A wire of length $$L$$ and mass per unit length $$6.0 \times 10^{-3}$$ kg m$$^{-1}$$ is put under tension of 540 N. Two consecutive frequencies that it resonates at are: 420 Hz and 490 Hz. Then $$L$$ in meters is:

We begin by noting that a stretched string fixed at both ends can vibrate in normal modes whose frequencies are given by the well-known formula

$$f_n=\dfrac{n v}{2L},$$

where $$f_n$$ is the frequency of the $$n^{\text{th}}$$ harmonic, $$v$$ is the speed of transverse waves on the string, and $$L$$ is the length of the string.

Before using the formula, we must first find the wave speed $$v$$ on the string. The speed of a wave on a stretched string is determined by the tension $$T$$ and the mass per unit length $$\mu$$ through the relation

$$v=\sqrt{\dfrac{T}{\mu}}.$$

We are given

$$T=540\ \text{N}, \qquad \mu = 6.0\times10^{-3}\ \text{kg m}^{-1}.$$

Substituting these values, we have

$$v=\sqrt{\dfrac{540}{6.0\times10^{-3}}}.$$

Inside the square root, the denominator is converted to a simpler decimal:

$$6.0\times10^{-3}=0.006.$$

So,

$$v=\sqrt{\dfrac{540}{0.006}}.$$

Carrying out the division first,

$$\dfrac{540}{0.006}=90\,000.$$

Taking the square root gives

$$v=\sqrt{90\,000}=300\ \text{m s}^{-1}.$$

Now, we are told that two consecutive resonant frequencies are $$420\ \text{Hz}$$ and $$490\ \text{Hz}$$. Let the lower of these, $$420\ \text{Hz}$$, correspond to harmonic number $$n$$. Then the higher one, $$490\ \text{Hz}$$, must correspond to harmonic number $$n+1$$. Using the formula for successive harmonics, we may write

$$f_{n+1}-f_n=\dfrac{v}{2L}.$$

The left-hand side is simply the numerical difference of the two given frequencies:

$$f_{n+1}-f_n = 490-420 = 70\ \text{Hz}.$$

Therefore,

$$70 = \dfrac{v}{2L}.$$

We have already found $$v=300\ \text{m s}^{-1}$$, so we substitute this value:

$$70 = \dfrac{300}{2L}.$$

Simplify the fraction in the numerator first:

$$\dfrac{300}{2L} = \dfrac{150}{L}.$$

Hence the equation becomes

$$70 = \dfrac{150}{L}.$$

To isolate $$L$$, we cross-multiply:

$$70L = 150.$$

Now, solve for $$L$$ by dividing both sides by 70:

$$L = \dfrac{150}{70}.$$

Reduce the fraction by dividing numerator and denominator by 10:

$$L = \dfrac{15}{7}\ \text{m}.$$

Carrying out the division for a decimal value,

$$L \approx 2.142857\ \text{m}.$$

Among the given choices, the closest value is 2.1 m, which corresponds to Option A.

Hence, the correct answer is Option A.