Physics (4300005) - Summer 2024 Solution

Table of Contents

Question 1(a) [3 marks]
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Define derived physical quantities and give three examples with their S.I. unit and symbol.

Answer: Derived physical quantities are those which are obtained by multiplication or division of fundamental physical quantities.

Table: Examples of Derived Physical Quantities

Derived Quantity	S.I. Unit	Symbol
Force	Newton (N)	F
Energy	Joule (J)	E
Electric Current	Ampere (A)	I

Mnemonic: “FEI: Force-Energy-Current derive from fundamentals”

Question 1(b) [4 marks]
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The length of a metal rod is 64.522 cm at 12°C temperature and 64.576 cm at 90°C temperature. Find the coefficient of linear expansion of the metal rod.

Answer: Formula: α = (L₂ - L₁)/[L₁ × (T₂ - T₁)]

Calculation:

Initial length (L₁) = 64.522 cm
Final length (L₂) = 64.576 cm
Initial temperature (T₁) = 12°C
Final temperature (T₂) = 90°C

α = (64.576 - 64.522)/[64.522 × (90 - 12)] α = 0.054/(64.522 × 78) α = 0.054/5032.716 α = 1.073 × 10⁻⁵ /°C

Mnemonic: “Change in Length over Original Length times Change in Temperature”

Question 1(c) [7 marks]
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Explain with figure: The principle, construction and working of a vernier calliper.

Answer: Principle: Vernier caliper works on the principle of vernier scale, which allows measurements with accuracy greater than the main scale.

Construction:

graph TD
    A[Vernier Caliper] --> B[Main Scale]
    A --> C[Vernier Scale]
    A --> D[Fixed Jaw]
    A --> E[Movable Jaw]
    A --> F[Depth Probe]
    A --> G[Locking Screw]

Working:

Zero error check: Close jaws and note if zero of vernier coincides with zero of main scale
External measurement: Place object between fixed and movable jaws
Reading process: Note main scale reading + (coinciding vernier division × least count)
Least count = (Smallest division on main scale)/(Number of divisions on vernier scale)

Diagram:

Mnemonic: “Main Scale Reading Plus Vernier Division Times Least Count”

Question 1(c) OR [7 marks]
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Explain with figure: The principle, construction and working of a micrometre screw gauge.

Answer: Principle: Micrometer screw gauge works on the principle of screw motion - rotational motion is converted into linear motion.

Construction:

graph TD
    A[Micrometer Screw Gauge] --> B[Frame]
    A --> C[Anvil]
    A --> D[Spindle]
    A --> E[Sleeve/Main Scale]
    A --> F[Thimble/Circular Scale]
    A --> G[Ratchet]
    A --> H[Lock Nut]

Working:

Zero error check: Close anvil and spindle, note if zero of circular scale aligns with reference line
Measurement process: Place object between anvil and spindle
Reading: Main scale reading + (Circular scale reading × Least count)
Least Count = Pitch/Number of divisions on circular scale

Diagram:

Mnemonic: “PST: Pitch divided by Scale gives Thimble’s least count”

Question 2(a) [3 marks]
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Find the diameter of a sphere if pitch of micrometer screw gauge is 1 mm and there are 100 divisions on circular scale. The edge of circular scale lies between 7 and 8 mm of the main scale and 65th division of the circular scale coincides with the horizontal line of the main scale.

Answer: Formula: Diameter = Main scale reading + (Circular scale reading × Least count)

Calculation:

Main scale reading = 7 mm
Circular scale reading = 65 divisions
Least count = Pitch/Number of divisions = 1/100 = 0.01 mm

Diameter = 7 + (65 × 0.01) = 7 + 0.65 = 7.65 mm

Mnemonic: “MSR + (CSR × LC) gives the final measurement”

Question 2(b) [4 marks]
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Explain phase difference and coherence.

Answer: Phase Difference: The difference in phase angle between two waves of the same frequency.

Table: Phase Difference Characteristics

Phase Difference	Interference Type	Result
0° or 360°	Constructive	Maximum amplitude
180°	Destructive	Minimum amplitude

Coherence: Property of waves that have a constant phase relationship.

Types of Coherence:

Temporal coherence: Related to frequency stability
Spatial coherence: Related to wavefront uniformity

Mnemonic: “Constant Phase Relationship Creates Coherent waves”

Question 2(c) [7 marks]
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Explain capacitor, its capacitance and the effect of dielectric material on the capacitance of parallel plate capacitor.

Answer: Capacitor: Device that stores electric charge and electrical energy in an electric field.

Capacitance: Ratio of charge stored to potential difference applied.

Formula: C = Q/V

Parallel Plate Capacitor: Capacitance formula: C = ε₀A/d

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

Effect of Dielectric:

Increases capacitance by K times (K = dielectric constant)
New formula: C = Kε₀A/d

Diagram:

Mnemonic: “KIDS: K Increases Dielectric Storage”

Question 2(a) OR [3 marks]
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If the lengths of two cylinders are (6.52±0.01) cm and (4.48±0.02) cm respectively. Find the difference in their length with percentage error.

Answer: Calculation:

Length of first cylinder (L₁) = 6.52 ± 0.01 cm
Length of second cylinder (L₂) = 4.48 ± 0.02 cm
Difference in length (ΔL) = L₁ - L₂ = 6.52 - 4.48 = 2.04 cm

Absolute error in difference = √[(0.01)² + (0.02)²] = √(0.0001 + 0.0004) = √0.0005 = 0.022 cm

Percentage error = (Absolute error/Measured value) × 100 = (0.022/2.04) × 100 = 1.08%

Mnemonic: “Add errors in quadrature for difference calculations”

Question 2(b) OR [4 marks]
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Explain the types of interference with relevant figures.

Answer: Types of Interference:

Table: Interference Types

Type	Phase Difference	Result	Wave Amplitude
Constructive	0°, 360°, 720°…	Reinforcement	Maximum
Destructive	180°, 540°, 900°…	Cancellation	Minimum

Constructive Interference: When crest meets crest or trough meets trough.

Destructive Interference: When crest meets trough.

Diagram:

Mnemonic: “Crest + Crest = Constructive, Crest + Trough = Destructive”

Question 2(c) OR [7 marks]
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Derive the expression for potential due to point charge with necessary figure.

Answer: Potential at a point due to point charge:

Formula development:

Definition: Work done per unit charge to bring a test charge from infinity to that point
Expression: V = W/q₀ = ∫(F·dr)

Step-by-step derivation:

Force between charges (Coulomb’s law): F = (1/4πε₀) × (Qq/r²)
Work done moving test charge: W = ∫(F·dr)
For radial motion: W = (Q/4πε₀) × ∫(1/r²)dr from ∞ to r
Integrating: W = (Q/4πε₀) × [-1/r]ᵣ∞
Final result: V = W/q₀ = (1/4πε₀) × (Q/r)

Final formula: V = (1/4πε₀) × (Q/r)

Diagram:

Mnemonic: “POD: Potential Over Distance equals charge over r”

Question 3(a) [3 marks]
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Explain in brief charging by friction and induction methods.

Answer: Charging by Friction: Process of charging by rubbing two different materials together.

Steps in friction charging:

Electrons transfer from one material to another
Material losing electrons becomes positively charged
Material gaining electrons becomes negatively charged

Charging by Induction: Process of charging without direct contact.

Steps in induction charging:

Bring charged body near a neutral conductor
Redistribution of charges in neutral body
Ground the conductor and remove ground
Remove the charged body

Mnemonic: “FTEE: Friction Transfers Electrons Easily”

Question 3(b) [4 marks]
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A tuning fork vibrates at frequency of 256 Hz. If its velocity is 340 m/s, find (a) wavelength and (b) distance travelled by it in 50 oscillations.

