What is the speed of sound in air at 0°C?

Correct Option: A. Solution: The speed of sound in air at 0°C is 331 m/s.

What is the speed of sound in air at 22°C?

Correct Option: B. Solution: The speed of sound in air at 22°C is 344 m/s.

Sound

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

Understand the concept of sound wave speed and its calculation.
Explain the phenomenon of reverberation and methods to reduce it.
Define loudness of sound and identify the factors it depends on.
Describe the application of ultrasound in cleaning processes.
Analyze how ultrasound can detect defects in metal blocks.

CBSE Revision Notes & Quick Summary for Last-Minute Study

Sound Waves and Their Properties

Key Concepts

Speed of Sound: The speed of sound varies in different media.
- Example: In air, the speed is approximately 346 m/s.
Frequency: The number of oscillations per unit time, measured in hertz (Hz).
- Formula:
  - v = f × λ
  - Where v = speed of sound, f = frequency, λ = wavelength.
Wavelength: The distance between successive crests or troughs of a wave.
Amplitude: The maximum disturbance in the medium, affecting loudness.
- Example: A louder sound has a larger amplitude.

Sound Wave Characteristics

Pitch: How the brain interprets frequency; higher frequency = higher pitch.
Loudness: Determined by amplitude; larger amplitude = louder sound.

Waveforms

Low Pitch Sound:
- Longer wavelength, lower frequency.
High Pitch Sound:
- Shorter wavelength, higher frequency.

Reflection of Sound Waves

Sound waves can reflect off surfaces, similar to light waves.
- Diagram: Illustrates sound waves reflecting off a wall.

Ultrasound Applications

Ultrasound for Cleaning: Uses high-frequency sound waves to remove dirt.
Defect Detection: Ultrasound can identify flaws in materials by analyzing wave reflections.

Important Formulas

Concept	Formula	Unit
Frequency	f = v / λ	Hz
Speed of Sound	v = f × λ	m/s
Time Period	T = 1 / f	s

Summary of Sound Properties

Speed in Different States:
- Solids: e.g., Aluminium (6420 m/s), Steel (5960 m/s)
- Liquids: e.g., Water (1531 m/s), Ethanol (1207 m/s)
- Gases: e.g., Air (346 m/s), Hydrogen (1284 m/s)

Diagrams

Ultrasound Testing Setup: Shows how ultrasound waves enter a metal block and detect flaws.
Waveforms: Illustrate the difference between soft and loud sounds based on amplitude.

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Frequency and Pitch: Students often confuse frequency with pitch. Remember, frequency is the number of oscillations per unit time, while pitch is how the brain interprets frequency.
Ignoring Amplitude's Role in Loudness: Many students forget that loudness is determined by amplitude, not frequency. Ensure you understand that a higher amplitude corresponds to a louder sound.
Confusing Time Period and Frequency: Students may mix up the definitions of time period (T) and frequency (f). Recall that frequency is the inverse of the time period: f = 1/T.
Neglecting the Medium's Effect on Sound Speed: Some students overlook that sound travels at different speeds in different media. Be aware of the speed of sound in solids, liquids, and gases.

Tips for Success

Practice Calculations: Regularly practice problems involving the speed of sound, frequency, and wavelength to solidify your understanding.
Visualize Concepts: Use diagrams to visualize sound wave properties, such as amplitude and frequency, to better grasp their relationships.
Review Definitions: Make sure you can clearly define key terms like loudness, pitch, frequency, and amplitude.
Understand Wave Behavior: Familiarize yourself with how sound waves reflect and refract, as this is often tested in exams.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

Echo

Reverberation

Resonance

Diffraction

Correct Answer: B

Solution:

Reverberation is the persistence of sound in a hall due to multiple reflections.

The wavelength is halved.

The wavelength remains unchanged.

The wavelength is doubled.

The wavelength is quadrupled.

Correct Answer: A

Solution:

The speed of sound

V

is related to its frequency

f

and wavelength

\lambda

by the equation

V = f \lambda

. If the frequency is doubled and the speed remains constant, the wavelength must be halved to maintain the equation.

