Question

The intensity level of sound $A$ is $30 \,dB$ greater than of $B$. How many times more intense is the sound $A$ than $B$?

• A. 10
• B. 100
• C. 1000
• D. 2

Answer 1

Answer: (C)

The intensity level in sound is given by
$L=\log _{10} \frac{I}{I_{0}}$
where $I_{0}$ is initial intensity.
Given, $ L_{A}=30+L_{B} $
$\therefore \log _{10} \frac{I_{A}}{I_{0}}=30+\log _{10} \frac{I_{B}}{I_{0}}$
$\Rightarrow \log _{10} \frac{I_{A}}{I_{B}}=3 $
$\Rightarrow \frac{I_{A}}{I_{B}}=10^{3}$
$\Rightarrow I_{A}=1000 I_{B}$
Hence, sound $A$ is $1000$ times more intense than sound $B$.