Question

A small sphere falls from rest in a liquid. Due to viscosity heat is produced. Find the relation between the rate of production of heat and the radius of the sphere at terminal velocity.

• A. $\frac{dQ}{dt}\propto r^{-5}$
• B. $\frac{dQ}{dt}\propto r^{\frac{3}{2}}$
• C. $\frac{dQ}{dt}\propto r^5$
• D. $\frac{dQ}{dt}\propto r^{-\frac{3}{2}}$

Answer 1

The correct option is C $\frac{dQ}{dt}\propto r^5$

In such case, terminal velocity is given as
$v_t=\frac{2r^{2}g}{9\eta }({\rho }_s-{\rho }_l)$
and viscous force is given as
$F_v=6\pi \eta r{v}_t$

Here, we need to find rate of production of heat, which is, power.

As viscous force is the only dissipative force.
Hence,
$\frac{dQ}{dt}=F_v{v}_t=\left(6\pi \eta r{v}_t\right)v_t$
$=6\pi \eta r{v}_t^2$

$=6\pi \eta r{\left[\frac{2}{9}\frac{r^{2}g}{\eta }\right({\rho }_s-{\rho }_l\left)\right]}^2$

$=\frac{8\pi g^2}{27\eta }({\rho }_s-{\rho }_l{)}^2{r}^5$

$\therefore \frac{dQ}{dt}\propto r^5$

Hence, $\left(C\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question?}\\ \text{This question is aimed at recalling the application of stoke's laws and terminal velocity.}\end{array}$