10th Class Physics Notes 2026 — Chapter 13 – Federal Board

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Class 10 Physics — Chapter 13: Sound

Board: Federal Board

1. Multiple Choice Questions (MCQs)

Q1. What are the three main parts of the human ear?

A. Outer ear, Middle ear, Inner ear
B. Inner ear, Cochlea, Eardrum
C. Middle ear, Eardrum, Auditory nerve
D. Outer ear, Auditory nerve, Cochlea

Correct Answer: A. Outer ear, Middle ear, Inner ear

Q2. Sound produced by piano and violin gives __________ waveform on an oscilloscope.

A. Simple
B. Harmonic
C. Complex
D. Symmetric

Correct Answer: C. Complex

Q3. Which is NOT true for sound?

A. Sound is produced by a vibrating body.
B. Sound travels due to variations of pressure in air.
C. Sound transfers energy along with matter of the medium.
D. Sound is a compressional and longitudinal wave.

Correct Answer: C. Sound transfers energy along with matter of the medium.

Sound waves transfer energy without transferring matter.

Q4. Sound travels faster in:

A. Rubber
B. Air
C. Water
D. Steel

Correct Answer: D. Steel

Sound generally travels faster in solids because of their greater elastic properties.

Q5. Speed of sound in air is \(332\ \text{m s}^{-1}\). What is its speed in a vacuum?

A. Equal to \(332\ \text{m s}^{-1}\)
B. Greater than \(332\ \text{m s}^{-1}\)
C. Less than \(332\ \text{m s}^{-1}\)
D. Zero

Correct Answer: D. Zero

Sound requires a material medium to propagate and cannot travel through a vacuum.

Q6. Sound travels from water to an iron rod, then into air and back into water. The speed of sound will successively:

A. Increase, decrease, increase
B. Decrease, increase, decrease
C. Increase, increase, increase
D. Decrease, decrease, decrease

Correct Answer: A. Increase, decrease, increase

The speed increases when sound travels from water to iron, decreases from iron to air, and increases from air to water.

Q7. What will happen to the speed of sound if its frequency is doubled? It will become:

A. Half
B. Double
C. Four times
D. Remain the same

Correct Answer: D. Remain the same

The speed of sound depends on the properties of the medium, not on frequency.

Q8. Infrasonic waves have a frequency:

A. Greater than \(20\ \text{Hz}\)
B. Less than \(20\ \text{Hz}\)
C. Exactly \(20\ \text{Hz}\)
D. \(20\ \text{kHz}\)

Correct Answer: B. Less than \(20\ \text{Hz}\)

Q9. For an echo, the minimum distance of a person from an obstacle is:

A. 17 m
B. 34 m
C. 0.1 m
D. Any distance above 50 m

Correct Answer: A. 17 m

Q10. The property of waves directly related to the loudness of sound is:

A. Frequency
B. Wavelength
C. Speed
D. Amplitude

Correct Answer: D. Amplitude

Q11. Sound entering the ear is amplified by:

A. Ossicles
B. Cochlea
C. Eardrum
D. Brain

Correct Answer: A. Ossicles

Q12. A silent whistle is used to train dogs. When a trainer blows the whistle, human beings cannot hear it, but dogs can. Its possible frequency is:

A. Greater than \(20\ \text{Hz}\)
B. Equal to \(20\ \text{Hz}\)
C. Less than \(20,000\ \text{Hz}\)
D. Greater than \(20,000\ \text{Hz}\)

Correct Answer: D. Greater than \(20,000\ \text{Hz}\)

This is in the ultrasonic range, above the normal human hearing limit.

