
The power of the sound from the speaker of a radio is 20 milliWatt. By turning the knob of the volume control the power of the sound is increased to 400 milliWatt. The power increase in decibels as compared to the original power is?
A) $13\,db$
B) $10\,db$
C) $20\,db$
D) $800\,db$
Answer
123.9k+ views
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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