A) \[10:8\]
B) \[9:1\]
C) \[4:1\]
D) \[2:1\]
Correct Answer: C
Solution :
\[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{a_{1}^{2}}{a_{2}^{2}}=\frac{9}{1}\] \[\Rightarrow \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{9}{1}\] \[\Rightarrow \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{1}\] Then, \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}\] \[=\frac{{{(3+1)}^{2}}}{{{(3-1)}^{2}}}\] \[=\frac{{{(4)}^{2}}}{{{(2)}^{2}}}=\frac{16}{4}=\frac{4}{1}\] Thus, \[{{I}_{\max }}:{{I}_{\min }}=4:1\]
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