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# Multiple choice question for engineering

## Set 1

1. In a magnetic material, always there exist magnetic dipoles as well as monopoles. State True/False.
a) True
b) False

Answer: b [Reason:] A magnetic material possesses only magnetic dipoles. The absence of magnetic monopoles is indicated by the equation Div(B) = 0.

2. Find the total flux in a material of flux density 15 units in an area of 24 units.
a) 3.6
b) 7.2
c) 9.6
d) 5.4

Answer: a [Reason:] The total flux in a material is given by φ = ∫ B.dS. Put B= 15 and ∫dS = 0.24. On substituting, we get φ = 15 x 0.24 = 3.6 units.

3. Find the magnetic field intensity of a material with flux density of 24 units in air(in 106 order)
a) 19.09
b) 21
c) 25
d) 26.78

Answer: a [Reason:] The magnetic flux density is given by B = μ H. To get H, put B = 24 and μ = 4∏ x 10-7. Thus H = 24/4∏ x 10-7 = 19.09 x 106 units.

4. Find the magnetic flux density when the vector potential is a position vector.
a) 1
b) 0
c) -1
d) ∞

Answer: b [Reason:] The magnetic flux density is given by B = Curl(A) and A = x i + y j + z k. The curl of the position vector A is i(0) – j(0) + k(0) = 0. Thus the flux density is also zero.

5. When the electric field travels in +x direction and the EM wave is travelling the –y direction, then the magnetic field will be travelling in which direction?
a) +z direction
b) –z direction
c) Either +z or –z direction
d) Does not travel

Answer: c [Reason:] The electric field and magnetic field will always travel perpendicular to each other and the EM wave will travel perpendicular to both these fields. In the given condition when E travels in +x direction and wave in –y direction, then the H field that is perpendicular to both components will be travelling in either +z or –z direction.

6. Inductance is present in semiconductor. State True/False.
a) True
b) False

Answer: b [Reason:] Inductance property exists only for pure conductors like coil, solenoid, toroid etc. It is not present in semiconductors.

7. Electromagnetic waves are longitudinal in nature. State True/False.
a) True
b) False

Answer: b [Reason:] Light and other electromagnetic radiations are transverse in nature as they travel at the same speed through a vacuum, such as through space. Such waves vibrate at right angles to the direction of propagation.

8. Identify the devices that do not use electromagnetic energy.
a) Television
b) Washing machine
c) Microwave oven
d) Mobile phones

Answer: b [Reason:] Television and mobile phones use the electromagnetic waves as signals. Microwave ovens generate electromagnetic waves (microwaves) for heating the food. Washing machine does not use any EM wave for its operation.

9. Find the flux density of a conductor in the square of the centre of the loop having current 3.14A and radius is 1.414m in air.
a) 8π x 10-7
b) 4π x 10-7
c) 6π x 10-7
d) 2π x 10-7

Answer: a [Reason:] The magnetic flux density of a conductor in the square of centre of the loop is given by B = 2√2 μo I/π R. Put I = 3.14 and R = 1.414, we get B = 2 x 1.414 x 4π x 10-7 x 3.14/π x 1.414 = 8π x 10-7 units.

10. Which type of flux will increase the inductance?
a) Series aiding
b) Series opposing
c) Shunt aiding
d) Shunt opposing

Answer: a [Reason:] The series aiding flux will give maximum inductance to a circuit compared to any other fluxing techniques. This is because all the individual and mutual inductances will get added.

## Set 2

1. The expression for refractive index is given by
a) N = v/c
b) N = c/v
c) N = cv
d) N = 1/cv

Answer: b [Reason:] The refractive index is defined as the ratio of the velocity of light in a vacuum to its velocity in a specified medium. It is given by n = c/v. It is constant for a particular material.

2. Numerical aperture is expressed as the
a) NA = sin θa
b) NA = cos θa
c) NA = tan θa
d) NA = sec θa

Answer: a [Reason:] The numerical aperture is the measure of how much light the fiber can collect. It is the sine of the acceptance angle, the angle at which the light must be transmitted in order to get maximum reflection. Thus it is given by NA = sin θa.

3. For total internal reflection to occur, which condition must be satisfied?
a) N1 = N2
b) N1 > N2
c) N1 < N2
d) N1 x N2=1

Answer: b [Reason:] The refractive of the transmitting medium should be greater than that of the receiving medium. In other words, the light must flow from denser to rarer medium, for total internal reflection to occur.

