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) ?

Answer

7

Explanation

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`