A metal block of area `0.10 m^2` is connected to a 0.010 kg mass via a string that passes over an ideal pulley(considered massless and frictionless) as in Fig 10.15 A liquid with a film thickness of 0.30 mm is placed between the block and the table . When released the block moves to the right with a constant speed of `0.085 ms^-1`. FInd the coefficienyt of viscosity of the liquid.

Explanation

The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m. Thus the shear force F is `F=T=mg=0.010 kg times 9.8 ms^-2=9.8 times 10^-2 N`
Shear stress on the fluid =`F//A=(9.8 times 10^-2)/0.10 N//m^2`
Strain rate `=v/l=0.08/(0.30 times 10^-3)`
`n=("stress")/("strainrate") s^-1`
`=((9.8 times 10^-2 N)(0.30 times 10^-3 m))/((0.085 ms^-1)(0.10 m^2))`
`=3.46 times 10^-3 Pas`