A) 49 : 1
B) 7 : 1
C) 4 : 1
D) 8 : 1
Correct Answer: C
Solution :
\[{{l}_{1}}=\frac{l}{4}\] \[{{l}_{2}}=\frac{1}{4}({{l}_{1}})\] \[=\frac{1}{16}l\] \[\Rightarrow \] \[\frac{{{l}_{\max }}}{{{l}_{\min }}}=\frac{{{({{A}_{1}}+{{A}_{2}})}^{2}}}{{{({{A}_{1}}-{{A}_{2}})}^{2}}}=\frac{{{(\sqrt{{{l}_{1}}}+\sqrt{{{l}_{2}}})}^{2}}}{{{(\sqrt{{{l}_{1}}}-\sqrt{{{l}_{2}}})}^{2}}}\] \[=\frac{(\sqrt{l}/2+\sqrt{l}/4}{{{(\sqrt{l}/2-\sqrt{l}/4)}^{2}}}\] \[=\frac{\sqrt{l}/2\,{{\left( 1+\frac{1}{2} \right)}^{2}}}{\sqrt{l}/2\left( 1-\frac{1}{2} \right)}\] \[=\frac{{{(3/2)}^{2}}}{{{(1/2)}^{2}}}=\frac{9}{1}\]You need to login to perform this action.
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