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A person speaking normally produces a sound intensity of $40\, dB$ at a distance of $1\, m$. If the threshold intensity for reasonable audibility is $20\,dB$, the maximum distance at which he can be heard clearly is ..... $m$
$4$
$5$
$10$
$20$
Solution
We have, $\beta=10 \log _{10}\left(\frac{\mathrm{I}}{\mathrm{I}_{0}}\right)$
Where $I_{0}=$ threshold intensity of sound
$=10^{-12} \mathrm{W} / \mathrm{m}^{2}$
i.e., $40=10 \log _{10}\left(\frac{I_{1}}{I_{0}}\right) \ldots .(i)$
and $20=10 \log _{10}\left(\frac{I_{2}}{I_{0}}\right) \ldots(ii)$
$\frac{(i)}{(ii)} \Rightarrow \frac{40}{20}=\log _{10}\left(\frac{I_{1}}{I_{2}}\right)$
$2=\log _{10}\left(\frac{I_{1}}{I_{2}}\right) \quad$ or $\frac{I_{1}}{I_{2}}=10^{2}$
$\therefore \frac{\mathrm{r}_{2}^{2}}{\mathrm{r}_{1}^{2}}=10^{2}\left(\text { since } \mathrm{I} \propto \frac{1}{\mathrm{r}^{2}}\right)$
$\mathrm{r}_{2}^{2}=10^{2} \mathrm{r}_{1}^{2}$ or $\mathrm{r}_{2}=10 \mathrm{r}_{1}=10 \times 1=10 \mathrm{m}$
