Question

Neglecting the density of air, the terminal velocity obtained by a raindrop of radius 0.3 mm falling through the air of viscosity $1.8\times {10}^{-5}Ns/m^2$ will be $({\rho }_{water}={10}^{3}kg/m^3)$

• A. 10.9 m/s
• B. 8.3 m/s
• C. 9.2 m/s
• D. 7.6 m/s

Answer 1

The correct option is A

10.9 m/s

The terminal velocity of the spherical raindrop of radius r is given by

$v_t=\frac{2r^2\rho g}{9\eta }$

Where $\rho$ is the density of water and $\eta$ the viscosity of air.

Substituting r = 0.3 mm $=0.3\times {10}^{-3}m,\rho ={10}^{3}kg/m^3,g=9.8m{s}^{-2}$ and $\eta =1.8\times {10}^{-5}Ns/m^2$

We get $v_t=\frac{2\times (0.3{)}^2\times {10}^{-3}\times 9.8}{9\times 1.8\times {10}^{-5}}=10.9m/s$