A spherical iron ball of radius 10   c m is coated with a layer of ice of uniform thickness that melts…

We know that the volume of a sphere of radius $r$ is $\frac{4}{3}{\mathrm{\pi r}}^3.$

Given the radius of the spherical iron ball is $R=10 cm$ and the thickness of the ice is $x cm.$

Volume of ice: $V=\frac{4}{3}\pi {\left(R+x\right)}^3-\frac{4}{3}\pi R^3$

$\Rightarrow V=\frac{4}{3}\pi {\left(10+x\right)}^3-\frac{4}{3}\pi {\left(10\right)}^3$

Differentiating with respect to $t,$ we get

$\frac{dV}{dt}=4\pi {\left(10+x\right)}^2\frac{dx}{dt}-0$

Given $\frac{dV}{dt}=50 c{m}^3/min$

$\Rightarrow 50=4\pi {\left(10+x\right)}^2\frac{dx}{dt}$

$\Rightarrow \frac{dx}{dt}=\frac{50}{4\pi {\left(10+x\right)}^2}$

$\Rightarrow {\left.\frac{dx}{dt}\right|}_{x=5}=\frac{50}{4\pi {\left(15\right)}^2}=\frac{1}{18\pi } cm/min.$