Physics: Model Question Sets and Predicted Questions

This chapter does three things. It reads the shape of the Grade 11 Physics paper (3 hours, full marks 75, Group A 11\times1, Group B 8\times5, Group C 3\times8) and shows how the 128 teaching hours are spread over the 26 units. It then sets two complete model papers to that grid, each fully solved. Finally it lists twenty-four predicted questions in ranked order, each with the reason it is predicted and a short model answer.

4.1 Pattern and weight analysis of the Grade 11 Physics paper

4.1.1 The grid

Part	Questions	Marks	What to expect
Group A	11 multiple-choice	11\times1 = 11	one mark each; drawn from the whole syllabus
Group B	8 short answer	8\times5 = 40	a mix of definition, short derivation and numerical
Group C	3 long answer	3\times8 = 24	derivation plus numerical, often in parts such as [3+3+2]

All questions are compulsory. Time per mark is 180/75 = 2.4 minutes: about 12 minutes for each Group B question and 19 for each Group C question.

4.1.2 Where the syllabus hours sit

Area	Units	Hours	Share	Expected presence in the paper
Mechanics	1–8	45	35 %	the largest block: a Group C question and three or four Group B questions
Heat and Thermodynamics	9–13	26	20 %	two Group B questions, sometimes a Group C part
Waves and Optics	14–18	15	12 %	one or two Group B questions; mirror, lens or prism numerical
Electricity and Magnetism	19–23	27	21 %	two Group B questions and often a Group C question
Modern Physics	24–26	15	12 %	one Group B question; Group A items

4.1.3 Unit by unit: hours and the form the question usually takes

Unit	Hours	Typical form	Priority
1 Physical quantities	3	dimensional analysis, significant figures	medium
2 Vectors	4	resultant, resolution, dot and cross product	high
3 Kinematics	5	equations of motion, projectile, relative velocity	very high
4 Dynamics	6	momentum, impulse, friction, torque and equilibrium	very high
5 Work, energy, power	6	work–energy theorem, collisions, conservation	very high
6 Circular motion	6	centripetal force, conical pendulum, vertical circle, banking	high
7 Gravitation	10	g variation, satellites, escape velocity	very high
8 Elasticity	5	Young modulus, Hooke's law, elastic energy	high
9 Heat and temperature	3	zeroth law, thermometry	low
10 Thermal expansion	4	\alpha,\beta,\gamma relations, Dulong and Petit	medium
11 Quantity of heat	6	calorimetry, latent heat, Newton's cooling	high
12 Rate of heat flow	5	conductivity, Stefan–Boltzmann	medium
13 Ideal gas	8	gas laws, kinetic theory, rms speed	high
14 Curved mirrors	2	mirror formula, image numerical	medium
15 Refraction at plane surfaces	4	apparent depth, critical angle	high
16 Prisms	3	n formula, minimum deviation	high
17 Lenses	3	lens formula, power, combination	medium
18 Dispersion	3	dispersive power, spectrum	medium
19 Electric charges	3	Coulomb's law, conservation	medium
20 Electric field	3	field of point charge, field lines	medium
21 Potential	4	potential, potential energy	medium
22 Capacitor	7	combinations, energy, parallel plate	very high
23 DC circuits	10	Kirchhoff, internal resistance, bridge	very high
24 Nuclear physics	6	decay law, binding energy	high
25 Solids	3	band theory, semiconductors	medium
26 Recent trends	6	applications, descriptive	medium

Two points follow. Nothing is safe to skip: Group A is spread over all 26 units. And the heavy units are the safest investment: Mechanics (45 h), Electricity (27 h) and Heat (26 h) together make up three-quarters of the teaching time.

4.2 Model Set 1

Time: 3 hours. Full marks: 75. Attempt all the questions. Figures in the margin indicate full marks. Use g = 10\ \text{m s}^{-2} unless stated. This is an original practice paper, built to the CDC pattern.

Group A — Rewrite the correct option of each question in your answer sheet. (11\times1 = 11)

