Class 10 Physics Chapter 7 Sound

Key Concept: Loudness level and amplitude relationship

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Given the relationship $L \propto a^2$, where $L$ is loudness and $a$ is amplitude. If amplitude increases by a factor of 4, loudness becomes $(4)^2 = 16$ times the original value. From the definition, each 10-fold increase in loudness corresponds to a 10 dB increase in loudness level. Here, $16 = 1.6 \times 10^1$, so the increase in loudness level is approximately 10 * log$_{10}(16)$ \approx 12 dB. Both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Amplitude, Compression and Rarefaction

c) 1.50 m
[Solution Description]
The wavelength ($\lambda$) of a longitudinal wave is defined as the distance between two consecutive compressions or rarefactions. Here, the distance between two consecutive compressions is given as 1.5 m. Therefore, the wavelength is simply equal to this distance.
$\lambda = 1.5$ m

Key Concept: Relation Between Surface Area and Amplitude

b) The drum produces a louder sound due to larger surface area
[Solution Description]
The amplitude of vibration is greater for a body with a larger surface area because it can displace more air molecules. Since the drum has a much larger surface area than the tuning fork, its amplitude of vibration will be greater for the same energy input, resulting in a louder sound.

Key Concept: Displacement-Time Graph

b) 0.125 Hz
[Solution Description]
The frequency $f$ of the wave is given by the formula:
$f = \frac{1}{T}$
Given that the time period $T = 8 \, \text{s}$, substituting the value gives:
$f = \frac{1}{8} \, \text{Hz} = 0.125 \, \text{Hz}$

Key Concept: Displacement-Time Graph

c) Frequency increases, pitch increases.
[Solution Description]
In a displacement-time graph, an increase in the number of waves within the same time interval means that the time period ($T$) of the wave has decreased. Since frequency ($f$) is inversely proportional to the time period ($f = 1/T$), a decrease in $T$ leads to an increase in frequency. Higher frequency corresponds to higher pitch. Therefore, the sound becomes shriller.

Key Concept: Decibel Scale

a) whisper
[Solution Description]
From the given table in the syllabus:
- 0 dB: Minimum audible sound
- 10 dB: Leaves rustling
- 20 dB: Recording studio
- 30 dB: Whisper
- 50 dB: Normal conversation
Thus, a sound level of 30 dB corresponds to a whisper.

Key Concept: Amplitude and Loudness

d) Loudness increases with increase in amplitude
[Solution Description]
Loudness is directly proportional to the square of the amplitude of the sound wave. A higher amplitude means greater energy carried by the wave, resulting in a louder sound.

Key Concept: Ultrasound in Medicine

b) Imaging internal organs
[Solution Description]
Ultrasound is widely used in medical imaging for visualizing internal body structures such as organs, tissues, and blood flow. It works by emitting high-frequency sound waves that reflect off tissues and are captured to create images.

Key Concept: Displacement-Time Graph

b) $\frac{\pi}{4}$ rad/s
[Solution Description]
The angular frequency ($\omega$) is related to the time period ($T$) by the equation:
$\omega = \frac{2\pi}{T}$
Given the time period $T = 8$ s, substituting the values:
$\omega = \frac{2\pi}{8} = \frac{\pi}{4} \text{ rad/s}$

Key Concept: Speed of Sound

d) Solid
[Solution Description]
The speed of sound depends on the medium it travels through. Solids have closely packed molecules, allowing sound to propagate faster (~5000 m/s). Liquids (~1500 m/s) and gases (~330 m/s) have progressively slower speeds due to less dense molecular arrangements. A vacuum cannot transmit sound at all.

Key Concept: Wave Properties

c) The frequency increases
[Solution Description]
As per the syllabus, when the number of waves in the same time interval increases, the time period decreases, which means the frequency (and pitch) increases.

Key Concept: Compression, Rarefaction, Wave Propagation

b) 1.5 meters
[Solution Description]
The distance between two consecutive compressions in a longitudinal wave is equal to one wavelength ($\lambda$). Therefore, if the distance between two consecutive compressions is given as 1.5 meters, the wavelength of the sound wave is 1.5 meters.

Key Concept: Definition of Pitch

b) The pitch decreases
[Solution Description]
Pitch is determined by the frequency of sound. Lower frequency results in lower pitch, as given in the syllabus. So, if frequency decreases, the pitch also decreases.

Key Concept: Cloud Computing

c) Eliminates need for physical hardware maintenance
[Solution Description]
The primary advantage of IaaS is that it eliminates the need for organizations to purchase and maintain physical hardware. Instead, they can provision virtualized computing resources (servers, storage, networking) over the internet on demand, paying only for what they use. This reduces capital expenses and allows for flexible scaling.

Key Concept: Wavelength and Frequency Relationship

c) 220 Hz
[Solution Description]
To find the frequency, we use the formula:
$v = f \lambda$
Rearranging for $f$:
$f = \frac{v}{\lambda} = \frac{330 \, \text{m/s}}{1.5 \, \text{m}} = 220 \, \text{Hz}$
The frequency of the wave is 220 Hz.

Key Concept: Quality, Wave Form

c) Because they produce different wave forms.
[Solution Description]
The quality or timbre of a sound depends on its wave form, which is determined by the presence of overtones or harmonics. Even if two sounds have the same pitch and loudness, their wave forms differ due to the unique combination of frequencies produced by different instruments.

Key Concept: Cybersecurity

d) Multi-factor authentication system
[Solution Description]
Multi-factor authentication (MFA) provides the strongest protection among these options because it requires multiple verification methods (something you know like password, something you have like phone, and/or something you are like fingerprint). While firewalls protect networks and antivirus protects against malware, MFA directly prevents unauthorized users from gaining access even if they obtain one authentication factor.

Key Concept: Ultrasound Imaging, Sonar Technology

b) 3000 meters
[Solution Description]
The distance can be calculated using the formula for sonar:
$\text{Distance} = \frac{\text{Speed} \times \text{Time}}{2}$
Since the sound travels to the iceberg and back, we divide by 2.
Given speed $= 1500 \text{ m/s}$ and time $= 4 \text{ s}$,
$\text{Distance} = \frac{1500 \times 4}{2} = \frac{6000}{2} = 3000 \text{ meters}$

Key Concept: Wave Parameters

d) 0.01 s
[Solution Description] The time period ($T$) is the reciprocal of the frequency ($f$). Given $f = 100 \, \text{Hz}$, we calculate:
$T = \frac{1}{f} = \frac{1}{100} = 0.01 \, \text{s}$

Key Concept: Propagation of Sound

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that sound waves are longitudinal waves due to compressions and rarefactions. This is correct because sound waves propagate through air by creating regions of high pressure (compressions) and low pressure (rarefactions).
The Reason states that in longitudinal waves, particles vibrate parallel to the direction of wave propagation. This is also correct as longitudinal waves are characterized by particle oscillations along the direction of wave travel.
Moreover, the Reason correctly explains the Assertion since it describes why compressions and rarefactions occur in longitudinal waves. Therefore, both statements are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Propagation of Sound

c) Longitudinal waves
[Solution Description]
Sound travels in air as longitudinal waves, where particles of the medium vibrate back and forth in the direction of the wave propagation.

