A) \[2:1\]
B) \[3:1\]
C) \[6:1\]
D) \[9:1\]
Correct Answer: D
Solution :
We have \[\frac{{{l}_{\max }}}{{{\operatorname{l}}_{\min }}}={{\left( \frac{\sqrt{{{l}_{1}}}+\sqrt{{{l}_{2}}}}{\sqrt{{{l}_{1}}}-\sqrt{{{l}_{2}}}} \right)}^{2}}=\frac{9}{1}\] \[(\because \,{{l}_{1}}=4/2)\]You need to login to perform this action.
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