Q. A spherical iron ball of radius 10 cm, coated with a layer of ice of uniform thickness, melts at a rate of 100$\pi\ {cm}^{3}/min$.The rate at which the thickness of decreases when the thickness of ice is 5 cm, is
A) 1cm/min
B) 2 cm/min
C) $\frac{1}{376}cm/min$
D) 5 cm/min
Explanation
(a)
Given, $\frac{dV}{dt} = 100\pi cm^{3}/min$
$\therefore\ \frac{d}{dt}\left( \frac{4}{3}\pi r^{3} \right) = 100\pi\ \Rightarrow 3r^{2}\frac{dr}{dt} = \frac{300\pi}{4\pi}$
$\Rightarrow \left( \frac{dr}{dt} \right)_{r = 5} = \frac{300}{4 \times 3 \times 25} = 1cm/\min$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.