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# Multiple choice question for engineering

## Set 1

1. Calculate the emf in a material with flux linkage of 3.5t2 at 2 seconds.
a) 3.5
b) -7
c) -14
d) 28

Answer: c [Reason:] The emf induced in a material with flux linkage is given by Vemf = -dλ/dt. On substituting λ= 3.5t2, we get emf = -7t. At time t = 2sec, the emf will be -14 units.

2. Find the emf induced in a coil of 60 turns with a flux rate of 3 units.
a) -60
b) -180
c) 60
d) 180

Answer: b [Reason:] The emf induced is the product of the turns and the flux rate. Thus Vemf = -Ndφ/dt. On substituting N = 60 and dφ/dt = 3, we get emf as -60 x 3 = -180 units.

3. Find the electric field intensity of a charge 2.5C with a force of 3N.
a) -7.5
b) 7.5
c) 2.5/3
d) 3/2.5

Answer: d [Reason:] The electric field intensity is the electric force per unit charge. It is given by E = F/q. On substituting F = 2.5 and q = 3, we get E = 3/2.5 units.

4. The electric field intensity of a field with velocity 10m/s and flux density of 2.8 units is
a) 0.28
b) 28
c) 280
d) 10/2.8

Answer: b [Reason:] The electric field is the product of the velocity and the magnetic flux density given by E = v x B. On substituting v = 10 and B = 2.8, we get E = 10 x 2.8 = 28 units.

5. The line integral of the electric field intensity is
a) Mmf
b) Emf
c) Electric potential
d) Magnetic potential

Answer: b [Reason:] From the Maxwell first law, the transformer emf is given by the line integral of the electric field intensity. Thus the emf is given by ∫ E.dl.

6. Which of the following relations is correct?
a) MMF = ∫ B.dl
b) MMF = ∫ H.dl
c) EMF = ∫ E.dl
d) EMF = ∫ D.dl

Answer: c [Reason:] The emf induced in a material is given by the line integral of the electric field intensity. Thus EMF = ∫ E.dl is the correct relation.

7. For static fields, the curl of E will be
a) Rotational
b) Irrotational
c) Solenoidal
d) Divergent

Answer: b [Reason:] For static fields, the charges will be constant and the field is constant. Thus curl of the electric field intensity will be zero. This implies the field is irrotational.

8. The line integral of which parameter is zero for static fields?
a) E
b) H
c) D
d) B

Answer: a [Reason:] The field is irrotational for static fields. Thus curl of E is zero. From Stokes theorem, the line integral of E is same as the surface integral of the curl of E. Since it is zero, the line integral of E will also be zero.

9. The magnitude of the conduction current density for a magnetic field intensity of a vector yi + zj + xk will be
a) 1.414
b) 1.732
c) -1.414
d) -1.732

Answer: b [Reason:] From the Ampere circuital law, the curl of H is the conduction current density. The curl of H = yi + zj + xk is –i – j – k. Thus conduction current density is –i – j – k. The magnitude will be √(1 + 1 + 1) = √3 = 1.732 units.

10. The charge density of a field with a position vector as electric flux density is given by
a) 0
b) 1
c) 2
d) 3

Answer: d [Reason:] The Gauss law for electric field states that the divergence of the electric flux density is the charge density. Thus Div(D) = ρ. For D as a position vector, the divergence of the position vector D will be always 3. Thus the charge density is also 3.

## Set 2

1. The given equation satisfies the Laplace equation.
V = x2 + y2 – z2. State True/False.
a) True
b) False

Answer: a [Reason:] Grad (V) = 2xi + 2yj – 4zk. Div (Grad (V)) = Del2(V) = 2+2-4 = 0. It satisfies the Laplacian equation. Thus the statement is true.

2. In free space, the Poisson equation becomes
a) Maxwell equation
b) Ampere equation
c) Laplace equation

Answer: c [Reason:] The Poisson equation is given by Del2(V) = -ρ/ε. In free space, the charges will be zero. Thus the equation becomes, Del2(V) = 0, which is the Laplace equation.

3. If Laplace equation satisfies, then which of the following statements will be true?
a) Potential will be zero
b) Current will be infinite
c) Resistance will be infinite
d) Voltage will be same

Answer: b [Reason:] Laplace equation satisfying implies the potential is not necessarily zero due to subsequent gradient and divergence operations following. Finally, if potential is assumed to be zero, then resistance is zero and current will be infinite.

