Category: Class 12 Physics Objective Questions

Physics Multiple Choice Questions on “Displacement Current”.

1. Which one of the following current flows in the gap between the capacitor plates?
a) Displacement current
b) Conduction current
c) Resistive current
d) Total current
Answer: a
Clarification: Displacement current (I_D) is the electric current that flows in the gap between the plates of the capacitor during its charging, which originates due to the time-varying electric field in the space between the two plates of the capacitor. The expression for displacement current is given as:
I_D = ε_o (frac {dPhi_{varepsilon}}{dt})

2. Identify the expression of Ampere-Maxwell’s Circuital law.
a) ∮( overset{rightharpoonup}{B}.overset{rightharpoonup}{dl}) = μ₀ (I_C-I_D)
b) ∮( overset{rightharpoonup}{B}.overset{rightharpoonup}{dl}) = μ₀ (I_D-I_C)
c) ∮( overset{rightharpoonup}{B}.overset{rightharpoonup}{dl}) = μ₀ (I_C+I_D)
d) ∮( overset{rightharpoonup}{B}.overset{rightharpoonup}{dl}) = μ₀ (I_C I_D)
Answer: c
Clarification: The Ampere-Maxwell Law tells you that this quantity is proportional to the enclosed current and rate of change of electric flux through any surface bounded by the path of integration. The expression for Ampere-Maxwell’s law is given as:
∮( overset{rightharpoonup}{B}.overset{rightharpoonup}{dl}) = μ₀(I_C+I_D)

3. Find the true statement.
a) Displacement current and conduction current are never equal
b) The current that flows through connection wires is called conduction current
c) During charging of the capacitor, in the connection wires, conduction current is discontinuous and displacement current is continuous
d) During charging of the capacitor, in the gap between the capacitor plates, conduction current is continuous and displacement current is discontinuous
Answer: b
Clarification: The true statement is ➔ the current that flows through the connection wires is called conduction current. All the other statements are not valid. Displacement current and conduction current can numerically be equal. During charging of the capacitor, in the connection wires, conduction current is continuous and displacement current is discontinuous. Similarly, during the charging of the capacitor, in the gap between the capacitor plates, conduction current is discontinuous and the displacement current is continuous.

4. Maxwell modified Ampere’s Circuital Law.
a) True
b) False
Answer: a
Clarification: Yes, Maxwell modified Ampere’s Law. Ampere’s law is true just for steady currents. Maxwell found the shortcoming in Ampere’s law and he modified Ampere’s law to incorporate time-varying electric fields. Maxwell was the one responsible for the correction of the Ampere’s circuital law by the addition of displacement current. He said that we’ve to think about not only the present existing outside the capacitor but also the present referred to as displacement current that existed between the plates of the capacitor.

5. A parallel plate capacitor with plate area A and separation between the plates d, is charged by a constant current I. Consider a plane surface of area A/4 parallel to the plates and drawn between the plates. What is the displacement current through this area?
a) I
b) (frac {I}{4})
c) 4I
d) (frac {I}{2})
Answer: b
Clarification: Electric field between the plates is given as:
E=(frac {q}{Avarepsilon_o}=frac {It}{Avarepsilon_o})
So, the electric flux through the area (frac {A}{4}) is given by:
Φ_E=((frac {A}{4}))E=(frac {It}{4varepsilon_o})
Then the displacement current will be:
I_D = ε_o(frac {dPhi_E}{dt})
I_D = ε_o(frac {d}{dt} (frac {It}{4epsilon_o})=frac {I}{4})
Therefore, the displacement current through this area is (frac {I}{4}).

Physics Multiple Choice Questions on “Photoelectric Effect”.

1. Which photon is more energetic: A red one or a violet one?
a) Both
b) Red
c) Violet
d) Neither
Answer: c
Clarification: Violet photon has more energy because the energy of a photon is given as:
E = hv
Since energy is directly proportional to wavelength, i.e. v_voilet > v_red, the violet photon is more energetic than the red photon.

