Question

If the volume of a sphere increases at the rate of $2\pi c{m}^3/s$, then the rate of increase of its radius (in $cm/s$ ), when the volume is $288\pi c{m}^3$, is

• A. $\frac{1}{36}$
• B. $\frac{1}{72}$
• C. $\frac{1}{18}$
• D. $\frac{1}{9}$

Answer 1

Answer: (B)

Given, $\frac{dV}{dt}=2\pi c{m}^3/s$
$\because$ Volume of sphere, $V=\frac{4}{3}\pi r^3$
On differentiating w.r.t. $t$, we get
$\frac{dV}{dt}=\frac{4}{3}\pi \times 3r^2\frac{dr}{dt}$
$\Rightarrow 2\pi =4\pi r^2\frac{dr}{dt}$
$\Rightarrow \frac{dr}{dt}=\frac{1}{2r^2}=\frac{1}{2\times 6^2}=\frac{1}{72}cm/s$
$\left[\because V=288\pi =\frac{4}{3}\pi r^3\Rightarrow 216=r^3\Rightarrow r=6\right]$