Solution:
Solution:
Increase in pressure $$= \left| \frac{mg}{Area} = \frac{mg}{a}\right| $$
$$ B = - \frac{dP}{\frac{dV}{V}} $$
$$ \frac{dV}{V} = - \frac{dP}{B} $$
$$\frac{dV}{V} = - \frac{mg}{Ka} $$
$$ v = \frac{4}{3}\pi R^{3} $$
$$ \frac{dV}{V} = \frac{3dR}{R} $$
$$ 3 \frac{dR}{R} = - \frac{mg}{Ka} $$
$$ \left|\frac{dR}{R}\right| = \frac{mg}{3Ka} $$
