Question

A sphere increases its volume at the rate of $\pi cc/s$. The rate at which its surface area increases when the radius is $1cm$ is

• A. $2\pi sq.cm/s$
• B. $\pi sq.cm/s$
• C. $\frac{3\pi }{2}sq.cm/s$
• D. $\frac{\pi }{2}sq.cm/s$

Answer 1

Answer: (A)

Given rate of increase volume of sphere is
$\frac{dV}{dt}=\pi$...(i)
We know that,
Volume of sphere; $V=\frac{4}{3}\pi r^3$
$\Rightarrow \frac{dV}{dt}=\frac{4}{3}\pi \cdot 3r^2\cdot \frac{dr}{dt}=4\pi r^2\frac{dr}{dt}$
$\Rightarrow \pi =4\pi r^2\frac{dr}{dt};$ [from Eq. (i)]
$\Rightarrow \frac{dr}{dt}=\frac{1}{4r^2}$...(ii)
Also, we know that, surface of sphere,
$S=4\pi r^2$
$\Rightarrow \frac{dS}{dt}=8\pi r\frac{dr}{dt}=8\pi r\cdot \frac{1}{4r^2}$ [from Eq.(ii)]
$\frac{dS}{dt}=\frac{2\pi }{r}[\because$ given $r=1]$
$\frac{dS}{dt}=2\pi$
So, rate of increase in surface $\equiv 2\pi sqcm/s$