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.

A metal block of area 0.10 m^2 is connected to a 0.010 kg mass via a s

1.	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.
Answer» Solution :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`