A) 9 : 4
B) 3 : 2
C) 25 : 1
D) 5 : 1
Correct Answer: C
Solution :
\[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{9}{4}\] \[\therefore \]\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{2}\] \[\therefore \]\[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}=\frac{{{(3+2)}^{2}}}{{{(3-2)}^{2}}}=\frac{2.5}{1}\] Hence, the correction option is [c].
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