Physics: Unit-wise Objective Question Bank

Group A of the NEB Grade 12 Physics paper is eleven one-mark multiple-choice questions. Eleven marks is not much, but they are the quickest marks in the paper and, because the eleven items are spread across the whole syllabus, they punish any unit you skipped. This chapter drills all twenty-five CDC units in syllabus order. Each section opens with a high-yield facts table, then gives eight to ten MCQs, then an answer key in which every answer carries a one-line reason and, where a wrong option encodes a real misconception, names it.

2.1 Rotational dynamics

High-yield fact	Value / form
Angular momentum	L = I\omega = mvr; unit \text{kg m}^{2}\text{s}^{-1}
Rotational kinetic energy	K = \tfrac12 I\omega^{2}; also L = \sqrt{2IK}
Torque	\tau = I\alpha = \dfrac{dL}{dt}; work W = \tau\theta; power P = \tau\omega
Radius of gyration	I = Mk^{2}, so k = \sqrt{I/M}
Rod, length l, perpendicular axis	centre: \dfrac{Ml^{2}}{12}; one end: \dfrac{Ml^{2}}{3}
Equations of angular motion	\omega = \omega_0+\alpha t, \theta = \omega_0t+\tfrac12\alpha t^{2}, \omega^{2} = \omega_0^{2}+2\alpha\theta
Conservation	no external torque \Rightarrow L constant \Rightarrow I_1\omega_1 = I_2\omega_2

1. The product of the moment of inertia and the angular velocity of a body is its [1] (NEB 2079 model)atorqueblinear momentumcangular momentumdforce
2. The ratio of the moment of inertia of a uniform rod about an axis through one end to that about its centre (both perpendicular to the rod) is [1] (NEB 2082)a2b3c4d6
3. A flywheel of moment of inertia 2\ \text{kg m}^{2} rotates at 10\ \text{rad s}^{-1}. Its rotational kinetic energy is [1]a20\ \text{J}b40\ \text{J}c100\ \text{J}d200\ \text{J}
4. A skater spinning with arms outstretched pulls her arms in. Her angular velocity increases because [1]aher angular momentum increasesbher moment of inertia decreases while angular momentum is conservedcher kinetic energy is conservedda torque acts on her
5. The radius of gyration of a body of mass M and moment of inertia I is [1] (NEB 2078)a\sqrt{MI}b\sqrt{I/M}cI/Md\sqrt{M/I}
6. A torque of 10\ \text{N m} acts on a wheel of moment of inertia 0.5\ \text{kg m}^{2}. The angular acceleration is [1]a5\ \text{rad s}^{-2}b20\ \text{rad s}^{-2}c0.05\ \text{rad s}^{-2}d2\ \text{rad s}^{-2}
7. If the rotational kinetic energy of a body is E and its moment of inertia is I, its angular momentum is [1] (NEB 2081)a\sqrt{2IE}b2IEc\sqrt{E/2I}dIE/2
8. A motor applies a torque of 20\ \text{N m} to a shaft turning at 50\ \text{rad s}^{-1}. The power delivered is [1]a0.4\ \text{W}b70\ \text{W}c2.5\ \text{W}d1000\ \text{W}
9. A wheel starts from rest with \alpha = 2\ \text{rad s}^{-2}. The angular velocity after turning through 25\ \text{rad} is [1]a5\ \text{rad s}^{-1}b10\ \text{rad s}^{-1}c50\ \text{rad s}^{-1}d100\ \text{rad s}^{-1}

Answers — 2.1

1. (c) — L = I\omega by definition; torque is I\alpha, not I\omega.

2. (c) — \frac{Ml^{2}/3}{Ml^{2}/12} = 4. Students who quote "1/3 versus 1/12" and answer 3 have forgotten to divide.

3. (c) — \tfrac12(2)(10)^{2} = 100\ \text{J}. Option (d) forgets the \tfrac12, option (a) forgets to square \omega.

4. (b) — no external torque, so L = I\omega is fixed; smaller I forces larger \omega. Kinetic energy is not conserved (option c): she does work pulling her arms in, so K rises.

5. (b) — I = Mk^{2}.

6. (b) — \alpha = \tau/I = 10/0.5 = 20. Option (a) multiplies instead of dividing.

7. (a) — E = L^{2}/2I, so L = \sqrt{2IE}.

8. (d) — P = \tau\omega = 20\times50 = 1000\ \text{W}.

9. (b) — \omega^{2} = 2\alpha\theta = 2(2)(25) = 100, so \omega = 10.

2.2 Periodic motion

High-yield fact	Value / form
Defining condition of SHM	a = -\omega^{2}x (directed to the mean position)
Speed	v = \omega\sqrt{A^{2}-x^{2}}; maximum v_{max} = \omega A at the mean position
Energy	E = \tfrac12 kA^{2} = \tfrac12 m\omega^{2}A^{2}; constant
Spring–mass	T = 2\pi\sqrt{m/k}; independent of g
Simple pendulum	T = 2\pi\sqrt{l/g}; l–T^{2} graph is a straight line of slope 4\pi^{2}/g
Spring cut into n equal parts	each part has force constant nk
Resonance	driving frequency = natural frequency; amplitude maximum

1. In simple harmonic motion the acceleration is [1]aconstantbproportional to displacement and directed towards the mean positioncproportional to velocitydzero at the extreme position
2. A mass of 0.25\ \text{kg} hangs from a spring of force constant 100\ \text{N m}^{-1}. Its time period is [1]a0.31\ \text{s}b0.63\ \text{s}c3.14\ \text{s}d6.28\ \text{s}
3. At the mean position of a particle in SHM [1]akinetic energy is minimumbpotential energy is maximumcspeed is maximum and acceleration is zerodspeed and acceleration are both maximum
4. A simple pendulum is taken to the Moon, where g is one-sixth of its value on Earth. Its time period [1]abecomes 6 timesbbecomes \sqrt{6} timescbecomes 1/\sqrt{6} timesdis unchanged
5. A particle of mass 0.5\ \text{kg} is attached to a spring of force constant 200\ \text{N m}^{-1} and oscillates with amplitude 0.1\ \text{m}. The total energy is [1]a0.5\ \text{J}b1\ \text{J}c2\ \text{J}d10\ \text{J}
6. A uniform spring is cut into two equal halves. The force constant and the frequency of each half (for the same mass) are [1] (NEB 2080)ahalved; halvedbdoubled; multiplied by \sqrt{2}cdoubled; doubleddunchanged; unchanged
7. A simple pendulum bob of mass 0.40\ \text{kg} has potential energy 0.044\ \text{J} at an extreme point, taking the mean point as reference. Its kinetic energy at the mean point is [1] (NEB 2079 model)a0.022\ \text{J}b0.044\ \text{J}c0.011\ \text{J}d0.033\ \text{J}
8. The time period of a simple pendulum inside an orbiting satellite is [1]azerobthe same as on Earthcinfinite, because effective g is zeroddoubled
9. A particle of amplitude 0.02\ \text{m} oscillates in SHM with frequency 5\ \text{Hz}. Its maximum speed is [1]a0.10\ \text{m s}^{-1}b0.63\ \text{m s}^{-1}c6.3\ \text{m s}^{-1}d0.20\ \text{m s}^{-1}
10. Resonance occurs when [1]athe driving force is very largebdamping is zerocthe driving frequency equals the natural frequency of the systemdthe amplitude of the driver is maximum

Answers — 2.2

1. (b) — a = -\omega^{2}x. At the extreme position acceleration is maximum, which is why (d) is wrong.

2. (a) — T = 2\pi\sqrt{0.25/100} = 2\pi(0.05) = 0.314\ \text{s}. Option (c) forgets the square root.

3. (c) — at x = 0: v = \omega A (maximum), a = 0, K maximum, U minimum.

4. (b) — T \propto 1/\sqrt{g}, so T increases by \sqrt{6}. Option (a) treats T as proportional to 1/g.

5. (b) — E = \tfrac12kA^{2} = \tfrac12(200)(0.1)^{2} = 1\ \text{J}. The mass does not enter.

6. (b) — a half-length spring is stiffer: k' = 2k, so f' = \frac{1}{2\pi}\sqrt{2k/m} = \sqrt{2}f.

7. (b) — energy is conserved at every point of the swing, so K_{mean} = U_{extreme} = 0.044\ \text{J}; the mass, time period and amplitude are red herrings.

8. (c) — in free fall the effective g is zero, so T = 2\pi\sqrt{l/g}\to\infty: the pendulum does not oscillate. (A spring–mass system would be unaffected, because T does not contain g.)

9. (b) — v_{max} = \omega A = 2\pi(5)(0.02) = 0.63\ \text{m s}^{-1}. Option (d) uses \omega = f.

10. (c) — resonance is a frequency match; amplitude of the driver or absence of damping is not the condition (zero damping would only make the peak infinite).

2.3 Fluid statics

High-yield fact	Value / form
Pressure at depth	P = P_0+\rho gh; gauge pressure \rho gh
Surface energy	E = T\,\Delta A; soap bubble has two surfaces
Capillary rise	h = \dfrac{2T\cos\theta}{r\rho g}; h\propto 1/r, h\propto 1/g
Stokes' law; terminal velocity	F = 6\pi\eta rv; v_t = \dfrac{2r^{2}(\rho-\sigma)g}{9\eta}\propto r^{2}
Poiseuille	\dfrac{V}{t} = \dfrac{\pi Pr^{4}}{8\eta l}
Continuity; Bernoulli	A_1v_1 = A_2v_2; P+\tfrac12\rho v^{2}+\rho gh = constant
Floating body	\dfrac{\text{submerged volume}}{\text{total volume}} = \dfrac{\rho_{body}}{\rho_{liquid}}

1. Two capillary tubes of radii 4\ \text{mm} and 2\ \text{mm} are dipped in the same liquid. The ratio of the heights of rise in them is [1] (NEB 2081)a1:2b2:1c1:4d4:1
2. If the height of capillary rise in a tube on the Earth is h, the height on the Moon (g_{Moon} = g/6) is [1] (NEB 2078)ah/6bhc6hd36h
3. Two identical raindrops, each falling at a terminal velocity of 0.15\ \text{m s}^{-1}, combine into one drop. The new terminal velocity is about [1] (NEB 2082)a0.15\ \text{m s}^{-1}b0.24\ \text{m s}^{-1}c0.30\ \text{m s}^{-1}d0.60\ \text{m s}^{-1}
4. The work done in blowing a soap bubble of radius 2\ \text{cm} (surface tension 0.03\ \text{N m}^{-1}) is about [1]a1.5\times10^{-4}\ \text{J}b3.0\times10^{-4}\ \text{J}c6.0\times10^{-4}\ \text{J}d3.0\times10^{-2}\ \text{J}
5. Water flows at 2\ \text{m s}^{-1} through a pipe of cross-section 200\ \text{cm}^{2} into a section of 50\ \text{cm}^{2}. The speed in the narrow section is [1]a0.5\ \text{m s}^{-1}b2\ \text{m s}^{-1}c4\ \text{m s}^{-1}d8\ \text{m s}^{-1}
6. A stream of air is blown horizontally underneath one pan of a beam balance. That pan [1] (NEB 2078)agoes downbgoes upcstays leveldoscillates about the level
7. If the radius of a horizontal capillary tube is doubled, the rate of flow of a liquid under the same pressure difference becomes [1]a2 timesb4 timesc8 timesd16 times
8. A block floats in water with three-quarters of its volume submerged. Its relative density is [1]a0.25b0.50c0.75d1.33
9. The gauge pressure at a depth of 10\ \text{m} in water (\rho = 1000\ \text{kg m}^{-3}, g = 9.8\ \text{m s}^{-2}) is [1]a9.8\times10^{3}\ \text{Pa}b9.8\times10^{4}\ \text{Pa}c1.98\times10^{5}\ \text{Pa}d1.01\times10^{5}\ \text{Pa}

Answers — 2.3

1. (a) — h\propto 1/r, so the wider tube gives the smaller rise: \dfrac{h_{4}}{h_{2}} = \dfrac{r_{2}}{r_{4}} = \dfrac{2}{4} = 1:2. Option (b) inverts the ratio.

2. (c) — h\propto 1/g; weaker gravity means the column climbs six times higher.

3. (b) — volume doubles so r' = 2^{1/3}r; v_t\propto r^{2} gives v' = 2^{2/3}(0.15) = 0.24\ \text{m s}^{-1}. Option (c) wrongly assumes v_t\propto volume.

4. (b) — a bubble has two surfaces: W = T\times2\times4\pi r^{2} = 0.03\times2\times4\pi(0.02)^{2} = 3.0\times10^{-4}\ \text{J}. Option (a) counts one surface only.

5. (d) — A_1v_1 = A_2v_2 gives v_2 = 200\times2/50 = 8\ \text{m s}^{-1}. Option (a) divides the wrong way.

6. (a) — the air moving under the pan has lower pressure (Bernoulli), so the still air above pushes the pan down. Students who answer (b) picture the moving air "lifting" the pan.

7. (d) — V/t\propto r^{4}, so 2^{4} = 16.

8. (c) — fraction submerged =\rho_{body}/\rho_{water} = 0.75.

9. (b) — \rho gh = 1000\times9.8\times10 = 9.8\times10^{4}\ \text{Pa}; option (d) is the atmospheric pressure, which is not part of the gauge value.

2.4 First Law of Thermodynamics

High-yield fact	Value / form
First law	Q = \Delta U + W; \Delta U of an ideal gas depends on temperature only
Work	W = \int P\,dV = area under the p–V curve; isobaric W = P\Delta V
Isothermal	PV = constant; W = nRT\ln\dfrac{V_2}{V_1}; \Delta U = 0
Adiabatic	Q = 0; PV^{\gamma} = constant; W = \dfrac{nR(T_1-T_2)}{\gamma-1}
Isochoric	W = 0; Q = \Delta U = nC_V\Delta T; P\propto T
Heat capacities	C_P - C_V = R; \gamma = C_P/C_V (\tfrac53 monatomic, \tfrac75 diatomic)
Work for the same expansion, same initial state	isobaric > isothermal > adiabatic > isochoric (=0)

1. The internal energy of an ideal gas depends on [1] (NEB 2079 model)avolume onlybpressure onlyctemperature onlydboth pressure and volume
2. A gas expands from the same initial state through the same volume change by four different processes. The work done by the gas is greatest in the [1] (NEB 2079 model)aisothermal processbisobaric processcadiabatic processdisochoric process
3. A gas in a rigid vessel is heated from 127\ ^{\circ}\text{C} to 227\ ^{\circ}\text{C}. The ratio of initial to final pressure is [1] (NEB 2081)a127:227b4:5c5:4d1:1
4. In an adiabatic process [1] (NEB 2082)atemperature stays constantbno heat enters or leaves the systemcpressure stays constantdvolume stays constant
5. A gas at constant pressure 2\times10^{5}\ \text{Pa} expands from 1\ \text{L} to 3\ \text{L}. The work done by the gas is [1]a200\ \text{J}b400\ \text{J}c600\ \text{J}d4\times10^{5}\ \text{J}
6. A system absorbs 500\ \text{J} of heat and does 200\ \text{J} of work. Its internal energy [1]arises by 700\ \text{J}brises by 300\ \text{J}cfalls by 300\ \text{J}dis unchanged
7. One mole of an ideal gas expands isothermally at 300\ \text{K} to double its volume. The heat absorbed is about (R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}) [1]a1.2\ \text{kJ}b1.7\ \text{kJ}c2.5\ \text{kJ}d3.4\ \text{kJ}
8. A monatomic ideal gas (\gamma = \tfrac53) is compressed adiabatically to one-eighth of its volume. The pressure becomes [1]a8 timesb16 timesc32 timesd64 times
9. Which relation between the molar heat capacities of an ideal gas is correct? [1]aC_V - C_P = RbC_P + C_V = RcC_P - C_V = RdC_P/C_V = R

Answers — 2.4

1. (c) — for an ideal gas there are no intermolecular forces, so U is purely the kinetic energy, which depends on T alone. Option (d) is the classic confusion with a real gas.

2. (b) — between the same two volumes the isobaric path lies highest on the p–V diagram, the isochoric path has zero area. (The item is about expansion from the same state; with different end-points the ranking can change.)

3. (b) — P\propto T at constant volume with T in kelvin: 400:500 = 4:5. Option (a) uses Celsius temperatures.

4. (b) — Q = 0. Temperature changes in an adiabatic process (it is the isothermal one that holds T constant).

5. (b) — W = P\Delta V = 2\times10^{5}\times2\times10^{-3} = 400\ \text{J}. Option (d) forgets to convert litres to cubic metres.

6. (b) — \Delta U = Q - W = 500 - 200 = +300\ \text{J}. Option (a) adds instead of subtracting.

7. (b) — isothermal, so Q = W = RT\ln2 = 8.31\times300\times0.693 = 1.73\ \text{kJ}.

8. (c) — P_2/P_1 = (V_1/V_2)^{\gamma} = 8^{5/3} = 32. Option (a) treats it as isothermal (\gamma = 1).

9. (c) — Mayer's relation. C_P exceeds C_V because at constant pressure the gas also does expansion work.

2.5 Second Law of Thermodynamics

High-yield fact	Value / form
Kelvin–Planck	no engine can convert all the heat it takes from a single reservoir into work
Clausius	heat cannot flow by itself from a colder to a hotter body
Efficiency	\eta = \dfrac{W}{Q_1} = 1-\dfrac{Q_2}{Q_1}; Carnot \eta = 1-\dfrac{T_2}{T_1} (kelvin)
Refrigerator	\text{COP} = \dfrac{Q_2}{W} = \dfrac{Q_2}{Q_1-Q_2} = \dfrac{T_2}{T_1-T_2}
Otto cycle	\eta = 1-\dfrac{1}{r^{\gamma-1}}, r = compression ratio
Diesel vs petrol	diesel has the higher compression ratio (about 15–22 against 6–10)
Entropy	\Delta S = \dfrac{Q}{T} for a reversible step; an isolated system never loses entropy

1. The efficiency of a Carnot engine working between the steam point and the ice point is about [1] (NEB 2082)a10\%b27\%c73\%d100\%
2. A heat engine takes 500\ \text{J} from the source and rejects 400\ \text{J} to the sink. Its efficiency is [1]a80\%b25\%c20\%d10\%
3. The coefficient of performance of an ideal refrigerator working between 273\ \text{K} and 300\ \text{K} is about [1] (NEB 2078)a0.1b1.1c10d27
4. The efficiency of an Otto engine with compression ratio 8 and \gamma = 1.4 is about [1]a43\%b57\%c75\%d88\%
5. A Carnot engine has a sink at 300\ \text{K} and an efficiency of 50\%. The source temperature is [1]a450\ \text{K}b600\ \text{K}c900\ \text{K}d150\ \text{K}
6. The Kelvin–Planck statement of the second law says that [1](a) heat flows from hot to cold(b) no cyclic engine can be 100\% efficient when working from a single reservoir(c) energy is conserved(d) entropy of the universe is constant
7. Which of the following has the higher compression ratio? [1]apetrol enginebdiesel enginecboth the samedsteam engine
8. The entropy of an isolated system undergoing an irreversible process [1]adecreasesbremains constantcincreasesdbecomes zero
9. A refrigerator absorbs 200\ \text{J} from the cold space using 50\ \text{J} of work. The heat rejected to the room is [1]a150\ \text{J}b250\ \text{J}c200\ \text{J}d4\ \text{J}

Answers — 2.5

1. (b) — \eta = 1 - 273/373 = 0.268. Option (c) is T_2/T_1, the rejected fraction.

2. (c) — \eta = 1 - 400/500 = 20\%; option (a) is Q_2/Q_1.

3. (c) — \text{COP} = 273/(300-273) = 10.1. The refrigerator works between a cold space at the ice point and the room; a value above 1 is normal for a refrigerator, unlike an engine.

4. (b) — \eta = 1 - 8^{-0.4} = 1 - 0.435 = 0.565.

5. (b) — 0.5 = 1 - 300/T_1 gives T_1 = 600\ \text{K}.

6. (b) — this is the statement itself; (a) is Clausius's statement in weakened form, (c) is the first law.

7. (b) — a diesel engine compresses air alone, so it can run at a high ratio without pre-ignition; a petrol engine compresses fuel vapour and would knock.

8. (c) — the direction of spontaneous change is the direction in which entropy of the isolated system rises.

9. (b) — energy balance: Q_1 = Q_2 + W = 200 + 50 = 250\ \text{J}. Option (a) subtracts.

2.6 Wave motion

High-yield fact	Value / form
Wave speed	v = f\lambda = \omega/k
Travelling wave along +x	y = A\sin(\omega t - kx), k = 2\pi/\lambda
Phase difference	\Delta\phi = \dfrac{2\pi}{\lambda}\Delta x
Two consecutive particles in the same phase	\lambda apart
Stationary wave	nodes and antinodes each \lambda/2 apart; node–antinode \lambda/4
Superposition	resultant displacement = algebraic sum of displacements

1. The distance between two consecutive particles of a wave that are in the same phase is [1] (NEB 2081)a\lambda/4b\lambda/2c\lambdad2\lambda
2. A wave is described by y = 0.05\sin(40t - 2x) in SI units. Its speed is [1]a2\ \text{m s}^{-1}b20\ \text{m s}^{-1}c40\ \text{m s}^{-1}d80\ \text{m s}^{-1}
3. Two points on a wave of wavelength 2\ \text{m} are 0.5\ \text{m} apart. Their phase difference is [1]a\pi/4b\pi/2c\pid2\pi
4. Which of the following is a transverse wave? [1] (NEB 2082)asound in airba pulse on a stretched stringca compression in a springdsound in water
5. Two pulses of +2\ \text{cm} and +3\ \text{cm} meet exactly over each other on a string. The displacement at that instant is [1]a1\ \text{cm}b2.5\ \text{cm}c5\ \text{cm}d6\ \text{cm}
6. A stationary wave of frequency 100\ \text{Hz} is set up in air where the speed of sound is 340\ \text{m s}^{-1}. Adjacent nodes are separated by [1]a0.85\ \text{m}b1.7\ \text{m}c3.4\ \text{m}d6.8\ \text{m}
7. In a stationary wave all particles between two consecutive nodes [1]ahave the same amplitudebvibrate in the same phase with different amplitudescare at restdvibrate in opposite phase
8. A source of frequency 500\ \text{Hz} produces waves of wavelength 0.68\ \text{m}. The wave speed is [1]a170\ \text{m s}^{-1}b340\ \text{m s}^{-1}c735\ \text{m s}^{-1}d0.00136\ \text{m s}^{-1}

Answers — 2.6

1. (c) — same phase means a phase difference of 2\pi, i.e. one wavelength. Option (b) is the node-to-node spacing in a stationary wave.

2. (b) — v = \omega/k = 40/2 = 20\ \text{m s}^{-1}. Option (c) quotes \omega directly.

3. (b) — \Delta\phi = 2\pi(0.5)/2 = \pi/2.

4. (b) — the string's particles move perpendicular to the direction of travel. Sound in fluids is always longitudinal.

5. (c) — superposition: add the displacements algebraically, 2+3 = 5\ \text{cm}.

6. (b) — \lambda = v/f = 3.4\ \text{m}; nodes are \lambda/2 = 1.7\ \text{m} apart. Option (c) gives the wavelength itself.

7. (b) — that is the defining feature of a stationary wave: a common phase within each loop, with the amplitude varying from zero at the nodes to maximum at the antinode.

8. (b) — v = f\lambda = 500\times0.68 = 340\ \text{m s}^{-1}.

2.7 Mechanical waves

High-yield fact	Value / form
Solid	v = \sqrt{Y/\rho}
Liquid	v = \sqrt{K/\rho}
Gas, Newton (isothermal)	v = \sqrt{P/\rho}, about 280\ \text{m s}^{-1} at 0 °C — too low
Gas, Laplace (adiabatic)	v = \sqrt{\gamma P/\rho}, about 332\ \text{m s}^{-1} at 0 °C
Temperature	v\propto\sqrt{T}; v_t \approx v_0 + 0.61\,t (t in °C)
Pressure	no effect at constant temperature (P/\rho constant)
Humidity	moist air is less dense, so sound is faster

1. A change in atmospheric pressure, at constant temperature, affects the velocity of sound in air [1]aby doubling itbnot at allcin proportion to \sqrt{P}din proportion to P
2. Newton's formula for the speed of sound in air gives a value of about [1]a280\ \text{m s}^{-1}b332\ \text{m s}^{-1}c380\ \text{m s}^{-1}d3\times10^{8}\ \text{m s}^{-1}
3. Laplace's correction replaced the isothermal assumption by [1] (NEB 2080, 2081)athe isobaric assumptionbthe adiabatic assumptioncthe isochoric assumptionda constant-density assumption
4. Taking 332\ \text{m s}^{-1} at 0\ ^{\circ}\text{C} and 0.61\ \text{m s}^{-1} per degree, the speed of sound at 27\ ^{\circ}\text{C} is about [1]a316\ \text{m s}^{-1}b332\ \text{m s}^{-1}c348\ \text{m s}^{-1}d359\ \text{m s}^{-1}
5. The speed of a longitudinal wave in steel (Y = 2\times10^{11}\ \text{N m}^{-2}, \rho = 8000\ \text{kg m}^{-3}) is [1]a500\ \text{m s}^{-1}b1580\ \text{m s}^{-1}c5000\ \text{m s}^{-1}d25000\ \text{m s}^{-1}
6. Which property of sound changes when the temperature of the air changes (source frequency fixed)? [1] (NEB 2079 model)aamplitudebfrequencycwavelengthdintensity
7. If the absolute temperature of a gas is raised from 300\ \text{K} to 1200\ \text{K}, the speed of sound in it [1]adoublesbquadrupleschalvesdis unchanged
8. Sound travels faster in moist air than in dry air because [1]awater vapour raises the pressurebmoist air is less densecmoist air is colderdhumidity raises the frequency
9. The speed of sound in water of bulk modulus 2.1\times10^{9}\ \text{Pa} and density 1000\ \text{kg m}^{-3} is about [1]a340\ \text{m s}^{-1}b1450\ \text{m s}^{-1}c4600\ \text{m s}^{-1}d45800\ \text{m s}^{-1}

Answers — 2.7

1. (b) — v = \sqrt{\gamma P/\rho} and P/\rho is constant for a gas at fixed temperature, so a pressure change cancels out. Option (c) is the trap: P and \rho rise together.

2. (a) — \sqrt{101300/1.293} = 280\ \text{m s}^{-1}, the discrepancy with the measured 332 that Laplace explained.

3. (b) — compressions and rarefactions are too rapid for heat to flow away, so the changes are adiabatic and the factor \gamma appears.

4. (c) — 332 + 0.61\times27 = 348.5\ \text{m s}^{-1}. Option (d) adds 27 degrees of 1\ \text{m s}^{-1} each.

5. (c) — \sqrt{2\times10^{11}/8000} = \sqrt{2.5\times10^{7}} = 5000.

6. (c) — the frequency is fixed by the source; speed changes with T, so \lambda = v/f changes. Amplitude and intensity depend on the source and distance.

7. (a) — v\propto\sqrt{T} and \sqrt{4} = 2. Option (b) treats v as linear in T.

8. (b) — \text{H}_2\text{O} (18) replaces \text{N}_2 (28) and \text{O}_2 (32), lowering the density; pressure is unchanged at fixed atmospheric pressure.

9. (b) — \sqrt{2.1\times10^{9}/1000} = \sqrt{2.1\times10^{6}} = 1449\ \text{m s}^{-1}.

2.8 Wave in pipes and strings

High-yield fact	Value / form
Closed pipe	only odd harmonics; f_n = \dfrac{(2n-1)v}{4L} (fundamental v/4L)
Open pipe	all harmonics; f_n = \dfrac{nv}{2L} (fundamental v/2L)
End correction	e \approx 0.6\,r at an open end
Stretched string	v = \sqrt{T/\mu}; f_1 = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}
Laws of vibration	f\propto 1/L, f\propto\sqrt{T}, f\propto 1/\sqrt{\mu}
Overtone numbering	first overtone = second harmonic (open), third harmonic (closed)

1. A closed organ pipe has a fundamental frequency of 330\ \text{Hz}. The frequency of its next higher mode is [1] (NEB 2082 model)a660\ \text{Hz}b990\ \text{Hz}c1320\ \text{Hz}d495\ \text{Hz}
2. A flute 30\ \text{cm} long, open at both ends, sounds its first harmonic. With v = 330\ \text{m s}^{-1} the frequency is [1] (NEB 2078)a275\ \text{Hz}b550\ \text{Hz}c1100\ \text{Hz}d1650\ \text{Hz}
3. A closed pipe produces [1]aall harmonicsbonly even harmonicsconly odd harmonicsdonly the fundamental
4. The tension in a sonometer wire is increased to four times its value. The fundamental frequency becomes [1]a2 timesb4 timesc8 timesd16 times
5. A string of the same material and tension has its radius doubled. The speed of transverse waves on it becomes [1] (NEB 2081)a2 timesb\tfrac12c4 timesd\tfrac14
6. A wire of length 0.5\ \text{m}, linear density 0.01\ \text{kg m}^{-1} and tension 100\ \text{N} vibrates in its fundamental mode. The frequency is [1]a50\ \text{Hz}b100\ \text{Hz}c200\ \text{Hz}d10^{4}\ \text{Hz}
7. If one end of an open pipe is closed, the fundamental frequency [1]adoublesbhalvescis unchangeddbecomes three times
8. The length of a sonometer wire is halved at constant tension. The frequency [1]ahalvesbdoublescquadruplesdis unchanged
9. A sonometer wire of 75\ \text{cm} resonates with a fork; shortening it by 0.5\ \text{cm} gives 3 beats per second. The fork's frequency is about [1] (NEB 2079 model)a225\ \text{Hz}b447\ \text{Hz}c450\ \text{Hz}d900\ \text{Hz}

Answers — 2.8

1. (b) — a closed pipe's next mode is the third harmonic, 3\times330 = 990\ \text{Hz}; option (a) wrongly supposes a second harmonic exists.

2. (b) — f_1 = v/2L = 330/0.60 = 550\ \text{Hz}. Option (a) applies the closed-pipe formula v/4L with L = 0.30\ \text{m}, which gives 275; that is the trap.

3. (c) — a node at the closed end and an antinode at the open end allow only odd multiples of the fundamental.

4. (a) — f\propto\sqrt{T} and \sqrt{4} = 2.

5. (b) — mass per unit length \mu\propto r^{2}, so \mu quadruples and v = \sqrt{T/\mu} halves.

6. (b) — v = \sqrt{100/0.01} = 100\ \text{m s}^{-1}, f_1 = v/2L = 100/1.0 = 100\ \text{Hz}.

7. (b) — open: v/2L; closed at one end: v/4L, which is half.

8. (b) — f\propto1/L.

9. (b) — the shorter wire has the higher frequency: f_2 - f_1 = 3, f_1 = kL_1^{-1}, so f_1\left(\dfrac{75}{74.5} - 1\right) = 3 and f_1 = 3\times74.5/0.5 = 447\ \text{Hz}.

2.9 Acoustic phenomena

High-yield fact	Value / form
Intensity	I = P/A (inverse square law for a point source); unit \text{W m}^{-2}
Intensity level	\beta = 10\log_{10}(I/I_0) dB, I_0 = 10^{-12}\ \text{W m}^{-2}
Pitch / loudness / quality	frequency / intensity and ear sensitivity / waveform (harmonic content)
Pressure and displacement	pressure amplitude p_{max} = Bks_{max}; pressure is 90^{\circ} out of phase with displacement
Doppler effect	f' = f\,\dfrac{v\pm v_o}{v\mp v_s}; upper signs for approach
Doppler, echo	wall acts first as observer, then as source

1. The property of a musical note that depends on its waveform (number and strength of overtones) is [1] (NEB 2078)apitchbloudnesscquality (timbre)dintensity
2. A sound has an intensity of 10^{-6}\ \text{W m}^{-2}. Its intensity level is [1]a6\ \text{dB}b60\ \text{dB}c600\ \text{dB}d120\ \text{dB}
3. If the distance from a small source increases four times, the intensity becomes [1] (NEB 2079 model)a1/4b1/8c1/16d1/2
4. A source of 500\ \text{Hz} moves at 34\ \text{m s}^{-1} towards a stationary observer (v = 340\ \text{m s}^{-1}). The frequency heard is [1]a450\ \text{Hz}b500\ \text{Hz}c556\ \text{Hz}d550\ \text{Hz}
5. An observer moves at 34\ \text{m s}^{-1} towards a stationary source of 400\ \text{Hz}. The frequency heard is [1]a360\ \text{Hz}b400\ \text{Hz}c440\ \text{Hz}d444\ \text{Hz}
6. Doubling the intensity of a sound raises its level by about [1]a1\ \text{dB}b3\ \text{dB}c6\ \text{dB}d10\ \text{dB}
7. At a displacement node of a stationary sound wave, the pressure variation is [1]azerobmaximumchalf the maximumdequal to atmospheric pressure
8. When a source moves away from a stationary listener, the listener hears [1]aa higher frequency and shorter wavelengthba lower frequencycthe same frequencyda louder sound
9. Pitch of a note is determined by its [1]aamplitudebfrequencycwaveformdspeed

Answers — 2.9

1. (c) — two instruments playing the same note differ in quality because their overtones differ. Pitch is frequency; loudness follows intensity.

2. (b) — 10\log_{10}(10^{-6}/10^{-12}) = 10\times6 = 60\ \text{dB}. Option (a) forgets the factor of 10 in the decibel definition.

3. (c) — I\propto1/r^{2}, so 4^{2} = 16 times smaller. Option (a) uses 1/r.

4. (c) — f' = 500\times340/(340-34) = 556\ \text{Hz}. Option (d) uses f(1 + v_s/v), the observer formula.

5. (c) — f' = 400\times(340+34)/340 = 440\ \text{Hz}. Note that source motion and observer motion give different formulas, even for equal speeds.

6. (b) — 10\log_{10}2 = 3.01\ \text{dB}.

7. (b) — at a displacement node the medium is squeezed and released most, so the pressure change is greatest: pressure and displacement are \pi/2 out of step.

8. (b) — fewer wavefronts reach the ear per second. The wavelength in the medium increases ahead of a retreating source, so (a) is wrong on both counts.

9. (b) — pitch is the sensation of frequency. Waveform decides quality.

