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A closed tube is filled with `AB=2m` `BC=4cm` water `(rho=10^(3)kg//m^(3))`. It rotating about an axis shown in figure with an angular velocity `omega=2 rad//s`. Find, `p_(A)-p_(C)`. |
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Answer» Correct Answer - A::B::C::D Pressure decreases in moving towards the axis of rotation and increases in moving away from the axis `(Delta p= +-(rho omega^(2)x^2)/(2))` ` :. p_(A)gtp_(B)` and `p_(B)gtp_(C)` `p_(A)-p_(C)=(p_(A)-p_(B))+(p_(B)-p_(C))` `=(+(rho omega^(2)x_(1)^(2))/(2))+((-rho omega^(2)x_(2)^(2))/(2))` Here, `x_(1)=AB=2m` and `x_(2)=BC=4m` Substituting teh values we have, `p_(A)-p_(C)=((10^(3))(2)^(2)(2)^(2))/(2)-((10^(3))(2)^(2)(4)^(2))/(2)` `=-2.4xx10^(3) N//m^(2)`. |
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