Question

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

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

Answer 1

The correct option is D $0.0005$

$A=(0.1{)}^2=0.01{\text{m}}^2\eta =0.01\text{Poise =}0.001\text{kg/m-s}$
$dv=0.1\text{m/s}$ and $F=0.002\text{N}$
According to Newton's law of viscosity,
$F=\eta A\frac{dv}{dx}$
$\therefore dx=\frac{\eta Adv}{F}=\frac{0.001\times 0.01\times 0.1}{0.002}=0.0005\text{m}$