How many times is $90\text{ dB}$ sound more intense than $40\text{ dB}$ sound?
How many times is $90\text{ dB}$ sound more intense than $40\text{ dB}$ sound?
- 5
- 50
- 500
- $10^5$
Solution
Here, $90 - 40 = 10 \log \frac{I_1}{I_0} - 10 \log \frac{I_2}{I_0}$
or $50 = 10 \log \left(\frac{I_1}{I_2}\right)$
$\therefore \frac{I_1}{I_2} = 10^5$
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