Key Concept: Audible and Inaudible Sounds

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Humans can hear sounds in the frequency range of 20 Hz to 20,000 Hz. Sounds with frequencies above 20,000 Hz are ultrasonic and are not audible to humans. A dog whistle produces sounds at frequencies higher than 20,000 Hz, which is why humans cannot hear it. Therefore, both Assertion and Reason are true, and Reason explains why the Assertion is correct.

Key Concept: Sound Production

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Sound is indeed produced by vibrating bodies. When an object vibrates, it causes the surrounding air particles to vibrate, creating pressure waves. These waves travel through the medium (air, water, or solid) and reach our ears, where they are interpreted as sound. The reason correctly explains why sound is produced by vibrating bodies.

Key Concept: Sound Perception, Sound Production

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that a blindfolded person can identify the closest player based on the intensity of the sound produced by their footsteps. This is true because the intensity of sound, which is perceived as loudness, is higher when the source is closer to the listener. The reason given is that the intensity of sound decreases with distance from the source, which is also true and explains why the person can identify the closest player. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Amplitude, Loudness, Sound Production

d) 240 dB

[Solution Description]

The loudness of a sound wave is directly proportional to the square of its amplitude. Given that the first wave has an amplitude of 2 cm and produces a sound of 60 dB, we can calculate the loudness of the second wave with an amplitude of 4 cm. Since the amplitude doubles, the loudness increases by a factor of $2^2 = 4$. Therefore, the loudness of the second wave is $60 \times 4 = 240$ dB.

Key Concept: Vibrating Objects

a) Striking a metal plate

[Solution Description]

In Activity 10.1, striking a metal plate produces sound, and touching the plate allows one to feel the vibrations. This demonstrates that sound is produced by a vibrating object. Therefore, the correct answer is striking a metal plate.

Key Concept: Vibration and Sound Production

c) The sound stops completely

[Solution Description]

When the metal plate is struck, it vibrates and produces sound. However, when it is held tightly, the vibrations are stopped, and consequently, the sound production also stops. This is because sound is produced by vibrations, and stopping the vibrations stops the sound.

Key Concept: Vibrations in Water

b) Waves form on the surface of the water.

[Solution Description]

When a metal dish filled with water is struck at its edge, it vibrates and produces sound. These vibrations create waves on the surface of the water, which can be observed visually. This indicates that the dish's vibrations are responsible for the sound produced.

Key Concept: Sound is Produced by a Vibrating Body

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that when a metal plate is struck, it produces sound. This is true because striking the plate causes it to vibrate, and these vibrations produce sound waves that we can hear. The reason given is that sound is produced due to the vibrations of the metal plate. This is also true because sound is indeed produced by vibrating objects. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Key Concept: Sound is Produced by a Vibrating Body

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

When a rubber band is plucked, it starts vibrating. The to and fro motion of the rubber band is called vibration. These vibrations create sound waves in the air, which our ears can detect as sound. Therefore, both the assertion and the reason are true, and the reason correctly explains why plucking the rubber band produces sound.

Key Concept: Sound Production

b) Waves form on the surface of the water

[Solution Description]

When the metal dish is struck with a spoon, it vibrates. These vibrations are transferred to the water, causing waves to form on its surface. This demonstrates that sound is produced by vibrating objects.

Key Concept: Vibrating Parts of Musical Instruments

b) Stretched membrane

[Solution Description]

The tabla is a percussion instrument that produces sound through the vibration of its stretched membrane. When the membrane is struck, it vibrates, producing sound waves.

Key Concept: Vibration in String Instruments

b) The stretched string

[Solution Description]

When a string of a sitar is plucked, the string vibrates, producing sound. However, the sound we hear is not only from the string but also from the vibration of the entire instrument. The body of the instrument amplifies the sound produced by the vibrating string.

Key Concept: Membrane Instruments

b) Stretched membrane

[Solution Description]

The tabla is a membrane instrument. When struck, the stretched membrane vibrates to produce sound. According to Table 10.1 in the syllabus, the vibrating part of a tabla is the stretched membrane.

