Ch 11  ·  Q–
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Class 9 Science Exercise NCERT Solutions Olympiad Board Exam
Chapter 11

Sound

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

17 Questions
40–55 min Ideal time
Q1 Now at
Q1
NUMERIC3 marks
What is sound and how is it produced?
📘 Concept & Theory Concept Used

Sound is one of the most common forms of energy that we experience in our daily life. It enables us to communicate, enjoy music, receive warning signals, and understand our surroundings. Scientifically, sound is a form of mechanical energy that is produced due to the vibration of objects and travels through a material medium in the form of longitudinal waves.

Every sound that we hear originates from a vibrating source. These vibrations transfer energy from one particle of the medium to the next without transporting the particles themselves over long distances. Since sound requires particles to carry vibrations, it cannot travel through a vacuum.

🗺️ Solution Roadmap Step-by-step Plan
  1. Define sound.

  2. Explain that sound is produced by vibrating objects.

  3. Describe how vibrations travel through a medium.

  4. Explain the formation of compressions and rarefactions.

  5. Give a simple real-life example.

📊 Graph / Figure Graph / Figure
Production & Propagation of Sound Vibrating Source Sound Wave Receiver (Ear) Compressions and rarefactions carry sound energy through a material medium.
✏️ Solution Complete Solution
Step-by-step Solution  ·  1 step
  1. Sound is a form of mechanical energy that produces the sensation of hearing. It travels through a material medium such as air, water, or solids in the form of longitudinal waves.

    Sound is produced when an object vibrates. A vibrating object moves to and fro about its mean position. During this motion, it transfers energy to the surrounding particles of the medium.

    The production of sound occurs in the following steps:

    1. A source such as a guitar string, tuning fork, loudspeaker diaphragm, or drum membrane begins to vibrate.
    2. These vibrations disturb the particles of the surrounding medium.
    3. The particles nearest to the vibrating source move back and forth, creating regions where particles come closer together called compressions, and regions where particles move farther apart called rarefactions.
    4. The disturbance is passed from one particle to the next, transferring energy throughout the medium.
    5. When these sound waves reach our ears, they make the eardrum vibrate. The brain interprets these vibrations as sound.

    Thus, sound is produced due to the vibrations of an object and propagates through a medium as successive compressions and rarefactions.

    Examples:

    • When a guitar string is plucked, it vibrates and produces sound.
    • When a bell is struck, its metal body vibrates and emits sound.
    • The vocal cords in our throat vibrate when we speak, producing our voice.
    • The membrane of a drum vibrates when hit, generating sound waves.
🎯 Exam Significance Exam Significance
  • This is one of the most frequently asked conceptual questions in CBSE and State Board examinations.
  • It forms the foundation for understanding sound waves, reflection of sound, echo, and ultrasound.
  • Competitive examinations often ask why sound cannot travel in vacuum and why vibration is essential for sound production.
  • Understanding this concept helps students answer assertion-reason, MCQ, case-study, and application-based questions.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. Sound is a form of mechanical energy.

  2. Every sound originates from a vibrating object.

  3. Sound requires a material medium for propagation.

  4. Sound travels through successive compressions and rarefactions.

  5. Sound cannot travel through vacuum because there are no particles to transmit vibrations.

  6. The vibrations reaching our eardrum are interpreted by the brain as sound.

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1 / 17  ·  6%
Q2 →
Q2
NUMERIC3 marks
Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
📘 Concept & Theory Concept Used

Sound is a longitudinal mechanical wave that propagates through a material medium. In a longitudinal wave, the particles of the medium vibrate back and forth in the same direction as the wave travels. These vibrations create alternate regions of high pressure and low pressure called compressions and rarefactions, respectively.

Compressions and rarefactions are responsible for the transmission of sound energy from the source to the listener. Without these alternating regions, sound cannot propagate through the medium.

🗺️ Solution Roadmap Step-by-step Plan
  1. Explain the forward motion of the vibrating source.

  2. Describe the formation of compression.

  3. Explain the backward motion of the vibrating source.

  4. Describe the formation of rarefaction.

  5. Conclude that alternate compressions and rarefactions form a sound wave.

📊 Graph / Figure Graph / Figure
Formation of Compressions and Rarefactions VIBRATION Source Compression (C) Rarefaction (R) Compression (C) Rarefaction (R) Compression (C) Rarefaction (R) Direction of Sound Wave Propagation
Explanation of the Diagram
  • The vibrating source pushes nearby air particles during its forward motion, creating a compression (C).
  • When the source moves backward, the air particles spread apart, producing a rarefaction (R).
  • Successive compressions and rarefactions travel through the air as a longitudinal sound wave.
  • The air particles only oscillate back and forth about their mean positions, whereas the sound energy moves forward.
✏️ Solution Complete Solution
Step-by-step Solution  ·  1 step
  1. A vibrating object continuously moves to and fro about its mean position. As it vibrates, it disturbs the surrounding air particles and produces alternating regions of high and low pressure.

    The formation of compressions and rarefactions takes place in the following steps:

    1. Suppose a vibrating object moves forward.
    2. During this forward motion, it pushes the air particles in front of it closer together.
    3. The air particles become crowded, increasing the pressure and density of that region. This region is called a compression (C).
    4. Next, the vibrating object moves backward.
    5. As it moves backward, the air particles spread farther apart.
    6. The pressure and density decrease in this region, forming a rarefaction (R).
    7. The object continues vibrating repeatedly, producing alternate compressions and rarefactions.
    8. These regions travel through the medium from one particle to another, carrying sound energy without transporting the particles themselves over long distances.

    Therefore, a sound wave consists of a continuous sequence of compressions and rarefactions produced by the vibrations of the source.

🎯 Exam Significance Exam Significance
  • This is one of the most important diagram-based questions in the Sound chapter.
  • CBSE frequently asks students to explain compressions and rarefactions with a labelled diagram.
  • Understanding this concept is essential for learning wavelength, frequency, speed of sound, echo, and reflection of sound.
  • Competitive examinations often test the difference between longitudinal and transverse waves using this concept.
  • Case-study and assertion-reason questions are frequently based on the formation of compressions and rarefactions.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. Sound is a longitudinal wave.

  2. Compression is a region of high pressure and high particle density.

  3. Rarefaction is a region of low pressure and low particle density.

  4. Alternate compressions and rarefactions constitute a sound wave.

  5. Sound energy propagates through the medium, while the particles merely oscillate about their mean positions.

  6. Sound cannot travel without a material medium because compressions and rarefactions cannot form in a vacuum.

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2 / 17  ·  12%
Q3 →
Q3
NUMERIC3 marks
Why is sound wave called a longitudinal wave?
📘 Concept & Theory Concept Used

Waves are classified according to the direction in which the particles of the medium vibrate relative to the direction of wave propagation. Based on this criterion, waves are of two main types:

  • Longitudinal waves: The particles of the medium vibrate parallel to the direction of wave propagation.
  • Transverse waves: The particles of the medium vibrate perpendicular to the direction of wave propagation.

