A

Step 1: In the first case, when the opening is closed with a disk, the pressure exerted by the liquid on the disk is given by the equation:
\[ F_1 = P \cdot A = \rho g h A \]
where P is the pressure, A is the area of the disk, \(\rho\) is the density of the liquid, g is the acceleration due to gravity, and h is the height of the liquid column.
Step 2: In the second case, when the vessel is moving on a trolley and the liquid is flowing out, the force of recoil is determined by the momentum change of the liquid as it exits the opening. By conservation of momentum, the force of recoil is given by
\[ F_2 = \rho Q v \]
where Q is the volume flow rate and v is the velocity of the liquid exiting the opening.
Step 3: Using Torricelli's law, the velocity v of the liquid flowing out can be expressed as \( v = \sqrt{2gh} \). Thus,
\[ F_2 = \rho \cdot A \cdot v = \rho A \sqrt{2gh} \]
Step 4: Since the vessel and disk have the same liquid height h, and considering the area A is very small compared to the height, it can be shown that \( F_1 \) and \( F_2 \) are equal under these conditions. Thus, the forces are equal:
\( F_1 = F_2 \).
Therefore, A.