A)
B)
C)
D)
Correct Answer: C
Solution :
\[{{X}_{L}}=2\pi fL\] Þ \[{{X}_{L}}\propto f\]\[\Rightarrow \frac{1}{{{X}_{L}}}\propto \frac{1}{f}\] i.e., graph between \[\frac{1}{{{X}_{L}}}\]and f will be a hyperbola.
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