Air column of `20cm` length in a resonance tube resonates with a certain tuning fork when sounded at its upper open end. The lower end of the tube is closed and adjustable by changing the quantity of mercury filled inside the tube. The temperature of the air is `27^(@)C`. The change in length of the air column required, if the temperature falls to `7^(@)C` and the same tuning fork is again sounded at the upper open end is nearly

Options

A. `1 mm`
B. `7 mm`
C. `5 mm`
D. `13 mm`

Explanation

To solve the problem, we need to find the change in the length of the air column in a resonance tube when the temperature changes from \(27^\circ C\) to \(7^\circ C\). We will use the relationship between the speed of sound in air, temperature, and the length of the air column.

### Step-by-Step Solution:

1. **Understand the relationship**: The speed of sound \(v\) in air is given by the formula:
\[ v = \sqrt{\frac{\gamma R T}{M}} \]
where \(T\) is the absolute temperature in Kelvin. For our purposes, we will use the simplified relationship that shows the speed of sound is proportional to the square root of the temperature:
\[ v \propto \sqrt{T} \]

2. **Convert temperatures to Kelvin**:
- Initial temperature \(T_1 = 27^\circ C = 27 + 273 = 300 \, K\)
- Final temperature \(T_2 = 7^\circ C = 7 + 273 = 280 \, K\)

3. **Relate the lengths of the air column**: Since the frequency remains constant (as the same tuning fork is used), we can say:
\[ \frac{L_1}{L_2} = \frac{v_1}{v_2} \]
where \(L_1\) and \(L_2\) are the lengths of the air column at temperatures \(T_1\) and \(T_2\) respectively.

4. **Express the lengths in terms of temperature**:
\[ \frac{L_1}{L_2} = \frac{\sqrt{T_1}}{\sqrt{T_2}} \implies L_2 = L_1 \cdot \sqrt{\frac{T_2}{T_1}} \]

5. **Substitute the known values**:
\[ L_1 = 20 \, cm, \quad T_1 = 300 \, K, \quad T_2 = 280 \, K \]
\[ L_2 = 20 \cdot \sqrt{\frac{280}{300}} \]

6. **Calculate \(L_2\)**:
\[ L_2 = 20 \cdot \sqrt{\frac{280}{300}} = 20 \cdot \sqrt{\frac{28}{30}} = 20 \cdot \sqrt{\frac{14}{15}} \approx 20 \cdot 0.935 = 18.7 \, cm \]

7. **Find the change in length**:
\[ \Delta L = L_1 - L_2 = 20 \, cm - 18.7 \, cm = 1.3 \, cm \]

### Final Answer:
The change in length of the air column required is approximately \(1.3 \, cm\).