Question

A small spherical body of radius $r$ is falling under gravity in a viscous medium and due to friction, the medium gets heated. When the body attains terminal velocity, then the rate of heating is proportional to:

• A. $r$
• B. $r^{3/2}$
• C. $r^5$
• D. $r^{1/2}$

Answer 1

The correct option is C $r^5$

Radius of the spherical body $=r$
$\text{Rate of heat production}=\text{Power dissipated}$
& $\text{Power dissipated}=F\times v...\left(i\right)$
Hence, rate of heat production is,
$\frac{dH}{dt}=\left(6\pi r\eta \right)v^2$
$[\because F=6\pi r\eta v\text{is the viscous force}]$
$\Rightarrow \frac{dH}{dt}=\left(6\pi \eta r\right){\left[\frac{2}{9}\frac{(\sigma -\rho )r^{2}g}{\eta }\right]}^2...\left(ii\right)$
$\sigma \&\rho$ are the density of solid body and liquid respectively.
$v=v_T$, when terminal velocity is achieved.
From Eq.$\left(ii\right)$, rate of heat production,
$=\frac{dH}{dt}\propto r^5$
Hence, the correct option is $\left(c\right)$