Physics: Unit-wise Objective Question Bank

Group A of the NEB Grade 11 Physics paper is eleven one-mark multiple-choice questions — eleven marks, and the only eleven marks in the paper you can secure in under fifteen minutes. This chapter drills all twenty-six syllabus units in order. Each section opens with a high-yield facts table (the constants and formulae the objective questions keep returning to), then gives eight 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 Physical Quantities

High-yield fact	Value / form
Force	[MLT^{-2}], newton
Work, energy, torque	[ML^{2}T^{-2}]
Power	[ML^{2}T^{-3}], watt
Pressure, stress, all elastic moduli, energy density	[ML^{-1}T^{-2}], pascal
Planck constant, angular momentum	[ML^{2}T^{-1}]
Gravitational constant G	[M^{-1}L^{3}T^{-2}]
Dimensionless quantities	strain, refractive index, relative density, angle, \mu, Poisson's ratio, dielectric constant
Vernier least count	\text{L.C.} = 1\ \text{MSD} - 1\ \text{VSD}
Screw gauge least count	pitch \div number of circular-scale divisions
Error in Z = A^{p}B^{q}/C^{r}	\frac{\Delta Z}{Z} = p\frac{\Delta A}{A} + q\frac{\Delta B}{B} + r\frac{\Delta C}{C}

1. The dimensional formula of the universal gravitational constant G is [1]a[M^{-1}L^{3}T^{-2}]b[ML^{3}T^{-2}]c[M^{-1}L^{2}T^{-2}]d[ML^{2}T^{-2}]
2. Which pair does not have the same dimensions? [1]awork and torquebimpulse and momentumcstress and pressuredforce and power
3. The number of significant figures in 0.002340 is [1]a7b6c4d3
4. Ten divisions of a vernier scale coincide with nine divisions of the main scale, and 1\ \text{MSD} = 1\ \text{mm}. The least count is [1]a0.1\ \text{mm}b0.01\ \text{mm}c1\ \text{mm}d0.05\ \text{mm}
5. Which quantity is dimensionless yet is given a unit? [1]aplane anglebstraincrefractive indexdrelative density
6. Dimensional analysis cannot be used to find [1]awhether an equation is dimensionally correctbthe value of a dimensionless constantcthe conversion factor between two systems of unitsdthe dimensions of a derived quantity
7. If Z = A^{2}B^{3}/C and the percentage errors in A, B and C are 1\%, 2\% and 3\%, the percentage error in Z is [1]a6\%b8\%c10\%d11\%
8. The dimensional formula [ML^{-1}T^{-2}] belongs to [1]asurface tensionbpressurecforce constantdlinear momentum

Answers — 2.1

1. (a) — from F = Gm_1m_2/r^2, G = Fr^2/m^2 gives [MLT^{-2}][L^2][M^{-2}] = [M^{-1}L^{3}T^{-2}].

2. (d) — force is [MLT^{-2}], power is [ML^{2}T^{-3}]. Distractor (a) tempts students who think "torque is not energy so its dimensions must differ"; they are identical, only the physical meaning differs.

3. (c) — leading zeros are never significant, the trailing zero after the decimal point is; so 2, 3, 4, 0 = four.

4. (a) — 10\ \text{VSD} = 9\ \text{MSD}, so 1\ \text{VSD} = 0.9\ \text{mm} and \text{L.C.} = 1 - 0.9 = 0.1\ \text{mm}. Option (b) is the screw-gauge least count, quoted from memory.

5. (a) — angle is length/length, so dimensionless, but SI still names the radian for it.

6. (b) — the method matches powers of M, L, T; a pure number carries none, so it survives undetermined.

7. (d) — 2(1) + 3(2) + 1(3) = 11\%. Option (a) comes from adding the errors without their powers; (c) from forgetting the C term.

8. (b) — pressure is force per area. Surface tension and force constant are both [MT^{-2}], momentum is [MLT^{-1}].

2.2 Vectors

High-yield fact	Value / form
Resultant (parallelogram law)	R = \sqrt{P^{2}+Q^{2}+2PQ\cos\theta}
Direction of resultant	\tan\alpha = \dfrac{Q\sin\theta}{P+Q\cos\theta}
Maximum, minimum resultant	P+Q at \theta = 0^{\circ}; |P-Q| at \theta = 180^{\circ}
Scalar product	\vec A\cdot\vec B = AB\cos\theta — a scalar
Vector product	|\vec A\times\vec B| = AB\sin\theta — a vector, \perp to both
Unit-vector rules	\hat i\cdot\hat i = 1, \hat i\times\hat i = 0, \hat i\times\hat j = \hat k
Resolution	A_x = A\cos\theta, A_y = A\sin\theta

1. Two equal vectors of magnitude P act at 120^{\circ} to each other. Their resultant has magnitude [1]a2PbP\sqrt{3}cPdzero
2. Forces of 3\ \text{N} and 4\ \text{N} act at right angles. The resultant is [1]a1\ \text{N}b5\ \text{N}c7\ \text{N}d12\ \text{N}
3. The minimum number of coplanar vectors of unequal magnitude whose resultant can be zero is [1]a2b3c4d5
4. If \vec A\cdot\vec B = 0 and neither vector is zero, the angle between them is [1]a0^{\circ}b45^{\circ}c90^{\circ}d180^{\circ}
5. \hat i\times\hat j is equal to [1]a0b1c\hat kd-\hat k
6. The vector product of two parallel vectors is [1]amaximumbzerocunitydinfinite
7. A vector of magnitude 10 units makes 60^{\circ} with the x-axis. Its x-component is [1]a5 unitsb5\sqrt{3} unitsc10 unitsd8.66 units
8. Which of the following is a scalar quantity? [1]atorquebimpulsecworkdangular velocity

Answers — 2.2

1. (c) — R = \sqrt{P^2+P^2+2P^2\cos120^{\circ}} = \sqrt{2P^2-P^2} = P. Option (a) is the answer for 0^{\circ}; students who forget that \cos120^{\circ} is negative pick it.

2. (b) — \sqrt{3^2+4^2}=5. Option (c) adds the magnitudes, the commonest vector error of all.

3. (b) — two unequal vectors can never cancel; three can, if they close a triangle.

4. (c) — AB\cos\theta = 0 with A,B \ne 0 forces \cos\theta = 0.

5. (c) — right-hand rule, in cyclic order \hat i\to\hat j\to\hat k. Option (d) is \hat j\times\hat i, the order reversed.

6. (b) — \sin 0^{\circ} = 0. Option (a) is true of the scalar product; that swap is the classic confusion.

7. (a) — A_x = 10\cos60^{\circ} = 5. Option (d) is 10\sin60^{\circ}, sine and cosine swapped.

8. (c) — work is \vec F\cdot\vec s, a scalar product. The other three are vectors.

2.3 Kinematics

High-yield fact	Value / form
Equations of motion	v = u+at, s = ut+\tfrac12at^{2}, v^{2} = u^{2}+2as
Distance in the n^{\text{th}} second	s_n = u + \dfrac{a}{2}(2n-1) — a distance, unit metre
Slope and area	x–t slope = v; v–t slope = a, v–t area = s
Projectile	T = \dfrac{2u\sin\theta}{g}, H = \dfrac{u^{2}\sin^{2}\theta}{2g}, R = \dfrac{u^{2}\sin2\theta}{g}
Maximum range	R_{\max} = u^{2}/g at \theta = 45^{\circ}; \theta and 90^{\circ}-\theta give equal range
Speed at the top	u\cos\theta, not zero
Horizontal projection	t = \sqrt{2h/g}, independent of u
Relative velocity	\vec v_{AB} = \vec v_A - \vec v_B

1. From the velocity–time graph above, the total displacement of the body in 10\ \text{s} is [1]a100\ \text{m}b120\ \text{m}c140\ \text{m}d200\ \text{m}
2. The area under an acceleration–time graph gives [1]adisplacementbchange in velocitycaverage velocitydjerk
3. At the highest point of its trajectory, the velocity of a projectile is [1]azerobvertical and equal to u\sin\thetachorizontal and equal to u\cos\thetadequal to u
4. A stone is dropped from a height and another is thrown horizontally from the same height at the same instant. They [1]areach the ground togetherbthe dropped one lands firstcthe thrown one lands firstdit depends on the mass
5. For a freely falling body starting from rest, the distances covered in the first, second and third seconds are in the ratio [1]a1:2:3b1:3:5c1:4:9d1:1:1
6. Two cars move towards each other at 10\ \text{m s}^{-1} and 20\ \text{m s}^{-1}. The speed of one relative to the other is [1]a10\ \text{m s}^{-1}b15\ \text{m s}^{-1}c30\ \text{m s}^{-1}d200\ \text{m s}^{-1}
7. A body is projected with speed u at 30^{\circ}. Another has the same range with the same speed at [1]a45^{\circ}b60^{\circ}c75^{\circ}d120^{\circ}
8. The quantity s_n = u + \frac{a}{2}(2n-1) has the SI unit [1]a\text{m s}^{-1}b\text{m s}^{-2}c\text{m}d\text{m s}

Answers — 2.3

1. (c) — triangle A = \tfrac12(4)(20) = 40, rectangle B = 4\times20 = 80, triangle C = \tfrac12(2)(20) = 20; total 140\ \text{m}. Option (d) is 20\times10, the area of the whole rectangle — the error of treating a varying velocity as constant.

2. (b) — area = \int a\,dt = \Delta v. Option (a) is the area under a velocity–time graph; mixing the two graphs up is the standard slip.

3. (c) — only the vertical component vanishes.

4. (a) — vertical motion is identical and independent of the horizontal motion; t = \sqrt{2h/g} contains no u.

5. (b) — s_n \propto (2n-1), giving 1:3:5. Option (c) is the ratio of total distances after 1, 2, 3 s.

6. (c) — opposite directions, so v_{AB} = 10-(-20) = 30\ \text{m s}^{-1}. Option (a) subtracts the speeds, which is right only for the same direction.

7. (b) — complementary angles give equal range, and 90^{\circ}-30^{\circ} = 60^{\circ}.

8. (c) — it is the distance travelled during one particular second, so a metre. The "per second" in the wording tempts (a).

