Question

A small sphere of radius $r$ falls from rest in a viscous liquid. Heat is produced due to friction between the liquid and the sphere. Choose the correct relation between rate of production of heat and the radius of the sphere, when terminal velocity is attained.

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

Answer 1

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

From Stokes's law,
$F_v=6\pi \eta r{V}_T$
where,
$\eta$ = coefficient of viscosity of liquid
$V_T$= Terminal velocity of sphere

The rate of production of heat is equal to the power generated by viscous force.
$\frac{dQ}{dt}=\text{Power}=F_v{V}_T=\left(6\pi \eta r{V}_T\right)V_T$
$\Rightarrow \frac{dQ}{dt}=6\pi \eta r{V}_T^2=6\pi \eta r{\left[\frac{2}{9}\frac{r^2}{\eta }g\right({\rho }_s-\rho \left)\right]}^2$
where, ${\rho }_s=$ density of sphere
$\rho =$ density of liquid
$\Rightarrow \frac{dQ}{dt}\propto r^5$
Hence, option (c) is the correct answer.