Two models of a windowpane are made. In one model, two identical glass panes of thickness 3 mm are separated with an air gap of 3 mm. This composite system is fixed in the window of a room The other model consist of a single glass pane of thickness 3 mm, the temperature difference being the same as for the first model. the ratio of the heat flow fot the double pane to that for the single pane is
(`K_("glass")=2.5 xx10^(-4)cal//s.m.^(@)C` and `K_("air")=6.2 xx10^(-6)cal//s.m.^(@)C`).

Options

A. `1//20`
B. `1//70`
C. `1//100`
D. `1//50`

Explanation

To solve the problem, we need to calculate the ratio of heat flow through two models of a windowpane: one with two glass panes separated by an air gap and the other with a single glass pane. We will use the concept of thermal resistance to find the heat flow.

### Step-by-Step Solution:

1. **Identify the Thermal Resistance for Each Model:**
- For the double pane model (Model A):
- There are two glass panes, each with a thickness of 3 mm, and an air gap of 3 mm.
- The thermal resistance due to the glass panes (R_G) can be calculated as:
\[ R_G = \frac{d}{K_G \cdot A} \]
where \(d\) is the thickness of the glass pane, \(K_G\) is the thermal conductivity of glass, and \(A\) is the area.
- The total thermal resistance for the double pane model is:
\[ R_A = 2R_G + R_A = 2 \left( \frac{3 \times 10^{-3}}{K_G \cdot A} \right) + \left( \frac{3 \times 10^{-3}}{K_{air} \cdot A} \right) \]

- For the single pane model (Model B):
- There is only one glass pane of thickness 3 mm.
- The thermal resistance is:
\[ R_B = \frac{3 \times 10^{-3}}{K_G \cdot A} \]

2. **Calculate the Heat Flow:**
- The heat flow \(I\) through a material is given by:
\[ I = \frac{\Delta T}{R} \]
where \(\Delta T\) is the temperature difference and \(R\) is the thermal resistance.
- For Model A:
\[ I_A = \frac{\Delta T}{R_A} \]
- For Model B:
\[ I_B = \frac{\Delta T}{R_B} \]

3. **Find the Ratio of Heat Flow:**
- The ratio of heat flow for the double pane to that for the single pane is:
\[ \frac{I_A}{I_B} = \frac{R_B}{R_A} \]

4. **Substituting the Values of Thermal Resistance:**
- Substitute the expressions for \(R_A\) and \(R_B\):
\[ \frac{I_A}{I_B} = \frac{\frac{3 \times 10^{-3}}{K_G \cdot A}}{2 \left( \frac{3 \times 10^{-3}}{K_G \cdot A} \right) + \left( \frac{3 \times 10^{-3}}{K_{air} \cdot A} \right)} \]

5. **Simplifying the Expression:**
- Cancel out common terms:
\[ = \frac{1}{2 + \frac{K_G}{K_{air}}} \]
- Substitute the given values:
- \(K_G = 2.5 \times 10^{-4} \, \text{cal/s.m.°C}\)
- \(K_{air} = 6.2 \times 10^{-6} \, \text{cal/s.m.°C}\)
- Thus:
\[ = \frac{1}{2 + \frac{2.5 \times 10^{-4}}{6.2 \times 10^{-6}}} \]

6. **Calculating the Final Ratio:**
- Calculate \(\frac{2.5 \times 10^{-4}}{6.2 \times 10^{-6}} \approx 40.32\)
- Therefore:
\[ \frac{I_A}{I_B} = \frac{1}{2 + 40.32} = \frac{1}{42.32} \approx \frac{1}{50} \]

### Conclusion:
The ratio of the heat flow for the double pane to that for the single pane is approximately \( \frac{1}{50} \).

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