2021 JUPEB Physics

PHY 001: MECHANICS AND PROPERTIES OF MATTER

Question 1

(a)(i) Define the following terms: Young modulus, Bulk modulus, and Shear modulus. [3 Marks]

(a)(ii) Two parallel oppositely directed forces, each 3500 N, are applied tangentially to the upper and lower faces of a cubical metal block 25 cm on a side. Find the angle of shear and displacement of the upper surface relative to the lower surface if the shear modulus for the metal is 80 GPa. [3 Marks]

(b) An airplane is travelling horizontally at a speed of 60 m/s and drops a crate of emergency supplies. To avoid damage, the maximum vertical speed of the crate on landing is 30 m/s. Assuming air resistance is negligible:

• (i) Calculate the maximum height of the airplane when the crate is dropped.

• (ii) Calculate the time taken for the crate to reach the ground from this height.

• (iii) If the airplane is travelling at the maximum permitted height, calculate the horizontal distance travelled by the crate after it is released. [4 Marks]

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Question 2

(a) Define the following: (i) terminal velocity, (ii) viscosity, (iii) surface tension. [3 Marks]

(b) State the following: (i) Stokes’ law, (ii) law of conservation of angular momentum. [3 Marks]

© A mass of 2 kg attached to the end of a vertical wire of length 2 m and diameter 2 mm extended the wire by 1 mm. Calculate the:

• (i) Young’s modulus of the wire

• (ii) Energy stored in the wire [4 Marks]

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PHY 002: HEAT, WAVES AND OPTICS

Question 3

(a) Define the following terms and state their units:

• (i) Specific heat capacity [2 Marks]

• (ii) Specific latent heat [2 Marks]

(b) A plane progressive wave is represented by the equation:

y = 0.3 sin(20πx/3 + 100πt) m

Calculate: (i) amplitude; (ii) frequency; (iii) wavelength; (iv) velocity of the wave. [4 Marks]

© The length of a test-tube is 15.0 cm. Calculate the two lowest frequencies for the sound emitted when the open end of the tube is blown. Take the speed of sound in air as 330 m/s. [2 Marks]

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Question 4

(a) Define the following: (i) Optical path, (ii) Coherent sources. [2 Marks]

(b)(i) A sound wave has an amplitude of 8.5 μm at a particular distance from the source. Calculate the amplitude when this distance is increased by 90%.

(b)(ii) State TWO uses of ultrasonics. [4 Marks]

© A parallel beam of sodium light is incident normally on a diffraction grating. The angle between the two first-order spectra on either side of the normal is 38°. If the wavelength of the light is 0.859 μm, find the:

• (i) Number of rulings per mm on the grating

• (ii) Greatest number of bright images obtained [4 Marks]

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PHY 003: ELECTRICITY AND MAGNETISM

Question 5

(a) Explain the following terms: (i) Electric flux, (ii) Equipotential surface. [2 Marks]

(b)(i) State the condition for resonance in RLC circuits. [1 Mark]

(b)(ii) A series RLC circuit with R = 425 Ω, L = 1.25 H, and C = 3.50 μF is connected to a maximum voltage supply V = 150 V, with ω = 377 s⁻¹. Determine the impedance and maximum current in the circuit. [5 Marks]

© The maximum power dissipated in a 5 kΩ resistor is 15 W. What is the maximum current? [2 Marks]

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Question 6

(a) Explain the statement: “The capacitance of a capacitor is 5 farads.” [2 Marks]

(b) A capacitor consisting of 2 parallel plates separated by oil of dielectric constant 1.5, each plate having an effective area of 1000 cm² and spaced 0.1 cm apart, is connected across a constant voltage source of 5000 V. Calculate:

• (i) The capacitance [1 Mark]

• (ii) The charge [1 Mark]

• (iii) The energy stored [1 Mark]

• (iv) If the voltage was doubled while capacitance remained constant, what would be the new energy stored? [2 Marks]

© Two charges Q₁ = +200 μC and Q₂ = −100 μC are separated by 100 cm, with Q₁ at the origin. How much work must be done to transfer Q₃ = +500 μC from x = 80 cm to x = 20 cm? [3 Marks]

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PHY 004: MODERN PHYSICS

Question 7

(a)(i) State three differences between intrinsic and extrinsic semiconductors. [3 Marks]

(a)(ii) Explain the term “doping” in semiconducting material. [2 Marks]

(b)(i) State Einstein’s photoelectric equation and define all terms. [2 Marks]

(b)(ii) Calculate the mass-energy equivalence of a neutron in eV.

