The volume of a sphere is increasing at the rate of $3$ cubic centimetre per second. Find the rate of increasing of its surface area, when the radius is $2cm$.

We know that,

Volume of sphere$=\frac{4}{3}\pi r^3$ (If radius r)

$i.e.V=\frac{4}{3}\pi r^3$

Differentiating equation with respect to $t$ and we get,

$\frac{dV}{dt}=\frac{4}{3}\pi .3r^2\frac{dr}{dt}$

$\Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.......\left(1\right)$

Given that, $\frac{dV}{dt}=3c.m{.}^3/\sec .$ and $r=2c.m.$

By equation $\left(1\right)$ to and we get,

$3=4\pi {\left(2\right)}^2\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{3}{16\pi }c.m./\sec .$

Now, let $S$ be the surface area

Then,

$S=4\pi r^2$

When

$\frac{dS}{dt}=8\pi r\frac{dr}{dt}$

$\frac{dS}{dt}=8\pi \left(2\right)\times \frac{3}{16\pi }$

$\frac{ds}{dt}=3c.m{.}^2/\sec$

Hence, this is the answer.