Key Concept: Effect of Distance on Loudness, Real-World Application

b) 68 dB
[Solution Description]
The loudness decreases with the square of the distance. If the distance increases from 10 m to 40 m (a factor of 4), the intensity decreases by $4^2 = 16$. The decibel level is calculated as $L = 10 \log_{10}(I/I_0)$. The new intensity is $\frac{I}{16}$, so the change in decibels is $10 \log_{10}(\frac{1/16}{1}) = -12 \, \text{dB}$. Thus, the new loudness is $80 - 12 = 68 \, \text{dB}$.

Key Concept: Amplitude Ratio, Loudness Comparison

c) 16:25
[Solution Description]
Given amplitude ratio $\frac{a_1}{a_2} = \frac{4}{5}$.
Loudness ratio $\frac{L_1}{L_2} = \left(\frac{4}{5}\right)^2 = \frac{16}{25}$.
The difference in loudness level (dB) is:
$10 \log_{10}\left(\frac{25}{16}\right) \approx 10 \times 0.1938 \approx 1.938 \text{ dB}.$
Therefore, the ratio of loudness levels is not straightforward—it’s a difference of ~1.94 dB. However, for MCQ options, the closest proportional ratio is derived from $\left(\frac{4}{5}\right)^2 = \frac{16}{25}$.

Key Concept: Sound Source Identification

c) Tuning forks produce single frequency, pianos produce multiple frequencies
[Solution Description]
A tuning fork produces sound with a single frequency, while a piano produces multiple frequencies (harmonics), resulting in different wave forms.

Key Concept: Amplitude

d) The maximum displacement of particles from their mean position
[Solution Description]
The amplitude ($a$) is the maximum displacement of particles from their mean position. In a displacement-distance graph, it is represented by the highest point on the y-axis.

Key Concept: Waveform of Monotone Sound

d) Simple sine wave
[Solution Description]
A monotone sound consists of a single frequency, and its waveform is represented as a simple sine wave without any harmonics or distortions.

Key Concept: Sonar

c) Sonar
[Solution Description]
Sonar (Sound Navigation and Ranging) is a technique used in submarines and ships to detect underwater objects by emitting sound waves and analyzing the echoes received back from them.

Key Concept: Wavelength, Speed, Frequency

b) 0.50 m
[Solution Description]
The relationship between speed ($v$), frequency ($f$), and wavelength ($\lambda$) is given by:
$v = f \times \lambda$
Rearranging to solve for $\lambda$:
$\lambda = \frac{v}{f} = \frac{340}{680} = 0.5$ m

Key Concept: Loudness and Amplitude Relationship

c) Becomes four times.
[Solution Description]
Loudness is directly proportional to the square of the amplitude. If the amplitude is doubled ($a \rightarrow 2a$), loudness becomes $(2)^2 = 4$ times the original loudness.

Key Concept: High Frequency, Wave Representation

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that increasing the number of cycles in the same time interval decreases frequency, which is incorrect. More cycles in the same time interval imply higher frequency (since frequency = cycles per second). The reason states that frequency is inversely proportional to time period, which is true ($f = \frac{1}{T}$). Thus, the Assertion is false, but the Reason is true.

Key Concept: Mathematical Relation, Musical Instruments

b) Frequency halves, pitch becomes lower
[Solution Description]
The frequency of a vibrating string is inversely proportional to its length ($f \propto \frac{1}{L}$). If the length ($L$) is doubled ($2L$), the frequency ($f$) becomes half of the original frequency ($\frac{f}{2}$). Since pitch depends on frequency, the pitch also decreases, making the sound graver.

Key Concept: Loudness vs Intensity

d) Loudness varies with the listener's ear sensitivity, but intensity does not.
[Solution Description]
Loudness is subjective and depends on the sensitivity of the listener's ears, whereas intensity is an objective measure of sound energy per unit area. Therefore, loudness varies between listeners, but intensity remains the same for a given sound.

Key Concept: Amplitude and Loudness

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that if the amplitude of a sound wave is doubled, its loudness increases by four times. From the syllabus, we know that the loudness $L$ is directly proportional to the square of the amplitude $a$, i.e., $L \propto a^2$. Therefore, if the amplitude is doubled ($2a$), the new loudness becomes $(2a)^2 = 4a^2$, which is four times the original loudness ($a^2$). Thus, the Assertion is true.
The Reason states that the loudness of sound is directly proportional to the square of the amplitude. This is explicitly mentioned in the syllabus as $L \propto a^2$. Hence, the Reason is also true and correctly explains why the Assertion is true, as it provides the mathematical relationship between loudness and amplitude.

Key Concept: Frequency and Time Period Relationship

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The relationship between frequency ($f$) and time period ($T$) is given by $f = \frac{1}{T}$. Here, the assertion states that the frequency is 5 Hz. Using the formula, the time period $T$ can be calculated as $T = \frac{1}{f} = \frac{1}{5} = 0.2$ s. Thus, both the assertion and reason are true, and the reason correctly explains the assertion using the formula.

Key Concept: Decibel Scale, Multiplicative Factor

b) 6 dB
[Solution Description]
The loudness decreases proportionally to the square of the amplitude reduction ($L \propto a^2$). Reducing the amplitude to half means the new loudness is $(1/2)^2 = 1/4$ of the original intensity. The change in decibels is calculated as $\Delta L = 10 \log_{10}(1/4) \approx -6$ dB. Therefore, the loudness decreases by approximately 6 dB.

Key Concept: Sound production in a homemade guitar

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the pitch of the sound produced by the rubber bands can be changed by adjusting their tension, which is true as per the syllabus. The reason explains that higher tension increases the frequency of vibration, leading to a higher pitch, which is also true and correctly explains the assertion. Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Amplitude, Frequency, Wavelength

b) Wavelength remains same, loudness increases
[Solution Description]
Step 1: Calculate the initial wavelength using the formula $v = f \lambda$.
$330 = 500 \times \lambda \implies \lambda = \frac{330}{500} = 0.66 \, \text{m}$
Step 2: Doubling the amplitude affects only the loudness (energy) of the sound, not the wavelength or speed.
Step 3: Therefore, the wavelength remains unchanged ($0.66 \, \text{m}$), but the loudness increases.

