Let the length of the other rod be \( L \).

Since the difference in their lengths remains the same at all temperatures, the expansion of each rod must be identical.

For the first rod:
\[
\Delta L_1 = 40 \times \alpha_1 \times \Delta T
\]

For the second rod:
\[
\Delta L_2 = L \times \alpha_2 \times \Delta T
\]

Since \(\Delta L_1 = \Delta L_2\):
\[
40 \times \alpha_1 = L \times \alpha_2
\]

Substitute \(\alpha_1 = 6 \times 10^{-6}/^\circ \text{C}\) and \(\alpha_2 = 4 \times 10^{-6}/^\circ \text{C}\):
\[
40 \times 6 \times 10^{-6} = L \times 4 \times 10^{-6}
\]

Dividing both sides by \(4 \times 10^{-6}\):
\[
L = \frac{40 \times 6}{4} = 60 \, \text{cm}
\]

So, the length of the other rod is \( 60 \, \text{cm} \).

We can solve this problem using the concept of heat exchange. Let \( m_h \) be the mass of hot water at 100°C, and \( m_c \) be the mass of cold water at 10°C. The total mass of water is given as 20 kg, so:

\[
m_h + m_c = 20 \, \text{kg}
\]

We also know that the final temperature of the mixture is 35°C. The heat gained by cold water must equal the heat lost by hot water:

\[
\text{Heat gained by cold water} = \text{Heat lost by hot water}
\]

Let the specific heat of water be \( c = 4200 \, \text{J/kg}^\circ \text{C} \). Using the formula for heat change:

\[
m_c c (35 - 10) = m_h c (100 - 35)
\]

Since the specific heat \( c \) is the same for both, it cancels out:

\[
m_c (35 - 10) = m_h (100 - 35)
\]
\[
m_c \times 25 = m_h \times 65
\]

We also know \( m_h + m_c = 20 \), so \( m_c = 20 - m_h \). Substituting this into the equation:

\[
(20 - m_h) \times 25 = m_h \times 65
\]
\[
500 - 25 m_h = 65 m_h
\]
\[
500 = 90 m_h
\]
\[
m_h = \frac{500}{90} \approx 5.56 \, \text{kg}
\]

Thus, the mass of hot water needed is approximately 5.56 kg.

Given:
- Mass of water \( m_w = 0.2 \, \text{kg} \)
- Initial temperature of water \( T_w = 80^\circ \text{C} \)
- Final temperature \( T_f = 60^\circ \text{C} \)
- Specific heat of water \( c_w = 4200 \, \text{J/kg}^\circ \text{C} \)
- Mass of solid \( m_s = 0.2 \, \text{kg} \)
- Initial temperature of solid \( T_s = 20^\circ \text{C} \)
- Specific heat of solid \( c_s \) (to be found)

Heat lost by water:
\[
Q_{\text{lost (water)}} = m_w c_w (T_w - T_f) = 0.2 \times 4200 \times (80 - 60) = 16800 \, \text{J}
\]

Heat gained by solid:
\[
Q_{\text{gained (solid)}} = m_s c_s (T_f - T_s) = 0.2 \times c_s \times (60 - 20) = 8 c_s
\]

From the heat exchange equation:
\[
16800 = 8 c_s
\]
\[
c_s = \frac{16800}{8} = 2100 \, \text{J/kg}^\circ \text{C}
\]

Since the specific heat of water is \( 4200 \, \text{J/kg}^\circ \text{C} \), we see that the specific heat of the solid is:

\[
c_s = \frac{4200}{2} = 2100 \, \text{J/kg}^\circ \text{C}
\]

Thus, the specific heat of the solid is indeed half of the specific heat of water.

In Part b to c temperature remains constant even when heat is supplied

Similarly in part d to e temperature remains constant

So state is changing in part b to c and in d to e.

Given:

- Diameter ratio = 1:2
- Density ratio = 2:1
- Specific heat ratio = 1:3

Step 1: Mass ratio
Mass is proportional to \( \text{density} \times \text{volume} \).
Volume is proportional to the cube of the diameter, so the volume ratio is \( (1:2)^3 = 1:8 \).

Thus, mass ratio = \( \text{density} \times \text{volume} = \frac{2 \times 1}{1 \times 8} = 1:4 \).

Step 2: Thermal capacity ratio
Thermal capacity = mass × specific heat.

So, thermal capacity ratio = \( \frac{1 \times 1}{4 \times 3} = 1:12 \).

Final Answer:
The ratio of the thermal capacities is 1:12.

To find the resulting temperature when two liquids A and B are mixed, we can use the principle of conservation of heat energy: the heat lost by the hotter liquid (A) equals the heat gained by the cooler liquid (B).

