Category: Class 12 Physics Objective Questions

Physics MCQs for Class 12 on “Magnetisation and Magnetic Intensity”.

1. Identify the expression for magnetic intensity from the following.
a) H = (frac {B_o}{mu_o})
b) H = B_o × μ_o
c) H = (frac {2B_o}{mu_o})
d) H = (frac {B_o}{2mu_o})
Answer: a
Clarification: Magnetic intensity in vacuum is defined as the ratio of applied magnetic field (B_o) to the permeability of free space (μ_o). Magnetic intensity is denoted by H. The expression is given by:
H = (frac {B_o}{mu_o})

2. Pick out the expression for intensity of magnetization from the following.
a) I = (frac {M^2}{V})
b) I = (frac {M}{V^2})
c) I = (frac {V}{M})
d) I = (frac {M}{V})
Answer: d
Clarification: Intensity of magnetization is defined as the magnetic moment developed per unit volume, when a magnetic specimen is subjected to the magnetizing field. Intensity of magnetization is denoted by I. The expression is given by:
I=(frac {M}{V})

3. Which of the following is another term for magnetization?
a) Magnetic neutrality
b) Magnetic polarization
c) Magnetic power
d) Magnetic moment
Answer: b
Clarification: Magnetization is also termed as magnetic polarization. Magnetic polarization is basically a balance between the magnetic flux density in a space which is devoid of matter and the magnetic flux density in a space with matter, i.e. in a material.

4. Both magnetization and magnetic intensity have the same unit.
a) True
b) False
Answer: a
Clarification: Both magnetization and magnetic intensity have the same unit ➔ A/m.
Magnetic intensity (H) = (frac {B_o}{mu_o})
B = (frac {current}{length}) (ampere circuital law) ➔ unit of B₀ = (frac {A}{m}); μ₀ is dimensionless as it is a constant. Therefore, the unit of H is the same as B₀.
Magnetization (I) = (frac {M}{V})
I = (frac {Magnetic , moment}{Volume}) ➔ Unit of magnetic moment = Am²; Unit of volume = m³; I = (frac {Am^2}{m^3} = frac {A}{m})
Thus, the units of magnetization and magnetic intensity is the same.

5. Find the false statement.
a) Magnetic intensity is a vector quantity
b) Induced magnetization is a process where you can magnetize a non-magnetic material
c) Magnetic intensity and intensity of magnetization are the same.
d) Total intensity is the measurement from the magnetometer after a model of the earth’s normal magnetic field is removed
Answer: c
Clarification: The statement magnetic intensity and intensity of magnetization are the same is the false one. They are not the same. When a magnet is entering a magnetic field, then the poles of the magnet experiences certain forces. Magnetic intensity refers to the measure of these forces. But, the intensity of magnetization explains the change in the magnetic moment of a magnet as a function of volume. All the other statements are valid.

Physics MCQs for Class 12,

Physics Multiple Choice Questions on “Alternating Current – LC Oscillations”.

1. LED lights of chargers glow even after it is switched off. Which of the following causes this situation?
a) LC Oscillations
b) Phasors
c) Power of the AC circuit
d) Eddy currents
Answer: a
Clarification: This situation is caused due to LC Oscillations. These circuits consist capacitors and inductors, thus the energy keeps oscillating in the circuit even after the electric connection is disconnected. The combination of a capacitor and an inductor constitutes an LC oscillator circuit.

2. Find the true statement.
a) When a resistor is connected to an inductor, the electric current in the circuit undergoes LC oscillations
b) When a resistor is connected to a capacitor, the electric current in the circuit undergoes LC oscillations
c) When a charged capacitor is connected to an inductor, the electric current in the circuit and charge on the capacitor undergoes LC oscillations
d) When a charged capacitor is connected to an inductor, only the electric current in the circuit undergoes LC oscillations
Answer: c
Clarification: When we are connecting a charged capacitor to an inductor, the electric current in the circuit and charge on the capacitor undergoes LC oscillations. All the others are incorrect statements regarding LC Oscillations.

