Question

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

Case I:
Drag force $F_1+F_2$
$=[{\eta }_0\frac{v}{h/2}+{\eta }_0\frac{v}{h/2}]A=\frac{4{\eta }_{0}vA}{h}$
Case II:
Drag force on the plate
$[\eta \frac{v}{h-y}+\eta \frac{v}{y}]A=\frac{\eta vha}{y(h-y)}$
In both case, drag force are equal
$\Rightarrow \frac{4{\eta }_{0}vA}{h}=\frac{\eta vha}{y(h-y)}$
$\Rightarrow \eta =\frac{4{\eta }_{0}y}{h}(1-\frac{y}{h})$.