Question

A small spherical drop fall from rest in viscous liquid. Due to friction, heat is produced. The correct relation between the rate of production of heat and the radius of the spherical drop at terminal velocity will be

• A. $\frac{dH}{dt}\propto \frac{1}{r^5}$
• B. $\frac{dH}{dt}\propto r^4$
• C. $\frac{dH}{dt}\propto \frac{1}{r^4}$
• D. $\frac{dH}{dt}\propto r^5$

Answer 1

Answer: (D)

$\frac{dH}{dt}\propto r^5$

Viscous force acting on spherical drop
$F_v=6\pi \eta v$
$\therefore$ Terminal velocity, $v=\frac{2}{9}\frac{r^2(\sigma -\rho )g}{\eta }$
where, $\eta =$ coefficient of viscosity of liquid
$\sigma =$ density of material of spherical drop
$\rho =$ density of liquid
Power imparted by viscous force $=$ Rate of production of heat
$\rho =\frac{dH}{dt}=F_v\cdot v=6\pi \eta v^2$
$=6\pi \eta \cdot {\left(\frac{2}{9}\frac{r^2(\sigma -\rho )g}{\eta }\right)}^2$
$\Rightarrow \frac{dH}{dt}\propto r^5$