Question

If the intensity of sound is doubled, by how many decibels does the sound level increase? (In dB)
A. $1$
B. $2$
C. $3$
D. $10$

Answer 1

Here, we have to determine the increase in the sound level when the intensity of sound is doubled to that of the original intensity of sound. We must know the concept of the sound level and use the definition properly and calculate the answer.

Complete step by step answer:
Let us consider the original intensity of sound as ${I_0}$ and the new intensity of sound be $I$. We know that the sound level is defined as the logarithm of intensity as shown below mathematically.
$\beta (dB) = 10{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)$ …. $\left( 1 \right)$
Where, $\beta$ is a sound intensity level in decibels.
Now, when the intensity of sound is doubled to that of the original intensity of sound, we have
$I = 2{I_0}$
Now, we put this value in equation $\left( 1 \right)$ and we get:
$\beta (dB) = 10{\log _{10}}\left( {\dfrac{{2{I_0}}}{{{I_0}}}} \right)$
$\Rightarrow \beta (dB) = 10{\log _{10}}\left( 2 \right)$
$\Rightarrow \beta (dB) = 10 \times 0.3010$.....…. (On placing the value of ${\log _{10}}(2)$)
Now, we have to calculate the value and find the required answer such that:
$\therefore {\text{Sound level (}}\beta {\text{) = }}3.010{\text{ decibels}}$

Thus, the correct answer is option C.

Note: Some more information on sound level intensity: These sound intensity levels are quoted in decibels rather than watt meters per second. The choice of such units in the scientific literature is based on the sound that is perceived by us. It is measured more accurately by the logarithmic of intensity rather than directly by intensity.