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(B) The pressure at point $A$ is due to the liquid of density $\rho$ at a depth $h$ from the surface of the upper liquid.$P_A = P_{atm} + \rho gh$The

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$3\rho gh$

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(B) The pressure at point $A$ is due to the liquid of density $\rho$ at a depth $h$ from the surface of the upper liquid.$P_A = P_{atm} + \rho gh$The pressure at point $B$ is due to the liquid of density $\rho$ (height $2h$) and the liquid of density $2\rho$ (height $h$).$P_B = P_{atm} + \rho g(2h) + (2\rho)gh = P_{atm} + 2\rho gh + 2\rho gh = P_{atm} + 4\rho gh$The difference in pressure between point $B$ and point $A$ is:$P_B - P_A = (P_{atm} + 4\rho gh) - (P_{atm} + \rho gh) = 3\rho gh$

$A$ cylinder stands vertically in two immiscible liquids of densities $\rho$ and $2\rho$. The heights of the two liquids are shown in the figure. Find the difference in pressure at point $A$ and point $B$.

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