1. Use the formula for resonant frequencies of a stretched wire

For a string fixed at both ends, $f_n = \frac{n}{2L}v$ where $n=1,2,3,\dots$ and wave speed $v = \sqrt{\frac{T}{\mu}}$

Here,

• $\mu = 6.0\times 10^{-3}\ \text{kg m}^{-1}$
• $T = 540\ \text{N}$

2. Find the wave speed

$v = \sqrt{\frac{540}{6.0\times 10^{-3}}}$

$v = \sqrt{90000} = 300\ \text{m/s}$

3. Use the fact that consecutive resonant frequencies differ by the fundamental frequency

Given consecutive frequencies are $420\ \text{Hz}$ and $490\ \text{Hz}$.

So, $f_{n+1} - f_n = \frac{v}{2L}$

Thus, $490 - 420 = \frac{300}{2L}$

$70 = \frac{300}{2L} = \frac{150}{L}$

4. Solve for $L$

$L = \frac{150}{70} = \frac{15}{7} \approx 2.14\ \text{m}$

5. Match with the options

$L \approx 2.1\ \text{m}$

So the correct option is B.

6. Comparison with stored answer

Stored correct answer: B

This matches our derived answer.