Sound propagates through air, water, and solids by causing the particles of the medium to move back and forth in the same direction as the wave travels. Therefore, sound is a longitudinal wave.

🗺️ Solution Roadmap Step-by-step Plan
  1. State the definition of a longitudinal wave.

  2. Explain the motion of particles in a sound wave.

  3. Describe the formation of compressions and rarefactions.

  4. Conclude why sound is classified as a longitudinal wave.

✏️ Solution Complete Solution
Step-by-step Solution  ·  1 step
  1. Sound waves are called longitudinal waves because the particles of the medium vibrate parallel to the direction in which the sound wave travels.

    The propagation of a sound wave takes place in the following steps:

    1. A vibrating source such as a tuning fork, loudspeaker, or guitar string produces vibrations.
    2. These vibrations disturb the nearby particles of the medium.
    3. The particles move to and fro about their mean positions in the same direction as the sound wave travels.
    4. This back-and-forth motion creates alternate regions of compressions (high pressure) and rarefactions (low pressure).
    5. The compressions and rarefactions move forward through the medium, carrying sound energy from one place to another.
    6. Although the sound wave travels forward, the particles themselves only oscillate about their mean positions and do not travel with the wave.

    Since the direction of particle vibration is the same as the direction of wave propagation, a sound wave is called a longitudinal wave.

🎯 Exam Significance Exam Significance
  • This is one of the most frequently asked conceptual questions in CBSE and State Board examinations.
  • Students should clearly distinguish between longitudinal and transverse waves.
  • Competitive examinations often ask the direction of particle vibration in sound waves.
  • This concept forms the basis for understanding compressions, rarefactions, wavelength, and wave propagation.
  • It is commonly tested through MCQs, assertion-reason questions, and case-study-based questions.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. Sound is a longitudinal mechanical wave.

  2. Particles of the medium vibrate parallel to the direction of wave propagation.

  3. Sound waves consist of alternate compressions and rarefactions.

  4. The particles only oscillate about their mean positions; they do not travel with the wave.

  5. The wave transfers energy from one place to another through the medium.

  6. Sound cannot travel through a vacuum because there are no particles to vibrate.

← Q2
3 / 17  ·  18%
Q4 →
Q4
NUMERIC3 marks
Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
📘 Concept & Theory Concept Used

Sound possesses several important characteristics, namely loudness, pitch, and quality (timbre). While loudness depends on the amplitude of vibration and pitch depends on the frequency of vibration, the quality or timbre of sound depends on the waveform produced by the source.

Even if two sounds have the same loudness and pitch, they may still sound different because their waveforms are different. This unique nature of the waveform is called the quality (timbre) of sound.

🗺️ Solution Roadmap Step-by-step Plan
  1. Recall the characteristics of sound.

  2. Identify the characteristic responsible for distinguishing voices.

  3. Explain why every person's voice is unique.

  4. Conclude with the correct answer.

📊 Graph / Figure Graph / Figure
Quality (Timbre) Helps Identify a Voice Every voice has a unique harmonic footprint, allowing us to recognize who is speaking. Friend Speaking UNIQUE TIMBRE WAVE Listener Different sound waveforms help our brains immediately distinguish different voices.
✏️ Solution Complete Solution
Step-by-step Solution  ·  2 steps
  1. The characteristic of sound that helps us identify our friend's voice is quality or timbre.

    The identification of a person's voice can be understood in the following steps:

    1. Every person has vocal cords that differ slightly in length, thickness, and tension.
    2. The shape of the throat, mouth cavity, tongue, and nasal cavity also varies from one person to another.
    3. These differences produce a unique waveform even if two people speak with the same pitch and loudness.
    4. This unique waveform gives each voice its own characteristic quality or timbre.
    5. Our ears and brain recognize this distinctive quality, allowing us to identify familiar voices even without seeing the speaker.

    Therefore, even in a dark room where visual identification is not possible, we can easily recognize our friend's voice because of its unique quality (timbre).

  2. Important Note
  3. Characteristic Depends On Purpose
    Loudness Amplitude Distinguishes loud and soft sounds.
    Pitch Frequency Distinguishes shrill and deep sounds.
    Quality (Timbre) Waveform and harmonics Distinguishes one person's voice or one musical instrument from another.
🎯 Exam Significance Exam Significance
  • This is a frequently asked conceptual question in CBSE and State Board examinations.
  • Students should clearly distinguish between loudness, pitch, and quality of sound.
  • Competitive examinations often ask which characteristic is responsible for identifying a person's voice.
  • The concept is also useful for solving MCQs, assertion-reason questions, and case-study-based questions.
  • Understanding timbre helps explain why different musical instruments produce distinct sounds even when playing the same note.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. Quality or timbre is the characteristic that distinguishes one sound from another.

  2. Every person's voice has a unique waveform.

  3. Quality depends on the construction of the vocal cords and resonating cavities.

  4. Voices can be recognized even if they have the same loudness and pitch.

  5. Loudness depends on amplitude, whereas pitch depends on frequency.

  6. Timbre is responsible for identifying familiar voices and different musical instruments.

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4 / 17  ·  24%
Q5 →
Q5
NUMERIC3 marks
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
📘 Concept & Theory Concept Used

During a thunderstorm, lightning and thunder are produced at the same instant. Lightning is an electrical discharge in the atmosphere, while thunder is the sound produced due to the rapid expansion of air heated by the lightning. Although both events occur simultaneously, they reach an observer at different times because light and sound travel at vastly different speeds.

Light travels extremely fast, whereas sound travels comparatively slowly through air. Hence, the flash is seen almost instantly, while the sound takes a noticeable amount of time to reach our ears.

🗺️ Solution Roadmap Step-by-step Plan
  1. State that lightning and thunder are produced simultaneously.

  2. Compare the speeds of light and sound.

  3. Explain why the flash reaches us first.

  4. Conclude why thunder is heard after a few seconds.

✏️ Solution Complete Solution
Step-by-step Solution  ·  4 steps
  1. Lightning (flash) and thunder are produced at the same time during a thunderstorm. However, we see the flash first and hear the thunder a few seconds later because the speed of light is much greater than the speed of sound.

