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Communication — JEE physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceJEE physicsCommunication1 Mark
MCQ
A $500 \mathrm{~Hz}$ modulating voltage fed into an FM generator produces a frequency deviation of $2.25 \mathrm{kHz}$. If amplitude of the voltage is kept constant but frequency is raised to $6 \mathrm{kHz}$ then the new deviation will be
  • A
    $4.5 \mathrm{kHz}$
  • ✓
    $54 \mathrm{kHz}$
  • C
    $27 \mathrm{kHz}$
  • D
    $15 \mathrm{kHz}$
✓

Answer

Correct option: B.
$54 \mathrm{kHz}$
$m_f=\frac{\delta}{f_m}=\frac{2250}{500}=4.5$
$\therefore$ New deviation $=2\left(m_f f_m\right)=2 \times 4.5 \times 6=54 \mathrm{kHz}$.

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