A pressurizing question of pressure

Question

faiqraees · Answer

attachment

faiqraees · Answer

@ganeshie8

faiqraees · Answer

@phi

irishboy123 · Answer

i'd go B

using the idea of **heads of pressure**

so from Bernoulli's Eqn, $ P + \frac{1}{2}\rho v^2 + \rho gh = const$

and here $v = 0$ so we can say, using $\tilde p = \dfrac{P}{\rho g}$, that $ \tilde p + h = const$ 

so for instance $ \tilde p_a = 760 ~~ mm ~ ~ Hg $

only B matches poss $(x,y, \tilde  p)$ combo's

attaching a drawing

faiqraees · Answer

$\color{#0cbb34}{\text{Originally Posted by}}$ @IrishBoy123
i'd go B

using the idea of **heads of pressure**

so from Bernoulli's Eqn, $P + \frac{1}{2}\rho v^2 + \rho gh = const$

and here $v = 0$ so we can say, using $\tilde p = \dfrac{P}{\rho g}$, that $\tilde p + h = const$ 

so for instance $\tilde p_a = 760 ~~ mm ~ ~ Hg$

only B matches poss $(x,y, \tilde  p)$ combo's

attaching a drawing
$\color{#0cbb34}{\text{End of Quote}}$
Can you justify your value of the variable of Pa? Why is it 760? Does any other value make any difference?

faiqraees · Answer

And can you also explain how did you get the equation for pressure for both sides?

jhonyy9 · Answer

i agree the @IrishBoy123 answer

faiqraees · Answer

I know his answer is correct I am asking how did he come up to it

irishboy123 · Answer

A) we assumed that $\tilde p$, the pressure in the air, is constant.  the question tells you that it is P and $P \to \tilde p$ under the approach i suggested.  

[BTW not trying to confuse things but it's not actually constant. it should be higher at the lower end, **but** air is a gas and really not very dense compared to mercury, so it's a decent assumption to make.  but as i say, if that is confusing, ignore it and just assume that the pressure in the air void is everywhere the same.]

that assumption allows us to apply Bernoulli's. as mentioned, throughout the continua of mercury on left and right.  in each case the boundary pressures are $\tilde p$ and $\tilde p_a$.


B) in terms of the numbers, we are told in the question that $ \tilde p_a = 760 ~~ mm ~ ~ Hg $

however, we end up with the equation $x = 50-y$ and the $\tilde p_a$'s cancel out.  so you would got the same x and y for any external air pressure, which makes sense

but $\tilde p_a$ only relevant when you want to calculate $\tilde p$ itself as $\tilde p = \tilde p_a + x$.

hope that makes it clearer......