1. The dimensions of impulse are [1]a[MLT^{-2}]b[MLT^{-1}]c[ML^{2}T^{-2}]d[ML^{-1}T^{-2}]
2. A car starts from rest and reaches 20\ \text{m s}^{-1} in 5\ \text{s} with uniform acceleration. The distance covered is [1]a20\ \text{m}b50\ \text{m}c100\ \text{m}d200\ \text{m}
3. A body of mass 2\ \text{kg} has kinetic energy 100\ \text{J}. Its speed is [1]a5\ \text{m s}^{-1}b7\ \text{m s}^{-1}c10\ \text{m s}^{-1}d50\ \text{m s}^{-1}
4. At a depth equal to half the Earth's radius, the acceleration due to gravity is (g = 9.8\ \text{m s}^{-2} at the surface) [1]a2.45\ \text{m s}^{-2}b4.9\ \text{m s}^{-2}c9.8\ \text{m s}^{-2}d19.6\ \text{m s}^{-2}
5. A spring of force constant 200\ \text{N m}^{-1} is stretched by 0.10\ \text{m}. The energy stored is [1]a0.5\ \text{J}b1.0\ \text{J}c2.0\ \text{J}d10\ \text{J}
6. A 1\ \text{m} metal rod (\alpha = 1.2\times10^{-5}\ \text{K}^{-1}) is heated through 100\ \text{K}. Its increase in length is [1]a0.12\ \text{mm}b1.2\ \text{mm}c12\ \text{mm}d120\ \text{mm}
7. An object is placed 40\ \text{cm} from a concave mirror of focal length 20\ \text{cm}. The image distance is [1]a20\ \text{cm}b40\ \text{cm}c60\ \text{cm}d80\ \text{cm}
8. The critical angle for a medium of refractive index 2 with respect to air is [1]a30^{\circ}b45^{\circ}c60^{\circ}d90^{\circ}
9. The equivalent resistance of 6\ \Omega and 3\ \Omega in parallel is [1]a1\ \Omegab2\ \Omegac4.5\ \Omegad9\ \Omega
10. The charge on a 5\ \mu\text{F} capacitor at 20\ \text{V} is [1]a4\ \mu\text{C}b25\ \mu\text{C}c100\ \mu\text{C}d400\ \mu\text{C}
11. A radioactive sample has a half-life of 10 minutes. The fraction left after 30 minutes is [1]a1/2b1/4c1/8d1/16

Group B — Short answer questions. (8\times5 = 40)

12. A stone is thrown horizontally at 15\ \text{m s}^{-1} from the top of a 80\ \text{m} cliff. Find the time to reach the ground, the horizontal distance covered, and the speed at impact. [5]
13. State Newton's second law in terms of momentum. A 0.2\ \text{kg} ball hits a wall at 20\ \text{m s}^{-1} and rebounds at 15\ \text{m s}^{-1}; the contact lasts 0.01\ \text{s}. Find the impulse and the average force. [5]
14. A uniform plank of weight 200\ \text{N} and length 4\ \text{m} rests on supports at its two ends. A person of weight 100\ \text{N} stands 1\ \text{m} from the left end. Find the reactions at the supports. [5]
15. Define work done by a variable force. A force F = (3x + 2)\ \text{N} acts on a body as it moves from x = 0 to x = 4\ \text{m}. Find the work done. [5]
16. Define specific latent heat of vaporization. How much heat is released when 200\ \text{g} of steam at 100^{\circ}\text{C} condenses and the water cools to 50^{\circ}\text{C}? (L_v = 2.26\times10^{6}\ \text{J kg}^{-1}, c_w = 4200\ \text{J kg}^{-1}\text{K}^{-1}) [5]
17. An object is 15\ \text{cm} from a concave lens of focal length 10\ \text{cm}. Find the position and magnification of the image and describe it. [5]
18. Capacitors of 4\ \mu\text{F} and 8\ \mu\text{F} are connected in series across a 12\ \text{V} battery. Find the equivalent capacitance, the charge and the p.d. across each. [5]
19. State Einstein's mass–energy relation. Find the energy equivalent to 1\ \text{g} of matter (c = 3\times10^{8}\ \text{m s}^{-1}) and explain what mass defect means. [5]

Group C — Long answer questions. (3\times8 = 24)

20. (a) State the law of conservation of linear momentum and derive it from Newton's laws. [3](b) A 2\ \text{kg} body moving at 6\ \text{m s}^{-1} collides elastically head-on with a 4\ \text{kg} body at rest. Find the velocities after collision and verify that kinetic energy is conserved. [5] [8]
21. (a) Derive an expression for the period of a conical pendulum. [4](b) A bob of mass 0.5\ \text{kg} on a string 1.0\ \text{m} long moves in a horizontal circle with the string at 60^{\circ} to the vertical. Find the period [2] and the tension. [2] [8]
22. (a) Define e.m.f. and internal resistance and show that I = \dfrac{E}{R + r}. [3](b) Three cells, each of e.m.f. 1.5\ \text{V} and internal resistance 0.5\ \Omega, are joined in series to a 3.0\ \Omega resistor. Find the current and the terminal p.d. [3](c) Find the power in the resistor and the efficiency of the circuit. [2] [8]

4.2.1 Solutions to Model Set 1

Solutions — Group A

1. (b) — impulse = F\Delta t = change of momentum, [MLT^{-1}].

2. (b) — a = 20/5 = 4; s = \tfrac12at^{2} = \tfrac12(4)(25) = 50\ \text{m}.

3. (c) — v = \sqrt{2K/m} = \sqrt{200/2} = 10\ \text{m s}^{-1}.

4. (b) — g_d = g(1 - d/R) = 9.8\times\tfrac12 = 4.9.

5. (b) — U = \tfrac12kx^{2} = \tfrac12(200)(0.01) = 1.0\ \text{J}.

6. (b) — \Delta l = l\alpha\Delta\theta = 1\times1.2\times10^{-5}\times100 = 1.2\times10^{-3}\ \text{m}.

