
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
149.4k+ 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.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electrical Field of Charged Spherical Shell - JEE

Charging and Discharging of Capacitor

Ideal and Non-Ideal Solutions Raoult's Law - JEE

The time period of an artificial satellite in a circular class 11 physics JEE_Main

Class 11 JEE Main Physics Mock Test 2025

Other Pages
JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Main Chemistry Question Paper with Answer Keys and Solutions

A pressure of 100 kPa causes a decrease in volume of class 11 physics JEE_Main

A boy wants to throw a ball from a point A so as to class 11 physics JEE_Main

Select incorrect statements A Zero acceleration of class 11 physics JEE_Main

Assertion On a rainy day it is difficult to drive a class 11 physics JEE_Main
