A metal plate of area 0.10 `m^(2)` is connected to a 0.01 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), A liquid with a film thickness of 0.3 mm is placed between the plate and the table. When released the plate moves to the right with a constant speed of 0.085 `ms^(-1)`. Find the coefficient of viscosity of the liquid.

Explanation

Here `A=0.10m^(2),m=0.01kg`
`dx=0.3mm=0.3xx10^(-3)m,dv=0.085ms^(-1)`
The metal plate moves towards right du to the tension T in the string which is equal to the weight of the suspended mass m. Assuming that the mass m moves with uniform velocity or zero acceleration, then the force of viscosity will be,
`F=T=mg=0.01xx9.8`
`=9.8xx10^(-2)N`
Taking velocity gradient to be uniform, then
`eta=(F)/(A).(dx)/(dv)=(9.8xx10^(-2)xx0.3xx10^(-3))/(0.10xx0.085)`
`=3.45xx10^(-3)PaS`