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