Question

A small sphere of density $\rho$ falls from rest into a viscous liquid of density $\sigma$ and viscosity $\eta$. Due to friction, heat is produced. Which of the follwing options correctly represents the relation between the rate of production of heat $H$ and the radius of the sphere $r$ at terminal velocity?

• A. $\frac{12\pi (\rho -\sigma )g^2{r}^5}{27\eta }$
• B. $\frac{12\pi (\rho -\sigma {)}^2{g}^2{r}^5}{27\eta }$
• C. $\frac{8\pi (\rho -\sigma )g^2{r}^5}{27\eta }$
• D. $\frac{8\pi (\rho -\sigma {)}^2{g}^2{r}^5}{27\eta }$

Answer 1

The correct option is D $\frac{8\pi (\rho -\sigma {)}^2{g}^2{r}^5}{27\eta }$

The forces acting on the sphere are:
Weight of the sphere $\left(F_w\right)=\frac{4}{3}\pi r^3\rho g$ (acting downwards $\downarrow$),
Buoyant force $\left(F_b\right)=\frac{4}{3}\pi r^3\sigma g$ (acting upwards$\uparrow$),
and
Viscous force $\left(F_t\right)=6\pi \eta rv$ (acting upwards$\uparrow$) where $v$ is the instantaneous velocity of sphere.

The sphere attains terminal velocity $v_t$ when the resultant force on it is zero i.e. $F_w=F_b+F_t$
$\frac{4}{3}\pi r^3\rho g=\frac{4}{3}\pi r^3\sigma g+6\pi \eta r{v}_t$.
Solving above equation, we get the terminal velocity
$v_t=\frac{2r^2(\rho -\sigma )g}{9\eta }.....\left(1\right)$
The rate of heat generation is equal to the rate of work done by the viscous force which in turn is equal to its power. Thus,
$H=\frac{dQ}{dt}=F_t\times v_t=\left(6\pi \eta r{v}_t\right)v_t.....\left(2\right)$
Using $\left(1\right)$ in$\left(2\right)$ we get
$H=\frac{dQ}{dt}=\frac{8\pi (\rho -\sigma {)}^2{g}^2{r}^5}{27\eta }$
Thus, option (d) is the correct answer.