JEE Physics Communication System MCQs Set 06

Answer: (c) 9 V

Explanation:
1. \( \mu=\frac{V_m}{V_c} \), so \( V_m=0.75 imes12=9 \) V.

Teacher's Note:
1. Modulation index is the ratio of message to carrier amplitude.

Answer: (b) \( \frac23 \)

Explanation:
1. \( \mu=\frac{A_{max}-A_{min}}{A_{max}+A_{min}}=\frac{8}{12}=\frac23 \).

Teacher's Note:
1. Max = Ac(1+mu), min = Ac(1-mu).

Answer: (d) 45.5 km

Explanation:
1. \( d=\sqrt{2Rh_T}+\sqrt{2Rh_R} \).
2. \( \sqrt{2\times6.4\times10^6\times32}=20.2 \) km and \( \sqrt{2\times6.4\times10^6\times50}=25.3 \) km, total 45.5 km.

Teacher's Note:
1. Each antenna has its own horizon distance.

Answer: (a) 1010 kHz and 990 kHz

Explanation:
1. Side bands are at \( f_c\pm f_m=1000\pm10 \) kHz.
2. So they are 1010 kHz and 990 kHz.

Teacher's Note:
1. Upper and lower side bands lie at fc plus and minus fm.

Answer: (a) to reduce the time lag between transmission and reception of the information signal

Explanation:
1. Radio waves travel at the speed of light whether modulated or not, so modulation does not shorten the time lag.
2. The other three are genuine reasons (antenna size, fractional bandwidth, selectivity and multiplexing).

Teacher's Note:
1. Modulation lets us use short antennas and many channels.

Answer: (b) \( 10^{31} \)

Explanation:
1. Photon energy \( =\frac{hc}{\lambda}=\frac{6.63\times10^{-34}\times3\times10^8}{500}=4\times10^{-28} \) J.
2. \( n=\frac{10^4}{4\times10^{-28}}=2.5\times10^{31} \) per second.

Teacher's Note:
1. Number per second = power / photon energy.

Answer: (b) \( \omega_c,\ \omega_c+\omega_m \) and \( \omega_c-\omega_m \)

Explanation:
1. Expanding, \( E_m\cos\omega_mt\cos\omega_ct=\frac{E_m}2[\cos(\omega_c+\omega_m)t+\cos(\omega_c-\omega_m)t] \).
2. So the carrier and the two side bands are present.

Teacher's Note:
1. Product of cosines gives sum and difference frequencies.

Answer: (b) \( A(1+\sin\omega_mt)\sin\omega_ct \) and 1

Explanation:
1. In AM the carrier amplitude is varied: \( [A+m(t)]\sin\omega_ct=A(1+\sin\omega_mt)\sin\omega_ct \).
2. The modulation index is \( \frac{A}{A}=1 \).

Teacher's Note:
1. Modulation index = message amplitude / carrier amplitude.

Answer: (c) Electromagnetic waves of frequencies higher than about 30 MHz cannot penetrate the ionosphere

Explanation:
1. Waves above about 30 MHz do penetrate the ionosphere (they are not reflected), which is why satellite communication uses such frequencies.
2. The other statements are correct, so (c) is the incorrect one.

Teacher's Note:
1. Sky wave range is about 3 to 30 MHz.

Answer: (b) 5 MHz and 25 MHz

Explanation:
1. The ionosphere reflects frequencies up to about 30 MHz.
2. Of the choices, 5 to 25 MHz is in the sky wave band.

Teacher's Note:
1. Above about 30 MHz waves escape; below that they are reflected.

Answer: (c) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal

Explanation:
1. In AM the carrier amplitude follows the message amplitude.
2. In FM the carrier frequency (not amplitude) follows the message.

Teacher's Note:
1. AM varies amplitude, FM varies frequency.

Answer: (a) 80 km

Explanation:
1. \( d=\sqrt{2Rh}=\sqrt{2\times6.4\times10^6\times500}=8\times10^4 \) m = 80 km.

Teacher's Note:
1. Only the height of the radar matters for an object on the ground.

Answer: (d) 0.014 N/C, 360 m

Explanation:
1. \( E_0=cB_0=3\times10^8\times4.82\times10^{-11}=0.0145 \) N/C.
2. \( \lambda=\frac cf=\frac{3\times10^8}{8.3\times10^5}=361 \) m.

Teacher's Note:
1. E = cB for an electromagnetic wave.

Answer: (b) 150 Hz, 50 Hz and 2/3

Explanation:
1. The carrier term has \( 300\pi=2\pi\times150 \), so \( f_c=150 \) Hz; the side bands are at 100 Hz and 200 Hz, so \( f_m=50 \) Hz.
2. Each side band has amplitude \( \frac{\mu A_c}{2}=10 \) with \( A_c=30 \), so \( \mu=\frac23 \).

Teacher's Note:
1. Side band amplitude is mu Ac / 2.