A hollow pipe of length `0.8m` is closed at one end. At its open end a `0.5 m` long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is `50 N` and the speed of sound is `320 ms^(-1)`, the mass of the string is

Options

A. (a)`5g`
B. (b)10g
C. (c)20g
D. (d)40g

Explanation

To solve the problem, we need to find the mass of the string given that it vibrates in its second harmonic and resonates with the fundamental frequency of a hollow pipe closed at one end.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a hollow pipe of length \( L_2 = 0.8 \, \text{m} \) closed at one end.
- The string of length \( L_1 = 0.5 \, \text{m} \) is vibrating in its second harmonic.
- The tension in the string is \( T = 50 \, \text{N} \).
- The speed of sound in air is \( v = 320 \, \text{m/s} \).

2. **Determine the Frequencies**:
- The fundamental frequency of the pipe (closed at one end) is given by:
\[ f_2 = \frac{v}{4L_2} \]
- The second harmonic frequency of the string is given by:
\[ f_1 = \frac{2}{L_1} \sqrt{\frac{T}{\mu}} \]
- Here, \( \mu \) is the mass per unit length of the string.

3. **Setting the Frequencies Equal**:
- Since the string resonates with the fundamental frequency of the pipe, we have:
\[ f_1 = f_2 \]
- Therefore:
\[ \frac{2}{L_1} \sqrt{\frac{T}{\mu}} = \frac{v}{4L_2} \]

4. **Substituting Known Values**:
- Substitute \( L_1 = 0.5 \, \text{m} \), \( L_2 = 0.8 \, \text{m} \), \( T = 50 \, \text{N} \), and \( v = 320 \, \text{m/s} \):
\[ \frac{2}{0.5} \sqrt{\frac{50}{\mu}} = \frac{320}{4 \times 0.8} \]
- Simplifying the right side:
\[ \frac{320}{3.2} = 100 \]
- Thus, we have:
\[ 4 \sqrt{\frac{50}{\mu}} = 100 \]

5. **Solving for \( \mu \)**:
- Squaring both sides:
\[ 16 \frac{50}{\mu} = 10000 \]
- Rearranging gives:
\[ \frac{50}{\mu} = 625 \quad \Rightarrow \quad \mu = \frac{50}{625} = 0.08 \, \text{kg/m} \]

6. **Finding the Mass of the String**:
- The mass \( m \) of the string can be calculated using:
\[ m = \mu \times L_1 \]
- Substituting \( \mu = 0.08 \, \text{kg/m} \) and \( L_1 = 0.5 \, \text{m} \):
\[ m = 0.08 \times 0.5 = 0.04 \, \text{kg} = 40 \, \text{g} \]

### Final Answer:
The mass of the string is \( 40 \, \text{g} \).