Question

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)

• A. \(\frac{dr}{dt}+k = 0\)
• B. \(\frac{dr}{dt}-k = 0\)
• C. \(\frac{dr}{dt}+kr = 0\)
• D. \(\frac{dr}{dt}-kr = 0\)

Hint:

Volume loss rate is proportional to surface area. Differentiate V = (4/3) pi r^3.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The raindrop is a sphere of radius \(r\). It loses volume at a rate proportional to its surface area, so \(\dfrac{dV}{dt}\) is negative.

Step 2: Key Formula or Approach:
\(V = \dfrac43\pi r^3\) and surface area \(S = 4\pi r^2\).

Step 3: Detailed Explanation:
Rate of change of volume:
\[ \frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt} \]
Evaporation rate is proportional to surface area, with a minus sign because volume decreases:
\[ \frac{dV}{dt} = -k\,(4\pi r^2) \]
Equate:
\[ 4\pi r^2\frac{dr}{dt} = -k\,4\pi r^2 \Rightarrow \frac{dr}{dt} = -k \]
\[ \frac{dr}{dt} + k = 0 \]
The radius shrinks at a constant rate. Options (C) and (D) say that the rate depends on \(r\), which is not the case since the \(r^2\) factors cancel. Option (B) has the wrong sign because the drop gets smaller.

Final Answer:
The differential equation is \(\dfrac{dr}{dt} + k = 0\), option (A). \[ \boxed{\frac{dr}{dt}+k=0 \text{ (A)}} \]