Communicating Vessels #2

Classical Mechanics Level 3

There are 2 similar communicating vessels, each contains 2 different liquids that aren't miscible in each other. The liquids are separated by a valve and at the same height $H$ . Open the valve.

The specific gravity of the vessel 1 is bigger than the specific gravity of the vessel 2 (or $d_{1} > d_{2}$ . Neglect the capacity of the part connecting 2 vessels altogether.

Add in a vessel another liquid (liquid 3), which isn't miscible in the other 2 liquids, with a specific gravity of $d_{3}$ until the first two liquids have the same liquid level. Find the liquid level of liquid 3.

Part 1 Part 3

\frac{H(d_{1}+d_{2}-d_{3})}{(d_{1}-d_{2}+d_{3})}

\frac{H(d_{1}-d_{2})}{d_{3}}

\frac{H}{2}

\frac{H(d_{1}-d_{2}-d_{3})}{(d_{1}+d_{2}+d_{3})}

d_{1} d_{2} d_{3} + H

None of the others are correct.

\frac{(d_{1}-d_{2}-d_{3})(d_{1}+d_{2}+d_{3})}{H}

1 solution

Anirudh Bhardwaj
Sep 21, 2019

See the figure.. Let solve according to it. After stopper is removed in fig 2. (H-x)d1g = (H+x)d2g - solve for x = $\frac{d1-d2}{d1+d2}$ H

H1 for orange liquid = $\frac{2d2H}{d1+d2}$ H2 for pink = $\frac{2d1H}{d1+d2}$

From figure 4 When liquid d3 (green) is added => d2 falls x and d1 rises x such that their final heights are equal i.e H2 - x = H1 +x solve for x = $\frac{d1-d2}{d1+d2}$ H

Final heights , orange liquid = H1 + x = H = pink liquid

now solve for green liquid

d3 y g + d2 H g = d1 H g

y = $\frac{d1-d2}{d3}$ H

Communicating Vessels #2

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