4. Suppose the potential function is a step function. The equation that gets satisfied is
a) Laplace equation
b) Poisson equation
c) Maxwell equation
d) Ampere equation

Answer: a [Reason:] Step is a constant function. The Laplace equation Div(Grad(step)) will become zero. This is because gradient of a constant is zero and divergence of zero vector will be zero.

5. Calculate the charge density when a potential function x2 + y2 + z2 is in air(in 10-9 order)
a) 1/6π
b) 6/2π
c) 12/6π
d) 10/8π

Answer: a [Reason:] The Poisson equation is given by Del2(V) = -ρ/ε. To find ρ, put ε = 8.854 x 10-12 in air and Laplacian of the function is 2 + 2 + 2 = 6. Ρ = 6 x 10-9/36π = 1/6π units.

6. The function V = exsin y + z does not satisfy Laplace equation. State True/False.
a) True
b) False

Answer: b [Reason:] Grad (V) = exsin y i + ex cos y j + k. Div(Grad(V)) = exsin y – exsin y + 0= 0.Thus Laplacian equation Div(Grad(V)) = 0 is true.

7. Poisson equation can be derived from which of the following equations?
a) Point form of Gauss law
b) Integral form of Gauss law
c) Point form of Ampere law
d) Integral form of Ampere law

Answer: a [Reason:] The point of Gauss law is given by, Div (D)= ρv. On putting D= ε E and E=- Grad (V) in Gauss law, we get Del2 (V)= -ρ/ε, which is the Poisson equation.

8. Find the charge density from the function of flux density given by 12x – 7z.
a) 19
b) -5
c) 5
d) -19

Answer: c [Reason:] From point form of Gauss law, we get Div (D) = ρv Div (D) = Div(12x – 7z) = 12-7 = 5, which the charge density ρv. Thus ρv = 5 units.

9. Find the electric field of a potential function given by 20 log x + y at the point (1,1,0).
a) -20 i – j
b) -i -20 j
c) i + j
d) (i + j)/20

Answer: a [Reason:] The electric field is given by E = -Grad(V). The gradient of the given function is 20i/x + j. At the point (1,1,0), we get 20i + j. The electric field E = -(20i + j) = -20i – j.

10. When a material has zero permittivity, the maximum potential that it can possess is
a) ∞
b) -∞
c) Unity
d) Zero

Answer: d [Reason:] Permittivity is zero, implies that the ability of the material to store electric charges is zero. Thus the electric field and potential of the material is also zero.

## Set 3

1. The best definition of polarisation is
a) Orientation of dipoles in random direction
b) Electric dipole moment per unit volume
c) Orientation of dipole moments
d) Change in polarity of every dipole

Answer: b [Reason:] The polarisation is defined mathematically as the electric dipole moment per unit volume. It is also referred to as the orientation of the dipoles in the direction of applied electric field.

2. Calculate the polarisation vector of the material which has 100 dipoles per unit volume in a volume of 2 units.
a) 200
b) 50
c) 400
d) 0.02

Answer: a [Reason:] Polarisation vector P = N x p, where N = 100 and p = 2. On substituting we get P = 200 units.

3. Polarizability is defined as the
a) Product of dipole moment and electric field
b) Ratio of dipole moment to electric field
c) Ratio of electric field to dipole moment
d) Product of dielectric constant and dipole moment

Answer: b [Reason:] Polarizability is a constant that is defined as the ratio of elemental dipole moment to the electric field strength.

4. Calculate the energy stored per unit volume in a dielectric medium due to polarisation when P = 9 units and E = 8 units.
a) 1.77
b) 2.25
c) 36
d) 144

Answer: c [Reason:] The energy stored per unit volume in a dielectric medium is given by, W = 0.5 X PE = 0.5 X 9 X 8 = 36 units.

5. Identify which type of polarisation depends on temperature.
a) Electronic
b) Ionic
c) Orientational
d) Interfacial

Answer: c [Reason:] The electronic, ionic and interfacial polarisation depends on the atoms which are independent with respect to temperature. Only the orientational polarisation is dependent on the temperature and is inversely proportional to it.

6. Calculate the polarisation vector in air when the susceptibility is 5 and electric field is 12 units.
a) 3
b) 2
c) 60
d) 2.4

Answer: c [Reason:] The polarisation vector is given by, P = ε0 x χe x E, where χe = 5 and ε0 = 12. On substituting, we get P = 1 x 5 x 12 = 60 units.

