The end correction of a resonance column is 1 cm. If the shortest length resonating with the tunning fork is 10 cm, the next resonating length should be :

Options

A. 40 cm
B. 36 cm
C. 28 cm
D. 32 cm

Answer

To find the next resonating length in a resonance column, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- End correction (e) = 1 cm
- Shortest resonating length (l1) = 10 cm

2. **Understanding the Resonance Condition**:
- The effective length of the resonance column when considering end correction is given by:
\[ L = l + e \]
- For the first resonance (fundamental frequency), we have:
\[ L_1 = l_1 + e = 10 \, \text{cm} + 1 \, \text{cm} = 11 \, \text{cm} \]
- This length corresponds to \(\frac{\lambda}{4}\) (where \(\lambda\) is the wavelength).

3. **Calculate the Wavelength**:
- From the resonance condition:
\[ L_1 = \frac{\lambda}{4} \]
- Therefore, we can express \(\lambda\) as:
\[ \lambda = 4L_1 = 4 \times 11 \, \text{cm} = 44 \, \text{cm} \]

4. **Finding the Next Resonating Length (l2)**:
- The next resonating length corresponds to the third harmonic (first overtone), which is given by:
\[ L_2 = \frac{3\lambda}{4} \]
- Substituting the value of \(\lambda\):
\[ L_2 = \frac{3 \times 44 \, \text{cm}}{4} = 33 \, \text{cm} \]
- Now, we need to account for the end correction again:
\[ l_2 = L_2 - e = 33 \, \text{cm} - 1 \, \text{cm} = 32 \, \text{cm} \]

5. **Final Answer**:
- The next resonating length \(l_2\) is **32 cm**.