    The explanation is as follows:

    1. Lightning produces an intense flash of light and generates thunder simultaneously.
    2. The speed of light in air is approximately \[ 3 \times 10^{8}\ \text{m/s} \] which is extremely high.
    3. The speed of sound in air at room temperature is approximately \[ 340\ \text{m/s} \] which is much slower than the speed of light.
    4. Because light travels almost instantaneously over distances encountered on Earth, the flash reaches our eyes almost immediately.
    5. The sound waves travel much more slowly through the air, so they require a few seconds to reach our ears.
    6. Therefore, we observe the lightning first and hear the thunder after a short delay.

    Hence, the delay between seeing lightning and hearing thunder is due to the enormous difference between the speeds of light and sound, not because they are produced at different times.

  2. Important Scientific Fact
  3. The time gap between the flash and the thunder can be used to estimate how far away the lightning occurred.
  4. Since \[ \text{Distance}=\text{Speed}\times\text{Time}, \] the approximate distance of the lightning is \[\text{Distance} \approx 340 \times (\text{time delay})\ \text{metres}\]
  5. For example, if thunder is heard 5 seconds after the flash, \[ \text{Distance}=340\times5=1700\ \text{m}=1.7\ \text{km}. \]
🎯 Exam Significance Exam Significance
  • This is a very common conceptual question in CBSE and State Board examinations.
  • Students should remember that lightning and thunder are produced simultaneously.
  • Competitive examinations frequently ask students to compare the speeds of light and sound.
  • The concept is also used in numerical problems involving the speed of sound and estimation of the distance of lightning.
  • This topic is often tested through MCQs, assertion-reason questions, and case-study-based questions.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. Lightning and thunder are produced at the same instant.

  2. Light travels much faster than sound.

  3. Speed of light: \[ 3\times10^{8}\ \text{m/s} \]

  4. Speed of sound in air: \[ \approx340\ \text{m/s} \]

  5. The flash reaches our eyes almost instantly, whereas thunder takes time to reach our ears.

  6. The delay between lightning and thunder can be used to estimate the distance of the lightning strike.

← Q4
5 / 17  ·  29%
Q6 →
Q6
NUMERIC3 marks
A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as \[ 344\,\text{m s}^{-1}. \]
📘 Concept & Theory Concept Used
p> The wavelength of a sound wave is the distance between two consecutive compressions or two consecutive rarefactions. It depends on the speed of sound in the medium and the frequency of the sound.

Frequency and wavelength are inversely proportional. Therefore:

  • Lower frequency corresponds to a longer wavelength.
  • Higher frequency corresponds to a shorter wavelength.

The relationship between speed, wavelength, and frequency is given by

\[ v=\lambda\nu \]

🗺️ Solution Roadmap Step-by-step Plan
  1. Write the given values.

  2. Use the wave equation.

  3. Calculate the wavelength for 20 Hz.

  4. Calculate the wavelength for 20 kHz.

  5. State the final answer with proper SI units.

✏️ Solution Complete Solution
Step-by-step Solution  ·  5 steps
  1. Given
  2. Lower frequency: \[ \nu_{1}=20\,\text{Hz} \]

    Higher frequency: \[ \nu_{2}=20\,\text{kHz}=20\times10^{3}\,\text{Hz}=20000\,\text{Hz} \]

    Speed of sound: \[ v=344\,\text{m s}^{-1} \]

  3. Formula
  4. \[ v=\lambda\nu \]

    Rearranging the formula,

    \[ \lambda=\frac{v}{\nu} \]

  5. Wavelength corresponding to 20 Hz
  6. Substitute the given values: \[ \begin{aligned} \lambda_{1} &=\frac{344}{20}\\&=17.2\,\text{m}\end{aligned} \]
  7. Wavelength corresponding to 20 kHz
  8. Substitute the given values: \[ \begin{aligned} \lambda_{2} &=\frac{344}{20000}\\&=0.0172\,\text{m} \end{aligned} \]
  9. Converting into centimetres,\[0.0172\,\text{m}=1.72\,\text{cm}\]
🎯 Exam Significance Exam Significance
  • This is a standard numerical based on the wave equation \[ v=\lambda\nu. \]
  • Students should remember that frequency and wavelength are inversely proportional.
  • CBSE frequently asks numericals involving the calculation of wavelength, frequency, or speed of sound.
  • Competitive examinations often test unit conversions such as \[ 20\,\text{kHz}=20000\,\text{Hz}. \]
  • This concept is useful for solving MCQs, assertion-reason questions, and application-based numericals.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. The wave equation is \[ v=\lambda\nu. \]

  2. Wavelength is calculated using \[ \lambda=\frac{v}{\nu}. \]

  3. Lower frequency produces a longer wavelength.

  4. Higher frequency produces a shorter wavelength.

  5. The human hearing range extends from \[ 20\,\text{Hz} \] to \[ 20\,\text{kHz}. \]

  6. For sound travelling at \[ 344\,\text{m s}^{-1}, \] the wavelength varies from 17.2 m to 0.0172 m (1.72 cm).

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Q7 →
Q7
NUMERIC3 marks
Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
📘 Concept & Theory Concept Used

The time taken by a sound wave to travel through a medium depends on the distance travelled and the speed of sound in that medium.

Sound travels much faster in solids than in gases because the particles in solids are closely packed and have stronger intermolecular forces. Therefore, sound reaches the other end of an aluminium rod much earlier than it reaches through the surrounding air.

The relation between distance, speed, and time is

\[ \text{Time}=\frac{\text{Distance}}{\text{Speed}} \]

🗺️ Solution Roadmap Step-by-step Plan
  1. Write the given speeds of sound in air and aluminium.

  2. Assume the length of the rod is \(L\).

  3. Write the expressions for time taken in both media.

  4. Take the ratio of the two times.

  5. Substitute the values and simplify.

✏️ Solution Complete Solution
Step-by-step Solution  ·  9 steps
  1. Given
  2. Let the length of the aluminium rod be \[ L. \]

    Speed of sound in air: \[ v_{\text{air}}=344\,\text{m s}^{-1} \]

    Speed of sound in aluminium: \[ v_{\text{Al}}=6420\,\text{m s}^{-1} \]

