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To solve the problem, we need to analyze the pressure changes in the system.

### Given dataTo solve the problem, we need to analyze the pressure changes in the system.

### Given 
1. **Pressure change in air**: \(\Delta P = 0.7 \, \text{kPa} = 700 \, \text{Pa}\)
2. **Drop in brine level**: \(\Delta h = 5 \, \text{mm} = 0.005 \, \text{m}\)
3. **Specific gravity (SG) of brine**: \(SG_{\text{brine}} = 1.1\)
4. **Specific gravity (SG) of mercury**: \(SG_{\text{mercury}} = 13.56\)

We are asked to determine the ratio of \(A_2/A_1\) (the areas in the two columns).

### Approach:

1. **Pressure equation**:
   The pressure difference caused by the height change in the brine can be related to the pressure change in the air. The system is a U-tube manometer, so we can write the pressure balance for the fluid columns.

   The pressure drop due to the brine in the right column is:
   \[
   \Delta P_{\text{brine}} = \rho_{\text{brine}} g \Delta h
   \]
   Where:
   - \(\rho_{\text{brine}}\) is the density of brine, calculated from its specific gravity: 
     \[
     \rho_{\text{brine}} = SG_{\text{brine}} \times \rho_{\text{water}} = 1.1 \times 1000 \, \text{kg/m}^3 = 1100 \, \text{kg/m}^3
     \]
   - \(g\) is the acceleration due to gravity (\(9.81 \, \text{m/s}^2\)).

   Now calculate the pressure drop:
   \[
   \Delta P_{\text{brine}} = 1100 \times 9.81 \times 0.005 = 53.955 \, \text{Pa}
   \]

2. **Mercury-brine balance**:
   The height change in the mercury side will be the opposite, and it should relate to the area difference. The pressure change due to mercury is:
   \[
   \Delta P_{\text{mercury}} = \rho_{\text{mercury}} g h_{\text{mercury}}
   \]
   Using the specific gravity of mercury:
   \[
   \rho_{\text{mercury}} = 13.56 \times 1000 = 13560 \, \text{kg/m}^3
   \]
   Since the system is static, the pressure change due to the height drop in the brine and the corresponding rise in mercury should balance out with the pressure change in the air.

3. **Relating areas and pressures**:
   The height drop in the brine (right side) is related to the height rise in mercury (left side) using the area ratio:
   \[
   A_1 h_{\text{mercury}} = A_2 \Delta h_{\text{brine}}
   \]
   Therefore:
   \[
   \frac{A_2}{A_1} = \frac{h_{\text{mercury}}}{\Delta h_{\text{brine}}}
   \]

4. **Final relation**:
   The pressure changes are equal, so:
   \[
   \rho_{\text{mercury}} g h_{\text{mercury}} = \Delta P_{\text{brine}}
   \]
   Solving for \(h_{\text{mercury}}\):
   \[
   h_{\text{mercury}} = \frac{\Delta P_{\text{brine}}}{\rho_{\text{mercury}} g}
   = \frac{53.955}{13560 \times 9.81}
   \approx 0.0004 \, \text{m} = 0.4 \, \text{mm}
   \]

Thus, the area ratio is:
\[
\frac{A_2}{A_1} = \frac{0.4 \, \text{mm}}{5 \, \text{mm}} = 0.08
\]

So, the ratio \(A_2/A_1 = 0.08\).

Would you like further details on this process or have any other questions?

### Related questions:
1. How does the specific gravity affect pressure changes in fluids?
2. What happens if the brine height change were doubled?
3. How would the result change if a different liquid were used instead of mercury?
4. Why does the air pressure only affect one side of the system?
5. How can you experimentally measure the pressure changes in such a system?

### Tip:
Always ensure unit consistency when working with pressures and heights in hydrostatic systems to avoid calculation errors.

Consider the system shown in Fig. P2. If a change of 0.7 kPa in the pressure of air causes the brine–mercury interface in the right column to drop by 5 mm in the brine level in the right column while the pressure in the brine pipe remains constant, determine the ratio of A2/A1.

Solve the problem of determining the area ratio A2/A1 in a brine–mercury system using principles of hydrostatics. Includes detailed calculation of pressure drop due to brine and mercury columns, using specific gravity and fluid pressure formulas.

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