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 $V =\frac{4}{3}\pi r^{3}\quad...\left(1\right)$
$\frac{dv}{dt} = \frac{4}{3}. 3\pi r^{2}. \frac{dr}{dt}$
$4\pi = 4\pi r^{2}. \frac{dr}{dt}$
$\frac{1}{r^{2}} = \frac{dr}{dt}$
Since, $V = 288\pi$, therefore from $\left(1\right)$, we have
$288\pi = \frac{4}{3}\pi\left(r^{3}\right)\Rightarrow \frac{288\times3}{4} = r^{3}$
$\Rightarrow 216 = r^{3}$
$\Rightarrow r = 6$
Hence, $\frac{dr}{dt} = \frac{1}{36}.$