A closed organ pipe has a frequency `'n'`. If its length is doubled and radius is halved , its frequency nearly becomes .

Options

A. halved
B. doubled
C. trebled
D. quadrupled

Explanation

To solve the problem, we need to analyze how the frequency of a closed organ pipe changes when its length is doubled and its radius is halved.

### Step-by-Step Solution:

1. **Understanding the Frequency of a Closed Organ Pipe**:
- The fundamental frequency (f) of a closed organ pipe is given by the formula:
\[ f = \frac{v}{\lambda} \]
- For a closed organ pipe, the wavelength (\(\lambda\)) is related to the length (L) of the pipe by:
\[ \lambda = 4L \]
- Therefore, the frequency can be expressed as:
\[ f = \frac{v}{4L} \]
- Here, \(v\) is the speed of sound in air.

2. **Initial Frequency**:
- Let the initial length of the organ pipe be \(L_1\) and the initial frequency be \(f_1 = n\).
- Thus, we have:
\[ f_1 = \frac{v}{4L_1} \]

3. **Changing the Length**:
- If the length of the pipe is doubled, the new length \(L_2\) becomes:
\[ L_2 = 2L_1 \]

4. **Calculating the New Frequency**:
- The new frequency \(f_2\) can be calculated using the new length:
\[ f_2 = \frac{v}{4L_2} = \frac{v}{4(2L_1)} = \frac{v}{8L_1} \]
- Now, substituting \(L_1\) from the initial frequency:
\[ f_2 = \frac{1}{2} \cdot \frac{v}{4L_1} = \frac{1}{2} f_1 \]
- Since \(f_1 = n\):
\[ f_2 = \frac{n}{2} \]

5. **Effect of Halving the Radius**:
- The frequency of a closed organ pipe is not directly dependent on the radius in the fundamental mode. Therefore, halving the radius does not affect the frequency calculation in this case.

6. **Final Result**:
- The new frequency of the closed organ pipe after doubling the length and halving the radius is:
\[ f_2 \approx \frac{n}{2} \]

### Conclusion:
The frequency of the closed organ pipe, when its length is doubled and its radius is halved, becomes nearly \(\frac{n}{2}\).