Q95 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
A spherical raindrop evaporates at a rate proportional to its surface area. The differential equation involving the rate of change of its radius $r$ with time '$t$' is $\ldots$ (where $k$ is a positive constant)
- Correct answerA$\dfrac{dr}{dt} + k = 0$
- B$\dfrac{dr}{dt} - k = 0$
- C$\dfrac{dr}{dt} + kr = 0$
- D$\dfrac{dr}{dt} - kr = 0$
Solution
Let $V$ be the volume and $S$ be the surface area of the spherical raindrop of radius $r$.
$V = \dfrac{4}{3}\pi r^3$ and $S = 4\pi r^2$
The rate of evaporation is the rate of decrease of volume, which is given to be proportional to the surface area.
$\dfrac{dV}{dt} \propto -S$
$\dfrac{dV}{dt} = -k S$ (where $k$ is a positive constant of proportionality)
Substituting the expressions for $V$ and $S$:
$\dfrac{d}{dt}\left(\dfrac{4}{3}\pi r^3\right) = -k(4\pi r^2)$
$4\pi r^2 \dfrac{dr}{dt} = -k(4\pi r^2)$
Since $r \neq 0$, dividing both sides by $4\pi r^2$ gives:
$\dfrac{dr}{dt} = -k$
$\dfrac{dr}{dt} + k = 0$
Answer: $\dfrac{dr}{dt} + k = 0$
$V = \dfrac{4}{3}\pi r^3$ and $S = 4\pi r^2$
The rate of evaporation is the rate of decrease of volume, which is given to be proportional to the surface area.
$\dfrac{dV}{dt} \propto -S$
$\dfrac{dV}{dt} = -k S$ (where $k$ is a positive constant of proportionality)
Substituting the expressions for $V$ and $S$:
$\dfrac{d}{dt}\left(\dfrac{4}{3}\pi r^3\right) = -k(4\pi r^2)$
$4\pi r^2 \dfrac{dr}{dt} = -k(4\pi r^2)$
Since $r \neq 0$, dividing both sides by $4\pi r^2$ gives:
$\dfrac{dr}{dt} = -k$
$\dfrac{dr}{dt} + k = 0$
Answer: $\dfrac{dr}{dt} + k = 0$
Hence, the correct answer is option: 'A'.