(Given: 1 amu = 1.66 × 10⁻²⁷ kg; mass of neutron = 1.67493 × 10⁻²⁷ kg; c = 3 × 10⁸ m/s; e = 1.6 × 10⁻¹⁹ C) [3 Marks]

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Question 8

(a)(i) State FOUR shortcomings of Bohr’s atomic theory. [2 Marks]

(a)(ii) State any TWO properties each of alpha (α), beta (β), and gamma (γ) rays emitted by radioactive atoms. [3 Marks]

(b) The wavelength of the Balmer first line is 6563 Å. Calculate the wavelength of the second line. [5 Marks]

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ANSWERS

PHY 001

Q1(a)(i) — Definitions

• Young’s Modulus: The ratio of tensile (longitudinal) stress to tensile strain. It measures a material’s resistance to stretching or compression along an axis.

• Bulk Modulus: The ratio of volumetric stress to volumetric strain. It measures a substance’s resistance to uniform compression.

• Shear Modulus: The ratio of shear stress to shear strain. It measures a material’s rigidity when subjected to tangential forces.

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Q1(a)(ii) — Shear Angle and Displacement

Given: F = 3500 N, side = 0.25 m, G = 80 × 10⁹ Pa

Area: A = 0.25 × 0.25 = 0.0625 m²

Shear stress: τ = F/A = 3500 / 0.0625 = 56,000 Pa

Angle of shear:

$$\theta = \frac{\tau}{G} = \frac{56{,}000}{80 \times 10^9} = 7 \times 10^{-7} \text{ rad}$$

Displacement:

$$\Delta x = h \cdot \theta = 0.25 \times 7 \times 10^{-7} = 1.75 \times 10^{-7} \text{ m}$$

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Q1(b) — Airplane Drop

(i) Maximum height: Using v² = u² + 2gh (vertical motion, u = 0):

$$30^2 = 2 \times 9.8 \times h \Rightarrow h = \frac{900}{19.6} \approx \boxed{45.92 \text{ m}}$$

(ii) Time to reach ground: Using v = u + gt:

$$t = \frac{30}{9.8} \approx \boxed{3.06 \text{ s}}$$

(iii) Horizontal distance:

$$x = v_x \times t = 60 \times 3.06 \approx \boxed{183.6 \text{ m}}$$

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Q2© — Wire Extension

Given: m = 2 kg, L = 2 m, d = 2 mm → r = 0.001 m, extension e = 0.001 m

(i) Young’s Modulus:

• F = mg = 2 × 9.8 = 19.6 N

• A = πr² = π × (0.001)² = 3.14 × 10⁻⁶ m²

• Stress = 19.6 / 3.14 × 10⁻⁶ = 6.24 × 10⁶ Pa

• Strain = 0.001 / 2 = 0.0005

$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{6.24 \times 10^6}{0.0005} = \boxed{1.248 \times 10^{10} \text{ Pa}}$$

(ii) Energy stored:

$$E = \frac{1}{2} \times F \times e = 0.5 \times 19.6 \times 0.001 = \boxed{9.8 \times 10^{-3} \text{ J}}$$

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PHY 002

Q3(b) — Wave Equation

From y = 0.3 sin(20πx/3 + 100πt):

| Quantity | Working | Result |

|—|---|—|

| Amplitude | A directly from equation | 0.3 m |

| Frequency | ω = 100π → f = ω/2π | 50 Hz |

| Wavelength | k = 20π/3 → λ = 2π/k | 0.3 m |

| Wave speed | v = fλ | 15 m/s |

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Q3© — Test Tube Frequencies

A test tube blown at the open end behaves as a closed pipe (closed at bottom, open at top). Only odd harmonics are present:

$$f_n = \frac{(2n-1)v}{4L}, \quad n = 1, 2, 3...$$

• 1st (fundamental): f₁ = 330 / (4 × 0.15) = 550 Hz

• 2nd (3rd harmonic): f₂ = 3 × 550 = 1650 Hz

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Q4(b)(i) — Amplitude with Increased Distance

Sound amplitude is inversely proportional to distance (A ∝ 1/r):

New distance = 1.9r (increased by 90%)

$$A_2 = A_1 \times \frac{r}{1.9r} = \frac{8.5}{1.9} \approx \boxed{4.47 \text{ μm}}$$

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Q4© — Diffraction Grating

(i) Rulings per mm:

The angle between the two first-order beams = 38°, so the diffraction angle θ = 19°.

