Question
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 0.3mmhas a terminal velocity of air 1ms−1. The viscosity of air is 18×10−3poise. What is the viscous force on it?
A. 101.78×10−2dyne
B. 101.37×10−5dyne
C. 16.95×10−5dyne
D. 16.95×10−4dyne
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=6πηrv
Where, F is the viscous force on rain drop by air, η viscosity of air in poise, r is the radius of rain drop in cm and v is the velocity of rain drop in cms−1.
Complete step by step solution:
Given, The Radius of raindrop, r=0.3mm (or) r=0.03cm
The Velocity of raindrop, v=1ms−1 (or) v=100cms−1
The Viscosity of air, η=18×10−5poise
The Stoke's law is given by,
F=6πηrv...............................(1)
Substitute the values of π, η, r and v in the equation (1),
F=6×3.14×(18×10−3poise)×(0.03cm)×(100cms−1) F=1.01736dyne F=101.736×10−2dyne F≃101.78×10−2dyne
Hence, the force of viscosity on rain drop, F≃101.78×10−2dyne
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 cm and cms−1 respectively. Because, the unit of the viscous force in the below options are in dyne. Another notation for the unit dyne is cgs, 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.