Two identical cylindrical vessels are kept on the ground and each...

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[ρ ( A ) {x}_{1}\dfrac{{x}_{1}}{2} + {ρAx}_{2}\dfrac{{x}_{2}}{2}\right]g - \left[ρA\left(\dfrac{{x}_{1} + {x}_{2}}{2}\right) \times\left(\dfrac{{x}_{1} + {x}_{2}}{4}\right) \times 2\right]g$</p><p>$= ρAg$\left[{x}_{1}^{2}2 + {x}_{2}^{2}2 - \dfrac{{\left({x}_{1} + {x}_{1}\right)}^{2}}{4}\right] = \dfrac{ρAg{\left({x}_{1} - {x}_{2}\right)}^{2}}{4}$