A person is taking in a small room and the sound intensity level is $60{\rm{ dB}}$ everywhere within the room. If there are eight people talking simultaneously in the room, what is the sound intensity level?
A. ${\rm{60 dB}}$
B. ${\rm{69 dB}}$
C. ${\rm{74 dB}}$
D. ${\rm{81 dB}}$
Answer
623.1k+ views
Hint:Hint: In the solution we use the basic relation between change in sound intensity levels and the intensities caused by different sources to find the new sound intensity level from the given information.
Complete Step by Step Answer:Initial sound intensity level ${B_1} = 60{\rm{ dB}}$
Let the intensity of the one person be ${I_1} = I$
Therefore, the intensity caused by 8 persons will be ${I_2} = 8I$
The change in sound intensity levels is given by the relation
\[\begin{array}{c}
{B_2} - {B_1} = 10\log \left( {\dfrac{{{I_2}}}{{{I_1}}}} \right)\\
{B_2} = {B_1} + 10\log \left( {\dfrac{{{I_2}}}{{{I_1}}}} \right)
\end{array}\]
Substituting the values in the above equation gives
$\begin{array}{c}
{B_2} = 60{\rm{ dB}} + 10\log \left( {\dfrac{{8I}}{I}} \right)\\
= 60 + 10\log \left( 8 \right)\\
= 60 + 10\log \left( {{2^3}} \right)\\
= 60 + 10 \times 3 \times \log 2\\
= 60 + 10 \times 3 \times 0.3\\
= 60 + 9\\
= 69{\rm{ dB}}
\end{array}$
Therefore, the correct answer is option B, that is ${\rm{69 dB}}$.
Note:Make sure that logarithm used in the equation is to the base of 10 and not natural logarithm as it has a significant effect on the results.
Complete Step by Step Answer:Initial sound intensity level ${B_1} = 60{\rm{ dB}}$
Let the intensity of the one person be ${I_1} = I$
Therefore, the intensity caused by 8 persons will be ${I_2} = 8I$
The change in sound intensity levels is given by the relation
\[\begin{array}{c}
{B_2} - {B_1} = 10\log \left( {\dfrac{{{I_2}}}{{{I_1}}}} \right)\\
{B_2} = {B_1} + 10\log \left( {\dfrac{{{I_2}}}{{{I_1}}}} \right)
\end{array}\]
Substituting the values in the above equation gives
$\begin{array}{c}
{B_2} = 60{\rm{ dB}} + 10\log \left( {\dfrac{{8I}}{I}} \right)\\
= 60 + 10\log \left( 8 \right)\\
= 60 + 10\log \left( {{2^3}} \right)\\
= 60 + 10 \times 3 \times \log 2\\
= 60 + 10 \times 3 \times 0.3\\
= 60 + 9\\
= 69{\rm{ dB}}
\end{array}$
Therefore, the correct answer is option B, that is ${\rm{69 dB}}$.
Note:Make sure that logarithm used in the equation is to the base of 10 and not natural logarithm as it has a significant effect on the results.
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