GREPhysics.NET: GR9677 #30

GR9677 #30

Problem

GREPhysics.NET Official Solution

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Advanced Topics $\Rightarrow$ }Fluid Mechanics

Equipotential leads to equipressure in a fluid. Thus, take the pressure at the base of the dark fluid and set it equal to the pressure (of the lighter-colored fluid) at a horizontal-line across on the right-hand side of the U:

$P_{dark}=\rho_4 g (5) = P_{light}= \rho_1 g (h_2 - (h_1-5))=\rho_1 g (h_2-h_1+5)$ . The initial total height of the columns is 40, thus after the darker liquid is added, the total height is 45. Plug $h_1+h_2=45$ into the equation above to get $h_1=15$ , $h_2=30$ , and therefore $h_2/h_1=2/1$ , as in choice (C).

(Ah, one should remember that the fluid pressure at a point is due to all the water on top of it, thus $P=\rho g h$ , where $h$ is the height of the water on top of the point.)

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Comments

phys2718
2008-10-06 13:44:57

This problem doesn't make sense physically. When you add the denser liquid to the water it will form at the bottom of the tube rather than the top. Just try adding water to oil.

HaveSpaceSuit
2008-10-16 19:27:18

The question says that the liquid is immiscible, so it can't mix with the water. Thus, it just acts as a weight on the top of the water.

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hoyas08
2008-06-22 16:53:38

I think one can also do this problem conceptually with only a bit of math. Since the liquid is 4 times denser than water, adding 5 cm of liquid to the left must be balanced by having 20 cm more water on the right side than on the left side at final equilibrium. This is the case when you push 10 cm of water from the left side over to the right, giving you:

$h_1$ = 20 cm - 10 cm + 5 cm = 15 cm
$h_2$ = 20 cm + 10 cm = 30 cm

$\frac{h_2}{h_1}$ = $\frac{2}{1}$

Saved me some algebra, at least.

cpstrehlow
2008-11-06 16:02:56

I solved this problem by equating 5 cm of immiscible liquid (four times as dense as water), with 20 cm of water. adding twenty cm of water to one side gives 40 cm on one side and 20 on the other. to balance out, there must be 30 cm on each side. knowing that h1 must be 15, it takes no algebra to see that the ratio is 2.

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grae313
2007-10-31 12:14:00

You can also equate the total pressures on each side using $P = \rho gh$ (this is exactly the same as what you did except this uses the base of the tube as the equipotential line, I found this more intuitive)

$(4)g(5) + (1)g(h_1 - 5) = (1)g(h_2)$

and then use the face that $h_1 + h_2 = 45$ to arrive at the same answer

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antithesis
2007-10-04 15:20:32

Technically, there is also the pressure of the air, but both sides are open, so it cancels on both sides.

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You are replying to:

You can also equate the total pressures on each side using $P = \rho gh$ (this is exactly the same as what you did except this uses the base of the tube as the equipotential line, I found this more intuitive) $(4)g(5) + (1)g(h_1 - 5) = (1)g(h_2)$ and then use the face that $h_1 + h_2 = 45$ to arrive at the same answer

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