JEE Main & Advanced Sample Paper JEE Main Sample Paper-23

  • 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 \[=3.0\times {{10}^{8}}m{{s}^{-1}},\pi =3.14\])

    A)  \[9.42\times {{10}^{4}}\] m                

    B)  \[18.8\times {{10}^{4}}\] m

    C)  \[4.5\times {{10}^{4}}\] m                  

    D)  None

    Correct Answer: A

    Solution :

    In an LC-circuit, the impedance of circuit is given by \[Z={{X}_{L}}-{{X}_{C}}\] When \[{{X}_{L}}={{X}_{C}}\] then Z = 0. In this situation, the amplitude s 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 \[=\frac{velocity}{wavelength}\] \[\Rightarrow \,\lambda =\frac{c}{f}\,=\frac{3\times {{10}^{8}}}{318.7}\,=9.42\,\times {{10}^{4}}m\]


You need to login to perform this action.
You will be redirected in 3 sec spinner