A) 10
B) 100
C) 1000
D) 2
Correct Answer: C
Solution :
The intensity level in sound is given by \[L={{\log }_{10}}\frac{I}{{{I}_{0}}}\] where\[{{I}_{0}}\]is initial intensity. Given, \[{{L}_{A}}=30+{{L}_{B}}\] \[\therefore \] \[{{\log }_{10}}\frac{{{I}_{A}}}{{{I}_{0}}}=30+{{\log }_{10}}\frac{{{I}_{B}}}{{{I}_{0}}}\] Since, \[1\,dB=\frac{1}{10}bel\] \[\therefore \] \[{{\log }_{10}}\frac{{{I}_{A}}}{{{I}_{0}}}-{{\log }_{10}}\frac{{{I}_{B}}}{{{I}_{0}}}=3\,bel\] \[\Rightarrow \] \[{{\log }_{10}}\frac{{{I}_{A}}}{{{I}_{B}}}=3\] \[\Rightarrow \] \[\frac{{{I}_{A}}}{{{I}_{B}}}={{10}^{3}}\] \[\Rightarrow \] \[{{I}_{A}}=1000{{I}_{B}}\] Hence, sound A is 1000 times more intense than sound B.
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