Select Page
Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

# Multiple choice question for engineering

## Set 1

1. The resonant circuit in a waveguide refers to the
a) Tank circuit
b) RC circuit
c) Bridge circuit
d) Attenuator circuit

Answer: a [Reason:] The resonant circuit refers to the tank circuit. It is parallel combination of an inductor and a capacitor.

2. When a waveguide is terminated, the mode of the guided termination will be
a) zero
b) non-zero
c) infinite
d) does not exist

Answer: b [Reason:] A waveguide mode with a guided termination is represented by TEmnp and TMmnp. Here m and n are the orders of the waveguide and p is the order of the resonant cavity. It is always a non-zero value.

3. The cut off frequency of a waveguide with resonant cavity is given by
a) V√((m/a)2 + (n/b)2 + (p/d)2)/2
b) V√((m/a)2 + (n/b)2 + (p/d)2)
c) 2V√((m/a)2 + (n/b)2 + (p/d)2)
d) V√((m/a) + (n/b) + (p/d))/2

Answer: a [Reason:] The cut off frequency of the waveguide of dimensions a x b with the resonant cavity of dimension d is given by fc = √((m/a)2 + (n/b)2 + (p/d)2)/2. Here m and n are the orders of the waveguide, p is the order of the cavity and v is the velocity.

4. The power of a wave in a transmission line, when the current and the resistance are 5A and 120 ohm respectively is
a) 3000
b) 4000
c) 2000
d) 1500

Answer: a [Reason:] The power of a line is given by P = I2R. On substituting for I = 5 and R = 120, we get P = 52 x 120 = 3000 units.

5. The power of a wave having a magnetic field intensity of 2.5 units is
a) 1.17 kilo watt
b) 1.17 mega watt
c) 1.17 watt
d) 11.7 watt

Answer: a [Reason:] The power of a wave is given by P = 0.5 η H2, where η is the intrinsic impedance and H is the magnetic field intensity. On substituting for η = 377 and H = 2.5, we get P = 0.5 x 377 x 2.52 = 1.17 kilo watts.

6. The power of a wave having an electric field strength of 12.8 units is
a) 0.217
b) 0.721
c) 0.127
d) 0.172

Answer: a [Reason:] The power of a wave is given by P = 0.5 E2/η, where E is the electric field intensity and η is the intrinsic impedance. On substituting for E = 12.8 and η = 377, we get P = 0.5 x 12.82/377 = 0.217 units.

7. The form or mode of propagation is determined by which factors?
a) Type of excitation device
b) Location of excitation device
c) Type and location of the excitation device
d) Waveguide characteristics

Answer: c [Reason:] The form and the mode of propagation of the wave in a waveguide is determined by the type and location of the excitation device.

8. The phase of the wave after the installation of the guided terminations will be
a) 0
b) 45
c) 90
d) 180

Answer: a [Reason:] Using a guided termination, the guides serves as a reflector. If the distance between the exciter and the wall is properly adjusted, the transmitted and the reflected wave will be in phase.

9. The exciter of the waveguide in a transmission line is the
a) Transmitter
c) Transponder
d) Antenna

Answer: d [Reason:] The waveguide is usually excited by the antenna rod. The reflection depends on the phase of excitation and the antenna current.

10. A waveguide imitates which type of filter characteristics?
a) Low pass filter
b) High pass filter
c) Band pass filter
d) Band reject filter

Answer: b [Reason:] A waveguide passes only high frequency waves (in GHz range) and attenuates low frequencies. This is the characteristic of a high pass filter.

## Set 2

1. Which of the following parameters does not exist in the transmission line equation?
a) R
b) Zo
c) ZL
d) Propagation constant

Answer: a [Reason:] The transmission line equation consists of secondary parameters only, which are derived from the primary parameters. The propagation constant, load impedance and the characteristic impedance are related in the transmission line equation.

2. For an infinite transmission line, the characteristic impedance is given by 50 ohm. Find the input impedance.
a) 25
b) 100
c) 2500
d) 50

Answer: d [Reason:] From the transmission line equation, the infinite line will have an input impedance same as that of the characteristic impedance. Thus Zin = Zo for l->∞. This shows that the line will be matched. The input impedance for the given case is 50 ohm.

3. The best transmission length for effective transmission of power is
a) L = λ/4
b) L = λ/8
c) L = λ/2
d) L = ∞

Answer: b [Reason:] Maximum transmission of power will occur, when the transmission line is matched. This implies that the input and characteristic impedances are the same. This condition is possible for l = λ/8 and l = ∞. Since l = ∞ is not feasible, the best option is l = λ/8.

