250+ TOP MCQs on Extrinsic Semiconductor | Class12 Physics

Physics Multiple Choice Questions on “Extrinsic Semiconductor”.

1. On doping germanium metal, with a little amount of indium, what does one get?
a) Intrinsic semiconductor
b) Insulator
c) n-type semiconductor
d) p-type semiconductor
Answer: d
Clarification: Indium impurity in germanium produces p-type semiconductors. A trivalent impurity added to germanium produces a p-type semiconductor. Trivalent impurities such as boron, indium, and gallium are called acceptor impurity.

2. In a pure semiconductor crystal, if current flows due to breakage of crystal bonds, then what is the semiconductor is called?
a) Acceptor
b) Donor
c) Intrinsic semiconductor
d) Extrinsic semiconductor
Answer: c
Clarification: Pure semiconductors are called intrinsic semiconductors. The number of electrons in the conduction band will be equal to the number of holes in the valence band. Intrinsic semiconductors are also called undoped and i-type semiconductors.

3. Which of the following, when added as an impurity, into the silicon, produces n-type semiconductor?
a) Phosphorous
b) Aluminum
c) Magnesium
d) Sulfur
Answer: a
Clarification: As phosphorous is pentavalent, it produces n-type semiconductor when added to silicon. They are called donor impurities. By adding phosphorus, extra valence electrons are added that become unbonded from individual atoms.

4. In n-type semiconductors, which one is the majority charge carriers?
a) Holes
b) Protons
c) Neutrons
d) Electrons
Answer: d
Clarification: In n-type semiconductors, electrons are majority charge carriers. It is made by adding an impurity to a pure semiconductor. This is the opposite scenario of p-type semiconductors where electrons are the minority charge carriers.

5. A small impurity is added to germanium to get a p-type semiconductor. Identify the impurity?
a) Bivalent substance
b) Trivalent substance
c) Pentavalent substance
d) Monovalent substance
Answer: b
Clarification: A trivalent impurity added to germanium produces a p-type semiconductor. Trivalent impurities such as boron, indium, and gallium are called acceptor impurity. These can be added to germanium in order to obtain a p-type semiconductor.

6. Boron when added as an impurity, into the silicon, produces n-type semiconductor.
a) True
b) False
Answer: b
Clarification: When a pentavalent impurity is added as an impurity to silicon, it produces n-type semiconductor. In n-type semiconductors, electrons are majority charge carriers, whereas the holes are minority charge carriers.

7. Identify the property which is not characteristic for a semiconductor?
a) At a very low temperature, it behaves like an insulator
b) At higher temperatures, two types of charge carriers will cause conductivity
c) The charge carriers are electrons and holes in the valence band at higher temperatures
d) The semiconductor is electrically neutral
Answer: c
Clarification: In a semiconductor, electrons are the charge carriers in the conduction band and holes are the charge carriers in the valence band at higher temperatures. The other statements are not valid.

8. The n-type semiconductor is which of the following?
a) Positively charged
b) Negatively charged
c) Neutral
d) Positive or negative depending upon doping materials
Answer: c
Clarification: Semiconductors maintain their electrical neutrality even after doping. This is achieved by adding an impurity to a pure semiconductor in order to obtain an n-type semiconductor.

9. The dominant contribution to current comes from holes in case of which of the following?
a) Metals
b) Intrinsic semiconductors
c) p-type extrinsic semiconductors
d) n-type extrinsic semiconductors
Answer: c
Clarification: Holes are the majority charge carriers in p-type extrinsic semiconductors. Trivalent impurities such as boron, indium, and gallium are called acceptor impurity. Also, in p-type semiconductors, electrons are the minority charge carriers.

10. In a p-type semiconductor, germanium is doped with which of the following?
a) Gallium
b) Copper
c) Phosphorous
d) Nitrogen
Answer: a
Clarification: Substances such as gallium, boron, and aluminum are all trivalent atoms. These are called acceptor impurities and they produce p-type semiconductors. Therefore, germanium is doped with gallium in a p-type semiconductor.

250+ TOP MCQs on Continuous Charge Distribution | Class12 Physics

Physics Multiple Choice Questions on “Continuous Charge Distribution”.

