A metal block of base area $0.2\; m ^{2}$ is connected to a $0.02\; kg$ mass via a string that passes over an ideal pulley as shown in figure. A liquid film of thickness $0.6\; mm$ is placed between the block and the table. When released the block moves to the right with a constant speed of $0.17\; m / s$. The co-efficient of viscosity of the liquid is
NEET 2017, Medium
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Shear force, $F=T=m g=0.01\; kg \times 10\; m s ^{-2}=0.1\; N$
Shear stress on the fluid $=\frac{F}{A}=\frac{0.1\; N }{0.1 \;m ^{2}}$
Strain rate $=\frac{v}{l}=\frac{0.085 \;m {s}^{-1}}{0.3 \times 10^{-3}\; m }$
Coefficient of viscosity, $\eta=\frac{\text { Shear stress }}{\text { Strain rate }}$
$=\left(\frac{0.1\; N }{0.1 \;m ^{2}}\right) \times \frac{\left(0.3 \times 10^{-3}\; m \right)}{0.085\; ms ^{-1}}$
$=3.5 \times 10^{-3} Pa\cdot s$
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