Category: Engineering Physics Objective Questions

Engineering Physics Multiple Choice Questions on “Significance of Rotational Motion”.

1. A flywheel has a mass of 25kg has a radius of 0.2m. What force should be applied tangentially to the rim of the flywheel so that it acquires an angular acceleration of 2rad/s²?
a) 2N
b) 5N
c) 3N
d) 1N
Answer: b
Clarification: M = 25kg, R = 0.2m, α = 2rad/s²
Moment of inertia of the flywheel about its axis,
I = 1/2 MR²=1/2×25×0.2²=0.5kgm²
As torque, τ = F×R=I×α
Therefore, Force F = Iα/R=(0.5×2)/0.2=5N.

2. A torque of 10Nm is applied to a flywheel of mass 1kg and radius of gyration 50cm. What is the resulting angular acceleration?
a) 4 rad/s²
b) 2 rad/s²
c) 5 rad/s²
d) 6 rad/s²
Answer: a
Clarification: τ=10Nm, M = 10kg, K = 0.50m
As τ=I×α = MK² α
α=τ/(MK² )=10/(10×0.50²)=4 rad/s².

3. The angular momentum of a body is 3.14 is and its rate of revolution is 10 cycles per second. Calculate the moment of inertia of the body about the axis of rotation.
a) 5 kgm²
b) 0
c) 31.4 kgm²
d) 0.5 kgm²
Answer: d
Clarification: L = 3.14 Js v = 10 rps
ω=2πv=2×3.14×10/rad
As L = Iω
I = L/ω = 3.14/(2×3.14×10)=0.5kgm².

4. The angular velocity of a planet revolving in an elliptical orbit around the sun increases, when it comes closer to the sun. This is due to ______________
a) Conservation of momentum
b) Newton’s second law of motion
c) Conservation of angular momentum
d) Law of angular velocity
Answer: c
Clarification: The angular velocity of a planet revolving in an elliptic orbit around the sun increases when it comes closer to the sun increases when it comes closer to the sun because its moment of inertia about the axis through the sun decreases. When it goes far away from the sun, its moment of inertia decreases so as to conserve angular momentum.

5. An ice skater or a ballet dancer can decrease her angular velocity by folding her arms and bringing the stretched leg close to the other leg.
a) True
b) False
Answer: b
Clarification: An ice skater can increase her angular velocity by folding her arm and bringing the leg closer to the other. When she stretches her hand outward, her moment of inertia increases and hence angular speed decreases to conserve angular momentum. When she folded her hands and brings the stretched legs close to the other leg, her moment of inertia decreases and hence her angular speed increases.

6. The centre of mass of a body should lie inside the body.
a) True
b) False
Answer: b
Clarification: The centre of mass of a body need not necessarily lie inside the body. The centre of mass of a ring lies in its hollow portion.

7. In a hand driven grinding machine, the handle is put _____________
a) Perpendicular to the stone
b) Near the circumference of the stone
c) A few distances above the stone
d) As far away from the stone
Answer: b
Clarification: For a given force, torque can be increased if the perpendicular distance of the point of application of the force from the axis of rotation is increased. Hence the handle put near the circumference produces maximum torque.

8. A faulty balance with unequal arms has its beam horizontal. The weights of the two pans are _____________
a) Equal
b) Unequal
c) Uncertain
d) Very large
Answer: b
Clarification: They are of unequal mass. Their masses are in the inverse ratio of the arms of the balance.

9. Which part of the ship is made heavy?
a) The whole ship should be light in weight
b) The top part
c) The bottom part
d) The sides
Answer: c
Clarification: The bottom of the ship is made heavy so that the centre of gravity remains low. This ensures the stability of its equilibrium.

10. Standing is not allowed in double decker bus.
a) True
b) False
Answer: a
Clarification: When the passengers stand in the upper deck, the centre of gravity of the loaded bus is raised this makes it less stable.

Engineering Physics Multiple Choice Questions on “Streamline and Turbulent Flow”.

1. A metal plate 5cm×5cm rest on a layer of castor oil 1mm thick whose coefficient of viscosity is 1.55Nsm^(-2). Find the horizontal force required to move the plate with a speed of 2 cms^(-1).
a) 0.775N
b) 0.0577N
c) 0.0775N
d) 0.577N
Answer: c
Clarification: F=ȠA dv/dx
F=1.55×25×10^-4×(2×10^-2)/(1×10^-3)=0.0775N.

