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.

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 \(-10^\circ \text{C}\).

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 \(K_5\).

The relationship between the coefficient of volume expansion (\( \beta \)) and the coefficient of linear expansion (\( \alpha \)) for a solid is:

\[
\beta = 3\alpha
\]

Given:
- Volume increase = 0.24%
- Temperature increase \( \Delta T = 40^\circ \text{C} \)

The coefficient of volume expansion \( \beta \) is given by:

\[
\beta = \frac{\text{Percentage increase in volume}}{\Delta T} = \frac{0.24}{40} = 0.006\% \, \text{per } ^\circ\text{C} = 6 \times 10^{-5} \, \text{per } ^\circ\text{C}
\]

Now, using \( \beta = 3\alpha \):

\[
\alpha = \frac{\beta}{3} = \frac{6 \times 10^{-5}}{3} = 2 \times 10^{-5} \, \text{per } ^\circ\text{C}
\]

Thus, the coefficient of linear expansion of the metal is \( 2 \times 10^{-5} \, \text{per } ^\circ\text{C} \).

Using the ideal gas equation \(P = \frac{nRT}{V}\), with total moles \(n = 0.2 + 0.3 = 0.5\text{ mol}\) and volume \(V = 200 \times 10^{-6}\text{ m}^3\). Calculating gives \(P = \frac{0.5 \times 8.3 \times 200}{2 \times 10^{-4}} = 4.15 \times 10^6\text{ Pa}\).

Using the relation \(\frac{C}{5} = \frac{F - 32}{9}\), let \(C = F = x\). This gives \(\frac{x}{5} = \frac{x - 32}{9} ⇒ 9x = 5x - 160 ⇒ x = -40\).

Steam has latent heat of vaporization, causing more severe burns than boiling water. Water also has an exceptionally high specific heat capacity compared to most other substances. Both statements are correct.

For isothermal process in A: \(P_A = 2P_0\). For adiabatic process in B: \(P_B = P_0 (2)^\gamma\). The ratio of final pressures is \(\frac{P_B}{P_A} = \frac{P_0 2^\gamma}{2 P_0} = 2^{\gamma - 1}\).