The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe \(L\) meter long. The length of the open pipe will be:

1.	\(L\)	2.	\(2L\)
3.	\(\dfrac{L}{2}\)	4.	\(4L\)

Three sound waves of equal amplitudes have frequencies of \((n-1),~n,\) and \((n+1).\) They superimpose to give beats. The number of beats produced per second will be:

A uniform rope, of length \(L\) and mass \(m_1,\) hangs vertically from a rigid support. A block of mass \(m_2\) is attached to the free end of the rope. A transverse pulse of wavelength \(\lambda_1\) is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is \(\lambda_2.\)${\mathrm{}}_{}$ The ratio \(\frac{\lambda_2}{\lambda_1}\) is:

\(4.0~\text{gm}\) of gas occupies \(22.4~\text{litres}\) at NTP. The specific heat capacity of the gas at a constant volume is  \(5.0~\text{JK}^{-1}\text{mol}^{-1}.\) If the speed of sound in the gas at NTP is \(952~\text{ms}^{-1},\) then the molar heat capacity at constant pressure will be:
(\(R=8.31~\text{JK}^{-1}\text{mol}^{-1}\)) 

The fundamental frequency of a closed organ pipe of a length \(20\) cm is equal to the second overtone of an organ pipe open at both ends. The length of the organ pipe open at both ends will be:

If \(n_1\), \(n_2\)_, and \(n_3\) are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency \(n\) of the string is given by: