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

We have, rate of decrease of the volume of spherical ball of salt at any instant is proportional to surface.

Let us assume that 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 $(S) = 4\pi {r^2}$

$\frac{{dV}}{{dt}} \propto S \Rightarrow \frac{d}{{dt}}\left( {\frac{4}{3}\pi {r^3}} \right) \propto 4\pi {r^2}$

$\Rightarrow$ $\frac{4}{3}\pi \cdot 3{r^2} \cdot \frac{{dr}}{{dt}} \propto 4\pi {r^2} \Rightarrow \frac{{dr}}{{dt}} \propto \frac{{4\pi {r^2}}}{{4\pi {r^2}}}$

$\Rightarrow$ $\frac{{dr}}{{dt}} = k \cdot 1\quad$ [where, $k$ is the constant of proportionality ]

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

Hence we can say that the radius of ball is decreasing at a constant rate.