In the arrangement shown, both the vessels A | vedclass

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(B) Let the cross-sectional area of the vessels be $S$. Let the height of the water level in vessel $A$ be $h_A$ and in vessel $B$ be $h_B$. Since the

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From $B$ to $A$

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(B) Let the cross-sectional area of the vessels be $S$. Let the height of the water level in vessel $A$ be $h_A$ and in vessel $B$ be $h_B$. Since the amount of water in $B$ is double that in $A$, we have $S \cdot h_B = 2(S \cdot h_A)$, which implies $h_B = 2h_A$.Let $P_A$ and $P_B$ be the pressures exerted by the water at the bottom of vessels $A$ and $B$ respectively. The pressure at the bottom is given by $P = P_{ext} + \rho gh$, where $P_{ext}$ is the pressure exerted by the piston.For the lever to be in equilibrium, the forces exerted by the pistons on the water must be equal (assuming the lever arm is balanced at the center). Let $F$ be the force exerted by each piston. Then $P_{ext} = F/S$ is the same for both vessels.Thus, $P_A = F/S + \rho gh_A$ and $P_B = F/S + \rho gh_B$.Since $h_B > h_A$, it follows that $P_B > P_A$.When the valve is opened, water flows from a region of higher pressure to a region of lower pressure. Therefore, water will flow from $B$ to $A$.

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