2.4 Dynamics

High-yield fact	Value / form
Newton's second law	\vec F = \dfrac{d\vec p}{dt} = m\vec a
Impulse	\vec J = \vec F_{av}\Delta t = m\vec v - m\vec u, unit N s = kg m s⁻¹
Conservation of momentum	m_1u_1+m_2u_2 = m_1v_1+m_2v_2 (no external force)
Torque	\tau = F\,d_\perp = rF\sin\theta, unit N m
Friction	\mu_s = f_{lim}/N, \mu_k = f_k/N, and \mu_k < \mu_s
Angle of repose	\mu_s = \tan\theta — equals the angle of friction \lambda
Body on a smooth incline	a = g\sin\theta, N = mg\cos\theta
Lift	apparent weight = m(g+a) going up, m(g-a) going down

1. Newton's second law of motion in its most general form is [1]aF = mvbF = dp/dtcF = ma onlydF = mg
2. The SI unit of impulse is the same as that of [1]aforcebenergyclinear momentumdpower
3. A rocket moves forward because [1]athe exhaust gases push against the airbof conservation of linear momentumcof the buoyancy of the gasesdthe Earth attracts it less
4. A body of mass 2\ \text{kg} at rest is acted on by a force of 10\ \text{N} for 3\ \text{s}. The impulse is [1]a6\ \text{N s}b15\ \text{N s}c30\ \text{N s}d60\ \text{N s}
5. In the free-body diagram above, the block is on the point of sliding. The coefficient of static friction is [1]a\sin\thetab\cos\thetac\tan\thetad1/\tan\theta
6. Which statement about action and reaction is correct? [1]athey act on the same body and cancelbthey act on different bodies and never cancelcthe action is always largerdthey act at different instants
7. A man of mass 60\ \text{kg} stands in a lift accelerating upwards at 2\ \text{m s}^{-2} (g = 10\ \text{m s}^{-2}). The reading of the weighing machine is [1]a480\ \text{N}b600\ \text{N}c720\ \text{N}d120\ \text{N}
8. The coefficient of kinetic friction is [1]agreater than the coefficient of static frictionbequal to itcless than itdalways equal to 1

Answers — 2.4

1. (b) — F = ma follows from F = dp/dt only when the mass is constant, which fails for a rocket.

2. (c) — impulse is the change in momentum, so N s = kg m s⁻¹. Option (a) tempts students who read "impulse" as "a big force".

3. (b) — the gas carries momentum backwards, so the rocket carries an equal momentum forwards. Option (a) is the classic error: a rocket works in vacuum, where there is nothing to push against.

4. (c) — J = F\Delta t = 10\times3 = 30\ \text{N s}.

5. (c) — on the point of sliding, f = \mu N, with f = mg\sin\theta and N = mg\cos\theta, so \mu = \tan\theta. This is the angle of repose.

6. (b) — they are equal and opposite but act on different bodies, so they can never be added together to cancel.

7. (c) — N = m(g+a) = 60(12) = 720\ \text{N}. Option (b) is the true weight, chosen by anyone who forgets the lift is accelerating; the machine reads the normal reaction, not the weight.

8. (c) — it always takes a larger force to start sliding than to keep it going.

2.5 Work, energy and power

High-yield fact	Value / form
Work	W = Fs\cos\theta = \vec F\cdot\vec s; variable force W = \int F\,dx
Power	P = W/t = \vec F\cdot\vec v; 1\ \text{hp} = 746\ \text{W}
Work–energy theorem	W_{net} = \Delta K
Kinetic energy	K = \tfrac12mv^{2} = \dfrac{p^{2}}{2m}, so p = \sqrt{2mK}
Spring	W = \tfrac12kx^{2}
1\ \text{kW h}	3.6\times10^{6}\ \text{J} — a unit of energy
Elastic collision	momentum and kinetic energy conserved; equal masses exchange velocities
Perfectly inelastic	v = \dfrac{m_1u_1+m_2u_2}{m_1+m_2}; KE is lost, momentum is not

1. The work done by the centripetal force on a body in uniform circular motion is [1]amaximumbzerocnegativedequal to mv^{2}/r
2. One kilowatt-hour is equal to [1]a1000\ \text{J}b3600\ \text{J}c3.6\times10^{6}\ \text{J}d3.6\times10^{3}\ \text{W}
3. If the momentum of a body is doubled, its kinetic energy becomes [1]ahalfbdoublecfour timesdunchanged
4. In a perfectly elastic head-on collision between two bodies of equal mass, one at rest, [1]aboth move togetherbthey exchange velocitiescboth stopdthe moving one rebounds
5. The work done by a conservative force around any closed path is [1]aalways positivebalways negativeczerodequal to the kinetic energy
6. A man of mass 60\ \text{kg} climbs a 5\ \text{m} staircase in 10\ \text{s} (g = 10\ \text{m s}^{-2}). His power output is [1]a30\ \text{W}b300\ \text{W}c3000\ \text{W}d600\ \text{W}
7. In an inelastic collision, the quantity that is not conserved is [1]alinear momentumbtotal energyckinetic energydmass
8. A body of mass m and kinetic energy K has linear momentum [1]a\sqrt{2mK}b2mKc\sqrt{mK}dmK^{2}

Answers — 2.5

1. (b) — the force is always perpendicular to the velocity, so \cos90^{\circ} = 0. Option (d) is the magnitude of the force, offered to anyone who reads "work" and reaches for the nearest formula.

2. (c) — 1000\ \text{W}\times3600\ \text{s}. Option (d) is the trap: kW h is energy, not power.

3. (c) — K = p^{2}/2m, so K \propto p^{2}. Option (b) comes from assuming K \propto p.

4. (b) — put m_1 = m_2 and u_2 = 0 in the elastic-collision result: v_1 = 0, v_2 = u_1. Option (a) describes a perfectly inelastic collision.

5. (c) — that is the definition of a conservative force.

6. (b) — W = mgh = 60\times10\times5 = 3000\ \text{J}, so P = 3000/10 = 300\ \text{W}. Option (c) is the work, not the power.

7. (c) — momentum and total energy always survive; the missing kinetic energy has become heat, sound and deformation.

8. (a) — invert K = p^2/2m.

2.6 Circular Motion

High-yield fact	Value / form
Angular relations	v = r\omega, a_t = r\alpha, \omega = 2\pi/T = 2\pi f
Centripetal acceleration	a_c = v^{2}/r = r\omega^{2}
Centripetal force	F_c = mv^{2}/r = mr\omega^{2}
Conical pendulum	\tan\theta = v^{2}/rg; T_p = 2\pi\sqrt{h/g}
Vertical circle	v_{top}^{\min} = \sqrt{gr}, v_{bottom}^{\min} = \sqrt{5gr}, T_2 - T_1 = 6mg
Banking	\tan\theta = v^{2}/rg; level road v_{\max} = \sqrt{\mu rg}
1\ \text{rpm}	2\pi/60\ \text{rad s}^{-1}

1. In uniform circular motion, which quantity stays constant? [1]avelocitybspeedcaccelerationdmomentum
2. The minimum speed at the highest point of a vertical circle of radius r is [1]a\sqrt{gr}b\sqrt{2gr}c\sqrt{5gr}dzero
3. The difference between the tensions at the lowest and the highest point of a vertical circle is [1]a2mgb3mgc5mgd6mg
4. A body completes 60 revolutions per minute. Its angular velocity is [1]a\pi\ \text{rad s}^{-1}b2\pi\ \text{rad s}^{-1}c60\ \text{rad s}^{-1}d120\pi\ \text{rad s}^{-1}
5. The angle of banking of a road of radius r for a design speed v is given by [1]a\tan\theta = rg/v^{2}b\tan\theta = v^{2}/rgc\sin\theta = v^{2}/rgd\tan\theta = v/rg
6. The period of a conical pendulum of vertical depth h is [1]a2\pi\sqrt{L/g}b2\pi\sqrt{h/g}c2\pi\sqrt{g/h}d2\pi\sqrt{L\cos\theta/h}
7. Centrifugal force is [1]athe reaction to the centripetal forceba pseudo-force that appears only in a rotating framecalways greater than the centripetal forceddirected towards the centre
8. A stone of mass 0.5\ \text{kg} moves in a circle of radius 2\ \text{m} at 4\ \text{m s}^{-1}. The centripetal force is [1]a1\ \text{N}b2\ \text{N}c4\ \text{N}d8\ \text{N}

Answers — 2.6

1. (b) — the direction of velocity changes continuously, so velocity, acceleration and momentum all change; only the magnitude of the velocity is fixed.

2. (a) — at the top, gravity alone supplies the centripetal force when the string goes slack: mg = mv^2/r.

3. (d) — subtracting the two force equations and using v_2^2 = v_1^2 + 4gr gives 6mg, independent of the speed.

4. (b) — 60\ \text{rpm} = 1\ \text{rev s}^{-1} = 2\pi\ \text{rad s}^{-1}. Option (c) is the trap: rpm read straight off as rad s⁻¹.

5. (b) — from N\sin\theta = mv^2/r and N\cos\theta = mg.

6. (b) — T_p = 2\pi\sqrt{L\cos\theta/g}, and L\cos\theta is exactly the depth h below the support.

7. (b) — it is not a real force and has no reaction partner. Option (a) is the commonest wrong answer: the reaction to the centripetal force acts on a different body.

8. (c) — F = mv^2/r = 0.5\times16/2 = 4\ \text{N}.

