Question

Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks

up moisture at a rate proportional to its surface area. Under these circumstances, what will be

the e‚ect on the drop’s radius.

Answer 1

Solution:

$V=\dfrac43\pi r^3 \\ \dfrac{dV}{dt}=\dfrac43.3\pi r^2 \dfrac{dr}{dt} \\=4\pi r^2 \dfrac{dr}{dt}\ ...(i)$

$A=4\pi r^2 ...(ii)$

It is given that $\dfrac{dV}{dt}=k.(Surface\ area)$

$\Rightarrow 4\pi r^2 \dfrac{dr}{dt}=k.(4\pi r^2)$ [using (i) and (ii)]

$\Rightarrow \dfrac{dr}{dt}=k$

It shows radius increases at a constant rate.