Q. A square plate of $0.1\ m$ side moves parallel to a second plate with a 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
B) 0.05
C) 0.005
D) 0.0005
Explanation
(d) $A = (0.1)^{2} = 0.01m^{2}$,
$\eta = 0.01\ Poise = 0.001\ decapoise$ (M.K.S. unit),
$dv = 0.1\ m/s$ and $F = 0.002\ N$
$F = \eta A\frac{dv}{dx}$
$\therefore dx = \frac{\eta Adv}{F} = \frac{0.001 \times 0.01 \times 0.1}{0.002} = 0.0005m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.