7. In isotropic materials, which of the following quantities will be independent of the direction?
a) Permittivity
b) Permeability
c) Polarisation
d) Polarizability

Answer: a [Reason:] Isotropic materials are those with radiate or absorb energy uniformly in all directions (eg. Isotropic antenna). Thus it is independent of the direction.

8. The total polarisation of a material is the
a) Product of all types of polarisation
b) Sum of all types of polarisation
c) Orientation directions of the dipoles
d) Total dipole moments in the material

Answer: b [Reason:] The total polarisation of a material is given by the sum of electronic, ionic, orientational and interfacial polarisation of the material.

9. In the given types of polarisation, which type exists in the semiconductor?
a) Electronic
b) Ionic
c) Orientational
d) Interfacial or space charge

Answer: d [Reason:] The interfacial type of polarisation occurs due to accumulation of charges at the interface in a multiphase material. This interface or junction is found in a semiconductor material.

10. Solids do not have which type of polarisation?
a) Ionic
b) Orientational
c) Interfacial
d) Electronic

Answer: c [Reason:] Solids possess permanent dipole moments. Moreover they do not have junction like semiconductors. Thus, solids neglect the interfacial and space charge polarisation. They possess only electronic, ionic and orientational polarisations.

## Set 4

1. The power of the electromagnetic wave with electric and magnetic field intensities given by 12 and 15 respectively is
a) 180
b) 90
c) 45
d) 120

Answer: b [Reason:] The Poynting vector gives the power of an EM wave. Thus P = EH/2. On substituting for E = 12 and H = 15, we get P = 12 x 15/2 = 90 units.

2. The power of a wave of with voltage of 140V and a characteristic impedance of 50 ohm is
a) 1.96
b) 19.6
c) 196
d) 19600

Answer: c [Reason:] The power of a wave is given by P = V2/2Zo, where V is the generator voltage and Zo is the characteristic impedance. on substituting the given data, we get P = 1402/(2×50) = 196 units.

3. The power reflected by a wave with incident power of 16 units is(Given that the reflection coefficient is 0.5)
a) 2
b) 8
c) 6
d) 4

Answer: d [Reason:] The fraction of the reflected to the incident power is given by the reflection coefficient. Thus Pref = R2xPinc. On substituting the given data, we get Pref = 0.52 x 16 = 4 units.

4. The power transmitted by a wave with incident power of 16 units is(Given that the reflection coefficient is 0.5)
a) 12
b) 8
c) 16
d) 4

Answer: a [Reason:] The fraction of the transmitted to the incident power is given by the reflection coefficient. Thus Pref = (1-R2) Pinc. On substituting the given data, we get Pref = (1- 0.52) x 16 = 12 units. In other words, it is the remaining power after reflection.

5. The incident and the reflected voltage are given by 15 and 5 respectively. The transmission coefficient is
a) 1/3
b) 2/3
c) 1
d) 3

Answer: b [Reason:] The ratio of the reflected to the incident voltage is the reflection coefficient. It is given by R = 5/15 = 1/3. To get the transmission coefficient, T = 1 – R = 1 – 1/3 = 2/3.

6. The current reflection coefficient is given by -0.75. Find the voltage reflection coefficient.
a) -0.75
b) 0.25
c) -0.25
d) 0.75

Answer: d [Reason:] The voltage reflection coefficient is the negative of the current reflection coefficient. For a current reflection coefficient of -0.75, the voltage reflection coefficient will be 0.75.

7. The attenuation is given by 20 units. Find the power loss in decibels.
a) 13.01
b) 26.02
c) 52.04
d) 104.08

Answer: a [Reason:] The attenuation refers to the power loss. Thus the power loss is given by 20 units. The power loss in dB will be 10 log 20 = 13.01 decibel.

8. The reflection coefficient is 0.5. Find the return loss.
a) 12.12
b) -12.12
c) 6.02
d) -6.02

Answer: c [Reason:] The return loss is given by RL = -20log R, where is the reflection coefficient. It is given as 0.5. Thus the return loss will be RL = -20 log 0.5 = 6.02 decibel.

9. The radiation resistance of an antenna having a power of 120 units and antenna current of 5A is
a) 4.8
b) 9.6
c) 3.6
d) 1.8

Answer: a [Reason:] The power of an antenna is given by Prad = Ia2 Rrad, where Ia is the antenna current and Rrad is the radiation resistance. On substituting the given data, we get Rrad = Prad/Ia2 = 120/52 = 4.8 ohm.

