2020 JUPEB Physics

PHY 001: MECHANICS AND PROPERTIES OF MATTER

Question 1

(a) Briefly explain the following:

• (i) Young’s modulus

• (ii) Stiffness of elastic material [2 marks]

(b) Figure 1 shows a graph of force against extension for a metal specimen. Calculate the:

• (i) Force constant of the specimen

• (ii) Work done in stretching the specimen up to the proportional limit [3 marks]

© Two particles of mass 0.20 kg and 0.30 kg are placed 0.15 m apart. A third particle of mass 0.050 kg is placed between them on the line joining the first two particles. Calculate:

• (i) The gravitational force acting on the third particle if it is placed 0.050 m from the 0.30 kg mass

• (ii) Where along the line should it be placed for no gravitational force to be exerted on it? [5 marks]

[Total = 10 marks]

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Question 2

(a)

• (i) Define Unit Vector

• (ii) State the work-energy theorem [3 marks]

(b) A certain satellite is in circular orbit about the Earth at an altitude of 550 km. If the satellite makes a revolution every 110 minutes, calculate:

• (i) Its orbital speed

• (ii) The centripetal acceleration (R_earth = 6.4 × 10⁶ m) [4 marks]

© State without proof Bernoulli’s equation for a tube flow with varying cross-sectional area. State each component and parameter with associated SI units. [3 marks]

[Total = 10 marks]

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PHY 002: HEAT, WAVES AND OPTICS

Question 3

(a) What is a Diffraction Grating? [2 marks]

(b) Distinguish between the following:

• (i) Coherent sources and incoherent sources

• (ii) Constructive interference and destructive interference [4 marks]

© Determine the angular positions of the first and second order maxima for light of wavelength 400 nm and 700 nm incident on a grating containing 10,000 lines/cm. [4 marks]

[Total = 10 marks]

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Question 4

(a)

• (i) List four characteristics of waves

• (ii) Define linear magnification

• (iii) Write down without proof the lens maker’s formula for a thin lens in air [4 marks]

(b) The particle displacement y of air molecules due to a sound wave is given by:

y = 0.008 cos(ωt) sin(kz)

where k = 4π/m and ω = 50π rad/s. Calculate:

• (i) The distance between two consecutive nodes

• (ii) The amplitude after 0.565 s [3 marks]

© A truck travels down a highway at 66 m/s with its horn emitting sound at 800 Hz. What frequency is heard by a passenger in a car travelling at 60 m/s in the same direction, as the car and truck approach each other? (Speed of sound = 350 m/s) [3 marks]

[Total = 10 marks]

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PHY 003: ELECTRICITY AND MAGNETISM

Question 5

A circuit is used to measure energy transferred from a battery to a variable resistor R. The battery has e.m.f. E and internal resistance r. The p.d. across R is V and current in the circuit is I.

(a)

• (i) Distinguish between the definitions of e.m.f. and p.d. with reference to the circuit

• (ii) State Kirchhoff’s second law

• (iii) Using Kirchhoff’s second law, determine an expression for current I [4 marks]

(b) The variation of p.d. V with current I across R is shown in fig. 2b. Use the graph to determine:

• (i) The e.m.f. E

• (ii) The internal resistance r [2 marks]

©

• (i) Using data from fig. 2b, calculate the energy transferred to R per unit time for a current of 1.6 A

• (ii) Use answers from (b)(i) and ©(i) to calculate the efficiency of the battery at 1.6 A [4 marks]

[Total = 10 marks]

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Question 6

(a)

• (i) State Fleming’s left hand rule

• (ii) State two factors upon which the magnitude of the force on a charge moving in a magnetic field depends [3 marks]

(b) An electron is accelerated by a 5.2 kV potential difference. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 4.0 cm? [3 marks]

© A step-up transformer’s primary coil has 400 turns and its secondary coil has 1200 turns. The primary coil is connected to an AC generator with e.m.f. of 150 V.

