A small spherical body of radius $r$ is falling under gravity in a viscous medium. Due to friction t

Question

A small spherical body of radius $r$ is falling under gravity in a viscous medium. Due to friction the medium gets heated. How does the rate of heating depend on radius of body when it attains terminal velocity is propotional to?\

Professor PC Thomas Classes · Answer

Rate of heat produced\ $=($ viscous force $F) 	imes($ velocity $v)$\ or $\frac{d Q}{d t}=(6 \pi \eta r v)(v)=6 \pi \eta r v^{2}$\ $=6 \pi \eta r\left[\frac{2}{9} \frac{(\sigma-ho) r^{2} g}{\eta}ight]^{2}$\ $=\left(\frac{6 \pi 	imes 4 g^{2} 	imes(\sigma-ho)^{2}}{81 	imes \eta}ight) 	imes r^{5}$\ $=($ constant $) 	imes r^{5}$ or Rate of heating $\propto r^{5}$.\ Hence the rate of heating is proportional to fifth power of radius of body when it attains terminal velocity.\