3. A capacitor of capacitance 5μF is charged to a potential difference of 20V. After that, it is connected across an inductor of inductance 0.5 mH. What is the current flowing in the circuit at a time when the potential difference across the capacitor is 10 V?
a) 1 A
b) 2 A
c) 5 A
d) 0.5 A
Answer: b
Clarification: Given: V₂ = 10 V; V₁ = 20 V; C = 5 μF; Inductance (I) = 0.5 mH
Initial charge on the capacitor (q₁) = C × V₁ = 5 × 10^-6 × 20 ……………….1
q₁ = 10^-4 C …………..B
The instantaneous charge on the capacitor as the capacitor discharges through the inductor ➔ q₂
q₂ = q₁cos (ωt) ➔ (frac {q_2}{q_2}) = cos (ωt) ………………..A
Also, q₂ = C × V₂ = 5 × 10^-6 × 10 …………………….2
q₂ = 0.5 × 10^-4 C
From 1 and 2 ➔ (frac {q_2}{q_2} = frac {V_2}{V_1}) ➔ (frac {q_2}{q_2}) = 0.5 = (frac {1}{2})
From equation A, we can equate as follows ➔ cos (ωt) = (frac {1}{2})
ωt = (frac {pi }{2}) rad ………………..3
For an LC circuit ➔ ω=(frac {1}{sqrt {LC}})
ω=20000(frac {rad}{s}) …………………….4
The current through the circuit is given as:
Current (I)=-(frac {dq}{dt})
Charge decreases with respect to time, so, (frac {dq}{dt}) obtained will be negative and this is why we add a negative sign to make a current positive.
Current = q₁ ω sinωt
Considering 3, 4 and B
Current = 10^-4 × 20000 × sin ((frac {pi }{2}))
Current = 2 A    [Sin((frac {pi }{2})) = 1]
Therefore, the current flowing through the circuit is 2 A.

4. LC oscillations only continues at a definite frequency.
a) True
b) False
Answer: b
Clarification: No, this statement is false. The process of LC oscillations, caused when we connect a charged capacitor to an inductor, continues at a definite frequency. But when the resistance in the LC circuit is zero, the LC oscillations continue at indefinite frequency. Therefore, indefinite frequency is also possible.

5. What is the most common application of LC oscillators?
a) Radio transmitters
b) Switches
c) Torch
d) Fans
Answer: a
Clarification: The most common application of LC oscillators is radio transmitters and receivers. LC oscillators have good phase noise characteristics as well as offer ease of implementation. Due to these factors, LC oscillators are most commonly used in radio-frequency circuits.

Physics Multiple Choice Questions on “Wave Optics – Diffraction”.

1. State the essential condition for diffraction of light to occur.
a) The size of the aperture must be less when compared to the wavelength of light
b) The size of the aperture must be more when compared to the wavelength of light
c) The size of the aperture must be comparable to the wavelength of light
d) The size of the aperture should not be compared to the wavelength of light
Answer: c
Clarification: Diffraction is the spreading out of waves as they pass through an aperture or around objects. The diffraction of light occurs when the size of the obstacle or the aperture is comparable to the wavelength of light.

2. What is the cause of diffraction?
a) Interference of primary wavelets
b) Interference of secondary wavelets
c) Reflection of primary wavelets
d) Reflection of secondary wavelets
Answer: b
Clarification: Diffraction occurs due to interference of secondary wavelets between different portions of a wavefront allowed to pass across a small aperture or obstacle. Interference can be either constructive or destructive. When interference is constructive, the intensity of the wave will increase.

3. What should be the order of the size of an obstacle or aperture for diffraction light?
a) Order of wavelength of light
b) Order of wavelength of obstacle
c) Order in ranges of micrometer
d) Order in ranges of nanometer
Answer: a
Clarification: The size of the obstacle or aperture should be of the order of the wavelength of light used. Therefore, the size of an obstacle is not of the order of obstacle, or micrometer or nanometer.

4. A small circular disc is placed in the path of light from a distant source. Identify the nature of the fringe produced.
a) Dual
b) Narrow
c) Dark
d) Bright
Answer: d
Clarification: Waves from the distant source are diffracted by the edge of the disc. These diffracted waves interfere constructively at the center of the shadow and produce a bright fringe. Therefore, the nature of the fringe produced is bright.

5. Single slit diffraction is completely immersed in water without changing any other parameter. How is the width of the central maximum affected?
a) Insignificant
b) Increases
c) Decreases
d) Becomes zero
Answer: c
Clarification: Wavelength of light in water decreases, so the width of the central maximum also decreases. This is the impact on the width of the central maximum when a single slit is completely immersed in water.

6. Diffraction is common in light waves.
a) True
b) False
Answer: b
Clarification: As the wavelength of light is much smaller than the size of the objects around us, so diffraction of light is not easily seen. Therefore, for diffraction of a wave, an obstacle or aperture of the size of the wavelength of light of the wave is needed. As the wavelength of light is of the order of 10-6m and (frac {obstacle}{aperture}) of this size are rare, diffraction is not common in light waves.

