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The impedance of coaxial cable, when its inductance is $0.40 μ H$ and capacitance is $1 \times 10^{- 11} F$ , can be

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$2 \times 10^{2} Ω$

$100 Ω$

$3 \times 10^{3} Ω$

$3 \times 10^{- 2} Ω$

answer is A.

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Detailed Solution

$U \sin g Z = \sqrt{\frac{L}{C}}, w e g e t Z = \sqrt{\frac{0.40 \times 10^{- 6}}{10^{- 11}}} = 2 \times 10^{2} Ω$

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