The heat absorbed during the fusion process is given by the formula:

\[
H = m \cdot L
\]

Where \( H \) is the heat absorbed, \( m \) is the mass, and \( L \) is the latent heat of fusion. Since both substances have the same mass and are being heated at the same rate (6 cal/s), the amount of heat absorbed depends on the duration of the phase change (the flat portion of the graph).

- For substance A: The time interval for fusion (flat portion of the graph) is from \( t = 3 \) to \( t = 5 \), so the time spent during fusion is \( 5 - 3 = 2 \) seconds.
- For substance B: The time interval for fusion is from \( t = 5 \) to \( t = 7 \), so the time spent during fusion is also \( 7 - 5 = 2 \) seconds.

However, the time during which each substance is being heated prior to the phase change is different, which affects the total heat absorbed by each. To calculate the heat absorbed for fusion alone (latent heat), we need to account for the heat absorbed during the temperature increase as well.

- For substance A: It takes 2 seconds to reach the fusion point, then 2 seconds to complete the fusion process.
- For substance B: It takes 3 seconds to reach the fusion point, then 2 seconds to complete the fusion.

Now, let's calculate the total heat absorbed for each substance:

\[
H_A = \text{Heat absorbed during heating phase} + \text{Heat absorbed during fusion}
\]

Substance A spends 2 seconds in the heating phase (at a rate of 6 cal/s), so it absorbs \( 2 \times 6 = 12 \) calories in the heating phase. For fusion, it absorbs heat for 2 seconds at the same rate:

\[
H_A = 12 + (2 \times 6) = 12 + 12 = 24 \, \text{cal}
\]

For substance B, it spends 3 seconds in the heating phase, absorbing \( 3 \times 6 = 18 \) calories. During the fusion, it also absorbs heat for 2 seconds:

\[
H_B = 18 + (2 \times 6) = 18 + 12 = 30 \, \text{cal}
\]

Now, the ratio of heat absorbed by substance A to substance B is:

\[
\frac{H_A}{H_B} = \frac{24}{30} = \frac{8}{5}
\]

Thus, the correct ratio of heat absorbed by substance A to substance B is indeed \(\frac{8}{5}\).

Let the length of the part made of metal A be \( L_A \) and that of metal B be \( L_B \), with \( L_A + L_B = 20 \) cm.

Given:
- Expansion of metal A's rod = 0.075 cm, so expansion per cm for metal A = \( \frac{0.075}{20} = 0.00375 \) cm.
- Expansion of metal B's rod = 0.045 cm, so expansion per cm for metal B = \( \frac{0.045}{20} = 0.00225 \) cm.

The combined expansion of the third rod is 0.060 cm:

\[
L_A \cdot 0.00375 + L_B \cdot 0.00225 = 0.060
\]

Since \( L_B = 20 - L_A \), substitute:

\[
L_A \cdot 0.00375 + (20 - L_A) \cdot 0.00225 = 0.060
\]

Expanding and solving:

\[
0.00375L_A + 0.045 - 0.00225L_A = 0.060
\]
\[
0.0015L_A = 0.015
\]
\[
L_A = \frac{0.015}{0.0015} = 10 \text{ cm}
\]

So, the portion made of metal A has length 10 cm.

To find \( x \), we need to compare the Z scale with the Fahrenheit scale.

1. Range on the Z scale:
\[
65^\circ Z - (-14^\circ Z) = 65 + 14 = 79^\circ Z
\]

2. Range on the Fahrenheit scale:
The boiling and freezing points of water are 212ºF and 32ºF, respectively, so:
\[
212^\circ F - 32^\circ F = 180^\circ F
\]

3. Relationship between scales:
A change of \( 79^\circ Z \) corresponds to \( 180^\circ F \). Therefore:
\[
x = \frac{180}{79}
\]

So, \( x = \frac{180}{79} \).