7. Determine the half angular width of the central maximum, if a wavelength of 1000 nm is observed when diffraction occurs from a single slit of 2 μm width.
a) 100^o
b) 30^o
c) 90^o
d) 150^o
Answer: b
Clarification: Half angular width of the central maximum is given by:
Sin θ = (frac {lambda }{d})
Sin θ = (frac {1000 times 10^{-9}}{2 times 10^{-6}})
Sin θ = 0.5.
Θ = 30^o

8. What will be the linear width of the central maximum on a screen that is kept 5 m away from the slit, if a light of wavelength 800 nm strikes a slit of 5 mm width.
a) 1.2 mm
b) 5.6 mm
c) 6.5 mm
d) 9.7 mm
Answer: a
Clarification: Linear width is given as:
β₀ = (frac {2Dlambda }{d})
β₀ = ( frac { ( 2 times 5 times 800 times 10^{-9} ) }{ (5 times 10^{-3} ) })
β₀ = 1.6 × 10^-3 m
β₀ = 1.6 mm
Therefore, the linear width of central maximum on a screen kept 5 m away from the slit is 1.6 mm.

Physics Multiple Choice Questions on “Nucleus – Radioactivity”.

1. Radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 years. The time, in years, after which one-fourth of the material remains?
a) 1080
b) 2430
c) 3240
d) 4860
Answer: a
Clarification: N = N₀e^{-(λ1 + λ2) t}.
4 = e^{(λ1 + λ2) t}.
t = (frac {(2 times 1620 times 810)}{(2430)}) = 1080 year.

2. Which of the following substances cannot be emitted by radioactive substances during their decay?
a) Protons
b) Neutrinos
c) Helium nuclei
d) Electrons
Answer: a
Clarification: Protons are not emitted by radioactive substances during their decay. They are positively charged subatomic particles found in the atomic nucleus. The others can be emitted by radioactive substances during their decay.

3. The electron emitted in β – radiation originates from where?
a) Inner orbits of atoms
b) Free electrons existing in nuclei
c) The decay of a neutron in nuclei
d) Photon escaping from the nucleus
Answer: c
Clarification: In β-emission, a neutron of nucleus decays into a proton and a β-particle.
The reaction given as:
₀n¹ ➔ ₁H² + _-1e⁰.

4. Find the probability that the nucleus of ₈₇Ra²²¹ undergoes decay after three half-lives, if its a radioactive substance which has a half-life of 6 days.
a) (frac {1}{6})
b) (frac {3}{2})
c) (frac {5}{6})
d) (frac {1}{2})
Answer: c
Clarification: After one-half life, (frac {N_0}{2}) sample remains and (frac {N_0}{2}) decays.
After three half-lives, (frac {N_0}{6}) sample remains and (frac {5N_0}{6}) decays.
Hence the probability that a nucleus undergoes decay after two half-lives is (frac {5}{6}).

5. A 300-day old radioactive substance shows an activity of 5000 dps, 150 days later its activity becomes 2500 dps. What was its initial activity?
a) 25000 dps
b) 20000 dps
c) 32000 dps
d) 5000 dps
Answer: b
Clarification: The expression is given as:
R₀ = 4R
4R = 4 × 5000
R₀ = 20000dps.

6. Emission of β-rays in radioactive decay results in the change of either mass or charge.
a) True
b) False
Answer: b
Clarification: No, this is a false statement. The emission of β-rays can result in changes in the charge of the nucleus but not the mass of the nucleus. Mass of the nucleus is always conserved in radioactive decay.

7. Which will be the unknown nucleus formed when ²²Ne₁₀ decays into two α-particles and an unknown nucleus?
a) Fluorine
b) Carbon
c) Neon
d) Oxygen
Answer: b
Clarification: The reaction is given as:
²²Ne₁₀ ➔ 2 ₂He⁴ + ₆X¹⁴.
Therefore, the unknown nucleus is ₆C¹⁴.

8. What is the half-time of a radioactive sample (in minutes), if its mean life is 200 s?
a) 0.69 min
b) 2 min
c) 2.57 min
d) 2.31 min
Answer: d
Clarification: T_1/2 = 0.693t
T_1/2 = 0.693 × 200s
T_1/2 = (frac {138.6}{60}) min
T_1/2 = 2.31 minutes

9. What will happen in a time of 7 hours, if a radioactive substance has an average life of 7 hours?
a) Half of the active nuclei decay
b) Less half of the active nuclei decay
c) More than half of the active nuclei decay
d) All active nuclei decay
Answer: c
Clarification: In one average life, i.e. at 7 hours, 63.2 % of the active nuclei will decay. Therefore, in a time of 7 hours, it can be considered that more than half of the active nuclei will decay.

10. A freshly prepared radioactive source of half-time 2h emits radiation of intensity which is 64 times the permissible safe level. Minimum time after which it would be possible to work safely with this source is which of the following?
a) 6 h
b) 12 h
c) 24 h
d) 20 h
Answer: b
Clarification: (frac {N}{N_0} = frac {1}{64}).
t = n T_1/2
t = 6 × 2hours
t = 12hours.