7. (b) — the object is at C (u = 2f = 40), so the image is also at C: v = 40\ \text{cm}.

8. (a) — \sin C = 1/2, so C = 30^{\circ}.

9. (b) — \dfrac{6\times3}{6+3} = 2\ \Omega.

10. (c) — Q = CV = 5\times20 = 100\ \mu\text{C}.

11. (c) — three half-lives: (\tfrac12)^{3} = 1/8.

Solutions — Group B (12–19)

12. Vertical fall: h = \tfrac12gt^{2}, so t = \sqrt{\dfrac{2\times80}{10}} = 4\ \text{s}. Horizontal distance = 15\times4 = 60\ \text{m}. At impact v_y = gt = 40\ \text{m s}^{-1}, v_x = 15; speed = \sqrt{15^{2} + 40^{2}} = \sqrt{1825} = 42.7\ \text{m s}^{-1}.

13. The rate of change of momentum of a body is proportional to the applied force and takes place in the direction of the force: F = \dfrac{\Delta p}{\Delta t}. Taking the rebound direction as positive, \Delta p = 0.2\,[15 - (-20)] = 7.0\ \text{N s} (impulse). F = \dfrac{7.0}{0.01} = 700\ \text{N}.

14. Let R_1 and R_2 be the reactions at the left and right ends. Moments about the right end: R_1(4) = 200(2) + 100(3) = 700, so R_1 = 175\ \text{N}. Vertical balance: R_1 + R_2 = 300, so R_2 = 125\ \text{N}.

15. For a variable force the work is the area under the F–x graph, W = \int F\,dx. W = \int_0^4(3x + 2)\,dx = \left[\tfrac32x^{2} + 2x\right]_0^4 = 24 + 8 = 32\ \text{J}.

16. The specific latent heat of vaporization is the heat needed to change 1\ \text{kg} of a liquid to vapour at its boiling point without change in temperature. Heat released = mL_v + mc\Delta\theta = 0.2(2.26\times10^{6}) + 0.2(4200)(50) = 4.52\times10^{5} + 4.2\times10^{4} = 4.94\times10^{5}\ \text{J}.

17. \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = -\dfrac{1}{10} - \dfrac{1}{15} = -\dfrac{5}{30}, so v = -6\ \text{cm} (virtual, on the same side as the object). m = -\dfrac{v}{u} = \dfrac{6}{15} = 0.4: erect, diminished, virtual.

18. C = \dfrac{4\times8}{4 + 8} = 2.67\ \mu\text{F}. In series each carries the same charge: Q = CV = 2.67\times12 = 32\ \mu\text{C}. V_1 = \dfrac{32}{4} = 8\ \text{V}; V_2 = \dfrac{32}{8} = 4\ \text{V} (sum 12\ \text{V} ✓).

19. E = mc^{2}: energy and mass are equivalent. E = 10^{-3}\times(3\times10^{8})^{2} = 9\times10^{13}\ \text{J}. Mass defect is the difference between the sum of the masses of the separate nucleons and the mass of the nucleus; it appears as the binding energy \Delta m\,c^{2}.

Solutions — Group C (20–22)

20. (a) The total linear momentum of an isolated system remains constant. For two bodies in collision, by Newton's third law F_{12} = -F_{21} for the same contact time \Delta t, so F_{12}\Delta t = -F_{21}\Delta t, i.e. \Delta p_1 = -\Delta p_2 and \Delta(p_1 + p_2) = 0. (b) Momentum: 2(6) = 2v_1 + 4v_2, so v_1 + 2v_2 = 6. Elastic: relative speed of approach equals relative speed of separation, 6 = v_2 - v_1. Adding: 3v_2 = 12, so v_2 = 4\ \text{m s}^{-1} and v_1 = -2\ \text{m s}^{-1} (the 2\ \text{kg} body rebounds). Kinetic energy before: \tfrac12(2)(36) = 36\ \text{J}. After: \tfrac12(2)(4) + \tfrac12(4)(16) = 4 + 32 = 36\ \text{J} ✓.

21. (a) (Diagram: bob of mass m on a string of length l at angle \theta to the vertical; tension T along the string; weight mg downward; circle of radius r = l\sin\theta.) Vertically: T\cos\theta = mg. Horizontally the component T\sin\theta provides the centripetal force: T\sin\theta = m\omega^{2}r = m\omega^{2}l\sin\theta, so T = m\omega^{2}l. Combining, \omega^{2} = \dfrac{g}{l\cos\theta} and the period T_p = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{l\cos\theta}{g}}. (b) T_p = 2\pi\sqrt{\dfrac{1.0\times0.5}{10}} = 2\pi\times0.2236 = 1.41\ \text{s}. Tension: T = \dfrac{mg}{\cos\theta} = \dfrac{0.5\times10}{0.5} = 10\ \text{N}.

22. (a) E.m.f. is the work done by the source per unit charge in carrying it round the complete circuit (equal to the terminal p.d. on open circuit). Internal resistance r is the resistance of the electrolyte and plates inside the cell. In a circuit with external resistance R, E = IR + Ir, so I = \dfrac{E}{R + r}. (b) In series, E_{tot} = 4.5\ \text{V} and r_{tot} = 1.5\ \Omega. I = \dfrac{4.5}{3.0 + 1.5} = 1.0\ \text{A}. Terminal p.d. V = E - Ir = 4.5 - 1.5 = 3.0\ \text{V}. (c) P = I^{2}R = 1^{2}\times3.0 = 3.0\ \text{W}. Efficiency = \dfrac{V}{E} = \dfrac{3.0}{4.5} = 66.7\ \%.

4.3 Model Set 2

Time: 3 hours. Full marks: 75. Attempt all the questions. Use g = 10\ \text{m s}^{-2} unless stated. This is an original practice paper, built to the CDC pattern.

Group A — Rewrite the correct option of each question in your answer sheet. (11\times1 = 11)

1. The dimensions of angular momentum are [1]a[MLT^{-1}]b[ML^{2}T^{-2}]c[ML^{2}T^{-1}]d[MLT^{-2}]
2. A stone falls freely from rest through 45\ \text{m}. The time taken is [1]a1.5\ \text{s}b2\ \text{s}c3\ \text{s}d4.5\ \text{s}
3. The momentum of a 0.5\ \text{kg} ball moving at 4\ \text{m s}^{-1} is [1]a0.5\ \text{kg m s}^{-1}b2\ \text{kg m s}^{-1}c8\ \text{kg m s}^{-1}d20\ \text{kg m s}^{-1}
4. The escape velocity from the Earth's surface is about [1]a7.9\ \text{km s}^{-1}b11.2\ \text{km s}^{-1}c15\ \text{km s}^{-1}d3\times10^{5}\ \text{km s}^{-1}
5. A force of 10\ \text{N} acts at 60^{\circ} to the displacement of 5\ \text{m}. The work done is [1]a50\ \text{J}b25\ \text{J}c43.3\ \text{J}d0
6. The heat needed to raise 2\ \text{kg} of water by 10\ \text{K} (c = 4200\ \text{J kg}^{-1}\text{K}^{-1}) is [1]a8.4\times10^{3}\ \text{J}b4.2\times10^{4}\ \text{J}c8.4\times10^{4}\ \text{J}d8.4\times10^{5}\ \text{J}
7. The speed of light in glass of refractive index 1.5 is [1]a1.5\times10^{8}\ \text{m s}^{-1}b2.0\times10^{8}\ \text{m s}^{-1}c3.0\times10^{8}\ \text{m s}^{-1}d4.5\times10^{8}\ \text{m s}^{-1}
8. The power of a convex lens of focal length 50\ \text{cm} is [1]a0.5\ \text{D}b1\ \text{D}c2\ \text{D}d5\ \text{D}
9. The field between two plates 2\ \text{cm} apart at a p.d. of 100\ \text{V} is [1]a50\ \text{V m}^{-1}b200\ \text{V m}^{-1}c5000\ \text{V m}^{-1}d20\,000\ \text{V m}^{-1}
10. The heat produced by 2\ \text{A} in a 5\ \Omega resistor in 10\ \text{s} is [1]a100\ \text{J}b200\ \text{J}c400\ \text{J}d1000\ \text{J}
11. After an \alpha-decay of ^{238}_{92}\text{U}, the mass number of the daughter nucleus is [1]a238b236c234d232

Group B — Short answer questions. (8\times5 = 40)

12. Two blocks of 3\ \text{kg} and 2\ \text{kg} lie on a smooth table, connected by a light string. A horizontal force of 10\ \text{N} pulls the 3\ \text{kg} block. Find the acceleration and the tension in the string. [5]
13. A 10\ \text{kg} block slides down a rough incline of 30^{\circ} with \mu_k = 0.2. Find its acceleration. [5]
14. Derive an expression for the escape velocity and evaluate it for g = 9.8\ \text{m s}^{-2} and R = 6.4\times10^{6}\ \text{m}. [5]
15. A load of 50\ \text{N} is hung from a steel wire 2\ \text{m} long and 0.5\ \text{mm} in radius (Y = 2\times10^{11}\ \text{N m}^{-2}). Find the extension. [5]
16. Define linear expansivity. A steel rail 12\ \text{m} long (\alpha = 1.2\times10^{-5}\ \text{K}^{-1}) is laid at 10^{\circ}\text{C}. Find the gap needed between rails for a rise to 40^{\circ}\text{C}, and say why gaps are left. [5]
17. An object is 30\ \text{cm} from a convex mirror of focal length 20\ \text{cm}. Find the image position and magnification. [5]
18. Charges +4\ \mu\text{C} and -4\ \mu\text{C} are 10\ \text{cm} apart. Find the electric field and the potential at the midpoint. [5]
19. Define activity of a radioactive source and state its SI unit. A source has an initial activity of 800\ \text{Bq} and half-life 2\ \text{h}; find its activity after 6\ \text{h}. Name one property that distinguishes \alpha from \gamma radiation. [5]

Group C — Long answer questions. (3\times8 = 24)

20. (a) State and prove the work–energy theorem for a constant force. [3](b) A pump raises 900\ \text{kg} of water per minute through 20\ \text{m}. Find the output power, and the input power if the efficiency is 75\ \%. [3](c) Define power and state its SI unit. [2] [8]
21. (a) Derive the minimum speeds at the top and at the bottom of a vertical circle for a body on a string. [5](b) For a 0.2\ \text{kg} body on a string of length 0.9\ \text{m}, find the minimum speeds at the top and bottom, and the tension at the bottom at that speed. [3] [8]
22. (a) State the laws of refraction and derive n = \dfrac{\text{real depth}}{\text{apparent depth}} for a nearly normal view. [4](b) A tank holds 12\ \text{cm} of water (n = 4/3) under 9\ \text{cm} of oil (n = 1.5). Find the apparent depth of the bottom as seen from above. [4] [8]

4.3.1 Solutions to Model Set 2

Solutions — Group A

1. (c) — L = mvr: [M][LT^{-1}][L] = [ML^{2}T^{-1}].

2. (c) — t = \sqrt{2h/g} = \sqrt{90/10} = 3\ \text{s}.

3. (b) — p = mv = 0.5\times4 = 2.

4. (b) — v_e = \sqrt{2gR} \approx 11.2\ \text{km s}^{-1}; (a) is the orbital speed near the surface.

5. (b) — W = Fs\cos\theta = 10\times5\times0.5 = 25\ \text{J}.

6. (c) — Q = mc\Delta\theta = 2\times4200\times10 = 8.4\times10^{4}\ \text{J}.

7. (b) — v = c/n = 3\times10^{8}/1.5 = 2\times10^{8}.

8. (c) — P = 1/f = 1/0.5 = 2\ \text{D}.

9. (c) — E = V/d = 100/0.02 = 5000\ \text{V m}^{-1}.

10. (b) — I^{2}Rt = 4\times5\times10 = 200\ \text{J}.

11. (c) — \alpha-emission removes 4 from the mass number: 238 - 4 = 234.

Solutions — Group B (12–19)

12. Treat the two blocks as one system: a = \dfrac{F}{m_1 + m_2} = \dfrac{10}{5} = 2.0\ \text{m s}^{-2}. The string alone accelerates the 2\ \text{kg} block: T = m_2a = 2\times2 = 4\ \text{N}.

13. Down the incline: mg\sin\theta - \mu mg\cos\theta = ma, so a = g(\sin30^{\circ} - 0.2\cos30^{\circ}) = 10(0.5 - 0.1732) = 3.27\ \text{m s}^{-2}.

14. To escape, the kinetic energy given must equal the work against gravity to infinity: \tfrac12mv_e^{2} = \dfrac{GMm}{R}. With GM = gR^{2}: v_e = \sqrt{2gR}. v_e = \sqrt{2\times9.8\times6.4\times10^{6}} = \sqrt{1.254\times10^{8}} = 1.12\times10^{4}\ \text{m s}^{-1} = 11.2\ \text{km s}^{-1}.

15. A = \pi r^{2} = \pi(0.5\times10^{-3})^{2} = 7.85\times10^{-7}\ \text{m}^{2}. \Delta l = \dfrac{FL}{AY} = \dfrac{50\times2}{7.85\times10^{-7}\times2\times10^{11}} = 6.4\times10^{-4}\ \text{m} = 0.64\ \text{mm}.

16. Linear expansivity is the fractional increase in length per unit original length per kelvin rise in temperature, \alpha = \dfrac{\Delta l}{l\Delta\theta}. \Delta l = 12\times1.2\times10^{-5}\times30 = 4.3\times10^{-3}\ \text{m} = 4.3\ \text{mm}. Gaps let the rails lengthen in summer; without them they would buckle.

17. With f = -20 and u = 30: \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = -\dfrac{1}{20} - \dfrac{1}{30} = -\dfrac{5}{60}, so v = -12\ \text{cm} (virtual, behind the mirror). m = -\dfrac{v}{u} = \dfrac{12}{30} = 0.4 (erect, diminished).

18. Each charge is 5\ \text{cm} from the midpoint. E_1 = \dfrac{kq}{r^{2}} = \dfrac{9\times10^{9}\times4\times10^{-6}}{(0.05)^{2}} = 1.44\times10^{7}\ \text{N C}^{-1}. Both fields point towards the negative charge, so they add: E = 2.88\times10^{7}\ \text{N C}^{-1}. Potential: V = k\left(\dfrac{4\times10^{-6}}{0.05} + \dfrac{-4\times10^{-6}}{0.05}\right) = 0.

