Q. Gas is being pumped into a spherical balloon at the rate of 30$ft^{3}/\min$.Then, the rate at which the radius increases when it reaches the value 15ft is
A) $\frac{1}{15\pi}$ft/min
B) $\frac{1}{30\pi}$ft/min
C) $\frac{1}{20}$ft/min
D) $\frac{1}{25}$ft/min
Explanation
(b)
Let $V = \frac{4}{3}\pi r^{3}\ \Rightarrow \ \frac{dV}{dt}4\pi r^{2}\frac{dr}{dt}$
$\Rightarrow \frac{dr}{dt} = \frac{30}{4 \times \pi \times 15 \times 15} = \frac{1}{30\pi}ft/\min\left\lbrack \therefore\frac{dV}{dt} = 30,\ r = 15 \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.