JEE Physics Communication System MCQs Set 14

Answer: (b) 3018 km²

Explanation:
1. Range \( d=\sqrt{2Rh} \), and the coverage area is \( \pi d^2 = 2\pi Rh \).
2. \( A = 2\pi\times6.4\times10^6\times75 = 3.02\times10^9 \) m² \( \approx 3018 \) km².

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (a) 500 Hz

Explanation:
1. Modulation index of FM \( m_f = \delta/f_m \).
2. \( \delta = 0.25\times2000 = 500 \) Hz.

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (b) 175 MHz

Explanation:
1. \( f_{MUF} = f_c\sec i \).
2. \( = 60/\cos70^\circ = 60/0.342 \approx 175 \) MHz.

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (a) 10

Explanation:
1. Carrier swing is twice the frequency deviation, so \( \delta = 100 \) kHz.
2. \( m_f = \delta/f_m = 100/10 = 10 \).

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (a) 150 W

Explanation:
1. \( P_t = P_c(1+m^2/2) = 1.5P_c = 900 \), so \( P_c = 600 \) W.
2. Sideband power total \( = 300 \) W, so each sideband carries 150 W.

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (b) 1503 kHz and 1497 kHz

Explanation:
1. Side bands are at \( f_c \pm f_m \).
2. \( 1500\pm3 = 1503 \) kHz and 1497 kHz.

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (d) 7 : 1

Explanation:
1. Helium atoms are excited first and transfer energy to neon atoms by collision.
2. A large excess of helium, about 7 : 1, gives efficient population inversion in neon.

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (b) stratosphere

Explanation:
1. In the troposphere and mesosphere the temperature falls with height.
2. In the stratosphere it rises with height because ozone absorbs ultraviolet radiation.

Answer: (c) 55 km

Explanation:
1. \( d = \sqrt{2Rh} \).
2. \( d = \sqrt{2\times6.4\times10^6\times240} = 5.54\times10^4 \) m \( \approx 55 \) km.

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (b) the period of revolution of the Moon is 27.3 days

Explanation:
1. A communication satellite must be geostationary, with a period of 24 hours.
2. The Moon's period is 27.3 days, so it cannot stay fixed over one place.

Answer: (c) \( \sqrt2:\sqrt5:\sqrt8 \)

Explanation:
1. Critical frequency \( f_c = 9\sqrt N \), so \( f_c\propto\sqrt N \).
2. The ratio is \( \sqrt2:\sqrt5:\sqrt8 \).

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (c) 208 kHz

Explanation:
1. \( 0.25 = 1 - 81N/f^2 \), so \( f^2 = 81N/0.75 \).
2. \( f^2 = 81\times4\times10^8/0.75 = 4.32\times10^{10} \), so \( f \approx 2.08\times10^5 \) Hz \( = 208 \) kHz.

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (c) sky wave propagation

Explanation:
1. Critical frequency \( f_c = 9\sqrt N = 9\times10^6 \) Hz \( = 9 \) MHz.
2. Since \( 5\text{ MHz} < f_c \), the wave is reflected by the ionosphere: sky wave propagation.

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (a) 80 km

Explanation:
1. \( d = \sqrt{2Rh} \).
2. \( d = \sqrt{2\times6.4\times10^6\times500} = \sqrt{6.4\times10^9} = 8\times10^4 \) m \( = 80 \) km.

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

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

Explanation:
1. Modulation allows a practical antenna size, reduces the fractional bandwidth and increases selectivity.
2. It does not reduce the time lag, as signals travel at the speed of light in either case.

Answer: (b) \( 2/3 \)

Explanation:
1. \( A_c + A_m = 10 \) and \( A_c - A_m = 2 \), so \( A_c = 6 \) V and \( A_m = 4 \) V.
2. \( \mu = A_m/A_c = 4/6 = 2/3 \).

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (c) 400 m

Explanation:
1. For an efficient antenna \( l = \lambda/4 \).
2. \( \lambda = 4l = 400 \) m.

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (b) 100 W

Explanation:
1. Total loss \( = 2\times5 = 10 \) dB, so \( 10\log(P_i/P_o) = 10 \).
2. \( P_i/P_o = 10 \), so \( P_o = 1000/10 = 100 \) W.

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (a) 1490 kHz and 1510 kHz

Explanation:
1. Side bands are \( f_c\pm f_m \) with \( f_c = 1500 \) kHz.
2. Lower \( = 1490 \) kHz, upper \( = 1510 \) kHz.

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.

Answer: (b) 9 V

Explanation:
1. \( \mu = V_m/V_c \).
2. \( V_m = 0.75\times12 = 9 \) V.

Teacher's Note:
1. Recall the standard relation first.
2. Check units before substituting.