COMEDK · Physics · 16. Waves and Sound
\(90 \mathrm{~dB}\) sound is \(x\) times more intense than \(40 \mathrm{~dB}\) sound, then \(x\) is
- A 5
- B 50
- C \(10^{5}\)
- D 500
Answer & Solution
Correct Answer
(C) \(10^{5}\)
Step-by-step Solution
Detailed explanation
The level of sound is given by \(L=10 \log \left(\frac{I}{I_{0}}\right) \mathrm{dB}\) Let for \(40 \mathrm{~dB}\) sound intensity be \(I\) and for \(90 \mathrm{~dB}\) sound intensity be \(I_{x}\), then \(40=10 \log \left(\frac{I}{I_{0}}\right)\)…
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