Q 12-08-158JEE MainJEE Main 2021 (25 Feb, Shift 1)Medium
A transmitting station releases waves of wavelength 960 m. A capacitor of $2.56\ \mu\text{F}$ is used in the resonant circuit. The self-inductance of coil necessary for resonance is $x\times10^{-8}$ H. Find $x$.
Numerical value type. Enter your answer.
Answer: 10
$f = \dfrac{c}{\lambda} = \dfrac{3\times10^8}{960} = 3.125\times10^5$ Hz.
$$L = \frac{1}{4\pi^2f^2C} = \frac{1}{39.48\times(3.125\times10^5)^2\times2.56\times10^{-6}} = \frac{1}{39.48\times2.5\times10^5} \approx 1.0\times10^{-7}\ \text{H}$$
So $L = 10\times10^{-8}$ H and $x = 10$.
Solution by Sreeraj P, M.Sc Physics