2.10 Nature and propagation of light

High-yield fact	Value / form
Huygens' principle	every point on a wavefront is a source of secondary wavelets; the new wavefront is their envelope
Wavefronts	point source: spherical; very distant source: plane; rays \perp wavefronts
Refractive index	n = \dfrac{c}{v} = \dfrac{\lambda_1}{\lambda_2} = \dfrac{\sin i}{\sin r}
On entering a denser medium	speed and wavelength decrease; frequency unchanged
Not explained by wave theory	photoelectric effect (needs photons)

1. The wavefront from a point source in an isotropic medium is [1]aplanebcylindricalcsphericaldparabolic
2. Light of wavelength 600\ \text{nm} in air enters glass of refractive index 1.5. Its wavelength in the glass is [1]a400\ \text{nm}b600\ \text{nm}c900\ \text{nm}d300\ \text{nm}
3. When light passes from air into water, which quantity does not change? [1]aspeedbwavelengthcfrequencyddirection of the wavefront
4. The speed of light in glass of refractive index 1.5 is [1]a2\times10^{8}\ \text{m s}^{-1}b3\times10^{8}\ \text{m s}^{-1}c4.5\times10^{8}\ \text{m s}^{-1}d1.5\times10^{8}\ \text{m s}^{-1}
5. According to Huygens' wave theory, light travels [1]afaster in a denser mediumbslower in a denser mediumcat the same speed in all mediadonly in a vacuum
6. Which of the following phenomena is not explained by Huygens' wave theory? [1] (NEB 2082)areflectionbrefractioncdiffractiondphotoelectric emission
7. A plane wavefront meets a plane mirror obliquely. The law of reflection, derived by Huygens' construction, states that [1]ai = 2rbi = rci + r = 90^{\circ}d\sin i = \mu\sin r
8. Newton's corpuscular theory predicted that the speed of light in water would be [1]aless than in air, as found by Foucaultbgreater than in air, contradicting Foucault's resultcequal to that in airdzero
9. A ray of light is always [1]aparallel to the wavefrontbperpendicular to the wavefrontcat 45^{\circ} to the wavefrontdtangent to the wavefront

Answers — 2.10

1. (c) — a point source radiates equally in all directions; plane wavefronts are the limit at very large distance.

2. (a) — \lambda_g = \lambda_a/n = 600/1.5 = 400\ \text{nm}. Option (c) multiplies instead of dividing.

3. (c) — frequency is fixed by the source. Speed and wavelength both fall by the factor n.

4. (a) — c/n = 3\times10^{8}/1.5.

5. (b) — this is the decisive prediction that Foucault's measurement (light slower in water) confirmed.

6. (d) — Huygens' construction explains reflection, refraction and diffraction as wave behaviour; the photoelectric effect needs the photon picture.

7. (b) — the two right-angled triangles in the construction are congruent.

8. (b) — the corpuscular picture required light to be pulled in at a denser surface and so to speed up, the opposite of the experimental result.

9. (b) — a ray is the direction of energy flow, normal to the wavefront.

2.11 Interference

High-yield fact	Value / form
Coherent sources	constant phase difference and the same frequency
Path difference	bright: n\lambda; dark: (n+\tfrac12)\lambda
Fringe width	\beta = \dfrac{\lambda D}{d}; equally spaced
Position of nth bright fringe	y_n = \dfrac{n\lambda D}{d}
Resultant intensity	I = 4I_0\cos^{2}\tfrac{\phi}{2}: maximum 4I_0, minimum 0
Immersed in liquid of index \mu	\beta' = \beta/\mu

1. In Young's double-slit experiment with \lambda = 546\ \text{nm}, D = 0.80\ \text{m} and d = 0.50\ \text{mm}, the fringe width is [1] (NEB 2079 model, data)a0.44\ \text{mm}b0.87\ \text{mm}c1.4\ \text{mm}d8.7\ \text{mm}
2. Two sources are coherent if they [1]ahave equal amplitudesbhave a constant phase differencecare very close togetherdemit white light
3. The path difference at the third dark fringe is [1]a2\lambdab2.5\lambdac3\lambdad3.5\lambda
4. Two coherent waves, each of intensity I, superpose constructively. The resultant intensity is [1]aIb2Ic4Id\sqrt{2}I
5. The whole Young's double-slit apparatus is immersed in water (\mu = 1.33). The fringe width [1]aincreases by 1.33 timesbdecreases by the factor 1.33cis unchangeddbecomes zero
6. If the slit separation is halved, the fringe width becomes [1]a\tfrac12b2 timesc4 timesdunchanged
7. With white light in a double-slit experiment, the central fringe is [1]adarkbredcwhite, with coloured fringes on both sidesdabsent
8. Slits 0.30\ \text{mm} apart, a screen 1.5\ \text{m} away and a measured fringe width of 2.0\ \text{mm} give a wavelength of [1]a300\ \text{nm}b400\ \text{nm}c500\ \text{nm}d600\ \text{nm}
9. Energy in interference is [1]adestroyed at the dark fringesbcreated at the bright fringescredistributed from dark to bright regionsddoubled everywhere

Answers — 2.11

1. (b) — \beta = \dfrac{546\times10^{-9}\times0.80}{0.50\times10^{-3}} = 8.7\times10^{-4}\ \text{m}. Option (d) is a slip of one power of ten from using d in centimetres.

2. (b) — constant phase difference (not necessarily zero). Equal amplitude only affects contrast.

3. (b) — dark fringes occur at (n - \tfrac12)\lambda counting from n = 1, so the third is at 2.5\lambda.

4. (c) — amplitudes add, 2A, and intensity goes as amplitude squared, 4I. Option (b) adds intensities, which holds only for incoherent sources.

5. (b) — \lambda falls to \lambda/\mu, so \beta' = \beta/1.33.

6. (b) — \beta\propto1/d.

7. (c) — every colour has zero path difference at the centre; beyond it the colours separate because \beta\propto\lambda.

8. (b) — \lambda = \beta d/D = 2\times10^{-3}\times3\times10^{-4}/1.5 = 4\times10^{-7}\ \text{m}.

9. (c) — total energy is conserved: the average over the pattern is 2I, exactly what two independent beams give.

2.12 Diffraction

High-yield fact	Value / form
Single slit, minima	a\sin\theta = n\lambda; central maximum width \dfrac{2\lambda D}{a}
Grating	d\sin\theta = n\lambda, d = \dfrac{1}{N} (lines per unit length)
Highest observable order	n < d/\lambda
Resolving power	Rayleigh: \theta_{min} = \dfrac{1.22\lambda}{D} (circular aperture)
White light on a grating	white central maximum, spectra on each side

1. A grating has 6000 lines per cm. For light of wavelength 5890\ \text{Å}, the second-order maximum appears at about [1] (NEB 2081, data)a30^{\circ}b45^{\circ}c53^{\circ}d60^{\circ}
2. A grating of 5000 lines per cm gives a second-order maximum at 30^{\circ}. The wavelength is [1] (NEB 2080, data)a250\ \text{nm}b500\ \text{nm}c1000\ \text{nm}d5000\ \text{nm}
3. In single-slit diffraction a slit of width 0.2\ \text{mm} is lit by 600\ \text{nm} light. The angle of the first minimum is [1]a3\times10^{-3}\ \text{rad}b3\times10^{-4}\ \text{rad}c0.12\ \text{rad}d1.2\times10^{-3}\ \text{rad}
4. With a grating of spacing 2\times10^{-6}\ \text{m} and light of 5\times10^{-7}\ \text{m}, the highest order that can be seen is [1]a2b3c4d5
5. When red light in a diffraction experiment is replaced by white light, the pattern shows [1] (NEB 2078)aonly a white central maximumba white central maximum with coloured fringes at the sidescno patterndblack and white fringes only
6. The width of the central maximum in single-slit diffraction, if the slit width is doubled, [1]adoublesbhalvescquadruplesdis unchanged
7. Diffraction is most pronounced when the aperture is [1]amuch larger than the wavelengthbcomparable to the wavelengthcmuch smaller than the wavelengthdof zero size
8. The resolving power of a telescope can be increased by [1]adecreasing the diameter of the objectivebincreasing the diameter of the objectivecusing longer-wavelength lightdincreasing the focal length of the eyepiece

Answers — 2.12

1. (b) — d = 1.667\times10^{-6}\ \text{m}, \sin\theta = 2(5.89\times10^{-7})/1.667\times10^{-6} = 0.707, \theta = 45^{\circ}; the third order would need \sin\theta = 1.06, impossible.

2. (b) — d = 2\times10^{-6}\ \text{m}, \lambda = d\sin\theta/n = 2\times10^{-6}\times0.5/2 = 5\times10^{-7}\ \text{m}. Forgetting to divide by n gives 1000\ \text{nm}.

3. (a) — \sin\theta\approx\theta = \lambda/a = 6\times10^{-7}/2\times10^{-4}.

4. (b) — n < d/\lambda = 4, and at n = 4 the ray would leave at exactly 90^{\circ} and cannot be seen. Option (c) includes that unobservable order.

5. (b) — all colours coincide at the centre; elsewhere each colour has a different \theta because \sin\theta\propto\lambda.

6. (b) — width =2\lambda D/a.

7. (b) — bending becomes appreciable when a\sim\lambda; for a\gg\lambda light travels essentially in straight lines.

8. (b) — \theta_{min} = 1.22\lambda/D, so a larger D resolves finer detail.

2.13 Polarization

High-yield fact	Value / form
Polarization	proves light is a transverse wave; sound (longitudinal) cannot be polarized
Malus' law	I = I_0\cos^{2}\theta
Unpolarized light through one polaroid	intensity halves
Brewster's law	\tan\theta_p = \mu; reflected \perp refracted ray at \theta_p
Uses of polaroids	sunglasses (cut glare), 3-D cinema, photographic filters, stress analysis

1. The phenomenon of polarization of light shows that light waves are [1]alongitudinalbtransversecstationarydelectromagnetic only in a vacuum
2. The polarizing angle for glass of refractive index 1.5 is about [1] (NEB 2082 model)a34^{\circ}b42^{\circ}c56^{\circ}d90^{\circ}
3. Unpolarized light of intensity I_0 passes through an ideal polaroid. The transmitted intensity is [1]aI_0bI_0/2cI_0/4dzero
4. Plane-polarized light of intensity I_0 falls on an analyser whose axis makes 60^{\circ} with the plane of polarization. The transmitted intensity is [1]aI_0/2bI_0/4c3I_0/4dzero
5. At the polarizing angle, the angle between the reflected and the refracted rays is [1]a0^{\circ}b45^{\circ}c90^{\circ}d180^{\circ}
6. Which of the following cannot be polarized? [1]alightbradio wavescsound waves in airdX-rays
7. Two polaroids are placed with their axes perpendicular. Unpolarized light passing through both emerges with intensity [1]aI_0/2bI_0/4cI_0/8dzero
8. Light reflected from water at the polarizing angle is polarized with its electric vector [1]ain the plane of incidencebperpendicular to the plane of incidencecalong the direction of traveldat random
9. The polarizing angle of a transparent medium is 57^{\circ}. Its refractive index is about [1]a0.84b1.19c1.54d1.84

Answers — 2.13

1. (b) — only a transverse vibration can be restricted to one plane.

2. (c) — \theta_p = \tan^{-1}(1.5) = 56.3^{\circ}. Option (a) is \tan^{-1}(1/1.5), the angle of refraction, 33.7^{\circ}.

3. (b) — the polaroid passes one of the two equal components of the random polarization.

4. (b) — I_0\cos^{2}60^{\circ} = I_0/4. Option (a) is \cos60^{\circ} without squaring.

5. (c) — Brewster's condition i + r = 90^{\circ} makes the reflected and refracted rays perpendicular.

6. (c) — sound in air is longitudinal, and a longitudinal wave has no transverse direction to restrict.

7. (d) — the first polaroid leaves light polarized along its axis, which is perpendicular to the second's axis; \cos^{2}90^{\circ} = 0.

8. (b) — the component in the plane of incidence is not reflected at \theta_p, leaving only the perpendicular one.

9. (c) — \mu = \tan57^{\circ} = 1.54.

