A Level Physics (9702) — MCQs & Practice Questions

A base unit is one of the seven independent SI units, such as the metre. A derived unit is built from base units by multiplication or division, such as the newton, 1 N = 1 kg m s⁻².

The smallest measurement the instrument can read — the value of one division on its scale. A metre rule has a least count of 1 mm; a vernier calliper, 0.1 mm; a screw gauge, 0.01 mm.

Random errors scatter either side of the true value, so averaging makes them partly cancel. A single reading may happen to be one of the extreme ones.

1. Convert: 12.5 cm = 0.125 m, 0.1 cm = 0.001 mdivide by 100[1]
2. Percentage uncertainty = (Δl / l) × 100 = (0.1 / 12.5) × 100same units top and bottom, so the conversion cancels[1]
3. = 0.8%[1]

0.125 ± 0.001 m, or 0.8%

1. 0.51 mm is anomalous — it lies well away from the others[1]
2. It should be excluded from the mean, but reported rather than deletedthe reporting half is the mark most often missed[1]
3. Mean = (0.42 + 0.43 + 0.42 + 0.43) / 4four readings, not five[1]
4. = 0.425 ≈ 0.43 mm[1]

(a) 0.51 mm, excluded but reported (b) 0.43 mm

A scalar has magnitude only, for example mass. A vector has both magnitude and direction, for example velocity.

Distance is the total path length, a scalar that only ever grows. Displacement is the straight line from start to finish, so returning to the starting point makes it zero.

The single force that has the same effect on the body as the two forces acting together.

1. Recognises the forces are perpendicular, so Pythagoras applies[1]
2. R = √(8.0² + 6.0²)[1]
3. R = 10.0 N[1]
4. Uses tan θ = 6.0 / 8.0[1]
5. θ = 36.9° north of easta direction with no reference line scores nothing[1]

10.0 N at 36.9° north of east

1. Replacing one vector by two components at right angles to each other[1]
2. Which together have the same effect as the original vector[1]
3. Uses W sin θ for the component along the slope[1]
4. = 250 × sin 20°[1]
5. = 85.5 N[1]
6. Uses W cos θ = 250 × cos 20°[1]
7. = 235 N[1]
8. The normal contact force from the surface of the slope[1]

(b) 85.5 N (c) 235 N

1. The current does not affect the crossing time: t = 30 / 1.2perpendicular components are independent[1]
2. t = 25 s[1]
3. Uses distance = 0.90 × 25[1]
4. = 22.5 m downstream[1]
5. Uses √(1.2² + 0.90²)[1]
6. = 1.5 m s⁻¹[1]

(a) 25 s (b) 22.5 m (c) 1.5 m s⁻¹

1. Selects v² = u² + 2asthe equation without t, since t is not given in part (a)[1]
2. 10.0² = 4.0² + 2 × a × 42 → 100 = 16 + 84a → a = 1.0 m s⁻²unit required for the mark[1]
3. Selects v = u + at (or s = ½(u+v)t)[1]
4. 10.0 = 4.0 + 1.0t → t = 6.0 s[1]

(a) 1.0 m s⁻² (b) 6.0 s

1. Uses s = ut + ½at² with u = 0"dropped from rest" is what tells you u = 0[1]
2. s = ½ × 9.81 × 3.2² = 50.2 ≈ 50 m[1]
3. Uses v = u + ator v² = u² + 2as[1]
4. v = 9.81 × 3.2 = 31.4 ≈ 31 m s⁻¹[1]
5. The actual speed would be lower / the stone would reach terminal velocitya statement about direction of change is enough[1]

(a) 50 m (b) 31 m s⁻¹ (c) the speed would be less

1. Acceleration = gradient = (20 − 0) / 40gradient of a velocity–time graph is acceleration[1]
2. = 0.50 m s⁻²[1]
3. Recognises distance = area under the graphthis is the mark most often missed[1]
4. Triangle ½ × 40 × 20 = 400; rectangle 60 × 20 = 1200; triangle ½ × 30 × 20 = 300all three areas needed[1]
5. Total = 1900 m[1]
6. Average speed = 1900 / 130 = 14.6 ≈ 15 m s⁻¹total distance ÷ total time, not the mean of the velocities[1]

(a) 0.50 m s⁻² (b) 1900 m (c) 15 m s⁻¹

1. Uses vertical motion with u_y = 0: 45 = ½ × 10 × t²"thrown horizontally" means the initial vertical velocity is zero[1]
2. t² = 9.0, t = 3.0 s[1]
3. Uses s_x = u_x t with constant horizontal velocity[1]
4. s_x = 15 × 3.0 = 45 m[1]
5. Horizontal and vertical motion are independent / gravity acts only vertically, so the vertical motion is unaffected by the horizontal velocity[1]

(a) 3.0 s (b) 45 m (c) the two components are independent

1. Displacement is the straight-line distance from the starting point to the finishing point[1]
2. …together with its direction / it is a vector quantitythe direction is required for the second mark[1]
3. Any curved or non-straight path, e.g. a runner going round a bend, a car following a winding roada full circular lap, where displacement is zero, also earns this[1]

Distance is the total length of the path travelled, a scalar. Displacement is the straight line from start to finish together with its direction, a vector.

Yes. A ball thrown vertically upward is momentarily at rest at the top of its flight, but gravity still acts, so its acceleration is g downward.

1. Uses a = (v − u)/t = (0 − 25)/5.0[1]
2. a = −5.0 m s⁻², a deceleration of 5.0 m s⁻²the negative sign or the word deceleration, not both required[1]
3. Uses s = ½(u + v)t or v² = u² + 2as[1]
4. s = ½(25 + 0) × 5.0 = 62.5 m[1]

deceleration 5.0 m s⁻², distance 62.5 m

For every action there is an equal and opposite reaction. The two forces are of the same type and act on different bodies.

By Newton's first law the passenger continues moving at the same velocity because no resultant force acts on them; the bus decelerates beneath them, so they move forward relative to it.

The product of mass and velocity, p = mv. A vector quantity. SI unit: kg m s⁻¹.