• (i) Calculate the e.m.f. of the secondary circuit

• (ii) Find the current in the primary circuit if secondary current is 5 A

• (iii) Determine the power drawn by the primary circuit

• (iv) Determine the power supplied by the secondary circuit [4 marks]

[Total = 10 marks]

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PHY 004: MODERN PHYSICS

Question 7

(a) State two advantages and two disadvantages of fusion power from the viewpoint of safety, pollution, and resources. [2 marks]

(b) The radioactive isotope ¹⁹⁸Au has a half-life of 64.8 h. A sample has an initial activity (t = 0) of 40.0 mCi. Calculate the number of nuclei that decay between t₁ = 10.0 h and t₂ = 12.0 h. (1 Ci = 3.7 × 10¹⁰ Bq) [3 marks]

© If the average energy released per fission is 208 MeV, calculate the total number of fission events required to operate a 100 W light bulb for 1.0 h. [2½ marks]

(d) Molybdenum has a work function of 4.20 eV. What is the stopping potential if the incident light has a wavelength of 180 nm? [2½ marks]

[Total = 10 marks]

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Question 8

(a) Define a photon. [2 marks]

(b) A photon of wavelength 0.0033 nm is incident on an isolated stationary electron of mass mₑ. The photon is deflected elastically by the electron. The wavelength of the deflected photon is 0.0038 nm. Calculate for the incident photon:

• (i) Its momentum

• (ii) Its energy [5 marks]

© The angle θ through which the photon is deflected is given by:

Δλ = (h / mₑc)(1 − cos θ)

• (i) State the phenomenon represented and identify all symbols

• (ii) Calculate angle θ using data from (b) above [3 marks]

[Total = 10 marks]

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ANSWERS

PHY 001

Q1(a) — Definitions

Young’s Modulus: The ratio of tensile stress to tensile strain within the elastic limit of a material. It quantifies a material’s resistance to elastic (recoverable) deformation under longitudinal stress.

Stiffness: The ability of a material or structure to resist deformation under an applied force. A stiffer material requires a greater force to produce the same extension.

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Q1(b) — Force-Extension Graph

(From the graph: at the proportional limit, F = 200 N, e = 0.3 × 10⁻³ m)

(i) Force constant:

$$k = \frac{F}{e} = \frac{200}{0.3 \times 10^{-3}} = \boxed{6.67 \times 10^5 \text{ N/m}}$$

(ii) Work done (area under graph up to proportional limit):

$$W = \frac{1}{2} \times F \times e = \frac{1}{2} \times 200 \times 0.3 \times 10^{-3} = \boxed{0.03 \text{ J}}$$

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Q1© — Gravitational Forces

G = 6.67 × 10⁻¹¹ N m²/kg²

The third particle (0.050 kg) is placed 0.050 m from the 0.30 kg mass, so it is 0.10 m from the 0.20 kg mass.

(i) Net gravitational force:

Force from 0.30 kg mass (attractive, pulling toward it):

$$F_1 = \frac{G \times 0.30 \times 0.050}{(0.050)^2} = \frac{6.67 \times 10^{-11} \times 0.015}{0.0025} = 4.002 \times 10^{-10} \text{ N}$$

Force from 0.20 kg mass (attractive, pulling away from 0.30 kg):

$$F_2 = \frac{G \times 0.20 \times 0.050}{(0.10)^2} = \frac{6.67 \times 10^{-11} \times 0.010}{0.01} = 6.67 \times 10^{-11} \text{ N}$$

Net force (toward 0.30 kg mass):

$$F_{net} = F_1 - F_2 = 4.002 \times 10^{-10} - 6.67 \times 10^{-11} = \boxed{3.335 \times 10^{-10} \text{ N}}$$

(ii) Position for zero net force:

Let x = distance from the 0.20 kg mass (then 0.15 − x from the 0.30 kg mass).

Setting forces equal:

$$\frac{G \times 0.20 \times 0.050}{x^2} = \frac{G \times 0.30 \times 0.050}{(0.15 - x)^2}$$

$$\frac{0.20}{x^2} = \frac{0.30}{(0.15 - x)^2}$$

Taking square roots:

$$\frac{\sqrt{0.20}}{x} = \frac{\sqrt{0.30}}{0.15 - x}$$

$$0.4472(0.15 - x) = 0.5477x$$

$$0.06708 = 0.9949x$$

$$\boxed{x \approx 0.0674 \text{ m from the 0.20 kg mass}}$$

(i.e., approximately 0.0826 m from the 0.30 kg mass)

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Q2(b) — Satellite Orbit

Orbital radius: r = R_earth + altitude = 6.4 × 10⁶ + 550 × 10³ = 6.95 × 10⁶ m

Period: T = 110 min = 6600 s

(i) Orbital speed:

$$v = \frac{2\pi r}{T} = \frac{2\pi \times 6.95 \times 10^6}{6600} = \boxed{6,614 \text{ m/s} \approx 6.61 \text{ km/s}}$$

(ii) Centripetal acceleration:

$$a = \frac{v^2}{r} = \frac{(6614)^2}{6.95 \times 10^6} = \boxed{6.30 \text{ m/s}^2}$$

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PHY 002

Q3© — Diffraction Grating Angular Positions

Grating spacing: d = 1/10,000 cm = 1 × 10⁻⁴ cm = 1 × 10⁻⁶ m

Using: d sin θ = nλ

For λ = 400 nm = 4 × 10⁻⁷ m:

| Order | sin θ | θ |

|—|---|—|

| n = 1 | 0.400 | 23.6° |

| n = 2 | 0.800 | 53.1° |

For λ = 700 nm = 7 × 10⁻⁷ m:

| Order | sin θ | θ |

|—|---|—|

| n = 1 | 0.700 | 44.4° |

| n = 2 | 1.400 | Not possible (sin θ > 1) |

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Q4(b) — Standing Sound Wave

From y = 0.008 cos(ωt) sin(kz), with k = 4π rad/m, ω = 50π rad/s

(i) Distance between consecutive nodes:

Nodes occur where sin(kz) = 0, i.e., kz = nπ

$$\text{Node separation} = \frac{\pi}{k} = \frac{\pi}{4\pi} = \boxed{0.25 \text{ m}}$$

(ii) Amplitude at t = 0.565 s:

The time-varying amplitude = 0.008 |cos(ωt)|:

$$= 0.008 \times |\cos(50\pi \times 0.565)|$$

$$= 0.008 \times |\cos(88.75 \text{ rad})|$$

$$88.75 \text{ rad} = 28.22\pi \Rightarrow \cos(0.22\pi) = \cos(39.6°) = 0.771$$

$$\boxed{A = 0.008 \times 0.771 \approx 6.17 \times 10^{-3} \text{ m}}$$

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Q4© — Doppler Effect

The truck (source) and car (observer) are moving toward each other (in the same direction, but the problem states they approach each other — the car is behind and catching up, or the truck approaches from ahead):

Using: $f_o = f_s \times \frac{v + v_o}{v - v_s}$

where v = 350 m/s, v_s = 66 m/s (source), v_o = 60 m/s (observer approaching source):

$$f_o = 800 \times \frac{350 + 60}{350 - 66} = 800 \times \frac{410}{284} = \boxed{1154.9 \text{ Hz}}$$

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PHY 003

Q5(b) — From Graph (fig. 2b)

(Graph shows V vs I for a battery circuit; V-intercept ≈ 6.0 V, line falls to about V = 1.0 V at I = 2.5 A)

(i) e.m.f.: The V-intercept (when I = 0):

$$\boxed{E = 6.0 \text{ V}}$$

(ii) Internal resistance: Magnitude of the slope:

$$r = \frac{\Delta V}{\Delta I} = \frac{6.0 - 1.0}{2.5 - 0} = \boxed{2.0 \text{ Ω}}$$

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Q5© — Power and Efficiency at I = 1.6 A

(i) Power transferred to R:

From graph at I = 1.6 A: V = E − Ir = 6.0 − (1.6 × 2.0) = 2.8 V

$$P = VI = 2.8 \times 1.6 = \boxed{4.48 \text{ W}}$$

(ii) Efficiency:

Total power from battery = EI = 6.0 × 1.6 = 9.6 W

$$\eta = \frac{P_{output}}{P_{total}} \times 100 = \frac{4.48}{9.6} \times 100 = \boxed{46.7\%}$$

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Q6(b) — Electron in Magnetic Field

Electron accelerated through V = 5200 V.

