Q. The rate of change of the surface area of a sphere of a sphere of radius$r,$when the radius is increasing at the rate of$\ 2\ cm/s$is proportional to
A) $\frac{1}{r}$
B) $\frac{1}{r^{2}}$
C) $r$
D) $r^{2}$
Explanation
(c)
Surface area of sphere $S = 4\pi r^{2}$ and $\frac{dr}{dt} = 2$
$\therefore\ \ \frac{dS}{dt} = 4\pi \times 2r\frac{dr}{dt} = 8\pi r \times 2 = 16\pi r$
$\Rightarrow \ \frac{dS}{dt} \propto r$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.