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Multiple choice question for engineering

Set 1

1. Identify which of the following is not a transmission line?
a) Telephone lines
b) Power transmission
c) Underground cables
d) Cavity resonators

View Answer

Answer: d [Reason:] The types of transmission line are telephone lines, power transmission lines, underground cables, coaxial cables, fibre optic cable transmission etc. Cavity resonators are not transmission lines, they are components that aid maximum transmission.

2. The open wire transmission line consists of
a) Conductor
b) Dielectric
c) Both conductor and dielectric
d) Either conductor or dielectric

View Answer

Answer: c [Reason:] The open wire is a common form of transmission line. The open wire consists of conductors. The conductors of such lines are considered to be parallel and separated by a dielectric.

3. Telephone lines and power lines are open wire transmission lines. State true/false.
a) True
b) False

View Answer

Answer: a [Reason:] The open wire lines are long and used for distant wired communication. Such lines are telephone and power lines.

4. The cable transmission line consists of
a) Conductors
b) Insulators
c) Insulated conductors
d) Insulated conductors with dielectric

View Answer

Answer: d [Reason:] The cable lines are underground transmission lines. The line consists of hundreds of individual paper insulated conductors twisted in pairs and combined inside a protective lead or plastic sheath, which is usually a solid dielectric.

5. The conductors lie perpendicularly with the dielectric in the cable line. State true/false.
a) True
b) False

View Answer

Answer: b [Reason:] In all the transmission lines, the conductors are considered to be parallel with a solid dielectric.

6. The coaxial cable consists of
a) Conductors
b) Dielectric
c) Conductor with dielectric
d) Two conductors with dielectric

View Answer

Answer: d [Reason:] The coaxial cable consists of a hollow conductor and the second conductor is located inside and coaxial with the tube. The dielectric may be solid or gaseous.

7. The coaxial cable are used in
a) Telephone cables transmission
b) Power transmission
c) Television signal transmission
d) Short wave transmission

View Answer

Answer: c [Reason:] Practically, the coaxial cables are employed in the transmission of the television signals from the dish antenna to the transponder.

8. Identify which is not a type of waveguide.
a) Rectangular
b) Circular
c) Cylindrical
d) Cavity resonator

View Answer

Answer: d [Reason:] Rectangular waveguide is a commonly used waveguide. Cylindrical and circular waveguides are the same. Cavity resonator is not a waveguide.

9. The range of frequencies handled by the waveguides is in
a) Hz
b) KHz
c) MHz
d) GHz

View Answer

Answer: d [Reason:] The waveguides are operated in the GHz range. In particular, the waveguides are active above 6 GHz. The range goes upto several tens of GHz. Beyond this range, the transmission is handled by optic fibre cables.

10. The range of frequencies operated by the coaxial cables is in
a) Hz
b) kHz
c) MHz
d) GHz

View Answer

Answer: c [Reason:] Coaxial cables are operated in the MHz range. The main application includes television cable line transmission.

11. The cut off frequency for waveguide operation is
a) 2 MHz
b) 6 GHz
c) 4 MHz
d) 6 MHz

View Answer

Answer: b [Reason:] The waveguides should be operated above the cut off frequency of 6 GHz. This will lead to effective power transmission. At a frequency below this, will lead to attenuation.

12. The highest means of communication is
a) Internet
b) Telephone
c) Television
d) Radio

View Answer

Answer: a [Reason:] Though all the means of communication use the radio frequencies, the internet communication is the fastest. It involves email, voice message, video message etc. The telephone, television and radio use transmission lines for communication through radio frequencies, whereas the internet uses both wired and wireless means. Also it employs optic fibre, which uses light as the medium, since it is faster than the radio signals.

Set 2

1. The phenomenon employed in the waveguide operation is
a) Reflection
b) Refraction
c) Total internal reflection
d) Adsorption

View Answer

Answer: c [Reason:] The waveguides use total internal reflection phenomenon to transmit the waves passing through it. Thus the acceptance angle and critical angle are important for effective transmission.

2. The dominant mode in waveguide is the mode which has
a) Highest frequency
b) Highest wavelength
c) Lowest phase constant
d) Highest attenuation

View Answer

Answer: b [Reason:] The dominant mode is the mode which has the minimum frequency or maximum wavelength available for propagation of the waves.

3. The modes are calculated from which parameter?
a) Frequency
b) Wavelength
c) Phase constant
d) V number

View Answer

Answer: d [Reason:] The modes are calculated from the V number of the waveguides. It is given by M= V2/2.

4. The circular waveguides use which function in the frequency calculation?
a) Laplace function
b) Schottky function
c) Bessel function
d) Transfer function

View Answer

Answer: c [Reason:] The circular or cylindrical waveguides use the Bessel function for the frequency calculation of a particular mode.

5. The scattering parameters are used to indicate the
a) Permittivity and permeability
b) Electric and magnetic field intensities
c) Reflection and transmission coefficients
d) Frequency and wavelength

View Answer

Answer: c [Reason:] The scattering matrix consists of the transmission coefficients in the main diagonal and the reflection coefficients in the opposite diagonal.

6. Which of the following two parameter models cannot be used to represent a transmission line?
a) H parameter model
b) T parameter model
c) ABCD parameter model
d) S parameter model

View Answer

Answer: a [Reason:] The T, ABCD and S parameter models are used in the transmission line modelling. The h parameter is not used for the same.

7. For the matched line, the parameters S12 and S21 are
a) 1
b) 0
c) -1
d) ∞

View Answer

Answer: b [Reason:] The parameters S12 and S21 are the reflection coefficients. For a matched line, the reflection coefficients are zero. Thus the parameters S12 and S21 are also zero.

8. The waveguides are materials with characteristics of
a) Low bulk resistivity
b) High bulk resistivity
c) High conductivity
d) Low conductivity

View Answer

Answer: a [Reason:] Generally, the waveguides are made of materials with low bulk resistivity like brass, copper, silver etc. But if the interior walls are properly plated, it is possible with poor conductivity materials too. It is even possible to make plastic waveguides.

9. The parameters S11 and S22 indicate the transmission coefficients. State true/false.
a) True
b) False

View Answer

Answer: a [Reason:] In a scattering matrix, the parameters S11 and S22 indicate the transmission coefficients and the parameters S21 and S12 indicate the reflection coefficients.

10. The waveguides increase the transmission of the electromagnetic waves. State true/false.
a) True
b) False

View Answer

Answer: a [Reason:] The waveguides aid in effective transmission of the electromagnetic power from the source antenna to the destination antenna.

Set 3

1. The point form of Ampere law is given by
a) Curl(B) = I
b) Curl(D) = J
c) Curl(V) = I
d) Curl(H) = J

View Answer

Answer: d [Reason:] Ampere law states that the line integral of H about any closed path is exactly equal to the direct current enclosed by that path. ∫ H.dl = I The point form will be Curl (H) = J.

2. The Ampere law is based on which theorem?
a) Green’s theorem
b) Gauss divergence theorem
c) Stoke’s theorem
d) Maxwell theorem

View Answer

Answer: c [Reason:] The proof of the Ampere’s circuital law is obtained from Stoke’s theorem for H and J only.

3. Electric field will be maximum outside the conductor and magnetic field will be maximum inside the conductor. State True/False.
a) True
b) False

View Answer

Answer: a [Reason:] At the conductor-free space boundary, electric field will be maximum and magnetic field will be minimum. This implies electric field is zero inside the conductor and increases as the radius increases and the magnetic field is zero outside the conductor and decreases as it approaches the conductor.

4. Find the magnetic flux density of a finite length conductor of radius 12cm and current 3A in air( in 10-6 order)
a) 4
b) 5
c) 6
d) 7

View Answer

Answer: b [Reason:] The magnetic field intensity is given by H = I/2πr, where I = 3A and r = 0.12. The magnetic flux density in air B = μ H, where μ = 4π x 10-7.Thus B = 4π x 10-7 x 3/2π x 0.12 = 5x 10-6 units.

5. Calculate the magnetic field intensity due to a toroid of turns 50, current 2A and radius 159mm.
a) 50
b) 75
c) 100
d) 200

View Answer

Answer: c [Reason:] The magnetic field intensity is given by H = NI/2πrm, where N = 50, I = 2A and rm = 1/2π. Thus H = 50 x 2/2π x 0.159 = 100 units.

6. Find the magnetic field intensity due to an infinite sheet of current 5A and charge density of 12j units in the positive y direction and the z component is above the sheet.
a) -6
b) 12k
c) 60
d) 6

View Answer

Answer: d [Reason:] The magnetic field intensity when the normal component is above the sheet is Hx = 0.5 K, where K = 12. Thus we get H = 0.5 x 12 = 6 units.

