Concept:A spherical balloon is filled with 3536 ? cubic meters of helium gas.The leakage rate of Helium gas from a balloon is \(\frac{{dV}}{{dt}} = - 70? ~\frac{{{m^3}}}{{min}}\)The volume of spherical balloon is, \(V = \frac{4}{3}? {r^3}\)The radius of the balloon decreases after the leakage after "t" time period is, \(\frac{{dV}}{{dt}} = 4? {r^2}\frac{{dr}}{{dt}}\).............(1)Calculation:Given:Vo = 3536? m3, t = 50 minThe rate of leakage from balloon is, \(\frac{{dV}}{{dt}} = - 70? \)After integration, we will get? V = -70?t + C At t = 0, ? V = Vo ? C = Vo? V = Vo -70?t Volume of balloon of after 50 minutes:V = 3536? - (70? × 50) V = 36? m3But the volume of balloon is, \(V = \frac{4}{3}? {r^3}\)Therefore, \( \frac{4}{3}? {r^3}=36\pi\)r = 3 m ..............(at 50 min)The radius of balloon decreases after the leakage after "t" time period is, \(\frac{{dr}}{{dt}} = \left[ {\frac{{dV}}{{dt}}} \right]\frac{1}{{4\pi {r^2}}}\)................(from eq.1)\( = - 70\pi \times \frac{1}{{4\pi \times 9}} = - \frac{{70}}{{36}}=-\frac{35}{18}\)(-ve sign indicates rate of decreases)