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A condenser of capacitor `0.144 pF` is used in a transmitte to transmit at wavelength `lambda`. If inductace of `1//pi^(2) mH` is used for resonance, what is the value of `lambda` (in m) ? |
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Answer» Correct Answer - 7 Here,`C = 0.144 pF = 0.144 xx 10^(-12) F` `L = (1)/(pi^(2)) mH = (10^(-3))/(pi^(2)) H, v = (1)/(2 pi sqrt(LC))` `lambda = (upsilon)/(v) = upsilon . 2 pi sqrt(LC)` `=3 xx 10^(8) xx 2 pi sqrt((10^(3))/(pi^(2)) xx 0.144 xx 10^(-12))` `lambda = 3 xx 10^(8) xx 2 pi xx (1.2 xx 10^(-8))/(pi) = 7.2 m` The correct answer is `7 m` |
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