Question

A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm³/min. When the thickness of ice is 15 cm, then the rate at which the thickness of ice decreases, is –

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

Answer 1

Answer: (C)

$\frac{1}{18\pi}$ cm/min

Answer 2

Given that, $\frac{dV}{dt}$ = 50 cm³/min

$\frac{d}{dt}$ $\left( \frac{4}{3}\pi r^{3} \right)$ = 50

̃ 3r² $\frac{dr}{dt}$ = $\frac{150}{4\pi}$ ̃ $\frac{dr}{dt}$ = $\frac{50}{4\pi r^{2}}$

̃ $\left( \frac{dr}{dt} \right)_{r = 15}$ = $\frac{50}{4\pi \times 225}$ = $\frac{1}{18\pi}$ cm/min.