  3. To Find
  4. Ratio of times taken by sound in air and aluminium: \[ \frac{t_{\text{air}}}{t_{\text{Al}}} \]
  5. Formula
  6. \[t=\frac{L}{v}\]
  7. Time taken through air
  8. \[t_{\text{air}}=\frac{L}{v_{\text{air}}}\]
  9. Time taken through aluminium
  10. \[t_{\text{Al}}=\frac{L}{v_{\text{Al}}}\]
  11. Calculate the ratio
  12. Dividing the two expressions, \[ \frac{t_{\text{air}}}{t_{\text{Al}}} = \frac{\left(\dfrac{L}{v_{\text{air}}}\right)} {\left(\dfrac{L}{v_{\text{Al}}}\right)} \]
  13. Since the length \(L\) is the same in both cases, it cancels out. \[ \frac{t_{\text{air}}}{t_{\text{Al}}} = \frac{v_{\text{Al}}}{v_{\text{air}}} \]
  14. Substituting the given values, \[ \begin{aligned} \frac{t_{\text{air}}}{t_{\text{Al}}} &= \frac{6420}{344}\\ &\approx 18.66 \end{aligned} \]
  15. Therefore,\[\boxed{t_{\text{air}}:t_{\text{Al}}=18.66:1}\]
💡 Answer Final Answer
Hence, sound takes about 18.7 times longer to travel through air than through aluminium.
🎯 Exam Significance Exam Significance
  • This is a frequently asked numerical based on the relation between speed, distance, and time.
  • Students should remember that sound travels fastest in solids, slower in liquids, and slowest in gases.
  • The cancellation of distance while taking ratios is an important mathematical step often tested in examinations.
  • Competitive examinations frequently ask conceptual questions comparing the speeds of sound in different media.
  • The concept is useful for MCQs, assertion-reason questions, and numerical problem-solving.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. Time taken by sound is given by \[ t=\frac{L}{v}. \]

  2. For the same distance, time is inversely proportional to the speed of sound.

  3. Sound travels much faster in aluminium than in air.

  4. The ratio of times is \[ \frac{t_{\text{air}}}{t_{\text{Al}}} = \frac{v_{\text{Al}}}{v_{\text{air}}}. \]

  5. The required ratio is \[ \boxed{18.7:1.} \]

  6. Solids are excellent conductors of sound because their particles are closely packed.

← Q6
7 / 17  ·  41%
Q8 →
Q8
NUMERIC3 marks
The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
📘 Concept & Theory concept Used

Frequency is the number of vibrations or oscillations made by a vibrating body in one second. Its SI unit is hertz (Hz).

By definition,

\[ 1\,\text{Hz} = 1\ \text{vibration per second} \]

Therefore, if the frequency of a source is known, the total number of vibrations during any time interval can be calculated using:

\[ \text{Number of vibrations}=\text{Frequency}\times\text{Time} \]

🗺️ Solution Roadmap Step-by-step Plan
  1. Write the given frequency.

  2. Convert one minute into seconds.

  3. Use the relation between frequency and number of vibrations.

  4. Calculate the total number of vibrations.

  5. Write the answer with the correct unit.

✏️ Solution Complete Solution
Step-by-step Solution  ·  4 steps
  1. Given
  2. Frequency of the source, \[ f=100\,\text{Hz} \]

    Time, \[ t=1\,\text{minute}=60\,\text{s} \]

  3. To Find
  4. Number of vibrations in one minute.
  5. Formula
  6. \[f=\frac{n}{t}\] where,
    • \(f\) = frequency
    • \(n\) = number of vibrations
    • \(t\) = time
    Rearranging the formula, \[n=f\times t\]
  7. Step-by-Step Calculation
  8. Substitute the given values:\[\begin{aligned}n&=100\times60\\&=6000\end{aligned}\]
💡 Answer Final Answer

Hence, the source of sound vibrates 6000 times in one minute.

🎯 Exam Significance Exam Significance
  • This is a basic numerical based on the definition of frequency.
  • Students should remember that 1 Hz means one vibration per second.
  • CBSE often asks direct numericals involving frequency, time, and number of vibrations.
  • Competitive examinations frequently test unit conversion from minutes to seconds.
  • This concept forms the foundation for solving more advanced wave-motion problems.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. Frequency is the number of vibrations per second.

  2. The SI unit of frequency is hertz (Hz).

  3. \[ 1\,\text{Hz}=1\ \text{vibration per second} \]

  4. The relation is \[ n=f\times t. \]

  5. Always convert time into seconds before performing calculations.

  6. A source of frequency \[ 100\,\text{Hz} \] makes \[ \boxed{6000} \] vibrations in one minute.

← Q7
8 / 17  ·  47%
Q9 →
Q9
NUMERIC3 marks
Does sound follow the same laws of reflection as light does? Explain.
📘 Concept & Theory Concept Used

Reflection is the phenomenon in which a wave bounces back into the same medium after striking a surface. Both light waves and sound waves undergo reflection. Although light is an electromagnetic wave and sound is a mechanical wave, the fundamental laws governing their reflection are the same.

A clear reflection of sound is obtained when sound strikes a large, hard, and smooth surface such as a wall, cliff, or the side of a building. This principle is used in echoes, megaphones, soundboards, auditoriums, and stethoscopes.

🗺️ Solution Roadmap Step-by-step Plan
  1. State whether sound follows the laws of reflection.

  2. Write the two laws of reflection.

  3. Explain each law briefly.

  4. Give a real-life example of reflection of sound.

  5. Conclude with the importance of reflected sound.

✏️ Solution Complete Solution
Step-by-step Solution  ·  1 step
  1. Yes. Sound follows the same laws of reflection as light.

    When sound waves strike a hard and smooth surface, they are reflected back into the same medium. The reflection of sound obeys the following two laws:

    1. First Law of Reflection
      The angle of incidence is equal to the angle of reflection.

      \[ \boxed{\angle i=\angle r} \]

    2. Second Law of Reflection
      The incident sound wave, the reflected sound wave, and the normal at the point of incidence all lie in the same plane.

    Therefore, the reflection of sound is completely analogous to the reflection of light. The only important difference is that sound requires a material medium for propagation, whereas light can travel through vacuum.

    Example: When a person shouts towards a large wall or a mountain, the sound waves strike the surface and are reflected back. If the reflected sound reaches the listener after a sufficient time interval, it is heard as an echo.

🎯 Exam Significance Exam Significance
  • This is one of the most frequently asked theory questions from the chapter "Sound".
  • Students should be able to state both laws of reflection accurately.
  • Diagram-based questions on reflection of sound are common in CBSE examinations.
  • Competitive examinations often ask the applications of reflected sound and the conditions required for good reflection.
  • The concept is frequently tested through MCQs, assertion-reason questions, and case-study-based questions.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. Sound obeys the same laws of reflection as light.

