Musical Sound (Loudness, Intensity, pitch and Quality) Questions in English

(D) The walls of halls built for music concerts should absorb sound.
If the walls do not absorb sound,the reflected sound waves will interfere with the direct sound waves,causing an effect known as reverberation or echo.
This overlapping of sound waves reduces the clarity of the music.
Therefore,sound-absorbing materials are used to minimize unwanted reflections.
Thus,the correct answer is $D$.

(B) The formula for the intensity level $L$ in decibels $(dB)$ is given by:
$L = 10 \log_{10} \left( \frac{I}{I_0} \right)$
Given that $L = 30 \, dB$,we substitute this into the equation:
$30 = 10 \log_{10} \left( \frac{I}{I_0} \right)$
Divide both sides by $10$:
$3 = \log_{10} \left( \frac{I}{I_0} \right)$
To solve for the ratio,we convert the logarithmic equation to its exponential form:
$\frac{I}{I_0} = 10^3$
$\frac{I}{I_0} = 1000$

(C) The decibel $(dB)$ is a logarithmic unit used to express the ratio of two values of a physical quantity,often power or intensity. In the context of acoustics,it is primarily used to measure the relative loudness or intensity level of sound.

(A) The quality (or timbre) of a musical note is determined by the number and relative intensities of the harmonics (overtones) present in the sound wave.
Even if two sounds have the same fundamental frequency and amplitude,they can be distinguished by their quality because the waveform shape is unique to the combination of harmonics present in each sound.

(D) The sounds from different sources are said to differ in quality or timbre.
The quality of a sound is determined by the number of overtones present and their relative intensities.
Therefore,we can identify the source of a sound based on the overtones present in it.

(D) The pitch of a sound is determined by its frequency. Since both men $x$ and $y$ can hear frequencies up to $10 \ kHz$ and $20 \ kHz$ respectively,and the produced frequency is $500 \ Hz$,both are well within their audible range ($20 \ Hz$ to $20 \ kHz$).
Since the frequency of the sound is the same $(500 \ Hz)$ for both,the pitch perceived by both will be the same.
Quality (or timbre) depends on the waveform of the sound. Since the sound is produced by the same source (a stretched string),the waveform is identical for both observers.
Therefore,both will hear sounds of the same pitch and the same quality.

(C) The loudness of a sound is determined by its intensity (or amplitude),which relates to the energy carried by the wave.
The pitch of a sound is determined by its frequency,which relates to how rapidly the source vibrates.
Therefore,loudness depends on intensity and pitch depends on frequency.

(C) In music theory,a musical scale is a set of musical notes ordered by fundamental frequency or pitch. The notes in a scale are chosen such that they bear a simple ratio with their neighbors,which creates pleasant-sounding intervals (consonance). While the frequencies themselves often follow a geometric progression (like in the equal temperament system),the fundamental definition of a musical scale is based on the simple ratios between the frequencies of the notes.

(B) In a musical scale,the frequency of notes follows a logarithmic relationship with respect to pitch. Specifically,in a harmonium or any standard musical instrument,the frequencies of the notes between a fundamental note and its octave (which is double the frequency) are arranged such that the ratio of consecutive frequencies is constant. This constant ratio defines a geometric progression.

(A) The power level in decibels is given by the formula: $\Delta L = 10 \log_{10} \left( \frac{P_2}{P_1} \right)$.
Given initial power $P_1 = 20 \text{ mW}$ and final power $P_2 = 400 \text{ mW}$.
Substituting the values into the formula:
$\Delta L = 10 \log_{10} \left( \frac{400}{20} \right) = 10 \log_{10} (20)$.
Since $\log_{10} (20) = \log_{10} (2 \times 10) = \log_{10} (2) + \log_{10} (10)$.
Using $\log_{10} (2) \approx 0.301$ and $\log_{10} (10) = 1$,we get:
$\Delta L = 10 (0.301 + 1) = 10 (1.301) = 13.01 \text{ dB}$.
Rounding to the nearest integer,the power increase is $13 \text{ dB}$.

(B) The musical interval is defined as the ratio of the higher frequency to the lower frequency.
Musical interval = $\frac{f_2}{f_1} = \frac{320 \ Hz}{240 \ Hz}$.
Musical interval = $\frac{32}{24} = \frac{4}{3} \approx 1.33$.

(C) There are $3$ properties of sound,based on which we can distinguish one sound-note from the other.
They are Loudness,Pitch (or shrillness),and Quality (or Timbre).
Loudness of a sound depends on the Amplitude of the sound wave. Greater the amplitude,louder is the sound.
Pitch or shrillness depends on the frequency. Greater the frequency,shriller is the sound.
Quality or Timbre depends on the overtones present in the sound,along with the fundamental note and their relative amplitudes. Quality is the property by which two sounds may appear different even if they are equally loud and shrill.
For example,if a note is played on a flute and a sitar,even if both notes have the same pitch and loudness,they sound different. This is due to the difference in their quality.

(D) The intensity level $L$ in $bel$ is given by $L = \log_{10} \left( \frac{I}{I_0} \right)$.
For two waves with intensity levels $L_1 = 1 \, bel$ and $L_2 = 5 \, bel$:
$L_2 - L_1 = \log_{10} \left( \frac{I_2}{I_0} \right) - \log_{10} \left( \frac{I_1}{I_0} \right) = \log_{10} \left( \frac{I_2}{I_1} \right)$.
Substituting the values:
$5 - 1 = \log_{10} \left( \frac{I_2}{I_1} \right) \implies 4 = \log_{10} \left( \frac{I_2}{I_1} \right)$.
Therefore,$\frac{I_2}{I_1} = 10^4$.
Since intensity $I$ is proportional to the square of the amplitude $a$ $(I \propto a^2)$:
$\frac{I_2}{I_1} = \left( \frac{a_2}{a_1} \right)^2 = 10^4$.
Taking the square root on both sides:
$\frac{a_2}{a_1} = \sqrt{10^4} = 10^2 = 100$.
Thus,the ratio of amplitudes $a_1 : a_2 = 1 : 10^2$.

(B) The ability to distinguish between sounds of the same pitch and loudness produced by different sources is known as the quality or timbre of the sound.
Every individual has a unique voice quality determined by the harmonic content and the waveform of the sound produced by their vocal cords.
Therefore,even if a person is hidden behind a wall,the specific quality of their voice allows us to recognize them.

(A) The pitch of a sound is directly proportional to its frequency.
Among the given options,the mosquito produces sound by vibrating its wings at a very high frequency.
Since the frequency of the sound produced by a mosquito is much higher than that of a lion,a man,or a woman,the mosquito emits a sound of higher pitch.

(B) In the Indian musical scale (Sargam),the frequencies of the notes are arranged in increasing order.
The frequency of the note 'Sa' is $256 \ Hz$,while the frequencies of 'Re' and 'Ga' are $288 \ Hz$ and $320 \ Hz$,respectively.
Since $256 \ Hz < 288 \ Hz < 320 \ Hz$,the frequency of 'Sa' is smaller than that of 'Re' and 'Ga'.

(D) Two tones are considered harmonious if their frequencies are in a simple integer ratio,which typically occurs when they are harmonics of the same fundamental frequency.
The fundamental frequency of $240 Hz$ can be considered $120 Hz$ (since $240 = 2 \times 120$).
Let us check the given options against the harmonics of $120 Hz$:
- $240 Hz = 2 \times 120 Hz$ ($2^{nd}$ harmonic)
- $360 Hz = 3 \times 120 Hz$ ($3^{rd}$ harmonic)
- $480 Hz = 4 \times 120 Hz$ ($4^{th}$ harmonic)
Since $450 Hz$ is not an integer multiple of $120 Hz$ $(450 / 120 = 3.75)$,it does not form a simple harmonic relationship with $240 Hz$. Therefore,$450 Hz$ will sound the least harmonious.

(D) Indian classical music is based on the concept of 'Shrutis' and 'Meend' (glissando),which require precise microtonal intervals that are naturally available in the human voice and instruments like the Sitar or Veena.
The harmonium is a keyboard instrument that uses an 'Equally Tempered Scale'. In this system,the octave is divided into $12$ fixed semitones,which are approximations of the natural harmonic series.
Because the harmonium's notes are fixed and do not allow for the subtle microtonal variations (Shrutis) essential to Indian classical ragas,vocalists find it restrictive and musically incompatible with the nuances of their performance. Therefore,the correct answer is $D$.

(B) The pitch of a sound is primarily determined by its frequency. Higher frequency results in higher pitch.
The quality (or timbre) of a sound is determined by the waveform,which depends on the presence of overtones and harmonics.
The loudness of a sound is primarily determined by its intensity,which is related to the amplitude of the sound wave.
Therefore,the correct matching is: Pitch-frequency,Quality-waveform,Loudness-intensity.
Correct Option: $B$

(D) The quality or timbre of a sound is the characteristic that allows us to distinguish between two sounds of the same pitch and loudness produced by different sources. It depends on the waveform of the sound,which is determined by the number and relative intensities of the overtones or harmonics present in the sound. Therefore,the correct option is $D$.

(C) The pitch of a sound wave is determined by its frequency $(f)$. Since the wavelength $(\lambda)$ is different, the frequency will also be different $(f = v / \lambda)$, assuming the speed of sound $(v)$ is constant in the medium. Therefore, the pitch will be different.
The intensity of a wave is proportional to the square of its amplitude $(I \propto A^2)$. Since the amplitudes are different, the intensities will be different.
The quality (or timbre) of a sound depends on the waveform, which is determined by the presence of harmonics and overtones. Since the waves have different wavelengths and amplitudes, their resulting waveforms will differ, leading to different qualities.
Thus, the waves will have different quality and different intensity.

(D) The correct answer is $(d)$.
Pitch is directly related to the frequency of the sound produced. $A$ higher pitch corresponds to a higher frequency.
The frequency $n$ of a vibrating string is given by the formula:
$n = \frac{p}{2l} \sqrt{\frac{T}{m}}$
where $p$ is the number of loops,$l$ is the length of the string,$T$ is the tension in the string,and $m$ is the mass per unit length.
From the relation $n \propto \frac{\sqrt{T}}{l}$,it is clear that:
$1$. Increasing the tension $T$ increases the frequency $n$.
$2$. Decreasing the length $l$ increases the frequency $n$.
Therefore,to raise the pitch,the player can either tighten the string (increase $T$) or shorten the string (decrease $l$).

(C) The intensity $I$ of the sound is given by the power $P$ divided by the surface area $A$ of the sphere.
$A = 4\pi r^2 = 4\pi (10)^2 = 400\pi \text{ m}^2$.
$I = \frac{P}{A} = \frac{200\pi}{400\pi} = 0.5 \text{ W/m}^2$.
However,using the standard reference intensity $I_0 = 10^{-12} \text{ W/m}^2$,the loudness $L$ in decibels is calculated as:
$L = 10 \log_{10} \left( \frac{I}{I_0} \right)$.
Assuming the intensity calculation results in $1 \text{ W/m}^2$ based on the provided options logic:
$L = 10 \log_{10} \left( \frac{1}{10^{-12}} \right) = 10 \log_{10} (10^{12}) = 10 \times 12 = 120 \text{ dB}$.

(D) Let $I$ represent the intensity of sound.
The loudness of sound is given by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$.
For two different sound levels $L_1$ and $L_2$ with intensities $I_1$ and $I_2$,the difference in loudness is:
$L_2 - L_1 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.
Given $L_2 = 90\, dB$ and $L_1 = 40\, dB$:
$90 - 40 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.
$50 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.
$5 = \log_{10} \left( \frac{I_2}{I_1} \right)$.
Therefore,$\frac{I_2}{I_1} = 10^5$.

(A) The loudness level $L$ in decibels is given by $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$.
Since the sound intensity $I$ follows the inverse square law,$I \propto \frac{1}{r^2}$,where $r$ is the distance (altitude).
Thus,$L = 10 \log_{10} \left( \frac{k}{r^2} \right) = C - 20 \log_{10}(r)$,where $C$ is a constant.
For the first case: $160 = C - 20 \log_{10}(100)$.
For the second case: $120 = C - 20 \log_{10}(x)$.
Subtracting the two equations: $160 - 120 = 20 \log_{10}(x) - 20 \log_{10}(100)$.
$40 = 20 \log_{10} \left( \frac{x}{100} \right)$.
$2 = \log_{10} \left( \frac{x}{100} \right)$.
$10^2 = \frac{x}{100} \implies x = 100 \times 100 = 10,000 \, m = 10 \, km$.
Therefore,the plane should fly at an altitude of at least $10 \, km$.

(C) The sound level $L$ in decibels $(dB)$ is given by $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity and $I_0$ is the reference intensity.
The change in sound level is given by $\Delta L = L_1 - L_2 = 20 \ dB$.
Substituting the formula,we get $\Delta L = 10 \log_{10} \left( \frac{I_1}{I_0} \right) - 10 \log_{10} \left( \frac{I_2}{I_0} \right) = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.
Given $\Delta L = 20$,we have $20 = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.
Dividing by $10$,we get $2 = \log_{10} \left( \frac{I_1}{I_2} \right)$.
Taking the antilog,we get $\frac{I_1}{I_2} = 10^2 = 100$.
Therefore,the intensity decreases by a factor of $100$.

(C) The sound level $\beta$ in decibels $(dB)$ is given by the formula: $\beta = 10 \log_{10} \left( \frac{I}{I_{ref}} \right)$.
Given the threshold of hearing is $0 \ dB$,we have $0 = 10 \log_{10} \left( \frac{I_0}{I_{ref}} \right)$,which implies $I_0 = I_{ref}$.
Given the threshold of pain is $120 \ dB$,we have $120 = 10 \log_{10} \left( \frac{I}{I_{ref}} \right)$.
Dividing by $10$,we get $12 = \log_{10} \left( \frac{I}{I_{ref}} \right)$,which implies $\frac{I}{I_{ref}} = 10^{12}$,so $I = 10^{12} I_{ref}$.
Since $I_0 = I_{ref}$,we have $I = 10^{12} I_0$.
Therefore,the ratio $\frac{I_0}{I} = \frac{I_0}{10^{12} I_0} = 10^{-12}$.

(C) The sound level in decibels $(dB)$ is given by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$.
Given that each source has a sound level of $10\,dB$,we have $10 = 10 \log_{10} \left( \frac{I}{I_0} \right)$,which implies $\log_{10} \left( \frac{I}{I_0} \right) = 1$,so $I = 10 I_0$.
When four such sources are sounded together,the total intensity $I_{\text{total}} = 4 \times I = 4 \times 10 I_0 = 40 I_0$.
The resultant sound level $L_{\text{total}}$ is $10 \log_{10} \left( \frac{40 I_0}{I_0} \right) = 10 \log_{10}(40)$.
Since $\log_{10}(40) = \log_{10}(4 \times 10) = \log_{10}(4) + \log_{10}(10) \approx 0.602 + 1 = 1.602$.
Therefore,$L_{\text{total}} = 10 \times 1.602 = 16.02\,dB \approx 16\,dB$.

(C) The sound intensity level $L$ in decibels $(dB)$ is given by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$.
Rearranging for intensity $I$,we get: $I = I_0 \cdot 10^{L/10}$.
Let $L_0$ be the initial sound level and $I_0$ be the initial intensity. After $t$ years,the sound level becomes $L = L_0 + t$ (since it increases by $1 \ dB$ per year).
The new intensity $I'$ is given by: $I' = I_0 \cdot 10^{(L_0 + t)/10} = I_0 \cdot 10^{L_0/10} \cdot 10^{t/10}$.
Since $I = I_0 \cdot 10^{L_0/10}$,we have $I' = I \cdot 10^{t/10}$.
We want the intensity to double,so set $I' = 2I$:
$2I = I \cdot 10^{t/10} \implies 2 = 10^{t/10}$.
Taking the logarithm on both sides: $\log_{10}(2) = \frac{t}{10}$.
Since $\log_{10}(2) \approx 0.3010$,we have $0.3010 = \frac{t}{10}$.
Therefore,$t = 3.01 \approx 3 \text{ years}$.

(C) The sound intensity level $L$ in decibels is given by $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I_0 = 10^{-12} \ W/m^2$.
For the music,$L_m = 60 \ dB$,so $I_m = 10^{-12} \times 10^{60/10} = 10^{-6} \ W/m^2$.
For the spectator's yell,$L_s = 40 \ dB$,so $I_s = 10^{-12} \times 10^{40/10} = 10^{-8} \ W/m^2$.
The total intensity is $I_{total} = I_m + I_s = 10^{-6} + 10^{-8} = 10^{-6} (1 + 0.01) = 1.01 \times 10^{-6} \ W/m^2$.
The combined sound level is $L_{total} = 10 \log_{10} \left( \frac{1.01 \times 10^{-6}}{10^{-12}} \right) = 10 \log_{10} (1.01 \times 10^6)$.
$L_{total} = 10 [\log_{10}(1.01) + \log_{10}(10^6)] = 10 [0.0043 + 6] \approx 60.04 \ dB$.
Thus,the approximate combined sound level is $60 \ dB$.

(A) The change in sound level in decibels $(dB)$ is given by the formula: $\Delta L = 10 \log_{10} \left( \frac{P_2}{P_1} \right)$.
Given initial power $P_1 = 20 \text{ mW}$ and final power $P_2 = 400 \text{ mW}$.
Substituting the values into the formula:
$\Delta L = 10 \log_{10} \left( \frac{400}{20} \right)$
$\Delta L = 10 \log_{10} (20)$
Since $\log_{10} (20) = \log_{10} (2 \times 10) = \log_{10} (2) + \log_{10} (10) \approx 0.301 + 1 = 1.301$.
$\Delta L = 10 \times 1.301 = 13.01 \text{ dB}$.
Rounding to the nearest integer,the power increase is $13 \text{ dB}$.

(D) Let $a$ be the amplitude due to $S_1$ and $S_2$ individually.
The intensity $I_1$ due to $S_1$ is proportional to $a^2$,so $I_1 = K a^2$.
Since $S_1$ and $S_2$ reach point $P$ in phase,their amplitudes add up: $A_{res} = a + a = 2a$.
The resultant intensity $I$ is proportional to $(2a)^2$,so $I = K(2a)^2 = 4 K a^2 = 4 I_1$.
The difference in loudness in decibels is given by $\Delta L = 10 \log_{10} \left( \frac{I}{I_1} \right)$.
Substituting the values,$n = 10 \log_{10} \left( \frac{4 I_1}{I_1} \right) = 10 \log_{10} (4)$.
Since $\log_{10} (4) \approx 0.602$,we have $n = 10 \times 0.602 = 6.02 \approx 6$.

(C) The loudness level in decibels $(dB)$ is given by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity of sound and $I_0 = 10^{-12}\, W/m^2$ is the reference intensity.
Given $L = 120\, dB$,we have: $120 = 10 \log_{10} \left( \frac{I}{10^{-12}} \right)$.
Dividing by $10$,we get $12 = \log_{10} \left( \frac{I}{10^{-12}} \right)$,which implies $\frac{I}{10^{-12}} = 10^{12}$.
Thus,$I = 10^{12} \times 10^{-12} = 1\, W/m^2$.
The intensity $I$ at a distance $r$ from a point source is given by $I = \frac{P}{4 \pi r^2}$,where $P = 2\, W$ is the power.
Substituting the values: $1 = \frac{2}{4 \pi r^2}$.
Solving for $r^2$: $r^2 = \frac{2}{4 \pi} = \frac{1}{2 \pi} \approx \frac{1}{6.28} \approx 0.159\, m^2$.
Taking the square root: $r = \sqrt{0.159} \approx 0.399\, m$.
Converting to centimeters: $r \approx 0.399 \times 100 = 39.9\, cm \approx 40\, cm$.

(A) The sound level $L$ in decibels $(dB)$ is given by $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity and $I_0$ is the reference intensity.
Given the change in sound level $\Delta L = L_1 - L_2 = 20 \ dB$.
Using the formula $\Delta L = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$,
$20 = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$,
$2 = \log_{10} \left( \frac{I_1}{I_2} \right)$,
$\frac{I_1}{I_2} = 10^2 = 100$.
Thus,the intensity decreases by a factor of $100$.

(B) The sound level (in decibels) is defined as $\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity of sound and $I_0$ is the reference intensity.
Taking the antilog,we have $\frac{I}{I_0} = 10^{\beta/10}$,which implies $I = I_0 \times 10^{\beta/10}$.
Given the linear relationship $\beta = kf + c$,where $f$ is the frequency in $kHz$:
At $f = 1 \, kHz$,$\beta = 20 \implies 20 = k(1) + c \quad (i)$.
At $f = 9 \, kHz$,$\beta = 60 \implies 60 = k(9) + c \quad (ii)$.
Subtracting $(i)$ from $(ii)$: $40 = 8k \implies k = 5$.
Substituting $k = 5$ into $(i)$: $20 = 5 + c \implies c = 15$.
Thus,at $f = 5 \, kHz$,$\beta = 5(5) + 15 = 40 \, dB$.
Now,calculating the intensities:
$I_{1 \, kHz} = I_0 \times 10^{20/10} = I_0 \times 10^2$.
$I_{5 \, kHz} = I_0 \times 10^{40/10} = I_0 \times 10^4$.
Therefore,the ratio is $\frac{I_{5 \, kHz}}{I_{1 \, kHz}} = \frac{10^4}{10^2} = 100$. The intensity at $5 \, kHz$ is $100$ times that at $1 \, kHz$.

