Q. Gas is being pumped into a spherical ballon at the rate of 30${ft}^{3}/min.$Then,the rate at which the radius increases when it reaches the value 15 ft, is
A) $\frac{1}{30\pi}$ft/min
B) $\frac{1}{15\pi}$ft/min
C) $\frac{1}{20}$ft/min
D) $\frac{1}{15}$ft/min
Explanation
(a)
Given that, $\frac{dv}{dt} = 30\ ft^{3}/$ min and $r = 15$ ft
Volume of spherical balloon
$V = \frac{4}{3}\pi r^{2}\ \ \Rightarrow \ \ \frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt}$
$\Rightarrow \ \ \ 30 = 4\pi r^{2}\frac{dr}{dt}$
$\Rightarrow \ \frac{dr}{dt} = \frac{30}{4 \times \pi \times 15 \times 15} = \frac{1}{30\pi}$ ft/min
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.