Given the relation between temperature scales \( A \) and \( B \):

\[
\frac{A - 42}{100} = \frac{B - 7}{220}
\]

To find the temperature at which both scales show the same reading, set \( A = B = x \):

\[
\frac{x - 42}{100} = \frac{x - 7}{220}
\]

Cross-multiplying:

\[
220(x - 42) = 100(x - 7)
\]

Expanding and solving for \( x \):

\[
220x - 9240 = 100x - 700
\]
\[
120x = 8540
\]
\[
x = \frac{8540}{120} = 12
\]

Thus, the temperature at which both scales read the same is 12.

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.

On heating an object its photographic expansion takes place. so, distance between any two point increases.

To find absolute zero in Fahrenheit, we use the relationship between Celsius and Fahrenheit:

\[
F = \frac{9}{5}C + 32
\]

Absolute zero in Celsius is \(-273.15^\circ C\). Substitute this into the formula:

\[
F = \frac{9}{5}(-273.15) + 32
\]
\[
F = -491.67 + 32 = -459.67
\]

Rounding to the nearest whole number, we get \(-460^\circ F\).

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

The change in length \( \Delta L \) due to temperature change is given by:

\[
\Delta L = L_0 \alpha \Delta T
\]

where:
- \( L_0 = 10 \, \text{cm} \)
- \( \alpha = 11 \times 10^{-6} / ^\circ \text{C} \)
- \( \Delta T = 20^\circ \text{C} - 19^\circ \text{C} = 1^\circ \text{C} \)

Substitute the values:

\[
\Delta L = 10 \times 11 \times 10^{-6} \times 1 = 11 \times 10^{-5} \, \text{cm}
\]

Thus, the bar will be \( 11 \times 10^{-5} \, \text{cm} \) shorter at 19ºC.

To find the final temperature when 5 g of ice at 0°C is mixed with 5 g of steam at 100°C, we need to consider the energy exchange between the ice and the steam.

We proceed step by step:

1. Heat required to melt the ice into water at 0°C:

\[
Q_1 = m_{\text{ice}} \times L_f = 5 \, \text{g} \times 80 \, \text{cal/g} = 400 \, \text{cal}
\]
This is the heat required to convert 5 g of ice at 0°C into 5 g of water at 0°C.

2. Heat released by steam as it condenses into water at 100°C:

\[
Q_2 = m_{\text{steam}} \times L_v = 5 \, \text{g} \times 540 \, \text{cal/g} = 2700 \, \text{cal}
\]
This is the heat released when 5 g of steam condenses into water at 100°C.

3. Heat required to raise the temperature of 5 g of water from 0°C to 100°C:

\[
Q_3 = m_{\text{water}} \times c \times \Delta T = 5 \, \text{g} \times 1 \, \text{cal/g°C} \times (100 - 0) = 5 \times 100 = 500 \, \text{cal}
\]

 Total heat available from the steam:

- The steam releases 2700 cal by condensing.
- The ice requires 400 cal to melt, and then 500 cal to be heated from 0°C to 100°C, totaling 900 cal.

Since the heat available from the steam (2700 cal) is more than the 900 cal required to melt the ice and raise its temperature to 100°C, the final temperature will be 100°C.

In conclusion, all the ice melts and the final temperature of the mixture is 100°C.

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.

To calculate the heat required to convert 40 g of ice at –20°C into water at 20°C, we need to consider three steps:

1. Heating ice from -20°C to 0°C:
\[
Q_1 = m \times c_{\text{ice}} \times \Delta T = 40 \, \text{g} \times 0.5 \, \text{cal/g°C} \times (0 - (-20)) = 40 \times 0.5 \times 20 = 400 \, \text{cal}
\]

2. Melting ice at 0°C (latent heat of fusion):
\[
Q_2 = m \times L_f = 40 \, \text{g} \times 80 \, \text{cal/g} = 3200 \, \text{cal}
\]

3. Heating water from 0°C to 20°C:
\[
Q_3 = m \times c_{\text{water}} \times \Delta T = 40 \, \text{g} \times 1 \, \text{cal/g°C} \times (20 - 0) = 40 \times 1 \times 20 = 800 \, \text{cal}
\]

Total heat required:

\[
Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 400 + 3200 + 800 = 4400 \, \text{cal}
\]

Thus, the total heat required is 4400 calories.

Since both gases \( A \) and \( B \) have the same temperature \( T \), pressure \( P \), and volume \( V \), each gas has the same number of moles, \( n \), by the ideal gas law:

\[
PV = nRT \Rightarrow n = \frac{PV}{RT}
\]

When gases \( A \) and \( B \) are mixed, the total number of moles becomes \( 2n \). Since temperature \( T \) and volume \( V \) remain the same, the total pressure \( P_{\text{mixture}} \) will be:

\[
P_{\text{mixture}} V = (2n)RT
\]

\[
P_{\text{mixture}} = 2P
\]

Answer: The pressure of the mixture is \( 2P \).