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:

Let us denote the density of the liquid by $$d$$ and the acceleration due to gravity by $$g$$. Each cylindrical vessel has the same base-area $$S$$, but the initial heights of the liquid columns are different: one is at height $$x_1$$ while the other is at height $$x_2$$.

For a uniform vertical column of liquid of height $$h$$, the centre of mass lies exactly at its mid-height $$\dfrac{h}{2}$$. Therefore, the gravitational potential energy $$U$$ of such a column is obtained from the general expression $$U = mgh_{\text{cm}}$$. Substituting the mass $$m = \rho V = d(S\,h)$$ and $$h_{\text{cm}}=\dfrac{h}{2}$$, we have

$$ U = (d\,S\,h)\,g\;\frac{h}{2} = \frac12\,d\,g\,S\,h^{2}. $$

We first compute the total gravitational potential energy of the system before the two vessels are connected.

For the vessel with height $$x_1$$ the energy is

$$U_1 = \frac12\,d\,g\,S\,x_1^{2},$$

and for the vessel with height $$x_2$$ the energy is

$$U_2 = \frac12\,d\,g\,S\,x_2^{2}.$$

Hence the initial total energy is

$$ U_{\text{initial}} = U_1 + U_2 = \frac12\,d\,g\,S\bigl(x_1^{2}+x_2^{2}\bigr). $$

Now we join the vessels by a tube of negligible volume fixed very near their bottoms. Because the liquid can now flow freely, it redistributes itself until both columns attain a common final height, which we shall call $$h_f$$.

The total volume of liquid is conserved. Initially the volume is

$$ V_{\text{total}} = Sx_1 + Sx_2 = S\,(x_1 + x_2). $$

After equilibrium the liquid occupies two cylinders, each of base area $$S$$, so the combined base area is $$2S$$. Therefore the common final height is obtained from

$$ 2S\,h_{f}=S\,(x_1 + x_2) \;\;\Longrightarrow\;\; h_{f} = \frac{x_1 + x_2}{2}. $$

For each vessel at this stage the liquid column has height $$h_f$$, so using the same energy formula we write

$$ U_{\text{one\,column\,(final)}} = \frac12\,d\,g\,S\,h_f^{2}. $$

Because there are two identical columns, the total final energy is

$$ U_{\text{final}} = 2\left(\frac12\,d\,g\,S\,h_f^{2}\right) = d\,g\,S\,h_f^{2}. $$

Next, we find the change in gravitational potential energy of the system:

$$ \Delta U = U_{\text{final}} - U_{\text{initial}} = d\,g\,S\,h_f^{2} - \frac12\,d\,g\,S\,(x_1^{2}+x_2^{2}). $$

We now substitute $$h_f = \dfrac{x_1 + x_2}{2}$$. First evaluate $$h_f^{2}$$:

$$ h_f^{2} = \left(\frac{x_1 + x_2}{2}\right)^{2} = \frac{(x_1 + x_2)^{2}}{4}. $$

Putting this into the expression for $$\Delta U$$ gives

$$ \Delta U = d\,g\,S\left[\frac{(x_1 + x_2)^{2}}{4} - \frac{1}{2}(x_1^{2}+x_2^{2})\right]. $$

To simplify the bracket, we write everything with the same denominator:

$$ \frac{(x_1 + x_2)^{2}}{4} - \frac{1}{2}(x_1^{2}+x_2^{2}) = \frac{(x_1 + x_2)^{2}}{4} - \frac{2(x_1^{2}+x_2^{2})}{4}. $$

Expanding $$ (x_1 + x_2)^{2} = x_1^{2} + 2x_1x_2 + x_2^{2} $$, we have

$$ \frac{x_1^{2} + 2x_1x_2 + x_2^{2} - 2x_1^{2} - 2x_2^{2}}{4} = \frac{-\,x_1^{2} + 2x_1x_2 - x_2^{2}}{4}. $$

Recognising the numerator as the negative of a perfect square,

$$ -\,x_1^{2} + 2x_1x_2 - x_2^{2} = -\,(x_1^{2} - 2x_1x_2 + x_2^{2}) = -\,(x_1 - x_2)^{2}. $$

Hence the entire bracket simplifies to

$$ \frac{-\,(x_1 - x_2)^{2}}{4} = -\,\frac{(x_1 - x_2)^{2}}{4}. $$

Substituting this back, the change in energy becomes

$$ \Delta U = d\,g\,S \left[-\,\frac{(x_1 - x_2)^{2}}{4}\right] = -\,\frac{1}{4}\,d\,g\,S\,(x_1 - x_2)^{2}. $$

The negative sign shows that energy has decreased. The magnitude of this decrease is

$$ \bigl|\Delta U\bigr| = \frac{1}{4}\,d\,g\,S\,(x_2 - x_1)^{2}, $$

which exactly matches Option D.

Hence, the correct answer is Option D.