2.7 Gravitation

High-yield fact	Value / form
G	6.67\times10^{-11}\ \text{N m}^{2}\text{kg}^{-2}
Surface value	g = GM/R^{2}, so M = gR^{2}/G
Altitude, depth	g_h = g\left(1-\dfrac{2h}{R}\right) for h\ll R; g_d = g\left(1-\dfrac{d}{R}\right), zero at the centre
Potential, potential energy	V = -GM/r (J kg⁻¹), U = -GMm/r
Orbital speed	v_o = \sqrt{GM/(R+h)}
Escape velocity	v_e = \sqrt{2GM/R} = \sqrt{2gR} = \sqrt{2}\,v_o = 11.2\ \text{km s}^{-1}
Satellite energy	K = +\dfrac{GMm}{2r}, U = -\dfrac{GMm}{r}, E = -\dfrac{GMm}{2r}
Geostationary satellite	T = 24\ \text{h}, h \approx 36\,000\ \text{km}, equatorial, west to east

1. The value of g at the centre of the Earth is [1]amaximumbequal to the surface valueczerodinfinite
2. The escape velocity from the surface of the Earth is approximately [1]a8\ \text{km s}^{-1}b11.2\ \text{km s}^{-1}c22.4\ \text{km s}^{-1}d3\times10^{8}\ \text{m s}^{-1}
3. The escape velocity and the orbital velocity close to the Earth's surface are related by [1]av_e = v_obv_e = 2v_ocv_e = \sqrt{2}\,v_odv_e = v_o/\sqrt{2}
4. An astronaut inside an orbiting satellite feels weightless because [1]athere is no gravity at that heightbthe astronaut and the satellite are in free fall togethercthe satellite is beyond the atmospheredthe astronaut's mass has become zero
5. The total energy of a satellite of mass m in a circular orbit of radius r is [1]a+GMm/rb-GMm/rc-GMm/2rd+GMm/2r
6. Escape velocity from a planet does not depend on [1]athe mass of the planetbthe radius of the planetcthe mass of the escaping bodydthe density of the planet
7. The time period of a geostationary satellite is [1]a1\ \text{h}b12\ \text{h}c24\ \text{h}d365\ \text{days}
8. Gravitational potential at a point is [1]aalways positivebalways negativeczero everywheredpositive near a mass and negative far from it

Answers — 2.7

1. (c) — g_d = g(1-d/R) gives zero at d = R; only the shell inside your radius pulls you, and at the centre there is none.

2. (b) — v_e = \sqrt{2gR} = \sqrt{2\times9.8\times6.4\times10^{6}} \approx 11.2\ \text{km s}^{-1}.

3. (c) — v_e = \sqrt{2GM/R} and v_o = \sqrt{GM/R}.

4. (b) — gravity is still about 89\% of its surface value at 400\ \text{km}; what vanishes is the normal reaction. Option (a) is the single commonest misconception in this unit.

5. (c) — E = K+U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}; it is negative because the satellite is bound.

6. (c) — m cancels out of \tfrac12mv_e^2 = GMm/R.

7. (c) — it must match the Earth's rotation period to stay above the same point.

8. (b) — with V = 0 chosen at infinity and gravity always attractive, V = -GM/r is negative everywhere.

2.8 Elasticity

High-yield fact	Value / form
Hooke's law	F = kx; k = YA/L, unit N m⁻¹
Young modulus	Y = \dfrac{FL}{Ae}, unit Pa = \text{N m}^{-2}
Bulk modulus	B = -\dfrac{pV}{\Delta V}; compressibility K = 1/B
Shear modulus	\eta = \dfrac{F/A}{\theta}
Poisson's ratio	\sigma = \dfrac{\text{lateral strain}}{\text{longitudinal strain}}; limits -1 to 0.5; real solids 0.2–0.4
Elastic energy	U = \tfrac12Fe; energy density u = \tfrac12\times stress \times strain
Only modulus for fluids	bulk modulus

1. The SI unit of Young modulus is the same as that of [1]aforcebpressurecstraindenergy
2. Strain has [1]athe unit metrebthe unit newtoncno unit and no dimensionsdthe unit pascal
3. Which elastic modulus is defined for liquids and gases? [1]aYoung modulusbshear moduluscbulk modulusdall three
4. The theoretical limits of Poisson's ratio are [1]a0 to 1b-1 to 0.5c-0.5 to 1d0 to \infty
5. The energy stored per unit volume of a stretched wire is [1]astress \times strainb\tfrac12\times stress \times strainc\tfrac12 Y\times stressd\tfrac12\times stress / strain
6. Breaking stress of a wire depends on [1]aits length onlybits area of cross-section onlycthe material onlydboth its length and its area
7. Steel is said to be more elastic than rubber because for the same stress steel [1]astretches morebstretches lesscbreaks soonerdis heavier
8. Hooke's law is obeyed only up to the [1]abreaking pointbyield pointcproportional limitdultimate stress

Answers — 2.8

1. (b) — Y is a stress divided by a dimensionless strain, so it carries the unit of stress, the pascal.

2. (c) — it is a ratio of two lengths (or two volumes). Option (d) is the unit of stress; swapping stress and strain is the classic slip.

3. (c) — fluids have no fixed shape, so they cannot sustain a shear or a tensile stress.

4. (b) — practical materials sit in 0.2–0.4; the negative half of the range belongs to rare auxetic materials.

5. (b) — from U = \tfrac12Fe divided by the volume AL.

6. (c) — breaking stress is a material property; breaking force is what depends on the area.

7. (b) — "more elastic" means a larger Y, i.e. a greater stress is needed per unit strain. Students routinely pick (a) because rubber "stretches more and springs back".

8. (c) — beyond the proportional limit the stress–strain graph is no longer a straight line, even though the wire may still return to its original length up to the elastic limit.

2.9 Heat and Temperature

High-yield fact	Value / form
Mean translational KE per molecule	\bar E_k = \tfrac32 k_BT — depends on T only
Boltzmann constant	k_B = 1.38\times10^{-23}\ \text{J K}^{-1}
Scale conversion	\dfrac{C}{100} = \dfrac{F-32}{180} = \dfrac{K-273.15}{100}
Zeroth law	if A \sim C and B \sim C then A \sim B — this is what defines temperature
Thermometric formula	t = \dfrac{X_t - X_0}{X_{100} - X_0}\times100\ ^{\circ}\text{C}
Triple point of water	273.16\ \text{K} = 0.01\ ^{\circ}\text{C}

1. The zeroth law of thermodynamics gives a physical basis for the concept of [1]aheatbinternal energyctemperaturedentropy
2. The mean translational kinetic energy of a gas molecule depends only on [1]aits massbthe absolute temperaturecthe pressuredthe volume
3. The temperature at which the Celsius and Fahrenheit readings are numerically equal is [1]a0^{\circ}b32^{\circ}c-40^{\circ}d100^{\circ}
4. Mercury is preferred to water as a thermometric liquid mainly because [1]ait is cheapbit expands uniformly and does not wet glasscit is transparentdit has a high specific heat capacity
5. Heat flows between two bodies in contact because of a difference in their [1]atotal thermal energiesbmassesctemperaturesdspecific heat capacities
6. The thermometric property used in a constant-volume gas thermometer is [1]avolumebpressureclength of a mercury threaddelectrical resistance
7. A body at 27\ ^{\circ}\text{C} has an absolute temperature of [1]a300.15\ \text{K}b273\ \text{K}c27\ \text{K}d246\ \text{K}
8. Two bodies are in thermal equilibrium when they have the same [1]aheat contentbtemperaturecmassdinternal energy

Answers — 2.9

1. (c) — it guarantees that "same temperature" is a consistent labelling of bodies.

2. (b) — \bar E_k = \tfrac32k_BT contains no mass term, so at the same temperature a heavy and a light molecule have the same mean translational energy (the heavy one simply moves more slowly).

3. (c) — set C = F in C/100 = (F-32)/180; both scales read -40.

4. (b) — plus a wide liquid range and good visibility. Option (d) is wrong twice over: mercury's specific heat capacity is low, which is an advantage.

5. (c) — a large cold lake holds far more thermal energy than a matchstick flame, yet heat still flows from the flame to the lake.

6. (b) — the volume is held fixed by definition, so pressure is the quantity that varies.

7. (a) — K = C + 273.15. Option (b) is 273 + 0, a careless subtraction.

8. (b) — equal temperature is the definition; equal heat content is neither necessary nor sufficient.

2.10 Thermal Expansion

High-yield fact	Value / form
Coefficients	\beta = 2\alpha, \gamma = 3\alpha, so \alpha:\beta:\gamma = 1:2:3; unit K⁻¹
Linear expansion	l_2 = l_1[1+\alpha\Delta\theta]
Liquid expansivities	\gamma_r = \gamma_a + \gamma_g = \gamma_a + 3\alpha_g
Dulong and Petit	\gamma = \dfrac{h_\theta - h_0}{h_0\,\theta} — gives the real expansivity, no vessel correction
Anomalous expansion of water	contracts from 0 to 4\ ^{\circ}\text{C}; maximum density at 4\ ^{\circ}\text{C}
Pendulum clock	\Delta t = \tfrac12\alpha\Delta\theta\times86400\ \text{s per day}

1. The ratio \alpha:\beta:\gamma for an isotropic solid is [1]a1:1:1b1:2:3c3:2:1d1:3:2
2. When a metal plate with a circular hole is heated, the hole [1]abecomes smallerbbecomes largercstays the same sizedbecomes oval
3. The real expansivity of a liquid is related to the apparent expansivity by [1]a\gamma_r = \gamma_a - \gamma_gb\gamma_r = \gamma_a + \gamma_gc\gamma_r = \gamma_a\gamma_gd\gamma_r = \gamma_a/\gamma_g
4. Water has its maximum density at [1]a0\ ^{\circ}\text{C}b4\ ^{\circ}\text{C}c100\ ^{\circ}\text{C}d-4\ ^{\circ}\text{C}
5. Dulong and Petit's method is preferred for a liquid because it [1]ais quickerbneeds no correction for the expansion of the vesselcuses less liquiddworks at any pressure
6. The SI unit of the coefficient of linear expansion is [1]a\text{m K}^{-1}b\text{K}^{-1}c\text{m}^{-1}d\text{m}^{3}\text{K}^{-1}
7. A brass rod 1\ \text{m} long (\alpha = 1.9\times10^{-5}\ \text{K}^{-1}) is heated through 100\ \text{K}. Its increase in length is [1]a0.19\ \text{mm}b1.9\ \text{mm}c19\ \text{mm}d1.9\ \text{cm}
8. A pendulum clock with a metal rod is taken from winter to summer. It will [1]again timeblose timeckeep correct timedstop

Answers — 2.10

1. (b) — one, two and three dimensions of the same linear stretch.

2. (b) — every linear dimension of the plate, including the diameter of the hole, is multiplied by (1+\alpha\Delta\theta). Option (a) is the standard misconception: students picture the metal "closing in".

3. (b) — the liquid's true expansion is the overflow plus the expansion of the vessel that held it.

