A metal plate of area `0.10m^(2)` is connected to a `0.01 kg` mass via a string that passes over an ideal pulley (considered to be friction-less), as shown in the figure. A liquid with a film thickness of `3.0 mm` is placed between the plate and the table. When released the plate moves to the right with a constant speed of `0.085ms^(-1)`. Find the coefficient of viscosity of the liquid.

Explanation

Here, `A= 0.10 m^(2) , m=0.01 kg`
`l= 0.3 xx 10^(-3) m = upsilon = 0.085 ms^(-1)`
The metal plate will move towards right due to tension T in the string, which is equal to the weight of the mass suspended at the end of string . If F is the viscous force involved, and there is no ac celeration of the mass, then
`F = T = mg = 0.01 xx 9.80 = 9.8 xx 10^(-2)N`
Assuming that the velocity gradient is uniform, then `eta = (Fl)/(upsilon A) = (9.8 xx 10^(-2) xx 0.3 xx 10^(-3))/(0.085 xx (0.10))`
`= 3.45 xx 10^(-3) Pa-s` .