Question

The terminal velocity of a rain drop is $30{{cm} \mathord{\left/ {\vphantom {{cm} s}} \right.} s}$. If the viscosity of air is $1.8 \times {10^{ - 5}}N{m^{ - 2}}$. The radius of rain drop is
(A) $1\mu m$
(B) $0.5mm$
(C) $0.05mm$
(D) $1mm$

Answer 1

Stokes’ law demonstrates that the frictional drag force is directly proportional to the weight of the object. The mass of the raindrop is the product of its volume and density. So using the formula $F = 6\pi \eta rv$ and equating it to the weight of the raindrop we can find the radius.

Formula Used: The following formulas are used to solve this question.
$m = Volume \times density$
$F = 6\pi \eta rv$ where force $F$ is acting on the interface between the fluid and the particle, $\eta$ is the dynamic viscosity, $r$ is the radius of the spherical object and $v$ is the flow velocity relative to the object.

Complete step by step answer
For the given rain drop, it is assumed that it is falling at terminal velocity in the same air density.
An object’s terminal velocity is the maximum velocity that it achieves when falling through a fluid. Terminal velocity occurs when a falling body’s rate of velocity increase slows down to a certain velocity which stays constant for the rest of its journey down. In other words the body or mass reaches zero acceleration.
When an object is at terminal velocity, the drag force opposing the motion is equal to the weight of the object pulling it down through the air.
According to Stokes Law,
Viscous drag force is given by, $F = 6\pi \eta rv$ where force $F$ is acting on the interface between the fluid and the particle, $\eta$ is the dynamic viscosity, $r$ is the radius of the spherical object and $v$ is the flow velocity relative to the object.
Weight of the particle is equal to $mg$ where $m$ is the mass and $g$ is the acceleration due to gravity of the particle.
For a spherical raindrop with radius $r$, mass is given by,
$m = Volume \times density$.
Let density be $\rho$, thus $m = \dfrac{4}{3}\pi {r^3}\rho$.
$\therefore Weight = \dfrac{4}{3}\pi {r^3}\rho g$
Now since weight is equal to the drag force, thus we can write,
$\dfrac{4}{3}\pi {r^3}\rho g = 6\pi \eta rv$
From here we can cancel the similar terms and write,
${r^2} = \dfrac{{9\eta v}}{{2\rho g}}$
The radius is thus given by, $r = \sqrt {\dfrac{{9\eta v}}{{2\rho g}}}$
Assigning the values given in the question, $v = 0.3m{s^{ - 1}}$, $\eta = 1.8 \times {10^{ - 5}}N{m^{ - 2}}$ and density of water $\rho = 1000{{kg} \mathord{\left/ {\vphantom {{kg} {{m^3}}}} \right.} {{m^3}}}$ and acceleration due to gravity $g = 10m{s^{ - 2}}.$
$r = \sqrt {\dfrac{{9 \times 1.8 \times {{10}^{ - 5}}N{m^{ - 2}} \times 0.3}}{{2 \times 1000 \times 10}}}$
$\Rightarrow r = \sqrt {27 \times 0.9 \times {{10}^{ - 8}}}$
From further simplification, we find the value of the radius.
Thus the radius is $r = 0.49 \times {10^{ - 4}}m \approx 0.05mm$.

$\therefore$ The correct answer is Option C.

Note
Stokes law is the basis of the falling-sphere viscometer, in which the fluid is stationary in a vertical glass tube. A sphere of known size and density is allowed to descend through the liquid. If correctly selected, it reaches terminal velocity, which can be measured by the time it takes to pass two marks on the tube. Electronic sensing can be used for opaque fluids.