A) Zero
B) 54 dB
C) 64 dB
D) 44 dB
Correct Answer: B
Solution :
\[I\propto \frac{1}{{{r}^{2}}}\Rightarrow \frac{{{I}_{2}}}{{{I}_{1}}}=\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{{{2}^{2}}}{{{(40)}^{2}}}=\frac{1}{400}\]Þ \[{{I}_{1}}=400{{I}_{2}}\] Intensity level at point 1, \[{{L}_{1}}=10{{\log }_{10}}\left( \frac{{{I}_{1}}}{{{I}_{0}}} \right)\] and intensity at point 2, \[{{L}_{2}}=10{{\log }_{10}}\left( \frac{{{I}_{2}}}{{{I}_{0}}} \right)\] \ \[{{L}_{1}}-{{L}_{2}}=10\log \frac{{{I}_{1}}}{{{I}_{2}}}=10{{\log }_{10}}(400)\] Þ \[{{L}_{1}}-{{L}_{2}}=10\times 2.602=26\] \[{{L}_{2}}={{L}_{1}}-26=80-26=54\ dB\]
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