A) \[\frac{1}{100}\]
B) \[\frac{1}{10}\]
C) \[\frac{10}{1}\]
D) \[\frac{3}{2}\]
Correct Answer: D
Solution :
\[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{100}{1}\] Now \[\frac{{{\operatorname{I}}_{max}}}{{{\operatorname{I}}_{min}}}={{\left( \frac{\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}}+1}{\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}}-1} \right)}^{2}}={{\left( \frac{\sqrt{100}+1}{\sqrt{100}-1} \right)}^{2}}=\frac{121}{81}=\frac{3}{2}\]
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