4. (b) — which is why a lake freezes from the top down and fish survive the winter.

5. (b) — the two liquid columns balance in a single U-tube, so the glass cancels out.

6. (b) — a fractional change in length per kelvin; the metres cancel.

7. (b) — \Delta l = l\alpha\Delta\theta = 1\times1.9\times10^{-5}\times100 = 1.9\times10^{-3}\ \text{m}.

8. (b) — the rod lengthens, so T = 2\pi\sqrt{l/g} increases and the clock ticks too slowly.

2.11 Quantity of Heat

High-yield fact	Value / form
Heat and phase change	Q = mc\Delta\theta; Q = mL
Water	c_w = 4200\ \text{J kg}^{-1}\text{K}^{-1}
Latent heats of water	L_f = 3.34\times10^{5}\ \text{J kg}^{-1}, L_v = 2.26\times10^{6}\ \text{J kg}^{-1}
Newton's law of cooling	\dfrac{d\theta}{dt} = -k(\theta-\theta_0) — small excess temperature only
Working form	\dfrac{\theta_1-\theta_2}{t} = k\left(\dfrac{\theta_1+\theta_2}{2} - \theta_0\right)
Water equivalent	W = mc/c_w, measured in kg
Triple point	the single (p,T) at which solid, liquid and vapour coexist

1. The SI unit of specific heat capacity is [1]a\text{J kg}^{-1}b\text{J kg}^{-1}\text{K}^{-1}c\text{J K}^{-1}d\text{J}
2. The specific latent heat of fusion of ice is about [1]a3.34\times10^{5}\ \text{J kg}^{-1}b2.26\times10^{6}\ \text{J kg}^{-1}c4200\ \text{J kg}^{-1}d336\ \text{J kg}^{-1}
3. Newton's law of cooling holds well only when [1]athe body is very hotbthe excess of temperature over the surroundings is smallcthe body is in vacuumdthe surroundings are also cooling
4. While ice at 0\ ^{\circ}\text{C} is melting, the heat supplied [1]araises the temperaturebbreaks the bonds of the solid lattice at constant temperaturecis lost to the surroundingsddoes nothing
5. The heat needed to melt 100\ \text{g} of ice at 0\ ^{\circ}\text{C} into water at 0\ ^{\circ}\text{C} is [1]a3.34\times10^{3}\ \text{J}b3.34\times10^{4}\ \text{J}c3.34\times10^{5}\ \text{J}d4.2\times10^{4}\ \text{J}
6. The unit of water equivalent is [1]ajoulebkilogramc\text{J K}^{-1}dkelvin
7. Steam at 100\ ^{\circ}\text{C} causes a worse burn than water at 100\ ^{\circ}\text{C} because [1]asteam is hotterbsteam releases its latent heat of vaporisation on condensingcsteam has a higher specific heat capacitydsteam moves faster
8. At the triple point of water [1]aonly ice and water coexistbonly water and vapour coexistcice, water and vapour coexistdwater boils and freezes at different pressures

Answers — 2.11

1. (b) — energy per kilogram per kelvin. Option (a) is the unit of specific latent heat; the two are swapped constantly.

2. (a) — option (b) is L_v; option (d) is the old value in cal g⁻¹ misread as SI.

3. (b) — for large excess temperatures the loss follows Stefan's T^4 law instead.

4. (b) — the temperature stays fixed during a phase change, which is why "heat always raises temperature" is false.

5. (b) — Q = mL_f = 0.1\times3.34\times10^{5} = 3.34\times10^{4}\ \text{J}. Option (c) forgets to convert 100\ \text{g} to 0.1\ \text{kg}.

6. (b) — it is the mass of water that would absorb the same heat for the same rise, so it is a mass.

7. (b) — each kilogram of steam gives up 2.26\times10^{6}\ \text{J} before its temperature even starts to fall.

8. (c) — a single point on the phase diagram, at 273.16\ \text{K} and 611.66\ \text{Pa}.

2.12 Rate of heat flow

High-yield fact	Value / form
Conduction	H = \dfrac{kA(\theta_1-\theta_2)}{L}; k in W m⁻¹ K⁻¹
Thermal resistance	R = L/kA
Black body	absorptance a = 1, emissivity e = 1
Kirchhoff's law of radiation	a good absorber is an equally good emitter
Stefan–Boltzmann law	E = \sigma T^{4}; P_{net} = e\sigma A(T^{4}-T_0^{4})
Stefan constant	\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}
Wien's displacement law	\lambda_m T = b = 2.898\times10^{-3}\ \text{m K}
Searle's method	good conductors; Lee's disc for bad conductors

1. The SI unit of thermal conductivity is [1]a\text{W m}^{-1}\text{K}^{-1}b\text{W m}^{-2}\text{K}^{-4}c\text{J s}^{-1}d\text{W m}^{-2}
2. Heat transfer across the vacuum between the Sun and the Earth takes place by [1]aconductionbconvectioncradiationdall three
3. If the absolute temperature of a black body is doubled, the power it radiates becomes [1]a2 timesb4 timesc8 timesd16 times
4. Wien's displacement law states that [1]a\lambda_m \propto Tb\lambda_m T = constantcE \propto T^{4}d\lambda_m \propto T^{4}
5. According to Kirchhoff's law of radiation, a good absorber is [1]aa poor emitterba good emitterca good reflectorda good transmitter
6. The value of Stefan's constant is [1]a5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}b1.38\times10^{-23}\ \text{J K}^{-1}c6.67\times10^{-11}d2.898\times10^{-3}\ \text{m K}
7. Searle's method is used to measure the thermal conductivity of [1]agood conductorsbbad conductorscliquidsdgases
8. A body radiates most when it is [1]apolished and silverybdull and blackcwhite and smoothdtransparent

Answers — 2.12

1. (a) — from H = kA\Delta\theta/L, k has units \text{W}\times\text{m}/(\text{m}^2\text{K}). Option (b) is Stefan's constant.

2. (c) — radiation needs no material medium.

3. (d) — E \propto T^{4}, so 2^{4} = 16.

4. (b) — the peak wavelength shifts inversely with temperature; option (a) is the sign of the relation reversed.

5. (b) — at a given temperature the emissivity equals the absorptance.

6. (a) — option (d) is Wien's constant, the usual mix-up in this unit.

7. (a) — a good conductor needs a long bar and a large temperature gradient; a bad conductor needs the thin disc of Lee's apparatus.

8. (b) — a dull black surface has e close to 1; a polished surface reflects and barely emits.

2.13 Ideal gas

High-yield fact	Value / form
Ideal gas equation	pV = nRT = NkT; R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}
Kinetic pressure	p = \tfrac13\rho\,\overline{c^{2}}, i.e. pV = \tfrac13mN\overline{c^{2}}
Mean translational energy	\overline{E} = \tfrac12 m\overline{c^{2}} = \tfrac32 kT
Boltzmann constant	k = R/N_A = 1.38\times10^{-23}\ \text{J K}^{-1}
Root-mean-square speed	c_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3kT}{m}} = \sqrt{\dfrac{3p}{\rho}}
Mayer's relation	C_p - C_v = R
Ratio of heat capacities	\gamma = 1.67 monatomic, 1.40 diatomic
Dulong and Petit for solids	C \approx 3R \approx 25\ \text{J mol}^{-1}\text{K}^{-1}

1. The value of the universal gas constant in SI units is [1]a8.314\ \text{J mol}^{-1}\text{K}^{-1}b0.0821\ \text{J mol}^{-1}\text{K}^{-1}c1.38\times10^{-23}d6.022\times10^{23}
2. According to kinetic theory, the pressure exerted by an ideal gas is [1]a\rho\overline{c^{2}}b\tfrac13\rho\overline{c^{2}}c\tfrac12\rho\overline{c^{2}}d\tfrac23\rho\overline{c^{2}}
3. If the absolute temperature of a gas is made four times, the rms speed of its molecules becomes [1]a2 timesb4 timesc8 timesd16 times
4. Mayer's relation for an ideal gas is [1]aC_p + C_v = RbC_p - C_v = RcC_p C_v = RdC_p/C_v = R
5. The ratio \gamma = C_p/C_v for a monatomic ideal gas is [1]a1.40b1.33c1.67d1.00
6. The mean kinetic energy of a molecule of an ideal gas at temperature T is [1]akTb\tfrac32kTc\tfrac32RTd\tfrac12kT
7. At the same temperature, hydrogen and oxygen molecules have the same [1]arms speedbmean translational kinetic energycmomentumddensity
8. The Boltzmann constant is equal to [1]aRN_AbR/N_AcN_A/RdR + N_A

Answers — 2.13

1. (a) — option (b) is R in \text{L atm mol}^{-1}\text{K}^{-1} with the wrong unit label attached.

2. (b) — the factor \tfrac13 comes from \overline{u^2} = \tfrac13\overline{c^2}, the three equivalent directions.

3. (a) — c_{rms} \propto \sqrt{T}, and \sqrt{4} = 2. Option (b) is the trap of reading the relation as linear.

4. (b) — the extra heat at constant pressure is exactly the work R\,\Delta T done in expanding.

5. (c) — three translational degrees of freedom only.

6. (b) — per molecule; option (c) is the value per mole, the commonest confusion here.

7. (b) — \overline{E} = \tfrac32 kT has no mass in it. Their rms speeds differ by \sqrt{32/2} = 4.

8. (b) — the gas constant shared out among the molecules of one mole.