7. Find the magnetic field intensity due to an infinite sheet of current 5A and charge density of 12j units in the positive y direction and the z component is below the sheet.
a) 6
b) 0
c) -6
d) 60k

View Answer

Answer: c [Reason:] The magnetic intensity when the normal component is below the sheet is Hy = -0.5 K, where K = 12.Thus we get H = -0.5 x 12 = -6 units.

8. Find the current density on the conductor surface when a magnetic field H = 3cos x i + zcos x j A/m, for z>0 and zero, otherwise is applied to a perfectly conducting surface in xy plane.
a) cos x i
b) –cos x i
c) cos x j
d) –cos x j

View Answer

Answer: b [Reason:] By Ampere law, Curl (H) = J. The curl of H will be i(-cos x) – j(0) + k(-z sin x) = -cos x i – zsin x k. In the xy plane, z = 0. Thus Curl(H) = J = -cos x i.

9. When the rotational path of the magnetic field intensity is zero, then the current in the path will be
a) 1
b) 0
c) ∞
d) 0.5

View Answer

Answer: b [Reason:] By Ampere law, Curl(H) = J. The rotational path of H is zero, implies the curl of H is zero. This shows the current density J is also zero. The current is the product of the current density and area, which is also zero.

10. Find the magnetic field intensity when the current density is 0.5 units for an area up to 20 units.
a) 10
b) 5
c) 20
d) 40

View Answer

Answer: a [Reason:] We know that ∫ H.dl = I. By Stoke’s law, we can write Curl(H) = J. In integral form, H = ∫ J.ds, where J = 0.5 and ds is defined by 20 units. Thus H = 0.5 x 20 = 10 units.

Set 4

1. The intrinsic impedance of a wave with electric and magnetic field of 10 and 8 respectively is
a) 1.6
b) 1.11
c) 1.25
d) 0.8

View Answer

Answer: b [Reason:] The intrinsic impedance of a wave is given by η = √(E/H). on substituting for E = 10 and H = 8, we get η = 1.11 units.

2. The intrinsic impedance is defined as the ratio of the magnetic field to the electric field of the electromagnetic wave. State true/false
a) True
b) False

View Answer

Answer: b [Reason:] The intrinsic impedance is defined as the ratio of the electric field intensity to the magnetic field intensity. It is denoted by η. It is a complex quantity.

3. The intrinsic impedance in free space is
a) 60π
b) 12π
c) 6π
d) 120π

View Answer

Answer: d [Reason:] The intrinsic impedance of the free space is the ratio of the permittivity to the permeability. For air medium, the value is 120π or 377 ohm.

4. The propagation constant is a complex quantity. State true/false.
a) True
b) False

View Answer

Answer: a [Reason:] The propagation constant is a complex quantity. It is given by γ = α + jβ, where α is the attenuation constant, a real value and β is the phase constant, a complex value.

5. Calculate the propagation constant of a wave with impedance and admittance given by 32 and 12 respectively.
a) 19.6
b) 17.6
c) 15.6
d) 13.6

View Answer

Answer: a [Reason:] The propagation constant is given by γ = √(ZY), where Z is the impedance and Y is the admittance. On substituting for Z = 32 and Y = 12, we get γ = 19.6 units.

6. The intrinsic angle is 250, find the loss angle.
a) 12.5
b) 50
c) 25
d) 75

View Answer

Answer: b [Reason:] The loss angle is twice of the intrinsic angle. It is given by δ = 2θn. On substituting for θn = 25, the loss angle δ = 2(25) = 50 degree.

7. The standing wave ratio of the wave with maximum and minimum electric field intensities of 12 and 4 is
a) 12
b) 4
c) 3
d) 48

View Answer

Answer: c [Reason:] The SWR is defined as the ratio of maximum electric field intensities to the minimum field intensities. It is given by S = EMAX/EMIN. Thus for the given data, S = 12/4 = 3.

8. The standing wave ratio of a wave travelling through two media having intrinsic impedances of 3 and 2 is
a) 23
b) 32
c) 15
d) 5

View Answer

Answer: b [Reason:] The standing wave ratio is given by the ratio of the intrinsic impedance of medium 1 to the intrinsic impedance of medium 2. Thus S = η1/η2. On substituting for η1 = 3 and η2 = 2, we get S = 3/2.

9. The reflection coefficient of a wave travelling through two media having electric intrinsic impedances of 3 and 5 respectively is
a) 2
b) 8
c) 53
d) 1/4

View Answer

Answer: d [Reason:] The reflection coefficient in terms for intrinsic impedance is R= η2-η1/η2+η1. On substituting the given data, we get R = 5 – 3/5 + 3 = 2/8 = ¼.

