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A tube of uniform cross section is used to siphon water from a vessel `V` as shown in the figure. The pressure over the open end of water in the vessel is atmospheric pressure `(P_(0))`. The height of the tube above and below the water level in the vessel are `h_(1)` and `h_(2)`, respectively. Given `h_(1) = h_(2) = 3.0 m`, the gauge pressure of water in the highest level `CD` of the tube will beA. `3.0xx10^(4)N//m^(2)`B. `5.9xx10^(4)N//m^(2)`C. `5.9xx10^(4)N//m^(2)`D. `1.5xx10^(4)N//m^(2)` |
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Answer» Correct Answer - B Putting `h_(1)=h_(2)=3.0m` in eqn iii above `P_(1)=` pressure at level `CD` `=P_(atm)-(rhog)(3+3)=1.0xx10^(5)-6rhog` The gauge pressurre at level `CD=6rhog=6xx10^(3)xx9.8N//m^(2)` `=5.9xx10^(4)N//m^(2)` |
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