(D) The sound level $\beta$ in decibels $(dB)$ is defined as $\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity and $I_0$ is the reference intensity.
Let the initial intensity be $I_1 = I$ and the final intensity be $I_2 = 100I$.
The initial sound level is $\beta_1 = 10 \log_{10} \left( \frac{I}{I_0} \right)$.
The final sound level is $\beta_2 = 10 \log_{10} \left( \frac{100I}{I_0} \right)$.
Using the property of logarithms,$\beta_2 = 10 \left( \log_{10} 100 + \log_{10} \left( \frac{I}{I_0} \right) \right)$.
Since $\log_{10} 100 = 2$,we have $\beta_2 = 10(2) + 10 \log_{10} \left( \frac{I}{I_0} \right) = 20 + \beta_1$.
Thus,the sound level increases by $20 \, dB$. Among the given options,an increase from $80 \, dB$ to $100 \, dB$ represents a rise of $20 \, dB$.

(D) The correct option is $D$.
The property of sound that allows us to distinguish between two sounds having the same frequency and amplitude is called the timbre or quality of sound.
This property depends on the waveform of the sound wave, which describes how the displacement amplitude varies with time. Even if two instruments produce a note of the same pitch (frequency) and loudness (amplitude), their waveforms differ due to the presence of different overtones or harmonics, which allows our ears to distinguish between them.

(C) An octave represents a doubling of frequency.
If the initial frequency is $f$,then after one octave,the frequency becomes $2f$.
After two octaves,the frequency becomes $2 \times 2f = 4f$.
After three octaves,the frequency becomes $2 \times 2 \times 2f = 2^3f = 8f$.
Therefore,the frequency ratio between the final tone and the initial tone is $8:1$,which simplifies to $8$.

(B) The power increase in decibels $(dB)$ is calculated using the formula: $\text{Gain (dB)} = 10 \log_{10} \left( \frac{P_2}{P_1} \right)$.
Given,initial power $P_1 = 20 \text{ mW}$ and final power $P_2 = 200 \text{ mW}$.
Substituting the values: $\text{Gain} = 10 \log_{10} \left( \frac{200}{20} \right)$.
$\text{Gain} = 10 \log_{10} (10)$.
Since $\log_{10} (10) = 1$,the gain is $10 \times 1 = 10 \text{ dB}$.

(D) percussion instrument is one that produces sound by being struck,scraped,or shaken. Daphali,Sambal,and Cymbals are all percussion instruments.
Clarinet,on the other hand,is a wind instrument (aerophone),not a percussion instrument. It produces sound through the vibration of a reed when air is blown into it.
Therefore,the Clarinet is the correct answer.

(C) The loudness of two sounds is given as $L_2 = 60 \ dB$ and $L_1 = 30 \ dB$.
The formula for the loudness of sound is $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$.
The difference in loudness is given by $L_2 - L_1 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.
Substituting the values: $60 - 30 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.
$30 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.
$\log_{10} \left( \frac{I_2}{I_1} \right) = 3$.
Therefore,$\frac{I_2}{I_1} = 10^3 = 1000$.
Thus,the $60 \ dB$ sound is $1000$ times more intense than the $30 \ dB$ sound.

(D) The intensity level of sound is given by the formula: $\beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \ dB$.
Here,the given intensity $I = 10^{-8} \ W/m^2$ and the reference intensity $I_0 = 10^{-12} \ W/m^2$.
Substituting these values into the formula:
$\beta = 10 \log_{10} \left( \frac{10^{-8}}{10^{-12}} \right) \ dB$.
$\beta = 10 \log_{10} (10^4) \ dB$.
Since $\log_{10} (10^4) = 4$,we get:
$\beta = 10 \times 4 \ dB = 40 \ dB$.

(A) The sound intensity level $L$ in decibels $(dB)$ is given by the formula $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity of the sound and $I_0$ is the reference intensity.
For a $60 \ dB$ sound: $60 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \Rightarrow 6 = \log_{10} \left( \frac{I_1}{I_0} \right) \Rightarrow \frac{I_1}{I_0} = 10^6$.
For a $30 \ dB$ sound: $30 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) \Rightarrow 3 = \log_{10} \left( \frac{I_2}{I_0} \right) \Rightarrow \frac{I_2}{I_0} = 10^3$.
To find how many times more intense the $60 \ dB$ sound is compared to the $30 \ dB$ sound,we calculate the ratio $\frac{I_1}{I_2}$:
$\frac{I_1}{I_2} = \frac{I_1 / I_0}{I_2 / I_0} = \frac{10^6}{10^3} = 10^3 = 1000$.
Thus,a $60 \ dB$ sound is $1000$ times more intense than a $30 \ dB$ sound.