Category: Engineering Physics Objective Questions

250+ TOP MCQs on Light and Answers

Engineering Physics Multiple Choice Questions on “Light”.

1. How is light emitted?
a) Klystron valves
b) Radioactive decay of the nucleus
c) Transmission of high energy electrons from an excited state to a lower state
d) Acceleration of electrons
Answer: c
Clarification: An electron is always stable in the ground state. When in an excited state, it tends to jump to a lower state. In this process, it releases a photon of energy equal to the energy difference. When the frequency of these photons lies in the visible region, we get light.

2. Who gave the quantum theory of light?
a) Albert Einstein
b) Heisenberg
c) Michaelson Morley
d) Max Planck
Answer: d
Clarification: Max Planck proposed the quantum theory of light which states that the light consists of small particles in the form of discrete bundles of energy called photons.

3. Huygen’s wave theory of light considered light to be ____________
a) Made up of particles
b) Longitudinal Waves
c) Transverse Waves
d) Of dual nature
Answer: b
Clarification: Huygen’s Wave Theory of Light considered Light to be a periodic disturbance that is transmitted through a medium in the form of longitudinal waves. This theory explained reflection, refraction, diffraction, and interference.

4. Michaelson-Morley experiment ruled out the existence of __________
a) Luminiferous Ether
b) Wave nature of light
c) Particle nature of light
d) Corpuscles
Answer: a
Clarification: Earlier it was believed, that space is filled up of a medium called Ether. It was believed, that light has different speed in different directions. However, by the Michaelson-Morley experiment, it was shown that the light has the same speed in all the directions and that ether does not exist.

5. Which theory was first to consider that light requires no medium for its propagation?
a) Huygen’s Wave Theory of Light
b) Maxwell’s Electromagnetic Theory
c) Quantum theory of light
d) Corpuscular theory
Answer: b
Clarification: In 1873, Maxwell introduced the Electromagnetic Theory which assumed the light to be an electromagnetic wave, thus it requires no medium for its propagation. It discarded the presence of ether.

6. What is the frequency of a photon having energy 2.1 X 10-30 J?
a) 3.1 X 1013 Hz
b) 4.1 X 1013 Hz
c) 5.1 X 1013 Hz
d) 6.1 X 1013 Hz
Answer: b
Clarification: Here, E = 2.1 X 10-30 J
Frequency, v = ?
We know, Energy = hv, where h is the Planck’s Constant
Therefore, v = E/h
= 2.1 X 10-30 J/6.6 X 10-34 Js
= 3.1 X 1013 Hz.

7. The wavelength of a photon is 5000 Å. The energy of the photon is ___
a) 3.47 eV
b) 2.41 eV
c) 6.78 eV
d) 2.12 eV
Answer: b
Clarification: We know, E = hc/λ
= 6.6 X 10-34 X 3 X 108/5 X 10-7
= 3.86 X 10-19 J
= 2.41 eV.

8. An electron makes a transition from n = 7 state to n = 3 state in the hydrogen atom. What is the frequency of the emitted photon?
a) 2 X 1014 Hz
b) 3 X 1014 Hz
c) 4 X 1014 Hz
d) 5 X 1014 Hz
Answer: b
Clarification: The electron makes a transition from n = 7 to n = 3.
ΔE=(R(frac{1}{n_1^2}-frac{1}{n_2^2}))
= 19.77 X 10-20J
Frequency, v = ΔE/h
= 3 X 1014 Hz.

9. The mass of a photon with wavelength 3.6 Å is _____________
a) 3.139 X 10-33 kg
b) 4.139 X 10-33 kg
c) 5.139 X 10-33 kg
d) 6.139 X 10-33 kg
Answer: d
Clarification: Here, wavelength = 3.6 X 10-10 m
Velocity of photon = velocity of light
Mass, m = h/λv = 6.139 X 10-33 kg.

10. Which phenomenon is shown by the following diagram?
engineering-physics-questions-answers-light-q10
a) Scattering of light
b) Dispersion of light
c) Primary Rainbow Formation
d) Secondary Rainbow formation
Answer: d
Clarification: In the case of secondary rainbow formation, the light rays undergo internal reflection twice inside the drop. The intensity of light, in this case is, reduced. Also, the order of the colors is reversed as shown in the figure.

250+ TOP MCQs on Optical Fibre and Answers

Engineering Physics Multiple Choice Questions on “Optical Fibre”.

1. What is the principle of fibre optical communication?
a) Frequency modulation
b) Population inversion
c) Total internal reflection
d) Doppler Effect
Answer: c
Clarification: In optical fibres, the light entering the fibre does not encounter any new surfaces, but repeatedly they hit the same surface. The reason for confining the light beam inside the fibres is the total internal reflection.

