The velocity distribution for the flow of a Newtonian fluid between two wide, parallel plates is given by the equation

`u=(3V)/2[1-(y/h)^(2)]`
where V is the mean velocity. The fluid has coefficient of viscosity `eta` Answer the following 3 questions for this situation.
Shear stress acting on the bottom wall is

Options

A. `tau_("bottom wall")=eta((3V)/h)`
B. `tau_("bottom wall")=eta((3V)/(2h))`
C. `tau_("bottom wall")=eta((6V)/h)`
D. `tau_("bottom wall")=eta((V)/h)`

Explanation

For this type of parallel flow the shearing stress is given as `ta=eta(du)/(dy)....(i)`
For the given distribution
`(du)/(dy)=-(3Vy)/(h^(2))....(ii)`
(a) Along the bottom wall so that (from eq. (ii))
`(du)/(dy)=(3V)/h` and therefore the shearing stress is `tau_("bottom wall")=eta((3V)/h)`
(b) Along the plane where `y=h//2` it follows from equation (ii) that
`(du)/(dy)=(-3Vy)/(h^(2))` and thus the shearing stress is `|tau|=eta(3V)/(2h)`.
(C)
Rate of volume flow
`2int_(0)^(h) (3V)/2(1-(y^(2))/(h^(2)))dy.l=2Vlh`