Q13. The pitch of sound depends upon:

A. Frequency
B. Amplitude
C. Quality
D. Displacement

Correct Answer: A. Frequency

Q14. Two sound waves having the same loudness and pitch can be distinguished by a characteristic of sound called:

A. Loudness
B. Pitch
C. Quality
D. Intensity

Correct Answer: C. Quality (or timbre)

Q15. The sensation of sound persists in our brain for about:

A. 10 s
B. 1 s
C. 0.1 s
D. 0.01 s

Correct Answer: C. 0.1 s

 

2. Constructed Response Questions

Q1. Mark the frequency regions for Infrasonic, Human Audible Frequency and Ultrasound.

  • Infrasonic Region: Frequencies below \(20\ \text{Hz}\), starting from \(0\ \text{Hz}\).

  • Human Audible Frequency Region: \(20\ \text{Hz}\) to \(20,000\ \text{Hz}\).

  • Ultrasound Region: Frequencies above \(20,000\ \text{Hz}\).

Examples from the given frequency scale:

Frequency

Region

0 Hz

Infrasonic

10 Hz

Infrasonic

20 Hz

Audible

1,000 Hz

Audible

3,500 Hz

Audible

13,000 Hz

Audible

20,000 Hz

Upper limit of audible range

40,000 Hz

Ultrasound

55,000 Hz

Ultrasound

80,000 Hz

Ultrasound

Q2. Label the major parts of the Human Hearing System.

1. Outer Ear

  • Pinna

  • Auditory canal (ear canal)

2. Middle Ear

  • Eardrum (tympanic membrane)

  • Ear ossicles (malleus, incus and stapes)

  • Eustachian tube

3. Inner Ear

  • Cochlea

  • Semicircular canals

  • Auditory nerve

 

3. Short Response Questions

Q1. Analyse the process of sound transmission from the outer ear to the brain.

Sound waves collected by the outer ear vibrate the eardrum. The ossicles (malleus, incus and stapes) amplify these vibrations. The cochlea converts mechanical vibrations into nerve impulses, which travel through the auditory nerve to the brain.

Q2. What are common sources of harmful noise? Assess their impact on human health.

Common sources of harmful noise include heavy traffic, industrial machinery, construction tools and loud music.

Prolonged exposure to noise can cause hearing impairment, chronic stress, hypertension, sleep disruption and decreased productivity.

Q3. How do animals use infrasound for communication? Compare this with human communication methods.

Elephants and whales use low-frequency infrasound (below \(20\ \text{Hz}\)) for long-distance communication through air and water.

Humans mainly communicate through audible sounds produced by vocal cords, within the approximate frequency range of \(20\ \text{Hz}\) to \(20\ \text{kHz}\), and through technology such as radio communication.

Q4. Evaluate the advantages and limitations of ultrasonic cleaning.

Advantages:

  • Reaches micro-crevices without scrubbing.

  • Can clean delicate jewellery and medical instruments.

Limitations:

  • Requires specialized cleaning fluids.

  • Cannot clean all soft materials.

  • Involves higher initial equipment costs.

Q5. Why does sound travel faster in solids than in liquids and gases?

Solids have much higher elastic moduli and tightly packed molecules. These properties allow disturbances to transfer rapidly between particles.

Consequently, the speed of sound is generally highest in solids compared with liquids and gases.

The speed of sound can be represented by:

\[ v=\sqrt{\frac{E}{\rho}} \]

Where:

  • \(v\) = Speed of sound

  • \(E\) = Elastic modulus

  • \(\rho\) = Density

Q6. How can two plastic glasses connected by a stretched string provide a better way to communicate than merely shouting through the air?

Sound travels efficiently through the taut solid string because mechanical vibrations are transferred between the connected glasses. The vibrations lose less energy than sound spreading spherically through open air.

Q7. How can we distinguish between two sounds having the same loudness?

We can distinguish between them by their quality (timbre) or pitch (frequency).

Quality depends on the waveform and overtones produced by an instrument. It allows us to distinguish between sounds produced by instruments such as a violin and a piano.

Q8. During a cricket match, you see a batsman striking the ball but hear the sound slightly later. Explain this time difference.

Light travels extremely fast, at approximately \(3\times10^8\ \text{m/s}\), so it reaches our eyes almost instantaneously.

