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 :

Solution :
$$\eqalign{ & {\text{Volume}}\,{\text{of}}\,{\text{spherical}}\,{\text{balloon}}\, = V = \frac{4}{3}\pi {r^3} \cr & \Rightarrow 4500\pi = \frac{{4\pi {r^3}}}{3}{\kern 1pt} \,\left( \because \right.\,{\text{Given,volume}}\,\left. { = 4500\pi {m^3}} \right) \cr & {\text{Differentiating both the sides, w}}{\text{.r}}{\text{.t }}'t'{\text{ we get,}} \cr & \frac{{dV}}{{dt}} = 4\pi {r^2}\left( {\frac{{dr}}{{dt}}} \right) \cr & {\text{Now,}}\,{\text{it}}\,{\text{is}}\,{\text{given}}\,{\text{that}}\,\frac{{dV}}{{dt}} = 72\pi \cr & \therefore {\text{After }}49{\text{ min, Volume}} \cr & \, = \left( {4500 - 49 \times 72} \right)\pi \cr & = \left( {4500 - 3528} \right)\pi \cr & = 972\pi {m^3} \cr & \Rightarrow V = 972\pi {m^3}\,\,\,\therefore 972\pi = \frac{4}{3}\pi {r^3} \cr & \Rightarrow {r^3} = 3 \times 243 \cr & \Rightarrow {r^3} = 3 \times {3^5} \cr & \Rightarrow {r^3} = {3^6} \cr & \Rightarrow {r^3} = {\left( {{3^2}} \right)^3} \cr & \Rightarrow r = 9 \cr & {\text{Also,we}}\,{\text{have}}\,\frac{{dV}}{{dt}} = 72\pi \cr & \therefore 72\pi = 4\pi \times 9 \times 9\left( {\frac{{dr}}{{dt}}} \right) \Rightarrow \frac{{dr}}{{dt}} = \left( {\frac{2}{9}} \right) \cr} $$