A wire is stretched betwween two rigid supports under a tension of 500 N. At a frequency of 360 Hz, the wire resonates. The next higher frequency at which resonance of wire takes place is 420 Hz. What will be the length of wire if mass per unit length of wire is `8 xx 10^(-3) kg m^(-1)`.

Answer

To find the length of the wire, we will use the information provided about the frequencies at which the wire resonates and the mass per unit length.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Tension (T) = 500 N
- Frequency (f1) = 360 Hz (first resonance frequency)
- Frequency (f2) = 420 Hz (next higher resonance frequency)
- Mass per unit length (μ) = \(8 \times 10^{-3} \, \text{kg/m}\)

2. **Use the Formula for Frequency of a Stretched Wire:**
The frequency of a stretched wire is given by the formula:
\[ f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \]
where \(n\) is the harmonic number, \(L\) is the length of the wire, \(T\) is the tension, and \(\mu\) is the mass per unit length.

3. **Set Up the Equations for Both Frequencies:**
For the first frequency (360 Hz):
\[ 360 = \frac{n}{2L} \sqrt{\frac{500}{8 \times 10^{-3}}} \]
For the second frequency (420 Hz):
\[ 420 = \frac{n+1}{2L} \sqrt{\frac{500}{8 \times 10^{-3}}} \]

4. **Calculate the Square Root Term:**
First, calculate \(\sqrt{\frac{500}{8 \times 10^{-3}}}\):
\[ \sqrt{\frac{500}{8 \times 10^{-3}}} = \sqrt{62500} = 250 \]

5. **Substitute the Square Root Value into the Equations:**
For the first frequency:
\[ 360 = \frac{n}{2L} \cdot 250 \quad \Rightarrow \quad n = \frac{360 \cdot 2L}{250} = \frac{720L}{250} = 2.88L \]
For the second frequency:
\[ 420 = \frac{n+1}{2L} \cdot 250 \quad \Rightarrow \quad n + 1 = \frac{420 \cdot 2L}{250} = \frac{840L}{250} = 3.36L \]

6. **Set Up the Equation Relating n and L:**
From the two equations:
\[ n = 2.88L \quad \text{and} \quad n + 1 = 3.36L \]
Substitute \(n\) from the first equation into the second:
\[ 2.88L + 1 = 3.36L \]
Rearranging gives:
\[ 1 = 3.36L - 2.88L \quad \Rightarrow \quad 1 = 0.48L \]

7. **Solve for L:**
\[ L = \frac{1}{0.48} \approx 2.08 \, \text{m} \]

### Final Answer:
The length of the wire is approximately **2.08 meters**.