
A person is talking in a small room and the sound intensity level is 60 dB everywhere in the room. If eight people are talking simultaneously in the room, what is the sound intensity level?
A. 60 dB
B. 69 dB
C. 74 dB
D. 81 dB
Answer
567k+ views
Hint: First find the intensity of the eight person and then use the formula of finding the change in the intensity level, which helps in finding the required result.
$ \Rightarrow {B_2} - {B_1} = 10\log \left( {\dfrac{{{I_8}}}{I}} \right)$
Complete step by step solution:
Let ${B_1}$and${B_2}$is the initial intensity when 1 person is talking and intensity after 8 persons talking simultaneously respectively.
It is given that the person is talking in a small room and the sound intensity level$\left( {{B_1}} \right)$ is $60$ dB everywhere in the room.
Let $I$ be the initial intensity due to 1 person,
Therefore, the intensity due to 8 persons$\left( {{I_8}} \right)$ $ = 8I$
Change in the sound intensity level is given using the formula:
$ \Rightarrow {B_2} - {B_1} = 10\log \left( {\dfrac{{{I_8}}}{I}} \right)$
Substitute 60 as the value of${B_1}$, $8I$ as the value of ${I_8}$ in the above formula:
$ \Rightarrow {B_2} - 60 = 10\log \left( {\dfrac{{8I}}{I}} \right)$
$ \Rightarrow {B_2} - 60 = 10\log \left( 8 \right)$
8 can also expressed as${2^3}$ :
$ \Rightarrow {B_2} - 60 = 10\log \left( {{2^3}} \right)$
$ \Rightarrow {B_2} - 60 = 10\log \left( {{2^3}} \right)$
We know that:
$\log \left( {{a^b}} \right) = b\log a$
After applying the above relation, we have
$ \Rightarrow {B_2} - 60 = 10 \times 3\log 2$
The value of $\log 2 \approx 0.3$, so
$ \Rightarrow {B_2} - 60 = 10 \times 3 \times 0.3$
$ \Rightarrow {B_2} - 60 = 10 \times 0.9$
$ \Rightarrow {B_2} - 60 = 9$
$ \Rightarrow {B_2} = 9 + 60$
$ \Rightarrow {B_2} = 69$ dB
We have found the value of the ${B_2} = 69$dB. It means that, when 8 people are talking simultaneously in a small room, then the intensity level is $69$dB everywhere in the room.
Therefore, the option $\left( B \right)$ is correct.
Note: Remember that if the initial intensity of one person is $I$, then the intensity of eight persons is eight times the intensity of one person. That is, the intensity of eight-person is $8I$.
$ \Rightarrow {B_2} - {B_1} = 10\log \left( {\dfrac{{{I_8}}}{I}} \right)$
Complete step by step solution:
Let ${B_1}$and${B_2}$is the initial intensity when 1 person is talking and intensity after 8 persons talking simultaneously respectively.
It is given that the person is talking in a small room and the sound intensity level$\left( {{B_1}} \right)$ is $60$ dB everywhere in the room.
Let $I$ be the initial intensity due to 1 person,
Therefore, the intensity due to 8 persons$\left( {{I_8}} \right)$ $ = 8I$
Change in the sound intensity level is given using the formula:
$ \Rightarrow {B_2} - {B_1} = 10\log \left( {\dfrac{{{I_8}}}{I}} \right)$
Substitute 60 as the value of${B_1}$, $8I$ as the value of ${I_8}$ in the above formula:
$ \Rightarrow {B_2} - 60 = 10\log \left( {\dfrac{{8I}}{I}} \right)$
$ \Rightarrow {B_2} - 60 = 10\log \left( 8 \right)$
8 can also expressed as${2^3}$ :
$ \Rightarrow {B_2} - 60 = 10\log \left( {{2^3}} \right)$
$ \Rightarrow {B_2} - 60 = 10\log \left( {{2^3}} \right)$
We know that:
$\log \left( {{a^b}} \right) = b\log a$
After applying the above relation, we have
$ \Rightarrow {B_2} - 60 = 10 \times 3\log 2$
The value of $\log 2 \approx 0.3$, so
$ \Rightarrow {B_2} - 60 = 10 \times 3 \times 0.3$
$ \Rightarrow {B_2} - 60 = 10 \times 0.9$
$ \Rightarrow {B_2} - 60 = 9$
$ \Rightarrow {B_2} = 9 + 60$
$ \Rightarrow {B_2} = 69$ dB
We have found the value of the ${B_2} = 69$dB. It means that, when 8 people are talking simultaneously in a small room, then the intensity level is $69$dB everywhere in the room.
Therefore, the option $\left( B \right)$ is correct.
Note: Remember that if the initial intensity of one person is $I$, then the intensity of eight persons is eight times the intensity of one person. That is, the intensity of eight-person is $8I$.
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