Question

A spherical iron ball $10 \, \text{cm}$ in radius is coated with a layer of ice of uniform thickness that melts at a rate of $50 \, \text{cm}^3/\text{min}$. When the thickness of ice is $15 \, \text{cm}$, then the rate at which the thickness of ice decreases is:

• A. $\frac{5}{6\pi} \, \text{cm/min}$
• B. $\frac{1}{54\pi} \, \text{cm/min}$
• C. $\frac{1}{18\pi} \, \text{cm/min}$
• D. $\frac{1}{36\pi} \, \text{cm/min}$

Hint:

For problems involving spherical volumes, note that the rate of change of volume depends on the square of the radius, especially when the shape is uniformly shrinking or melting.

Answer 1

The Correct Option is C

$$\text{Given: } \frac{dV}{dt} = 50 \, \text{cm}^3/\text{min},$$
$$\frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 50,$$
$$3r^2 \frac{dr}{dt} = \frac{150}{4\pi} \quad \Rightarrow \quad \frac{dr}{dt} = \frac{50}{4\pi r^2}.$$
\text{Substitute } r = 15: $$\left( \frac{dr}{dt} \right)_{r = 15} = \frac{50}{4\pi \times 225} = \frac{1}{18\pi} \, \text{cm/min}.$$ Final Answer: $$\boxed{\frac{1}{18\pi} \, \text{cm/min}}$$