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

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

It is given that, a spherical iron ball of $10 \mathrm{cm}$ radius is coated with a layer of ice of uniform thickness, let the thickness is '$x$' $\mathrm{cm}$, then volume of the ball is $V=\frac{4}{3}\pi {\left(10+x\right)}^3 c{m}^3.$

On differentiating w.r.t. ' $t$', we get

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

where $t$ is time in $\min$.

It is given, the $\frac{dV}{dt}=-50 {\mathrm{cm}}^3/\min ,$ when $x$ is $5 \mathrm{cm}$, then

$-50=4\pi {\left(10+5\right)}^2\frac{dx}{dt}$  [from Eq. $\left(\mathrm{i}\right)$]

$\Rightarrow \frac{dx}{dt}=-\frac{50}{4\pi \left(225\right)}=-\frac{1}{18\pi } \mathrm{cm}/\min .$

Negative sign indicates the thickness of ice layer decreases with time.

Thus, the thickness of ice layer decreases at a rate $\frac{1}{18\pi } \mathrm{cm}/\min .$

Hence, option $\left(D\right)$ is correct.