To solve the problem with the updated parameters, let's rework the calculations based on the new values.

Updated Given Data:

1. Gauge pressure of air: $P_{\text{air}} = 30 \, \text{kPa}$
2. Height of water column $h_2$: $h_2 = 50 \, \text{cm} = 0.50 \, \text{m}$
3. Height of the pipe above the water surface $h_1$: $h_1 = 8 \, \text{cm} = 0.08 \, \text{m}$
4. Length $L_1$ and $L_2$ of the fluid columns in the manometer: $L_1 = L_2 = 6 \, \text{cm} = 0.06 \, \text{m}$
5. Specific Gravity (SG) of the manometer fluid: SG = 2.4
6. Density of water $\rho_{\text{water}}$: Approximately $1000 \, \text{kg/m}^3$
7. Acceleration due to gravity $g$: $9.81 \, \text{m/s}^2$

Key Equations:

1. Manometer fluid density $\rho_{\text{gage}}$: $\rho_{\text{gage}} = \text{SG} \times \rho_{\text{water}} = 2.4 \times 1000 \, \text{kg/m}^3 = 2400 \, \text{kg/m}^3$

2. Pressure at the water surface inside the container: $P_{\text{surface}} = P_{\text{air}} = 30 \, \text{kPa} = 30 \times 10^3 \, \text{Pa}$

3. Pressure difference due to the manometer fluid (using the manometer readings $L_1$ and $L_2$): The difference in pressure indicated by the manometer is: $\Delta P_{\text{gage}} = \rho_{\text{gage}} \times g \times (L_2 - L_1)$ However, since $L_1 = L_2$, there is no differential pressure contribution from the manometer fluid itself. The difference will be zero due to the symmetric setup: $\Delta P_{\text{gage}} = 0$

4. Pressure at the base of the water column $h_2$: $P_{\text{base}} = P_{\text{surface}} + \rho_{\text{water}} \times g \times h_2$ Substituting values: $P_{\text{base}} = 30 \times 10^3 \, \text{Pa} + 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.50 \, \text{m}$ $P_{\text{base}} = 30000 \, \text{Pa} + 4905 \, \text{Pa}$ $P_{\text{base}} \approx 34905 \, \text{Pa}$

5. Pressure in the pipe (at height $h_1$ above the water surface): $P_{\text{pipe}} = P_{\text{base}} - \rho_{\text{water}} \times g \times h_1$ Substituting the values: $P_{\text{pipe}} = 34905 \, \text{Pa} - 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.08 \, \text{m}$ $P_{\text{pipe}} = 34905 \, \text{Pa} - 784.8 \, \text{Pa}$ $P_{\text{pipe}} \approx 34120.2 \, \text{Pa}$

Final Answer:

The pressure in the pipe is approximately 34.12 kPa.

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Would you like any further explanations or have any additional questions? Here are some related questions to consider:

1. What would happen if the lengths $L_1$ and $L_2$ were different, introducing a differential pressure?
2. How does the specific gravity of the manometer fluid affect the pressure reading in this scenario?
3. If the pipe's height $h_1$ were further increased, how would that influence the pressure?
4. How would the results change if the fluid in the container were oil with a different density?
5. Can this manometer setup be used for highly compressible fluids like gases? Why or why not?

Tip: Always ensure that you account for the specific gravity when dealing with manometric fluids, as it directly influences the pressure readings by altering the density factor in the calculations.