  2. First law: \[ \angle i=\angle r \]

  3. The incident wave, reflected wave, and normal lie in the same plane.

  4. Hard, smooth, and large surfaces produce better reflection of sound.

  5. Reflection of sound is responsible for echoes, SONAR, stethoscopes, megaphones, and auditorium acoustics.

  6. Unlike light, sound requires a material medium for propagation.

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9 / 17  ·  53%
Q10 →
Q10
NUMERIC2 marks
When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remain the same. Do you hear an echo sooner on a hotter day?
📘 Concept & Theory Concept Used

An echo is the repetition of a sound caused by the reflection of sound waves from a distant obstacle such as a wall, hill, or building. The time taken to hear an echo depends on two factors:

  • The distance between the source of sound and the reflecting surface.
  • The speed of sound in the medium.

The speed of sound in air is not constant. It increases with an increase in the temperature of the air. Therefore, for the same distance, sound takes less time to travel on a hotter day than on a colder day.

The empirical relation between the speed of sound and temperature is:

\[ v=331+0.6T \]

where \(v\) is the speed of sound in \(\text{m s}^{-1}\) and \(T\) is the temperature in \(^\circ\text{C}\).

🗺️ Solution Roadmap Step-by-step Plan
  1. Recall how an echo is produced.

  2. State the effect of temperature on the speed of sound.

  3. Relate speed with the time taken to hear an echo.

  4. Draw the conclusion.

✏️ Solution Complete Solution
Step-by-step Solution  ·  3 steps
  1. Yes. An echo is heard sooner on a hotter day.

    The explanation is given below:

    1. When a person produces a sound, the sound waves travel to the reflecting surface.
    2. After striking the surface, the sound waves are reflected back towards the listener.
    3. The total distance travelled by the sound is twice the distance between the source and the reflecting surface.
    4. On a hotter day, the temperature of air is higher.
    5. As the temperature increases, the speed of sound in air also increases according to the relation:

      \[ v=331+0.6T \]

    6. Since the distance remains unchanged but the speed increases, the time taken by the sound to travel to the obstacle and back decreases.
    7. Hence, the reflected sound reaches the listener earlier.

    Therefore, although the distance between the source and the reflecting surface remains the same, the echo is heard earlier on a hotter day because sound travels faster in warm air.

  2. Scientific Explanation
  3. The time taken by sound is given by\[t=\frac{\text{Distance}}{\text{Speed}}\]
  4. Since the distance is constant,
    • If speed increases, time decreases.
    • If speed decreases, time increases.
🎯 Exam Significance Exam Significance
  • This is one of the most frequently asked conceptual questions related to echo.
  • Students should know that the speed of sound increases with temperature.
  • The empirical relation \[ v=331+0.6T \] is commonly asked in objective and competitive examinations.
  • The concept is useful for solving numericals involving echo, speed of sound, and temperature.
  • CBSE often asks application-based and assertion-reason questions based on this concept.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. An echo is produced due to the reflection of sound.

  2. The speed of sound increases with an increase in temperature.

  3. \[ v=331+0.6T \] gives the approximate speed of sound in air.

  4. For the same distance, \[ t=\frac{d}{v}, \] so a higher speed means a shorter time.

  5. Therefore, an echo is heard earlier on a hotter day.

  6. Temperature affects the speed of sound but does not change the distance travelled by the echo.

← Q9
10 / 17  ·  59%
Q11 →
Q11
NUMERIC3 marks
Give two practical applications of the reflection of sound waves.
📘 Concept & Theory Concept Used

Reflection of sound is the phenomenon in which sound waves bounce back into the same medium after striking a hard, smooth, and large surface. This property of sound is widely used in science, engineering, medicine, navigation, and architecture.

The reflected sound can be used to determine the distance of an object, improve the audibility of sound, or direct sound waves towards a desired location. Many modern technologies are based on this principle.

🗺️ Solution Roadmap Step-by-step Plan
  1. Recall the concept of reflection of sound.

  2. Identify practical devices or situations where reflected sound is useful.

  3. Explain the working principle of each application.

  4. Conclude with the importance of reflection of sound in everyday life.

✏️ Solution Complete Solution
Step-by-step Solution  ·  2 steps
  1. Reflection of sound has numerous practical applications. Two important applications are explained below.

    SONAR (Sound Navigation and Ranging)

    SONAR is used in ships and submarines to measure the depth of the sea and to detect underwater objects such as rocks, submarines, icebergs, and shoals of fish.

    It works on the principle of reflection of sound. A pulse of ultrasonic sound is transmitted into the water. When it strikes an underwater object, it gets reflected back as an echo. The time taken by the echo is measured to determine the distance of the object.

    \[ \boxed{\text{Distance}=\frac{v\times t}{2}} \]

    where

    • \(v\) = speed of sound in water
    • \(t\) = total time taken by the echo
    Design of Auditoriums and Concert Halls

    Large auditoriums, theatres, conference halls, and lecture rooms are carefully designed so that reflected sound reaches every listener clearly.

    Curved ceilings, sound reflectors, and specially designed walls distribute sound uniformly throughout the hall. At the same time, sound-absorbing materials are used to reduce excessive reflections and prevent echoes or reverberation.

    This improves the clarity, loudness, and quality of sound for the audience.

  2. Other Common Applications of Reflection of Sound
    • Stethoscope used by doctors.
    • Megaphones and loudhailers.
    • Hearing aids.
    • Soundboards in musical instruments.
    • Whispering galleries.
    • Ultrasonic imaging and industrial testing.
🎯 Exam Significance Exam Significance
  • This is one of the most important application-based theory questions from the chapter "Sound".
  • CBSE frequently asks students to explain the working principle of SONAR.
  • Students should remember that SONAR uses ultrasonic waves and the principle of reflection of sound.
  • Competitive examinations often ask the formula used to calculate the distance of an object using echo.
  • Knowledge of auditorium acoustics is useful for answering case-study and assertion-reason questions.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. Reflection of sound occurs when sound strikes a hard and smooth surface.

  2. SONAR is based on the reflection of ultrasonic sound waves.

  3. The distance of an underwater object is calculated using \[ \boxed{\text{Distance}=\frac{v\times t}{2}} \]

  4. Auditoriums are designed to distribute reflected sound uniformly and reduce echoes.

  5. Reflection of sound is also used in stethoscopes, megaphones, hearing aids, and whispering galleries.

  6. Reflection of sound has numerous applications in medicine, navigation, architecture, and engineering.

← Q10
11 / 17  ·  65%
Q12 →
Q12
NUMERIC3 marks
A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, \[ g=10\,\text{m s}^{-2} \] and speed of sound \[ =340\,\text{m s}^{-1}. \]
📘 Concept & Theory Concept Used

This problem involves two successive motions:

  1. The stone falls freely from the top of the tower to the pond.
  2. After the splash is produced, the sound travels upward from the pond to the observer at the top.

