In a resonance tube experiment , the first resonance is obtained for `10 cm` of air column and the sound for `32 cm`. The end correction for this apparatus is

Options

A. `0.5 cm`
B. `1.0 cm`
C. `1.5 cm`
D. `2 cm`

Explanation

To find the end correction in a resonance tube experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- First resonance (L1) = 10 cm
- Second resonance (L2) = 32 cm

2. **Understand the Resonance Conditions**:
- At the first resonance, the length of the air column (L1) plus the end correction (e) is equal to one-fourth of the wavelength (λ/4).
\[ L1 + e = \frac{\lambda}{4} \]

- At the second resonance, the length of the air column (L2) plus the end correction (e) is equal to three-fourths of the wavelength (3λ/4).
\[ L2 + e = \frac{3\lambda}{4} \]

3. **Set Up the Equations**:
- From the first resonance:
\[ e = \frac{\lambda}{4} - L1 \]
- From the second resonance:
\[ e = \frac{3\lambda}{4} - L2 \]

4. **Equate the Two Expressions for e**:
\[ \frac{\lambda}{4} - L1 = \frac{3\lambda}{4} - L2 \]

5. **Rearranging the Equation**:
\[ L2 - L1 = \frac{3\lambda}{4} - \frac{\lambda}{4} \]
\[ L2 - L1 = \frac{2\lambda}{4} = \frac{\lambda}{2} \]

6. **Substituting the Values**:
- Substitute L1 = 10 cm and L2 = 32 cm:
\[ 32 - 10 = \frac{\lambda}{2} \]
\[ 22 = \frac{\lambda}{2} \implies \lambda = 44 \text{ cm} \]

7. **Finding the End Correction**:
- Substitute λ back into one of the equations for e. Using the first resonance equation:
\[ e = \frac{\lambda}{4} - L1 \]
\[ e = \frac{44}{4} - 10 = 11 - 10 = 1 \text{ cm} \]

### Final Answer:
The end correction (e) for this apparatus is **1 cm**.

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