300+ REAL TIME Engineering Physics Objective Questions & Answers 2023

250+ TOP MCQs on Quantum Number and Answers

Engineering Physics Multiple Choice Questions on “Quantum Number”.

1. Quantum Numbers are solutions of _____________
a) Heisenberg’s Uncertainty Principle
b) Einstein’s mass energy relation
c) Schrodinger’s Wave Equation
d) Hamiltonian Operator

Answer: c
Clarification: When the wave function for an atom is solved using the Schrodinger Wave Equation, the solutions obtained are called the Quantum Number which are basically n, l and m.

2. Which quantum numbers gives the shell to which the electron belongs?
a) n
b) l
c) m
d) s

Answer: a
Clarification: The principal quantum number, n, gives the shell to which the electron belongs. The energy of the shell is dependent on ‘n’.

3. What is the maximum number of electrons in a shell?
a) n
b) 2n
c) n²
d) 2n²

Answer: d
Clarification: The total number of electrons in a shell are given by: 2n² while the number of orbitals present in a shell is given by n², which is half the total number of electrons in that shell.

4. Which of the following is the correct expression for the orbital angular momentum?
a) (sqrt{l+1})
b) (sqrt{n(l+1)})
c) (sqrt{l(l+1)})
d) (sqrt{m(l+1)})

Answer: c
Clarification: The orbital angular momentum of an electron is given by the expression: (sqrt{l(l+1)}). Thus, when the azimuthal quantum number, l, is zero the angular momentum is zero as well.

5. Which of the following quantum number gives the shape of atomic orbital of sub-shell?
a) n
b) l
c) m
d) s

Answer: b
Clarification: The Azimuthal quantum number, l, helps in determining the shape of the atomic orbital of sub-shell. It also gives the sub-shell to which the electron belongs.

6. What is the range of Azimuthal Quantum Number, l?
a) 0 to n
b) 0 to s
c) 0 to n-1
d) 0 to s-1

Answer: c
Clarification: The value of Azimuthal Quantum number, l, varies from 0 to n-1. It helps us in identifying to which subshell the electron belongs.

7. The total value of the magnetic quantum number are _______________
a) 2n
b) 2l
c) 2n + 1
d) 2l + 1

Answer: d
Clarification: The magnetic quantum number denotes the orientation of electrons in an atom. The total values of m_e are 2l + 1. They vary from –l to +l.

8. How many values does the spin quantum number have?
a) 2
b) 2l
c) 2n
d) 2m_e

Answer: a
Clarification: The spin quantum number have only two values: +1/2 and -1/2. It is not a solution of the Schrodinger wave equation.

9. Which of the following can be the quantum numbers for an orbital?
a) n = 4, l = 4, m = 3
b) n = 2, l = 3, m = 1
c) n = 3, l = 2, m = -1
d) n = 3, l = 0, m = -3

Answer: c
Clarification: In the given options, option c is the correct option because in this the value of l is between 0 – n-1 and value of m is between –l to +l.

10. An electron makes a transition from n = 5 state to n = 2 state in the hydrogen atom. What is the frequency of the emitted photon?
a) 4.9 X 10¹⁴ Hz
b) 5.9 X 10¹⁴ Hz
c) 6.9 X 10¹⁴ Hz
d) 7.9 X 10¹⁴ Hz

Answer: c
Clarification: The electron makes a transition from n = 5 to n = 2.
ΔE=(R(frac{1}{n_1^2}frac{-1}{n_2^2}))
= 45.774 X 10^-20J
Frequency, v = ΔE/h
= 6.9 X 10¹⁴ Hz.

11. The subshell dz² has no nodal plane.
a) True
b) False

Answer: a
Clarification: The subshell dz² has a ring in x-y plane and is based on the z-plane. Thus the probability of finding an electron is never zero for this sub-shell.

12. Nodes are the plane where the probability of finding an electron is 1.
a) True
b) False

Answer: b
Clarification: Nodal planes are described as the planes where the probability of finding an electron is not equal to zero. The total number of nodes in an orbital is n -1.