1. How many electrons must be added to an isolated spherical conductor of radius 20cm to produce an electric field 1000N/C just outside the surface?
a) 2.77*1020
b) 2.77*1010
c) 1.77*1010
d) 5.4*1010

Answer: b
Clarification: The electric field E = k (frac {Q}{r^2}) where k=9*109 N m2 C-2 and Q is the charge of the sphere, r is the radius of the sphere. Therefore Q = (frac {E*r^2}{k}=frac {1000*0.2^2}{9*10^9}) = 4.44*10-9. Number of electrons required for this amount of charge is = (frac {4.44*10^{-9}}{1.602*10^{-19}}) = 2.77*1010.

2. A circular annulus of inner radius r and outer radius R has a uniform charge density a. What will be the total charge on the annulus?
a) a*(R2-r2)
b) π*a*(R2-r2)
c) a*(R-r)
d) π*a* R2

Answer: b
Clarification: Total surface area of the annulus = π*(R2-r2) because it has outer radius R and inner radius r. We know surface charge density is the amount of charge stored on the unit surface area. In this case, surface charge density is a. Therefore total charge on the annulus = π*a*(R2-r2).

3. What is the dimension of linear charge density?
a) [I T L-1]
b) [I T-1 L]
c) [I T L]
d) [I-1 T-1 L]

Answer: a
Clarification: Linear charge density λ=(frac {Amount , of, charge}{Total , length}). The dimension of electric charge = [I T] and the dimension of length is [L]. Hence the dimension of linear-charge-density = [I T L-1]. In the case of surface charge density, the dimension is [I T L-2] because it means charge stored on the unit surface area.

4. A charge is distributed along an infinite curved line in space with linear charge distribution λ. What will be the amount of force on a point charge q kept at a certain distance from the line?
a) q∫(frac {lambda}{r^2}hat{r})dl
b) q∫(frac {lambda}{r^2}hat{r})dr
c) q∫(frac {lambda}{r^3}hat{r})dl
d) q∫(frac {lambda}{r^2})

Answer: a
Clarification: Let the point charge is situated at a distance r from a small part dl on the line. The charge stored in stat small part=λ.dl. The force due to that small part will be directed towards the unit vector (hat{r}). Therefore, force on that charge due to the entire linear charge distribution=q∫(frac {lambda}{r^2}hat{r})dl.

5. The term ‘volume charge density’ is valid for a solid conducting sphere.
a) True
b) False

Answer: b
Clarification: We know that in the case of any conducting material, the charge is always distributed on its outer surface instead of distribution throughout the material. Therefore, the word volume charge density doesn’t make any sense in the case of the conducting sphere. Instead, we can use the term surface charge density.

6. A solid non conducting sphere of radius 1m carries a total charge of 10 C which is uniformly distributed throughout the sphere. Determine the charge density of the sphere.
a) 10 C/m3
b) 4.76 C/m3
c) 0.1 C/m3
d) 2.38 C/m3

Answer: d
Clarification: Volume of the sphere = (frac {4}{3})πr3 where r is the radius of the sphere. Therefore, the charge density = (frac {total , charge}{frac {4}{3}pi r^3}). Now substituting the values, charge density = (frac {10}{frac {4}{3}pi r^3}) = 2.38 C/m3. But if the sphere is conducting, we have to consider the surface charge density.

250+ TOP MCQs on Electric Current | Class12 Physics

Physics Multiple Choice Questions on “Electric Current”.

1. What is the SI unit of current?
a) Coulomb (C)
b) Ampere (A)
c) Farad (F)
d) Newton (N)
Answer: b
Clarification: The SI unit of current is Ampere (A). The current through a wire is called one ampere if one coulomb of charge flows through the wire per second.
1 Ampere = (frac {1 , Coulomb}{1 , Second}) or 1A = 1Cs-1.

2. What is the number of electrons that constitutes a current of one Ampere?
a) 2.25 × 1018
b) 2.25 × 10-18
c) 6.25 × 1018
d) 6.25 × 10-18
Answer: c
Clarification: Number of electrons, n = (frac {Current (I) times time (t)}{Charge , of , 1 , electron (e)})
= (frac {(1 times 1)}{1.6 times 10^{-19}})
= 6.25 × 1018.

3. Give the number of electrons passing through a wire per minute. The current flowing through it is 500mA.
a) 1.875 × 1020
b) 6.875 × 1020
c) 1.875 × 10-20
d) 6.875 × 10-20
Answer: a
Clarification: Number of electrons, n = (frac {It}{e}).
= (frac {(500 times 10^{-3} times 60)}{(1.6 times 10^{-19})})
= 1.875 × 1020.

4. Which of these is a correct definition of conventional current?
a) Current that flows from lower potential to higher potential
b) The current which remains static
c) Current constituted by the flow of ions
d) Current that flows from higher potential to lower potential
Answer: d
Clarification: The current that flows from a point at the higher (positive) potential to a point at lower (negative) potential is called conventional current. The direction of motion of positive charges is taken as the direction of electric current.

5. Which type of current is flowing through a circuit?
a) Static current
b) Conventional current
c) Electronic current
d) Potential current
Answer: c
Clarification: The carriers of electric current are electrons. The current in a circuit is due to the flow of electrons. Therefore, the direction of the conventional current is opposite to the direction of flow of electrons.

6. In electrolytes and ionic crystals, the current carriers are free electrons.
a) True
b) False
Answer: b
Clarification: In electrolytes and ionic crystals, the current carriers are positive and negative ions. Whereas in most of the substances like metals, the carriers of electric current are electrons.

7. Identify the dimensional formula of electric current.
a) [M L T-2]
b) [M0 L0 T0 A1]
c) [M L3 T0]
d) [M L2 T-3]
Answer: b
Clarification: The flow of charge per unit time is defined as current. The fundamental unit of current is Ampere (named after Andre Marie Ampere).
Hence the dimensional formula of electric current is [M0 L0 T0 A1].

8. Which type of a physical quantity is electric current?
a) Scalar quantity
b) Vector quantity
c) Bipolar quantity
d) Thermodynamic quantity
Answer: a
Clarification: The electric current is a scalar quantity. Laws of ordinary algebra are used to add electric currents and the law of vector algebra are not applicable.

9. Which of these is a correct definition of electronic current?
a) Current that flows from lower potential to higher potential
b) The current which remains static
c) Current constituted by the flow of ions
d) Current that flows from higher potential to lower potential
Answer: a
Clarification: The current that flows from a point at the lower (negative) potential to a point at higher (positive) potential is called electronic current. Electronic current is produced by the movement of negatively charged electrons.

10. What is the study of electric charges in motion called?
a) Charge mobility
b) Electronic mobility
c) Static electricity
d) Current electricity
Answer: d
Clarification: The study of motion or dynamics of charges constitute an electric current. Therefore, the study of electric charges in motion is called current electricity.

250+ TOP MCQs on Motion in a Magnetic Field | Class12 Physics

Physics Multiple Choice Questions on “Motion in a Magnetic Field”.

1. A positive charge is moving vertically upwards. When it enters a region of magnetic field directed towards north, what is the direction of the force on the charge?
a) Up
b) Down
c) Left
d) Right
Answer: c
Clarification: According to Fleming’s left-hand rule, the magnetic force will act towards left. When the positive charge enters a region of magnetic field directed towards north, the magnetic force will act towards left.

2. An electron moving with a velocity of 15 ms-1 enters a uniform magnetic field of 0.2 T, along a direction parallel to the field. What would be its trajectory in this field?
a) Elliptical
b) Straight path
c) Helical
d) Circular
Answer: b
Clarification: The electron will continue to follow its straight path because a parallel magnetic field does not exert any force on the electron. So, there won’t be any change in its trajectory when the electron enters a uniform magnetic field.

3. In a certain arrangement, a proton does not get deflected while moving through a magnetic field region. Under what condition is it possible?
a) F = 0
b) F = 180 N
c) F = -180 N
d) F = 3600 N
Answer: a
Clarification: Magnetic force on a proton is Fm = qvBsin(θ).
A proton moving parallel or antiparallel to a magnetic field does not experience any force. Hence, the proton does not get deflected while moving through a magnetic field region.

4. Which of the following will experience a maximum force, when projected with the same velocity perpendicular to the magnetic field : (i) α-particle, and (ii) β-particle?
a) Both α-particle and β-particle
b) None
c) β-particle
d) α-particle
Answer: d
Clarification: F = qvB.
For α-particle, q = 2e, Fα = 2evB
For β-particle, q = e, Fβ = evB.
Thus, the α-particle will experience maximum force.

5. A charged particle in a plasma trapped in a magnetic bottle leaks out after a millisecond. What is the total work done by the magnetic field during the time the particle is trapped?
a) Maximum
b) Minimum
c) Zero
d) Depends on the strength of the magnetic field
Answer: c
Clarification: Work done is zero. Since a magnetic field exerts a force perpendicular to the direction of motion of the charged particle, no work is done by it on the charged particle. Therefore, the total work done by the magnetic field during the time the particle is trapped is zero.

6. Force on a charge ‘q’ moving with a velocity ‘v’ in a magnetic field is given by F = qvB. State true or false.
a) True
b) False
Answer: b
Clarification: Force on a charge ‘q’ moving with a velocity ‘v’ in a magnetic field is given by:
F = qvB sinθ.
The direction of the force is given by Fleming’s left-hand rule.

7. A proton enters a magnetic field of flux density 5 T with a velocity of 5 × 107 ms-1 at an angle of 30o with the field. Find the force on the proton.
a) 0.2 × 10-11 N
b) 2 × 10-11 N
c) 20 × 10-11 N
d) 200 × 10-11 N
Answer: b
Clarification: F = qvB sinθ.
F = 1.6 × 10-19 × 5 × 107 × 5 × sin30o
F = 2 × 10-11 N.

250+ TOP MCQs on Electromagnetic Induction – Magnetic Flux | Class12 Physics

Physics Multiple Choice Questions on “Electromagnetic Induction – Magnetic Flux”.

1. Pick out the SI unit of magnetic flux.
a) Ampere
b) Tesla meter
c) Weber
d) Maxwell
Answer: c
Clarification: The magnetic flux through a surface is the surface integral of the normal component of the magnetic field flux density passing through that surface. The SI unit of magnetic flux is weber (Wb). One weber is the flux produced when a uniform magnetic field of one tesla acts normally over an area of 1 cm2.

2. Identify the correct dimensions of magnetic flux.
a) [M3 L2 A-1 T-2]
b) [M L2 A-1 T-2]
c) [M2 L2 A-1 T-2]
d) [M L2 A-1 T2]
Answer: b
Clarification: Magnetic flux = Magnetic field × Area.
Flux = (frac { ( [ M L T^{-2} ].[ L^2 ])}{(A L) })
Flux = (frac { [M L^2 T^{-2} ] }{ [ A ] })
Flux = [M L2 A-1 T-2].

3. Which of the following is a unit of magnetic flux?
a) Maxwell
b) Gauss
c) Tesla
d) Ampere
Answer: a
Clarification: The CGS unit of magnetic flux is Maxwell (Mx). One Maxwell is the flux produced when a uniform magnetic field of one gauss acts normally over an area of 1 cm2.

4. Find out the correct relation from the following.
a) 1 Wb = 1 G cm
b) 1 Wb = 1 Tm
c) 1 Wb = 1 G cm2
d) 1 Wb = 108 Maxwell
Answer: d
Clarification: 1 Wb = 1 T × 1 m2
Wb = 104 G × 104 cm2.
1 Wb = 108 Maxwell.

5. When is the magnetic flux said to be positive?
a) θ = 180o
b) θ = 360o
c) θ = 0o
d) θ = 90o
Answer: c
Clarification: A normal to a plane can be drawn from either side. If the normal drawn to a plane points out in the direction of the field, then θ = 0o and the flux are taken as positive. If the flux is positive and increasing the voltage will be negative.

6. The total number of magnetic lines of force crossing the surface placed in a magnetic field normally is called the magnetic induction.
a) True
b) False
Answer: b
Clarification: The magnetic flux through any surface place in a magnetic field is the total number of magnetic lines of force crossing this surface normally. It is measured as the product of the component of the magnetic field normal to the surface and the surface area.

7. Calculate the magnetic flux when the magnetic field is perpendicular to the surface area.
a) Minimum
b) Maximum
c) Zero
d) Depends on the surface area
Answer: b
Clarification: Magnetic flux = B ΔS cos 00
Magnetic flux = B ΔS.
It follows that magnetic flux linked with a surface is maximum when the direction of the magnetic field is perpendicular to the surface area.

8. Which type of physical quantity is magnetic flux?
a) Scalar
b) Vector
c) Isotropic
d) Isentropic
Answer: a
Clarification: The magnetic flux is measured as the product of the component of the magnetic field normal to the surface and the surface area.
Magnetic flux is a scalar quantity.