Key Concept: Echolocation

c) Bats and dolphins
[Solution Description]
Bats and dolphins are examples of animals that use echolocation to navigate and hunt. They emit sound waves and interpret the returning echoes to determine the location and distance of objects or prey.

Key Concept: Sound Wave Properties

b) Wave B
[Solution Description]
Since pitch is determined by frequency, the wave with the higher frequency has the higher pitch. Here, wave B has a frequency of 400 Hz, which is higher than wave A's 200 Hz.

Key Concept: Frequency and Time Period

b) 0.002 seconds
[Solution Description]
The relationship between frequency ($f$) and time period ($T$) is given by:
$f = \frac{1}{T}$
Given $f = 500$ Hz, we can find the time period as:
$T = \frac{1}{f} = \frac{1}{500} = 0.002 \text{ seconds}$

Key Concept: Amplitude, Loudness

c) 4 times
[Solution Description]
The loudness ($L$) of a sound wave is directly proportional to the square of its amplitude ($a$). Mathematically, $L \propto a^2$.
Given:
Original amplitude ($a_1$) = 5 cm
New amplitude ($a_2$) = 10 cm
Step 1: Calculate the ratio of loudness.
$\frac{L_2}{L_1} = \frac{a_2^2}{a_1^2} = \frac{(10)^2}{(5)^2} = \frac{100}{25} = 4$
Thus, the new sound will be 4 times louder than the original sound.

Key Concept: Loudness and Amplitude

d) It becomes 9 times.
[Solution Description]
The loudness of sound is directly proportional to the square of the amplitude of the wave ($L \propto a^2$). When the amplitude is tripled, the new loudness becomes $(3)^2 = 9$ times the original loudness.

Key Concept: Amplitude and Loudness

c) It quadruples
[Solution Description]
The loudness ($L$) of a sound is proportional to the square of its amplitude ($a$):
$L \propto a^2$
If the amplitude is doubled ($2a$), the new loudness ($L'$) becomes:
$L' \propto (2a)^2 = 4a^2$
Thus, the loudness increases by a factor of 4.

Key Concept: Dependence of Loudness on Surface Area, Energy and Amplitude

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The loudness of sound depends on the amplitude of vibration, which is determined by the energy imparted to the vibrating body. A larger surface area allows for more air particles to be displaced, increasing the amplitude and hence the loudness of the sound. The Reason correctly explains the Assertion because it links the increased surface area to greater amplitude, which directly affects loudness. Therefore, both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Pitch and Frequency

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that a sound with 200 Hz has a higher pitch than a sound with 500 Hz. From the syllabus, we know that higher frequency corresponds to higher pitch and lower frequency corresponds to lower pitch. Therefore, the Assertion is false because 200 Hz is less than 500 Hz, so it should have a lower pitch. The Reason states that pitch is directly proportional to the frequency, which is true as per the syllabus. Thus, the Assertion is false, but the Reason is true.

Key Concept: Distance from Source

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that loudness decreases with increasing distance, which is true based on the syllabus. The reason explains this phenomenon by stating the inverse square law relationship between intensity and distance, which is also correct. Moreover, the reason correctly explains why the assertion is valid because the decrease in loudness is directly linked to the decrease in intensity due to distance following $I \propto \frac{1}{d^2}$.

Key Concept: Pitch and Frequency

b) High pitch
[Solution Description]
The frequency of a sound wave determines its pitch. A sound of high frequency is said to have a high pitch.

Key Concept: Longitudinal Waves, Compression, Rarefaction

c) Wavelength
[Solution Description]
In a longitudinal wave, the distance between two consecutive compressions or two consecutive rarefactions is equal to one wavelength ($\lambda$). This distance represents the spatial period of the wave.

Key Concept: Wavelength Definition

c) Wavelength
[Solution Description]
The wavelength ($\lambda$) in a longitudinal wave is the distance between two consecutive compressions or rarefactions.

Key Concept: Comparison of Loudness

c) 4 times louder.
[Solution Description]
Given amplitudes: $A_A = 5 \, \text{cm}, \, A_B = 10 \, \text{cm}$. The ratio of loudness is $\frac{L_B}{L_A} = \left( \frac{A_B}{A_A} \right)^2 = \left( \frac{10}{5} \right)^2 = 4$. Sound B is 4 times louder than sound A.

Key Concept: Pitch and Frequency, Waveform

c) Quality (Timbre)
[Solution Description]
The quality (or timbre) of the sound changes because the waveform is altered by the presence of additional frequencies (harmonics). However, the fundamental frequency (pitch) remains the same as it is still 256 Hz.

Key Concept: Amplitude, Graphical Representation

b) Amplitude = 7 cm; 4 times louder
[Solution Description]
Amplitude is defined as the maximum displacement of a vibrating particle from its mean position.
Step 1: Determine the amplitude.
Here, the maximum displacement = 7 cm, so amplitude ($a_1$) = 7 cm.
Given another wave's amplitude ($a_2$) = 3.5 cm.
Step 2: Compare the loudness using the relation $L \propto a^2$.
$\frac{L_1}{L_2} = \frac{a_1^2}{a_2^2} = \frac{7^2}{3.5^2} = \frac{49}{12.25} = 4$
Thus, the first wave is 4 times louder than the second wave.

Key Concept: Longitudinal vs transverse waves

c) Particles vibrate parallel to the direction of wave propagation.
[Solution Description]
In a longitudinal wave, the particles of the medium vibrate in the same direction as the wave propagation. Transverse waves have vibrations perpendicular to the direction of propagation.

Key Concept: Factors Affecting Loudness of Sound

b) The drum with $1.2 \, \text{m}^2$ surface area
[Solution Description]
The loudness of sound is directly proportional to the surface area of the vibrating body. A larger surface area results in greater energy transfer, leading to a louder sound. Here, the second drum has a larger surface area ($1.2 \, \text{m}^2$) compared to the first drum ($0.5 \, \text{m}^2$). Therefore, the second drum will produce a louder sound when struck with the same force.

Key Concept: Amplitude and Loudness Ratio

b) Becomes one-fourth of its original value
[Solution Description]
The loudness of a sound wave is proportional to the square of its amplitude. If the amplitude is halved ($a' = \frac{a}{2}$), then the new loudness $L'$ becomes:
$L' \propto \left(\frac{a}{2}\right)^2 = \frac{a^2}{4}$
Thus, the loudness reduces to $\frac{1}{4}$ of its original value.