Physics Multiple Choice Questions on “Electric Flux”.

1. What is the dimension of electric flux?
a) [M L³ T^-3 I^-1]
b) [M L² T^-3 I^-1]
c) [M L³ T^-3 I¹]
d) [M L³ T³ I^-1]
Answer: a
Clarification: Electric flux=electric field intensity* area. The dimension of field intensity is [M L T^-3 I^-1] and the dimension of the area is [L²]. Therefore, the dimension of flux = [M L³ T^-3 I^-1]. This can also be justified that flux=potential*length. By putting the dimensions of potential and length, we can get the same result.

2. What is the unit of electric flux?
a) V/m
b) N/m
c) V*m
d) N/Coulomb
Answer: c
Clarification: Flux is equal to the product of field intensity and area of the surface. But field intensity multiplied by a length gives the unit of electric potential as E=-(frac {dV}{dx}). Therefore, flux also means electric potential multiplied by length. This gives us the unit V*m. There is another unit of flux N*m²*C^-1.

3. Which one is the correct expression of electric flux?
a) ∫ (vec{E}.dvec{s})
b) ∫ (vec{E^2}.dvec{s})
c) ∫ (vec{E}^{-1}.dvec{s})
d) ∫ (vec{E}.dvec{l})
Answer: a
Clarification: Electric flux is defined as the number of field lines crossing perpendicularly through a surface area. The number of electric field lines crossing through the unit cross-section area is known as electric field intensity (E). Therefore field lines crossing through small area ds are E.ds. Taking the integral gives the flux in the entire surface.

4. Electric flux will be maximum if the angle between the field lines and area vector is ______
a) 45 degree
b) 135 degree
c) 90 degree
d) 0 degree
Answer: d
Clarification: We know that the electric field and area both are vector quantity and the electric flux is expressed as (vec{E}.vec{s}) (taking dot product). But if the angle between the two vectors is θ then the formula becomes E.s.cosθ.Cosθ will be maximum if theta is zero, in all other cases, the value of Cosθ is less than 1. Therefore flux will be the maximum if the angle is 0 degrees.

5. Flux linked to a surface is said to be positive if the flux lines are coming out of the surface. The statement is ______
a) True
b) False
Answer: a
Clarification: Depending on the direction of electric flux lines i.e. electric field intensity, we can differentiate between positive and negative flux. If the flux lines are going inside a surface, the flux is said to be negative. But if the flux lines are coming out of the surface, the flux is said to be positive.

6. If a charge is placed outside a closed surface, flux due to that charge inside the surface will be ________
a) Positive
b) May be positive or negative, depending on the nature of the charge
c) Negative
d) Zero
Answer: d
Clarification: According to Gauss’s principle, if there is no charge bound inside a surface, net electric flux coming out of the surface will always be 0. In this case, the charge is kept outside the surface, so it will generate no field lines and hence no flux inside the surface. The situation will be the same if an electric dipole is placed inside a surface as dipole has equal positive and negative charges; the total charge inside the surface becomes 0.

7. Two separate charges q and 10q are placed inside two different spheres. In which case, the electric flux will be greater?
a) Flux will be same in both the cases
b) 1^st sphere
c) 2^nd sphere
d) No flux in any of the spheres
Answer: c
Clarification: The electric field line is directly proportional to the charge bound inside a sphere. We know that, field lines are the measure of electric flux i.e. number of field lines crossing through a surface are know an electric flux. Therefore flux will be lesser in the case of q charge and will be 10 times in case of a 10q charge.

8. Which of the following law explains the relation between the charge inside a surface and electric flux?
a) Gauss’s Law
b) Coulomb’s Law
c) Faraday’s Law
d) Pascal’s Law
Answer: a
Clarification: Gauss’s Law gives the relation between electric flux and charges inside a surface. It states that electric flux coming out from a closed surface is equal to (frac {1}{varepsilon}) times the charge inside the surface. Coulomb’s Law explains the force between two charges and Pascal’s Law is related to fluids.

9. Flux coming out from a balloon of radius 10 cm is 1.0*10³ N.m².C-1^-1. If the radius of the balloon is doubled, the flux coming out from the balloon will be _______
a) 0.5 times
b) 2 times
c) Same
d) 4 times
Answer: c
Clarification: If the radius of the balloon is 2 times the initial value, the surface area of the balloon will be 4 times the initial value because of surface area=4π(radius)². But in this case, flux coming out from the balloon is dependent only upon the charge inside the balloon, not on the area of the surface. As charge enclosed in the balloon is the same in both the cases, the flux will also be the same.