A) 25:1
B) 5:l
C) 9:4
D) 25:16
Correct Answer: C
Solution :
\[\frac{{{\operatorname{I}}_{max}}}{{{\operatorname{I}}_{min}}}=\frac{25}{1}\] \[{{\left[ \frac{{{A}_{1}}+{{A}_{2}}}{{{A}_{1}}-{{A}_{2}}} \right]}^{2}}=\frac{25}{1}\] \[\frac{{{A}_{1}}+{{A}_{2}}}{{{A}_{1}}-{{A}_{2}}}=\frac{5}{1}\] \[\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{3}{2}\] \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)}^{2}}=\frac{9}{4}\]
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