A) \[\sqrt{{{\operatorname{r}}_{1}}\,\,{{r}_{2}}}\]
B) \[{{\operatorname{r}}_{1}}+\,{{r}_{2}}\]
C) \[{{\operatorname{r}}_{1}}-\,{{r}_{2}}\]
D) \[\frac{{{\operatorname{r}}_{1}}+\,{{r}_{2}}}{2}\]
Correct Answer: C
Solution :
\[\frac{{{\operatorname{I}}_{max}}}{{{\operatorname{I}}_{min}}}=\frac{{{\left( \sqrt{{{\operatorname{I}}_{1}}}+\sqrt{{{\operatorname{I}}_{2}}} \right)}^{2}}}{{{\left( \sqrt{{{\operatorname{I}}_{1}}}-\sqrt{{{\operatorname{I}}_{2}}} \right)}^{2}}}=\frac{9}{1}\] \[\frac{\sqrt{{{\operatorname{I}}_{1}}}+\sqrt{{{\operatorname{I}}_{2}}}}{\sqrt{{{\operatorname{I}}_{1}}}-\sqrt{{{\operatorname{I}}_{2}}}}=\frac{3}{1}\] \[3\sqrt{{{\operatorname{I}}_{1}}}-3\sqrt{{{\operatorname{I}}_{2}}}=\sqrt{{{\operatorname{I}}_{1}}}+\sqrt{{{\operatorname{I}}_{2}}}\] \[2\sqrt{{{\operatorname{I}}_{1}}}=4\sqrt{{{\operatorname{I}}_{1}}}\] \[\frac{{{\operatorname{I}}_{1}}}{{{\operatorname{I}}_{2}}}=\frac{4}{1}\] \[\frac{K{{A}^{2}}_{1}}{K{{A}^{2}}_{2}}=\frac{4}{1}\] \[\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{2}{1}\]
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