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 :
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Official solution
Volume of spherical balloon \[ = V = {4 \over 3}\pi {r^3}\]
\[ \Rightarrow 4500\pi = {{4\pi {r^3}} \over 3}\]
( as Given, volume \[ = 4500\pi {m^3}\] )
Differentiating both the sides, w.r.t′t′ we get,
\[{{dV} \over {dt}} = 4\pi {r^2}\left( {{{dr} \over {dt}}} \right)\]
Now, it is given that \[{{dV} \over {dt}} = 72\pi \]
\[\therefore\] After 49 min, Volume -
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 49 \times 72} \right)\pi \]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 3528} \right)\pi \]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 972\pi {m^3}\]
\[ \Rightarrow V = 972\,\,\pi {m^3}\]
\[\therefore\] \[972\pi = {4 \over 3}\pi r{}^3\]
\[ \Rightarrow {r^3} = 3 \times 243 = 3 \times {3^5} = {3^6} = {\left( {{3^2}} \right)^3} \Rightarrow r = 9\]
Also, we have \[{{dV} \over {dt}} = 72\pi \]
\[\therefore\] \[72\pi = 4\pi \times 9 \times 9\left( {{{dr} \over {dt}}} \right) \Rightarrow {{dr} \over {dt}} = \left( {{2 \over 9}} \right)\]