A horizontal force of 1 N is required to move a metal plate of area [tex]\sf 10^{-2}[/tex] m² with a velocity of [tex]\sf 2 \times 10^{-2}[/tex] m/s, when it rests on a layer of oil [tex]\sf 1.5 \times 10^{-3}[/tex] m thick. Find the coefficient of viscosity of oil.
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Answer
Given
To Find
Formula Used
[tex]\bf F=\eta A\dfrac{dv}{dx}\\[/tex]
where ,
F denotes force
η denotes coefficient of viscosity
A denotes area
dv denotes velocity
dx denotes thickness
Solution
[tex]\bf F=\eta A\dfrac{dv}{dx}\\\\\implies 1=\eta \times 10^{-2}\times \dfrac{2 \times 10^{-2}}{1.5 \times 10^{-3}}\\\\ \implies \bf \eta = \dfrac{1.5 \times 10^{-3}}{2 \times 10^{-4}}\\\\\implies \bf \eta = \dfrac{1.5 \times 10}{2}\\\\\implies \bf \eta =\dfrac{15}{2}\\\\\implies \bf \eta =7.5[/tex]
So , Coefficient of viscosity of oil , η = 7.5
GiveN :
To FinD :
SolutioN :
Use Newton's Law of Viscosity :
[tex]\longrightarrow {\boxed{\sf{F \: = \: \eta A \dfrac{dv}{dx}}}} \\ \\ \longrightarrow \sf{1 \: = \: \eta \: \times \: 10^{-2} \bigg\lgroup \dfrac{2 \: \times \: 10^{-2}}{1.5 \: \times \: 10^{-3}} \bigg\rgroup} \\ \\ \longrightarrow \sf{1 \: = \: \eta \: \times \: 10^{-2} \bigg( 1.33 \: \times \: 10^{-2 \: + \: 3} \bigg)} \\ \\ \longrightarrow \sf{1 \: = \: \eta \: \times \: 10^{-2} \: \times \: 1.33 \: \times \: 10^1} \\ \\ \longrightarrow \sf{1 \: = \: \eta \: \times \: 1.33 \: \times \: 10^{-2 \: + \: 1}} \\ \\ \longrightarrow \sf{1 \: = \: \eta \: \times \: 1.33 \: \times \: 10^{-1}} \\ \\ \longrightarrow \sf{\eta \: = \: \dfrac{1}{1.33 \: \times \: 10^{-1}}} \\ \\ \longrightarrow \sf{\eta \: = \: 0.75 \: \times \: 10} \\ \\ \longrightarrow \sf{\eta \: = \: 7.5 \: poise \: (approx.)}[/tex]