Question 1 - Fill in the blanks/MCQs [14 marks]

Answer:

Question	Answer	Question	Answer
(1)	(a) Si	(8)	(b) 0.5 Hz
(2)	(a) 1.50	(9)	(a) 300000 km/s
(3)	(b) greater than	(10)	(b) solid
(4)	(c) 4	(11)	(a) crest and trough
(5)	(d) Total internal reflection	(12)	(b) monochromatic
(6)	(d) frequency	(13)	(a) Single mode
(7)	(a) Coulomb	(14)	(b) 45°

Mnemonic: "Silicon Glass Bridge Optic Frequency Coulomb Hz Solid Crest Mono Single 45"

Question 2(A) - Attempt any two [6 marks]

Question 2(A)(1) [3 marks]

Differentiate between accuracy and precision.

Answer:

Parameter	Accuracy	Precision
Definition	Closeness to true value	Consistency of repeated measurements
Focus	Correctness	Reproducibility
Error Type	Systematic error	Random error
Example	Hitting bullseye	Hitting same spot repeatedly

Accuracy: How close measurement is to actual value
Precision: How close repeated measurements are to each other

Mnemonic: "Accurate Aims Actual, Precise Repeats Reliably"

Question 2(A)(2) [3 marks]

Determine the diameter of a sphere measured by micrometer screw, main scale reading is 5 mm and 50th division of circular scale is coinciding with base line. The least count of this instrument is 0.01 mm.

Answer:

Given:
Main Scale Reading (MSR) = 5 mm
Circular Scale Reading (CSR) = 50 divisions
Least Count (LC) = 0.01 mm

Formula: Total Reading = MSR + (CSR × LC)
Total Reading = 5 + (50 × 0.01)
Total Reading = 5 + 0.5 = 5.5 mm

Diameter of sphere = 5.5 mm

Mnemonic: "Main Scale Reading + Circular × Least Count"

Question 2(A)(3) [3 marks]

Calculate the amount of electric charge stored on either plate of a capacitor of capacitance 4 µF when connected across 12 volt battery.

Answer:

Given:
Capacitance (C) = 4 µF = 4 × 10⁻⁶ F
Voltage (V) = 12 V

Formula: Q = CV
Q = 4 × 10⁻⁶ × 12
Q = 48 × 10⁻⁶ C
Q = 48 µC

Electric charge stored = 48 µC

Mnemonic: "Charge equals Capacitance times Voltage"

Question 2(B) - Attempt any two [8 marks]

Question 2(B)(1) [4 marks]

Draw a sketch of micrometer screw gauge with proper nomenclature.

Answer:

                     Ratchet
                        |
                        v
    +------------------+
    |                  |
    |   Thimble Scale  |
    |        ||        |
    +--------||--------+
             ||
             ||  Spindle
             ||
    +--------||--------+
    |    Main Scale    |
    |  0   5   10  15  |
    +--||----------||--+
       ||          ||
    Anvil         Frame

Main Components:

Frame: U-shaped structure providing support
Anvil: Fixed jaw for placing object
Spindle: Movable screw mechanism
Thimble Scale: Circular scale with 50 divisions
Main Scale: Linear scale in mm
Ratchet: For consistent pressure application

Mnemonic: "Frame Anvil Spindle Thimble Main Ratchet"

Question 2(B)(2) [4 marks]

Explain the zero, positive and negative errors for vernier calipers with proper diagram and list necessary steps to remove these types of errors.

Answer:

Types of Errors:

Error Type	Condition	Reading
Zero Error	Zero line of vernier doesn't coincide with main scale zero	Non-zero reading when jaws closed
Positive Error	Vernier zero is right of main scale zero	Add correction
Negative Error	Vernier zero is left of main scale zero	Subtract correction

Diagram:

Zero Error:
Main Scale:  |0|1|2|3|4|5|
Vernier:      |0|1|2|3|4|

Positive Error:
Main Scale:  |0|1|2|3|4|5|
Vernier:       |0|1|2|3|4|

Negative Error:
Main Scale:  |0|1|2|3|4|5|
Vernier:     |0|1|2|3|4|

Steps to Remove Errors:

Check zero error before measurement
Apply correction to final reading
Clean jaws regularly to prevent debris
Handle carefully to avoid mechanical damage

Mnemonic: "Check Clean Correct Carefully"

Question 2(B)(3) [4 marks]

In an experiment of finding the periodic time of a simple pendulum, the observations are 1.96 s, 1.98 s, 2.00 s, 2.02 s, 2.04 s. Calculate absolute error, mean absolute error, relative error and percentage error.

