Neglecting the density of air, the terminal velocity obtained by a raindrop of radius `0.3 mm` falling through the air of viscosity `1.8 xx10^(-5) N//m^(2)` will be

Options

A. `10.9ms^(-1)`
B. `7.48ms^(-1)`
C. `3.7ms^(-1)`
D. `12.8ms^(-1)`

Answer

A

Explanation

The terminal velocity of the spherical raindrop of radius r is given by
`v_(t)=(2r^(2)rhog)/(9eta)` where `rho` is the density of water and `eta` the viscosity of air. Substituting r = 0.3 mm
`=0.3xx10^(-3)m, rho =10^(3)kg//m^(3), g=9.8m//s^(2) and eta=1.8xx10^(-3)"N s/m"^(2)`, we get
`v_(t)=(2xx(0.3)^(2)xx10^(-3)xx9.8)/(9xx1.8xx10^(-5))`
`=10.9"m s"^(-1)`