Question

A football is inflated by pumping air in it. When it acquires spherical shape its radius increases at the rate of \( 0.02 \, \text{cm/s} \). The rate of increase of its volume when the radius is \( 10 \, \text{cm} \) is:

• A. \( \pi \, \text{cm}^3/\text{s} \)
• B. \( 4 \pi \, \text{cm}^3/\text{s} \)
• C. \( 6 \pi \, \text{cm}^3/\text{s} \)
• D. \( 8 \pi \, \text{cm}^3/\text{s} \)

Hint:

The rate of change of volume for a sphere can be found by differentiating its volume formula with respect to time.

Answer 1

The Correct Option is C

Step 1: The volume of a sphere is given by: \[ V = \frac{4}{3} \pi r^3. \] Step 2: Differentiating with respect to time, we get: \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}. \] Step 3: Given \( \frac{dr}{dt} = 0.02 \, \text{cm/s} \) and \( r = 10 \, \text{cm} \), substitute to find the rate of change of volume: \[ \frac{dV}{dt} = 4 \pi (10)^2 (0.02) = 6 \pi \, \text{cm}^3/\text{s}. \]
Final Answer: \[ \boxed{6 \pi \, \text{cm}^3/\text{s}} \]