For a liquid in a U-tube with different temperatures in each arm, the expansion of the liquid in each arm is affected by the temperature difference.

Let:
- \( l_1 \) and \( l_2 \) be the heights of the liquid columns at temperatures \( t_1 \) and \( t_2 \), respectively.
- \( \beta \) be the coefficient of volume expansion of the liquid.

Since the pressure at the same horizontal level in both arms must be equal, we have:
\[
l_1 (1 + \beta t_1) = l_2 (1 + \beta t_2)
\]

Rearranging, we get:
\[
l_1 + l_1 \beta t_1 = l_2 + l_2 \beta t_2
\]

Solving for \( \beta \):
\[
\beta = \frac{l_1 - l_2}{l_2 t_1 - l_1 t_2}
\]

Thus, the coefficient of volume expansion of the liquid is:
\[
\beta = \frac{l_1 - l_2}{l_2 t_1 - l_1 t_2}
\]

If the portion AB is slant, it indicates that both pressure (P) and volume (V) are changing. In thermodynamic terms, this suggests that AB could represent a compression or expansion, process where both pressure decreases and volume decreases.

Given that you mentioned that the correct answer is "the liquid state of matter," it is likely that AB represents the compression of a liquid.

In many thermodynamic diagrams, when a liquid is compressed or expands slightly, both its pressure and volume can change, though not as dramatically as in gases. This phase corresponds to the liquid region of a substance's phase diagram, where both pressure and volume are decreasing, which is why AB represents the liquid state.

To summarize:
- AB is slant: Both pressure and volume are decreasing.
- Liquid state: Liquids are not perfectly incompressible; hence, both pressure and volume can change, but the changes in volume are relatively small compared to gases.

This slant line indicates the behaviour of a liquid as it is compressed, leading to the conclusion that AB represents the liquid state of matter.

The root mean square (r.m.s.) speed \( v_{\text{rms}} \) of gas molecules is given by the formula:

\[
v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}
\]

where:
- \( k_B \) is the Boltzmann constant,
- \( T \) is the temperature (assumed to be room temperature here),
- \( m \) is the mass of one molecule of the gas.

Rearranging to find \( m \):

\[
m = \frac{3 k_B T}{v_{\text{rms}}^2}
\]

For a diatomic gas like \( \text{H}_2 \), using the given \( v_{\text{rms}} = 1930 \, \text{m/s} \), we calculate that this speed corresponds to the molecular mass of hydrogen (\( \text{H}_2 \)). Therefore, the gas is identified as \( \text{H}_2 \).

Since the gases are mixed at constant volume \( V \) and temperature \( T \), we can apply the ideal gas law for each gas:

For each gas, we have:
\[
P = \frac{nRT}{V}
\]

Since both gases have the same pressure \( P \), volume \( V \), and temperature \( T \), they contribute equally to the total pressure when mixed.

After mixing, the total pressure of the mixture is the sum of the partial pressures of each gas:
\[
P_{\text{total}} = P + P = 2P
\]

Thus, the pressure of the mixture is 2P.

To solve this, we use the relationship from Charles's Law, which states that for a gas at constant pressure:

\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\]

where:
- \( V_1 \) is the initial volume at temperature \( T_1 \),
- \( V_2 \) is the final volume at temperature \( T_2 \).

Given:
- Initial temperature \( T_1 = 0^\circ \text{C} = 273 \, \text{K} \),
- \( V_2 = 2V_1 \) (double the initial volume).

Now, substituting into Charles's Law:

\[
\frac{V_1}{273} = \frac{2V_1}{T_2}
\]

Solving for \( T_2 \):

\[
T_2 = 2 \times 273 = 546 \, \text{K}
\]

Converting \( T_2 \) back to Celsius:

\[
T_2 = 546 - 273 = 273^\circ \text{C}
\]

So, the required temperature is 273°C.

To find the ratio of the slopes of curves \( B \) to \( A \) for gases with masses \( m \) and \( 3m \) at constant volume:

1. For an ideal gas at constant volume, \( P = \frac{nRT}{V} \).
2. Since \( n = \frac{\text{mass}}{\text{molar mass}} \), the pressure \( P \propto \frac{\text{mass} \cdot T}{M} \).
3. So, slope \( \propto \frac{\text{mass}}{M} \) for each gas.

