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NEET Sample Paper NEET Sample Test Paper-89

  • question_answer
    The tuning circuit of a radio receiver has a resistance of \[50\,\Omega ,\]an inductance of 10mH and a variable capacitance C. A 1 MHz radio wave produces a potential difference of 0.1 mV. The value of capacitance to produce the resonance will be

    A) 2.5 pF  

    B)        5.0 pF

    C) 25 pF   

    D)        50 pF

    Correct Answer: A

    Solution :

    [a] The resonant frequency \[{{f}_{r}}=\frac{1}{2\pi \sqrt{LC}}\] \[\Rightarrow \]\[{{10}^{6}}=\frac{1}{2\pi \sqrt{10\times {{10}^{-3}}\times C}}\] \[\Rightarrow \]\[C=\frac{1}{4{{\pi }^{2}}\times 10\times {{10}^{-3}}\times {{10}^{12}}}\] \[=2.5\times {{10}^{-12}}F=2.5pF\]


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