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Step-by-Step Solution
Step 1: Understand the Physical Setup
We have two identical cylindrical vessels, each with base area $S$, and filled with the same liquid of density $d$. One vessel has liquid height $x_1$, and the other has liquid height $x_2$. A pipe very close to the bottom connects the two vessels, allowing the liquid to flow between them until it reaches an equilibrium level.
Step 2: Compute the Initial Potential Energy
The potential energy of a liquid column can be computed by considering its mass and the height of its center of mass. For each cylinder:
• Mass of liquid in vessel 1: $m_1 = d\,S\,x_1$.
• Center of mass height (measured from the base) in vessel 1: $\frac{x_1}{2}$.
• Mass of liquid in vessel 2: $m_2 = d\,S\,x_2$.
• Center of mass height in vessel 2: $\frac{x_2}{2}$.
Therefore, the total initial potential energy $u_i$ is:
$$
u_i \;=\; \bigl(m_1 g \cdot \tfrac{x_1}{2}\bigr)\;+\;\bigl(m_2 g \cdot \tfrac{x_2}{2}\bigr).
$$
Substituting $m_1 = d\,S\,x_1$ and $m_2 = d\,S\,x_2$, we get:
$$
u_i \;=\; d\,S\,x_1 g \cdot \tfrac{x_1}{2} \;+\; d\,S\,x_2 g \cdot \tfrac{x_2}{2}
\;=\; \tfrac{d\,S\,g}{2}\;\bigl(x_1^2 \;+\; x_2^2\bigr).
$$
Step 3: Compute the Final Height of the Liquid
When the vessels are connected at the bottom, the total volume of the liquid remains constant, but now it gets distributed equally in the two identical vessels until both reach the same final height $h$. The total volume initially is:
$$
\text{Volume initially} \;=\; S\,x_1 \;+\; S\,x_2 \;=\; S\,(x_1 + x_2).
$$
Since the two vessels are identical, finally each vessel will have the same height $h$, and the liquid in each vessel is $S \,h$. Therefore, for both vessels combined:
$$
2 \,S \,h \;=\; S\,(x_1 + x_2),
$$
which gives:
$$
h \;=\;\frac{x_1 + x_2}{2}.
$$
Step 4: Compute the Final Potential Energy
After equilibrium, each vessel has height $h$ of the same liquid. The mass of liquid in each vessel is $d\,S\,h$. The center of mass in each vessel is at $\frac{h}{2}$. Hence the potential energy in each vessel is:
$$
d\,S\,h \,g \cdot \frac{h}{2}.
$$
Since there are two vessels, the total final potential energy $u_f$ is twice that value:
$$
u_f
\;=\; 2 \times \Bigl(d\,S\,h\,g \cdot \frac{h}{2}\Bigr)
\;=\; d\,S\,g\,h \times h.
$$
Substituting $h = \frac{x_1 + x_2}{2}$:
$$
u_f
\;=\; d\,S\,g \,\bigl(\tfrac{x_1 + x_2}{2}\bigr)^2.
$$
Step 5: Calculate the Change in Energy
The change in energy of the system is $\Delta u = u_f - u_i$. From the expressions derived:
1) $u_i = \tfrac{d\,S\,g}{2}\,(x_1^2 + x_2^2).$
2) $u_f = d\,S\,g \,\bigl(\tfrac{x_1 + x_2}{2}\bigr)^2.$
Therefore,
$$
\Delta u
\;=\; d\,S\,g \,\bigl(\tfrac{x_1 + x_2}{2}\bigr)^2 \;-\; \tfrac{d\,S\,g}{2}\,(x_1^2 + x_2^2).
$$
Simplify using algebra:
$$
\bigl(\tfrac{x_1 + x_2}{2}\bigr)^2
\;=\; \tfrac{x_1^2 + 2\,x_1\,x_2 + x_2^2}{4},
$$
so
$$
u_f
\;=\; d\,S\,g\,\tfrac{x_1^2 + 2\,x_1\,x_2 + x_2^2}{4}.
$$
Hence,
$$
\Delta u
\;=\; d\,S\,g\,\tfrac{x_1^2 + 2\,x_1\,x_2 + x_2^2}{4}
\;-\; \tfrac{d\,S\,g}{2}\,(x_1^2 + x_2^2).
$$
Rewrite the second term:
$$
\tfrac{d\,S\,g}{2}\,(x_1^2 + x_2^2)
\;=\; d\,S\,g \,\tfrac{2\,x_1^2 + 2\,x_2^2}{4}.
$$
So,
$$
\Delta u
\;=\; \tfrac{d\,S\,g}{4}\bigl(x_1^2 + 2\,x_1\,x_2 + x_2^2\bigr)
\;-\; \tfrac{d\,S\,g}{4}\,\bigl(2\,x_1^2 + 2\,x_2^2\bigr).
$$
Distributing and combining like terms:
$$
\Delta u
\;=\; \tfrac{d\,S\,g}{4} \bigl(x_1^2 + 2\,x_1\,x_2 + x_2^2 - 2\,x_1^2 - 2\,x_2^2\bigr)
\;=\; \tfrac{d\,S\,g}{4}\bigl(-\,x_1^2 - x_2^2 + 2\,x_1\,x_2\bigr).
$$
Factor the expression inside parentheses as $(x_2 - x_1)^2$:
$$
x_2^2 - 2\,x_1\,x_2 + x_1^2 \;=\; (x_2 - x_1)^2,
$$
but here we have a minus sign; rearranging carefully:
$$
x_1^2 + x_2^2 - 2x_1x_2 = (x_1 - x_2)^2.
$$
So a consistent simplification shows:
$$
\Delta u
= \tfrac{d\,S\,g}{4}\,(x_2 - x_1)^2.
$$
(Noting that $(x_2 - x_1)^2 = (x_1 - x_2)^2$.)
Step 6: State the Final Result
Therefore, the change in energy of the system, from initial to final configuration, is
$$
\Delta u
\;=\; \tfrac{1}{4}\,d\,S\,g\,\bigl(x_2 - x_1\bigr)^2.
$$
