Class 9 Science Chapter 10 Important Questions with Answers PDF
Why does a tabla sound different from a flute even when both play the same musical note? How can a ship detect an underwater object, and why might a person hear two tapping sounds through a steel fence? Chapter 10, Sound Waves: Characteristics and Applications, explores these questions through sound-producing vibrations, wave propagation, frequency, amplitude, reflection and the applications of sound beyond human hearing.
Table of Content
These Class 9 Science Chapter 10 Important Questions include solved MCQs, experiment-based problems, diagrams, numerical calculations and detailed explanations from the Exploration textbook for 2026–27. Students can practise important questions from Chapter 10 of Class 9 Science covering sound intensity, pitch, timbre, noise pollution, Indian musical instruments, echoes, SONAR and echolocation. For additional chapter-wise practice, visit Class 9 Science Important Questions.
These chapter 10 Class 9 Science important questions cover different skills rather than repeatedly asking for definitions. Students will interpret observations, analyse sound graphs, compare explanations, calculate wave quantities and apply scientific ideas to real situations.
Class 9 Science Chapter 10 Important Questions MCQ
Q1. A singer records a song using a microphone. Which energy conversion occurs inside the microphone?
(a) Electrical energy into heat energy
(b) Sound energy into electrical energy
(c) Light energy into sound energy
(d) Chemical energy into mechanical energy
Answer: (b) Sound energy into electrical energy.
Explanation: Sound waves make the microphone's diaphragm vibrate. These vibrations are converted into an electrical signal. When the recording is played through a speaker, the speaker converts electrical energy into sound energy through a vibrating diaphragm.
Q2. A student marks one coil of a slinky and pushes its end forward and backward. The compression travels towards the opposite end, but the marked coil remains near its original position. What does this show?
(a) The medium moves completely from one end to the other.
(b) Only the marked coil carries energy.
(c) The disturbance travels while particles oscillate around their mean positions.
(d) The marked coil is not involved in wave propagation.
Answer: (c) The disturbance travels while particles oscillate around their mean positions.
Explanation: In a longitudinal mechanical wave, the medium's particles vibrate parallel to the direction of propagation. The disturbance and energy travel through the medium, but the particles do not continuously move forward with the wave.
Q3. A sound wave is represented by a graph of air density against distance. What does the highest point of the graph indicate?
(a) Minimum air density
(b) Maximum air density within a compression
(c) Zero particle movement
(d) The end of the sound wave
Answer: (b) Maximum air density within a compression.
Explanation: On a density–distance graph, a crest represents maximum density. A trough represents minimum density within a rarefaction. These are features of the graphical representation of longitudinal sound waves, not evidence that the air particles themselves follow a transverse path.
Q4. Consider the following statements about an electric bell inside a bell jar.
Assertion (A): The bell becomes difficult to hear after most of the air is removed, even though it continues vibrating.
Reason (R): Sound requires a material medium for propagation.
(a) Both A and R are true, and R explains A.
(b) Both A and R are true, but R does not explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer: (a) Both A and R are true, and R explains A.
Explanation: The bell continues producing vibrations. However, removing air reduces the particles available to transmit the sound disturbance. This establishes why sound cannot propagate through a perfect vacuum.
Q5. A musician plays a note of fundamental frequency 240 Hz and then another note of frequency 480 Hz. What is the relationship between these notes?
(a) The second note is one octave higher.
(b) The notes have identical frequencies.
(c) The second note has half the frequency.
(d) The second note must be twice as loud.
Answer: (a) The second note is one octave higher.
Explanation: An octave separates notes whose fundamental frequencies have a ratio of 2:1. Doubling frequency raises a note by one octave but does not automatically increase loudness.
Class 9 Science Chapter 10 Very Short Answer Questions
Q6. Why is a metal taal described as sonorous?
Answer: Sonority is the property of certain metals to produce a ringing sound when struck. A taal produces sound through the vibrations created when its metal parts are struck together.
Q7. What does amplitude mean in the density representation of a sound wave?
Answer: Amplitude is the maximum change in the density of the medium above or below its average density. Greater density amplitude generally indicates more energy carried by the wave, when other conditions are comparable.
Q8. What is Jingrwai Iawbei, mentioned in the chapter?
Answer: Jingrwai Iawbei is a tune-naming tradition associated with Kongthong village in Meghalaya. A mother composes a special tune for her child, which can be sung or whistled as a form of identification.
Q9. Why are decibels used when discussing loud sounds and noise pollution?