4. Find the refractive index of a medium having a velocity of 1.5 x 108.
a) 0.5
b) 5
c) 0.2
d) 2

Answer: d [Reason:] The refractive index is given by the ratio of the speed of light to the velocity in a particular medium. It is given by n = c/v. On substituting for v = 1.5 x 108 and c = 3 x 108, we get n = 3/1.5 = 2. The quantity has no unit.

5. The refractive index of water will be
a) 1
b) 2.66
c) 5
d) 1.33

Answer: d [Reason:] The velocity of light in water as medium will be 2.25 x 108. On substituting for the speed of light, we get refractive index as n = 3/2.25 = 1.33(no unit).

6. The refractive index of air is unity. State True/False.
a) True
b) False

Answer: a [Reason:] The velocity of light in the air medium and the speed of light are both the same. Since light travels at maximum velocity in air only. Thus the refractive index n = c/v will be unity.

7. The numerical aperture of a coaxial cable with core and cladding indices given by 2.33 and 1.4 respectively is
a) 3.73
b) 0.83
c) 3.46
d) 1.86

Answer: d [Reason:] The numerical aperture is given by NA = √(n12 – n22), where n1 and n2 are the refractive indices of core and cladding respectively. On substituting for n1 = 2.33 and n2 = 1.4, we get NA = √(2.332-1.42) = 1.86.

8. Find the acceptance angle of a material which has a numerical aperture of 0.707 in air.
a) 30
b) 60
c) 45
d) 90

Answer: c [Reason:] The numerical aperture is given by NA = n sin θa, where n is the refractive index. It is unity in air. Thus NA = sin θa. To get θ= sin-1(NA), put NA = 0.707, thus θa = sin-1(0.707) = 45 degree.

9. The numerical aperture of a material with acceptance angle of 60 degree in water will be
a) 1.15
b) 2.15
c) 5.21
d) 1.52

Answer: a [Reason:] The numerical aperture is given by NA = n sin θa, where n is the refractive index. It is 1.33 for water medium. Given that the acceptance angle is 60, we get NA = 1.33 sin 60 = 1.15.

10. The core refractive index should be lesser than the cladding refractive index for a coaxial cable. State True/False
a) True
b) False

Answer: b [Reason:] The light should pass through the core region only, for effective transmission. When light passes through cladding, losses will occur, as cladding is meant for protection. Thus core refractive index must be greater than the cladding refractive index.

11. The refractive index is 2.33 and the critical angle is 350. Find the numerical aperture.
a) 2
b) 1.9
c) 2.33
d) 12

Answer: b [Reason:] The numerical aperture is given by NA = n cos θc, where θc is the critical angle and n is the refractive index. On substituting for n = 2.33 and θc = 35, we get NA = 2.33 cos 35 = 1.9(no unit).

12. Choose the optical fibre material from the given materials.
a) Glass
b) Plastic
c) Silica
d) Quartz

Answer: c [Reason:] Silica is the most dominant optical fibre material. This is because of its hardness, flexibility, melting point. Also it is an easily available material.

## Set 3

1. The electric flux density and electric field intensity have which of the following relation?
a) Linear
b) Nonlinear
c) Inversely linear
d) Inversely nonlinear

Answer: a [Reason:] The electric flux density is directly proportional to electric field intensity. The proportionality constant is permittivity. D=ε E. It is clear that both are in linear relationship.

2. The electric field intensity is the negative gradient of the electric potential. State True/False.
a) True
b) False

Answer: a [Reason:] V = -∫E.dl is the integral form. On differentiating both sides, we get E = -Grad (V). Thus the electric field intensity is the negative gradient of the electric potential.

3. Find the electric potential for an electric field 3units at a distance of 2m.
a) 9
b) 4
c) 6
d) 3/2

Answer: c [Reason:] The electric field intensity is the ratio of electric potential to the distance. E = V/d. To get V = E X d = 3 X 2 = 6units.

4. Find the potential at a point (4, 3, -6) for the function V = 2x2y + 5z.
a) 96
b) 66
c) 30
d) -66

Answer: b [Reason:] The electric potential for the function V = 2x2y + 5z at the point (4, 3, -6) is given by V = 2(4)2(3) + 5(-6) = 96-30 = 66 units.

5. Find the electric flux density surrounding a material with field intensity of 2xyz placed in transformer oil ( εr = 2.2) at the point P(1,2,3) is
(in 10-10 units)
a) 2.1
b) 2.33
c) 2.5
d) 2.77

Answer: c [Reason:] D = εE, where ε = εo εr. The flux density is given by, D = 8.854 X 10-12 X 2.2 X 2(1)(2)(3) = 2.33 X 10-10 units.