Sound travels much more slowly, at approximately \(340\ \text{m/s}\) in air. Therefore, we hear the sound after seeing the batsman strike the ball.

Q9. When a pendulum vibrates, we do not hear its sound. Why?

A pendulum generally vibrates at a frequency below \(20\ \text{Hz}\), which lies in the infrasonic region.

Since the normal human audible range is approximately \(20\ \text{Hz}\) to \(20\ \text{kHz}\), the human ear cannot detect these vibrations as sound.

Q10. Two students are talking in the corridor of your school. You can hear them in your classroom but cannot see them. How?

Sound waves have relatively long wavelengths compared with light waves. They can diffract (bend) around corners and through doorways.

Light waves have extremely short wavelengths and generally travel in straight lines, so the students may not be visible.

Q11. What steps would you take to stop echo and reverberation effects in a large room?

We can reduce echo and reverberation by installing sound-absorbing materials, such as:

  • Thick carpets

  • Heavy curtains

  • Acoustic foam panels

  • Upholstered furniture

Soft, porous and irregular surfaces absorb sound energy and reduce unwanted reflections.

 

4. Long Response Questions

Q1. Explain the production of sound waves with examples.

Definition and Mechanism

Sound is a form of energy produced by vibrating bodies. When an object vibrates back and forth, it disturbs the surrounding air molecules. This creates periodic pressure variations called compressions and rarefactions, which travel through the medium as longitudinal mechanical waves.

Examples of Sound Production

1. Tuning Fork

Striking a tuning fork causes its tines to move rapidly back and forth. As a tine moves outward, it compresses the adjacent air, creating a high-pressure region. As it moves inward, it creates a rarefaction, which is a low-pressure region.

2. Vocal Cords

Air forced from the lungs passes through the vocal cords in the larynx, causing them to vibrate and produce speech sounds.

3. Stretched String (Guitar/Violin)

Plucking a tight string creates vibrations. These vibrations cause the surrounding air molecules and the instrument’s sound box to produce longitudinal pressure waves.

4. Drums (Membranes)

Striking a drum skin causes the membrane to move up and down, generating longitudinal pressure waves in the surrounding air.

 

Q2. Justify why sound waves cannot travel in a vacuum. Design an experiment that demonstrates this principle.

Theoretical Justification

Sound is a mechanical wave, meaning it relies on particle-to-particle interactions to propagate energy.

A vacuum is devoid of matter (atoms or molecules). Therefore, there are no particles to transfer pressure oscillations from one point to another.

Bell-Jar Experiment

Experimental arrangement

       +-----------------------+
       |     Electric Bell     |
       |       /=======\       |
       |      |  ( o )  |      |
       |       \=======/       |
       |          ||           |
       |      [Glass Jar]      |
       +-----------||----------+
                   ||
              To Vacuum Pump

Apparatus:

  • Electric bell

  • Glass bell jar

  • Vacuum pump

  • Battery

  • Connecting wires

Procedure:

  1. Suspend an electric bell inside an airtight glass bell jar connected to a vacuum pump.

  2. Turn on the bell circuit. The hammer strikes the gong, and the ringing sound is clearly audible outside.

  3. Start the vacuum pump to gradually remove air from inside the jar.

Observation:

As air is pumped out, the intensity of the sound diminishes. Eventually, when a near-total vacuum is created, no sound is heard despite seeing the hammer strike the gong.

When air is allowed back into the jar, the sound becomes audible again.

Conclusion:

Sound requires a material medium to propagate. Therefore, sound cannot travel through a vacuum.

 

Q3. What is the nature of sound waves? How is sound propagated? Explain.

Nature of Sound Waves

  • Mechanical: Sound requires a material medium (solid, liquid or gas) to travel.

  • Longitudinal: Particles of the medium oscillate parallel to the direction of wave propagation.

  • Pressure Waves: Sound consists of alternating high-density regions called compressions and low-density regions called rarefactions.

Propagation of Sound

Direction of Wave Propagation  ---->

|||||||||  | | | |  |||||||||  | | | |  |||||||||
Compressions  Rarefactions  Compressions

Propagation Process

  1. Source Vibration: A vibrating object moves forward, pushing air molecules together to form a compression zone of high pressure and density.

  2. Restoring Motion: When the object swings back, it creates a rarefaction zone of low pressure.

  3. Chain Reaction: Air molecules do not travel from the source to the listener. Instead, they oscillate locally around their equilibrium positions, transmitting energy continuously to adjacent molecules.

 

Q4. Discuss how changes in amplitude and frequency affect the loudness and pitch of sound waves.

Amplitude and Loudness

Loudness: Loudness is the degree of auditory sensation produced by a sound wave. It is directly linked to the amplitude of the wave.

The relationship is:

\[ \text{Loudness}\propto A^2 \]
  • High Amplitude: Produces greater energy and larger air displacement, resulting in a loud sound.

  • Low Amplitude: Produces smaller energy and minor air displacement, resulting in a faint sound.

Frequency and Pitch

Pitch: Pitch is the sensation that allows a sound to be perceived as high or low. It directly depends on frequency.

\[ \text{Pitch}\propto f \]
  • High Frequency: Produces rapid vibrations and a high-pitched or shrill sound, such as a whistle or a woman’s voice.

  • Low Frequency: Produces slower vibrations and a low-pitched or deep sound, such as a bass drum or a man’s voice.

 

Q5. What is the quality of sound? How do the shape and material of a sound source influence the waveform of the sound it produces?

Quality (Timbre)

Quality is the characteristic of sound that enables a listener to distinguish between two sounds having the same loudness and pitch but produced by different instruments.

Influence of Shape and Material on Waveform

1. Fundamental Frequency and Overtones

Pure tones are simple sine waves, such as those produced by a tuning fork. Musical instruments produce a fundamental frequency along with multiple secondary frequencies called harmonics or overtones.

2. Material Influence

Wood, metal and plastic vibrate differently because of variations in elasticity and internal damping.

For example, a wooden acoustic guitar absorbs higher harmonics faster than a brass trumpet.

3. Shape and Geometry

The shape of a resonant cavity determines which harmonic frequencies are amplified or reduced. For example, the body of a violin and the soundboard of a piano influence their resulting sounds.

4. Resulting Waveform

The combination of fundamental and overtone waves produces a unique complex waveform. This gives each instrument its characteristic sound quality.

 

Q6. Explain the speed of sound in different media. How does the speed of sound differ in solids, liquids and gases? Explain why, and discuss real-world implications.

Speed of Sound Formula

The velocity of a mechanical wave depends on the elastic properties and density of the medium.

\[ v=\sqrt{\frac{E}{\rho}} \]

Where:

  • \(v\) = Speed of sound

  • \(E\) = Elastic modulus

  • \(\rho\) = Density

Comparison Across Different States

Phase

Approximate Speed

Explanation

Solids

3,000–6,000 m/s

Strong intermolecular forces and high elastic modulus.

Liquids

1,400–1,600 m/s

Moderately bound molecules; less elastic than solids.

Gases

330–350 m/s

Weak intermolecular forces and high compressibility.

Real-World Implications

1. Railway Track Detection

A person can detect an approaching train by placing an ear against steel rails, sometimes before hearing it through the air.

2. SONAR and Marine Life

Submarines use the relatively high speed of sound propagation in water (approximately \(1,500\ \text{m/s}\)) for navigation and target tracking.

3. Acoustic Design

Sound travel delays must be considered and compensated for in large outdoor public-address (PA) systems.

 

Q7. What is the reflection of sound? Differentiate between echo and reverberation.

Reflection of Sound

When a sound wave encounters the boundary of a different medium or a reflecting surface, it bounces back into the original medium. This phenomenon is called reflection of sound.

