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)  \[18\,\times 8.\,{{10}^{4}}m\]  

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 \[\text{=}\,\frac{\text{velocity}}{\text{wavelength}}\] \[\Rightarrow \]               \[\lambda =\frac{c}{f}=\frac{3\times {{10}^{8}}}{3184.7}\] \[\Rightarrow \]               \[\lambda =9.42\times {{10}^{4}}m\]