Consider two liquids A and B in a U-shaped tu | vedclass

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(B) In static equilibrium,the pressure at the same horizontal level in a continuous fluid is the same.Let the pressure at the interface of liquid $A$

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$h_{A}=\frac{h_{B}}{2}$

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(B) In static equilibrium,the pressure at the same horizontal level in a continuous fluid is the same.Let the pressure at the interface of liquid $A$ and liquid $B$ in the left arm be $P$. For the left arm,the pressure at this level is $P_0 + \rho_A g h_A$,where $P_0$ is the atmospheric pressure.For the right arm,the pressure at the same horizontal level is $P_0 + \rho_B g h_B$.Equating the pressures:$P_0 + \rho_A g h_A = P_0 + \rho_B g h_B$$\rho_A h_A = \rho_B h_B$Given that the density of liquid $A$ is twice the density of liquid $B$,i.e.,$\rho_A = 2\rho_B$.Substituting this into the equation:$(2\rho_B) h_A = \rho_B h_B$$2 h_A = h_B$$h_A = \frac{h_B}{2}$

Consider two liquids $A$ and $B$ in a $U$-shaped tube in static equilibrium as shown in the figure. If the density of the liquid $A$ is twice the density of liquid $B$,then the relation between $h_A$ and $h_B$ is

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