Physics Class 12 CBSE Solved Question Paper 2024 – Section C

Answer: Einstein’s photoelectric equation is K.E. = hν − ϕ. Millikan proved its validity by experimentally verifying the linear relationship between the frequency of incident light and the kinetic energy of the ejected electrons.
Solution: Einstein’s Photoelectric Equation is given by:K.E. = hν − ϕ,

where:

• K.E. is the maximum kinetic energy of the ejected electrons,
• h is Planck’s constant,
• ν is the frequency of the incident light,
• ϕ is the work function of the material.

Millikan’s experiment involved shining light of varying frequencies onto a metal surface and measuring the stopping potential required to halt the emitted electrons. He observed that the kinetic energy of the ejected electrons depended linearly on the frequency of the incident light, and this experimental data matched Einstein’s equation perfectly, validating the photoelectric equation.

Answer: The threshold frequency is the minimum frequency of incident radiation required to provide the energy needed to overcome the work function of the material and cause photoelectric emission.
Solution: The threshold frequency ν₀ is the minimum frequency of incident light required to eject electrons from a material. This is because the energy of the incident photon must be greater than or equal to the work function ϕ of the material to overcome the attractive forces holding the electron to the surface.According to Einstein’s equation:

hν₀ = ϕ.

If the frequency ν of the incident light is less than ν₀, the photon’s energy hν is not sufficient to eject electrons, and no photoelectric emission occurs.

Answer: The electric flux is a measure of the number of electric field lines passing through a surface, and its dimensions are [ML³T⁻³I⁻¹].

Solution: Electric Flux Φₑ is a measure of the number of electric field lines passing through a given surface. It is mathematically defined as:

Φₑ = E (vector) ⋅ A (vector) = E A cosθ

where:

• E (vector) is the electric field,
• A (vector) is the area vector
• θ is the angle between the electric field and the area.

Dimensions of Electric Flux: The electric field has dimensions of [MLT⁻³I⁻¹] and area has dimensions of [L²], so the dimensions of electric flux are [ML³T⁻³I⁻¹].

Answer: The electric flux through the surface is 8 × 10⁻³ Nm²/C.
Solution: The electric flux Φₑ through a surface is given by:Φₑ = E(vector) A(vector)

where A(vector)’s the area vector, whose magnitude is A = side² and direction is given by the unit normal vector n̂.

The area of the square is:

A = (1 cm)² = 1 × 10⁻⁴ m².

The dot product E(vector).A(vector)’s:

Φₑ = E A cosθ = E A (Ê ⋅ n̂),

where:

E = 100 N/C, A = 1 × 10⁻⁴ m², Ê = î, and n̂ = 0.8 î + 0.6 k̂.

The dot product Ê ⋅ n̂ is (1⋅0.8) + (0⋅0.6) = 0.8.

Thus, the electric flux Φₑ is:

Φₑ = E A⋅(0.8) = (100) × (1 × 10⁻⁴) × (0.8) = 8 × 10⁻³ Nm²/C.

Answer: Lenz’s law states that the induced current opposes the change in magnetic flux, which is a manifestation of the law of conservation of energy.
Solution: Lenz’s Law states that the direction of the induced current in a closed loop is such that it opposes the change in magnetic flux that produced it. Mathematically, it is represented as:ε = − dΦB/dt,

where ε is the induced emf and ΦB is the magnetic flux.

This law ensures that the induced current creates a magnetic field that opposes the change in the original magnetic field, thus conserving energy. For example, if the magnetic flux through a loop increases, the induced current will flow in such a direction as to create a magnetic field opposing the increase. Similarly, if the flux decreases, the induced current will try to increase it, thus opposing the change.

Lenz’s law is consistent with the law of conservation of energy because it prevents a system from generating energy out of nothing.

Answer: The emf induced between the center and the end of the rod is approximately 377 V.
Solution: The emf induced in a rotating rod in a magnetic field is given by the formula:ε = ½ B ω L²,

where:

• B = 2 T (magnetic field),
• ω = 2πf = 2π×60 = 120π rad/s (angular velocity),
• L = 1 m (distance from center to the end of the rod, which is half the total length).

Substituting the values:

ε = ½ × 2 × 120π × (1)² = 120π V ≈ 377 V.

Answer: Ampere’s circuital law relates the integral of the magnetic field around a closed loop to the current passing through the loop.
Solution:

Answer: The net magnetic field is 1 × 10⁻⁵ T and is directed towards the wire carrying 10 A.
Solution: The magnetic field due to a long straight wire at a distance r from the wire is given by:B = μ₀ I / 2πr,

where I is the current, r is the distance from the wire, and μ₀ = 4π × 10⁻⁷ T⋅m/A.

Since the point is midway between the two wires, the distance from each wire to the point is r = 0.2/2 = 0.1 m.

For the first wire (carrying 5 A):

B₁ = 4π × 10⁻⁷ × 5 / 2π × 0.1 = 1 × 10⁻⁵ T.

