1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# 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 the radius is 15 cm.

Open in App
Solution

## $\text{Let}\mathit{\text{r}}\text{be the radius and}\mathit{\text{V}}\text{be the volume of the spherical balloon at any time}\mathit{\text{t}}\text{. Then,}\phantom{\rule{0ex}{0ex}}\text{V=}\frac{4}{3}\text{π}{\mathit{\text{r}}}^{3}\phantom{\rule{0ex}{0ex}}\text{⇒}\frac{dV}{dt}{\text{=4πr}}^{2}\frac{dr}{dt}\phantom{\rule{0ex}{0ex}}\text{⇒}\frac{dr}{dt}\text{=}\left(\frac{1}{4\mathrm{\pi }{r}^{2}}\right)\frac{dV}{dt}\phantom{\rule{0ex}{0ex}}\text{⇒}\frac{dr}{dt}\text{=}\frac{900}{4\mathrm{\pi }{\left(15\right)}^{2}}\text{}\left[\because r=15\mathrm{cm}\mathrm{and}\frac{dV}{dt}=900{\mathrm{cm}}^{3}/\mathrm{sec}\right]\phantom{\rule{0ex}{0ex}}\text{⇒}\frac{dr}{dt}\text{=}\frac{900}{900\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}\text{⇒}\frac{dr}{dt}\text{=}\frac{1}{\mathrm{\pi }}\text{cm/sec}$

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Formation of Differential Equation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program