10 dB

20 dB

50 dB

100 dB

Correct Answer: B

Solution:

The increase in decibels is calculated using the formula

\Delta L = 10 \log_{10}(\text{factor})

. For a factor of 100,

\Delta L = 10 \log_{10}(100) = 10 \times 2 = 20 \text{ dB}

Detecting defects in metal blocks

Measuring temperature

Calculating speed of light

Determining gravitational force

Correct Answer: A

Solution:

Ultrasound is used to detect defects in metal blocks by analyzing the reflection of sound waves.

Pitch

Loudness

Quality

Intensity

Correct Answer: C

Solution:

The quality of sound, also known as timbre, helps to distinguish different voices.

0.1 m

1 m

10 m

0.01 m

Correct Answer: A

Solution:

The wavelength

\lambda

can be calculated using the formula

v = f \lambda

, where

v

is the speed of sound and

f

is the frequency. Given

v = 343 \text{ m/s}

and

f = 3430 \text{ Hz}

\lambda = \frac{v}{f} = \frac{343}{3430} = 0.1 \text{ m}

Sound waves with frequencies below 20 Hz

Sound waves with frequencies above 20 kHz

Sound waves with frequencies between 20 Hz and 20 kHz

Sound waves with frequencies exactly at 20 kHz

Correct Answer: B

Solution:

Sound waves with frequencies above 20 kHz are termed 'ultrasonic'.

The wavelength increases.

The wavelength decreases.

The wavelength remains constant.

The wavelength becomes zero.

Correct Answer: A

Solution:

As the speed of sound

V

increases with temperature, for a constant frequency

f

, the wavelength

\lambda = \frac{V}{f}

increases.

513 m

1026 m

171 m

684 m

Correct Answer: A

Solution:

The total distance traveled by the sound is

342 \times 3 = 1026\, \text{m}

. The distance to the mountain is half of this, so

\frac{1026}{2} = 513\, \text{m}

Detecting flaws in metal blocks

Listening to music

Amplifying speech in a megaphone

Measuring the speed of a car

Correct Answer: A

Solution:

Ultrasonic waves are used in non-destructive testing methods to detect flaws in metal blocks. This is because they can penetrate materials and reflect off internal defects, allowing for detailed internal inspection. Therefore, the correct answer is option a.

22600 Hz

2260 Hz

226 Hz

22.6 Hz

Correct Answer: A

Solution:

Frequency is calculated using the formula

v = f \lambda

. Here,

v = 339

m/s and

\lambda = 0.015

m. Thus,

f = \frac{v}{\lambda} = \frac{339}{0.015} = 22600

Hz.

Ultrasound waves reflect off defects within the metal.

Ultrasound waves increase the temperature of the metal.

Ultrasound waves change the color of the metal.

Ultrasound waves can only travel through perfect metal structures.

Correct Answer: A

Solution:

Ultrasound waves reflect off defects within the metal, allowing for the detection of flaws as the reflected waves can be captured and analyzed.

10 Hz - 10 kHz

20 Hz - 20 kHz

30 Hz - 30 kHz

40 Hz - 40 kHz

Correct Answer: B

Solution:

The audible range of hearing for average human beings is from 20 Hz to 20 kHz.

Reflection of waves at material boundaries

Refraction of waves through different media

Diffraction of waves around obstacles

Interference of waves from different sources

Correct Answer: A

Solution:

Ultrasound waves detect internal flaws based on the reflection of waves at material boundaries. When ultrasound waves encounter a defect, they reflect back differently compared to uniform areas, allowing detection of irregularities.

Temperature of the medium

Density of the medium

Frequency of the sound wave

Elasticity of the medium

Correct Answer: C

Solution:

The speed of sound in a medium is primarily affected by the medium's temperature, density, and elasticity. The frequency of the sound wave does not affect the speed of sound.

680 m

1360 m

340 m

170 m

Correct Answer: A

Solution:

The total distance travelled by the sound is

340 \times 4 = 1360

m. Since this is the distance to the mountain and back, the distance to the mountain is

\frac{1360}{2} = 680

Reverberation is the absorption of sound by walls.

