A ring is dropped into a quiet pond and waves move in circles at a speed of 2 cm/second. At the instant, when the radius of the circular wave is 12 cm, how fast is the enclosed area increasing?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 48\pi\ \mathrm{cm^{2}/sec} \)
The area enclosed by the circular wave is \( A=\pi r^{2} \). Differentiating with respect to time (chain rule):
\( \displaystyle \frac{dA}{dt}=2\pi r\,\frac{dr}{dt} \)
With r = 12 cm and \( \frac{dr}{dt} \) = 2 cm/sec:
\( \displaystyle \frac{dA}{dt}=2\pi(12)(2) \)
\( \displaystyle =48\pi\ \mathrm{cm^{2}/sec} \)
Why the other options are incorrect
Option (1) \( 24\pi\ \mathrm{cm^{2}/sec} \) — Incorrect. 24π = 2π(12) leaves out the factor \( \frac{dr}{dt} \) = 2.
Option (2) \( 24\pi\ \mathrm{cm^{3}/sec} \) — Incorrect. Both the value (factor 2 missing) and the unit are wrong; area changes in cm2/sec.
Option (4) \( 48\pi\ \mathrm{cm^{3}/sec} \) — Incorrect. The value is right but the unit is not: area changes in cm2/sec, not cm3/sec.
NCERT (2026-27 edition):
Rate of Change of Quantities