Key Concept: Displacement-time graph, Displacement-distance graph, Wavelength, Frequency

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The displacement-time graph represents the oscillation of a single particle over time and provides information about the amplitude and time period (or frequency) of the wave. However, it does not provide direct information about the spatial characteristics like wavelength. On the other hand, the displacement-distance graph captures the positions of multiple particles at a specific instant, allowing us to measure the wavelength as the distance between two consecutive points in the same phase. Therefore, while both graphs represent aspects of wave motion, only the displacement-distance graph can be used to determine the wavelength. This makes the Assertion true. The Reason is also true because it correctly describes the fundamental difference between the two graphs, and it logically explains why the Assertion holds. Hence, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Amplitude Calculation

c) 10 cm
[Solution Description]
Given $L \propto a^2$, if loudness increases by 4 times, then $4L \propto (ka)^2$, where $k$ is the factor by which amplitude changes. Solving, $k^2 = 4 \Rightarrow k = 2$. So, the new amplitude is $2 \times 5\,\text{cm} = 10\,\text{cm}$.

Key Concept: Quality/Timbre

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that two sounds with identical pitch and loudness can be differentiated by their timbre, which aligns with the syllabus stating that quality (timbre) distinguishes such sounds. The Reason explains that this is because timbre depends on the wave form, which varies uniquely for different sources. Since both statements are correct and the Reason correctly explains the Assertion, the answer is option a).

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 increases with its frequency, which is correct because higher frequency corresponds to a higher pitch. The reason states that frequency is inversely proportional to the time period ($f = \frac{1}{T}$), which is also true. Moreover, this relationship explains why higher frequency results in higher pitch, as a shorter time period means more waves per unit time, leading to a perceived increase in pitch. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Membrane Instruments

c) Tension and size of the membrane
[Solution Description]
In membrane instruments like the tabla, the pitch of the sound depends on the tension and size of the membrane. A tighter and smaller membrane produces a higher pitched sound, while a looser and larger membrane results in a lower pitched sound. Thus, the correct factor affecting pitch is the tension and size of the membrane.

Key Concept: Material Selection, Safety Precautions

d) To prevent snapping and ensure consistent sound production
[Solution Description]
Properly stretching and securing the rubber bands ensures consistent tension, which is crucial for producing stable and predictable sound frequencies. Additionally, improper stretching can lead to snapping, posing safety risks.
Secure fixation prevents sudden release of tension, ensuring both functionality and safety during the experiment.

Key Concept: Propagation of Sound

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that sound cannot travel in a vacuum, which is true because sound waves require a medium (like air, water, or solids) to propagate. The reason given is that sound requires a medium for propagation, which directly explains why sound cannot travel in a vacuum. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Frequency Definition

a) Hertz
[Solution Description]
Frequency is denoted by $f$ and its S.I. unit is second$^{-1}$ or hertz (symbol Hz). Thus, the correct answer is hertz.

Key Concept: Loudness and Pitch

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The loudness of a sound wave is related to the amplitude of the wave. The energy carried by a wave is proportional to the square of its amplitude, so an increase in amplitude results in an increase in loudness. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Amplitude Comparison

c) 4:1
[Solution Description]
The loudness is proportional to the square of the amplitude. For wave A ($a_1 = 5 \, \text{cm}$) and wave B ($a_2 = 10 \, \text{cm}$), the loudness ratio is:
$\frac{L_B}{L_A} = \frac{(10)^2}{(5)^2} = \frac{100}{25} = 4$

Key Concept: String Instruments

c) The pitch increases
[Solution Description]
The pitch of sound produced by a string instrument depends on the tension and thickness of the string. Higher tension increases the frequency of vibration of the string, resulting in a higher pitch.

Key Concept: Sound Production in Musical Toy

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The pitch of sound is determined by its frequency. When the rubber bands are tightened, their tension increases, leading to faster vibrations and thus higher frequency. This results in a higher pitch. Therefore, both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion.

Key Concept: Unit of time period

c) Second (s)
[Solution Description]
The S.I. unit of time period is the second (\text{s}), as it measures the duration of one vibration.

Key Concept: Frequency and Time Period

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that if the frequency of a sound wave is doubled, its time period is halved. This is true because frequency ($f$) and time period ($T$) are inversely proportional to each other as given by the equation $f = \frac{1}{T}$. Therefore, doubling the frequency will indeed halve the time period. The reason correctly explains the assertion using the fundamental relationship between frequency and time period.

Key Concept: Frequency, Wavelength, Wave speed

b) 200 Hz
[Solution Description]
To find the frequency ($f$), we rearrange the wave equation $v = f \lambda$ to solve for $f$:
$f = \frac{v}{\lambda}$
Given:
$v = 500 \, \text{m/s}$
$\lambda = 2.5 \, \text{m}$
Calculation:
$f = \frac{500}{2.5}$
$f = 200 \, \text{Hz}$
Therefore, the frequency of the wave is $200 \, \text{Hz}$.

Key Concept: Waveform of Sound

b) 500 Hz
[Solution Description]
The distance between two consecutive compressions is equal to the wavelength ($\lambda$) of the sound wave. Given $\lambda = 0.66 \, \text{m}$ and $v = 330 \, \text{m s}^{-1}$, the frequency ($f$) can be calculated using the formula:
$f = \frac{v}{\lambda} = \frac{330}{0.66} = 500 \, \text{Hz}.$

Key Concept: Visual Representation of Pitch

b) A graph with more waves in the same time interval.
[Solution Description]
A higher-pitched sound corresponds to a higher frequency. In a displacement-time graph, higher frequency means more waves (or cycles) are plotted in the same time interval compared to a lower-frequency sound.

Key Concept: Monotone Sound and Tuning Fork

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that a tuning fork produces sound of a single frequency, which is true as per the syllabus definition of a monotone sound. The reason states that the sound is called a monotone due to its simple sine waveform. While the waveform is indeed a sine wave, the term "monotone" specifically refers to the sound being of a single frequency, not just the waveform shape. Thus, the reason does not correctly explain why the sound is called a monotone. Therefore, both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Frequency, Time Period, Wave Characteristics

b) 6.8 m
[Solution Description]
Step 1: Given time period $T = 0.02$ s.
Step 2: Calculate frequency using $f = \frac{1}{T} = \frac{1}{0.02} = 50$ Hz.
Step 3: Given wave speed $v = 340$ m/s.
Step 4: Use the formula $v = f \lambda$ to find wavelength $\lambda$.
Step 5: Substitute values $\lambda = \frac{v}{f} = \frac{340}{50} = 6.8$ m.

Key Concept: Frequency, Pitch

b) Sound B has a higher pitch than sound A
[Solution Description]
Pitch depends on the frequency of the sound wave. Higher frequency corresponds to higher pitch, and lower frequency corresponds to lower pitch.
Sound B has a higher frequency (800 Hz) than sound A (200 Hz), so sound B is shriller (higher pitch) compared to sound A.

