2023 JUPEB Physics

PHY 001: MECHANICS AND PROPERTIES OF MATTER

1. (a) Mention the THREE types of modulus.

(b) Briefly explain the following: [5 Marks]

• i. Elasticity

• ii. Plasticity

• iii. Ductility

• iv. Malleability

© A particle is moving along the x-axis such that its position is given by: [4 Marks]

x = 4t² − 16t + 12

• i. Find the instantaneous velocity when t = 5s

• ii. Find the instantaneous acceleration when t = 5s

• iii. At what time is the particle stationary?

[4½ Marks]

[Total = 10 Marks]

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2. (a) i. State the Archimedes’ Principle.

• ii. List FOUR characteristics of pressure in a fluid. [3 Marks]

(b) Distinguish between a compressible fluid and an incompressible fluid. [2 Marks]

© Two divers M and N are at a depth 60 m and 80 m respectively below the water surface of a sea. The pressure on M is P₁ and the pressure on N is P₂. If the atmospheric pressure is equivalent to 20 m of water, find the value of P₂/P₁. [2 Marks]

(d) Differentiate between conservative force and non-conservative force, giving one example each.

[3 Marks]

[Total = 10 Marks]

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PHY 002: HEAT, WAVES AND OPTICS

3. (a) State Huygens’ principle. [1 Mark]

(b) i. Explain the term “electromagnetic spectrum”. List any THREE of its components.

• ii. In a Young’s slit experiment, the separation between the first and the fifth bright fringes is 2.5 mm when the wavelength used is 4.5 × 10⁻⁷ m. The distance from the slits to the screen is 0.9 m. Calculate the separation of the two slits. [4 Marks]

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© i. Define internal energy.

• ii. State the second law of thermodynamics. [2 Marks]

(d) i. Using the first law of thermodynamics, write expressions for adiabatic and isochoric processes.

• ii. State FOUR factors which affect heat loss by convection. [3 Marks]

[Total = 10 Marks]

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4. (a) State TWO similarities and TWO differences between image formed by a converging mirror and a converging lens. [2 Marks]

(b) i. Mention any TWO uses each of plane mirror, concave and convex mirror.

• ii. Copy and complete the table below for the image formed by a concave mirror for different positions of the object.

| Position of Object | Position of Image | Size of Image | Nature of Image |

|—|---|—|---|

| At infinity | | | |

| At C | | | |

| Between C and F | | | |

| At F | | | |

| Between F and P | | | |

[8 Marks]

[Total = 10 Marks]

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PHY 003: ELECTRICITY AND MAGNETISM

5. (a) i. Explain what is meant by relative permittivity.

• ii. State TWO physical desirable properties in a material considered for dielectric in a capacitor. [3 Marks]

(b) A long magnet is removed from the centre of a coil of 30 turns. The speed of the magnet is controlled to maintain an induced e.m.f. of 80 μV across the coil. Removing the magnet in this way takes 3 minutes. Calculate the change of flux through the coil. [3 Marks]

© An electron enters the region of a uniform electric field as shown in Figure 1 below, with v₀ = 3.00 × 10⁶ m/s and E = 200 N/C. The horizontal length of such plates is l = 0.100 m.

• i. Find the acceleration of the electron while it is in the electric field.

• ii. Assuming the electron enters the field at time t = 0, find the time at which it leaves the field. (mass of an electron = 1.67 × 10⁻³¹ kg, charge of an electron = 1.60 × 10⁻¹⁹ C)

[4 Marks]

[Total = 10 Marks]

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6. (a) How can the motion of a moving charged particle be used to distinguish between a magnetic field and an electric field? [2 Marks]

(b) Consider the circuit diagram in Figure 2. Calculate the current in each resistor.

(Circuit diagram: Two batteries — 11.0 V with 20.0 Ω resistor, and 12.0 V with 17.0 Ω resistor — connected in a network with a 10.0 Ω resistor)

[5 Marks]

© A proton is moving in a circular orbit of radius 14 cm in a uniform 0.35 T magnetic field perpendicular to the velocity of the proton. Determine the speed of the proton. [3 Marks]

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PHY 004: MODERN PHYSICS

7. (a) Calculate the de Broglie wavelength for a particle moving with a speed of 2.30 × 10⁸ m/s if the particle is: [3 Marks]

• i. an electron

• ii. a proton

• iii. a 200 g bullet

(b) i. Give the properties of α, β, and γ radiation in terms of charge, mass, ionizing effect, and field effect.

• ii. Explain the effect of temperature and pressure on the rate of disintegration of a radioactive nucleus.

[7 Marks]

[Total = 10 Marks]

© i. State Heisenberg’s Uncertainty Principle.

• ii. List FOUR practical applications of X-rays. [3 Marks]

(d) A photon has a wavelength of 1Å. Calculate the:

• i. energy of the photon in electron volts

• ii. momentum of the photon

(1Å = 10⁻¹⁰ m) [3 Marks]

(e) i. Define binding energy.

• ii. Calculate the average binding energy per nucleon of ¹⁵⁶N which has a mass number of 56.0930 u. (mass of proton = 1.007825 u; mass of neutron = 1.008665 u; 1u = 931.5 MeV)

SECTION B: PHYSICS ESSAY — COMPLETE SOLUTIONS

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PHY 001: MECHANICS AND PROPERTIES OF MATTER

Question 1

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(a) THREE Types of Modulus

1. Young’s Modulus (E) — ratio of tensile stress to tensile strain

2. Bulk Modulus (K) — ratio of volumetric stress to volumetric strain

3. Shear (Rigidity) Modulus (G) — ratio of shear stress to shear strain

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(b) Brief Explanations

i. Elasticity:

The property of a material that enables it to regain its original shape and size after the removal of an applied force (deforming force).

ii. Plasticity:

The property of a material that causes it to permanently deform and not return to its original shape after the deforming force is removed.

iii. Ductility:

The property of a material that allows it to be drawn into thin wires without breaking when a tensile force is applied (e.g., copper, gold).

iv. Malleability:

The property of a material that allows it to be rolled or hammered into thin sheets without breaking (e.g., aluminium, lead).

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© x = 4t² − 16t + 12

i. Instantaneous velocity when t = 5s

$$v = \frac{dx}{dt} = 8t - 16$$

At t = 5:

$$v = 8(5) - 16 = 40 - 16 = \boxed{24 \text{ m/s}}$$

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ii. Instantaneous acceleration when t = 5s

$$a = \frac{dv}{dt} = 8 \text{ m/s}^2$$

Acceleration is constant = 8 m/s² (independent of t)

$$\boxed{a = 8 \text{ m/s}^2}$$

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iii. Time when particle is stationary (v = 0)

$$8t - 16 = 0$$

$$8t = 16$$

$$\boxed{t = 2 \text{ s}}$$

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Question 2

(a) i. Archimedes’ Principle

“When a body is wholly or partially immersed in a fluid, it experiences an upthrust (buoyant force) equal to the weight of the fluid displaced.”

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(a) ii. FOUR Characteristics of Pressure in a Fluid

1. Pressure at a point in a fluid acts equally in all directions.

2. Pressure increases with depth (P = ρgh).

3. Pressure depends on the density of the fluid.

4. Pressure acts perpendicular to any surface in contact with the fluid.

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(b) Compressible vs Incompressible Fluid

| Compressible Fluid | Incompressible Fluid |

|—|---|

| Density changes with applied pressure | Density remains constant regardless of pressure |

| Volume decreases under pressure | Volume does not change under pressure |

| Example: gases (air, steam) | Example: liquids (water, oil) |

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© Find P₂/P₁

Atmospheric pressure = 20 m of water

Pressure at depth h: P = P_atm + ρgh → in metres of water: P = (20 + h)

• P₁ (diver M at 60 m) = 20 + 60 = 80 m of water

• P₂ (diver N at 80 m) = 20 + 80 = 100 m of water

$$\frac{P_2}{P_1} = \frac{100}{80} = \boxed{1.25}$$

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(d) Conservative vs Non-Conservative Force

| Conservative Force | Non-Conservative Force |

|—|---|

| Work done is independent of path | Work done depends on path |

| Total mechanical energy is conserved | Energy is lost (usually as heat) |

| Work done in a closed loop = 0 | Work done in a closed loop ≠ 0 |

| Example: Gravitational force, elastic spring force | Example: Friction force, air resistance |

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PHY 002: HEAT, WAVES AND OPTICS

Question 3

(a) Huygens’ Principle

“Every point on a wavefront acts as a source of secondary wavelets that spread out in the forward direction with the same speed as the wave. The new wavefront is the tangent (envelope) to all these secondary wavelets.”

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(b) i. Electromagnetic Spectrum

Definition: The electromagnetic spectrum is the complete range of all electromagnetic waves arranged in order of increasing frequency (or decreasing wavelength). These waves all travel at the speed of light (3 × 10⁸ m/s) in a vacuum.

THREE components:

1. Radio waves

2. X-rays

3. Visible light (gamma rays, microwaves, infrared, ultraviolet also acceptable)

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(b) ii. Young’s Double Slit — Separation of Slits

Given:

• Fringe separation between 1st and 5th bright fringe: distance spans 4 fringe widths

• Total distance = 2.5 mm = 2.5 × 10⁻³ m

• λ = 4.5 × 10⁻⁷ m

• D = 0.9 m

Fringe width:

$$\beta = \frac{2.5 \times 10^{-3}}{4} = 6.25 \times 10^{-4} \text{ m}$$

Formula: $\beta = \frac{\lambda D}{d}$

$$d = \frac{\lambda D}{\beta} = \frac{4.5 \times 10^{-7} \times 0.9}{6.25 \times 10^{-4}}$$

$$d = \frac{4.05 \times 10^{-7}}{6.25 \times 10^{-4}}$$

$$\boxed{d = 6.48 \times 10^{-4} \text{ m} \approx 0.648 \text{ mm}}$$

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© i. Internal Energy

Internal energy is the total sum of the kinetic and potential energies of all the molecules (particles) making up a substance. It includes translational, rotational, and vibrational energies of molecules.

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© ii. Second Law of Thermodynamics

“Heat cannot spontaneously flow from a colder body to a hotter body without the expenditure of external work.”

OR (Kelvin-Planck statement): “It is impossible to construct a heat engine that operates in a cycle and converts all its heat input into work with no other effect.”

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(d) i. First Law Expressions

First Law: ΔU = Q − W

Adiabatic process (no heat exchange, Q = 0):

$$\Delta U = -W \quad \Rightarrow \quad W = -\Delta U$$

Isochoric process (constant volume, W = 0 since W = PΔV and ΔV = 0):

$$\Delta U = Q$$

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(d) ii. FOUR Factors Affecting Heat Loss by Convection

1. Temperature difference between the surface and the surrounding fluid — greater difference increases convection.

2. Surface area — larger area allows more heat loss.

3. Density and viscosity of the fluid — less dense, less viscous fluids convect more easily.

4. Velocity/flow of the fluid — faster fluid movement (forced convection) increases heat loss.

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Question 4

(a) Similarities and Differences: Converging Mirror vs Converging Lens

TWO Similarities:

1. Both can produce real and inverted images when the object is beyond the focal point.

2. Both produce virtual, erect, and magnified images when the object is between the focal point and the optical centre/pole.

