- ✓$0.5$ and $9\, kHz$
- B$0.3$ and $9\, kHz$
- C$0.5$ and $10\, kHz$
- D$0.4$ and $10\, kHz$
$m=\frac{A_{m}}{A_{c}}=\frac{2}{4}=0.5$
and $(a)$ carrier wave frequency is given by
$=2 \pi \mathrm{f}_{\mathrm{c}}=2 \times 10^{4} \pi$
$\mathrm{f}_{\mathrm{c}}=1 \mathrm{kHz}$
lower side band frequency $\Rightarrow \mathrm{f}_{\mathrm{c}}-\mathrm{f}_{\mathrm{m}}$
$\Rightarrow 10 \mathrm{kHz}-1 \mathrm{kHz}=9 \mathrm{kHz}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Statement $(I)$ : When currents vary with time, Newton's third law is valid only if momentum carried by the electromagnetic field is taken into account.
Statement $(II)$ : Ampere's circuital law does not depend on Biot-Savart's law.
In the light of the above statements, choose the correct answer from the options given below:
