Question

Gas is being pumped into a spherical balloon at the rate of 30 cm³/min. The rate at which the radius increases when it reaches the value 15 cm, is ___________.

Answer 1

Let r be the radius and V be the volume of the balloon at time t.

It is given that, $\frac{dV}{dt}$ = 30 cm³/min

Now,

Volume of the spherical balloon, V = $\frac{4}{3}\mathrm{\pi }{r}^3$

$V=\frac{4}{3}\mathrm{\pi }{r}^3$

Differentiating both sides with respect to t, we get

$\frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}\mathrm{\pi }{r}^3\right)$

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

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

When r = 15 cm and $\frac{dV}{dt}$ = 30 cm³/min, we get

$30=4\mathrm{\pi }\times {\left(15\right)}^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{30}{4\mathrm{\pi }\times {\left(15\right)}^2}$

$\Rightarrow \frac{dr}{dt}=\frac{1}{30\mathrm{\pi }}$ cm/min

Thus, the radius of the balloon is increasing at the rate of $\frac{1}{30\mathrm{\pi }}$ cm/min.


Gas is being pumped into a spherical balloon at the rate of 30 cm³/min. The rate at which the radius increases when it reaches the value 15 cm, is $\underline{\frac{1}{30\mathrm{\pi }}\mathrm{cm}/\min }$.