Properties of Matter 4 Question 14
16. A small sphere falls from rest in a viscous liquid. Due to friction, heat is produced. Find the relation between the rate of production of heat and the radius of the sphere at terminal velocity.
(2004, 2M)
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Answer:
Correct Answer: 16. $\frac{d Q}{d t} \propto r^{5}$
Solution:
- Terminal velocity $v _T=\frac{2 r^{2} g}{9 \eta}\left(\rho _S-\rho _L\right)$
and viscous force $F=6 \pi \eta r v _T$
Rate of production of heat (power) : as viscous force is the only dissipative force.
Hence,
$$ \begin{aligned} \frac{d Q}{d t}=F v _T & =\left(6 \pi \eta r v _T\right)\left(v _T\right)=6 \pi \eta r v _T^{2} \\ & =6 \pi \eta r[{{\frac{2}{9} \frac{r^{2} g}{\eta}\left(\rho _S-\rho _L\right)}}]^{2} \\ & =\frac{8 \pi g^{2}}{27 \eta}\left(\rho _S-\rho _L\right)^{2} r^{5} \text { or } \frac{d Q}{d t} \propto r^{5} \end{aligned} $$
