
Decibel \[\left( {db} \right)\] is a unit of loudness of sound. It is defined in a manner such that when amplitude of sound is multiplied by a factor of $\sqrt {10} $, the decibel level increases by \[10\] units. Loud music of \[70dB\] is being played at a function. To reduce the loudness to a level of \[30dB\] , the amplitude of the instrument playing music to be reduced by a factor of
A. $0$
B. $10\sqrt {10} $
C. $100$
D. $100\sqrt {100} $
Answer
448.5k+ views
Hint: In this question, we can solve the equation \[dB = 10\log {a_0}^2 - 10\log {a_1}^2\]. After then we can assume that the amplitude is changed by the factor $n$. Now, we can solve the above equation for $n$.
Complete step by step solution: -
We know that the loudness of sound is given by equation
$dB = 10\log \left( {\dfrac{{{I_0}}}{{{I_1}}}} \right) $
$\Rightarrow dB = 10\log {I_0} - 10\log {I_1} $
We know that intensity is directly proportional to the square of the amplitude i.e.
$I \propto {a^2}$
So,
\[dB = 10\log {a_0}^2 - 10\log {a_1}^2\]
According to the question, if amplitude of sound is multiplied by a factor of$\sqrt {10} $, the decibel level increases by \[10\] units. So,
$d{B^1} = 10\log {\left( {\sqrt {10} } \right)^2} - 10\log {a_1}^2 $
$\Rightarrow d{B^1} = 10\log 10 - 10\log {a_1}^2 $
$\Rightarrow d{B^1} = 10 - 10\log {a_1}^2 $
$\Rightarrow d{B^1} = 10 - dB $
Where \[dB = 10\log {a_1}^2$
Now, if \[70dB\] is being played at a function and if it is reduced to a level of\[30dB\] , then let the amplitude of the instrument playing music be reduced by a factor of $n$.
So,
$d{B^1} = 10\log \left( {\dfrac{{{a_1}^2}}{{{n^2}}}} \right) $
$\Rightarrow d{B^1} = 10\log {a_1}^2 - 10\log {n^2} $
$\Rightarrow d{B^1} = dB - 20\log n $
$\Rightarrow d{B^1} - dB = 20\log n$
$\Rightarrow 70 - 30 = 20\log n $
$\Rightarrow 40 = 20\log n $
$\Rightarrow 2 = \log n $
$\Rightarrow n = {10^2} $
$\Rightarrow n = 100 $
So, loud music of \[70dB\] is being played at a function. To reduce the loudness to a level of\[30dB\] , the amplitude of the instrument playing music to be reduced by a factor of $100$ .
Hence, option C is correct.
Additional information: -
Decibel is a logarithmic unit which is used to measure the loudness. It is used in electronics, signals and communications. Decibel is a logarithmic way of describing ratios of power, sound pressure, voltage, intensity, etc. Generally, it is used to measure the loudness of the sound. The level $0dB$ occurs when the intensity of the sound is equal to the reference level of the sound.
Note:
In this question, we have kept in mind that \[d{B^1}\] is the difference in decibel. We have to remember the calculations of logarithmic also such as $\log 10 = 1$ and $\log 1 = 0$.
Complete step by step solution: -
We know that the loudness of sound is given by equation
$dB = 10\log \left( {\dfrac{{{I_0}}}{{{I_1}}}} \right) $
$\Rightarrow dB = 10\log {I_0} - 10\log {I_1} $
We know that intensity is directly proportional to the square of the amplitude i.e.
$I \propto {a^2}$
So,
\[dB = 10\log {a_0}^2 - 10\log {a_1}^2\]
According to the question, if amplitude of sound is multiplied by a factor of$\sqrt {10} $, the decibel level increases by \[10\] units. So,
$d{B^1} = 10\log {\left( {\sqrt {10} } \right)^2} - 10\log {a_1}^2 $
$\Rightarrow d{B^1} = 10\log 10 - 10\log {a_1}^2 $
$\Rightarrow d{B^1} = 10 - 10\log {a_1}^2 $
$\Rightarrow d{B^1} = 10 - dB $
Where \[dB = 10\log {a_1}^2$
Now, if \[70dB\] is being played at a function and if it is reduced to a level of\[30dB\] , then let the amplitude of the instrument playing music be reduced by a factor of $n$.
So,
$d{B^1} = 10\log \left( {\dfrac{{{a_1}^2}}{{{n^2}}}} \right) $
$\Rightarrow d{B^1} = 10\log {a_1}^2 - 10\log {n^2} $
$\Rightarrow d{B^1} = dB - 20\log n $
$\Rightarrow d{B^1} - dB = 20\log n$
$\Rightarrow 70 - 30 = 20\log n $
$\Rightarrow 40 = 20\log n $
$\Rightarrow 2 = \log n $
$\Rightarrow n = {10^2} $
$\Rightarrow n = 100 $
So, loud music of \[70dB\] is being played at a function. To reduce the loudness to a level of\[30dB\] , the amplitude of the instrument playing music to be reduced by a factor of $100$ .
Hence, option C is correct.
Additional information: -
Decibel is a logarithmic unit which is used to measure the loudness. It is used in electronics, signals and communications. Decibel is a logarithmic way of describing ratios of power, sound pressure, voltage, intensity, etc. Generally, it is used to measure the loudness of the sound. The level $0dB$ occurs when the intensity of the sound is equal to the reference level of the sound.
Note:
In this question, we have kept in mind that \[d{B^1}\] is the difference in decibel. We have to remember the calculations of logarithmic also such as $\log 10 = 1$ and $\log 1 = 0$.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Who is Mukesh What is his dream Why does it look like class 12 english CBSE

Who was RajKumar Shukla Why did he come to Lucknow class 12 english CBSE

The word Maasai is derived from the word Maa Maasai class 12 social science CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE
