Two sounds differ by 30 dB.The intensity of the louder sound I_L,compared with the softer I_S,is I_L/I_S.The value of the ratio is A)1000 B)30 C)9 D)100 E)300

Question

Quizplus · Accepted Answer

The difference in decibels (dB) between two sounds is given by $10 \log_{10}(\frac{I_1}{I_2})$, where $I_1$ and $I_2$ are the intensities of the two sounds. Given a 30 dB difference, we have $30 = 10 \log_{10}(\frac{I_1}{I_2})$. Solving for $\frac{I_1}{I_2}$, we get $\frac{I_1}{I_2} = 10^{30/10} = 10^3 = 1000$.

Two Sounds Differ by 30 DB