Reverberation is the persistence of sound due to multiple reflections.

Reverberation is the transmission of sound through a vacuum.

Reverberation is the increase in sound speed in a denser medium.

Correct Answer: B

Solution:

Reverberation is defined as the persistence of sound in a space after the original sound is produced, due to multiple reflections from surfaces such as walls and ceilings.

Air

Water

Iron

Helium

Correct Answer: C

Solution:

Sound travels fastest in solids due to the close proximity of particles which facilitates quicker transmission of sound waves. Among the given options, iron is a solid and allows sound to travel faster than in air, water, or helium.

0.77 mm

0.85 mm

0.65 mm

0.95 mm

Correct Answer: A

Solution:

The wavelength

\lambda

is given by

\lambda = \frac{V}{f}

, where

V = 1540 \text{ m/s}

and

f = 2 \times 10^6 \text{ Hz}

. Thus,

\lambda = \frac{1540}{2 \times 10^6} = 0.77 \text{ mm}

0.68 m

1.36 m

0.34 m

2.00 m

Correct Answer: A

Solution:

The wavelength

\lambda

of a sound wave is given by the formula

\lambda = \frac{v}{f}

, where

v

is the speed of sound and

f

is the frequency. Substituting the given values,

\lambda = \frac{340}{500} = 0.68

Echo

Reverberation

Diffraction

Refraction

Correct Answer: B

Solution:

Reverberation is the persistence of sound in a hall due to repeated reflection.

Transverse wave

Longitudinal wave

Electromagnetic wave

Mechanical wave

Correct Answer: B

Solution:

Sound is classified as a longitudinal wave because it propagates through compressions and rarefactions in a medium.

513 m

1026 m

171 m

684 m

Correct Answer: A

Solution:

The total distance traveled by the sound is

342 \times 3 = 1026

m. Since this distance is to the surface and back, the distance to the surface is

\frac{1026}{2} = 513

686 m

343 m

171.5 m

1372 m

Correct Answer: A

Solution:

The total distance traveled by the sound is

v \times t = 343 \text{ m/s} \times 4 \text{ s} = 1372 \text{ m}

. Since this distance is to the cliff and back, the distance to the cliff is

\frac{1372}{2} = 686 \text{ m}

Because particles oscillate perpendicular to the wave direction.

Because particles oscillate parallel to the wave direction.

Because it travels faster in solids.

Because it can travel in a vacuum.

Correct Answer: B

Solution:

Sound waves are longitudinal because the particles of the medium oscillate parallel to the direction of wave propagation.

Pitch

Loudness

Quality

Intensity

Correct Answer: C

Solution:

The quality of sound helps in identifying a friend's voice.

0.01 s

1 s

10 s

0.1 s

Correct Answer: A

Solution:

The time period

T

is the reciprocal of the frequency

f

. Thus,

T = \frac{1}{f} = \frac{1}{100} = 0.01

Pitch

Loudness

Quality

Intensity

Correct Answer: C

Solution:

The quality or timbre of sound helps in identifying different voices.

0.5 m

1 m

2 m

0.25 m

Correct Answer: A

Solution:

The wavelength

\lambda

can be calculated using the formula

v = f \lambda

, where

v

is the speed of sound and

f

is the frequency. Given

v = 340 \text{ m/s}

and

f = 680 \text{ Hz}

\lambda = \frac{v}{f} = \frac{340}{680} = 0.5 \text{ m}

Air

Water

Iron

Glass

Correct Answer: C

Solution:

Sound travels fastest in solids due to their dense molecular structure. Among the options, iron is a solid and has a higher speed of sound compared to liquids and gases. The speed of sound in iron is approximately 5950 m/s, which is faster than in air, water, or glass.

Ultrasonic

Infrasonic

Supersonic

Subsonic

Correct Answer: B

Solution:

Sound waves with frequencies below the audible range are termed 'infrasonic'.

10 Hz

15 kHz

25 kHz

5 Hz

Correct Answer: C

Solution:

Ultrasonic frequencies are those above the audible range of 20 kHz.

