Two vibrating strings of the same material but of lengths Land \(2 \mathrm{~L}\) have radii \(2 \mathrm{r}\) and \(\mathrm{r}\) respectively. They are stretched under the same tension. Both the string vibrate in their fundamental modes. The one of length \(\mathrm{L}\) with frequency \(\mathrm{U}_{1}\) and the other with frequency \(\mathrm{U}_{2}\). What is the ratio $\left(\mathrm{U}_{1} / \mathrm{U}_{2}\right)$ ? (A) 8 (B) 2 (C) 1 (D) 4

We are given two vibrating strings with the following properties: String 1: Length L, radius 2r, fundamental frequency U1 String 2: Length 2L, radius r, fundamental frequency U2 Both strings are of the same material and are stretched under the same tension.

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

The linear mass density, \(\mu\), is defined as the mass per unit length: \[ \mu = \frac{m}{L} \] For a string with radius \(r\) and length \(L\), its volume is given by: \[ V=r^2 L\pi \] Mass can be expressed as a product of volume and density: \[ m = \rho V \] Substituting volume into the mass equation, we get: \[ m = \rho\pi r^2 L \] Now, we can express the linear mass density in terms of radius: \[ \mu = \frac{m}{L} = \rho\pi r^2 \]

To find the fundamental frequencies of both strings, we need to apply the formula we derived earlier: For String 1, use the values \(L\), \(2r\) for radius, and the same tension \(T\). \[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\rho\pi(2r)^2}} \] For String 2, use the values \(2L\), \(r\) for radius, and the same tension \(T\). \[ f_2 = \frac{1}{4L} \sqrt{\frac{T}{\rho\pi r^2}} \]

We need to find the ratio of the fundamental frequencies: \[ \frac{f_1}{f_2}=\frac{\left(\frac{1}{2L} \sqrt{\frac{T}{\rho\pi(2r)^2}}\right)}{\left(\frac{1}{4L} \sqrt{\frac{T}{\rho\pi r^2}}\right)} \] Simplify: \[ \frac{f_1}{f_2}=\frac{\left(\frac{1}{2L} \sqrt{\frac{T}{4\rho\pi r^2}}\right)}{\left(\frac{1}{4L} \sqrt{\frac{T}{\rho\pi r^2}}\right)} \] Cancel out common factors, and the ratio becomes: \[ \frac{f_1}{f_2}=\frac{2 \sqrt{\frac{1}{4}}}{\sqrt{1}}=\frac{1}{2} \] However, the question asks for the ratio \(\frac{U_{1}}{U_{2}}\), which is the inverse of the ratio \(\frac{f_1}{f_2}\). So, taking the inverse of the ratio we calculated: \[ \frac{U_{1}}{U_{2}}=\frac{1}{\frac{1}{2}}=2 \] The ratio U1 / U2 is 2, so the correct answer is (B) 2.