Worked examples that train every level of thinking

Density is the mass per unit volume of a substance, ρ = m/V (1)

SI unit: kg/m³ (1)

Water is less dense than milk (1.000 < 1.030 g/cm³), so it adds less mass for the same volume (1)

the mixture has less mass in each cm³, so its density falls towards 1.000 g/cm³ (1)

Mass of milk: A = 326.0 − 120.0 = 206.0 g; B = 323.0 − 120.0 = 203.0 g (1); ρ_A = 206.0/200 = 1.030 g/cm³ and ρ_B = 203.0/200 = 1.015 g/cm³ (1)

ρ_A = 1.030 × 1000 = 1030 kg/m³ (1)

200 cm³ of pure milk would have a mass of 200 × 1.030 = 206 g, but B's sample has only 203 g: 3 g missing (1); each cm³ of water that replaces milk makes the mass 1.030 − 1.000 = 0.030 g smaller, so volume of water = 3 ÷ 0.030 (1)

= 100 cm³: half the sample is water, so the claim is false (1)

The whole difference between pure milk (206 g) and B's watered milk (203 g) is only 3 g, and the accepted range 1.028–1.034 g/cm³ is only 1.2 g wide for 200 cm³; a scale reading to the nearest 10 g can only give milk masses such as 200 g or 210 g, i.e. densities of 1.00 or 1.05 g/cm³ (1)

its 10 g step is more than three times the 3 g difference, so it cannot reliably tell good milk from watered milk: keep the electronic balance (1)

Any three of: zero (tare) the balance and use the same clean, dry cylinder each time; measure exactly 200 cm³ of well-stirred milk, reading the bottom of the meniscus at eye level; weigh, subtract the cylinder's mass and calculate ρ = m/V; accept the milk only if ρ is between 1.028 and 1.034 g/cm³, otherwise reject it or test again; record the farmer's name, the date and the result in a book; repeat a doubtful test with a second sample

any three, (1) each

MA = load ÷ effort (1)

VR = distance moved by effort ÷ distance moved by load (1)

Some of the work done by the effort is wasted overcoming friction in the pulley bearings and the rope (1)

and in lifting the lower pulley block and rope, so the useful work output is less than the work input (1)

MA = 500/150 ≈ 3.33 (1); VR = 4 (number of supporting rope sections) (1)

η = MA/VR × 100 = 3.33/4 × 100 ≈ 83 % (1)

100 N: MA = 2, η = 2/4 × 100 = 50 %; efficiency rises with load (50 %, 67 %, 75 %, 80 %, 83 %) (1); the effort always rises by 25 N per 100 N of load, so effort = (load + w)/4 (1)

w = 4 × 50 − 100 = 4 × 150 − 500 = 100 N: the lower block weighs 100 N; this fixed extra weight matters less as the load grows, so η increases (1)

Any three of: effort: carrying puts 500 N on a worker's back on every trip (injury risk); B needs an effort of at least 500 N, about a worker's whole body weight, too hard to pull steadily 40 times a day; C needs only 150 N; time: with C the rope must be pulled 4 × 6 = 24 m per bag, slower than B; energy: work input with C = 150 × 24 = 3600 J per bag for 3000 J of useful work, a small cost; safety and cost: C needs a strong support beam but is cheap compared with injuries. Judgement: C is best

any three, (1) each

With the 100 N lower block and little friction: effort = (1000 + 100)/n; for 4 sections 275 N (just within 300 N, no margin for friction) (1); choose 6 sections (an upper and a lower block of three pulleys each): effort ≈ 1100/6 ≈ 183 N, well below 300 N, leaving room for friction and the slightly heavier lower block (1)

design details: fix the upper block to a strong beam, attach the wheelbarrow with a hook and safety catch, guide the rope, and test with a 1000 N load before use; rope must be pulled 6 m per metre lifted (1)

Renewable: solar (or wind, small hydro, biogas) (1)

Non-renewable: diesel/petrol (or coal, natural gas) (1)

Sunlight is intermittent: there is none at night and little on cloudy days, but the fridge runs 24 hours (1)

Batteries store energy in the day for use at night; a generator produces electricity whenever fuel is burned, so it needs no storage (1)

E = Pt = 0.150 kW × 24 h = 3.6 kWh (1); useful energy per panel = 0.80 × 1.5 = 1.2 kWh, so 3.6 ÷ 1.2 = 3 panels (1)

Diesel: 3.6 × 0.40 = 1.44 L; 1.44 × 1600 = 2304 RWF per day (1)

Diesel is cheaper at first (1200 < 3000); at year 2 the costs are almost equal (3182 vs 3200), and by year 4 solar is far cheaper (3400 vs 5164), so solar becomes cheaper just after year 2, in the period between year 2 and year 4 (1); saving = 11 110 − 4600 = 6510 thousand RWF ≈ 6.5 million RWF (1)

The jump of 600 thousand is the replacement of the batteries, which wear out after about 5 years (1)

Creditworthy points (any 3): solar is much cheaper over 10 years (saves about 6.5 million RWF); no fuel has to be carried on bad roads, so the fridge does not stop when deliveries fail; solar produces no CO₂ or fumes; solar is quiet and needs little maintenance; diesel is cheaper to install and works in any weather; vaccines are spoiled if power fails, so reliability is the key criterion; judgement: solar (possibly with a small generator as backup)

Justified judgement using three criteria (1 + 1 + 1)

Creditworthy points (any 3): solar panels (at least 3 × 300 W, facing the sun, tilted, no shade) → charge controller (stops overcharging/deep discharge) → battery bank (stores energy for night) → inverter if the fridge is a.c. (d.c. solar fridge avoids inverter losses) → fridge; fuse/circuit breaker for protection; extra panel and larger battery for several days of autonomy; use a solar direct-drive fridge with an ice bank or cool boxes with ice packs; keep a small generator as backup

