In a sphere the rate of change of volume with respect to (w.r.t) time is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Surface area times the rate of change of radius w.r.t. time
For a sphere of radius r, \( V=\frac{4}{3}\pi r^{3} \). Differentiating with respect to t by the chain rule:
\( \displaystyle \frac{dV}{dt}=\frac{4}{3}\pi\cdot 3r^{2}\frac{dr}{dt} \)
\( \displaystyle =4\pi r^{2}\,\frac{dr}{dt} \)
Since the surface area of the sphere is \( S=4\pi r^{2} \),
\( \displaystyle \frac{dV}{dt}=S\cdot\frac{dr}{dt} \)
Why the other options are incorrect
Option (1) \( \pi \) times the rate of change of radius w.r.t. time. — Incorrect. The factor is \( 4\pi r^{2} \), not \( \pi \).
Option (3) Surface area times the rate of change of diameter w.r.t. time — Incorrect. The diameter D = 2r changes twice as fast, \( \frac{dD}{dt}=2\frac{dr}{dt} \), so \( S\cdot\frac{dD}{dt}=2\frac{dV}{dt} \).
Option (4) \( \displaystyle\frac{1}{3}\pi \) times the rate of change of radius w.r.t. time — Incorrect. The \( \frac{1}{3} \) cancels with the 3 from differentiating \( r^{3} \); the factor is \( 4\pi r^{2} \).