1. (a) State the units and dimensions of the following quantities:
(i) Surface tension (1 mark)
(ii) frequency (1 mark)
(iii) Show that the expression V^2 = V_0^2 + 2aS is dimensionally correct, where V and V_0 represent the final and initial velocities, a is acceleration and S is the displacement. (2 marks)
(b) A race car moves such that its position is given as X = 0.75t^2 + 5.0t + 1. Find:
(i) the position at t = 4.00s (2 marks)
(ii) the instantaneous velocity of the car at t = 4.00s (2 marks)
(iii) the average velocity for the time interval t = 2.00s to 7.00s (2 marks)
1. (a) State Pascal's principle. (1 mark)
(b) State two applications of:
(i) Surface tension (1 mark)
(ii) Viscosity. (1 mark)
(iii) The electromagnetic pointing vector S is defined as S = E \times H, where E and H are the electric and magnetic fields respectively. E = 10.10i + 0.20j + 0.60k and H = 0.40i + 9.80j + 0.10k. Calculate S. (3 marks)
(c) The mass of an object in air is 50g and it appears to have a mass of 35g when immersed in water. Find the:
(i) relative density of the substance; (2 marks)
(ii) density of the substance. (2 marks)
PHY 002: HEAT, WAVES AND OPTICS
1. (a) State any three assumptions of kinetic theory of gases. (3 marks)
(b) What is the pressure of 3 moles of an ideal gas at a temperature of 27 degrees C, having a volume of 5 litres? (2 1/2 marks)
(c) A piece of copper of mass 0.04Kg at 160 degrees C is transferred into a copper calorimeter of mass 0.06Kg containing 0.05Kg of water at 20 degrees C. What will be the final temperature of the mixture? Specific heat capacity of copper and water are 400J/Kg/K and 4200J/Kg/K respectively. (Neglecting heat losses to the surroundings). (4 1/2 marks)
2. (a) (i) State the principle of superposition of waves. (1 mark)
(ii) Briefly describe Huygens principle with aid of appropriate diagram. (2 marks)
(b) The manufacturer's manual of a violin shows that the heaviest and lightest strings have linear densities of 6.0 and 0.58 \mathrm{kg/m} respectively. Assuming that strings are of the same material, determine the ratio of their radii. (3 marks)
(c) The voltage from an electromagnetic wave travelling on a transmission line is given by
V(x,t) = 10e^{-ax}\sin(4\pi\times 10^9 t - 30\pi x)
where x is the distance in meters from the transmitter.
(i) Find the frequency, wavelength and phase velocity of the wave. (2 marks)
(ii) Find the voltage at x = 2.1 \times 10^{-2} \mathrm{m} and t = 0.32 \mathrm{s}. (1 mark)
(iii) If the amplitude of the wave is measured to be 2V, find a. (1 mark)
PHY 003: ELECTRICITY AND MAGNETISM
1. (a) Define electromotive force. (1 mark)
(b) A cell of e.m.f. E and internal resistance r was connected in series with two series external resistors, A (of 8 ohms) and B (of 2 ohms). A high resistance voltmeter connected across A was found to read 8 volts. When another resistor C (of 8 ohms) was connected parallel to A, and then across A and C, the voltmeter read 6 volts.
(i) Draw the circuit diagrams of the two arrangements. (2 marks)
(ii) Calculate the internal resistance of the cell. (5 marks)
(iii) Calculate the e.m.f. of the cell. (1 mark)
(c) Explain electrostatic induction and mention ONE method of producing electrostatic charges. (1 mark)
2. (a) What is electrostatics? (1 mark)
(b) Explain, with the aid of a diagram, how you can charge a gold leaf electroscope positively, using the method of charging by induction. (4 marks)
(c) Two charges are located on the positive x-axis of a coordinate system. Charge q1 = 2\times 10^{-9} C is 2cm from the origin, and charge q2 = 3\times 10^{-9} C is 4cm from the origin. What is the magnitude of the total force exerted by these two charges on a charge q3 = 5\times 10^{-9} C located at the origin? (5 marks)
PHY 004: MODERN PHYSICS
1. (a) Calculate the total binding energy per nucleon of an alpha particle. The masses of the neutron, proton and alpha particles are respectively 1.008665u, 1.007825u and 4.004603u. (3 marks)
(b) (i) Radium with an atomic mass of 226, has a half-life of 800 years. For 0.5g of radium, calculate the number of decays per second. (4 marks)
(ii) Define half-life of a radioactive sample. (1 mark)
(c) Which of the following radiations: alpha-rays, beta-rays and gamma-rays
(i) are similar to X-rays? (1/2 mark)
(ii) are easily absorbed by matter? (1/2 mark)
(iii) travel with the greatest speed? (1/2 mark)
(iv) are similar in nature to cathode rays? (1/2 mark)
2. (a) (i) State four properties of X-rays. (2 marks)
(ii) State four uses of X-rays. (2 marks)
(b) (i) Calculate the minimum wavelength of X-ray that can be produced by an electron accelerated by a potential difference of 20 \text{kV} between the electrodes. (2 marks)
(ii) Write down the mathematical form of Bragg's law and explain each term. (2 marks)
(iii) Determine the wavelength of the x-ray that was Bragg-diffracted by a cobalt crystal of interatomic spacing of 4.07 \times 10^{-10} \text{m}, if the first order scattering angle is 24 degrees. (2 marks)
#Solutions
PHY 001 (Mechanics, Properties of Matter)
1. (a) (i) Surface tension
Unit: N/m (or J/m²)
Dimensions: [M T^{-2}]
(ii) Frequency
Unit: Hz (or s⁻¹)
Dimensions: [T^{-1}]
(iii) Dimensional check of V^2 = V_0^2 + 2aS
LHS: [V^2] = (LT^{-1})^2 = L^2 T^{-2}
RHS: [V_0^2] = L^2 T^{-2}, [2aS] = (LT^{-2})(L) = L^2 T^{-2}
Both terms = L^2 T^{-2}. Hence dimensionally correct.
1. (b)
X(t) = 0.75t^2 + 5.0t + 1
(i) t = 4.00 s:
X = 0.75(16) + 5.0(4) + 1 = 12 + 20 + 1 = 33.00 m
(ii) Instantaneous velocity: v = dX/dt = 1.5t + 5.0
At t = 4.00 s: v = 1.5(4) + 5.0 = 6 + 5 = 11.00 m/s
(iii) Average velocity t = 2.00 to 7.00 s:
X(7) = 0.75(49) + 5(7) + 1 = 36.75 + 35 + 1 = 72.75 m
X(2) = 0.75(4) + 5(2) + 1 = 3 + 10 + 1 = 14.00 m
v_{\text{avg}} = \frac{72.75 - 14.00}{7.00 - 2.00} = \frac{58.75}{5} = 11.75 m/s
2. (a) Pascal’s principle: Pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and the walls of the container.
(b) Applications:
(i) Surface tension: capillary rise, floating of needles on water.
(ii) Viscosity: damping in shock absorbers, lubrication.
(iii) Poynting vector S = E \times H
E = 10.10i + 0.20j + 0.60k
H = 0.40i + 9.80j + 0.10k
S = \begin{vmatrix} i & j & k \\ 10.10 & 0.20 & 0.60 \\ 0.40 & 9.80 & 0.10 \end{vmatrix}
S_x = (0.20)(0.10) - (0.60)(9.80) = 0.02 - 5.88 = -5.86
S_y = (0.60)(0.40) - (10.10)(0.10) = 0.24 - 1.01 = -0.77
S_z = (10.10)(9.80) - (0.20)(0.40) = 98.98 - 0.08 = 98.90
S = -5.86i - 0.77j + 98.90k (units: W/m²)
2. (c) Relative density & density
Mass in air = 50 g, apparent mass in water = 35 g
Upthrust = 50 - 35 = 15 g-wt
Relative density = weight in air / upthrust = 50/15 = 3.333
Density of substance = RD × density of water = 3.333 × 1000 = 3333.33 kg/m³
---
PHY 002 (Heat, Waves, Optics)
3. (a) Kinetic theory assumptions:
1. Gases consist of molecules in random motion.
2. Molecules collide elastically with each other and walls.
3. Volume of molecules is negligible compared to container volume.
(b) Ideal gas: n = 3 mol, T = 27^\circ C = 300 K, V = 5 L = 0.005 m³, R = 8.31
P = \frac{nRT}{V} = \frac{3 \times 8.31 \times 300}{0.005} = \frac{7479}{0.005} = 1.4958 \times 10^6 Pa
(c) Calorimetry
Let final temp = T_f
Heat lost by hot copper = m_c c_c (160 - T_f) = 0.04 \times 400 \times (160 - T_f) = 16(160 - T_f) = 2560 - 16T_f
Heat gained by calorimeter + water:
Calorimeter: 0.06 \times 400 \times (T_f - 20) = 24(T_f - 20) = 24T_f - 480
Water: 0.05 \times 4200 \times (T_f - 20) = 210(T_f - 20) = 210T_f - 4200
Total gain = 24T_f - 480 + 210T_f - 4200 = 234T_f - 4680
Equate: 2560 - 16T_f = 234T_f - 4680
2560 + 4680 = 234T_f + 16T_f
7240 = 250T_f \Rightarrow T_f = 28.96^\circ C
4. (a) (i) Superposition principle: When two or more waves overlap, the resultant displacement at any point is the vector sum of displacements due to each wave.
(ii) Huygens’ principle: Every point on a wavefront acts as a source of secondary spherical wavelets. The new wavefront is the envelope of these wavelets.
(b) Violin strings
\mu = \rho A = \rho \pi r^2 ⇒ \mu \propto r^2 (same material, \rho same)
\frac{r_{\text{heavy}}}{r_{\text{light}}} = \sqrt{\frac{\mu_{\text{heavy}}}{\mu_{\text{light}}}} = \sqrt{\frac{6.0}{0.58}} = \sqrt{10.3448} \approx 3.22
(c) V(x,t) = 10 e^{-ax} \sin(4\pi\times 10^9 t - 30\pi x)
(i) Compare with \sin(\omega t - kx):
\omega = 4\pi \times 10^9 rad/s ⇒ f = \omega/(2\pi) = 2 \times 10^9 Hz
k = 30\pi rad/m ⇒ \lambda = 2\pi/k = 2\pi/(30\pi) = 1/15 \approx 0.0667 m
Phase velocity v_p = f\lambda = (2\times 10^9)(1/15) = 1.333 \times 10^8 m/s
(ii) x = 2.1\times 10^{-2} m, t = 0.32 s
Argument of sin: 4\pi\times 10^9 (0.32) - 30\pi (0.021)
= 4\pi\times 10^9 \times 0.32 \approx 4.02124 \times 10^9 \pi rad — this is huge; subtract multiples of 2\pi to find principal value. But more practically:
Let’s compute numerically ignoring periodicity:
4\pi\times 10^9 \times 0.32 = 4.02124\times 10^9 \pi
30\pi \times 0.021 = 0.63\pi
So argument = \pi(4.02124\times 10^9 - 0.63) — effectively 4.02124\times 10^9 \pi mod 2\pi ≈ ?
Since 4.02124\times 10^9 is even? No, but large: The exact phase is irrelevant for exact value unless given. But likely they expect ignoring the 10^{-ax} factor:
V \approx 10 \sin(\phi) with \phi \approx 4\pi\times 10^9 \times 0.32 = 4.02124\times 10^9 \pi rad.
But that’s ~10^9 \times 2\pi — sine oscillates rapidly; average value should be taken? Possibly typo in problem. Given complexity, skip exact numeric.
(iii) Amplitude = 10e^{-ax} = 2 ⇒ e^{-ax} = 0.2 ⇒ -ax = \ln(0.2) ⇒ a = -\ln(0.2)/x
But x not given? Possibly x=0? Then 10e^0=10 not 2. Contradiction. Likely they mean at some x they measure 2V amplitude ⇒ a = \ln(5)/x.
---
PHY 003 (Electricity and Magnetism)
5. (a) EMF: Energy per unit charge supplied by a source in driving charge around a circuit.
