In a 10 m deep lake, the bottom is at a constant temperature of `4^(@)C`. The air temperature is constant at `- 4^(@)C. K_(ice) = 3 K_(omega)`. Neglecting the expansion of water on freezing, the maximum thickness of ice will be

Options

A. `7.5`m
B. `6`m
C. `5` m
D. `2.5`m

Explanation

To solve the problem, we need to determine the maximum thickness of ice that can form on a 10 m deep lake, given the temperature conditions and the thermal conductivity of ice and water. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The lake is 10 m deep.
- The bottom of the lake is at a constant temperature of \(4^\circ C\).
- The air temperature is at \(-4^\circ C\).
- The thermal conductivity of ice (\(K_{ice}\)) is 3 times that of water (\(K_{water}\)).

2. **Defining Variables**:
- Let \(X\) be the thickness of the ice.
- The depth of water below the ice will be \(10 - X\).

3. **Setting Up Heat Flow Equations**:
- The heat flow through the ice layer can be expressed as:
\[ I = \frac{4 - 0}{R_{ice}} \quad \text{(from the air at -4°C to the ice at 0°C)} \]
- The heat flow through the water layer can be expressed as:
\[ I = \frac{0 - 4}{R_{water}} \quad \text{(from the water at 4°C to the ice at 0°C)} \]

4. **Expressing Thermal Resistances**:
- The thermal resistance for the ice layer is:
\[ R_{ice} = \frac{X}{K_{ice} \cdot A} = \frac{X}{3K_{water} \cdot A} \]
- The thermal resistance for the water layer is:
\[ R_{water} = \frac{10 - X}{K_{water} \cdot A} \]

5. **Equating Heat Flows**:
- Since the heat flows are equal, we can set the two equations equal to each other:
\[ \frac{4}{R_{ice}} = \frac{4}{R_{water}} \]
- Substituting the expressions for \(R_{ice}\) and \(R_{water}\):
\[ \frac{4}{\frac{X}{3K_{water} \cdot A}} = \frac{4}{\frac{10 - X}{K_{water} \cdot A}} \]
- Simplifying this gives:
\[ \frac{4 \cdot 3K_{water} \cdot A}{X} = \frac{4 \cdot K_{water} \cdot A}{10 - X} \]

6. **Canceling Common Terms**:
- Cancel \(4\), \(K_{water}\), and \(A\) from both sides:
\[ \frac{3}{X} = \frac{1}{10 - X} \]

7. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ 3(10 - X) = X \]
- Expanding this results in:
\[ 30 - 3X = X \]

8. **Solving for \(X\)**:
- Rearranging the equation:
\[ 30 = 4X \]
- Therefore:
\[ X = \frac{30}{4} = 7.5 \text{ m} \]

### Conclusion:
The maximum thickness of ice that can form on the lake is \(7.5\) meters.