Question

If the volume of a spherical ball is increasing at the rate of $4\pi$ cc/sec, then the rate of increase of its radius (in cm/ sec), when the volume is $288\pi$ cc, is :

• A. $\frac{1}{6}$
• B. $\frac{1}{9}$
• C. $\frac{1}{36}$
• D. $\frac{1}{24}$

Answer 1

Answer: (C)

Volume of sphere $V=\frac{4}{3}\pi r^3...\left(1\right)$
$\frac{dv}{dt}=\frac{4}{3}.3\pi r^2.\frac{dr}{dt}$
$4\pi =4\pi r^2.\frac{dr}{dt}$
$\frac{1}{r^2}=\frac{dr}{dt}$
Since, $V=288\pi$, therefore from $\left(1\right)$, we have
$288\pi =\frac{4}{3}\pi \left(r^3\right)\Rightarrow \frac{288\times 3}{4}=r^3$
$\Rightarrow 216=r^3$
$\Rightarrow r=6$
Hence, $\frac{dr}{dt}=\frac{1}{36}.$