Question

A metal block of Film 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 speed of $ 0.085 \,m\, s^{-1} $ . The coefficient of viscosity of the liquid is (Take $ g = 10\, m \,s^{-2} $ )

• A. $ 2.5 \times 10^{-3} Pa\,s $
• B. $ 3.5 \times 10^{-3} Pa\,s $
• C. $ 4.5 \times 10^{-3} Pa\,s $
• D. $ 6.5 \times 10^{-3} Pa\,s $

Answer 1

Answer: (B)

Here, $ m = 0.01 \,kg, l= 0.3\, mm = 0.3 \times 10^{-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 \times 10 \, m \, s^{-2} = 0.1 \, N $
Shear stress on the fluid = $ \frac{F}{A}=\frac{0.1\,N}{0.1\,m^{2}} $
Strain rate $ =\frac{v}{l}=\frac{0.085\,m\,s^{-1}}{0.3\times10^{-3}\,m} $
Coefficient of viscosity, $ \eta=\frac{\text{Shear stress}}{\text{Strain rate}} $
$ =\left(\frac{0.1\,N}{0.1\, m^{2}}\right) \times \frac{\left(0.3\times10^{-3}\,m\right)}{0.085\, m \, s^{-1}} $
$ =3.5\times10^{-3} \, Pa\,s $