A metal block of area `0.10m^(2)` is connected to a 0.010 kg mass via a string that passes over an ideal pulley (cosidered massless and frictionless). As in figure. A liquid with a film thickness of 0.30 mm is placed b etween the block and the table. when released the block moves to the right with a constant speed of `0.085ms^(-1)`. find the coefficient of viscosity of the liquid.

Explanation

As the metal plate moves with constant velocity.
`mg-T=0impliesT=mg`
If `F` is the vicous force on the plate then
`F=T=mg=0.01xx9.8=9.8xx10^(-2)N`
By Newton's law of viscosity,
`F=etaA(dv)/(dy)`
where `(dv)/(dy)=0.085/(0.3xx10^(-3))=28.3s`
`implies 9.8xx10^(-2)=etaxx00.10xx28.3`
`implies eta=3.8xx10^(-3)Ns//m^(2)`