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Physics Question on Viscosity

A square plate of 0.1 m side moves parallel to a second plate with a velocity of 0.1 m/s, both plates being immersed in water. If the viscous force is 0.002 N and the coefficient of viscosity is 0.01 Poise, distance between the plates in metre is:

A

0.1

B

0.05

C

0.005

D

0.0005

Answer

0.0005

Explanation

Solution

Given: Coefficient of viscosity η=0.01\eta =0.01 poise =10−2poise=10−3kg/ms={{10}^{-2}}poise={{10}^{-3}}kg/ms Viscous force =0.002N=2×10−3N=0.002N=2\times {{10}^{-3}}N (∵ 1 poise =110 kg/ms)(\because \,1\,poise\,=\frac{1}{10}\,kg/ms) Length of square plate =0.1 m Area of square plate A=0.1×0.1=10−2m2A=0.1\times 0.1={{10}^{-2}}{{m}^{2}} Velocity =0.1m/s=10−1m/s=0.1m/s={{10}^{-1}}m/s From relation, the viscous force is given by F=ηAdvdxF=\eta A\frac{dv}{dx} or dx=ηAvFdx=\frac{\eta Av}{F} =10−3×10−2×10−12×10−3=10−32=\frac{{{10}^{-3}}\times {{10}^{-2}}\times {{10}^{-1}}}{2\times {{10}^{-3}}}=\frac{{{10}^{-3}}}{2} =0.5×10−3=0.0005 m=0.5\times {{10}^{-3}}=0.0005\,m