Find electron speed using energy conservation:

$$eV = \frac{1}{2}m_e v^2 \Rightarrow v = \sqrt{\frac{2eV}{m_e}} = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 5200}{9.11 \times 10^{-31}}}$$

$$v = \sqrt{1.827 \times 10^{15}} = 4.275 \times 10^7 \text{ m/s}$$

Find B using circular motion: evB = mv²/r → B = mv/er

$$B = \frac{m_e v}{er} = \frac{9.11 \times 10^{-31} \times 4.275 \times 10^7}{1.6 \times 10^{-19} \times 0.04} = \boxed{6.09 \times 10^{-3} \text{ T} \approx 6.1 \text{ mT}}$$

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Q6© — Transformer

N_p = 400, N_s = 1200, E_p = 150 V, I_s = 5 A

(i) Secondary e.m.f.:

$$\frac{E_s}{E_p} = \frac{N_s}{N_p} \Rightarrow E_s = 150 \times \frac{1200}{400} = \boxed{450 \text{ V}}$$

(ii) Primary current (using turns ratio for ideal transformer):

$$\frac{I_p}{I_s} = \frac{N_s}{N_p} \Rightarrow I_p = 5 \times \frac{1200}{400} = \boxed{15 \text{ A}}$$

(iii) Power drawn by primary:

$$P_p = E_p \times I_p = 150 \times 15 = \boxed{2250 \text{ W}}$$

(iv) Power supplied by secondary:

$$P_s = E_s \times I_s = 450 \times 5 = \boxed{2250 \text{ W}}$$

(Equal, as expected for an ideal transformer — 100% efficiency)

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PHY 004

Q7© — Fission Events for Light Bulb

Total energy needed: E = P × t = 100 × 3600 = 3.6 × 10⁵ J

Energy per fission: E_f = 208 MeV = 208 × 1.6 × 10⁻¹³ = 3.328 × 10⁻¹¹ J

$$N = \frac{3.6 \times 10^5}{3.328 \times 10^{-11}} = \boxed{1.08 \times 10^{16} \text{ fission events}}$$

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Q7(d) — Stopping Potential (Molybdenum)

Work function: φ = 4.20 eV; λ = 180 nm = 180 × 10⁻⁹ m

Energy of incident photon:

$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{180 \times 10^{-9}} = 1.104 \times 10^{-18} \text{ J} = 6.90 \text{ eV}$$

Einstein’s photoelectric equation: eV_s = E − φ

$$V_s = 6.90 - 4.20 = \boxed{2.70 \text{ V}}$$

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Q8(b) — Photon Momentum and Energy

λ = 0.0033 nm = 3.3 × 10⁻¹² m; h = 6.626 × 10⁻³⁴ J·s

(i) Momentum:

$$p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{3.3 \times 10^{-12}} = \boxed{2.01 \times 10^{-22} \text{ kg m/s}}$$

(ii) Energy:

$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.3 \times 10^{-12}} = \boxed{6.02 \times 10^{-14} \text{ J} \approx 375.9 \text{ keV}}$$

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Q8© — Compton Scattering

(i) Phenomenon: This is the Compton Effect — the scattering of a photon by a free electron, resulting in an increase in the photon’s wavelength.

Symbols:

• Δλ = change in wavelength (λ_scattered − λ_incident)

• h = Planck’s constant (6.626 × 10⁻³⁴ J·s)

• mₑ = rest mass of the electron (9.11 × 10⁻³¹ kg)

• c = speed of light (3 × 10⁸ m/s)

• θ = angle of photon deflection

(ii) Calculate θ:

$$\Delta\lambda = 0.0038 - 0.0033 = 0.0005 \text{ nm} = 5 \times 10^{-13} \text{ m}$$

Compton wavelength: $\frac{h}{m_e c} = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 3 \times 10^8} = 2.426 \times 10^{-12} \text{ m}$

$$5 \times 10^{-13} = 2.426 \times 10^{-12} \times (1 - \cos\theta)$$

$$1 - \cos\theta = \frac{5 \times 10^{-13}}{2.426 \times 10^{-12}} = 0.2061$$

$$\cos\theta = 0.7939 \Rightarrow \boxed{\theta \approx 37.4°}$$