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

\[\text{Rate of change of volume is } \frac{dV}{dt} = 50 \, \text{cm}^3/\text{min}.\]
\[\text{Differentiating the volume formula with respect to time: } \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 50,\]
\[\text{which simplifies to } 3r^2 \frac{dr}{dt} = \frac{150}{4\pi}, \text{ leading to } \frac{dr}{dt} = \frac{50}{4\pi r^2}.\]
\text{When the radius is } r = 15 \text{ cm}:\[\left( \frac{dr}{dt} \right)_{r = 15} = \frac{50}{4\pi \times 225} = \frac{1}{18\pi} \, \text{cm/min}.\] Final Answer:\[\boxed{\frac{1}{18\pi} \, \text{cm/min}}\]