Question

A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume is proportional to the surface. Prove that the radius is decreasing at a constant rate.

Answer 1

We have rate of decrease of the volume of spherical ball of salt at any instant is $\alpha$ surface . let the radius of the spherical ball of the salt be $r$.
$\therefore$ Volume of the ball (V) =$\frac{4}{3}\pi r^3$
and surface area (S) $=4\pi r^2$
$\because \frac{dV}{dT}\propto S$
$\Rightarrow \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)\propto 4\pi r^2$
$\Rightarrow \frac{4}{3}\pi \left(3r^2\right)\frac{dr}{dt}\propto 4\pi r^2$
$\Rightarrow \frac{dr}{dt}\propto \frac{4\pi r^2}{4\pi r^2}$
$\Rightarrow \frac{dr}{dt}=k.1$ [ where k is the propotionality constant ]
$\Rightarrow \frac{dr}{dt}=k$
Hence the radius of ball is decreasing at a constant rate.