InterviewSolution
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InterviewSolution
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A metal block of area 0.10 `m^(2)` is connected to a 0.02 kg mass via a string. The string passes over an ideal pulley (considered massless and frictionless) as shown in figure. A liquid with a flim of thickness 0.15 mm is placed between the plate and the table. When released the plate movees to the right with a constant speed of 0.075 m `s^(-1)`. What is the coefficient of viscosity of the liquid? |
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Answer» Area of the film, A = 0.10 `m^(2)` Thickness of the film = 0.15 mm x = 0.25 `xx 10^(-3)`m Relative velocity of plate v = 0.075 m `s^(-1)` Since the plate moves with constant speed, the tension developed in the string is equal to the viscous force F. But tension is equal to the weight of suspended mass m. `therefore` Viscous force F = weight of mass = 0.02 `xx` 9.8 = 0.196 N `F = eta (Av)/(x)` `rArr eta = (Fx)/(Av)` `=(0.196 xx 0.15 xx 10^(-3))/(0.10 xx 0.075)` `=(0.0196)/(5)` = 0.00392 Pa s |
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