A) 3 : 2
B) 5 : 1
C) 25 : 1
D) 9 : 1
Correct Answer: C
Solution :
\[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{a_{1}^{2}}{a_{2}^{2}}\] \[\therefore \] \[\sqrt{\frac{9}{4}}=\frac{{{a}_{1}}}{{{a}_{2}}}\] \[\Rightarrow \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{2}\] \[\therefore \] \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}={{\left( \frac{3+2}{3-2} \right)}^{2}}\] or \[{{I}_{\max }}:{{I}_{\min }}=25:1\]You need to login to perform this action.
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