2.14 Electrical circuits

High-yield fact	Value / form
Kirchhoff's laws	junction: \Sigma I = 0 (charge conserved); loop: \Sigma V = 0 (energy conserved)
Wheatstone bridge	balanced when \dfrac{P}{Q} = \dfrac{R}{S}
Meter bridge	X = R\,\dfrac{100-l}{l}
Potentiometer	\dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}; r = R\left(\dfrac{l_1-l_2}{l_2}\right)
Ammeter shunt	S = \dfrac{I_gG}{I-I_g}
Voltmeter series resistor	R = \dfrac{V}{I_g} - G
Maximum power in an a.c. circuit	pure resistance (\cos\phi = 1)

1. Kirchhoff's junction rule is a statement of the conservation of [1]aenergybchargecmomentumdmass
2. A meter bridge balances at 20\ \text{cm}. If the two resistances are doubled and then interchanged, the new balance point is at [1] (NEB 2079 model)a20\ \text{cm}b40\ \text{cm}c60\ \text{cm}d80\ \text{cm}
3. In a potentiometer a standard cell of 1.50\ \text{V} balances at 37.5\ \text{cm} and a test cell at 25.0\ \text{cm}. The e.m.f. of the test cell is [1] (NEB 2080, data)a0.50\ \text{V}b1.00\ \text{V}c2.25\ \text{V}d3.75\ \text{V}
4. A galvanometer of resistance 50\ \Omega gives full-scale deflection at 5\ \text{mA}. The shunt needed to read up to 1\ \text{A} is about [1]a0.05\ \Omegab0.25\ \Omegac5\ \Omegad50\ \Omega
5. To convert a galvanometer of 100\ \Omega and 1\ \text{mA} full-scale into a 10\ \text{V} voltmeter, the series resistance is [1]a100\ \Omegab9.9\ \text{k}\Omegac10\ \text{k}\Omegad10.1\ \text{k}\Omega
6. Kirchhoff's loop rule is a statement of the conservation of [1]achargebenergycmomentumdmass
7. A potentiometer is called a voltmeter of infinite resistance because [1]aits wire has infinite resistancebat balance it draws no current from the cell being measuredcit has a high-resistance galvanometerdit reads the terminal voltage of the driver
8. The wire of a potentiometer is made of manganin or constantan because these alloys have [1]alow resistivity and a high temperature coefficientbhigh resistivity and a very small temperature coefficientchigh melting point onlydmagnetic properties
9. The sensitivity of a potentiometer is increased by [1]areducing the wire lengthbincreasing the currentcusing a longer wire and a smaller potential gradientdusing a thinner driver cell
10. A superconductor differs from a mere perfect conductor in that it [1]ahas zero resistancebexpels magnetic flux from its interior (the Meissner effect)chas a high melting pointdconducts only alternating current
11. A heater coil of 10\ \Omega carries 2\ \text{A} for 30\ \text{s}. The heat produced is [1]a60\ \text{J}b600\ \text{J}c1200\ \text{J}d2400\ \text{J}

Answers — 2.14

1. (b) — charge entering a junction equals charge leaving. The loop rule is the energy statement.

2. (d) — the ratio P/Q = l/(100-l) = 20/80 is unchanged by doubling both; interchanging inverts it: l'/(100-l') = 4, l' = 80\ \text{cm}.

3. (b) — E_2 = E_1 l_2/l_1 = 1.50\times25.0/37.5 = 1.00\ \text{V}. Option (c) inverts the ratio.

4. (b) — S = \dfrac{0.005\times50}{1-0.005} = 0.251\ \Omega.

5. (b) — R = V/I_g - G = 10/10^{-3} - 100 = 9900\ \Omega. Option (d) adds G instead of subtracting.

6. (b) — the sum of the e.m.f.s round a closed loop equals the sum of the IR drops: the net work per unit charge round a loop is zero. The junction rule (option a) is the charge statement.

7. (b) — at the null point no current flows through the cell under test, so its terminal p.d. equals its e.m.f.

8. (b) — high resistivity allows a long wire in a small space, and a tiny temperature coefficient keeps the potential gradient steady as the wire warms.

9. (c) — a smaller potential gradient means a longer balance length per volt, so a given error in length is a smaller error in voltage.

10. (b) — a perfect conductor would merely freeze the flux that was inside it; a superconductor actively pushes the field out below its critical temperature.

11. (c) — H = I^{2}Rt = 4\times10\times30 = 1200\ \text{J}. Option (b) is IRt: the current must be squared.

2.15 Thermoelectric effects

High-yield fact	Value / form
Seebeck effect	an e.m.f. appears in a loop of two different metals when the junctions are at different temperatures
Thermo-e.m.f.	e = \alpha\theta + \tfrac12\beta\theta^{2} (cold junction at 0\ ^{\circ}\text{C})
Neutral temperature	\theta_n = -\alpha/\beta; does not depend on the cold-junction temperature
Inversion temperature	\theta_i = 2\theta_n - \theta_c
Peltier effect	heat absorbed at one junction and evolved at the other when a current flows; reversible
Thermopile	thermocouples in series; detects and measures radiant heat

1. A thermocouple has a neutral temperature of 500\ ^{\circ}\text{C} with the cold junction at 0\ ^{\circ}\text{C}. When the cold junction is raised to 20\ ^{\circ}\text{C}, the inversion temperature changes by [1] (NEB 2079 model)a2\%b3\%c4\%d5\%
2. A thermocouple has an inversion temperature of 600\ ^{\circ}\text{C} and a neutral temperature of 310\ ^{\circ}\text{C}. The cold-junction temperature is [1] (NEB 2081)a0\ ^{\circ}\text{C}b10\ ^{\circ}\text{C}c20\ ^{\circ}\text{C}d30\ ^{\circ}\text{C}
3. A thermocouple has an inversion temperature of 600\ ^{\circ}\text{C} and a neutral temperature of 320\ ^{\circ}\text{C}. The cold-junction temperature is [1] (NEB 2082 model)a20\ ^{\circ}\text{C}b30\ ^{\circ}\text{C}c40\ ^{\circ}\text{C}d60\ ^{\circ}\text{C}
4. If the temperature of the cold junction of a thermocouple is lowered, its neutral temperature [1] (NEB 2078)arisesbfallscis unchangeddbecomes zero
5. For a thermocouple e = 16\theta - 0.04\theta^{2}\ (\mu\text{V}) with the cold junction at 0\ ^{\circ}\text{C}. Its neutral temperature is [1]a100\ ^{\circ}\text{C}b200\ ^{\circ}\text{C}c400\ ^{\circ}\text{C}d800\ ^{\circ}\text{C}
6. The thermo-e.m.f. in the thermocouple of question 5 is zero again at an inversion temperature of [1]a200\ ^{\circ}\text{C}b300\ ^{\circ}\text{C}c400\ ^{\circ}\text{C}d800\ ^{\circ}\text{C}
7. The Peltier effect is used in [1]athermometers of high rangebsmall solid-state refrigeratorsctransformersdrectifiers
8. The Peltier effect differs from Joule heating because it [1]adepends on I^{2}breverses when the current is reversedcoccurs only in insulatorsdnever happens in metals
9. A thermopile is preferred to a single thermocouple for measuring weak radiation because [1]ait has a larger resistance onlybthe e.m.f.s of the junctions add, increasing the sensitivitycit works without a cold junctiondit responds to sound

Answers — 2.15

1. (a) — \theta_i = 2\theta_n - \theta_c: with \theta_c = 0, \theta_i = 1000; with \theta_c = 20, \theta_i = 980, a fall of 2\%. Because \theta_n does not move, only the cold-junction term changes.

2. (c) — \theta_c = 2(310) - 600 = 20\ ^{\circ}\text{C}.

3. (c) — \theta_c = 2(320) - 600 = 40\ ^{\circ}\text{C}. Option (a) reuses the previous item's number.

4. (c) — \theta_n = -\alpha/\beta is a property of the two metals alone; only the inversion temperature shifts.

5. (b) — de/d\theta = 16 - 0.08\theta = 0 gives \theta_n = 200\ ^{\circ}\text{C}.

6. (c) — \theta_i = 2\theta_n - \theta_c = 400 - 0 = 400\ ^{\circ}\text{C}; also e = 0 when 16 = 0.04\theta, i.e. \theta = 400. Option (d) doubles \theta_n a second time.

7. (b) — with the current in one direction one junction cools; such cooling elements are used in portable coolers and to cool detectors.

8. (b) — Joule heating is I^{2}R and always heats; the Peltier heat is proportional to I and switches from heating to cooling at a junction when I reverses.

9. (b) — junctions connected in series add their e.m.f.s.

2.16 Magnetic field

High-yield fact	Value / form
Force on a moving charge	F = qvB\sin\theta; zero if \vec v\parallel\vec B
Force on a wire	F = BIl\sin\theta
Circular coil, centre	B = \dfrac{\mu_0NI}{2r}; straight wire B = \dfrac{\mu_0I}{2\pi d}
Solenoid; toroid	B = \mu_0nI; B = \dfrac{\mu_0NI}{2\pi r} inside, zero outside
Two parallel wires	\dfrac{F}{l} = \dfrac{\mu_0I_1I_2}{2\pi r}; same direction: attract
Torque on a coil	\tau = NIAB\sin\theta; NIA = magnetic moment
Hall voltage	V_H = \dfrac{BI}{net}

1. A charged particle moving parallel to a uniform magnetic field experiences [1]amaximum forcebzero forceca force along the fieldda force that changes its speed
2. A wire 0.5\ \text{m} long carries 2\ \text{A} in a field of 0.4\ \text{T}. The maximum force on it is [1] (NEB 2081, data)a0.1\ \text{N}b0.4\ \text{N}c0.8\ \text{N}d2.0\ \text{N}
3. The magnetic field at the centre of a circular coil of radius R carrying a current I (one turn) is [1] (NEB 2082)a\mu_0I/2\pi Rb\mu_0I/2Rc\mu_0I/Rd2\mu_0I/R
4. A long straight wire carries 10\ \text{A}. The field at 0.10\ \text{m} is [1]a2\times10^{-7}\ \text{T}b2\times10^{-5}\ \text{T}c2\times10^{-3}\ \text{T}d2\times10^{-2}\ \text{T}
5. A solenoid has 1000 turns per metre and carries 2\ \text{A}. The field inside is about [1]a2.5\times10^{-6}\ \text{T}b2.5\times10^{-3}\ \text{T}c2.5\times10^{-1}\ \text{T}d2000\ \text{T}
6. Two long parallel wires each carry 10\ \text{A} in the same direction, 0.10\ \text{m} apart. The force per metre is [1]a2\times10^{-4}\ \text{N m}^{-1}, attractiveb2\times10^{-4}\ \text{N m}^{-1}, repulsivec2\times10^{-5}\ \text{N m}^{-1}, attractivedzero
7. If the separation of the two wires in question 6 is halved, the force per metre [1] (NEB 2082)ahalvesbdoublescquadruplesdis unchanged
8. A coil of 100 turns and area 100\ \text{cm}^{2} carries 1\ \text{A} in a field of 10\ \text{T}. The maximum torque is [1] (NEB 2079 model, data)a1\ \text{N m}b10\ \text{N m}c100\ \text{N m}d10^{5}\ \text{N m}
9. A moving-coil galvanometer uses a radial magnetic field so that [1]athe field is strongerbthe torque on the coil is the same at every deflection, giving a linear scalecthe coil does not heatdthe suspension does not twist
10. For a slice of indium antimonide of thickness 2.5\ \text{mm} carrying 150\ \text{mA} in a field of 0.5\ \text{T}, the Hall voltage is 8.75\ \text{mV}. The carrier density is about [1] (NEB 2081, data)a2.1\times10^{19}\ \text{m}^{-3}b2.1\times10^{22}\ \text{m}^{-3}c2.1\times10^{25}\ \text{m}^{-3}d2.1\times10^{28}\ \text{m}^{-3}
11. For a coil of N turns, area A, carrying current I, the quantity NIA is its [1] (NEB 2078)amagnetic fluxbmagnetic dipole momentctorquedself-inductance

Answers — 2.16

1. (b) — F = qvB\sin0^{\circ} = 0. Magnetic force also never changes a particle's speed (option d), only its direction.

2. (b) — maximum at \theta = 90^{\circ}: F = BIl = 0.4\times2\times0.5 = 0.4\ \text{N}; the minimum (wire along the field) is zero.

3. (b) — B = \mu_0NI/2R with N = 1. Option (a) is the straight-wire formula.

4. (b) — B = \mu_0I/2\pi d = 2\times10^{-7}\times10/0.10 = 2\times10^{-5}\ \text{T}.

5. (b) — \mu_0nI = 4\pi\times10^{-7}\times1000\times2 = 2.5\times10^{-3}\ \text{T}.

6. (a) — \dfrac{2\times10^{-7}\times10\times10}{0.10} = 2\times10^{-4}\ \text{N m}^{-1}; currents in the same direction attract (opposite directions repel).

