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A liquid is flowing through a horizontal channel. The speed of flow `(v)` depends on height `(y)` from the floor as `v = v_(0)[2((y)/(h))-((y)/(h))^(2)]`. Where `h` is the height of liquid in the channel and `v_(0)` is the speed of the top layer. Coefficient of viscosity is `eta`. Then the shear stress that the liquid exerts on the floor is. ` `A. `(2etav_(0))/(h)`B. `(3etav_(0))/(h)`C. `(4etav_(0))/(h)`D. `(2etav_(0))/(3h)` |
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Answer» Correct Answer - A Shear stress is tangential force applied by the liquid on unit area of the floor: Velocity gradient `= (dv)/(dy) = (2v_(0))/(h) - (2v_(0))/(h_(2))y` At `y = 0, (dv)/(dy) = (2v_(0))/(h)` `:.` Viscous force per unit area `= eta(dv)/(dy) = (2etav_(0))/(h)` |
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