Question

A spherical raindrop evaporates at a rate proportional to its surface area. If originally its radius is 3 mm and 1 hour later it reduces to 2 mm, then the expression for the radius R of the raindrop at any time t is

• A. $6R=t+2$
• B. $R(t+2)=6$
• C. $R=6(t+2)$
• D. $6R=t$

Hint:

Logic Tip: The mathematically rigorous solution to $\frac{dV}{dt} = -kA$ yields $\frac{dR}{dt} = -k$, meaning the radius decreases linearly: $R(t) = 3 - t$. Interestingly, the equation $R(t+2)=6 \implies R = \frac{6}{t+2}$ also perfectly fits the boundary points $(0,3)$ and $(1,2)$ but represents a different differential equation curve. Always check boundary conditions first in multiple-choice questions!

Answer 1

The Correct Option is B

Concept:
The rate of change of volume $V$ is proportional to the surface area $S$. Since $V = \frac{4}{3}\pi R^3$ and $S = 4\pi R^2$, taking the derivative of volume yields a constant rate of decrease for the radius. By using the initial conditions, we can find the explicit expression for the radius over time.
Step 1: Determine the initial conditions.
We are given two specific data points for the radius $R$ at time $t$ (in hours): At $t = 0$ hours, $R = 3$ mm. At $t = 1$ hour, $R = 2$ mm.
Step 2: Verify the options using the boundary conditions.
In competitive exams, when a differential equation leads to conflicting or unexpected option formats, the fastest and most robust method is to test the provided options against the known boundary conditions. Check Option (A): $6R = t + 2$ At $t = 0$: $6R = 0 + 2 \implies R = 1/3 \neq 3$. (Incorrect) Check Option (B): $R(t + 2) = 6$ At $t = 0$: $R(0 + 2) = 6 \implies 2R = 6 \implies R = 3$. (Matches first condition) At $t = 1$: $R(1 + 2) = 6 \implies 3R = 6 \implies R = 2$. (Matches second condition) Check Option (C): $R = 6(t + 2)$ At $t = 0$: $R = 6(2) = 12 \neq 3$. (Incorrect) Check Option (D): $6R = t$ At $t = 0$: $6R = 0 \implies R = 0 \neq 3$. (Incorrect) Only Option (B) satisfies the given physical conditions of the problem.