It follows the law of reflection:

\[ \text{Angle of incidence}=\text{Angle of reflection} \]

Difference Between Echo and Reverberation

Feature

Echo

Reverberation

Definition

A clear, distinct repetition of sound after reflection from a distant surface.

Prolonged sound caused by multiple overlapping reflections in an enclosed space.

Minimum Distance

At least 17 metres in air at \(20^\circ\text{C}\).

Less than 17 metres.

Time Delay

Reflected sound arrives at least 0.1 s after the original sound.

Reflections arrive within 0.1 s of the original sound.

Perception

Heard as a separate, distinct repetition.

Heard as a continuous, unclear extension of sound.

 

Q8. Explain the phenomenon of an echo as the reflection of sound waves. Design an experiment to measure the time delay of an echo, and discuss how this can be used to determine distances in various applications, such as SONAR.

Phenomenon of Echo

An echo occurs when a sound wave strikes a hard obstacle, such as a cliff or wall, and reflects back to the observer.

The persistence of hearing in humans is approximately \(0.1\ \text{s}\). For a distinct echo, the reflected sound must reach the ear at least \(0.1\ \text{s}\) after the original sound.

The distance can be calculated using:

\[ d=\frac{v\times t}{2} \]

Where:

  • \(d\) = Distance to the reflecting surface

  • \(v\) = Speed of sound

  • \(t\) = Total time taken by sound to travel to the obstacle and back

Experiment: Measuring the Time Delay of an Echo

  1. Stand at a known distance \(S=50\ \text{m}\) from a large, flat, vertical wall.

  2. Use a stopwatch and clap two wooden blocks together sharply and loudly.

  3. Start the timer upon clapping and stop it when the reflected echo is heard. Let the total time be \(t\).

  4. The speed of sound can be calculated using:

\[ v=\frac{2S}{t} \]

Applications: SONAR and Depth Detection

SONAR (Sound Navigation and Ranging):

A transmitter on a ship emits ultrasonic pulses towards the ocean floor.

Calculation:

The receiver detects the reflected sound after time \(t\). If the speed of sound in ocean water is approximately \(1,500\ \text{m/s}\), the depth of the seabed is:

\[ d=\frac{v\times t}{2} \]

This principle is used to measure underwater distances and sea depth.

 

Q9. Differentiate between noise and music. Explain why noise is a nuisance.

Differences Between Noise and Music

Feature

Music

Noise

Waveform

Regular, periodic and continuous vibrations.

Irregular, non-periodic and abrupt vibrations.

Auditory Perception

Pleasant and soothing to human ears.

Unpleasant, irritating and disturbing to the ears.

Frequency Pattern

Defined musical frequencies and harmonics.

Random mixtures of frequencies without an orderly pattern.

Examples

Flute, violin and singing.

Heavy traffic, drilling machines and explosions.

Why Noise Is a Nuisance

Excessive noise disrupts physiological and psychological balance. It can cause:

  • Sleep disturbance

  • Stress

  • Elevated blood pressure

  • Severe headaches

  • Lack of concentration

  • Permanent noise-induced hearing loss (NIHL)

 

Q10. Justify the importance of acoustic protection in environments such as schools and hospitals. How can design elements and materials contribute to creating sound-friendly spaces?

Importance in Critical Spaces

1. Schools

High noise levels interfere with speech clarity, causing student fatigue and reducing learning performance.

2. Hospitals

Quiet environments are important for patient healing, blood pressure stabilization and restful recovery.

Design Elements and Materials

1. Absorptive Surface Linings

Acoustic ceiling tiles, porous plaster and wall panels absorb incident sound energy.

2. Soft Interior Furnishings

Carpeting, heavy drapes and upholstered chairs minimize reverberation.

3. Structural Isolation

Double-glazed windows, cavity walls and rubberized door gaskets help block exterior noise transmission.

