Find the speed of the water coming out of the container in the given figure when pressure at $P_0$ is $1\ \ \mathrm{atm}$.

enter image description here

Solution in the book :-

By equation of continuity,
$Av_0 = av$ where $A, v_0$ are the volume and area at top of the tank.

$v_0 = av/A$

Since $A << a \ \ \therefore v_0 \approx 0$

$\color{#A28}{P_0 = P ,\text{Because both are exposed to air}}$

By Bernoulli's equation,

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

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

$v = \sqrt{2gh}$

In my attempt, I did not take $P = P_0$ instead I took $P = P_0 + \rho_{water} h (0 - -h) = P_0 + \rho gh$, for which I got different answer, which is wrong I know for sure.

I did not get the part in purple, I am heavily confused why we have to take those two pressure equal just because the things are exposed to air.

  • With height the pressure increases.
  • The water inside is pushing water near hole out so that pressure should also be taken in account.

Why we neglected these two factors is beyond my understanding, please help.


The pressure inside the container, far from the hole, is $P_0+\rho g h$. But the pressure at the hole is atmospheric, because (as the solution says) the water here is exposed to the atmosphere. The pressure in the liquid changes from $P_0+\rho g h$ in the region surrounding the hole to $P_0$ at the hole. If the hole is small, this region of change is small.

If there were not such a pressure difference the water would not be pushed sideways out of the hole.

  • $\begingroup$ So if the hole was big (say 10 cm or m ) the pressure at the hole will be $P_0+\rho g h$ $\endgroup$ – A---B Jan 13 '17 at 21:23
  • $\begingroup$ No. Whatever the size of the hole, the pressure exactly at the hole is $P_0$. But if the hole is small, the pressure will be $P_0+\rho gh$ quite close to the hole, whereas if the hole is big you would have to go further into the liquid, further from the hole, to find a point where the pressure is equal to $P_0+\rho gh$. $\endgroup$ – sammy gerbil Jan 13 '17 at 21:27
  • $\begingroup$ ahh, I got it. The pressure just inside the hole is $\displaystyle P_0 + \rho_{water} gh$ and just outside hole is $P_0$ and pressure difference is $\displaystyle\rho_{water} gh$. right ? $\endgroup$ – A---B Jan 13 '17 at 21:30
  • $\begingroup$ Yes, that is it. $\endgroup$ – sammy gerbil Jan 13 '17 at 21:33

Pressure of the atmosphere at the top and at the small hole is the same, so we dont need to factor this in the calculation of the speed. Atmospheric pressure is not significantly different at these two heights. Second objection, I dont see where did we neglect the pressure of the water at the exit hole? enter image description here

  • $\begingroup$ We neglected the water pressure because we only took the pressure of the atmosphere. $\endgroup$ – A---B Jan 13 '17 at 21:25
  • $\begingroup$ ? no we did not take the pressure of the atmosphere, it would be in the final result in some way if we did. $\endgroup$ – Žarko Tomičić Jan 13 '17 at 21:35
  • $\begingroup$ Anyway I hope the answer is clear enough I even drew it.... :-) $\endgroup$ – Žarko Tomičić Jan 13 '17 at 21:37
  • $\begingroup$ Yes you are correct, thanks for clearing it, I upvoted your answer. Also can I just use energy conservation instead of Bernoulli's horrendous equation ? $\endgroup$ – A---B Jan 13 '17 at 21:39
  • 1
    $\begingroup$ It is the same thing so...you must I guess...and the equation is actually simple enough in this case. Instead of using mass you use densitiy...so it is actually energy per unit volume. Other than that if you check it out its he same thing... $\endgroup$ – Žarko Tomičić Jan 13 '17 at 21:42

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