A metal plate of area `0.10m^(2)` is connected to a `0.01 kg` mass via a string that passes over an ideal pulley (considered to be friction-less), as shown in the figure. A liquid with a film thickness of `3.0 mm` is placed between the plate and the table. When released the plate moves to the right with a constant speed of `0.085ms^(-1)`. Find the coefficient of viscosity of the liquid.

Explanation

The metal block moves to the right because of the tension in the string. The tenstion T is equal in magnitude to the weight of the suspended mass m. Thus, the shear force F is
`F=T=mg=0.010kgxx9.8ms^(-2)=9.8xx10^(-2)N`
Shear stress on the fluid `=F//A=(9.8xx10^(-2))/(0.10)N//m^(2)`
straight rate `=(v)/(l)=(0.085)/(0.30xx10^(-3))`
`eta=("Stress")/("Strain rate")s^(-1)`
`=((9.8xx10^(-2))N)(0.30xx10^(3)m)/((0.085ms^(-1))(0.10m^(2)))`
`=3.46xx10^(3)Pa` s