13. The probability of finding an electron is uniform in every direction is in which orbital?
a) s
b) p
c) d
d) p

Answer: a
Clarification: For an s-orbital, the probability of finding an electron is uniform in every direction. For p orbitals, the probability of finding an electron is along one direction only.

250+ TOP MCQs on Polarization and Answers

Engineering Physics Multiple Choice Questions on “Polarization”.

1. A window which can transmit all the incident light without any reflection is called ___________
a) Polarized Window
b) Malus Window
c) Brewster Window
d) Non-reflecting window
Answer: c
Clarification: Brewster’s window is the practical use of Brewster law. When light is incident on a clean glass plate, most of the incident light (>> 92%) is transmitted while a small fraction is reflected.

2. What should be the phase difference between the two plane-polarized waves, vibrating at right angles to each other, to produce circularly polarized light?
a) π/6
b) π/2
c) π/4
d) π/3
Answer: b
Clarification: The circularly polarized light is produced when the phase difference between the two rays is π/2 and the angle of incidence is π/4. In that case, the equation of the polarized light turns out to be x² + y² = a², which is the equation of the circle.

3. Sound waves can be polarized.
a) True
b) False
Answer: b
Clarification: Longitudinal waves cannot be polarized as their direction is the same as its vibration. Thus, it’s intensity will not change as it passes through a rotating analyzer.

4. Which of the following is a uniaxial crystal?
a) Borax
b) Mica
c) Quartz
d) Selenite
Answer: c
Clarification: Uniaxial crystals are those in which there is only one optic axis. Calcite, quartz, ice are a few examples. Borax, mica, and selenite are biaxial crystals which have two optic axes.

5. If the phase difference between two rays is π/2 and the angle of incidence is not equal to π/4, the emergent light is __________
a) Linearly Polarized
b) Elliptically Polarized
c) Circularly Polarized
d) Non-Polarized
Answer: b
Clarification: In this case, the equation of the emergent light becomes the equation of an ellipse, which is (frac{x^2}{a^2}+frac{y^2}{b^2}) =1. Thus, the emergent light is elliptically polarized and the plane of the ellipse is normal to the direction of propagation.

6. The velocity of light in water is 2.2 X 10⁸ m/s. What is the polarizing angle of incidence?
a) 47.23
b) 51.02
c) 53.74
d) 65.36
Answer: c
Clarification: Refractive index of water = Speed of light in space/Speed of light in water
= 3/2.2
= 1.3636
Using Brewster’s Law, Angle of incidence = tan^-1μ
= tan^-11.3636
= 53.74.

7. An electromagnetic beam has an intensity 28 W/m² and is linearly polarized vertically. What is the intensity of the transmitted beam, if the angle of incidence on the polaroid is 45o with the vertical?
a) 10 Wm^-1
b) 12 Wm^-1
c) 14 Wm^-1
d) 16 Wm^-1
Answer: c
Clarification: Here, I_o = 28 Wm^-2, θ = 45
Using Malus law, I = I_ocos²θ
I = 28 X cos²45
I = 14 Wm^-1.

8. What should be the thickness of quarter-wave plate for a light of wavelength 5890 Å if µ_e = 1.553 and µ_o = 1.544?
a) 1.33 X 10^-3 cm
b) 1.43 X 10^-3 cm
c) 1.53 X 10^-3 cm
d) 1.63 X 10^-3 cm
Answer: d
Clarification: t = (frac{lambda}{4(μ_E-μ_0)})
Λ = 5890 Å = 5890 X 10^-8 cm, µ_e = 1.553 and µ_o = 1.544
Therefore, t = 5890a X 10^-8/4 X 0.009
= 1.63 X 10^-3 cm.