6. If potential V = 20/(x2 + y2). The electric field intensity for V is
40(x i + y j)/(x2 + y2)2. State True/False.
a) True
b) False

Answer: a [Reason:] E = -Grad (V) = -Grad(20/(x2 + y2)) = -(-40x i /(x2 + y2)2 – 40(y j)/(x2 + y2)2) = 40(x i + y j)/(x2 + y2)2. Thus the statement is true.

7. Find the potential of the function V = 60cos θ/r at the point P(3, 60, 25).
a) 20
b) 10
c) 30
d) 60

Answer: b [Reason:] Given V = 60cos θ/r. For r = 3m and θ = 60, we get V = 60cos 60/3 = 20cos 60 = 10 units.

8. Find the work done moving a charge 2C having potential V = 24volts is
a) 96
b) 24
c) 36
d) 48

Answer: d [Reason:] The work done is the product of charge and potential. W = Q X V = 2 X 24 = 48 units.

9. If the potential is given by, V = 10sin θ cosφ/r, find the density at the point P(2, π/2, 0)
(in 10-12 units)
a) 13.25
b) 22.13
c) 26.31
d) 31.52

Answer: b [Reason:] Since V is given find out E.E = -Grad(V) = – Grad(10sin θ cosφ/r). From E, we can easily compute D. D = εE = 8.854 X 10-12 X 5/2 = 22.13 units.

10. If V = 2x2y + 20z – 4/(x2 + y2), find the density at A(6, -2.5, 3) in nC/m2.
a) 0.531i – 0.6373j – 0.177k
b) 0.6373i – 0.177j -0.531k
c) 0.177i – 0.6373j – 0.531k
d) 0.531i – 0.177j – 0.6373k

Answer: a [Reason:] Find E from V, E = -Grad (V). We get E at A(6,-2.5,3) as 59.97i – 71.98j -20k. Thus D = εE = 8.854 X 10-12 X (59.97i – 71.98j -20k) = (0.531i – 0.6373j – 0.177k) nC/m2.

## Set 4

1. Find the resistivity of a material having resistance 20kohm, area 2 units and length of 12m.
a) 6666.6
b) 3333.3
c) 1200
d) 2000

Answer: b [Reason:] The resistance of a material is given by R = ρL/A. To get ρ, put R = 20 x 103, A = 2 and L = 12. We get ρ = 3333.3 units.

2. A resistor value of colour code orange violet orange will be
a) 37 kohm
b) 37 Mohm
c) 48 kohm
d) 48 Mohm

Answer: a [Reason:] Orange refers to number 3. Violet refers to number 7. The third colour code orange refers to 103. Thus the resistor value will be 37 kilo ohm.

3. A infinite resistance is considered as a/an
a) Closed path(short circuit)
b) Open path
c) Not defined

Answer:b [Reason:] When there exists infinite resistance in a path, the current flowing will ideally be zero. This is possible only for an open path/circuit.

4. Find the time constant in a series R-L circuit when the resistance is 4 ohm and the inductance is 2 H.
a) 0.25
b) 0.2
c) 2
d) 0.5

Answer: d [Reason:] The time constant for an R-L series circuit will be τ = L/R. Put R = 4 and L = 2. We get τ = 2/4 = 0.5 second.

5. Find the time constant for a R-C circuit for resistance R = 24 kohm and C = 16 microfarad.
a) 1.5 millisecond
b) 0.6 nanosecond
c) 384 millisecond
d) 384 microsecond

Answer: c [Reason:] The time constant for R-C circuit is τ = RC. Put R = 24 kilo ohm and C = 16 micro farad. We get τ = 24 x 103 x 16 x 10-6 = 0.384 = 384 millisecond.

6. Find the capacitance when charge is 20 C has a voltage of 1.2V.
a) 32.67
b) 16.67
c) 6.67
d) 12.33

Answer: b [Reason:] Capacitance is related to Q and V as C = Q/V. Put C = 20C and V = 1.2V, we get Q = 20/1.2 = 16.67 farad.

7. Calculate the capacitance of two parallel plates of area 2 units separated by a distance of 0.2m in air(in picofarad)
a) 8.84
b) 88.4
c) 884.1
d) 0.884

Answer: b [Reason:] Capacitance is given by, C = εo A/d. Put A = 2, d = 0.2, εo = 8.854 x 10-12, we get C = 8.841 x 10-11 = 88. 41 pF.

