question_answer

The inductance of the oscillatory circuit of a radio 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\[3.0\times {{10}^{8}}\,m/s,\,\pi =3.14\])

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

Correct Answer: A

Solution :

Key Idea: When inductive reactance is equal to capacitive reactance circuit is in resonance.                          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 cycles/s. Also frequency \[=\frac{velocity}{wavelength}\] \[\Rightarrow \] \[\lambda =\frac{c}{f}=\frac{3\times {{10}^{8}}}{3184.7}\] \[\Rightarrow \] \[\lambda =9.42\times {{10}^{4}}\,m\]