Key Concept: Vibrating Parts

d) Entire body

[Solution Description]

The Ghatam is a percussion instrument made of clay. When it is struck, the entire body of the Ghatam vibrates to produce sound.

Key Concept: Vocal Cords, Airflow

c) Louder sound without changing pitch

[Solution Description]

Increasing the airflow through the vocal cords while keeping the tension constant results in an increase in the amplitude of the vibrations. This leads to a louder sound but does not affect the pitch. The pitch remains unchanged because it depends on the tension and length of the vocal cords, not the airflow.

Key Concept: Vocal Cord Length

b) 15 mm

[Solution Description]

In women, the vocal cords are approximately 15 mm long. This length is shorter than in men (\~20 mm) and longer than in children.

Key Concept: Sound Production in Humans

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The voice box, or larynx, is indeed located at the upper end of the windpipe, as stated in the syllabus. This makes the Assertion true. The Reason states that the vocal cords vibrate when air passes through them, producing sound. This is also true according to the syllabus, which explains that air forced through the slit between the vocal cords causes them to vibrate, thereby producing sound. Furthermore, the Reason correctly explains why the Assertion is true because the location of the voice box allows it to function effectively in sound production.

Key Concept: Vocal cord length, Pitch differentiation

c) The length of the vocal cords differs, with men having longer cords than women and children.

[Solution Description]

The length of the vocal cords varies among men, women, and children. Men typically have longer vocal cords (about 20 mm), women have shorter ones (about 15 mm), and children have even shorter vocal cords. This difference in length affects the pitch of the voice, with longer cords producing lower pitches and shorter cords producing higher pitches.

Key Concept: Variation in Voice

b) Men have longer vocal cords

[Solution Description]

Men typically have longer vocal cords (about 20 mm) compared to women (about 15 mm). Longer vocal cords vibrate at a lower frequency, producing a deeper voice.

Key Concept: Vocal cord length, Frequency

b) 270 Hz

[Solution Description]

The frequency of sound produced by vocal cords is inversely proportional to their length. Given that the woman's vocal cords are 15 mm long and produce a frequency of 180 Hz, and the child's vocal cords are 10 mm long, let’s denote the frequency produced by the child as $f_c$.

Using the inverse proportionality:

$\frac{f_w}{f_c} = \frac{L_c}{L_w}$

where $f_w = 180$ Hz, $L_w = 15$ mm, and $L_c = 10$ mm.

Solving for $f_c$:

$f_c = f_w \times \frac{L_w}{L_c} = 180 \times \frac{15}{10} = 270 \text{ Hz}$

Thus, the child's voice would have an approximate frequency of 270 Hz.

Key Concept: Role of vocal cord length in pitch

b) The woman's voice has a higher pitch because her vocal cords are shorter.

[Solution Description]

The pitch of a sound depends on the frequency of vibration of the vocal cords. Shorter vocal cords vibrate at a higher frequency, producing a higher pitch. Since the woman's vocal cords are shorter (15 mm) compared to the man's (20 mm), her voice will have a higher pitch.

Key Concept: Sound Medium, Eardrum, Frequency

b) Loudness increases, pitch remains the same

[Solution Description]

The loudness of sound is directly related to the amplitude of the sound wave. Increasing the amplitude by a factor of 3 will increase the loudness significantly. However, the pitch of the sound is determined by the frequency, which remains unchanged at 500 Hz. Therefore, the loudness increases, but the pitch remains the same.

Key Concept: Sound Propagation, Medium Requirement

c) No sound is heard

[Solution Description]

Sound requires a medium (such as air) to propagate. When a vacuum is created inside the tumbler, there is no medium for sound to travel through. Therefore, no sound will be heard from outside the tumbler.

Key Concept: Sound Needs a Medium for Propagation

b) The sound becomes fainter

[Solution Description]

When air is sucked out of the tumbler, the medium for sound propagation decreases. As a result, the sound becomes fainter because sound needs a medium (like air) to travel. If all the air is removed, creating a vacuum, no sound can be heard.

Key Concept: Sound in liquids

c) Sound requires a medium to travel, and liquids can act as a medium

[Solution Description]

The experiment involves placing a bell inside water and listening to the sound produced by shaking it. When the ear is placed gently on the water surface, the sound of the bell can be heard. This indicates that sound waves can travel through liquids like water.