1. States conservation of momentum: total before = total after[1]
2. Before: p = 2.0 × 3.0 + 4.0 × 0 = 6.0 kg m s⁻¹[1]
3. After: combined mass = 6.0 kg, so 6.0 = 6.0 × vthey stick together, so they share one velocity[1]
4. v = 1.0 m s⁻¹ in the original directiondirection expected for full marks[1]

1.0 m s⁻¹ in the direction of the original motion

1. Resultant force = 15 − 6.0 = 9.0 Nfriction opposes, so it subtracts[1]
2. Uses F = ma[1]
3. a = F/m = 9.0 / 3.0[1]
4. a = 3.0 m s⁻²unit required[1]

3.0 m s⁻²

1. Uses p = mv[1]
2. p = 1200 × 15 = 18 000 kg m s⁻¹the stationary car contributes nothing[1]
3. Applies conservation: total after = 18 000 kg m s⁻¹[1]
4. Combined mass 2000 kg, so v = 18 000 / 2000 = 9.0 m s⁻¹[1]
5. Kinetic energy is not conserved[1]
6. Inelastic collisionobjects sticking together is always inelastic[1]

(a) 1.8 × 10⁴ kg m s⁻¹ (b) 9.0 m s⁻¹ (c) not conserved; inelastic

1. Total momentum before is zeronothing is moving[1]
2. 0 = 0.020 × 400 + 4.0 × v[1]
3. v = −8.0 / 4.0 = −2.0 m s⁻¹, i.e. 2.0 m s⁻¹ backwardsdirection required[1]
4. The rifle's momentum must equal the bullet's in size and be opposite in direction[1]
5. Since p = mv is fixed, a larger m gives a smaller v[1]

2.0 m s⁻¹ backwards

1. Takes the initial direction as positive, so the rebound velocity is −8.0 m s⁻¹the sign is the whole question[1]
2. Δp = m(v − u) = 0.15 × (−8.0 − 12) = −3.0 kg m s⁻¹a change of 20 m s⁻¹, not 4[1]
3. Uses F = Δp/Δt[1]
4. F = −3.0 / 0.050 = −60 N, i.e. 60 N away from the wall[1]

60 N, directed away from the wall

The total momentum of a system before an interaction equals the total momentum after it, provided no resultant external force acts on the system.

It increases the time over which the ball is brought to rest. Since F = Δp/Δt and the change in momentum is fixed, a longer time means a smaller force on the hands.

1. Uses Δp = m(v − u) with u = 0the ball starts at rest on the tee[1]
2. Δp = 0.045 × 60 = 2.7 kg m s⁻¹[1]
3. Converts the time: 0.50 ms = 5.0 × 10⁻⁴ s[1]
4. Uses F = Δp/Δt[1]
5. F = 2.7 / 5.0 × 10⁻⁴ = 5400 N[1]
6. Applies conservation of momentum to club and ball together[1]
7. Before: 0.30 × 70 = 21 kg m s⁻¹the ball contributes nothing[1]
8. After: 21 = 0.30 × v + 2.7[1]
9. v = 18.3 / 0.30 = 61 m s⁻¹[1]

(a) 2.7 kg m s⁻¹ (b) 5400 N (c) 61 m s⁻¹

1. At the centre of gravity, which for a uniform beam is its midpoint, 1.0 m from the left end"uniform" is the word that tells you this[1]
2. Takes moments about the pivot; the beam's weight acts 1.0 − 0.50 = 0.50 m to the right of it[1]
3. Clockwise = anticlockwise: 40 × 0.50 = W × 0.50the load acts 0.50 m to the left of the pivot[1]
4. W = 40 N[1]

(a) at the midpoint, 1.0 m from the left end (b) W = 40 N

1. Passengers upstairs raise the centre of gravity of the bus[1]
2. A higher centre of gravity means a smaller tilt is needed before the vertical line through it falls outside the base / wheelbasethis is the key reasoning mark[1]
3. So the bus topples at a smaller angle / is less stable[1]

1. Recognises that only the perpendicular component turns the nutor equivalently uses the perpendicular distance[1]
2. Perpendicular component = 12 sin 30° = 6.0 N[1]
3. Moment = 6.0 × 0.25 = 1.5 N munit required[1]

1.5 N m

The turning effect of a force about a pivot, equal to force × perpendicular distance from the pivot to the line of action. SI unit: the newton metre, N m.

They are different quantities. Work is force acting along a displacement; a moment is force acting across a distance from a pivot. Sharing base units does not make them the same thing.

ΣF = 0 — no resultant force, so it does not accelerate. Σ moments = 0 — no resultant turning effect, so it does not rotate.

1. Left distance = 50 − 20 = 30 cm = 0.30 mdistance from the pivot, not from the end of the rule[1]
2. Anticlockwise moment = 3.0 × 0.30 = 0.90 N m[1]
3. Right distance = 70 − 50 = 20 cm = 0.20 m; for balance W₂ × 0.20 = 0.90principle of moments[1]
4. W₂ = 4.5 Nunit required[1]

4.5 N

A scalar has magnitude only, for example mass. A vector has both magnitude and direction, for example velocity.

Distance is the total path length, a scalar that only ever grows. Displacement is the straight line from start to finish, so returning to the starting point makes it zero.

The single force that has the same effect on the body as the two forces acting together.

1. Recognises the forces are perpendicular, so Pythagoras applies[1]
2. R = √(8.0² + 6.0²)[1]
3. R = 10.0 N[1]
4. Uses tan θ = 6.0 / 8.0[1]
5. θ = 36.9° north of easta direction with no reference line scores nothing[1]

10.0 N at 36.9° north of east

1. Replacing one vector by two components at right angles to each other[1]
2. Which together have the same effect as the original vector[1]
3. Uses W sin θ for the component along the slope[1]
4. = 250 × sin 20°[1]
5. = 85.5 N[1]
6. Uses W cos θ = 250 × cos 20°[1]
7. = 235 N[1]
8. The normal contact force from the surface of the slope[1]

(b) 85.5 N (c) 235 N

1. The current does not affect the crossing time: t = 30 / 1.2perpendicular components are independent[1]
2. t = 25 s[1]
3. Uses distance = 0.90 × 25[1]
4. = 22.5 m downstream[1]
5. Uses √(1.2² + 0.90²)[1]
6. = 1.5 m s⁻¹[1]

