Two models of a windowpane are made, two identical glass panes of thickness `3 mm` are separated with an air gap of `3mm`. This composite system is fixed in the window of a room. The other model consists of a single glass pane of thickness `6mm`, the temperature difference being the same as for first model. the ratio of the heat flow for the double pane to that for the single pane is `(K_("glass") = 2.5 xx 10^(-4) cal//s.m.^(@)C` and `K_("air") = 6.2 xx 10^(-6)cal//s.m.^(@)C)`

Options

A. `1//20`
B. `1//70`
C. `31//1312`
D. `31//656`

Explanation

To solve the problem of comparing the heat flow through two models of windowpanes, we will follow these steps:

### Step 1: Understand the Heat Flow Equation
The heat flow (I) through a material is given by the formula:
\[ I = \frac{\Delta T}{R} \]
where:
- \( \Delta T \) is the temperature difference across the material,
- \( R \) is the thermal resistance of the material.

### Step 2: Define the Two Models
- **Model 1**: Two glass panes of thickness \( 3 \, \text{mm} \) each, separated by an air gap of \( 3 \, \text{mm} \).
- **Model 2**: A single glass pane of thickness \( 6 \, \text{mm} \).

### Step 3: Calculate the Thermal Resistance for Model 1
For Model 1, the total thermal resistance \( R_1 \) is the sum of the resistances of the two glass panes and the air gap:
\[ R_1 = R_{\text{glass1}} + R_{\text{air}} + R_{\text{glass2}} \]

Where:
- \( R_{\text{glass}} = \frac{L}{K \cdot A} \) (for each glass pane),
- \( R_{\text{air}} = \frac{L}{K_{\text{air}} \cdot A} \).

Given:
- Thickness of each glass pane \( L = 3 \, \text{mm} = 0.003 \, \text{m} \),
- Conductivity of glass \( K_{\text{glass}} = 2.5 \times 10^{-4} \, \text{cal/s.m.}^\circ C \),
- Conductivity of air \( K_{\text{air}} = 6.2 \times 10^{-6} \, \text{cal/s.m.}^\circ C \).

Calculating each resistance:
1. For one glass pane:
\[ R_{\text{glass}} = \frac{0.003}{2.5 \times 10^{-4} \cdot A} \]
Therefore, for two glass panes:
\[ R_{\text{glass1}} + R_{\text{glass2}} = 2 \cdot \frac{0.003}{2.5 \times 10^{-4} \cdot A} = \frac{0.006}{2.5 \times 10^{-4} \cdot A} \]

2. For the air gap:
\[ R_{\text{air}} = \frac{0.003}{6.2 \times 10^{-6} \cdot A} \]

Combining these:
\[ R_1 = \frac{0.006}{2.5 \times 10^{-4} \cdot A} + \frac{0.003}{6.2 \times 10^{-6} \cdot A} \]

### Step 4: Calculate the Thermal Resistance for Model 2
For Model 2, the thermal resistance \( R_2 \) is:
\[ R_2 = R_{\text{glass}} = \frac{0.006}{2.5 \times 10^{-4} \cdot A} \]

### Step 5: Find the Ratio of Heat Flow
Using the heat flow equations:
\[ \frac{I_1}{I_2} = \frac{R_2}{R_1} \]

Substituting the values:
\[ \frac{I_1}{I_2} = \frac{\frac{0.006}{2.5 \times 10^{-4} \cdot A}}{\frac{0.006}{2.5 \times 10^{-4} \cdot A} + \frac{0.003}{6.2 \times 10^{-6} \cdot A}} \]

### Step 6: Simplify the Expression
The area \( A \) cancels out:
\[ \frac{I_1}{I_2} = \frac{1}{1 + \frac{0.003 \cdot 2.5 \times 10^{-4}}{0.006 \cdot 6.2 \times 10^{-6}}} \]

Calculating the fraction:
\[ \frac{0.003 \cdot 2.5 \times 10^{-4}}{0.006 \cdot 6.2 \times 10^{-6}} = \frac{0.00000075}{0.0000000372} \approx 20.16 \]

Thus:
\[ \frac{I_1}{I_2} = \frac{1}{1 + 20.16} = \frac{1}{21.16} \]

### Final Ratio
The final ratio of heat flow for the double pane to that for the single pane is approximately:
\[ \frac{I_1}{I_2} \approx \frac{1}{21.16} \]