To find the equivalent thermal conductivity (\(K_{\text{eq}}\)) of the compound slab with two materials in series, we use the formula for thermal resistances in series.

Formula:
The total thermal resistance \(R_{\text{total}}\) for two materials in series is:

\[
R_{\text{total}} = R_1 + R_2 = \frac{L_1}{K_1 A} + \frac{L_2}{K_2 A}
\]

Where \(L_1 = L_2\) (equal thickness) and \(A\) is the cross-sectional area (same for both). The equivalent conductivity \(K_{\text{eq}}\) is given by:

\[
R_{\text{total}} = \frac{2L}{K_{\text{eq}} A}
\]

Given:
- Thickness of each layer, \(L_1 = L_2 = L/2\)
- Thermal conductivities \(K_1 = K\) and \(K_2 = 2K\)

Substitute into the resistance formula:

\[
R_{\text{total}} = \frac{L/2}{K A} + \frac{L/2}{2K A} = \frac{L}{2KA} + \frac{L}{4KA} = \frac{3L}{4KA}
\]

Now, equate this to the total resistance for the equivalent conductivity:

\[
\frac{2L}{K_{\text{eq}} A} = \frac{3L}{4KA}
\]

Solving for \(K_{\text{eq}}\):

\[
K_{\text{eq}} = \frac{4K}{3}
\]

Thus, the equivalent thermal conductivity of the slab is:

\[
K_{\text{eq}} = \frac{4K}{3}
\]

To find the temperature at the interface of the rectangular slab, we use the concept of  thermal conduction through two materials in series.

Given:
- Thermal conductivity of iron, \(K_{\text{iron}} = 0.2 \, \text{W/mK}\)
- Thermal conductivity of brass, \(K_{\text{brass}} = 0.3 \, \text{W/mK}\)
- Temperature at one side of the iron slab, \(T_{\text{iron}} = 100^\circ C\)
- Temperature at one side of the brass slab, \(T_{\text{brass}} = 0^\circ C\)
- Thicknesses of both slabs are equal.

Since the slabs are in series and have equal thicknesses, the **heat flux** through both materials is the same. Using the formula for heat conduction, the temperature at the interface \(T_{\text{interface}}\) can be calculated using the ratio of thermal conductivities:

\[
\frac{T_{\text{iron}} - T_{\text{interface}}}{T_{\text{interface}} - T_{\text{brass}}} = \frac{K_{\text{brass}}}{K_{\text{iron}}}
\]

Substitute the known values:

\[
\frac{100 - T_{\text{interface}}}{T_{\text{interface}} - 0} = \frac{0.3}{0.2} = 1.5
\]

Now solve for \(T_{\text{interface}}\):

\[
100 - T_{\text{interface}} = 1.5 T_{\text{interface}}
\]

\[
100 = 2.5 T_{\text{interface}}
\]

\[
T_{\text{interface}} = \frac{100}{2.5} = 40^\circ C
\]

Thus, the temperature at the interface is \(40^\circ C\).

In lakes, temperature stratification occurs, where different layers of water have distinct temperatures.

- Surface Temperature (2°C): The surface of the lake cools down and can freeze when temperatures drop, resulting in water that is less dense.

- Bottom Temperature (4°C): Water reaches its maximum density at 4°C. Below this temperature, water becomes less dense, causing it to rise. Therefore, in many lakes, the bottom water remains at around 4°C, even when the surface is colder.

This stratification helps maintain aquatic life during cold seasons, as the bottom layer remains relatively stable and can support organisms.

To find the temperature difference (ΔT) between the two ends of the conductor, we can use the formula for heat conduction:

\[
Q = \frac{K \cdot A \cdot \Delta T}{L}
\]

Where:
- \( Q = 6000 \, \text{J/s} \)
- \( K = 200 \, \text{J} \, \text{m}^{-1} \, \text{K}^{-1} \)
- \( A = 0.75 \, \text{m}^2 \)
- \( L = 1 \, \text{m} \)

Rearranging to solve for ΔT:

\[
\Delta T = \frac{Q \cdot L}{K \cdot A}
\]

\[
\Delta T = \frac{6000 \cdot 1}{200 \cdot 0.75} = 40 \, \text{K}
\]

The temperature difference between the two ends is 40 K or 40 °C

Given:

- Layer 1: Thermal conductivity \(K_1\)
- Layer 2: Thermal conductivity \(K_2\)
- Thickness of each layer: \(d\)

Formula for Equivalent Thermal Conductivity

The equivalent thermal conductivity for two parallel layers of the same thickness can be given by:

\[
K_{eq} = \frac{K_1 + K_2}{2}
\]

Derivation:

1. Resistance of Each Layer:
The thermal resistance for each layer can be expressed as:
\[
R_1 = \frac{d}{K_1 A}, \quad R_2 = \frac{d}{K_2 A}
\]
where \(A\) is the cross-sectional area.