(a) 25 s (b) 22.5 m (c) 1.5 m s⁻¹

1. Volume = 0.20 × 0.10 × 0.050 = 1.0 × 10⁻³ m³converting every length to metres first[1]
2. Density = 0.80 / 1.0 × 10⁻³ = 800 kg m⁻³[1]
3. Weight = mg = 0.80 × 10 = 8.0 Npressure needs force, and the force here is the weight[1]
4. Largest face area = 0.20 × 0.10 = 0.020 m²largest face gives the lowest pressure[1]
5. Pressure = 8.0 / 0.020 = 400 Pa[1]

(a) 800 kg m⁻³ (b) 400 Pa

1. Uses p = ρgh[1]
2. p = 1000 × 10 × 12 = 1.2 × 10⁵ Paunit required[1]
3. No effect / the pressure is unchanged[1]
4. Pressure in a liquid depends only on depth, density and g — not on the shape or width of the containerthe reasoning mark; the statement alone scores 1 of 2[1]

(a) 1.2 × 10⁵ Pa (b) no change — pressure depends only on depth

1. Extension = 12.0 − 8.0 = 4.0 cm = 0.040 mextension, not total length — the mark most often lost on this topic[1]
2. k = F/x = 5.0 / 0.040 = 125 N m⁻¹[1]
3. New extension = 8.0 / 125 = 0.064 m = 6.4 cm[1]
4. New length = 8.0 + 6.4 = 14.4 cmadding the natural length back on[1]

(a) 125 N m⁻¹ (b) 14.4 cm

Mass per unit volume, ρ = m/V. SI unit: kg m⁻³.

Broad feet spread the camel's weight over a larger area, so the pressure on the sand is smaller and it does not sink.

The extension of a spring is directly proportional to the load applied, provided the limit of proportionality is not exceeded.

1. Uses p = ρgh[1]
2. p = 800 × 10 × 1.5 = 12 000 Paunit required[1]
3. Rearranges p = F/A to F = pA[1]
4. F = 12 000 × 2.0 = 24 000 N[1]

p = 1.2 × 10⁴ Pa, F = 2.4 × 10⁴ N

Work = force × distance moved in the direction of the force, W = Fs cos θ. It is zero when the force is perpendicular to the motion, since cos 90° = 0.

Energy cannot be created or destroyed, only transferred from one store to another. The total energy of a closed system remains constant.

No. Energy is transferred to heat and sound in the ball and the floor at each bounce, so less remains as gravitational potential energy. The total is unchanged.

1. Uses E = mghthe useful energy is the gravitational potential energy gained[1]
2. E = 300 × 10 × 12 = 36 000 J[1]
3. Uses P = E/t[1]
4. P = 36 000 / 40 = 900 Wunit required[1]

900 W

1. Uses KE = ½mv²[1]
2. KE = ½ × 1200 × 20² = 240 000 Jsquare the velocity before multiplying[1]
3. Gain in KE = 240 000 J since it started from rest[1]
4. P = 240 000 / 8.0 = 30 000 W = 30 kW[1]

2.4 × 10⁵ J and 30 kW

A crystalline solid has a regular repeating lattice and a sharp melting point. An amorphous solid has no long-range order and softens over a range of temperature.

In elastic deformation the atoms are pulled slightly further apart but keep the same neighbours, so the material returns to its original length. In plastic deformation planes of atoms slip permanently over one another, so it does not.

In a semiconductor, thermal energy lifts more electrons across the small band gap into the conduction band, so there are more charge carriers. In a metal the number of carriers is fixed, and heating makes the ions vibrate more so the electrons collide with them more often.

1. σ = F/A = 38 / 3.5 × 10⁻⁷[1]
2. σ = 1.09 × 10⁸ Pa[1]
3. ε = e/L = 1.8 × 10⁻³ / 2.0convert mm to m[1]
4. ε = 9.0 × 10⁻⁴no units[1]
5. E = σ/ε = 1.2 × 10¹¹ Paabout right for copper[1]

σ = 1.09 × 10⁸ Pa, ε = 9.0 × 10⁻⁴, E = 1.2 × 10¹¹ Pa

1. Copper: straight line at first, then curving over into a long plastic region before breaking[1]
2. Copper reaches a maximum stress and then extends considerably before failure[1]
3. Glass: straight line all the way[1]
4. Glass breaks abruptly at the end of the straight line, with no plastic region[1]
5. Copper is ductile — it has a long plastic region and can be drawn into wire[1]
6. Glass is brittle — it breaks with almost no plastic deformation[1]
7. The gradient is the Young modulus[1]
8. It measures the stiffness of the material and is independent of the sample dimensions[1]

copper ductile with a long plastic region; glass brittle; gradient = Young modulus

1. A range of energies that electrons in the solid are not allowed to have[1]
2. It separates the valence band from the conduction band[1]
3. Conductor: no gap — the bands overlap or the upper band is part-filled[1]
4. Semiconductor: a small gap, about 1 eV, which some electrons cross at room temperature[1]
5. Insulator: a large gap of several eV that electrons cannot cross at ordinary temperatures[1]

no gap / small gap / large gap

Wavelength is the distance between two neighbouring points that are in phase, for example crest to crest. Frequency is the number of complete waves passing a fixed point each second.

In a transverse wave the vibration is perpendicular to the direction of travel, for example light. In a longitudinal wave it is parallel to the direction of travel, for example sound.

The frequency stays the same. The speed decreases and, since v = fλ, the wavelength decreases in proportion.

1. Uses λ = length ÷ number of waves[1]
2. λ = 3.0 / 15 = 0.20 m[1]
3. Uses v = fλ[1]
4. v = 12 × 0.20 = 2.4 m s⁻¹[1]

λ = 0.20 m, v = 2.4 m s⁻¹

1. The waves spread out after passing through the gap[1]
2. This is diffraction[1]
3. The wavelength and frequency are unchanged; only the shape of the wavefront changes[1]
4. The waves spread out more[1]
5. Because the gap width is closer to the wavelengthmaximum spreading when gap ≈ λ[1]
6. Converts 25 mm = 0.025 m[1]
7. Uses v = fλ[1]
8. v = 8.0 × 0.025 = 0.20 m s⁻¹[1]

(c) 0.20 m s⁻¹

1. The waves change direction at the boundary[1]
2. They bend towards the normal[1]
3. The waves travel more slowly in the shallow water[1]
4. The end of each wavefront entering the shallow region slows first[1]
5. So the wavefront pivots, changing the direction of travel[1]
6. The direction does not change[1]
7. Because the whole wavefront slows at the same instant, so there is nothing to pivot aboutthe wavelength still shortens[1]