Given:
- Temperature of liquid A = 75°C
- Temperature of liquid B = 15°C
- Masses of A and B are in the ratio 2:3, so let the masses of A and B be \( 2m \) and \( 3m \), respectively.
- Specific heats of A and B are in the ratio 3:4, so let the specific heats of A and B be \( 3c \) and \( 4c \), respectively.

Let the final temperature of the mixture be \( T \).

Step 1: Heat lost by liquid A (hotter liquid):
\[
Q_A = \text{mass of A} \times \text{specific heat of A} \times (\text{initial temp of A} - T)
\]
\[
Q_A = 2m \times 3c \times (75 - T) = 6mc(75 - T)
\]

Step 2: Heat gained by liquid B (cooler liquid):
\[
Q_B = \text{mass of B} \times \text{specific heat of B} \times (T - \text{initial temp of B})
\]
\[
Q_B = 3m \times 4c \times (T - 15) = 12mc(T - 15)
\]

Step 3: Apply the conservation of heat:
Heat lost by A = Heat gained by B
\[
6mc(75 - T) = 12mc(T - 15)
\]

 Step 4: Simplify and solve for \( T \):
Cancel out \( mc \) from both sides:
\[
6(75 - T) = 12(T - 15)
\]
Expand both sides:
\[
450 - 6T = 12T - 180
\]
Combine like terms:
\[
450 + 180 = 12T + 6T
\]
\[
630 = 18T
\]
Solve for \( T \):
\[
T = \frac{630}{18} = 35 \, ^\circ\text{C}
\]

Final Answer:
The resulting temperature is 35°C.

Here’s a short solution:

Given

- Mass of water = 54 gm
- Initial temperature of water = 30°C, Final temperature = 90°C
- Latent heat of steam = 530 cal/gm
- Specific heat of water = 1 cal/gm°C

Step 1: Heat gained by water to reach 90°C:

\[
Q_{\text{gained}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T = 54 \times 1 \times (90 - 30) = 3240 \, \text{cal}
\]

Step 2: Heat lost by steam:

\[
Q_{\text{lost}} = m_s \times (530 + 10) = 540 m_s
\]

Step 3: Equating heat gained and heat lost:

\[
3240 = 540 m_s \quad \Rightarrow \quad m_s = \frac{3240}{540} = 6 \, \text{gm}
\]

Step 4: Total mass of the mixture:

\[
m_{\text{mixture}} = 54 \, \text{gm} + 6 \, \text{gm} = 60 \, \text{gm}
\]

Final Answer:
The mass of the mixture is 60 grams.

For an ideal gas, at constant volume, the pressure \( P \) is directly proportional to the temperature \( T \) (Gay-Lussac's Law):

\[
P \propto T
\]

In the graph, each line represents a pressure-temperature relationship at different volumes. A steeper slope indicates a smaller volume (since \( P = \frac{nRT}{V} \) and a higher \( V \) gives a less steep line).

From the figure:
1. Lines \( 1 \) and \( 2 \) have the same slope, indicating \( V_1 = V_2 \).
2. Lines \( 3 \) and \( 4 \) also have the same slope, indicating \( V_3 = V_4 \).
3. The slope of lines \( 3 \) and \( 4 \) is steeper than that of lines \( 1 \) and \( 2 \), meaning \( V_2 > V_3 \).

Thus, the correct answer is \( V_1 = V_2 \), \( V_3 = V_4 \), and \( V_2 > V_3 \).

To solve this, we can use Gay-Lussac's Law, which states:

\[
\frac{P_1}{T_1} = \frac{P_2}{T_2}
\]

Given:
- Initial pressure \( P_1 = 870 \, \text{mm Hg} \),
- Final pressure \( P_2 = 1800 \, \text{mm Hg} \),
- Initial temperature \( T_1 = 17^\circ \text{C} = 17 + 273 = 290 \, \text{K} \).

Rearrange to find \( T_2 \):

\[
T_2 = \frac{P_2 \times T_1}{P_1}
\]

Substitute the values:

\[
T_2 = \frac{1800 \times 290}{870}
\]

Calculating this:

\[
T_2 = 600 \, \text{K}
\]

So, the temperature at which the pressure becomes 1800 mm of Hg is  600 K.

To make the volume of an ideal gas four times its initial volume, we can use Boyle's Law, which states:

\[
P \propto \frac{1}{V} \quad \text{(at constant temperature)}
\]

If the initial pressure is \( P \) and the initial volume is \( V \), we want the final volume \( V_{\text{final}} = 4V \).

According to Boyle's Law:

\[
P_{\text{final}} \times V_{\text{final}} = P \times V
\]

Substituting \( V_{\text{final}} = 4V \):

\[
P_{\text{final}} \times 4V = P \times V
\]

\[
P_{\text{final}} = \frac{P}{4}
\]

So, to make the volume four times, we quarter the pressure at constant temperature.