9. A tube of length 20cm containing a 10% sugar solution rotates the plane of polarization by 13.2°. What is the specific rotation of sugar solution?
a) 66°
b) 55
c) 44
d) 33
Answer: a
Clarification: The specific rotation is given by S = 10θ/lC
Here, θ = 13.2°, C = 0.1 g/cc and l = 20 cm
Therefore, S = 10 X 13.2/20 X 0.1
= 66°.

10. Unpolarized light is incident on a plane glass surface. What should be the angle of incidence such that the reflect and refracted rays are perpendicular to each other?
a) 90°
b) 45°
c) 57°
d) 60°
Answer: c
Clarification: Using Brewster’s Law, µ = tani_p
In this case, i + r is equal to π/2.
For glass, µ = 1.5
Therefore, i_p = tan^-1µ = 57°.

11. A plate which induces the desired amount of phase difference between two rays is known as ___________
a) Polaroid
b) Phasor plates
c) Retardation Plates
d) Quartz plates
Answer: c
Clarification: Sometimes, it is required to induce a certain phase difference between E-ray and O-ray. For that, a plate with a specific thickness is chosen to induce that phase difference. This plate is known as the retardation plate as it retards the motion of one of the rays.

12. What does the following graph show?

a) Linearly Polarized Light
b) Unpolarized Light
c) Partially Polarized Light
d) Circular Polarized Light
Answer: a
Clarification: The graph shows a linearly polarized light. The vibrations are restricted in the x-y plane. Such a polarized light is obtained by using a polarizer/Polaroid and the phenomenon is called polarization of light.

250+ TOP MCQs on Production of Plane Polarised Light and Answers

Engineering Physics Questions & Answers for Exams focuses on “Production of Plane Polarised Light”.

1. The direction of polarization of an electromagnetic wave is considered to be the direction of __________
a) Electric Field
b) Magnetic field
c) Resultant of Electric field and Magnetic Field
d) Perpendicular to both Electric and Magnetic Field
Answer: a
Clarification: The direction of polarization of an electromagnetic wave is taken as the direction of the electric field vector because many common electromagnetic wave detectors respond to the electric forces on electrons in material not magnetic forces.

2. The second polaroid placed after the ray has been polarized is called as __________
a) Polarizer
b) Analyzer
c) Brewster’s Window
d) Multi Polaroid
Answer: c
Clarification: The second polaroid is placed after the ray has already been polarized to analyze whether the ray is polarized completely or not. Thus, it is called the polarizer.

3. An unpolarized light can be polarized by __________
a) Reflection
b) Refraction
c) Diffraction
d) Interference
Answer: a
Clarification: An unpolarized light can be polarized by reflection. The extent of polarization depends on the angle of incidence. The reflected light is completely polarized for a specific angle, called the polarizing angle.

5. For a glass surface, the value of polarizing angle is _________
a) 45.7°
b) 52.7°
c) 57.5°
d) 67.8°
Answer: c
Clarification: Polarizing angle is the angle for which the reflected light would be completely plane polarized. For glass, it is 57.5°. Thus, if a light is incident on glass at 57.5° the reflected light would be completely polarized.

6. The refractive index of a material is 1.75. What is the polarizing angle?
a) 47.23°
b) 51.02°
c) 53.74°
d) 60.25°
Answer: d
Clarification: Using Brewster’s Law, Angle of incidence = tan^-1μ
= tan^-11.752
= 60.25.

7. A polarized light of Intensity 40 Wm² is incident on the analyzer at a certain angle. If the transmitted beam has an Intensity of 20 Wm², the angle made with the vertical is _____________
a) 30°
b) 45°
c) 60°
d) 75°
Answer: b
Clarification: Here, I_o = 40 Wm^-2, I = 20 Wm^-2
Using Malus law, I = I_ocos²θ
20 = 40 X cos²θ
cos²θ = ¹⁄₂
θ = 45°.

