Concept:

A spherical balloon is filled with 3536 π cubic meters of helium gas.

The leakage rate of Helium gas from a balloon is \(\frac{{dV}}{{dt}} = - 70π ~\frac{{{m^3}}}{{min}}\)

The volume of spherical balloon is, \(V = \frac{4}{3}π {r^3}\)

The radius of the balloon decreases after the leakage after "t" time period is, \(\frac{{dV}}{{dt}} = 4π {r^2}\frac{{dr}}{{dt}}\).............(1)

Calculation:

Given:

V_o = 3536π m³, t = 50 min

The rate of leakage from balloon is, \(\frac{{dV}}{{dt}} = - 70π \)

After integration, we will get

∴ V = -70πt + C 

At t = 0, ⇒ V = V_o ∴ C = V_o

⇒ V = V_o -70πt 

Volume of balloon of after 50 minutes:

V = 3536π - (70π × 50) 

V = 36π m³

But the volume of balloon is, \(V = \frac{4}{3}π {r^3}\)

Therefore, \( \frac{4}{3}π {r^3}=36\pi\)

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)