Using the grating equation: d sin θ = nλ (n = 1):

$$d = \frac{\lambda}{\sin\theta} = \frac{0.859 \times 10^{-6}}{\sin 19°} = \frac{0.859 \times 10^{-6}}{0.3256} = 2.638 \times 10^{-6} \text{ m}$$

$$N = \frac{1}{d} = \frac{1}{2.638 \times 10^{-3} \text{ mm}} \approx \boxed{379 \text{ rulings/mm}}$$

(ii) Greatest number of bright images:

Maximum order when sin θ = 1:

$$n_{max} = \frac{d}{\lambda} = \frac{2.638 \times 10^{-6}}{0.859 \times 10^{-6}} \approx 3$$

Total bright images = 2n + 1 (both sides + central) = 7 images

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PHY 003

Q5(b)(ii) — RLC Impedance

• X_L = ωL = 377 × 1.25 = 471.25 Ω

• X_C = 1/(ωC) = 1/(377 × 3.5 × 10⁻⁶) = 757.86 Ω

$$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{425^2 + (471.25 - 757.86)^2}$$

$$= \sqrt{180{,}625 + 82{,}142} = \boxed{512.6 \text{ Ω}}$$

$$I_{max} = \frac{V_{max}}{Z} = \frac{150}{512.6} \approx \boxed{0.293 \text{ A}}$$

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Q5© — Maximum Current in Resistor

$$P = I^2 R \Rightarrow I = \sqrt{\frac{P}{R}} = \sqrt{\frac{15}{5000}} = \sqrt{0.003} \approx \boxed{0.0548 \text{ A}}$$

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Q6(b) — Parallel Plate Capacitor

Given: εᵣ = 1.5, A = 1000 cm² = 0.1 m², d = 0.1 cm = 0.001 m, V = 5000 V

(i) Capacitance:

$$C = \frac{\varepsilon_r \varepsilon_0 A}{d} = \frac{1.5 \times 8.85 \times 10^{-12} \times 0.1}{0.001} = \boxed{1.327 \times 10^{-9} \text{ F}}$$

(ii) Charge:

$$Q = CV = 1.327 \times 10^{-9} \times 5000 = \boxed{6.635 \times 10^{-6} \text{ C}}$$

(iii) Energy stored:

$$E = \frac{1}{2}CV^2 = 0.5 \times 1.327 \times 10^{-9} \times (5000)^2 = \boxed{0.01659 \text{ J}}$$

(iv) Energy if V doubles (C constant):

Since E ∝ V², doubling V quadruples energy:

$$E_{new} = 4 \times 0.01659 = \boxed{0.0664 \text{ J}}$$

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PHY 004

Q7(b)(ii) — Mass-Energy of Neutron

$$E = mc^2 = 1.67493 \times 10^{-27} \times (3 \times 10^8)^2 = 1.5074 \times 10^{-10} \text{ J}$$

$$E \text{ (in eV)} = \frac{1.5074 \times 10^{-10}}{1.6 \times 10^{-19}} \approx \boxed{942.1 \text{ MeV}}$$

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Q8(b) — Balmer Series Second Line

Using the Balmer formula: $\frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{n^2}\right)$

First line (n = 3): Find R:

$$\frac{1}{6563} = R\left(\frac{1}{4} - \frac{1}{9}\right) = R \times \frac{5}{36}$$

$$R = \frac{36}{5 \times 6563} = 1.097 \times 10^{-3} \text{ Å}^{-1}$$

Second line (n = 4):

$$\frac{1}{\lambda_2} = R\left(\frac{1}{4} - \frac{1}{16}\right) = R \times \frac{3}{16}$$

$$\lambda_2 = \frac{5 \times 6563 \times 16}{36 \times 3} = \frac{524,040}{108} \approx \boxed{4856 \text{ Å}}$$