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# . 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.

Jitender Singh IIT Delhi
6 years ago
Ans:
Let radius of balloon to be ‘r’
Volume ‘V’:
$V = \frac{4}{3}\pi r^{3}$
$\frac{\partial V}{\partial t} = 4\pi r^{2}.\frac{\partial r}{\partial t} = 900 \frac{cm^{3}}{sec}$
Surface Area ‘S’:
$S = 4\pi r^{2}$
$\frac{\partial S}{\partial t} = 8\pi r.\frac{\partial r}{\partial t} = \frac{2}{r}.4\pi r^{2}.\frac{\partial r}{\partial t}$
$\frac{\partial S}{\partial t} = \frac{2}{30}.4\pi (r)^{2}.\frac{\partial r}{\partial t} = \frac{2}{15}.900 = 120\frac{cm^{2}}{sec}$
$8\pi r.\frac{\partial r}{\partial t} = 60$
$\frac{\partial r}{\partial t} = \frac{120}{8\pi .15} = \frac{1}{\pi }\frac{cm}{sec}$
Thanks & Regards
Jitender Singh
IIT Delhi