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: Work with the radius directly:
For a sphere, $\frac{dV}{dr} = S$. So the chain rule gives $\frac{dV}{dt} = S\,\frac{dr}{dt}$.

Step 2: Apply the given proportionality:
Evaporation rate is proportional to $S$: $\frac{dV}{dt} = -kS$.
Compare: $S\frac{dr}{dt} = -kS$. Cancelling $S$ (it is not zero for a drop that exists) gives $\frac{dr}{dt} = -k$.

Step 3: Conclusion:
The radius decreases linearly with time. So the equation is $\frac{dr}{dt} + k = 0$.

Final Answer:
Option (A). \[ \boxed{\frac{dr}{dt}+k=0 \text{ (A)}} \]