Therefore, the total time after which the splash is heard is the sum of:

\[ \boxed{\text{Total Time}=\text{Time of Fall}+\text{Time Taken by Sound}} \]

🗺️ Solution Roadmap Step-by-step Plan
  1. Calculate the time taken by the stone to fall using the equation of motion.

  2. Calculate the time taken by the sound to travel from the pond to the top.

  3. Add both times to obtain the total time.

  4. Write the final answer with the correct unit.

✏️ Solution Complete Solution
Step-by-step Solution  ·  9 steps
  1. Given
  2. Height of the tower, \[ h=500\,\text{m} \]

    Acceleration due to gravity, \[ g=10\,\text{m s}^{-2} \]

    Speed of sound, \[ v=340\,\text{m s}^{-1} \]

  3. To Find
  4. Time after which the splash is heard at the top of the tower.
  5. Calculate the time taken by the stone to reach the pond
  6. Since the stone is dropped from rest,\[u=0\]
  7. Using the second equation of motion,\[s=ut+\frac{1}{2}gt^{2}\]
  8. Substituting the given values, \[ \begin{aligned} 500&=0+\frac{1}{2}\times10\times t_{1}^{2}\\ &=5t_{1}^{2}\\ \Rightarrow t_{1}^{2}&=100\\ t_{1} &=10\,\text{s} \end{aligned} \]
  9. Calculate the time taken by the sound to travel upward
  10. The splash occurs at the bottom of the tower, and the sound travels upward through a distance of 500 m.
  11. Using the relation,\[\text{Time}=\frac{\text{Distance}}{\text{Speed}}\]
  12. \[ \begin{aligned} t_{2}&=\frac{500}{340}\\ &=1.47\,\text{s} \end{aligned} \]
  13. Calculate the total time
  14. Total time, \[ \begin{aligned} t&=t_{1}+t_{2}\\ &=10+1.47\\ &11.47\,\text{s} \end{aligned} \]
💡 Answer Final Answer

Hence, the splash is heard approximately 11.47 seconds after the stone is dropped.

🎯 Exam Significance Exam Significance
  • This is one of the most important numerical questions from the chapter "Sound".
  • It combines concepts of free fall and propagation of sound in a single problem.
  • Students should remember to calculate the falling time and sound travel time separately before adding them.
  • Competitive examinations frequently ask similar multi-step numericals involving motion and sound.
  • The problem strengthens the application of equations of motion and the relation \[ \text{Time}=\frac{\text{Distance}}{\text{Speed}}. \]
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. The stone first falls under gravity and then the sound travels upward.

  2. Use \[ s=ut+\frac{1}{2}gt^{2} \] to calculate the falling time.

  3. Use \[ t=\frac{d}{v} \] to calculate the time taken by sound.

  4. The total time is the sum of both time intervals.

  5. The splash is heard after \[ \boxed{11.47\ \text{seconds}}. \]

  6. Always solve such problems step by step instead of combining both motions into a single equation.

← Q11
12 / 17  ·  71%
Q13 →
Q13
NUMERIC3 marks
A sound wave travels at a speed of \(339\,\text{m s}^{-1}\). If its wavelength is \(1.5\,\text{cm}\), what is the frequency of the wave? Will it be audible?
📘 Concept & Theory Concept Used

Every sound wave is characterized by three important quantities: speed, wavelength, and frequency. These quantities are related by the wave equation

\[ v=\lambda\nu \]

where

  • \(v\) = speed of sound
  • \(\lambda\) = wavelength
  • \(\nu\) = frequency

Before using the formula, all quantities must be expressed in SI units. Since the wavelength is given in centimetres, it must first be converted into metres.

The normal hearing range of the human ear is

\[ 20\,\text{Hz}\ \text{to}\ 20\,000\,\text{Hz} \]

Sounds having frequencies greater than \(20\,000\,\text{Hz}\) are called ultrasonic (ultrasound) and cannot be heard by humans.

🗺️ Solution Roadmap Step-by-step Plan
  1. Write the given values.

  2. Convert wavelength from centimetres to metres.

  3. Apply the wave equation.

  4. Calculate the frequency.

  5. Compare the obtained frequency with the audible range.

✏️ Solution Complete Solution
Step-by-step Solution  ·  7 steps
  1. Given
  2. Speed of sound, \[ v=339\,\text{m s}^{-1} \]

    Wavelength, \[ \lambda=1.5\,\text{cm} \]

    Converting centimetres into metres,

    \[ 1.5\,\text{cm} =\frac{1.5}{100}\,\text{m} =1.5\times10^{-2}\,\text{m} \]

  3. To Find
  4. Frequency of the sound wave \((\nu)\) and whether it is audible.
  5. Formula
  6. \[v=\lambda\nu\] Rearranging, \[\nu=\frac{v}{\lambda}\]
  7. Step-by-Step Calculation
  8. Substitute the given values: \[ \begin{aligned} \nu &=\frac{339}{1.5\times10^{-2}}\\ &=\frac{339}{0.015}\\ &=22600\,\text{Hz} \end{aligned} \]
  9. Checking Whether the Sound is Audible
  10. \[ 20\,\text{Hz} \leq \nu \leq 20\,000\,\text{Hz} \]
  11. Since\[22600\,\text{Hz}>20000\,\text{Hz},\]
  12. the sound lies beyond the audible range of the human ear.
🎯 Exam Significance Exam Significance
  • This is a standard numerical based on the wave equation \[ v=\lambda\nu. \]
  • Students should always convert wavelength into SI units before substituting values.
  • Questions involving the audible and ultrasonic frequency ranges are frequently asked in CBSE and State Board examinations.
  • Competitive examinations often test unit conversion, wave equations, and the classification of sound based on frequency.
  • This concept forms the basis for understanding ultrasonic applications such as SONAR and medical ultrasonography.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. The wave equation is \[ v=\lambda\nu. \]

  2. Convert wavelength into metres before applying the formula.

  3. The calculated frequency is \[ \boxed{2.26\times10^{4}\,\text{Hz}}. \]

  4. The audible range of the human ear is \[ 20\,\text{Hz}\text{ to }20\,000\,\text{Hz}. \]

  5. Frequencies greater than \[ 20\,000\,\text{Hz} \] are called ultrasonic waves.

  6. The given sound is an ultrasonic wave and cannot be heard by humans.

← Q12
13 / 17  ·  76%
Q14 →
Q14
NUMERIC3 marks
What is reverberation? How can it be reduced?
📘 Concept & Theory Concept Used

When a sound is produced inside a room or hall, it is reflected repeatedly from the walls, ceiling, floor, and other hard surfaces. If these reflected sound waves continue to reach our ears even after the original sound has stopped, the sound appears to persist for a short duration. This phenomenon is called reverberation.

