Plus Two Physics Previous Year Question Papers and Answers PDF HSSlive: Complete Guide (2010-2024)

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Why HSSlive Plus Two Physics Previous Year Question Papers PDFs Are Essential

Physics requires both conceptual clarity and systematic practice. HSSlive.co.in offers the most reliable collection of Plus Two Physics question papers that:

• Help you master the exact Kerala Higher Secondary Board examination pattern
• Reveal frequently tested topics and concepts from past papers
• Develop effective time management strategies
• Build confidence through targeted practice
• Identify your strengths and weak areas in different chapters

How to Download Plus Two Physics Previous Year Question Papers and Answers PDF from HSSlive

Quick Access Guide:

1. Visit the official HSSlive website: www.hsslive.co.in
2. Navigate to “Previous Question Papers” or “Question Bank” section
3. Select “Plus Two” from the class options
4. Choose “Physics” from the subject list
5. Download the PDF files for different years (2010-2024)

Pro Tip: Create a dedicated folder to organize your HSSlive Physics PDFs by year for structured revision.

Kerala Plus Two Physics Exam Pattern (Important for HSSlive PDF Users)

Understanding the exact question paper structure will help you extract maximum value from HSSlive PDFs:

Section	Question Type	Marks per Question	Number of Questions
Part A	Very Short Answer	1 mark	8 questions
Part B	Short Answer	2 marks	10 questions
Part C	Short Essay	3 marks	9 questions
Part D	Long Essay	5 marks	3 questions
Total		60 marks	30 questions

15 Plus Two Physics Previous Year Question Papers with Answers (HSSlive PDF Collection)

 

Plus Two Physics Previous Year Question Papers with Answers (2010-2024)

1. March 2024 Physics Question Paper with Answers

Question 1: What is the SI unit of electric field intensity? (1 mark) Answer: Newton per coulomb (N/C) or Volt per meter (V/m)

Question 2: A moving coil galvanometer of resistance 100 Ω gives full-scale deflection for a current of 1 mA. How will you convert it into a voltmeter of range 0-5 V? (3 marks) Answer:

• For a voltmeter, we need to connect a high resistance in series with the galvanometer.
• If R is the required resistance, then:
  • Current for full-scale deflection = 1 mA = 0.001 A
  • Voltage range = 5 V
  • Galvanometer resistance = 100 Ω
• Using V = I(R + G), where G is galvanometer resistance
• 5 = 0.001(R + 100)
• R + 100 = 5000
• R = 4900 Ω
• Therefore, a resistance of 4900 Ω must be connected in series with the galvanometer.

Question 3: Explain the working principle of a cyclotron with necessary diagram. (5 marks) Answer:

• Cyclotron is a device used to accelerate charged particles to high energies.
• Working principle: It works on the principle of Lorentz force and the fact that the time period of a charged particle in a magnetic field is independent of its velocity and radius of orbit.

Diagram: [A circular device with two D-shaped electrodes (Dees) with a gap between them, placed in a magnetic field perpendicular to the plane of Dees]

Working:

• The two D-shaped hollow conductors (Dees) are connected to an AC source.
• A uniform magnetic field is applied perpendicular to the plane of Dees.
• Charged particles (like protons) are released at the center.
• The particles move in a semicircular path inside the Dees due to the magnetic field.
• As particles reach the gap between Dees, the electric field accelerates them.
• The frequency of AC source is adjusted to match the cyclotron frequency: f = qB/2πm
• With each crossing of the gap, particles gain energy and move in larger semicircular paths.
• Eventually, they emerge as a high-energy beam.

Limitations:

• Cannot accelerate electrons (relativistic effects)
• Cannot accelerate neutral particles
• Upper limit to energy due to relativistic effects

2. March 2023 Physics Question Paper with Answers

Question 1: Write the expression for radius of nth Bohr orbit in hydrogen atom. (1 mark) Answer: rₙ = n²h²ε₀/πmZe²

Question 2: State Faraday’s laws of electromagnetic induction. (2 marks) Answer: First Law: Whenever there is a change in magnetic flux linked with a closed circuit, an emf is induced in the circuit. Second Law: The magnitude of induced emf is directly proportional to the rate of change of magnetic flux linkage. Mathematically, ε = -dΦ/dt

Question 3: A ray of light traveling in a medium of refractive index n₁ is incident on a plane surface separating it from another medium of refractive index n₂. Derive Snell’s law of refraction. (5 marks) Answer:

• Consider a ray of light traveling from medium 1 (with refractive index n₁) to medium 2 (with refractive index n₂).
• Let the angle of incidence be i and the angle of refraction be r.
• According to Fermat’s principle of least time, light follows the path that takes the least time.
• If v₁ and v₂ are the velocities of light in the two media, then:
  • n₁ = c/v₁ and n₂ = c/v₂ (where c is speed of light in vacuum)
• For minimum time of travel:
  • sin i/v₁ = sin r/v₂
• Substituting refractive indices:
  • sin i/sin r = v₁/v₂ = n₂/n₁
• Therefore, n₁sin i = n₂sin r, which is Snell’s law of refraction.

