A) \[\frac{f}{3}\]
B) f
C) \[\frac{4f}{3}\]
D) \[\frac{3f}{4}\]
Correct Answer: D
Solution :
\[\frac{1}{f}=({{\mu }_{g}}-1)\frac{2}{R}=\frac{1}{R},\left( {{\mu }_{g}}=\frac{3}{2} \right),R=f\] \[\frac{1}{{{f}^{1}}}=({{\mu }_{W}}-1)\frac{2}{R}=-\frac{2}{3R}=-\frac{2}{3f}\] \[\frac{1}{{{f}_{eq}}}=\frac{1}{f}+\frac{1}{f}+\frac{1}{{{f}_{1}}}=\frac{1}{R}+\frac{1}{R}-\frac{2}{3R}\] \[\frac{1}{{{f}_{eq}}}=\frac{1}{f}+\frac{1}{f}=\frac{2}{3f}\] \[\frac{1}{{{f}_{eq}}}=\frac{2}{f}-\frac{2}{3f}\] \[\frac{1}{{{f}_{eq}}}=\frac{4}{3f}\] \[{{f}_{eq}}=\frac{3f}{4}\]You need to login to perform this action.
You will be redirected in
3 sec