The volume of spherical balloon is increasing at the rate of $4\, cm^3/sec$. The rate of increase of its surface area, when the radius is 3 cm will be :-

$2.66\, cm^2/sec$

The correct answer is Option (2) → $2.66\, cm^2/sec$

Volume of sphere: $V = \frac{4}{3}\pi r^3$

Surface area of sphere: $S = 4\pi r^2$

Given: $\frac{dV}{dt} = 4 \, cm^3/sec$

Differentiate $V = \frac{4}{3}\pi r^3$ w.r.t. $t$:

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

Substitute values: $4 = 4\pi (3^2)\frac{dr}{dt}$

$4 = 36\pi \frac{dr}{dt}$

$\frac{dr}{dt} = \frac{1}{9\pi}$

Now, differentiate $S = 4\pi r^2$:

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

At $r=3$: $\frac{dS}{dt} = 8\pi (3)\cdot \frac{1}{9\pi}$

$\frac{dS}{dt} = \frac{24}{9} = \frac{8}{3} \, cm^2/sec$

Final Answer: $\frac{8}{3}\, cm^2/sec$