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: \[ \boxed{\frac{1}{18\pi} \, \text{cm/min}} \]