Question

90 dB sound is x times more intense than 40 dB sound, then x is

• A. 5
• B. 50
• C. $10^5$
• D. 500

Answer 1

Answer: (C)

Sound level, $L=10\log \left(\frac{I}{I_0}\right)$
Let intensity of 40 dB sound be $I$, then for 90 dB sound, intensity will be $I_x$.
$\therefore 40dB=\left(10dB\right)\log \left(\frac{I}{I_0}\right)$ ....(i)
and $90dB=\left(10dB\right)\log \left(\frac{I_x}{I_0}\right)$.....(ii)
or, $\log \left(\frac{I}{I_0}\right)=4$ and $9=\log \left(\frac{I}{I_0}\right)+\log x$.
or ,$9=4+\log xor,\log x=5$.
$\therefore x=10^5$