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Question: A rain drop of radius \(0 \cdot 3\,mm\) falls through air with a terminal velocity of \(1\,m\,{s^{ -...

A rain drop of radius 0⋅3 mm0 \cdot 3\,mm falls through air with a terminal velocity of 1 m s−11\,m\,{s^{ - 1}}. the viscosity of air is 18×10−5 poise18 \times {10^{ - 5}}\,{\text{poise}}. What is the viscous force on the raindrop ?
A. 1⋅018×10−2 dyne1 \cdot 018 \times {10^{ - 2}}\,{\text{dyne}}
B. 2⋅018×10−2 dyne2 \cdot 018 \times {10^{ - 2}}\,{\text{dyne}}
C. 3⋅018×10−2 dyne3 \cdot 018 \times {10^{ - 2}}\,{\text{dyne}}
D. 4⋅018×10−2 dyne4 \cdot 018 \times {10^{ - 2}}\,{\text{dyne}}

Explanation

Solution

In order to find the viscous force acting on the raindrop when it falls through the air we can use the Stokes law.
According to this law, the viscous force or the resistive force that retards the motion of a sphere in a viscous fluid will be directly proportional to the velocity of the sphere, radius of the sphere, and the viscosity of the fluid through which it travels. In equation form, it can be written as,
F=6πηrvF = 6\pi \eta rv
Where η\eta represents the viscosity, r represents the radius of the sphere and v represents the velocity.

Complete step by step answer:
It is given that radius of the raindrop is 0⋅3 mm0 \cdot 3\,mm
That is,
r=0⋅3 mmr = 0 \cdot 3\,mm
⇒r=0⋅3×10−1 cm\Rightarrow r = 0 \cdot 3 \times {10^{ - 1}}\,cm
The terminal velocity or the maximum velocity attained by the raindrop when it falls through the air is given as 1 m s−11\,m\,{s^{ - 1}}.
That is,
v=1ms−1v = 1m{s^{ - 1}}
⇒v=100cms−1\Rightarrow v = 100cm{s^{ - 1}}
The viscosity of air is 18×10−5poise18 \times {10^{ - 5}}poise. Viscosity is represented using the symbol η\eta .
Therefore,
⇒η=18×10−5poise\Rightarrow \eta = 18 \times {10^{ - 5}}poise
We need to find the viscous force on the raindrop. For this, we can use Stokes law.
According to this law, the viscous force or the resistive force that retards the motion of a sphere in a viscous fluid will be directly proportional to the velocity of the sphere, the radius of the sphere, and the viscosity of the fluid through which it travels. In equation form, it can be written as,
⇒F=6πηrv\Rightarrow F = 6\pi \eta rv
Where η\eta represents the viscosity, r represents the radius of the sphere and vv represents the velocity.
We know value of π\pi is3⋅143 \cdot 14
Now let us substitute all the given values in the above equation. Then we get,
⇒F=6×3⋅14×18×10−5×0⋅03×100\Rightarrow F = 6 \times 3 \cdot 14 \times 18 \times {10^{ - 5}} \times 0 \cdot 03 \times 100
∴F=1⋅018×10−2dyne\therefore F = 1 \cdot 018 \times {10^{ - 2}}{\text{dyne}}

Thus, the correct answer is option A.

Note:
Stokes law is applied only in a streamline flow that is when the flow is steady. If the spherical body that is falling moves so fast in the fluid then the flow will not be steady then stokes is not applicable.