Question

A square plate 0.1 m side moves parallel to second plate with a velocity of 0.1 $m{s}^{-1}$, both plates being immersed in water. If the viscous force is 0.002 N and the coefficient of viscosity 0.001 poise, distance between the plates is

• A. 0.1 m
• B. 0.05 m
• C. 0.005 m
• D. 0.0005 m

Answer 1

The correct option is D 0.0005 m

Given data,

$Area=a^2=1\times 0.1=0.1m^2$

$\eta =0.01poise$

$\eta =\frac{0.01}{10}=0.001decapoise$

$dv=0.1m/s$

$F=0.002N$

We know that

As $F=\eta A\frac{dv}{dx}$

or $dx=\eta \frac{Adv}{F}$

$=\frac{0.001\times (0.1{)}^2\times 0.1}{0.002}=0.0005m$