2. At what speed will the velocity head of stream of water be 40 cm?
a) 2.80cms^-1
b) 280 cms^-1
c) 28 cms^-1
d) 0.280 cms^-1
Answer: b
Clarification: h=v²/2g
v=√2gh=√(2×980×40)=280cms^-1.

3. It is easier to cut an apple with a blunt knife than a sharp knife.
a) True
b) False
Answer: b
Clarification: The area of a sharp edge is much less than the area of a blunt edge. For the same total force, the effective force per unit area is more for the sharp edge than the blunt one. Hence a sharp knife cuts easily than a blunt knife.

4. It is easier to swim in sea water than in river water.
a) True
b) False
Answer: a
Clarification: The sea water has many salts dissolved in it. So the density of sea water is greater than that of river water. Consequently, the sea water exerts greater upthrust on the swimmer than the river water. Hence it is easier to swim in sea water than in river water.

5. If there were no gravity, which of the following will not be there for a fluid?
a) Viscosity
b) Surface tension
c) Pressure
d) Archimedes’ upward thrust
Answer: d
Clarification: Archimedes’ upward thrust will be absent for a fluid if there were no gravity.

6. Motion of a liquid in a tube is best described by ___________
a) Bernoulli theorem
b) Poiseuille’s equation
c) Stoke’s law
d) Archimedes’ principle
Answer: b
Clarification: Poiseuille’s formula gives the volume of a liquid flowing out per second through a horizontal capillary tube of length l, radius r, under a pressure difference p applied across its ends.
Q=V/t=(πpr⁴)/8Ƞl.

7. Critical velocity of the liquid ___________
a) Decrease when radius decreases
b) Increases when radius increases
c) Decreased when density increases
d) Increases when density increases
Answer: c
Clarification:
v_c=(R_e Ƞ)/pD
Critically velocity decreases when density ρ increases or diameter D increases.

8. A steel ball is dropped in oil, then ___________
a) The ball attains constant velocity after sometime
b) The ball stops
c) The speed of ball will keep on increasing
d) The speed of the ball will decrease
Answer: a
Clarification: The ball attains constant velocity after falling through some distance in oil when the weight of ball gets balanced by upthrust and the upward viscous force.

9. An aeroplane gets its upward lift due to a phenomenon described by the ___________
a) Archimedes’ principle
b) Bernoulli’s principle
c) Buoyancy principle
d) Pascal law
Answer: b
Clarification: An aeroplane gets dynamic upward lift in accordance with Bernoulli’s principle.

10. The rate of flow of liquid through an orifice of a tank does not depend upon ___________
a) The size of the orifice
b) Density of liquid
c) The height of fluid column
d) Acceleration due to gravity
Answer: d
Clarification: The rate of flow of liquid through an orifice depends on the size of the orifice, atomising surface area, liquid characteristic. It does not depend on acceleration due to gravity.

11. The velocity of efflux of a liquid through an orifice in the bottom of the tank does not depend upon ___________
a) Size of orifice
b) Height of liquid
c) Acceleration due to gravity
d) Quantity
Answer: a
Clarification: Velocity of efflux, v=√2gh
Clearly, it does not depend on the size of the orifice.

12. Assertion: Smaller drops of liquid resist deforming forces better than the larger drops.
Reason: Excess pressure inside a drop is directly proportional to its surface area.
a) Both assertion and reason are true and the reason is the correct explanation of the assertion
b) Both assertion and reason are true but the reason is not the correct explanation of the assertion
c) Assertion is true but the reason is false
d) Assertion and reason is false
Answer: c
Clarification: The assertion is true but the reason is false. The excess pressure inside a liquid drop,
p=2σ/R
The excess pressure is large in a small drop due to which it can resist the deforming forces.

13. A body of mass 15kg is dropped into the water. If the apparent weight of the body is 107N, then the applied thrust will be ___________
a) 40 N
b) 80 N
c) 60 N
d) 100 N
Answer: a
Clarification: Apparent weight = Weight – Upthrust
Therefore, Upthrust = Weight – apparent weight
Upthrust = 15 × 9.8 – 107 = 147 – 107 = 40 N.

Engineering Physics Multiple Choice Questions on “Oscillatory Motion – 1”.

1. The motion of the earth about its axis is periodic and simple harmonic.
a) True
b) False
Answer: b
Clarification: The earth takes 24 hours to complete its rotation about its axis, but the concept of to and fro motion is absent, and hence the rotation of the earth is periodic and not simple harmonic.

2. An object of mass 0.2kg executes simple harmonic motion along the x-axis with a frequency of (25/π)Hz. At the position x = 0.04, the object has kinetic energy of 0.5J and potential energy 0.4J. The amplitude of oscillation is?
a) 6cm
b) 4cm
c) 8cm
d) 2cm
Answer: a
Clarification: Total energy,
E=2π² mv² A²
0.5+0.4=2π²×0.2×(25/π)² A²
A²=0.9/(0.4×25²)
A=3/(2×25)=3/50 m=6cm.

3. A spring of force constant 800N/m has an extension of 5cm. The work done in extending it from 5cm to 15cm is?
a) 8J
b) 16J
c) 24J
d) 32J
Answer: a
Clarification: At x₁ = 5 cm,
U₁=1/2×k(x₁)²=1/2×800×0.05²=1J
At x₂=15cm,
U₂=1/2×k(x₂)²=1/2×800×0.15²=9J
W=U₂-U₁=9-1=8J.

4. A simple pendulum is attached to the roof of a lift. If the time period of oscillation, when the lift is stationary is T, then the frequency of oscillation when the lift falls freely, will be ___________
a) Zero
b) T
c) 1/T
d) ∞
Answer: a
Clarification: In a freely falling lift,
g=0
v=1/2π×√(g/l)=1/2π×√(0/l)=0.

5. There is a simple pendulum hanging from the ceiling of a lift. When the lift is standstill, the time period of the pendulum is T. If the resultant acceleration becomes g/4, then the new time period of the pendulum is?
a) 0.8T
b) 0.25T
c) 2T
d) 4T
Answer: c
Clarification: T=2×√(l/g)
T^‘=2π×√((l/g)/4)
T^‘=2T.

6. A lightly damped oscillator with a frequency v is set in motion by a harmonic driving force of frequency v’. When v’ is lesser than v, then the response of the oscillator is controlled by ___________
a) Spring constant
b) Inertia of the mass
c) Oscillator frequency
d) Damping coefficient
Answer: a
Clarification: Frequency of driving force is lesser than frequency v of a damped oscillator. The vibrations are nearly in phase with the driving force and response of the oscillator is controlled by spring constant.

7. Statement: In simple harmonic motion, the velocity is maximum, when the acceleration is minimum.
Reason: Displacement and velocity in simple harmonic motion is differ in phase by π/2.
a) Both statement and reason are true and the reason is the correct explanation of the statement
b) Both statement and reason are true but the reason is not the correct explanation of the statement
c) Statement is true, but the reason is false
d) Statement and reason are false
Answer: b
Clarification: Both statement and reason are true but the reason is not the correct explanation of the statement, In fact, the phase difference between velocity and acceleration is π/2.

8. What is the time period of a pendulum hanged in a satellite? (T is the time period on earth)
a) Zero
b) T
c) Infinite
d) T/√6
Answer: c
Clarification: In a satellite, g= 0
T=2π√(l/g)=2π√(l/0)=∞.

9. Which of the following functions represents a simple harmonic oscillation?
a) sinωt-cosωt
b) sinωt+sin2ωt
c) sinωt-sin2ωt
d) sin² ωt
Answer: a
Clarification: y=sinωt-cosωt
dy/dt=ωcosωt+ωsinωt
(d² y)/dt² =-ω² sinωt+ω² cosωt
=-ω² (sinωt-cosωt)
a=-ω² y that is a∝y
This satisfies the condition of simple harmonic motion.

10. The displacement of a simple harmonic motion doing oscillation when kinetic energy = potential energy (amplitude = 4cm) is?
a) 2√2cm
b) 2cm
c) 1/√2 cm
d) √2 cm
Answer: a
Clarification: When kinetic energy = potential energy,
y=a/√2=4/√2=2√2 cm.

Engineering Physics Multiple Choice Questions on “Superposition of Waves”.

1. Which is different from others by units?
a) Phase difference
b) Mechanical equivalent
c) Loudness of sound
d) Poisson’s ratio
Answer: d
Clarification: Of all the given quantities, Poisson’s ratio has no units.

2. The waves, in which the particles of the medium vibrate in a direction perpendicular to the direction of wave motion, is known as __________
a) Transverse waves
b) Longitudinal waves
c) Propagated waves
d) Magnetic waves
Answer: a
Clarification: In transverse waves, particles of the medium vibrate in a direction perpendicular of the wave

3. The waves produced by a motor boat sailing in water are __________
a) Transverse
b) Longitudinal
c) Longitudinal and transverse
d) Stationary
Answer: c
Clarification: Transverse waves are generated on the water surface. Inside water longitudinal waves longitudinal waves are produced due to vibrations of the rudder.

4. For a wave propagating in a medium, identify the property that is independent of the others.
a) Velocity
b) Wavelengths
c) Frequency
d) All these depend on each other
Answer: c
Clarification: Wave velocity=frequency×wavelength
Frequency remains unchanged while velocity and wavelength and interdependent.

5. A boat at anchor is rocked by waves, whose crests are 100m apart and velocity is 25m/s. The boat bounces up once in every __________
a) 2500s
b) 75s
c) 4s
d) 0.25s
Answer: c
Clarification: ʎ=100m, v=25m/s
T=ʎ/v=100/25=4s.

6. A transverse wave passes through a string with the equation
y=10sinπ(0.02x-2t)
Where x is in metres and t in seconds. What is the maximum velocity of the particles in wave motion?
a) 63m/s
b) 78m/s
c) 100m/s
d) 121m/s
Answer: a
Clarification: y=10sinπ(0.02x-2t)
u=dy/dt=-20πcosπ(0.02x-2t)
u_max=20π=63m/s.

7. If equation of a sound wave is
y=0.0015sin⁡(62.8x+314t)
Then its wavelength will be __________
a) 0.1unit
b) 0.2units
c) 0.3unit
d) 2unit
Answer: a
Clarification: y=0.0015sin⁡(62.8x+314t)
y=asin⁡(2π/ʎ x+2π/T t)
2π/ʎ=62.8
ʎ = (2×3.14)/62.8=0.1unit.

8. The plane wave is described by the equation
y=3cos⁡(x/4-10t-π/2),
Where c is in meters and t in seconds. The maximum velocity of the particles of the medium due to this is?
a) 30m/s
b) 3π/2 m/s
c) 3/4 m/s
d) 40m/s
Answer: a
Clarification: y=3cos⁡(x/4-10t-π/2)
u=dy/dt=30sin⁡(x/4-10t-π/2)
u_max=30m/s.

9. The equation of wave is given by
y=10sin⁡(2πt/30+α)
If the displacement is 5cm at t=0, then the total phase at t=7.5s will be __________
a) π/3rad
b) π/2rad
c) 2π/5rad
d) 2π/3rad
Answer: d
Clarification: y=10sin⁡(2πt/30+α)
At t=0, 5=10sinα or sin α=1/2
α=π/6
Total phase at t=7.5s,
φ=(2π×7.5)/300+π/6=π/2+π/6=2π/3.

10. If two sound waves having a phase difference of 60°, then they will have a path difference of __________
a) ʎ/6
b) ʎ/3
c) ʎ
d) 3ʎ
Answer: a
Clarification: ∆φ=60°=π/3
∆x=ʎ/2π×∆φ=ʎ/2π×π/3=ʎ/6.

Engineering Physics Multiple Choice Questions on “Black Body Radiation”.

1. As the wavelength of the radiation decreases, the intensity of the black body radiations ____________
a) Increases
b) Decreases
c) First increases then decrease
d) First decreases then increase
Answer: c
Clarification: In the case of Black Body radiations, as the body gets hotter the wavelength of the emitted radiation decreases. However, the intensity first increases up to a specific wavelength than starts decreasing, as the wavelength continues to decrease.

2. The radiations emitted by hot bodies are called as ________________
a) X-rays
b) Black-body radiation
c) Gamma radiations
d) Visible light
Answer: b
Clarification: The phenomenon of black—body radiations was given by Max Planck. He stated that hot bodies emit radiation over a wide range of wavelengths. An ideal body is the one that emits and absorbs radiation of all frequencies. Such a body called a Black Body and the radiations are called Black body radiations.

3. An iron rod is heated. The colors at different temperatures are noted. Which of the following colors shows that the iron rod is at the lowest temperature?
a) Red
b) Orange
c) White
d) Blue
Answer: a
Clarification: As the body gets hotter, the frequency of the emitted radiation keeps on increasing. Blue color has the highest frequency out of red, orange and white. Thus, as the iron rod gets heated first it would become red, then orange, then white and then finally blue.

4. A black body is defined as a perfect absorber of radiations. It may or may not be a perfect emitter of radiations.
a) True
b) False
Answer: b
Clarification: A black body is defined as the one which is a perfect absorber as well as a perfect emitter of radiations. Such a body would absorb all the radiations falling on it and would emit all of them when heated.

5. From the figure, what’s the relation between T₁, T₂, and T₃?

a) T₁ > T₂ > T₃
b) T₃ > T₂ > T₂
c) T₃ > T₁ > T₂
d) T₂ > T₁ > T₃
Answer: b
Clarification: We already know, as the temperature of the body is higher, the intensity of the black body radiations would be higher. Thus, from the graph, the radiations with temperature T₃ has the highest intensity followed by the one with temperature T₂ and then T₁. Thus, T₃> T₂ > T₁.

6. Electromagnetic wave theory of light could not explain Black Body radiations.
a) True
b) False
Answer: a
Clarification: According to electromagnetic theory, the absorption and the emission should be continuous. As the wavelength keeps decreasing, the intensity of the emitted radiations should keep increasing to infinity. Such is not the case with Black Body Radiations.

7. The unit of absorptive power is _______________
a) T
b) Ts^-1
c) Ts
d) No unit
Answer: d
Clarification: Absorptive power can be defined as the ratio of energy absorbed per unit area upon energy incident per unit time per unit area. For a black body, it’s absorptive power is equal to one.

8. For an object other than a black body, it’s emissivity, e is _______________
a) 1
b) 0 < e < 1
c) e > 1
d) e = 0
Answer: b
Clarification: Emissivity is the ratio of emissive power of any object and the emissive power of the black body having the same temperature and surface area as the object. Thus, for a black body, it is equal to 1. For any other object, it is less than 1.

9. What relation between emissivity, e, and Absorptive Power, a, is given by Kirchhoff’s law?
a) e < a
b) e > a
c) e = a
d) no specific relation
Answer: c
Clarification: Kirchhoff’s law states that for any object the emissivity is always equal to absorptive power. For a black body, both of them are equal to one.

10. What is the relation between the Energies as shown in the figure?

a) E_r = 0
b) E_a = 0
c) E_t = E_i
d) E_i = E_r
Answer: a
Clarification: As a black body is a perfect absorber, the reflected energy and the transmitted energy should be zero. Also, the energy of the incident radiation should be equal to the energy absorbed.

Engineering Physics Multiple Choice Questions on “Fermi-Dirac Distribution”.

1. Fermi-Dirac statistics is for the ________
a) Distinguishable particles
b) Symmetrical Particles
c) Particles with half integral spin
d) Particles with integral spin
Answer: c
Clarification: The Maxwell-Boltzmann statistics is for the distinguishable particles, which are basically the classical particles like atoms and molecules.

2. The difference between fermions and bosons is that bosons do not obey ______
a) Aufbau principle
b) Pauli’s Exclusion Principle
c) Hund’s Rule of Maximum Multiplicity
d) Heisenberg’s Uncertainty Principle
Answer: b
Clarification: The particles that follow Pauli’s Exclusion Principle are defined as Fermions while that don’t are called bosons. Bosons have an integral spin number.

3. The Maxwell-Boltzmann law is given by the expression n_i = ________
a) (frac{g}{e^{alpha+beta E}} )
b) (frac{g}{e^{alpha+beta E}-1} )
c) (frac{g}{e^{alpha+beta E}+1} )
d) (frac{g}{e^{alpha+beta E}+k} )
Answer: c
Clarification: The correct expression for the Maxwell-Boltzmann law is n_i = (frac{g}{e^{alpha+beta E}+1} ), where α depends on the volume and the temperature of the gas and β is equal to 1/kT.

4. The wave function of fermions is not _________
a) Continuous
b) Single Valued
c) Symmetric
d) Differentiable
Answer: c
Clarification: The particles which have antisymmetric wave function having half odd integral spin number and obey Pauli’s principle are called fermions.

5. Fermi-Dirac statistics cannot be applied to ________
a) Electrons
b) Photons
c) Fermions
d) Protons
Answer: b
Clarification: Fermi-Dirac Statistics can be applied to particles having half odd integral spin number and obey Pauli’s principle which are electrons, fermions and protons. Photon has an integral spin number.

