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

1. \(8.0~\text{JK}^{-1}\text{mol}^{-1}\)  2. \(7.5~\text{JK}^{-1}\text{mol}^{-1}\)
3. \(7.0~\text{JK}^{-1}\text{mol}^{-1}\) 4. \(8.5~\text{JK}^{-1}\text{mol}^{-1}\)
Subtopic:  Speed of Sound |
From NCERT
NEET - 2015
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
Links
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

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. \( 155~\text{Hz} \)
2. \( 205~\text{Hz} \)
3. \( 10.5~\text{Hz} \)
4. \( 105~\text{Hz} \)
Subtopic:  Standing Waves |
 78%
From NCERT
NEET - 2015
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

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:

1. \(80\) cm 2. \(100\) cm
3. \(120\) cm 4. \(140\) cm
Subtopic:  Standing Waves |
 78%
From NCERT
NEET - 2015
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
Links
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

advertisementadvertisement

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. \( n=n_1+n_2+n_3\)

Subtopic:  Standing Waves |
 77%
From NCERT
AIPMT - 2014
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

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= \(340~\text{m/s}\)
1. \(4\)
2. \(5\)
3. \(7\)
4. \(6\)

Subtopic:  Standing Waves |
 68%
From NCERT
AIPMT - 2014
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

If we study the vibration of a pipe open at both ends, then which of the following statements is not true:
1. Odd harmonics of the fundamental frequency will be generated.
2. All harmonics of the fundamental frequency will be generated.
3. Pressure change will be maximum at both ends.
4. The open end will be an antinode.
Subtopic:  Standing Waves |
 58%
From NCERT
AIPMT - 2013
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
Links
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

advertisementadvertisement

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:

1. \(246~\text{Hz}\) 2. \(240~\text{Hz}\)
3. \(260~\text{Hz}\) 4. \(254~\text{Hz}\)
Subtopic:  Beats |
 77%
From NCERT
AIPMT - 2013
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
Links
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

A wave traveling in the +ve \(x\text-\)direction having maximum displacement along \(y\text-\)direction as \(1~\text{m}\), wavelength \(2\pi~\text{m}\) and frequency of \(\frac{1}{\pi}~\text{Hz}\), is represented by:
1. \(y=\sin (2 \pi x-2 \pi t)\)
2. \(y=\sin (10 \pi x-20 \pi t)\)
3. \(y=\sin (2 \pi x+2 \pi t)\)
4. \( y=\sin (x-2 t)\)

Subtopic:  Wave Motion |
 87%
From NCERT
AIPMT - 2013
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
Links
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

The length of a wire between the two ends of a sonometer is \(100\) cm. What should be the positions of two bridges below the wire so that the three segments of the wire have their fundamental frequencies in the ratio \(1:3:5\)?
1. \(\frac{1500}{23} \mathrm{~cm}, \frac{500}{23} \mathrm{~cm} \)
2. \(\frac{1500}{23} \mathrm{~cm}, \frac{300}{23} \mathrm{~cm} \)
3. \(\frac{300}{23} \mathrm{~cm}, \frac{1500}{23} \mathrm{~cm} \)
4. \(\frac{1500}{23} \mathrm{~cm}, \frac{2000}{23} \mathrm{~cm}\)
Subtopic:  Standing Waves |
From NCERT
NEET - 2013
Please attempt this question first.
Hints
Please attempt this question first.

advertisementadvertisement

Two sources \(\text{P}\) and \(\text{Q}\) produce notes of frequency \(660~\text{Hz}\) each. A listener moves from \(\text{P}\) to \(\text{Q}\) with a speed of \(1\) m/s. If The speed of sound is \(330\) m/s, then number of beats heard by the listener per second will be:
1. \(4\)
2. \(8\)
3. \(2\)
4. zero
Subtopic:  Doppler's Effect (OLD NCERT) |
From NCERT
NEET - 2013
Please attempt this question first.
Hints
Please attempt this question first.