A thin plate of large area is placed midway in a gap of height `h` filled with oil of viscosity and the plate is pulled at constant velocity `v `by applying the same drag force on the plate. If a lighter oil of viscosity `eta` is then substituted in the gap. it is found that for the velocity `v`, and the same drag force as previous case the plate is located unsymmetrically in the gap but parallel to the walls. Find `eta` in terms of distance from nearer wall to the plane `y`.

Answer

`(1-y/h)`

Explanation

Case I:
Drag force `F_(1) +F_(2)`
`[eta_(0)v/(h//2)+eta_(0)v/(h//2)]A`

Case II:
Drag force on the plane
`=[eta-v/(h-y)+etav/y]A=(etavha)/(y(h-y))`
In both cases drag force are equal
`implies(4eta_(0)vA)/h=(etavha)/(y(h-a))`
`implies eta=(4eta_(0))/h(1-y/h)`