TWO Differences:

| Converging Mirror (Concave) | Converging Lens |

|—|---|

| Works by reflection of light | Works by refraction of light |

| Image forms on the same side as the object | Image forms on the opposite side from the object (for real image) |

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(b) i. Uses of Mirrors

Plane Mirror:

1. Used in dressing/bathrooms to see one’s reflection

2. Used in periscopes and kaleidoscopes

Concave Mirror:

1. Used in shaving/makeup mirrors (magnified upright image)

2. Used as reflectors in torches, car headlamps, and satellite dishes

Convex Mirror:

1. Used as driving/rear-view mirrors in vehicles (wide field of view)

2. Used in supermarkets and road junctions as security/surveillance mirrors

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(b) ii. Completed Table — Concave Mirror

| Position of Object | Position of Image | Size of Image | Nature of Image |

|—|---|—|---|

| At infinity | At F (focal point) | Highly diminished (point) | Real, Inverted |

| At C | At C | Same size as object | Real, Inverted |

| Between C and F | Beyond C | Magnified (enlarged) | Real, Inverted |

| At F | At infinity | Infinitely large | Real, Inverted |

| Between F and P | Behind the mirror (same side as object) | Magnified (enlarged) | Virtual, Erect |

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PHY 003: ELECTRICITY AND MAGNETISM

Question 5

(a) i. Relative Permittivity

Relative permittivity (εᵣ), also called the dielectric constant, is the ratio of the permittivity of a material (ε) to the permittivity of free space (ε₀):

$$\varepsilon_r = \frac{\varepsilon}{\varepsilon_0}$$

It indicates how much more charge a capacitor can store when a dielectric material is inserted compared to vacuum.

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(a) ii. TWO Physical Desirable Properties of Dielectric in a Capacitor

1. High dielectric constant (εᵣ) — to increase the capacitance.

2. High dielectric strength — ability to withstand large electric fields without breaking down (conducting).

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(b) Change of Flux Through the Coil

Given:

• N = 30 turns

• EMF (ε) = 80 μV = 80 × 10⁻⁶ V

• Time (t) = 3 minutes = 180 s

Faraday’s Law:

$$\varepsilon = N \frac{\Delta\Phi}{\Delta t}$$

$$\Delta\Phi = \frac{\varepsilon \cdot \Delta t}{N} = \frac{80 \times 10^{-6} \times 180}{30}$$

$$\Delta\Phi = \frac{14.4 \times 10^{-3}}{30}$$

$$\boxed{\Delta\Phi = 4.8 \times 10^{-4} \text{ Wb}}$$

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© Electron in Uniform Electric Field

Given:

• v₀ = 3.00 × 10⁶ m/s

• E = 200 N/C

• l = 0.100 m

• mₑ = 9.11 × 10⁻³¹ kg (standard value; note: question gives 1.67 × 10⁻³¹ which is closer to proton — using standard electron mass)

• q = 1.60 × 10⁻¹⁹ C

i. Acceleration of electron:

$$F = qE = 1.60 \times 10^{-19} \times 200 = 3.20 \times 10^{-17} \text{ N}$$

$$a = \frac{F}{m} = \frac{3.20 \times 10^{-17}}{9.11 \times 10^{-31}}$$

$$\boxed{a = 3.51 \times 10^{13} \text{ m/s}^2}$$

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ii. Time to leave the field:

The electron travels horizontal distance l = 0.100 m at constant horizontal velocity v₀:

$$t = \frac{l}{v_0} = \frac{0.100}{3.00 \times 10^6}$$

$$\boxed{t = 3.33 \times 10^{-8} \text{ s}}$$

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Question 6

(a) Distinguishing Magnetic and Electric Fields Using a Moving Charged Particle

• In an electric field, a charged particle experiences a force parallel (or anti-parallel) to the field direction, regardless of whether the particle is moving or stationary.

• In a magnetic field, a charged particle experiences a force only when it is moving, and the force is always perpendicular to both the velocity and the magnetic field (F = qv × B).

Therefore, if a stationary charge is undeflected but a moving charge is deflected in a direction perpendicular to its motion, the field is magnetic. If the charge is deflected regardless of motion, the field is electric.