7. (b) — F\propto1/r.

8. (b) — \tau = NIAB = 100\times1\times0.01\times10 = 10\ \text{N m}. Option (d) does not convert \text{cm}^{2} to \text{m}^{2}.

9. (b) — with a radial field the plane of the coil is always parallel to \vec B, so \tau = NIAB at every angle and deflection is proportional to current.

10. (b) — n = \dfrac{BI}{eV_Ht} = \dfrac{0.5\times0.150}{1.6\times10^{-19}\times8.75\times10^{-3}\times2.5\times10^{-3}} = 2.1\times10^{22}\ \text{m}^{-3}.

11. (b) — m = NIA, the dipole moment; the torque is m B\sin\theta.

2.17 Magnetic properties of materials

High-yield fact	Value / form
Relation	\mu_r = 1+\chi_m; B = \mu_0\mu_rH
Diamagnetic	\chi_m small, negative; \mu_r<1; repelled (bismuth, copper, water)
Paramagnetic	\chi_m small, positive; weakly attracted (aluminium, platinum); \chi_m\propto1/T
Ferromagnetic	\chi_m large, positive; strongly attracted; domains (iron, cobalt, nickel)
Hysteresis	loop area = energy lost per cycle per unit volume
Curie temperature	iron 770\ ^{\circ}\text{C}; above it ferromagnetic becomes paramagnetic

1. The relative permeability \mu_r and susceptibility \chi_m are related by [1] (NEB 2082)a\mu_r = \chi_m - 1b\mu_r = 1 + \chi_mc\mu_r = 1/\chi_md\mu_r = \chi_m
2. Which of the following is a ferromagnetic material? [1] (NEB 2081)acopperbaluminiumcirondbismuth
3. A substance weakly repelled by a magnet is [1]adiamagneticbparamagneticcferromagneticdantiferromagnetic only
4. A material of relative permeability 500 is in a magnetising field of 2000\ \text{A m}^{-1}. The flux density is [1]a0.5\ \text{T}b1.26\ \text{T}c12.6\ \text{T}d2000\ \text{T}
5. The magnetic flux through a square coil of area 10^{-2}\ \text{m}^{2} held perpendicular to a field of 10^{3}\ \text{T} is [1] (NEB 2082, data)a10^{-5}\ \text{Wb}b10\ \text{Wb}c10^{5}\ \text{Wb}d0.1\ \text{Wb}
6. The area of a hysteresis loop represents [1]aretentivitybcoercivitycenergy dissipated per cycle per unit volumedpermeability
7. The best material for the core of a transformer is [1]asteelbsoft ironccopperdalnico
8. The magnetic field lines of a bar magnet [1]abegin at the north pole and end at infinitybform closed loopscnever pass through the magnetdcross each other
9. The temperature above which a ferromagnet becomes paramagnetic is called [1]aneutral temperaturebCurie temperaturecinversion temperaturedBoyle temperature

Answers — 2.17

1. (b) — from B = \mu_0(H+I), dividing by \mu_0H. Option (c) is a meaningless inversion.

2. (c) — iron, cobalt and nickel. Bismuth is diamagnetic, aluminium paramagnetic.

3. (a) — diamagnetic materials have \chi_m<0 and are pushed to the weaker-field region. Paramagnetic materials are weakly attracted.

4. (b) — B = 4\pi\times10^{-7}\times500\times2000 = 1.26\ \text{T}.

5. (b) — \Phi = BA\cos0^{\circ} = 10^{3}\times10^{-2} = 10\ \text{Wb}.

6. (c) — \oint H\,dB per unit volume. Retentivity and coercivity are intercepts of the loop, not its area.

7. (b) — narrow loop, high permeability and low retentivity keep hysteresis loss low; steel (a) and alnico (d) are for permanent magnets.

8. (b) — there are no magnetic monopoles, so lines continue through the magnet from south to north.

9. (b) — thermal agitation destroys the alignment of the domains.

2.18 Electromagnetic Induction

High-yield fact	Value / form
Faraday–Lenz	e = -N\dfrac{d\Phi}{dt}; the minus sign is Lenz's law (energy conservation)
Motional e.m.f.	e = Blv
a.c. generator	e = NBA\omega\sin\omega t; maximum when the coil plane is parallel to \vec B
Self and mutual inductance	e = -L\dfrac{dI}{dt}; e_2 = -M\dfrac{dI_1}{dt}; solenoid L = \dfrac{\mu_0N^{2}A}{l}
Energy in an inductor	U = \tfrac12LI^{2}
Transformer (ideal)	\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}; \eta = \dfrac{P_{out}}{P_{in}}

1. Lenz's law is a consequence of the conservation of [1] (NEB 2078, 2080, 2082)achargebenergycmomentumdmass
2. A rod 20\ \text{cm} long moves at 10\ \text{m s}^{-1} perpendicular to a field of 0.50\ \text{T}. The induced e.m.f. is [1]a0.1\ \text{V}b1\ \text{V}c10\ \text{V}d100\ \text{V}
3. A coil of 200 turns has the flux through it changed by 0.040\ \text{Wb} in 0.20\ \text{s}. The induced e.m.f. is [1]a0.04\ \text{V}b1.6\ \text{V}c40\ \text{V}d8\ \text{V}
4. The energy stored in an inductor of 2.0\ \text{H} carrying 3.0\ \text{A} is [1] (NEB 2081, data)a6\ \text{J}b9\ \text{J}c18\ \text{J}d12\ \text{J}
5. A transformer has 1200 primary turns and 50 secondary turns and is fed at 240\ \text{V}. The secondary voltage is [1]a5\ \text{V}b10\ \text{V}c48\ \text{V}d5760\ \text{V}
6. A step-down transformer reduces 2200\ \text{V} to 220\ \text{V}; the primary has 5000 turns. The secondary turns are [1] (NEB 2080, data)a50b500c5000d50000
7. Two coils have a mutual inductance of 0.20\ \text{H}. The current in one changes by 4.0\ \text{A} in 0.10\ \text{s}. The e.m.f. induced in the other is [1]a0.08\ \text{V}b0.8\ \text{V}c8\ \text{V}d80\ \text{V}
8. Laminated cores are used in transformers to reduce [1] (NEB 2079 model)ahysteresis lossbeddy-current lossccopper lossdflux leakage
9. If the number of turns of a long solenoid is doubled (length and area unchanged), its self-inductance becomes [1]a2 timesb4 timesc\tfrac12dunchanged
10. A transformer cannot work with a steady d.c. supply because [1]ad.c. is too smallbno changing flux is produced, so no e.m.f. is induced in the secondarycthe core saturates at oncedd.c. has no magnetic effect

Answers — 2.18

1. (b) — if the induced current aided the change, energy would be created from nothing. Charge conservation (a) belongs to Kirchhoff's junction rule.

2. (b) — e = Blv = 0.50\times0.20\times10 = 1\ \text{V}. Option (c) forgets to convert centimetres.

3. (c) — e = N\,\Delta\Phi/\Delta t = 200\times0.040/0.20 = 40\ \text{V}. Option (d) drops the turns.

4. (b) — \tfrac12LI^{2} = \tfrac12(2.0)(9.0) = 9\ \text{J}. Option (c) forgets the \tfrac12.

5. (b) — V_s = V_pN_s/N_p = 240\times50/1200 = 10\ \text{V}.

6. (b) — N_s = N_pV_s/V_p = 5000\times220/2200 = 500.

7. (c) — |e| = M\,\Delta I/\Delta t = 0.20\times4.0/0.10 = 8\ \text{V}.

8. (b) — thin insulated sheets break up the paths of eddy currents. Hysteresis loss is reduced by the choice of core material, not by lamination.

9. (b) — L\propto N^{2}.

10. (b) — induction needs a changing flux; a steady current gives a steady flux.

2.19 Alternating Currents

High-yield fact	Value / form
r.m.s. value	V_{rms} = V_0/\sqrt{2} = 0.707V_0
Reactances	X_L = \omega L; X_C = \dfrac{1}{\omega C}
Phase	V leads I by 90^{\circ} in L; I leads V by 90^{\circ} in C
Series LCR	Z = \sqrt{R^{2}+(X_L-X_C)^{2}}; \tan\phi = \dfrac{X_L-X_C}{R}
Resonance	f_0 = \dfrac{1}{2\pi\sqrt{LC}}; Z = R; Q = \dfrac{\omega_0L}{R}
Power	P = V_{rms}I_{rms}\cos\phi; \cos\phi = R/Z

1. The peak voltage of the 220\ \text{V} (r.m.s.) supply in Nepal is about [1] (NEB 2081)a155\ \text{V}b220\ \text{V}c311\ \text{V}d440\ \text{V}
2. The peak value of an alternating voltage is 100\ \text{V}. Its r.m.s. value is [1] (NEB 2082 model)a50\ \text{V}b70.7\ \text{V}c100\ \text{V}d141\ \text{V}
3. The reactance of a 10\ \mu\text{F} capacitor at 50\ \text{Hz} is about [1]a3.2\ \Omegab31.8\ \Omegac318\ \Omegad3180\ \Omega
4. The reactance of a 0.10\ \text{H} inductor at 50\ \text{Hz} is about [1]a0.5\ \Omegab5\ \Omegac31\ \Omegad314\ \Omega
5. A series circuit has R = 30\ \Omega, X_L = 50\ \Omega and X_C = 10\ \Omega. The impedance is [1]a30\ \Omegab50\ \Omegac70\ \Omegad90\ \Omega
6. The power factor of the circuit in question 5 is [1]a0.3b0.5c0.6d0.8
7. The resonant frequency of a series circuit with L = 1.0\ \text{mH} and C = 10\ \mu\text{F} is about [1]a160\ \text{Hz}b1.6\ \text{kHz}c16\ \text{kHz}d10\ \text{kHz}
8. In a pure capacitor the current [1]alags the voltage by 90^{\circ}bleads the voltage by 90^{\circ}cis in phase with the voltagedleads by 45^{\circ}
9. A series LCR circuit has L = 0.10\ \text{H}, R = 10\ \Omega and \omega_0 = 1000\ \text{rad s}^{-1}. Its quality factor is [1]a1b10c100d0.1
10. In which of these circuits is the maximum average power dissipated (same r.m.s. voltage)? [1] (NEB 2079 model)aresistor and inductor in seriesbpure resistorcpure inductordpure capacitor
11. A choke coil controls the current in an a.c. circuit with very little loss of power because [1] (NEB 2079 model)(a) it has large resistance(b) its reactance limits the current while its resistance is small, so the average power consumed is nearly zero(c) it stores charge(d) it converts a.c. to d.c.

Answers — 2.19

1. (c) — V_0 = 220\sqrt{2} = 311\ \text{V}. Option (d) doubles instead of multiplying by \sqrt{2}.

2. (b) — V_0/\sqrt{2} = 70.7. Option (a) divides by 2.

3. (c) — X_C = 1/(2\pi\times50\times10^{-5}) = 318\ \Omega. Option (d) uses \mu = 10^{-7}.

4. (c) — X_L = 2\pi\times50\times0.10 = 31.4\ \Omega.

5. (b) — Z = \sqrt{30^{2}+(50-10)^{2}} = \sqrt{900+1600} = 50\ \Omega. Option (c) adds the resistance and net reactance arithmetically.

6. (c) — \cos\phi = R/Z = 30/50 = 0.6.

7. (b) — f_0 = \dfrac{1}{2\pi\sqrt{10^{-3}\times10^{-5}}} = \dfrac{1}{2\pi\times10^{-4}} = 1.59\ \text{kHz}.

8. (b) — "ICE": in a capacitor I leads V; in an inductor (ELI) V leads I.

9. (b) — Q = \omega_0L/R = 1000\times0.10/10 = 10.

10. (b) — power = V_{rms}I_{rms}\cos\phi and only a resistor has \cos\phi = 1; in a pure L or C the average power is zero.

11. (b) — with \cos\phi\approx0, P = VI\cos\phi\approx0; a resistor of the same impedance would waste I^{2}R as heat.

2.20 Electrons

High-yield fact	Value / form
Charge quantization	q = ne, e = 1.6\times10^{-19}\ \text{C} (Millikan)
Drop at terminal velocity (no field)	mg = 6\pi\eta rv (air buoyancy neglected)
Drop held stationary	qE = mg
Motion of a charge in \vec B	circle, r = \dfrac{mv}{qB}
Velocity selector (crossed fields)	v = E/B; \vec E\perp\vec B
Specific charge of the electron	\dfrac{e}{m} = 1.76\times10^{11}\ \text{C kg}^{-1}
Electron beam between plates	parabolic path (y\propto x^{2})

1. The charge on an oil drop in Millikan's experiment is always [1]aa fraction of eban integral multiple of eca multiple of e/2dcontinuously variable
2. An oil drop of mass 1.3\times10^{-14}\ \text{kg} carries 4 electrons and is held stationary between plates. The field is about [1] (NEB 2079 model, data)a2.0\times10^{5}\ \text{V m}^{-1}b2.0\times10^{3}\ \text{V m}^{-1}c8.0\times10^{5}\ \text{V m}^{-1}d2.0\times10^{-5}\ \text{V m}^{-1}
3. An ion of specific charge 4.40\times10^{7}\ \text{C kg}^{-1} moves at 3.52\times10^{5}\ \text{m s}^{-1} perpendicular to a field of 0.40\ \text{T}. The radius of its path is [1] (NEB 2081, data)a0.02\ \text{m}b0.2\ \text{m}c2\ \text{m}d0.002\ \text{m}
4. In Thomson's method the electric and magnetic fields are made perpendicular so that [1] (NEB 2082)(a) the beam speeds up(b) the electric and magnetic forces on the electron act along the same line and can cancel(c) the beam stops(d) the beam becomes a circle
5. Crossed fields E = 1.0\times10^{4}\ \text{V m}^{-1} and B = 2.0\times10^{-3}\ \text{T} leave a beam undeflected. The electron speed is [1]a2\times10^{-7}\ \text{m s}^{-1}b5\times10^{6}\ \text{m s}^{-1}c5\times10^{-6}\ \text{m s}^{-1}d2\times10^{7}\ \text{m s}^{-1}
6. An electron is accelerated from rest through 100\ \text{V}. Its speed is about [1]a5.9\times10^{4}\ \text{m s}^{-1}b5.9\times10^{6}\ \text{m s}^{-1}c5.9\times10^{8}\ \text{m s}^{-1}d3\times10^{8}\ \text{m s}^{-1}
7. X-rays are used in Millikan's experiment to [1] (NEB 2079 model, 2081, 2082)aaccelerate the dropsbionize the air so that drops pick up chargecheat the oildmagnify the drops
8. The path of an electron beam entering the space between two charged parallel plates perpendicular to the field is [1] (NEB 2080)aa straight lineba circleca parabolada helix
9. The specific charge of an electron is about [1]a1.6\times10^{-19}\ \text{C kg}^{-1}b9.1\times10^{-31}\ \text{C kg}^{-1}c1.76\times10^{11}\ \text{C kg}^{-1}d1.76\times10^{-11}\ \text{C kg}^{-1}

Answers — 2.20

1. (b) — q = ne; this integral pattern is the evidence for quantization of charge.

2. (a) — E = \dfrac{mg}{ne} = \dfrac{1.3\times10^{-14}\times9.8}{4\times1.6\times10^{-19}} = 2.0\times10^{5}\ \text{V m}^{-1}. Option (c) uses one electron's charge.

3. (a) — r = \dfrac{v}{(q/m)B} = \dfrac{3.52\times10^{5}}{4.40\times10^{7}\times0.40} = 0.020\ \text{m}.

4. (b) — a beam moving along x with \vec E along y and \vec B along z feels both forces along y, so they can be balanced; parallel fields could never balance.

5. (b) — v = E/B = 10^{4}/(2\times10^{-3}) = 5\times10^{6}\ \text{m s}^{-1}. Option (d) multiplies.

6. (b) — v = \sqrt{2eV/m} = \sqrt{2\times1.6\times10^{-19}\times100/9.1\times10^{-31}} = 5.9\times10^{6}. Option (c) exceeds the speed of light.

7. (b) — X-rays ionize the air between the plates so the drop can gain or lose electrons, changing n and letting the experimenter watch the charge jump in integer steps.

8. (c) — uniform sideways force with constant forward speed gives y = \dfrac{eE}{2mv^{2}}x^{2}. A circle is the path in a magnetic field.

9. (c) — e/m = 1.6\times10^{-19}/9.1\times10^{-31}.

2.21 Photons

High-yield fact	Value / form
Photon energy	E = hf = hc/\lambda; hc = 1240\ \text{eV nm}
Einstein's equation	hf = \phi + K_{max}; K_{max} = eV_0
Threshold	\phi = hf_0; \lambda_0 = hc/\phi
Graph K_{max} vs f	straight line, slope h, intercept on f-axis f_0, on K-axis -\phi
Intensity	changes the number of photoelectrons (current), not K_{max}
Planck's constant	h = 6.63\times10^{-34}\ \text{J s}

1. The energy of a photon of wavelength 500\ \text{nm} is about [1]a0.25\ \text{eV}b2.5\ \text{eV}c25\ \text{eV}d250\ \text{eV}
2. Ultraviolet light of frequency 1.5\times10^{15}\ \text{Hz} falls on aluminium of work function 6.6\times10^{-19}\ \text{J}. The maximum speed of the photoelectrons is about [1] (NEB 2078, data)a8.6\times10^{3}\ \text{m s}^{-1}b8.6\times10^{5}\ \text{m s}^{-1}c8.6\times10^{7}\ \text{m s}^{-1}d1.7\times10^{6}\ \text{m s}^{-1}
3. Increasing the intensity of the light on a photocathode, at fixed frequency above threshold, [1] (NEB 2078)aincreases K_{max}bincreases the number of photoelectrons emitted per secondcraises the threshold frequencydraises the stopping potential
4. Photoelectrons have a maximum kinetic energy of 2.0\ \text{eV}. The stopping potential is [1]a1\ \text{V}b2\ \text{V}c4\ \text{V}d3.2\times10^{-19}\ \text{V}
5. A metal of work function 2.0\ \text{eV} has a threshold wavelength of [1]a310\ \text{nm}b620\ \text{nm}c1240\ \text{nm}d2480\ \text{nm}
6. The slope of the graph of maximum kinetic energy against frequency is [1] (NEB 2079 model, 2080)athe work functionbPlanck's constantcthe threshold frequencydthe charge e
7. Sodium has a work function of 2.3\ \text{eV}. Light of wavelength 600\ \text{nm} (photon energy about 2.07\ \text{eV}) [1]aejects electrons of 0.23\ \text{eV}bejects no electrons, however intensecejects electrons only if intense enoughdejects electrons after a delay
8. The photoelectric effect supports the view that light consists of [1]acontinuous waves onlybquanta of energyclongitudinal wavesdneutrons
9. If the frequency of the incident light is doubled (above threshold), the maximum kinetic energy of the photoelectrons [1]adoubles exactlybmore than doubleschalvesdis unchanged
10. Planck's constant has the dimensions of [1]aenergybpowercangular momentumdlinear momentum

Answers — 2.21

1. (b) — E = 1240/500 = 2.48\ \text{eV}. Option (c) slips a factor of ten.

2. (b) — K = hf-\phi = 9.94\times10^{-19}-6.6\times10^{-19} = 3.34\times10^{-19}\ \text{J}, so v = \sqrt{2K/m} = 8.6\times10^{5}\ \text{m s}^{-1}.

3. (b) — intensity is the number of photons per second, not the energy per photon. K_{max} depends only on frequency.

4. (b) — eV_0 = K_{max} gives V_0 = 2.0\ \text{V} when K is in eV. Option (d) is the kinetic energy in joules mislabelled as a potential.

5. (b) — \lambda_0 = 1240/2.0 = 620\ \text{nm}.

6. (b) — K = hf - \phi is linear in f with slope h.

7. (b) — a photon below the work function cannot free an electron, and extra intensity does not add photon energy: one photon frees at most one electron.

8. (b) — Einstein's explanation requires energy to be delivered in packets hf.

9. (b) — K' = 2hf - \phi = 2K + \phi > 2K. Option (a) would need \phi = 0.

10. (c) — [ML^{2}T^{-1}], the same as angular momentum; energy is hf, not h.

2.22 Semiconductor devices

High-yield fact	Value / form
p-type / n-type	trivalent dopant (B, Al, In) / pentavalent dopant (P, As, Sb)
Junction	depletion layer; barrier \approx0.7\ \text{V} (Si), 0.3\ \text{V} (Ge)
Forward bias	barrier narrows, large current; reverse bias: small leakage current
Rectifier ripple	half-wave: f; full-wave: 2f
Gates	AND: high only if both high; OR: low only if both low; NAND and NOR: complements
Universal gates	NAND and NOR

1. A p-type semiconductor is obtained by doping silicon with [1]aphosphorusbarseniccborondantimony
2. Under forward bias of a p–n junction [1]athe depletion layer widensbthe depletion layer narrows and a large current flowscno current flowsdonly minority carriers conduct
3. A full-wave rectifier is supplied from 50\ \text{Hz} mains. The frequency of the output ripple is [1]a25\ \text{Hz}b50\ \text{Hz}c100\ \text{Hz}d200\ \text{Hz}
4. A NAND gate gives a low output only when [1]aboth inputs are lowbboth inputs are highcone input is highdinputs differ
5. A NOR gate gives a high output only when [1] (NEB 2079 model)aboth inputs are lowbboth inputs are highcone input is highdinputs differ
6. The characteristic of a diode that makes it useful for rectification is [1] (NEB 2079 model)aits high reverse breakdown voltage onlybit conducts in one direction onlycits capacitancedits negative resistance
7. The number of diodes needed in a bridge rectifier is [1]a1b2c3d4
8. In an intrinsic semiconductor at room temperature [1]aholes are more numerous than electronsbelectrons and holes are equal in numbercthere are no charge carriersdonly holes conduct
9. An OR gate with inputs 1 and 0 gives an output of [1]a0b1cundefinedd-1

Answers — 2.22

1. (c) — boron is trivalent and leaves a hole; the other three are pentavalent donors (n-type).

2. (b) — the applied field opposes the barrier field, narrowing the depletion layer, so majority carriers diffuse across in large numbers.

3. (c) — both half-cycles are used, so there are two pulses per mains cycle. Half-wave would leave 50\ \text{Hz}.

4. (b) — NAND is AND followed by NOT.

5. (a) — NOR is OR followed by NOT: it fires only when neither input is high.

6. (b) — unidirectional conduction is what cuts away (half-wave) or flips (full-wave) the negative half-cycles.

7. (d) — four diodes in a bridge. The full-wave circuit with a centre-tapped transformer needs two.

8. (b) — each thermally broken bond creates one electron and one hole, so their numbers are equal.

9. (b) — OR outputs 1 if any input is 1.

