A tube of uniform cross-section is used to siphon water from a vessel, as shown in the figure. Assuming that the cross – section of the vessel is very much greater than that of the tube, find the maximum value of height  $h_1$ (in m), for which the siphon will work. [Take $h_2=3m,$ atmospheric pressure  $P_{atm}={10}^{5}N{m}^{-2},$ density of water $\rho =1000kg-m^{-3}$ and acceleration due to gravity  $g=10m{s}^{-2}$ ]

 ${\text{P}}_{\text{A}}{\text{=1×10}}^{\text{5}}{\text{Nm}}^{\text{-2}}{\text{,h}}_{\text{2}}\text{=3m}$
Let p be the density of the water and ${\text{P}}_{\text{atm}}$  be the atmospheric pressure at A and B openings. Applying Bernoulli’s theorem for stream line motion,
${\text{P}}_{\text{A}}{\text{=P}}_{\text{1}}\text{+}\frac{\text{1}}{\text{2}}{\text{ρv}}_{\text{1}}^{\text{2}}{\text{+ρgh}}_{\text{1}}\text{,}$  where ${\text{p}}_{\text{1}}{\text{,v}}_{\text{1}}$  are pressure and velocity at point C, at a height ${\text{h}}_1$  below the topmost point on the tube,
Since the tube is of uniform cross-section and the liquid is incompressible,
Flow speed at C = flow speed at B 
From Torricelli’s theorem between A and B,
 $$\begin{gathered} {\text{V}}_{\text{B}}\text{=}\sqrt{{\text{2gh}}_{\text{2}}}{\text{=v}}_{\text{1}} \\ \therefore {\text{P}}_1-P_A-\frac{1}{2}\rho \times 2g{h}_2-\rho g{h}_1 \\ -P_A-pg(h_1+h_g) \end{gathered}$$
 Now pressure at C should be greater than or equal to zero (for negative pressure, water could not reach the )
For maximum ${\text{h}}_{\text{1}}{\text{,p}}_{\text{1}}$  should be minimum, equal to zero.
 $\therefore {\text{h}}_{\text{1}}\text{=}\frac{{\text{P}}_{\text{A}}}{\rho \text{g}}\text{-}{h}_2\text{=}\frac{{\text{1×10}}^{\text{5}}}{{\text{10}}^{\text{3}}\text{×10}}\text{-3=7m}$