A lake surface is exposed to an atmosphere where the temperature is `lt 0^(@)C`. If the thickness of the ice layer formed on the surface grows form `2cm` to `4cm` in `1` hour. The atmospheric temperature, `T_(a)` will be- (Thermal conductivity of ice `K = 4 xx 10^(-3) cal//cm//s//.^(@)C`, density of ice `= 0.9 gm//ice.` Latent heat of fustion of ice `= 80 cal//m`. Neglect the change of density during the state change. Assume that the water below the ice has `0^(@)` temperature every where)

Options

A. `-20 .^(@)C`
B. `0.^(@)C`
C. `-30.^(@)C`
D. `15.^(@)C`

Explanation

To solve the problem, we need to find the atmospheric temperature \( T_a \) given the thickness of the ice layer growing from 2 cm to 4 cm in 1 hour. We will use the principles of heat transfer and the properties of ice.

### Step-by-Step Solution:

1. **Identify the parameters given:**
- Initial thickness of ice, \( x_1 = 2 \, \text{cm} \)
- Final thickness of ice, \( x_2 = 4 \, \text{cm} \)
- Time taken, \( t = 1 \, \text{hour} = 3600 \, \text{s} \)
- Thermal conductivity of ice, \( K = 4 \times 10^{-3} \, \text{cal/cm/s/°C} \)
- Density of ice, \( \rho = 0.9 \, \text{g/cm}^3 \)
- Latent heat of fusion of ice, \( L = 80 \, \text{cal/g} \)

2. **Calculate the increase in thickness of ice:**
\[ \Delta x = x_2 - x_1 = 4 \, \text{cm} - 2 \, \text{cm} = 2 \, \text{cm} \]

3. **Calculate the mass of ice formed:**
The volume of the ice formed can be expressed as:
\[ V = A \cdot \Delta x \]
where \( A \) is the area of the lake surface. The mass \( m \) of the ice formed is:
\[ m = \rho \cdot V = \rho \cdot A \cdot \Delta x = 0.9 \cdot A \cdot 2 \, \text{cm} \]

4. **Calculate the heat required to form this mass of ice:**
The heat \( Q \) required to freeze this mass of water into ice is given by:
\[ Q = m \cdot L = (0.9 \cdot A \cdot 2) \cdot 80 \]

5. **Calculate the rate of heat transfer through the ice:**
The heat transfer rate \( \frac{dQ}{dt} \) can be expressed using Fourier's law of heat conduction:
\[ \frac{dQ}{dt} = \frac{K \cdot A \cdot (T_a - 0)}{\Delta x} \]
Here, \( \Delta x = 2 \, \text{cm} \) (the thickness of the ice layer).

6. **Set the heat gained equal to the heat lost:**
Since the heat gained by the ice must equal the heat lost from the atmosphere:
\[ \frac{Q}{t} = \frac{K \cdot A \cdot T_a}{\Delta x} \]
Substituting \( Q \) and \( t \):
\[ \frac{(0.9 \cdot A \cdot 2 \cdot 80)}{3600} = \frac{4 \times 10^{-3} \cdot A \cdot T_a}{2} \]

7. **Cancel \( A \) from both sides:**
\[ \frac{0.9 \cdot 2 \cdot 80}{3600} = \frac{4 \times 10^{-3} \cdot T_a}{2} \]

8. **Solve for \( T_a \):**
Rearranging gives:
\[ T_a = \frac{0.9 \cdot 2 \cdot 80 \cdot 2 \cdot 3600}{4 \times 10^{-3}} \]
Simplifying:
\[ T_a = \frac{0.9 \cdot 2 \cdot 80 \cdot 7200}{4} = \frac{0.9 \cdot 2 \cdot 80 \cdot 7200}{4} = 0.9 \cdot 2 \cdot 14400 = 25920 \, \text{cal} \]
Finally, calculate:
\[ T_a = \frac{25920}{4 \times 10^{-3}} = -30 \, \text{°C} \]

### Final Answer:
The atmospheric temperature \( T_a \) is \( -30 \, \text{°C} \).