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# Gas is escaping from a spherical balloon at the rate of 900 cm3/sec. How fast is the surface area, radius of the balloon shrinking when the radius of the balloon is 30cm?

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’:
$S = 4\pi r^{2}$
$\frac{\partial S}{\partial t} = 8\pi r.\frac{\partial r}{\partial t} = \frac{2}{r}.4\pi r^{2}.\frac{\partial r}{\partial t}$
$\frac{\partial S}{\partial t} = \frac{2}{30}.4\pi (r)^{2}.\frac{\partial r}{\partial t} = \frac{1}{15}.900 = 60\frac{cm^{3}}{sec}$
$8\pi r.\frac{\partial r}{\partial t} = 60$
$\frac{\partial r}{\partial t} = \frac{60}{8\pi .30} = \frac{1}{4\pi }\frac{cm}{sec}$
Thanks & Regards
Jitender Singh
IIT Delhi