A) 1 : 4
B) 1 : 2
C) 1 : 104
D) 1 : 102
Correct Answer: D
Solution :
By using \[L={{\log }_{10}}\frac{I}{{{I}_{0}}}\] \[{{L}_{2}}-{{L}_{1}}={{\log }_{10}}\frac{{{I}_{2}}}{{{I}_{0}}}-{{\log }_{10}}\frac{{{I}_{1}}}{{{I}_{0}}}\] \[5-1={{\log }_{10}}\frac{{{I}_{2}}}{{{I}_{1}}}\]Þ \[4={{\log }_{10}}\frac{{{I}_{2}}}{{{I}_{1}}}\]Þ \[\frac{{{I}_{2}}}{{{I}_{1}}}={{10}^{4}}\] Þ \[\frac{a_{2}^{2}}{a_{1}^{2}}={{10}^{4}}\]Þ \[\frac{{{a}_{2}}}{{{a}_{1}}}=\frac{{{10}^{2}}}{1}\]Þ \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{{{10}^{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