For gas \( A \) with mass \( m \), let the slope be \( S \propto \frac{m}{M} \).
For gas \( B \) with mass \( 3m \), slope \( S_B \propto \frac{3m}{M} \).

Thus, the ratio of slopes \( \frac{S_B}{S_A} = \frac{3m/M}{m/M} = 3:1 \).

To analyze the \( PV \) versus \( T \) graph for gases of equal mass, we can use the ideal gas law and the fact that for equal masses of different gases, the number of moles \( n \) varies inversely with the molar mass \( M \) of each gas.

1. **Ideal Gas Law**:
\[
PV = nRT
\]
For a given temperature \( T \), \( PV \propto n \). Thus, \( PV \) will be larger for gases with more moles (i.e., smaller molar mass).

2. **Order of Molar Masses**:
- \( H_2 \): Molar mass = 2 g/mol
- \( He \): Molar mass = 4 g/mol
- \( O_2 \): Molar mass = 32 g/mol

Since equal masses are used, the number of moles \( n \) will be highest for \( H_2 \), followed by \( He \), and lowest for \( O_2 \).

3. **Conclusion**:
- Line with the highest \( PV \) value corresponds to the gas with the highest number of moles, i.e., \( H_2 \) (line **C**).
- The middle line corresponds to \( He \) (line **B**).
- The line with the lowest \( PV \) value corresponds to \( O_2 \) (line **A**).

Thus, **C corresponds to \( H_2 \)**, **B to \( He \)**, and **A to \( O_2 \)**.

Given that the gas obeys the law \( VP^2 = \text{constant} \).

1. Initially:
\[
VP^2 = k
\]

2. When the volume changes from \( V \) to \( 2V \):
\[
(2V)P'^2 = k
\]
where \( P' \) is the new pressure.

Since \( VP^2 = (2V)P'^2 \), we can relate the pressures as:
\[
P'^2 = \frac{P^2}{2}
\]
\[
P' = \frac{P}{\sqrt{2}}
\]

3. Use the ideal gas law initially and finally:
\[
PV = nRT
\]
\[
P' \cdot 2V = nRT'
\]

Substitute \( P' = \frac{P}{\sqrt{2}} \):
\[
\frac{P}{\sqrt{2}} \cdot 2V = nRT'
\]

4. Simplify:
\[
\sqrt{2} PV = nRT'
\]

Since \( PV = nRT \):
\[
\sqrt{2} \cdot nRT = nRT'
\]
\[
T' = \sqrt{2} T
\]

So, the new temperature is \( \sqrt{2} T \).

To determine which line corresponds to each type of gas (monoatomic, diatomic, polyatomic), we can use the fact that the specific heat at constant pressure \( C_p \) varies with the degrees of freedom of each gas. Since \( Q = n C_p \Delta T \), for a given \( Q \), the slope of the \( Q \)-\( \Delta T \) line is inversely proportional to \( C_p \).

1. Monoatomic gas (M): \( C_p = \frac{5}{2} R \).
2. Diatomic gas (D): \( C_p = \frac{7}{2} R \).
3. Polyatomic gas (P): \( C_p \) is higher than both monoatomic and diatomic due to additional rotational degrees of freedom.

Since the slope is inversely related to \( C_p \):
- Line with the lowest slope (shallowest) corresponds to the monoatomic gas (M).
- Line with a medium slope corresponds to the diatomic gas (D).
- Line with the steepest slope corresponds to the polyatomic gas (P).

Thus, lines  a, b, and c correspond to  P, D, and M, respectively.

For an ideal gas, the relationship between \( C_p \) and \( C_v \) is:

\[
C_p - C_v = R
\]

where \( R \approx 2 \, \text{cal/mole-K} \).

Check each option:

1. If \( C_v = 3 \) and \( C_p = 5 \):
\[
C_p - C_v = 5 - 3 = 2 = R
\]
This matches the expected result.

2. Other sets will not satisfy \( C_p - C_v = 2 \) as accurately.

Thus, \( C_v = 3 \) and \( C_p = 5 \) is the most reliable set.