2022 JUPEB physics

1. (a) (i) Define the following terms: Young modulus, Bulk modulus, and Shear modulus.

(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.

(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 from the airplane.

2. (a) Define the following:

(i) terminal velocity, (ii) viscosity, (iii) surface tension.

(b) State the following:

(i) Stokes’ law, (ii) law of conservation of angular momentum.

(i) Young’s modulus of the wire,

(ii) energy stored in the wire.

PHY 002: HEAT, WAVES AND OPTICS

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

(i) specific heat capacity

(ii) specific latent heat

(b) A plane progressive wave is represented by the equation y = 0.3 sin(20πx/3 + 100πt) m. Calculate the following:

(i) amplitude; (ii) frequency; (iii) wavelength; (iv) velocity of the wave.

4. (a) Define the following:

(i) Optical path (ii) Coherent sources

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

(ii) State TWO uses of ultrasonics.

© 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.

PHY 003: ELECTRICITY AND MAGNETISM

5. (a) Explain the following terms:

(i) Electric flux (ii) Equipotential surface.

(b) (i) State the condition for resonance in RLC circuits.

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

6. (a) Explain the statement: “The capacitance of a capacitor is 5 farads.”

(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 5000V. Calculate:

(i) the capacitance of the capacitor,

(ii) the charge of the capacitor,

(iii) the energy stored in the capacitor, and

(iv) If the voltage source was doubled, while the capacitance was kept constant, what would be the new amount of energy stored?

© Two charges Q₁ = +200 μC and Q₂ = −100μC are separated by a distance of 100 cm, with Q₁ located at the origin. How much work must be done to transfer a charge Q₃ = +500μC from a point x = 80 cm to a point x = 20 cm?

PHY 004: MODERN PHYSICS

7. (a) (i) State three differences between intrinsic and extrinsic semiconductors.

(ii) Explain the term “doping” in semiconducting material.

(b) (i) State Einstein’s photoelectric equation and define the terms in the equation.

(ii) Calculate the mass-energy equivalence of a neutron in eV. (Take 1 amu as 1.66 × 10⁻²⁷ kg; mass of neutron = 1.67493 × 10⁻²⁷ kg; c = 3 × 10⁸ m/s, e = 1.6 × 10⁻¹⁹ C).

8. (a) (i) State FOUR shortcomings of Bohr’s atomic theory.

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

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

ANSWERS

PHY 001: MECHANICS AND PROPERTIES OF MATTER

1(a)(i) Definitions:

Young’s Modulus (E): The ratio of tensile stress to tensile strain within the elastic limit. E = (F/A)/(ΔL/L). Unit: Pa (N/m²).

Bulk Modulus (K): The ratio of volumetric stress to volumetric strain. K = −P/(ΔV/V). Unit: Pa.

Shear Modulus (G): The ratio of shear stress to shear strain. G = (F/A)/θ. Unit: Pa.

1(a)(ii)

Given: F = 3500 N, side = 25 cm = 0.25 m, G = 80 GPa = 80 × 10⁹ Pa

Area = (0.25)² = 0.0625 m²

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

Shear strain (angle of shear) θ = stress/G = 56,000 / (80 × 10⁹) = 7 × 10⁻⁷ rad

Displacement = θ × L = 7 × 10⁻⁷ × 0.25 = 1.75 × 10⁻⁷ m

1(b)

Given: horizontal speed = 60 m/s, max vertical landing speed = 30 m/s, g = 10 m/s²

(i) Using v² = u² + 2gh, u = 0 (dropped):

h = v²/2g = (30)²/(2×10) = 900/20 = 45 m

(ii) v = u + gt → t = v/g = 30/10 = 3 s

(iii) Horizontal distance = horizontal speed × time = 60 × 3 = 180 m

2(a)

(i) Terminal velocity: The constant maximum velocity reached by a falling body when the drag force equals the gravitational force.

(ii) Viscosity: The property of a fluid that resists the relative motion between its layers; a measure of internal friction.

(iii) Surface tension: The property of a liquid surface that causes it to behave like a stretched elastic membrane, due to cohesive forces among molecules.

2(b)

(i) Stokes’ Law: The drag force on a sphere moving through a viscous fluid is F = 6πηrv, where η is viscosity, r is the radius, and v is the velocity.

(ii) Law of conservation of angular momentum: The total angular momentum of a system remains constant if no external torque acts on it.

2©

Given: m = 2 kg, L = 2 m, d = 2 mm → r = 1 mm = 1×10⁻³ m, ΔL = 1 mm = 1×10⁻³ m, g = 10 m/s²

A = πr² = π × (10⁻³)² = 3.142 × 10⁻⁶ m²

F = mg = 2 × 10 = 20 N

(i) E = (F × L)/(A × ΔL) = (20 × 2)/(3.142 × 10⁻⁶ × 10⁻³) = 40/(3.142 × 10⁻⁹) = 1.27 × 10¹⁰ Pa

(ii) Energy = ½ × F × ΔL = ½ × 20 × 10⁻³ = 0.01 J

PHY 002: HEAT, WAVES AND OPTICS

3(a)

(i) Specific heat capacity: The amount of heat required to raise the temperature of 1 kg of a substance by 1 K (or 1°C). Unit: J kg⁻¹ K⁻¹.