Answer: Formulas:

Wavelength (λ) = Velocity (v) / Frequency (f)
Distance (d) = Number of oscillations (n) × Wavelength (λ)

Calculation: (a) Wavelength (λ) = v/f = 340/256 = 1.328 m

(b) Distance (d) = n × λ = 50 × 1.328 = 66.4 m

Mnemonic: “VFD: Velocity, Frequency and Distance are connected”

Question 3(c) [7 marks]
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Write the principle and construction of a bimetallic thermometer with a labelled diagram. Also mention its advantages and disadvantages.

Answer: Principle: Different metals expand differently when heated, causing the strip to bend.

Construction:

graph TD
    A[Bimetallic Thermometer] --> B[Fixed End]
    A --> C[Bimetallic Strip]
    A --> D[Pointer]
    A --> E[Scale]
    A --> F[Protective Case]
    C --> G[Metal with Higher Expansion]
    C --> H[Metal with Lower Expansion]

Working:

Temperature change causes different expansion rates
Bimetallic strip bends toward metal with lower expansion coefficient
Pointer movement indicates temperature

Diagram:

Advantages:

Simple, robust design
No power supply needed
Wide temperature range

Disadvantages:

Less accurate than other types
Slow response time
Subject to mechanical wear

Mnemonic: “BEDS: Bimetallic Elements Deform with Stress”

Question 3(a) OR [3 marks]
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Explain work done on a point charge in an electric field.

Answer: Work Done on Point Charge: The work done to move a point charge q in an electric field E.

Formula: W = q(Vₐ - Vᵦ) = qΔV

Where:

q = charge being moved
Vₐ = potential at initial position
Vᵦ = potential at final position
ΔV = potential difference

Key properties:

Work is independent of path taken
Work is positive when moving against electric field
Work is negative when moving along electric field

Mnemonic: “PEW: Potential difference × Electric charge = Work”

Question 3(b) OR [4 marks]
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What will be the distance travelled by a sound wave in 75 vibrations if its speed is 0.33 km/s and frequency is 660 Hz.

Answer: Formulas:

Wavelength (λ) = Velocity (v) / Frequency (f)
Distance (d) = Number of vibrations (n) × Wavelength (λ)

Calculation:

Convert velocity: v = 0.33 km/s = 330 m/s
Wavelength: λ = v/f = 330/660 = 0.5 m
Distance: d = n × λ = 75 × 0.5 = 37.5 m

Mnemonic: “FVW: Frequency into Velocity gives Wavelength”

Question 3(c) OR [7 marks]
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Write the principle and construction of a Mercury thermometer with a labelled diagram. Also mention its advantages and disadvantages.

Answer: Principle: Mercury thermometer works on the principle of thermal expansion of mercury when heated.

Construction:

graph TD
    A[Mercury Thermometer] --> B[Glass Bulb]
    A --> C[Capillary Tube]
    A --> D[Scale]
    A --> E[Mercury]
    A --> F[Vacuum/Nitrogen Space]
    A --> G[Safety Bulb]

Working:

Mercury expands when heated
Expansion causes mercury to rise in capillary
Height of mercury column indicates temperature

Diagram:

Advantages:

High accuracy
Wide temperature range (-38°C to 357°C)
Linear expansion of mercury
Good visibility of mercury thread

Disadvantages:

Mercury is toxic
Fragile glass construction
Cannot be used below -38°C
Slow response to temperature changes

Mnemonic: “MELT: Mercury Expands Linearly with Temperature”

Question 4(a) [3 marks]
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The electric force between two positive ions of equal magnitude separated by distance 5×10⁻¹⁰ m from eachother is 3.7 × 10⁻⁹ N. How many electrons would have been removed from each atom.

Answer: Formula: F = (1/4πε₀) × (q₁q₂/r²)

Calculation:

F = 3.7 × 10⁻⁹ N
r = 5 × 10⁻¹⁰ m
q₁ = q₂ = ne (n = number of electrons, e = electron charge)
1/4πε₀ = 9 × 10⁹ Nm²/C²
e = 1.6 × 10⁻¹⁹ C

3.7 × 10⁻⁹ = (9 × 10⁹) × (n²e²/(5 × 10⁻¹⁰)²) 3.7 × 10⁻⁹ = (9 × 10⁹) × (n² × (1.6 × 10⁻¹⁹)²/25 × 10⁻²⁰) Solving: n = 1 (1 electron removed from each atom)

Mnemonic: “FACE: Force Affects Charge Equally”

Question 4(b) [4 marks]
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State Snell’s law and derive its formula.

Answer: Snell’s Law: The ratio of sine of angle of incidence to sine of angle of refraction is constant for a given pair of media.

Formula: (sin i)/(sin r) = n₂/n₁ = constant

Derivation steps:

Light travels at different speeds in different media
When light passes from one medium to another, it changes direction
Using Fermat’s principle of least time
Ratio of speeds equals ratio of refractive indices
Final formula: n₁sin i = n₂sin r

Diagram:

Mnemonic: “SINIS: SIN I over SIN R equals refractive index ratio”

Question 4(c) [7 marks]
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Explain any three applications of Ultrasonic waves.

Answer: Applications of Ultrasonic Waves:

Table: Ultrasonic Applications

Application	Principle	Use
Medical Imaging	Reflection from tissues	Visualize internal organs
NDT (Non-Destructive Testing)	Reflection from defects	Find flaws in materials
Cleaning	Cavitation effect	Clean jewelry, surgical instruments

1. Medical Imaging (Sonography):

Frequencies: 1-10 MHz
Principle: Pulse-echo technique
Uses: Fetal imaging, organ scanning, blood flow measurement

2. Industrial NDT:

Detects cracks, voids, and flaws in materials
Quality control in manufacturing
Thickness measurement of materials

3. Ultrasonic Cleaning:

Creates microscopic bubbles (cavitation)
Removes contaminants from surfaces
Used for jewelry, optical components, surgical instruments

Mnemonic: “MIC: Medical, Industrial, Cleaning applications”

Question 4(a) OR [3 marks]
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Obtain the equivalent capacitance for series and parallel combinations of 3 capacitors having capacitances 5 µF, 10 µF and 15 µF respectively.

Answer: Parallel Combination: Cₚ = C₁ + C₂ + C₃ = 5 + 10 + 15 = 30 µF

Series Combination: 1/Cₛ = 1/C₁ + 1/C₂ + 1/C₃ 1/Cₛ = 1/5 + 1/10 + 1/15 1/Cₛ = 0.2 + 0.1 + 0.067 = 0.367 Cₛ = 1/0.367 = 2.72 µF

Mnemonic: “ASAP: Add for Series, Add inverse for Parallel”

Question 4(b) OR [4 marks]
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Explain the construction of an optical fibre with a neat diagram.

Answer: Construction of Optical Fiber:

Components:

Core: Light transmission medium
Cladding: Outer layer with lower refractive index
Buffer coating: Protective plastic covering

Parameters:

Core diameter: 8-50 μm (single mode), 50-100 μm (multimode)
Cladding diameter: 125-140 μm
Core refractive index > Cladding refractive index

Diagram:

Mnemonic: “CBC: Core-Buffer-Cladding from inside out”

Question 4(c) OR [7 marks]
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Explain production of ultrasonic waves by magnetostriction method.

Answer: Magnetostriction Method: The process of generating ultrasonic waves using the property of ferromagnetic materials to change dimensions when placed in a magnetic field.

Principle: Ferromagnetic materials change length when magnetized, producing mechanical vibrations that create ultrasonic waves.

