question_answer

Two cylindrical vessels of equal cross-sectional area A contain water upto heights \[{{h}_{1}}\]and \[{{h}_{2}}\].The vessels are interconnected so that the levels in them become equal. Calculate the work done by the force of gravity during the process. The density of water is \[\rho \].

Answer:

                Initial levels of water in two cylindrical vessels are at A and B.   As total volume of water is constant, therefore, after interconnection, the height of water in each vessel becomes  \[\frac{{{h}_{1}}+{{h}_{2}}}{2}\], Fig. 4(HT).7. In the left vessel, level falls through      \[AC=\left( {{h}_{1}}-\frac{{{h}_{1}}+{{h}_{2}}}{2} \right)=\frac{{{h}_{1}}-{{h}_{2}}}{2}\] In the right vessel, level rises through  \[BD=\frac{{{h}_{1}}-{{h}_{2}}}{2}\] Mass of water moved \[(m)=\rho A\left( \frac{{{h}_{1}}-{{h}_{2}}}{2} \right)\] Work done = \[mgh=\left[ \rho A\frac{\left( {{h}_{1}}-{{h}_{2}} \right)}{2} \right]g\left( \frac{{{h}_{1}}-{{h}_{2}}}{2} \right)=\rho Ag{{\left( \frac{{{h}_{1}}-{{h}_{2}}}{2} \right)}^{2}}\]