Category: Class 12 Physics Objective Questions

Physics Question Papers for Class 12 on “Reflection of Light by Spherical Mirrors”.

1. A concave mirror is held in water. What should be the change in the focal length of the mirror?
a) Halved
b) Doubled
c) Remains the same
d) Increases exponentially
Answer: c
Clarification: No change. The focal length of a concave mirror does not depend on the nature of the medium. Therefore, the focal length of the mirror, even after holding it in water, will remain the same.

2. Two concave mirrors have the same focal length but the aperture of one is larger than that of the other. Which mirror forms the sharper image and why?
a) Plane
b) Concave
c) Convex
d) Prism
Answer: b
Clarification: The concave mirror with a smaller aperture forms the sharper image because it is free from spherical aberration. So, even though both have the same focal length, the change of aperture affects the formation of the image.

3. A man stands in front of a mirror of special shape. He finds that his image has a very small head, a fat body, and legs of normal size. What can we say about the shapes of the three parts of the mirror?
a) Convex, Concave, Plane
b) The plane, Concave, Convex
c) Concave, Convex, Plane
d) Convex, Plane, Concave
Answer: a
Clarification: The upper part of the mirror is convex. The middle part of the mirror is concave. The lower part of the mirror is plane. The image formed by a plane mirror will not have any changes, whereas that will not be the case for concave and convex mirrors.

4. Can the relative refractive index of a medium w.r.t. another medium be less than unity?
a) Insufficient data
b) Indefinite
c) No
d) Yes
Answer: d
Clarification: Yes. The refractive index of a medium concerning another medium can be less than unity. For example, the relative refractive index of water concerning glass is less than unity.

5. Which of the following causes refraction of light?
a) Change in the density of light from one medium to another
b) Change in viscosity of light from one medium to another
c) Change in the speed of light from one medium to another
d) Change in direction of light from one medium to another
Answer: c
Clarification: Light travels at different speeds in a different medium. The bending or refraction of light occurs due to the change in the speed of light as it passes from one medium to another. Other statements are not true regarding the refraction of light.

6. Frequency and wavelength change when light passes from a rarer to a denser medium.
a) True
b) False
Answer: b
Clarification: When light passes from a rarer to a denser medium, the wavelength of light changes but the frequency remains unchanged. Therefore, both frequency and wavelength do not change when light passes from a rarer to a denser medium.

7. The magnification of ‘X’ is more than unity. Identify X.
a) Convex mirror
b) Concave mirror
c) Plane mirror
d) Prism
Answer: b
Clarification: If the magnification of the mirror is one, then it is a plane mirror. If magnification is more than one, the mirror is concave. If magnification is less than one, then the mirror is convex. So the X required in this question is a concave mirror.

8. What happens to the shining of diamond if it is dipped in a transparent oil?
a) Reduces
b) Increases
c) Remains the same
d) Becomes colorless
Answer: a
Clarification: As the critical angle for diamond – oil interface is greater than that for the diamond – air interface, the shining of diamond reduces when it is dipped in transparent oil.

9. Calculate its focal length if the radii of curvature of a double convex lens are 10 cm, and 20 cm and its refractive index is 1.75.
a) 50 cm
b) 13 cm
c) 8.9 cm
d) 8 cm
Answer: c
Clarification: (frac {1}{f}) = (μ-1)( [ frac {1}{R_1} – frac {1}{R_2} ] )
(frac {1}{f}) = (1.75 – 1)( [ frac {1}{10} – frac {1}{(-20)} ] )
(frac {1}{f} = frac {9}{80})
f = +8.89 cm ≈ 8.9 cm.

10. If the radius of curvature of each face of the biconcave lens, made of glass of refractive index 1.25 is 50 cm, then what is the focal length of the lens in air?
a) 30 cm
b) -30 cm
c) 5 cm
d) -5 cm
Answer: b
Clarification: (frac {1}{f}) = (μ-1)( [ frac {1}{R_1} – frac {1}{R_2} ] )
(frac {1}{f}) = (1.25 – 1)( [ frac {1}{-50} – frac {1}{50} ] )
(frac {1}{f} = frac {-1}{100})
f = -100 cm
Therefore, the focal length of the lens in air is -100 cm.

Physics Question Papers for Class 12,

Physics Multiple Choice Questions on “Einstein’s Photoelectric Equation : Energy Quantum of Radiation”.

1. Two beams, one of red light and the other of blue light, of the same intensity are incident on a metallic surface to emit photoelectrons. Which emits electrons of greater frequency?
a) Both
b) Red light
c) Blue light
d) Neither
Answer: c
Clarification: From Einstein’s photoelectric equation, we have
(frac {1}{2}) mv² = h(v – v₀).
Since the frequency of blue light is greater than that of the red light, blue light emits electrons of greater kinetic energy.

2. If the intensity of incident radiation in a photo-cell is increased, how does the stopping potential vary?
a) Increases
b) Remains the same
c) Decreases
d) Infinite
Answer: b
Clarification: There is no effect on stopping potential. The intensity of incident radiation is independent of stopping potential. Therefore, even if the incident radiation in a photo-cell is increased, the stopping potential remains unchanged.