 

Q11. Analyse the effects of noise pollution on the environment and human health. What strategies can be implemented to reduce these effects, and how can communities balance development with acoustic quality?

Impact on Environment and Health

1. Human Health

Noise pollution can cause:

  • Cardiovascular diseases

  • Chronic anxiety

  • Sleep disorders

  • Hearing loss

  • Reduced cognitive performance

2. Environment and Wildlife

Noise pollution disrupts animal navigation, mating calls and hunting behaviours. It can affect songbirds and marine mammals that rely on sound and echolocation.

Mitigation Strategies and Community Balancing

1. Engineering Controls

Install sound barriers along highways, use mufflers on exhausts and design quieter machinery.

2. Urban Zoning

Keep industrial zones and airports away from residential areas where possible.

3. Green Buffers

Plant dense belts of trees along roadways to help reduce noise.

 

Q12. Humans can hear frequencies from approximately 20 Hz to 20,000 Hz. Analyse how sounds outside this range (infrasound and ultrasound) are used in nature and technology. What are the implications of these frequencies for human hearing?

Utilization of Infrasound (Below 20 Hz)

1. Nature

Elephants, whales and rhinos use low-frequency sounds for communication over long distances. Some animals can also detect early seismic waves preceding earthquakes.

2. Technology

Infrasound is used for:

  • Monitoring volcanic eruptions

  • Detecting nuclear tests

  • Studying atmospheric phenomena

Utilization of Ultrasound (Above 20,000 Hz)

1. Nature

  • Bats navigate and hunt insects through echolocation.

  • Dolphins explore underwater environments using echolocation.

2. Technology

Ultrasound is used in:

  • Medical ultrasonography for imaging organs and fetuses

  • Non-destructive testing of metals

  • Ultrasonic cleaning

  • SONAR depth measurement

Implications for Human Hearing

Although humans cannot normally hear these frequencies, extremely high-intensity infrasound can cause dizziness or resonance in internal organs, while high-intensity ultrasound can cause localized thermal heating in tissues.

 

Q13. Explain how sound waves are converted into electrical signals by the eardrum and auditory nerves.

Step-by-Step Transduction Process

Sound Waves
     |
     v
Eardrum (Mechanical Vibration)
     |
     v
Ossicles (Amplification)
     |
     v
Oval Window / Cochlear Fluid
     |
     v
Hair Cells in Organ of Corti
     |
     v
Electrical Impulses
(Auditory Nerve)
     |
     v
Brain

1. Collection and Mechanical Vibration

Sound waves pass through the ear canal and strike the eardrum (tympanic membrane), causing it to vibrate in response to pressure variations.

2. Mechanical Amplification

Three small bones—the malleus, incus and stapes (ossicles)—amplify these mechanical vibrations and push against the oval window of the inner ear.

3. Fluid Movement and Hair Cell Deflection

Vibrations enter the fluid-filled cochlea, generating fluid waves that displace microscopic hair cells in the Organ of Corti.

4. Bioelectric Conversion and Nerve Transmission

Bending of the hair cells triggers membrane depolarization, converting mechanical movement into electrical nerve impulses. These impulses travel through the auditory nerve to the brain for interpretation.

 

5. Numerical Response Questions

Q1. A tourist clapped his hands near a cliff and heard the echo after 5 s. What is the distance of the cliff from the student if the speed of sound is 344 m/s?

Given:

  • Time interval for echo, \(t=5\ \text{s}\)

  • Speed of sound, \(v=344\ \text{m/s}\)

Formula:

Since the sound wave travels to the cliff and back, the total distance travelled is \(2d\).

\[ 2d=v\times t \]
\[ d=\frac{v\times t}{2} \]

Calculation:

\[ d=\frac{344\times5}{2} \]
\[ d=\frac{1720}{2} \]
\[ d=860\ \text{m} \]

Result: The distance of the cliff is 860 m.

 

Q2. Speed of sound in seawater is 1500 m/s. If its wavelength is 45 cm, find the frequency of these sound waves.