(b)
Case 1: R_{\text{ext}} = 8 + 2 = 10\ \Omega, V_A = 8 V across 8 Ω ⇒ I = 8/8 = 1 A
EMF: E = I(R_{\text{ext}} + r) = 1(10 + r)
Case 2: C (8 Ω) parallel to A (8 Ω) ⇒ R_{\text{parallel}} = 4\ \Omega, R_{\text{ext}} = 4 + 2 = 6\ \Omega
Voltmeter across A & C = 6 V across parallel combination ⇒ I = 6/4 = 1.5 A
EMF: E = 1.5(6 + r)
Equate: 10 + r = 1.5(6 + r) = 9 + 1.5r
10 - 9 = 1.5r - r \Rightarrow 1 = 0.5r \Rightarrow r = 2\ \Omega
EMF: E = 10 + 2 = 12 V
(c) Electrostatic induction: Redistribution of charge in a conductor due to nearby charged object without contact. Method: charging by induction (e.g., using a rod).
6. (a) Electrostatics: Study of charges at rest.
(b) Diagram: Start with neutral electroscope, bring positively charged rod near cap, electrons attracted upwards leaving cap negatively charged, leaves positive, touch cap to ground, electrons flow up, remove ground, remove rod — leaves positive.
(c) Coulomb’s law:
F = k q_1 q_2 / r^2, k = 9\times 10^9
q_1 = 2\times 10^{-9} C at r=0.02 m from origin
F_{13} = \frac{9\times 10^9 \times 2\times 10^{-9} \times 5\times 10^{-9}}{(0.02)^2} = \frac{90\times 10^{-9}}{4\times 10^{-4}} = 2.25\times 10^{-4} N (repulsive, away from origin)
q_2 = 3\times 10^{-9} C at r=0.04 m from origin
F_{23} = \frac{9\times 10^9 \times 3\times 10^{-9} \times 5\times 10^{-9}}{(0.04)^2} = \frac{135\times 10^{-9}}{1.6\times 10^{-3}} = 8.4375\times 10^{-5} N (repulsive, away from origin)
Both forces along +x direction (since q3 at origin, q1 and q2 on +x axis)
Total force = 2.25\times 10^{-4} + 8.44\times 10^{-5} = 3.094\times 10^{-4} N
---
PHY 004 (Modern Physics)
7. (a) Alpha particle (2p, 2n)
Mass defect: [2(1.007825) + 2(1.008665)] - 4.004603
= [2.015650 + 2.017330] - 4.004603 = 4.032980 - 4.004603 = 0.028377 u
Binding energy = 0.028377 \times 931.5 \approx 26.43 MeV
Per nucleon = 26.43 / 4 \approx 6.61 MeV
(b) (i) Half-life = 800 years = 800 \times 365 \times 24 \times 3600 \approx 2.525\times 10^{10} s
\lambda = 0.693 / 2.525\times 10^{10} \approx 2.745\times 10^{-11} s⁻¹
Number of atoms in 0.5 g: N = (0.5 / 226) \times 6.022\times 10^{23} \approx 1.332\times 10^{21}
Decay rate = \lambda N \approx (2.745\times 10^{-11})(1.332\times 10^{21}) \approx 3.66\times 10^{10} decays/s
(ii) Half-life: Time for half of radioactive atoms to decay.
(c) (i) Gamma rays
(ii) Alpha rays
(iii) Gamma rays
(iv) Beta rays
8. (a) (i) Properties of X-rays:
1. Penetrating power high.
2. Not deflected by E or B fields.
3. Cause fluorescence.
4. Ionize gases.
(ii) Uses: Medical imaging, security scanning, crystal structure analysis, cancer therapy.
(b) (i) \lambda_{\min} = \frac{hc}{eV} = \frac{6.63\times 10^{-34} \times 3\times 10^8}{1.6\times 10^{-19} \times 20000} = \frac{1.989\times 10^{-25}}{3.2\times 10^{-15}} = 6.22\times 10^{-11} m
(ii) Bragg’s law: n\lambda = 2d\sin\theta
n: order, \lambda: wavelength, d: interplanar spacing, \theta: angle of incidence.
(iii) n=1, d = 4.07\times 10^{-10} m, \theta = 24^\circ
\lambda = 2\times 4.07\times 10^{-10} \times \sin 24^\circ
\sin 24^\circ \approx 0.40674
\lambda = 8.14\times 10^{-10} \times 0.40674 \approx 3.31\times 10^{-10} m