2.14 Reflection at curved mirror

High-yield fact	Value / form
Focal length	f = R/2
Mirror formula	\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}
Magnification	m = -\dfrac{v}{u} = \dfrac{h_i}{h_o}
Sign convention	distances measured from the pole, against the incident light are negative; for a concave mirror f is negative, for a convex mirror f is positive
Convex mirror	image always virtual, erect and diminished, whatever the object distance
Concave, object at C	image at C, real, inverted, same size
Concave, object between F and P	image virtual, erect, magnified — the shaving mirror

1. The focal length of a spherical mirror of radius of curvature R is [1]aRbR/2c2RdR/4
2. The image formed by a convex mirror is always [1]areal, inverted and magnifiedbvirtual, erect and diminishedcreal, erect and diminisheddvirtual, inverted and magnified
3. The mirror formula is [1]a\dfrac1v - \dfrac1u = \dfrac1fb\dfrac1v + \dfrac1u = \dfrac1fc\dfrac1u - \dfrac1v = \dfrac1fd\dfrac1v + \dfrac1u = \dfrac2f
4. The linear magnification produced by a mirror is [1]am = v/ubm = -v/ucm = u/vdm = -u/v
5. A concave mirror forms an image of the same size as the object when the object is placed at [1]athe focusbthe polecthe centre of curvaturedinfinity
6. An object is placed 15\ \text{cm} from a concave mirror of focal length 10\ \text{cm}. The image distance is [1]a6\ \text{cm} in frontb30\ \text{cm} in frontc30\ \text{cm} behindd6\ \text{cm} behind
7. In the figure above, the object lies between C and F, so the image is [1]avirtual and diminishedbreal, inverted and magnified, beyond Ccreal and at Fdvirtual and magnified
8. A convex mirror is used as a driver's rear-view mirror because it [1]aforms a real imagebgives a wide field of view with an erect imagecmagnifies distant vehiclesdremoves chromatic aberration

Answers — 2.14

1. (b) — paraxial rays from infinity cross the axis half way to the centre of curvature.

2. (b) — the reflected rays always diverge, so the image is caught behind the mirror; no object position changes this.

3. (b) — the plus sign is the mirror formula, the minus sign belongs to the lens formula. Option (a) is the single most costly mix-up in the whole optics section.

4. (b) — the minus sign is what tells you a real image is inverted. Option (a) is the lens magnification.

5. (c) — u = -R gives v = -R and m = -1.

6. (b) — with u = -15\ \text{cm}, f = -10\ \text{cm}: 1/v = 1/f - 1/u = -1/10 + 1/15 = -1/30, so v = -30\ \text{cm}, i.e. 30\ \text{cm} in front of the mirror and real.

7. (b) — read the ray diagram: the two construction rays meet in front of the mirror, past C.

8. (b) — the diminished virtual image packs a wider scene into the same glass, at the price of making cars look further away than they are.

2.15 Refraction at plane surfaces

High-yield fact	Value / form
Snell's law	\dfrac{\sin i}{\sin r} = {}_1n_2 = \dfrac{n_2}{n_1} = \dfrac{v_1}{v_2}
Absolute refractive index	n = c/v; n_{water} = 1.33, n_{glass} = 1.5
Unchanged on refraction	frequency (colour); wavelength and speed both change
Real and apparent depth	n = \dfrac{\text{real depth}}{\text{apparent depth}}; shift = t\left(1-\dfrac1n\right)
Critical angle	\sin C = \dfrac{1}{n}; glass C \approx 41.8^{\circ}, water C \approx 48.8^{\circ}
Total internal reflection	needs denser \to rarer and i > C
Lateral shift	d = \dfrac{t\sin(i-r)}{\cos r}

1. The refractive index of a medium is defined as [1]av/cbc/vccvdc - v
2. When light passes from air into glass, the quantity that does not change is its [1]aspeedbwavelengthcfrequencyddirection
3. The critical angle for a medium of refractive index n is given by [1]a\sin C = nb\sin C = 1/nc\tan C = nd\cos C = 1/n
4. For glass of refractive index 1.5, the critical angle is about [1]a30^{\circ}b41.8^{\circ}c48.8^{\circ}d60^{\circ}
5. A pond is 4\ \text{m} deep. To an observer looking vertically down, its apparent depth (n = 4/3) is [1]a5.33\ \text{m}b3\ \text{m}c2\ \text{m}d4\ \text{m}
6. A mirage in a desert is caused by [1]adiffractionbdispersionctotal internal reflection in layers of air of different densitydinterference
7. A ray entering an optically denser medium bends [1]aaway from the normalbtowards the normalcalong the normaldnot at all
8. Total internal reflection can occur when light travels from [1]aair to glassbglass to air at any anglecglass to air at an angle greater than the critical angledair to water at grazing incidence

Answers — 2.15

1. (b) — light slows down in a medium, so n > 1 always. Option (a) inverts the ratio and would make n < 1.

2. (c) — the source sets the frequency; the medium sets the speed, and therefore the wavelength through v = f\lambda.

3. (b) — put r = 90^{\circ} in Snell's law.

4. (b) — C = \sin^{-1}(1/1.5) = 41.8^{\circ}. Option (c) is the value for water, a frequent swap.

5. (b) — apparent depth = 4/(4/3) = 3\ \text{m}. Option (a) multiplies by n instead of dividing.

6. (c) — hot air near the ground is rarer, so rays curve and are totally internally reflected.

7. (b) — into the slower medium, the wavefront turns towards the normal.

8. (c) — both conditions are needed; option (b) drops the angle condition, option (a) has the media the wrong way round.

2.16 Refraction through prisms

High-yield fact	Value / form
Angle relations	A = r_1 + r_2 and \delta = i_1 + i_2 - A
At minimum deviation	i_1 = i_2, r_1 = r_2 = A/2, ray inside is parallel to the base
Prism formula	n = \dfrac{\sin\left(\dfrac{A+\delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}
Thin prism	\delta = (n-1)A
Grazing emergence condition	maximum A for emergence is 2C

1. For a prism, the relation between the angle of the prism and the angles of refraction is [1]aA = r_1 - r_2bA = r_1 + r_2cA = i_1 + i_2dA = \delta + i_1
2. At the position of minimum deviation, the ray inside the prism is [1]aperpendicular to the basebparallel to the basecalong the basedat the critical angle
3. The deviation produced by a thin prism is [1]anAb(n-1)Ac(n+1)AdA/(n-1)
4. For a prism of angle 60^{\circ} with a minimum deviation of 30^{\circ}, the refractive index is [1]a1.33b1.414c1.50d1.73
5. A thin prism of angle 5^{\circ} and refractive index 1.5 produces a deviation of [1]a2.5^{\circ}b5^{\circ}c7.5^{\circ}d1.5^{\circ}
6. A ray will emerge from the second face of a prism only if the angle of the prism is [1]aless than Cbless than or equal to 2Ccgreater than 2Cdexactly 90^{\circ}
7. The deviation produced by a prism is [1]ai_1 + i_2 - Abi_1 - i_2 + AcA - i_1 - i_2di_1 + i_2 + A
8. As the angle of incidence on a prism is increased from a small value, the deviation [1]aincreases steadilybdecreases steadilycfirst decreases to a minimum then increasesdstays constant

Answers — 2.16

1. (b) — from the quadrilateral formed by the two normals and the two faces.

2. (b) — the symmetry r_1 = r_2 is exactly what makes the internal ray parallel to the base.

3. (b) — the (n-1) factor; option (a) forgets that a prism of n = 1 would deviate nothing.

4. (b) — n = \sin45^{\circ}/\sin30^{\circ} = 0.7071/0.5 = 1.414.

5. (a) — \delta = (1.5-1)\times5^{\circ} = 2.5^{\circ}.

6. (b) — if A > 2C then r_2 > C at the second face and the ray is totally internally reflected instead.

7. (a) — the two deviations at the two faces, less the prism angle.

8. (c) — the deviation curve is U-shaped, which is why \delta_m exists at all.

2.17 Lenses

High-yield fact	Value / form
Lens formula	\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} (note the minus)
Magnification	m = \dfrac{v}{u}
Lens maker's formula	\dfrac1f = (n-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)
Power	P = \dfrac{1}{f\ \text{in metres}}, in dioptre (D); convex +, concave -
Lenses in contact	\dfrac{1}{F} = \dfrac{1}{f_1}+\dfrac{1}{f_2}, P = P_1 + P_2
Lenses separated by d	\dfrac{1}{F} = \dfrac{1}{f_1}+\dfrac{1}{f_2}-\dfrac{d}{f_1f_2}
Simple microscope	M = 1 + \dfrac{D}{f} with D = 25\ \text{cm}

1. The SI unit of the power of a lens is [1]awattbdioptrecmetredlumen
2. The lens formula is [1]a\dfrac1v + \dfrac1u = \dfrac1fb\dfrac1v - \dfrac1u = \dfrac1fc\dfrac1u - \dfrac1v = \dfrac1fd\dfrac1f = \dfrac1v\dfrac1u
3. A convex lens of focal length 20\ \text{cm} is placed in contact with a concave lens of focal length 30\ \text{cm}. The power of the combination is [1]a+1.67\ \text{D}b-1.67\ \text{D}c+8.33\ \text{D}d-0.6\ \text{D}
4. When a convex glass lens is immersed in water, its focal length [1]aincreasesbdecreasescstays the samedbecomes negative
5. The magnification produced by a lens is [1]a-v/ubv/ucu/vdf/u
6. An object is 15\ \text{cm} from a convex lens of focal length 10\ \text{cm}. The image is formed at [1]a6\ \text{cm}b30\ \text{cm}c-30\ \text{cm}d25\ \text{cm}
7. The angular magnification of a simple microscope for an image at the near point is [1]aD/fb1 + D/fcD/f - 1df/D
8. A lens of focal length -50\ \text{cm} has a power of [1]a-2\ \text{D}b+2\ \text{D}c-0.02\ \text{D}d-50\ \text{D}

Answers — 2.17

1. (b) — one dioptre is one reciprocal metre of focal length.

2. (b) — the minus sign distinguishes it from the mirror formula.

3. (a) — 1/F = 1/20 - 1/30 = 1/60, so F = 60\ \text{cm} = 0.6\ \text{m} and P = 1/0.6 = +1.67\ \text{D}. Option (c) adds the powers as 5 + 3.33 with the wrong sign on the concave lens.

4. (a) — the lens maker's formula uses the relative index n_{glass}/n_{water} \approx 1.13, so (n-1) shrinks and f grows.

5. (b) — no minus sign here; a negative m still comes out automatically from the signs of u and v.

6. (b) — u = -15, f = +10: 1/v = 1/10 - 1/15 = 1/30, so v = +30\ \text{cm}, real and on the far side.

7. (b) — the +1 appears only for the near-point setting; D/f is the relaxed-eye value.

8. (a) — P = 1/(-0.5\ \text{m}) = -2\ \text{D}. Option (d) forgets to convert centimetres to metres.

