Radius of a spherical balloon is increasing at the rate of 0.5 cm/sec; then the rate of increase of its volume when radius is 10 cm is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(200\pi\ \text{cm}^3/\text{sec}\)
Volume of a sphere, differentiated with respect to time (rate of change):
\( \displaystyle V = \frac{4}{3}\pi r^3 \)
\( \displaystyle \frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt} \)
With r = 10 cm and dr/dt = 0.5 cm/sec:
\( \displaystyle \frac{dV}{dt} = 4\pi(100)(0.5) = 200\pi\ \text{cm}^3/\text{sec} \)