To find the rise in temperature of the block, we can follow these steps:

1. Calculate the potential energy of the block before it falls:
\[
\text{Potential energy} = m \times g \times h
\]
where:
- \( m = 20 \, \text{kg} \) (mass of the block)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 20 \, \text{m} \) (height)

\[
\text{Potential energy} = 20 \times 10 \times 20 = 4000 \, \text{J}
\]

2. Energy converted to heat:
Since 75% of the energy goes into raising the temperature of the block:
\[
\text{Heat energy} = 0.75 \times 4000 = 3000 \, \text{J}
\]

3. Use the heat energy to calculate the rise in temperature:
The formula for heat energy is:
\[
Q = m \times c \times \Delta T
\]
where:
- \( Q = 3000 \, \text{J} \) (heat energy)
- \( m = 20 \, \text{kg} \)
- \( c = 100 \, \text{J/kg°C} \) (specific heat of the block)
- \( \Delta T \) is the rise in temperature (to be found)

Rearranging the formula to solve for \( \Delta T \):
\[
\Delta T = \frac{Q}{m \times c} = \frac{3000}{20 \times 100} = \frac{3000}{2000} = 1.5^\circ \text{C}
\]

Thus, the rise in temperature of the block is 1.5°C.

In the provided graph, the temperature of the material increases with heat input. The flat portions of the graph (AB and CD) represent phase changes where heat is absorbed but the temperature remains constant. Specifically:

- Segment AB corresponds to the latent heat of fusion (solid to liquid).
- Segment CD corresponds to the latent heat of vaporization (liquid to gas).

The length of CD is given as twice the length of AB, meaning the heat input required for vaporization is twice that of fusion.

However, the false conclusion stated in the answer is that "the latent heat of fusion is twice the latent heat of vaporization." This contradicts the graph because CD represents vaporization and is twice the length of AB, implying that the latent heat of vaporization is actually twice the latent heat of fusion, not the other way around.

Hence, the false conclusion is that the latent heat of fusion is larger than the latent heat of vaporization. The correct interpretation from the graph is that the latent heat of vaporization is twice the latent heat of fusion.

To find the water equivalent of the calorimeter, we can use the principle of heat exchange:

\[
\text{Heat lost by hotter water} = \text{Heat gained by colder water} + \text{Heat gained by the calorimeter}
\]

Let \( m_1 = 0.2 \, \text{kg} \), \( T_1 = 30^\circ \text{C} \) (colder water), \( m_2 = 0.1 \, \text{kg} \), \( T_2 = 60^\circ \text{C} \) (hotter water), and the final temperature \( T_f = 35^\circ \text{C} \). Let \( W_c \) be the water equivalent of the calorimeter, and \( c = 4200 \, \text{J/kg}^\circ \text{C} \) be the specific heat capacity of water.

Heat lost by hotter water:
\[
Q_{\text{lost}} = m_2 c (T_2 - T_f) = 0.1 \times 4200 \times (60 - 35) = 0.1 \times 4200 \times 25 = 10500 \, \text{J}
\]

Heat gained by colder water:
\[
Q_{\text{gained (water)}} = m_1 c (T_f - T_1) = 0.2 \times 4200 \times (35 - 30) = 0.2 \times 4200 \times 5 = 4200 \, \text{J}
\]

Heat gained by calorimeter:
\[
Q_{\text{gained (calorimeter)}} = W_c \times c \times (T_f - T_1) = W_c \times 4200 \times 5
\]

Using heat balance equation:
\[
Q_{\text{lost}} = Q_{\text{gained (water)}} + Q_{\text{gained (calorimeter)}}
\]
\[
10500 = 4200 + W_c \times 4200 \times 5
\]
\[
10500 - 4200 = 21000 W_c
\]
\[
6300 = 21000 W_c
\]
\[
W_c = \frac{6300}{21000} = 0.3 \, \text{kg}
\]

Since the water equivalent is in terms of mass and \( c = 4200 \, \text{J/kg}^\circ \text{C} \), the water equivalent in \( \text{J/K} \) is:
\[
W_c \times c = 0.3 \times 4200 = 1260 \, \text{J/K}
\]

So, the water equivalent of the calorimeter is 1260 J/K.

1. Heat gained by ice to reach 0°C:
\[
Q_{\text{ice heating}} = 10 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 20°C = 100 \, \text{cal}
\]

2. Heat gained by ice to melt:
\[
Q_{\text{ice melting}} = 10 \, \text{g} \cdot 80 \, \text{cal/g} = 800 \, \text{cal}
\]

Total heat gained by ice:
\[
Q_{\text{total ice}} = 100 \, \text{cal} + 800 \, \text{cal} = 900 \, \text{cal}
\]

3. Heat lost by water:
\[
Q_{\text{water cooling}} = 10 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (50 - T) = 500 - 10T \, \text{cal}
\]

4. Set heat gained equal to heat lost:
\[
500 - 10T = 900 \Rightarrow -10T = 400 \Rightarrow T = -40°C
\]

Since this temperature is below 0°C, all the ice will not melt.

