A metal block of area 0.10 m2 is connected to a 0.01 kg mass via a string that passes over a massless and frictionless 0.01 kg 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 ms−1. The coefficient of viscosity of the liquid is (Take g = 10 ms−2)

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Here, m=0.01kg,l=0.3mm=0.3×10−3m,g=10ms−2,v=0.085ms−1,A=0.1m2.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×10ms−2=0.1NShear stress on the fluid = AF​= 0.1m20.1N​Strain rate = lv​=0.3×10−3m0.085ms−1​Coefficient of viscosity, η= StrainrateShear stress​=(0.1 m20.1N​)×0.085ms−10.3×10−3m​=3.5×10−3Pas

Question Text	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 0.01 kg 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 $m s^{- 2}$ )
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