A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness that melts at a rate of $$50 \; cm^3/min$$. When the thickness of ice is 5 cm, then the rate (in cm/min) at which the thickness of ice decreases, is:

Let us denote by $$r = 10\;{\rm cm}$$ the fixed radius of the iron ball and by $$y(t)$$ the thickness of the surrounding ice at time $$t$$ minutes. Therefore, the outer radius of the ice sphere is $$R(t)=r+y(t)=10+y(t)$$.

The volume of just the ice layer is obtained by subtracting the iron volume from the total volume of the ice-covered sphere. Using the standard volume formula for a sphere, $$V=\dfrac{4}{3}\pi(\text{radius})^{3}$$, we have

$$ V(t)=\frac{4}{3}\pi\Bigl[R(t)^{3}-r^{3}\Bigr] =\frac{4}{3}\pi\Bigl[(10+y)^{3}-10^{3}\Bigr]. $$

Given data state that the ice is melting at a rate of $$50\;{\rm cm^{3}/min}$$, so the volume is decreasing. Hence $$\dfrac{dV}{dt}=-50\;{\rm cm^{3}/\min}$$ (negative sign because the volume is going down).

To relate this to the change in thickness, we differentiate the expression for $$V(t)$$ with respect to time. First rewrite the inner function: $$R(t)=10+y$$, so $$\dfrac{dR}{dt}=\dfrac{dy}{dt}$$ because the iron ball itself does not change size.

Differentiate: using the chain rule and the fact that $$\dfrac{d}{dt}(R^{3})=3R^{2}\dfrac{dR}{dt}$$, we get

$$ \frac{dV}{dt}= \frac{4}{3}\pi\bigl[3R^{2}\tfrac{dR}{dt}\bigr] =4\pi R^{2}\frac{dR}{dt}. $$

Since $$\dfrac{dR}{dt}=\dfrac{dy}{dt}$$, we can write

$$ \frac{dV}{dt}=4\pi R^{2}\frac{dy}{dt}. $$

We are interested in the instant when the ice thickness is $$y=5\;{\rm cm}$$, so the outer radius is

$$ R=10+5=15\;{\rm cm}. $$

Substituting $$\dfrac{dV}{dt}=-50$$ and $$R=15$$ into the differentiated formula:

$$ -50 = 4\pi (15)^{2}\frac{dy}{dt}. $$

Simplify the numerical factor:

$$ -50 = 4\pi \times 225 \times \frac{dy}{dt} = 900\pi \,\frac{dy}{dt}. $$

Now isolate $$\dfrac{dy}{dt}$$:

$$ \frac{dy}{dt}=\frac{-50}{900\pi} =\frac{-1}{18\pi}\;{\rm cm/min}. $$

The negative sign confirms that the thickness is decreasing. The rate at which the thickness decreases, asked as a positive quantity, is

$$ \bigl|\tfrac{dy}{dt}\bigr|=\frac{1}{18\pi}\;{\rm cm/min}. $$

Hence, the correct answer is Option D.