Question

A thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids, whose densities are \[{{\rho }_{1}}\] and \[{{\rho }_{2}}\] fill half the circle. The angle \[\theta \] between the radius vector passing through the common interface and the vertical is:
A. \[\theta ={{\tan }^{-1}}\left[ \dfrac{\pi }{2}\left( \dfrac{{{\rho }_{1}}-{{\rho }_{2}}}{{{\rho }_{1}}-{{\rho }_{2}}} \right) \right]\]
B. \[\theta ={{\tan }^{-1}}\left[ \dfrac{\pi }{2}\left( \dfrac{{{\rho }_{1}}-{{\rho }_{2}}}{{{\rho }_{1}}+{{\rho }_{2}}} \right) \right]\]
C. \[\theta ={{\tan }^{-1}}\pi \left( \dfrac{{{\rho }_{1}}}{{{\rho }_{2}}} \right)\]
D. None of the above.

Answer 1

Hint: To find the \[\theta \] we will have to find the heights in relation with \[{{\rho }_{1}}\]and \[{{\rho }_{2}}\]. Using this fact, we will calculate the heights of liquids filled in the circle. Later, using trigonometric ratios we will calculate \[\theta \] for a given arrangement.

Formula Used:
\[P=\rho gh\]

Complete step by step answer:

Let ac be the diameter of the circle as shown in the figure above. From point a to b be filled with liquid of density \[{{\rho }_{1}}\] and from point b to c be filled with liquid of density \[{{\rho }_{2}}\] . That makes point b as our interface making angle \[\theta \]with the vertical as shown in the figure.
So, now let us find the pressure to the right side of the point 1 say \[{{P}_{1}}\] with height \[{{h}_{1}}\]
Therefore,
\[{{P}_{1}}\]= \[{{\rho }_{1}}g{{h}_{1}}\]
Similarly, let us find pressure to the other side. But, this has both liquid.
Therefore,
\[{{P}_{2}}={{\rho }_{1}}g{{h}_{2}}+{{\rho }_{2}}g{{h}_{3}}\]
Now we know that \[{{P}_{1}}\] = \[{{P}_{2}}\]
Therefore after equating both
We get,
\[{{\rho }_{1}}g{{h}_{1}}\] = \[{{\rho }_{1}}g{{h}_{2}}+{{\rho }_{2}}g{{h}_{3}}\] ………………….. (A)
Now, we know that the vertical line from origin to point O is radius say R.
So, using trigonometry to find \[{{h}_{1}}\] in terms of \[\theta \]
Therefore,
\[{{h}_{1}}=R\left( 1-\sin \theta \right)\] …………… (1)
Similarly the other heights can be found as
\[{{h}_{2}}=R(1-\cos \theta )\] ……………. (2)
\[{{h}_{3}}=R(\cos \theta +\sin \theta )\] ………………. (3)
Now, after substituting (1), (2) and (3) in equation (A)
We get,
\[{{\rho }_{1}}gR\left( 1-\sin \theta \right)={{\rho }_{1}}gR(1-\cos \theta )+{{\rho }_{2}}gR(\cos \theta +\sin \theta )\]
Now, dividing both sides by g and R
We get,
\[{{\rho }_{1}}\left( 1-\sin \theta \right)={{\rho }_{1}}(1-\cos \theta )+{{\rho }_{2}}(\cos \theta +\sin \theta )\]
\[{{\rho }_{1}}-{{\rho }_{1}}\sin \theta ={{\rho }_{1}}-{{\rho }_{1}}\cos \theta +{{\rho }_{2}}\cos \theta +{{\rho }_{2}}\sin \theta \]
Now, let us subtract \[{{\rho }_{1}}\] from both sides and rearrange the terms
We get,
\[(-{{\rho }_{2}}-{{\rho }_{1}})\sin \theta =({{\rho }_{2}}-{{\rho }_{1}})\cos \theta \]
Now, multiplying both sides by -1
We get,
\[({{\rho }_{1}}+{{\rho }_{2}})\sin \theta =({{\rho }_{1}}-{{\rho }_{2}})\cos \theta \]
Therefore,
\[\tan \theta =\dfrac{({{\rho }_{1}}-{{\rho }_{2}})}{({{\rho }_{1}}+{{\rho }_{2}})}\]
\[\therefore \theta ={{\tan }^{-1}}\left[ \dfrac{({{\rho }_{1}}-{{\rho }_{2}})}{({{\rho }_{1}}+{{\rho }_{2}})} \right]\]
We do not see this answer in the options.
Therefore, the correct answer is option D.

Note:
Our answer might look similar to option B, but there is a \[\dfrac{\pi }{2}\] in multiplication with our answer which will change our answer by 90 degrees. So students must be careful. Pressure in a liquid depends only on the height of the object submerged in liquid and the density of liquid. Pressure is equally divided in all directions in any liquid. So if a force is applied at an appointment, it will transmit the same in all directions. Therefore, in our case equal pressure is applied on both liquids from the empty half. Which makes the pressure at lowest point 1 equal.