1. Radio, microwave, infrared, visible, ultraviolet, X-ray, gammatwo marks for the full correct order; one for a mostly correct order[2]
2. 3.0 × 10⁸ m s⁻¹[1]
3. Gamma rays have a much higher frequency, so each photon carries much more energy[1]
4. They are ionising — able to remove electrons from atoms, which can damage or mutate cellsthe word "ionising" is what the mark scheme looks for[1]

Order as listed; 3.0 × 10⁸ m s⁻¹; gamma is ionising because of its far higher photon energy

1. Converts MHz to Hz: 96.0 × 10⁶ = 9.60 × 10⁷ Hzthe prefix is where this question is won or lost[1]
2. Rearranges c = fλ to λ = c/f[1]
3. λ = 3.0 × 10⁸ / 9.60 × 10⁷ = 3.1 m[1]

3.1 m

1. Infrared use: remote controls / thermal imaging / heatersany one[1]
2. Infrared danger: skin burns[1]
3. Ultraviolet use: sterilising / security marking / detecting forgeriesany one[1]
4. Ultraviolet danger: skin cancer / eye damage[1]

They are all transverse waves, and they all travel at 3.0 × 10⁸ m s⁻¹ in a vacuum. (Also acceptable: none requires a medium; all carry energy.)

Microwaves pass through the atmosphere with little absorption, so the signal reaches the satellite and returns. Infrared is strongly absorbed by the atmosphere.

1. X-rays pass through soft tissue but are absorbed by the denser bone[1]
2. Producing a shadow image on the detector[1]
3. Gamma rays are ionising and can kill living cells[1]
4. A focused beam destroys the tumour cells[1]
5. But healthy cells would also be damaged or mutated, so exposure elsewhere must be minimised[1]
6. Ultraviolet is ionising enough to kill bacteria[1]
7. So it sterilises surfaces and instruments[1]
8. Staff must wear eye protection / avoid skin exposure, since UV causes eye damage and skin cancer[1]

Sources with the same frequency and a constant phase difference. Without that, the phase relationship changes randomly and no stable pattern of fringes is seen.

Light is a transverse wave, so its vibrations are perpendicular to the direction of travel and one plane can be selected. Sound is longitudinal, vibrating along the direction of travel, so there is no plane to select.

Diffraction is greatest when the gap is comparable to the wavelength. Sound wavelengths are around a metre, similar to a doorway, while visible light is about 5 × 10⁻⁷ m — far too small to spread noticeably.

1. d = 1 / (300 × 10³) = 3.33 × 10⁻⁶ m300 per mm is 3 × 10⁵ per metre[1]
2. Uses d sin θ = nλ with n = 1[1]
3. sin θ = 590 × 10⁻⁹ / 3.33 × 10⁻⁶ = 0.177[1]
4. θ = 10.2°[1]
5. Highest order: n ≤ d/λ = 5.6, so n = 5sin θ cannot exceed 1[1]

θ = 10.2° for the first order; 5 orders visible

1. The two paths to the centre are equal in length[1]
2. So the path difference is zero, the waves arrive in phase and interfere constructively[1]
3. Uses x = λD/a[1]
4. x = (5.9 × 10⁻⁷ × 2.4) / 4.0 × 10⁻⁴[1]
5. x = 3.5 × 10⁻³ m, about 3.5 mm[1]
6. The fringe spacing doubles[1]
7. Because x is inversely proportional to a[1]
8. The fringes become more widely separated and easier to measure[1]

(b) 3.5 mm (c) spacing doubles

1. A laser beam is split down two perpendicular arms and recombined[1]
2. The arms are set so the returning beams cancel by destructive interference[1]
3. A tiny change in one arm's length upsets the cancellation, so light appears at the detectora fraction of a wavelength is enough[1]
4. Events such as black hole mergers emit almost no light[1]
5. But they do radiate gravitational waves, which also pass through intervening dust and gas unimpeded[1]

Sources with the same frequency and a constant phase difference. Without that, the phase relationship changes randomly and no stable pattern of fringes is seen.

Light is a transverse wave, so its vibrations are perpendicular to the direction of travel and one plane can be selected. Sound is longitudinal, vibrating along the direction of travel, so there is no plane to select.

Diffraction is greatest when the gap is comparable to the wavelength. Sound wavelengths are around a metre, similar to a doorway, while visible light is about 5 × 10⁻⁷ m — far too small to spread noticeably.

1. d = 1 / (300 × 10³) = 3.33 × 10⁻⁶ m300 per mm is 3 × 10⁵ per metre[1]
2. Uses d sin θ = nλ with n = 1[1]
3. sin θ = 590 × 10⁻⁹ / 3.33 × 10⁻⁶ = 0.177[1]
4. θ = 10.2°[1]
5. Highest order: n ≤ d/λ = 5.6, so n = 5sin θ cannot exceed 1[1]

θ = 10.2° for the first order; 5 orders visible

1. The two paths to the centre are equal in length[1]
2. So the path difference is zero, the waves arrive in phase and interfere constructively[1]
3. Uses x = λD/a[1]
4. x = (5.9 × 10⁻⁷ × 2.4) / 4.0 × 10⁻⁴[1]
5. x = 3.5 × 10⁻³ m, about 3.5 mm[1]
6. The fringe spacing doubles[1]
7. Because x is inversely proportional to a[1]
8. The fringes become more widely separated and easier to measure[1]

(b) 3.5 mm (c) spacing doubles

1. A laser beam is split down two perpendicular arms and recombined[1]
2. The arms are set so the returning beams cancel by destructive interference[1]
3. A tiny change in one arm's length upsets the cancellation, so light appears at the detectora fraction of a wavelength is enough[1]
4. Events such as black hole mergers emit almost no light[1]
5. But they do radiate gravitational waves, which also pass through intervening dust and gas unimpeded[1]

The rate of flow of electric charge, I = Q/t. Its unit is the ampere (A), where one ampere is one coulomb per second.

The current through a metallic conductor is directly proportional to the potential difference across it, provided the temperature stays constant.

The metal ions vibrate more strongly, so the moving electrons collide with them more often. Each collision impedes the flow, so the resistance rises.