Key Concept: Sound Propagation in Solids

c) Faster than in air

[Solution Description]

Sound travels faster in solids than in gases (like air) because the particles in solids are closer together, allowing vibrations to transfer more quickly. In wood, the speed of sound is significantly higher than in air due to its higher density and elasticity.

Key Concept: Sound Propagation in Liquids

b) Whales communicating underwater

[Solution Description]

Sound can travel through liquids, such as water. Whales and dolphins communicate underwater by producing sound waves that travel through the water.

Key Concept: Sound Propagation in Air

c) The loudness decreases

[Solution Description]

Sound requires a medium to propagate. In this case, air acts as the medium. When air is gradually removed from the container, the loudness of the sound decreases because there is less air to carry the sound waves. If all the air is removed, creating a vacuum, the sound will not be heard at all as sound cannot travel through a vacuum.

Key Concept: Sound Needs a Medium

c) Solids

[Solution Description]

Sound travels fastest through solids because the particles are closely packed together, allowing vibrations to pass more quickly compared to liquids or gases like air or water.

Key Concept: Sound Transmission in Liquids

c) Sound can travel through liquids

[Solution Description]

The experiment demonstrates that sound can travel through liquids. When the bell is shaken underwater, it produces vibrations that travel through the water as sound waves. Placing the ear on the water surface allows the sound waves to reach the ear, indicating that liquids are a medium for sound propagation.

Key Concept: Sound Transmission in Liquids

c) Sound can travel through water

[Solution Description]

Sound is a mechanical wave that requires a medium to travel. Liquids, such as water, provide a medium for sound waves to propagate. The activity involving the bell ringing underwater demonstrates that sound can indeed travel through water. Therefore, sound can travel through water.

Key Concept: Sound Propagation through Solids

c) Metal rods allow vibrations to propagate

[Solution Description]

Sound travels through a metal rod because metals are solid materials that allow vibrations to propagate efficiently. The particles in the metal vibrate and transfer energy from one end to the other, enabling sound to travel through it.

Key Concept: Toy Telephone Experiment

c) Sound can travel through strings.

[Solution Description]

The toy telephone experiment shows that sound can travel through strings. When one person speaks into one cup, the vibrations from their voice travel through the string to the other cup. This allows the second person to hear the sound, demonstrating that strings (a type of solid) can transmit sound waves.

Key Concept: Vacuum Effect on Sound

c) There is less medium for sound to travel

[Solution Description]

As air is removed from a container, the medium for sound propagation decreases. With less air, fewer molecules are available to vibrate and transmit sound, making the sound fainter. If all the air is removed, creating a vacuum, sound cannot be heard at all.

Key Concept: Practical Observation of Sound Propagation

b) The sound stops completely when all the air is removed.

[Solution Description]

When air is sucked out of the tumbler, the sound of the ringing phone becomes fainter because the medium (air) that carries the sound is being removed. When all the air is removed, the sound stops completely, demonstrating that sound requires a medium to travel.

Key Concept: Sound production, vocal cords, frequency

c) The voice will become shriller.

[Solution Description]

The pitch of a sound is directly related to its frequency. If the vocal cords vibrate at a higher frequency, the pitch of the voice will increase, making the voice sound shriller. On the other hand, if the frequency is lower, the pitch will decrease, making the voice sound deeper.

Key Concept: Sound Needs a Medium for Propagation

c) The medium (air) carrying the sound waves decreases

[Solution Description]

When air is sucked out of the tumbler, the medium (air) that carries the sound waves decreases. As the amount of air reduces, the sound becomes fainter because there are fewer particles to transmit the sound. In a vacuum, where there is no air, sound cannot travel at all.

Key Concept: Eardrum function, Sound transmission

b) The grains jump up and down due to the vibrations of the rubber sheet caused by sound waves.

[Solution Description] When sound waves enter the tin-can model, they cause the rubber sheet (acting as the eardrum) to vibrate. These vibrations are transferred to the grains placed on the rubber sheet, causing them to jump up and down. This happens because the sound waves create pressure changes in the air, which make the rubber sheet vibrate.

