
A raindrop of radius $0.3\;mm$has a terminal velocity of air $1\;m{s^{ - 1}}$. The viscosity of air is $18 \times {10^{ - 3}}\;poise$. What is the viscous force on it?
A. $101.78 \times {10^{ - 2}}\;dyne$
B. $101.37 \times {10^{ - 5}}\;dyne$
C. $16.95 \times {10^{ - 5}}\;dyne$
D. $16.95 \times {10^{ - 4}}\;dyne$
Answer
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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,
$F = 6\pi \eta rv$
Where, $F$ is the viscous force on rain drop by air, $\eta $ viscosity of air in $poise$, $r$ is the radius of rain drop in $cm$ and $v$ is the velocity of rain drop in $cm{s^{ - 1}}$.
Complete step by step solution:
Given, The Radius of raindrop, $r = 0.3\;mm$ (or) $r = 0.03\;cm$
The Velocity of raindrop, $v = 1\;m{s^{ - 1}}$ (or) $v = 100\;cm{s^{ - 1}}$
The Viscosity of air, $\eta = 18 \times {10^{ - 5}}\;poise$
The Stoke's law is given by,
$F = 6\pi \eta rv\;...............................\left( 1 \right)$
Substitute the values of $\pi $, $\eta $, $r$ and $v$ in the equation (1),
$
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, $F \simeq 101.78 \times {10^{ - 2}}\;dyne$
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 $cm$ and $cm{s^{ - 1}}$ 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.
Formula used:
The Stoke's law is given by,
$F = 6\pi \eta rv$
Where, $F$ is the viscous force on rain drop by air, $\eta $ viscosity of air in $poise$, $r$ is the radius of rain drop in $cm$ and $v$ is the velocity of rain drop in $cm{s^{ - 1}}$.
Complete step by step solution:
Given, The Radius of raindrop, $r = 0.3\;mm$ (or) $r = 0.03\;cm$
The Velocity of raindrop, $v = 1\;m{s^{ - 1}}$ (or) $v = 100\;cm{s^{ - 1}}$
The Viscosity of air, $\eta = 18 \times {10^{ - 5}}\;poise$
The Stoke's law is given by,
$F = 6\pi \eta rv\;...............................\left( 1 \right)$
Substitute the values of $\pi $, $\eta $, $r$ and $v$ in the equation (1),
$
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, $F \simeq 101.78 \times {10^{ - 2}}\;dyne$
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 $cm$ and $cm{s^{ - 1}}$ 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.
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