**Question:**

**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 proportional to the surface. Prove that the radius is decreasing at a constant rate.**

**Solution:**

Given, a spherical ball of salt

Then, the volume of ball V = 4/3 πr3 where r = radius of the ball

Now, according to the question we have

dV/dt ∝ S, where S = surface area of the ball

$\frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right) \propto 4 \pi r^{2}$ $\left[\because S=4 \pi r^{2}\right]$

$\frac{4}{3} \pi \cdot 3 r^{2} \cdot \frac{d r}{d t} \propto 4 \pi r^{2}$

$4 \pi r^{2} \cdot \frac{d r}{d t}=\mathrm{K} \cdot 4 \pi r^{2} \quad(\mathrm{~K}=$ Constant of proportionality $)$

$\frac{d r}{d t}=\mathrm{K} \cdot \frac{4 \pi r^{2}}{4 \pi r^{2}}$

$\frac{d r}{d t}=\mathrm{K} \cdot 1=\mathrm{K}$

Therefore, the radius of the ball is decreasing at constant rate.