Air is pumped into a spherical balloon at the rate of \(72\pi\) cm3/s. The rate at which the radius of the balloon is increasing when the radius is 6 cm is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{2}\) cm/s
Let r be the radius and V the volume of the balloon at time t. Then
\( \displaystyle V = \frac{4}{3}\pi r^{3} \)
Differentiating with respect to t by the chain rule:
\( \displaystyle \frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt} \)
With \(\frac{dV}{dt} = 72\pi\) and r = 6:
\( 72\pi = 4\pi \times 36 \times \displaystyle\frac{dr}{dt} \)
\( \displaystyle \frac{dr}{dt} = \frac{72\pi}{144\pi} \)
\( \displaystyle = \frac{1}{2} \) cm/s