1. Converts the area: 0.50 mm² = 0.50 × 10⁻⁶ m²the step most often dropped[1]
2. Uses ρ = RA/L[1]
3. ρ = (0.068 × 0.50 × 10⁻⁶) / 2.0[1]
4. ρ = 1.7 × 10⁻⁸ Ω mcopper[1]
5. Doubling the length doubles the resistance: 0.136 Ω[1]

ρ = 1.7 × 10⁻⁸ Ω m; R = 0.14 Ω

1. The graph passes through the origin[1]
2. It is a straight line at low potential difference[1]
3. It then curves towards the V axis, so the gradient falls[1]
4. A larger current heats the filament[1]
5. The ions vibrate more and the electrons collide with them more often[1]
6. So the resistance increases and the current no longer rises in proportion[1]
7. R = V/I = 6.0 / 0.50 = 12 Ω[1]
8. P = VI = 6.0 × 0.50 = 3.0 W[1]

(c) 12 Ω and 3.0 W

1. Total resistance = 20 + 25 = 45 Ω[1]
2. Uses I = V/R[1]
3. I = 9.0 / 45 = 0.20 A[1]
4. Uses V = IR for the thermistor[1]
5. V = 0.20 × 25 = 5.0 V[1]
6. The potential difference across the thermistor decreases[1]
7. Its resistance falls as it warms, so it takes a smaller share of the supply voltage[1]

(a) 0.20 A (b) 5.0 V (c) it falls

1. Uses I = P/V = 2300 / 230converting kW to W[1]
2. I = 10 A[1]
3. 13 A fuse[1]
4. It is the next standard rating above the normal working current of 10 A — a 5 A fuse would blow in normal usethe justification is a separate mark[1]
5. Energy = 2.3 kW × 0.25 h = 0.575 kWh15 minutes is 0.25 hours[1]
6. Cost = 0.575 × 25 = 14.4 rupees[1]

(a) 10 A (b) 13 A, the next rating above 10 A (c) about 14 rupees per day

1. Green and yellow[1]
2. Connected to the metal case[1]
3. A fault would send a very large current from the live wire through the case to earth[1]
4. This large current melts the fuse[1]
5. Which disconnects the live supply, so the case cannot give a shockthe fuse must be in the live wire for this to work[1]

1. Converts to kilowatts: 60 W = 0.060 kW[1]
2. E = 0.060 × 8.0 = 0.48 kWh[1]
3. Converts hours to seconds: 8.0 × 3600 = 28 800 s[1]
4. E = 60 × 28 800 = 1.73 × 10⁶ J[1]

0.48 kWh, or 1.7 × 10⁶ J

When the fuse blows it must disconnect the appliance from the dangerous side of the supply. A fuse in the neutral would leave the appliance connected to the live wire and still dangerous.

The energy transferred by an appliance of power 1 kilowatt operating for 1 hour. It is a unit of energy, not power.

1. Uses E = Pt with power in kW and time in hours[1]
2. E = 3.0 × 2.5 = 7.5 kWh[1]
3. Total lamp power = 5 × 12 = 60 W = 0.060 kW[1]
4. Uses E = 0.060 × 6.0[1]
5. = 0.36 kWh[1]
6. Total = 7.5 + 0.36 = 7.86 kWh[1]
7. Cost = 7.86 × 22 = 173 rupees[1]
8. I = P/V = 3000 / 230 = 13.0 A[1]
9. A 13 A fuse is marginal — a higher-rated fuse or a dedicated circuit is neededaccept 13 A with a comment, or a stated higher rating[1]

(a) 7.5 kWh (b) 0.36 kWh (c) about 173 rupees (d) 13 A

1. 1/R = 1/8.0 + 1/8.0 = 0.25, so R = 4.0 Ωone mark for the method, one for inverting correctly[2]
2. Total = 3.0 + 4.0 = 7.0 Ω[1]
3. I = V/R = 9.0 / 7.0 = 1.3 Aaccept 1.29[1]
4. Uses V = IR for the 3.0 Ω resistor[1]
5. V = 1.29 × 3.0 = 3.9 V[1]

(a) 4.0 Ω (b) 7.0 Ω (c) 1.3 A (d) 3.9 V

1. Its resistance decreases[1]
2. The thermistor now takes a smaller share of the supply voltage[1]
3. So a larger share appears across the fixed resistor[1]
4. The output voltage therefore increases[1]

1. Each lamp receives the full mains voltage[1]
2. So each operates at its correct brightness[1]
3. If one lamp fails, the circuit through the others is unbroken[1]
4. So the rest continue to work, and each can be switched independently[1]

1. The total resistance halves[1]
2. Because the second branch gives the current an additional path, so more current flows for the same potential difference[1]

It halves. The second branch gives the current an additional path, so more current flows for the same potential difference.

1. There is only one path for the charge to follow[1]
2. So the same charge passes through each component every second[1]

There is only one path for the charge to follow, so the same charge passes through each component every second.

1. Uses 1/R = 1/R₁ + 1/R₂[1]
2. 1/R = 1/4.0 + 1/12 = 0.25 + 0.0833 = 0.3333[1]
3. R = 3.0 Ωinverting is a separate mark[1]
4. Total resistance = 6.0 + 3.0 = 9.0 Ω[1]
5. Uses I = V/R[1]
6. I = 12 / 9.0 = 1.33 A[1]
7. P.d. across the parallel section = 1.33 × 3.0 = 4.0 Vboth branches share this[1]
8. Uses I = V/R for the 12 Ω branch[1]
9. I = 4.0 / 12 = 0.33 A[1]

(a) 3.0 Ω (b) 1.33 A (c) 0.33 A

1. 92 protonsthe proton number[1]
2. 235 − 92 = 143 neutrons[1]
3. 92 electrons, since the atom is neutral[1]
4. Same number of protons (92); different number of neutrons (146 instead of 143)[1]
5. Chemical behaviour depends on the electrons[1]
6. Both have 92 electrons arranged identically, so they react in the same way[1]

(a) 92 p, 143 n, 92 e (b) same Z, different A (c) identical electron arrangement

1. Most alpha particles passed straight through[1]
2. So the atom is mostly empty space[1]
3. Some were deflected through large angles[1]
4. So there is a concentrated positive charge repelling them[1]
5. A very few were reflected almost straight back[1]
6. So the nucleus is very small and contains most of the atom's mass[1]

1. In that model the positive charge is spread thinly throughout the atom[1]
2. So the repulsive force on an alpha particle anywhere would be small[1]
3. It could not produce the large-angle deflections or backscattering that were observed[1]

The nucleon number A is the total number of protons and neutrons in the nucleus. The proton number Z is the number of protons alone.

