Three different liquids are filled in a U-tube as shown in the figure. Their densities are \( \rho_1 \), \( \rho_2 \), and \( \rho_3 \), respectively. From the figure, we may conclude that:
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In pressure equilibrium problems, remember that the pressure at the same level in different liquids must be equal for the system to be in equilibrium.
Step 1: Problem Overview
A U-tube contains three liquids at different heights. The liquid columns' pressures at the U-tube's base must be balanced for equilibrium.
Step 2: Pressure at Equilibrium
For the U-tube to be balanced, pressure at the same level must be equal on both sides. The pressure at the bottom of each liquid column is calculated using:
\[\nP = \rho g h\n\]
Where:
- \( \rho \) is the liquid's density,
- \( g \) is gravitational acceleration,
- \( h \) is the liquid column's height.
Step 3: Applying Pressure Balance
On the U-tube's left side, the liquid column with density \( \rho_1 \) and height \( h \) exerts pressure:
\[\nP_1 = \rho_1 g h\n\]
On the right side, there are two columns: one with density \( \rho_2 \) and height \( h \), and another with density \( \rho_3 \) and height \( \frac{h}{2} \). The total pressure on the right side is:
\[\nP_2 = \rho_2 g h + \rho_3 g \frac{h}{2}\n\]
For equilibrium:
\[\nP_1 = P_2\n\]
Substituting:
\[\n\rho_1 g h = \rho_2 g h + \rho_3 g \frac{h}{2}\n\]
Step 4: Equation Simplification
We can cancel \( g h \):
\[\n\rho_1 = \rho_2 + \frac{\rho_3}{2}\n\]
Multiply by 2:
\[\n2 \rho_1 = 2 \rho_2 + \rho_3\n\]
Rearrange:
\[\n\rho_3 = 2 (\rho_1 - \rho_2)\n\]
Step 5: Conclusion
Therefore, the correct relationship is:
\[\n\boxed{(C)} \, \rho_3 = 2 (\rho_2 - \rho_1)\n\]