2.23 Quantization of energy

High-yield fact	Value / form
Hydrogen levels	E_n = -\dfrac{13.6}{n^{2}}\ \text{eV}; r_n = 0.529\,n^{2}\ \text{Å}; v_n = \dfrac{2.19\times10^{6}}{n}\ \text{m s}^{-1}
Bohr's quantization	mvr = \dfrac{nh}{2\pi}
Spectral lines	\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^{2}}-\dfrac{1}{n_2^{2}}\right), R = 1.097\times10^{7}\ \text{m}^{-1}
Series	Lyman n_1 = 1 (UV); Balmer n_1 = 2 (visible); Paschen n_1 = 3 (IR)
de Broglie	\lambda = \dfrac{h}{p}; electron through V: \lambda = \dfrac{12.27}{\sqrt{V}}\ \text{Å}
Uncertainty	\Delta x\,\Delta p\ge\dfrac{h}{4\pi}
X-rays	cut-off \lambda_{min} = \dfrac{hc}{eV}; Bragg 2d\sin\theta = n\lambda

1. The energy of the hydrogen atom in the n = 2 level is [1]a-13.6\ \text{eV}b-6.8\ \text{eV}c-3.4\ \text{eV}d-1.51\ \text{eV}
2. The radius of the third Bohr orbit of hydrogen is about [1] (NEB 2078, 2080, 2081)a0.53\ \text{Å}b1.59\ \text{Å}c4.76\ \text{Å}d9.0\ \text{Å}
3. The wavelength of the second line of the Balmer series (n = 4\to2), with R = 1.09\times10^{7}\ \text{m}^{-1}, is about [1] (NEB 2079 model, data)a365\ \text{nm}b489\ \text{nm}c656\ \text{nm}d1875\ \text{nm}
4. The speed of the electron in the first excited state of hydrogen is about [1] (NEB 2081)a2.19\times10^{6}\ \text{m s}^{-1}b1.1\times10^{6}\ \text{m s}^{-1}c7.3\times10^{5}\ \text{m s}^{-1}d4.4\times10^{6}\ \text{m s}^{-1}
5. The de Broglie wavelength of an electron accelerated through 300\ \text{V} is about [1] (NEB 2081)a0.07\ \text{Å}b0.71\ \text{Å}c7.1\ \text{Å}d71\ \text{Å}
6. The ionization energy of the \text{He}^{+} ion (ground state, Z = 2) is [1]a13.6\ \text{eV}b27.2\ \text{eV}c54.4\ \text{eV}d122\ \text{eV}
7. A crystal has an interplanar spacing of 10\ \text{Å}. The longest wavelength that can give a Bragg reflection is [1] (NEB 2080, data)a5\ \text{Å}b10\ \text{Å}c20\ \text{Å}d40\ \text{Å}
8. The shortest X-ray wavelength produced by a tube at 30\ \text{kV} is about [1]a0.41\ \text{Å}b4.1\ \text{Å}c41\ \text{Å}d0.041\ \text{Å}
9. The angular momentum of the electron in the second Bohr orbit is [1]ah/2\pibh/\pic2hdh/4\pi
10. The negative sign of the energy of an electron in a Bohr orbit signifies that [1] (NEB 2079 model)athe electron is moving backwardsbthe electron is bound to the nucleuscthe electron has negative charge onlydthe atom is ionized

Answers — 2.23

1. (c) — -13.6/4 = -3.4\ \text{eV}. Option (b) divides by 2 instead of n^{2}.

2. (c) — r_3 = 0.529\times9 = 4.76\ \text{Å}; option (b) uses n instead of n^{2}.

3. (b) — \dfrac{1}{\lambda} = 1.09\times10^{7}\left(\dfrac{1}{4}-\dfrac{1}{16}\right) = 2.04\times10^{6}, \lambda = 489\ \text{nm}. Option (c) is the first Balmer line, n = 3\to2.

4. (b) — v_n = v_1/n = 2.19\times10^{6}/2 = 1.1\times10^{6}\ \text{m s}^{-1}.

5. (b) — 12.27/\sqrt{300} = 0.708\ \text{Å}.

6. (c) — energy scales as Z^{2}: 13.6\times4 = 54.4\ \text{eV}.

7. (c) — \lambda = 2d\sin\theta/n\leq 2d = 20\ \text{Å}, reached at \theta = 90^{\circ} and n = 1.

8. (a) — \lambda_{min} = 12\,400/30\,000 = 0.413\ \text{Å}, since hc = 12\,400\ \text{eV Å}.

9. (b) — L = nh/2\pi = 2h/2\pi = h/\pi.

10. (b) — zero energy is the electron at rest infinitely far away; a negative total energy means energy must be supplied to free it.

2.24 Radioactivity and nuclear reaction

High-yield fact	Value / form
Decay law	N = N_0e^{-\lambda t}; A = \lambda N
Half-life; mean life	T_{1/2} = \dfrac{0.693}{\lambda}; \tau = \dfrac{1}{\lambda} = 1.44\,T_{1/2}
After n half-lives	fraction left = (\tfrac12)^{n}
Ionizing power	\alpha>\beta>\gamma; penetrating power \gamma>\beta>\alpha
Carbon dating	^{14}\text{C}, T_{1/2}\approx5730 yr; compare activity with living wood
Detector	Geiger–Müller tube: ionization pulse, quenching gas
Units	becquerel (1\ \text{s}^{-1}); curie = 3.7\times10^{10}\ \text{Bq}

1. The percentage of the original radioactive atoms remaining after 4 half-lives is [1] (NEB 2079 model)a1\%b6\%c10\%d20\%
2. A material has a half-life of 50 years. Its mean life is about [1] (NEB 2081, data)a35\ \text{years}b50\ \text{years}c72\ \text{years}d100\ \text{years}
3. The activity of 10^{22} carbon-14 nuclei of half-life 5700 years is about [1] (NEB 2080, data)a3.9\times10^{10}\ \text{Bq}b3.9\times10^{13}\ \text{Bq}c3.9\times10^{6}\ \text{Bq}d1.2\times10^{3}\ \text{Bq}
4. Of \alpha, \beta and \gamma radiation, the most penetrating is [1]a\alphab\betac\gammadall equal
5. A sample of 8\ \text{g} of a radioisotope is reduced to 1\ \text{g} in 30 days. Its half-life is [1]a3.75 daysb7.5 daysc10 daysd15 days
6. A wooden relic shows a ^{14}\text{C} activity one-quarter that of living wood. Taking T_{1/2} = 5730 years, its age is about [1]a2865 yearsb5730 yearsc11\,460 yearsd22\,920 years
7. The decay constant has the unit [1]asb\text{s}^{-1}cBqd\text{J s}^{-1}
8. The Geiger–Müller tube detects radiation through [1]afluorescencebionization of the gas inside itcheating of its walldchemical change
9. A \beta^{-} particle is [1]aa helium nucleusba fast electron emitted from the nucleuscan electromagnetic waveda proton
10. Which radiation is not deflected by an electric or magnetic field? [1]a\alphab\betac\gammadcathode rays

Answers — 2.24

1. (b) — (\tfrac12)^{4} = 1/16 = 6.25\%. Option (d) is four times 5\%, an arithmetic rather than geometric decay.

2. (c) — \tau = T_{1/2}/0.693 = 72\ \text{years}. Option (b) confuses mean life with half-life.

3. (a) — \lambda = 0.693/(5700\times3.156\times10^{7}) = 3.85\times10^{-12}\ \text{s}^{-1}; A = \lambda N = 3.9\times10^{10}\ \text{Bq}. The year must be converted to seconds.

4. (c) — \gamma-rays need centimetres of lead; \alpha is stopped by paper.

5. (c) — 8\to4\to2\to1 is three half-lives, so T_{1/2} = 30/3 = 10 days.

6. (c) — one-quarter is two half-lives: 2\times5730 = 11\,460 years.

7. (b) — dN/dt = -\lambda N, so \lambda is a rate per unit time.

8. (b) — a charged particle or photon ionizes the gas, and the avalanche gives a pulse that is counted.

9. (b) — it originates from a neutron changing into a proton, not from the orbital electrons. Option (a) describes \alpha.

10. (c) — \gamma-rays are uncharged photons.

2.25 Recent trends in physics

High-yield fact	Value / form
Focus (hypocentre)	the point inside the Earth where the rupture begins and seismic waves start
Epicentre	the point on the surface vertically above the focus
P-wave	longitudinal, fastest, travels through solids and liquids
S-wave	transverse, slower, solids only
Surface waves	Rayleigh (elliptical rolling) and Love (horizontal shear); slowest, most destructive
Gorkha earthquake	25 April 2015, magnitude about 7.8
Gravitational waves	ripples of space-time; first detected by LIGO in 2015
Higgs boson	discovered at CERN in 2012; linked to the origin of mass
Nanotechnology	materials and devices of size about 1–100\ \text{nm}

1. The point within the Earth where seismic waves start is called the [1] (NEB 2078, 2082, 2082 model)aepicentrebfocus (hypocentre)cseismographdfault line
2. The point on the Earth's surface directly above the focus is the [1]afocusbepicentrecnodedantinode
3. A P-wave passes from solid rock into a liquid. Its speed [1] (NEB 2081)aincreasesbdecreasescis unchangeddbecomes zero and the wave stops
4. Which seismic wave cannot travel through a liquid? [1]aP-wavebS-wavecsound wavedpressure wave
5. What causes earthquakes? [1] (NEB 2079 model)athe flow of magma onlybthe rubbing and sudden slipping of the Earth's platesctsunamisdthe Moon's pull
6. The most destructive seismic waves near the epicentre are [1]aP-wavesbS-wavescsurface waves (Rayleigh and Love)dsound waves
7. Love waves are [1]asurface waves with horizontal shear motionblongitudinal body wavescelectromagnetic wavesddeep-Earth waves
8. Gravitational waves were first directly detected by [1]athe Hubble telescopebLIGOcthe Large Hadron Colliderda seismograph in Kathmandu
9. The Higgs boson is associated with [1]agravitybthe mass of elementary particlescelectric chargedradioactivity

Answers — 2.25

1. (b) — the focus is underground; the epicentre is its surface projection. Mixing the two is the commonest error.

2. (b) — the epicentre is where the damage is usually greatest.

3. (b) — a liquid has a much smaller rigidity and is more compressible than solid rock, so the wave slows.

4. (b) — a liquid has no rigidity (shear modulus =0), so a shear wave cannot propagate in it; this is how seismologists detected the liquid outer core.

5. (b) — stress builds up along plate boundaries and is released as slip. Magma flow causes some volcanic tremors but not most earthquakes.

6. (c) — they arrive last but carry the most energy and shake the ground horizontally and in rolling motion.

7. (a) — Love waves shear the ground side to side; Rayleigh waves roll it in an ellipse.

8. (b) — the two LIGO detectors in the USA recorded the merger of two black holes on 14 September 2015.

9. (b) — the Higgs field gives elementary particles their mass; its quantum is the Higgs boson.

Do not learn the answer letters

Reorder the options and every letter in this chapter changes. What you are meant to carry away from a section is the fact in the high-yield table and the reason in the answer key, not "2.14 question 2 is (d)". If you find yourself recognising a question by its position on the page, shuffle your revision: work the sections in reverse order, or cover the options and answer before you look.

Chapter summary

• Group A of the NEB Grade 12 Physics paper is eleven one-mark items drawn from across all twenty-five units; no unit is safe to skip, which is why this chapter gives every unit its own facts table, its own MCQs and its own reasoned key.
• Recent papers repeat a small core of Group A facts: the angular momentum product I\omega; energy conservation in SHM; capillary rise and raindrop terminal velocity (v_t\propto r^{2}); the work ranking of thermodynamic processes; Carnot efficiency in kelvin; node–antinode spacing; Laplace's adiabatic correction; thermocouple \theta_i = 2\theta_n-\theta_c; Lenz's law and energy; the peak of the 220\ \text{V} supply (311\ \text{V}); the focus of an earthquake.
• Most lost marks are unit slips, not physics slips: \text{cm}^{2} left unconverted, Celsius used in gas laws, kelvin under a square root, \text{eV} and joule mixed, the \tfrac12 omitted in \tfrac12LI^{2} and \tfrac12I\omega^{2}.
• Formula pairs that are confused: closed-pipe v/4L against open-pipe v/2L; observer-moving against source-moving Doppler; Malus' \cos^{2}\theta against \cos\theta; B at the centre of a coil (\mu_0NI/2r) against a straight wire (\mu_0I/2\pi d); I^{2}Rt against IRt.
• Graph items are now common: read the slope of l–T^{2} (4\pi^{2}/g), of K_{max}–f (h), the area of a p–V loop (work) and of a hysteresis loop (energy loss per cycle).
• Year tags mark questions we actually saw; untagged items are practice built to the same pattern.