Key Concept: Wavelength and Frequency Relationship

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the wavelength of a sound wave decreases if its frequency increases, keeping the speed constant. This is true because wavelength and frequency are inversely related when the speed is constant, as shown by the wave equation $v = \lambda f$. The reason correctly explains this relationship using the wave equation, which directly links wavelength and frequency under constant speed conditions. Therefore, both the assertion and reason are true, and the reason is the correct explanation for the assertion.

Key Concept: Low frequency and pitch

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
From the given syllabus, it is clear that pitch depends on the frequency of sound. The higher the frequency, the higher the pitch, and vice versa. Therefore, the Assertion (A) correctly states that a sound with 1 Hz (lower frequency) has a lower pitch compared to a sound with 2 Hz (higher frequency). The Reason (R) also correctly explains the relationship between pitch and frequency. Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Unit of Wavelength

d) Meter (m)
[Solution Description]
The S.I. unit of wavelength is meter, denoted by \$m\$.

Key Concept: Relationship Between Loudness and Amplitude

c) 1:9
[Solution Description]
Given the ratio of amplitudes is $1:3$. Since loudness is proportional to the square of the amplitude ($L \propto a^2$), the ratio of loudness is $(1)^2 : (3)^2 = 1:9$.

Key Concept: Frequency Definition

d) The number of vibrations per second
[Solution Description]
Frequency measures the number of complete vibrations or cycles of a wave that occur per unit time (usually one second). It is not related to amplitude, wavelength, or energy directly.

Key Concept: Displacement-Distance Graph

b) 4 cm
[Solution Description]
The displacement-distance graph represents the positions of particles at a fixed time. The maximum displacement corresponds to the amplitude of the wave. At a distance of $\frac{\lambda}{2}$ (1.5 m here), the particle would be at its maximum displacement, which is equal to the amplitude. Therefore, the displacement is 4 cm.

Key Concept: Higher Frequency = Higher Pitch, Wave Patterns

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that a sound wave with a frequency of $512 \, \text{Hz}$ will have a higher pitch than one with $256 \, \text{Hz}$, which is correct because pitch is directly proportional to frequency. The Reason explains that the pitch is determined by frequency and that higher frequencies correspond to shorter time periods between wave cycles, which is also true. Moreover, the Reason correctly explains why higher frequency leads to higher pitch, as shorter time periods imply more cycles per second, resulting in a higher perceived pitch. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Pitch

c) Tension of the string
[Solution Description]
The pitch of sound in a guitar depends on the thickness, length, and tension of the string. Thinner, shorter, and tighter strings produce higher pitch sounds.

Key Concept: Frequency Definition

c) Hertz (Hz)
[Solution Description]
The unit of frequency is hertz, denoted by Hz. It represents the number of complete vibrations per second.

Key Concept: Decibel Scale

b) 20 dB
[Solution Description]
A 10-fold increase in loudness corresponds to 10 dB. Similarly, a 100-fold increase corresponds to $10 \, \text{dB} \times \log_{10}(100) = 20 \, \text{dB}$ since $\log_{10}(100) = 2$.

Key Concept: Effect of distance on loudness

b) One-fourth of its original value
[Solution Description]
The intensity ($I$) of sound is inversely proportional to the square of the distance ($d$) from the source:
$I \propto \frac{1}{d^2}$.
If the distance doubles ($d \rightarrow 2d$), the new intensity ($I'$) will be:
$I' = \frac{I}{(2)^2} = \frac{I}{4}$.
Therefore, the intensity becomes one-fourth of its original value.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Sound is a mechanical wave that needs a medium (solid, liquid, or gas) to propagate. In vacuum, there are no particles to transmit the sound vibrations, which is why sound cannot travel in vacuum. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Energy Conversion in Sound

d) Kinetic energy and potential energy interconvert
[Solution Description]
Sound is a form of energy where particles of the medium vibrate. During these vibrations, kinetic energy converts into potential energy and vice versa. This interconversion allows sound to propagate as a wave.

Key Concept: Tuning Fork Frequency

b) $320 \, \text{s}^{-1}$
[Solution Description]
Common frequencies of tuning forks include $256 \, \text{s}^{-1}$, $288 \, \text{s}^{-1}$, $320 \, \text{s}^{-1}$, $384 \, \text{s}^{-1}$, $480 \, \text{s}^{-1}$, and $512 \, \text{s}^{-1}$. Among the options, $320 \, \text{s}^{-1}$ is one of these standard frequencies.

Key Concept: Characteristics of Sound

c) Amplitude
[Solution Description]
Loudness is determined by the amplitude of the sound wave, which is the maximum displacement of particles from their mean position.

Key Concept: Propagation of Sound, Loudness

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The relationship between loudness ($L$) and amplitude ($a$) of a sound wave is given by:
$L \propto a^2$
If the initial amplitude is $a_1$ and the new amplitude is $a_2$, then the ratio of loudness is:
$\frac{L_2}{L_1} = \left(\frac{a_2}{a_1}\right)^2$
For the loudness to double ($L_2 = 2L_1$), the above equation becomes:
$2 = \left(\frac{a_2}{a_1}\right)^2$
Taking the square root of both sides gives:
$\sqrt{2} = \frac{a_2}{a_1}$
Therefore, the amplitude must be increased by $\sqrt{2}$ times to double the loudness, confirming the Assertion is true. The Reason correctly explains the Assertion as it states the proportionality between loudness and the square of amplitude.

Key Concept: Nature of Sound

d) Sound is produced by vibrating bodies and needs a medium to travel.
[Solution Description]
Sound requires a medium to propagate and cannot travel in a vacuum. It travels fastest in solids, slower in liquids, and slowest in gases. The statement that aligns with these facts is correct.

Key Concept: Nature of Sound

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the speed of sound is higher in solids than in liquids and gases because particles in solids are more closely packed. This is true as the syllabus mentions that sound travels fastest in solids (e.g., $5000 \, \text{m/s}$ in iron) compared to liquids ($1500 \, \text{m/s}$ in water) and gases ($330 \, \text{m/s}$ in air).
The reason states that in solids, the vibrations are transmitted faster due to strong intermolecular forces. This is also true according to the syllabus, which explains that closer packing and stronger bonds in solids allow quicker energy transfer, leading to higher speed of sound. Moreover, the reason correctly explains why the assertion is true.

Key Concept: Quality of Sound

c) Their quality
[Solution Description]
Quality (or timbre) is the characteristic that distinguishes two sounds of the same pitch and loudness based on their waveform or overtones. Even if the loudness and pitch are identical, the quality differs due to variations in their waveforms.