2. If the wavelength of electromagnetic radiation is doubled, what will happen to the energy of photons?
a) Remains the same
b) Doubled
c) Halved
d) Infinite
Answer: c
Clarification: Energy of a photon is given as:
E = hv = (frac {hc}{lambda})
The energy of the photon reduces to one-half when the wavelength of radiation is doubled.

3. What happens to the wavelength of a photon after it collides with an electron?
a) Increases
b) Decreases
c) Remains the same
d) Infinite
Answer: a
Clarification: A photon transfers a part of its energy to the colliding electron, so its energy decreases, and consequently wavelength increases. This is because the energy of a photon is inversely proportional to the wavelength of a photon.

4. Why are alkali metals most suited as photo-sensitive metals?
a) High frequency
b) Zero rest mass
c) High work function
d) Low work function
Answer: d
Clarification: Alkali metals have low work functions. Even visible radiation can eject out electrons from them. So alkali metals are the most suitable photo-sensitive metals.

5. Which radiations will be most effective for the emission of electrons from a metallic surface?
a) Microwaves
b) X rays
c) Ultraviolet
d) Infrared
Answer: c
Clarification: Ultraviolet rays are most effective for photoelectric emission because they have the highest frequency when compared to the other electromagnetic waves, and hence the most energetic among all of them.

6. Photoelectric emission is possible at all frequencies.
a) True
b) False
Answer: b
Clarification: No. Photoelectric emission is possible only if the energy of the incident photon is greater than the work function (W₀ = hv₀) of the metal. Hence, the frequency v of the incident radiation must be greater than the threshold frequency v₀.

7. Two metals A and B have work functions 4 eV and 10 eV respectively. Which metal has a higher threshold wavelength?
a) Metal A
b) Metal B
c) Both
d) Neither
Answer: a
Clarification: According to the equation, W₀ = hv₀ = (frac {hc}{lambda_0}), the work function is inversely proportional to the wavelength.
So metal A with lower work function has a higher threshold wavelength.

8. Give the unit of work function.
a) Electron volt
b) Joule
c) Hertz
d) Watt
Answer: a
Clarification: One electron volt is the kinetic energy gained by an electron when it is accelerated through a potential difference of 1 volt.
Therefore, 1 eV = 1.602 × 10^-19 J.

9. What is the frequency of a photon whose energy is 66.3 eV?
a) 196 × 10¹⁶ Hz
b) 1336 × 10¹⁶ Hz
c) 1.6 × 10¹⁶ Hz
d) 16 × 10¹⁶ Hz
Answer: c
Clarification: Frequency can be calculated as follows:
V = (frac {E}{h})
V = (frac {(66.3 times 1.6 times 10^{-19})}{(6.63 times 10^{-34})})
V = 1.6 × 10¹⁶Hz

10. Calculate the energy of a photon of wavelength 6600 angstroms.
a) 30 × 10^-19 J
b) 3 × 10^-19 J
c) 300 × 10^-19 J
d) 3000 × 10^-19 J
Answer: b
Clarification: Given: λ = 6600 angstrom = 6600 × 10^-10 m.
Energy of a photon = (frac {hc}{lambda })
E = (frac {(6.6 times 10^{-34} times 3 times 10^8)}{(6600 times 10^{-10})})
E = 3 × 10^-19 J

Physics Quiz for IIT JEE Exam on “Classification of Metals, Conductors and Semiconductors”.

1. The manifestation of the band structure in solids is due to which of the following?
a) Heisenberg’s uncertainty principle
b) Pauli’s exclusion principle
c) Bohr’s correspondence principle
d) Boltzmann’s law
Answer: b
Clarification: The electronic configuration of atoms and consequently their manifestation of the band structure of solids can be well explained based on Pauli’s exclusion principle.

2. The energy band gap is maximum in which of the following?
a) Metals
b) Superconductors
c) Insulators
d) Semiconductors
Answer: c
Clarification: The bandgap is maximum in insulators. This makes it difficult for electrons to move to the conduction band. This is contrary to metals and superconductors, which has minimum band gaps facilitating the movement of electrons.