9. Calculate the magnetic flux produced when the magnetic field is parallel to the surface area.
a) Maximum
b) Minimum
c) Zero
d) Depends on the magnetic field
Answer: c
Clarification: Magnetic flux = B ΔS cos 90o
Magnetic flux = 0.
It follows that magnetic flux linked with a surface is zero when the direction of the magnetic field is parallel to the surface area.

10. When is the magnetic flux said to be negative?
a) θ = 180o
b) θ = 360o
c) θ = 90o
d) θ = 0o
Answer: a
Clarification: If the normal points in the opposite direction of the field, then θ = 180o and the flux is taken as negative. If the magnetic field is pointing opposite the direction of the surface area then the value of this flux is negative.

250+ TOP MCQs on Electromagnetic Spectrum | Class12 Physics

Physics Multiple Choice Questions on “Electromagnetic Spectrum”.

1. On what basis is the classification of electromagnetic waves done?
a) Electromagnetic spectrum
b) Electric field
c) Magnetic field
d) Propagation constant
Answer: a
Clarification: The classification of electromagnetic waves done according to the frequency called the electromagnetic spectrum. The basic difference various type of electromagnetic waves lies in their wavelength or frequency since all of them travel through vacuum at the same speed and also, the waves differ in their mode of interaction with matter.

2. Which among the following has a frequency range of 500 kHz to 1000 MHz?
a) Microwaves
b) Infrared Waves
c) Radio Waves
d) Gamma Rays
Answer: c
Clarification: Radio Waves are generally in the frequency range ➔ 500 kHz to 1000 MHz. Radio waves are used for long-distance communication, such as in television, mobiles, and radios. These devices receive radio waves and convert them to mechanical vibrations in the speaker to create sound waves.

3. What is the full form of UHF?
a) Under heavy frequency
b) Ultra-high frequency
c) Ultra heavy frequency
d) Under high frequency
Answer: b
Clarification: UHF stands for Ultra high frequency. Cellular phones used radio waves to transmit voice communication in the Ultra high-frequency Band. The UHF band extends from 900 MHz to 5 x 109 Hz or 5000 MHz. Radio waves are produced by oscillating circuits having an inductor and capacitor.

4. Microwaves are produced by klystrons.
a) True
b) False
Answer: a
Clarification: Yes, microwaves are produced by special vacuum tubes such as klystrons and magnetrons. Microwaves are high-frequency radio waves of frequency range 1 G Hz to 300 G Hz. Microwaves are used in radar systems and the study of atomic and molecular structure.

5. Which among the following is an application of microwaves?
a) Solar cells
b) Ovens
c) Radiotherapy
d) Burglar alarms
Answer: b
Clarification: Microwave ovens are an application of microwaves. In microwave ovens, the frequency of microwave produced is matched with the natural frequency of water molecules so that resonance occurs and water molecules in the material vibrate at the higher amplitude and transfer energy to nearby food molecules, results in heating the food.

6. Identify the electromagnetic wave which is also known as heatwaves.
a) Radio Waves
b) Gamma Waves
c) X-Rays
d) Infrared Waves
Answer: d
Clarification: Infrared waves are heat radiations also known as heatwaves. These waves are produced by hot bodies and molecules. They do heating because water molecules present in most of the materials readily absorb infrared waves and their thermal motion increases, so they heat themselves and also heat their surroundings.

7. Pick out the electromagnetic wave which is highly harmful to humans.
a) Radio waves
b) Ultraviolet Rays
c) Microwaves
d) Infrared Waves
Answer: b
Clarification: Ultraviolet (UV) radiations in large quantities are highly harmful to humans. These rays in solar radiation on reaching earth are absorbed by the ozone layer in the atmosphere. UV rays are produced by special lamps such as mercury and from arc lamps and by very hot bodies like the sun.

8. Stephen is thinking of going to the hospital for checking whether he has tuberculosis since he believes that he may have some early symptoms of the disease. His friend suggests him to do an x-ray checkup to clear up the confusion. Is the suggestion made by Stephen’s friend going to help him?
a) True
b) False
Answer: b
Clarification: No, the suggestion is not going to help Stephen. Even though X-rays have higher energy than light waves and can pass through the body, it is not of great value in the early diagnosis of pathological conditions of bones and joints like tuberculosis. This is because, in the early stages of the disease, the damage to the lungs may not yet have become sufficiently marked in the X-ray to be detectable, so people who have active tuberculosis can be missed.

