Question

The rate of change of volume of spherical balloon at any instant is directly proportional to its surface area. If initially its radius is 3 cm, after 2 minutes its radius becomes 9 cm, then radius of balloon after 4 minutes is

• A. 12 cm
• B. 14 cm
• C. 15 cm
• D. 18 cm

Hint:

If $dV/dt \propto S$ for a sphere, the radius increases at a constant rate ($dr/dt = \text{const}$).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This is an application of differential equations to a physical process (growth of a balloon).
Step 2: Key Formula or Approach:
1. Volume \( V = \frac{4}{3} \pi r^3 \), Surface Area \( S = 4\pi r^2 \).
2. Given \( \frac{dV}{dt} = kS \).
Step 3: Detailed Explanation:
Differentiate volume with respect to time: \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \).
Substitute into the given relation: \( 4\pi r^2 \frac{dr}{dt} = k(4\pi r^2) \implies \frac{dr}{dt} = k \).
Integrate: \( r = kt + C \).
At \( t = 0 \), \( r = 3 \implies 3 = k(0) + C \implies C = 3 \).
At \( t = 2 \), \( r = 9 \implies 9 = k(2) + 3 \implies 2k = 6 \implies k = 3 \).
The equation for radius is \( r = 3t + 3 \).
Now find \( r \) at \( t = 4 \):
\[ r = 3(4) + 3 = 12 + 3 = 15 \text{ cm} \] Step 4: Final Answer:
The radius after 4 minutes is 15 cm.