Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density d. The area of the base of both vessels is S but the height of liquid in one vessel is x1 and in the other, x2 . When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is:
Solution
$${u_i} = \left[ {dS{x_1}.{{{x_1}} \over 2} + dS{x_2}.{{{x_2}} \over 2}} \right]g\left\{ {dS{x_1} \to m,\,{{{x_1}} \over 2} \to h(C.O.M)} \right\}$$
<br><br>Here total volume remains same.
<br><br>$\therefore$ V<sub>i</sub> = V<sub>f</sub>
<br><br>$\Rightarrow$ S(x<sub>1</sub> + x<sub>2</sub>) = S(h + h)
<br><br>$\Rightarrow$ h = ${{{x_1} + {x_2}} \over 2}$
<br><br>u<sub>f</sub> = (dSh)g${h \over 2} \times 2$
<br><br>$\Rightarrow$ $${u_f} = \left[ {dS\left( {{{{x_1} + {x_2}} \over 2}} \right) \times \left( {{{{x_1} + {x_2}} \over 4}} \right) \times 2} \right]g$$<br><br>$\therefore$ $${u_i} - {u_f} = dsg\left[ {{{x_1^2} \over 2} + {{x_2^2} \over 2} - {{{{\left( {{x_1} + {x_2}} \right)}^2}} \over 4}} \right]$$<br><br>$= dsg{{{{\left( {{x_1} - {x_2}} \right)}^2}} \over 4}$
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
This question is part of PrepWiser's free JEE Main question bank. 183 more solved questions on Properties of Solids and Liquids are available — start with the harder ones if your accuracy is >70%.