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.

If `h_(1)=h_(2) = 3.0 m`, the maximum value of `h_(1)`, for which the siphon will work will be

Options

A. `3.0m`
B. `6.0m`
C. `7.2m`
D. `4.8m`

Answer

C

Explanation

We apply Bernoulli's theorem for level `A` and for the highest level, `CD` (labelled by subscript `1`) to get
`P_(A)=P_(1)+1/2rhov_(1)^(2)+rhogh_(1)`…….i
Since the tune has uniform cross section and water is incompressible
`v_(1)=v_(B)=sqrt(2gh_(2))`………..ii
From eqn i and
`P_(1)=P_(A)-1/2rho[sqrt(2gh_(2))]^(2)-rhogh_(1)`
`=P_(atm)-rhog(h_(1)+h_(2))`...........iii
The minimum value of `P_(1)=0(P_(1)` cannot be negatie because then no water will reach the level). Hence putting `P_(1)=0`.
`(h_(1))_("max")=P_(atm)/(rhog)-h_(2)=(1.0xx10^(5))/(1xx10^(3)xx9.8)-3.0`
`=10.2m-3.0m=7.2m`