A wire of length 30 cm, stretched between rigid supports, has its $$n^{th}$$ and $$(n+1)^{th}$$ harmonics at 400 Hz and 450 Hz, respectively. If tension in the string is 2700 N, its linear mass density is _____ kg m$$^{-1}$$.

We are given that the length of the wire is $$L = 30$$ cm $$= 0.3$$ m, the $$n^{th}$$ harmonic frequency is $$f_n = 400$$ Hz, the $$(n+1)^{th}$$ harmonic frequency is $$f_{n+1} = 450$$ Hz, and the tension is $$T = 2700$$ N.

For a string fixed at both ends, the frequency of the $$n^{th}$$ harmonic is $$f_n = n \cdot f_1$$, and the difference between consecutive harmonics gives the fundamental frequency as $$f_{n+1} - f_n = f_1 = 450 - 400 = 50 \text{ Hz}$$. Hence, $$n = \frac{f_n}{f_1} = \frac{400}{50} = 8$$, so we are dealing with the 8th and 9th harmonics. The fundamental frequency for a string of length $$L$$ is $$f_1 = \frac{v}{2L}$$, from which $$v = 2Lf_1 = 2 \times 0.3 \times 50 = 30 \text{ m/s}$$.

The wave speed on a string is related to tension and linear mass density by $$v = \sqrt{\frac{T}{\mu}}$$, yielding $$\mu = \frac{T}{v^2} = \frac{2700}{(30)^2} = \frac{2700}{900} = 3 \text{ kg/m}$$.

The answer is 3.