They have the same number of protons and therefore the same number of electrons in the same arrangement. Chemical behaviour is decided by the electrons, not by the number of neutrons.

1. Almost all the alpha particles passed straight through[1]
2. A very small fraction were deflected through large angles, some straight back[1]
3. Passing straight through shows the atom is mostly empty space[1]
4. Large deflections show a concentrated region of positive charge[1]
5. Which repels the positive alpha particle[1]
6. Backward scattering shows that region is also very massive, and very small[1]
7. So each alpha particle meets at most one nucleus[1]
8. A thicker foil would cause multiple scattering and the result could not be interpreted[1]

Its direction of motion is changing continuously, so its velocity is changing. Velocity is a vector, and a changing velocity is an acceleration.

The tension in the string. If it breaks there is no longer a resultant force, so the stone flies off along the tangent in a straight line, not outwards along the radius.

No external torque acts, so angular momentum L = Iω is conserved. Pulling her arms in moves mass closer to the axis, reducing the moment of inertia, so the angular velocity must increase.

1. Uses ω = 2π/T[1]
2. ω = 2π / 5800 = 1.08 × 10⁻³ rad s⁻¹[1]
3. Uses v = rω[1]
4. v = 7.0 × 10⁶ × 1.08 × 10⁻³ = 7.6 × 10³ m s⁻¹[1]
5. a = ω²r = (1.08 × 10⁻³)² × 7.0 × 10⁶ = 8.2 m s⁻²close to g, as expected in low orbit[1]

ω = 1.08 × 10⁻³ rad s⁻¹, v = 7.6 km s⁻¹, a = 8.2 m s⁻²

1. Weight mg acting downwards[1]
2. Tension T also acting downwards, towards the centreat the top both point the same way[1]
3. At minimum speed the string just goes slack, so T = 0 and gravity alone provides the force[1]
4. mg = mv²/r, so v² = gr = 9.81 × 0.80the mass cancels[1]
5. v = 2.8 m s⁻¹[1]
6. At the bottom the tension acts up and the weight down: T − mg = mv²/r[1]
7. T = 0.25 × 6.0² / 0.80 + 0.25 × 9.81[1]
8. T = 11.25 + 2.45 = 13.7 N[1]

(b) 2.8 m s⁻¹ (c) 13.7 N

1. Uses KE = ½Iω²[1]
2. = ½ × 0.45 × 120² = 3240 J[1]
3. Uses L = Iω[1]
4. = 0.45 × 120 = 54 kg m² s⁻¹[1]
5. No external torque, so angular momentum is conserved: 54 = 0.60 ω[1]
6. ω = 90 rad s⁻¹kinetic energy has fallen — the collision is inelastic[1]

3240 J, 54 kg m² s⁻¹, 90 rad s⁻¹

Gravitational field strength is the gravitational force per unit mass at a point, g = F/m. Since the gravitational force on a mass is F = mg and Newton's second law gives F = ma, equating them gives a = g — so a freely falling mass accelerates at the numerical value of the field strength.

Potential is defined as the work done per unit mass in bringing a mass from infinity, where the potential is defined as zero. Because gravity is always attractive, it does positive work pulling the mass inwards, so the work done on the mass is negative. Every finite distance therefore has a negative potential, rising to zero only at infinity.

1. g = GM/r² = (6.67 × 10⁻¹¹)(7.35 × 10²²)/(1.74 × 10⁶)²Using the surface radius as r.[1]
2. g = 4.902 × 10¹² / 3.028 × 10¹² = 1.62 N kg⁻¹About one sixth of Earth's, which is the well-known figure and a good check.[1]
3. Escape velocity: ½mv² = GMm/r, so v = √(2GM/r)The kinetic energy must equal the depth of the potential well.[1]
4. v = √(2 × 4.902 × 10¹² / 1.74 × 10⁶) = √(5.635 × 10⁶) = 2370 m s⁻¹About 2.4 km s⁻¹, far less than Earth's 11.2 km s⁻¹.[1]

g = 1.62 N kg⁻¹, escape velocity 2.37 km s⁻¹

1. (a) The orbit must be in the plane of the equatorAny other plane would make the satellite drift north and south.[1]
2. and the satellite must travel west to east, in the same sense as the Earth's rotationOtherwise it would move relative to the ground even with a 24-hour period.[1]
3. (b) T = 24 h = 86 400 s. Using T² = 4π²r³/(GM), r³ = GMT²/(4π²)Rearranging Kepler's third law for r.[1]
4. r³ = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(86400)²/(4π²) = 7.53 × 10²²Careful with the square of the period.[1]
5. r = 4.22 × 10⁷ m as requiredTaking the cube root.[1]
6. (c) Height = 4.22 × 10⁷ − 6.37 × 10⁶ = 3.58 × 10⁷ mSubtracting the Earth's radius, since r was measured from the centre.[1]
7. (d) Equating GMm/r² with mv²/r cancels the satellite mass m from both sides, so the orbit depends only on r and the Earth's mass.The mark is for identifying the cancellation, not merely asserting independence.[1]

(a) equatorial, west to east; (b) 4.22 × 10⁷ m; (c) 3.58 × 10⁷ m; (d) m cancels

Field follows an inverse square, so it is g/9. Potential follows an inverse first power, so it is one third of the surface value. The two scale differently, which is why they must never be treated interchangeably.

1. Uses E = mcΔθ with Δθ = 70 °C88 − 18, not 88[1]
2. E = 1.5 × 4200 × 70 = 4.41 × 10⁵ J[1]
3. Uses t = E/P with P = 2000 Wconverting kW to W[1]
4. t = 441000 / 2000 = 220 saccept 220.5 s[1]
5. Energy is lost to the surroundings / the container is also heatedeither reason accepted[1]

(a) 4.4 × 10⁵ J (b) 220 s (c) heat lost to surroundings

1. Temperature rises from −10 °C to 0 °C[1]
2. Horizontal section at 0 °C while the ice melts[1]
3. Temperature rises from 0 °C to 100 °C[1]
4. Longer horizontal section at 100 °C while the water boilsthe boiling plateau must be longer than the melting one[1]
5. During the flat sections the energy supplied breaks bonds between particles rather than raising kinetic energythe explanation mark[1]

1. Heating makes particles vibrate more and push further apart, so each metal expands[1]
2. The two metals expand by different amounts for the same temperature risethis is the essential point[1]
3. The strip bends toward the metal that expands less[1]
4. Used in a thermostat / fire alarm / oven switchany valid use[1]

Temperature is the average kinetic energy of the particles, measured in °C or K. Heat is the total thermal energy of all the particles, measured in joules, and depends on how many there are.

