Find the rate of change of volume of a sphere with respect to its surface area when radius is $5 m$.

$\frac{5}{2}m$

The correct answer is Option (2) → $\frac{5}{2}m$

Let $r$ be the radius of the sphere and V and S be the volume and surface area of the sphere respectively, then

$V=\frac{4}{3}πr^3$

Differentiating w.r.t. r, we get

$\frac{dV}{dr}=4πr^2$   ...(i)

Also $S=4πr^2⇒\frac{dS}{dr}=8πr$   ...(ii)

Now, $\frac{dV}{dS}=\frac{\frac{dV}{dr}}{\frac{dS}{dr}}=\frac{4πr^2}{8πr}=\frac{r}{2}$   (Using (i) and (ii))

So, $\left[\frac{dV}{dS}\right]_{r=5}=\frac{5}{2}m^3/m^2$ i.e. $\frac{5}{2}m$