Question 64 of 192

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 ${x}_{1}$ and in the other ${x}_{2}$. 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 :

A$gdS\left({x}_{2}^{2}+{x}_{1}^{2}\right)$B$gdS{\left({x}_{2}+{x}_{1}\right)}^{2}$C$\dfrac{3}{4}gdS{\left({x}_{2}-{x}_{1}\right)}^{2}$D$\dfrac{1}{4}gdS{\left({x}_{2}-{x}_{1}\right)}^{2}$

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Answer: (D) $\dfrac{1}{4}gdS{\left({x}_{2}-{x}_{1}\right)}^{2}$

Initial height of liquid in container's of same cross section are ${x}_{1}$ and ${x}_{2}$ respectively. Now value is opened find loss in potential energy when water level be become same
loss in $PE={U}_{i}-{U}_{f}$
$=\left[\rho (A){x}_{1}\dfrac{{x}_{1}}{2}+{\rho Ax}_{2}\dfrac{{x}_{2}}{2}\right]g-\left[\rho A\left(\dfrac{{x}_{1}+{x}_{2}}{2}\right)\times \left(\dfrac{{x}_{1}+{x}_{2}}{4}\right)\times 2\right]g$
$=\rho Ag\left[\dfrac{{x}_{1}^{2}}{2}+\dfrac{{x}_{2}^{2}}{2}-\dfrac{{\left({x}_{1}+{x}_{1}\right)}^{2}}{4}\right]=\dfrac{\rho Ag{\left({x}_{1}-{x}_{2}\right)}^{2}}{4}$