Answer: Decibels (dB) express sound levels on a logarithmic scale. They are useful for comparing sound levels and discussing potentially harmful exposure, although perceived loudness also depends on the listener.
Q10. State the law of reflection of sound involving the angles of incidence and reflection.
Answer: The angle of incidence equals the angle of reflection, measured from the normal to the reflecting surface. The incident sound direction, reflected sound direction and normal lie in the same plane.
Class 9 Science Chapter 10 Short Answer Questions
Q11. Two students are 250 m apart along a steel fence. One student taps the fence, and the other hears sound through the steel and surrounding air. Calculate the approximate difference in arrival times.
Use the following speeds:
Sound in steel = 5000 m/s
Sound in air = 340 m/s
Solution:
Step 1: Calculate the time through steel.
Time = Distance ÷ Speed
= 250 ÷ 5000
= 0.05 s
Step 2: Calculate the time through air.
= 250 ÷ 340
≈ 0.735 s
Step 3: Calculate the difference.
= 0.735 − 0.05
≈ 0.685 s
Answer: The sound travelling through air arrives approximately 0.685 seconds after the sound travelling through steel.
Explanation: Sound propagates much faster through steel than through air. Since the arrival-time difference is greater than 0.1 s, the student may distinguish the two sounds if both are sufficiently audible.
Q12. A teacher stretches a balloon sheet over a container and sprinkles rice grains on it. When a loud sound is produced nearby, the grains begin moving. Why does this happen without touching the container?
Answer:
The loud sound produces pressure and density disturbances in the surrounding air.
These disturbances reach the stretched sheet and cause it to vibrate. The vibrating sheet transfers energy to the rice grains, making them move or jump.
The observation demonstrates that sound carries energy through a material medium.
The movement of the grains does not mean that air particles travel all the way from the sound source to the grains. Instead, energy is transferred through the medium as the wave propagates.
Q13. A sound wave appears as an up-and-down curve on a graph. Does this mean the particles in air move up and down while sound travels horizontally? Explain.
Answer:
No. A curved graph of sound does not necessarily represent the actual path followed by air particles.
In the chapter's density–distance graph:
The horizontal axis shows distance.
The vertical axis shows air density.
Crests represent maximum density.
Troughs represent minimum density.
The curve displays how density changes with distance.
Actual air particles oscillate back and forth parallel to the direction of sound propagation, making sound in air a longitudinal wave.
Q14. A flute and a tabla play notes that are perceived to have the same pitch and loudness. Why can listeners still recognise which instrument is playing?
Answer:
The instruments have different timbres, or characteristic sound qualities.
A musical note usually contains a fundamental frequency together with higher frequencies called overtones.
The combination and relative strengths of these frequencies vary with an instrument's materials, shape and vibrating parts.
A flute uses a vibrating air column, whereas the tabla uses a vibrating membrane. Their different vibration patterns produce recognisably different sounds even when pitch and loudness are similar.
Q15. Thunder travels 1720 m through air. Compare the approximate travel time at 22°C and at 0°C, using sound speeds of 344 m/s and 331 m/s, respectively.
Solution:
At 22°C:
Time = 1720 ÷ 344
= 5 s
At 0°C:
Time = 1720 ÷ 331
≈ 5.20 s
Extra time at 0°C:
= 5.20 − 5.00
≈ 0.20 s
Answer: The thunder takes about 0.20 seconds longer to travel the same distance at 0°C.
Explanation: The chapter explains that sound generally travels faster in warmer air. This calculation assumes all other relevant conditions remain unchanged.
Q16. A sound wave travels through air at 340 m/s with a frequency of 680 Hz. Calculate its wavelength and time period.
Solution:
Given:
Speed = 340 m/s
Frequency = 680 Hz
Step 1: Find wavelength.
Wave speed = Frequency × Wavelength
Wavelength = Speed ÷ Frequency
= 340 ÷ 680
= 0.5 m
Step 2: Find time period.
Time period = 1 ÷ Frequency
= 1 ÷ 680
≈ 0.00147 s
Answer:
Wavelength = 0.5 m
Time period ≈ 0.00147 s
Q17. Why might a newly built auditorium produce unclear speech even when the microphones and speakers are working properly?
Answer:
Excessive reverberation may make the speech unclear.
When sound repeatedly reflects from hard walls, ceilings and floors, it may persist after the speaker has finished producing it. Overlapping reflections can make words difficult to distinguish.
This problem can be reduced through acoustic design, including sound-absorbing panels, curtains and upholstered seating.
However, some reverberation is useful in performance spaces. The goal is to control the reflections rather than eliminate them completely.
