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14.Waves and Sound
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A sound absorber attenuates (decreases) the sound level by $20\, dB$. The intensity decreases by a factor of
A
$100$
B
$1000$
C
$10000$
D
$10$
Solution
$\mathrm{L}_{1}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}\right) ; \quad \mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}\right)$
$\therefore \mathrm{L}_{1}-\mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}\right)-10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}\right)$
or $\quad \Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right) \quad$ or $\quad 20 \mathrm{dB}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$
or $10^{2}=\frac{I_{1}}{I_{2}}$ or $I_{2}=\frac{I_{1}}{100}$
Standard 11
Physics
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