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. \(\frac{L}{2}\)
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:

The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is \(20~\text{cm}\), the length of the open organ pipe is:
1. \(13.2~\text{cm}\)
2. \(8~\text{cm}\)
3. \(12.5~\text{cm}\)
4. \(16~\text{cm}\)

A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 ms^-1 at an angle of ${60}^0$ with the source-observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 ms^-1), is: 
              

1. 100 Hz

2. 103 Hz

3. 106 Hz

4. 97 Hz

\(4.0~\text{gm}\) 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:
(\(R=8.31~\text{JK}^{-1}\text{mol}^{-1}\)) 

A string is stretched between fixed points separated by \(75.0~\text{cm}\). It is observed to have resonant frequencies of \(420~\text{Hz}\) and \(315~\text{Hz}\). There are no other resonant frequencies between these two. The lowest resonant frequency for this string is:
1. \( 155 \mathrm{~Hz} \)
2. \( 205 \mathrm{~Hz} \)
3. \( 10.5 \mathrm{~Hz} \)
4. \( 105 \mathrm{~Hz}\)

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:
1. \( \frac{1}{n}=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}\)
2. \( \frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n_1}}+\frac{1}{\sqrt{n_2}}+\frac{1}{\sqrt{n_3}}\)
3. \( \sqrt{n}=\sqrt{n_1}+\sqrt{n_2}+\sqrt{n_3}\)
4. \( n=n_1+n_2+n_3\)

The number of possible natural oscillations of the air column in a pipe closed at one end of length \(85\) cm whose frequencies lie below \(1250\) Hz are:(velocity of sound= $\mathrm{}$\(340~\text{m/s}\)) 
1. \(4\)
2. \(5\)
3. \(7\)
4. \(6\)