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.

Determine the velocity `v_(B)` of the water issuing out at `B`.

Options

A. `v_(B)=sqrt(2gh_(2))`
B. `v_(B)=sqrt(2h(h_(1)+h_(2))`
C. `v_(B)=sqrt(2h(h_(2)-h_(1))`
D. `v_(B)=sqrt(gh_(2))`

Answer

A

Explanation

As long as water fills the tube (as shown in the figure) and points `A` and `B` are open to the atmosphere, the velocity at `B` will be given by Torricelli's theorem.

Hence, `v_(B)=sqrt(2gh_(2))` where `h_(2)` is the difference in the levels `A` and `B`.