The fundamental frequency (\( f \)) of a vibrating string is given by: \[ f = \frac{v}{2L}, \] where: \item \( f \) is the frequency, \item \( v \) is the wave velocity, \item \( L \) is the length of the string.
Step 1: Relation between frequencies and lengths. For the initial string length (\( L \)) and frequency (\( f_1 = 120 \, \text{Hz} \)): \[ f_1 = \frac{v}{2L}. \] For the new string length (\( L' \)) and frequency (\( f_2 = 180 \, \text{Hz} \)): \[ f_2 = \frac{v}{2L'}. \] Taking the ratio of the two frequencies: \[ \frac{f_2}{f_1} = \frac{L}{L'}. \] Substitute \( f_1 = 120 \, \text{Hz} \) and \( f_2 = 180 \, \text{Hz} \): \[ \frac{180}{120} = \frac{L}{L'}. \] Simplify: \[ \frac{3}{2} = \frac{L}{L'}. \] Rearrange to solve for \( L' \): \[ L' = \frac{2}{3} L. \]
Step 2: Substitute the initial length. Given \( L = 90 \, \text{cm} \): \[ L' = \frac{2}{3} \cdot 90 = 60 \, \text{cm}. \]
Final Answer: The new string length is: \[ \boxed{60 \, \text{cm}}. \]
Two loudspeakers (\(L_1\) and \(L_2\)) are placed with a separation of \(10 \, \text{m}\), as shown in the figure. Both speakers are fed with an audio input signal of the same frequency with constant volume. A voice recorder, initially at point \(A\), at equidistance to both loudspeakers, is moved by \(25 \, \text{m}\) along the line \(AB\) while monitoring the audio signal. The measured signal was found to undergo \(10\) cycles of minima and maxima during the movement. The frequency of the input signal is _____________ Hz.
(Speed of sound in air is \(324 \, \text{m/s}\) and \( \sqrt{5} = 2.23 \)) 
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.