Question

The power of sound from speakers of radio is \[10W\] , the power of sound from the speaker of radio is \[400W\] by increasing the volume of radio. The power increased in dB as compared to original power is nearly
A. 8
B. 12
C. 13
D. 16

Answer 1

Hint: Use the expression for the decibel difference between two level Intensities and substitute the values as given in the Question to get the answer.

Complete step by step answer:
As we know that, power is proportional to intensity, the decibel difference between two levels with intensities \[{L_1}\] and \[{L_2}\] is given by
Here,
\[P\propto I\]
\[{L_1} = 10\log (\dfrac{{{I_1}}}{{{I_0}}})\] and \[{L_2} = 10\log (\dfrac{{{I_2}}}{{{I_0}}})\]
So Now, we have
\[{L_2} - {L_1} = 10\log (\dfrac{{{I_2}}}{{{I_1}}}) = 10\log (\dfrac{{{P_2}}}{{{P_1}}})\]
\[\Rightarrow 10\log (\dfrac{{400}}{{10}})\]
\[\Rightarrow 10\log (40)\]
\[\Rightarrow 10\log (4 \times 10)\]
\[\Rightarrow 10(0.60 + 1)\]
\[\Rightarrow 16dB\]
So, the Correct Answer is Option D

Additional Information:
Sound power level (SWL)- It is a logarithmic measure of the power of a sound relative to a reference value.
In a small reflective (reverberant) room a sound source will produce a higher sound pressure level than in a large acoustically dead (absorptive) room or an open space.
For acoustic purposes, sound pressure levels and sound power are quantified in decibels that are a logarithmic ratio. The reference value is 10-12 in the case of a sound power level.
Sound Intensity Level (Li) is the logarithmic ratio of the measured sound intensity to a reference sound intensity in decibels.
The Sound Intensity Level Formula
\[10\log (\dfrac{I}{{{I_0}}})dB\]
where is the reference sound intensity of 10-12 W/m² and I is the sound intensity in W/m² and If we double distance from the source, then there will be decrease of 6dB in the sound level

Note:If we have multiple sound sources, the total power emitted is the sum of the power from each source.
And if the two sources sound simultaneously with equal power, then the sound power level would double
\[{L_W} = 10\log (\dfrac{{2W}}{{{W_0}}})\]
\[ = 10\log (\dfrac{W}{{{W_0}}}) + 10\log (2)\]
\[ = 10\log (\dfrac{W}{{{W_0}}}) + 3dB\]
And there would be an increase of \[3dB\].