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Multiple choice question for engineering

Set 1

1. The waveguide is employed in the transmission lines, when operated at the range of
a) Hz
b) KHz
c) MHz
d) GHz

View Answer

Answer: d [Reason:] Waveguides are employed for effective transmission, when the lines carry electromagnetic waves in the GHz range.

2. The cut off frequency for a waveguide to operate is
a) 3 MHz
b) 3 GHz
c) 6 MHz
d) 6 GHz

View Answer

Answer: d [Reason:] The cut off frequency of the waveguide is 6 GHz. This is the frequency at which the waveguide will start to operate.

3. In rectangular waveguides, the dimensions a and b represent the
a) Broad wall dimensions
b) Broad wall and side wall dimension respectively
c) Side wall and broad wall dimension respectively
d) Side wall dimensions

View Answer

Answer: b [Reason:] In rectangular waveguide, the a parameter is the broad wall dimension of the waveguide and the b parameter is the side wall dimension of the waveguide. Always, a > b in a waveguide.

4. The Bessel function is denoted by
a) Jn(ha)
b) Jm(ha)
c) Jn(hb)
d) Jm(hb)

View Answer

Answer: a [Reason:] The Bessel function is used in the circular waveguides. Normally Jn(ha) = 0. Here n is the order of the Bessel function.

5. In a waveguide, always which condition holds good?
a) phase velocity = c
b) phase velocity greater than c
c) phase velocity lesser than c
d) group velocity = c

View Answer

Answer: b [Reason:] In air medium, the phase velocity is assumed to be the speed of light. For waveguides, the phase velocity is always greater than the speed of the light.

6. The group wavelength is greater than the wavelength at any point. State true/false.
a) True
b) False

View Answer

Answer: a [Reason:] In a waveguide, the phase velocity is greater than the velocity of light. Thus the group velocity will be less. This implies the group wavelength will be greater than the wavelength at any point.

7. Find the group wavelength of a wave, given that the group phase constant is 6.28 units.
a) 2
b) 3.14
c) 6.28
d) 1

View Answer

Answer: d [Reason:] The group wavelength is given by λg = 2π/βg, where βg is the group wavelength of the wave. On substituting for βg = 6.28, we get group wavelength as unity.

8. The phase velocity of a wave with frequency of 15 radian/sec and group phase constant of 2 units is
a) 30
b) 15
c) 7.5
d) 2/15

View Answer

Answer: c [Reason:] The phase velocity of a wave is given by Vp = ω/βg. on substituting for ω = 15 and βg = 2, we get phase velocity as 15/2 = 7.5 units.

9. The modes in a material having a V number of 20 is
a) 20
b) 400
c) 200
d) 100

View Answer

Answer: c [Reason:] The relation between the modes and the V number is given by m = v2/2. Given that v = 20, we get m = 202/2 = 200 modes.

10. The number of modes is given as 50, find the V number.
a) 5
b) 10
c) 50
d) 2500

View Answer

Answer: b [Reason:] The relation between the modes and the V number is given by m = v2/2. Given that m = 50, we get v2 = 2 x 50 = 100. The V number is 10.

Set 2

1. The reflection coefficient of a wave with transmission coefficient 0.35 is
a) 1.35
b) 0.65
c) 0.35
d) 0.7

View Answer

Answer: b [Reason:] The reflection coefficient is the reverse of the transmission coefficient. Thus T + R = 1. On substituting for T = 0.35, the reflection coefficient R will be 1 – 0.35 = 0.65 (no unit).

2. The incident wave amplitude is 24 units. Find the reflected wave amplitude if the reflection coefficient is 0.6.
a) 14.4
b) 16.6
c) 13.3
d) 11.1

View Answer

Answer: a [Reason:] The reflection coefficient is the ratio of the reflected amplitude to the incident amplitude. Thus R = Er/Ei. On substituting for Ei = 24 and R = 0.6, we get Er = R Ei = 0.6 X 24 = 14.4 units.

3. Find the reflection coefficient of the wave passing through two media having intrinsic impedances of 4 and 9 respectively.
a) 0.5
b) 1
c) 0.38
d) 0.1

View Answer

Answer: c [Reason:] The reflection coefficient in terms of intrinsic impedances is given by R = η2 – η1/η2 + η1. On substituting for η1 = 4 and η2 = 9, we get R = 9 – 4/9 + 4 = 5/13 = 0.38.