Q18. Why can infrasonic waves be useful for monitoring natural events even though humans cannot normally hear them?
Answer:
Infrasonic waves have frequencies below 20 Hz, which are generally outside the human audible range.
Some large natural events, such as volcanic eruptions and severe storms, produce low-frequency disturbances.
These waves can travel considerable distances and may be detected by sensitive instruments.
Scientists can analyse them to monitor environmental activity without relying on human hearing.
Class 9 Science Chapter 10 Long Answer Questions
Q19. Explain how sound travels from a person speaking to another person's brain. Include sound production, wave propagation and hearing.
Answer:
Sound travels from a speaker to a listener through the following processes:
1. Sound Production: Air passing through the larynx causes the vocal cords to vibrate. The tongue, lips and mouth shape these vibrations into speech.
2. Wave Propagation: Vibrations create alternating compressions and rarefactions in the surrounding air. These disturbances carry sound energy towards the listener.
3. Particle Motion: Air particles vibrate back and forth around their mean positions. They transfer energy without travelling continuously from the speaker to the listener.
4. Eardrum and Middle Ear: Sound waves make the eardrum vibrate. Three tiny bones in the middle ear transmit and amplify these vibrations.
5. Cochlea and Brain: The cochlea converts vibrations into electrical signals. The auditory nerve carries these signals to the brain, which interprets them as speech.
Conclusion: Hearing involves sound production, wave propagation, vibration transfer and interpretation by the brain.
Q20. Design an investigation to compare the sounds produced by a thick rubber band and a thin rubber band. Explain what should be kept constant and how the results should be interpreted.
Answer:
This investigation examines how the physical properties of a vibrating rubber band affect the sound produced.
Aim
To compare the vibration frequencies and sounds produced by two rubber bands of different thicknesses.
Materials
A cardboard box
One thin rubber band
One thick rubber band
A ruler
A sound-frequency measuring application, if available
Step 1: Arrange the Setup
Stretch one rubber band across the cardboard box.
Pluck it gently and observe its sound.
Step 2: Record the Sound
Use a suitable application to record the approximate dominant frequency.
Repeat the observation several times.
Step 3: Change the Rubber Band
Replace it with the second rubber band.
Keep the vibrating length and stretching conditions as comparable as possible.
Step 4: Repeat Measurements
Pluck the second band in the same manner and record its frequency.
Step 5: Compare Results
A thin band may vibrate at a higher frequency than a thick band under comparable conditions, but the result also depends on tension, length and material.
Variables to Consider
Scientific Interpretation
If the thin rubber band vibrates more frequently, its sound will generally have a higher pitch.
If the difference is small or unexpected, the student should check whether the bands had comparable tension and length.
Conclusion
The investigation demonstrates that sound depends on the characteristics of a vibrating source. Reliable experiments require controlled conditions and repeated observations.
Q21. Two sound waves, A and B, travel through the same air at 340 m/s. The distance between successive compressions is 2 cm for Wave A and 1 cm for Wave B. Compare their wavelengths, frequencies and audible classification.
Answer:
The two waves travel at the same speed but have different wavelengths.
Step 1: Identify the Wavelengths
Wave A:
2 cm = 0.02 m
Wave B:
1 cm = 0.01 m
Step 2: Use the Wave Equation
Wave speed = Frequency × Wavelength
Frequency = Speed ÷ Wavelength
Step 3: Calculate Wave A's Frequency
Frequency A = 340 ÷ 0.02
= 17,000 Hz
Step 4: Calculate Wave B's Frequency
Frequency B = 340 ÷ 0.01
= 34,000 Hz
Step 5: Compare the Frequencies
Wave B has twice the frequency of Wave A.
This happens because Wave B's wavelength is half the wavelength of Wave A.
Step 6: Classify the Sounds
Wave A has a frequency of 17,000 Hz, which lies within the conventional human audible range, although hearing at high frequencies varies among individuals.
Wave B has a frequency of 34,000 Hz, which is ultrasonic.
Comparison Table
Conclusion
For waves travelling at the same speed, frequency and wavelength are inversely related.
Q22. Compare a whispering gallery and a modern auditorium to explain how the reflection of sound can be used or controlled in architecture.
Answer:
Sound reflection plays an important role in the design of buildings such as whispering galleries and auditoriums.
1. Reflection of Sound: When sound strikes a surface, some of its energy is reflected. The angle of incidence equals the angle of reflection.
2. Whispering Gallery: The Whispering Gallery at Gol Gumbaz in Vijayapura, Karnataka, demonstrates how curved surfaces and repeated reflections can make faint sounds audible at distant locations.
