A balloon which always remains spherical is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon is increasing when its radius is 15 cm.

$\frac{1}{\pi}$

The correct answer is Option (1) → $\frac{1}{\pi}$

Let r be the radius of the spherical balloon and V be its volume at any time t, then

$V =\frac{4}{3}πr^3$   …(i)

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

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

But $\frac{dV}{dt}=900\, cm^3/sec$  (given)

∴ From (ii), $900=4\pi r^2\frac{dr}{dt}⇒\frac{dr}{dt}=\frac{225}{πr^2}$

When $r=15cm,\frac{dr}{dt}=\frac{225}{\pi ×15^2}=\frac{1}{\pi}$

Hence, the radius of the balloon is increasing at the rate of $\frac{1}{\pi}\,cm/sec$ when its radius is 15 cm.