Multiple choice question for engineering
1. The charge within a conductor will be
Answer: c [Reason:] No charges exist in a conductor. An illustration for this statement is that, it is safer to stay inside a car rather than standing under a tree during lightning. Since the car has a metal body, no charges will be possessed by it to get ionized by the lightning.
2. For a conservative field which of the following equations holds good?
a) ∫ E.dl = 0
b) ∫ H.dl = 0
c) ∫ B.dl = 0
d) ∫ D.dl = 0
Answer: a [Reason:] A conservative field implies the work done in a closed path will be zero. This is given by ∫ E.dl = 0.
3. Find the electric field if the surface density at the boundary of air is 10-9.
Answer: c [Reason:] It is the conductor-free space boundary. At the boundary, E = ρ/εo. Put ρ = 10-9 and εo = 10-9/36π. We get E = 36π units.
4. Find the flux density at the boundary when the charge density is given by 24 units.
Answer: b [Reason:] At the boundary of a conductor- free space interface, the flux density is equal to the charge density. Thus D = ρv = 24 units.
5. Which component of the electric field intensity is always continuous at the boundary?
Answer: a [Reason:] At the boundary of the dielectric-dielectric, the tangential component of the electric field intensity is always continuous. We get Et1 = Et2.
6. The normal component of which quantity is always discontinuous at the boundary?
Answer: b [Reason:] The normal component of an electric flux density is always discontinuous at the boundary for a dielectric-dielectric boundary. We get Dn1 = Dn2, when we assume the free surface charge exists at the interface.
7. The electric flux density of a surface with permittivity of 2 is given by 12 units. What the flux density of the surface in air?
Answer: b [Reason:] The relation between electric field and permittivity is given by Dt1/Dt2 = ε1/ε2. Put Dt1 = 12, ε1 = 2 and ε2 =1, we get Dt2 = 12 x 1/ 2 = 6 units.
8. The electric field intensity of a surface with permittivity 3.5 is given by 18 units. What the field intensity of the surface in air?
Answer: c [Reason:] The relation between flux density and permittivity is given by En1/En2 = ε2/ ε1. Put En1 = 18, ε1 = 3.5 and ε2 = 1. We get En2 = 18 x 3.5 = 63 units.
9. A wave incident on a surface at an angle 60 degree is having field intensity of 6 units. The reflected wave is at an angle of 30 degree. Find the field intensity after reflection.
Answer: c [Reason:] By Snell’s law, the relation between incident and reflected waves is given by, E1 sin θ1 = E2 sin θ2. Thus 6 sin 60 = E2 sin 30. We get E2 = 6 x 1.732 = 10.4 units.
10. Find the permittivity of the surface when a wave incident at an angle 60 is reflected by the surface at 45 in air.
Answer: d [Reason:] From the relations of the boundary conditions of a dielectric-dielectric interface, we get tan θ1/tan θ2 = ε1/ε2. Thus tan 60/tan 45 = ε1/1. We get ε1 = tan 60 = 1.73.
1. Brewster angle is valid for which type of polarisation?
c) S polarised
d) P polarised
Answer: b [Reason:] The parallel polarisation of the electromagnetic waves is possible only when the transmission occurs at the Brewster angle.
2. The Brewster angle is expressed as
d) Tan (n)
Answer: c [Reason:] The tangent of the Brewster angle is the ratio of the refractive indices of the second medium to that of the first medium. It is given by tan θb= n2/n1. Thus the Brewster angle will be θb = tan-1(n2/n1).
3. The refractive index of a material with permittivity 16 is given by
Answer: c [Reason:] The refractive index is the square root of the permittivity. Thus n = √ε. Given that ε = 16, we get refractive index as n = 4. It has no unit.
4. The reflection coefficient in the wave propagation when it is transmitted with the Brewster angle is
Answer: a [Reason:] Brewster angle propagation refers to complete transmission. The wave transmitted at the Brewster angle will be completely transmitted without reflection. Thus the reflection coefficient will be zero.
5. The transmission coefficient of a wave propagating in the Brewster angle is
Answer: b [Reason:] The transmission coefficient is the reverse of the reflection coefficient. At Brewster angle, the reflection will be zero. Thus the transmission is T = 1-R. Since R = 0, T = 1. It is to be noted that T and R lies in the range of 0 to 1.
6. A circularly polarised wave transmitted at the Brewster angle will be received as linearly polarised wave. State True/False
Answer: a [Reason:] The Brewster angle is said to be the polarisation angle. When a circularly polarised wave is incident at the Brewster angle, the resultant wave will be linearly polarised.
7. An elliptically polarised wave transmitted at the Brewster angle will be received as an elliptically polarised wave. State True/False
Answer: b [Reason:] Any polarised wave transmitted at the Brewster angle will be linearly polarised. It can be a parallel, perpendicular, circular or elliptical polarisation. The resultant wave is always linearly polarised. This is the reason why the Brewster angle is called polarisation angle.
