
Which of the following statements are incorrect? (This question has multiple correct options)
${\text{A}}{\text{.}}$ Wave pulses in string are transverse waves
${\text{B}}{\text{.}}$ Sound waves in air are transverse waves of compression and rarefaction
${\text{C}}{\text{.}}$ The speed of sound in air at 20${}^0{\text{C}}$ is twice that at 5${}^0{\text{C}}$
${\text{D}}{\text{.}}$ A 60 dB sound has twice the intensity of a 30 dB sound
Answer
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Hint- Here, we will proceed by stating whether the wave pulses in string and sound waves are longitudinal or transverse waves. Then, we will be comparing speed of sound in air at 20${}^0{\text{C}}$ and 5${}^0{\text{C}}$. Finally, we will find out the intensities of the sound corresponding to sound intensity levels of 60 dB and 30 dB.
Complete step-by-step solution -
Formula Used- $\beta = 10{\log _{10}}\left( {\dfrac{{\text{I}}}{{{{\text{I}}_0}}}} \right)$.
${\text{A}}{\text{.}}$ In general, a wave is defined as a disturbance in a medium that transports both energy and momentum.
String waves are an example of transverse waves because the string moves up and down at right angles to the horizontal motion of the wave. (There are also longitudinal waves like sound waves where the medium and the wave move along the same direction).
So, option A is correct.
${\text{B}}{\text{.}}$ Sound is a mechanical wave that results from the back and forth vibration of the particles of the medium through which the sound wave is moving. If a sound wave is moving from left to right through air, then particles of air will be displaced both rightward and leftward as the energy of the sound wave passes through it. The motion of the particles is parallel (and anti-parallel) to the direction of the energy transport. This is what characterizes sound waves in air as longitudinal waves.
So, option B is incorrect.
${\text{C}}{\text{.}}$ Speed of sound in air at 20${}^0{\text{C}}$ = 343.26 m/s
Speed of sound in air at 5${}^0{\text{C}}$ = 334.33 m/s
Clearly, the speed of sound in air at 20${}^0{\text{C}}$ is not twice that at 5${}^0{\text{C}}$
So, option C is also incorrect.
${\text{D}}{\text{.}}$ As we know that the sound intensity level $\beta $ in decibels (dB) of a sound having intensity I in W/${{\text{m}}^2}$ is given by
$\beta = 10{\log _{10}}\left( {\dfrac{{\text{I}}}{{{{\text{I}}_0}}}} \right)$ where ${{\text{I}}_0} = {10^{ - 12}}$ W/${{\text{m}}^2}$ and is the reference intensity.
For 60 dB sound, put $\beta = 60$ dB and ${{\text{I}}_0} = {10^{ - 12}}$ W/${{\text{m}}^2}$
\[
60 = 10{\log _{10}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}}} \right) \\
\Rightarrow {\log _{10}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}}} \right) = \dfrac{{60}}{{10}} \\
\Rightarrow {\log _{10}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}}} \right) = 6 \\
\]
Using the formula ${\log _{10}}\left( a \right) = b \Rightarrow a = {10^b}$ in the above equation, we get
\[
\Rightarrow \dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}} = {10^6} \\
\Rightarrow {{\text{I}}_1} = {10^6} \times {10^{ - 12}} \\
\Rightarrow {{\text{I}}_1} = {10^{6 - 12}} \\
\Rightarrow {{\text{I}}_1} = {10^{ - 6}} \\
\]
So, the intensity of 60 dB sound is \[{10^{ - 6}}\] W/${{\text{m}}^2}$
Similarly, for 30 dB sound, put $\beta = 30$ dB and ${{\text{I}}_0} = {10^{ - 12}}$ W/${{\text{m}}^2}$
\[
30 = 10{\log _{10}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}}} \right) \\
\Rightarrow {\log _{10}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}}} \right) = \dfrac{{30}}{{10}} \\
\Rightarrow {\log _{10}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}}} \right) = 3 \\
\]
Using the formula ${\log _{10}}\left( a \right) = b \Rightarrow a = {10^b}$ in the above equation, we get
\[
\Rightarrow \dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}} = {10^3} \\
\Rightarrow {{\text{I}}_1} = {10^3} \times {10^{ - 12}} \\
\Rightarrow {{\text{I}}_1} = {10^{3 - 12}} \\
\Rightarrow {{\text{I}}_1} = {10^{ - 9}} \\
\]
So, the intensity of 30 dB sound is \[{10^{ - 9}}\] W/${{\text{m}}^2}$
Since, ${{\text{I}}_1} \ne 2{{\text{I}}_2}$
Clearly, the intensity of a 60 dB sound is not equal to twice the intensity of a 30 dB sound.