19. Activity is the number of nuclei disintegrating per second; the SI unit is the becquerel (1\ \text{Bq} = 1 decay per second). 6\ \text{h} = 3 half-lives, so A = 800\times\tfrac18 = 100\ \text{Bq}. Distinction: \alpha particles are helium nuclei, strongly ionizing and stopped by paper; \gamma rays are electromagnetic waves, weakly ionizing and very penetrating.

Solutions — Group C (20–22)

20. (a) For a constant net force F, v^{2} = u^{2} + 2as and F = ma. Multiply by m/2: \tfrac12mv^{2} - \tfrac12mu^{2} = mas = Fs = W. The net work done equals the change in kinetic energy. (b) Output power = \dfrac{mgh}{t} = \dfrac{900\times10\times20}{60} = 3000\ \text{W}. Input = \dfrac{3000}{0.75} = 4000\ \text{W}. (c) Power is the rate of doing work, P = W/t; the unit is the watt (\text{J s}^{-1}).

21. (a) (Diagram: circle of radius r with the body at the top and bottom, tension T towards the centre, weight mg downward.) At the top: T_t + mg = \dfrac{mv_t^{2}}{r}. The least speed is when T_t = 0: v_t = \sqrt{gr}. Between bottom and top the body rises 2r; energy conservation gives \tfrac12mv_b^{2} = \tfrac12mv_t^{2} + mg(2r), so v_b^{2} = gr + 4gr = 5gr and v_b = \sqrt{5gr}. At the bottom, T_b - mg = \dfrac{mv_b^{2}}{r} = 5mg, so T_b = 6mg. (b) v_t = \sqrt{10\times0.9} = 3.0\ \text{m s}^{-1}; v_b = \sqrt{5\times10\times0.9} = \sqrt{45} = 6.7\ \text{m s}^{-1}. T_b = 6mg = 6\times0.2\times10 = 12\ \text{N}.

22. (a) Laws: the incident ray, the refracted ray and the normal lie in one plane; \dfrac{\sin i}{\sin r} is constant for a given pair of media (Snell). (Diagram: object O at depth d in the denser medium; a ray OA at a small angle r to the normal refracts at A into air at angle i; the eye traces it back to the apparent position I at depth d'.) With x the horizontal distance of A from the normal, for small angles \sin r \approx \tan r = x/d and \sin i \approx \tan i = x/d'. Hence n = \dfrac{\sin i}{\sin r} = \dfrac{x/d'}{x/d} = \dfrac{d}{d'}, i.e. n = \dfrac{\text{real depth}}{\text{apparent depth}}. (b) The apparent depth is the sum of the apparent thicknesses of each layer: \dfrac{9}{1.5} + \dfrac{12}{4/3} = 6 + 9 = 15\ \text{cm} (the real depth is 21\ \text{cm}).

4.4 Predicted Questions — ranked

The ranking is by teaching hours of the unit and how standard the question is. Each prediction states its reason. Tier A are the questions a paper on this syllabus almost has to contain in some form; Tier B are likely; Tier C are possible and worth a pass if time remains. These are judgements, never guarantees.

4.4.1 Tier A — revise first

P1. Equations of motion by the graphical method, and a numerical (unit 3, [5\text{ or }8]; Mechanics). Why: kinematics is the foundation of Mechanics (45 h), and the graphical derivation is named in the syllabus. Outline: v = u + at from the slope; s as the area under the v–t graph; v^{2} = u^{2} + 2as by eliminating t. Example: u = 5, a = 2, t = 6 gives v = 17\ \text{m s}^{-1}, s = 66\ \text{m}. P2. Projectile motion: time of flight, maximum height, range (unit 3, [5\text{ or }8]). Why: named in the syllabus and the standard application of vectors and kinematics. Outline: T = \dfrac{2u\sin\theta}{g}, H = \dfrac{u^{2}\sin^{2}\theta}{2g}, R = \dfrac{u^{2}\sin2\theta}{g}; maximum range at 45^{\circ}; R equal for \theta and 90^{\circ} - \theta. P3. Conservation of linear momentum, with an elastic or inelastic collision (unit 4/5, [5\text{ or }8]). Why: named in both units, and gives a clean numerical. Outline: proof from Newton's third law; elastic: v_1 = \dfrac{m_1 - m_2}{m_1 + m_2}u, v_2 = \dfrac{2m_1}{m_1 + m_2}u; inelastic: v = \dfrac{m_1u}{m_1 + m_2}, kinetic energy lost. P4. Laws of solid friction and angle of repose (unit 4, [5]). Why: the syllabus names the laws and their verification. Outline: four laws; \mu = \tan\theta at the angle of friction; incline with and without friction. P5. Work–energy theorem and conservation of energy (unit 5, [5\text{ or }8]). Why: a six-hour unit with a standard proof. Outline: W = \Delta K; K + U constant for a freely falling body; check at two heights. P6. Satellite: orbital speed, period, geostationary orbit (unit 7, [8]). Why: gravitation has the largest unit weight (10 h) and the satellite topics fill half the unit. Outline: v = \sqrt{GM/(R+h)}; T = 2\pi\sqrt{(R+h)^{3}/GM}; energy K = GMm/2r, U = -GMm/r, E = -GMm/2r; geostationary conditions. P7. Variation of g with altitude and depth; escape velocity (unit 7, [5]). Why: named in the syllabus; clean derivations. Outline: g_h = g\left(\dfrac{R}{R+h}\right)^{2}; g_d = g\left(1 - \dfrac{d}{R}\right); v_e = \sqrt{2gR} = 11.2\ \text{km s}^{-1}. P8. Young modulus, stress, strain and elastic energy (unit 8, [5]). Why: a numerical type with exact formulae. Outline: Y = \dfrac{FL}{A\Delta l}; U = \tfrac12F\Delta l; energy per volume \tfrac12\times\text{stress}\times\text{strain}. P9. Combination of capacitors and energy stored (unit 22, [5]). Why: a seven-hour unit with direct numericals. Outline: series \dfrac{1}{C} = \sum\dfrac{1}{C_i}; parallel C = \sum C_i; U = \tfrac12CV^{2}. P10. Kirchhoff's laws with a two-loop or bridge numerical; internal resistance (unit 23, [5\text{ or }8]). Why: DC circuits carry 10 hours, the joint-largest unit weight. Outline: \sum I = 0 and \sum E = \sum IR; I = \dfrac{E}{R + r}; Wheatstone balance \dfrac{P}{Q} = \dfrac{R}{S}. P11. Calorimetry and latent heat numerical (unit 11, [5]). Why: a recurring numerical type and a six-hour unit. Outline: heat lost = heat gained; list each stage: warm, melt, warm, boil. P12. Kinetic theory: P = \tfrac13\rho\overline{c^{2}} and rms speed (unit 13, [5\text{ or }8]). Why: the standard derivation of an eight-hour unit. Outline: momentum change 2mu_x; time 2l/u_x; \overline{u_x^{2}} = \tfrac13\overline{c^{2}}; c_{rms} = \sqrt{3RT/M}.

