Question

A spherical balloon is filled with 3536 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 70 π cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 50 minutes after the leakage began is
1. 1/9
2. 1/2
3. 2/9
4. 35/18

Answer 1

Correct Answer - Option 4 : 35/18

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)

\( = - 70\pi \times \frac{1}{{4\pi \times 9}} = - \frac{{70}}{{36}}=-\frac{35}{18}\)

(-ve sign indicates rate of decreases)