Answer
Verified
377.1k+ views
Hint: Study the definition of decibel. Use the formula to find the intensity of the sound. From there find the noise level due to the said number of students.Noise is defined as the unpleasant sound that causes disturbance.
Formula used:
The formula to find the sound energy in units of decibel is given by,
\[\beta = 10{\log _{10}}\dfrac{{{S_1}}}{{{S_2}}}\]
where, is the sound energy in units of \[dB\] \[{S_1}\] and \[{S_2}\] are the intensity of sound from two sources.
Complete step by step answer:
We have given here that a class full of \[50\] students are making noise in absence of the teacher is \[50\,dB\].The output of each student is the same sound energy per second. So, let the intensity of sound by each student is \[I\] with respect to some source \[{I_0}\]. Hence, the sound energy produced by \[50\] students is \[50I\]. Hence the sound energy produced by \[100\] students will be, \[100I\].
The mathematical expression for the sound energy in units of decibel is given by,
\[\beta = 10{\log _{10}}\dfrac{{{S_1}}}{{{S_2}}}\]
where, $\beta$ is the sound energy in units of \[dB\] \[{S_1}\] and \[{S_2}\]are the intensity of sound from two sources.
Now, in units of sound energy decibel the noise by \[50\] students is,
\[50 = 10{\log _{10}}\dfrac{{50I}}{{{I_0}}}\]..…(i)
Now, let the sound produced by \[100\] students in units of decibel is \[\beta \]. So, the noise produced by \[100\] students will be,
\[\beta = 10{\log _{10}}\dfrac{{100I}}{{{I_0}}}\]……(ii)
So, Subtracting equation(ii) from equation(i) we get,
\[\beta - 50 = 10{\log _{10}}\dfrac{{100I}}{{{I_0}}} - 10{\log _{10}}\dfrac{{50I}}{{{I_0}}}\]
Upon simplifying we will have,
\[\beta - 50 = 10 \times {\log _{10}}\dfrac{{100I}}{{50I}}\]
\[\Rightarrow \beta = 50 + 10{\log _{10}}2\]
\[\Rightarrow \beta = 50 + 3.01\]
\[\therefore \beta = 53.01\]
Hence, the noise output by \[100\] students in units of decibel in absence of teacher is \[53.01\,dB\].
Note: The noise output in units of bel is not proportional to the energy output by the students. It is so since the logarithm curve follows the curve of exponent here the curve is of the form \[y = {10^x}\]. So, it is not linear so the increase in noise is less than the increase in intensity of sound.
Formula used:
The formula to find the sound energy in units of decibel is given by,
\[\beta = 10{\log _{10}}\dfrac{{{S_1}}}{{{S_2}}}\]
where, is the sound energy in units of \[dB\] \[{S_1}\] and \[{S_2}\] are the intensity of sound from two sources.
Complete step by step answer:
We have given here that a class full of \[50\] students are making noise in absence of the teacher is \[50\,dB\].The output of each student is the same sound energy per second. So, let the intensity of sound by each student is \[I\] with respect to some source \[{I_0}\]. Hence, the sound energy produced by \[50\] students is \[50I\]. Hence the sound energy produced by \[100\] students will be, \[100I\].
The mathematical expression for the sound energy in units of decibel is given by,
\[\beta = 10{\log _{10}}\dfrac{{{S_1}}}{{{S_2}}}\]
where, $\beta$ is the sound energy in units of \[dB\] \[{S_1}\] and \[{S_2}\]are the intensity of sound from two sources.
Now, in units of sound energy decibel the noise by \[50\] students is,
\[50 = 10{\log _{10}}\dfrac{{50I}}{{{I_0}}}\]..…(i)
Now, let the sound produced by \[100\] students in units of decibel is \[\beta \]. So, the noise produced by \[100\] students will be,
\[\beta = 10{\log _{10}}\dfrac{{100I}}{{{I_0}}}\]……(ii)
So, Subtracting equation(ii) from equation(i) we get,
\[\beta - 50 = 10{\log _{10}}\dfrac{{100I}}{{{I_0}}} - 10{\log _{10}}\dfrac{{50I}}{{{I_0}}}\]
Upon simplifying we will have,
\[\beta - 50 = 10 \times {\log _{10}}\dfrac{{100I}}{{50I}}\]
\[\Rightarrow \beta = 50 + 10{\log _{10}}2\]
\[\Rightarrow \beta = 50 + 3.01\]
\[\therefore \beta = 53.01\]
Hence, the noise output by \[100\] students in units of decibel in absence of teacher is \[53.01\,dB\].
Note: The noise output in units of bel is not proportional to the energy output by the students. It is so since the logarithm curve follows the curve of exponent here the curve is of the form \[y = {10^x}\]. So, it is not linear so the increase in noise is less than the increase in intensity of sound.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE