In a resonance tube experiment, two successive resonances are heard at 15 cm 48 cm. End correction will be

Options

A. `1.5 cm`
B. 3 cm
C. `2.5` cm
D. 1 cm

Answer

To find the end correction in a resonance tube experiment where two successive resonances are heard at lengths of 15 cm and 48 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the lengths of the resonances**:
- Let the first resonance length be \( L_1 = 15 \) cm.
- Let the second resonance length be \( L_2 = 48 \) cm.

2. **Understand the relationship between the lengths and end correction**:
- For a closed organ pipe, the resonant lengths can be expressed as:
- First resonance: \( L_1 + e \) (where \( e \) is the end correction)
- Second resonance: \( L_2 + e \)

3. **Use the formula for resonances in a closed pipe**:
- The resonances occur at odd multiples of \( \frac{\lambda}{4} \):
- First resonance corresponds to \( \frac{\lambda}{4} \): \( L_1 + e \)
- Second resonance corresponds to \( \frac{3\lambda}{4} \): \( L_2 + e \)

4. **Set up the equations**:
- From the first resonance:
\[ L_1 + e = \frac{\lambda}{4} \]
- From the second resonance:
\[ L_2 + e = \frac{3\lambda}{4} \]

5. **Subtract the first equation from the second**:
\[ (L_2 + e) - (L_1 + e) = \frac{3\lambda}{4} - \frac{\lambda}{4} \]
This simplifies to:
\[ L_2 - L_1 = \frac{2\lambda}{4} = \frac{\lambda}{2} \]

6. **Substitute the values of \( L_1 \) and \( L_2 \)**:
\[ 48 \text{ cm} - 15 \text{ cm} = \frac{\lambda}{2} \]
\[ 33 \text{ cm} = \frac{\lambda}{2} \]
Therefore,
\[ \lambda = 66 \text{ cm} \]

7. **Now, substitute \( \lambda \) back to find \( e \)**:
- From the first resonance equation:
\[ L_1 + e = \frac{\lambda}{4} \]
Substitute \( L_1 = 15 \) cm and \( \lambda = 66 \) cm:
\[ 15 + e = \frac{66}{4} = 16.5 \]
Therefore,
\[ e = 16.5 - 15 = 1.5 \text{ cm} \]

### Final Answer:
The end correction \( e \) is **1.5 cm**.