A spherical balloon is filled with $4500 \pi$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $72 \pi$ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is

JEE Main · 2012 · 29 Apr 2012 · Offline · Application of Derivatives

A spherical balloon is filled with

4500 π

cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of

72 π

cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is

Answer: (C)

After

49

minutes the volume of gas left is

V = 4500 π - 49 (72 π) = 972 π

cubic metres. From

V = \frac{4}{3} π r^{3}

this gives

r^{3} = 729

, so

r = 9

m at that instant. Differentiating

V = \frac{4}{3} π r^{3}

gives

\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}

, and the gas is leaving at

\frac{d V}{d t} = - 72 π

, so

\frac{d r}{d t} = \frac{- 72 π}{4 π ( 81 )} = - \frac{2}{9}

. The radius is shrinking at a rate of

\frac{2}{9}

metres per minute.

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option