A) 0.1
B) 0.05
C) 0.005
D) 0.0005
Correct Answer: D
Solution :
\[A={{(0.1)}^{2}}=0.01{{m}^{2}}\], \[\eta =0.01\,\ Poise=0.001\ \ decapoise\] (M.K.S. unit), dv = 0.1 m/s and F = 0.002 N \[F=\eta A\frac{dv}{dx}\] \[\therefore \ \ dx=\frac{\eta Adv}{F}=\frac{0.001\times 0.01\times 0.1}{0.002}=0.0005m\].
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