Modern Physics (DI01000061) - Winter 2024 Solution

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

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Answer:

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

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

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

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Differentiate between accuracy and precision.

Answer:

• 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]

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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]

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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]

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

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Draw a sketch of micrometer screw gauge with proper nomenclature.

Answer:

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]

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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:

Diagram:

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]

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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]

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

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Define: Electric flux, Electric field, Potential Difference

Answer:

• 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]

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

Answer:

Circuit Diagram:

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]

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Define: Infrasonic sound, Audible Sound, Ultrasonic sound

Answer:

• 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]

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

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Prove C = ε₀A/d for parallel plate capacitor.

Answer:

Diagram:

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]

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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]

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

Answer:

Construction:

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]

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

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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]

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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:

• 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]

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Compare: Ordinary light and LASER

Answer:

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]

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

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Demonstrate the structure of an optical fiber with necessary diagram.

Answer:

Optical Fiber Structure:

Components:

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]

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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]

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Explain P-type and N-type semiconductors.

Answer:

N-type Semiconductor:

P-type Semiconductor:

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]

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

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

Answer:

Energy Band Diagram:

• 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]

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Explain OR and AND logic gates with necessary truth table.

Answer:

OR Gate:

AND Gate:

Symbols:

• 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]

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Describe the use of Zener diode as a voltage regulator.

Answer:

Circuit Diagram:

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]

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

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

Answer:

Center-tap Full Wave Rectifier:

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:

Advantages: Better efficiency, lower ripple, better transformer utilization

Mnemonic: “Full wave uses Full cycle, Better efficiency Better output”

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

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

Answer:

Forward Bias Characteristics:

Reverse Bias Characteristics:

I-V Characteristic Curve:

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]

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

Answer:

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

Construction:

Materials Used:

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”