2. Total Resistance in Parallel:
The total resistance for two resistors in parallel is:
\[
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}
\]
Substituting the resistances, we get:
\[
\frac{1}{R_{eq}} = \frac{K_1 A}{d} + \frac{K_2 A}{d}
\]
Simplifying:
\[
\frac{1}{R_{eq}} = \frac{A}{d} \left(K_1 + K_2\right)
\]

3. Total Conductivity:
Now, the equivalent conductivity can be expressed as:
\[
K_{eq} = \frac{d}{A} \cdot \frac{1}{R_{eq}} = \frac{d}{A} \cdot \frac{d}{A (K_1 + K_2)} = \frac{K_1 + K_2}{2}
\]

Conclusion:
Thus, for two parallel layers of materials with equal thickness, the correct equivalent thermal conductivity is:

\[
K_{eq} = \frac{K_1 + K_2}{2}
\]

Convection is the process by which heat is transferred through the movement of fluids, including gases like air. In the atmosphere, convection occurs when warm air rises and cool air sinks.

Here's a short explanation of how this works:

1. Heating the Surface: The sun heats the Earth's surface, which in turn warms the air above it.
2. Rising Warm Air: As the air warms, it becomes less dense and rises.
3. Cooling and Sinking: Once the warm air rises, it cools down at higher altitudes, becomes denser, and eventually sinks back down.
4. Cycle Continuation: This cycle of rising warm air and sinking cool air creates convection currents, which distribute heat throughout the atmosphere, influencing weather patterns and temperature distribution.

Overall, convection plays a crucial role in regulating the Earth's climate and weather systems.

The temperature gradient is the rate at which temperature changes with respect to distance. It's given as \( 80^\circ \text{C/m} \), and the length of the rod is \( 0.5 \, \text{m} \).

To find the temperature difference across the rod, we use the formula:

\[
\Delta T = \text{Temperature gradient} \times \text{Length}
\]

Substitute the values:

\[
\Delta T = 80^\circ \text{C/m} \times 0.5 \, \text{m} = 40^\circ \text{C}
\]

Now, the temperature at the hotter end is given as \( 30^\circ \text{C} \), so the temperature at the colder end is:

\[
T_{\text{colder end}} = T_{\text{hotter end}} - \Delta T
\]

Substituting the values:

\[
T_{\text{colder end}} = 30^\circ \text{C} - 40^\circ \text{C} = -10^\circ \text{C}
\]

Therefore, the temperature at the colder end of the rod is \(-10^\circ \text{C}\).

Given that the thermal resistance is the same for both rods, we have the equation:

\[
\frac{L_1}{K_1} = \frac{L_2}{K_2}
\]

So, the ratio of lengths is:

\[
\frac{L_1}{L_2} = \frac{K_1}{K_2}
\]

If the ratio of the thermal conductivities \(K_1:K_2 = 5:4\), the ratio of the lengths will be the same:

\[
\frac{L_1}{L_2} = \frac{5}{4}
\]

Therefore, the correct ratio of the lengths is \(5:4\).

No heat flows through the central rod \(K_5\) if the system is balanced like a Wheatstone bridge. The condition for this is:

\[
\frac{K_1}{K_2} = \frac{K_3}{K_4}
\]

This ensures that the temperature difference between points B and D is zero, preventing any heat flow through \(K_5\).

1. Heat released by \(x\) gm steam at \(100^\circ C\):
\[
Q_{\text{steam}} = x \times 540
\]

2. Heat required to convert \(y\) gm ice at \(0^\circ C\) to water at \(100^\circ C\):
\[
Q_{\text{ice}} = y \times 80 + y \times 100 = y \times 180
\]

3. Equating:
\[
x \times 540 = y \times 180
\]

\[
\frac{y}{x} = 3
\]

So, \( y : x = 3 : 1 \).