So, option D is also incorrect.
Therefore, the statements given in options B, C and D are all incorrect.
Note- The speed of sound in air at higher temperature is more than that at lower temperature because higher temperature means molecules in the air will be having higher energy and will move faster as compared to the molecules in air at lower temperature.
Complete step-by-step solution -
Formula Used- $\beta = 10{\log _{10}}\left( {\dfrac{{\text{I}}}{{{{\text{I}}_0}}}} \right)$.
${\text{A}}{\text{.}}$ In general, a wave is defined as a disturbance in a medium that transports both energy and momentum.
String waves are an example of transverse waves because the string moves up and down at right angles to the horizontal motion of the wave. (There are also longitudinal waves like sound waves where the medium and the wave move along the same direction).
So, option A is correct.
${\text{B}}{\text{.}}$ Sound is a mechanical wave that results from the back and forth vibration of the particles of the medium through which the sound wave is moving. If a sound wave is moving from left to right through air, then particles of air will be displaced both rightward and leftward as the energy of the sound wave passes through it. The motion of the particles is parallel (and anti-parallel) to the direction of the energy transport. This is what characterizes sound waves in air as longitudinal waves.
So, option B is incorrect.
${\text{C}}{\text{.}}$ Speed of sound in air at 20${}^0{\text{C}}$ = 343.26 m/s
Speed of sound in air at 5${}^0{\text{C}}$ = 334.33 m/s
Clearly, the speed of sound in air at 20${}^0{\text{C}}$ is not twice that at 5${}^0{\text{C}}$
So, option C is also incorrect.
${\text{D}}{\text{.}}$ As we know that the sound intensity level $\beta $ in decibels (dB) of a sound having intensity I in W/${{\text{m}}^2}$ is given by
$\beta = 10{\log _{10}}\left( {\dfrac{{\text{I}}}{{{{\text{I}}_0}}}} \right)$ where ${{\text{I}}_0} = {10^{ - 12}}$ W/${{\text{m}}^2}$ and is the reference intensity.
For 60 dB sound, put $\beta = 60$ dB and ${{\text{I}}_0} = {10^{ - 12}}$ W/${{\text{m}}^2}$
\[
60 = 10{\log _{10}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}}} \right) \\
\Rightarrow {\log _{10}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}}} \right) = \dfrac{{60}}{{10}} \\
\Rightarrow {\log _{10}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}}} \right) = 6 \\
\]
Using the formula ${\log _{10}}\left( a \right) = b \Rightarrow a = {10^b}$ in the above equation, we get
\[
\Rightarrow \dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}} = {10^6} \\
\Rightarrow {{\text{I}}_1} = {10^6} \times {10^{ - 12}} \\
\Rightarrow {{\text{I}}_1} = {10^{6 - 12}} \\
\Rightarrow {{\text{I}}_1} = {10^{ - 6}} \\
\]
So, the intensity of 60 dB sound is \[{10^{ - 6}}\] W/${{\text{m}}^2}$
Similarly, for 30 dB sound, put $\beta = 30$ dB and ${{\text{I}}_0} = {10^{ - 12}}$ W/${{\text{m}}^2}$
\[
30 = 10{\log _{10}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}}} \right) \\
\Rightarrow {\log _{10}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}}} \right) = \dfrac{{30}}{{10}} \\
\Rightarrow {\log _{10}}\left( {\dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}}} \right) = 3 \\
\]
Using the formula ${\log _{10}}\left( a \right) = b \Rightarrow a = {10^b}$ in the above equation, we get
\[
\Rightarrow \dfrac{{{{\text{I}}_1}}}{{{{10}^{ - 12}}}} = {10^3} \\
\Rightarrow {{\text{I}}_1} = {10^3} \times {10^{ - 12}} \\
\Rightarrow {{\text{I}}_1} = {10^{3 - 12}} \\
\Rightarrow {{\text{I}}_1} = {10^{ - 9}} \\
\]
So, the intensity of 30 dB sound is \[{10^{ - 9}}\] W/${{\text{m}}^2}$
Since, ${{\text{I}}_1} \ne 2{{\text{I}}_2}$
Clearly, the intensity of a 60 dB sound is not equal to twice the intensity of a 30 dB sound.
So, option D is also incorrect.
Therefore, the statements given in options B, C and D are all incorrect.
Note- The speed of sound in air at higher temperature is more than that at lower temperature because higher temperature means molecules in the air will be having higher energy and will move faster as compared to the molecules in air at lower temperature.
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