In a resonance apparatus, the first and second resonating lengths of air column are `15cm` and `48cm` respectively. The end correction for this apparatus is

Options

A. `6cm`
B. `3cm`
C. `1.5cm`
D. `2cm`

Explanation

To find the end correction for the resonance apparatus, we can use the formula for end correction, which is given by:

\[ e = \frac{L_2 - 3L_1}{2} \]

where:
- \( L_1 \) is the first resonating length,
- \( L_2 \) is the second resonating length,
- \( e \) is the end correction.

### Step-by-Step Solution:

1. **Identify the given values**:
- First resonating length, \( L_1 = 15 \, \text{cm} \)
- Second resonating length, \( L_2 = 48 \, \text{cm} \)

2. **Substitute the values into the formula**:
\[ e = \frac{L_2 - 3L_1}{2} \]
Substituting the values:
\[ e = \frac{48 \, \text{cm} - 3 \times 15 \, \text{cm}}{2} \]

3. **Calculate \( 3L_1 \)**:
\[ 3L_1 = 3 \times 15 \, \text{cm} = 45 \, \text{cm} \]

4. **Subtract \( 3L_1 \) from \( L_2 \)**:
\[ L_2 - 3L_1 = 48 \, \text{cm} - 45 \, \text{cm} = 3 \, \text{cm} \]

5. **Divide by 2 to find the end correction**:
\[ e = \frac{3 \, \text{cm}}{2} = 1.5 \, \text{cm} \]

Thus, the end correction for the apparatus is \( 1.5 \, \text{cm} \).

### Final Answer:
The end correction is \( 1.5 \, \text{cm} \).