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

• A. \(h_A=\dfrac{h_B}{\sqrt{2}}\)
• B. \(h_A=\dfrac{h_B}{2}\)
• C. \(h_A=\dfrac{h_B}{3}\)
• D. \(h_A=\dfrac{h_B}{\sqrt{3}}\)

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.

Answer 1

The Correct Option is B

Step 1: State the fundamental U-tube equilibrium principle.
In a U-tube in static equilibrium, pressure at the same horizontal level is equal on both sides (Pascal's principle).
Step 2: Write pressure expressions for each column.
$P_A = \rho_A g h_A$ and $P_B = \rho_B g h_B$.
Step 3: Apply the equilibrium condition $P_A = P_B$.
$\rho_A g h_A = \rho_B g h_B$. Factor $g$ cancels: $\rho_A h_A = \rho_B h_B$.
Step 4: Use the given density relationship $\rho_A = 2\rho_B$.
\[ 2\rho_B h_A = \rho_B h_B \Rightarrow 2h_A = h_B \Rightarrow h_A = \frac{h_B}{2} \]
Step 5: Interpret physically.
Denser liquid A needs a shorter column to exert the same pressure. Doubling density halves the required height, consistent with $P = \rho g h$.
Step 6: State the final answer.
\[ \boxed{h_A = \frac{h_B}{2}} \]