Question

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

Answer 1

The correct answer is \(\frac{1}{π} cm/s.\)
The volume of a sphere \((V)\) with radius \((r)\) is given by,
\(v=\frac{4}{3}πr^3\)
∴Rate of change of volume \((V)\) with respect to time \((t)\) is given by
\(\frac{dv}{dt}=\frac{dv}{dr}.\frac{dr}{dt}\) [By chain rule]
\(=\frac{dv}{dt}(\frac{4}{3}πr^3).\frac{dr}{dt}\)
\(=4πr^2\frac{dr}{dt}\)
It is given that \(\frac{dv}{dt}=900cm^3/s\)
\(∴ 900=4πr^2.\frac{dr}{dt}\)
\(\frac{dr}{dt}=\frac{900}{4πr^2}=\frac{225}{r^2}\)
Therefore, when radius=15 cm
\(\frac{dr}{dt}=\frac{225}{\pi(15)^2}=\frac{1}{\pi}\)
Hence, the rate at which the radius of the balloon increases when the radius is \(15cm\) is  \(\frac{1}{π} cm/s.\)