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 be

Options

A. `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)`

Answer

B

Explanation

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)`