4. When the length of the transmission line is same as that of the wavelength, then which condition holds good?
a) Zin = Zo
b) Z = Zo
c) ZL = Zo
d) Zin = ZL

Answer: d [Reason:] When the transmission line has a length same as that of the wavelength of the wave propagating through it, the input impedance will be same as the load impedance. This is the case where the wave is not amplified. The transmission line acts as a buffer.

5. The input impedance of a half wave transmission line with a load impedance of 12.5 ohm is
a) 25
b) 50
c) 6.25
d) 12.5

Answer: d [Reason:] For a half wave transmission line L = λ/2, the input and the load impedances will be the same. Thus for the given data, the input impedance will be 12.5 ohm.

6. The condition for a quarter wave transformer is
a) Zo2 = Zin ZL
b) Zo = Zin ZL
c) ZL2 = Zin Zo
d) Zo = Zin

Answer: a [Reason:] The quarter wave transformer represents L = λ/4. In this case, the characteristic impedance is the geometric mean of the input and load impedances. Thus Zo2 = Zin ZL is the required condition.

7. Find the characteristic impedance of a quarter wave with input and load impedances given by 50 and 25 respectively.
a) 50
b) 25
c) 75
d) 35.35

Answer: d [Reason:] For a quarter line wave, the characteristic impedance is the geometric mean of input and load impedances. Thus Zo2 = Zin ZL. On substituting for Zin = 50 and ZL = 25, we get Zo2 = 50 x 25. The characteristic impedance will be 35.35 ohm.

8. Find the load impedance in a quarter line transformer with characteristic impedance of 75 ohm and input impedance of 200 ohm.
a) 28.125
b) 12.285
c) 52.185
d) 85.128

Answer: a [Reason:] For a quarter line wave, the characteristic impedance is the geometric mean of input and load impedances. Thus Zo2 = Zin ZL. On substituting for Zo = 75 and Zin = 200, we get ZL = Zo2/Zin = 752/200 = 28.125 ohm.

9. The reflection coefficient of a perfectly matched transmission line is
a) 1
b) -1
c) 0
d) ∞

Answer: c [Reason:] In a perfectly matched line, maximum power transfer will occur. Losses will be minimal. This implies unity transmission coefficient and zero reflection coefficient.

10. The purpose of the transmission line equation is to
a) Find primary parameters
b) Find secondary parameters
c) Find the reflection cofficient
d) Impedance matching

Answer: d [Reason:] The transmission line equation is useful in finding the length of the line which gives maximum power transfer. Thus it is useful for impedance matching.

11. The quarter wave transformer can be considered as a
a) Impedance inverter
b) Impedance doubler
c) Impedance tripler

Answer: a [Reason:] A quarter wave transformer may be considered as an impedance inverter as it can transform a low impedance into a high impedance and vice-versa.

12. Which transmission line is called as one to one transformer?
a) L = λ
b) L = λ/2
c) L = λ/4
d) L = λ/8

Answer: b [Reason:] The half wave transformer line repeats its terminating impedance. In other words, when l = λ/2, Zin = ZL. Thus it is considered to be one to one transformer.

## Set 3

1. Which of the following parameters is not a primary parameter?
a) Resistance
b) Attenuation constant
c) Capacitance
d) Conductance

Answer: b [Reason:] The primary parameters of a transmission line are the resistance, inductance, capacitance and conductance. The attenuation, phase and propagation constant are secondary parameters. Thus the odd one out is the attenuation constant.

2. The networks in which the R, L, C parameters are individually concentrated or lumped at discrete points in the circuit are called
a) Lumped
b) Distributed
c) Parallel
d) Paired

Answer: a [Reason:] The networks in which the R, L, C parameters are individually concentrated or lumped at discrete points in the circuit are called lumped networks. These networks can be identified definitely as representing a particular parameter. An example is the filters.

3. The lines having R, L, C distributed along the circuit are called
a) Lumped
b) Distributed
c) Parallel
d) Paired

Answer: b [Reason:] In distributed lines, the primary parameters are distributed along the circuit with each elemental length having its own values and the concentration of the individual parameters is not possible. An example is the transmission of power.

4. Which primary parameter is uniformly distributed along the length of the conductor?
a) G
b) C
c) L
d) R

Answer: d [Reason:] The resistance is a primary parameter that is uniformly distributed along the length of the conductor. It depends on the cross section area and the length of the conductor.

5. The primary parameter that is associated with the magnetic flux linkage is
a) R
b) L
c) C
d) G

Answer: b [Reason:] When the conductors carry current, the conductor will be surrounded and linked by magnetic flux. The flux linkages per ampere of current gives rise to the effect of inductance. It is denoted by L.

6. The primary parameter that is associated with the electric charges is
a) G
b) R
c) C
d) L

Answer: c [Reason:] Conductors separated by insulating dielectrics in order to store electric charges, gives rise to the capacitance effect. The capacitance is distributed in the whole conductor length.

7. The leakage current in the transmission lines is referred to as the
a) Resistance
c) Conductance
d) Polarisation

Answer: c [Reason:] The dielectrics or insulators of the open wire line may not be perfect and a leakage current will flow. This leakage conductance exists between the conductors.

8. Find the receiving impedance of a transmission line having a voltage of 24V and a conduction current of 1.2A is
a) 25.2
b) 22.8
c) 28.8
d) 20

Answer: d [Reason:] By Ohm’s law, the impedance is the ratio of the voltage to the current. On substituting for V = 24 and I = 1.2, we get Z = V/I = 24/1.2 = 20 units.

9. The characteristic impedance of a transmission line with impedance and admittance of 16 and 9 respectively is
a) 25
b) 1.33
c) 7
d) 0.75

Answer: b [Reason:] The characteristic impedance is given by Zo = √(Z/Y), where Z is the impedance and Y is the admittance. On substituting for Z = 16 and Y = 9, we get the characteristic impedance as √(16/9) = 1.33 units.

10. The propagation constant of a transmission line with impedance and admittance of 9 and 16 respectively is
a) 25
b) 144
c) 12
d) 7

Answer: c [Reason:] The propagation constant is given by γ = √(ZY), where Z is given by 9 and Y is 16. On substituting the given values, the propagation constant will be γ = √(ZY) = √(9 x 16) = 12 units.

11. Find the characteristic impedance expression in terms of the inductance and capacitance parameters.
a) Zo = √(LC)
b) Zo = LC
c) Zo = √(L/C)
d) Zo = L/C

Answer: c [Reason:] The characteristic impedance is given by the square root of the ratio of the inductance to the capacitance. Thus Zo = √(L/C) is the required expression.

12. When a transmission line has a load impedance same as that of the characteristic impedance, the line is said to be
a) Parallel
b) Perpendicular
c) Polarized
d) Matched

Answer: d [Reason:] When a transmission line load impedance is same as that of the characteristic impedance, the line is said to be matched. In such cases, full transmission of power will occur, with minimal losses.

## Set 4

1. In transverse magnetic waves, which of the following is true?
a) E is parallel to H
b) H is parallel to wave direction
c) H is transverse to wave direction
d) E is transverse to H

Answer: c [Reason:] In transverse magnetic waves, the magnetic field strength is transverse to the wave direction. They are also called E waves.

2. The dominant mode in the TM waves is
a) TM01
b) TM10
c) TM20
d) TM11

Answer: d [Reason:] The modes TM10, TM01 and TM20 does not exist in any waveguide. The TM11 mode is the dominant mode in the waveguide.

3. The modes in a waveguide having a V number of 20 is
a) 400
b) 200
c) 100
d) 40

Answer: b [Reason:] The number of modes in a waveguide is given by m = V2/2. On substituting for V = 20, we get m = 400/2 = 200 modes.

4. The v number of a waveguide having 120 modes is
a) 15.5
b) 18
c) 24
d) 12

Answer: a [Reason:] The number of modes in a waveguide is given by m = V2/2. On substituting for m = 120, we get V = √(2 x 120) = 15.5.

5. The intrinsic impedance of a TM wave will be
a) Greater than 377 ohm
b) Equal to 377 ohm
c) Lesser than 377 ohm
d) Does not exist

Answer: c [Reason:] The intrinsic impedance of the transverse magnetic wave is given by ηTM = η √(1-(fc/f)2). Here the term √(1-(fc/f)2) is always lesser than unity. Thus the intrinsic impedance of the TM wave is lesser than 377 ohms.

6. The modes TMmo and TMon are called
a) Generate modes
b) Degenerate modes
c) Dominant modes
d) Evanescent modes

Answer: d [Reason:] The modes TMmo and TMon does not exist. These modes are said to be evanescent mode.

7. Two modes with same cut off frequency are said to be
a) Generate modes
b) Dominant modes
c) Degenerate modes
d) Regenerate modes

Answer: c [Reason:] Two modes with same cut off frequency are called as degenerate modes. These modes have same field distribution.

8. Does the mode TM30 exists?
a) Yes
b) No

Answer: b [Reason:] Modes in the format of TMmo and TMon does not exist. The given mode is in the form of TMmo, which is does not exist. It is an evanescent mode.