3. March 2022 Physics Question Paper with Answers

Question 1: What is the phase difference between voltage and current in a purely inductive AC circuit? (1 mark) Answer: π/2 or 90° (voltage leads current by 90°)

Question 2: What is meant by binding energy of a nucleus? Calculate the binding energy per nucleon of ₂He⁴. Given: mass of proton = 1.00783 u, mass of neutron = 1.00867 u, mass of ₂He⁴ = 4.00260 u, 1 u = 931.5 MeV/c². (3 marks) Answer:

• Binding energy is the energy required to break a nucleus into its constituent nucleons.
• For ₂He⁴: 2 protons and 2 neutrons
• Total mass of constituents = 2(1.00783) + 2(1.00867) = 2.01566 + 2.01734 = 4.033 u
• Mass of ₂He⁴ nucleus = 4.00260 u
• Mass defect = 4.033 – 4.00260 = 0.0304 u
• Binding energy = 0.0304 × 931.5 = 28.3176 MeV
• Binding energy per nucleon = 28.3176/4 = 7.0794 MeV per nucleon

Question 3: With a neat diagram, explain the working of a full-wave rectifier. Draw its input and output waveforms. (5 marks) Answer:

• A full-wave rectifier converts both halves of AC input to DC output.
• Center-tapped transformer full-wave rectifier uses two diodes.
• Bridge rectifier uses four diodes arranged in a bridge configuration.

Diagram: [Shows a bridge rectifier with four diodes arranged in a bridge with AC input and DC output connections]

Working:

• During positive half-cycle of input: Diodes D₁ and D₃ conduct, D₂ and D₄ are reverse biased.
• During negative half-cycle: Diodes D₂ and D₄ conduct, D₁ and D₃ are reverse biased.
• Current through the load is always in the same direction.
• Output is pulsating DC with frequency twice that of input AC.

Input waveform: [Sinusoidal wave] Output waveform: [All positive half-cycles showing pulsating DC]

Advantages:

• Higher efficiency (81% vs. 40.6% for half-wave)
• Less ripple
• Better transformer utilization

4. March 2021 Physics Question Paper with Answers

Question 1: What happens to the resistance of a conductor when its temperature increases? (1 mark) Answer: The resistance of a metallic conductor increases with increase in temperature.

Question 2: Draw the circuit diagram of a common emitter amplifier using an NPN transistor. Explain its working. (3 marks) Answer: Diagram: [Shows NPN transistor in CE configuration with biasing resistors, coupling capacitors, and load resistor]

Working:

• Base is connected to input signal via coupling capacitor C₁
• Collector is connected to output via coupling capacitor C₂
• Biasing resistors R₁ and R₂ provide proper DC biasing
• When input signal increases base current, collector current increases
• Small change in base current produces large change in collector current
• Phase shift of 180° between input and output
• Voltage gain (Av) = -βRL/re (where re is AC emitter resistance)
• Provides high voltage gain, medium current gain, and medium input impedance

Question 3: State Gauss’s law in electrostatics. Using this law, derive an expression for the electric field due to an infinitely long straight uniformly charged wire. (5 marks) Answer: Gauss’s law states that the total electric flux through any closed surface is equal to 1/ε₀ times the total charge enclosed by the surface. Mathematically: ∮E⃗.dA⃗ = q/ε₀

Derivation for electric field due to charged wire:

• Consider an infinitely long straight wire with uniform linear charge density λ.
• Choose a cylindrical Gaussian surface of radius r and length l around the wire.
• Due to symmetry, electric field is radially outward and has the same magnitude at all points on the cylindrical surface.
• Electric flux through the curved surface = E(2πrl)
• Flux through the flat ends is zero as E⃗ is parallel to these surfaces.
• Total charge enclosed = λl
• By Gauss’s law: E(2πrl) = λl/ε₀
• Therefore: E = λ/(2πε₀r)
• The electric field varies inversely with distance from the wire.

