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The correct option is (a) SURFACE wave

To explain I would say: Ground waves are also known as Surface waves. Ground waves are vertically polarized and transmitting and receiving antennas are PLACED closely and the wave follows the curvature of the EARTH.

Right choice is (b) False

To explain: If the WAVE is horizontally polarized, then the electric FIELD is short CIRCUITED by the earth. So the ground wave is always vertically polarized and VERTICAL antennas are USED.

Right CHOICE is (a) D LAYER

To ELABORATE: During NIGHT time D layer disappears and F1, F2 layers combine TOGETHER to form an F layer. D layer is present at 70km from the earth surface and is the nearest layer to earth surface. E layer lies at 110km, f1 layer at 220km and F2 at 250 to 400km.

The CORRECT option is (b) 15km

The explanation is: The LOWEST layer of the ATMOSPHERE is troposphere. So it ranges up to 15km from the SURFACE of the EARTH. The gas components remain almost constant in this region.

Right option is (a) 2

To explain: During night time D LAYER is not PRESENT. F1 and F2 layers get combined to FORM an F layer during night time. Therefore only two layers NAMELY E and F are present at night time.

Correct answer is (c) IONOSPHERE

To EXPLAIN: Ionosphere is most widely used for long DISTANCE communication and it lies at 90 to 400km from EARTH surface. D LAYER is not present at night hours.

The correct CHOICE is (a) Appleton layer

Best explanation: E layer is called as Kennelly Heaviside layer and F layer is called as Appleton layer. D layer is also KNOWN as absorption layer for SHORT WAVE signal at HF.

Right CHOICE is (a) Tropopause

Explanation: At Tropopause starts after the top of troposphere and ends at stratosphere. Above critical height called Tropopause the temperature remains CONSTANT THROUGHOUT belt and increases thereby. ENTIRE belt of troposphere is called Region of change.

The correct ANSWER is (a) Horizontal

To explain I would say: Any horizontal component of electric field which is in contact with EARTH is short CIRCUITED by earth. USUALLY ground wave propagation is DONE by vertical antennas so it is vertically polarized.

Right CHOICE is (b) 0

To explain I would say: The reflection factor is given by the ratio of Reflected wave to incident wave.

\(ρ=\frac{reflected \,wave}{incident \,wave}=\frac{E_r}{E_i} \)

No reflections MEANS ER = 0 =>ρ = 0

The correct OPTION is (a) \(\frac{4πh_t h_r}{λd^2}E_0\)

The explanation: Space wave propagation reflects frequencies above 30MHz. The FIELD strength due to space wave propagation is \(E=\frac{2E_0}{d} sin⁡ \frac{2πh_t h_r}{λd} ≈ \frac{4πh_t h_r}{λd^2}E_0\)

Here E0 is the field strength due to LOS at a unit distance which depends on transmits power, HT and HR is the HEIGHT of transmitting and receiving antenna and d is the distance between two antennas.

Correct answer is (a) Outer ATMOSPHERE

To elaborate: Outer atmosphere is ALSO known as G region. It lies above 400km from the earth surface. E layer is called as Kennelly Heaviside layer and F layer is called as Appleton layer. D layer is also known as ABSORPTION layer for short wave signal at HF.

The correct answer is (a) Surface wave

Explanation: GROUND waves are also known as Surface waves are USED for low frequencies and broadcasting. TROPOSPHERIC waves are used for MF and HF signals. Ionospheric waves are used for long DISTANCE communications.

Correct choice is (a) F layer

The explanation: F layer is the TOP layer in the ionosphere and is also PRESENT at the night time. So it is used for the long distance sky wave propagation and also its IONIZATION density is HIGH. D layer is not present during night hours.

The CORRECT choice is (a) 50 to 400km

For explanation: SKY WAVE propagation also known as ionosphere propagation REFLECTS the waves with frequency 2 to 30MHz. It is present at 50 to 400km from earth surface.

Right answer is (a) \(θ=1/2 tan^{-1}⁡(\frac{σ}{ωϵ})\)

The explanation is: The wave tilt of the surface wave DEPENDS on the CONDUCTIVITY of EARTH, permittivity and ANGULAR frequency of the waves. Expression for tilt angle is \(θ=1/2 tan^{-1}⁡(\frac{σ}{ωϵ}). \)

Right OPTION is (b) It is the ANGLE of incident at which no REFLECTIONS occur

The explanation is: Brewster angle is defined as the angle of incident at which no reflections occurs.

\(θ=tan^{-1}⁡\sqrt{\frac{ϵ_{r2}}{ϵ_{r1}}} \)

Correct option is (d) D layer

The best explanation: D layer is ALSO known as absorption layer for short wave SIGNAL at HF. It reflects SIGNALS up to VLF and LF. Its absorption increases with solar activity.

Right option is (b) E and F layers

The BEST I can explain: During NIGHT TIME D layer is not present. F1 and F2 layers get combined to FORM an F layer during night time. So E and F layers are present at night time.

The correct ANSWER is (b) 15KM

For EXPLANATION I WOULD say: The Troposphere portion of earth extends up to 15km from the earth surface. IONOSPHERE from 50 to 400km and Outer atmosphere extends above 400km.

The correct OPTION is (b) Sky wave

To explain I would say: In Sky wave or IONOSPHERIC wave propagation waves are reflected from the ionosphere layers DEPENDING upon different FREQUENCIES.

Right option is (b) 2 MHz to 30MHz

For explanation I WOULD say: GROUND wave propagation also known as SURFACE wave propagation is used up to 2MHz. SKY wave propagation is used at 2MHz to 30MHz.

Right choice is (a) \(1.8×10^6\FRAC{σ}{f(MHZ)}\)

For explanation: DISSIPATION factor for the DIELECTRIC is Df = \(1.8×10^6\frac{σ}{f(MHz)}\), σ is the conductivity of earth. This is used to calculate the phase constant for the surface wave given by \(b=tan^{-1}(\frac{ϵ+1}{D_f})\)

The CORRECT answer is (B) False

For explanation: As the wave front tilts more and more towards the surface then the electric field gets short CIRCUITED. As it gets reduced with tilt, at some point it is COMPLETELY attenuated.

Correct answer is (a) Surface WAVE

Explanation: GROUND waves are also KNOWN as Surface waves as the wave propagates close to the surface of earth. Space waves are known as TROPOSPHERIC waves and SKY waves as Ionospheric waves.

Right option is (a) 20- 70km

Explanation: STRATOSPHERE is present at 20 to 70km from earth SURFACE. TROPOSPHERE is up to 15KM and above 400km is the outer atmosphere.

The correct ANSWER is (a) VHF

For EXPLANATION I WOULD say: Space WAVE propagation reflects frequencies at 30 to 300MHz range. So it propagates at VHF band.

MF- 300 KHz – 3MHz

EHF- 30GHz-300GHz

The correct OPTION is (a) True

Easy explanation: Brewster angle is DEFINED as the angle of incident at which no REFLECTIONS OCCURS.

\(θ=tan^{-1}⁡\sqrt{\frac{ϵ_{r2}}{ϵ_{r1}}}\)

At θ =90, tan θ=∞, which doesn’t exist in REALITY.

Right choice is (a) Reflected WAVE to INCIDENT wave

Explanation: The REFLECTION factor is given by the ratio of Reflected wave to incident wave.

\(ρ=\frac{reflected \,wave}{incident\, wave}=\frac{E_r}{E_i} \)

ρ=1 It means that surface is perfect REFLECTING MEDIUM.

The correct ANSWER is (a) R < 0.1

To explain: Earth is CLASSIFIED as electrically SMOOTH and electrically rough.

If the ROUGHNESS of earth R < 0.1, then it is electrically smooth.

If the roughness of earth R > 10, then it is electrically rough.

