
as force on the plate due to viscosity is from upper as well as lower portion of the oil, equal from each part,
Then, $\mathrm{F}=2 \eta \mathrm{A} \frac{\mathrm{dv}}{\mathrm{dz}}=2 \times \eta \times(0.5) \frac{0.5}{1.25 \times 10^{-2}}$
$\Rightarrow \eta=2.5 \times 10^{-2} \mathrm{kg}-\mathrm{sec} / \mathrm{m}^{2}$