The two metals expand by different amounts for the same temperature rise, so the strip curves toward the metal that expands less.

The energy supplied is used to break the bonds holding the particles in the lattice, not to increase their kinetic energy. Since temperature measures kinetic energy, it does not change.

1. Recognises two stages: melting, then warminga single-stage answer cannot score more than two[1]
2. Melting: E₁ = mL = 0.20 × 3.34 × 10⁵ = 6.68 × 10⁴ J[1]
3. Warming: E₂ = mcΔθ = 0.20 × 4200 × 20[1]
4. E₂ = 1.68 × 10⁴ J[1]
5. Total = 6.68 × 10⁴ + 1.68 × 10⁴ = 8.36 × 10⁴ Jadding the two stages[1]

8.36 × 10⁴ J

1. Brownian motion[1]
2. The smoke particles are bombarded by air molecules[1]
3. The bombardment is uneven / random, so the resultant force keeps changing directionthe mark that separates a full answer[1]
4. Air molecules are very smalltoo small to see individually[1]
5. They are moving rapidly and randomly[1]

1. The pressure increases[1]
2. The particles gain kinetic energy and move faster[1]
3. They collide with the walls more frequently[1]
4. Each collision exerts a greater force, so the total force per unit area risesboth "more often" and "harder" are needed for full marks[1]

1. Uses p₁V₁ = p₂V₂temperature constant, so Boyle's law applies[1]
2. 3.0 × 10⁵ × 2.0 = 1.0 × 10⁵ × V₂[1]
3. V₂ = 6.0 cm³lower pressure, larger bubble — check the direction[1]

6.0 cm³

In a gas the particles are far apart with large spaces between them, which can be reduced. In a liquid the particles are already touching, so there is no space to remove.

Raise its temperature, so the particles move faster and collide harder and more often. Or reduce its volume, so the same particles strike a smaller area more frequently.

1. Particles move rapidly and randomly, colliding with the walls[1]
2. Each collision exerts a small force on the wall[1]
3. Pressure is the total force per unit area of wall[1]
4. Uses p₁V₁ = p₂V₂, stating that the temperature is constant[1]
5. 2.0 × 10⁵ × 0.50 = p₂ × 0.20[1]
6. p₂ = 5.0 × 10⁵ Pa[1]
7. The pressure would increase[1]
8. The particles gain kinetic energy, so they hit the walls harder and more often[1]

(b) 5.0 × 10⁵ Pa (c) pressure rises

For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume. Both the mass and the temperature must be constant.

The particles gain kinetic energy and move faster, so they strike the walls harder and more frequently. With the same wall area, the force per unit area increases.

The gas does work pushing back the surroundings, so W is negative. With Q = 0, ΔU is negative, the internal energy falls and the temperature drops.

1. Convert: 17 °C = 290 K, 137 °C = 410 K[1]
2. Constant pressure, so V₁/T₁ = V₂/T₂Charles's law[1]
3. 0.015 / 290 = V₂ / 410[1]
4. V₂ = 0.015 × 410 / 290[1]
5. V₂ = 0.021 m³a 41% rise in kelvin gives a 41% rise in volume[1]

0.021 m³

1. Uses 1 − T_c/T_h with kelvin[1]
2. 1 − 300/500 = 0.40, so 40%[1]
3. Maximum work = 0.40 × 6000 = 2400 J[1]
4. Heat rejected = 6000 − 2400[1]
5. = 3600 Jthis cannot be avoided — the second law requires it[1]
6. Actual efficiency = 1800/6000 = 0.30, or 30%[1]
7. Lower than the maximum because the real engine is not reversible[1]
8. Friction, turbulence and heat lost to the surroundings all reduce the work obtained[1]

(a) 40% (b) 2400 J of work, 3600 J rejected (c) 30%

1. The increase in internal energy equals the heat supplied plus the work done on the system[1]
2. ΔU = Q + W[1]
3. W = +250 J (work done ON the gas), Q = −100 J (heat lost)the signs are the difficult part[1]
4. ΔU = −100 + 250 = +150 J[1]
5. The temperature rises[1]
6. Internal energy has increased, and for a gas that means greater mean kinetic energy[1]

ΔU = +150 J, so the temperature rises

For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume. Both the mass and the temperature must be constant.

The particles gain kinetic energy and move faster, so they strike the walls harder and more frequently. With the same wall area, the force per unit area increases.

The gas does work pushing back the surroundings, so W is negative. With Q = 0, ΔU is negative, the internal energy falls and the temperature drops.

1. Convert: 17 °C = 290 K, 137 °C = 410 K[1]
2. Constant pressure, so V₁/T₁ = V₂/T₂Charles's law[1]
3. 0.015 / 290 = V₂ / 410[1]
4. V₂ = 0.015 × 410 / 290[1]
5. V₂ = 0.021 m³a 41% rise in kelvin gives a 41% rise in volume[1]

0.021 m³

1. Uses 1 − T_c/T_h with kelvin[1]
2. 1 − 300/500 = 0.40, so 40%[1]
3. Maximum work = 0.40 × 6000 = 2400 J[1]
4. Heat rejected = 6000 − 2400[1]
5. = 3600 Jthis cannot be avoided — the second law requires it[1]
6. Actual efficiency = 1800/6000 = 0.30, or 30%[1]
7. Lower than the maximum because the real engine is not reversible[1]
8. Friction, turbulence and heat lost to the surroundings all reduce the work obtained[1]

(a) 40% (b) 2400 J of work, 3600 J rejected (c) 30%

1. The increase in internal energy equals the heat supplied plus the work done on the system[1]
2. ΔU = Q + W[1]
3. W = +250 J (work done ON the gas), Q = −100 J (heat lost)the signs are the difficult part[1]
4. ΔU = −100 + 250 = +150 J[1]
5. The temperature rises[1]
6. Internal energy has increased, and for a gas that means greater mean kinetic energy[1]

ΔU = +150 J, so the temperature rises

The acceleration must be proportional to the displacement from a fixed equilibrium point, and it must be directed towards that point. Together these give a = −ω²x.

Speed is greatest at the equilibrium position, because all the energy is kinetic there and the net force is zero. Acceleration is greatest at maximum displacement, because the restoring force is largest there — although the body is momentarily at rest.

1. ω = 2π/T = 2π/0.80 = 7.854 rad s⁻¹Angular frequency comes first; everything else is built on it.[1]
2. v_max = ωx₀ = 7.854 × 0.040The amplitude must be converted to metres.[1]
3. v_max = 0.314 m s⁻¹Occurring at the equilibrium position.[1]
4. a_max = ω²x₀ = 61.68 × 0.040 = 2.47 m s⁻²Occurring at maximum displacement, where the speed is zero.[1]

v_max = 0.314 m s⁻¹, a_max = 2.47 m s⁻²

1. (a) T = 2π√(L/g) = 2π√(1.2/9.81) = 2π√0.12232The standard pendulum formula.[1]
2. T = 2.20 sClose to 2 s, as expected for a metre-scale pendulum.[1]
3. (b) ω = 2π/2.20 = 2.856 rad s⁻¹; v_max = ωx₀ = 2.856 × 0.050Converting 5.0 cm to 0.050 m.[1]
4. v_max = 0.143 m s⁻¹At the lowest point of the swing.[1]
5. (c) The period is unchanged, because T = 2π√(L/g) contains no amplitude termIsochronism — the property that made pendulum clocks possible.[1]
6. The total energy is quadrupled, since E ∝ x₀² and the amplitude has doubledThe squared dependence is the point of the part.[1]
7. (d) Critical damping returns the system to equilibrium in the shortest time without overshooting — used in car suspension or door closers.Definition plus a named application.[1]

(a) 2.20 s; (b) 0.143 m s⁻¹; (c) period unchanged, energy ×4; (d) fastest return without overshoot

Resonance occurs when the frequency of a driving force equals the natural frequency of the system, producing a maximum amplitude and the most efficient transfer of energy. Increasing the damping makes the peak lower and broader, and shifts it slightly to a lower frequency.

Electrons are transferred from the cloth onto the rod. The rod gains electrons and becomes negative; the cloth is left with a positive charge.

A region in which a charge experiences a force. The direction of the field is the direction of the force on a small positive test charge.

The rod induces a separation of charge in the paper, drawing the opposite charge to the near side. That near charge is closer to the rod, so its attraction outweighs the repulsion of the far side.

1. Converts 25 mm = 0.025 m[1]
2. Uses E = V/d = 5000 / 0.025[1]
3. E = 2.0 × 10⁵ V m⁻¹[1]
4. F = QE = 3.2 × 10⁻¹⁹ × 2.0 × 10⁵ = 6.4 × 10⁻¹⁴ N[1]

E = 2.0 × 10⁵ V m⁻¹, F = 6.4 × 10⁻¹⁴ N

1. Friction between the tanker and the air, and between the fuel and the tank walls, transfers electrons[1]
2. The tanker is insulated by its rubber tyres, so the charge cannot escape and accumulates[1]
3. The charge raises the potential of the tanker[1]
4. Eventually a spark jumps to a nearby earthed object[1]
5. The spark could ignite the fuel vapour and cause an explosion[1]
6. The wire provides a conducting path to earth[1]
7. Electrons flow along it until the tanker is at the same potential as the earth[1]
8. So charge leaks away steadily instead of building to a spark[1]

1. The lines are parallel and equally spaced, showing a uniform field[1]
2. They run from the positive plate to the negative plate[1]
3. The force is upward, towards the positive plate[1]
4. Because the drop is negative, so the force is opposite to the field direction[1]
5. Increase the potential difference across the plates[1]
6. Move the plates closer together, or increase the charge on the dropany one[1]

(b) upward, towards the positive plate

Capacitance is the charge stored per unit potential difference, C = Q/V, measured in farads. The plates carry equal and opposite charges, +Q and −Q, so the net charge on the device is zero — what is stored is the separation of charge.

The potential difference rises linearly from zero as charge accumulates, so charge is not delivered at the full voltage throughout. The average potential difference during charging is V/2, giving W = ½QV. Graphically it is the triangular area under the V–Q line.

1. τ = RC = 47 000 × 220 × 10⁻⁶Both quantities converted to base units first.[1]
2. τ = 10.34 sOhms × farads gives seconds, which is a useful check.[1]
3. V = V₀e^(−t/RC) = 9.0 × e^(−15/10.34) = 9.0 × e^(−1.451)Substituting into the discharge equation.[1]
4. V = 9.0 × 0.2344 = 2.11 VJust under 1.5 time constants, so a little above the 23% mark.[1]

τ = 10.3 s, V = 2.11 V

1. (a) Q = CV = 100 × 10⁻⁶ × 20 = 2.0 × 10⁻³ CStraight substitution.[1]
2. W = ½CV² = ½ × 100 × 10⁻⁶ × 400 = 0.020 JUsing the squared-voltage form.[1]
3. (b) τ = RC = 25 000 × 100 × 10⁻⁶ = 2.5 sConsistent units give seconds.[1]
4. I₀ = V₀/R = 20/25 000 = 8.0 × 10⁻⁴ AAt t = 0 the full voltage is across the resistor.[1]
5. (c) 5.0 = 20e^(−t/2.5), so e^(−t/2.5) = 0.25Dividing by the initial voltage first.[1]
6. −t/2.5 = ln 0.25 = −1.386, so t = 3.47 sThis is two half-lives, since the voltage fell to a quarter — a useful check.[1]
7. (d) The exponential approaches zero asymptotically without ever reaching it, since e^(−t/RC) is never exactly zero for finite t. In practice it is treated as discharged after about 5RC.The mark is for the asymptotic argument plus the practical convention.[1]

(a) 2.0 mC, 0.020 J; (b) 2.5 s, 0.80 mA; (c) 3.47 s; (d) exponential never reaches zero

In parallel: 6 + 6 = 12 µF. In series: 1/C = 1/6 + 1/6 = 1/3, so C = 3 µF. The series result is smaller because the charge must pass through both, so the same charge is stored for a larger total voltage — which by C = Q/V means less capacitance.