What will be the approximate terminal velocity of a rain drop of diamteter `1.8xx 10^(-3)m,` when density of rain water `~~10^(3)kgm^(-3)` and the co-efficient of viscosity of air `~~1.8xx10^(-5)Nsm^(-2)` ? (Neglect buoyancy of air. )

Options

A. `49 ms^(-1)`
B. `98 ms^(-1)`
C. `392 ms^(-1)`
D. `980 ms^(-1)`

Answer

D

Explanation

Here, diameter of a rain drop = `1.8 xx 10^(-3)m`
`r = (1.8 xx 10^(-3))/(2) m = 0.9 xx 10^(-3) m`
Density of rain water, ` rho = 10^3 kg m^(-3)`
Coefficient of viscosity of air, `eta = 1.8 xx 10^(-5) N m^(-2)`
When a raindrop falls in air (viscous medium), then it accelerates initially due to gravity. The viscous drag force, `F = 6pi eta rv ` where v = terminal velocity
In equilibrium, F=mg
or `6pi etarv = 4/3 pi r^3 rhog " or " v = (2)/(9 eta) r^2 rho g `
` rArr v = (2 xx 0.9 xx 0.9 xx 10^(-6) xx 10^3 xx 9.8)/(9 xx 1.8 xx 10^(-5)) = 98 ms^(-1)`