Q. The rate of change of the surface area of the sphere of radius r when the radius is increasing at the rate of 2 cm/s is proportional to
A) $\frac{1}{r^{2}}$
B) $\frac{1}{r}$
C) $r^{2}$
D) $r$
Explanation
(c)
Let surface area of sphere,
$S = \frac{4}{3}\pi r^{3}\ \ \ \ \ \ \ \ \Rightarrow \frac{dS}{dt} = 4\pi r^{2}\frac{dr}{dt}$
$\Rightarrow \ \frac{dS}{dt} = 4\pi r^{2}(2) = 8\pi r^{2}\ \ \ \ \ \therefore\frac{dS}{dt} \propto r^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.