MCQ
How many times more intense is $90\,\, dB$ sound than $40\,\, dB$ sound?
- A$5$
- B$50$
- C$500$
- ✓$10^5$
✓
Answer
Correct option: D.
$10^5$
d
Let $I$ represents the intensity of sound.
Let $I$ represents the intensity of sound.
Loudness of sound $\quad L=10 \log _{10} \frac{I}{I_{o}}$
$\Longrightarrow L_{2}-L_{1}=10 \log _{10} \frac{I_{2}}{I_{1}}$
$90-40=10 \log _{10} \frac{I_{2}}{I_{1}}$
$5=\log _{10} \frac{I_{2}}{I_{1}} \Longrightarrow \frac{I_{2}}{I_{1}}=10^{5}$
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