Key Concept: Pitch and Frequency, String Instruments

c) The pitch becomes double
[Solution Description]
The frequency $f$ of a vibrating string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length of the string, $T$ is the tension, and $\mu$ is the linear mass density. If the tension increases to $4T$, the new frequency $f'$ becomes $f' = \frac{1}{2L} \sqrt{\frac{4T}{\mu}} = 2f$. Thus, the pitch doubles.

Key Concept: Longitudinal vs Transverse waves

d) Transverse waves that propagate along the string
[Solution Description]
According to the syllabus notes, sound waves produced in strings are transverse waves because the particles of the medium vibrate perpendicular to the direction of wave propagation.

Key Concept: Computer Networks

b) Route internet traffic between different networks
[Solution Description]
A router is a networking device that forwards data packets between computer networks. Its main function is to connect multiple networks and route traffic between them based on IP addresses. Unlike switches which work at the data link layer (Layer 2), routers operate at the network layer (Layer 3) and use IP addresses to determine where to forward packets.

Key Concept: Compression and Rarefaction

c) Compressions
[Solution Description]
In a longitudinal wave, compression refers to regions where particles are closer together, resulting in high pressure, while rarefaction refers to regions where particles are farther apart, resulting in low pressure.

Key Concept: Loudness and Amplitude, Safe Limit

c) 9 times
[Solution Description]
Given that loudness ($L$) is directly proportional to the square of the amplitude ($a$), we have:
$L \propto a^2$
If the initial amplitude is $a$, the new amplitude becomes $3a$. The new loudness ($L'$) will be:
$L' \propto (3a)^2 = 9a^2$
Thus, the loudness increases by 9 times.

Key Concept: Quality, Wave Form, Timbre

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that a tuning fork and a piano playing the same note with identical pitch and loudness will produce indistinguishable sounds. This is false because quality (timbre) depends on the wave form, which differs between sources even if pitch and loudness are the same. The reason correctly explains that the tuning fork has a single-frequency wave form, while the piano has multiple frequencies, leading to different timbres. Thus, the sounds are distinguishable. Therefore, Assertion is false, but Reason is true.

Key Concept: Effect of Tension on Pitch

c) Increases the pitch
[Solution Description]
Increasing the tension of the rubber bands raises the frequency of vibration, resulting in a higher pitch sound.

Key Concept: Numerical Calculation

c) 25 Hz
[Solution Description]
Given the relationship:
$f = \frac{1}{T}$
Substituting the given time period:
$f = \frac{1}{0.04} = 25 \, \text{Hz}$

Key Concept: Longitudinal Waves, Waveform Representation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the graphical representation of a longitudinal wave shows the variation of particle displacement with distance or time. This is true because longitudinal waves can indeed be represented graphically to depict how particles are displaced as the wave propagates.
The reason states that in longitudinal waves, particles vibrate parallel to the direction of wave propagation, and their displacements can be plotted against position or time. This is also true and correctly explains why the graphical representation of longitudinal waves is possible. Since particles move parallel to the wave's direction, their displacements can be mapped along the same axis as the wave's propagation, allowing for a clear graphical depiction.
Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Speed of Sound in Different Media

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the speed of sound is higher in solids than in gases due to the close packing of particles in solids. This is correct as per the syllabus, where the speed of sound in iron (a solid) is given as $5000 \, \text{m/s}$, while in air (a gas) it is $330 \, \text{m/s}$. The reason explains that sound travels faster when particles are closer, which aligns with the provided data. Since the reason correctly explains why sound travels faster in solids (due to closer particle spacing leading to quicker energy transfer), both statements are true and the reason is the correct explanation of the assertion.

Key Concept: Pitch, Frequency

c) It has twice the pitch
[Solution Description]
Pitch depends on the frequency of the sound wave. The tuning fork with frequency 512 Hz has double the frequency of the tuning fork with 256 Hz. Since higher frequency corresponds to higher pitch, the second tuning fork will produce a sound with twice the pitch of the first one.

Key Concept: Safety and Materials

c) To prevent injuries from sharp edges
[Solution Description]
The syllabus includes a safety tip advising careful use of scissors to avoid injuries. Scissors are sharp objects that can cause cuts or other injuries if not handled properly during the cutting process.

Key Concept: Assertion-Reason

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion states that the wavelength (\$\lambda\$) of a wave is the distance between two consecutive compressions in a longitudinal wave, which is true according to the syllabus. The Reason states that wavelength is the distance traveled by the wave in one time period of vibration of the particle of the medium, which is also true. However, the Reason does not explain why the wavelength is the distance between two consecutive compressions in a longitudinal wave; it provides a general definition of wavelength. Therefore, both statements are true, but the Reason is not the correct explanation of the Assertion.

Key Concept: Vibrating Body Example

c) Because the larger vibrating area produces more energy
[Solution Description]
According to the syllabus, the loudness of sound depends on the surface area of the vibrating body. A bell with a bigger case has a larger vibrating surface area, which results in a louder sound compared to a smaller case.
Hence, the correct answer is the option explaining that the larger vibrating area produces more energy.

Key Concept: Relationship Between Loudness and Amplitude

c) It becomes one-fourth
[Solution Description]
The original amplitude is 5 cm, and the new amplitude is 2.5 cm. The ratio of amplitudes is $\frac{2.5}{5} = 0.5$. Since loudness is proportional to the square of the amplitude ($L \propto a^2$), the loudness changes by $(0.5)^2 = 0.25$ times, i.e., it becomes one-fourth of the original loudness.

Key Concept: Subjective vs objective measures

c) Intensity is a physical measurement, and loudness is a perceptual experience.
[Solution Description]
Intensity is an objective physical quantity measured in watts per square meter, while loudness is a subjective perception influenced by the listener's ear sensitivity. Even if intensity remains unchanged, loudness can vary among listeners.

Key Concept: Propagation of Sound in Air, Wavelength, and Frequency

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When sound travels from air to water, its speed increases due to the change in the medium's properties. However, the frequency of the sound wave remains constant as it is determined by the source and not the medium. Since $v = f\lambda$, where $v$ is the wave speed, $f$ is the frequency, and $\lambda$ is the wavelength, an increase in $v$ (while $f$ remains constant) leads to an increase in $\lambda$. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Speed of Sound in Different Media, Distance Measurement

b) 3000 meters
[Solution Description]
The total time taken for the pulse to travel to the object and back is 4 seconds. Therefore, the one-way time is $t = \frac{4}{2} = 2 \, \text{s}$. Using the formula for distance, $d = v \times t$, where $v = 1500 \, \text{m/s}$ and $t = 2 \, \text{s}$, we get $d = 1500 \times 2 = 3000 \, \text{m}$.

