In this how they have written the equation .

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Applying the equation of continuity, $(2A)v=2(A)u$ - ie the velocities $v,u$ in the vertical and horizontal sections are equal.

The total volume of liquid is $V=(2A)H+2(AH)=4AH$ so the total mass is $M=\rho V$. The initial height of liquid in the vertical tube is $H$. When this height has fallen by $x$ a volume of $2Ax$ has moved into the horizontal sections. The height of the centre of this section of liquid above the horizontal section is $H-\frac12x$. The loss of GPE is therefore $mg(H-\frac12x)$ where $m=2Ax\rho $.

The whole of the fluid is moving with speed $v$ after this liquid falls. The loss in GPE equals the gain in the KE of the fluid so

$(2Ax\rho)g(H-\frac12x)=\frac12Mv^2=\frac12(4AH\rho)v^2$

$x(H-\frac12x)g=Hv^2$.

When $x=\frac12H$ then

$Hv^2=\frac12H(H-\frac14H)g$

$v=\sqrt{\frac38gH}$.

Differentiating wrt $x$ we get

$(H-\frac12x)g+x(-\frac12)g=(H-x)g=2Hv\frac{dv}{dx}=2H\frac{dx}{dt}\frac{dv}{dx}=2Ha$

where $a=\frac{dv}{dt}$ is acceleration.

When $x=0$ the initial acceleration of the liquid is $a=\frac12g$.

The mass of liquid to the right of P is $m'=\rho A\frac{H}{2}$. The initial acceleration of this liquid is $a=\frac12g$ so the resultant force on it is $m'a=\frac14\rho AHg$. The force on this liquid comes from the difference in pressure at the two ends, which is the gauge pressure $P$ since the right side of this liquid is at atmospheric pressure. Therefore

$PA= \frac14\rho AHg$

$P=\frac14\rho Hg$.

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I thought that we can only apply it for two points .