1. Initial setup

Let the liquid density be $d$, base area of each cylinder be $S$, and initial heights be $x_1$ and $x_2$.

Since the vessels are identical and connected near the bottom, at equilibrium the liquid levels must become equal.

2. Final common height

Total initial volume: $V_i = Sx_1 + Sx_2 = S(x_1+x_2)$

If the final common height is $h$, then total final volume is: $V_f = 2Sh$

By conservation of volume, $2Sh = S(x_1+x_2)$ $h = \frac{x_1+x_2}{2}$

3. Gravitational potential energy of a liquid column

For a liquid column of height $x$ and base area $S$:

• Volume $= Sx$
• Mass $= dSx$
• Its center of mass is at height $x/2$

So its gravitational potential energy is: $U = (dSx)g\left(\frac{x}{2}\right) = \frac{1}{2}gdSx^2$

4. Initial energy

For the two vessels initially, $U_i = \frac{1}{2}gdSx_1^2 + \frac{1}{2}gdSx_2^2$ $U_i = \frac{1}{2}gdS(x_1^2 + x_2^2)$

5. Final energy

After equilibrium, each vessel has height $h = \frac{x_1+x_2}{2}$

So final energy is: $U_f = 2\left(\frac{1}{2}gdSh^2\right) = gdSh^2$

Substitute $h$: $U_f = gdS\left(\frac{x_1+x_2}{2}\right)^2 = \frac{1}{4}gdS(x_1+x_2)^2$

6. Change in energy

The decrease in energy is: $\Delta U = U_i - U_f$

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

Factor out $\frac{1}{4}gdS$: $\Delta U = \frac{1}{4}gdS\left[2(x_1^2+x_2^2) - (x_1+x_2)^2\right]$

Expand: $2(x_1^2+x_2^2) - (x_1^2+2x_1x_2+x_2^2) = x_1^2 + x_2^2 - 2x_1x_2$

$= (x_2-x_1)^2$

Hence, $\Delta U = \frac{1}{4}gdS(x_2-x_1)^2$

This is the loss in gravitational potential energy of the system.

7. Option matching

Thus the required change in energy is: $\boxed{\frac{1}{4}gdS(x_2-x_1)^2}$

So the correct option is C.