For the second wire (carrying 10 A):

B₂ = 4π × 10⁻⁷ × 10 / 2π × 0.1 = 2 × 10⁻⁵ T.

The magnetic fields are in opposite directions because the currents are in the same direction. Hence, the net magnetic field is:

Bₙₑₜ = B₂ − B₁ = 2 × 10⁻⁵ − 1 × 10⁻⁵ = 1 × 10⁻⁵ T.

The direction of the net magnetic field is towards the wire carrying 10 A.

Answer: (i) Infrared radiation; (ii) Ultraviolet radiation.
Solution:

(i) Heat waves are typically referred to as infrared radiation. Infrared waves are emitted by hot objects and are responsible for heat transfer through radiation.

(ii) Ultraviolet (UV) radiation is absorbed by the ozone layer in the atmosphere. The ozone layer effectively blocks harmful UV radiation, particularly the shorter wavelength UV-B and UV-C rays, which can damage living tissues.

Answer:

• Infrared production: Heating objects (e.g., heaters, infrared LEDs).
• Infrared detection: Thermocouples, infrared cameras.
• Ultraviolet production: Sun, mercury vapor lamps, electrical discharges.
• Ultraviolet detection: UV photodiodes, photographic plates.

Solution:

(i) Production of Infrared Radiation: Infrared radiation is produced by any object with a temperature above absolute zero. Specifically, it can be produced by heating objects (such as electric heaters) or through infrared LEDs and lasers.

(ii) Detection of Infrared Radiation: Infrared radiation can be detected using thermocouples, bolometers, or infrared cameras that convert infrared energy into a visible image or signal.

(iii) Production of Ultraviolet Radiation: Ultraviolet radiation is produced by the sun, gas-discharge lamps (e.g., mercury vapor lamps), and certain chemical reactions, including fluorescence and electrical discharges in gases.

(iv) Detection of Ultraviolet Radiation: Ultraviolet radiation can be detected using UV-sensitive photodiodes, photoelectric cells, or photographic plates coated with UV-sensitive chemicals.

Answer: The p-n junction diode’s unidirectional conductivity, threshold voltage, and reverse breakdown voltage make it suitable for rectification.
Solution:

• Unidirectional Conductivity: The diode allows current to flow easily in the forward direction but blocks current flow in the reverse direction, essential for converting AC to DC in rectifier circuits.
• Threshold Voltage: In forward bias, the diode conducts only when the applied voltage exceeds a certain threshold (typically 0.7 V for silicon diodes).
• Reverse Breakdown Voltage: The diode blocks current in reverse bias until the reverse breakdown voltage is reached, protecting the circuit in normal operating conditions.

Answer: In a full-wave rectifier, both diodes conduct in alternate half-cycles, ensuring continuous current flow in the same direction across the load resistor, converting the AC signal into pulsating DC.
Solution:Full-Wave Rectifier Circuit:

A full-wave rectifier uses two diodes and a center-tapped transformer to convert the entire AC waveform (both positive and negative halves) into pulsating DC.

Working:

• During the positive half of the AC input, diode D₁ is forward biased and conducts, allowing current to flow through the load resistor Rₗ. Diode D₂ is reverse biased and does not conduct.
• During the negative half of the AC input, diode D₂ is forward biased and conducts, allowing current to flow through Rₗ in the same direction as before. Diode D₁ is reverse biased and does not conduct.

Thus, the output across Rₗ is a pulsating DC that contains both halves of the input AC signal. A capacitor can be used to smooth the output.

Answer:

• (a) A doped semiconductor is neutral because the dopant atoms introduce no net charge.
• (b) In a p-n junction, drift and diffusion currents cancel out at equilibrium, so no net current flows.
• (c) The reverse current is limited by the minority carriers and remains constant despite changes in reverse voltage.

Solution:

• (a) A doped semiconductor is electrically neutral: When a semiconductor is doped, either donor atoms (n-type doping) or acceptor atoms (p-type doping) are introduced, which provide free electrons or holes, respectively. However, these atoms themselves are electrically neutral (donor atoms are neutral before ionization), and they introduce no net charge to the material. The overall number of positive and negative charges remains balanced, maintaining electrical neutrality.
• (b) In a p-n junction under equilibrium, there is no net current: At equilibrium, a depletion region forms at the junction where the electric field opposes the diffusion of charge carriers. The drift current and the diffusion current balance each other. Thus, there is no net movement of charge carriers, and hence no net current flows across the junction.
• (c) In a diode, the reverse current is practically not dependent on the applied voltage: In reverse bias, the current is due to minority carriers (electrons in the p-region and holes in the n-region). The number of minority carriers is very small and relatively constant. Even as the reverse voltage increases, the reverse current remains nearly constant because the supply of minority carriers is limited. This reverse saturation current is determined by temperature and material properties, not the applied voltage.