A small amount of reverberation improves the richness of sound in auditoriums and concert halls. However, excessive reverberation makes speech and music unclear because the reflected sounds overlap with the original sound.

🗺️ Solution Roadmap Step-by-step Plan
  1. Define reverberation.

  2. Explain why reverberation occurs.

  3. Describe its harmful effects.

  4. List practical methods to reduce reverberation.

✏️ Solution Complete Solution
Step-by-step Solution  ·  2 steps
  1. Reverberation is the persistence of sound in an enclosed space due to multiple reflections of sound waves from walls, ceilings, floors, and other hard surfaces, even after the original sound has stopped.

    The phenomenon occurs in the following steps:

    1. A sound is produced inside a room or hall.
    2. The sound waves strike the walls, ceiling, floor, and furniture.
    3. The sound waves are reflected repeatedly from these surfaces.
    4. The reflected waves continue to reach the listener's ears for a short time after the original sound has ceased.
    5. As a result, the sound appears prolonged or blurred. This effect is known as reverberation.

    For example, in an empty classroom, auditorium, or large hall, a person's voice may continue to be heard for a short time due to repeated reflections.

  2. Reduction of Reverberation
  3. Reverberation can be reduced by using materials that absorb sound instead of reflecting it.

    • Cover the floor with carpets, rugs, or mats to absorb reflected sound.
    • Hang thick curtains, draperies, or fabric coverings on windows and walls.
    • Use cushioned chairs and upholstered furniture that absorb sound energy.
    • Install acoustic panels, cork sheets, foam panels, or fibre boards on walls and ceilings.
    • Provide false ceilings or wooden acoustic panels to reduce multiple reflections.
    • Design auditoriums with properly shaped walls and ceilings so that sound is distributed uniformly without excessive reflection.
🎯 Exam Significance Exam Significance
  • This is one of the most frequently asked theory questions from the chapter "Sound".
  • Students should know the difference between echo and reverberation.
  • CBSE often asks practical methods used to reduce reverberation in auditoriums.
  • Competitive examinations frequently include MCQs on sound-absorbing materials and auditorium acoustics.
  • This concept is also useful in assertion-reason and case-study-based questions.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. Reverberation is the persistence of sound caused by multiple reflections.

  2. Hard and smooth surfaces increase reverberation.

  3. Excessive reverberation reduces the clarity of speech and music.

  4. Soft and porous materials absorb sound and reduce reverberation.

  5. Carpets, curtains, acoustic tiles, foam panels, and cushioned seats are commonly used to reduce reverberation.

  6. Proper acoustic design of auditoriums ensures clear and uniform distribution of sound.

← Q13
14 / 17  ·  82%
Q15 →
Q15
NUMERIC3 marks
What is the loudness of sound? What factors does it depend on?
📘 Concept & Theory Concept Used

Sound possesses three important characteristics: loudness, pitch, and quality (timbre). Among these, loudness enables us to distinguish between loud and soft sounds.

Loudness is a physiological sensation perceived by the human ear, whereas intensity is a measurable physical quantity. In general, a sound with greater intensity is perceived as louder.

The loudness of a sound mainly depends on the amplitude of vibration of the source. A larger amplitude means more energy is carried by the sound wave, making the sound appear louder.

🗺️ Solution Roadmap Step-by-step Plan
  1. Define loudness.

  2. Differentiate loudness from intensity.

  3. Explain the main factor affecting loudness.

  4. List the other factors affecting loudness with suitable explanations.

✏️ Solution Complete Solution
Step-by-step Solution  ·  2 steps
  1. Loudness of Sound
  2. Loudness is the sensation produced in the ear that enables us to distinguish between a loud sound and a soft sound. It represents how strong or weak a sound is perceived by a listener.

    Loudness is directly related to the intensity of sound reaching the ear. A sound having greater energy is generally heard as louder than a sound having lower energy.

    For example, the sound of a drum is louder than the sound of a flute because the drum produces vibrations with much larger amplitudes.

  3. Factors Affecting Loudness
    1. Amplitude of Vibration (Most Important Factor)

      Loudness mainly depends on the amplitude of the vibrating source.

      • Greater amplitude → greater energy → louder sound.
      • Smaller amplitude → less energy → softer sound.
    2. Distance from the Source

      As the distance from the source increases, the sound energy spreads over a larger area. Consequently, the intensity and loudness decrease.

    3. Sensitivity of the Ear

      Different individuals may perceive the same sound differently because the sensitivity of human ears varies from person to person.

    4. Nature of the Medium

      The medium through which sound travels also affects loudness. Sound generally travels more efficiently through solids than through gases because solids transmit sound energy more effectively.

    5. Intensity of the Sound Wave

      A sound wave carrying greater energy has higher intensity and is perceived as louder by the human ear.

🎯 Exam Significance Exam Significance
  • This is one of the most frequently asked theory questions from the chapter "Sound".
  • Students should clearly distinguish between loudness, pitch, and quality of sound.
  • CBSE often asks the factor responsible for loudness in MCQs and short-answer questions.
  • Competitive examinations frequently test the relationship between amplitude and loudness.
  • The concept is also useful in assertion-reason and case-study-based questions involving sound waves.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. Loudness is the sensation that distinguishes loud sounds from soft sounds.

  2. The most important factor affecting loudness is the amplitude of vibration.

  3. Greater amplitude produces a louder sound, while smaller amplitude produces a softer sound.

  4. Loudness also depends on the distance from the source, the nature of the medium, the intensity of sound, and the sensitivity of the listener's ear.

  5. Loudness depends mainly on amplitude, whereas pitch depends on frequency.

  6. Loudness is a physiological sensation, while intensity is a measurable physical quantity.

← Q14
15 / 17  ·  88%
Q16 →
Q16
NUMERIC3 marks
How is ultrasound used for cleaning?
📘 Concept & Theory Concept Used

Ultrasound refers to sound waves having frequencies greater than \[ 20,000\,\text{Hz} \; (20\,\text{kHz}). \] These high-frequency sound waves possess enough energy to clean objects that have tiny holes, narrow grooves, and inaccessible parts where ordinary cleaning methods cannot reach.

Ultrasonic cleaning works on the principle of cavitation. In this process, ultrasonic waves passing through a cleaning liquid produce millions of microscopic bubbles. These bubbles grow and collapse rapidly, generating tiny shock waves that remove dirt and other impurities from the surface of the object.

🗺️ Solution Roadmap Step-by-step Plan
  1. State what ultrasound is.

  2. Explain the principle of cavitation.

  3. Describe the cleaning process step by step.

  4. Mention common applications of ultrasonic cleaning.

✏️ Solution Complete Solution
Step-by-step Solution  ·  2 steps
  1. Ultrasound is widely used for cleaning delicate and complex objects by the process of cavitation.

