Communicating Vessels #3

Classical Mechanics Level 5

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. What is the condition of $d_{3}$ so that the liquid level of the required liquid 3 (part 2) is the same as the difference of the liquid level of the 2 vessels in part 1?

Part 1 Part 2

\frac{d_{3}}{d_{1}}=\frac{d_{2}}{d_{3}}

None of the other answers are correct.

d_{1}=d_{3}

d_{2}=d_{3}+d_{1}

d_{2}=d_{3}

\frac{d_{1}}{d_{2}}=\frac{d_{3}}{d_{1}}

d_{1}=d_{3}=d_{2}

1 solution

Kumudesh Ghosh
May 26, 2020

Because d1 = d3

Such a helpful solution!

Tin Le - 1 year ago

I want a clear solution, I don't understand. @Aryan Sanghi , @Siddharth Chakravarty , @Kumudesh Ghosh

Vinayak Srivastava - 10 months, 3 weeks ago

From the first part of the question, you get the difference in the liquid levels is $\frac{H(d_{1}-d_{2})}{d_{1}}$ . And from part 2, you get the answer for the liquid level of d3 as $\frac{H(d_{1}-d_{2})}{d_{3}}$ . So these two levels have to be equal as per the question, so equating them we get $\frac{H(d_{1}-d_{2})}{d_{1}}$ = $\frac{H(d_{1}-d_{2})}{d_{3}}$ So cancelling the terms and then doing invertendo, we get d1=d3. If you want a comprehensive solution, I will post it here, just let me know.

Siddharth Chakravarty - 10 months, 3 weeks ago

Ok, thank you!

Vinayak Srivastava - 10 months, 3 weeks ago

Communicating Vessels #3

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