8. What angle should the beam made with the vertical, so that after passing through the analyzer, it’s intensity should become 0.86 times the initial intensity?
a) 30°
b) 45°
c) 60°
d) 75°
Answer: c
Clarification: Here, I = 0.86 I_o
Using Malus law, I = I_ocos²θ
0.86 I_o = I_o X cos²θ
cos²θ = 0.86
θ = 60°.

9. The sunlight as received by the observer is __________
a) Unpolarized
b) Partially Polarized
c) Polarized
d) Varying Polarizability
Answer: c
Clarification: The incident sunlight, when enters the earth’s atmosphere, is scattered by the particles present in the earth’s atmosphere. Due to this, the sunlight gets polarized.

10. If ray R₁ is incident at the polarizing angle, which of the following rays are unpolarized?

a) R₁
b) R₂
c) Both R₂ and R₃
d) Both R₁ and R₃
Answer: d
Clarification: The incident ray is incident at an angle 57.5° which is the polarizing angle of the glass. Thus the reflected light is completely plane polarized, while the other two are not.

11. At the angle of polarization, p, the angle between the reflected and the refracted beam is __________
a) p
b) π
c) π /2
d) p/2
Answer: c
Clarification: If the angle of incidence is equal to p, the angle made by the refracted beam with the x-axis would be π /2 – p. The refracted beam would make an angle p with the x-axis. Thus, the angle between the two rays would be π /2.

12. Which of the two rays are polarized?

a) X
b) Y
c) Both X and Y
d) Neither X and Y
Answer: b
Clarification: The incident sunlight is unpolarized. On encountering the molecules of earth’s atmosphere, it changes its direction. Due to the scattering of light, the incident sunlight gets polarized.

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250+ TOP MCQs on Bragg’s Law and Answers

Engineering Physics Multiple Choice Questions on “Bragg’s Law”.

1. The correct expression for Bragg’s law is nλ = ___________
a) dsinθ
b) dcosθ
c) 2dsinθ
d) 2dcosθ
Answer: c
Clarification: Bragg’s law gives us the relation between the wavelength of the incident X-ray and the angle of incidence. The expression is: nλ = 2dsinθ.

2. Peaks of scattered intensity is observed when the path difference would be equal to __________
a) λ
b) λ/2
c) 3 λ/2
d) No peaks are observed at any condition
Answer: a
Clarification: In the Bragg’s law experiment, the x-ray radiations are scattered by the crystal lattice. The peaks of scattered intensity are observed when the path difference is equal to the integral multiple of wavelength. Also, the incident angle should be equal to the reflecting angle.

3. Bragg’s law is used in which process?
a) X-ray production
b) Gamma-ray production
c) X-ray crystallography
d) X-ray scan
Answer: c
Clarification: Bragg’s law is a special case of Laue diffraction. It is used in X-ray crystallography, the basis of which is to identify the crystal lattice.

4. When X-ray are incident on an atom, they make an electronic cloud move.
a) True
b) False
Answer: a
Clarification: X-ray is an electromagnetic wave. So, just like any other electromagnetic wave, when X-ray is incident on an atom, it makes the electronic cloud around that atom to move.

5. The scattering of waves in Bragg’s law experiment is due to __________
a) Einstein’s scattering
b) Rayleigh scattering
c) Newton scattering
d) Inelastic scattering
Answer: b
Clarification: In Bragg’s law experiment, the phenomenon is observed due to the Rayleigh scattering, i.e., the movement of charges is re-radiated with the wave of same frequency.

6. Bragg’s law is a special case of Laue diffraction.
a) True
b) False
Answer: a
Clarification: Laue diffraction is the one used in X-ray crystallography. Bragg’s law is only a special case of Laue diffraction with the condition that the incident angle be equal to scattering angle and the wavelength is integer times the path difference.

7. Bragg’s law was proposed in __________
a) 1903
b) 1913
c) 1893
d) 1853
Answer: b
Clarification: Lawrence Bragg and William Henry Bragg proposed Bragg’s law in 1913. It was based on the fact that surprising patterns are observed when the X-rays are reflected by a crystalline solid.

8. If the angle of incidence is 30°, then the wavelength for first-order spectrum is equal to __________
a) d
b) 2d
c) d/2
d) d/3
Answer: a
Clarification: We know, nλ = 2dsinθ.
As, θ = 30° and n = 1, we get
λ = d.

9. If X-ray of wavelength 100 Å is incident on an atom at an angle of 90°, then what should be the value of d for first-order sepctrum?
a) 30 Å
b) 40 Å
c) 50 Å
d) 60 Å
Answer: c
Clarification: We know, nλ = 2dsinθ.
Here, λ = 10^-8 m, n = 1, sinθ = 1.
Therefore, 10^-8 = 2 X d X 1
d = 50 Å.

10. What should be the value of X?

a) θ
b) θ/2
c) 2θ
d) θ/3
Answer: c
Clarification: As we can see in the following figure, the angle X is between the incident light and the diffracted light. This angle should be 2θ, as the angle made by both with horizontal plane would be θ.

250+ TOP MCQs on Hall Effect and Answers

Engineering Physics Interview Questions and Answers for freshers focuses on “Hall Effect”.

1. When does a normal conductor become a superconductor?
a) At normal temperature
b) At Curie temperature
c) At critical temperature
d) Never
Answer: c
Clarification: The temperature at which a normal conductor loses its resistivity and becomes a superconductor is known as transition temperature or critical temperature.

2. In which of the following does the residual resistivity exist?
a) Impure metal at high temperature
b) Pure metal at low temperature
c) Pure metal at high temperature
d) Impure metal at low temperature
Answer: d
Clarification: When the temperature is reduced to 0K, the resistivity of the impure metal doesn’t become zero, because there exist some impurities which gives rise to minimum resistivity known as residual resistivity.

3. Meissner effect occurs in superconductors due to which of the following properties?
a) Diamagnetic property
b) Magnetic property
c) Paramagnetic property
d) Ferromagnetic property
Answer: a
Clarification: A diamagnetic material has a tendency to expel magnetic lines of forces. Since the superconductor also expels magnetic lines of force it behaves as a perfect diamagnet. This behaviour is first observed by Meissner and is hence called Meissner effect.

4. What happens when a large value a.c. current is passed through superconductors?
a) Conductivity increases
b) Superconducting property is destroyed
c) It acts as a magnet
d) It becomes resistant
Answer: b
Clarification: When a large value of a.c. current is applied to a superconducting material it induces some magnetic field in the material and because of this magnetic field, the superconducting property of the material is destroyed.

5. How is persistent current produced in supermagnets?
a) By passing ac current
b) By magnetising it
c) By passing dc current
d) By increasing the resistance
Answer: c
Clarification: When dc current of large magnitude is once induced in a super conducting ring then the current persists in the ring even after the removal of the field. This current is called persistent current. This is due to diamagnetic property. The magnetic flux inside the ring will be trapped in it and hence current persists.

6. Superconductors can be used as a memory or storage elements in computers.
a) True
b) False
Answer: a
Clarification: Since the current in superconducting ring can flow without any change in its value, it can be used as a memory or storage element in computers.

7. Superconducting tin has a critical temperature of 3.7K at zero magnetic field and a critical field at 0.0306 Tesla at 0K. Find the critical field at 2K.
a) 0.0306 Tesla
b) 7.4 Tesla
c) 0.02166 Tesla
d) 0 Tesla
Answer: c
Clarification: Critical field, H_c = H₀ [1-T²/(T_c)²)]
Critical college = 0.02166 Tesla.

8. Calculate the critical current for a wire of lead having a diameter of 1mm at 4.2 K. Calculate temperature for lead is 7.18 K and H_c = 6.5×10⁴ A/m. Critical field is 42.758×10³ A/m.
a) 3.5593 A
b) 27.3 A
c) 46.67 A
d) 134.26 A
Answer: d
Clarification: Critical current I_c = 2πrH_c
I_c = 134.26 A.