513 meters

Correct Answer: A

Solution:

The total distance traveled by the sound is

V \times t = 342 \times 3 = 1026

meters. Since this distance includes the journey to the cliff and back, the distance to the cliff is

\frac{1026}{2} = 513

meters.

331 m/s

344 m/s

346 m/s

340 m/s

Correct Answer: C

Solution:

The speed of sound in air at 25°C is approximately 346 m/s.

Medical imaging

Breaking kidney stones

Detecting flaws in metal blocks

Cooking food

Correct Answer: D

Solution:

Ultrasound is commonly used in medical imaging, breaking kidney stones, and detecting flaws in metal blocks. It is not typically used for cooking food.

0.308 mm

0.154 mm

0.308 m

0.154 m

Correct Answer: A

Solution:

The wavelength

\lambda

is calculated using the formula

\lambda = \frac{v}{f}

. Here,

v = 1540 \text{ m/s}

and

f = 5 \times 10^6 \text{ Hz}

. Thus,

\lambda = \frac{1540}{5 \times 10^6} = 0.000308 \text{ m} = 0.308 \text{ mm}

331 m/s

344 m/s

346 m/s

340 m/s

Correct Answer: A

Solution:

The speed of sound in air at 0°C is 331 m/s.

333 m/s

340 m/s

350 m/s

360 m/s

Correct Answer: A

Solution:

The time taken for the sound to travel to the wall and back is 0.6 s. Therefore, the distance traveled by the sound is 200 meters (100 meters to the wall and 100 meters back). The speed of sound

v

is calculated as

v = \frac{\text{distance}}{\text{time}} = \frac{200}{0.6} = 333 \text{ m/s}

Color of the medium

Temperature of the medium

Shape of the medium

Thickness of the medium

Correct Answer: B

Solution:

The speed of sound depends primarily on the temperature of the medium.

0.29 seconds

0.25 seconds

0.35 seconds

0.30 seconds

Correct Answer: A

Solution:

The time taken

t

for the echo to return is given by

t = \frac{2d}{V}

, where

d = 50 \text{ m}

and

V = 343 \text{ m/s}

. Thus,

t = \frac{2 \times 50}{343} \approx 0.29 \text{ seconds}

Pitch

Loudness

Quality

Frequency

Correct Answer: B

Solution:

Loudness is the physiological response of the ear to the intensity of sound.

400 Hz, Yes

500 Hz, Yes

400 Hz, No

500 Hz, No

Correct Answer: B

Solution:

The frequency

f

of a wave is given by the equation

f = \frac{V}{\lambda}

, where

V

is the speed of sound and

\lambda

is the wavelength. Substituting the given values:

f = \frac{340}{0.85} = 400 \text{ Hz}

. Since 400 Hz is within the audible range of 20 Hz to 20 kHz, it will be audible.

12,000 times

200 times

1,200 times

20,000 times

Correct Answer: A

Solution:

Frequency is the number of oscillations per second. In one minute (60 seconds), it vibrates 200 Hz \times 60 = 12,000 times.

331 m/s

344 m/s

346 m/s

340 m/s

Correct Answer: B

Solution:

The speed of sound in air at 22°C is 344 m/s.

Frequency

Amplitude

Wavelength

Speed of sound

Correct Answer: B

Solution:

Loudness of sound is determined by its amplitude. The greater the amplitude, the louder the sound.

0.4 m

0.5 m

0.3 m

0.2 m

Correct Answer: A

Solution:

The wavelength

\lambda

can be calculated using the formula

\lambda = \frac{V}{f}

, where

V = 340 \text{ m/s}

is the speed of sound and

f = 850 \text{ Hz}

is the frequency. Thus,

\lambda = \frac{340}{850} = 0.4 \text{ m}

100 times

6000 times

60 times

3600 times

Correct Answer: B

Solution:

The sound wave vibrates 100 times per second. In one minute (60 seconds), it vibrates

100 \times 60 = 6000

times.

Detecting defects in metal blocks

Breaking kidney stones

Cooking food

Medical imaging

Correct Answer: C

Solution:

Ultrasound is used for detecting defects in metal blocks, breaking kidney stones, and medical imaging, but not for cooking food.

