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\[n\propto \sqrt{I}\] \[\therefore \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\sqrt{\frac{I}{4l}}\] \[\Rightarrow \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{2}\] \[\frac{{{l}_{max}}}{{{\operatorname{I}}_{min}}}={{\left( \frac{{{a}_{1}}+{{a}_{2}}}{{{a}_{1}}-{{a}_{2}}} \right)}^{2}}\] \[={{\left( \frac{1+2}{1-2} \right)}^{2}}\] \[=\frac{9}{1}\] \[\therefore \]\[{{\operatorname{I}}_{max}}=9I,{{\operatorname{I}}_{min}}=I\]
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