3. At absolute zero, Si acts as which of the following?
a) Non-metal
b) Metal
c) Insulator
d) Superconductor
Answer: c
Clarification: At absolute zero, the element, silicon (Si) acts as an insulator due to the absence of free electrons in the conduction band. The other options are not valid.

4. A piece of copper and another of germanium are cooled from room temperature to 77 K. what will impact on the resistance of each of them?
a) Each of these decreases
b) Copper strip increases and that of germanium decreases
c) Copper strip decreases and that of germanium increases
d) Each of these increases
Answer: c
Clarification: With the decrease of temperature, the resistance of copper, which is a metallic conductor, will decrease, whereas the resistance of germanium, which is a semiconductor will increase.

5. In semiconductors at room temperature, which of the following is likely to happen?
a) The valence band is partially empty and the conduction band is partially filled
b) The valence band is filled and the conduction band is partially filled
c) The valence band is filled
d) The conduction band is empty
Answer: a
Clarification: In semiconductors at room temperature, the valence band is partially empty and the conduction band is partially filled. The other options are not suitable according to the room temperature condition of semiconductors.

6. At absolute zero temperature, a semiconductor acts as a conductor.
a) True
b) False
Answer: b
Clarification: At absolute zero temperature, a semiconductor does act as an insulator. When the electron gains enough energy to participate in conduction, the electron is at a low energy state. Therefore, the semiconductors pretend like an insulator.

7. Which of the following is true regarding insulators?
a) The valence band is partially filled with electrons
b) The conduction band is partially filled with electrons
c) The conduction band is filled with electrons and valence band empty
d) The conduction band is empty and valence band is filled with electrons
Answer: a
Clarification: In insulators, the conduction band is empty and the valence band is filled with electrons. The other options do not correspond to the concept of insulators, but rather conductors and semiconductors.

8. Crystalline solids are which of the following?
a) Anisotropic
b) Isotropic
c) Amorphous
d) Unipotential
Answer: a
Clarification: Crystalline solids are anisotropic as they show different physical properties in a different direction. Therefore, they are neither isotropic, amorphous, nor unit potential in nature.

9. Which of the following is an amorphous solid?
a) Glass
b) Diamond
c) Salt
d) Sugar
Answer: a
Clarification: Glass is an amorphous solid. It is a non-crystalline solid in which the atoms and molecules are not organized in a definite lattice pattern. The other options are crystalline.

10. At which temperature, a pure semiconductor behaves slightly as a conductor?
a) Low temperature
b) Room temperature
c) High temperature
d) Supercritical temperature
Answer: c
Clarification: A pure semiconductor behaves slightly as a conductor at high temperatures. Their resistivity increase as temperature increases. Therefore, at high temperatures, semiconductors slightly act as a conductor.

Physics Quiz for IIT JEE Exam,

Physics Multiple Choice Questions on “Van de Graff Generator”.

1. What is the use of a Van de Graff generator?
a) Van de Graff generator is used to create a large amount of current
b) Van de Graff generator is used to create a small amount of voltage
c) Van de Graff generator is used to create a large amount of static electricity
d) Van de Graff generator is used to create a small amount of resistance
Answer: c
Clarification: Van de Graff generator is used to create a large amount of static electricity. A Van de Graff generator uses static electricity and a moving belt to charge a large metal sphere to a very high voltage. As the belt moves, electrons move from the rubber belt to the silicon roller, causing the belt to become positively charged and the roller to become negatively charged. As a result, it builds up positive charge.

2. Find the true statement.
a) A Van de Graff generator produces large voltage and less current
b) A Van de Graff generators produces large resistance and less voltage
c) A Van de Graff generators produces large current and large resistance
d) A Van de Graff generators produces large current and less voltage
Answer: a
Clarification: A Van de Graff generators produces large voltage and less current. A Van de Graff generator is an electrostatic generator which creates very high electric potentials. It produces very high voltage direct current electricity at low current levels.

3. What is the order of potential difference built up by the Van de Graff generator?
a) Potential difference of the order of hundreds
b) Potential difference of the order of several million volts
c) Potential difference of the order of thousands
d) Potential difference of the order of tens
Answer: b
Clarification: A Van de Graff generator, by means of a moving belt and suitable brushes, transfers charge continuously to a large spherical conducting shell. As a result, a potential difference of the order of several million volts is built up and this can be used for accelerating charged particles.

