Question:
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
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
Show Hint
In a U-tube at static equilibrium, pressures at the same horizontal level are equal:
\[
\rho_1gh_1=\rho_2gh_2
\]
A denser liquid has a smaller height column for the same pressure.
Updated On: Jun 25, 2026
- \(h_A=\dfrac{h_B}{\sqrt{2}}\)
- \(h_A=\dfrac{h_B}{2}\)
- \(h_A=\dfrac{h_B}{3}\)
- \(h_A=\dfrac{h_B}{\sqrt{3}}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understand the condition of static equilibrium.
In a U-shaped tube containing two immiscible liquids, the pressure at the same horizontal level must be equal.
So, at the common horizontal level shown in the figure, \[ P_A=P_B \]
Step 2: Write pressure due to liquid columns.
Pressure due to a liquid column is given by \[ P=\rho gh \] For liquid \(A\), \[ P_A=\rho_A g h_A \] For liquid \(B\), \[ P_B=\rho_B g h_B \] Since pressures are equal, \[ \rho_A g h_A=\rho_B g h_B \]
Step 3: Use the given density relation.
Given that density of liquid \(A\) is twice the density of liquid \(B\).
Therefore, \[ \rho_A=2\rho_B \] Substituting in \[ \rho_A g h_A=\rho_B g h_B, \] we get \[ 2\rho_B g h_A=\rho_B g h_B \] Canceling \(\rho_B g\) from both sides, \[ 2h_A=h_B \]
Step 4: Find the relation between \(h_A\) and \(h_B\).
From \[ 2h_A=h_B, \] we get \[ h_A=\frac{h_B}{2} \]
Step 5: Final conclusion.
Hence, the required relation is \[ \boxed{h_A=\frac{h_B}{2}} \]
In a U-shaped tube containing two immiscible liquids, the pressure at the same horizontal level must be equal.
So, at the common horizontal level shown in the figure, \[ P_A=P_B \]
Step 2: Write pressure due to liquid columns.
Pressure due to a liquid column is given by \[ P=\rho gh \] For liquid \(A\), \[ P_A=\rho_A g h_A \] For liquid \(B\), \[ P_B=\rho_B g h_B \] Since pressures are equal, \[ \rho_A g h_A=\rho_B g h_B \]
Step 3: Use the given density relation.
Given that density of liquid \(A\) is twice the density of liquid \(B\).
Therefore, \[ \rho_A=2\rho_B \] Substituting in \[ \rho_A g h_A=\rho_B g h_B, \] we get \[ 2\rho_B g h_A=\rho_B g h_B \] Canceling \(\rho_B g\) from both sides, \[ 2h_A=h_B \]
Step 4: Find the relation between \(h_A\) and \(h_B\).
From \[ 2h_A=h_B, \] we get \[ h_A=\frac{h_B}{2} \]
Step 5: Final conclusion.
Hence, the required relation is \[ \boxed{h_A=\frac{h_B}{2}} \]
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