Q. A spherical balloon is expanding. If the radius is increasing at the rate of 2 cm/min, the rate at which the volume increase (in cubic centimeters per minute)when the radius is 5 cm, is
A) $10\pi$
B) $100\pi$
C) $200\pi$
D) $50\pi$
Explanation
(c)
Let volume of sphere $V = \frac{4}{3}\pi r^{3}$
$\Rightarrow \ \frac{dV}{dt} = 4\pi r^{2}\ \ \frac{dr}{dt} = 4\pi r^{2}.(2)\ \ \ \lbrack\because\frac{dr}{dt} = 2\rbrack$
$\therefore\ \frac{dV}{dt} = 8\pi(5)^{2} = 200\pi\ cm^{3}/\min{\ \lbrack\because r = 5cm\rbrack}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.