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Question

A square plate of 0.1 m side moves parallel to a second plate with a relative 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 m is

A
0.1
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B
0.05
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C
0.005
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D
0.0005
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Solution

The correct option is D 0.0005
A=(0.1)2=0.01 m2η=0.01 Poise = 0.001 kg/m-s
dv=0.1 m/s and F=0.002 N
According to Newton's law of viscosity,
F=ηAdvdx
dx=ηAdvF =0.001×0.01×0.10.002 =0.0005 m

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