8. Compute the capacitance between two concentric shells of inner radius 2m and the outer radius is infinitely large.
a) 0.111 nF
b) 0.222 nF
c) 4.5 nF
d) 5.4 nF

Answer: b [Reason:] The concentric shell with infinite outer radius is considered to be an isolated sphere. The capacitance C = 4πε/(1/a – 1/b). If b->∞, then C = 4πεa. Put a = 2m, we get C = 4π x 8.854 x 10-12 x 2 = 0.222 nF.

9. The capacitance of a material refers to
a) Ability of the material to store magnetic field
b) Ability of the material to store electromagnetic field
c) Ability of the material to store electric field
d) Potential between two charged plates

Answer: c [Reason:] The capacitance of a material is a measure of the ability of the material to store electric field. It is the ratio of charge stored to the voltage across the parallel plates.

10. A cable of core radius 1.25cm and impregnated paper insulation of thickness 2.13cm and relative permittivity 3.5. Compute the capacitance of the cable/km(in nF)
a) 195.7
b) 179.5
c) 157.9
d) 197.5

Answer: a [Reason:] Capacitance between coaxial cylinders of inner radius 1.25cm and outer radius 1.25 + 2.13 = 3.38cm will be C = 2πεL/ ln(b/a). Put b = 3.38, a = 1.25 and L = 1000m, we get C = 1.957 x 10-7 = 195.7 nF.

## Set 5

1. The resultant electric field of a wave with Ex = 3 and Ey = 4 will be
a) 7
b) 1
c) 25
d) 5

Answer: d [Reason:] The resultant electric field of two electric components Ex and Ey is E = √(Ex2 + Ey2). On substituting for Ex = 3 and Ey = 4, we get E = 5 units.

2. In S polarisation, the electric field lies in the plane perpendicular to that of the interface. State True/False
a) True
b) False

Answer: a [Reason:] In the EM wave propagation, the electric and magnetic fields are perpendicular to each other. The S polarised wave is similar to the transverse magnetic (TM) wave, the electric field lies in the plane perpendicular to that of the interface.

3. In P polarisation, the electric field lies in the same plane as the interface. State True/False.
a) True
b) False

Answer: a [Reason:] In the EM wave propagation, the electric and magnetic fields are perpendicular to each other. The P polarised wave is similar to the transverse electric (TE) wave, the magnetic field lies in the plane perpendicular to that of the interface or the electric field lies in the same plane as the interface.

4. The group delay of a wave with phase constant 2.5 units and frequency of 1.2 radian/sec is
a) 3.7
b) 1.3
c) 3
d) 2.08

Answer: d [Reason:] The group delay is given by td = β/ω. On substituting for β = 2.5 and ω = 1.2, we get the group delay as td = 2.5/1.2 = 2.08 units.

5. The Brewster angle is valid for which type of polarisation?
a) S polarised
b) P polarised
c) Elliptical
d) Linear

Answer: b [Reason:] The Brewster angle is valid for perpendicular polarisation. The P polarised wave is also a type of perpendicular polarisation. In P polarisation, the electric field lies in the plane of the interface.

6. Find the reflection coefficient of a wave with an incident electric field of 5 units and reflected electric field of 2 units.
a) 2.5
b) 0.4
c) 0.8
d) 1.2

Answer: b [Reason:] The reflection coefficient is the ratio of the reflected electric field to the incident electric field. Thus τ = Er/Ei. On substituting for Ei = 5 and Er = 2, we get τ = 2/5 = 0.4(no unit).

7. The transmission coefficient of a wave with incident and transmitted electric field of 5 and 5 respectively is
a) 0
b) 1
c) 10
d) 5

Answer: b [Reason:] The transmission coefficient is the ratio of the transmitted electric field to the incident electric field. Thus T = Et/Ei. On substituting for Et = 5 and Ei = 5, we get T = 5/5 = 1(no unit). Simply, when the incident and transmitted field are same, no reflection occurs and the transmission is unity.

8. Find the relative permittivity of the medium having a refractive index of 1.6
a) 0.4
b) 2.56
c) 3.2
d) 4.8

Answer: b [Reason:] The refractive index is the square root of the relative permittivity. It is given by n = √εr. To get εr, put n = 1.6. We get εr = n2 = 1.62 = 2.56(no unit).

9. Calculate the transmission coefficient of a wave with a reflection coefficient of 0.6
a) 0.6
b) 1
c) 0
d) 0.4

Answer: d [Reason:] The transmission coefficient is the reverse of the reflection coefficient. Thus T + τ = 1. On substituting for τ = 0.6, we get T = 0.4. It has no unit.

10.The phase constant of a wave propagation with frequency of 35 radian/sec and time delay of 7.5 sec is
a) 4.66
b) 262.5
c) 46.6
d) 26.25