    The cleaning process takes place in the following steps:

    1. The object to be cleaned is immersed in a tank containing a suitable cleaning liquid.
    2. Ultrasonic waves of very high frequency are passed through the liquid.
    3. These ultrasonic waves produce millions of tiny bubbles in the liquid.
    4. The bubbles rapidly collapse, producing minute shock waves and high-pressure jets.
    5. These shock waves remove dust, grease, oil, dirt, and other impurities even from very small holes, cracks, and narrow grooves.
    6. The object becomes thoroughly cleaned without causing any damage to its surface.

    Therefore, ultrasound provides an efficient and non-destructive method for cleaning delicate objects that are difficult to clean by ordinary methods.

  2. Common Applications of Ultrasonic Cleaning
    • Cleaning surgical and medical instruments.
    • Cleaning jewellery such as rings, necklaces, and watches.
    • Cleaning electronic components and circuit boards.
    • Cleaning laboratory equipment.
    • Cleaning precision machine parts and automobile components.
🎯 Exam Significance Exam Significance
  • This is a frequently asked application-based question from the chapter "Sound".
  • Students should remember that ultrasonic cleaning works on the principle of cavitation.
  • CBSE examinations often ask the uses of ultrasound in medicine and industry.
  • Competitive examinations frequently include MCQs on ultrasonic cleaning and its applications.
  • This concept is also useful for case-study and assertion-reason questions.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. Ultrasound has a frequency greater than \[ 20\,000\,\text{Hz}. \]

  2. Ultrasonic cleaning works on the principle of cavitation.

  3. Microscopic bubbles formed in the cleaning liquid collapse rapidly and remove dirt.

  4. Ultrasound can clean tiny holes, cracks, and narrow grooves that are difficult to clean manually.

  5. It is widely used for cleaning jewellery, surgical instruments, laboratory equipment, and electronic components.

  6. Ultrasonic cleaning is efficient, safe, and non-destructive.

← Q15
16 / 17  ·  94%
Q17 →
Q17
NUMERIC3 marks
Explain how defects in a metal block can be detected using ultrasound.
📘 Concept & Theory Concept Used

Ultrasound consists of sound waves having frequencies greater than \[ 20,000\,\text{Hz}. \] These waves can penetrate solids and are reflected whenever they encounter a boundary between two materials having different properties, such as a crack, cavity, or air gap inside a metal block.

The detection of internal defects is based on the reflection of ultrasonic waves. Since ultrasonic waves have very short wavelengths, they can detect even very small cracks that are invisible from the outside.

🗺️ Solution Roadmap Step-by-step Plan
  1. State the principle used for defect detection.

  2. Explain how ultrasonic waves are transmitted into the metal.

  3. Describe what happens when no defect is present.

  4. Explain how a defect reflects the ultrasonic waves.

  5. Conclude how the position of the defect is determined.

✏️ Solution Complete Solution
Step-by-step Solution  ·  2 steps
  1. Defects inside a metal block are detected using the reflection of ultrasonic waves.

    The detection process takes place in the following steps:

    1. An ultrasonic transmitter sends high-frequency ultrasonic waves into the metal block.
    2. The ultrasonic waves travel through the metal.
    3. If the metal block is free from defects, the waves travel to the opposite surface and are reflected back to the receiver.
    4. If the block contains an internal crack, cavity, air gap, or other defect, the ultrasonic waves are reflected from that defect before reaching the opposite surface.
    5. The reflected waves (echoes) are received by the ultrasonic detector.
    6. The time taken by the echo to return is measured.
    7. Using the speed of ultrasound in the metal, the position and approximate size of the defect are determined.

    Thus, ultrasound makes it possible to detect internal defects without cutting or damaging the metal block. This technique is known as Non-Destructive Testing (NDT).

  2. Applications of Ultrasonic Defect Detection
    • Inspection of railway tracks.
    • Testing aircraft wings and engine components.
    • Inspection of bridges and steel structures.
    • Testing pressure vessels and pipelines.
    • Quality control in manufacturing industries.
🎯 Exam Significance Exam Significance
  • This is one of the most important application-based questions from the chapter "Sound".
  • Students should remember that ultrasonic testing works on the principle of reflection of sound.
  • CBSE examinations frequently ask the industrial uses of ultrasound.
  • Competitive examinations often test the concept of Non-Destructive Testing (NDT) and ultrasonic inspection.
  • This topic is useful for MCQs, assertion-reason questions, and case-study-based questions.
🔑 Key Takeaways Key Takeaways
Key Takeaways  ·  6 points
  1. Ultrasound is used to detect internal cracks and defects in metal blocks.

  2. The technique is based on the reflection of ultrasonic waves.

  3. Defects such as cracks, cavities, and air gaps reflect ultrasonic waves before they reach the opposite surface.

  4. The time taken by the reflected echo helps determine the location of the defect.

  5. This method is called Non-Destructive Testing (NDT) because the object is tested without causing any damage.

  6. Ultrasonic testing is widely used in industries, aviation, railways, pipelines, and bridge inspection.

← Q16
17 / 17  ·  100%
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Chapter Complete!

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NCERT Class 9 Sound Exercise Solutions
NCERT Class 9 Sound Exercise Solutions — Complete Notes & Solutions · academia-aeternum.com
Sound is a fascinating topic that connects physics to our everyday experiences — from hearing a friend’s voice to enjoying music or detecting objects with sonar. In NCERT Class 9 Science Chapter 11 “SOUND”, students explore how sound is produced, travels through different media, and reflects to create echoes and reverberations. The textbook exercises help learners apply these concepts through numerical problems, reasoning questions, and real-life applications like ultrasound, SONAR, and sound…
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    Sound — Learning Resources

    📄 Detailed Notes
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    ✔️ True / False
    💬 Q&A Discussion

    Frequently Asked Questions

    Longitudinal waves are waves in which particle movement is parallel to wave direction, like sound waves in air.

    Human ears cannot detect frequencies higher than 20,000 Hz, which is the range of ultrasonic sound.

    Audible sounds are within human hearing range (20 Hz–20,000 Hz); sounds outside this range are inaudible to us.

    Greater amplitude means louder sound, while smaller amplitude results in a softer sound to our ears.

    Using oscilloscopes, ripple tanks, or computer-based simulations, sound waves can be displayed as graphs.

    Sound requires particles to transfer vibrations. In vacuum, where no particles exist, sound cannot travel.

    Devices like SONAR send out ultrasound waves and measure their reflections to detect objects and measure distances underwater.

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