10 dB

20 dB

5 dB

15 dB

Correct Answer: A

Solution:

The decibel level is a logarithmic measure of sound intensity. An increase in loudness by a factor of 10 corresponds to an increase of 10 decibels. Therefore, the correct answer is option a.

Particles of the medium move perpendicular to the direction of wave propagation.

Particles of the medium move parallel to the direction of wave propagation.

Sound waves can travel in a vacuum.

Sound waves have a fixed frequency and amplitude.

Correct Answer: B

Solution:

Sound waves are longitudinal waves because the particles of the medium move parallel to the direction of wave propagation, creating compressions and rarefactions.

Travel faster than light in tissues.

Penetrate bones without reflection.

Reflect off tissues to create images.

Be absorbed completely by soft tissues.

Correct Answer: C

Solution:

Ultrasound waves are used in medical imaging because they reflect off different tissues and these reflections are used to create images of internal body structures.

0.68 m

0.34 m

1.36 m

0.17 m

Correct Answer: A

Solution:

The wavelength

\lambda

is given by the formula

\lambda = \frac{v}{f}

, where

v

is the speed of sound and

f

is the frequency. Substituting the given values,

\lambda = \frac{340 \text{ m/s}}{500 \text{ Hz}} = 0.68 \text{ m}

Polished marble

Glass panels

Compressed fibreboard

Metal sheets

Correct Answer: C

Solution:

Compressed fibreboard is a sound-absorbent material that helps reduce excessive reverberation by absorbing sound waves, unlike polished marble, glass panels, or metal sheets which reflect sound.

172 Hz

688 Hz

344 Hz

86 Hz

Correct Answer: A

Solution:

The frequency

v

is calculated using the formula

v = \frac{V}{\lambda}

, where

V

is the speed of sound and

\lambda

is the wavelength. Thus,

v = \frac{344}{2} = 172

Hz.

Detecting flaws in metal blocks

Measuring air temperature

Recording sound in a studio

Amplifying sound in a concert

Correct Answer: A

Solution:

Ultrasound is used in detecting flaws in metal blocks by sending high-frequency sound waves through the metal and analyzing the reflected waves to identify any irregularities.

257.25 m

514.5 m

171.5 m

343 m

Correct Answer: A

Solution:

The distance

d

to the mountain can be calculated using the formula

d = \frac{V \times t}{2}

, where

V

is the speed of sound and

t

is the time for the echo. Substituting the given values:

d = \frac{343 \times 1.5}{2} = 257.25 \text{ m}

It has a higher speed than audible sound.

It can penetrate through bones.

It has a frequency higher than the audible range, allowing for detailed imaging.

It is less expensive than other imaging techniques.

Correct Answer: C

Solution:

Ultrasound has a frequency higher than the audible range, which allows it to create detailed images of internal body structures without ionizing radiation, making it suitable for medical imaging.

It decreases

It remains the same

It increases

It fluctuates

Correct Answer: C

Solution:

The speed of sound increases with an increase in the temperature of the medium.

Air

Water

Iron

Glass

Correct Answer: C

Solution:

Sound travels fastest in solids. Among the given options, sound travels fastest in iron.

Sound waves cause particles to move perpendicular to the wave direction.

Sound waves cause particles to move parallel to the wave direction.

Sound waves do not cause any particle movement.

Sound waves cause particles to move in a circular motion.

Correct Answer: B

Solution:

Sound waves are called longitudinal waves because the particles of the medium move parallel to the direction of wave propagation.

343 m

857.5 m

1715 m

686 m

Correct Answer: D

Solution:

The total distance traveled by the sound is

343 \times 5 = 1715

m. Since this is the distance to the cliff and back, the distance to the cliff is

\frac{1715}{2} = 857.5

Air

Water

Steel

Ethanol

Correct Answer: C

Solution:

Solution:

The speed of sound in air varies with temperature; it increases as the temperature increases.