(ii) Specific latent heat: The amount of heat required to change the state of 1 kg of a substance without a change in temperature. Unit: J kg⁻¹.

3(b)

y = 0.3 sin(20πx/3 + 100πt)

Standard form: y = A sin(kx + ωt)

(i) Amplitude A = 0.3 m

(ii) ω = 100π → f = ω/2π = 100π/2π = 50 Hz

(iii) k = 20π/3 → λ = 2π/k = 2π/(20π/3) = 6/10 = 0.3 m

(iv) v = fλ = 50 × 0.3 = 15 m/s

3©

Closed tube (open at one end): resonates at odd harmonics only.

λ = 4L/n where n = 1, 3, 5…

L = 15.0 cm = 0.15 m

1st lowest frequency (fundamental): f₁ = v/4L = 330/(4×0.15) = 330/0.6 = 550 Hz

2nd lowest frequency (3rd harmonic): f₂ = 3v/4L = 3×550 = 1650 Hz

4(a)

(i) Optical path: The product of the refractive index of a medium and the geometric path length travelled by light through it. Optical path = n × d.

(ii) Coherent sources: Sources that emit light waves of the same frequency, same wavelength, and a constant phase difference.

4(b)(i)

Intensity ∝ 1/r², and Amplitude ∝ 1/r.

New distance = r + 0.9r = 1.9r

A₂/A₁ = r/1.9r = 1/1.9

A₂ = 8.5/1.9 = 4.47 μm

(ii) Two uses of ultrasonics:

Medical imaging (ultrasound scanning).
Detection of cracks and flaws in metal structures (non-destructive testing).

4©

Angle between two first-order spectra = 38° → angle of diffraction θ = 38°/2 = 19°

λ = 0.859 μm = 0.859 × 10⁻⁶ m

(i) dsinθ = nλ → d = λ/sinθ = (0.859 × 10⁻⁶)/sin19° = (0.859 × 10⁻⁶)/0.3256 = 2.638 × 10⁻⁶ m

Rulings per mm = 1/(2.638 × 10⁻³) = 379 rulings/mm

(ii) Maximum order: n_max = d/λ = (2.638 × 10⁻⁶)/(0.859 × 10⁻⁶) = 3.07 → n_max = 3

Greatest number of bright images = 2n_max + 1 (including central) = 7

PHY 003: ELECTRICITY AND MAGNETISM

5(a)

(i) Electric flux: The total number of electric field lines passing perpendicularly through a given surface area. Φ = E·A·cosθ. Unit: N·m²/C.

(ii) Equipotential surface: A surface on which every point is at the same electric potential, so no work is done in moving a charge along it.

5(b)(i)

Condition for resonance in RLC circuit: Inductive reactance equals capacitive reactance.

X_L = X_C, i.e., ωL = 1/ωC

5(b)(ii)

R = 425 Ω, L = 1.25 H, C = 3.50 μF = 3.50×10⁻⁶ F, ω = 377 s⁻¹

X_L = ωL = 377 × 1.25 = 471.25 Ω

X_C = 1/ωC = 1/(377 × 3.50×10⁻⁶) = 1/0.001320 = 757.6 Ω

Z = √[R² + (X_L − X_C)²] = √[425² + (471.25 − 757.6)²]

= √[180,625 + (−286.35)²]

= √[180,625 + 81,976]

= √262,601 = 512.4 Ω

I_max = V_max/Z = 150/512.4 = 0.293 A

5©

P = I²R → I = √(P/R) = √(15/5000) = √(0.003) = 0.0548 A ≈ 54.8 mA

6(a)

A capacitance of 5 farads means that the capacitor stores 5 coulombs of charge for every 1 volt of potential difference applied across its plates. That is, C = Q/V → Q = 5 × V.

6(b)

ε₀ = 8.85×10⁻¹² F/m, k = 1.5, A = 1000 cm² = 0.1 m², d = 0.1 cm = 0.001 m, V = 5000 V

(i) C = kε₀A/d = (1.5 × 8.85×10⁻¹² × 0.1)/0.001 = (1.3275×10⁻¹²)/0.001 = 1.3275×10⁻⁹ F ≈ 1.33 nF

(ii) Q = CV = 1.3275×10⁻⁹ × 5000 = 6.64×10⁻⁶ C ≈ 6.64 μC

(iii) E = ½CV² = ½ × 1.3275×10⁻⁹ × (5000)² = ½ × 1.3275×10⁻⁹ × 25×10⁶ = 1.66×10⁻² J ≈ 16.6 mJ

(iv) New V = 10,000 V; C constant:

E_new = ½CV² = ½ × 1.3275×10⁻⁹ × (10,000)² = ½ × 1.3275×10⁻⁹ × 10⁸ = 6.64×10⁻² J ≈ 66.4 mJ

6©

k_e = 9×10⁹ N·m²/C², Q₁ = +200×10⁻⁶ C at x=0, Q₂ = −100×10⁻⁶ C at x=100 cm = 1 m

Potential at x = 80 cm = 0.8 m:

V_A = k_e[Q₁/0.8 + Q₂/0.2]

= 9×10⁹[(200×10⁻⁶)/0.8 + (−100×10⁻⁶)/0.2]

= 9×10⁹[0.25 − 0.5]

= 9×10⁹ × (−0.25) = −2.25×10⁹ V

Potential at x = 20 cm = 0.2 m:

V_B = k_e[Q₁/0.2 + Q₂/0.8]

= 9×10⁹[(200×10⁻⁶)/0.2 + (−100×10⁻⁶)/0.8]

= 9×10⁹[1.0 − 0.125]

= 9×10⁹ × 0.875 = 7.875×10⁹ V

Work done = Q₃(V_B − V_A) = 500×10⁻⁶ × (7.875×10⁹ − (−2.25×10⁹))

= 500×10⁻⁶ × 10.125×10⁹

= 5.0625×10⁶ J ≈ 5.06 MJ

PHY 004: MODERN PHYSICS

7(a)(i) Three differences between intrinsic and extrinsic semiconductors:

| Intrinsic | Extrinsic |

|—|---|

| Pure semiconductor (e.g., pure Si or Ge) | Semiconductor doped with impurity atoms |

| Equal numbers of holes and electrons | Unequal numbers of holes and electrons |

| Lower conductivity | Higher conductivity |

7(a)(ii) Doping:

Doping is the deliberate introduction of a small amount of impurity atoms (trivalent or pentavalent) into a pure semiconductor to increase its electrical conductivity and control the type of charge carrier (electrons or holes).

7(b)(i) Einstein’s Photoelectric Equation:

hf = φ + ½mv²_max

Where:

h = Planck’s constant
f = frequency of incident radiation
φ (or hf₀) = work function of the metal (minimum energy to eject an electron)
½mv²_max = maximum kinetic energy of the emitted photoelectron
m = mass of electron, v_max = maximum velocity of ejected electron

7(b)(ii)

Mass of neutron = 1.67493 × 10⁻²⁷ kg

1 amu = 1.66 × 10⁻²⁷ kg → E per amu = 931.5 MeV

Mass in amu = (1.67493 × 10⁻²⁷)/(1.66 × 10⁻²⁷) = 1.00dot ≈ 1.00899 amu

E = mc² = 1.67493×10⁻²⁷ × (3×10⁸)²

= 1.67493×10⁻²⁷ × 9×10¹⁶

= 1.50744×10⁻¹⁰ J

Convert to eV: E = 1.50744×10⁻¹⁰ / 1.6×10⁻¹⁹ = 9.42 × 10⁸ eV = 942 MeV

8(a)(i) Four shortcomings of Bohr’s atomic theory:

It fails to explain the spectra of multi-electron atoms.
It cannot account for the fine structure (splitting) of spectral lines.
It does not explain the Zeeman effect (splitting of lines in a magnetic field).
It violates the Heisenberg uncertainty principle by assuming definite electron orbits.

8(a)(ii) Properties of α, β, and γ rays:

Alpha (α):

Consists of helium nuclei (2 protons, 2 neutrons); charge = +2e.
Low penetrating power (stopped by a sheet of paper).

Beta (β):

Consists of fast-moving electrons (or positrons); charge = −e (or +e).
Moderate penetrating power (stopped by a few mm of aluminium).

Gamma (γ):

Electromagnetic radiation of very short wavelength; no charge, no mass.
Highest penetrating power (requires several cm of lead to attenuate).

8(b) Wavelength of Balmer second line:

Balmer series: 1/λ = R[1/2² − 1/n²]

First line (n=3): 1/λ₁ = R[1/4 − 1/9] = R × 5/36

Second line (n=4): 1/λ₂ = R[1/4 − 1/16] = R × 3/16

λ₂/λ₁ = (5/36)/(3/16) = (5×16)/(36×3) = 80/108 = 20/27

λ₂ = λ₁ × 20/27 = 6563 × 20/27 = 131,260/27 = 4861.5 Å

2022 JUPEB physics

ANSWERS

PHY 001: MECHANICS AND PROPERTIES OF MATTER

PHY 002: HEAT, WAVES AND OPTICS

PHY 003: ELECTRICITY AND MAGNETISM

PHY 004: MODERN PHYSICS

Inside The Group:

Contact form

2022 JUPEB physics

ANSWERS

PHY 001: MECHANICS AND PROPERTIES OF MATTER

PHY 002: HEAT, WAVES AND OPTICS

PHY 003: ELECTRICITY AND MAGNETISM

PHY 004: MODERN PHYSICS

You may like these posts

Contact form