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

The correct option is C $\frac{1}{36}$

Given, volume $V=288\pi$ cc, $\frac{dV}{dt}=4\pi$ cc/sec

volume of sphere

$V=\frac{4}{3}\pi r^3$
$288\pi =\frac{4}{3}\pi r^3$
$\Rightarrow r=6$

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

$4\pi =4\pi 6^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{1}{36}$ cm/sec