In a U-shaped tube, water and oil are separated by a movable membrane. What is the ratio of the heights $\frac{h1}{h2}$ (density of the oil $ρ_{oil}$ = 0.92 $\frac{g}{cm^3}$)?

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I tried solving by saying that the pressure at the membrane should be the same so

$$P_{H_2O} = P_{oil}$$ $$=> P_0 + \rho_{H_2O} \cdot g \cdot h_1 = P_0 + \rho_{oil} \cdot g \cdot h_2$$ where $P_0$ is the atmospheric pressure.

then i got

$$\frac{h_1}{h_2} = \frac{\rho_1}{\rho_2} = 0.92$$

But my friend argues that the Net force on the membrane should be equal and he got the answer like following,

$$A_1 \cdot P_1 = A_2 \cdot P_2$$

After substituting the formula and values he got, $$=> \frac{h_1}{h_2} = 0.92 \frac{D^2}{d^2}$$

So all i want to ask is, which method is correct?? Thanks.


Your friend is correct in that the net force on the membrane should be zero. And the force from each side of the membrane is the pressure times the area.

The area of the membrane is the same on each side. It is not the case that one side of the membrane is of size $D$ and the other side is of size $d$. So his formula is incorrect.

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  • $\begingroup$ if the area of membrane is the same, $A_1 = A_2$, that will give us $P_1 = P_2$ but that's how i solved the question. So, is it safe to say, I'm correct?? $\endgroup$ – rndflas Jan 20 '15 at 23:40
  • $\begingroup$ @rndflas Correct in this case but only because the areas are equal. Incorrect in the general sense -- pressures don't have to be equal. The forces do. $\endgroup$ – tpg2114 Jan 20 '15 at 23:42
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    $\begingroup$ @mdflas, you're correct if we can consider the bottom tube to be "small", that is $d \ll h_{1}$. $\endgroup$ – BowlOfRed Jan 20 '15 at 23:45

$P_1 =\gamma_{water} \cdot h_1$

$P_2= \gamma_{oil} \cdot h_2$

$\gamma_{water} \cdot h_1 \cdot A = \gamma_{oil} \cdot h_2 \cdot A$

$\gamma_{water} \cdot h_1 = \gamma_{oil} \cdot h_2 $

$\frac{\gamma_{water} \cdot h_1} {\gamma_{oil} \cdot h_2} = 1$

$\frac{h_1} {h_2} = \frac{\gamma_{oil}} {\gamma_{water}} = 0.92$

Sorry to say your friend is incorrect because the diameter of concern would be at the interface. There is only one diameter there.

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