The radius of a sphere is increasing at the rate of 0.2 cm/sec.

(b) $180 \pi \mathrm{cm}^{3} / \mathrm{sec}$

Let $r$ be the radius and $V$ be the volume of the sphere at any time $t .$ Then,

$\mathrm{V}=\frac{4}{3} \pi r^{3}$

$\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t}$

$\Rightarrow \frac{d V}{d t}=4 \pi(15)^{2} \times 0.2$

$\Rightarrow \frac{d V}{d t}=180 \pi \mathrm{cm}^{3} / \mathrm{sec}$