5. March 2020 Physics Question Paper with Answers

Question 1: Define self-inductance. Write its SI unit. (1 mark) Answer: Self-inductance is the property of a coil due to which it opposes any change in current flowing through it by inducing an emf in itself. SI unit is henry (H).

Question 2: Draw the block diagram of a simple amplitude modulation (AM) transmitter and explain the function of each block. (3 marks) Answer: Block diagram of AM transmitter consists of:

1. Microphone: Converts sound signals to electrical signals
2. Audio amplifier: Amplifies the weak audio signal
3. Carrier oscillator: Generates high-frequency carrier wave
4. Modulator: Mixes audio signal with carrier wave to produce AM wave
5. Power amplifier: Amplifies the modulated signal
6. Antenna: Radiates the signal as electromagnetic waves

Functions:

• Microphone: Transducer that converts sound to electrical signals
• Audio amplifier: Increases amplitude of audio signals for proper modulation
• Carrier oscillator: Generates stable high-frequency sine wave
• Modulator: Performs actual modulation process by varying amplitude of carrier
• Power amplifier: Boosts power of modulated signal
• Antenna: Converts electrical energy to electromagnetic waves for transmission

Question 3: What is a potentiometer? Explain with a neat circuit diagram how a potentiometer is used to compare the emf of two cells. (5 marks) Answer: A potentiometer is a device used to measure potential difference, emf, and for comparing emfs without drawing current from the source.

Diagram: [Shows a potentiometer setup with a long wire, battery, galvanometer, and two cells to be compared]

Working principle:

• The potentiometer wire AB of uniform cross-section is connected to a battery through a rheostat.
• The wire has a potential gradient along its length.
• To compare emfs of cells E₁ and E₂:
  • Connect cell E₁ with its positive terminal to point A through a galvanometer
  • Find the null point J₁ where no current flows through galvanometer
  • Length AJ₁ = l₁
  • Repeat with cell E₂ to find null point J₂
  • Length AJ₂ = l₂
• At null points: E₁ = K×l₁ and E₂ = K×l₂ (where K is potential gradient)
• Therefore: E₁/E₂ = l₁/l₂

Advantages:

• No current is drawn from the cell at balance point
• High accuracy in measurement
• Can compare EMFs of cells with different internal resistances

6. March 2019 Physics Question Paper with Answers

Question 1: How does the energy gap in a semiconductor vary with doping? (1 mark) Answer: The energy gap in a semiconductor decreases with doping.

Question 2: State Brewster’s law. What is the relation between Brewster’s angle and refractive index of the medium? (2 marks) Answer: Brewster’s law states that when unpolarized light is incident on a transparent dielectric medium at the polarizing angle, the reflected light is completely plane polarized perpendicular to the plane of incidence.

Relation: tan(ip) = n₂/n₁ Where ip is Brewster’s angle, n₁ is refractive index of first medium, and n₂ is refractive index of second medium. If the first medium is air, then tan(ip) = n, where n is the refractive index of the second medium.

Question 3: With a neat diagram, explain the principle and working of a moving coil galvanometer. Derive the expression for the torque acting on the coil. (5 marks) Answer: Principle: A moving coil galvanometer works on the principle that when a current-carrying coil is placed in a magnetic field, it experiences a torque.

Diagram: [Shows rectangular coil suspended between pole pieces of a horseshoe magnet with a soft iron core]

Construction:

• Rectangular coil of fine insulated copper wire wound on a metallic frame
• Suspended by phosphor bronze strip between pole pieces of a horseshoe magnet
• Cylindrical soft iron core inside the coil to provide radial magnetic field
• Light mirror attached to suspension to detect rotation
• Phosphor bronze strip provides restoring torque

Working:

• When current passes through the coil, it experiences a torque
• Coil rotates until magnetic torque equals restoring torque
• Deflection is proportional to current

Derivation of torque:

• For a coil with N turns, area A, carrying current I in magnetic field B:
• Force on each vertical side = BIl (where l is length of side)
• Torque on coil = F×w = BIl×w = BIA (where w is width of coil)
• Total torque for N turns = NBIA
• At equilibrium, τ = Cθ (where C is torsional constant and θ is deflection)
• Therefore, NBIA = Cθ or θ = (NBIA)/C
• θ ∝ I, showing deflection is proportional to current

7. March 2018 Physics Question Paper with Answers

Question 1: Define mobility of charge carriers in a conductor. Write its SI unit. (1 mark) Answer: Mobility is defined as the drift velocity per unit electric field. SI unit is m²/Vs.

