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)

$10^5$

Answer 2

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 $I_x$.
$\therefore \, 40 dB = \left(10 dB\right) \log\left(\frac{I}{I_{0}}\right)$ ....(i)
and $90 dB = \left(10 dB\right) \log\left(\frac{I_{x}}{I_{0}}\right)$.....(ii)
or, $\log \left(\frac{I}{I_{0}}\right)= 4$ and $9 = \log \left(\frac{I}{I_{0}}\right) + \log x$.
or ,$9= 4 + \log x or , \log x = 5$.
$\therefore \, x = 10^{5}$