10. The reflection coefficient of a wave travelling through two media having magnetic intrinsic impedances of 2 and 1 respectively is
a) 12
b) 13
c) 14
d) 15

View Answer

Answer: b [Reason:] The reflection coefficient in terms for magnetic intrinsic impedance is R = η1 – η2/η1 + η2. On substituting the given data, we get R = 2 – 1/2 + 1 = 1/3.

11. The transmission coefficient of a wave travelling through two media having electric intrinsic impedances of 3 and 2 respectively is
a) 22
b) 89
c) 45
d) 14

View Answer

Answer: c [Reason:] The transmission coefficient in terms for intrinsic impedance is R = 2η2/η2 + η1. On substituting the given data, we get R = 2 x 2/2 + 3 = 4/5.

12. The transmission coefficient of a wave travelling through two media having electric intrinsic impedances of 3 and 4 respectively is
a) 23
b) 89
c) 25
d) 67

View Answer

Answer: d [Reason:] The transmission coefficient in terms for intrinsic impedance is R = 2η1/η2 + η1. On substituting the given data, we get R = 2 x 34 + 3 = 6/7.

Set 5

1. Biot Savart law in magnetic field is analogous to which law in electric field?
a) Gauss law
b) Faraday law
c) Coulomb’s law
d) Ampere law

View Answer

Answer: c [Reason:] Biot Savart law states that the magnetic flux density H = I.dl sinθ/4πr2, which is analogous to the electric field F = q1q2/4πεr2, which is the Coulomb’s law.

2. Which of the following cannot be computed using the Biot Savart law?
a) Magnetic field intensity
b) Magnetic flux density
c) Electric field intensity
d) Permeability

View Answer

Answer: c [Reason:] The Biot Savart law is used to calculate magnetic field intensity. Using which we can calculate flux density and permeability by the formula B = μH.

3. Find the magnetic field of a finite current element with 2A current and height 1/2π is
a) 1
b) 2
c) 1/2
d) 1/4

View Answer

Answer: a [Reason:] The magnetic field due to a finite current element is given by H = I/2πh. Put I = 2 and h = 1/2π, we get H = 1 unit.

4. Calculate the magnetic field at a point on the centre of the circular conductor of radius 2m with current 8A.
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] The magnetic field due to a point in the centre of the circular conductor is given by H = I/2a. Put I = 8A and a = 2m, we get H = 8/4 = 2 units.

5. The current element of the solenoid of turns 100, length 2m and current 0.5A is given by,
a) 100 dx
b) 200 dx
c) 25 dx
d) 50 dx

View Answer

Answer: c [Reason:] The current element of the solenoid is given by NI dx/L. Put N = 100, I = 0.5 and L = 2 to get, I dx = 100 x 0.5 x dx/2 = 25 dx.

6. Find the magnetic field intensity at the centre O of a square of the sides equal to 5m and carrying 10A of current.
a) 1.2
b) 1
c) 1.6
d) 1.8

View Answer

Answer: d [Reason:] The magnetic field is given by H = 4I/√2πω. Put I = 10 and ω = 5m. Thus H = 4 x 10/√2π(5) = 1.8 unit.

7. Find the magnetic flux density when a point from a finite current length element of current 0.5A and radius 100nm.
a) 0
b) 0.5
c) 1
d) 2

View Answer

Answer: c [Reason:] The magnetic flux density is B = μH, where H is given by I/2πr. Put μ = 4π x 10-7, I = 0.5 and r = 10-7, we get B = 4π x 10-7 x 0.5/2π x 10-7 = 1 unit.

8. In a static magnetic field only magnetic dipoles exist. State True/False.
a) True
b) False

View Answer

Answer: a [Reason:] From Gauss law for magnetic field, we get divergence of the magnetic flux density is always zero (ie, Div(B) = 0). This implies the non-existence of magnetic monopole.

9. The magnetic field intensity will be zero inside a conductor. State true/false.
a) True
b) False

View Answer

Answer: b [Reason:] Electric field will be zero inside a conductor and magnetic field will be zero outside the conductor. In other words, the conductor boundary, E will be maximum and H will be minimum.

10. Find the magnetic field when a circular conductor of very high radius is subjected to a current of 12A and the point P is at the centre of the conductor.
a) 1
b) ∞
c) 0
d) -∞

View Answer

Answer: c [Reason:] The magnetic field of a circular conductor with point on the centre is given by I/2a. If the radius is assumed to be infinite, then H = 12/2(∞) = 0.

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