A raindrop of the radius \[0.3mm\] has a terminal velocity \[1\dfrac{m}{s}\] in the air. The viscosity of air is \[18 \times {10^{ - 5}}poise\]. The viscous force on the airdrop is
A. \[16.695 \times {10^{ - 7}}N\]
B. \[1.695 \times {10^{ - 8}}N\]
C. \[1.017 \times {10^{ - 12}}N\]
D. \[101.73 \times {10^{ - 9}}N\]

Hint:Poisson's ratio is defined as the ratio of the change in a sample width per unit width to the change in the length per unit length due to strain. Where strain is the force which determines the strength of an object when they are stretched or deformed.
Viscosity is the measure of the extent to which fluid flow is redistricted, and its unit is newton-second per square meter, which is usually expressed as Pascal-second. The viscosity of liquids decreases rapidly with an increase in temperature. Viscous force is the force acting between the layers of flowing fluid. It is the rate at which the fluid velocity changes its space.
Use the viscous force formula \[F = 6\pi \eta r{V_T}\], where \[{V_T}\] is the terminal velocity \[\eta \] is the poisson's ratio.

Complete step by step answer:
Terminal velocity \[{V_T} = 1\dfrac{m}{s}\]
Poisson's ratio \[\eta = 18 \times {10^{ - 5}}poise = 18 \times {10^{ - 6}}decapoise\]
The radius of the rain droplet \[r = 0.3mm = 0.3 \times {10^{ - 3}}m\]
 Viscous force on the raindrops is given by the formula \[F = 6\pi \eta r{V_T}\]
Now substitute the values in the formula,
\[
  F = 6\pi \eta r{V_T} \\
   = 6 \times \dfrac{{22}}{7} \times 18 \times {10^{ - 6}} \times 0.3 \times {10^{ - 3}} \times 1 \\
   = \dfrac{{6 \times 22 \times 18 \times 3}}{{7 \times 10}} \times {10^{ - 9}} \\
   = 101.73 \times {10^{ - 9}}N \\
 \]
Hence the Viscous force on the air droplet is\[ = 101.73 \times {10^{ - 9}}N\]
Option D is correct.

Note:Students must note that the viscosity is the same as the friction as it also opposes the motion of the fluid, and it is non-conservative.