9. The boundary between the Fresnel and Fraunhofer zones having a length of 12 units and a wavelength of 3 units is
a) 96
b) 48
c) 192
d) 36

Answer: a [Reason:] The Fresnel- Fraunhofer boundary is related by the wavelength as R = 2L2/λ. On substituting for L = 12 and λ = 3, we get R = 2 x 122/3 = 96 units.

10. The reflection coefficient, when a resonant cavity is placed between the waveguide is
a) 0
b) 1
c) -1
d) Infinite

Answer: b [Reason:] When the waveguide is shorted by conducting plates, the reflection coefficient will be unity. This will lead to the occurrence of standing waves.

11. The distance between two terminated plates is given by the
a) Guided wavelength
b) 2(guided wavelength)
c) Guided wavelength/2
d) (Guided wavelength)/4

Answer: c [Reason:] The distance between the terminating plates is given by Vmin = λg/2, where λg is the guided wavelength.

12. Find the guided wavelength if the distance between the two conducting plates in the waveguide is 2 cm.
a) 4cm
b) 2cm
c) 1cm
d) 0.5cm

Answer: a [Reason:] The distance between the terminating plates is given by Vmin = λg/2, where λg is the guided wavelength. On substituting for Vmin = 2cm, we get λg = 2Vmin = 2 x 0.02 = 4cm.

## Set 5

1. The divergence theorem converts
a) Line to surface integral
b) Surface to volume integral
c) Volume to line integral
d) Surface to line integral

Answer: b [Reason:] The divergence theorem is given by, ∫∫ D.ds = ∫∫∫ Div (D) dv. It is clear that it converts surface (double) integral to volume(triple) integral.

2. The triple integral is used to compute volume. State True/False
a) True
b) False

Answer: a [Reason:] The triple integral, as the name suggests integrates the function/quantity three times. This gives volume which is the product of three independent quantities.

3. The volume integral is three dimensional. State True/False
a) True
b) False

Answer: a [Reason:] Volume integral integrates the independent quantities by three times. Thus it is said to be three dimensional integral or triple integral.

4. Find the charged enclosed by a sphere of charge density ρ and radius a.
a) ρ (4πa2)
b) ρ(4πa3/3)
c) ρ(2πa2)
d) ρ(2πa3/3)

Answer: b [Reason:] The charge enclosed by the sphere is Q = ∫∫∫ ρ dv. Where, dv = r2 sin θ dr dθ dφ and on integrating with r = 0->a, φ = 0->2π and θ = 0->π, we get Q = ρ(4πa3/3).

5. Evaluate Gauss law for D = 5r2/4 i in spherical coordinates with r = 4m and θ = π/2 as volume integral.
a) 600
b) 588.9
c) 577.8
d) 599.7

Answer: b [Reason:] ∫∫ D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed. The divergence of D given is, Div(D) = 5r and dv = r2 sin θ dr dθ dφ. On integrating, r = 0->4, φ = 0->2π and θ = 0->π/4, we get Q = 588.9.

6. Compute divergence theorem for D = 5r2/4 i in spherical coordinates between r = 1 and r = 2 in volume integral.
a) 80 π
b) 5 π
c) 75 π
d) 85 π

Answer: c [Reason:] D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed. The divergence of D given is, Div(D) = 5r and dv = r2 sin θ dr dθ dφ. On integrating, r = 1->2, φ = 0->2π and θ = 0->π, we get Q = 75 π.

7. Compute the Gauss law for D = 10ρ3/4 i, in cylindrical coordinates with ρ = 4m, z = 0 and z = 5, hence find charge using volume integral.
a) 6100 π
b) 6200 π
c) 6300 π
d) 6400 π

Answer: d [Reason:] Q = D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed. The divergence of D given is, Div(D) = 10 ρ2 and dv = ρ dρ dφ dz. On integrating, ρ = 0->4, φ = 0->2π and z = 0->5, we get Q = 6400 π.

8. Using volume integral, which quantity can be calculated?
a) area of cube
b) area of cuboid
c) volume of cube
d) distance of vector

Answer: c [Reason:] The volume integral gives the volume of a vector in a region. Thus volume of a cube can be computed.

9. Compute the charge enclosed by a cube of 2m each edge centered at the origin and with the edges parallel to the axes. Given D = 10y3/3 j.
a) 20
b) 70/3
c) 80/3
d) 30

Answer: c [Reason:] Div(D) = 10y2 ∫∫∫Div (D) dv = ∫∫∫ 10y2 dx dy dz. On integrating, x = -1->1, y = -1->1 and z = -1->1, we get Q = 80/3.

10. Find the value of divergence theorem for the field D = 2xy i + x2 j for the rectangular parallelepiped given by x = 0 and 1, y = 0 and 2, z = 0 and 3.
a) 10
b) 12
c) 14
d) 16