Answer:

Observations: 1.96, 1.98, 2.00, 2.02, 2.04 s

Mean value = (1.96 + 1.98 + 2.00 + 2.02 + 2.04) ÷ 5 = 2.00 s

Absolute errors: |xi - mean|
|1.96 - 2.00| = 0.04 s
|1.98 - 2.00| = 0.02 s
|2.00 - 2.00| = 0.00 s
|2.02 - 2.00| = 0.02 s
|2.04 - 2.00| = 0.04 s

Mean absolute error = (0.04 + 0.02 + 0.00 + 0.02 + 0.04) ÷ 5 = 0.024 s

Relative error = Mean absolute error ÷ Mean value = 0.024 ÷ 2.00 = 0.012

Percentage error = Relative error × 100 = 0.012 × 100 = 1.2%

Results: Mean absolute error = 0.024 s, Relative error = 0.012, Percentage error = 1.2%

Mnemonic: "Mean Absolute Relative Percentage"

Question 3(A) - Attempt any two [6 marks]

Question 3(A)(1) [3 marks]

Define: Electric flux, Electric field, Potential Difference

Answer:

Term	Definition	Unit	Formula
Electric Flux	Number of electric field lines passing through a surface	Nm²/C	Φ = E·A
Electric Field	Force per unit positive charge	N/C	E = F/q
Potential Difference	Work done per unit charge between two points	Volt	V = W/q

Electric flux: Measure of field lines penetrating surface
Electric field: Region where electric force acts on charges
Potential difference: Energy difference per unit charge

Mnemonic: "Flux Field Force, Work Watts Volts"

Question 3(A)(2) [3 marks]

Derive the formula for equivalent capacitance when three different capacitors are connected in series with necessary circuit diagram.

Answer:

Circuit Diagram:

    +----||----||----||----+
    |    C1    C2    C3    |
    |                      |
    +----------V-----------+

Derivation:

Same charge Q flows through each capacitor
Voltage divides: V = V₁ + V₂ + V₃
For each capacitor: V₁ = Q/C₁, V₂ = Q/C₂, V₃ = Q/C₃
Total voltage: V = Q/C₁ + Q/C₂ + Q/C₃ = Q(1/C₁ + 1/C₂ + 1/C₃)
For equivalent: V = Q/Cs
Therefore: 1/Cs = 1/C₁ + 1/C₂ + 1/C₃

Formula: 1/Cs = 1/C₁ + 1/C₂ + 1/C₃

Mnemonic: "Series Sums reciprocals, Same charge Splits voltage"

Question 3(A)(3) [3 marks]

Define: Infrasonic sound, Audible Sound, Ultrasonic sound

Answer:

Sound Type	Frequency Range	Characteristics	Applications
Infrasonic	Below 20 Hz	Inaudible to humans	Earthquake detection
Audible	20 Hz to 20 kHz	Audible to humans	Communication, music
Ultrasonic	Above 20 kHz	Inaudible to humans	Medical imaging, SONAR

Infrasonic: Low frequency sounds below human hearing
Audible: Normal hearing range for humans
Ultrasonic: High frequency sounds above human hearing

Mnemonic: "Infra-Below, Audible-Between, Ultra-Above"

Question 3(B) - Attempt any two [8 marks]

Question 3(B)(1) [4 marks]

Prove C = ε₀A/d for parallel plate capacitor.

Answer:

Diagram:

    +--------+  +--------+
    |   +Q   |  |   -Q   |
    |        |  |        |
    | Plate1 |  | Plate2 |
    |   A    |  |   A    |
    +--------+  +--------+
         <---d--->

Derivation:

Electric field between plates: E = σ/ε₀ = Q/(ε₀A)
Potential difference: V = E × d = Qd/(ε₀A)
Capacitance definition: C = Q/V
Substituting: C = Q ÷ [Qd/(ε₀A)] = ε₀A/d

Final Formula: C = ε₀A/d

Where:

ε₀: Permittivity of free space
A: Area of plates
d: Distance between plates

Mnemonic: "Capacitance equals epsilon-zero Area over distance"

Question 3(B)(2) [4 marks]

List the characteristics of electric field lines.

Answer:

Key Characteristics:

Direction: From positive to negative charge
Density: Indicates field strength
Continuous: Never break in free space
Non-intersecting: No two lines cross
Perpendicular: To conductor surface
Closed loops: Only around changing magnetic fields
Tangent: Gives field direction at any point
Uniform spacing: In uniform field regions

Properties:

Start from positive charges
End at negative charges
Higher density means stronger field
Never intersect each other

Mnemonic: "Positive to Negative, Dense means Strong, Never cross, Always perpendicular"

Question 3(B)(3) [4 marks]

Describe working and construction of magnetostriction method used for production of ultrasonic waves.

Answer:

Construction:

    Oscillator -> Coil -> Nickel Rod -> Horn
                    |       |          |
                   AC    Vibrates   Amplifies

Components:

Nickel rod: Magnetostrictive material
Coil: Electromagnet around rod
AC oscillator: High frequency current source
Horn: Sound amplifier and transmitter

Working Principle:

AC current flows through coil
Magnetic field changes rapidly
Nickel rod expands and contracts
Mechanical vibrations produced
Ultrasonic waves generated

Applications: Medical imaging, cleaning, welding

Mnemonic: "AC Coil Makes Nickel vibrate, Creates Ultrasonic"

Question 4(A) - Attempt any two [6 marks]

Question 4(A)(1) [3 marks]

A radio station broadcasts its radio signals at 9.26 × 10⁷ Hz. Find the wavelength if the waves travel at a speed of 3.00 × 10⁸ m/s.

Answer:

Given:
Frequency (f) = 9.26 × 10⁷ Hz
Speed (c) = 3.00 × 10⁸ m/s

Formula: c = fλ
Therefore: λ = c/f

λ = (3.00 × 10⁸) ÷ (9.26 × 10⁷)
λ = 3.24 m

Wavelength = 3.24 m

Mnemonic: "Speed equals frequency times wavelength"

Question 4(A)(2) [3 marks]

State the Snell's law and explain refractive index of media.

Answer:

Snell's Law: n₁ sin θ₁ = n₂ sin θ₂

Where:

n₁, n₂: Refractive indices of media 1 and 2
θ₁, θ₂: Angles of incidence and refraction

Refractive Index:

Type	Definition	Formula
Absolute	Speed of light in vacuum to medium	n = c/v
Relative	Ratio of speeds in two media	n₂₁ = v₁/v₂

Higher refractive index: Denser medium, slower light
Lower refractive index: Rarer medium, faster light

Mnemonic: "Snell Says Sine ratio constant, Dense slows Down light"

Question 4(A)(3) [3 marks]

Compare: Ordinary light and LASER

Answer:

Property	Ordinary Light	LASER
Coherence	Incoherent	Coherent
Color	Polychromatic	Monochromatic
Direction	Divergent	Parallel beam
Intensity	Low	Very high
Phase	Random	Fixed phase relationship
Wavelength	Multiple wavelengths	Single wavelength

Key Differences:

LASER: Coherent, monochromatic, parallel, intense
Ordinary: Incoherent, polychromatic, divergent, less intense

Mnemonic: "LASER: Coherent Monochromatic Parallel Intense"

Question 4(B) - Attempt any two [8 marks]

Question 4(B)(1) [4 marks]

Demonstrate the structure of an optical fiber with necessary diagram.

Answer:

Optical Fiber Structure:

    |<------ Core ------>|
    |                    |
    +--------------------+  <- Cladding
    |   Higher n₁        |
    |                    |  <- Lower n₂
    +--------------------+
    |                    |
    |   Protective       |  <- Jacket
    |   Coating          |
    +--------------------+

Components:

Component	Material	Function	Refractive Index
Core	Glass/Plastic	Light transmission	Higher (n₁)
Cladding	Glass	Total internal reflection	Lower (n₂)
Jacket	Plastic	Protection	-

Working Principle:

Light enters core at acceptance angle
Total internal reflection at core-cladding boundary
Light travels in zigzag path through core
n₁ > n₂ ensures light confinement

Mnemonic: "Core Cladding Jacket, Higher Lower Protection"

Question 4(B)(2) [4 marks]

List applications of LASER in engineering and medical field.

Answer:

Engineering Applications:

Cutting and welding: Precision metal cutting
3D printing: Laser sintering
Measurement: Distance and surveying
Communication: Optical fiber systems
Material processing: Surface hardening
Barcode scanning: Retail and inventory

Medical Applications:

Surgery: Precise tissue cutting
Eye treatment: Corrective surgery
Cancer treatment: Tumor destruction
Diagnostics: Spectroscopy
Dentistry: Cavity treatment
Skin treatment: Cosmetic procedures

Advantages: Precision, non-contact, sterile, minimal damage

Mnemonic: "Engineering: Cut Weld Measure Communicate, Medical: Surgery Eye Cancer Diagnose"

Question 4(B)(3) [4 marks]

Explain P-type and N-type semiconductors.

Answer:

N-type Semiconductor:

Property	N-type
Dopant	Phosphorus, Arsenic (5 valence electrons)
Majority carriers	Electrons
Minority carriers	Holes
Charge	Negative

P-type Semiconductor:

Property	P-type
Dopant	Boron, Aluminum (3 valence electrons)
Majority carriers	Holes
Minority carriers	Electrons
Charge	Positive

Formation Process:

N-type: Pentavalent atoms donate electrons
P-type: Trivalent atoms accept electrons, create holes
Doping: Controlled addition of impurities
Conductivity: Increases due to free carriers

Mnemonic: "N-type Negative electrons, P-type Positive holes"

Question 5(A) - Attempt any two [6 marks]

Question 5(A)(1) [3 marks]

Classify conductors, semiconductors and insulators based on energy band gap.

