Answer
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Hint: We will calculate the increase in intensity in power by using the sound intensity formula. Just remember, the sound intensity level is the level of the intensity of a sound relative to a reference value. Also, it is the logarithmic expression that is relative to a reference value of sound intensity.
Complete step by step solution:
We are given that
${P_1} = 20\,mW$(Power of the sound from the speaker)
${P_2} = 400\,mW$(Power of the sound increased)
As we all know that intensity is the power per unit area.
Now, For a given source, \[P\,\alpha \,I\].
Where $P$ is the power and $I$ is the intensity in the sound.
If ${L_1}$and ${L_2}$ are the initial and final level of loudness, then
${L_1} = 10\log (\dfrac{{{I_1}}}{{{I_0}}})$
And ${L_2} = 10\log (\dfrac{{{I_2}}}{{{I_0}}})$
$ \Rightarrow {L_2} - {L_1} = 10\log (\dfrac{{{I_2}}}{{{I_1}}})$
Therefore, an increase in loudness of sound can be calculated as
$\Delta L = 10\log (\dfrac{{{P_2}}}{{{P_1}}})$
\[ \Rightarrow \Delta L = 10\log (\dfrac{{400}}{{20}})\]
$ \Rightarrow \,\Delta L = 10\log (20)$
$\therefore \,$ $\Delta L = 13dB$
Therefore, the increase in power as compared to the original power is$13\,db$.
Therefore, option (A) is the correct option.
Additional Information:
As we know, the Sound intensity level is the level of intensity of sound which is relative to a reference value. It is denoted by ${L_I}$ and is expressed as
${L_I} = \dfrac{1}{2}\ln (\dfrac{I}{{{I_0}}})\,Np$
Or ${L_I} = {\log _{10}}(\dfrac{I}{{{I_0}}})\,B$
Or ${L_I} = 10\log (\dfrac{I}{{{I_0}}})\,dB$
Where ${L_I}$ is the sound intensity level, $I$ is the intensity of sound, and ${I_0}$ is the reference sound intensity.
Here,$Np$, $B$, and $dB$ are the units in which sound intensity level could be measured and these are abbreviated as $neper$, $bel$, and $decibel$.
Note: As we know, ${I_0}$ is the reference sound intensity and its value in the air is given by
${I_0} = 1\,pW/{m^2}$(Pico watts per meter square).
It is the least value of the sound intensity but it is hearable by the human ear at room conditions.
Complete step by step solution:
We are given that
${P_1} = 20\,mW$(Power of the sound from the speaker)
${P_2} = 400\,mW$(Power of the sound increased)
As we all know that intensity is the power per unit area.
Now, For a given source, \[P\,\alpha \,I\].
Where $P$ is the power and $I$ is the intensity in the sound.
If ${L_1}$and ${L_2}$ are the initial and final level of loudness, then
${L_1} = 10\log (\dfrac{{{I_1}}}{{{I_0}}})$
And ${L_2} = 10\log (\dfrac{{{I_2}}}{{{I_0}}})$
$ \Rightarrow {L_2} - {L_1} = 10\log (\dfrac{{{I_2}}}{{{I_1}}})$
Therefore, an increase in loudness of sound can be calculated as
$\Delta L = 10\log (\dfrac{{{P_2}}}{{{P_1}}})$
\[ \Rightarrow \Delta L = 10\log (\dfrac{{400}}{{20}})\]
$ \Rightarrow \,\Delta L = 10\log (20)$
$\therefore \,$ $\Delta L = 13dB$
Therefore, the increase in power as compared to the original power is$13\,db$.
Therefore, option (A) is the correct option.
Additional Information:
As we know, the Sound intensity level is the level of intensity of sound which is relative to a reference value. It is denoted by ${L_I}$ and is expressed as
${L_I} = \dfrac{1}{2}\ln (\dfrac{I}{{{I_0}}})\,Np$
Or ${L_I} = {\log _{10}}(\dfrac{I}{{{I_0}}})\,B$
Or ${L_I} = 10\log (\dfrac{I}{{{I_0}}})\,dB$
Where ${L_I}$ is the sound intensity level, $I$ is the intensity of sound, and ${I_0}$ is the reference sound intensity.
Here,$Np$, $B$, and $dB$ are the units in which sound intensity level could be measured and these are abbreviated as $neper$, $bel$, and $decibel$.
Note: As we know, ${I_0}$ is the reference sound intensity and its value in the air is given by
${I_0} = 1\,pW/{m^2}$(Pico watts per meter square).
It is the least value of the sound intensity but it is hearable by the human ear at room conditions.
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