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Chapter 15: Problem 16

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density \(\rho .\) The area of either base is \(A\), but in one vessel the liquid height is \(h_{1}\) and in the other \(h_{2}\). Find the work done by gravity in equalizing the levels when the two vessels are connected.

Short Answer

Expert verified
The work done by gravity in equalizing the levels when the two vessels are connected is \(\frac{1}{2} \rho g A (h_{1}-h_{2})^2\).

Step by step solution

01

Identify given values

Two identical cylindrical vessels have base area \(A\) and liquids of densities \(\rho\) with heights \(h_{1}\) and \(h_{2}\). The gravitational force \(g\) is also present.
02

Derive formula

Work done \(W\) due to gravity is \(Force × Distance\), and here force on a tiny amount of liquid at height \(x\) is \(\Delta m × g = \rho × A × \Delta x × g\). The distance the liquid moves is \(h_{1} - x\), if \(h_{1} > h_{2}\). So \(\Delta W = \rho × A × \Delta x × g × (h_{1} - x)\). When integrated, we get the total work done \(W = \int_{h_{2}}^{h_{1}} \rho g A x (h_{1} - x) dx\).
03

Calculate the total work

By solving the integration, we get \(W = \frac{1}{2} \rho g A (h_{1}-h_{2})^2\).

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