Key Concept: Sound Medium

b) Gas

[Solution Description]

Sound requires a medium to travel, which can be a gas, liquid, or solid. Sound cannot travel through a vacuum because there are no particles to carry the vibrations. Therefore, the correct medium for sound to travel through can be any of these three states of matter.

Key Concept: Structure of Ear, Eardrum Damage

c) The person will experience hearing impairment

[Solution Description]

The eardrum is crucial for hearing as it transmits sound vibrations to the inner ear. Damage to the eardrum can impair its ability to vibrate properly, leading to hearing loss or difficulty in hearing.

Key Concept: Eardrum Function, Sound Vibrations

c) The grains will jump higher and more vigorously

[Solution Description]

When sound enters the ear, it causes the eardrum to vibrate. In the tin-can model, the stretched rubber represents the eardrum. Louder sounds produce stronger vibrations, causing the grains of cereal to jump higher and more vigorously.

Key Concept: Outer Ear Structure

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description] The outer ear indeed has a funnel-like shape, which helps in directing sound waves into the ear canal. However, the amplification of sound waves primarily occurs due to the resonance in the ear canal and the mechanical action of the ossicles in the middle ear, not solely due to the funnel-like structure of the outer ear. Therefore, while the assertion is true, the reason provides an incorrect explanation for the function of the outer ear.

Key Concept: Outer Ear Structure

b) To amplify and direct sound waves into the ear canal

[Solution Description]

The outer ear's funnel-like structure helps to collect sound waves and direct them into the ear canal. This design amplifies the sound and channels it towards the eardrum, which then vibrates to transmit the sound to the inner ear.

Key Concept: Eardrum Damage

b) It can damage the eardrum

[Solution Description]

Inserting sharp objects into the ear can damage the eardrum, which is a delicate membrane. A damaged eardrum can impair its ability to vibrate properly in response to sound waves, leading to hearing loss or other auditory issues.

Key Concept: Eardrum Function

c) To transmit sound vibrations to the inner ear

[Solution Description]

The eardrum, also known as the tympanic membrane, vibrates when sound waves hit it. These vibrations are then transmitted to the inner ear, where they are converted into signals that the brain can interpret as sound. Therefore, the primary function of the eardrum is to transmit sound vibrations to the inner ear.

Key Concept: Inner ear: Nerve signals to the brain

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The eardrum, being a thin stretched membrane, vibrates when sound waves hit it. These vibrations are then transmitted to the inner ear, which converts them into nerve signals that are sent to the brain. This is how we perceive sound. Both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Hearing Process, Eardrum Damage

b) It can damage the eardrum

[Solution Description]

Inserting sharp objects into the ear canal can damage the eardrum. The eardrum is a thin membrane that plays a crucial role in hearing by transmitting sound vibrations to the inner ear. Damage to the eardrum can impair this transmission process, leading to hearing loss.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The eardrum is a thin membrane that vibrates when sound waves hit it. Sound travels through the air as vibrations, and when these vibrations reach the eardrum, they cause it to move back and forth. This movement of the eardrum sends vibrations to the inner ear, which then transmits signals to the brain for interpretation. Thus, both the assertion and reason are true, and the reason correctly explains why the eardrum vibrates.

Key Concept: Eardrum Sensitivity

b) Because the rubber sheet vibrates due to sound

[Solution Description]

When sound is produced near a stretched rubber sheet (which acts like an eardrum), the rubber sheet vibrates due to the sound waves. These vibrations cause the grains placed on the rubber sheet to jump up and down. This demonstrates how the eardrum vibrates in response to sound.

Key Concept: Eardrum and Sound

c) The membrane's vibrations cause the grains to move

[Solution Description]

When sound waves enter the tin-can model, they cause vibrations in the air inside the can. These vibrations are transferred to the stretched rubber membrane, making it vibrate. The grains placed on the membrane also vibrate due to these movements, causing them to jump up and down.

Key Concept: Function of Eardrum

d) To transmit sound vibrations

[Solution Description]

The eardrum acts as a thin membrane that vibrates when sound waves hit it. These vibrations are then transmitted to the inner ear, where they are converted into electrical signals for the brain to interpret. Thus, its primary function is to transmit sound vibrations.

Key Concept: Frequency

c) 5 Hz

[Solution Description]

Frequency is calculated as the number of oscillations per second. Here, the object completes 50 oscillations in 10 seconds. So, the frequency $f$ is given by:

$f = \frac{\text{Number of oscillations}}{\text{Time}} = \frac{50}{10} = 5 \, \text{Hz}$

The frequency of the object is 5 Hz.

Key Concept: Frequency, Pitch

c) Pitch doubles and becomes higher

[Solution Description]

The pitch of a sound is directly related to its frequency. If the frequency is doubled, the pitch also doubles. Since the original frequency is 500 Hz, doubling it results in:

$\text{New frequency} = 2 \times 500 = 1000 \, \text{Hz}$

Thus, the pitch becomes higher as the frequency increases to 1000 Hz.

Key Concept: Decibel Level

d) Average factory.

[Solution Description]

Decibel (\$dB\$) is a unit used to measure the loudness of sound. According to the syllabus, normal conversation is around 60 dB, busy traffic is 70 dB, average factory noise is 80 dB, and above 80 dB, the noise becomes physically painful. Among the given options, average factory noise has the highest decibel level.

Key Concept: Amplitude, Loudness

c) Factor of 4

[Solution Description]

The loudness of sound is proportional to the square of the amplitude. Let the initial amplitude be $A_1 = 5$ units and the final amplitude be $A_2 = 10$ units. The ratio of loudness is given by:

$\text{Loudness Ratio} = \left(\frac{A_2}{A_1}\right)^2 = \left(\frac{10}{5}\right)^2 = 2^2 = 4$

Therefore, the loudness increases by a factor of 4.

Key Concept: Amplitude and Displacement

c) Amplitude is 3 cm

[Solution Description]

The displacement of the thermocol ball is equal to the amplitude of vibration of the tumbler. Therefore, if the ball is displaced by 3 cm, the amplitude of vibration is also 3 cm.

Key Concept: Frequency, Time Period

b) 0.02 seconds

[Solution Description]

The time period ($T$) is the reciprocal of the frequency ($f$). Given $f = 50$ Hz, we can calculate the time period using the formula:

$T = \frac{1}{f} = \frac{1}{50} = 0.02 \text{ seconds}$

Therefore, the time period of the oscillation is 0.02 seconds.

Key Concept: Amplitude, Time Period, and Frequency of a Vibration

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the time period ($T$) of a vibrating object is inversely proportional to its frequency ($f$). This is true because the relationship between time period and frequency is given by $T = \frac{1}{f}$.

The reason states that the frequency of a vibrating object is defined as the number of oscillations per second. This is also true because frequency is indeed defined as the number of complete oscillations or vibrations per unit time (second).

Since both the assertion and the reason are true, and the reason correctly explains why the assertion is true, the correct answer is option a).

Key Concept: Time Period, Frequency

a) Time period: 0.25 s, Frequency: 4 Hz

[Solution Description]

To find the time period ($T$), we use the formula $T = \frac{\text{Total Time}}{\text{Number of Oscillations}}$. Given that the pendulum completes 60 oscillations in 15 seconds, we substitute the values into the formula:

$T = \frac{15}{60} = 0.25 \text{ seconds}$

Next, to find the frequency ($f$), we use the formula $f = \frac{1}{T}$. Substituting the value of $T$:

$f = \frac{1}{0.25} = 4 \text{ Hz}$

Therefore, the time period is 0.25 seconds and the frequency is 4 Hz.

Key Concept: Audible Frequencies

b) 1,000 Hz

[Solution Description]

The range of audible frequencies for the human ear is roughly from 20 Hz to 20,000 Hz. Among the given options, 1,000 Hz falls within this range and is therefore audible.

Key Concept: Frequency of Oscillation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The frequency of an object is defined as the number of oscillations it completes in one second. If the frequency is 50 Hz, it means the object completes 50 oscillations in one second. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Frequency

b) 5 Hz

[Solution Description]

Frequency is calculated as the number of oscillations per second. Here, the number of oscillations is $15$, and the time taken is $3$ seconds. So, frequency = $\frac{15}{3} = 5$ Hz.

