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Question: The power of the sound from the speaker of a radio is 20 milliWatt. By turning the knob of the volum...

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) 13db13\,db
B) 10db10\,db
C) 20db20\,db
D) 800db800\,db

Explanation

Solution

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
P1=20mW{P_1} = 20\,mW(Power of the sound from the speaker)
P2=400mW{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, PαIP\,\alpha \,I.
Where PP is the power and II is the intensity in the sound.
If L1{L_1}and L2{L_2} are the initial and final level of loudness, then
L1=10log(I1I0){L_1} = 10\log (\dfrac{{{I_1}}}{{{I_0}}})
And L2=10log(I2I0){L_2} = 10\log (\dfrac{{{I_2}}}{{{I_0}}})
L2L1=10log(I2I1)\Rightarrow {L_2} - {L_1} = 10\log (\dfrac{{{I_2}}}{{{I_1}}})
Therefore, an increase in loudness of sound can be calculated as
ΔL=10log(P2P1)\Delta L = 10\log (\dfrac{{{P_2}}}{{{P_1}}})
ΔL=10log(40020)\Rightarrow \Delta L = 10\log (\dfrac{{400}}{{20}})
ΔL=10log(20)\Rightarrow \,\Delta L = 10\log (20)
\therefore \, ΔL=13dB\Delta L = 13dB
Therefore, the increase in power as compared to the original power is13db13\,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 LI{L_I} and is expressed as
LI=12ln(II0)Np{L_I} = \dfrac{1}{2}\ln (\dfrac{I}{{{I_0}}})\,Np
Or LI=log10(II0)B{L_I} = {\log _{10}}(\dfrac{I}{{{I_0}}})\,B
Or LI=10log(II0)dB{L_I} = 10\log (\dfrac{I}{{{I_0}}})\,dB
Where LI{L_I} is the sound intensity level, II is the intensity of sound, and I0{I_0} is the reference sound intensity.
Here,NpNp, BB, and dBdB are the units in which sound intensity level could be measured and these are abbreviated as neperneper, belbel, and decibeldecibel.

Note: As we know, I0{I_0} is the reference sound intensity and its value in the air is given by
I0=1pW/m2{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.