A metal block of area `0.10 m^(2)` is connected to a 0.01 kg mass via a string that passes over a massless and frictionless pulley as shown in figure. A liquid with a film thickness of 0.3 mm is placed between the block and the table. When released the block moves to the right with a constant of `0.085 m s^(-1)`. The coefficient of viscosity of the liquid is

Options

A. `2.5 xx 10^(-3) `Pa s
B. `3.5xx 10^(-3)` Pa s
C. `4.5xx 10^(-3)` Pa s
D. `6.5 xx10^(-3)` Pa s

Answer

B

Explanation

Here, m=0.01 kg, l=0.3 mm `=0.3xx10^(-3)` m,
`g=10 m s^(-2), v=0.085 m s^(-1), A=0.1 m^(2)`
The metal block moves to the right due to tension T of the string which is equal to the weight of the mass suspended at the end of the string.
Thus,
Shear force, `F=T=mg=0.01 kg xx 10 m s^(-2)=0.1 N`
Shear stress on the fluid`=(F)/(A)=(0.1 N)/(0.1 m^2)`
Strain rate`=(v)/(l)=(0.085 m s^(-1))/(0.3xx10^(-3))`
Coefficient of viscosity, `eta=("Shear stress")/("Strain rate")`
`=((0.1 N)/(0.1 m^(2)))xx((0.3xx10^(-3)m))/(0.085 m s^(-1))=3.5xx10^(-3) Pa s`