4.4.2 Tier B — likely

P13. Conical pendulum and motion in a vertical circle (unit 6, [5\text{ or }8]). Why: named in the syllabus; clean derivations. Outline: T = 2\pi\sqrt{l\cos\theta/g}; v_{top} = \sqrt{gr}, v_{bottom} = \sqrt{5gr}, T_{bottom} = 6mg. P14. Banking of roads (unit 6, [5]). Why: named in the syllabus. Outline: \tan\theta = \dfrac{v^{2}}{rg}; with friction v^{2} = rg\,\dfrac{\mu + \tan\theta}{1 - \mu\tan\theta}. P15. Moment, torque and equilibrium of a beam (unit 4, [5]). Why: a standard numerical (two supports, a hanging load). Outline: take moments about one support; sum of forces zero. P16. Triangle, parallelogram law and resolution of vectors; scalar and vector products (unit 2, [5]). Why: four hours, all formulae exact. Outline: R = \sqrt{P^{2} + Q^{2} + 2PQ\cos\theta}; \tan\alpha = \dfrac{Q\sin\theta}{P + Q\cos\theta}; \vec A\cdot\vec B = AB\cos\theta, |\vec A\times\vec B| = AB\sin\theta. P17. Thermal expansion: \gamma = 3\alpha and the apparent expansion of a liquid (unit 10, [5]). Why: named relations in the syllabus. Outline: \beta = 2\alpha, \gamma = 3\alpha; \gamma_r = \gamma_a + \gamma_g. P18. Newton's law of cooling and Stefan–Boltzmann law (units 11/12, [5]). Why: named in the syllabus; formula-based. Outline: rate \propto excess temperature; P = \sigma AT^{4}; net loss \sigma A(T^{4} - T_0^{4}). P19. Refraction: apparent depth, critical angle, total internal reflection (unit 15, [5]). Why: four hours; standard numericals. Outline: n = \dfrac{\text{real}}{\text{apparent}}; \sin C = 1/n; optical fibre as application. P20. Prism: n in terms of A and D_m (unit 16, [5]). Why: the prism formula is the syllabus's single derivation of its unit. Outline: D = i_1 + i_2 - A; n = \dfrac{\sin\frac{A + D_m}{2}}{\sin\frac{A}{2}}; thin prism D = (n - 1)A.

4.4.3 Tier C — possible

P21. Radioactive decay law and half-life (unit 24, [5]). Why: the one derivation of Modern Physics; a standard numerical. Outline: N = N_0e^{-\lambda t}; T_{1/2} = 0.693/\lambda. P22. Binding energy and mass defect (unit 24, [5]). Why: a numerical type with a fixed constant (1\ \text{u} = 931.5\ \text{MeV}). Outline: B = \Delta m\times931.5\ \text{MeV}; binding energy per nucleon. P23. Band theory: conductor, semiconductor, insulator; n-type and p-type (unit 25, [5]). Why: the whole content of a three-hour unit. Outline: band diagrams; doping with Group V (donor) or Group III (acceptor) atoms. P24. Dispersive power and the pure spectrum (unit 18, [5]). Why: a standard definition with a one-line numerical. Outline: \omega = \dfrac{n_v - n_r}{n_y - 1}; the pure spectrum is formed with two prisms or a lens and a screen.

