Question

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

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

Answer 1

Answer: (B)

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