Now the force exerted by the liquid on the plate is:

There will be backward viscous force acting on both the sides, first from liquid above the plate, and second from liquid below the plate. In this case $dx$ for one side liquid is 2 cm and for another side liquid, $dx$ is 4cm. so we have F_v=\left(\mu
A\dfrac{dv}{dx}\right)_1+\left(\mu A\dfrac{dv}{dx}\right)_2 \\\Rightarrow
F_v= 0.8 \times 0.01 \times \dfrac{1}{0.02} +0.8 \times 0.01 \times
\dfrac{1}{0.04}= 0.6N