The problem involves a U-shaped tube with mercury and a liquid of unknown density. Here's the step-by-step solution:

Data:

• Cross-sectional area: $A = 1.5 \, \text{cm}^2$
• Volume of mercury: $V_{\text{Hg}} = 50 \, \text{cm}^3$
• Density of mercury: $\rho_{\text{Hg}} = 13.6 \, \text{g/cm}^3$ (relative density)
• Height difference of mercury: $h_{\text{Hg}} = 2.75 \, \text{cm}$
• A liquid of unknown density is added.

Step-by-Step Solution:

1. Height of Mercury Column in the Tube Before Adding the Liquid:

   • Mercury initially fills both sides symmetrically.
   • The height of mercury in the tube is: $h_{\text{Hg,initial}} = \frac{V_{\text{Hg}}}{A} = \frac{50}{1.5} = 33.33 \, \text{cm}$

2. Effect of the Added Liquid:

   • When the unknown liquid is added, it pushes mercury down on one side.
   • The height difference in mercury columns is $h_{\text{Hg}} = 2.75 \, \text{cm}$.

3. Pressure Balance in the U-Tube:

   • At equilibrium, the pressures at the same horizontal level in both arms are equal.
   • Pressure due to the unknown liquid: $P_{\text{liquid}} = \rho_{\text{liquid}} g h_{\text{liquid}}$
   • Pressure due to displaced mercury: $P_{\text{Hg}} = \rho_{\text{Hg}} g h_{\text{Hg}}$

   Since the pressures are equal: $\rho_{\text{liquid}} g h_{\text{liquid}} = \rho_{\text{Hg}} g h_{\text{Hg}}$

   Simplify by canceling $g$: $\rho_{\text{liquid}} h_{\text{liquid}} = \rho_{\text{Hg}} h_{\text{Hg}}$

4. Determine the Height of the Liquid Column:

   • The height of the liquid column added is equal to the height of mercury displaced on the other side: $h_{\text{liquid}} = h_{\text{Hg}} = 2.75 \, \text{cm}$

5. Solve for Relative Density:

   • Substitute the known values: $\rho_{\text{liquid}} = \rho_{\text{Hg}} \frac{h_{\text{Hg}}}{h_{\text{liquid}}}$

   • With $\rho_{\text{Hg}} = 13.6$: $\rho_{\text{liquid}} = 13.6 \cdot \frac{2.75}{2.75} = 13.6$

Final Answer:

The relative density of the unknown liquid is approximately 1.0 (water's density).

Would you like further clarification or a breakdown of another part? Let me know!

Related Questions:

1. How does the cross-sectional area affect the equilibrium in a U-tube?
2. Why is the pressure at the same horizontal level in a fluid always equal?
3. What role does the density of mercury play in solving this problem?
4. How would the solution change if the liquid added had a higher density?
5. Can you calculate the absolute density of the unknown liquid in $\text{g/cm}^3$?

Tip: Always set up the pressure balance equation carefully when solving fluid dynamics problems involving equilibrium.