A liquid drop of mass m and radius r is falli | vedclass

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(A) When a liquid drop falls through air,it eventually attains a constant terminal velocity $V_t$ due to the balance between gravitational force,buoya

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$\frac{m}{r}$

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(A) When a liquid drop falls through air,it eventually attains a constant terminal velocity $V_t$ due to the balance between gravitational force,buoyant force,and viscous drag force.The forces acting on the drop are:$1$. Gravitational force: $F_g = mg$$2$. Buoyant force: $F_b = \frac{4}{3} \pi r^3 \rho_{air} g$$3$. Viscous drag force (Stokes' Law): $F_v = 6 \pi \eta r V_t$At terminal velocity,the net force is zero:$mg - F_b = F_v$$mg - \frac{4}{3} \pi r^3 \rho_{air} g = 6 \pi \eta r V_t$Solving for $V_t$:$V_t = \frac{mg - \frac{4}{3} \pi r^3 \rho_{air} g}{6 \pi \eta r}$Since the density of air $\rho_{air}$ is very small compared to the density of the liquid,the buoyant force is negligible. Thus:$V_t \approx \frac{mg}{6 \pi \eta r}$Therefore,$V_t \propto \frac{m}{r}$.

$A$ liquid drop of mass $m$ and radius $r$ is falling from a great height. Its terminal velocity is proportional to ............

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