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250+ TOP MCQs on Dimensions of a Physical Quantity and Answers

Engineering Physics Multiple Choice Questions on “Dimensions of a Physical Quantity”.

1. Which of the following is used for measuring long time intervals?
a) Electrical oscillators
b) Atomic clocks
c) Decay of elementary particles
d) Radioactive dating
Answer: d
Clarification: Radioactive dating is used to measure long time intervals by finding the ratio of the number of radioactive atoms that have undergone decay to the number of atoms left undecayed. Carbon dating is used to estimate the age of fossils, uranium dating is used to estimate the age of rocks.

2. The average life of an Indian is 56 years. Find the number of times the human heart beats in the life of an Indian, if the heat beats once in 0.8 s.
a) 20.4×10⁹ times
b) 2.5×10⁹ times
c) 2.2×10⁹ times
d) 6.1×10⁹ times
Answer: c
Clarification:
Average life of an Indian = 56 years = 56×365.25×24×60×60 s
Period of heart beat = 0.8 s
Total number of heart beats in 56 years = (56×365.25×24×60×60)/0.8 = 2.2×10⁹ times.

3. How many dimensions of the world are there?
a) 7
b) 3
c) 2
d) 4
Answer: a
Clarification: All the derived quantities can be expressed in terms of some combination of the seven fundamental or base quantities. We call these seven fundamental quantities as the seven dimensions of the world.

4. What is the dimensional formula for Gravitational constant?
a) ML² T^(-3)
b) ML^(-1) T^(-2)
c) ML^(-1) T^(-1)
d) M^(-1) L³ T^(-2)
Answer: d
Clarification: F = G(m₁ m₂)/r²
G = ([F][r²])/([m₁ ][m₂]) = (MLT^(-2) L²)/MM = M^(-1) L³ T^(-2).

5. Which of the following is a dimensional constant?
a) e
b) Area
c) Specific gravity
d) Gravitational constant
Answer: d
Clarification: The physical quantities which posses dimensions and have constant values are called dimensional constants. Hence Gravitational constant is a dimensional constant.

6. If the unit of force is 1kN, unit of length 1km and the unit of time is 100s, what will be the unit of mass?
a) 1000 kg
b) 10⁴ kg
c) 100 kg
d) 10⁵ kg
Answer: b
Clarification: M = (MLT^(-2) T²)/L = (FT²)/L
M = (1000 N ×10⁴ s²)/(1000 m) = 10⁴ kg.

7. Which of the following is a systematic error?
a) Constant error
b) Least count error
c) Gross error
d) Personal error
Answer: d
Clarification: The errors which occur in one direction, either positive or negative, are called systematic error. Personal errors arise due to improper setting of instruments. Hence it comes under systematic error.

8. All physical quantities have dimensions.
a) True
b) False
Answer: b
Clarification: All physical quantities do not have dimensions. Like angle, strain and relative density are examples for dimension less quantity.

9. An instrument cannot be precise without being accurate.
a) False
b) True
Answer: a
Clarification: An instrument can be precise without being accurate but the instrument cannot be accurate without being precise.

10. Can the diameter of a thread be measure by using a scale?
a) Yes
b) No
Answer: a
Clarification: The thread is wound on a meter scale such that its turns are close together. Thickness of the thread coil is measure and the number of turns made by the thread is counted. Diameter is then given by dividing the thickness by a number of turns.

11. Calculate the de-Broglie wavelength of an electron which has been accelerated from rest on application of potential of 400volts.
a) 0.1653 Amstrong
b) 0.5125 Amstrong
c) 0.6135 Amstrong
d) 0.2514 Amstrong
Answer: c
Clarification: de-Broglie wavelength = h/√(2×m×e×V)
De-Broglie wavelength = (6.625×10^(-14))/√(2×9.11×10^(-31)×1.6×10^(-19)×400)
Wavelength = 0.6135 Amstrong.