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  $0.3mm$ 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 $(\mathrm{Take} \mathrm{g} =10 \mathrm{m} {\mathrm{s}}^{-2})$ 

 

Here, $\mathrm{m}=0.01 \mathrm{kg}, \mathrm{l}=0.3\mathrm{mm} =0.3\times {10}^{-3} \mathrm{m}, \mathrm{g}=10 \mathrm{m} {\mathrm{s}}^{-2}, \mathrm{v}=0.085\mathrm{m} {\mathrm{s}}^{-2}, \mathrm{A}=0.1 {\mathrm{m}}^2$

The metal block moves to right due to tension $T$ of the string which is equal to the weight of mass suspended at the end of the string .

Thus,

Shear force,$\mathrm{F}= \mathrm{T}= \mathrm{mg}=0.01\mathrm{kg} \times 10 \mathrm{m} {\mathrm{s}}^{-2} =0.1 \mathrm{N}$

Shear stress on the fluid $= \frac{\mathrm{F}}{A} =\frac{0.1 \mathrm{N}}{0.1 {\mathrm{m}}^2}$

Strain rate $=\frac{\mathrm{v}}{\mathrm{l}}=\frac{0.085 \mathrm{m} {\mathrm{s}}^{-1}}{0.3\times {10}^{-3}\mathrm{m}}$

Coefficient of viscosity, $\eta =\frac{\mathrm{Shear} \mathrm{stress} }{\mathrm{Strain} \mathrm{rate}}=\left(\frac{0.1\mathrm{N}}{0.1{\mathrm{m}}^2}\right) \times \frac{0.3\times {10}^{-3}\mathrm{m}}{0.085 \mathrm{m} {\mathrm{s}}^{-1}} =3.5\times {10}^{-3} \mathrm{Pa} \mathrm{s}$