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. \(16~\text{cm}\)

An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is \(50\) cm. The next larger length of the column resonating with the same tuning fork 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:
1. \(\sqrt{\frac{m_1+m_2}{m_2}}\)
2. \(\sqrt{\frac{m_2}{m_1}}\)
3. \(\sqrt{\frac{m_1+m_2}{m_1}}\)
4. \(\sqrt{\frac{m_1}{m_2}}\)

\(4.0~\text{gm}\) 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. \(6\)

If we study the vibration of a pipe open at both ends, then which of the following statements is not true:

A source of unknown frequency gives \(4\) beats/s when sounded with a source of known frequency of \(250~\text{Hz}\). The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency of \(513~\text{Hz}\). The unknown frequency will be: