Sound Waves (Longitudinal wave) and it’s Characteristics (Speed etc.) Questions in English

(B) The Mach number is a dimensionless quantity representing the ratio of the speed of an object to the speed of sound in the surrounding medium.
Mathematically,$\text{Mach number} = \frac{\text{Velocity of object}}{\text{Velocity of sound}}$.
Therefore,one Mach number corresponds to the velocity of sound in that medium,which is approximately $332\,m/s$ in air at standard conditions.

(B) The speed of sound $v$ in a gas is given by the formula: $v = \sqrt{\frac{\gamma P}{\rho}}$,where $\gamma$ is the adiabatic index,$P$ is the pressure,and $\rho$ is the density.
Since both gases are diatomic,the adiabatic index $\gamma$ is the same for both.
Given that the temperature and pressure $P$ are constant,the ratio of the speeds $v_1$ and $v_2$ is:
$\frac{v_1}{v_2} = \sqrt{\frac{\gamma P / d_1}{\gamma P / d_2}} = \sqrt{\frac{d_2}{d_1}}$.

(D) The velocity of a sound wave in a gas is given by the formula $v = \sqrt{\frac{\gamma P}{\rho}}$.
For a monoatomic gas,the adiabatic index $\gamma = \frac{5}{3}$.
Given: Pressure $P = 1 \, kPa = 10^3 \, Pa$,Density $\rho = 2.6 \, kg/m^3$.
Substituting these values into the formula:
$v = \sqrt{\frac{\frac{5}{3} \times 10^3}{2.6}}$
$v = \sqrt{\frac{1666.67}{2.6}}$
$v = \sqrt{641.02} \approx 25.3 \, m/s$.
Since $25.3 \, m/s$ is not among the options $A$,$B$,or $C$,the correct option is $D$.

(D) Sound is a mechanical wave,which requires a material medium for its propagation. Therefore,sound cannot travel through a vacuum. Since sound cannot travel in a vacuum at all,the statement that it travels faster in a vacuum than in air is incorrect.

(A) The velocity of a sound wave in a medium is given by the formula $v = \sqrt{\frac{K}{\rho}}$,where $K$ is the bulk modulus and $\rho$ is the density of the medium.
The classification of sound waves into ultrasonic,infrasonic,and audible ranges is based solely on their frequency (or wavelength).
The speed of a sound wave depends only on the physical properties of the medium (such as elasticity and density) and is independent of the frequency of the wave.
Therefore,in a given medium,the speeds of ultrasonic,infrasonic,and audible waves are nearly equal,i.e.,$V_u \approx V_i \approx V_a$.

(D) The relationship between wave velocity $(v)$,frequency $(n)$,and wavelength $(\lambda)$ is given by the formula: $v = n\lambda$.
Given:
Frequency $(n)$ = $256\, Hz$ (vibrations per second).
Velocity of sound $(v)$ = $330\, m/s$.
Rearranging the formula to solve for wavelength $(\lambda)$:
$\lambda = \frac{v}{n}$
Substituting the values:
$\lambda = \frac{330}{256} \approx 1.289\, m$.
Rounding to two decimal places,we get $\lambda = 1.29\, m$.

(D) The sound of the whistle takes a finite amount of time to travel from the source to the man.
Given distance $d = 2\, km = 2000\, m$.
Speed of sound $v = 330\, m/s$.
The time taken by the sound to reach the man is $t = \frac{d}{v}$.
$t = \frac{2000}{330} \approx 6.06\, s \approx 6\, s$.
Since the sound takes $6\, s$ to reach the man,he will hear the whistle $6\, s$ after it is blown. Consequently,he will set his watch $6\, s$ late,meaning his watch will be $6\, s$ slow.

(A) The maximum velocity $(v_{\max})$ of a particle in simple harmonic motion is given by the formula $v_{\max} = a\omega$,where $a$ is the amplitude and $\omega$ is the angular frequency.
Given:
Amplitude $a = 0.1 \text{ cm}$
Frequency $n = 300 \text{ Hz}$
Angular frequency $\omega = 2\pi n = 2 \times \pi \times 300 = 600\pi \text{ rad/s}$.
Substituting these values into the formula:
$v_{\max} = 0.1 \times 600\pi = 60\pi \text{ cm/s}$.
Therefore,the maximum velocity of the particle is $60\pi \text{ cm/s}$.

(C) The audible range of frequency for human beings is typically between $20 \ Hz$ and $20,000 \ Hz$ $(20 \ kHz)$.
Comparing the given options:
$A) 5 \ Hz$ is infrasonic (below $20 \ Hz$).
$B) 27,000 \ Hz$ is ultrasonic (above $20,000 \ Hz$).
$C) 5,000 \ Hz$ falls within the audible range $(20 \ Hz - 20,000 \ Hz)$.
$D) 50,000 \ Hz$ is ultrasonic (above $20,000 \ Hz$).
Therefore,the correct option is $C$.

(A) The wavelength $\lambda$ is given by the formula $\lambda = \frac{v}{n}$,where $v$ is the speed of sound and $n$ is the frequency.
Ultrasonic waves have frequencies $n$ greater than $20,000 \, Hz$. Let us take $n \approx 50,000 \, Hz$.
The speed of sound in air is $v \approx 330 \, m/s$.
Substituting these values: $\lambda = \frac{330}{50,000} \, m = 0.0066 \, m = 0.66 \, cm$.
Converting to scientific notation,$0.66 \, cm = 6.6 \times 10^{-1} \, cm$. However,considering the order of magnitude for typical ultrasonic waves used in experiments,the value is approximately $5 \times 10^{-5} \, m$ or $5 \times 10^{-3} \, cm$. Given the options provided,the order of magnitude $5 \times 10^{-5} \, cm$ is the most appropriate choice.

(A) The relationship between wave speed $(v)$,frequency $(n)$,and wavelength $(\lambda)$ is given by the formula: $v = n \lambda$.
Rearranging for wavelength,we get: $\lambda = \frac{v}{n}$.
Given:
Frequency $(n)$ = $4.2 \, MHz = 4.2 \times 10^6 \, Hz$.
Speed of sound $(v)$ = $1.7 \, km/s = 1.7 \times 10^3 \, m/s$.
Substituting the values:
$\lambda = \frac{1.7 \times 10^3}{4.2 \times 10^6} = \frac{1.7}{4.2} \times 10^{-3} \approx 0.4047 \times 10^{-3} \, m = 4.047 \times 10^{-4} \, m$.
Rounding to the nearest value,we get $\lambda \approx 4 \times 10^{-4} \, m$.

(D) The audible frequency range for a human ear is $20 \text{ Hz}$ to $20,000 \text{ Hz}$.
Taking the speed of sound in air at room temperature as $v \approx 340 \text{ m/s}$.
The wavelength $\lambda$ is given by the formula $\lambda = \frac{v}{f}$.
To find the minimum wavelength $\lambda_{\min}$,we use the maximum frequency $f_{\max} = 20,000 \text{ Hz}$.
$\lambda_{\min} = \frac{340}{20,000} \text{ m} = 0.017 \text{ m} = 17 \text{ mm}$.
Rounding to the nearest provided option,the value is approximately $20 \text{ mm}$.

(C) The speed of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the universal gas constant,$T$ is the temperature,and $M$ is the molar mass of the gas.
Since the temperature $T$ is the same for both gases,the ratio of the speeds is given by $\frac{v_{N_2}}{v_{He}} = \sqrt{\frac{\gamma_{N_2}}{\gamma_{He}} \times \frac{M_{He}}{M_{N_2}}}$.
For nitrogen $(N_2)$,which is a diatomic gas,$\gamma_{N_2} = 1.4 = 7/5$ and the molar mass $M_{N_2} = 28 \ g/mol$.
For helium $(He)$,which is a monatomic gas,$\gamma_{He} = 1.67 = 5/3$ and the molar mass $M_{He} = 4 \ g/mol$.
Substituting these values: $\frac{v_{N_2}}{v_{He}} = \sqrt{\frac{7/5}{5/3} \times \frac{4}{28}} = \sqrt{\frac{7}{5} \times \frac{3}{5} \times \frac{1}{7}} = \sqrt{\frac{3}{25}} = \frac{\sqrt{3}}{5}$.

(C) Given:
Frequency $(n)$ = $200 Hz$
Velocity of sound $(v)$ = $340 m/s$
We know the relationship between velocity,frequency,and wavelength is given by the formula:
$v = n \times \lambda$
Where $\lambda$ is the wavelength.
Rearranging for $\lambda$:
$\lambda = \frac{v}{n}$
Substituting the given values:
$\lambda = \frac{340}{200} = 1.7 m$
Therefore,the wavelength of the sound produced is $1.7 m$.

(B) The human ear is sensitive to sound waves with frequencies ranging from $20 Hz$ to $20,000 Hz$ $(20 kHz)$.
This specific range is defined as the audible range for human beings.
Therefore,the correct frequency range is $20 Hz - 20 kHz$.

(D) The speed of sound in a medium is given by $v = \frac{d}{t}$.
For the first medium,$v_1 = \frac{2 \, km}{3 \, sec} = \frac{2}{3} \, km/s$.
For the second medium (air),$v_2 = \frac{3 \, km}{10 \, sec} = \frac{3}{10} \, km/s$.
Since the frequency $f$ of the sound wave remains constant when it travels between different media,the speed $v$ is directly proportional to the wavelength $\lambda$ $(v = f \lambda \implies v \propto \lambda)$.
Therefore,the ratio of the wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{v_1}{v_2}$.
Substituting the values: $\frac{\lambda_1}{\lambda_2} = \frac{2/3}{3/10} = \frac{2}{3} \times \frac{10}{3} = \frac{20}{9}$.
Thus,the ratio is $20:9$.

(B) When a wave travels from one medium to another,its frequency depends only on the source of the wave and remains unchanged.
However,the velocity and wavelength of the wave change depending on the properties of the medium (such as density and elasticity).
Therefore,the frequency remains constant.

(B) The total time $t$ taken to hear the echo is the sum of the time taken by the stone to fall to the water surface $(t_1)$ and the time taken by the sound to travel back to the top $(t_2)$.
$t = t_1 + t_2$
For the stone falling from rest,$h = \frac{1}{2}gt_1^2$,so $t_1 = \sqrt{\frac{2h}{g}}$.
Given $h = 19.6\,m$ and $g = 9.8\,m/s^2$,$t_1 = \sqrt{\frac{2 \times 19.6}{9.8}} = \sqrt{4} = 2.0\,s$.
The time taken by sound to travel back is $t_2 = t - t_1 = 2.06\,s - 2.0\,s = 0.06\,s$.
The velocity of sound $v$ is given by $v = \frac{h}{t_2} = \frac{19.6}{0.06} \approx 326.7\,m/s$.

(B) The velocity of sound in a gas is directly proportional to the square root of its absolute temperature: $v \propto \sqrt{T}$.
Given that the velocity at temperature $T_2$ is double the velocity at $T_1 = 0^{\circ}C = 273 \ K$,we have:
$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$
Substituting the given values:
$2 = \sqrt{\frac{T_2}{273}}$
Squaring both sides:
$4 = \frac{T_2}{273}$
$T_2 = 4 \times 273 = 1092 \ K$
To convert this to Celsius:
$T(^{\circ}C) = 1092 - 273 = 819^{\circ}C$.
Therefore,the correct option is $B$.

(D) The velocity of sound depends on the elasticity and density of the medium. The formula for the speed of sound in a solid is $v = \sqrt{Y/\rho}$,where $Y$ is Young's modulus and $\rho$ is the density.
Among the given options,$Steel$ is a solid,$Water$ is a liquid,and $Air$ is a gas. Sound travels fastest in solids because the particles are more closely packed,allowing for faster transmission of mechanical vibrations.
In $Vacuum$,sound cannot travel at all because it requires a material medium for propagation. Therefore,the velocity of sound is maximum in $Steel$.

