https://file.aiquickdraw.com/mathbot/file/1731000165879g13znlyg.jpg

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

Un tubo en forma de U tiene una sección de 1.5 cm² y contiene inicialmente 50 cm³ de Hg. A un brazo se le agrega un líquido desconocido y se observa que el desnivel de Hg en los brazos es de 2.75 cm. Determine la densidad relativa del líquido desconocido (densidad relativa del mercurio 13.6).

Solve a fluid mechanics problem involving a U-shaped tube with mercury and an unknown liquid. Learn how to calculate the relative density of the unknown liquid by analyzing the pressure equilibrium and displacement caused by adding the liquid. Suitable for Grades 11-12 students.

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