A) \[3:1\]
B) \[4:1\]
C) \[9:1\]
D) \[10:1\]
Correct Answer: B
Solution :
\[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{(a+b)}^{2}}}{{{(a-b)}^{2}}}\] \[\frac{9}{1}=\frac{{{(a+b)}^{2}}}{{{(a-b)}^{2}}}\] or \[\frac{3}{1}=\frac{a+b}{a-b}\] or \[3a-3b=a+b\] or\[2a=4b\]or \[a=2b\] \[\therefore \] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{a}^{2}}}{{{b}^{2}}}=\frac{4}{1}\]You need to login to perform this action.
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