A) 10
B) 200
C) 100
D) 400
Correct Answer: C
Solution :
Let \[{{L}_{1}}=10\log \left( \frac{{{I}_{1}}}{{{I}_{0}}} \right)\] and \[{{L}_{2}}=10\log \left( \frac{{{I}_{2}}}{{{I}_{0}}} \right)\] Therefore, difference of sound levels \[\Delta L={{L}_{2}}-{{L}_{1}}=10\log \left( \frac{{{I}_{2}}}{{{I}_{1}}} \right)\] Now, \[40=10\log \left( \frac{{{I}_{2}}}{{{I}_{1}}} \right)=20\log \left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)\] \[[\because \,I\propto {{P}^{2}}]\] or \[\frac{40}{20}=\log \left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)\]or \[\frac{{{P}_{2}}}{{{P}_{1}}}={{(10)}^{2}}=100\]
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