Question 2: Draw the energy band diagrams of (i) conductors, (ii) semiconductors, and (iii) insulators. Explain the differences. (3 marks) Answer: Diagram: [Shows three band diagrams with different energy gaps]

Conductors:

• Valence band and conduction band overlap or partially filled conduction band
• No energy gap between valence and conduction bands
• Electrons move freely to conduction band at room temperature

Semiconductors:

• Small energy gap (about 1 eV) between valence and conduction bands
• At room temperature, some electrons can gain enough energy to cross the gap
• Conductivity increases with temperature

Insulators:

• Large energy gap (more than 3 eV) between valence and conduction bands
• Electrons cannot gain enough energy to cross the gap at room temperature
• Poor conductors of electricity

Question 3: State the laws of radioactive decay. Derive the relation N = N₀e^(-λt) where the symbols have their usual meanings. (5 marks) Answer: Laws of radioactive decay:

1. Radioactive decay is a random process.
2. The rate of decay is proportional to the number of radioactive nuclei present.
3. Decay is independent of physical and chemical conditions.

Derivation:

• If N is the number of undecayed nuclei at time t
• Rate of decay: -dN/dt = λN (where λ is decay constant)
• Rearranging: dN/N = -λdt
• Integrating both sides from t=0 (N=N₀) to t=t (N=N):
• ∫(dN/N) = -λ∫dt
• ln(N/N₀) = -λt
• N = N₀e^(-λt)

This equation shows exponential decay of radioactive nuclei with time.

8. March 2017 Physics Question Paper with Answers

Question 1: Which electromagnetic wave has the shortest wavelength? (1 mark) Answer: Gamma rays

Question 2: Draw the circuit diagram of a Zener diode as a voltage regulator and explain its working. (3 marks) Answer: Diagram: [Shows Zener diode connected in reverse bias with a series resistor Rs between input and output]

Working:

• Zener diode is connected in reverse bias parallel to the load.
• Series resistor Rs limits the current.
• When input voltage increases beyond Zener voltage:
  • Zener diode breaks down and conducts
  • Extra voltage appears across Rs
  • Voltage across Zener (and load) remains constant
• When input voltage decreases:
  • Current through Zener decreases
  • Voltage across load remains constant until input reaches Zener voltage
• If input falls below Zener voltage, regulation fails

Question 3: Describe Young’s double-slit experiment to demonstrate interference of light waves. Derive the expression for fringe width. (5 marks) Answer: Young’s double-slit experiment:

• Monochromatic light passes through two narrow slits S₁ and S₂ close to each other.
• The slits act as coherent sources.
• Interference pattern of alternate bright and dark fringes forms on the screen.

Diagram: [Shows light source, double slit, and interference pattern on screen]

Derivation of fringe width:

• Let d be the separation between slits, D be the distance to screen
• Path difference for point P on screen: δ = S₂P – S₁P = d sinθ
• For small angles: sinθ ≈ tanθ ≈ y/D (where y is distance from central fringe)
• Path difference: δ = d×y/D
• For bright fringe: δ = nλ (n = 0, 1, 2, …)
• Therefore: d×y/D = nλ
• Position of nth bright fringe: yn = nλD/d
• Distance between consecutive bright fringes (fringe width): β = λD/d

9. March 2016 Physics Question Paper with Answers

Question 1: What is space wave propagation? (1 mark) Answer: Space wave propagation is a mode of electromagnetic wave transmission in which radio waves travel in a straight line from transmitting antenna to receiving antenna, including direct waves and ground-reflected waves. It is limited to line of sight and used for VHF and UHF communication.

Question 2: A capacitor of capacitance 5 μF is charged to a potential of 100 V. Calculate the energy stored in the capacitor. (2 marks) Answer:

• Energy stored in a capacitor: U = ½CV²
• Given: C = 5 μF = 5 × 10^-6 F, V = 100 V
• U = ½ × 5 × 10^-6 × 100²
• U = 2.5 × 10^-2 J or 0.025 J

Question 3: Explain the principle, construction, and working of a transformer. Derive the relation between input and output voltages in terms of number of turns in primary and secondary coils. (5 marks) Answer: Principle: Transformer works on the principle of mutual induction – when current in one coil changes, an emf is induced in the nearby coil.

