A) 0.1
B) 0.05
C) 0.005
D) 0.0005
Correct Answer: D
Solution :
Given: Coefficient of viscosity \[\eta =0.01\]poise \[={{10}^{-2}}poise={{10}^{-3}}kg/ms\] Viscous force \[=0.002N=2\times {{10}^{-3}}N\] \[(\because \,1\,poise\,=\frac{1}{10}\,kg/ms)\] Length of square plate =0.1 m Area of square plate \[A=0.1\times 0.1={{10}^{-2}}{{m}^{2}}\] Velocity \[=0.1m/s={{10}^{-1}}m/s\] From relation, the viscous force is given by \[F=\eta A\frac{dv}{dx}\] or \[dx=\frac{\eta Av}{F}\] \[=\frac{{{10}^{-3}}\times {{10}^{-2}}\times {{10}^{-1}}}{2\times {{10}^{-3}}}=\frac{{{10}^{-3}}}{2}\] \[=0.5\times {{10}^{-3}}=0.0005\,m\]
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