Question

A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to:

• A. $r^5$
• B. $r^2$
• C. $r^3$
• D. $r^4$

Answer 1

The correct option is A $r^5$

A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to

Let r is the radius of sphere and $v_t$ is its terminal speed. Then the weight of sphere is balanced by the buoyant force and viscous force such that:

Weight,

$w=mg$

$\because \rho =\frac{m}{V}$

$m=\frac{4}{3}\pi r^3\rho g........\left(1\right)$

$so,$

$w=\frac{4}{3}\pi r^3\rho$

Buoyant force,

$F_B=\frac{4}{3}\pi r^3\sigma g..........\left(2\right)$

Where, $\sigma$ is density of water.

Viscous force, $F=6\pi \eta rvt..............\left(3\right)$

Where, $\eta$is viscosity.

From equation (1) (2) and (3)

$w=F_B+F_v$

$\frac{4}{3}\pi r^3\rho g=\frac{4}{3}\pi r^3\sigma g+6\pi \eta rvt$

$v_t=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }.......\left(4\right)$

The rate of production of heat when the sphere attains its terminal velocity is

equal to work done by the viscous forces.

$W=\frac{dQ}{dt}=F_v\times v_t$

$W=6\pi \eta r{v}_t^2$

$W=6\pi \eta r{\left(\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }\right)}^2$

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