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

Explanation

Area of block `A=0.10m^(2)`
Thickness of film of liquid,
`l=0.3mm=0.3xx10^(-3)m`
Speed of block `v=0.085ms^(-2)`
The coefficient of viscosity `eta=?`
Mass of sphere `m=0.01kg`
Tension in the string F=mg
`=0.01xx9.8`
`=9.8xx10^(-2)N`
Shear strain (Modulus of rigidity)
`eta=("shear stress")/("strain rate")=(FcancelA)/(vcancelt)`
`=(Fl)/(Av)`
`=(9.8xx10^(-2)xx0.3xx10^(-3))/(0.1xx0.085)`
`=345.88xx10^(-5)`
`thereforeeta=3.46xx10^(-3)Pas`
`eta=(2)/(9)xx(r^(2)g)/(v_(t))(rho-sigma)`
`=(2xx(2xx10^(-3))^(2)xx9.8(8.9xx10^(3)-1.5xx10^(3)))/(9xx6.5xx10^(-2))`
`=9.917xx10^(-6+3+2)`
`=9.9xx10^(-1)`
`=0.99kgm^(-1)s^(-1)`