The length of a sonometer wire is 0.75 m and density `9xx 10^(3) Kg//m^(3)`. It can bear a stress of `8.1 xx 10^(8) N//m^(2)` without exceeding the elastic limit . The fundamental that can be produced in the wire , is

Options

A. 200 Hz
B. 150 Hz
C. 600 Hz
D. 450Hz

Answer

To solve the problem of finding the fundamental frequency of a sonometer wire, we can follow these steps:

### Step 1: Understand the formula for fundamental frequency
The fundamental frequency \( f \) of a vibrating string can be expressed as:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) = length of the wire
- \( T \) = tension in the wire
- \( \mu \) = mass per unit length of the wire

### Step 2: Calculate the tension \( T \)
Given that the wire can bear a stress \( \sigma \) of \( 8.1 \times 10^8 \, \text{N/m}^2 \), we can express tension \( T \) in terms of stress and area \( A \):
\[ T = \sigma \cdot A \]
However, we do not have the area directly, but we will see that it cancels out later.

### Step 3: Calculate the mass per unit length \( \mu \)
The mass per unit length \( \mu \) can be expressed as:
\[ \mu = \frac{m}{L} = \frac{\rho \cdot V}{L} = \frac{\rho \cdot A \cdot L}{L} = \rho \cdot A \]
where \( \rho = 9 \times 10^3 \, \text{kg/m}^3 \) is the density of the wire.

### Step 4: Substitute \( T \) and \( \mu \) into the frequency formula
Substituting \( T \) and \( \mu \) into the frequency formula:
\[ f = \frac{1}{2L} \sqrt{\frac{\sigma \cdot A}{\rho \cdot A}} = \frac{1}{2L} \sqrt{\frac{\sigma}{\rho}} \]
Notice that the area \( A \) cancels out.

### Step 5: Plug in the values
Now we can plug in the values:
- \( L = 0.75 \, \text{m} \)
- \( \sigma = 8.1 \times 10^8 \, \text{N/m}^2 \)
- \( \rho = 9 \times 10^3 \, \text{kg/m}^3 \)

Calculating:
\[ f = \frac{1}{2 \times 0.75} \sqrt{\frac{8.1 \times 10^8}{9 \times 10^3}} \]
\[ = \frac{1}{1.5} \sqrt{\frac{8.1 \times 10^8}{9 \times 10^3}} \]
\[ = \frac{1}{1.5} \sqrt{9 \times 10^4} \]
\[ = \frac{1}{1.5} \times 300 \]
\[ = 200 \, \text{Hz} \]

### Conclusion
The fundamental frequency that can be produced in the wire is \( 200 \, \text{Hz} \).

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