Two square metal plates, each of side 10 cm are immersed in water. One plate moves parallel to the other with a velocity of 5 cm `s^(-1)`. If the viscous force is 150 dyne, what is their distance of separation ? Given `eta_("water")` = .001 PI

Explanation

Area of the plates, A = 10 `xx` 10
= 100 `cm^(2)`
= 0.01 `m^(2)`
Viscous force = 150 dyne
= 150 `xx 10^(-5)` N
`eta = 0.001` PI
Relative velocity, v = 5 cm `s^(-1)`
= 0.05 m `s^(-1)`
Let the separation between the plates be l,
then viscous force, `F = eta A(v)/(l)`
`150 xx 10^(-5) N = (0.001 xx 0.01 xx 0.05)/(l)`
`rArr l = (0.05)/(150)`m
`=(0.01)/(30)`m
= 0.033 cm