Explanation

(c) Surface area, $A = 4\ \pi\ r^{2}$ or $r = (A/4\pi)^{1/2}$
Volume $V = \frac{4}{3}\ \pi r^{3}$
Or $\ = \frac{4}{3}\pi(A/4\pi)^{3/2} = kA^{3/2}$
Where $\frac{4\pi}{3} \times \frac{1}{(4\pi)^{3/2}} = k =$ constant
Using Boyle’s law, we have
$p_{1}V_{1} = p_{2}\ V_{2}$
Or $p_{2} = \frac{p_{1}V_{1}}{V_{2}} = \frac{(10 + h)kA_{1}^{3/2}}{kA_{2}^{3/2}}$
Or $p_{2} = (10 + h)\left( \frac{A_{1}}{A_{2}} \right)^{3/2}$
$= (10 + h)\left( \frac{1}{4} \right)^{3/2} = \frac{10 + h}{8}$
As $p_{2} = 10$ m of water, so
$10 = \frac{10 + h}{8}$ or $80 = 10 + h$
Or $h = 70\ $m