A small ball of radius r is falling in a viscous liquid under gravity. Find the dependency of rate of heat produced in terms of radius r after the drop attains terminal velocity

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Solution

Rate of heat produced $= P = f . v$
$f = 6 π η r v o .$
$v = 2 v^{2} \frac{(p - σ) g}{9 η}$
Substituting
$P = 6 π η r v . v = 6 π η v (v)^{2}$
$= 6 π η v {[\frac{2 r^{2} (p - σ) g}{q n}]}^{2}$
$=^{2} t π η v \frac{4}{9 \times 9} v^{4} (p - σ)^{2} g^{2}$
$= \frac{8 v^{5} (p - σ)^{2} g^{2}}{27 η}$
$P =$ density of solid
$σ =$ density of liq.
$η =$ coeff of viscosity

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A small ball of radius r is falling in a viscous liquid under gravity. Find the dependency of rate of heat produced in terms of radius r after the drop attains terminal velocity

Rate of heat produced =P=f.vf=6πηrvo.v=2v2(p−σ)g9ηSubstituting P=6πηrv.v=6πηv(v)2=6πηv[2r2(p−σ)gqn]2=2tπηv49×9v4(p−σ)2g2=8v5(p−σ)2g227ηP= density of solidσ= density of liq.η= coeff of viscosity