The surface of a spherical balloon is increasing at the rate of 2 cm^…

The correct answer is Option (1) → $6\,cm^3/s$

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

$S = 4πг^2$   ...(i)

and $V =\frac{4}{3}πr^3$   ...(ii)

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

$\frac{dS}{dt}=4π.2r\frac{dr}{dt}=8πr\frac{dr}{dt}$ but $\frac{dS}{dt}=2\,cm^2/sec$ (given)

$⇒ 2 = 8πr\frac{dr}{dt}⇒\frac{dr}{dt}=\frac{1}{4πr}$   …(iii)

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

$\frac{dV}{dt}=\frac{4}{3}π. 3r^2\frac{dr}{dt} = 4πr^2 ×\frac{1}{4πr}$  (using (iii))

$⇒\frac{dV}{dt}=r$

When $r=6\,cm, \frac{dV}{dt}=6$

Hence, the volume is increasing at the rate of $6\, cm^3/sec$.