6. The distribution function is given by ________
a) (frac{1}{Aefrac{E}{nT}+1})
b) (frac{1}{Aefrac{E}{nT}-1})
c) (frac{1}{efrac{E}{nT}+A})
d) (frac{1}{efrac{E}{nT}-A})
Answer: a
Clarification: The distribution function is given by (frac{1}{Aefrac{E}{nT}+1}), where A = (e^{frac{-E_F}{nT}}) The energy E_F is called the Fermi energy and is constant for a given system.

7. At T > 0K, the probability of a state with E > E_F filled is zero.
a) True
b) False
Answer: b
Clarification: At T > 0 K, the probability that a state with E > E_F is filled is ¹⁄₂. Hence, fermi energy is the energy at which the probability of occupation is ¹⁄₂ at any temperature above 0 K.

8. The expression for mean energy is given by ________
a) (frac{3}{5}NE_F)
b) (frac{2}{5}NE_F)
c) (frac{3}{5}E_F)
d) (frac{2}{5}E_F)
Answer: a
Clarification: The expression for the mean energy of an electron at absolute zero is given by (frac{3}{5}NE_F). The expression (frac{3}{5}E_F) is called the zero point energy.

9. For all the quantum states with energy greater than Fermi energy to be empty in a Fermi-Dirac system, the temperature should be ______
a) 273 K
b) 373 K
c) 0 K
d) 100 K
Answer: c
Clarification: We know that the Fermi-Dirac distribution is given by:
F_FD(E) = ( frac{1}{efrac{E-E_F}{nT}+1})
For all the quantum states with energy greater than Fermi energy to be empty,
F_FD(E) = 0, for E > E_F and F_FD(E) = 1, for E < E_F
Therefore, for E < E_F ( frac{1}{efrac{E-E_F}{nT}+1}=1)
(efrac{E-E_F}{nT}= 0)
As, E < E_F, E- E_F < 0. Therefore, to satisfy the given statement,
T = 0 K
Thus, we can define Fermi energy as the energy of the uppermost occupied level at 0 K.

10. What is the relationship between T1 and T2?

a) T1 > T2
b) T1 < T2
c) T1 = T2
d) Insufficient Information
Answer: b
Clarification: The given figure shows the variation of Fermi-Dirac distribution with energy E. In this case, T2 > T1, according to the expression: F_FD(E) =( frac{1}{efrac{E-E_F}{nT}+1}).

11. The density of silver is 10.5 g/cm³ and its atomic weight is 108. If each atom contributes one electron for conduction, what is the fermi energy?
a) 2.12 eV
b) 3.31 eV
c) 4.69 eV
d) 5.51 eV
Answer: d
Clarification: Fermi Energy, (E_F = frac{h^2}{8m} (frac{3N}{pi v})^{2∕3})
Here, N /V = 10.5 X 6.02 X 10²³/108
= 5.85 X 10²⁸ m^-3
Therefore, (E_F = frac{6.626 X 6.626}{8 X 9.1} times 10^{-19} times frac{3 times 5.585}{pi}^{2∕3})
= 8.816 X 10^-19J
= 5.51 eV.

12. The Fermi energy of a material is 3.45 eV. What is the zero-point energy of the material?
a) 1.02 eV
b) 2.07 eV
c) 3.45 eV
d) 4.16 eV
Answer: b
Clarification: As we know, the fermi energy of the material is 3.45 eV.
Now, zero-point energy =(frac{3}{5}E_F)
= 3/5 X 3.45 eV
= 2.07 eV.

13. The average energy of one electron silver is 3.306 eV. What is the fermi-energy of Silver at 0 K?
a) 2.32 eV
b) 3.78 eV
c) 4.12 eV
d) 5.51 eV
Answer: d
Clarification: We know, Average Energy, E =(frac{3}{5}NE_F)
Here, E = 3.306 eV, N = 1
Therefore, we get: E_F = 5E/3
= 5.51 eV.

14. In Fermi-Dirac Statistics, one energy state can be occupied by more than one particle.
a) True
b) False
Answer: b
Clarification: In Fermi-Dirac statistics, one energy state can be occupied by only one particle. Therefore, when one state is filled, the particle will go on to another state.

15. Which of the following is the curve for Fermi-Dirac statistics?

a) X
b) Y
c) Z
d) None
Answer: c
Clarification: As seen the curve the order of the ration of f(K) with E/kT is in the order X > Y > Z. Here, Y is Maxwell-Boltzmann Statistics, X is Bose-Einstein Statistics and Z is Fermi-Dirac Statistics.