The volume of a spherical balloon is increasing at the rate of $2{ \mathrm{cm}}^3/\sec .$ When its radius is $4 \mathrm{cm},$ the rate of change of its surface area (in $\mathrm{cm}{}^2/\sec$) is

Let $V$ be the volume & $S$ be the surface area of spherical balloon having radius $r$.

Given:$\frac{dV}{dt}=2{\mathrm{cm}}^3/\sec$

To find:${\left(\frac{dS}{dt}\right)}_{r=4}$

$\text{As, volume} V=\frac{4}{3}\pi r^3$

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

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

$\therefore \frac{dr}{dt}=\frac{1}{2{\mathrm{\pi r}}^2}$

Now,

$S=4{\mathrm{\pi r}}^2$

$\Rightarrow \frac{dS}{dt}=4\pi \times 2r\frac{dr}{dt}$

$\therefore {\left(\frac{dS}{dt}\right)}_{r=4}=4\mathrm{\pi }\times 2\times 4\times \frac{1}{2\mathrm{\pi }{\left(4\right)}^2}=1{\mathrm{cm}}^2/\mathrm{s}$

Hence, option $1$ is correct.