8. Find the Brewster angle of a wave transmitted from a medium of permittivity 4 to a medium of permittivity 2.
Answer: a [Reason:] The Brewster angle is given by θb = tan-1(n2/n1), where n = √ε. Thus we can express the formula in terms of permittivity as θb = tan-1√ (ε 2/ε 1). Here ε1 = 4 and ε2 = 2. Thus we get θb = tan-1√ (2/4) =
tan-1(0.707) = 35.26 degree.
9. Find the ratio of refractive index of medium 2 to that of medium 1, when the Brewster angle is 60 degree.
Answer: d [Reason:] The tangent of the Brewster angle is the ratio of the medium 2 permittivity to the medium 1 permittivity. Thus tan θb = (n2/n1). Given that θb = 60 degree, the ratio n2/n1 will be tan 60 = 1.732.
10. The Brewster angle is the angle of
Answer: a [Reason:] The Brewster angle is the angle of incidence at which complete transmission of the electromagnetic wave occurs.
1. Standing waves occurs due to
a) Impedance match
b) Impedance mismatch
Answer: b [Reason:] Impedance mismatches result in standing waves along the transmission line. It shows the variation of the wave amplitudes due to mismatching.
2. Standing wave ratio is defined as the
a) Ratio of voltage maxima to voltage minima
b) Ratio of current maxima to current minima
c) Product of voltage maxima and voltage minima
d) Product of current maxima and current minima
Answer: a [Reason:] SWR is defined as the ratio of the partial standing wave’s amplitude at an antinode (maximum) to the amplitude at a node (minimum) along the line. It is given by S = VMAX/VMIN.
3. Given that the reflection coefficient is 0.6. Find the SWR.
Answer: b [Reason:] The relation between reflection coefficient and SWR is given by S = 1 + R/1 – R. On substituting for R = 0.6, we get S = 1 + 0.6/1 – 0.6 = 1.6/0.4 = 4.
4. The maxima and minima voltage of the standing wave are 6 and 2 respectively. The standing wave ratio is
Answer: b [Reason:] The ratio of voltage maxima to voltage minima is given by the standing wave ratio SWR. Thus S = VMAX/VMIN. On substituting the given data, we get S = 6/2 = 3.
5. Find the standing wave ratio, when a load impedance of 250 ohm is connected to a 75 ohm line.
Answer: d [Reason:] The standing wave ratio is the ratio of the load impedance to the characteristic impedance. Thus S = ZL/Zo. On substituting for ZL = 250 and Zo = 75, we get S = 250/75 = 3.33.
6. Find the reflection coefficient of the wave with SWR of 3.5.
Answer: a [Reason:] The reflection coefficient in terms of the SWR is given by R = S – 1/S + 1. On substituting for S = 3.5, we get 3.5 – 1/3.5 + 1 = 0.55.
7. The range of the standing wave ratio is
a) 0 < S < 1
b) -1 < S < 1
c) 1 < S < ∞
d) 0 < S < ∞
Answer: c [Reason:] The standing wave ratio is given by S = 1 – R/1 + R. Thus the minimum value of S is 1. It can extend upto infinity for long lines. Thus the range is 1 < S < ∞.
8. For matched line, the standing wave ratio will be
Answer: d [Reason:] In a matched line, maximum transmission occurs. The reflection will be zero. The standing wave ratio S = 1 – R/1 + R. For R = 0, the SWR is unity for matched line.
9. The maximum impedance of a 50 ohm transmission line with SWR of 3 is
Answer: c [Reason:] The maximum impedance is given by the product of the characteristic impedance and the SWR. Thus Zmax = S Zo. On substituting for S = 3 and Zo = 50, we get ZMAX = 3 X 50 = 150 units.
10. The minimum impedance of a 75 ohm transmission line with a SWR of 2.5 is
Answer: d [Reason:] The minimum impedance in terms of SWR is given by ZMIN = Zo/S. Substituting the given data for S = 2.5 and Zo = 75, we get Zmin = 75/2.5 = 30.
11. The standing wave ratio of short circuited and open circuited lines will be
Answer: d [Reason:] The transmission line will reflect high power when it is short or circuited. This will lead to high reflection coefficient. Thus the standing wave ratio will be infinity for these extreme cases.
12. The current reflection coefficient of a line with voltage reflection coefficient of 0.65 is given by
Answer: c [Reason:] The current reflection coefficient at any point on the line is the negative of the voltage reflection coefficient at that point, i.e, -R. Given that the voltage reflection coefficient is 0.65, thus the current reflection coefficient is -0.65.
1. Which of the following are conductors?
Answer: c [Reason:] Normally, metals are said to be good conductors. Here mercury is the only metal (which is in liquid form). The other options are insulators.
2. Find the range of band gap energy for conductors.
a) >6 eV
b) 0.2-0.4 eV
c) 0.4-2 eV
d) 2-6 eV
Answer: b [Reason:] Conductors are materials with least band gap energy. The smallest range in this group is 0.2-0.4 eV.
3. Conduction in metals is due to
a) Electrons only
b) Electrons and holes
c) Holes only
d) Applied electric field
Answer: a [Reason:] Conduction in metals is only due to majority carriers, which are electrons. Electrons and holes are responsible for conduction in a semiconductor.