3. Modern Auditoriums: Auditoriums use carefully designed surfaces to distribute sound evenly and maintain speech clarity.
4. Reverberation: Repeated reflections can cause sound to persist. Excessive reverberation makes speech unclear, although some reverberation improves musical performances.
5. Sound Control: Curtains, acoustic panels and upholstered seats absorb sound and help control unwanted reflections.
Conclusion: Whispering galleries demonstrate unusual acoustic effects, while auditoriums use controlled reflection and absorption to improve listening conditions.
Q23. A research vessel sends an ultrasonic pulse towards an underwater object. The echo returns after 0.8 seconds. If sound travels at 1500 m/s in water, calculate the object's distance and explain how SONAR uses this principle.
Solution:
Step 1: Identify the Given Values
Speed of sound = 1500 m/s
Echo return time = 0.8 s
Step 2: Understand the Travel Path
The signal travels from the vessel to the object.
After reflection, it travels back to the vessel.
Therefore, the measured time includes both journeys.
Step 3: Calculate the Total Distance
Total distance = Speed × Time
= 1500 × 0.8
= 1200 m
Step 4: Calculate the One-Way Distance
Distance to object = 1200 ÷ 2
= 600 m
Answer: The underwater object is approximately 600 metres away along the sound path.
How SONAR Works
SONAR stands for Sound Navigation and Ranging.
A transmitter sends sound pulses into water.
The pulses travel through water and reach underwater objects.
The sound reflects from the objects.
A receiver detects the returning echoes.
The travel time is used to calculate the distance.
Applications
SONAR is used to detect submarines, locate underwater wreckage, measure water depth and investigate the sea floor.
Important Point
The return time must be divided by two when calculating the distance to a reflecting object.
Q24. Study the illustration of a sound wave containing alternating compressions and rarefactions. Explain how the density graph represents the wave and why the particles do not travel with the disturbance.
Answer:
The illustration contains two related representations of the same sound wave.
1. Particle Representation
Closely spaced particles represent compressions.
Widely spaced particles represent rarefactions.
2. Density Graph
The vertical axis represents the density of the medium.
The horizontal axis represents distance.
3. Crests
A crest on the density graph represents a region of maximum density within a compression.
4. Troughs
A trough represents a region of minimum density within a rarefaction.
5. Wavelength
The distance between two successive compressions or two successive rarefactions is one wavelength.
On the density graph, this is the horizontal distance between consecutive crests or consecutive troughs.
6. Direction of Particle Motion
The particles vibrate back and forth parallel to the wave's direction of propagation.
7. Why the Particles Do Not Travel with the Wave
A particle passes its disturbance to neighbouring particles and continues oscillating around its mean position.
As a result, the energy and disturbance move forward, but the medium does not undergo continuous bulk motion along with the wave.
Conclusion
The illustration demonstrates how sound carries energy through density variations in a material medium.
Q25. A school is organising a music festival in its auditorium. Explain how sound-wave concepts can help the organisers improve sound quality while protecting students' hearing.
Answer:
Sound-wave concepts can help organisers improve music quality while protecting students' hearing.
1. Choose Suitable Equipment: Microphones convert sound into electrical signals, while speakers convert electrical signals into sound. Good equipment helps reproduce music and speech clearly.
2. Maintain Sound Quality: Pitch determines how high or low a sound is perceived, while timbre helps listeners distinguish musical instruments. The sound system should reproduce these characteristics accurately.
3. Control Sound Levels: Excessively loud sound can damage hearing, especially during prolonged exposure. Organisers should monitor sound levels and avoid unnecessary volume.
4. Reduce Reverberation: Acoustic panels, curtains and upholstered seating can reduce unwanted reflections and improve speech clarity.
5. Position Speakers and Protect Listeners: Speakers should distribute sound evenly. Students should be able to move away from loud areas, and people working near loudspeakers may require hearing protection.
Conclusion: Proper sound equipment, controlled volume and good acoustic design can create an enjoyable and safer music festival.
Important Concepts, Formulas and Preparation Tips for Class 9 Science Chapter 10
The Class 9 Science Chapter 10 important questions cover sound production, wave characteristics, reflection, SONAR and ultrasound. Revise these essential concepts, formulas and applications for effective exam preparation.
Preparation Tips: Practise sound-wave diagrams, experimental observations, numerical calculations and real-life applications. Focus on understanding concepts rather than memorising answers.
