Category: Class 12 Physics Objective Questions

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.

Physics Multiple Choice Questions on “Current Electricity – Meter Bridge”.

1. Two resistances are connected in two gaps of Meter Bridge. The balance is 20cm from the zero end. A resistance of 15 ohms is connected in series with the smaller of the two. The null point shifts to 40cm. What is the value of the bigger resistance?
a) 9
b) 18
c) 27
d) 36
Answer: d
Clarification: Let P be the smaller resistance and Q be the bigger resistance.
First case → (frac {P}{Q} = frac {20}{80} = frac {1}{4})
Second case → (frac {(P+15)}{Q} = frac {40}{60} = frac {2}{3})
Comparing both → (frac {P}{(P + 15)} = frac {1}{4} , times , frac {3}{2} = frac {3}{8})
8P = 3P + 45 → 5P = 45 → P = 9 ohms
Therefore, substituting in (frac {P}{Q} = frac {1}{4}) → (frac {9}{Q} = frac {1}{4}) → Q = 36 ohms.

2. What is the effect on null deflection of galvanometer, when the radius of the wire is tripled?
a) No change
b) Becomes half
c) Reduces by (frac {1}{3})
d) Thrice the original value
Answer: a
Clarification: For a balanced Meter Bridge (frac {P}{Q} = frac {x}{(100-x)}). From this, we can understand that the null deflection of galvanometer does not depend on the radius of the wire. So, even if the radius of the wire is tripled, the null deflection of the galvanometer undergoes no change.

3. A resistance of 5 ohms is connected across the gap of a Meter Bridge and an unknown resistance, greater than 5 ohms, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 50 cm. Neglecting any correction, what is the unknown resistance? The length of the wire is 150 cm.
a) 3 ohms
b) 10 ohms
c) 7 ohms
d) 5 ohms
Answer: b
Clarification: Let × be the unknown resistance.
First case → (frac {5}{x} = frac {l}{(150 – l)})
750 – 5l = xl ……….. 1
Second case → (frac {x}{5} = frac {(l + 50)}{(100 – l)})
100 – xl = 5l + 250
xl = 100x – 5l – 250 ………… 2
Comparing 1 and 2
750 – 5l = 100x – 5l – 250
100x = 750 + 250
100x = 1000
x = 10 ohms
Therefore, the unknown resistance is 10 ohms.

4. The sensitivity of the meter bridge is at the peak when all resistors have the same order.
a) True
b) False
Answer: a
Clarification: Yes, the sensitivity of the meter bridge is at the peak when all resistors have the same order. The sensitivity can be increased by keeping the current in the galvanometer high and this can be achieved by reducing the values of the resistors used in the Meter Bridge.

5. Which error is removed when the known and unknown resistances are interchanged in a Meter Bridge?
a) Percentage error
b) Measurement error
c) End error
d) Index error
Answer: c
Clarification: End error is removed when the known and unknown resistances are interchanged in a Meter Bridge. This additional length has a resistance known as end resistance. So, when an end error arises, it can be removed by interchanging the known and unknown resistances and taking the mean of the resistances determined.

Physics Multiple Choice Questions on “Magnetic Properties of Materials”.

1. Identify the expression for magnetic induction from the following.
a) B = μ_o(H+I)
b) B = μ_o(H×I)
c) B = μ_o(H-I)
d) B = μ_o((frac {H}{I}))
Answer: a
Clarification: Magnetic induction (B) is defined as the number of magnetic lines of induction crossing per unit area normally through the magnetic substance, and is given by:
B = μ_o(H+I)
The SI unit of magnetic induction is Tesla (T).

2. Which among the following is true about magnetic susceptibility?
a) It is the ratio of magnetic intensity to intensity of magnetization
b) The SI unit of magnetic susceptibility is Tesla (T)
c) It is the ratio of intensity of magnetization to magnetic intensity
d) It is the ratio of magnetic moment to volume
Answer: c
Clarification: Magnetic susceptibility of a magnetic substance is defined as the ratio of the intensity of magnetization (I) to the magnetic intensity (H). The expression is given by:
χ_m=(frac {I}{H})
Magnetic susceptibility measures the aptness of a magnetic substance to acquire magnetism. It is a unit less constant of magnetic substance.

3. Pick out the unit of magnetic permeability.
a) TmA
b) (frac {Tm}{A})
c) (frac {T}{mA})
d) (frac {Am}{T})
Answer: b
Clarification: Magnetic permeability (μ) of a magnetic substance is defined as the ratio of its magnetic induction (B) to the magnetic intensity (H). The expression for magnetic permeability is given as:
μ=(frac {B}{H})
Magnetic permeability is the ability of a magnetic substances to permit magnetic field lines to pass through it. The SI unit for this quantity is (frac {Tm}{A}).

4. Relative magnetic permeability is unit less.
a) True
b) False
Answer: a
Clarification: Yes, relative magnetic permeability is a unit less constant of magnetic substance. Relative magnetic permeability (μ_r) of a magnetic substance is defined as the ratio of its magnetic permeability to the permeability of free space. The expression for it is given as:
μ_r=(frac {μ}{mu_o})

5. The relative permeability of a medium is 0.050. What is its magnetic susceptibility?
a) 500
b) 501
c) 49.9
d) 499
Answer: d
Clarification: Given: relative permeability (μ_r) = 500
Required equation ➔ μ_r = 1+χ_m
χ_m = μ_r – 1
χ_m = 500 – 1
χ_m = 499
Therefore, the value of magnetic susceptibility is 499.

6. The magnetic susceptibility of a paramagnetic material at -80⁰C is 0.0085, then what is its value at -180⁰C?
a) 0.015
b) 0.016
c) 0.017
d) 0.018
Answer: c
Clarification: Given: χ_m1 = 0.0085; T₁ = -80⁰ C = -80 + 273 = 193 K; T₂ = -180⁰ C = -180 + 273 = 93 K
χ_m ∝ (frac {1}{T})
(frac {chi_{m2}}{chi_{m1}} , = , frac {T_1}{T_2})
(frac {chi_{m2}}{0.0085} , = , frac {193}{93})
χ_m2 = (frac {193 , times , 0.0085}{93})
χ_m2 = 0.017

7. A material has a permeability of 0.1 H/m when the magnetic intensity is 70 A/m. What will be the magnetic induction inside the material?
a) 7 T
b) 0.7 T
c) 70 T
d) 0.07
Answer: a
Clarification: Given: permeability (μ_o) = 0.1 H/m; Magnetic intensity (H) = 70 A/m
Required equation ➔ Magnetic induction (B) = μ_o × H
B = 0.1 × 70
B = 7 T
Therefore, the magnetic induction inside the material is 7 T.

8. Ferromagnetic materials does not show hysteresis.
a) True
b) False
Answer: b
Clarification: Ferromagnetic materials do show hysteresis. When a ferromagnetic material is magnetized in one direction, it will not relax back to zero magnetization when the imposed magnetizing field is removed. The phenomenon of hysteresis in ferromagnetic materials is the result of two effects ➔ rotation of magnetization and changes in size or number of magnetic domains.

9. Find the correct combination regarding relative permeability and magnetic susceptibility of a paramagnetic substance.
a) μ_r > 1, χ < 0
b) μ_r < 1, χ > 0
c) μ_r < 1, χ < 0
d) μ_r > 1, χ > 0
Answer: d
Clarification: The correct combination for paramagnetic substances is given as follows:
μ_r > 1, χ > 0
Paramagnetic materials have a small and positive value of magnetic susceptibility (χ). The relative permeability (μ_r) of paramagnetic materials is just greater than 1.