Answer:

Material	Energy Band Gap	Characteristics	Examples
Conductor	No gap (0 eV)	Valence and conduction bands overlap	Copper, Silver
Semiconductor	Small gap (1-3 eV)	Moderate band gap	Silicon, Germanium
Insulator	Large gap (>3 eV)	Wide band gap	Glass, Rubber

Energy Band Diagram:

Conductor    Semiconductor    Insulator
   
   CB            CB              CB
   --            --              --
   VB            VB              VB
   
No Gap        Small Gap       Large Gap

CB: Conduction Band
VB: Valence Band
Gap determines electrical conductivity

Mnemonic: "No gap Conducts, Small gap Semi, Large gap Insulates"

Question 5(A)(2) [3 marks]

Explain OR and AND logic gates with necessary truth table.

Answer:

OR Gate:

A	B	Y = A + B
0	0	0
0	1	1
1	0	1
1	1	1

AND Gate:

A	B	Y = A · B
0	0	0
0	1	0
1	0	0
1	1	1

Symbols:

OR Gate:     A ----\
                   >---- Y
             B ----/

AND Gate:    A ----\
                   &---- Y
             B ----/

OR: Output HIGH when any input is HIGH
AND: Output HIGH when all inputs are HIGH

Mnemonic: "OR: Any high makes high, AND: All high makes high"

Question 5(A)(3) [3 marks]

Describe the use of Zener diode as a voltage regulator.

Answer:

Circuit Diagram:

    Vin ----[Rs]----+----Vout
                     |
                   [Zener]
                     |
                    GND

Working Principle:

Forward bias: Acts like normal diode
Reverse bias: Breaks down at Zener voltage
Voltage regulation: Maintains constant Vout = Vz
Series resistor: Limits current through Zener

Characteristics:

Zener voltage: Constant breakdown voltage
Current range: Wide operating range
Temperature stability: Good voltage stability
Power rating: Must not exceed maximum power

Applications: Power supplies, voltage references, protection circuits

Mnemonic: "Zener Zealously maintains Voltage despite Variations"

Question 5(B) - Attempt any two [8 marks]

Question 5(B)(1) [4 marks]

Explain full wave rectifier with necessary circuit and draw input and output waveforms.

Answer:

Center-tap Full Wave Rectifier:

    AC Input ----+----[D1]----+---- Positive Output
                 |             |
            Transformer     Load (RL)
                 |             |
                 +----[D2]----+---- Common

Working:

Positive half cycle: D1 conducts, D2 off
Negative half cycle: D2 conducts, D1 off
Both halves: Current flows through load in same direction

Waveforms:

Input:     /\  /\  /\  /\
          /  \/  \/  \/  \
         /                \

Output:   /\    /\    /\
         /  \  /  \  /  \
        /    \/    \/    \

Advantages: Better efficiency, lower ripple, better transformer utilization

Mnemonic: "Full wave uses Full cycle, Better efficiency Better output"

Question 5(B)(2) [4 marks]

Demonstrate forward and reverse characteristics of P-N junction diode.

Answer:

Forward Bias Characteristics:

Voltage Range	Current	Behavior
0 to 0.3V (Si)	Very small	Cut-in voltage
Above 0.7V	Exponential increase	Conducting

Reverse Bias Characteristics:

Voltage Range	Current	Behavior
0 to breakdown	Reverse saturation	Leakage current
Breakdown voltage	Sharp increase	Avalanche breakdown

I-V Characteristic Curve:

       I(mA)
        |
        |    Forward
       /|     bias
      / |
     /  |
----+---+----V(V)
  -20   0.7  
    |
    |Reverse
    |bias

Key Points:

Forward: Low resistance, high current
Reverse: High resistance, low current
Cut-in voltage: 0.7V for Silicon, 0.3V for Germanium

Mnemonic: "Forward Flow, Reverse Resist"

Question 5(B)(3) [4 marks]

Write the principle of LED and explain its construction and working.

Answer:

Principle: Electroluminescence - Direct conversion of electrical energy to light energy

Construction:

    Light Output
        ↑
    +-------+
    | P-type|  <- Anode
    +-------+
    |Junction|
    +-------+
    | N-type|  <- Cathode
    +-------+

Materials Used:

Color	Material	Wavelength
Red	GaAs	700 nm
Green	GaP	550 nm
Blue	GaN	470 nm

Working:

Forward bias: Electrons and holes recombine at junction
Energy release: Photons emitted during recombination
Light color: Depends on band gap energy
Efficiency: High electrical to optical conversion

Applications: Displays, indicators, lighting, optical communication

Mnemonic: "LED: Light Emitting Diode, Electrons and holes Dance to make Light"