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JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-II

  • question_answer
    The carrier frequency of a transmitter is provided by a tank circuit of a coil of inductance \[49\mu H\] and a capacitance of\[2.5nF\]. It is modulated by an audio signal of\[12kHz\]. The frequency range occupied by the side bands is:                                                                                                                                 [JEE Online 15-04-2018 (II)]

    A) \[18kHz-30kHz\]              

    B) \[63kHz-75kHz\]              

    C) \[442kHz-466kHz\]         

    D) \[13482kHz-13494kHz\]

    Correct Answer: C

    Solution :

    \[w=\frac{1}{\sqrt{LC}}\] \[=\frac{1}{\sqrt{49\times {{10}^{-6}}\times \frac{2.5}{10}\times {{10}^{-9}}}}\] \[=\frac{1}{7\times 5\times {{10}^{-8}}}=\frac{{{10}^{8}}}{7\times 5}=w\] \[=\frac{{{10}^{8}}}{7\times 5}=2\pi \times f=2\times \frac{22}{7}\times f\] \[\frac{{{10}^{8}}}{22\times 10}=f\] \[\frac{{{10}^{7}}}{22}=f\] \[\frac{{{10}^{4}}}{22}kHz=f\] \[f=454.54kHz\] For frequency range \[454.54\pm 12kHz\] \[442kHz-466kHz\]


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