3. If the frequency of the incident radiation is equal to the threshold frequency, what will be the value of the stopping potential?
a) 0
b) Infinite
c) 180 V
d) 1220 V
Answer: a
Clarification: From Einstein’s photoelectric equation, we have
(frac {1}{2}) mv² = h(v – v₀).
When v = v₀,
hv₀ = hv₀ + eV₀
Therefore, V₀ = 0.

4. How does retarding potential vary with the frequency of light causing photoelectric effect?
a) Infinite
b) Zero
c) Decreases
d) Increases
Answer: d
Clarification: The stopping potential is directly proportional to the frequency of light.
Hence, the stopping potential increases with an increase in the frequency of the incident light.

5. If the intensity of the radiation incident on a photo-sensitive plate is doubled, how does the stopping potential change?
a) Increases
b) Decreases
c) No effect
d) Infinite
Answer: c
Clarification: No effect. This is because the intensity of radiation incident on a photo-sensitive plate is independent on stopping potential. So, the stopping potential remains the same, even if the intensity is doubled.

6. The stopping potential in photoelectric emission does not depend upon the frequency of the incident radiation.
a) True
b) False
Answer: b
Clarification: The stopping potential depends on the frequency of incident radiation. Above the threshold frequency, the stopping potential is directly proportional to the frequency of incident radiation.

7. The maximum kinetic energy of a photoelectron is 3 eV. What is the stopping potential?
a) 0 V
b) 3 V
c) 9 V
d) 12 V
Answer: b
Clarification: Stopping potential is given as:
V₀ = (frac {K_{max}}{e})
V₀ = (frac {3eV}{e})
V₀ = 3 V
Therefore, the stopping potential is 3 V.

8. The stopping potential in an experiment on the photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?
a) 1.5 eV
b) 3 eV
c) 4.5 eV
d) 6 eV
Answer: a
Clarification: K_max = eV₀
K_max = 1.6 × 10^-19 C × 1.5 V
K_max = 24 × 10^-19 J
It can also be converted into units of electron volts that will be:
K_max = 1.5 eV

9. For a photosensitive surface, the work function is 3.3 × 10^-19 J. Calculate the threshold frequency.
a) 15 × 10¹⁴ Hz
b) 25 × 10¹⁴ Hz
c) 5 × 10¹⁴ Hz
d) 55 × 10¹⁴ Hz
Answer: c
Clarification: Threshold frequency is given as:
v₀ = (frac {W_0}{h})
v₀ = (frac {W_0}{h})
v₀ = (frac {(3.3 times 10^{-19})}{(6.6 times 10^{-34})})
v₀ = 5 × 10¹⁴ Hz.

10. Calculate the kinetic energy of a photoelectron (in eV) emitted on shining light of wavelength 6.2 × 10^-6 m on a metal surface. The work function of the metal is 0.1 eV.
a) 10 eV
b) 0.1 eV
c) 0.01 eV
d) 1000 eV
Answer: b
Clarification: K_max = (( frac {hc}{lambda })) – W₀
K_max = ( [ frac {(6.6 times 10^{-34} times 3 times 10^8 )}{(6.2 times 10^{-6} times 1.6 times 10^{-19}) }] ) – 0.1.
K_max = 0.2 – 0.1
K_max = 0.1 eV

Physics Multiple Choice Questions on “Semiconductor Diode”.

1. Why is there a sudden increase in current in Zener diode?
a) Due to the rupture of ionic bonds
b) Due to rupture of covalent bonds
c) Due to viscosity
d) Due to potential difference
Answer: b
Clarification: The sudden increase in current in a Zener diode is due to the rupture of the many covalent bonds present. Therefore, the Zener diode should be connected in reverse bias.

2. What is the semiconductor diode used as?
a) Oscillator
b) Amplifier
c) Rectifier
d) Modulator
Answer: c
Clarification: Semiconductor diode can be used as a rectifier. The function of a rectifier is that it converts an alternating current into direct current by allowing the current to pass through in one direction.

3. What is an oscillator?
a) An amplifier with a large gain
b) An amplifier with negative feedback
c) An amplifier with positive feedback
d) An amplifier with no feedback
Answer: c
Clarification: An oscillator is considered as an amplifier with positive feedback. It converts direct current from a power supply to an alternating current signal. It produces an alternating waveform without any input.

4. What is rectification?
a) Process of conversion of ac into dc
b) Process of conversion of low ac into high ac
c) Process of conversion of dc into ac
d) Process of conversion of low dc into high dc
Answer: a
Clarification: Rectification is the process of conversion of alternating current into direct current. The conversion first powers to alternating current then use a transformer to change the voltage, and finally rectifies power back to direct current.

5. What is a Zener diode used as?
a) Oscillator
b) Regulator
c) Rectifier
d) Filter
Answer: b
Clarification: Zener diode can be used as a voltage regulator. They can also be used as shunt regulators to regulate the voltage across small circuits. Zener diodes are always operated in a reverse-biased condition.