2.18 Dispersion

High-yield fact	Value / form
Angular dispersion	\theta = \delta_V - \delta_R = (n_V - n_R)A
Dispersive power	\omega = \dfrac{n_V - n_R}{n_Y - 1} = \dfrac{\delta_V-\delta_R}{\delta_Y} — dimensionless, no unit
Cause of dispersion	n depends on wavelength; violet has the largest n, so it deviates most
Deviation without dispersion	\omega_1\delta_1 + \omega_2\delta_2 = 0
Achromatic doublet	\dfrac{\omega_1}{f_1} + \dfrac{\omega_2}{f_2} = 0
Chromatic aberration	coloured fringes; cured by an achromatic combination
Spherical aberration	blurring from marginal rays; reduced by stops or parabolic surfaces

1. Dispersive power has the unit [1]adioptrebradiancno unitdmetre
2. Which colour is deviated most by a glass prism? [1]aredbyellowcgreendviolet
3. Dispersion of white light by a prism occurs because [1]athe speed of light in glass depends on wavelengthbglass absorbs some colourscthe prism has two facesdlight is transverse
4. The condition for an achromatic combination of two thin lenses in contact is [1]a\omega_1 f_1 = \omega_2 f_2b\dfrac{\omega_1}{f_1} + \dfrac{\omega_2}{f_2} = 0cf_1 = f_2d\omega_1 = \omega_2
5. For a thin prism of angle 5^{\circ} with n_V = 1.52 and n_R = 1.50, the angular dispersion is [1]a0.02^{\circ}b0.1^{\circ}c1^{\circ}d2.5^{\circ}
6. The defect by which a lens forms coloured images of a white object is called [1]aspherical aberrationbchromatic aberrationcastigmatismdcoma
7. Spherical aberration in a lens can be reduced by [1]ausing a combination of two different glassesbusing a stop to block marginal rayscincreasing the aperturedusing monochromatic light
8. Dispersive power of a prism material depends on [1]athe angle of the prismbthe material of the prism onlycthe angle of incidencedthe size of the prism

Answers — 2.18

1. (c) — it is a ratio of two angles (or of two refractive-index differences).

2. (d) — violet has the shortest wavelength and the largest n.

3. (a) — a wavelength-dependent speed means a wavelength-dependent n, hence a wavelength-dependent deviation. Option (b) is the classic wrong picture: the prism does not create or filter the colours.

4. (b) — the two dispersions cancel while some net power survives.

5. (b) — \theta = (1.52-1.50)\times5^{\circ} = 0.02\times5 = 0.1^{\circ}. Option (a) drops the factor A.

6. (b) — colour is in the name.

7. (b) — spherical aberration is a geometry problem, not a colour problem, so an achromatic doublet (option a) is the answer to the other defect.

8. (b) — \omega contains only refractive indices; A cancels out of the ratio.

2.19 Electric Charges

High-yield fact	Value / form
Coulomb's law	F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^{2}}
Constant	k = \dfrac{1}{4\pi\varepsilon_0} = 9\times10^{9}\ \text{N m}^{2}\text{C}^{-2}
Permittivity of free space	\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^{2}\text{N}^{-1}\text{m}^{-2}
In a medium	F_{med} = \dfrac{F_{vac}}{\varepsilon_r}
Quantisation	q = ne, e = 1.6\times10^{-19}\ \text{C}
1\ \text{C}	6.25\times10^{18} electronic charges

1. The value of 1/4\pi\varepsilon_0 in SI units is [1]a9\times10^{9}b8.85\times10^{-12}c6.67\times10^{-11}d1.6\times10^{-19}
2. The force between two point charges is inversely proportional to [1]arbr^{2}cr^{3}d\sqrt{r}
3. If the distance between two charges is halved, the force between them becomes [1]ahalfbdoublecfour timesdone quarter
4. When a charged body is brought near an uncharged conductor, the near face of the conductor acquires [1]athe same kind of chargebthe opposite kind of chargecno chargedtwice the charge
5. Two charges of 1\ \mu\text{C} each are placed 1\ \text{m} apart in air. The force between them is [1]a9\times10^{-3}\ \text{N}b9\times10^{3}\ \text{N}c9\times10^{-9}\ \text{N}d9\ \text{N}
6. The charge on a body is always an integral multiple of e. This is called [1]aconservation of chargebquantisation of chargecadditivity of chargedinduction
7. Coulomb's law in a medium of relative permittivity \varepsilon_r gives a force that is [1]a\varepsilon_r times the vacuum valueb1/\varepsilon_r times the vacuum valuecunchangedd\varepsilon_r^{2} times the vacuum value
8. One coulomb of charge is equivalent to [1]a1.6\times10^{-19} electronsb6.25\times10^{18} electronsc6.022\times10^{23} electronsd9\times10^{9} electrons

Answers — 2.19

1. (a) — option (b) is \varepsilon_0 itself, which is what the 9\times10^{9} is built from.

2. (b) — the inverse-square law, like gravitation.

3. (c) — F \propto 1/r^2, so halving r multiplies F by 4. Option (b) treats the law as 1/r.

4. (b) — charging by induction; the far face takes the like charge.

5. (a) — F = 9\times10^{9}\times(10^{-6})^{2}/1^{2} = 9\times10^{-3}\ \text{N}. Option (b) drops the micro prefix.

6. (b) — no free charge smaller than e has ever been isolated.

7. (b) — the medium screens the field, so the force is always reduced.

8. (b) — 1/1.6\times10^{-19}.

2.20 Electric field

High-yield fact	Value / form
Field strength	\vec E = \vec F/q; unit \text{N C}^{-1} = \text{V m}^{-1}
Point charge	E = \dfrac{kQ}{r^{2}}
Gauss's law	\oint \vec E\cdot d\vec A = \dfrac{q_{enc}}{\varepsilon_0}; flux in \text{N m}^{2}\text{C}^{-1}
Inside a conductor	E = 0; all excess charge sits on the surface
Charged conducting surface	E = \dfrac{\sigma}{\varepsilon_0}; isolated non-conducting sheet E = \dfrac{\sigma}{2\varepsilon_0}
Infinite line charge	E = \dfrac{\lambda}{2\pi\varepsilon_0 r} \propto \dfrac1r
Field lines	start on +, end on -, never cross, never form closed loops in electrostatics

1. The SI unit of electric field strength is [1]a\text{N C}^{-1} onlyb\text{V m}^{-1} onlycboth \text{N C}^{-1} and \text{V m}^{-1}d\text{J C}^{-1}
2. The electric field inside a charged hollow conductor in equilibrium is [1]azerob\sigma/\varepsilon_0cmaximum at the centreduniform and non-zero
3. The electric field due to an infinitely long straight charged wire varies as [1]a1/rb1/r^{2}c1/r^{3}dindependent of r
4. The total electric flux through a closed surface enclosing a charge q is [1]aq\varepsilon_0bq/\varepsilon_0c\varepsilon_0/qdq/4\pi\varepsilon_0
5. Two electric lines of force can never intersect because [1]athey are parallelbthe field would then have two directions at one pointcthey repeldthey are imaginary
6. The electric field at 3\ \text{m} from a point charge of 1\ \mu\text{C} is [1]a1000\ \text{N C}^{-1}b3000\ \text{N C}^{-1}c9000\ \text{N C}^{-1}d300\ \text{N C}^{-1}
7. The SI unit of electric flux is [1]a\text{N C}^{-1}b\text{N m}^{2}\text{C}^{-1}c\text{C m}^{-2}d\text{V}
8. Just outside the surface of a charged conductor with surface charge density \sigma, the field is [1]a\sigma/\varepsilon_0b\sigma/2\varepsilon_0czerod\sigma\varepsilon_0

Answers — 2.20

1. (c) — \text{N C}^{-1} and \text{V m}^{-1} are the same combination of base units. Option (d) is the unit of potential.

2. (a) — free charges rearrange until the interior field vanishes; this is electrostatic shielding.

3. (a) — a cylindrical Gaussian surface has area \propto r, so E \propto 1/r; only a point charge gives 1/r^2.

4. (b) — Gauss's law, and it does not depend on the shape of the surface or where the charge sits inside it.

5. (b) — the tangent at a point gives the unique direction of \vec E there.

6. (a) — E = 9\times10^{9}\times10^{-6}/9 = 1000\ \text{N C}^{-1}.

7. (b) — field strength times area.

8. (a) — the factor of two that appears for a thin non-conducting sheet is absent here, because all the charge lies on one face and there is no field inside the metal.

2.21 Potential, potential difference and potential energy

High-yield fact	Value / form
Potential	V = W/q, a scalar; unit volt = \text{J C}^{-1}
Point charge	V = \dfrac{kQ}{r} — note 1/r, not 1/r^{2}
Field from potential	E = -\dfrac{dV}{dx}
Potential energy of a pair	U = \dfrac{kq_1q_2}{r}
Electronvolt	1\ \text{eV} = 1.6\times10^{-19}\ \text{J}
Equipotential surface	V constant; no work to move a charge along it; always perpendicular to \vec E
Charged conducting sphere	inside, V = V_{surface} = kQ/R and E = 0

1. Electric potential is a [1]avector quantitybscalar quantityctensorddimensionless number
2. The potential at a distance r from a point charge Q varies as [1]a1/rb1/r^{2}crdr^{2}
3. One electronvolt equals [1]a1.6\times10^{-19}\ \text{J}b1\ \text{J}c1.6\times10^{19}\ \text{J}d9\times10^{9}\ \text{J}
4. The relation between the electric field and the potential gradient is [1]aE = +dV/dxbE = -dV/dxcE = V xdE = V/x^{2}
5. The work done in moving a charge from one point to another on an equipotential surface is [1]amaximumbzerocqVdinfinite
6. Inside a charged hollow conducting sphere, the potential is [1]azerobequal to the value on the surfacecinfinitedgreater than on the surface
7. The potential at 0.3\ \text{m} from a charge of 3\ \text{nC} is [1]a9\ \text{V}b90\ \text{V}c900\ \text{V}d0.9\ \text{V}
8. A potential difference of 200\ \text{V} is applied across two plates 2\ \text{mm} apart. The field between them is [1]a100\ \text{V m}^{-1}b1\times10^{4}\ \text{V m}^{-1}c1\times10^{5}\ \text{V m}^{-1}d400\ \text{V m}^{-1}

Answers — 2.21

1. (b) — which is why potentials add algebraically while fields must be added as vectors. That single fact makes most potential problems easier than the matching field problem.

