# 8. A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Given =  $\frac{dV}{dt} = 900 \ cm^{3}/s$
To find =   $\frac{dr}{dt}$   at r = 15 cm
Solution:-

Volume of sphere(V) =  $\frac{4}{3}\pi r^{3}$
$\frac{dV}{dt} = \frac{dV}{dr}.\frac{dr}{dt} = \frac{d(\frac{4}{3}\pi r^{3})} {dr}.\frac{dr}{dt} = \frac{4}{3}\pi\times 3r^{2} \times \frac{dr}{dt}$

$\frac{dV}{dt}= 4 \pi r^{2} \times \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^{2}} = \frac{900}{4\pi \times(15)^{2}} = \frac{900}{900\pi} = \frac{1}{\pi} \ cm/s$
Hence,  the rate at which the radius of the balloon increases when the radius is 15 cm is  $\frac{1}{\pi} \ cm/s$

## Related Chapters

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-