4. The reflection coefficient of a wave travelling through two media having permittivities 4 and 9 respectively is
a) 0
b) 0.5
c) 0.25
d) 0.2

View Answer

Answer: d [Reason:] The reflection coefficient in terms of permittivity is given by R = √ε2 – √ε1/√ε2 + √ε1. On substituting for ε1 = 4 and ε2 = 9, we get R = 3 – 2/3 + 2 = 1/5 = 0.2.

5. Calculate the transmission coefficient, when the incident and transmitted amplitudes are 10 and 7 respectively.
a) 17
b) 3
c) 10/7
d) 0.7

View Answer

Answer: d [Reason:] The transmission coefficient is defined as the ratio of the transmitted amplitude to the incident amplitude. Thus T = Et/Ei. On substituting for Ei = 10 and Et = 7, we get T = 7/10 = 0.7.

6. The transmission coefficient in a wave travelling through two media having intrinsic impedances of 5.5 and 1.33 is
a) 0.389
b) 0.55
c) 0.133
d) 0.42

View Answer

Answer: a [Reason:] The transmission coefficient is terms of the intrinsic impedance is given by T = 2η2/η1 + η2. On substituting for η1 = 5.5 and η2 = 1.33, we get T = 2 x 1.33/1.33 + 5.5 = 2.66/6.83 = 0.389.

7. The transmission coefficient in a wave travelling through two media having permittivities 4 and 1 is
a) 1/4
b) 3/2
c) 3/4
d) 2/3

View Answer

Answer: d [Reason:] The transmission coefficient in terms of the permittivity is given by T= 2√ε2/√ε1 + √ε2. On substituting for ε1 = 4 and ε2 = 1, we get T= 2(1)/1 + 2 = 2/3.

8. The reflection coefficient of a transmission line having characteristic and load impedances as 50 and 30 ohm respectively is
a) 1/4
b) 1/8
c) 1/2
d) 3/4

View Answer

Answer: a [Reason:] The reflection coefficient of a transmission line is given by, R = ZL – Zo/ZL + Zo, where ZL and Zo is the load and characteristic impedances respectively. On substituting ZL = 30 and Zo = 50, the reflection coefficient R = 50 – 30/50 + 30 = 20/80 = 1/4.

9. In matched line, the transmission coefficient is
a) 0
b) 1
c) -1
d) Infinity

View Answer

Answer: b [Reason:] In matched line, the maximum power is transferred from transmitter to receiver. Such reflection coefficient will be zero and transmission coefficient is unity.

10. Find the power reflected in a transmission line, when the reflection coefficient and input power are 0.45 and 18V respectively.
a) 3.645
b) 6.453
c) 4.563
d) 5.463

View Answer

Answer: a [Reason:] The ratio of the reflected to incident amplitudes gives the reflection. Similarly, the ratio of reflected to incident power gives square of the reflection coefficient. Thus Prefl = R2Pinc. On substituting for R = 0.45 and Pinc = 18, we get Prefl = 0.452 x 18 = 3.645 units.

11. The transmitted power in a transmission line, when the reflection coefficient and the incident power are 0.6 and 24V respectively, is
a) 15.36
b) 51.63
c) 15.63
d) 51.36

View Answer

Answer: a [Reason:] The transmitted power in terms of the reflection coefficient and the incident power is Ptr = (1-R2)Pinc, where R = 0.6 and Pinc = 24, from the given data. Thus Ptr = (1- 0.62) x 24 = 15.36 units.

12. The reflection coefficient of a short circuit transmission line is -1. State True/False.
a) True
b) False

View Answer

Answer: a [Reason:] For a short circuit line, the losses are maximum due to heavy current flow. This leads to less transmission and more attenuation. Thus the reflection coefficient is negative. R = -1 for short circuit lines.

Set 3

1. The Snell’s law can be derived from which type of incidence?
a) Incidence angle
b) Reflected angle
c) Refracted angle
d) Oblique incidence

View Answer

Answer: d [Reason:] The oblique incidence refers to the interface between dielectric media. Consider a planar interface between two dielectric media. A plane wave is incident at an angle from medium 1 and reflected from medium 2. The interface plane defines the boundary between the media. This is the oblique medium.