Key Concept: Effect of Amplitude on Loudness

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the loudness of sound increases with an increase in the amplitude of vibration. This is true because, according to the syllabus, loudness is proportional to the square of the amplitude of vibration. When the amplitude increases, the loudness increases by a factor equal to the square of the amplitude change. For example, if the amplitude doubles, the loudness increases by a factor of four. The reason given is also correct, as it explains that loudness is directly proportional to the square of the amplitude of vibration. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Amplitude, Loudness, Frequency

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

According to the syllabus, loudness of sound is proportional to the square of the amplitude of vibration. This means that if the amplitude is doubled, the loudness increases by a factor of 4 (since $2^2 = 4$). Therefore, the Assertion is true because doubling the amplitude indeed increases the loudness. The Reason is also true as it correctly states the relationship between loudness and amplitude. Furthermore, the Reason provides the correct explanation for the Assertion since it explains why the loudness increases when the amplitude is doubled.

Key Concept: Loudness and Amplitude

c) The loudness becomes four times greater

[Solution Description]

The loudness of sound is proportional to the square of the amplitude of the vibration producing the sound. If the amplitude is doubled, the loudness increases by a factor of $2^2 = 4$. Therefore, the loudness becomes four times greater.

Key Concept: Amplitude, Loudness

c) 9

[Solution Description]

The loudness of sound is proportional to the square of the amplitude. Let the initial amplitude be $A_1 = 5$ units and the final amplitude be $A_2 = 15$ units. The factor by which the loudness increases is given by:

$\text{Factor} = \left(\frac{A_2}{A_1}\right)^2 = \left(\frac{15}{5}\right)^2 = 3^2 = 9$

Therefore, the loudness increases by a factor of 9.

Key Concept: Frequency

c) 50 Hz

[Solution Description]

Frequency is defined as the number of oscillations per second. If an object oscillates 50 times in one second, its frequency is $50$ Hz.

Key Concept: Frequency and Pitch

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the pitch of a sound produced by a whistle is higher than that of a drum because the whistle vibrates with a higher frequency. This is true as the whistle indeed has a higher frequency of vibration compared to the drum. The reason states that the pitch of a sound is directly determined by its frequency, with higher frequencies resulting in higher pitches. This is also true as frequency is directly proportional to the pitch of the sound. Moreover, the reason correctly explains the assertion, making it the correct explanation.

Key Concept: Frequency and Sound Characteristics

c) The drum has a lower frequency and therefore a lower pitch.

[Solution Description]

The drum typically produces a low-pitched sound due to its low frequency of vibration, whereas the whistle produces a high-pitched sound due to its high frequency of vibration. Therefore, the frequency of the whistle is higher than that of the drum, resulting in a higher pitch for the whistle compared to the drum.

Key Concept: Amplitude, Frequency, and Pitch

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that a bird chirp has a higher pitch than a lion\'s roar because it has a higher frequency. This is true as the pitch of a sound is directly related to its frequency. Higher frequency results in a higher pitch, which is why a bird chirp sounds sharper compared to a lion\'s roar. The reason provided is also correct as the pitch of a sound is indeed determined by its frequency. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Frequency, Pitch, Shrillness

c) The sound with 1000 Hz has a higher pitch and is more shrill

[Solution Description]

The pitch and shrillness of a sound are determined by its frequency. A higher frequency corresponds to a higher pitch and greater shrillness. Therefore, the sound with a frequency of 1000 Hz will have a higher pitch and be more shrill compared to the sound with a frequency of 500 Hz.

Key Concept: Frequency and Pitch

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The frequency of vibration determines the pitch of a sound. A lion's roar has a low frequency, which results in a low-pitched sound. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Amplitude and Loudness

c) It becomes four times

[Solution Description]

The loudness of sound is proportional to the square of the amplitude. If the amplitude is doubled, the loudness increases by a factor of $2^2 = 4$. Therefore, the loudness becomes four times greater than before.

