Question 1(a) [3 marks]#
Define accuracy and precision.
Answer:
- Accuracy: Closeness of a measured value to the true value
- Precision: Consistency or repeatability of measurement values
Mnemonic: “Accuracy Aims at Truth, Precision Produces Repeatability”
Question 1(b) [4 marks]#
Derive SI unit of work and Velocity using fundamental physical units.
Answer:
Table: Derivation of Work and Velocity Units
| Physical Quantity | Formula | SI Unit Derivation | SI Unit |
|---|---|---|---|
| Work (W) | W = F × d | W = [Force] × [Distance] = [kg·m/s²] × [m] = [kg·m²/s²] | Joule (J) |
| Velocity (v) | v = d/t | v = [Distance]/[Time] = [m]/[s] | m/s |
- Work: When a force (kg·m/s²) acts through a distance (m), we get kg·m²/s² = Joule
- Velocity: When an object covers distance (m) in time (s), we get m/s
Mnemonic: “Work Forces Distance, Velocity Distances Time”
Question 1(c) [7 marks]#
What is Least Count of instrument. State equation of Least count of Vernier calipers. Explain measurement by vernier calipers with neat and clean diagram.
Answer:
Least Count: Smallest measurement that can be directly measured using a measuring instrument.
Equation for Least Count of Vernier Caliper: Least Count = 1 Main Scale Division - 1 Vernier Scale Division or Least Count = Value of 1 MSD / Number of VSD
Diagram: Vernier Caliper
┌────────┐
│ │
┌────┘ ┌───┘
│ │
│ ┌─────┘
│ │
─┼───┼───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬
0 1 2 3 4 5 6 7 8 9 10
│ │ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┴───┴───┘
0 5 10 15 20 25 30 35 40 45
Vernier Scale
Measurement Process:
Step 1: Close the jaws of caliper around the object
Step 2: Note the main scale reading just before the zero of vernier scale
Step 3: Find which vernier division exactly coincides with a main scale division
Step 4: Add the vernier reading to the main scale reading: Total = MSR + (VC × LC)
Main Scale Reading (MSR): Value on main scale just before vernier zero
Vernier Coincidence (VC): Division number where vernier line aligns with main scale line
Least Count (LC): Usually 0.02 mm or 0.001 inch
Mnemonic: “Main plus Matched makes Measurement”
Question 1(c) OR [7 marks]#
What is Least Count of instrument. State equation of Least count of micrometer screw. Explain the positive and negative error in micrometer screw with neat and clean diagram.
Answer:
Least Count: Smallest measurement that can be directly measured using a measuring instrument.
Equation for Least Count of Micrometer Screw: Least Count = Pitch of screw / Number of divisions on circular scale
Diagram: Micrometer Screw Gauge
┌─────────────────┐
│ │
│ ┌───────┐ │
│ │ │ │
└────┤ ├────┘
│ │
└───────┘
0 5 10 15 20 25
────────────────────
│
V
┌───────┐
│0 5 │ ← Circular Scale
└───────┘
Positive Error: When zero of circular scale is above the reference line. The measured reading will be more than the actual value.
Negative Error: When zero of circular scale is below the reference line. The measured reading will be less than the actual value.
Error Correction:
- For positive error: Actual Reading = Observed Reading - Zero Error
- For negative error: Actual Reading = Observed Reading + Zero Error
Mnemonic: “Positive Produces Plus, Negative Needs Addition”
Question 2(a) [3 marks]#
Write characteristics of electric lines of force.
Answer:
Table: Characteristics of Electric Field Lines
| Characteristic | Description |
|---|---|
| Direction | Always from positive to negative charge |
| Shape | Straight lines for uniform fields, curved for non-uniform fields |
| Density | Proportional to field strength |
| Path | Never intersect each other |
| Nature | Start from positive and end at negative charges |
Mnemonic: “Direction, Density, Never Cross, Start-End”
Question 2(b) [4 marks]#
Calculate the equivalent capacitance for both series and parallel connection of capacitors having capacitance of values 9 μF, 12 μF & 15 μF.