Correct order with functions and one cloudy-week measure (1 + 1 + 1)

For a body in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point (1)

Conditions: the resultant force is zero and the resultant moment about any point is zero (1)

Taking moments about the axle (the pivot): F × 1.4 = W × d, so F = Wd/1.4 (1)

Loading near the wheel makes d small, so the moment of the load and hence the effort F needed are smaller (1)

Moments about the axle: F × 1.4 = 600 × 0.40 (1); F ≈ 171 N (1)

Vertical forces balance: R = 600 − 171 ≈ 429 N (1)

F/d is constant ≈ 429 N/m for all readings except one (86/0.20, 129/0.30, 171/0.40, 257/0.60), and the line passes through the origin, so F ∝ d (1); gradient = W/1.4, so W = 429 × 1.4 ≈ 600 N, as stated (1)

The reading at d = 0.50 m is anomalous: it should be 600 × 0.50/1.4 ≈ 214 N, not 230 N (1)

Creditworthy points (any 3): wheelbarrow: wheel carries about 70% of the weight when loaded near the wheel, but with one wheel it can tip sideways on uneven ground; two-wheeled cart: wider base, more stable on level ground, carries more, but too wide for narrow paths and hard to control on steep slopes; head loading: no equipment and fits any path, but the whole 600 N is carried by the person with a high centre of gravity, so it is tiring and risky, needing several trips; judgement: wheelbarrow, loaded low and near the wheel, is the best compromise on narrow hill paths

Justified judgement using stability and effort (1 + 1 + 1)

Creditworthy points (any 3): place the pivot close to the tank, e.g. 0.5 m from the end under the tank, leaving 2.5 m for the effort arm; effort = 1500 × 0.5/2.5 = 300 N (less if only one edge is lifted, since part of the weight rests on the ground); put a flat plate under the pivot block so it does not sink; push down at the end of the pole; safety: empty the tank first, keep feet and hands clear and slide the support in with a stick, check the pole has no cracks, lift slowly with two people, chock the tank

Workable lever design with calculation and two precautions (1 + 1 + 1)

P = hρg (above atmospheric pressure) (1)

Pressure = normal force per unit area, P = F/A (unit Pa) (1)

Saline flows only if the pressure it gives at the needle is greater than the blood pressure in the vein (1)

when h is small, hρg is less than the vein pressure, so the flow stops or reverses (1)

P = hρg = 1.0 × 1020 × 9.8 (1)

≈ 1.0 × 10⁴ Pa (10 kPa) (1)

Each 0.2 m rise in h adds the same 15 drops/min, so the graph of rate against h is a straight line (1); extrapolating back, the rate is zero at h₀ = 0.40 − 0.20 = 0.20 m (1)

P_vein = h₀ρg = 0.20 × 1020 × 9.8 ≈ 2.0 kPa (1)

Creditworthy points: A gives a changing height so the rate varies, and ties up a nurse for hours (1); B is free, steady and gives enough pressure (about 15 kPa, well above 2 kPa), but the rate must be set with the roller clamp and checked (1); C is the most precise and has alarms but is expensive, needs electricity and maintenance (1); verdict: B is most suitable for a remote post, C where precise dosing and power are available (1)

any 3

Creditworthy points: drops needed = 1000 × 20 = 20 000 in 480 min (1), so aim for ≈ 42 drops per minute (1); hang the bag at a fixed height (about 1 m above the arm), expel air bubbles from the tube, then adjust the roller clamp while counting drops in the drip chamber for 1 minute with a watch (1); re-check the rate and the volume left every hour, because the rate falls as the bag empties and the patient moves (1)

any 3

Intensity: the sound power passing through unit area at right angles to the wave, I = P/A (W/m²) (1)

Intensity level: β = 10 log₁₀(I/I₀) in dB, with I₀ = 1.0 × 10⁻¹² W/m² (1)

With the window open, sound travels straight through the gap in the air with almost no loss of energy (1)

when it is closed, the sound must pass through glass and walls, which are much denser than air: most of the sound energy is reflected at their surfaces and some is absorbed, so much less energy per second reaches the room (1)

I = I₀ × 10^80/10 = 1.0 × 10⁻⁴ W/m² (1); point source: I ∝ 1/r², so at 40 m I is 1/16 of the value at 10 m (1)

β = 80 − 10 log₁₀16 ≈ 68 dB (1)

The measured level falls by 3 dB for each doubling of distance, not 6 dB (74 dB at 40 m, not 68 dB) (1); −3 dB means the intensity halves, so I ∝ 1/r (1)

the road is a long line of many cars, so the sound spreads out as a cylinder, not a sphere (1)

Creditworthy points: A — from 20 m to 80 m is two doublings, only 77 → 71 dB (−6 dB) for a line source, and moving a classroom is costly (1); B — about 15 dB reduction for all rooms and the playground, but expensive and it may block air flow and light (1); C — large reduction indoors, but closed windows can make crowded classrooms hot and stuffy unless there is other ventilation (1); verdict with reason, e.g. B gives the biggest benefit for the whole school: −15 dB cuts the intensity to about 1/30, while moving the classroom (−6 dB) only cuts it to 1/4 (1)

any 3

Creditworthy points: mark a grid of points (e.g. every 10 m) on a plan of the compound and measure the level at each, at the same height and with the same phone (1); measure at the same time of day, taking the average of several readings (or over 1 minute) to allow for changing traffic (1); for the trees, measure at equal distances from the road behind the trees and in an open area beside them (1); calibrate or compare the phones against one reference meter, and draw contour lines of equal dB on the plan (1)

any 3