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Question: A raindrop of radius \(0.3\;mm\)has a terminal velocity of air \(1\;m{s^{ - 1}}\). The viscosity of ...

A raindrop of radius 0.3  mm0.3\;mmhas a terminal velocity of air 1  ms11\;m{s^{ - 1}}. The viscosity of air is 18×103  poise18 \times {10^{ - 3}}\;poise. What is the viscous force on it?
A. 101.78×102  dyne101.78 \times {10^{ - 2}}\;dyne
B. 101.37×105  dyne101.37 \times {10^{ - 5}}\;dyne
C. 16.95×105  dyne16.95 \times {10^{ - 5}}\;dyne
D. 16.95×104  dyne16.95 \times {10^{ - 4}}\;dyne

Explanation

Solution

Hint: A raindrop tends to fall towards the ground, with certain velocity. Hence, due to the viscosity of air surrounded by the raindrop generates a drag force opposite to the movement of the raindrop. That force of viscosity can be obtained with the help of Stoke's law. It gives the detailed relation between the radius of the drop, the velocity of raindrop, viscosity of air and the viscous force on rain drop.

Formula used:
The Stoke's law is given by,
F=6πηrvF = 6\pi \eta rv
Where, FF is the viscous force on rain drop by air, η\eta viscosity of air in poisepoise, rr is the radius of rain drop in cmcm and vv is the velocity of rain drop in cms1cm{s^{ - 1}}.

Complete step by step solution:
Given, The Radius of raindrop, r=0.3  mmr = 0.3\;mm (or) r=0.03  cmr = 0.03\;cm
The Velocity of raindrop, v=1  ms1v = 1\;m{s^{ - 1}} (or) v=100  cms1v = 100\;cm{s^{ - 1}}
The Viscosity of air, η=18×105  poise\eta = 18 \times {10^{ - 5}}\;poise

The Stoke's law is given by,
F=6πηrv  ...............................(1)F = 6\pi \eta rv\;...............................\left( 1 \right)
Substitute the values of π\pi , η\eta , rr and vv in the equation (1),
F=6×3.14×(18×103  poise)×(0.03  cm)×(100  cms1) F=1.01736  dyne F=101.736×102  dyne F101.78×102  dyne  F = 6 \times 3.14 \times \left( {18 \times {{10}^{ - 3}}\;poise} \right) \times \left( {0.03\;cm} \right) \times \left( {100\;cm{s^{ - 1}}} \right) \\\ F = 1.01736\;dyne \\\ F = 101.736 \times {10^{ - 2}}\;dyne \\\ F \simeq 101.78 \times {10^{ - 2}}\;dyne \\\
Hence, the force of viscosity on rain drop, F101.78×102  dyneF \simeq 101.78 \times {10^{ - 2}}\;dyne

Thus, the option (A) is the correct answer.

Note: In this question, Stoke's law used to find the viscous force on the raindrop which is due to the air viscosity. In the above solution, the unit of radius of rain drop and velocity of rain drop should be in cmcm and cms1cm{s^{ - 1}} respectively. Because, the unit of the viscous force in the below options are in dynedyne. Another notation for the unit dynedyne is cgscgs, which is centi-gram second. Hence, the length unit should be in centimetre. If the values entered in metre will cause error in the result.