A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm³/min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is 

$\frac{1}{18 \pi}$ cm/min

Let at any timet, h cm be the thickness of ice. Then,

V = Volume of ice = $\frac{4}{3} \pi(10+h)^3-\frac{4}{3} \pi \times 10^3$

$\Rightarrow \frac{d V}{d t}=4 \pi(10+h)^2 \frac{d h}{d t}$

$\Rightarrow -50=4 \pi(10+5)^2 \times \frac{d h}{d t}$                $\left[\begin{array}{l}∵ \frac{d v}{d t}=-50 \mathrm{~cm}^3 / \mathrm{min} \\ \text { and } h=5\end{array}\right]$

$\Rightarrow \frac{d h}{d t}=-\frac{1}{18 \pi}$ cm/min