Two tuning forks `n_1 " and "n_2` when sounded together produces 5 beats per second . If the fork `n_1` is in resonance with a closed tube of length 8cm and other `n_2` is resonance with an open tube of length 16.5cm the frequency `n_1` will be

Options

A. 165Hz
B. 140Hz
C. 140Hz
D. 145Hz

Explanation

To solve the problem, we need to find the frequency \( n_1 \) of the tuning fork that is in resonance with a closed tube of length 8 cm. We also know that two tuning forks \( n_1 \) and \( n_2 \) produce 5 beats per second when sounded together, and that \( n_2 \) is in resonance with an open tube of length 16.5 cm.

### Step-by-step Solution:

1. **Understanding the Beat Frequency**:
- The beat frequency is given as 5 beats per second. This means that the difference in frequencies of the two tuning forks is:
\[ |n_1 - n_2| = 5 \text{ Hz} \]

2. **Finding the Frequency of \( n_1 \)**:
- The frequency of a closed organ pipe (which resonates with \( n_1 \)) is given by the formula:
\[ n_1 = \frac{v}{4L} \]
- Here, \( L = 8 \text{ cm} = 0.08 \text{ m} \). Therefore:
\[ n_1 = \frac{v}{4 \times 0.08} = \frac{v}{0.32} \]

3. **Finding the Frequency of \( n_2 \)**:
- The frequency of an open organ pipe (which resonates with \( n_2 \)) is given by the formula:
\[ n_2 = \frac{v}{2L} \]
- Here, \( L = 16.5 \text{ cm} = 0.165 \text{ m} \). Therefore:
\[ n_2 = \frac{v}{2 \times 0.165} = \frac{v}{0.33} \]

4. **Setting Up the Equation**:
- We know from the beat frequency that:
\[ n_1 - n_2 = 5 \]
- Substituting the expressions for \( n_1 \) and \( n_2 \):
\[ \frac{v}{0.32} - \frac{v}{0.33} = 5 \]

5. **Finding a Common Denominator**:
- The common denominator for \( 0.32 \) and \( 0.33 \) is \( 0.32 \times 0.33 \). Thus, we can rewrite the equation:
\[ \frac{v \cdot 0.33 - v \cdot 0.32}{0.32 \times 0.33} = 5 \]
- Simplifying the numerator:
\[ v(0.33 - 0.32) = 5 \cdot 0.32 \cdot 0.33 \]
- This simplifies to:
\[ v(0.01) = 5 \cdot 0.32 \cdot 0.33 \]

6. **Calculating the Velocity \( v \)**:
- Solving for \( v \):
\[ v = \frac{5 \cdot 0.32 \cdot 0.33}{0.01} = 5 \cdot 32 \cdot 33 = 5280 \text{ cm/s} \]

7. **Finding \( n_1 \)**:
- Now substitute \( v \) back into the equation for \( n_1 \):
\[ n_1 = \frac{5280}{0.32} = 165 \text{ Hz} \]

### Final Answer:
The frequency \( n_1 \) is \( 165 \text{ Hz} \).