Question

A thin plate $AB$of large area $A$ is placed symmetrically in a small gap of height h filled with water of viscosity ${\eta }_0$, and the plate has a constant velocity $v$ by applying a force F as shown in the figure.
If the gap is filled with some other liquid of viscosity $0.75{\eta }_0$,
the minimum distance (in cm) from top wall that the plate should be placed in the gap, so that the plate can again be pulled with the same constant velocity $v$ by applying the same force F is (Take
$h=20cm$) (Answer upto two digit after the decimal point)

Answer 1

case 1: plate is placed symmetrically
$\frac{\partial u}{\partial y}=\frac{v}{\frac{h}{2}}=\frac{2v}{h}$
$viscousforce=\frac{{\eta }_{0}2v}{h}A$
since the plate has two sides, the total force
$F_1=\frac{4{\eta }_{0}vA}{h}$
case 2: let the plate is at y distance from top
$F_2=\frac{\eta vA}{y}+\frac{\eta vA}{h-y}=\frac{\eta vAh}{y(h-y)}{F}_1=F_2\Rightarrow \frac{\eta vAh}{y(h-y)}=\frac{4{\eta }_{0}Av}{h}$
$\frac{3}{16}{h}^2=y(h-y)$solving, we get
$y=\frac{h}{4},\frac{3h}{4}$Hence, for minimum distance from top, $y=\frac{20}{4}=5cm$