Key Concept: Audible Frequency Range, Inaudible Sounds

d) The sound is inaudible to both humans and dogs.

[Solution Description]

The human audible frequency range is $20$ to $20,000 Hz$. A sound of $15 Hz$ is below the lower limit of human hearing, making it inaudible to humans.

Key Concept: Animal Hearing

b) Dogs have a broader hearing range

[Solution Description] Dogs have a broader hearing range compared to humans, extending beyond $20,000 \text{ Hz}$. This allows them to hear high-frequency sounds that are inaudible to humans.

Key Concept: Inaudible Sounds

d) 25,000 Hz

[Solution Description]

Sounds with frequencies less than $20 \text{ Hz}$ or greater than $20,000 \text{ Hz}$ are inaudible to the human ear. Therefore, a frequency of $25,000 \text{ Hz}$ is inaudible.

Key Concept: Measures to Limit Noise Pollution

c) Planting trees around buildings

[Solution Description]

Planting trees along roads and around buildings can help reduce noise pollution by cutting down on the sounds reaching the residents.

Key Concept: Definition of Noise and Music, Conceptual Complexity

b) Singing is music; shouting is noise

[Solution Description]

When students sing together in harmony, the sound produced is musical because it is pleasing to the ear. However, if they shout loudly and chaotically, the sound becomes unpleasant and is classified as noise. This demonstrates the difference between organized musical sounds and disorganized noise.

Key Concept: Noise Pollution, Control Measures

c) Planting trees along the roads

[Solution Description]

To control noise pollution in residential areas, it is essential to implement measures that directly address the sources of noise. Planting trees along roads can significantly reduce the sound reaching the residents by acting as a natural sound barrier. This method is both effective and environmentally friendly.

Key Concept: Sound Production

b) Stretched string

[Solution Description]

The Veena is a stringed musical instrument. According to Table 10.1, the sound in a Veena is produced by the vibration of its stretched string.

Key Concept: Noise Pollution

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that continuous exposure to loud sounds from traffic and construction sites can lead to temporary or permanent hearing impairment. This is true because prolonged exposure to high decibel levels can damage the auditory system. The reason states that noise pollution causes health issues like hypertension, anxiety, and hearing loss due to excessive or unwanted sounds in the environment. This is also true as noise pollution is known to have various adverse health effects. Moreover, the reason correctly explains the assertion because hearing impairment is one of the direct consequences of noise pollution, which includes exposure to loud sounds from traffic and construction sites.

Key Concept: Audible Range

b) 1000 Hz

[Solution Description]

The audible range for humans is from 20 Hz to 20,000 Hz. Among the given options, 1000 Hz falls within this range, making it audible to humans.

Key Concept: Health Effects of Noise Pollution

c) Diabetes

[Solution Description]

Noise pollution can cause several health issues such as hypertension, anxiety, lack of sleep, and hearing impairment. However, it does not directly cause diabetes, which is related to insulin regulation.

Key Concept: Control of Noise Pollution

c) Using earplugs in noisy areas

[Solution Description]

To control noise pollution, measures such as reducing the volume of sound systems, using earplugs, and maintaining vehicles properly can be effective.

Key Concept: Measures to Control Noise Pollution

c) Planting trees along roads and around buildings

[Solution Description]

Effective measures to control noise pollution in residential areas include minimizing the use of automobile horns, running TV and music systems at low volumes, and planting trees along roads and around buildings to reduce sound levels.

Key Concept: Noise Pollution, Home Appliances

c) 78 dB

[Solution Description]

The sound level reduction is calculated as follows: First, convert the original sound level to intensity: $I = 10^{-12} \times 10^{80/10} = 10^{-4}$ W/m². Reducing this by 20% gives $I_{new} = 0.8 \times 10^{-4} = 8 \times 10^{-5}$ W/m\u00B2. Now, convert back to decibels: $L_{new} = 10 \log_{10}(8 \times 10^{-5}/10^{-12}) = 10 \log_{10}(8 \times 10^7) \approx 79.03$ dB.