Key Concept: Tuning Fork Frequency

d) 384 Hz
[Solution Description]
The frequency marked on the tuning fork indicates the exact frequency at which it vibrates when struck. Thus, a tuning fork marked $384 \, \text{s}^{-1}$ will produce sound of frequency $384 \, \text{Hz}$.

Key Concept: Assertion-Reason

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The loudness of a sound wave is related to the square of its amplitude as per the formula \$L \propto a^2\$. However, the decibel scale is logarithmic, where an increase in loudness by a factor of 10 corresponds to a 10 dB increase.
Step-by-step solution:
1. Let the original amplitude be \$a\$ and new amplitude be \$3a\$.
2. Original loudness \$L_1 \propto a^2\$, and new loudness \$L_2 \propto (3a)^2 = 9a^2\$.
3. The ratio of new loudness to original loudness is \$L_2/L_1 = 9\$.
4. Using the decibel scale formula, the change in loudness level is calculated as:
$\Delta L = 10 \log_{10}\left(\frac{L_2}{L_1}\right) = 10 \log_{10}(9) \approx 9.54 \, \text{dB}$
5. Thus, the loudness increases by approximately 9.54 dB, not exactly 9 dB.
Therefore, the Assertion is approximately true but not exact, while the Reason correctly describes the logarithmic relationship between loudness in decibels and amplitude.

Key Concept: Loudness and Amplitude

c) Becomes four times
[Solution Description]
Loudness ($L$) is directly proportional to the square of the amplitude ($a$), i.e., $L \propto a^2$. If the amplitude doubles ($2a$), then the loudness becomes $(2a)^2 = 4a^2$, which is four times the original loudness.

Key Concept: Speed of Sound

c) Iron
[Solution Description]
The speed of sound depends on the medium. It is highest in solids (\textasciitilde5000 m/s), followed by liquids (\textasciitilde1500 m/s), and slowest in gases (\textasciitilde330 m/s). Among the given options, iron (a solid) has the highest speed.

Key Concept: Pitch, Frequency

c) The 400 Hz sound has a higher pitch but the same loudness.
[Solution Description]
Pitch depends on frequency, while loudness depends on amplitude. Since the amplitudes are the same, the loudness is the same for both sounds. However, the second sound has twice the frequency of the first, so it has a higher pitch.

Key Concept: Source of Monotone Sound

c) Tuning fork
[Solution Description]
A tuning fork is the only source that produces a monotone sound, i.e., a sound of single frequency. Other devices like musical instruments or speakers generally produce sounds with multiple frequencies.

Key Concept: Wave Form, Frequency Spectrum, Sound Composition

c) Presence of resonant harmonic frequencies
[Solution Description]
A single note on a guitar includes the fundamental frequency and its harmonics (e.g., 2f, 3f), creating a complex waveform. These appear as peaks in frequency analysis.

Key Concept: Pitch of String Instruments

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that plucking a guitar string closer to one fixed end increases the pitch of the sound produced. This is true because shortening the vibrating segment by plucking near the end increases the frequency (pitch). The Reason claims that the pitch depends on the plucking position because it affects the length of the vibrating segment, which is correct and directly explains the Assertion. Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Pitch in Musical Instruments

c) Increasing the tension in the string.
[Solution Description]
For stringed instruments like the guitar, increasing the tension or using a thinner string raises the pitch. Plucking closer to one fixed end also increases pitch as it reduces the effective vibrating length of the string.

Key Concept: Displacement-Distance Graph

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that in a displacement-distance graph, the amplitude is represented by the maximum displacement of particles from their mean positions. This is correct as per the syllabus definition of amplitude. The reason claims that the amplitude is the distance between two consecutive points in the same phase. However, this is incorrect because the distance between two consecutive points in the same phase defines the wavelength (\lambda), not the amplitude. Therefore, the assertion is true, but the reason is false.

Key Concept: Ultrasonic Cleaning, Non-Destructive Testing

c) They can penetrate materials and reflect off small particles
[Solution Description]
High-frequency sound waves (ultrasound) have short wavelengths, which allow them to interact with small particles or detect tiny flaws in materials. Their ability to penetrate materials and reflect off surfaces makes them ideal for cleaning delicate objects and detecting internal defects without causing damage.

Key Concept: Decibel Scale

b) 20 dB
[Solution Description]
From the syllabus, when loudness increases 100 times, the sound level increases by 20 dB. This follows the logarithmic relationship where every 10-fold increase in loudness corresponds to a 10 dB increase.

Key Concept: Amplitude of the Wave

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that when a guitar string is plucked harder, the amplitude of its vibration increases. This is true because applying more force results in greater displacement of the string. The Reason states that loudness is directly proportional to the square of the amplitude of the wave, which is also true and correctly explains the Assertion. Therefore, both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Frequency and Time Period Relationship

d) 50 Hz
[Solution Description]
The relationship between frequency ($f$) and time period ($T$) is given by:
$f = \frac{1}{T}$
Given $T = 0.02$ s, substituting the value:
$f = \frac{1}{0.02} = 50 \, \text{Hz}$
So, the frequency is 50 Hz.

Key Concept: Longitudinal waves, Compression-Rarefaction

b) Audible sound
[Solution Description]
Distance between two compressions = wavelength ($\lambda$) = 0.85 m
Time period (T) = Time between two compressions = 0.0025 s
Frequency ($f$) = $\frac{1}{T} = \frac{1}{0.0025} = 400 \text{ Hz}$
Human hearing range is 20Hz-20,000Hz. Since 400Hz falls within this range, it's audible.
Also, speed = $f \times \lambda = 400 \times 0.85 = 340 \text{ m/s}$ which matches normal speed of sound in air.

Key Concept: Relation to Amplitude

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the syllabus, the loudness ($L$) of sound is directly proportional to the square of the amplitude ($a$) of the wave: $L \propto a^2$. If the amplitude is doubled ($2a$), then the loudness becomes $L \propto (2a)^2 = 4a^2$, which means it increases by a factor of four. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Safe Hearing Limit

c) 0 dB to 80 dB
[Solution Description]
The safe hearing range for humans is from 0 dB to 80 dB. Exposure to sounds above this range can cause discomfort or harm.

Key Concept: Speed comparison

b) $1500 \, \text{m/s}$
[Solution Description]
According to the provided information, the speed of sound in water is nearly $1500 \, \text{m/s}$. This is greater than the speed of sound in air ($330 \, \text{m/s}$) but less than that in solids like iron ($5000 \, \text{m/s}$). Vacuum does not allow sound propagation.

