A plate of area `10m^(2)` is made to move horizontally with a speed of `0.5 ms^(-1)` by applying a force over the surface a liquid. If the depth of the liquid is 2.5m, coefficient of viscosity of liquid is `10^(-3)Nsm^(-2)`, then calculate the horizontal force needed to move the plate.

Explanation

Given `Deltav=0.5 ms^(-1)`
`Deltax=2.5m`
velocity gradient `=(Deltav)/(Deltax)=(0.5)/(2.5)=0.2(ms^(-1))/(m)`

We know that |F| `=etaA((Deltav)/(Deltax))=10^(-3) xx 10^(2) xx 10^(-2)=10^(-3)N`
Hence `10^(-3)N` of force is needed to be applied, to keep the plate moving.