Answer:
For Series Connection: 1/Ceq = 1/C₁ + 1/C₂ + 1/C₃ 1/Ceq = 1/9 + 1/12 + 1/15 1/Ceq = 5/36 + 3/36 + 2.4/36 = 10.4/36 Ceq = 36/10.4 = 3.46 μF
For Parallel Connection: Ceq = C₁ + C₂ + C₃ Ceq = 9 + 12 + 15 = 36 μF
Mnemonic: “Series Sums Reciprocals, Parallel Puts Together”
Question 2(c) [7 marks]#
Explain coulombs inverse square law and derive its equation. Calculate coulomb force between two electrons separated by 10 meter. (e=1.66 x 10⁻¹⁹ C, K= 9 x 10⁹ Nm² C⁻²)
Answer:
Coulomb’s Law: The electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Equation Derivation: F ∝ q₁q₂ F ∝ 1/r² Combining: F ∝ q₁q₂/r² With constant: F = k(q₁q₂/r²)
Where k = 1/(4πε₀) = 9 × 10⁹ Nm²/C²
Diagram: Coulomb’s Law
q₁ q₂
●─────────●
←────r────→
F₁²→ ←F₂₁
Calculation: F = k(q₁q₂/r²) F = 9 × 10⁹ × [(1.66 × 10⁻¹⁹) × (1.66 × 10⁻¹⁹)] / (10)² F = 9 × 10⁹ × 2.76 × 10⁻³⁸ / 100 F = 9 × 2.76 × 10⁻³⁸⁻² × 10⁹ F = 2.48 × 10⁻³¹ N
Mnemonic: “Charges Multiply, Distance Squares, Force Declines”
Question 2(a) OR [3 marks]#
Explain electric field and and derive its unit.
Answer:
Electric Field: The region around a charge where another charge experiences a force.
Definition: Electric field at a point is the force experienced by a unit positive charge placed at that point.
E = F/q
Unit Derivation: E = F/q = [N]/[C] = [kg·m/s²]/[A·s] = [kg·m/(A·s³)] SI unit: N/C or V/m
Mnemonic: “Electric field Equals Force per Charge”
Question 2(b) OR [4 marks]#
Explain electric flux with neat figure and derive its unit.
Answer:
Electric Flux: Measure of the electric field passing through a given area.
Equation: ϕₑ = E·A·cosθ
Where:
- E is the electric field
- A is the area
- θ is the angle between E and the normal to the area
Diagram: Electric Flux
↑ n (normal)
│
│ θ
│/
───────┼─────→ E (electric field)
│
│
Surface Area A
Unit Derivation: ϕₑ = E·A·cosθ = [N/C]·[m²]·[dimensionless] = [N·m²/C] Since 1 N/C = 1 V/m, flux unit = V·m = N·m²/C
SI unit: N·m²/C or V·m
Mnemonic: “Flux Flows through Fields and Areas”
Question 2(c) OR [7 marks]#
Define capacitor and derive its unit. Give the formula of parallel plate capacitor and explain each term. Calculate the capacitance of a parallel plate capacitor having 20 cm x 20 cm square plates separated by a distance of 1.0 mm.
Answer:
Capacitor: A device that stores electric charge.
Definition: Capacitance is the ratio of charge stored to the potential difference applied. C = Q/V
Unit Derivation: C = Q/V = [C]/[V] = [A·s]/[J/C] = [A·s]/[N·m/C] = [A²·s⁴/(kg·m²)] = Farad (F)
Parallel Plate Capacitor Formula: C = ε₀εᵣA/d
Where:
- C is the capacitance
- ε₀ is the permittivity of free space (8.85 × 10⁻¹² F/m)
- εᵣ is the relative permittivity of dielectric
- A is the area of overlap of plates
- d is the distance between plates
Diagram: Parallel Plate Capacitor
┌───────────────┐ ┐
│ + + + + + + + │ │
└───────────────┘ │ d
┌───────────────┐ │
│ - - - - - - - │ │
└───────────────┘ ┘
Area A
Calculation: A = 20 cm × 20 cm = 0.2 m × 0.2 m = 0.04 m² d = 1.0 mm = 0.001 m εᵣ = 1 (air) ε₀ = 8.85 × 10⁻¹² F/m
C = ε₀εᵣA/d = 8.85 × 10⁻¹² × 1 × 0.04/0.001 = 354 × 10⁻¹² F = 354 pF
Mnemonic: “Capacitance Collects Charge between Closer Plates”
Question 3(a) [3 marks]#
Explain heat conduction in solid with example.