2. The Snell’s law is given by
a) N1 sin θi = N2 sin θt
b) N2 sin θi = N1 sin θt
c) sin θi = sin θt
d) N1 cos θi = N2 cos θt

View Answer

Answer: a [Reason:] The Snell law states that in an oblique medium, the product of the refractive index and sine of incidence angle in medium 1 is same as that of medium 2. Thus it is given by N1 sin θi = N2 sin θt.

3. Calculate the ratio of sine of incident angle to the sine of reflected angle when the refractive indices of medium 1 and 2 are given as 2.33 and 1.66 respectively.
a) 0.71
b) 1.4
c) 2
d) 3.99

View Answer

Answer: a [Reason:] The Snell law is given by N1 sin θi = N2 sin θt. To get sin θi/sin θt, the ratio is N2/N1. On substituting for N1 = 2.33 and N2 = 1.66, we get 1.66/2.33 = 0.71.

4. Find the ratio of the refractive index of medium 1 to that of medium 2, when the incident and reflected angles are given by 300 and 450 respectively.
a) 0.5
b) 1
c) 2
d) 4

View Answer

Answer: c [Reason:] The Snell law is given by N1 sin θi = N2 sin θt. For getting N1/N2, the ratio is sin θt/sin θi. On substituting for θi = 30 and θt = 45, we get sin 45/sin 30 = 2.

5. The refractive index of a medium with permittivity of 2 and permeability of 3 is given by
a) 3.56
b) 2.45
c) 3.21
d) 1.78

View Answer

Answer: b [Reason:] The refractive index is given by n = c √(με), where c is the speed of light. Given that relative permittivity and relative permeability are 2 and 3 respectively. Thus n = 3 x 108 √(2 x 4π x 10-7 x 3 x 8.854 x 10-12) = 2.45.

6. The critical angle is defined as the angle of incidence at which the total internal reflection starts to occur. State True/False.
a) True
b) False

View Answer

Answer: a [Reason:] The critical angle is the minimum angle of incidence which is required for the total internal reflection to occur. This is the angle that relates the refractive index with the angle of reflection in an oblique incidence medium.

7. The critical angle for two media of refractive indices of medium 1 and 2 given by 2 and 1 respectively is
a) 0
b) 30
c) 45
d) 60

View Answer

Answer: b [Reason:] The sine of the critical angle is the ratio of refractive index of medium 2 to that in medium 1. Thus sin θc = n2/n1. To get θc, put n1 = 2 and n2 = 1. Thus we get θc = sin-1(n2/n1) = sin-1(1/2) = 30 degree.

8. The critical angle for two media with permittivities of 16 and 9 respectively is
a) 48.59
b) 54.34
c) 60
d) 45

View Answer

Answer: a [Reason:] The sine of the critical angle is the ratio of refractive index of medium 2 to that in medium 1. Thus sin θc = n2/n1. Also n = √ε, thus sin θc = √ε2/√ε1. Put ε1 = 16 and ε2 = 9, we get θc = sin-1(3/4) = 48.59 degree.

9. The angle of incidence is equal to the angle of reflection for perfect reflection. State True/False.
a) True
b) False

View Answer

Answer: a [Reason:] For complete wave reflection, the angle of incidence should be same as the angle of the reflection. In such cases, the reflection coefficient is unity and the transmission coefficient is zero.

10. The angle of incidence of a wave of a wave with angle of transmission 45 degree and the refractive indices of the two media given by 2 and 1.3 is
a) 41.68
b) 61.86
c) 12.23
d) 27.89

View Answer

Answer: a [Reason:] The Snell law is given by N1 sin θi = N2 sin θt. To get θi, put N1 = 2, N2 = 1.3, θt = 45 degree. Thus we get θi = sin-1(1.3 sin 45)/2 = 41.68 degree.

11. The angle at which the wave must be transmitted in air media if the angle of reflection is 45 degree is
a) 45
b) 30
c) 60
d) 90

View Answer

Answer: a [Reason:] In air media, n1 = n2 = 1. Thus, sin θi=sin θt and the angle of incidence and the angle of reflection are same. Given that the reflection angle is 45, thus the angle of incidence is also 45 degree.

12. For a critical angle of 60 degree and the refractive index of the first medium is 1.732, the refractive index of the second medium is
a) 1
b) 1.5
c) 2
d) 1.66

View Answer

Answer: b [Reason:] From the definition of Snell law, sin θc = n2/n1. To get n2, put n1 = 1.732 and θc = 60. Thus we get sin 60 = n2/1.732 and n2 = 1.5.

Set 4

1. In a transverse electric magnetic wave, which of the following will be true?
a) E is transverse to H
b) E is transverse to wave direction
c) H is transverse to wave direction
d) E and H are transverse to wave direction

View Answer

Answer: d [Reason:] In the transverse electric magnetic wave (TEM wave), both the electric and magnetic field strengths are transverse to the wave propagation.

2. The cut off frequency of the TEM wave is
a) 0
b) 1 GHz
c) 6 GHz
d) infinity

View Answer

Answer: a [Reason:] The TEM waves have both E and H perpendicular to the guide axis. Thus its cut off frequency is zero.

3. Which component is non zero in a TEM wave?
a) Ex
b) Hz
c) Ez
d) Attenuation constant

View Answer

Answer: a [Reason:] In a TEM wave, the wave propagates along the guided axis. Thus the components Ez and Hz are zero. The attenuation is also zero. The non-zero component will be Ex.

4. TEM wave can propagate in rectangular waveguides. State true/false.
a) True
b) False

View Answer

Answer: b [Reason:] The rectangular waveguide does not allow the TEM wave. TEM mode can exist only in two conductor system and not in hollow waveguide in which the centre conductor does not exist.

5. The cut off wavelength in the TEM wave will be
a) 0
b) Negative
c) Infinity
d) 1/6 GHz

View Answer

Answer: c [Reason:] The cut off frequency in a TEM wave is zero. Thus the cut off wavelength will be infinity.

6. The guided wavelength of a TEM wave in a waveguide having a wavelength of 5 units is
a) 0
b) Infinity
c) 5
d) 1/5

View Answer

Answer: c [Reason:] The guided wavelength is same as the wavelength of the waveguide with a TEM wave. Thus the guided wavelength is 5 units.

7. The guided phase constant of a TEM wave in a waveguide with a phase constant of 2.8 units is
a) 2.8
b) 1.4
c) 0
d) Infinity

View Answer

Answer: a [Reason:] The guided phase constant is same as the phase constant of the waveguide. For the given data, the guided phase constant is 2.8 units.

8. Which type of transmission line accepts the TEM wave?
a) Copper cables
b) Coaxial cable
c) Rectangular waveguides
d) Circular waveguides

View Answer

Answer: b [Reason:] Hollow transmission lines support TE and TM waves only. The TEM wave is possible only in the coaxial cable transmission line, which is not hollow.

9. For a TEM wave to propagate in a medium, the medium has to be
a) Air
b) Insulator
c) Dispersive
d) Non dispersive

View Answer

Answer: d [Reason:] The medium in which the TEM waves propagate has to be non- dispersive. This implies the phase velocity and the characteristic impedance has to be constant over a wide band.

10. Stripline and parallel plate waveguides support the TEM wave. State true/false.
a) True
b) False

View Answer

Answer: a [Reason:] The stripline and parallel plate waveguides are not hollow and the dielectric is lossless. The medium is non dispersive. Thus the statement is true.

Set 5

1. When the phase angle between the Ex and Ey component is 00 or 1800, the polarisation is
a) Elliptical
b) Circular
c) Linear
d) Perpendicular

View Answer

Answer: c [Reason:] The phase angle between the Ex and Ey component is 00 and 1800 for linearly polarised wave. The wave is assumed to be propagating in the z direction.

2. The magnitude of the Ex and Ey components are same in which type of polarisation?
a) Linear
b) Circular
c) Elliptical
d) Perpendicular

View Answer

Answer: b [Reason:] In circular polarisation, the magnitude of the Ex and Ey components are the same. This is a form of the elliptical polarisation in which the major and minor axis are the same.

3. When the Ex and Ey components of a wave are not same, the polarisation will be
a) Linear
b) Elliptical
c) Circular
d) Parallel

View Answer

Answer: b [Reason:] In elliptical polarisation, the magnitude of Ex and Ey components are not same. This is due to the variation in the major and minor axes of the waves representing its magnitude.

4. Identify the polarisation of the wave given, Ex = Exo cos wt and Ey = Eyo sin wt. The phase difference is +900.
a) Left hand circularly polarised
b) Right hand circularly polarised
c) Left hand elliptically polarised
d) Right hand elliptically polarised

View Answer

Answer: c [Reason:] The magnitude of the Ex and Ey components are not same. Thus it is elliptical polarisation. For +90 phase difference, the polarisation is left handed. In other words, the rotation is in clockwise direction. Thus the polarisation is left hand elliptical.

5. Identify the polarisation of the wave given, Ex = 2 cos wt and Ey = sin wt. The phase difference is -900.
a) Left hand circularly polarised
b) Right hand circularly polarised
c) Left hand elliptically polarised
d) Right hand elliptically polarised

View Answer

Answer: d [Reason:] The magnitude of the Ex and Ey components are not same. Thus it is elliptical polarisation. For -90 phase difference, the polarisation is right handed. In other words, the rotation is in anti-clockwise direction. Thus the polarisation is right hand elliptical.

6. Identify the polarisation of the wave given, Ex = 2 cos wt and Ey = 2 sin wt. The phase difference is +900.
a) Left hand circularly polarised
b) Right hand circularly polarised
c) Left hand elliptically polarised
d) Right hand elliptically polarised

View Answer

Answer: a [Reason:] The magnitude of the Ex and Ey components are the same. Thus it is circular polarisation. For +90 phase difference, the polarisation is left handed. In other words, the rotation is in clockwise direction. Thus the polarisation is left hand circular.

7. Identify the polarisation of the wave given, Ex = cos wt and Ey = sin wt. The phase difference is -900.
a) Left hand circularly polarised
b) Right hand circularly polarised
c) Left hand elliptically polarised
d) Right hand elliptically polarised

View Answer

Answer: b [Reason:] The magnitude of the Ex and Ey components are the same. Thus it is circular polarisation. For -90 phase difference, the polarisation is right handed. In other words, the rotation is in anti-clockwise direction. Thus the polarisation is right hand circular.

8. For a non-zero Ex component and zero Ey component, the polarisation is
a) Parallel
b) Perpendicular
c) Elliptical
d) Circular

View Answer

Answer: a [Reason:] When the Ex is non-zero and the Ey is zero, the polarisation is parallel. The parallel polarisation is classified under the linear polarisation type.

9. Identify the polarisation of the wave given that, Ex = 2 cos wt and Ey = cos wt.
a) Elliptical
b) Circular
c) Parallel
d) Linear

View Answer

Answer: d [Reason:] The magnitude of the Ex and Ey components are not the same. Thus it cannot be circular polarisation. For a phase difference of 0, the polarisation is linear. In other words, the waves are in phase. Thus the polarisation is linear.

10. The Snell law is applicable for perpendicular polarisation and the Brewster law is applicable for parallel polarisation. State True/False.
a) True
b) False

View Answer

Answer: a [Reason:] The Snell law is calculated from the oblique incidence media. Thus it is applicable for perpendicular polarisation. The Brewster law is applicable for perpendicular polarisation.

11. When the polarisation of the receiving antenna is unknown, to ensure that it receives atleast half the power, the transmitted wave should be
a) Linearly polarised
b) Elliptically polarised
c) Circularly polarised
d) Normally polarised

View Answer

Answer: c [Reason:] The polarisation of the transmitting and receiving antenna has to be the same. This is the condition for maximum power transfer to occur. This is possible only when the polarisation is circular.

12. Identify the polarisation of the wave given that, Ex = 2 sin wt and Ey = 3 sin wt.
a) Linear
b) Elliptical
c) Circular
d) Parallel

View Answer

Answer: a [Reason:] The magnitude of the Ex and Ey components are not the same. Thus it cannot be circular polarisation. For a phase difference of 0, the polarisation is linear. In other words, the waves are in phase. Thus the polarisation is linear.