A closed organ pipe in `3^(rd)` harmonic and an open organ pipe in `5^(th)` harmonic mode are in resonance with a tuning fork P. The ratio of lengths of closed to open organ pipe will be

Options

A. `3 : 10`
B. `10 : 3`
C. `5 : 4`
D. `10 : 5`

Explanation

To solve the problem, we need to find the ratio of the lengths of a closed organ pipe (Lc) to an open organ pipe (Lo) when they are in resonance with a tuning fork. The closed organ pipe is in its 3rd harmonic, and the open organ pipe is in its 5th harmonic.

### Step-by-step Solution:

1. **Understand the Harmonics**:
- For a closed organ pipe, the harmonics are given by the formula:
\[ f_n = \frac{nV}{4L_c} \]
where \( n \) is the harmonic number, \( V \) is the speed of sound in air, and \( L_c \) is the length of the closed organ pipe.
- For the 3rd harmonic, \( n = 3 \):
\[ f_1 = \frac{3V}{4L_c} \]

2. **Open Organ Pipe Harmonics**:
- For an open organ pipe, the harmonics are given by:
\[ f_n = \frac{nV}{2L_o} \]
where \( L_o \) is the length of the open organ pipe.
- For the 5th harmonic, \( n = 5 \):
\[ f_2 = \frac{5V}{2L_o} \]

3. **Setting the Frequencies Equal**:
- Since both pipes are resonating with the same tuning fork, their frequencies are equal:
\[ f_1 = f_2 \]
- Therefore, we can write:
\[ \frac{3V}{4L_c} = \frac{5V}{2L_o} \]

4. **Canceling the Speed of Sound**:
- We can cancel \( V \) from both sides of the equation:
\[ \frac{3}{4L_c} = \frac{5}{2L_o} \]

5. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ 3 \cdot 2L_o = 5 \cdot 4L_c \]
- Simplifying this:
\[ 6L_o = 20L_c \]

6. **Finding the Ratio**:
- Rearranging gives:
\[ \frac{L_c}{L_o} = \frac{6}{20} = \frac{3}{10} \]

### Final Answer:
The ratio of lengths of the closed organ pipe to the open organ pipe is:
\[ \frac{L_c}{L_o} = \frac{3}{10} \]