2. (a) — one power of r less than the field, because potential is the integral of the field.

3. (a) — the energy an electron gains falling through one volt.

4. (b) — the field points from high to low potential, hence the minus sign.

5. (b) — V does not change, so W = q\Delta V = 0.

6. (b) — E = 0 inside means V cannot change, so it holds the surface value. Option (a) confuses zero field with zero potential.

7. (b) — V = 9\times10^{9}\times3\times10^{-9}/0.3 = 27/0.3 = 90\ \text{V}.

8. (c) — E = V/d = 200/0.002 = 1\times10^{5}\ \text{V m}^{-1}. Option (a) forgets to convert millimetres.

2.22 Capacitor

High-yield fact	Value / form
Capacitance	C = Q/V; unit farad = \text{C V}^{-1}
Parallel-plate	C = \dfrac{\varepsilon_0 A}{d}, with dielectric C = \dfrac{\varepsilon_0\varepsilon_r A}{d}
In series	\dfrac{1}{C} = \dfrac{1}{C_1}+\dfrac{1}{C_2}+\dots — total is less than the smallest
In parallel	C = C_1+C_2+\dots
Energy stored	U = \tfrac12CV^{2} = \tfrac12QV = \dfrac{Q^{2}}{2C}
Dielectric inserted, battery connected	V fixed, C up, so Q and U both rise
Dielectric inserted, battery removed	Q fixed, C up, so V and U both fall
\varepsilon_r of vacuum	exactly 1

1. The SI unit of capacitance is the [1]acoulombbvoltcfaraddhenry
2. The capacitance of a parallel-plate capacitor is [1]a\varepsilon_0 A db\varepsilon_0 A/dc\varepsilon_0 d/AdA/\varepsilon_0 d
3. Three capacitors of 3\ \mu\text{F} each are joined in series. The equivalent capacitance is [1]a9\ \mu\text{F}b3\ \mu\text{F}c1\ \mu\text{F}d0.33\ \mu\text{F}
4. The energy stored in a capacitor of capacitance C charged to a potential difference V is [1]aCVb\tfrac12CVc\tfrac12CV^{2}dCV^{2}
5. A 2\ \mu\text{F} capacitor is charged to 100\ \text{V}. The energy stored is [1]a0.01\ \text{J}b0.02\ \text{J}c1\ \text{J}d200\ \text{J}
6. A dielectric slab is inserted into a parallel-plate capacitor that stays connected to a battery. The charge on the plates [1]aincreasesbdecreasescstays the samedbecomes zero
7. The relative permittivity of free space is [1]a0b1c8.85\times10^{-12}d9\times10^{9}
8. Capacitors are connected in parallel to obtain [1]aa smaller capacitanceba larger capacitanceca larger working voltageda smaller charge

Answers — 2.22

1. (c) — one farad is one coulomb per volt, an enormous unit, which is why real capacitors are in μF and pF.

2. (b) — bigger plates and a smaller gap both raise C.

3. (c) — 1/C = 3/3, so C = 1\ \mu\text{F}. Option (a) applies the parallel rule; capacitor combination rules are the exact mirror image of the resistor rules, and that inversion is the commonest error in this unit.

4. (c) — the \tfrac12 comes from integrating V\,dq as the capacitor charges up.

5. (a) — U = \tfrac12\times2\times10^{-6}\times100^{2} = 0.01\ \text{J}.

6. (a) — V is held fixed by the battery and C has increased, so Q = CV must rise. With the battery removed the answer would be (c).

7. (b) — by definition; 8.85\times10^{-12} is the absolute permittivity \varepsilon_0.

8. (b) — parallel adds plate area in effect. Series is used to share a high voltage.

2.23 DC Circuits

High-yield fact	Value / form
Ohm's law	V = IR; resistivity \rho = RA/L, unit \Omega\ \text{m}
Drift	I = nAv_de; drift speed is only about 10^{-4}\ \text{m s}^{-1}
Series / parallel	R_s = R_1+R_2; \dfrac{1}{R_p} = \dfrac{1}{R_1}+\dfrac{1}{R_2}
Internal resistance	V = E - Ir; E = I(R+r)
Power	P = VI = I^{2}R = \dfrac{V^{2}}{R}
Wheatstone bridge, balanced	\dfrac{P}{Q} = \dfrac{R}{S}; no current through the galvanometer
Metre bridge	\dfrac{R}{S} = \dfrac{l}{100-l}

1. The SI unit of resistivity is [1]a\Omegab\Omega\ \text{m}c\Omega\ \text{m}^{-1}d\Omega^{-1}\text{m}^{-1}
2. The drift velocity of free electrons in a metallic conductor carrying a normal current is of the order of [1]a10^{8}\ \text{m s}^{-1}b10^{6}\ \text{m s}^{-1}c10^{-4}\ \text{m s}^{-1}d10^{-10}\ \text{m s}^{-1}
3. Three resistors of 6\ \Omega each are joined in parallel. The equivalent resistance is [1]a18\ \Omegab6\ \Omegac3\ \Omegad2\ \Omega
4. The terminal potential difference of a cell of emf E and internal resistance r delivering a current I is [1]aE + IrbE - IrcIr - EdE/Ir
5. The balance condition of a Wheatstone bridge with arms P, Q, R, S is [1]aPQ = RSbP/Q = R/ScP/R = S/QdP + Q = R + S
6. In the network shown, the resistance between A and B is [1]a2\ \Omegab4\ \Omegac9\ \Omegad11\ \Omega
7. Two bulbs are rated 100\ \text{W} and 60\ \text{W} at the same voltage. Compared with the 60\ \text{W} bulb, the 100\ \text{W} bulb has [1]aa higher resistanceba lower resistancecthe same resistancedzero resistance
8. The current in a conductor is given by [1]aI = nAv_debI = nAv_d/ecI = nA/v_dedI = Av_de/n

Answers — 2.23

1. (b) — from \rho = RA/L, the units are \Omega\times\text{m}^2/\text{m}.

2. (c) — slower than a walking ant; the signal travels near the speed of light, which is what option (a) is fishing for.

3. (d) — 6/3 = 2\ \Omega; n equal resistors in parallel give R/n.

4. (b) — the cell loses Ir inside itself while discharging.

5. (b) — the ratio of adjacent arms on the two sides.

6. (b) — 3\ \Omega in parallel with 6\ \Omega is 18/9 = 2\ \Omega; add the series 2\ \Omega to get 4\ \Omega. Option (d) adds all three in series.

7. (b) — P = V^2/R at fixed V, so more power means less resistance. Students reliably answer (a) because "brighter feels like more resistance".

8. (a) — number density times area times drift speed times charge per carrier.

2.24 Nuclear physics

High-yield fact	Value / form
Nuclear radius	R = R_0A^{1/3}, R_0 = 1.2\times10^{-15}\ \text{m}
Nuclear density	about 2.3\times10^{17}\ \text{kg m}^{-3}, the same for all nuclei
Mass–energy	1\ \text{u} = 931.5\ \text{MeV} = 1.66\times10^{-27}\ \text{kg}
Mass defect	\Delta m = [Zm_p + (A-Z)m_n] - M_{nucleus}
Binding energy per nucleon	peaks near 8.8\ \text{MeV} at A \approx 56 (iron)
Fission of U-235	about 200\ \text{MeV} released per fission
Pair production	needs at least 1.02\ \text{MeV}
Isotopes / isobars / isotones	same Z / same A / same N

1. The radius of a nucleus of mass number A is proportional to [1]aAbA^{1/3}cA^{2/3}dA^{2}
2. Nuclear density is [1]aproportional to Abproportional to A^{1/3}cnearly the same for all nucleidinversely proportional to A
3. The energy equivalent of one atomic mass unit is [1]a931.5\ \text{MeV}b1.6\times10^{-19}\ \text{J}c13.6\ \text{eV}d200\ \text{MeV}
4. The difference between the sum of the masses of the free nucleons and the mass of the nucleus is called the [1]abinding energybmass defectcpacking fractiondatomic mass
5. The binding energy per nucleon is maximum for nuclei of mass number about [1]a2b56c138d235
6. The energy released in the fission of one uranium-235 nucleus is about [1]a2\ \text{MeV}b20\ \text{MeV}c200\ \text{MeV}d2000\ \text{MeV}
7. Isotopes of an element have the same [1]amass numberbatomic numbercnumber of neutronsdnuclear radius
8. The minimum photon energy needed for electron–positron pair production is [1]a0.51\ \text{MeV}b1.02\ \text{MeV}c931.5\ \text{MeV}d13.6\ \text{eV}

Answers — 2.24

1. (b) — volume is proportional to A, so radius goes as the cube root.

2. (c) — mass \propto A and volume \propto A, so the ratio is constant; that constancy is the evidence that nuclear matter is incompressible.

3. (a) — from E = mc^{2} with m = 1.66\times10^{-27}\ \text{kg}.

4. (b) — the binding energy is that defect multiplied by c^{2}; options (a) and (b) are the classic swap, and marks are lost every year for naming the energy when the question asked for the mass.

5. (b) — iron and nickel, which is why both fusion below and fission above release energy.

6. (c) — the number worth memorising for reactor numericals.

7. (b) — same Z, different A.

8. (b) — twice the rest energy of an electron, 2\times0.51\ \text{MeV}. Option (a) forgets that two particles must be made.