The Class 9 Science Chapter 10 Important Questions PDF, once available, will help students revise solved questions, wave diagrams and numerical problems from Sound Waves: Characteristics and Applications for the 2026–27 academic session.
Additional Study Materials for Class 9 Science Chapter 10 – Sound Waves: Characteristics and Applications
CBSE Class 9 Science Chapter-wise Important Questions
The chapter-wise important questions for CBSE Class 9 Science cover essential concepts from the Exploration textbook for 2026–27. These resources support conceptual understanding, scientific reasoning and examination preparation across different chapters.
Other Important Related Links for CBSE Class 9 Science
FAQs on Important Questions and Answers for Class 9 Science Exploration Chapter 10 – Sound Waves: Characteristics and Applications |2026–27
1. Why do some Class 9 Science Chapter 10 important questions ask students to explain experiments instead of just defining sound?
Experiments such as the vibrating rubber band, tuning fork and slinky help students connect scientific definitions with visible observations. Explaining what happens, why it happens and what the result proves develops a deeper understanding of sound production, propagation and energy transfer.
2. How can students tell whether a sound-wave diagram shows particle displacement or air density?
Students should first examine the axes and labels. A density–distance graph shows changes in the medium's density, with crests representing maximum density and troughs minimum density. A displacement diagram represents particle motion. The two diagrams may look similar but describe different physical quantities.
3. Can two instruments produce the same musical note but still sound different?
Yes. Instruments can have the same fundamental frequency but different combinations and strengths of overtones. These differences create timbre, which allows listeners to distinguish a flute, tabla, sitar or other instrument even when notes have similar pitch and loudness.
4. What does the syaahi on a tabla or mridangam contribute to its sound?
The syaahi is the black patch applied to the drum membrane. It modifies the membrane's vibration pattern and helps produce the instrument's characteristic tones. The chapter connects this acoustic design with Indian percussion instruments and C. V. Raman's work on musical acoustics.
5. How is the pitch of a sound related to its frequency, and why is loudness different?
Pitch describes how high or low a sound is perceived, and it generally increases with frequency. Loudness concerns how strong a sound seems to the listener and is influenced by sound intensity, amplitude and hearing sensitivity. Increasing frequency does not automatically make a sound louder.
6. Why does a sound become weaker as we move farther from its source?
As sound spreads away from its source, the energy is distributed over an increasingly large area. Consequently, the sound intensity generally decreases with distance. Obstacles, absorption and environmental conditions can also affect how strongly a sound is heard.
7. How can students use Class 9 Chapter 10 Science important questions to improve numerical accuracy?
Students should identify the relevant quantities, choose the correct relation and check units before calculating. They must pay particular attention to echo and SONAR questions, where the recorded time covers the outward and return journeys. Comparing the final result with the physical situation helps detect errors.
8. Why does the speed of sound change with air temperature, but not normally with the frequency of the sound source?
The speed of sound in air depends on the medium's physical conditions, including temperature. When the source frequency changes in the same medium under unchanged conditions, the wavelength adjusts while the speed remains approximately constant. The chapter uses this relationship to connect sound speed, frequency and wavelength.
9. What is special about the Whispering Gallery at Gol Gumbaz?
The Whispering Gallery at Gol Gumbaz in Vijayapura, Karnataka, is known for its unusual acoustic effects. Sound reflects around the dome, allowing faint sounds to be heard at distant points. It is a useful real-world example of how building shape affects sound reflection.
10. Why is too much reverberation a problem for speech but sometimes desirable for music?
Excessive reverberation makes reflected sounds overlap, reducing the clarity of spoken words. A controlled amount can make musical performances sound fuller. Auditoriums therefore use a combination of reflecting and absorbing surfaces to achieve suitable acoustics.
11. How are microphones, speakers and hearing aids connected through sound energy?
A microphone converts sound-induced vibrations into electrical signals. A speaker uses electrical signals to create vibrations that produce sound. A hearing aid generally includes a microphone, electronic processing or amplification and a receiver that delivers sound to the ear.
12. How do scientists use infrasonic waves to study earthquakes, storms or volcanoes?
Some large natural events produce low-frequency sound waves. Sensitive instruments can detect and analyse these waves even when they are not audible to humans. The information can help scientists monitor environmental activity and investigate distant events.
13. What is the difference between a bat's echolocation and SONAR used by ships?
Both rely on the reflection of sound waves. Bats produce ultrasonic calls and interpret echoes to locate insects or obstacles. SONAR equipment sends sound pulses through water and analyses reflected signals to determine the position or distance of underwater objects. The underlying principle is similar, but the systems and environments differ.
