1. Set up the radius of the ice-coated ball

Let the radius of the iron ball be $10$ cm. If the ice thickness at time $t$ is $x$ cm, then the outer radius is $R=10+x.$

When the thickness is $5$ cm, $R=10+5=15\text{ cm.}$

2. Volume of the ice layer

The volume of ice is the difference of volumes of two spheres: $V=\frac{4}{3}\pi R^3-\frac{4}{3}\pi (10)^3.$

Since the iron ball radius is constant, differentiating with respect to time gives $\frac{dV}{dt}=4\pi R^2\frac{dR}{dt}.$

But $R=10+x$, so $\frac{dR}{dt}=\frac{dx}{dt}.$

Thus, $\frac{dV}{dt}=4\pi R^2\frac{dx}{dt}.$

3. Use the given melting rate

The ice melts at a rate of $50\text{ cm}^3/\text{min}$, so the ice volume is decreasing: $\frac{dV}{dt}=-50.$

At $R=15$ cm, $-50=4\pi(15)^2\frac{dx}{dt}.$

So, $-50=4\pi(225)\frac{dx}{dt}$ $-50=900\pi\frac{dx}{dt}.$

Therefore, $\frac{dx}{dt}=-\frac{50}{900\pi}=-\frac{1}{18\pi}.$

4. Interpret the result

The thickness is decreasing, so the rate at which it decreases is the positive magnitude: $\left|\frac{dx}{dt}\right|=\frac{1}{18\pi}\text{ cm/min}.$

5. Check options

• A: $\dfrac{1}{18\pi}$ ✅
• B: $\dfrac{1}{36\pi}$ ❌
• C: $\dfrac{1}{54\pi}$ ❌
• D: $\dfrac{5}{6\pi}$ ❌

Hence, the correct option is A.