Construction:

• Primary and secondary coils wound on a soft iron core
• Laminated core to reduce eddy current losses
• Primary connected to AC input, secondary to load

Diagram: [Shows transformer with primary and secondary coils on a closed core]

Working:

• AC in primary produces changing magnetic flux
• Changing flux induces emf in secondary by mutual induction
• Power is transferred from primary to secondary

Derivation of voltage relation:

• If Φ is the magnetic flux in core, then:
  • Induced emf in primary: εₚ = -Nₚ(dΦ/dt)
  • Induced emf in secondary: εₛ = -Nₛ(dΦ/dt)
• Taking ratio: εₛ/εₚ = Nₛ/Nₚ
• Since εₚ ≈ Vₚ and εₛ ≈ Vₛ (ignoring resistance):
  • Vₛ/Vₚ = Nₛ/Nₚ
• This is the transformer equation relating input and output voltages

10. March 2015 Physics Question Paper with Answers

Question 1: What is the order of wavelength of X-rays? (1 mark) Answer: 10^-10 m or 0.1 nm or 1 Å (Angstrom)

Question 2: A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 40 cm. Calculate the power of the combination. (2 marks) Answer:

• Power of convex lens: P₁ = 1/f₁ = 1/0.2 = +5 D
• Power of concave lens: P₂ = 1/f₂ = 1/(-0.4) = -2.5 D
• Power of combination: P = P₁ + P₂ = 5 + (-2.5) = 2.5 D

Question 3: What is nuclear fission? Explain the working of a nuclear reactor with a neat diagram. (5 marks) Answer: Nuclear fission is the process in which a heavy nucleus splits into lighter nuclei with the release of energy.

Nuclear reactor components:

1. Fuel rods (enriched uranium or plutonium)
2. Moderator (heavy water, graphite)
3. Control rods (cadmium, boron)
4. Coolant (water, liquid sodium)
5. Radiation shield (concrete, lead)
6. Pressure vessel

Diagram: [Shows cross-section of nuclear reactor with components labeled]

Working:

• Nuclear fission of U-235 releases energy and neutrons
• Neutrons are slowed down by moderator
• Controlled chain reaction is maintained
• Control rods absorb excess neutrons
• Heat generated is carried away by coolant
• Heat is used to generate steam for electricity production
• Radiation shield protects environment from radiation

11. March 2014 Physics Question Paper with Answers

Question 1: What is the principle of digital communication? (1 mark) Answer: The principle of digital communication is to convert analog signals into digital form (binary bits – 0 and 1) for transmission and then reconvert to analog form at the receiver.

Question 2: Define intensity of electric field at a point. Derive an expression for the intensity of electric field at a point on the axial line of an electric dipole. (3 marks) Answer: Electric field intensity is the force experienced by a unit positive charge placed at that point.

Derivation for electric field on axial line of dipole:

• Consider dipole with charges +q and -q separated by distance 2a
• Point P is at distance r from center of dipole on axial line
• Distance from +q to P = r-a
• Distance from -q to P = r+a
• Electric field due to +q at P: E₁ = kq/(r-a)²
• Electric field due to -q at P: E₂ = kq/(r+a)²
• Net field: E = E₁ – E₂ = kq/(r-a)² – kq/(r+a)²
• Simplifying for r>>a: E = 2kp/r³
• Where p = 2aq is dipole moment

Question 3: Describe Rutherford’s alpha-particle scattering experiment. What are the conclusions drawn from this experiment? (5 marks) Answer: Rutherford’s experiment:

• A thin gold foil was bombarded with alpha particles
• Scattering of alpha particles was observed using zinc sulfide screen
• Most alpha particles passed through with little or no deflection
• A few particles were deflected through large angles
• Very few bounced back

Diagram: [Shows experimental setup with alpha source, gold foil, and detector]

Observations:

• Most alpha particles passed straight through
• Some were deflected at small angles
• Very few (1 in 20,000) bounced back

Conclusions:

1. Most of the atom is empty space
2. The entire positive charge and most of the mass is concentrated in a very small region called the nucleus
3. The size of nucleus is very small compared to the size of atom
4. The atom must contain negatively charged particles to balance the positive charge
5. Strong electric forces exist within the nucleus

12. March 2013 Physics Question Paper with Answers

Question 1: What is meant by ‘demodulation’ in communication systems? (1 mark) Answer: Demodulation is the process of extracting the original message signal from the modulated carrier wave at the receiver end of a communication system.

Question 2: Explain Hall effect. Derive the expression for Hall voltage. (3 marks) Answer: Hall effect is the production of a voltage difference across a conductor when placed in a magnetic field perpendicular to the current flow.