Chapter summary

• The Physics paper is 75 marks: 11\times1, 8\times5 and 3\times8; Mechanics (45 h), Electricity (27 h) and Heat (26 h) make three-quarters of the teaching time.
• Two complete model sets, built to the grid, are solved in full; each answer shows the law, the substitution with units and the final result.
• Recurring numerical types: equations of motion and projectiles; collisions and impulse; g variation and satellites; Y = FL/A\Delta l; calorimetry; capacitor combinations; internal resistance and Kirchhoff; half-life.
• The twenty-four predicted questions are ranked by unit hours and standardness; they are judgements, not leaks.
• In the hall: formula first, then substitution with units, then the answer; draw the diagram before the algebra; convert to kelvin, metres and radians before calculating.

Practice questions

Group A — Multiple choice (1 mark each)

1. The tension in a conical-pendulum string of mass m at angle \theta to the vertical is [1]amg\sin\thetabmg\cos\thetacmg/\cos\thetadmg/\sin\theta
2. The maximum range of a projectile launched at 20\ \text{m s}^{-1} (g = 10) is [1]a20\ \text{m}b40\ \text{m}c80\ \text{m}d200\ \text{m}
3. Two 6\ \mu\text{F} capacitors in parallel give [1]a3\ \mu\text{F}b6\ \mu\text{F}c12\ \mu\text{F}d36\ \mu\text{F}
4. The minimum speed at the top of a vertical circle of radius 1.6\ \text{m} (g = 10) is [1]a2\ \text{m s}^{-1}b4\ \text{m s}^{-1}c8\ \text{m s}^{-1}d16\ \text{m s}^{-1}

Answers to Group A

1. (c) — T\cos\theta = mg.

2. (b) — R_{max} = u^{2}/g = 400/10 = 40\ \text{m}.

3. (c) — parallel capacitances add: 6 + 6 = 12.

4. (b) — \sqrt{gr} = \sqrt{16} = 4\ \text{m s}^{-1}.

Group B — Short answer (5 marks each)

1. A 60\ \text{kg} person stands on a lift accelerating upward at 2\ \text{m s}^{-2}. Find the apparent weight (g = 10). [5]
2. Find the heat needed to raise 0.5\ \text{kg} of ice at 0^{\circ}\text{C} to water at 20^{\circ}\text{C} (L_f = 3.36\times10^{5}\ \text{J kg}^{-1}). [5]
3. Find the equivalent resistance between the ends when 2\ \Omega is in series with the parallel pair 3\ \Omega and 6\ \Omega. [5]
4. A sample has 5\ \text{g} left after 12 days from 40\ \text{g}. Find its half-life. [5]

Answers to Group B

1. R - mg = ma, so R = m(g + a) = 60\times12 = 720\ \text{N} (apparent weight, compared with 600\ \text{N} at rest).

2. Melting: 0.5\times3.36\times10^{5} = 1.68\times10^{5}\ \text{J}. Warming: 0.5\times4200\times20 = 4.2\times10^{4}\ \text{J}. Total 2.10\times10^{5}\ \text{J}.

3. Parallel: \dfrac{3\times6}{9} = 2\ \Omega; total 2 + 2 = 4\ \Omega.

4. 40/5 = 8 = 2^{3}, so 3T_{1/2} = 12 days and T_{1/2} = 4 days.

Group C — Long answer (8 marks each)

1. Derive the expression for the energy stored in a charged capacitor and find the energy released when a 20\ \mu\text{F} capacitor charged to 50\ \text{V} is discharged. [8]
2. Derive the relation between the focal length and radius of curvature of a concave mirror and find the image formed by an object 30\ \text{cm} from a concave mirror of f = 10\ \text{cm}. [8]

Answers to Group C

1. Charging from q to q + dq against the p.d. V = q/C needs dW = \dfrac{q}{C}\,dq. Total work W = \int_0^Q\dfrac{q}{C}\,dq = \dfrac{Q^{2}}{2C} = \tfrac12CV^{2}, stored as electric field energy. U = \tfrac12(20\times10^{-6})(50)^{2} = 0.025\ \text{J}, which is released as heat or light on discharge.

2. A ray parallel to the axis strikes the mirror at D; the radius CD is the normal, so the angle of incidence and reflection are equal, \theta. The reflected ray cuts the axis at F. In \triangle CDF, \angle FCD = \angle CDF = \theta (since DA \parallel CP), so CF = FD; for a small aperture FD \approx FP, hence CP = CF + FP = 2FP and f = R/2. Numerical: \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{10} - \dfrac{1}{30} = \dfrac{2}{30}, so v = 15\ \text{cm} (real, inverted); m = -\dfrac{15}{30} = -0.5 (diminished).