A sonometer wire has a length of 0.75 m and density of the material of the wire is `10^(4) kg//m^(3)`. If the wire can bear a stress of `7.5 xx 10^(8) N//m^(2)` without exceeding the elastic limit, what fundamental frequency can be produced in the wire ?

Explanation

To find the fundamental frequency produced by the sonometer wire, we can follow these steps:

### Step 1: Identify the given values
- Length of the wire, \( L = 0.75 \, \text{m} \)
- Density of the material, \( \rho = 10^4 \, \text{kg/m}^3 \)
- Maximum stress the wire can bear, \( \sigma = 7.5 \times 10^8 \, \text{N/m}^2 \)

### Step 2: Calculate the area of cross-section using the stress formula
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