9. Monica wanted to take photographs of a monument. But since the surrounding is filled with smoke, she is not able to take good photos. Which one of the following electromagnetic waves can be used in this situation to help Monica?
a) X-rays
b) Gamma rays
c) Ultraviolet rays
d) Infrared waves
Answer: d
Clarification: Infrared waves can be used to take photographs during conditions of smoke, fog, etc. as these waves are scattered less than visible rays and hence travel longer distances through the atmosphere. So, using infrared waves can help Monica out.

10. Which is the frequency range of gamma rays from the following?
a) 1 × 1018 to 3 × 1022 Hz
b) 3 × 10-18 to 5 × 1022 Hz
c) 3 × 1018 to 5 × 1022 Hz
d) 3 × 10-18 to 5 × 10-22 Hz
Answer: c
Clarification: The frequency range of Gamma rays is 3 × 1018 to 5 × 1022 Hz. These rays have a wavelength of 6 × 10-13 to 10-10m. Gamma rays are produced in nuclear reactions and are also emitted by radioactive nuclei.

11. A radio can tune to any station in the frequency range from 2.5 MHz to 15 MHz bands. What is the corresponding wavelength band?
a) 30 m to 130 m
b) 40 m to 140 m
c) 20 m to 120 m
d) 10 m to 110 m
Answer: c
Clarification: Maximum wavelength in the band will correspond to the lowest frequency:
c=fminλmax
λmax=(frac {c}{f_{min}}=frac {3times 10^8}{2.5times 10^6})=120 m
Minimum wavelength in the band will correspond to the highest frequency:
c=fmaxλmin
λmin=(frac {c}{f_{max}} =frac {3times 10^8}{15 times 10^6 })=20 m
Therefore, the corresponding wavelength band is 20 m to 120 m.

12. The speeds of microwaves, infrared waves, and ultraviolet waves are Vm, Vi, and Vu respectively. Identify the correct combination showing the different waves in vacuum.
a) Vm > Vi > Vu
b) Vm = Vi = Vu
c) Vm < Vi < Vu
d) Vm > Vi < Vu
Answer: b
Clarification: The correct combination is ➔ Vm = Vi = Vu. This is because, in vacuum, all the electromagnetic waves in question will travel at the same speed. The speed with which they travel in vacuum is the speed of light. (c = 3 × 108 m/s).

13. Electromagnetic radiation has an energy of 15 keV. Then this radiation belongs to which region of electromagnetic waves?
a) X rays
b) Infrared rays
c) Gamma rays
d) UV rays
Answer: a
Clarification: Given: Energy (E) = 15 keV
The required equation ➔ λ(in Å)=( frac {hc}{E(in , eV)})
λ=(frac {12400}{15 times 10^3})
λ=0.826 Å=0.826 × 10-10 m
X rays cover wavelengths ranging from above 10-8 m to 10-13 m. Therefore, this electromagnetic radiation belongs to the region of X rays.

14. Pick out the correct increasing order of energy of electromagnetic waves from the following.
a) Einfrared < Emicro < Evisible < Eultravioletgamma
b) Emicro < Einfrared < Evisible < Eultraviolet < Egamma
c) Emicro < Einfrared < Evisible < Egamma < Eultraviolet
d) Emicro < Einfrared < Eultraviolet < Evisible < Egamma
Answer: b
Clarification: The energy of electromagnetic waves is directly proportional to the frequency of the electromagnetic waves. So the order of frequency is given as:
vmicro < vinfrared < vvisible < vultraviolet < vgamma
Since E=hv ➔ E ∝ v
The order of energy is as follows:
Emicro < Einfrared < Evisible < Eultraviolet < Egamma

15. Which of the following cannot travel in vacuum?
a) Radio waves
b) Gamma Waves
c) Infrared Waves
d) Infrasonic waves
Answer: d
Clarification: Radio waves, gamma waves, and infrared waves are electromagnetic waves and due to this they do not need any material medium to travel and hence, can travel in vacuum. Whereas, infrasonic waves are mechanical waves and so, they need a material medium to travel. Therefore, infrasonic waves cannot travel in vacuum.