10. The transmission coefficient is given by 0.65. Find the return loss of the wave.
a) 9.11
b) 1.99
c) 1.19
d) 9.91

Answer: a [Reason:] The transmission coefficient is the reverse of the reflection coefficient, i.e, T + R = 1. When T = 0.65, we get R = 0.35. Thus the return loss RL = -20log R = -20log 0.35 = 9.11 decibel.

11. The return loss is given as 12 decibel. Calculate the reflection coefficient.
a) 0.35
b) 0.55
c) 0.25
d) 0.75

Answer: c [Reason:] The return loss is given by RL = -20log R. The reflection coefficient can be calculated as R = 10(-RL/20), by anti logarithm property. For the given return loss RL = 12, we get R = 10(-12/20) = 0.25.

12. Find the transmission coefficient of a wave, when the return loss is 6 decibel.
a) 0.498
b) 0.501
c) 0.35
d) 0.65

Answer: a [Reason:] The return loss is given by RL = -20log R. The reflection coefficient can be calculated as R = 10(-RL/20), by anti logarithm property. For the given return loss RL = 6, we get R = 10(-6/20) = 0.501. The transmission coefficient will be T = 1 – R = 1-0.501 = 0.498.

## Set 5

1. Calculate the capacitance of a material in air with area 20 units and distance between plates is 5m.
a) 35.36pF
b) 3.536pF
c) 35.36nF
d) 3.536nF

Answer: a [Reason:] The capacitance of any material is given by, C = εA/d, where ε = εoεr is the permittivity in air and the material respectively. Thus C = 1 X 8.854 X 10-12 X 20/5 = 35.36pF.

2. The resistance of a material with conductivity 2millimho/m2, length 10m and area 50m is
a) 500
b) 200
c) 100
d) 1000

Answer: c [Reason:] The resistance is given by, R = ρL/A, where ρ is the resistivity, the inverse of conductivity. R = 10/(0.002 X 50) = 100 ohm.

3. Find the inductance of a coil with permeability 3.5, turns 100 and length 2m. Assume the area to be thrice the length.
a) 131.94mH
b) 94.131mH
c) 131.94H
d) 94.131H

Answer: a [Reason:] The inductance is given by L = μ N2A/l, where μ= μoμr is the permeability of air and the material respectively. N = 100 and Area = 3 X 2 = 6. L = 4π X 10-7 X 1002 X 6/2 = 131.94mH.

4. Find the current density of a material with resistivity 20 units and electric field intensity 2000 units.
a) 400
b) 300
c) 200
d) 100

Answer: d [Reason:] The current density is given by J = σ E, where σ is the conductivity. Thus resistivity ρ = 1/σ. J = E/ρ = 2000/20 = 100 units.

5. Find the current in a conductor with resistance 2 ohm, electric field 2 units and distance 100cm.
a) 1A
b) 10mA
c) 10A
d) 100mA

Answer: a [Reason:] We know that E = V/d. To get potential, V = E X d = 2 X 1 = 2 volts. From Ohm’s law, V = IR and current I = V/R = 2/2 = 1A.

6. In electric fields, D= ε E. The correct expression which is analogous in magnetic fields will be
a) H = μ B
b) B = μ H
c) A = μ B
d) H = μ A

Answer: b [Reason:] In electric fields, the flux density is a product of permittivity and field intensity. Similarly, for magnetic fields, the magnetic flux density is the product of permeability and magnetic field intensity, given by B= μ H.

7. Find the force on a conductor of length 12m and magnetic flux density 20 units when a current of 0.5A is flowing through it.
a) 60
b) 120
c) 180
d) 200

Answer: b [Reason:] The force on a conductor is given by F = BIL, where B = 20, I = 0.5 and L = 12. Force F = 20 X 0.5 x 12 = 120 N.

8. From the formula F = qE, can prove that work done is a product of force and displacement. State True/False
a) True
b) False

Answer: a [Reason:] We know that F = qE = qV/d and W = qV. Thus it is clear that qV = W and qV = Fd. On equating both, we get W = Fd, which is the required proof.

9. Calculate the power of a material with electric field 100 units at a distance of 10cm with a current of 2A flowing through it.
a) 10
b) 20
c) 40
d) 80