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# Multiple choice question for engineering

## Set 1

1. The wavelength of a line with a phase constant of 6.28 units is
a) 2
b) 1
c) 0.5
d) 3.14

Answer: b [Reason:] The wavelength and the phase constant are related by λ = 2π/β, where β is given as 6.28. On substituting for β, we get λ = 2π/6.28 = 1 unit.

2. The wavelength of a wave with a frequency of 6 GHz in air is
a) 50
b) 5
c) 0.5
d) 0.05

Answer: d [Reason:] The wavelength is given by the ratio of the velocity to the frequency of the wave. In air medium, the velocity can be assumed as the speed of light. On substituting for v and f, we get λ = v/f = 3×108/6×109 = 0.05 units.

3. The phase constant of a wave with a wavelength of 2 units is given by
a) 2
b) 3.14
c) 6.28
d) 1

Answer: b [Reason:] The phase constant is given by β = 2π/λ. On substituting for λ = 2, we get β = 2π/2 = 3.14 units.

4. The frequency of a wave travelling in a transmission line with velocity 4 x 108 and wavelength 3 units is
a) 0.75 GHz
b) 0.133 GHz
c) 7.5 GHz
d) 1.33 GHz

Answer: b [Reason:] The frequency and wavelength relation is given by f = v/λ. On substituting for v and λ, we get f = 4 x 108/3 = 0.133 GHz.

5. The velocity and phase constant relation is given by
a) V = ω/β
b) V = ωβ
c) V = β/ω
d) Vωβ = 1

Answer: a [Reason:] The velocity of a wave is the ratio of the frequency in radian/second to the phase constant. It is given by V = ω/β.

6. Find the phase constant of a wave travelling with a velocity of 1.2 x 108 and a frequency of 7.5 giga radian/sec
a) 62.5
b) 26.5
c) 56.2
d) 52.6

Answer: a [Reason:] The phase constant is given by β = ω/v, from the definition of phase constant and velocity. On substituting for ω = 7.5 x 109 and v = 1.2 x 108, we get the phase constant β = 7.5 x 109/1.2 x 108 = 62.5 units.

7. The electrical length in a transmission line refers to the
a) Product of attenuation constant and length
b) Ratio of attenuation constant and length
c) Product of phase constant and length
d) Ratio of phase constant and length

Answer: a [Reason:] The electrical length in a transmission line refers to the product of the attenuation constant α and the length of the line l. It is given by αl.

8. The unit of attenuation constant is
a) Decibel
b) Bel
c) Neper
d) No unit

Answer: c [Reason:] Attenuation constant is the measure of the power loss of the wave during its transmission. It is expressed in terms of neper and 1 neper= 8.686 decibel/m.

9. The attenuation constant causes phase distortion and the phase constant causes frequency distortion. State True/False.
a) True
b) False

Answer: b [Reason:] There are always some distortions, even in the perfect transmission line. This is due to the variation of the secondary parameters. The attenuation constant causes the frequency distortion, whereas the phase constant causes the phase distortion.

10. The propagation constant of a wave with attenuation and phase constant given by 2 and 3 respectively is
a) 2 – 3j
b) 3 – 2j
c) 2 + 3j
d) 3 + 2j

Answer: c [Reason:] The propagation constant is given by γ = α + jβ. Given that α = 2 and β = 3. Thus we get the propagation constant as γ = 2 + 3j.

11. The velocity of wave in the air medium is
a) 1 x 108
b) 1.5 x 108
c) 3 x 108
d) 1 x 109

Answer: c [Reason:] The light is travelling at its fastest speed in air medium. Thus the velocity of a wave in the air medium is assumed to have the speed of light. It is given by c = 3 x 108.

12. Identify the secondary parameter from the options given below.
a) Resistance
b) Conductance
c) Phase constant
d) Capacitance

Answer: c [Reason:] Primary parameters are directly observed from the circuit characteristics. Secondary parameters are derived or calculated from the primary parameters. R, L, C, G are primary parameters, whereas α, β, γ, Zo are secondary parameters.

## Set 2

1. For time varying currents, the field or waves will be
a) Electrostatic
b) Magneto static
c) Electromagnetic
d) Electrical

Answer: c [Reason:] For stationary charges, the field is electrostatic. For steady currents, the field is magneto static. But for time varying currents, the field or waves will be electromagnetic.