Answer:
Heat Conduction: Transfer of heat through a solid material without the movement of the material itself.
Process: Heat energy transfers from high temperature region to low temperature region through molecular vibrations.
Diagram: Heat Conduction
Hot Cold
↓ ↓
┌────────────────────────┐
│ >>>>>>>>>>>>>>>>>>>>>>> │
└────────────────────────┘
Heat flow →
Example: Metal spoon in hot tea gets heated up at the handle end through conduction.
Mnemonic: “Hot Energizes, Atoms Transfer, Conducts Outward”
Question 3(b) [4 marks]#
A person has fever 102. What is the temperature scale here? Convert the temperature in remaining two scales.
Answer:
Temperature Scale: 102°F (Fahrenheit)
Conversion Formulas:
- °C = (°F - 32) × 5/9
- K = °C + 273.15
Calculation: °C = (102 - 32) × 5/9 = 70 × 5/9 = 38.89°C K = 38.89 + 273.15 = 312.04 K
Table: Temperature Conversion
| Fahrenheit | Celsius | Kelvin |
|---|---|---|
| 102°F | 38.89°C | 312.04 K |
Mnemonic: “Fahrenheit First, Convert Celsius, Kelvin Comes last”
Question 3(c) [7 marks]#
Explain the principle of platinum resistance thermometer and list out its uses.
Answer:
Principle: The electrical resistance of platinum changes predictably and consistently with temperature, allowing for precise temperature measurement.
Working: Based on the relationship R = R₀[1 + α(T - T₀)], where R is resistance at temperature T, R₀ is resistance at reference temperature T₀, and α is temperature coefficient of resistance.
Diagram: Platinum Resistance Thermometer
┌───────────────┐
│ Indicator │
└───┬───────┬───┘
│ │
│ │
┌───┴───────┴───┐
│ Wheatstone │
│ Bridge │
└───┬───────┬───┘
│ │
│ │
┌───┴───────┴───┐
│ Platinum │
│ Resistance │
│ Coil │
└───────────────┘
Uses:
- Industrial process: Temperature monitoring in manufacturing
- Scientific research: Laboratory measurements requiring high precision
- Calibration: Standard for calibrating other thermometers
- Medical applications: Temperature monitoring in medical equipment
Mnemonic: “Platinum Provides Precise Proportional Resistance”
Question 3(a) OR [3 marks]#
Define specific heat and heat capacity. And write its units.
Answer:
Specific Heat: Amount of heat energy required to raise the temperature of 1 kg of substance by 1 K.
Heat Capacity: Amount of heat energy required to raise the temperature of an entire object by 1 K.
Table: Heat Capacity Terms
| Term | Formula | SI Unit |
|---|---|---|
| Specific Heat (c) | Q = mc∆T | J/(kg·K) |
| Heat Capacity (C) | Q = C∆T | J/K |
Mnemonic: “Specific for Substance, Capacity for Complete Object”
Question 3(b) OR [4 marks]#
Explain heat convection in fluid with example.
Answer:
Heat Convection: Transfer of heat through a fluid (liquid or gas) by the movement of the fluid itself.
Process: Hot fluid expands, becomes less dense, rises; cooler fluid descends, creating a continuous circulation pattern called convection current.
Diagram: Convection Current
↑ ↑ ↑
warm warm warm
^ ^ ^
| | |
┌──────────────────┐
│ heat source │
└──────────────────┘
Cool fluid
↓ ↓ ↓
Example: Boiling water in a pot - heated water rises to the top while cooler water sinks to the bottom.
Mnemonic: “Heat Rises, Cool Descends, Currents Circulate”
Question 3(c) OR [7 marks]#
Define coefficient of thermal conductivity. Derive its equation of coefficient of thermal conductivity for heat transfer in solids.
Answer:
Coefficient of Thermal Conductivity: The amount of heat transferred per unit time per unit area per unit temperature gradient.
Definition: The quantity of heat flowing per second through unit area when temperature gradient is unity.
