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A liquid of density `rho` is rotated by an angular speed `omega` as shown in figure,Using the concept of pressure equation, find a relation between `h_(1),h_(2),x_(1)` and `x_(2)`. |
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Answer» Writing pressure equation between points `A` and `B`, we have `p_(A)+rho gh_(1)-(rho omega^(2)x_(1)^(2))/(2)+(rho omega^(2)x_(2)^(2))/(2)-rho gh_(2)=p_(B)` Points `A` and `B` are open to atmosphere. Therefore, `p_(A) = p_(B) = p_(0) =` atmospheric pressure. Substituting the values in above equation we have, `rho gh_(1) - (rho omega^(2)x_(1)^(2))/(2) +(rho omega^(2) x_(2)^(2))/(2) -rho g h_(2) = 0` On simplifying we get, `(h_(2)-h_(1))=(omega^(2))/(2g)(x_(2)^(2)-x_(1)^(2))` This is the desired relation between `h_(1),h_(2),x_(1)` and `x_(2)`. |
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