Q. If the volume of a sphere is increasing at a constant rate, then the rate at which its radius is increasing is
A) A constant
B) proportional to the radius
C) Inversely proportional to the radius
D) Inversely proportional to the surface area
Explanation
(d)
Given , $\ \ \ \frac{dV}{dt} = k$ [say]…(i)
$\because\ \ \ V = \frac{4}{3}\pi r^{3}$
$\Rightarrow \ \ \frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt}$
$\ \therefore\ \ \ \ \ \ \frac{dr}{dt} = \frac{k}{4\pi r^{2}}$ [from Eq.(i)]
$\Rightarrow \ \ \$Rate of increasing radius is inversely
proportional to its surface area
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.