1. Given data

• Linear mass density: $\mu = 9.0 \times 10^{-4}\ \text{kg/m}$
• Tension: $T = 900\ \text{N}$
• One resonant frequency: $f_n = 500\ \text{Hz}$
• Next higher resonant frequency: $f_{n+1} = 550\ \text{Hz}$

We need to find the length $L$ of the wire.

2. Wave speed on the stretched wire

For a stretched string/wire, $v = \sqrt{\frac{T}{\mu}}$

Substitute the values: $v = \sqrt{\frac{900}{9.0\times 10^{-4}}}$

$\frac{900}{9.0\times 10^{-4}} = 1000\times 1000 = 10^6$

So, $v = \sqrt{10^6} = 1000\ \text{m/s}$

3. Use the resonance condition

For a wire fixed at both ends, resonant frequencies are $f_n = \frac{nv}{2L}, \quad n=1,2,3,\dots$

Hence the difference between successive resonant frequencies is $f_{n+1}-f_n = \frac{v}{2L}$

Given: $550 - 500 = 50\ \text{Hz}$

Therefore, $\frac{v}{2L} = 50$

Substitute $v=1000$ m/s: $\frac{1000}{2L} = 50$

$\frac{1000}{100} = L$

$L = 10\ \text{m}$

4. Final answer

The length of the wire is $\boxed{10\ \text{m}}$

5. Comparison with stored answer

Stored correct answer: $10$

Our derived answer matches the stored answer.