2. According to Faraday’s law, EMF stands for
a) Electromagnetic field
b) Electromagnetic force
c) Electromagnetic friction
d) Electromotive force

Answer: d [Reason:] The force in any closed circuit due to the change in the flux linkage of the circuit is called as electromotive force EMF. This phenomenon is called as Faraday’s law.

3. Calculate the emf when the flux is given by 3sin t + 5cos t
a) 3cos t – 5sin t
b) -3cos t + 5sin t
c) -3sin t – 5cos t
d) 3cos t + 5sin t

Answer: b [Reason:] The electromotive force is given by Vemf = -dλ/dt. Thus Vemf = -dλ/dt = -(3cos t – 5sin t) = -3cos t + 5sin t.

4. The induced voltage will oppose the flux producing it. State True/False.
a) True
b) False

Answer: a [Reason:] According to Lenz law, the induced voltage acts in such a way that it opposes the flux producing it. This is indicated by a negative sign.

5. Calculate the emf when a coil of 100 turns is subjected to a flux rate of 0.3 tesla/sec.
a) 3
b) 30
c) -30
d) -300

Answer: c [Reason:] The induced emf is given by Vemf = -dλ/dt = -Ndψ/dt. Thus emf will be -100 x 0.3 = -30 units.

6. Find the displacement current when the flux density is given by t3 at 2 seconds.
a) 3
b) 6
c) 12
d) 27

Answer: c [Reason:] The displacement current is given by Jd = dD/dt. Thus Jd = 3t2. At time t = 2, we get Jd = 3(2)2= 12A.

7. Find the force due to a current element of length 2cm and flux density of 12 tesla. The current through the element will be 5A.
a) 1 N
b) 1.2 N
c) 1.4 N
d) 1.6 N

Answer: b [Reason:] The force due to a current element is given by F = BI x L. Thus F = 12 x 5 x 0.02 = 1.2 units.

8. Which of the following statements is true?
a) E is the cross product of v and B
b) B is the cross product of v and E
c) E is the dot product of v and B
d) B is the dot product of v and E

Answer: a [Reason:] The electric field is the cross product of the velocity and the magnetic field intensity. This is given by Lorentz equation.

9. The time varying electric field E is conservative. State True/False.
a) True
b) False

Answer: b [Reason:] The time varying electric field E(t) is not a closed path. Thus the curl will be non-zero. This implies E(t) is not conservative and the statement is false.

10. When the conduction current density and displacement current density are same, the dissipation factor will be
a) Zero
b) Minimum
c) Maximum
d) Unity

Answer: d [Reason:] Dissipation factor refers to the tangent of loss angle. It is the ratio of conduction current density to displacement current density. When both are same, the loss tangent or the dissipation factor will be unity.

## Set 3

1. Gauss theorem uses which of the following operations?
b) Curl
c) Divergence
d) Laplacian

Answer: c [Reason:] The Gauss divergence theorem uses divergence operator to convert surface to volume integral. It is used to calculate the volume of the function enclosing the region given.

2. Evaluate the surface integral ∫∫ (3x i + 2y j). dS, where S is the sphere given by x2 + y2 + z2 = 9.
a) 120π
b) 180π
c) 240π
d) 300π

Answer: b [Reason:] We could parameterise surface and find surface integral, but it is wise to use divergence theorem to get faster results. The divergence theorem is given by ∫∫ F.dS = ∫∫∫ Div (F).dV Div (3x i + 2y j) = 3 + 2 = 5. Now the volume integral will be ∫∫∫ 5.dV, where dV is the volume of the sphere 4πr3/3 and r = 3units.Thus we get 180π.

3. The Gauss divergence theorem converts
a) line to surface integral
b) line to volume integral
c) surface to line integral
d) surface to volume integral

Answer: d [Reason:] The divergence theorem for a function F is given by ∫∫ F.dS = ∫∫∫ Div (F).dV. Thus it converts surface to volume integral.

4. The divergence theorem for a surface consisting of a sphere is computed in which coordinate system?
a) Cartesian
b) Cylindrical
c) Spherical
d) Depends on the function

Answer: d [Reason:] Seeing the surface as sphere, we would immediately choose spherical system, but it is wrong. The divergence operation is performed in that coordinate system in which the function belongs to. It is independent of the surface region.

