A) 25 : 16
B) 5 : 3
C) 16 : 1
D) 25 : 9
Correct Answer: C
Solution :
\[{{a}_{1}}:{{a}_{2}}=3:5\] Let \[{{a}_{1}}=3a\] and \[{{a}_{2}}=5a\] \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}\] \[=\frac{{{(3a+5a)}^{2}}}{{{(3a-5a)}^{2}}}\] \[=\frac{64}{4}=\frac{16}{1}\]
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