A) \[{{\lambda }_{1}}={{\lambda }_{2}}=4{{\lambda }_{3}}=9{{\lambda }_{4}}\]
B) \[4{{\lambda }_{1}}=2{{\lambda }_{2}}=2{{\lambda }_{3}}={{\lambda }_{4}}\]
C) \[{{\lambda }_{1}}=2{{\lambda }_{2}}=2\surd 2{{\lambda }_{3}}=3\surd 2{{\lambda }_{4}}\]
D) \[{{\lambda }_{1}}={{\lambda }_{2}}=2{{\lambda }_{3}}=3\surd 2{{\lambda }_{4}}\]
Correct Answer: A
Solution :
\[{{Z}_{1}}=1,{{Z}_{2}}=1,{{Z}_{3}}=2\operatorname{and}\,{{Z}_{4}}=3\] |
\[\frac{1}{\lambda }=R{{Z}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)\] |
Or \[\lambda =\frac{4}{3R{{Z}^{2}}}\] |
Or \[\lambda {{Z}^{2}}=\operatorname{constant}\] |
So \[{{\lambda }_{1}}{{\left( 1 \right)}^{2}}={{\lambda }_{2}}{{\left( 1 \right)}^{2}}={{\lambda }_{3}}{{\left( 2 \right)}^{2}}=\lambda a\left( {{3}^{2}} \right)\] |
Or \[{{\lambda }_{1}}={{\lambda }_{2}}=4{{\lambda }_{3}}=9{{\lambda }_{4}}\] |
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