Construction:

graph TD
    A[Magnetostriction Generator] --> B[AC Power Supply]
    A --> C[Coil/Solenoid]
    A --> D[Ferromagnetic Rod]
    A --> E[Acoustic Medium]
    A --> F[Cooling System]

Working Process:

AC current passes through solenoid
Alternating magnetic field produced
Ferromagnetic rod expands and contracts
Vibrations transmitted to medium
Ultrasonic waves generated

Diagram:

Advantages:

Simple construction
High power output
Suitable for liquids

Disadvantages:

Limited to frequencies below 100 kHz
Heating effects
Lower efficiency

Mnemonic: “FAME: Ferromagnetic Alternating Magnetic Effect”

Question 5(a) [3 marks]
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Explain in brief the three modes of heat transfer.

Answer: Three Modes of Heat Transfer:

Table: Heat Transfer Modes

Mode	Medium Requirement	Example
Conduction	Physical contact	Heat through metal rod
Convection	Fluid medium	Hot air rising
Radiation	No medium needed	Heat from sun

1. Conduction:

Transfer through direct molecular collision
No bulk movement of matter
Good in solids, especially metals

2. Convection:

Transfer through fluid movement
Requires density differences
Natural or forced convection

3. Radiation:

Transfer through electromagnetic waves
Works in vacuum
Depends on temperature and surface properties

Mnemonic: “CCR: Conduction Contact, Convection Current, Radiation Rays”

Question 5(b) [4 marks]
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Calculate the numerical aperture and acceptance angle of an optical fibre if the refractive indices of core and cladding of an optical fibre are 1.55 and 1.5 respectively.

Answer: Formulas:

Numerical Aperture (NA) = √(n₁² - n₂²)
Acceptance angle (θₐ) = sin⁻¹(NA)

Calculation:

Core refractive index (n₁) = 1.55
Cladding refractive index (n₂) = 1.5

NA = √(1.55² - 1.5²) NA = √(2.4025 - 2.25) NA = √0.1525 NA = 0.391

Acceptance angle (θₐ) = sin⁻¹(0.391) θₐ = 23.03°

Mnemonic: “CORE: Calculate Optical Refractive-index Exactly”

Question 5(c) [7 marks]
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Explain any three applications of optical fibres.

Answer: Applications of Optical Fibers:

Table: Major Optical Fiber Applications

Application	Advantage	Example
Communications	High bandwidth	Internet, phone networks
Medical	Flexibility, imaging	Endoscopy
Sensors	Immunity to EMI	Temperature sensing

1. Communication Networks:

Telecommunications and internet
Higher bandwidth than copper cables
Less signal attenuation over long distances
More secure against tapping

2. Medical Applications:

Endoscopy for minimally invasive procedures
Light delivery for photodynamic therapy
Dental procedures
Surgical illumination

3. Sensing Applications:

Temperature and pressure sensors
Strain gauges for structural monitoring
Chemical sensors
Gyroscopes for navigation

Mnemonic: “CMS: Communication, Medical, Sensing applications”

Question 5(a) OR [3 marks]
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Give a detailed explanation of specific heat.

Answer: Specific Heat: Amount of heat required to raise the temperature of 1 kg of a substance by 1 Kelvin (or 1°C).

Formula: Q = mc∆T

Where:

Q = Heat energy (J)
m = Mass (kg)
c = Specific heat capacity (J/kg·K)
∆T = Temperature change (K)

Units: J/kg·K or J/kg·°C

Significance:

Measures thermal inertia of materials
Higher specific heat means material requires more energy to heat up
Water has unusually high specific heat (4,186 J/kg·K)

Mnemonic: “STEM: Specific heat measures Temperature change per Energy and Mass”

Question 5(b) OR [4 marks]
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If the refractive indices of core and cladding of an optical fibre are 1.48 and 1.45 respectively. Calculate its acceptance angle and critical angle.

Answer: Formulas: