Physics: Model Question Sets and Predicted Questions

This chapter does three things. It reads the shape of the Grade 12 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 twenty-five CDC units, so that you can see where the marks are likely to sit. It then sets two complete model papers to that exact grid, each fully solved in the order a full-mark script would follow. Finally it lists thirty-nine predicted questions, ranked, every one with the reason it is predicted and a short model answer.

4.1 Pattern and weight analysis of the Grade 12 Physics paper

4.1.1 The grid

Part	Questions	Marks	What to expect
Group A	Q1 to Q11, multiple choice	11\times1 = 11	one mark each, spread over the whole syllabus; a definition, a unit, one substitution or a misconception
Group B	Q12 to Q19, short answer	8\times5 = 40	a definition or statement with a short derivation or numerical; marks split 2+3, 1+2+2 or 3+2
Group C	Q20 to Q22, long answer	3\times8 = 24	a full derivation with a numerical, often split 4+2+2, 3+3+2 or 5+3

All questions are compulsory; any choice is inside a question, between a question and its Or alternative. The written paper is 75 marks; the remaining 25 are internal (practical work and project). Time per mark is 180/75 = 2.4 minutes, which gives about 12 minutes for each Group B question and about 19 minutes for each Group C question, with roughly 15 minutes left for Group A and a final check.

4.1.2 Where the syllabus hours sit

The CDC publishes no per-unit marks grid, so the "expected marks" column below is our estimate: each area's share of the 128 teaching hours applied to the 75 written marks.

Area	Units	Hours	Share of hours	Expected marks (of 75)	Typical presence
Mechanics	1–3	22	17 %	about 13	two MCQs, one or two Group B, often a Group C (SHM or rotation)
Heat and Thermodynamics	4–5	12	9 %	about 7	one MCQ and one Group B, or a Group C alternative
Waves and Optics	6–13	26	20 %	about 15	three or four MCQs and two Group B (pipes, Doppler, interference, grating)
Electricity and Magnetism	14–19	35	27 %	about 20	two or three MCQs, two Group B and a Group C
Modern Physics	20–25	33	26 %	about 19	two or three MCQs, one or two Group B and a Group C

Two points follow. Electricity and Magnetism and Modern Physics together carry more than half the hours (68 of 128), so together they are worth about forty marks. And Waves and Optics is eight small units of two to five hours each: no single unit is heavy, but the area as a whole is worth as much as Mechanics plus Heat.

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

#	Unit	Hours	Typical form of the question	Priority
1	Rotational dynamics	7	MI of a rod; \tau = I\alpha; rotational KE; conservation of angular momentum	very high
2	Periodic motion	6	SHM equation and energy; spring–mass and simple pendulum derivations	very high
3	Fluid statics	9	capillary rise; Stokes' law and terminal velocity; Bernoulli and continuity	very high
4	First law of thermodynamics	6	C_p - C_v = R; isothermal work; PV^{\gamma} = constant	very high
5	Second law of thermodynamics	6	Carnot, Otto and Diesel efficiency; refrigerator COP; statements of the law	high
6	Wave motion	2	reading a progressive-wave equation; progressive against stationary	medium
7	Mechanical waves	4	Newton's formula and Laplace's correction; effect of temperature and humidity	high
8	Wave in pipes and strings	4	harmonics in open and closed pipes; end correction; string laws	very high
9	Acoustic phenomena	5	Doppler effect; intensity level in decibels	very high
10	Nature and propagation of light	3	Huygens' principle; reflection and refraction by wave theory	medium
11	Interference	2	YDSE fringe width derivation and numerical	high
12	Diffraction	3	grating equation; single-slit width; resolving power	high
13	Polarization	3	Brewster's law; polaroids; transverse nature of light	medium
14	Electrical circuits	6	Kirchhoff; Wheatstone and meter bridge; potentiometer; galvanometer conversion	very high
15	Thermoelectric effects	3	Seebeck and Peltier effects; neutral and inversion temperature	medium
16	Magnetic field	9	Biot–Savart and Ampère; force on charge and conductor; Hall effect; galvanometer	very high
17	Magnetic properties of materials	5	hysteresis loop; dia-, para- and ferromagnetism; \mu_r and \chi	medium
18	Electromagnetic induction	6	Faraday and Lenz; motional emf; AC generator; inductor energy; transformer	very high
19	Alternating currents	6	series LCR impedance, phase, resonance, power factor	very high
20	Electrons	4	Millikan's experiment; Thomson's e/m; electron in fields	high
21	Photons	3	Einstein's equation; stopping potential; Planck's constant from a graph	very high
22	Semiconductor devices	6	p–n junction; full-wave rectifier; logic-gate truth tables	high
23	Quantization of energy	8	Bohr's radius and energy; spectral series; de Broglie; X-rays and Bragg's law	very high
24	Radioactivity and nuclear reaction	6	decay law; half-life and mean life; carbon dating; GM tube	very high
25	Recent trends in physics	6	seismic waves and the 2015 Gorkha earthquake; gravitational waves; Higgs boson; nanotechnology	medium

Nothing is safe to skip, because the eleven MCQs are spread across all twenty-five units. But the thirteen "very high" units hold 81 of the 128 hours, and a student who has every standard derivation and numerical of those units ready has covered most of Groups B and C.

4.1.4 The derivations and numerical types that every textbook treats

Area	Derivations to have word-perfect	Numerical types
Mechanics	MI of a uniform rod (ML^{2}/12, ML^{2}/3); \tau = I\alpha; T = 2\pi\sqrt{m/k}; T = 2\pi\sqrt{l/g}; h = 2T\cos\theta/(r\rho g); terminal velocity; Bernoulli's equation	rotational KE and angular momentum; spring–mass period and energy; capillary height; terminal velocity; pipe flow with continuity
Heat	C_p - C_v = R; W = nRT\ln(V_2/V_1); PV^{\gamma} = constant	adiabatic temperature and pressure; Carnot, Otto efficiency; refrigerator COP
Waves and Optics	v = \sqrt{\gamma P/\rho}; harmonics in pipes; Doppler formulae; YDSE fringe width; Brewster's law	pipe and string frequencies with end correction; Doppler; fringe width; grating angle and highest order
Electricity and Magnetism	Wheatstone balance; potentiometer comparison and internal resistance; B of a coil and a solenoid; Hall voltage; \tfrac12LI^{2}; LCR impedance	shunt and series resistance; potentiometer lengths; solenoid field and radius of a charge; peak emf; LCR current, power and resonance
Modern Physics	Thomson's e/m; Millikan's q = mg/E; Einstein's equation; Bohr's radius and energy; N = N_0e^{-\lambda t}; Bragg's law	stopping potential and threshold wavelength; Balmer wavelength; de Broglie wavelength; carbon-dating age; \lambda_{min} of X-rays

Definitions that a five-mark question opens with again and again: moment of inertia and radius of gyration; SHM; surface tension and angle of contact; coefficient of viscosity; internal energy; adiabatic and isothermal processes; the Kelvin–Planck and Clausius statements; coherent sources; Brewster's angle; the ampere; magnetic susceptibility; Lenz's law; self-inductance; power factor; work function and stopping potential; excitation and ionization potential; half-life and mean life.

4.2 Model Question Paper 1

Area	Questions	Marks
Mechanics	1, 2, 12, 13, 20	20
Heat and Thermodynamics	3, 14 (and 20 Or)	6
Waves and Optics	4, 5, 6, 15, 16	13
Electricity and Magnetism	7, 8, 17, 18, 21	20
Modern Physics	9, 10, 11, 19, 22	16

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

1. A flywheel of moment of inertia 2\ \text{kg m}^{2} rotates at 10\ \text{rad s}^{-1}. Its kinetic energy of rotation is [1]a20\ \text{J}b100\ \text{J}c200\ \text{J}d10\ \text{J}
2. The length of a simple pendulum whose period is 2.0\ \text{s} at a place where g = 9.8\ \text{m s}^{-2} is about [1]a0.25\ \text{m}b0.50\ \text{m}c0.99\ \text{m}d1.98\ \text{m}
3. In an isothermal expansion of an ideal gas [1]athe internal energy does not changebno heat is exchangedcno work is donedthe temperature rises
4. The fundamental frequency of a closed pipe 0.85\ \text{m} long (neglect end correction) is [1]a50\ \text{Hz}b100\ \text{Hz}c200\ \text{Hz}d400\ \text{Hz}
5. In Young's double-slit experiment the slit separation is doubled and the slit-to-screen distance is halved. The fringe width becomes [1]aunchangedbone-fourthcfour timesdone-half
6. The Brewster angle for glass of refractive index \sqrt{3} is [1]a30^{\circ}b45^{\circ}c60^{\circ}d90^{\circ}
7. A charged particle moves parallel to a uniform magnetic field. It experiences [1]aa maximum forcebzero forceca force along the fieldda force that makes it move in a circle
8. The rms value of an alternating voltage of peak value 311\ \text{V} is [1]a156\ \text{V}b220\ \text{V}c311\ \text{V}d440\ \text{V}
9. A full-wave rectifier is fed from the 50\ \text{Hz} mains. The frequency of the ripple in its output is [1]a25\ \text{Hz}b50\ \text{Hz}c100\ \text{Hz}d200\ \text{Hz}
10. Photons of energy 3.5\ \text{eV} fall on a metal of work function 2.0\ \text{eV}. The stopping potential is [1]a1.5\ \text{V}b2.0\ \text{V}c3.5\ \text{V}d5.5\ \text{V}
11. The mean life \tau of a radioactive substance is related to its half-life T_{1/2} by [1]a\tau = 0.693\,T_{1/2}b\tau = T_{1/2}c\tau = 1.44\,T_{1/2}d\tau = 2T_{1/2}

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

12. (a) Define moment of inertia and radius of gyration. [2](b) A uniform disc of mass 2.0\ \text{kg} and radius 0.20\ \text{m} rotates about its own axis at 600 revolutions per minute. Find its angular momentum and its kinetic energy of rotation (I = \tfrac12MR^{2}). [3] [5]
13. (a) State Bernoulli's theorem and the equation of continuity. [2](b) Water flows through a horizontal pipe. At A the cross-section is 4.0\times10^{-3}\ \text{m}^{2} and the speed is 2.0\ \text{m s}^{-1}; at B the cross-section is 1.0\times10^{-3}\ \text{m}^{2}. Find the speed at B and the pressure difference between A and B. [3] [5]
14. (a) Derive Mayer's relation C_p - C_v = R for one mole of an ideal gas. [3](b) For a gas C_p = 29.1\ \text{J mol}^{-1}\text{K}^{-1}. Find C_v and \gamma. [2] [5] Or
15. (a) State the Kelvin–Planck and the Clausius statements of the second law of thermodynamics. [2](b) A Carnot engine works between 527^{\circ}\text{C} and 127^{\circ}\text{C} and absorbs 2000\ \text{J} from the source in each cycle. Find its efficiency, the work done per cycle and the heat rejected to the sink. [3] [5]
16. (a) State the laws of transverse vibration of a stretched string. [2](b) A sonometer wire 0.60\ \text{m} long has a mass of 3.0\ \text{g} and is stretched by a tension of 72\ \text{N}. Find the speed of the transverse wave, the fundamental frequency and the frequency of the second overtone. [3] [5]
17. (a) What is the Doppler effect? [1](b) Derive the apparent frequency heard by a stationary observer when the source moves towards him. [2](c) The siren of an ambulance has frequency 680\ \text{Hz}. Find the frequency heard by a stationary observer when the ambulance approaches at 20\ \text{m s}^{-1} and when it moves away at the same speed. [2] [5] Or
18. (a) What is a diffraction grating? Define the grating element. [1](b) Light of wavelength 600\ \text{nm} falls normally on a grating with 5000 lines per cm. Find the angle of the second-order maximum and the highest order that can be seen. [4] [5]
19. (a) State the principle of a potentiometer. [1](b) Two cells balance at 60\ \text{cm} and 80\ \text{cm} of a potentiometer wire. The first has emf 1.5\ \text{V}. Find the emf of the second. [2](c) A cell balances at 75\ \text{cm} on open circuit and at 60\ \text{cm} when it is shunted by an 8\ \Omega resistor. Find its internal resistance. [2] [5]
20. (a) State Faraday's laws of electromagnetic induction and Lenz's law. [2](b) A coil of 100 turns and area 0.020\ \text{m}^{2} rotates at 50 revolutions per second in a uniform field of 0.50\ \text{T}. Find the peak emf and the rms emf. [2](c) In which position of the coil is the emf maximum? [1] [5]
21. (a) State Bohr's postulates of the hydrogen atom. [2](b) Calculate the wavelength of the first line of the Balmer series (R = 1.097\times10^{7}\ \text{m}^{-1}). [2](c) What is the ionization potential of hydrogen in its ground state? [1] [5]

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

20. (a) Show that the vertical oscillation of a mass hung from a light spring is simple harmonic, and derive T = 2\pi\sqrt{m/k}. [4](b) A 0.50\ \text{kg} mass hung from a spring stretches it by 4.9\ \text{cm}. Find the force constant and the period of vertical oscillation. [2](c) If it oscillates with amplitude 3.0\ \text{cm}, find its maximum speed and its total energy. [2] [8] Or
21. (a) Derive PV^{\gamma} = constant for a reversible adiabatic change of an ideal gas. [4](b) Air (\gamma = 1.4) at 27^{\circ}\text{C} and 1.0\times10^{5}\ \text{Pa} is compressed suddenly to \tfrac{1}{32} of its volume. Find the final temperature and pressure. [3](c) Why is an adiabatic curve steeper than an isothermal curve through the same point? [1] [8]
22. (a) State Ampère's circuital law and use it to find the magnetic field inside a long solenoid. [4](b) A solenoid 0.50\ \text{m} long has 1000 turns and carries 2.0\ \text{A}. Find the field inside it. [2](c) An electron enters this field at 2.0\times10^{6}\ \text{m s}^{-1} at right angles to it. Find the radius of its path. [2] [8] Or
23. (a) Draw the phasor diagram of a series LCR circuit and derive its impedance. [3](b) A resistor of 40\ \Omega, an inductor of reactance 80\ \Omega and a capacitor of reactance 50\ \Omega are joined in series to a 220\ \text{V}, 50\ \text{Hz} supply. Find the impedance, the current and the phase angle. [2](c) Find the power factor and the power consumed. [2](d) Find the resonant frequency of the circuit. [1] [8]
24. (a) Write Einstein's photoelectric equation and use it to explain the laws of photoelectric emission. [3](b) Light of wavelength 400\ \text{nm} falls on a metal of work function 2.0\ \text{eV}. Find the maximum kinetic energy of the photoelectrons, the stopping potential and the threshold wavelength. [3](c) How is Planck's constant found from a graph of stopping potential against frequency? [2] [8]

4.3 Solutions to Model Paper 1

Answers — Model Paper 1, Group A

1. (b) — K = \tfrac12I\omega^{2} = \tfrac12(2)(10)^{2} = 100\ \text{J}; option (c) forgets the half.

2. (c) — l = gT^{2}/4\pi^{2} = 9.8\times4/39.48 = 0.99\ \text{m} (the "seconds pendulum").

3. (a) — for an ideal gas U depends only on T, so \Delta U = 0 and Q = W; "no heat exchanged" is adiabatic.

4. (b) — f = v/4L = 340/3.4 = 100\ \text{Hz}; option (c) uses the open-pipe formula v/2L.

5. (b) — \beta = \lambda D/d; D halves and d doubles, so \beta becomes \tfrac12\times\tfrac12 = \tfrac14 of its value.

6. (c) — Brewster's law \tan i_p = n = \sqrt{3}, so i_p = 60^{\circ}.

7. (b) — F = qvB\sin\theta with \theta = 0^{\circ} is zero; the particle goes straight on.

8. (b) — V_{rms} = V_0/\sqrt{2} = 311/1.414 = 220\ \text{V}, the Nepal mains value.

9. (c) — both half-cycles give a pulse, so there are two pulses per input cycle: 2\times50 = 100\ \text{Hz}.

10. (a) — eV_0 = h\nu - \phi = 3.5 - 2.0 = 1.5\ \text{eV}, so V_0 = 1.5\ \text{V}.

11. (c) — \tau = 1/\lambda and T_{1/2} = 0.693/\lambda, so \tau = T_{1/2}/0.693 = 1.44\,T_{1/2}.

Solutions — Model Paper 1, Group B (Q12 to Q15)

12. (a) The moment of inertia of a body about an axis is I = \sum m_ir_i^{2}, the sum of each particle's mass times the square of its distance from the axis; it measures the body's opposition to a change in its rotation (SI unit \text{kg m}^{2}). The radius of gyration K is the distance from the axis at which the whole mass could be put without changing the moment of inertia: I = MK^{2}. (b) I = \tfrac12MR^{2} = \tfrac12(2.0)(0.20)^{2} = 0.040\ \text{kg m}^{2}. \omega = 600\times2\pi/60 = 20\pi = 62.8\ \text{rad s}^{-1}.

L = I\omega = 0.040\times62.8 = 2.51\ \text{kg m}^{2}\,\text{s}^{-1}

K = \tfrac12I\omega^{2} = \tfrac12(0.040)(62.8)^{2} = 79.0\ \text{J}

13. (a) Bernoulli's theorem: for the steady, streamline flow of an ideal (incompressible, non-viscous) fluid, the total energy per unit volume is constant along a streamline: P + \tfrac12\rho v^{2} + \rho gh = constant. Continuity: the volume flowing per second is the same at every section, A_1v_1 = A_2v_2. (b) v_B = A_Av_A/A_B = (4.0\times10^{-3})(2.0)/(1.0\times10^{-3}) = 8.0\ \text{m s}^{-1}. The pipe is horizontal, so

P_A - P_B = \tfrac12\rho(v_B^{2} - v_A^{2}) = \tfrac12(1000)(64 - 4) = 3.0\times10^{4}\ \text{Pa}

The pressure is lower at the narrow section B, where the water is faster.

14. (a) Heat one mole at constant volume through dT: no work is done, so dQ = dU = C_v\,dT. Heat the same mole through the same dT at constant pressure: dQ = C_p\,dT = dU + P\,dV. For an ideal gas U depends only on T, so dU = C_v\,dT in both cases. From PV = RT at constant P, P\,dV = R\,dT. Hence C_p\,dT = C_v\,dT + R\,dT, and C_p - C_v = R. (b) C_v = C_p - R = 29.1 - 8.31 = 20.8\ \text{J mol}^{-1}\text{K}^{-1}; \gamma = C_p/C_v = 29.1/20.79 = 1.40 (a diatomic gas such as air). Or: (a) Kelvin–Planck: no engine working in a cycle can take heat from one reservoir and convert it completely into work. Clausius: heat cannot flow by itself from a colder body to a hotter body; a refrigerator needs work. (b) T_1 = 800\ \text{K}, T_2 = 400\ \text{K}. \eta = 1 - T_2/T_1 = 1 - 400/800 = 0.50 = 50\ \%. Work W = \eta Q_1 = 0.50\times2000 = 1000\ \text{J}; heat rejected Q_2 = Q_1 - W = 1000\ \text{J}.

15. (a) For a stretched string the fundamental frequency f = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}, so: (i) law of length f\propto1/l at fixed T and \mu; (ii) law of tension f\propto\sqrt{T} at fixed l and \mu; (iii) law of mass f\propto1/\sqrt{\mu} at fixed l and T. (b) \mu = 3.0\times10^{-3}/0.60 = 5.0\times10^{-3}\ \text{kg m}^{-1}.

v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{72}{5.0\times10^{-3}}} = \sqrt{14\,400} = 120\ \text{m s}^{-1}

f_1 = v/2l = 120/1.20 = 100\ \text{Hz}. The second overtone is the third harmonic: f_3 = 3f_1 = 300\ \text{Hz}.

Solutions — Model Paper 1, Group B (Q16 to Q19)

16. (a) The Doppler effect is the apparent change in the frequency of a wave when the source, the observer, or both move relative to the medium. (b) A source of frequency f moves towards the observer at v_s. In one period 1/f the wave travels v/f but the source moves v_s/f after it, so the wavelength in front is squeezed to \lambda' = (v - v_s)/f. The observer is at rest, so the waves reach him at speed v:

f' = \frac{v}{\lambda'} = \frac{v}{v - v_s}\,f

(c) Approaching: f' = \dfrac{340}{340 - 20}\times680 = \dfrac{340}{320}\times680 = 722.5\ \text{Hz}. Receding (replace v_s by -v_s): f'' = \dfrac{340}{360}\times680 = 642.2\ \text{Hz}. Or: (a) A diffraction grating is a plate with a large number of equally spaced, parallel, very narrow slits. The grating element d = a + b is the width of one slit plus one opaque space. (b) d = \dfrac{1}{5000}\ \text{cm} = 2.0\times10^{-6}\ \text{m}. For a maximum d\sin\theta = n\lambda:

\sin\theta_2 = \frac{2\times600\times10^{-9}}{2.0\times10^{-6}} = 0.60\;\Rightarrow\;\theta_2 = 36.9^{\circ}

The highest order needs \sin\theta\le1: n\le d/\lambda = 2.0\times10^{-6}/6.0\times10^{-7} = 3.33, so the highest order seen is n = 3.

17. (a) When a steady current flows through a wire of uniform cross-section, the potential difference across any length of it is proportional to that length: V\propto l. (b) \dfrac{E_2}{E_1} = \dfrac{l_2}{l_1}, so E_2 = 1.5\times\dfrac{80}{60} = 2.0\ \text{V}. (c) On open circuit E\propto l_1; when shunted the terminal p.d. V\propto l_2. Since E/V = (R + r)/R:

r = R\left(\frac{l_1}{l_2} - 1\right) = 8\left(\frac{75}{60} - 1\right) = 8\times0.25 = 2.0\ \Omega

18. (a) Faraday's first law: an emf is induced in a circuit whenever the magnetic flux linked with it changes. Second law: the induced emf equals the rate of change of flux linkage, \varepsilon = -N\,d\phi/dt. Lenz's law: the induced current flows in a direction that opposes the change producing it (the minus sign). (b) \omega = 2\pi\times50 = 314\ \text{rad s}^{-1}. \varepsilon_0 = NBA\omega = 100\times0.50\times0.020\times314 = 314\ \text{V}. \varepsilon_{rms} = 314/\sqrt{2} = 222\ \text{V}. (c) \varepsilon = NBA\omega\sin\omega t is maximum when \omega t = 90^{\circ}, that is, when the plane of the coil is parallel to the field: the flux through the coil is then zero but it is changing fastest.

19. (a) (i) The electron moves round the nucleus in certain circular orbits without radiating (stationary orbits). (ii) In an allowed orbit its angular momentum is a whole multiple of h/2\pi: mvr = nh/2\pi. (iii) It emits or absorbs radiation only when it jumps between orbits, with h\nu = E_2 - E_1. (b) The first Balmer line is the jump n = 3\to2:

\frac{1}{\lambda} = R\left(\frac{1}{2^{2}} - \frac{1}{3^{2}}\right) = 1.097\times10^{7}\times\frac{5}{36} = 1.524\times10^{6}\ \text{m}^{-1}

\lambda = 6.56\times10^{-7}\ \text{m} = 656\ \text{nm} (the red H_{\alpha} line). (c) The ground-state energy is -13.6\ \text{eV}, so 13.6\ \text{eV} removes the electron: the ionization potential is 13.6\ \text{V}.

Solutions — Model Paper 1, Group C (Q20 to Q22)

20. (a) A spring of force constant k is stretched by e when a mass m hangs at rest: ke = mg. Pull the mass down a further x. The upward spring force is now k(e + x), so the net (restoring) force is F = mg - k(e + x) = -kx. Then a = -\dfrac{k}{m}x: the acceleration is proportional to the displacement and directed towards the equilibrium position, so the motion is SHM with \omega^{2} = k/m. Hence

T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{e}{g}}

(b) k = mg/e = 0.50\times9.8/0.049 = 100\ \text{N m}^{-1}. T = 2\pi\sqrt{0.50/100} = 2\pi\times0.0707 = 0.444\ \text{s}. (c) \omega = \sqrt{k/m} = \sqrt{200} = 14.1\ \text{rad s}^{-1}. v_{max} = A\omega = 0.030\times14.1 = 0.424\ \text{m s}^{-1}. E = \tfrac12kA^{2} = \tfrac12(100)(0.030)^{2} = 0.045\ \text{J} (check: \tfrac12mv_{max}^{2} = \tfrac12(0.5)(0.424)^{2} = 0.045\ \text{J}). Or: (a) For one mole in an adiabatic change dQ = 0, so the first law gives 0 = C_v\,dT + P\,dV. Differentiating PV = RT: P\,dV + V\,dP = R\,dT, so dT = (P\,dV + V\,dP)/R. Substituting:

C_v(P\,dV + V\,dP) + RP\,dV = 0\;\Rightarrow\;C_vV\,dP + C_pP\,dV = 0

using C_v + R = C_p. Divide by C_vPV: \dfrac{dP}{P} + \gamma\dfrac{dV}{V} = 0. Integrating, \ln P + \gamma\ln V = constant, so PV^{\gamma} = constant. (b) TV^{\gamma-1} = constant: T_2 = T_1(V_1/V_2)^{\gamma-1} = 300\times32^{0.4} = 300\times4 = 1200\ \text{K} (927^{\circ}\text{C}), since 32^{0.4} = 2^{2}. P_2 = P_1(V_1/V_2)^{\gamma} = 1.0\times10^{5}\times32^{1.4} = 1.0\times10^{5}\times128 = 1.28\times10^{7}\ \text{Pa}, since 32^{1.4} = 2^{7}. (c) The slope of an isothermal is dP/dV = -P/V; of an adiabatic it is -\gamma P/V. Because \gamma > 1, the adiabatic is steeper.

21. (a) Ampère's law: the line integral of B round any closed loop equals \mu_0 times the current enclosed, \oint B\,dl = \mu_0I. Take a rectangular loop PQRS of length l along the axis, with PQ inside the solenoid and RS outside. The field inside is uniform and along PQ; outside it is nearly zero; on QR and SP the field is either zero or perpendicular to the path. So \oint B\,dl = Bl. With n turns per metre the loop encloses nl turns, each carrying I:

Bl = \mu_0nlI\;\Rightarrow\;B = \mu_0nI

(b) n = 1000/0.50 = 2000\ \text{m}^{-1}. B = 4\pi\times10^{-7}\times2000\times2.0 = 5.03\times10^{-3}\ \text{T}. (c) The magnetic force gives the centripetal force: evB = mv^{2}/r, so

r = \frac{mv}{eB} = \frac{9.1\times10^{-31}\times2.0\times10^{6}}{1.6\times10^{-19}\times5.03\times10^{-3}} = 2.26\times10^{-3}\ \text{m}

about 2.3\ \text{mm}. Or: (a) The same current I flows through R, L and C, so draw I as the reference phasor. V_R = IR is in phase with I; V_L = IX_L leads I by 90^{\circ}; V_C = IX_C lags I by 90^{\circ}. V_L and V_C are opposite, so their resultant is I(X_L - X_C) at right angles to V_R. By Pythagoras, V^{2} = (IR)^{2} + [I(X_L - X_C)]^{2}, so

Z = \frac{V}{I} = \sqrt{R^{2} + (X_L - X_C)^{2}},\quad\tan\phi = \frac{X_L - X_C}{R}

(b) Z = \sqrt{40^{2} + 30^{2}} = 50\ \Omega. I = 220/50 = 4.4\ \text{A}. \tan\phi = 30/40 = 0.75, so \phi = 36.9^{\circ}; since X_L > X_C the current lags the voltage. (c) Power factor \cos\phi = R/Z = 40/50 = 0.80. P = VI\cos\phi = 220\times4.4\times0.80 = 774\ \text{W} (check: I^{2}R = 4.4^{2}\times40 = 774\ \text{W}). (d) X_L = \omega L and X_C = 1/\omega C, so X_C/X_L = 1/(\omega^{2}LC) = \omega_0^{2}/\omega^{2}. Hence f_0 = f\sqrt{X_C/X_L} = 50\sqrt{50/80} = 39.5\ \text{Hz}.

22. (a) A photon of energy h\nu gives all its energy to one electron; part, the work function \phi, frees the electron and the rest is its maximum kinetic energy: h\nu = \phi + \tfrac12mv_{max}^{2}. This explains the laws: (i) emission needs h\nu\ge\phi, so there is a threshold frequency \nu_0 = \phi/h below which no electrons come out, however bright the light; (ii) the maximum KE rises linearly with frequency and does not depend on intensity; (iii) the number of electrons per second is proportional to the number of photons, that is, to the intensity; (iv) one photon acts on one electron at once, so emission is instantaneous. (b) Photon energy:

E = \frac{hc}{\lambda} = \frac{6.63\times10^{-34}\times3\times10^{8}}{4.0\times10^{-7}} = 4.97\times10^{-19}\ \text{J} = 3.11\ \text{eV}

K_{max} = 3.11 - 2.0 = 1.11\ \text{eV} (1.77\times10^{-19}\ \text{J}); stopping potential V_0 = 1.11\ \text{V}. Threshold wavelength \lambda_0 = hc/\phi = 1.989\times10^{-25}/(3.2\times10^{-19}) = 6.22\times10^{-7}\ \text{m} = 622\ \text{nm}. (c) Since eV_0 = h\nu - \phi, V_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}. Measure V_0 for several frequencies and plot V_0 against \nu: the graph is a straight line of slope h/e, so h = e\times\text{slope}. The intercept on the frequency axis gives \nu_0.

4.4 Model Question Paper 2

Area	Questions	Marks
Mechanics	1, 2, 12, 20	15
Heat and Thermodynamics	3, 14, 15	11
Waves and Optics	4, 5, 6, 7, 13, 16 (and 15 Or, 20 Or)	14
Electricity and Magnetism	8, 9, 17, 18, 21	20
Modern Physics	10, 11, 19, 22	15

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

1. A skater spinning with arms stretched out pulls them in, so that her moment of inertia is halved. Her angular velocity [1]ais halvedbis unchangedcis doubleddbecomes four times
2. The radius of a small steel ball falling through glycerine is doubled. Its terminal velocity becomes [1]atwicebfour timeschalfdeight times
3. The coefficient of performance of an ideal refrigerator working between -3^{\circ}\text{C} and 27^{\circ}\text{C} is [1]a0.1b9c10d1.11
4. The speed of sound in air is 332\ \text{m s}^{-1} at 0^{\circ}\text{C}. At 91^{\circ}\text{C} it is about [1]a332\ \text{m s}^{-1}b383\ \text{m s}^{-1}c443\ \text{m s}^{-1}d664\ \text{m s}^{-1}
5. The intensity level of a sound rises by 20\ \text{dB}. Its intensity has been multiplied by [1]a2b20c100d200
6. According to the wave theory of light, the speed of light in a denser medium is [1]agreater than in airbless than in aircthe same as in airdinfinite
7. Light of wavelength 600\ \text{nm} falls on a single slit 0.20\ \text{mm} wide. The width of the central maximum on a screen 2.0\ \text{m} away is [1]a3\ \text{mm}b6\ \text{mm}c12\ \text{mm}d24\ \text{mm}
8. To convert a galvanometer into an ammeter we connect [1]aa high resistance in seriesba low resistance in parallelca high resistance in parallelda low resistance in series
9. When the hot junction of a thermocouple is at the neutral temperature, the thermo-emf is [1]azerobmaximumcreverseddminimum but not zero
10. The output of a NAND gate is 0 only when [1]aboth inputs are 0bA = 0, B = 1cA = 1, B = 0dboth inputs are 1
11. The de Broglie wavelength of an electron accelerated from rest through 100\ \text{V} is about [1]a0.0123\ \text{nm}b0.123\ \text{nm}c1.23\ \text{nm}d12.3\ \text{nm}

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

12. (a) Derive the relation \tau = I\alpha between torque and angular acceleration for a rigid body. [2](b) A constant torque of 20\ \text{N m} acts for 10\ \text{s} on a wheel of moment of inertia 4.0\ \text{kg m}^{2} that starts from rest. Find the angular acceleration, the angular velocity reached, the angle turned through and the work done by the torque. [3] [5] Or
13. (a) State Stokes' law and derive an expression for the terminal velocity of a sphere falling through a viscous liquid. [3](b) A steel ball of radius 1.0\ \text{mm} and density 8000\ \text{kg m}^{-3} falls through glycerine of density 1250\ \text{kg m}^{-3} and viscosity 1.5\ \text{Pa s}. Find its terminal velocity. [2] [5]
14. (a) Distinguish between progressive and stationary waves. [2](b) A progressive wave is y = 0.02\sin(400\pi t - 2\pi x) in SI units. Find its amplitude, frequency, wavelength and speed. [2](c) Find the phase difference between two points 0.25\ \text{m} apart along the wave. [1] [5]
15. (a) Derive an expression for the work done by n moles of an ideal gas in an isothermal expansion from V_1 to V_2. [3](b) Two moles of an ideal gas at 300\ \text{K} expand isothermally to three times their volume. Find the work done and the heat absorbed. [2] [5]
16. (a) Draw the P–V diagram of the Otto cycle and name its four processes. [3](b) Find the efficiency of an Otto engine of compression ratio 8 (\gamma = 1.4). [2] [5] Or
17. (a) What are coherent sources? Why can two separate lamps not act as coherent sources? [2](b) In Young's experiment the slits are 0.50\ \text{mm} apart and the screen is 1.0\ \text{m} away. The fringe width is 1.2\ \text{mm}. Find the wavelength and the distance of the third bright fringe from the central fringe. [3] [5]
18. (a) What is polarization of light? How does it show that light is a transverse wave? [2](b) State Brewster's law and show that at the polarizing angle the reflected and refracted rays are at right angles. [2](c) Find the polarizing angle for water (n = 1.33). [1] [5]
19. (a) How is a galvanometer converted into a voltmeter? [1](b) A galvanometer of resistance 50\ \Omega gives full-scale deflection for 2.0\ \text{mA}. Find the resistance needed, and how it is connected, to convert it into (i) a voltmeter of range 0–10\ \text{V} and (ii) an ammeter of range 0–1.0\ \text{A}. [4] [5]
20. (a) What is the Hall effect? Derive an expression for the Hall voltage across a strip carrying current in a magnetic field. [3](b) A copper strip 0.50\ \text{mm} thick carries 10\ \text{A} in a field of 1.5\ \text{T} perpendicular to its face. Copper has 8.5\times10^{28} free electrons per \text{m}^{3}. Find the Hall voltage. [2] [5]
21. (a) Derive the law of radioactive decay N = N_0e^{-\lambda t}. [2](b) Define half-life and obtain its relation to the decay constant. [1](c) The ¹⁴C activity of an old piece of wood is 40\ \% of that of fresh wood. Find its age (half-life of ¹⁴C = 5730 years). [2] [5]

