\Delta}r}$
$\therefore\ \ \frac{\mathrm{\Delta}r}{r} = \frac{mg}{3Ka}$