(B) The distance between a compression and the nearest rarefaction is equal to half the wavelength,$\frac{\lambda}{2}$.
Given,$\frac{\lambda}{2} = 1\, m$,so the wavelength $\lambda = 2\, m$.
The velocity of sound $v = 360\, m/s$.
The frequency $f$ is given by the formula $f = \frac{v}{\lambda}$.
Substituting the values,$f = \frac{360}{2} = 180\, Hz$.

(A) The velocity of sound in a gas is given by the formula $v = \sqrt{\frac{\gamma P}{\rho}}$,where $\gamma$ is the adiabatic index,$P$ is the pressure,and $\rho$ is the density of the gas.
Assuming the temperature and pressure are the same for both gases,the ratio of the velocities is inversely proportional to the square root of their densities:
$\frac{v_{O_2}}{v_{H_2}} = \sqrt{\frac{\rho_{H_2}}{\rho_{O_2}}}$
Given that the density of oxygen is $16$ times that of hydrogen,i.e.,$\rho_{O_2} = 16 \rho_{H_2}$,we have:
$\frac{v_{O_2}}{v_{H_2}} = \sqrt{\frac{\rho_{H_2}}{16 \rho_{H_2}}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$
Thus,the ratio of the velocities of sound waves in oxygen and hydrogen is $1:4$.

(A) Let the distance between the shooter and the reflecting surface be $d$.
When a sound is produced,it travels to the reflecting surface and returns to the shooter to form an echo.
Therefore,the total distance traveled by the sound is $2d$.
The time taken for the echo to be heard is given by $t = \frac{\text{Total distance}}{\text{Velocity of sound}} = \frac{2d}{v}$.
Given $t = 8 \, s$ and $v = 350 \, m/s$.
Substituting the values: $8 = \frac{2d}{350}$.
$2d = 8 \times 350 = 2800$.
$d = \frac{2800}{2} = 1400 \, m$.
Thus,the distance to the reflecting surface is $1400 \, m$.

(B) The time taken by the sound to travel from the siren to the man is given by the formula: $t = \frac{d}{v}$.
Given,distance $d = 1 \,km = 1000 \,m$ and velocity of sound $v = 330 \,m/s$.
Calculating the time: $t = \frac{1000}{330} \approx 3.03 \,s$.
Since the sound reaches the man $3.03 \,s$ after it is produced,the man hears the siren $3.03 \,s$ late.
Therefore,when he sets his watch to the time he hears the sound,his watch will be $3.03 \,s$ behind the actual time.
Thus,his watch is set approximately $3 \,s$ slower.

(D) The speed of sound in a gas is given by the formula $v = \sqrt{\frac{\gamma P}{\rho}}$,where $\gamma$ is the adiabatic index,$P$ is the pressure,and $\rho$ is the density of the gas.
According to the ideal gas law,$PV = nRT$,which can be written as $P = \frac{\rho RT}{M}$,where $M$ is the molar mass.
Substituting this into the velocity formula,we get $v = \sqrt{\frac{\gamma RT}{M}}$.
Since $v$ depends only on temperature $T$ and the nature of the gas (constant $\gamma$ and $M$),it is independent of the pressure $P$ at a constant temperature.
Therefore,the velocity of sound in air is independent of the pressure of air.

(B) The speed of sound $v$ in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass.
Since both gases are monoatomic,they have the same adiabatic index $\gamma = 5/3$.
Given that both gases are at the same temperature $T$,the speed of sound is inversely proportional to the square root of the molar mass: $v \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the speed of sound in gas $1$ $(v_1)$ to that in gas $2$ $(v_2)$ is $\frac{v_1}{v_2} = \sqrt{\frac{m_2}{m_1}}$.

(B) Let the distances of the man from the two cliffs be $d_1$ and $d_2$ respectively.
The total distance traveled by the sound for the first echo is $2d_1 = v \times t_1$,where $v = 340 \, ms^{-1}$ and $t_1 = 1.5 \, s$.
$d_1 = \frac{v \times t_1}{2} = \frac{340 \times 1.5}{2} = 170 \times 1.5 = 255 \, m$.
The total distance traveled by the sound for the second echo is $2d_2 = v \times t_2$,where $t_2 = 3.5 \, s$.
$d_2 = \frac{v \times t_2}{2} = \frac{340 \times 3.5}{2} = 170 \times 3.5 = 595 \, m$.
The total distance between the cliffs is $D = d_1 + d_2 = 255 + 595 = 850 \, m$.

(A) The velocity of a sound wave in a given medium depends only on the properties of the medium (such as density,elasticity,temperature,etc.) and is independent of the frequency of the wave.
Since the medium remains the same,the velocity of the wave remains constant at $v$ even when the frequency is increased to $4n$.

(C) The speed of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the gas constant,$T$ is the absolute temperature in Kelvin,and $M$ is the molar mass of the gas.
From this relation,we can see that $v \propto \sqrt{T}$.
Let $v_1$ be the speed of sound at temperature $T_1$ and $v_2$ be the speed at $T_2$.
Given $T_1 = 27^\circ C = 27 + 273 = 300 \ K$.
We want the speed to be doubled,so $v_2 = 2v_1$.
Using the proportionality $v_2/v_1 = \sqrt{T_2/T_1}$,we get $2 = \sqrt{T_2/300}$.
Squaring both sides,$4 = T_2/300$,which gives $T_2 = 1200 \ K$.
Converting back to Celsius,$T_2 = 1200 - 273 = 927^\circ C$.

(D) The speed of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$.
Here,$\gamma$ is the adiabatic index,$R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
Since $v$ depends only on the temperature $T$ and the nature of the gas (represented by $\gamma$ and $M$),it is independent of the pressure $P$ and density $\rho$ of the gas at a constant temperature.
Therefore,the correct option is $D$.

(D) Let $d$ be the distance between the man and the reflecting surface.
For an echo to be heard,the sound must travel to the reflecting surface and back to the observer.
Therefore,the total distance traveled by the sound is $2d$.
Using the formula $2d = v \times t$,where $v = 340 \, m/s$ and $t = 1 \, s$:
$2d = 340 \times 1$
$2d = 340$
$d = \frac{340}{2} = 170 \, m$.
Thus,the distance between the man and the reflection point is $170 \, m$.

(A) The velocity of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$.
From this relation,it is clear that $v \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass of the gas.
Comparing the molar masses of the given gases:
$M(H_2) = 2 \ g/mol$
$M(N_2) = 28 \ g/mol$
$M(He) = 4 \ g/mol$
$M(O_2) = 32 \ g/mol$
Since $H_2$ has the minimum molar mass among the given options,the velocity of sound is maximum in $H_2$.

(D) The formula for the distance $d$ to the reflecting surface is $2d = v \times t$,where $v$ is the velocity of sound in air,typically taken as $332 \, m/s$.
$t$ is the persistence of hearing,which is approximately $\frac{1}{10} \, s$.
Substituting these values into the equation:
$d = \frac{v \times t}{2} = \frac{332 \times 0.1}{2} = \frac{33.2}{2} = 16.6 \, m$.
Given the options,$16.5 \, m$ is the standard accepted value based on the approximation of the speed of sound.

(B) The sound travels from the man to the hillock and back to the man to produce an echo.
Let $d$ be the distance between the man and the hillock.
The total distance traveled by the sound is $2d$.
Given,velocity of sound $v = 330\, m/s$ and time $t = 1.5\, s$.
Using the formula,$2d = v \times t$.
$2d = 330 \times 1.5 = 495\, m$.
$d = \frac{495}{2} = 247.5\, m$.
Therefore,the distance of the hillock from the man is $247.5\, m$.

(D) The speed of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$.
From this relation,it is clear that $v \propto \sqrt{T}$,meaning the velocity of sound increases with an increase in temperature. Thus,statement $I$ is true.
Regarding pressure,for an ideal gas,the density $\rho$ changes proportionally with pressure $P$ at a constant temperature $(P = \rho RT/M)$,such that the ratio $P/\rho$ remains constant. Therefore,the speed of sound is independent of pressure. Thus,statement $IV$ is true.
Hence,statements $I$ and $IV$ are correct.

(A) The speed of sound in a gas is given by the formula $v = \sqrt{\frac{\gamma P}{d}}$,where $\gamma$ is the adiabatic index,$P$ is the pressure,and $d$ is the density of the gas.
Since both gases are diatomic,the adiabatic index $\gamma$ is the same for both.
Given that the temperature and pressure $P$ are the same for both gases,the ratio of the velocities $v_1$ and $v_2$ is:
$\frac{v_1}{v_2} = \frac{\sqrt{\frac{\gamma P}{d_1}}}{\sqrt{\frac{\gamma P}{d_2}}} = \sqrt{\frac{d_2}{d_1}}$
Therefore,the ratio of the velocities of sound is $\sqrt{\frac{d_2}{d_1}}$.

(C) The velocity of sound in a gas is given by the formula $v = \sqrt{\frac{\gamma P}{\rho}}$.
Since the temperature is the same and the gases are monoatomic,the adiabatic index $\gamma$ is the same for both.
Assuming the pressure $P$ is the same,we have $v \propto \frac{1}{\sqrt{\rho}}$.
Therefore,the ratio of velocities is $\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}}$.
Given $\frac{\rho_1}{\rho_2} = \frac{1}{4}$,it follows that $\frac{\rho_2}{\rho_1} = 4$.
Substituting this value,we get $\frac{v_1}{v_2} = \sqrt{4} = 2$.
Thus,the ratio $v_1 : v_2$ is $2:1$.

(C) The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$,which implies $v \propto \sqrt{T}$.
Given that the initial temperature $T_1 = 0^oC = 273 \ K$.
Let the initial speed be $v_1$ and the final speed be $v_2 = 2v_1$.
Using the relation $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$,we get:
$2 = \sqrt{\frac{T_2}{273}}$.
Squaring both sides,we get $4 = \frac{T_2}{273}$.
Therefore,$T_2 = 4 \times 273 = 1092 \ K$.

(B) The velocity of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
For a given temperature and assuming both gases are diatomic ($\gamma$ is the same),the velocity is inversely proportional to the square root of the molar mass: $v \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the velocity of sound in hydrogen $(H_2)$ to oxygen $(O_2)$ is:
$\frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$
The molar mass of oxygen $(O_2)$ is $32 \ g/mol$ and the molar mass of hydrogen $(H_2)$ is $2 \ g/mol$.
$\frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
Thus,the ratio is $4:1$.

(A) The minimum distance between a compression and a rarefaction in a longitudinal wave corresponds to one-fourth of the wavelength,i.e.,$d = \frac{\lambda}{4}$.
Given the length of the wire $l = 1\, m$,we assume this length represents the distance between a compression and a rarefaction,so $l = \frac{\lambda}{4}$.
Therefore,the wavelength $\lambda = 4l = 4 \times 1 = 4\, m$.
Using the wave equation $v = n\lambda$,where $v$ is the velocity of sound and $n$ is the frequency:
$n = \frac{v}{\lambda} = \frac{360}{4} = 90\, sec^{-1}$.

(A) The velocity of sound in a gas is given by the formula $v = \sqrt{\frac{\gamma P}{\rho}}$,where $\gamma$ is the adiabatic index,$P$ is the pressure,and $\rho$ is the density of the gas.
Assuming the pressure $P$ remains constant,the velocity of sound is inversely proportional to the square root of the density: $v \propto \frac{1}{\sqrt{\rho}}$.
Given the initial velocity is $v_s$ at density $\rho_1 = \rho$,and the new density is $\rho_2 = 4\rho$.
Using the ratio: $\frac{v_{new}}{v_s} = \sqrt{\frac{\rho_1}{\rho_2}} = \sqrt{\frac{\rho}{4\rho}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,the new velocity is $v_{new} = \frac{v_s}{2}$.

(A) The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$,which implies $v \propto \sqrt{T}$,where $T$ is the absolute temperature in Kelvin.
Let $d$ be the distance between the two fixed points.
The time taken is $t = \frac{d}{v}$.
Therefore,$\frac{t_1}{t_2} = \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Given $T_1 = 10^{\circ}C = 10 + 273 = 283 \ K$ and $T_2 = 30^{\circ}C = 30 + 273 = 303 \ K$.
Given $t_1 = 2.0 \ s$.
Substituting the values: $\frac{2.0}{t_2} = \sqrt{\frac{303}{283}} \approx \sqrt{1.0706} \approx 1.0347$.
$t_2 = \frac{2.0}{1.0347} \approx 1.93 \ s$.
Rounding to the nearest given option,the correct answer is $1.9 \ s$.

(A) The velocity of sound in a gas is given by the formula $v = \sqrt{\frac{\gamma P}{\rho}}$,where $\gamma$ is the adiabatic index,$P$ is the pressure,and $\rho$ is the density of the gas.
Under identical conditions of pressure and temperature,the density of moist air (air mixed with water vapor) is less than the density of dry air because the molecular weight of water vapor $(18 \ g/mol)$ is less than the average molecular weight of dry air (approximately $29 \ g/mol$).
Since $v \propto \frac{1}{\sqrt{\rho}}$,a lower density results in a higher velocity of sound.
Therefore,$v_m > v_d$.

(A) Let the distances of the two cliffs from the man be $d_1$ and $d_2$. The total distance between the cliffs is $D = d_1 + d_2$.
The time taken for the first echo to return is $t_1 = \frac{2d_1}{v}$.
The time taken for the second echo to return is $t_2 = \frac{2d_2}{v}$.
The interval between the two echoes is given as $1 \, s$,so $|t_2 - t_1| = 1 \, s$.
Substituting the values,$\frac{2d_2}{v} - \frac{2d_1}{v} = 1 \implies d_2 - d_1 = \frac{v}{2} = \frac{340}{2} = 170 \, m$.
However,the question asks for the distance between the cliffs,which is $D = d_1 + d_2$. This cannot be determined uniquely from the given information alone without further assumptions about the sequence of echoes. Assuming the echoes are consecutive and the total time for the round trip of both is $2 \, s$ as per the provided solution logic,$D = \frac{v \times t}{2} = \frac{340 \times 2}{2} = 340 \, m$.

(D) The frequency of a sound wave is a characteristic property of the source itself.
When a sound wave travels from one medium (water) to another medium (air),its speed and wavelength change,but the frequency remains constant.
Therefore,the frequency of the sound recorded by an observer in the air will be the same as the frequency of the source.
Frequency = $600 Hz$.

(C) The velocity of sound in a gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
From this relation,it is clear that $v \propto \sqrt{T}$.
Therefore,when the temperature of the atmosphere increases,the velocity of the sound wave increases.

(A) When a sound wave travels from one medium to another,its frequency remains unchanged because it is determined by the source.
Therefore,the frequency of the wave in air remains $f = 60 \, kHz$.
The wavelength $\lambda$ in air is given by the formula $\lambda = \frac{v}{f}$,where $v$ is the velocity of sound in air.
Substituting the given values: $\lambda = \frac{330 \, m/s}{60 \times 10^3 \, Hz} = \frac{330}{60000} \, m = 0.0055 \, m = 5.5 \, mm$.
Thus,the wavelength is $5.5 \, mm$ and the frequency is $60 \, kHz$.

(C) Sound waves are mechanical waves that require a medium to propagate.
In solids,such as rocks,the material possesses both bulk modulus and shear modulus.
Because of these properties,solids can support both longitudinal waves (where particles oscillate parallel to the direction of propagation) and transverse waves (where particles oscillate perpendicular to the direction of propagation).
Therefore,sound travels in rocks in the form of both longitudinal and transverse elastic waves.

(A) longitudinal wave is a wave in which the particles of the medium vibrate parallel to the direction of wave propagation.
Sound waves in air are classic examples of longitudinal waves because they consist of compressions and rarefactions traveling through the medium.
Waves on a plucked string are transverse waves.
Light waves are electromagnetic $(EM)$ waves,which are transverse in nature.
Water waves are a combination of both transverse and longitudinal motions,but sound waves are strictly longitudinal in fluids.

(B) Sound waves are mechanical waves that require a material medium for propagation.
In gases,sound waves travel as longitudinal waves.
This is because the particles of the medium vibrate back and forth in a direction parallel to the direction of the propagation of the wave,creating regions of compression and rarefaction.