A) 5I and 3I
B) 9I and 3I
C) 4I and I
D) 9I and I
Correct Answer: D
Solution :
[d] \[{{I}_{1}}=I,{{I}_{2}}=4I\] \[{{I}_{\max }}={{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}={{\left( \sqrt{I}+\sqrt{4I} \right)}^{2}}={{\left( I+2I \right)}^{2}}={{\left( 3I \right)}^{2}}\] \[{{I}_{\max }}={{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}={{\left( \sqrt{I}-\sqrt{4I} \right)}^{2}}={{\left( I-2I \right)}^{2}}={{\left( -I \right)}^{2}}\] \[\frac{{{I}_{\max }}}{{{I}_{\max }}}=\frac{9}{1}\]
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