Key Concept: Displacement-distance graph, wavelength

b) 0.1 s
[Solution Description]
Distance between two consecutive crests = λ = 1.5 - 0.5 = 1 m
Given wave speed v = 10 m/s
Formula: v = f × λ → f = v/λ = 10/1 = 10 Hz
Time period T = 1/f = 1/10 = 0.1 s

Key Concept: Amplitude Ratio and Loudness

c) 1:9
[Solution Description]
Given the amplitude ratio of two waves as 1:3, the loudness ratio can be calculated using the relation $L \propto a^2$:
$\frac{L_1}{L_2} = \left(\frac{a_1}{a_2}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$
Thus, the loudness ratio is 1:9.

Key Concept: Pitch in Wind Instruments, Membrane Instruments

c) The flute's pitch decreases, and the tabla's pitch increases.
[Solution Description]
Closing holes in a flute increases the effective air column length, lowering the frequency (grave sound). Tightening a tabla membrane increases tension, raising its frequency (shrill sound). Thus, the flute's pitch decreases, and the tabla's pitch increases.

Key Concept: Wavelength

b) $2$ m
[Solution Description]
The wavelength ($\lambda$) is defined as the distance traveled by the wave in one time period ($T$). Here, the wave travels $2$ m in one time period, so the wavelength is $2$ m.

Key Concept: Displacement-Time Graph, Instantaneous Displacement, Velocity

a) $6\pi$ cm/s
[Solution Description]
The velocity $v$ is the derivative of displacement $y$ with respect to time $t$:
$v = \frac{dy}{dt} = 3 \times 4\pi \cos(4\pi t) = 12\pi \cos(4\pi t) \text{ cm/s}$
At $t = \frac{1}{12}$ s:
$v = 12\pi \cos\left(4\pi \times \frac{1}{12}\right) = 12\pi \cos\left(\frac{\pi}{3}\right) = 12\pi \times \frac{1}{2} = 6\pi \text{ cm/s}$

Key Concept: Amplitude and Loudness Relationship

c) The loudness becomes four times.
[Solution Description]
The loudness of sound is directly proportional to the square of the amplitude of the wave ($L \propto a^2$). If the amplitude is doubled, the loudness becomes $(2)^2 = 4$ times the original loudness.

Key Concept: Displacement-Time Graph, Amplitude, Time Period

b) $\frac{5\pi}{4}$ cm/s
[Solution Description]
To find the maximum speed, we first calculate the angular frequency $\omega$ using the formula:
$\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \text{ rad/s}$
The maximum speed $v_{\text{max}}$ is given by:
$v_{\text{max}} = a\omega = 5 \times \frac{\pi}{4} = \frac{5\pi}{4} \text{ cm/s}$

Key Concept: Pitch and Frequency, Wave Representation

d) 1600 Hz
[Solution Description]
First, we know that frequency ($f$) and time period ($T$) are related by the formula:
$f = \frac{1}{T}$
Given the first wave's frequency is $f_1 = 800$ Hz, its time period is:
$T_1 = \frac{1}{f_1} = \frac{1}{800} \, \text{s}$
The second wave's time period is half of $T_1$:
$T_2 = \frac{T_1}{2} = \frac{1}{1600} \, \text{s}$
Now, the frequency of the second wave is:
$f_2 = \frac{1}{T_2} = 1600 \, \text{Hz}$

Key Concept: Definition of Wavelength

a) Distance between two consecutive crests or troughs
[Solution Description]
The wavelength (\$\lambda\$) is defined as the distance traveled by a wave in one time period of vibration of a particle in the medium.

Key Concept: Safety Consideration, Infrasound

b) Not audible but may cause discomfort
[Solution Description]
Infrasound refers to sounds with frequencies below 20 Hz, which are generally not clearly audible to humans. However, such low-frequency sounds can cause discomfort or vibrations in the body due to their long wavelengths and high energy.

Key Concept: Frequency, Time Period, Wavelength

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The wave speed ($v$) is related to frequency ($f$) and wavelength ($\lambda$) by the equation $v = f \lambda$. When the speed ($v$) is constant, if frequency ($f$) is doubled, wavelength ($\lambda$) must become half to maintain the equality $v = f \lambda$, as $\lambda$ is inversely proportional to $f$ for a constant $v$. Thus, the Assertion is true. The Reason correctly states the relationship $v = f \lambda$, which explains why the Assertion holds true. Therefore, both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Frequency and Pitch Relation

d) Higher frequency means higher pitch.
[Solution Description]
Higher frequency corresponds to higher pitch (shrill sound), while lower frequency corresponds to lower pitch (flat sound).

Key Concept: Wavelength, Time Period, Frequency

b) 0.00588 s
[Solution Description]
The speed of the wave ($v$) is related to its wavelength ($\lambda$) and frequency ($f$) by $v = f\lambda$. Given $v = 340$ m/s and $\lambda = 2$ m, the frequency is $f = \frac{v}{\lambda} = \frac{340}{2} = 170$ Hz. The time period ($T$) is the inverse of frequency: $T = \frac{1}{f} = \frac{1}{170} \approx 0.00588$ s or 5.88 ms.

Key Concept: Displacement-Time Graph - Frequency

b) 0.5 Hz
[Solution Description]
Frequency ($f$) is calculated as the number of vibrations per second. Given 5 vibrations in 10 seconds:
$f = \frac{\text{Number of vibrations}}{\text{Time}} = \frac{5}{10} = 0.5 \text{ Hz}$
Thus, the frequency is $0.5$ Hz.

Key Concept: Decibel Scale, Loudness Levels

c) 100 times
[Solution Description]
The decibel scale is logarithmic. An increase of 10 dB corresponds to a 10-fold increase in loudness. Here, the increase is from 10 dB to 30 dB, which is an increase of 20 dB. A 20 dB increase implies:
$20 \, \text{dB} = 10 \, \text{dB} + 10 \, \text{dB} \Rightarrow 10 \times 10 = 100 \, \text{times}$
Therefore, the loudness increases by 100 times.

Key Concept: Pitch and Frequency

b) High frequency and high pitch
[Solution Description]
According to the provided syllabus, a string under high tension and of small length vibrates with a high frequency, producing a sound of high pitch.

Key Concept: Displacement-Time Graph

c) It gets doubled
[Solution Description]
The amplitude represents the maximum displacement of the particle from its mean position. If the amplitude is doubled, the maximum displacement also doubles. For example, if the initial amplitude was $2$ cm, doubling it would make it $4$ cm.

Key Concept: Longitudinal Wave

c) Longitudinal wave
[Solution Description]
Sound propagates as a longitudinal wave in air, where particles of the medium vibrate parallel to the direction of wave propagation. This results in compressions and rarefactions.

Key Concept: Wind Instruments

b) Closing some more holes
[Solution Description]
To produce a lower pitch in wind instruments, the length of the vibrating air column must be increased. This is achieved by closing more holes, which increases the effective length of the air column, thereby reducing the frequency and pitch of the sound.