2. What is the other name for a maximum external incident angle?
a) Optical angle
b) Total internal reflection angle
c) Refraction angle
d) Wave guide acceptance angle
Answer: d
Clarification: Only this rays which pass within the acceptance angle will be totally reflected. Therefore, light incident on the core within the maximum external incident angle can be coupled into the fibre to propagate. This angle is called a wave guide acceptance angle.

3. A single mode fibre has low intermodal dispersion than multimode.
a) True
b) False
Answer: a
Clarification: In both single and multimode fibres the refractive indices will be in step by step. Since a single mode has less dispersion than multimode, the single mode step index fibre also has low intermodal dispersion compared to multimode step index fibre.

4. How does the refractive index vary in Graded Index fibre?
a) Tangentially
b) Radially
c) Longitudinally
d) Transversely
Answer: b
Clarification: The refractive index of the core is maximum along the fibre axis and it gradually decreases. Here the refractive index varies radially from the axis of the fibre. Hence it is called graded index fibre.

5. Which of the following has more distortion?
a) Single step-index fibre
b) Graded index fibre
c) Multimode step-index fibre
d) Glass fibre
Answer: c
Clarification: When rays travel through longer distances there will be some difference in reflected angles. Hence high angle rays arrive later than low angle rays. Therefore the signal pulses are broadened thereby results in a distorted output.

6. In which of the following there is no distortion?
a) Graded index fibre
b) Multimode step-index fibre
c) Single step-index fibre
d) Glass fibre
Answer: a
Clarification: The light travels with different speeds in different paths because of the variation in their refractive indices. At the outer edge it travels faster than near the centre But almost all the rays reach the exit end at the same time due to the helical path. Thus, there is no dispersion in the pulses and hence the output is not a distorted output.

7. Which of the following loss occurs inside the fibre?
a) Radiative loss
b) Scattering
c) Absorption
d) Attenuation
Answer: b
Clarification: Scattering is a wavelength dependent loss. Since the glass used in the fabrication of fibres, the disordered structure of glass will make some vibrations in the refractive index inside the fibre. This causes Rayleigh scattering.

8. What causes microscopic bend?
a) Uniform pressure
b) Non-uniform volume
c) Uniform volume
d) Non-uniform pressure
Answer: d
Clarification: Micro-bends losses are caused due to non-uniformities inside the fibre. This micro-bends in fibre appears due to non-uniform pressures created during the cabling of fibre.

9. When more than one mode is propagating, how is it dispersed?
a) Dispersion
b) Inter-modal dispersion
c) Material dispersion
d) Waveguide dispersion
Answer: b
Clarification: When more than one mode is propagating through a fibre, then inter modal dispersion will occur. Since many modes are propagating, they will have different wavelengths and will take different time to propagate through the fibre.

10. A fibre optic telephone transmission can handle more than thousands of voice channels.
a) True
b) False
Answer: a
Clarification: Optical fibre has larger bandwidth hence it can handle a large number of channels for communication.

11. Which of the following is known as fibre optic back bone?
a) Telecommunication
b) Cable television
c) Delay lines
d) Bus topology
Answer: d
Clarification: Each computer on the network is connected to the rest of the computers by the optical wiring scheme called bus topology, which is an application known as fibre optic back bone.

12. Calculate the numerical aperture of an optical fibre whose core and cladding are made of materials of refractive index 1.6 and 1.5 respectively.
a) 0.55677
b) 55.77
c) 0.2458
d) 0.647852
Answer: a
Clarification: Numerical aperture = (sqrt{n1^2-n2^2})
Numerical aperture = 0.55677.

13. A step-index fibre has a numerical aperture of 0.26, a core refractive index of 1.5 and a core diameter of 100micrometer. Calculate the acceptance angle.
a) 1.47°
b) 15.07°
c) 2.18°
d) 24.15°
Answer: b
Clarification: sin i = (Numerical aperture)/n
sin i = 15.07°.

250+ TOP MCQs on Properties of Dielectric Materials and Answers

Engineering Physics Multiple Choice Questions on “Properties of Dielectric Materials”.

1. What is the process of producing electric dipoles inside the dielectric by an external electric field?
a) Polarisation
b) Dipole moment
c) Susceptibility
d) Magnetisation
Answer: a
Clarification: When an external magnetic field is applied to the dielectrics, the field exerts a force on each positive charges in its own direction while negative charges are pushed in the opposite direction. Consequently, an electric dipole is created in all the atoms. Thus the process of producing electric dipoles inside the dielectrics by an external electric field is called polarisation.

