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. Coulomb is the unit of which quantity?
a) Field strength
b) Charge
c) Permittivity
d) Force

View Answer

Answer: b [Reason:] The standard unit of charge is Coulomb. One coulomb is defined as the 1 Newton of force applied on 1 unit of electric field.

2. Coulomb law is employed in
a) Electrostatics
b) Magnetostatics
c) Electromagnetics
d) Maxwell theory

View Answer

Answer: a [Reason:] Coulomb law is applied to static charges. It states that force between any two point charges is proportional to the product of the charges and inversely proportional to square of the distance between them. Thus it is employed in electrostatics.

3. Find the force between 2C and -1C separated by a distance 1m in air(in newton).
a) 18 X 106
b) -18 X 106
c) 18 X 10-6
d) -18 X 10-6

View Answer

Answer: b [Reason:] F = q1q2/(4∏εor2) = -2 X 9/(10-9 X 12) = -18 X 109.

4. Two charges 1C and -4C exists in air. What is the direction of force?
a) Away from 1C
b) Away from -4C
c) From 1C to -4C
d) From -4C to 1C

View Answer

Answer: c [Reason:] Since the charges are unlike, the force will be attractive. Thus the force directs from 1C to -4C.

5. Find the force of interaction between 60 stat coulomb and 37.5 stat coulomb spaced 7.5cm apart in transformer oil(εr=2.2) in 10-4 N,
a) 8.15
b) 5.18
c) 1.518
d) 1.815

View Answer

Answer: d [Reason:] 1 stat coulomb = 1/(3 X 109) C F = (1.998 X 1.2488 X 10-16)/(4∏ X 8.854 X 10-12 X 2.2 X (7.5 X 10-2)2) = 1.815 X 10-4 N.

6. Find the force between two charges when they are brought in contact and separated by 4cm apart, charges are 2nC and -1nC, in μN.
a) 1.44
b) 2.44
c) 1.404
d) 2.404

View Answer

Answer: c [Reason:] Before the charges are brought into contact, F = 11.234 μN. After charges are brought into contact and then separated, charge on each sphere is, (q1 + q2)/2 = 0.5nC On calculating the force with q1 = q2 = 0.5nC, F = 1.404μN.

7. The Coulomb law is an implication of which law?
a) Ampere law
b) Gauss law
c) Biot Savart law
d) Lenz law

View Answer

Answer: b [Reason:] The Coulomb law can be formulated from the Gauss law, using the divergence theorem. Thus it is an implication of Gauss law.

8. Two small diameter 10gm dielectric balls can slide freely on a vertical channel. Each carry a negative charge of 1μC. Find the separation between the balls if the lower ball is restrained from moving.
a) 0.5
b) 0.4
c) 0.3
d) 0.2

View Answer

Answer: c [Reason:] F = mg = 10 X 10-3 X 9.81 = 9.81 X 10-2 N. On calculating r by substituting charges, we get r = 0.3m.

9. A charge of 2 X 10-7 C is acted upon by a force of 0.1N. Determine the distance to the other charge of 4.5 X 10-7 C, both the charges are in vacuum.
a) 0.03
b) 0.05
c) 0.07
d) 0.09

View Answer

Answer: d [Reason:] F = q1q2/(4∏εor2) , substituting q1, q2 and F, r2 = q1q2/(4∏εoF) = We get r = 0.09m.

10. For a charge Q1, the effect of charge Q2 on Q1 will be,
a) F1 = F2
b) F1 = -F2
c) F1 = F2 = 0
d) F1 and F2 are not equal

View Answer

Answer: b [Reason:] The force of two charges with respect with each other is given by F1 and F2. Thus F1 + F2 = 0 and F1 = -F2.

Set 2

1. Curl is defined as the angular velocity at every point of the vector field. State True/False.
a) True
b) False

View Answer

Answer: a [Reason:] Curl is defined as the circulation of a vector per unit area. It is the cross product of the del operator and any vector field. Circulation implies the angular at every point of the vector field. It is obtained by multiplying the component of the vector parallel to the specified closed path at each point along it, by the differential path length and summing the results.

2. The curl of curl of a vector is given by,
a) Div(Grad V) – (Del)2V
b) Grad(Div V) – (Del)2V
c) (Del)2V – Div(Grad V)
d) (Del)2V – Grad(Div V)

View Answer

Answer: b [Reason:] Curl (Curl V) = Grad (Div V) – (Del)2V is a standard result of the curl operation.

3. Which of the following theorem use the curl operation?
a) Green’s theorem
b) Gauss Divergence theorem
c) Stoke’s theorem
d) Maxwell equation

View Answer

Answer: c [Reason:] The Stoke’s theorem is given by ∫ A.dl = ∫Curl(A).ds, which uses the curl operation. There can be confusion with Maxwell equation also, but it uses curl in electromagnetics specifically, whereas the Stoke’s theorem uses it in a generalised manner. Thus the best option is c.

4. The curl of a curl of a vector gives a
a) Scalar
b) Vector
c) Zero value
d) Non zero value

View Answer

Answer: b [Reason:] Curl is always defined for vectors only. The curl of a vector is a vector only. The curl of the resultant vector is also a vector only.

5. Find the curl of the vector and state its nature at (1,1,-0.2)
F = 30 i + 2xy j + 5xz2 k
a) √4.01
b) √4.02
c) √4.03
d) √4.04

View Answer

Answer: d [Reason:] Curl F = -5z2 j + 2y k. At (1,1,-0.2), Curl F = -0.2 j + 2 k. |Curl F| = √(-0.22+22) = √4.04.

6. Is the vector is irrotational. E = yz i + xz j + xy k
a) Yes
b) No

View Answer

Answer: a [Reason:] Curl E = i(Dy(xy) – Dz(xz)) – j (Dx(xy) – Dz(yz)) + k(Dx(xz) – Dy(yz)) = i(x – x) – j(y – y) + k(z – z) = 0 Since the curl is zero, the vector is irrotational or curl-free.

7. Find the curl of A = (y cos ax)i + (y + ex)k
a) 2i – ex j – cos ax k
b) i – ex j – cos ax k
c) 2i – ex j + cos ax k
d) i – ex j + cos ax k

View Answer

Answer: b [Reason:] Curl A = i(Dy(y + ex)) – j (Dx(y + ex) – Dz(y cos ax)) + k(-Dy(y cos ax)) = 1.i – j(ex) – k cos ax = i – ex j – cos ax k.

8. Find the curl of the vector A = yz i + 4xy j + y k
a) xi + j + (4y – z)k
b) xi + yj + (z – 4y)k
c) i + j + (4y – z)k
d) i + yj + (4y – z)k

View Answer

Answer: d [Reason:] Curl A = i(Dy(y) – Dz(0)) – j (Dx(0) – Dz(yz)) + k(Dx(4xy) – Dy(yz)) = i + y j + (4y – z)k, which is option d.

9. Curl cannot be employed in which one of the following?
a) Directional coupler
b) Magic Tee
c) Isolator and Terminator
d) Waveguides

View Answer

Answer: d [Reason:] In the options a, b, c, the EM waves travel both in linear and angular motion, which involves curl too. But in waveguides, as the name suggests, only guided propagation occurs (no bending or curl of waves).

10. Which of the following Maxwell equations use curl operation?
a) Maxwell 1st and 2nd equation
b) Maxwell 3rd and 4th equation
c) All the four equations
d) None of the equations

View Answer

Answer: a [Reason:] Maxwell 1st equation, Curl (H) = J (Ampere law) Maxwell 2nd equation, Curl (E) = -D(B)/Dt (Faraday’s law) Maxwell 3rd equation, Div (D) = Q (Gauss law for electric field) Maxwell 4th equation, Div (B) = 0(Gauss law for magnetic field) It is clear that only 1st and 2nd equations use the curl operation.

Set 3

1. The real part of the propagation constant is the
a) Attenuation constant
b) Phase constant
c) Permittivity
d) Permeability

View Answer

Answer: a [Reason:] The propagation constant is given by γ = α + jβ. Here the real part is the attenuation constant and the imaginary part is the phase constant.

2. The phase constant of a wave is given by
a) ω√(LC)
b) ω√(L/C)
c) ω√(C/L)
d) ω√(1/LC)

View Answer

Answer: a [Reason:] The phase constant of a wave in a transmission line is given by β = ω√(LC), where L and C are the specifications of the line.

3. The cut off frequency of the dominant mode in a TE wave in the line having a and b as 2.5 cm and 1 cm respectively is
a) 4.5 GHz
b) 5 GHz
c) 5.5 GHz
d) 6 GHz

View Answer

Answer: d [Reason:] The dominant mode in TE is TE10. The cut off frequency will be mc/2a, where m = 1 and a = 0.025 are given. On substituting, we get the frequency as 1 x 3 x 108/2 x 0.025 = 6 GHz.

4. The cut off frequency of the TE01 mode will be
a) mc/2a
b) mc/2b
c) nc/2a
d) nc/2b

View Answer

Answer: d [Reason:] The cut off frequency consists of modes m and n. For m = 0, the dimension b will be considered. Thus the frequency is nc/2b, where c is the speed of the light.

5. The condition which will satisfy the dimensions of the waveguide is
a) a = b
b) a > b
c) a < b
d) ab = 0

View Answer

Answer: b [Reason:] The dimensions a and b represent the broad wall and the side wall dimensions respectively. The broad wall will be greater than the side wall. Thus the condition a>b is true.

6. The cut off wavelength of the TE10 mode having a broad wall dimension of 5cm is
a) 0.1
b) 1
c) 10
d) 0.01

View Answer

Answer: a [Reason:] The cut off wavelength of the waveguide is given by λc = 2a/m. on substituting for a = 0.05 and m = 1, we get λc = 2 x 0.05/1 = 0.1 units.

7. The broad wall dimension of a waveguide having a cut off frequency of 7.5 GHz is
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] The cut off frequency and the broad wall dimension are related by fc = mc/2a. On substituting for m = 1 and fc = 7.5 GHz, we get a = 0.02 or 2 cm.

8. The sin θ in the waveguide refers to the ratio of the
a) Frequency to wavelength
b) Wavelength to frequency
c) Cut off frequency to frequency
d) Frequency to cut off frequency

View Answer

Answer: c [Reason:] The ratio of the cut off frequency to the frequency at any point gives the sin θ in a waveguide.

9. Is the transmission of a frequency 5 GHz possible in waveguides?
a) Yes
b) No

View Answer

Answer: a [Reason:] The cut off frequency for waveguide operation is 6 GHz. Thus a wave of 5 GHz is not possible for transmission in a waveguide.

10. The dimension for a waveguide in dominant mode with a cut off wavelength of 2 units is
a) 2
b) 4
c) 6
d) 8

View Answer

Answer: b [Reason:] The cut off wavelength of a waveguide is given by λc = 2a/m. For the dominant mode, m = 1. Given that λc = 2, thus we get a = 4 units.

11. The waveguides are used in a transmission line for
a) Increasing transmission coefficient
b) Increasing reflection coefficient
c) Decreasing transmission coefficient
d) Decreasing reflection coefficient

View Answer

Answer: a [Reason:] The waveguides are used to increase the transmission efficiency of the waves travelling through it.

12. The attenuation coefficient of the wave having a resistance of 15 ohm in a 50 ohm line is
a) 15
b) 1.5
c) 0.15
d) 0.015

View Answer

Answer: c [Reason:] The attenuation coefficient of a wave with a resistance of R in a line of characteristic impedance Zo is α = R/2Zo. On substituting for R = 15 and Zo = 50, we get α = 15/(2 x 50) = 0.15 units.

Set 4

1. The cylindrical coordinate system is also referred to as
a) Cartesian system
b) Circular system
c) Spherical system
d) Space system

View Answer

Answer: b [Reason:] The cylindrical coordinates(r,φ,z) is also called as circular system and is used for systems with circular dimensions.

2. Transform the vector B=yi+(x+z)j located at point (-2,6,3) into cylindrical coordinates.
a) (6.325,-71.57,3)
b) (6.325,71.57,3)
c) (6.325,73.57,3)
d) (6.325,-73.57,3)

View Answer

Answer: a [Reason:] ρ = √(x2+y2) = √40 = 6.325 Φ = tan-1(y/x) = tan-1(-6/2) = -71.57 z = 3.

3. Cylindrical systems have the following scalar values respectively
a) 1, ρ ,1
b) 1, 1, 1
c) 0,1,0
d) 1,0,0

View Answer

Answer: a [Reason:] The range of radius is one to unity, that of plane angle is one to 360 degree and that of z plane is one to infinity. Thus the minimum scalar factor has to be 1, ρ , 1.

4. A charge located at point p (5,30⁰,2) is said to be in which coordinate system?
a) Cartesian system
b) Cylindrical system
c) Spherical system
d) Space system

View Answer

Answer: b [Reason:] The cylindrical system is of the form (ρ, φ, z), which relates the point given in the question.

5. Cylindrical system is employed in waveguides. State True/False.
a) True
b) False

View Answer

Answer: a [Reason:] Cylindrical systems are employed in circular waveguides, whereas Cartesian systems are employed in rectangular waveguides.

6. The pressure inside a piston cylinder is a variable of
a) Radius
b) Plane angle
c) Z plane distance
d) Constant, not a variable

View Answer

Answer: c [Reason:] Pressure varies up and down in a cylinder due to suction. Thus it is dependent on the z plane distance of the cylinder.

7. Charges filled inside a cylindrical will possess flux in which direction?
a) Upwards
b) Downwards
c) Laterally outwards
d) Inwards

View Answer

Answer: c [Reason:] The flux due to the charges will act outside the cylinder. Since the cylinder possesses curved surfaces, it will flow laterally outwards.

8. Rectangular waveguides dominate the circular waveguides. Find the reason.
a) Low cut-off frequency
b) Easy to design
c) More wave propagation
d) The statement is false

View Answer

Answer: b [Reason:] Due to linear design, the desired dimensions can be easily constructed using rectangular waveguides than circular ones.

9. Transform the spherical system B = (10/r)i + (10cos θ)j + k into cylindrical form at (5, π/2, -2)
a) 2.467i + j + 1.167k
b) 2.467i – j + 1.167k
c) 2.467i – j – 1.167k
d) 2.467i + j – 1.167k

View Answer

Answer: a [Reason:] The equivalent cylindrical form is given by, B = (10sin θ/r + rcos2θ)i + j + (10cos θ/r –r sin θ cos θ)k At (5, π/2, -2), r = √(52+-22) = √29 sin θ = 5/√29 and cos θ = -2/√29 Thus, B = 2.467i + j + 1.167k.

10. Convert the given rectangular coordinates A(2,3,1) into corresponding cylindrical coordinates
a) (3.21,56.31,1)
b) (3.21,57.31,0)
c) (3.61,57.31,0)
d) (3.61,56.31,1)

View Answer

Answer: d [Reason:] ρ = √(x2+y2) = √13 = 3.61 Φ = tan-1(y/x) = 56.31 z = 1 Thus, A = (3.61,56.31,1).

Set 5

1. A dielectric is always an insulator. But an insulator is not necessarily a dielectric. State True/False.
a) True
b) False

View Answer

Answer: a [Reason:] For a material to be dielectric, its permittivity should be very high. This is seen in insulators. For a material to be insulator, the condition is to have large band gap energy. However, this is not necessary for a dielectric.

2. Identify a good dielectric.
a) Iron
b) Ceramics
c) Plastic
d) Magnesium

View Answer

Answer: b [Reason:] Iron and magnesium are metals. Hence they need not be considered. Both ceramics and plastic are insulators. But dielectric constant is more for ceramics always. Hence ceramics is the best dielectric.

3. A dielectric can be made a conductor by
a) Compression
b) Heating
c) Doping
d) Freezing

View Answer

Answer: b [Reason:] On increasing the temperature, the free electrons in an insulator can be promoted from valence to conduction band. Gradually, it can act as a conductor through heating process. This condition is called dielectric breakdown, wherein the insulator loses its dielectric property and starts to conduct.

4. Find the dielectric constant for a material with electric susceptibility of 4.
a) 3
b) 5
c) 8
d) 16

View Answer

Answer: b [Reason:] The electric susceptibility is given by χe = εr – 1. For a susceptibility of 4, the dielectric constant will be 5. It has no unit.

5. For a dielectric which of the following properties hold good?
a) They are superconductors at high temperatures
b) They are superconductors at low temperatures
c) They can never become a superconductor
d) They have very less dielectric breakdown voltage

View Answer

Answer: b [Reason:] Superconductors are characterised by diamagnetism behaviour and zero resistivity, which true for a dielectric. They occur only at low temperature. Thus a dielectric can become a superconductor at low temperatures with very high dielectric breakdown voltage.

6. The magnetic field which destroys the superconductivity is called
a) Diamagnetic field
b) Ferromagnetic field
c) Ferrimagnetic field
d) Critical field

View Answer

Answer: d [Reason:] Critical field is that strong magnetic field which can destroy the superconductivity of a material. The temperature at which this occurs is called transition temperature.

7. The magnetic susceptibility in a superconductor will be
a) Positive
b) Negative
c) Zero
d) Infinity

View Answer

Answer: b [Reason:] Due to perfect diamagnetism in a superconductor, its magnetic susceptibility will be negative. This phenomenon is called Meissner effect.

8. The superconducting materials will be independent of which of the following?
a) Magnetic field
b) Electric field
c) Magnetization
d) Temperature

View Answer

Answer: b [Reason:] Superconducting materials depends only on the applied magnetic field, resultant magnetization at the temperature considered. It is independent of the applied electric field and the corresponding polarization.

9. Find the mean free path of an electron travelling at a speed of 18m/s in 2 seconds.
a) 9
b) 36
c) 0.11
d) 4.5

View Answer

Answer: b [Reason:] The mean free path is defined as the average distance travelled by an electron before collision takes place. It is given by, d = v x τc, where v is the velocity and τc is the collision time. Thus d = 18 x 2 = 36m.

10. Find the velocity of an electron when its kinetic energy is equal to one electron volt (in 105m/s).
Given charge of an electron e = 1.6 x 10-19 and mass of an electron m = 9.1 x 10-31.
a) 3.9
b) 4.9
c) 5.9
d) 6.9

View Answer

Answer: c [Reason:] When the kinetic energy and one electron volt are equal, we can equate mv2/2 = eV. Put e and m in the equation to get velocity v = 5.9 x 105 m/s.