Given:

  • Speed of sound, \(v=1500\ \text{m/s}\)

  • Wavelength, \(\lambda=45\ \text{cm}=0.45\ \text{m}\)

Formula:

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

Calculation:

\[ f=\frac{1500}{0.45} \]
\[ f=\frac{150000}{45} \]
\[ f=3333.33\ \text{Hz} \]

Result: The frequency of the sound waves is 3,333.3 Hz.

 

Q3. A sound wave in seawater travels at a speed of 1500 m/s. It is sent and detected back after reflection from the seabed in 8.5 s. Find the distance from the surface to the bottom of the sea.

Given:

  • Speed of sound in seawater, \(v=1500\ \text{m/s}\)

  • Total time taken, \(t=8.5\ \text{s}\)

Formula:

\[ d=\frac{v\times t}{2} \]

Calculation:

\[ d=\frac{1500\times8.5}{2} \]
\[ d=750\times8.5 \]
\[ d=6375\ \text{m} \]

Result: The distance from the surface to the bottom of the sea is 6,375 m.

 

Q4. A physician counts 69 heartbeats in 1 minute. Calculate the frequency and period of the heartbeats.

Given:

  • Number of heartbeats, \(N=69\)

  • Time taken, \(t=1\ \text{min}=60\ \text{s}\)

Formulae:

Frequency:

\[ f=\frac{N}{t} \]

Time period:

\[ T=\frac{1}{f} \]

Calculation:

1. Frequency

\[ f=\frac{69}{60} \]
\[ f=1.15\ \text{Hz} \]

2. Time Period

\[ T=\frac{1}{1.15} \]
\[ T\approx0.8696\ \text{s} \]

Result:

  • Frequency = 1.15 Hz

  • Time period = 0.8696 s

 

Q5. Find the range of wavelengths at a speed of 343 m/s for the audible frequency range (20 Hz to 20 kHz).

Given:

  • Speed of sound, \(v=343\ \text{m/s}\)

  • Lower frequency limit, \(f_1=20\ \text{Hz}\)

  • Upper frequency limit, \(f_2=20\ \text{kHz}=20,000\ \text{Hz}\)

Formula:

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

Calculation:

1. For the lower frequency limit, \(f_1=20\ \text{Hz}\):

\[ \lambda_1=\frac{343}{20} \]
\[ \lambda_1=17.15\ \text{m} \]

2. For the upper frequency limit, \(f_2=20,000\ \text{Hz}\):

\[ \lambda_2=\frac{343}{20000} \]
\[ \lambda_2=0.01715\ \text{m} \]

Result: The wavelength range is from 17.15 m to 0.01715 m.

 

Q6. During a thunderstorm, thunder is heard 3 seconds after the lightning flash. Find the distance of the clouds from the ground. (Speed of sound = 340 m/s)

Given:

  • Time delay, \(t=3\ \text{s}\)

  • Speed of sound, \(v=340\ \text{m/s}\)

Formula:

\[ d=v\times t \]

Calculation:

\[ d=340\times3 \]
\[ d=1020\ \text{m} \]

Result: The distance of the clouds from the ground is 1,020 m.

 

Q7. SONAR (Sound Navigation and Ranging System) sends an ultrasound signal towards the seabed. It is received back after 5.3 s. If the speed of sound in seawater is 1550 m/s, find the depth of the sea.

Given:

  • Total time taken, \(t=5.3\ \text{s}\)

  • Speed of sound in seawater, \(v=1550\ \text{m/s}\)

Formula:

Since the sound travels to the seabed and returns:

\[ d=\frac{v\times t}{2} \]

Calculation:

\[ d=\frac{1550\times5.3}{2} \]
\[ d=\frac{8215}{2} \]
\[ d=4107.5\ \text{m} \]

Result: The depth of the sea is 4,107.5 m.

 

Notes by Sir Dilawar
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