6. Forward biasing of p-n junction offers infinite resistance.
a) True
b) False
Answer: b
Clarification: No, this is a false statement. Forward biasing of p-n junction offers low resistance. In the case of an ideal p-n junction, the resistance offered is zero. So, forward biasing does not offer any resistance.

7. When a junction diode is reverse biased, what causes current across the junction?
a) Diffusion of charges
b) Nature of material
c) Drift of charges
d) Both drift and diffusion of charges
Answer: c
Clarification: The reverse current is mainly due to the drift of charges. It is due to the carriers like holes and free electrons passing through a square centimeter area that is perpendicular to the direction of flow.

8. Identify the condition for a transistor to act as an amplifier.
a) The emitter-base junction is forward biased and the base-collector junction is reverse biased
b) No bias voltage is required
c) Both junctions are forward biased
d) Both junctions are reverse biased
Answer: a
Clarification: In order to use a transistor as an amplifier the emitter-base junction is set up as forward biased and the base-collector junction is set up as reverse biased. This is the criteria for making a transistor function as an amplifier.

9. What can a p-n junction diode be used as?
a) Condenser
b) Regulator
c) Amplifier
d) Rectifier
Answer: d
Clarification: A junction diode can be used as a rectifier. The rectifier converts alternating current into direct current. During the positive half cycle, the diode is forward biased and allows electric current through it.

10. What is a transistor made up of?
a) Chip
b) Insulator
c) Semiconductor
d) Metal
Answer: c
Clarification: A transistor is a semiconductor device. Transistors can work either as an amplifier or as a switch. It is used to amplify or switch electronic signals. A transistor is a solid-state device made up of silicon and germanium.

Physics Problems for Schools on “Magnetic Field due to a Current Element & Biot-Savart Law”.

1. Give the SI unit of magnetic permeability of free space.
a) T A m^-2
b) T A^-2 m
c) T A^-1 m
d) T A m²
Answer: c
Clarification: Magnetic permeability of free space is a measure of the amount of resistance encountered when forming a magnetic field in a classical vacuum. The SI unit of permeability is weber ampere^-1 metre^-1 (Wb A^-1 m^-1) or tesla ampere^-1meter (T A^-1 m).

2. State the rule that is used to find the direction of field acting at a point near a current-carrying straight conductor.
a) Cork rule
b) The right-hand thumb rule
c) Swimming rule
d) Flemings rule
Answer: b
Clarification: Right-hand thumb rule can be used to find the direction of the magnetic field at a point near a current-carrying conductor. Right hand rule states that, if the thumb of the right hand is in the direction of the current flow then, the curl fingers show the direction of the magnetic field.

3. Give the dimensional formula for magnetic permeability of free space.
a) [M L T^-2 A^-2]
b) [M² L T^-2 A^-2]
c) [M L² T^-2 A^-2]
d) [M^-1 L T^-2 A^-2]
Answer: a
Clarification: Magnetic permeability = Magnetic flux density × [Magnetic field strength]^-1.
μ = [M L⁰ T^-2 A^-1] × [M⁰ L^-1 T⁰ A¹]^-1
μ = [M L T^-2 A^-2].

4. A wire placed along the north-south direction carries a current of 8 A from south to north. Find the magnetic field due to a 1 cm piece of wire at a point 200 cm north-east from the piece.
a) 14 × 10^-9 T
b) 1004 × 10^-9 T
c) 204.4 × 10^-9 T
d) 1.4 × 10^-9 T
Answer: d
Clarification: dB = (frac {(mu_0 I dl sin (theta))}{(4pi r^2)}).
dB = (frac {(4 pi , times , 10^{-7} , times , 8 , times , 1 , times , 10^{-2} , times , sin 45^o}{(4 pi , times , 22)})
dB = 1.4 × 10^-9 T.

5. Which of the following is not a point of similarity between Biot-Savart law and Coulomb’s law.
a) Both fields depend inversely on the square of the distance from the source to the point of observation
b) They are not a universal law
c) The principle of superposition does not apply to both
d) Both are long-range fields
Answer: c
Clarification: The principle of superposition applies to both fields. This is because the magnetic field is linearly related to its source, namely, the current element and the electrostatic field is related linearly to its source, the electric charge.

6. The magnetic field due to a current element is minimum in a plane passing through the element when it is perpendicular to its axis. State true or false.
a) True
b) False
Answer: b
Clarification: If θ = 90^o → sin θ = 1
→ Then dB is maximum.
The magnetic field due to a current element is maximum in a plane passing through the element and perpendicular to its axis.

7. Give the SI unit of the magnetic field from Biot-Savart law.
a) Ampere
b) Tesla
c) Weber
d) Gauss
Answer: b
Clarification: The SI unit of the magnetic field is the tesla (T). One tesla is 10⁷ times the magnetic field produced by a conducting wire of length one meter and carrying a current of one ampere at a distance of one meter from it and perpendicular to it.

Physics Problems for Schools,