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(b) Circuit Analysis — Current in Each Resistor

Given circuit (from Figure 2):

• Battery 1: 11.0 V with 20.0 Ω (top branch)

• Battery 2: 12.0 V with 17.0 Ω (middle branch)

• 10.0 Ω resistor (bottom branch, no battery)

Using Kirchhoff’s Voltage Law (KVL) — assign mesh currents I₁ (top loop) and I₂ (bottom loop):

Mesh 1 (top loop):

$$11.0 = 20I_1 + 10(I_1 - I_2)$$

$$11.0 = 30I_1 - 10I_2 \quad \text{...(1)}$$

Mesh 2 (bottom loop):

$$12.0 = 17I_2 + 10(I_2 - I_1)$$

$$12.0 = -10I_1 + 27I_2 \quad \text{...(2)}$$

Solving simultaneously:

From (1): $11 = 30I_1 - 10I_2$ → multiply by 2.7:

$$29.7 = 81I_1 - 27I_2 \quad \text{...(3)}$$

Add (2) + (3):

$$12.0 + 29.7 = -10I_1 + 81I_1$$

$$41.7 = 71I_1$$

$$I_1 = \frac{41.7}{71} = 0.587 \text{ A}$$

Substitute into (1):

$$11 = 30(0.587) - 10I_2$$

$$11 = 17.61 - 10I_2$$

$$10I_2 = 6.61$$

$$I_2 = 0.661 \text{ A}$$

Currents:

• 20 Ω resistor: I₁ = 0.587 A

• 17 Ω resistor: I₂ = 0.661 A

• 10 Ω resistor: I₁ − I₂ = 0.587 − 0.661 = −0.074 A (flows in direction of I₂)

$$\boxed{I_{20\Omega} = 0.587 \text{ A}, \quad I_{17\Omega} = 0.661 \text{ A}, \quad I_{10\Omega} = 0.074 \text{ A}}$$

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© Speed of Proton in Circular Orbit

Given:

• r = 14 cm = 0.14 m

• B = 0.35 T

• mₚ = 1.67 × 10⁻²⁷ kg

• q = 1.60 × 10⁻¹⁹ C

Using: $r = \frac{mv}{qB}$

$$v = \frac{qBr}{m} = \frac{1.60 \times 10^{-19} \times 0.35 \times 0.14}{1.67 \times 10^{-27}}$$

$$v = \frac{7.84 \times 10^{-21}}{1.67 \times 10^{-27}}$$

$$\boxed{v = 4.69 \times 10^6 \text{ m/s}}$$

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PHY 004: MODERN PHYSICS

Question 7

(a) de Broglie Wavelength: λ = h/mv

Given: v = 2.30 × 10⁸ m/s, h = 6.626 × 10⁻³⁴ J·s

i. Electron (mₑ = 9.11 × 10⁻³¹ kg):

$$\lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 2.30 \times 10^8}$$

$$\lambda = \frac{6.626 \times 10^{-34}}{2.095 \times 10^{-22}}$$

$$\boxed{\lambda_e = 3.16 \times 10^{-12} \text{ m}}$$

ii. Proton (mₚ = 1.67 × 10⁻²⁷ kg):

$$\lambda = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 2.30 \times 10^8}$$

$$\lambda = \frac{6.626 \times 10^{-34}}{3.841 \times 10^{-19}}$$

$$\boxed{\lambda_p = 1.72 \times 10^{-15} \text{ m}}$$

iii. 200 g bullet (m = 0.200 kg):

$$\lambda = \frac{6.626 \times 10^{-34}}{0.200 \times 2.30 \times 10^8}$$

$$\lambda = \frac{6.626 \times 10^{-34}}{4.60 \times 10^7}$$

$$\boxed{\lambda_{bullet} = 1.44 \times 10^{-41} \text{ m}}$$

(Negligibly small — confirming macroscopic objects show no wave behaviour)

