Question

A spherical balloon expands at a rate proportional to its surface area. Initially its radius is 2 cm and 5 minutes later it increases upto 7 cm, then surface area of spherical balloon after 12 minutes will be

• A. \(2480\) sq.cm.
• B. \(2460\) sq.cm.
• C. \(2464\) sq.cm.
• D. \(2400\) sq.cm.

Hint:

The volume rate is proportional to surface area, which makes the radius grow at a constant rate.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a sphere, \(V = \dfrac{4}{3}\pi r^3\) and the surface area is \(S = 4\pi r^2\). The statement says the volume grows at a rate proportional to the surface area.

Step 2: Form the differential equation.
\[ \frac{dV}{dt} = kS \Rightarrow 4\pi r^2\frac{dr}{dt} = k\cdot 4\pi r^2 \Rightarrow \frac{dr}{dt} = k \]
So the radius increases at a constant rate: \(r = kt + C\).

Step 3: Use the data.
At \(t = 0\), \(r = 2\), so \(C = 2\). At \(t = 5\), \(r = 7\), so \(5k = 5\) and \(k = 1\) cm per minute.

Step 4: Find the radius at 12 min.
\(r = 2 + 12 = 14\) cm.

Step 5: Surface area.
\[ S = 4\pi r^2 = 4\times\frac{22}{7}\times 196 = 4\times 22\times 28 = 2464\text{ cm}^2 \]

Final Answer:
The surface area after 12 minutes is 2464 square cm, option (C). \[ \boxed{2464\text{ sq. cm}} \]