5. Equilibrium Calculation:
Let \( x \) be the mass of ice that melts:
\[
500 = 70x + 100 \Rightarrow 400 = 70x \Rightarrow x \approx 5.71 \, \text{g}
\]

Result:
The final temperature of the system will be **0°C**.

To find the ratio of the masses of \( \text{N}_2 \) and \( \text{He} \) in vessels \( A \) and \( B \), we use the ideal gas law:

\[
PV = \frac{m}{M} RT
\]

where:
- \( m \) is the mass of the gas,
- \( M \) is the molar mass,
- \( P, V, R, T \) are pressure, volume, gas constant, and temperature, respectively.

For vessel \( A \) (containing \( \text{N}_2 \)):
\[
m_{\text{N}_2} = \frac{PVM_{\text{N}_2}}{RT}
\]

For vessel \( B \) (containing \( \text{He} \) at temperature \( 2T \)):
\[
m_{\text{He}} = \frac{PV M_{\text{He}}}{R \cdot 2T} = \frac{PVM_{\text{He}}}{2RT}
\]

Now, the ratio of the masses \( \frac{m_{\text{N}_2}}{m_{\text{He}}} \) is:

\[
\frac{m_{\text{N}_2}}{m_{\text{He}}} = \frac{\frac{PVM_{\text{N}_2}}{RT}}{\frac{PVM_{\text{He}}}{2RT}} = \frac{M_{\text{N}_2}}{M_{\text{He}}} \times 2
\]

Since \( M_{\text{N}_2} = 28 \) and \( M_{\text{He}} = 4 \):

\[
\frac{m_{\text{N}_2}}{m_{\text{He}}} = \frac{28}{4} \times 2 = 14
\]

Thus, the ratio of masses of \( \text{N}_2 \) to \( \text{He} \) is  14:1.

Thank you for the clarification. Given the correct answer, let's interpret the graph accordingly.

In this pressure (\( P \)) vs. temperature (\( T \)) graph:

1. **Parallel Lines**: When two lines have the same slope, it implies they have the same volume. This is because the slope in a \( P \)-\( T \) graph (for constant \( V \)) is given by \( \frac{nR}{V} \).

- Since lines **1** and **2** are parallel, we have \( V_1 = V_2 \).
- Similarly, lines **3** and **4** are parallel, so \( V_3 = V_4 \).

2. **Comparing Slopes**: The line with a steeper slope corresponds to a smaller volume, and the line with a shallower slope corresponds to a larger volume.

- Since lines 1 and 2 have a shallower slope compared to lines 3 and 4, we conclude that \( V_1 = V_2 > V_3 = V_4 \).

Final Answer:
- \( V_1 = V_2 \), \( V_3 = V_4 \), and \( V_2 > V_3 \).

The system shows two coaxial cylinders made of materials with different thermal conductivities (\(K_1\) and \(K_2\)) and radii \(r\) and \(3r\). For axial heat flow, the conductivities act in parallel.

To find the equivalent thermal conductivity (\(K_{\text{eq}}\)) of the entire system, we use the formula for thermal conductivities in parallel, similar to how resistances in parallel are treated for electrical systems:

\[
K_{\text{eq}} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}
\]

Where:
- \(A_1 = \pi r^2\) is the area of the inner cylinder.
- \(A_2 = \pi (3r)^2 - \pi r^2\  = 8\pi r^2\) is the area of the outer cylinder.

Substituting into the equation:

\[
K_{\text{eq}} = \frac{K_1 (\pi r^2) + K_2 (8\pi r^2)}{\pi r^2 + 8\pi r^2}
\]

Simplifying:

\[
K_{\text{eq}} = \frac{K_1 + 8K_2}{1 + 8} = \frac{K_1 + 8K_2}{9}
\]

Thus, the equivalent thermal conductivity is:

\[
K_{\text{eq}} = \frac{K_1 + 8K_2}{9}
\]