The CORRECT answer is (d) \(\frac{2h_t h_r}{d}\)

To ELABORATE: The path difference GIVEN by two-ray MODEL is \(\frac{2h_t h_r}{d}.\) and its phase difference is given as \(∅=\frac{2π}{λ} ∆x=\frac{4πh_t h_r}{d}.\)

The correct choice is (b) False

Explanation: LINE of sight provides a direct communication link from transmitter to receiver. If the HEIGHT of ANTENNAS is increased then the LOS is ALSO improved.

Correct option is (a) AEo/d

Explanation: The GROUND wave FIELD strength at a point \(E=\frac{AE_o}{d}V/m \)

Where Eo is field strength of wave at UNIT distance from TRANSMITTING antenna

A is factor of ground losses

D is distance of point from Transmitting antenna.

Right ANSWER is (a) (4πd/λ)^2

The BEST explanation: Spatial ATTENUATION coefficient reflects the decrease in the power DENSITY due to the spherical SPREAD of the wave. It is also known as the loss factor. Expression for loss factor is given by Ls=(4πd/λ)^2

The correct OPTION is (a) Vertically

Best explanation: If the wave is horizontally polarized, then the ELECTRIC field is SHORT CIRCUITED by the earth. So the ground wave is always vertically polarized and vertical ANTENNAS are used.

The correct choice is (a) Forward SCATTER PROPAGATION

For explanation: Forward scatter propagation or TROPOSPHERIC propagation occurs at frequencies above 300MHz through the fine layers or blobs in the troposphere. UFH and microwaves PROPAGATE much beyond the LOS through forward scattering in tropospheric irregularities.

Right answer is (a) TROPOSPHERE

Easy explanation: Troposphere is the lowest layer and nearest to EARTH surface in the atmosphere RANGING up to 15km. Stratosphere lies at 50 to 90km and up to 400km IONOSPHERE. Above 400km is the outer atmosphere.

Correct OPTION is (a) poles

The BEST explanation: The actual height of troposphere is least at poles and maximum at equator. At other COMPOSITIONS it is ALMOST REMAINS same. The entire belt of troposphere is called Region of change.

Right option is (d) Above 30MHz

For explanation: GROUND wave propagation also known as SURFACE wave propagation is USED up to 2MHZ. sky wave propagation is used at 2MHz to 30MHz. Space wave propagation REFLECTS frequencies above 30MHz.

The correct option is (B) Infinity

Explanation: When the ray APPEARS to be a straight line over a flat earth surface horizontally, its refractive INDEX is not CHANGING with respect to height. \(\frac{dM}{dh}=0\)

Effective radius is k times the ACTUAL radius and \(k=0.048/\frac{dM}{dh}=∞\)

Right option is (a) 1

For explanation: The reflection factor is defined as the RATIO of Reflected WAVE to incident wave.

\(ρ=\FRAC{reflected \,wave}{incident \,wave}=\frac{E_r}{E_i} \)

For a perfect REFLECTING surface, total incident waves are reflected completely. Er = Ei

\(ρ=\frac{E_r}{E_i}=1.\)

The correct answer is (a) \(P_D=\frac{P_t G_t}{4πd^2}\)

To ELABORATE: Power density is the power per unit AREA at a distance d from the transmitter and is GIVEN by \(P_D=\frac{P_t G_t}{4πd^2}\) where Pt is the transmitter power and Gt is the transmitter ANTENNA gain.

Power received by the antenna is given by \(P_r=\frac{P_t G_t G_r}{(4πd/λ)^2}\)

The correct option is (a) N=(μ-1)×10^-6

The best I can explain: The EXCESS refractive index over UNITY in MILLIONTHS is given by N=(μ-1)×10^-6

Where refractive index μ=0.0003% GREATER than unity for TROPOSPHERE.

Correct OPTION is (a) Leaky CAPACITOR

To EXPLAIN: Any horizontal component of electric field which is in contact with earth is short circuited by earth. So earth BEHAVES like a leaky capacity. It forms a resistor in shunt with a capacitor.