A wire having a linear mass density $$9.0 \times 10^{-4}$$ kg m$$^{-1}$$ is stretched between two rigid supports with a tension of 900 N. The wire resonates at a frequency of 500 Hz. The next higher frequency at which the same wire resonates is 550 Hz. The length of the wire is _________ m.

We start by recalling that a stretched wire fixed at both ends behaves like a stretched string. For such a string the resonant (natural) frequencies are given by the well-known formula

$$f_n \;=\; \frac{n\,v}{2L}, \qquad n = 1,2,3,\dots$$

Here $$f_n$$ is the $$n^{\text{th}}$$ resonant frequency, $$v$$ is the speed of transverse waves on the wire and $$L$$ is the length of the wire.

For two consecutive resonant frequencies, say $$f_n$$ and $$f_{n+1}$$, we have

$$f_{n+1} - f_n \;=\; \frac{(n+1)v}{2L} \;-\; \frac{n\,v}{2L} \;=\; \frac{v}{2L}.$$

This shows that the difference between any two successive resonant frequencies is always $$\dfrac{v}{2L}$$, independent of the harmonic number.

According to the data, the wire resonates at $$500\;\text{Hz}$$ and the next higher frequency is $$550\;\text{Hz}$$. Hence

$$f_{n+1} - f_n \;=\; 550 - 500 \;=\; 50\;\text{Hz}.$$

Using the relation for consecutive differences, we write

$$\frac{v}{2L} \;=\; 50.$$

To find $$v$$ we employ the formula for the speed of transverse waves on a stretched string:

$$v \;=\; \sqrt{\frac{T}{\mu}},$$

where $$T$$ is the tension and $$\mu$$ is the linear mass density of the wire.

The given tension is $$T = 900\;\text{N}$$ and the linear mass density is $$\mu = 9.0 \times 10^{-4}\;\text{kg m}^{-1}$$. Substituting these values, we obtain

$$v \;=\; \sqrt{\frac{900}{9.0 \times 10^{-4}}}.$$

Dividing inside the square root, we get

$$\frac{900}{9.0 \times 10^{-4}} \;=\; \frac{900}{0.0009} \;=\; 1\,000\,000.$$

Taking the square root,

$$v \;=\; \sqrt{1\,000\,000} \;=\; 1000\;\text{m s}^{-1}.$$

Now we substitute this value of $$v$$ into the difference equation $$\dfrac{v}{2L}=50$$:

$$\frac{1000}{2L} \;=\; 50.$$

Simplifying the numerator first:

$$\frac{1000}{2} \;=\; 500,$$

so the equation becomes

$$\frac{500}{L} \;=\; 50.$$

Cross-multiplying gives

$$500 \;=\; 50\,L.$$

Dividing both sides by $$50$$, we find

$$L \;=\; \frac{500}{50} \;=\; 10\;\text{m}.$$

So, the answer is $$10\;\text{m}$$.