To solve this problem, let's go through the details step-by-step.

1. Problem Analysis:

   • We have a U-shaped tube with a cross-sectional area $A = 1.5 \, \text{cm}^2$.
   • Initially, it contains 50 cm³ of mercury (Hg), which means that the height of mercury can be determined by: $h_{\text{Hg initial}} = \frac{\text{volume of Hg}}{\text{area}} = \frac{50 \, \text{cm}^3}{1.5 \, \text{cm}^2} = 33.33 \, \text{cm}$
   • A new, unknown liquid is added to one arm of the U-tube, causing the mercury to shift by 2.75 cm.
   • The relative density of mercury (Hg) is given as $13.6$.

2. Goal:

   • Determine the relative density (specific gravity) of the unknown liquid.

3. Method:

   • The difference in height on both sides of the mercury ($2.75 \, \text{cm}$) gives a clue to the pressure exerted by the unknown liquid.
   • In a U-tube in equilibrium, the pressure exerted by the height of the unknown liquid should equal the pressure difference caused by the displaced mercury.

4. Pressure Equilibrium:

   • Let $h_{\text{unknown}}$ be the height of the unknown liquid causing this displacement.
   • The pressure due to the unknown liquid on one side is given by: $P_{\text{unknown}} = \rho_{\text{unknown}} \cdot g \cdot h_{\text{unknown}}$
   • The pressure due to the displaced mercury height (2.75 cm) on the other side is: $P_{\text{Hg}} = \rho_{\text{Hg}} \cdot g \cdot 2.75 \, \text{cm}$
   • Since the system is in equilibrium: $\rho_{\text{unknown}} \cdot g \cdot h_{\text{unknown}} = \rho_{\text{Hg}} \cdot g \cdot 2.75 \, \text{cm}$

5. Calculations:

   • Let’s express the density of mercury in terms of its relative density and the unknown liquid density.
   • Since the relative density of mercury is $13.6$: $\rho_{\text{Hg}} = 13.6 \cdot \rho_{\text{water}}$
   • Rearranging the equilibrium equation, we get: $\rho_{\text{unknown}} = \frac{\rho_{\text{Hg}} \cdot 2.75}{h_{\text{unknown}}}$

   By solving