4. A Van de Graff generator can kill a person.
a) True
b) False
Answer: b
Clarification: Van de Graff generators typically produce a very small amount of current, i.e. the current produced is in microamperes. Therefore, an accidental shock from a Van de Graff generator may be startling and it may be painful, but it will not cause serious harm to most individuals, let alone kill, even at a high voltages.

5. When was Van de Graff generator invented and by whom?
a) 1944, Robin Van de Graff
b) 1932, Robert Van de Graff
c) 1933, Robin Van de Graff
d) 1933, Robert Van de Graff
Answer: d
Clarification: Van de Graff generator was invented by Robert Jemison Van de Graff on November 28, 1933. Robert Jemison invented the Van de Graff generator, which is a kind of high-voltage electrostatic generator that accelerates particles, while he was doing his PhD in Princeton University.

Physics Multiple Choice Questions on “Magnetic Force”.

1. What is the space around a current-carrying conductor, in which its magnetic effect can be experienced called?
a) Electric field
b) Magnetic pole
c) Magnetic field
d) Charge distribution
Answer: c
Clarification: The space around a current-carrying conductor, in which its magnetic effect can be experienced is called the magnetic field. When a current is passed through a conductor, it modifies the space around the conductor and forms a magnetic field.

2. Give the SI unit of the magnetic field.
a) Ampere
b) Tesla
c) Oersted
d) Weber
Answer: b
Clarification: The SI unit of the magnetic field is tesla, named after the great scientist Nikola tesla. 1 tesla is 10⁷ times the magnetic field produced by a conducting wire of length one metre and carrying a current of one ampere at a distance of one metre from it and perpendicular to it.

3. What is the force exerted by a stationary charge when it is placed in a magnetic field?
a) Zero
b) Maximum
c) Minimum
d) Depends on the strength of the magnetic field
Answer: a
Clarification: A stationary charge does not produce any magnetic field and it does not suffer any interaction against the external magnetic field. Hence the force exerted is zero.

4. What is the work done by the magnetic field on a moving charged particle?
a) Maximum
b) Minimum
c) Depends on the strength of the magnetic field
d) Zero
Answer: d
Clarification: As the magnetic force acts in a direction perpendicular to the direction of the velocity or the direction of motion of the charged particle, so the work done is zero.
W = F × dl × cos 90^°
W = 0.

5. The north pole of a magnet is brought near a stationary negatively charged conductor. What is the force experienced by it at the poles?
a) Maximum
b) Minimum
c) Zero
d) Depend on the nature of the conductor
Answer: c
Clarification: The north pole of a magnet will not experience any force. This is because a stationary charge does not produce any magnetic field. Therefore, the force experienced by the magnet at the poles is zero.

6. When a charge moves parallel or antiparallel to the direction of the magnetic field, it experiences a maximum force. State true or false.
a) True
b) False
Answer: b
Clarification: When a charge moves parallel or antiparallel to the direction of the magnetic field, it experiences a minimum (zero) force.
When θ = 0^o or 180^o,
F_m = qvBsinθ = qvB (0) = 0.

7. Identify the condition under which the force acting on a charge moving through a uniform magnetic field is minimum.
a) θ = 90^o
b) θ = 180^o
c) θ = 270^o
d) θ = 340^o
Answer: b
Clarification: When θ = 0^o or 180^o,
F_m = qvBsinθ = qvB (0) = 0.
So when a charge moves parallel or antiparallel to the direction of the magnetic field, it experiences a minimum force.

8. Identify the condition under which the force acting on a charge moving through a uniform magnetic field is maximum.
a) θ = 90^o
b) θ = 180^o
c) θ = 0^o
d) θ = 360^o
Answer: a
Clarification: When θ = 90^o → F_m is maximum.
Thus a charge experiences a maximum force when it moves perpendicular to the direction of the magnetic field. So, this is the condition when the force experienced is maximum.