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

20. (a) Derive the moment of inertia of a thin uniform rod of mass M and length L about an axis perpendicular to it through (i) its centre and (ii) one end. [4](b) A uniform rod 1.2\ \text{m} long and of mass 0.50\ \text{kg} is pivoted freely at one end and released from the horizontal position. Find its angular velocity and the speed of its free end when it passes through the vertical. [3](c) Find its radius of gyration about the pivot. [1] [8] Or
21. (a) Show that an open pipe gives all the harmonics while a closed pipe gives only the odd harmonics. [4](b) A pipe 30\ \text{cm} long and 4.0\ \text{cm} in diameter is closed at one end. Find its fundamental frequency, allowing for end correction (e = 0.3d). [2](c) Find the fundamental frequency when both ends are opened. [2] [8]
22. (a) Define self-inductance and derive the energy stored in an inductor carrying current I. [3](b) State the principle of a transformer. Why is its core laminated? [2](c) A step-down transformer with 4000 primary turns changes 2200\ \text{V} to 220\ \text{V} and supplies 11\ \text{A} to a load. Its efficiency is 90\ \%. Find the number of secondary turns, the output power and the primary current. [3] [8] Or
23. (a) State Kirchhoff's laws and use them to derive the balance condition of a Wheatstone bridge. [4](b) In a metre bridge an unknown resistance X is in the left gap and 6.0\ \Omega in the right gap. The balance point is 40\ \text{cm} from the left end. Find X. [2](c) X is a wire 1.0\ \text{m} long and 0.40\ \text{mm} in diameter. Find its resistivity. [2] [8]
24. (a) Describe J. J. Thomson's experiment for the specific charge of the electron, and derive e/m = E/(B^{2}r). [5](b) In such an experiment crossed fields E = 3.4\times10^{4}\ \text{V m}^{-1} and B = 2.0\times10^{-3}\ \text{T} leave the beam undeflected. Find the speed of the electrons, and the radius of their path when the electric field is switched off. [3] [8]

4.5 Solutions to Model Paper 2

Answers — Model Paper 2, Group A

1. (c) — no external torque, so I_1\omega_1 = I_2\omega_2; halving I doubles \omega.

2. (b) — v_t = 2r^{2}(\rho - \sigma)g/9\eta is proportional to r^{2}, so it becomes four times.

3. (b) — \beta = T_2/(T_1 - T_2) = 270/30 = 9; option (c) uses T_1 instead of T_2.

4. (b) — v\propto\sqrt{T}: 332\sqrt{364/273} = 332\times1.155 = 383\ \text{m s}^{-1}; option (d) doubles v instead of using the square root of kelvin temperatures.

5. (c) — \Delta L = 10\log(I_2/I_1) = 20, so I_2/I_1 = 10^{2} = 100.

6. (b) — Huygens' construction gives n = c/v with n > 1, so light is slower in glass or water; the corpuscular theory wrongly predicted faster.

7. (c) — the first minimum is at y = \lambda D/a = 6\times10^{-7}\times2/2\times10^{-4} = 6\ \text{mm}, and the central maximum is twice that, 12\ \text{mm}.

8. (b) — a low-resistance shunt in parallel carries most of the current; a series high resistance makes a voltmeter.

9. (b) — at the neutral temperature dE/d\theta = 0, so the emf is maximum; it is zero and reverses at the inversion temperature.

10. (d) — NAND is NOT-AND: Y = \overline{A\cdot B} is 0 only when A = B = 1.

11. (b) — \lambda = h/\sqrt{2meV} = 6.63\times10^{-34}/5.40\times10^{-24} = 1.23\times10^{-10}\ \text{m} = 0.123\ \text{nm}.

Solutions — Model Paper 2, Group B (Q12 to Q15)

12. (a) A particle of mass m_i at distance r_i from the axis has tangential acceleration a_i = r_i\alpha, so the tangential force on it is F_i = m_ir_i\alpha and its torque is F_ir_i = m_ir_i^{2}\alpha. Every particle of a rigid body has the same \alpha, so the total torque is \tau = (\sum m_ir_i^{2})\alpha = I\alpha. (b) \alpha = \tau/I = 20/4.0 = 5.0\ \text{rad s}^{-2}. \omega = \alpha t = 5.0\times10 = 50\ \text{rad s}^{-1}. \theta = \tfrac12\alpha t^{2} = \tfrac12(5.0)(100) = 250\ \text{rad}. W = \tau\theta = 20\times250 = 5000\ \text{J} (check: \tfrac12I\omega^{2} = \tfrac12(4.0)(2500) = 5000\ \text{J}). Or: (a) Stokes' law: the viscous drag on a small sphere of radius r moving slowly at speed v through a fluid of viscosity \eta is F = 6\pi\eta rv. When the sphere (density \rho) reaches terminal velocity, the forces balance: weight = upthrust + drag,

\tfrac43\pi r^{3}\rho g = \tfrac43\pi r^{3}\sigma g + 6\pi\eta rv_t\;\Rightarrow\;v_t = \frac{2r^{2}(\rho - \sigma)g}{9\eta}

(b) v_t = \dfrac{2(1.0\times10^{-3})^{2}(8000 - 1250)(9.8)}{9\times1.5} = \dfrac{0.1323}{13.5} = 9.8\times10^{-3}\ \text{m s}^{-1}, about 9.8\ \text{mm s}^{-1}.

13. (a)

Progressive wave	Stationary wave
the disturbance travels forward and carries energy	the pattern stays in place; no net energy transfer
every particle has the same amplitude	amplitude varies from zero (node) to maximum (antinode)
phase changes continuously along the wave	all particles between two nodes are in the same phase

(b) Compare with y = A\sin(\omega t - kx): A = 0.02\ \text{m}; \omega = 400\pi, so f = \omega/2\pi = 200\ \text{Hz}; k = 2\pi, so \lambda = 2\pi/k = 1.0\ \text{m}; v = f\lambda = 200\ \text{m s}^{-1}. (c) \Delta\phi = k\,\Delta x = 2\pi\times0.25 = \pi/2 rad (90^{\circ}).

14. (a) At constant T the pressure is P = nRT/V. The work done in a small expansion dV is P\,dV, so

W = \int_{V_1}^{V_2}\frac{nRT}{V}\,dV = nRT\ln\frac{V_2}{V_1} = 2.303\,nRT\log\frac{V_2}{V_1}

(b) W = 2\times8.31\times300\times\ln3 = 4986\times1.099 = 5.48\times10^{3}\ \text{J}. In an isothermal change of an ideal gas \Delta U = 0, so the heat absorbed is also Q = W = 5.48\ \text{kJ}.

15. (a) The ideal Otto cycle (petrol engine) on a P–V diagram: (i) adiabatic compression of the fuel–air mixture from V_1 to V_2; (ii) isochoric heating — the spark fires and pressure rises at constant volume, heat Q_1 taken in; (iii) adiabatic expansion — the power stroke; (iv) isochoric cooling — the exhaust valve opens, heat Q_2 given out. (Draw the closed loop of two adiabatic curves joined by two vertical lines, with arrows clockwise.) (b) \eta = 1 - \dfrac{1}{r^{\gamma-1}} = 1 - \dfrac{1}{8^{0.4}} = 1 - \dfrac{1}{2.297} = 1 - 0.435 = 0.565 = 56.5\ \%. Or: (a) Coherent sources emit waves of the same frequency with a constant phase difference. Two lamps emit light from millions of atoms that radiate independently in bursts of about 10^{-8}\ \text{s}, so their phase difference changes at random and no steady fringes form. (b) \lambda = \dfrac{\beta d}{D} = \dfrac{1.2\times10^{-3}\times0.50\times10^{-3}}{1.0} = 6.0\times10^{-7}\ \text{m} = 600\ \text{nm}. The third bright fringe is at y_3 = 3\beta = 3.6\ \text{mm} from the centre.

Solutions — Model Paper 2, Group B (Q16 to Q19)

16. (a) Polarization is the restriction of the vibrations of light to a single plane containing the direction of travel. When light passes through one polaroid and a second polaroid is rotated, the intensity falls to zero twice in each turn. A longitudinal wave would pass equally at every orientation, so light must be transverse. (b) Brewster's law: the tangent of the polarizing angle equals the refractive index, \tan i_p = n. By Snell's law n = \sin i_p/\sin r. Equating, \sin i_p/\cos i_p = \sin i_p/\sin r, so \sin r = \cos i_p = \sin(90^{\circ} - i_p) and i_p + r = 90^{\circ}. The reflected ray leaves at i_p and the refracted ray at r, so the angle between them is 180^{\circ} - (i_p + r) = 90^{\circ}. (c) i_p = \tan^{-1}1.33 = 53.1^{\circ}.

17. (a) A high resistance is connected in series with the galvanometer, so that full-scale current flows when the full-range voltage is applied. (b)(i) R = \dfrac{V}{I_g} - G = \dfrac{10}{2.0\times10^{-3}} - 50 = 5000 - 50 = 4950\ \Omega in series. (ii) The shunt and galvanometer have the same p.d.: I_gG = (I - I_g)S, so

S = \frac{I_gG}{I - I_g} = \frac{2.0\times10^{-3}\times50}{1.0 - 0.002} = 0.100\ \Omega

connected in parallel.

18. (a) When a strip carrying a current is placed in a magnetic field at right angles to it, a small voltage appears across the strip, perpendicular to both the current and the field: this is the Hall effect. Electrons of drift speed v_d feel a force ev_dB and pile up on one edge until the electric field E_H = V_H/w across the width w balances it: eE_H = ev_dB, so V_H = v_dBw. The current is I = neAv_d = new t v_d for a strip of thickness t, so v_d = I/(newt) and

V_H = \frac{BI}{net}

(b) V_H = \dfrac{1.5\times10}{8.5\times10^{28}\times1.6\times10^{-19}\times0.50\times10^{-3}} = \dfrac{15}{6.8\times10^{6}} = 2.2\times10^{-6}\ \text{V}, about 2.2\ \mu\text{V}.

19. (a) The rate of decay is proportional to the number of undecayed nuclei: dN/dt = -\lambda N. Separate and integrate from N_0 at t = 0: \int dN/N = -\lambda\int dt, so \ln(N/N_0) = -\lambda t and N = N_0e^{-\lambda t}. (b) Half-life is the time in which half the nuclei decay. Put N = N_0/2: e^{-\lambda T_{1/2}} = \tfrac12, so T_{1/2} = \ln2/\lambda = 0.693/\lambda. (c) Activity is proportional to N, so N/N_0 = 0.40. \lambda = 0.693/5730\ \text{yr}^{-1} and

t = \frac{1}{\lambda}\ln\frac{N_0}{N} = \frac{5730}{0.693}\ln2.5 = 8268\times0.916 = 7.58\times10^{3}\ \text{years}

Solutions — Model Paper 2, Group C (Q20 to Q22)

20. (a)(i) Take the axis through the centre O, perpendicular to the rod. The mass per unit length is M/L. An element dx at distance x from O has mass (M/L)\,dx and contributes (M/L)x^{2}\,dx:

I_C = \int_{-L/2}^{L/2}\frac{M}{L}x^{2}\,dx = \frac{M}{L}\left[\frac{x^{3}}{3}\right]_{-L/2}^{L/2} = \frac{ML^{2}}{12}

(ii) About one end, x runs from 0 to L: I_E = \dfrac{M}{L}\left[\dfrac{x^{3}}{3}\right]_0^L = \dfrac{ML^{2}}{3}. (b) The centre of mass falls L/2, and that loss of potential energy becomes rotational KE about the end: Mg\dfrac{L}{2} = \dfrac12\cdot\dfrac{ML^{2}}{3}\omega^{2}, so the mass cancels and

\omega = \sqrt{\frac{3g}{L}} = \sqrt{\frac{3\times9.8}{1.2}} = \sqrt{24.5} = 4.95\ \text{rad s}^{-1}

Speed of the free end v = \omega L = 4.95\times1.2 = 5.94\ \text{m s}^{-1}. (c) MK^{2} = ML^{2}/3, so K = L/\sqrt{3} = 1.2/1.732 = 0.69\ \text{m}. Or: (a) Open pipe: both ends are antinodes. The allowed patterns fit L = n\lambda/2 with n = 1, 2, 3, \ldots, so f_n = nv/2L: the frequencies are f_1, 2f_1, 3f_1, \ldots — all harmonics. Closed pipe: the closed end is a node and the open end an antinode, so the length holds an odd number of quarter-wavelengths, L = (2n - 1)\lambda/4, and f = (2n - 1)v/4L: the frequencies are f_1, 3f_1, 5f_1, \ldots — only odd harmonics. (b) e = 0.3\times4.0 = 1.2\ \text{cm}; effective length 0.30 + 0.012 = 0.312\ \text{m}. f = \dfrac{340}{4\times0.312} = 272\ \text{Hz}. (c) Open at both ends, the correction applies twice: 0.30 + 2(0.012) = 0.324\ \text{m}. f = \dfrac{340}{2\times0.324} = 525\ \text{Hz}.

21. (a) Self-inductance L is the flux linkage per unit current, N\phi = LI; equivalently, the emf induced in a coil when its current changes at 1\ \text{A s}^{-1} (\varepsilon = -L\,dI/dt). While the current grows, the source does work against the back emf at the rate P = \varepsilon I = LI\,dI/dt. So in time dt, dW = LI\,dI, and from 0 to I:

W = \int_0^I LI\,dI = \tfrac12LI^{2}

This is stored as energy in the magnetic field of the inductor. (b) A transformer works by mutual induction: an alternating current in the primary produces a changing flux in the core, and this flux induces an alternating emf in the secondary. The core is laminated — thin insulated sheets — to cut down eddy currents and the heat they waste. (c) N_s = N_p\dfrac{V_s}{V_p} = 4000\times\dfrac{220}{2200} = 400 turns. Output power = V_sI_s = 220\times11 = 2420\ \text{W}. Input power = 2420/0.90 = 2689\ \text{W}, so I_p = 2689/2200 = 1.22\ \text{A}. Or: (a) Junction law: the sum of currents entering a junction equals the sum leaving (\sum I = 0; conservation of charge). Loop law: round any closed loop, the sum of emfs equals the sum of the IR drops (\sum E = \sum IR; conservation of energy). In the bridge ABCD with P in AB, Q in BC, R in AD and S in DC, and the galvanometer between B and D, let no current flow through the galvanometer at balance. Then P and Q carry the same current I_1, and R and S the same current I_2. Loop ABDA: I_1P - I_2R = 0. Loop BCDB: I_1Q - I_2S = 0. Dividing: \dfrac{P}{Q} = \dfrac{R}{S}. (b) \dfrac{X}{6.0} = \dfrac{40}{60}, so X = 4.0\ \Omega. (c) A = \pi r^{2} = \pi(0.20\times10^{-3})^{2} = 1.26\times10^{-7}\ \text{m}^{2}. \rho = \dfrac{RA}{l} = \dfrac{4.0\times1.26\times10^{-7}}{1.0} = 5.0\times10^{-7}\ \Omega\,\text{m}.

22. (a) Apparatus: a discharge tube in which electrons from the cathode are accelerated towards a perforated anode and pass as a narrow beam between two parallel plates (electric field E) placed inside a region where a magnetic field B acts at right angles to both E and the beam; the beam strikes a fluorescent screen. Step 1 — find v: the fields are adjusted until the spot returns to the undeflected position. The two forces then balance, eE = evB, so v = E/B. Step 2 — find r: the electric field is switched off. The magnetic force alone bends the beam into a circular arc of radius r (found from the displacement of the spot and the geometry of the tube): evB = mv^{2}/r, so e/m = v/(Br). Result: substituting v = E/B,

\frac{e}{m} = \frac{E}{B^{2}r}

Thomson found the same value whatever gas or cathode metal was used, which showed the electron is a universal particle. The modern value is 1.76\times10^{11}\ \text{C kg}^{-1}. (b) v = E/B = 3.4\times10^{4}/2.0\times10^{-3} = 1.7\times10^{7}\ \text{m s}^{-1}.

r = \frac{mv}{eB} = \frac{9.1\times10^{-31}\times1.7\times10^{7}}{1.6\times10^{-19}\times2.0\times10^{-3}} = 4.8\times10^{-2}\ \text{m}

about 4.8\ \text{cm} (check: v/(Br) = 1.7\times10^{7}/(2.0\times10^{-3}\times0.0483) = 1.76\times10^{11}\ \text{C kg}^{-1}).

4.6 Thirty-nine predicted questions, ranked

How to read this list. Questions are ranked in three tiers. Tier 1 are the standard derivations and numerical types of the "very high" units in 4.1.3, which together hold most of the teaching hours: a paper on this syllabus can hardly avoid several of them. Tier 2 are likely: named syllabus bullets with a clean derivation or a fixed formula, in units of medium weight. Tier 3 are possible: descriptive bullets, short units and the recent-trends unit, worth a pass once Tiers 1 and 2 are secure. Each entry gives the unit, the reason, then a short model answer; the full answers are in Chapter 3 (sections 3.1 to 3.25) and should be learnt in full.

Tier 1 — drill first

1. Series LCR circuit: phasor diagram, impedance, phase angle, resonance and quality factor (unit 19). Why: a six-hour unit whose whole content builds to this one result, and it gives a many-part numerical. Answer: Z = \sqrt{R^{2} + (X_L - X_C)^{2}}, \tan\phi = (X_L - X_C)/R; at resonance X_L = X_C, f_0 = 1/(2\pi\sqrt{LC}), Z = R, current maximum; Q = \dfrac{1}{R}\sqrt{L/C}. For L = 0.5\ \text{H}, C = 20\ \mu\text{F}, R = 10\ \Omega: f_0 = 50.3\ \text{Hz}, Q = 15.8.
2. Moment of inertia of a uniform rod about its centre and about one end; radius of gyration (unit 1). Why: the one MI derivation named in the syllabus. Answer: integrate (M/L)x^{2}\,dx: ML^{2}/12 about the centre, ML^{2}/3 about an end; K = L/\sqrt{12} and L/\sqrt{3}.
3. Conservation of angular momentum, with a numerical (unit 1). Why: a named bullet with everyday examples (skater, diver, a man on a turntable). Answer: with no external torque dL/dt = 0, so I_1\omega_1 = I_2\omega_2; if I halves, \omega doubles and rotational KE L^{2}/2I doubles, the extra energy coming from the work done by the muscles.
4. SHM: show the spring–mass and simple-pendulum motions are SHM, and find the period (unit 2). Why: both are named bullets and both give a five-mark derivation. Answer: spring F = -kx, T = 2\pi\sqrt{m/k}; pendulum F = -mg\sin\theta\approx-(mg/l)x for small \theta, T = 2\pi\sqrt{l/g}.
5. Energy in SHM: KE, PE and total energy, with the graph against displacement (unit 2). Why: a short derivation and a frequent Group B part. Answer: K = \tfrac12m\omega^{2}(A^{2} - x^{2}), U = \tfrac12m\omega^{2}x^{2}, E = \tfrac12m\omega^{2}A^{2} constant; K = U at x = A/\sqrt{2}.
6. Capillary rise: derive h = 2T\cos\theta/(r\rho g) and calculate it (unit 3). Why: the standard derivation of surface tension, with a direct numerical. Answer: the upward pull 2\pi rT\cos\theta balances the weight \pi r^{2}h\rho g. For water in a 0.25\ \text{mm} radius tube (T = 0.072\ \text{N m}^{-1}, \theta = 0): h = 5.9\ \text{cm}.
7. Stokes' law and terminal velocity (unit 3). Why: a named bullet; the derivation is three lines and the numerical is common. Answer: 6\pi\eta rv_t = \tfrac43\pi r^{3}(\rho - \sigma)g, so v_t = 2r^{2}(\rho - \sigma)g/9\eta; uses: Millikan's experiment, raindrops, parachutes.
8. Bernoulli's equation with the equation of continuity; an application (unit 3). Why: a named bullet with a standard derivation from the work–energy theorem. Answer: P + \tfrac12\rho v^{2} + \rho gh = constant; A_1v_1 = A_2v_2; applications: venturimeter, lift on an aircraft wing, the atomiser, a roof blown off in a storm.
9. Mayer's relation C_p - C_v = R, and PV^{\gamma} = constant (unit 4). Why: the two classic derivations of a six-hour unit. Answer: at constant P the extra heat P\,dV = R\,dT does work; for an adiabatic change C_vV\,dP + C_pP\,dV = 0 gives PV^{\gamma} = constant and TV^{\gamma-1} = constant.
10. Isothermal and adiabatic work; a P–V numerical (unit 4). Why: the work integral is named in the syllabus. Answer: isothermal W = nRT\ln(V_2/V_1); adiabatic W = (P_1V_1 - P_2V_2)/(\gamma - 1); work is the area under the P–V curve.
11. Carnot engine and refrigerator: efficiency and COP (unit 5). Why: gives a clean numerical and the second-law statements. Answer: \eta = 1 - T_2/T_1; COP = T_2/(T_1 - T_2); between 27^{\circ}\text{C} and -3^{\circ}\text{C} the COP is 9.
12. Harmonics in open and closed pipes; end correction (unit 8). Why: the most standard derivation of Waves. Answer: open: f_n = nv/2L, all harmonics; closed: f = (2n - 1)v/4L, odd only; e = 0.3d per open end, so a closed pipe uses L + 0.3d and an open pipe L + 0.6d.
13. Transverse waves on a stretched string and the laws of vibration (unit 8). Why: named bullets with a direct sonometer numerical. Answer: v = \sqrt{T/\mu}, f_n = \dfrac{n}{2l}\sqrt{T/\mu}; f\propto1/l, f\propto\sqrt{T}, f\propto1/\sqrt{\mu}.
14. Doppler effect: source moving, observer moving, with a numerical (unit 9). Why: the main numerical of a five-hour unit. Answer: f' = f\dfrac{v + v_o}{v - v_s} (signs positive when each moves towards the other); a 680\ \text{Hz} source approaching at 20\ \text{m s}^{-1} is heard at 722.5\ \text{Hz}.
15. Young's double-slit experiment: derive the fringe width (unit 11). Why: the only derivation of the interference unit. Answer: path difference yd/D; bright for n\lambda, dark for (n + \tfrac12)\lambda; \beta = \lambda D/d.
16. Diffraction grating: the grating equation and the highest order (unit 12). Why: a clean numerical. Answer: d\sin\theta = n\lambda; n_{max} is the whole number below d/\lambda; 5000 lines per cm and 600\ \text{nm} give \theta_2 = 36.9^{\circ} and n_{max} = 3.
17. Potentiometer: comparison of emfs and internal resistance; why it beats a voltmeter (unit 14). Why: a named bullet and a recurring five-mark numerical. Answer: E_1/E_2 = l_1/l_2; r = R(l_1/l_2 - 1); at balance it draws no current from the cell, so it measures the true emf.
18. Conversion of a galvanometer into an ammeter and a voltmeter (unit 14). Why: a named bullet with the cleanest numerical in the unit. Answer: shunt S = I_gG/(I - I_g) in parallel; series R = V/I_g - G.
19. Biot–Savart law: field at the centre of a circular coil and near a long straight wire (unit 16). Why: a nine-hour unit, and the derivations are named in the syllabus. Answer: dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^{2}}; centre of coil B = \mu_0NI/2r; straight wire B = \mu_0I/2\pi d. For N = 50, I = 2.0\ \text{A}, r = 0.10\ \text{m}: B = 6.3\times10^{-4}\ \text{T}.
20. Ampère's law: long solenoid and toroid; force between parallel conductors and the definition of the ampere (unit 16). Why: named bullets, each a short derivation. Answer: solenoid B = \mu_0nI; toroid B = \mu_0NI/2\pi r; F/l = \mu_0I_1I_2/2\pi d; one ampere gives 2\times10^{-7}\ \text{N m}^{-1} between wires 1\ \text{m} apart in vacuum.
21. Force and torque on a rectangular coil; the moving-coil galvanometer (unit 16). Why: a named bullet that ends in the instrument. Answer: \tau = NBIA\sin\theta; with a radial field \tau = NBIA = k\theta, so I = (k/NBA)\theta; current sensitivity \theta/I = NBA/k.
22. Faraday and Lenz; motional emf; the AC generator (unit 18). Why: three named bullets, a derivation and a numerical. Answer: \varepsilon = -N\,d\phi/dt; a rod of length l moving at v across B gives \varepsilon = Blv (0.5\ \text{T}, 0.2\ \text{m}, 10\ \text{m s}^{-1} give 1.0\ \text{V}); a coil gives \varepsilon = NBA\omega\sin\omega t.
23. Photoelectric effect: Einstein's equation, stopping potential, and Planck's constant from a graph (unit 21). Why: the whole three-hour unit in one question. Answer: h\nu = \phi + eV_0; V_0 against \nu is a line of slope h/e and intercept \nu_0; intensity changes the current, not V_0.
24. Bohr's theory: radius and energy of the nth orbit; spectral series (unit 23). Why: the heaviest unit of Modern Physics (8 h). Answer: r_n = \dfrac{\varepsilon_0n^{2}h^{2}}{\pi me^{2}} = 0.529n^{2}\ \text{Å}; E_n = -\dfrac{me^{4}}{8\varepsilon_0^{2}n^{2}h^{2}} = -13.6/n^{2}\ \text{eV}; Lyman ends on n = 1 (UV), Balmer on n = 2 (visible), Paschen on n = 3 (IR).
25. Radioactive decay law, half-life, mean life, and carbon dating (unit 24). Why: named bullets with the clearest numerical in Modern Physics. Answer: N = N_0e^{-\lambda t}; T_{1/2} = 0.693/\lambda; \tau = 1/\lambda = 1.44\,T_{1/2}; a sample at 40\ \% of fresh ¹⁴C activity is about 7.6\times10^{3} years old.
26. Millikan's oil-drop experiment and Thomson's e/m (unit 20). Why: the two experiments of the electrons unit; one or the other is a natural long question. Answer: a balanced drop has qE = mg, so q = mgd/V, and every q is a whole multiple of 1.6\times10^{-19}\ \text{C} (a drop of 1.6\times10^{-15}\ \text{kg} balanced by 4.9\times10^{4}\ \text{V m}^{-1} carries 3.2\times10^{-19}\ \text{C} = 2e); Thomson: v = E/B, e/m = E/(B^{2}r).

Tier 2 — likely

27. Newton's formula for the speed of sound and Laplace's correction; effect of temperature, pressure and humidity (unit 7). Why: three named bullets in one question. Answer: Newton v = \sqrt{P/\rho} = 280\ \text{m s}^{-1} at STP, too low; Laplace: sound is adiabatic, v = \sqrt{\gamma P/\rho} = 331\ \text{m s}^{-1}; v\propto\sqrt{T}; pressure has no effect at constant T; moist air is less dense, so sound is faster.
28. Stationary waves: equation, nodes and antinodes (unit 6). Why: the basis of the pipes and strings questions. Answer: two equal waves in opposite directions give y = 2A\sin kx\cos\omega t; nodes where \sin kx = 0, at x = 0, \lambda/2, \lambda, \ldots; adjacent nodes are \lambda/2 apart.
29. Intensity, loudness, pitch and quality; the decibel (unit 9). Why: named bullets, often a Group B part with a one-line numerical. Answer: intensity is power per unit area; L = 10\log(I/I_0) with I_0 = 10^{-12}\ \text{W m}^{-2}; pitch depends on frequency, quality on the overtones present.
30. Huygens' principle; the laws of reflection and refraction from wave theory (unit 10). Why: the whole of a three-hour unit. Answer: every point on a wavefront is a source of secondary wavelets; the new wavefront is their common tangent; refraction gives \sin i/\sin r = v_1/v_2 = n.
31. Brewster's law and polaroids (unit 13). Why: named bullets with a short proof. Answer: \tan i_p = n and i_p + r = 90^{\circ}; polaroids are used in sunglasses, photography filters and 3-D films.
32. Wheatstone bridge, metre bridge and Kirchhoff's laws in a two-loop circuit (unit 14). Why: named bullets with a standard proof. Answer: P/Q = R/S at balance; metre bridge X = R\,l/(100 - l); in a two-loop circuit write one junction equation and two loop equations, then solve.
33. Self- and mutual inductance; energy stored in an inductor; transformer and its losses (unit 18). Why: named bullets. Answer: \varepsilon = -L\,dI/dt; U = \tfrac12LI^{2}; V_s/V_p = N_s/N_p; losses: copper, eddy current, hysteresis and flux leakage.
34. p–n junction; full-wave rectifier; truth tables of the basic gates (unit 22). Why: a six-hour unit whose items are drawn and tabulated. Answer: depletion layer and barrier potential (about 0.7\ \text{V} for Si); bridge or centre-tap rectifier with output ripple at 2f; NOT, OR, AND, NAND (\overline{AB}) and NOR (\overline{A+B}) tables.
35. de Broglie wavelength and the uncertainty principle; X-rays and Bragg's law (unit 23). Why: the second half of the heaviest Modern Physics unit. Answer: \lambda = h/p = h/\sqrt{2meV}, 0.123\ \text{nm} at 100\ \text{V}; \Delta x\,\Delta p\ge h/4\pi; \lambda_{min} = hc/eV (0.062\ \text{nm} at 20\ \text{kV}); Bragg 2d\sin\theta = n\lambda.

Tier 3 — possible

36. Seebeck and Peltier effects; neutral and inversion temperatures; thermopile (unit 15). Why: a three-hour unit with one definite graph. Answer: E = a\theta + \tfrac12b\theta^{2} is a parabola; E is maximum at the neutral temperature \theta_n = -a/b and zero again at the inversion temperature 2\theta_n (cold junction at 0^{\circ}\text{C}).
37. Hysteresis; dia-, para- and ferromagnetic materials; \mu_r = 1 + \chi (unit 17). Why: a five-hour unit, mostly descriptive. Answer: the B–H loop shows retentivity and coercivity, and its area is the energy lost per cycle per unit volume; soft iron (narrow loop) for transformer cores, steel (wide loop) for permanent magnets; \chi is small negative, small positive and large positive for dia-, para- and ferromagnets.
38. Damped and forced oscillations; resonance; entropy and the second law; resolving power (units 2, 5, 12). Why: short named bullets that usually come as one part of a longer question. Answer: damping makes the amplitude die away exponentially; resonance is maximum amplitude when the driving frequency equals the natural frequency (Barton's pendulums, the Tacoma bridge); \Delta S = Q/T (1\ \text{kg} of ice melting at 273\ \text{K}: 1.23\times10^{3}\ \text{J K}^{-1}); a telescope resolves \theta = 1.22\lambda/D.
39. Recent trends: P, S, Rayleigh and Love waves and the 2015 Gorkha earthquake; gravitational waves; Higgs boson; nanotechnology; GM tube and medical uses of radiation (units 24, 25). Why: a six-hour unit and two descriptive bullets of unit 24, usually asked as short descriptive parts. Answer: P waves are longitudinal and fastest; S waves are transverse and cannot pass through the liquid outer core; Love and Rayleigh waves travel along the surface and do most of the damage — the magnitude 7.8 Gorkha earthquake of 25 April 2015 ruptured eastward from Gorkha district to beneath Kathmandu; gravitational waves are ripples in space-time, first detected by LIGO in 2015; the Higgs boson, found at CERN in 2012, explains why particles have mass; a GM tube counts ionizing particles by the avalanche each one sets off; radiation is used to diagnose and to treat cancer but damages cells and DNA.

Chapter summary

• The grid never changes: eleven one-mark MCQs (Q1 to Q11), eight five-mark questions (Q12 to Q19) and three eight-mark questions (Q20 to Q22), all compulsory, 3 hours for 75 marks — about 2.4 minutes a mark.
• Marks follow teaching hours. Mechanics 22 h (about 13 marks), Heat 12 h (about 7), Waves and Optics 26 h (about 15), Electricity and Magnetism 35 h (about 20), Modern Physics 33 h (about 19).
• Thirteen units are "very high" priority and hold 81 of the 128 hours; their standard derivations (ML^{2}/12, T = 2\pi\sqrt{m/k}, h = 2T\cos\theta/r\rho g, C_p - C_v = R, PV^{\gamma} = constant, f = nv/2L, \beta = \lambda D/d, B = \mu_0nI, Z = \sqrt{R^{2} + (X_L - X_C)^{2}}, h\nu = \phi + eV_0, E_n = -13.6/n^{2}\ \text{eV}, N = N_0e^{-\lambda t}) are the Tier 1 predictions.
• Recurring numerical types: rotational KE and angular momentum; spring period and SHM energy; capillary height and terminal velocity; adiabatic and Carnot; pipe and string frequencies; Doppler; fringe width and grating; potentiometer, shunt and series resistance; solenoid field and the radius of a charge; LCR; stopping potential; Balmer wavelength; carbon-dating age.
• Two complete model papers are solved in full; sit each in three hours, mark yourself by the steps in the solution (law, substitution with units, answer), and then return to Chapters 2 and 3 for the units where you lost marks.

Sources and verification