>A Venturi tube without manometer is given with initial area $A$ and area in its throat as $a$, initial pressure as $P_1 = 3 \ \ \mathrm{atm}$ and $A = 5a$, Find velocity of water as it enter the tube.

by equation of continuity :-

$$Av_1 = av_2 \implies 5av_1 = av_2 \implies 5v_1 = v_2 \tag{1}$$

Since $\Delta F = 0$

Thus, $$P \propto \frac1A \implies P_1 A = P_2a \implies P_1\times 5a = P_2a \implies 5P_1 = P_2 \tag{2}$$

By Bernoulli's principle :-

$$\Delta P + \rho g\Delta h + \frac12\rho\Delta v^2 = 0$$

$$(P_2 - P_1) + \frac12\rho(v_2^2 - v_1^2) = 0$$

From 1 and 2,

$$(5P_1 - P_1) + \frac12\rho(25v_1^2 - v_1^2) = 0$$

$$(P_1) + \rho(3v_1^2) = 0$$

$$\sqrt{-\frac{P_1} {3\rho }} = v_1$$

Now this is impossible and I am a idiot to get to this result but even after staring at this for, I can't get spot the flaw. Surely some of you can do that.

I am grateful for any help .