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(b) i. Properties of α, β, γ Radiation

| Property | Alpha (α) | Beta (β) | Gamma (γ) |

|—|---|—|---|

| Charge | +2 (positive) | −1 (negative) | 0 (neutral) |

| Mass | 4 u (heavy) | ~1/1836 u (light) | 0 (massless photon) |

| Ionizing Effect | Strongly ionizing | Moderately ionizing | Weakly ionizing |

| Field Effect | Deflected by E and B fields (toward −ve plate) | Deflected (toward +ve plate) | Not deflected |

| Penetrating Power | Least (stopped by paper) | Moderate (stopped by Al) | Most (reduced by thick Pb) |

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(b) ii. Effect of Temperature and Pressure on Radioactive Disintegration

Temperature: Radioactive decay is a nuclear process (occurs in the nucleus). Temperature changes affect only the electron energy levels (chemical changes) and have no effect on the rate of radioactive disintegration.

Pressure: Similarly, pressure affects the electron cloud and intermolecular distances but cannot affect the nucleus. Therefore, pressure also has no effect on the rate of radioactive decay.

This distinguishes radioactive decay from ordinary chemical reactions, which are affected by both temperature and pressure.

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© i. Heisenberg’s Uncertainty Principle

“It is impossible to simultaneously determine with perfect accuracy both the position and the momentum (or velocity) of a particle.”

Mathematically:

$$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$

Where Δx = uncertainty in position, Δp = uncertainty in momentum.

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© ii. FOUR Practical Applications of X-rays

1. Medical diagnosis — used in radiography to detect bone fractures and internal injuries.

2. Cancer treatment (Radiotherapy) — high-energy X-rays destroy cancerous tumours.

3. Airport security — X-ray scanners inspect luggage for contraband items.

4. Crystallography — X-ray diffraction is used to determine the crystal structure of materials.

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(d) Photon with wavelength λ = 1 Å = 10⁻¹⁰ m

i. Energy of photon in electron volts:

$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{10^{-10}}$$

$$E = \frac{1.988 \times 10^{-25}}{10^{-10}} = 1.988 \times 10^{-15} \text{ J}$$

Converting to eV (1 eV = 1.6 × 10⁻¹⁹ J):

$$E = \frac{1.988 \times 10^{-15}}{1.6 \times 10^{-19}}$$

$$\boxed{E = 12,425 \text{ eV} \approx 12.4 \text{ keV}}$$

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ii. Momentum of the photon:

$$p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{10^{-10}}$$

KaTeX parse error: Undefined control sequence: \cdotp at position 1: \̲c̲d̲o̲t̲p̲

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(e) Binding Energy of ⁵⁶Fe (Iron-56)

(Note: The question states mass number 56 with 56.0930 u — this corresponds to Iron-56, ²⁶Fe₅₆ with 26 protons and 30 neutrons)

Given:

• Mass of nucleus = 56.0930 u (as given)

• mₚ = 1.007825 u

• mₙ = 1.008665 u

• 1 u = 931.5 MeV

• Z = 26 (protons), N = 30 (neutrons)

i. Binding Energy Definition:

Binding energy is the minimum energy required to completely separate all the nucleons (protons and neutrons) in a nucleus from each other, or equivalently, the energy released when nucleons combine to form the nucleus.

ii. Calculation:

Mass of constituents:

$$M_{constituents} = 26(1.007825) + 30(1.008665)$$

$$= 26.20345 + 30.25995 = 56.46340 \text{ u}$$

Mass defect:

$$\Delta m = 56.46340 - 56.0930 = 0.37040 \text{ u}$$

Total binding energy:

$$BE = \Delta m \times 931.5 = 0.37040 \times 931.5$$

$$BE = 345.13 \text{ MeV}$$

Average binding energy per nucleon:

$$BE/A = \frac{345.13}{56}$$

$$\boxed{BE/nucleon \approx 6.16 \text{ MeV/nucleon}}$$