2. Which of the following easily adapt itself to store electrical energy?
a) Passive dielectric
b) Superconductor
c) Active dielectric
d) Polar molecules
Answer: c
Clarification: When a dielectric is subjected to an external electric field, if the dielectric actively accepts electricity, then they are termed as active dielectrics. Thus active dielectrics are the dielectrics which can easily adapt itself to store the electrical energy in it.

3. Which of the following restricts the flow of electrical energy?
a) Superconductors
b) Passive dielectrics
c) Polar molecules
d) Active dielectric
Answer: b
Clarification: Passive dielectric acts as an insulator; conduction will not take place through this dielectrics. Thus passive dielectrics are the dielectrics that restrict the flow of electrical energy in it.

4. For non-polar molecules, there is no absorption or emission in the range of infrared.
a) True
b) False
Answer: a
Clarification: These molecules possess centre of symmetry and hence the centres of positive and negative charges coincide. Therefore the net charges and net dipole moment of these molecules will be zero and hence these non-polar molecules will not posses any dipole moment in it. Hence there is no absorption or emission in the range of infrared.

5. How does ionic polarisation occur?
a) Splitting of ions
b) Passing magnetic field
c) Displacement of cations and anions
d) Never occurs
Answer: c
Clarification: Ionic polarisation occurs due to the displacement of cations and anions from its original position in the opposite directions, in the presence of an electric field.

6. Polar molecules have permanent dipole moments even in the absence of an electric field.
a) False
b) True
Answer: b
Clarification: In the absence of an electric field the polar molecules posses some dipole moment. These dipoles are randomly arranged and they cancel each other. Hence the net dipole moment is very less.

7. Which of the following polarisations is very rapid?
a) Electronic polarisation
b) Ionic polarisation
c) Space charge polarisation
d) Orientation polarisation
Answer: a
Clarification: Electronic polarisation is very rapid and will complete at the instant the voltage is applied. The reason is that the electrons are very light particles. Therefore even for high frequency this kind of polarisation occurs.

8. Which of the following is the slowest polarisation method?
a) Ionic polarisation
b) Orientation polarisation
c) Electronic polarisation
d) Space charge polarisation
Answer: d
Clarification: Space charge polarisation is very slow because in this case, the ions have to diffuse over several interatomic distances. Also, this process occurs at a very low frequency.

9. When does a dielectric become a conductor?
a) At avalanche breakdown
b) At high temperature
c) At dielectric breakdown
d) In the presence of magnetic field
Answer: c
Clarification: When a dielectric is placed in an electric field and if the electric field is increased, when the electric field exceeds the critical field, the dielectric loses its insulating property and becomes conducting. This is called dielectric breakdown.

10. Which of the following breakdowns occur at a higher temperature?
a) Avalanche breakdown
b) Thermal breakdown
c) Electrochemical breakdown
d) Dielectric breakdown
Answer: b
Clarification: When a dielectric is subjected to an electric field, heat is generated. This generated heat is dissipated by the dielectric. In some cases, the generated heat will be very high compared to the heat dissipated. Under such conditions, the temperature inside the dielectric increases and heat may produce breakdown. This is thermal breakdown.

11. When mobility increases, insulation resistance decreases and dielectric becomes conducting.
a) True
b) False
Answer: a
Clarification: If the temperature is increased, the mobility of ions increases and hence electrochemical reaction may be induced to take place. Therefore when the mobility of ions is increased, insulation resistance decreases and hence dielectric becomes conducting.

12. Which of the following materials exhibit Ferro-electricity?
a) Iron
b) Platinum
c) Hydrogen
d) Rochelle salt
Answer: d
Clarification: When a dielectric exhibits electric polarisation even in the absence of an external field, it is known as ferro-elecricity and these materials are termed as Ferro-electrics. They are anisotropic crystals that exhibit spontaneous polarisation. Hence only Rochelle salt exhibits Ferro-electricity.

13. Calculate the electronic polarizability of an argon atom whose ɛr = 1.0024 at NTP and N = 2.7×1025 atoms/m3.
a) 0.0024 Fm2
b) 7.87 ×10-40 Fm2
c) 7.87 Fm2
d) 1.0024×10-40 Fm2
Answer: b
Clarification: Electronic polarisabilty αe = (ε0 (εr-1))/N
Electronic polarisability = 7.87 × 10-40 Fm2.

14. Calculate the dielectric constant of a material which when inserted in parallel condenser of area 10mm × 10mm and distance of separation of 2mm, gives a capacitance of 10-9 F.
a) 8.854×10-12
b) 100
c) 2259
d) 5354
Answer: c
Clarification: C = (εr ε0 A)/d
εr = Cd/(ε0 A) = 2259.

15. Find the capacitance of layer of A12 O3 that is 0.5μm thick and 2000mm2 of square area εr = 8.
a) 1000μF
b) 0.283μF
c) 16μF
d) 2.83μF
Answer: b
Clarification: C = (εr ε0 A)/d
Capacitance = 0.283μF.

250+ TOP MCQs on Transistors and Answers

Engineering Physics Multiple Choice Questions on “Transistors”.

1. BJT stands for __________
a) Bi-Junction Transfer
b) Blue Junction Transistor
c) Bipolar Junction Transistor
d) Base Junction Transistor
Answer: c
Clarification: BJT stands for Bipolar Junction Transistor. It was the first transistor to be invented. It is widely used in circuits.

2. The doped region in a transistor are ________
a) Emitter and Collector
b) Emitter and Base
c) Collector and Base
d) Emitter, Collector and Base
Answer: d
Clarification: There are three doped regions forming two p-n junctions between them. There are two types of transistors n-p-n transistor and p-n-p transistor.

3. Which region of the transistor is highly doped?
a) Emitter
b) Base
c) Collector
d) Both Emitter and Collector
Answer: a
Clarification: In a transistor, emitter is of moderate size and heavily doped. Collector is moderately doped and larger as compared to the emitter. Base is very thin and lightly doped.

4. Both the junctions in a transistor are forward biased.
a) True
b) False
Answer: b
Clarification: Emitter-base junction of the transistor is forwards biased while the collector-base junction of the transistor is reverse biased or vice versa depending on the condition desired.

5. Which junction is forward biased when transistor is used as an amplifier?
a) Emitter-Base
b) Emitter-Collector
c) Collector-Base
d) No junction is forward biased
Answer: a
Clarification: For Transistor to be used as an amplifier, the emitter-base junction is forward biased and the base-collector region is reverse biased. This state is called an active state.

6. If Ie is the current entering the emitter, Ib is the current leaving the base and Ic is the current leaving the collector in a p-n-p transistor used for amplification, what is the relation between Ie, Ib and Ic?
a) Ie < Ic
b) Ic < Ib
c) Ib < Ic
d) Ie < Ib + Ic
Answer: c
Clarification: The total current entering the emitter, Ie, goes to the base form where most of the current enters the collector and a very small fraction of the current leaves the base. Thus, Ib < Ic.

7. In the active state, the emitter-base junction has a higher resistance than the collector-base junction.
a) True
b) False
Answer: b
Clarification: Since the emitter-base junction is forward biased, their resistance is lower than the collector-base junction, which is reverse biased.

8. From the figure, what is βac when VCE is 10V and Ic is 4 mA?
engineering-physics-questions-answers-transistors-q8
a) 50
b) 100
c) 150
d) 200
Answer: c
Clarification: We know, βac = ΔIc/ΔIb
Now, at VCE = 10V, we read two values of Ic from the graph.
Then, ΔIb = 10 μA, ΔIc = 1.5 mA
Therefore, βac = 1.5 mA/10 μA
= 150.

9. In which state is the switch said to be on?
engineering-physics-questions-answers-transistors-q9
a) A
b) B
c) C
d) Neither region
Answer: c
Clarification: A is the Cut-off region, B is the Active Region and C is the saturation region. When the transistor is not conducting, it is said to be switched off and when it is Region C it is said to be switched on.

10. A low input to the transistor gives __________
a) Low output
b) High Output
c) Normal Output
d) No Output
Answer: b
Clarification: A low input to the transistor gives a high output and a high input gives a low output. The switching circuits are designed such a way that the transistor does not stay in the active state.

11. In a CE transmitter amplifier, if the amplification factor is 150 and the collector voltage is 4 V and resistance is 2 kΩ, what should be the value of RB, given that the dc base current is 10 times the signal current?
a) 5 kΩ
b) 10 kΩ
c) 15 kΩ
d) 20 kΩ
Answer: d
Clarification: Here, the output voltage is 4 V.
So, ic = 4/2000
= 2 mA
The signal current through the base, IB = ic/β
= 0.013 mA
The dc base current = 0.13 mA
Assuming here, that VBE = 1.4 V, we get
RB = (4.0-1.4)/0.13
= 20 kΩ.

12. From the output characteristics of a transistor, one cannot calculate __________
a) IB
b) VBE
c) Ic
d) VCE
Answer: b
Clarification: The output characteristics graph for a transistor gives us the relation between the collector current and the emitter voltage. It also gives us the value of base current. But it gives no information about the base-emitter voltage.

