Question

A metal block of Film area $0.10m^2$ is connected to a $0.01kg$ 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 $0.085m{s}^{-1}$ . The coefficient of viscosity of the liquid is (Take $g=10m{s}^{-2}$ )

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

Answer 1

Answer: (B)

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