2.25 Solids

High-yield fact	Value / form
Band gap	conductor: bands overlap; semiconductor E_g \approx 1\ \text{eV}; insulator E_g > 3\ \text{eV}
Silicon and germanium	E_g = 1.1\ \text{eV} and 0.7\ \text{eV}
n-type	pentavalent dopant (P, As, Sb); majority carriers electrons
p-type	trivalent dopant (B, Al, In); majority carriers holes
Mass-action law	n_e n_h = n_i^{2}
Temperature coefficient	semiconductors are negative; metals positive
Doped crystal	electrically neutral overall

1. The forbidden energy gap of silicon at room temperature is about [1]a0.7\ \text{eV}b1.1\ \text{eV}c3.2\ \text{eV}d5.4\ \text{eV}
2. Germanium is doped with a pentavalent impurity. The resulting semiconductor is [1]ap-typebn-typecintrinsicdan insulator
3. The resistance of a pure semiconductor with rising temperature [1]aincreasesbdecreasescstays constantdfirst rises then falls
4. In an extrinsic semiconductor at equilibrium [1]an_e = n_hbn_en_h = n_i^{2}cn_e + n_h = n_idn_e = n_i^{2}
5. A material with a forbidden gap greater than about 3\ \text{eV} behaves as [1]aa conductorba semiconductorcan insulatorda superconductor
6. An n-type semiconductor crystal is [1]anegatively chargedbpositively chargedcelectrically neutraldcharged only when heated
7. In the band diagram of a conductor [1]athe valence and conduction bands overlapbthe gap is about 1\ \text{eV}cthe gap is greater than 3\ \text{eV}dthere is no valence band
8. The majority charge carriers in a p-type semiconductor are [1]aelectronsbholescprotonsdnegative ions

Answers — 2.25

1. (b) — option (a) is germanium, the standard mix-up.

2. (b) — the fifth electron is loosely bound and becomes a free carrier; "n" is for the negative carrier, not for a negative crystal.

3. (b) — more electrons are thermally excited across the gap; semiconductors have a negative temperature coefficient of resistance, unlike metals.

4. (b) — the mass-action law holds for both types; option (a) is true only for an intrinsic sample.

5. (c) — see the figure: the gap is too wide for thermal excitation at room temperature.

6. (c) — each donor atom contributes one electron and one fixed positive ion core, so the crystal stays neutral. Option (a) is the most popular wrong answer in this unit.

7. (a) — there is no gap at all, so electrons move under the smallest field.

8. (b) — a trivalent dopant leaves a vacancy in the bonding, which behaves as a mobile positive carrier.

2.26 Recent trends in physics

High-yield fact	Value / form
Quark content	proton = uud, neutron = udd
Quark charges	up, charm, top = +\tfrac23e; down, strange, bottom = -\tfrac13e
Baryon / meson	three quarks / one quark and one antiquark
Leptons	electron, muon, tau and their neutrinos
Hubble's law	v = H_0d
Age of the universe	about 13.8 billion years
Cosmic microwave background	about 2.7\ \text{K}
Schwarzschild radius	R_s = \dfrac{2GM}{c^{2}}

1. The quark composition of a proton is [1]auudbuddcuuudddd
2. The charge on an up quark is [1]a+\tfrac13eb-\tfrac13ec+\tfrac23ed-\tfrac23e
3. A baryon is made of [1]atwo quarksbthree quarksca quark and an antiquarkdno quarks
4. Hubble's law states that the recession speed of a galaxy is [1]aindependent of distancebproportional to its distancecinversely proportional to its distancedproportional to its mass
5. The temperature of the cosmic microwave background radiation is about [1]a0\ \text{K}b2.7\ \text{K}c300\ \text{K}d5800\ \text{K}
6. The present estimate of the age of the universe is about [1]a4.5 billion yearsb13.8 billion yearsc100 billion yearsd1 million years
7. The Schwarzschild radius of a body of mass M is [1]aGM/c^{2}b2GM/c^{2}cGM/2c^{2}d2GM/c
8. The neutrino belongs to the family of [1]abaryonsbmesonscleptonsdgauge bosons

Answers — 2.26

1. (a) — \tfrac23+\tfrac23-\tfrac13 = +1; option (b) is the neutron, whose charges sum to zero.

2. (c) — the two-thirds family.

3. (b) — protons and neutrons are the familiar examples; option (c) defines a meson.

4. (b) — the linear law that is the observational basis of the expanding universe.

5. (b) — the cooled relic of the hot early universe.

6. (b) — option (a) is the age of the Earth, offered to catch a hurried reader.

7. (b) — set the escape velocity equal to c in v_e = \sqrt{2GM/R}.

8. (c) — it feels only the weak and gravitational interactions and has no quark content.

Estimating your way to an optics answer

Question 6 of section 2.17 — a 10\ \text{cm} convex lens with an object at 15\ \text{cm} — has options 6, 30, -30 and 25\ \text{cm}. You can eliminate three of them without touching the lens formula.

The object is between f and 2f, so the standard result applies at once: the image is real, inverted, magnified and lies beyond 2f, that is beyond 20\ \text{cm}. That kills 6\ \text{cm} (too near), kills -30\ \text{cm} (real means positive v for a lens) and leaves 30\ \text{cm} against 25\ \text{cm}. Now one line of arithmetic settles it: 1/v = 1/10 - 1/15 = 1/30.

Option 6\ \text{cm} is not an accident either — it is the answer to 1/v = 1/10 + 1/15, i.e. the mirror formula applied to a lens. The examiner put it there for exactly the student who has the two formulas the wrong way round.

Do not learn the answer letters

Reorder the options and every letter in this chapter changes. What you are meant to carry out of a section is the fact in the high-yield table and the reason in the answer key, not "2.14 question 3 is (b)". If you find yourself recognising a question by its position on the page rather than by its physics, shuffle your revision: work the sections in reverse order, or cover the options and answer from memory before you look.

Chapter summary

• Group A of the NEB Physics paper is eleven one-mark objective items drawn from across all twenty-six units, so no unit is safe to skip; the whole of this chapter is built one section per syllabus unit for exactly that reason.
• Every section opens with a high-yield facts table. Those tables are the smallest thing you can revise that still covers the objective paper: constants, defining formulae, standard values and the one or two named results that MCQs keep returning to.
• The answer keys give a reason, not a letter. Where an option encodes a standard misconception — mass for weight, the mirror formula for the lens formula, the resistor rule for the capacitor rule, zero field for zero potential, mass defect for binding energy — the key names it, because recognising your own error is worth more than recognising the right answer.
• Recurring traps to carry into the hall: unit prefixes (μ, m, k) left unconverted; \text{kW h} treated as a power; c_{rms} \propto \sqrt{T} read as linear; the \tfrac12 in capacitor energy; the sign convention in mirror and lens problems; \tfrac32kT per molecule versus \tfrac32RT per mole; weightlessness explained as "no gravity".
• Objective marks are the cheapest marks in the paper and the easiest to throw away. One minute per item is the budget; if an item resists for longer, mark it, move on, and come back after Group C.

Practice questions

Group A

1. The dimensional formula of surface tension is [1]a[MT^{-2}]b[MLT^{-2}]c[ML^{-1}T^{-2}]d[ML^{2}T^{-2}]
2. A body thrown vertically upwards has, at the highest point, [1]azero velocity and zero accelerationbzero velocity and acceleration g downwardscmaximum velocitydacceleration zero and velocity g
3. The work done in moving a charge along an equipotential surface is [1]aqVbzerocV/qdinfinite
4. The escape velocity of a body from the Earth's surface is \sqrt{2} times its [1]aorbital velocity near the surfacebterminal velocitycrms speedddrift velocity
5. Two resistors of 4\ \Omega and 12\ \Omega are joined in parallel. The equivalent resistance is [1]a16\ \Omegab8\ \Omegac3\ \Omegad48\ \Omega
6. The focal length of a plane mirror is [1]azerobinfinitecequal to the object distancedR/2 with R finite
7. The specific latent heat of vaporisation of water is about [1]a3.34\times10^{5}\ \text{J kg}^{-1}b2.26\times10^{6}\ \text{J kg}^{-1}c4200\ \text{J kg}^{-1}d1\times10^{3}\ \text{J kg}^{-1}
8. Which of the following is a scalar? [1]aelectric fieldbelectric potentialctorquedimpulse
9. The energy equivalent of a mass defect of 1\ \text{u} is [1]a931.5\ \text{MeV}b200\ \text{MeV}c1.02\ \text{MeV}d13.6\ \text{eV}
10. If the temperature of a black body is halved, the energy it radiates per second becomes [1]a1/2b1/4c1/8d1/16
11. A p-type semiconductor is obtained by doping silicon with [1]aphosphorusbarseniccborondantimony

Answers to Group A

1. (a) — surface tension is force per unit length, [MLT^{-2}]/[L].

2. (b) — the velocity is instantaneously zero but gravity never switches off.

3. (b) — \Delta V = 0 on an equipotential.

4. (a) — v_e = \sqrt{2}\,v_o.

5. (c) — 4\times12/16 = 3\ \Omega; the parallel value is always smaller than the smaller resistor.

6. (b) — a plane mirror is a spherical mirror of infinite radius, so f = R/2 \to \infty.

7. (b) — option (a) is the latent heat of fusion.

8. (b) — potential is energy per unit charge, with no direction.

9. (a) — the standard conversion.

10. (d) — E \propto T^{4}, so halving T gives (1/2)^{4}.

11. (c) — boron is trivalent; the other three are pentavalent and would give n-type.

Group B

1. State the sign convention used for spherical mirrors and use it to explain why the image in a convex mirror is always virtual, erect and diminished. Support your answer with a ray diagram. [5]
2. Distinguish between the real and the apparent expansivity of a liquid, derive the relation between them and explain why Dulong and Petit's method needs no correction for the vessel. [5]
3. Define electric field strength and electric potential. Show that E = -dV/dx, and explain why the potential inside a charged hollow conductor equals the surface value. [5]
4. Explain, using the band theory of solids, the difference between a conductor, a semiconductor and an insulator. Why does the resistance of a semiconductor fall with rising temperature while that of a metal rises? [5]

Group C

1. Derive an expression for the pressure exerted by an ideal gas on the basis of kinetic theory, hence show that the mean translational kinetic energy of a molecule is \tfrac32kT. A gas at 300\ \text{K} has a molar mass of 32\ \text{g mol}^{-1}; calculate the rms speed of its molecules. [8]
2. State Coulomb's law and define the farad. Derive an expression for the capacitance of a parallel-plate capacitor with a dielectric slab of thickness t and relative permittivity \varepsilon_r between the plates. A 2\ \mu\text{F} capacitor is charged to 200\ \text{V} and then connected across an uncharged 3\ \mu\text{F} capacitor; find the common potential difference and the energy lost. [8]