Q. The inductance of the oscillatory circuit of the ratio station is 10 mH and its capacitance is $0.25\ \mu F$. Taking the effect of resistance negligible, wavelength of the broadcasted waves will be (velocity of light $= 3.0 \times 10^{8}\ ms^{- 1},\ \pi = 3.14$ )
A) $9.42 \times 10^{4}\ m$
B) $18.8 \times 10^{4}\ m$
C) $4.5 \times 10^{4}\ m$
D) None of these
Explanation
(a) In an L – C circuit the impedance of circuit is
$Z = X_{L} - X_{C}$
When$X_{L} = X_{C}$ , then Z = 0. In this situation the amplitude of current in the circuit would be infinite. It will be condition of electrical resonance and frequency is given by
$f = \frac{1}{2\pi\sqrt{LC}}$
$= \frac{1}{2 \times 3.14 \times \sqrt{10 \times 10^{- 3} \times 0.25 \times 10^{- 6}}}$
$= 3184.7\ $cycle$\ s^{- 1}$
Also frequency $= \frac{velocity}{wavelength}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{c}{f} = \frac{3 \times 10^{8}}{3184.7}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = 9.42 \times 10^{4}\ $m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.