Derivation:
- Consider a rod with cross-sectional area A and length L
- Temperature difference between ends is ∆T
- Heat flow Q in time t
Heat current = Q/t Temperature gradient = ∆T/L Area = A
According to Fourier’s law: Q/t = k·A·(∆T/L)
Rearranging: k = (Q·L)/(t·A·∆T)
Where k is the coefficient of thermal conductivity.
Diagram: Thermal Conductivity
T₁ T₂
↓ ↓
┌────────────────────────┐
│ │ Area A
└────────────────────────┘
←───── L ─────→
Heat flow →
Unit: W/(m·K)
Mnemonic: “Heat Quantity Transfers Along Length Divided by Area and Temperature”
Question 4(a) [3 marks]#
Write the difference between transverse waves and longitudinal waves.
Answer:
Table: Transverse vs Longitudinal Waves
| Property | Transverse Waves | Longitudinal Waves |
|---|---|---|
| Particle motion | Perpendicular to wave direction | Parallel to wave direction |
| Medium displacement | Crests and troughs | Compressions and rarefactions |
| Examples | Light waves, water waves | Sound waves, seismic P-waves |
| Medium requirements | Can travel through solids | Can travel through solids, liquids, gases |
| Polarization | Can be polarized | Cannot be polarized |
Mnemonic: “Transverse Takes Perpendicular Path, Longitudinal Likes Linear Lanes”
Question 4(b) [4 marks]#
Calculate the wavelength of a wave having velocity 350 m/s and frequency 10 Hz.
Answer:
Wave Equation: v = fλ
Where:
- v is wave velocity (350 m/s)
- f is frequency (10 Hz)
- λ is wavelength (to be calculated)
Calculation: λ = v/f = 350/10 = 35 m
Mnemonic: “Velocity Values frequency times wavelength”
Question 4(c) [7 marks]#
Define Ultrasonic waves and write its characteristics. Write its four major applications of Ultrasonic wave.
Answer:
Ultrasonic Waves: Sound waves with frequencies higher than the upper audible limit of human hearing (above 20 kHz).
Characteristics:
- High frequency: Above 20 kHz
- Short wavelength: Enables detection of small objects
- Directional: Can be focused in a specific direction
- Non-ionizing: Safe for biological tissues
- Penetration: Can travel through various media
Diagram: Ultrasonic Wave
Amplitude
↑
│ /\ /\ /\
│ / \ / \ / \
───────┼─/────\──/────\──/────\──────→ Time
│/ \/ \/ \
│
Period < 50 μs (f > 20 kHz)
Applications:
- Medical: Diagnostic imaging, therapeutic procedures
- Industrial: Non-destructive testing, flaw detection
- Cleaning: Ultrasonic cleaning baths for precision parts
- Distance measurement: Sonar, parking sensors, level indicators
Mnemonic: “Ultrasonic Uses Sound to Sense, Scan, Sanitize”
Question 4(a) OR [3 marks]#
Explain the polarization of light with neat diagram.
Answer:
Polarization: The process of restricting the vibrations of light waves to a single plane.
Types:
- Linear polarization
- Circular polarization
- Elliptical polarization
Diagram: Light Polarization
Unpolarized Light Polarizer Polarized Light
↓ ↓ ↓
⊥↔↕⊢⊣|↖↗↘↙ ┌─────┐ ↔↔↔↔
⊥↔↕⊢⊣|↖↗↘↙ → │/////│ → ↔↔↔↔
⊥↔↕⊢⊣|↖↗↘↙ └─────┘ ↔↔↔↔
Multiple Allows only Single plane
vibration one plane vibration
planes
Mnemonic: “Polarizers Pick Particular Planes”
Question 4(b) OR [4 marks]#
If velocity of light in air is 3 x 10⁸ m/s and velocity of light in water is 2.25 x 10⁸ m/s. Calculate reflective index of water.
Answer:
Refractive Index Formula: n = c/v
Where:
- n is the refractive index
- c is the speed of light in vacuum (or air)
- v is the speed of light in medium
Calculation: n = 3 × 10⁸ / 2.25 × 10⁸ = 3/2.25 = 4/3 = 1.33
Mnemonic: “Slower Speeds Show higher index”
Question 4(c)(i) OR [4 marks]#
Define: velocity, wavelength and frequency of wave. And derive the relationship between wave velocity, wavelength and frequency.
Answer:
Wave Velocity (v): The speed at which a wave travels through a medium.
Wavelength (λ): The distance between two consecutive similar points on a wave.
Frequency (f): Number of complete wave cycles passing a point per unit time.
Diagram: Wave Parameters
Amplitude
↑
│ /\ /\ /\
│ / \ / \ / \
────┼─/────\──/────\──/────\─→ Distance
│/ \/ \/ \
│
↑ ↑ ↑
Wavelength (λ) Period (T)
Derivation:
- In time T (period), the wave travels a distance of one wavelength λ
- So, v = λ/T
- Since f = 1/T (frequency is inverse of period)
- Therefore, v = λf
Mnemonic: “Velocity Values frequency times wavelength”
Question 4(c)(ii) OR [3 marks]#
Write properties of light.
Answer:
Table: Properties of Light
| Property | Description |
|---|---|
| Propagation | Travels in straight lines in homogeneous medium |
| Speed | 3 × 10⁸ m/s in vacuum |
| Reflection | Bounces off surfaces following law of reflection |
| Refraction | Changes direction when passing between media |
| Dispersion | White light splits into component colors |
| Interference | Waves can superimpose to create patterns |
| Diffraction | Bends around obstacles and through small openings |
| Polarization | Can be restricted to vibrate in one plane |
| Dual nature | Exhibits both wave and particle properties |
Mnemonic: “Light Reflects, Refracts, Disperses, Interferes, Polarizes”
Question 5(a) [3 marks]#
Explain law of refraction of light for plane surface. And explain Snell’s law.
Answer:
Law of Refraction: When light passes from one medium to another, it changes direction at the boundary.
Snell’s Law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media.
n₁sin(θ₁) = n₂sin(θ₂)
Where:
- n₁ is the refractive index of first medium
- n₂ is the refractive index of second medium
- θ₁ is the angle of incidence
- θ₂ is the angle of refraction
Diagram: Refraction
┌─────────────
Normal │
↑ │ Medium 1 (n₁)
│ │
│ │ Incident ray
│ /│
│ / │
│ / │
│/θ₁ │
├────┼────────────────
│\θ₂ │
│ \ │
│ \ │
│ \│ Medium 2 (n₂)
│ Refracted ray
│
│
└─────────────
Mnemonic: “Sines Show Speeds in Separate Substances”
Question 5(b) [4 marks]#
A step index fiber has core refractive index of 1.30 and relative refractive index difference is Δ=0.02. Find numerical aperture.
Answer:
Numerical Aperture Formula: NA = √(n₁² - n₂²)
For step index fiber: NA = n₁√(2Δ)
Where:
- n₁ is the core refractive index
- Δ is the relative refractive index difference
Calculation: NA = 1.30 × √(2 × 0.02) NA = 1.30 × √0.04 NA = 1.30 × 0.2 NA = 0.26
Mnemonic: “Numerical Aperture Needs core And Delta”
Question 5(c) [7 marks]#
Explain Total internal reflection of light. And derive the equation of critical angle.
Answer:
Total Internal Reflection (TIR): The complete reflection of light at the boundary between two media when light travels from a denser medium to a rarer medium at an angle greater than the critical angle.
Conditions for TIR:
- Light must travel from denser to rarer medium
- Angle of incidence must exceed critical angle
Critical Angle: The angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°.
Derivation: Using Snell’s law: n₁sin(θ₁) = n₂sin(θ₂)
At critical angle (θc):
- θ₁ = θc
- θ₂ = 90°
- sin(90°) = 1
Therefore: n₁sin(θc) = n₂sin(90°) = n₂ × 1 = n₂
Rearranging: sin(θc) = n₂/n₁
Diagram: Total Internal Reflection
Medium 1 (n₁)
(Denser)
┌─────────────────
│ \ /
│ \θc /
│ \ /
│ \/
│ /\
│ / \
│ / \
│ / \
└─────────────────
Medium 2 (n₂)
(Rarer)
Mnemonic: “Critical Comes when Dense to Rare with Sine at Ratio”
Question 5(a) OR [3 marks]#
Explain numerical aperture and acceptance angle for fiber optic cable.
Answer:
Numerical Aperture (NA): Measure of the light-gathering ability of an optical fiber.
Acceptance Angle (θₐ): Maximum angle at which light can enter the fiber and still experience total internal reflection.
Relationship: NA = sin(θₐ)
Diagram: Numerical Aperture and Acceptance Angle
θₐ
/│\
Cladding /a│ \ Cladding
────────┼──┼──┼────────
│ │ │
Core │ │ │ Core
────────┼──┼──┼────────
│ │ │
Cladding│ │ │ Cladding
────────┴──┴──┴────────
Mnemonic: “Acceptance Angle Allows light, Numerical Aperture Names its Sine”
Question 5(b) OR [4 marks]#
Write full form LASER. Write its characteristics.
Answer:
LASER: Light Amplification by Stimulated Emission of Radiation
Table: Characteristics of LASER
| Characteristic | Description |
|---|---|
| Monochromatic | Single wavelength or color |
| Coherent | All waves in same phase |
| Highly directional | Travels in straight line with minimal divergence |
| High intensity | Concentrated energy in narrow beam |
| Collimated | Parallel rays with minimal spreading |
Mnemonic: “LASER Light: Mono, Coherent, Direct, Intense”
Question 5(c) OR [7 marks]#
Explain the construction of optical fiber cable in details. And Explain Step index and Graded index optical fiber.
Answer:
Optical Fiber Construction:
- Core: Central light-transmitting portion (glass or plastic)
- Cladding: Surrounds core, with lower refractive index than core
- Buffer Coating: Protective plastic coating
- Jacket: Outer protective covering
Diagram: Optical Fiber Structure
┌───────────────┐
│ │ ← Jacket
│ ┌─────────┐ │
│ │ │ │ ← Buffer Coating
│ │ ┌───┐ │ │
│ │ │ │ │ │
│ │ │ │ │ │
│ │ └───┘ │ │
│ │ ↑ │ │
│ └────┼────┘ │
│ │ │
└───────┼───────┘
↑
Core
Cladding
Step Index Fiber:
- Abrupt change in refractive index between core and cladding
- Light travels in zigzag path by total internal reflection
- Higher modal dispersion (signal spreading)
- Simpler construction
Graded Index Fiber:
- Gradual change in refractive index from center of core to cladding
- Light travels in helical path due to continuous refraction
- Lower modal dispersion
- More complex construction
Diagram: Step Index vs Graded Index Fiber
Step Index:
─────────────────────
/ \
/ ┌────────────┐ \
/ │ │ \
| │ Core │ |
\ │ │ /
\ └────────────┘ /
\ Cladding /
────────────────────
Graded Index:
─────────────────────
/ \
/ ┌──────────┐ \
/ / \ \
| | Core | |
\ \ / /
\ └──────────┘ /
\ Cladding /
────────────────────
Refractive Index Profile:
Step Index: Graded Index:
│ │
n₁ ─┤▄▄▄▄▄▄▄ ▄▄▄▄▄
│ │ ▄ ▄
│ │ ▄ ▄
n₂ ─┤ ▀▀▀▀▀▀▀ ▄ ▄
│ ▀▀▀▀▀▀▀▀▀▀▀
└───────→ r └───────→ r
Mnemonic: “Step Shows Sharp Shift, Graded Gradually Goes down”
Overall Exam Solution Summary#
This complete solution set covers all the questions in the Physics (4300005) Winter 2024 exam, including both main questions and OR alternatives. Key physics concepts covered include:
Measurement and Instruments:
- Vernier calipers and micrometer screw gauge
- Accuracy, precision, and error analysis
Electrostatics:
- Coulomb’s law and electric fields
- Capacitors and capacitance calculations
Heat and Thermometry:
- Heat transfer mechanisms
- Temperature scales and thermal conductivity
Wave Motion:
- Wave parameters and relationship
- Transverse vs longitudinal waves
- Ultrasonic waves and applications
Optics and Modern Physics:
- Refraction and Snell’s law
- Total internal reflection
- Optical fibers and LASER properties
Each solution follows a structured approach with:
- Clear definitions
- Relevant equations
- Simple diagrams
- Worked-out calculations
- Mnemonics for easy recall
These solutions are designed specifically for weak students who struggle with exams, using simplified explanations, visual aids, and memory techniques to help with comprehension and retention.