The volume of a spherical balloon is increasing at the rate of 3 cm^3…

The correct answer is Option (2) → $3\, cm^2/sec$

Let r be the radius of the spherical balloon at any time t, V be its volume and S its surface area at that instant, then

$V =\frac{4}{3}πr^3$   …(i)

and $S = 4πг^2$   …(ii)

Diff. (i) w.r.t. t, we get

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

But $\frac{dV}{dt}=3\, cm^3/sec$  (given)

$∴3=4πr^2\frac{dr}{dt}⇒\frac{dr}{dt}=\frac{3}{4πr^2}$   ...(iii)

Diff. (ii) w.r.t. t, we get

$\frac{dS}{dt}=4π.2r\frac{dr}{dt}= 8πг.\frac{3}{4 πr^2}$  (Using (iii))

$⇒\frac{dS}{dt}=\frac{6}{r}$.

When $r=2\, cm,\frac{dS}{dt}=\frac{6}{2}=3$

Hence, the surface area is increasing at the rate of $3\, cm^2/sec$ when its radius is 2 cm.