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:
Show Hint
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: Understanding the Problem
We have three different liquids in a U-tube, and the columns of these liquids are at different heights. The pressures exerted by the columns of liquids at the bottom of the U-tube must balance because the system is in equilibrium.
Step 2: Pressure Equilibrium
For the U-tube to be in equilibrium, the pressure at the same level on both sides must be equal. The pressure at the bottom of each liquid column is given by the formula:
\[
P = \rho g h
\]
Where:
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( h \) is the height of the liquid column.
Step 3: Applying the Pressure Balance
On the left side of the U-tube, the pressure due to the column of liquid with density \( \rho_1 \) and height \( h \) is:
\[
P_1 = \rho_1 g h
\]
On the right side of the U-tube, we have two columns: one with liquid of density \( \rho_2 \) and height \( h \), and the other with liquid of density \( \rho_3 \) and height \( \frac{h}{2} \). The total pressure on the right side is:
\[
P_2 = \rho_2 g h + \rho_3 g \frac{h}{2}
\]
For equilibrium:
\[
P_1 = P_2
\]
Substituting the expressions for \( P_1 \) and \( P_2 \):
\[
\rho_1 g h = \rho_2 g h + \rho_3 g \frac{h}{2}
\]
Step 4: Simplifying the Equation
We can cancel out \( g h \) from both sides:
\[
\rho_1 = \rho_2 + \frac{\rho_3}{2}
\]
Multiplying both sides by 2:
\[
2 \rho_1 = 2 \rho_2 + \rho_3
\]
Rearranging the equation:
\[
\rho_3 = 2 (\rho_1 - \rho_2)
\]
Step 5: Conclusion
Thus, the correct relationship is:
\[
\boxed{(C)} \, \rho_3 = 2 (\rho_2 - \rho_1)
\]