Sound

AI Learning Assistant

CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

CBSE Revision Notes & Quick Summary for Last-Minute Study

Sound Waves and Their Properties

Key Concepts

Sound Wave Characteristics

Waveforms

Reflection of Sound Waves

Ultrasound Applications

Important Formulas

Summary of Sound Properties

Diagrams

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

Multiple Choice Questions

Q1.What is the term for the persistence of sound in a hall due to multiple reflections?

Correct Answer: B

Q2.A sound wave travels through a medium and undergoes compressions and rarefactions. If the frequency of the sound wave is doubled, how does this affect the wavelength of the sound wave, assuming the speed of sound in the medium remains constant?

Correct Answer: A

Q3.If the loudness of a sound is increased by a factor of 100, by how many decibels does the sound level increase?

Correct Answer: B

Q4.Which of the following is a practical application of ultrasound?

Correct Answer: A

Q5.Which characteristic of sound allows you to identify a friend's voice in a dark room?

Correct Answer: C

Q6.If the speed of sound in air is 343 m/s and a sound wave has a frequency of 3430 Hz, what is its wavelength?

Correct Answer: A

Q7.Which of the following waves are termed as 'ultrasonic'?

Correct Answer: B

Q8.If the speed of sound in air increases with temperature, what happens to the wavelength of a sound wave of constant frequency as the air temperature increases?

Correct Answer: A

Q9.A person hears an echo 3 seconds after shouting near a mountain. If the speed of sound is 342 m/s, how far is the mountain?

Correct Answer: A

Q10.Which of the following applications uses ultrasonic waves?

Correct Answer: A

Q11.If a sound wave travels at a speed of 339 m/s and has a wavelength of 1.5 cm, what is its frequency?

Correct Answer: A

Q12.Which of the following best describes the reason why ultrasound can be used to detect flaws in metal blocks?

Correct Answer: A

Q13.What is the audible range of hearing for average human beings?

Correct Answer: B

Q14.Ultrasound waves are used to detect defects in metal blocks. What principle allows ultrasound waves to identify internal flaws?

Correct Answer: A

Q15.Which of the following factors does NOT affect the speed of sound in a medium?

Correct Answer: C

Q16.If a person claps near a mountain and hears the echo after 4 seconds, what is the distance between the person and the mountain? Assume the speed of sound is 340 m/s.

Correct Answer: A

Q17.Which of the following statements correctly describes the concept of reverberation?

Correct Answer: B

Q18.In which of the following media does sound travel the fastest at a given temperature?

Correct Answer: C

Q19.In an ultrasound imaging scenario, if the speed of sound in the tissue is 1540 m/s and the frequency of the ultrasound wave is 2 MHz, what is the wavelength of the ultrasound in the tissue?

Correct Answer: A

Q20.A sound wave travels through a medium with a speed of 340 m/s. If the frequency of the sound wave is 500 Hz, what is its wavelength?

Correct Answer: A

Q21.What is the phenomenon called when sound persists in a hall due to repeated reflection?

Correct Answer: B

Q22.Which type of wave is sound classified as?

Correct Answer: B

Q23.A sound wave is reflected from a surface and an echo is heard after 3 seconds. If the speed of sound is 342 m/s, what is the distance to the reflecting surface?

Correct Answer: A

Q24.A person hears an echo 4 seconds after shouting near a cliff. If the speed of sound is 343 m/s, what is the distance to the cliff?

Correct Answer: A

Q25.Why is sound wave called a longitudinal wave?

Correct Answer: B

Q26.Which of the following is a characteristic of sound that helps identify a friend by their voice?

Correct Answer: C

Q27.What is the time period of a sound wave if its frequency is 100 Hz?

Correct Answer: A

Q28.Which of the following is a characteristic of sound that allows you to recognize a friend's voice?

Correct Answer: C

Q29.A sound wave travels through a medium at a speed of 340 m/s. If the frequency of the wave is 680 Hz, what is the wavelength of the sound wave in the medium?

Correct Answer: A

Q30.Which of the following materials will allow sound to travel the fastest?

Correct Answer: C

Q31.What is the term for sound waves with frequencies below the audible range?