13. What is the expression for the Current Amplification factor?
a) (frac{Delta I_c}{Delta V_c})
b) (frac{Delta V_c}{Delta I_c})
c) ((frac{Delta I_C}{Delta I_B})_{V_{CE}})
d) ((frac{Delta I_C}{Delta I_B})_{V_{BE}})
Answer: c
Clarification: Amplification factor can be defined as the ratio of the change in collector current to the change in base current at a constant collector-emitter voltage when the transistor is in active state. The correct expression for the amplification factor is: ((frac{Delta I_C}{Delta I_B})_{V_{CE}}).

14. In the output characteristics, the resistance is the __________
a) Slope of the curve
b) Trace of the curve
c) Asymptote of the curve
d) Reciprocal of the slope of the curve
Answer: d
Clarification: The reciprocal of the slope of the linear part of the output characteristic gives the value of the output resistance, which is given by ((frac{Delta V_{CE}}{Delta I_C})_{I_B}.)

15. The output in an oscillator is __________
a) Discontinuous
b) Oscillating
c) Self-sustained
d) Spiked
Answer: c
Clarification: In an oscillator, we get ac output without any external input signal. Thus, the output in an oscillator is self-sustained. To attain this, an amplifier is taken, which uses a transistor.

250+ TOP MCQs on Applications of Nanomaterials and Answers

Engineering Physics Multiple Choice Questions on “Applications of Nanomaterials”.

1. Which property of Nanomaterials make them suitable to be used for elimination of pollutants?
a) High purity
b) Better thermal conductivity
c) Enhanced chemical activity
d) Small size
Answer: c
Clarification: Nanomaterials have enhanced chemical activity. Due to this, they can be used as catalysts to react with toxic gases such as CO and NO2 in automobile catalytic converters.

2. Nano crystalline materials synthesised by sol-gel technique results in a foam like structures called ___________
a) Gel
b) Aerosol
c) Foam
d) Aerogel
Answer: d
Clarification: The foam-like material formed is called aerogel. They are porous and extremely lightweight, yet they can load equivalent to 100 times their weight. They are used as insulation materials.

3. Which nanomaterial is used for cutting tools?
a) Fullerene
b) Aerogel
c) Tungsten Carbide
d) Gold
Answer: c
Clarification: Cutting tools made of Nano crystalline crystals like Tungsten carbide are much harder, much more wear-resistant and last longer than their conventional counterparts.

4. Aerogels can hold more energy than the separators in batteries.
a) True
b) False
Answer: a
Clarification: The energy density of a conventional battery is quite low. The Nano crystalline materials, as aerogels, have considerably more energy holding capacity.

5. A Carbon monoxide sensor made of zirconia uses which characteristic to detect any change?
a) Capacitance
b) Resistivity
c) Activity
d) Permeability
Answer: c
Clarification: Zirconia uses its chemical activity to sense the presence of CO. In case CO is present, the oxygen atoms in zirconia would react with the carbon in CO to partially reduce Zirconium oxide.

6. Which components of an automobile are envisioned to be coated with zirconia?
a) Spark plugs
b) Liners
c) Tyres
d) Brakes
Answer: b
Clarification: Zirconia is a hard and brittle ceramic. Its use in the coating of liners in an automobile is envisioned, as it can heat up the engine more effectively. Alumina can also be used for coating.

7. The main purpose of CNTs in fuel cells is __________
a) Production of energy
b) Active medium
c) Catalyst
d) Storage
Answer: d
Clarification: Carbon nanotubes are useful in fuel cells as well as in batteries, primarily for storage purposes of hydrogen. The tubes need to gold 6.5% hydrogen by weight.

8. Zirconia is a hard, brittle ___________
a) Metal
b) Non-metal
c) Composite
d) Ceramics
Answer: d
Clarification: Zirconia is a ceramic and even has been rendered as superplastic. Usage of these nano crystals have been encouraged instead of the ceramics.

9. Nanoscale aluminium oxide increases the _________
a) Conductivity
b) Resistance
c) Ductility
d) Stability
Answer: b
Clarification: Nanoscale aluminium oxide and titanium oxide are optically transparent and they greatly increase abrasion resistance of traditional coatings.

10. Quantum dots can be used in ________
a) Crystallography
b) Optoelectronics
c) Mechanics
d) Quantum physics
Answer: b
Clarification: Inorganic nanomaterials, like quantum dots, can be used in optoelectronics because of their interesting optical and electrical properties.