A) 25 : 1
B) 25 : 16
C) 9 : 4
D) 5 : 1
Correct Answer: C
Solution :
[c]: \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{(a+b)}^{2}}}{{{(a-b)}^{2}}}=\frac{25}{1}\] \[\frac{a+b}{a-b}=\frac{5}{1}\]or\[5a-5b-a+b\] or\[4a=6b\]or\[\frac{a}{b}=\frac{3}{2}\therefore \frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{a}^{2}}}{{{b}^{2}}}=\frac{9}{4}\]You need to login to perform this action.
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