Key Concept: Sources of Noise Pollution, Health Effects

a) Temporary hearing loss

[Solution Description]

The question focuses on the impact of continuous exposure to loud noise from industrial sources. According to the syllabus, prolonged exposure to high noise levels can cause temporary or permanent hearing impairment, sleep disturbances, and hypertension. Among these, hearing impairment is the most direct consequence of continuous exposure to loud noise. Therefore, the most likely health issue for residents living near the factory is hearing impairment.

Key Concept: Noise Pollution, Health Issues

c) Assertion is true, but Reason is false.

[Solution Description]

Noise pollution, especially when exposure is prolonged and at high decibel levels, can indeed cause permanent hearing impairment. The eardrum plays a vital role in the hearing process by transmitting sound vibrations to the inner ear. However, while noise pollution can damage the auditory system, it primarily affects the hair cells in the cochlea rather than directly damaging the eardrum. The assertion is true because noise pollution can lead to permanent hearing impairment, but the reason is false because it incorrectly attributes the damage to the eardrum.

Key Concept: Measures to Control Noise Pollution

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] Planting trees along roads and around buildings is indeed an effective measure to control noise pollution in residential areas. Trees act as natural sound barriers by absorbing and deflecting sound waves, which reduces the intensity of noise reaching the residents. This is a well-documented method for mitigating noise pollution in urban and residential settings. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Measures to Limit Noise Pollution

c) Planting trees along roads

[Solution Description]

Planting trees along roads and around buildings can significantly reduce noise pollution in residential areas. Trees act as natural sound barriers, absorbing and deflecting sound waves, thus reducing the harmful effects of noise on residents.

Key Concept: Noise Pollution Control

b) Reducing noise pollution

[Solution Description]

Planting trees in urban areas helps reduce noise pollution by acting as natural sound barriers. This helps in cutting down the sounds that reach the residents, thus reducing the harmful effects of noise pollution.

Key Concept: Effects of Noise Pollution

c) Sleep disturbances

[Solution Description]

The syllabus states that noise pollution can have harmful effects on residents, such as stress, hearing loss, and sleep disturbances. These effects are caused by prolonged exposure to excessive or unwanted sounds.

Key Concept: Hearing Impairment

a) Disease

[Solution Description]

Partial hearing impairment is generally the result of a disease, injury, or age. It is less commonly from birth and more often acquired later in life due to external factors.

Key Concept: Causes of Hearing Impairment

b) Disease or injury

[Solution Description]

Partial hearing impairment is generally the result of a disease, injury, or age. This information is directly provided in the syllabus.

Key Concept: Hearing Impairment and Speech Development

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Assertion (A) states that a child with congenital deafness is likely to have defective speech development. This is true because speech development heavily relies on the ability to hear and mimic sounds. Reason (R) explains that speech develops as a direct result of hearing, and hearing loss can impede this process. This is also true and directly explains why the assertion is correct. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Hearing Impairment, Speech Development

c) Inability to hear and mimic sounds

[Solution Description] Speech development depends on the ability to hear sounds and mimic them. A child with congenital hearing impairment cannot hear sounds clearly, making it difficult for them to develop speech naturally. This leads to delayed or defective speech development.

Key Concept: Hearing Impairment, Technological Support

c) Speech-to-text converter; textual communication

[Solution Description] A device that converts spoken language into text in real-time is known as a speech-to-text converter. This technology enhances communication by allowing individuals with total hearing impairment to read spoken words instead of relying on auditory input.

Key Concept: Sign Language

c) Sign language

[Solution Description]

Individuals with total hearing impairment often rely on sign language as their primary mode of communication. Sign language uses visual gestures and signs to convey meaning, allowing effective communication without the need for auditory input.

Key Concept: Hearing Impairment, Social Awareness

a) Cochlear implants

[Solution Description]

Technological devices such as hearing aids, cochlear implants, and assistive listening devices are designed to help the hearing-impaired communicate more effectively in everyday situations. Among these, cochlear implants are particularly effective for individuals with severe hearing loss.

Key Concept: Helping Hearing-Impaired Individuals

b) Learning sign language

[Solution Description]

Children with hearing impairments can effectively communicate by learning sign language. This helps them overcome the challenges posed by their hearing loss.