Q. 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 $cm^{2}/\min.$ 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{5}{54\pi}\$cm/min
C) $\frac{5}{18\pi}\$cm/min
D) $\frac{1}{36\pi}\$cm/min
Explanation
(c)
Given , $\frac{d}{dt}\left( \frac{4}{3}\pi r^{3} \right) = - 50$
$\Rightarrow \ \frac{dr}{dt} = - \frac{50}{4\pi r^{2}}$
$\Rightarrow \ \left( \frac{dr}{dt} \right)_{r = 15} = - \frac{50}{4\pi \times 225} = - \frac{1}{18\pi}cm/\min$
Hence , the thickness of ice decrease by
1/18$\pi\$cm/min
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.