4. Find the band gap energy when a light of wavelength 1240nm is incident on it.
Answer: a [Reason:] The band gap energy in electron volt when wavelength is given is, Eg = 1.24(μm)/λ = 1.24 x 10-6/1240 x 10-9 = 1eV.
5. Alternating current measured in a transmission line will be
a) Peak value
b) Average value
c) RMS value
Answer: c [Reason:] The instantaneous current flowing in a transmission line, when measured using an ammeter, will give RMS current value. This value is 70.7% of the peak value. This is because, due to oscillations in AC, it is not possible to measure peak value. Hence to normalise, we consider current at any time in a line will be the RMS current.
6. The current in a metal at any frequency is due to
a) Conduction current
b) Displacement current
c) Both conduction and displacement current
d) Neither conduction nor displacement current
Answer: a [Reason:] At any frequency, the current through the metal will be due to conduction current. Only at high frequencies and when medium is air, the conduction is due to displacement current. Thus in general the current in metal is due to conduction current, which depends on the mobility of the carriers.
7. For conductors, the free electrons will exist at
a) Valence band
b) Middle of valence and conduction band
c) Will not exist
d) Conduction band
Answer: d [Reason:] In conductors, the free electrons exist in the conduction band. Since the band gap energy is very low, less energy is required to transport the free electrons to the conduction band, as they are readily available to conduct.
8. The current flowing through an insulating medium is called
Answer: b [Reason:] A beam of electrons in a vacuum tube is called convection current. It occurs when current flows through an insulating medium like liquid, vacuum etc.
9. Find the conduction current density when conductivity of a material is 500 units and corresponding electric field is 2 units.
Answer: c [Reason:] The conduction current density is given by, J = σE
J = 500 X 2 = 1000 units.
10. Calculate the convection current when electron density of 200 units is travelling at a speed of 12m/s.
Answer: b [Reason:] The convection current density is given by, J = ρeV
J = 200 X 12= 2400 units.
1. Find the current when the charge is a time function given by q(t) = 3t + t2 at 2 seconds.
Answer: c [Reason:] The current is defined as the rate of change of charge in a circuit ie, I = dq/dt. On differentiating the charge with respect to time, we get 3 + 2t. At time t = 2s, I = 7A.
2. The continuity equation is a combination of which of the two laws?
a) Ohm’s law and Gauss law
b) Ampere law and Gauss law
c) Ohm’s law and Ampere law
d) Maxwell law and Ampere law
Answer: b [Reason:] I = ∫ J.ds is the integral form of Ohm’s law and Div (J) = dq/dt is the Gauss law analogous to D. Through these two equations, we get Div(J) = -dρ/dt. This is the continuity equation.
3. Calculate the charge density for the current density given 20sin x i + ycos z j at the origin.
Answer: b [Reason:] Using continuity equation, the problem can be solved. Div(J) =
– dρ/dt. Div(J) = 20cos x + cos z. At origin, we get 20cos 0 + cos 0 = 21. To get ρ, on integrating the Div(J) with respect to t, the charge density will be 21t.
4. Compute the conductivity when the current density is 12 units and the electric field is 20 units. Also identify the nature of the material.
a) 1.67, dielectric
b) 1.67, conductor
c) 0.6, dielectric
d) 0.6, conductor
Answer: c [Reason:] The current density is the product of conductivity and electric field intensity. J = σE. To get σ, put J = 12 and E = 20. σ = 12/20 = 0.6. Since the conductivity is less than unity, the material is a dielectric.
5. Find the electron density when convection current density is 120 units and the velocity is 5m/s.
Answer: c [Reason:] The convection current density is given by J = ρe x v. To get ρe, put J = 120 and v = 5. ρe = 120/5 = 24 units.
6. Calculate the electric field when the conductivity is 20 units, electron density is 2.4 units and the velocity is 10m/s. Assume the conduction and convection current densities are same.
Answer: d [Reason:] The conduction current density is given by J = σE and the convection current density is J = ρe v. When both are equal, ρe v = σE. To get E, put σ = 20, ρe = 2.4 and v = 10, E = 2.4 x 10/20 = 1.2 units.
7. Find the mobility of the electrons when the drift velocity is 23 units and the electric field is 11 units.
Answer: b [Reason:] The mobility is defined as the drift velocity per unit electric field. Thus μe = vd/E = 23/11 = 2.1 units.
8. Find the resistance of a cylinder of area 200 units and length 100m with conductivity of 12 units.
Answer: a [Reason:] The resistance is given by R = ρL/A = L/σA. Put L = 100, σ = 12 and A = 200, we get R = 100/(12 x 200) = 1/24 units.
9. Calculate the potential when a conductor of length 2m is having an electric field of 12.3units.
Answer: d [Reason:] The electric field is given by E = V/L. To get V, put E = 12.3 and L = 2.Thus we get V = E x L = 12.3 x 2 = 24.6 units.
10. On equating the generic form of current density equation and the point form of Ohm’s law, we can obtain V=IR. State True/False.
Answer: a [Reason:] The generic current density equation is J = I/A and the point form of Ohm’s law is J = σ E. On equating both and substituting E = V/L, we get V = IL/σ A = IR which is the Ohm’s law.