Question

Find the terminal velocity of a raindrop of radius 0.01mm. Coefficient of viscosity of air is $1.8 \times {10^{ - 5}}$ and its density is 1.2kg/m$^3$. Density of water$ = 1000kg/{m^3}$. Take g $ = 10m/{s^2}$. (Force of buoyancy due to air is neglected) .

Answer 1

Hint: This question can be solved using the general formula for terminal velocity of a fluid which is given as $v = \dfrac{2}{9}\dfrac{{{r^2}\left( {\rho - \sigma } \right)g}}{\eta }$. By substituting the value of radius r, coefficient of viscosity $\eta $and density of water $\rho$ and density of air $\sigma$ given in the question we find the terminal velocity of the raindrop v.

Complete step by step answer:
Given, the radius of raindrop $r = 0.01mm = 1 \times {10^{ - 5}}m$
The coefficient of viscosity of air $\eta = 1.8 \times {10^{ - 5}}$
Density of air $\sigma = 1.2kg/{m^3}$
Density of water $\rho = 1000kg/{m^3}$
We know that the terminal velocity e is given as$v = \dfrac{2}{9}\dfrac{{{r^2}\left( {\rho - \sigma } \right)g}}{\eta }$
Substituting the given values in this equation we get,
$v = \dfrac{2}{9} \times \dfrac{{{{({{10}^{ - 5}})}^2}\left( {1000 - 1.2} \right)}}{{1.8 \times {{10}^{ - 5}}}} = \dfrac{2}{9} \times \dfrac{{998.8}}{{1.8}} \times {10^{ - 5}} = 123.3 \times {10^{ - 5}}$

Therefore the terminal velocity of the raindrop is given as $123.3 \times {10^{ - 5}}$m/s.

Additional information:
Terminal velocity is defined as the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling prevents further acceleration. The coefficient of viscosity defines the resistance for the flow of in a medium or in other words it can be defined as the tendency of resistance to the flow of fluid. In general the viscosity of air is less than the viscosity of liquid.

Note: If the density of air is not mentioned in the question we take it to be negligible to find out the terminal velocity of the raindrop. Therefore the formula for terminal velocity becomes.