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A liquid flows through a horizontal tube. The velocities of the liquid in the two sections, which have areas of cross section `A_(1)` and `A_(2)` are `v_(1)` and `v_(2)` respectively. The difference in the levels of the liquid in the two vertical tubes is `h`. Then A. the volume of the liquid flowing through the tube in unit time is `A_(1)v_(1)`B. `v_(2)-v_(1)=sqrt(2gh)`C. `v_(2)^(2)-v_(1)^(2)=2gh`D. the energy per unit mass of the liquid is the same in both sections of the tube |
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Answer» Correct Answer - A::C::D `(P_(1))/(rho)+(v_(1)^(2))/2=(P_(2))/(rho)+(v_(2)^(2))/2` `P_(1)-P_(2)=(rho)/2(v_(2)^(2)-v_(1)^(2))` But `P_(1)-P_(2)=rhogh=(rho)/2(v_(2)^(2)-v_(1)^(2))` or `v_(2)^(2)-v_(1)^(2)=2gh` |
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