Diagram: [Shows conductor with current, magnetic field, and Hall voltage]

Derivation:

• Consider a conductor carrying current I along x-axis
• Magnetic field B applied along z-axis
• Electrons experience magnetic force F = e(v×B) along negative y-axis
• Electrons accumulate on one face, creating electric field EH
• At equilibrium: eEH = evB
• Hall field: EH = vB
• If w is width of conductor: Hall voltage VH = EHw = vBw
• Using drift velocity v = I/(neA) where n is electron density
• VH = IBw/(net) = IB/(ne) (where t is thickness)
• Hall coefficient: RH = 1/(ne)

Applications:

• Determination of carrier type and concentration
• Measurement of magnetic field
• Study of semiconductors

Question 3: Define mutual inductance between two coils. Derive an expression for the mutual inductance between two long co-axial solenoids. (5 marks) Answer: Mutual inductance is defined as the property of two coils due to which emf is induced in one coil when current changes in the nearby coil.

If Φ₂ is the magnetic flux linked with secondary due to current I₁ in primary: M = Φ₂/I₁

For two co-axial solenoids:

• Primary solenoid: n₁ turns per unit length, radius r₁, current I₁
• Secondary solenoid: n₂ turns per unit length, radius r₂, length l
• Magnetic field due to primary: B₁ = μ₀n₁I₁
• Magnetic flux through each turn of secondary: Φ = πr₁²B₁ = πr₁²μ₀n₁I₁
• Total flux linked with secondary: Φ₂ = n₂l × Φ = πr₁²μ₀n₁n₂lI₁
• Mutual inductance: M = Φ₂/I₁ = πr₁²μ₀n₁n₂l

If r₁ = r₂ = r (perfectly co-axial): M = μ₀n₁n₂πr²l

13. March 2012 Physics Question Paper with Answers

Question 1: What do you mean by polar and non-polar molecules? Give one example each. (1 mark) Answer: Polar molecules have a permanent dipole moment due to asymmetric charge distribution (e.g., H₂O, NH₃). Non-polar molecules have symmetric charge distribution with zero net dipole moment (e.g., H₂, CO₂).

Question 2: Derive lens maker’s formula for a thin lens. (3 marks) Answer: Consider a thin lens with refractive index μ and radii of curvature R₁ and R₂:

• For refraction at first surface (using spherical surface formula):
  • 1/v₁ – μ/u = (μ-1)/R₁
• For refraction at second surface:
  • 1/v – 1/v₁ = (1-μ)/R₂
• Eliminating v₁:
  • 1/v – μ/u = (μ-1)/R₁ + (1-μ)/R₂
  • 1/v – 1/u = (μ-1)(1/R₁ – 1/R₂)
• If focal length is f, then for object at infinity (u = ∞):
  • 1/f = (μ-1)(1/R₁ – 1/R₂)

This is the lens maker’s formula, relating focal length to refractive index and radii of curvature.

Question 3: Write Einstein’s photoelectric equation. Explain how this equation is verified experimentally. (5 marks) Answer: Einstein’s photoelectric equation: hν = W₀ + ½mv²max

Where:

• hν is energy of incident photon
• W₀ is work function of the metal
• ½mv²max is maximum kinetic energy of emitted photoelectron

Experimental verification:

• Millikan used a setup with vacuum tube containing emitter plate and collector plate
• A variable potential difference opposed photoelectric current
• Stopping potential (V₀) was measured for different frequencies of light
• From Einstein’s equation: eV₀ = hν – W₀
• Plot of V₀ vs. frequency is a straight line
• Slope gives h/e (Planck’s constant/electronic charge)
• Intercept on frequency axis gives threshold frequency
• Intercept on V₀ axis gives -W₀/e

Diagram: [Shows photoelectric effect experimental setup]

Results confirmed:

1. Maximum kinetic energy is linearly related to frequency
2. Threshold frequency exists below which no emission occurs
3. Calculated value of Planck’s constant matched with expected value

14. March 2011 Physics Question Paper with Answers

Question 1: What is wavefront? Name different types of wavefronts. (1 mark) Answer: A wavefront is the locus of all points in a medium which vibrate in the same phase. Types: Spherical wavefront, Cylindrical wavefront, and Plane wavefront.

Question 2: Two thin lenses of focal lengths f₁ and f₂ are kept in contact. Find the focal length and power of the combination. (2 marks) Answer:

• For lenses in contact, the focal length of combination is given by:
  • 1/F = 1/f₁ + 1/f₂
• Power of combination:
  • P = P₁ + P₂ = 1/f₁ +