Fundamental frequency of a sonometer wire is \(\mathrm{n}\). If the length and diameter of the wire are doubled keeping the tension same, the new fundamental frequency is...... (A) \((2 \mathrm{n} / \sqrt{2})\) (B) \(\\{\mathrm{n} /(2 \sqrt{2})\) (C) \(\sqrt{2 n}\) (D) \((\mathrm{n} / 4)\)

The formula for the fundamental frequency of a vibrating string is given by: \(f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}\) Here, \(f\) is the fundamental frequency, \(L\) is the length of the string, \(T\) is the tension, and \(\mu\) is the linear mass density.

Since the problem only provides us with the diameter change, we will first find the relationship between linear mass density and the diameter. Linear mass density (\(\mu\)) can be calculated as: \(\mu = \dfrac{m}{L}\) Where \(m\) is the mass of the wire, \(L\) is the length of the wire, and the mass of a cylinder (the wire) is given by: \(m = \pi r^2 h \rho\) Here \(r\) is the radius (half of the diameter), \(h\) is the height (which is equal to length in this case), and \(\rho\) is the density of the material. When we substitute the mass formula into the linear mass density formula, we will have: \(\mu = \dfrac{\pi r^2 L \rho}{L}\) This will simplify to: \(\mu = \pi r^2 \rho\)

Since the length and diameter are doubled, let the new length be \(2L\), and the new diameter be \(2d\), where \(d\) is the original diameter. The new radius is thus equal to \(r_{new} = 2r\), thus the new linear mass density (\(\mu_{new}\)) will be: \(\mu_{new} = \pi (2r)^2 \rho = 4\pi r^2 \rho\)

Substitute the new length and linear mass density into the fundamental frequency formula: \(f_{new} = \dfrac{1}{(2)(2L)} \sqrt{\dfrac{T}{4\pi r^2 \rho}}\) Factor out the ratio of the new frequency to the original frequency: \(\dfrac{f_{new}}{n} = \dfrac{1}{(2)(2L)} \sqrt{\dfrac{T}{(4\pi r^2 \rho)(2L)}}\) The original fundamental frequency can be written as: \(n = \dfrac{1}{2L} \sqrt{\dfrac{T}{\pi r^2 \rho}}\) We can now write the ratio of the new frequency to the old frequency as: \(\dfrac{f_{new}}{n} = \dfrac{1}{(2)(2L)} \sqrt{\dfrac{T}{(4\pi r^2 \rho)(2L)}} \times \dfrac{\sqrt{\dfrac{T}{\pi r^2 \rho}}}{\dfrac{1}{2L}}\) After simplification, we find: \(\dfrac{f_{new}}{n} = \dfrac{1}{4}\) Finally, we can find the new fundamental frequency: \(f_{new} = \dfrac{n}{4}\) The correct answer is (D) \(\dfrac{n}{4}\).