In a sphere, the rate of change of volume is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Surface area times of rate of change of radius.
\( \displaystyle V=\frac{4}{3}\pi r^3 \)
\( \displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt} \)
Since \( 4\pi r^2 \) is the surface area, the rate of change of volume is the surface area times the rate of change of the radius.
Why the other options are incorrect
Option (1) \( \pi \) times the rate of change of radius. — Incorrect. The factor is \( 4\pi r^2 \), not \( \pi \).
Option (2) \( \pi \) times the rate of change of diameter. — Incorrect. With d = 2r, \( \frac{dV}{dt}=2\pi r^2\frac{dd}{dt} \); the factor is \( 2\pi r^2 \), not \( \pi \).
Option (3) Surface area times the rate of change of diameter. — Incorrect. Since d = 2r, \( \frac{dd}{dt}=2\frac{dr}{dt} \), so this would be twice the correct value.