Find the rate of change of volume of a sphere with respect to its surface area when radius is $5 m$.
Answer & explanation
Correct answer: option 2
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. $\frac{5}{2}m$