Q175 / 180 · KCET 2008 · Physics
How many times more intense is a $60 \mathrm{~dB}$ sound than a $30 \mathrm{~dB}$ sound?
- Correct answerA1000
- B2
- C100
- D4
Solution
$\mathrm{L}=10 \log _{10} \frac{\mathrm{I}}{\mathrm{I}_{0}}$
$60=10 \log _{10} \frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}$
$\Rightarrow \frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}=10^{6}$
Similarly,
$$
\begin{aligned}
&30=10 \log _{10} \frac{\mathrm{I}_{2}}{\mathrm{I}_{0}} \\
&\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}=10^{3}
\end{aligned}
$$
Dividing Eq. (i) by Eq. (ii), we get
$$
\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=1000
$$
Hence, the correct answer is option: 'A'.