5. Find the Gauss value for a position vector in Cartesian system from the origin to one unit in three dimensions.
a) 0
b) 3
c) -3
d) 1

Answer: b [Reason:] The position vector in Cartesian system is given by R = x i + y j + z k. Div(R) = 1 + 1 + 1 = 3. By divergence theorem, ∫∫∫3.dV, where V is a cube with x = 0->1, y = 0->1 and z = 0->1. On integrating, we get 3 units.

6. The divergence theorem value for the function x2 + y2 + z2 at a distance of one unit from the origin is
a) 0
b) 1
c) 2
d) 3

Answer: d [Reason:] Div (F) = 2x + 2y + 2z. The triple integral of the divergence of the function is ∫∫∫(2x + 2y + 2z)dx dy dz, where x = 0->1, y = 0->1 and z = 0->1. On integrating, we get 3 units.

7. If a function is described by F = (3x + z, y2 − sin x2z, xz + yex5), then the divergence theorem value in the region 0<x<1, 0<y<3 and 0<z<2 will be
a) 13
b) 26
c) 39
d) 51

Answer: c [Reason:] Div (F) = 3 + 2y + x. By divergence theorem, the triple integral of Div F in the region is ∫∫∫ (3 + 2y + x) dx dy dz. On integrating from x = 0->1, y = 0->3 and z = 0->2, we get 39 units.

8. Find the divergence theorem value for the function given by (ez, sin x, y2)
a) 1
b) 0
c) -1
d) 2

Answer: b [Reason:] Since the divergence of the function is zero, the triple integral leads to zero. The Gauss theorem gives zero value.

9. For a function given by F = 4x i + 7y j +z k, the divergence theorem evaluates to which of the values given, if the surface considered is a cone of radius 1/2π m and height 4π2 m.
a) 1
b) 2
c) 3
d) 4

Answer: b [Reason:] Div (F) = 4 + 7 + 1 = 12. The divergence theorem gives ∫∫∫(12).dV, where dV is the volume of the cone πr3h/3, where r = 1/2π m and h = 4π2 m. On substituting the radius and height in the triple integral, we get 2 units.

10. Divergence theorem computes to zero for a solenoidal function. State True/False.
a) True
b) False

Answer: a [Reason:] The divergence theorem is given by, ∫∫ F.dS = ∫∫∫ Div (F).dV, for a function F. If the function is solenoidal, its divergence will be zero. Thus the theorem computes to zero.

## Set 4

1. Divergence theorem is based on
a) Gauss law
b) Stoke’s law
c) Ampere law
d) Lenz law

Answer: a [Reason:] The divergence theorem relates surface integral and volume integral. Div(D) = ρv, which is Gauss’s law.

2. The Gaussian surface for a line charge will be
a) Sphere
b) Cylinder
c) Cube
d) Cuboid

Answer: b [Reason:] A line charge can be visualized as a rod of electric charges. The three dimensional imaginary enclosed surface of a rod can be a cylinder.

3. The Gaussian surface for a point charge will be
a) Cube
b) Cylinder
c) Sphere
d) Cuboid

Answer: c [Reason:] A point charge is single dimensional. The three dimensional imaginary enclosed surface of a point charge will be sphere.

4. A circular disc of radius 5m with a surface charge density ρs = 10sinφ is enclosed by surface. What is the net flux crossing the surface?
a) 3
b) 2
c) 1
d) 0

Answer: d [Reason:] Q = ∫ ρsds = ∫∫ 10sinφ rdrdφ, on integrating with r = 0->5 and φ = 0->2π, we get Q = ψ = 0.

5. The total charge of a surface with densities 1,2,…,10 is
a) 11
b) 33
c) 55
d) 77

Answer: c [Reason:] Q = ∫∫D.ds. Since the data is discrete, the total charge will be summation of 1,2,…,10,i.e, 1+2+…+10 = 10(11)/2 = 55.

6. The work done by a charge of 10μC with a potential 4.386 is (in μJ)
a) 32.86
b) 43.86
c) 54.68
d) 65.68

Answer: b [Reason:] By Gauss law principles, W = Q X V = 10 X 10-6 X 4.386 = 43.86 X 10-6 joule.

7. The potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 109)
a) 12.74
b) 13.47
c) 12.47
d) 13.74

Answer: c [Reason:] The potential of a coaxial cylinder will be ρl ln(b/a)/2πε, where ρl = 1, b = 2m and a = 1m. We get V = 12.47 X 109 volts.

