Two identical cylindrical vessels with their bases at same level each contains a liquid of density d. The height of the liquid in one vessel is h1 and that in the other vessel is h2. The area of either vases is A. The work done by gravity in equalizing the levels when the two vessels are connected, is

The correct option is 0.25(h1-h2)²gAd.

Let's understand the physical system and compute the work done by gravity step-by-step when the two cylindrical vessels are connected to equalize their liquid levels.

1. Initial State:
We have two identical cylindrical vessels, each with a base area $A$. The density of the liquid is $d$.
In the first vessel, the liquid column has a height of $h_1$.
In the second vessel, the liquid column has a height of $h_2$.
Without loss of generality, let us assume $h_1>h_2$.

The mass of the liquid in the first vessel is:
$m_1=A{h}_{1}d$
The mass of the liquid in the second vessel is:
$m_2=A{h}_{2}d$

For a uniform column of liquid of height $h$, the center of mass lies at half the height, i.e., at $h/2$ above the base. Therefore, the initial potential energy of the liquid in each vessel is given by:
$U_1=m_{1}g\left(\frac{h_1}{2}\right)=\left(A{h}_{1}d\right)g\left(\frac{h_1}{2}\right)=\frac{1}{2}Adg{h}_1^2$
$U_2=m_{2}g\left(\frac{h_2}{2}\right)=\left(A{h}_{2}d\right)g\left(\frac{h_2}{2}\right)=\frac{1}{2}Adg{h}_2^2$

The total initial potential energy of the system is:
$U_i=U_1+U_2=\frac{1}{2}Adg\left(h_1^2+h_2^2\right)$

2. Final State:
When the vessels are connected at their bases, liquid flows from the vessel with the higher level to the one with the lower level until their heights are equalized at a common height $h$.
Since the total volume of the liquid is conserved and the vessel areas are equal:
$A{h}_1+A{h}_2=2Ah\Rightarrow h=\frac{h_1+h_2}{2}$

The final height of the liquid in each vessel is $h$. The mass of liquid in each vessel is now:
$m_f=Ahd$
The new center of mass for each vessel is at $h/2$ above the base. The final potential energy of the liquid in each vessel is:
$U_{\mathrm{f1}}=U_{\mathrm{f2}}=m_{f}g\left(\frac{h}{2}\right)=\frac{1}{2}Adg{h}^2$

The total final potential energy of the system is:
$U_f=U_{\mathrm{f1}}+U_{\mathrm{f2}}=Adg{h}^2$
Substituting $h=\frac{h_1+h_2}{2}$:
$U_f=Adg{\left(\frac{h_1+h_2}{2}\right)}^2=\frac{1}{4}Adg{\left(h_1+h_2\right)}^2$

3. Work Done by Gravity:
The work done by gravity ($W$) on the system is equal to the decrease in the potential energy of the system:
$W=U_i-U_f$
Let us calculate this difference:
$W=\frac{1}{2}Adg\left(h_1^2+h_2^2\right)-\frac{1}{4}Adg{\left(h_1+h_2\right)}^2$
Taking the common factor $\frac{1}{4}Adg$ out:
$W=\frac{1}{4}Adg\left[2\left(h_1^2+h_2^2\right)-{\left(h_1+h_2\right)}^2\right]$
Expand the terms inside the bracket:
$2h_1^2+2h_2^2-\left(h_1^2+h_2^2+2h_1{h}_2\right)$
$=h_1^2+h_2^2-2h_1{h}_2$
$={\left(h_1-h_2\right)}^2$

Substituting this back into the work equation gives:
$W=0.25{\left(h_1-h_2\right)}^{2}gAd$