8. Find the potential due to a charged ring of density 2 units with radius 2m and the point at which potential is measured is at a distance of 1m from the ring.
a) 18π
b) 24π
c) 36π
d) 72π

Answer: d [Reason:] The potential due to a charged ring is given by λa/2εr, where a = 2m and r = 1m. We get V = 72π volts.

9. Gauss law cannot be used to find which of the following quantity?
a) Electric field intensity
b) Electric flux density
c) Charge
d) Permittivity

Answer: d [Reason:] Permittivity is constant for a particular material(say permittivity of water is 1). It cannot be determined from Gauss law, whereas the remaining options can be computed from Gauss law.

10. Gauss law for magnetic fields is given by
a) Div(E) = 0
b) Div(B) = 0
c) Div(H) = 0
d) Div(D) = 0

Answer: b [Reason:] The divergence of magnetic flux density is always zero. This is called Gauss law for magnetic fields. It implies the non-existence of magnetic monopoles in any magnetic field.

## Set 5

1. Gauss law can be used to compute which of the following?
a) Permittivity
b) Permeability
d) Electric potential

Answer: c [Reason:] Gauss law relates the electric flux density and the charge density. Thus it can be used to compute radius of the Gaussian surface. Permittivity and permeability are constants for a particular material.

2. Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 1m.
a) 0
b) 1
c) 2
d) 3

Answer: a [Reason:] Since 1m does not enclose any cylinder (three Gaussian surfaces of radius 2m, 4m, 5m exists), the charge density and charge becomes zero according to Gauss law. Thus flux density is also zero.

3. Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 3m.
a) 3
b) 10/3
c) 11/3
d) 4

Answer: b [Reason:] The radius is 3m, hence it will enclose one Gaussian cylinder of R = 2m. By Gauss law, ψ = Q D(2πRL) = σ(2πRL), D(2π X 3) = σ(2π X 2), Thus D = 10/3 units.

4. Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ =-3 at R = 5m. Find the flux density at R = 4.5m.
a) 4/4.5
b) 3/4.5
c) 2/4.5
d) 1/4.5

Answer: c [Reason:] The Gaussian cylinder of R = 4.5m encloses sum of charges of two cylinders (R = 2m and R = 4m). By Gauss law, ψ = Q D(2πRL) = σ(2πRL), D(2π X 4.5) = Q1 + Q2 = σ1(2π X 2) + σ2(2π X 4), here σ1 = 5 and σ2 = -2. We get D = 2/4.5 units.

5. Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 6m.
a) 17/6
b) -17/6
c) 13/6
d) -13/6

Answer: d [Reason:] The radius R = 6m encloses all the three Gaussian cylinders. By Gauss law, ψ = Q D(2πRL) = σ(2πRL), D(2π X 6) = Q1 + Q2 + Q3 = σ1(2π X 2) + σ2(2π X 4) + σ3(2π X 5), here σ1 = 5, σ2 = -2 and σ3 = -3. We get D = -13/6 units.

6. Gauss law can be evaluated in which coordinate system?
a) Cartesian
b) Cylinder
c) Spherical
d) Depends on the Gaussian surface

Answer: d [Reason:] The Gauss law exists for all materials. Depending on the Gaussian surface of the material, we take the coordinate systems accordingly. Suppose if the material is a coaxial cable, the Gaussian surface is in the form of cylinder. Thus we take Cylinder/Circular coordinate system.

7. Gauss law cannot be expressed in which of the following forms?
a) Differential
b) Integral
c) Point
d) Stokes theorem

Answer: d [Reason:] Gauss law can be expressed in differential or point form as, Div (D)= ρv and in integral form as ∫∫ D.ds = Q = ψ . It is not possible to express it using Stoke’s theorem.

8. The tangential component of electric field intensity is always continuous at the interface. State True/False.
a) True
b) False

Answer: a [Reason:] Consider a dielectric-dielectric boundary, the electric field intensity in both the surfaces will be Et1 = Et2, which implies that the tangential component of electric field intensity is always continuous at the boundary.

9. The normal component of the electric flux density is always discontinuous at the interface. State True/False.
a) True
b) False

Answer: a [Reason:] In a dielectric-dielectric boundary, if a free surface charge density exists at the interface, then the normal components of the electric flux density are discontinuous at the boundary, which means Dn1 = Dn2.

10. With Gauss law as reference which of the following law can be derived?
a) Ampere law
c) Coulomb’s law
d) Ohm’s law