Secondary 3 Integrated Science Practice Paper: Forces, Pressure, Circuits, Acids and Salts, Density, Inheritance and Ecology

2026-11-13

Science – Secondary 3 – Practice Paper

Integrated science practice paper for Secondary 3: forces, pressure, work and power, circuits, acids and salts, density, inheritance and ecology, with full answers.

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Paper details　Secondary 3 integrated science practice paper with full answers: resultant force and acceleration, pressure and liquid pressure, series and parallel circuits, neutralisation and salt mass, density and particle model, genetic ratios, energy flow in a food chain, mean and percentage error.

Section A: Force, Pressure and Work (18 marks)

Data 1: a 2.5 kg block on a smooth horizontal surface is pulled by a horizontal force of 12 N from rest. A brick weighs 240 N and its base is a square of side 0.20 m. Take the density of water as 1000 kg/m3 and g as 10 m/s2.

#	Question	Answer and marking notes
1	(a) Find the acceleration of the block, and its speed and displacement after 3.0 s.	Answer: acceleration = 12 / 2.5 = 4.8 m/s2; speed = 4.8 x 3.0 = 14.4 m/s; displacement = 0.5 x 4.8 x 3.0 squared = 21.6 m. (Marking: 2 marks, 1 mark, 2 marks)
2	(b) Find the pressure the brick exerts on the ground when it rests on its base.	Answer: contact area = 0.20 x 0.20 = 0.040 m2; pressure = 240 / 0.040 = 6000 Pa (6.0 kPa). (Marking: area 2 marks, pressure 3 marks)
3	(c) Find the liquid pressure at a depth of 1.5 m in the pool.	Answer: pressure = 1000 x 10 x 1.5 = 15000 Pa (15 kPa). (Marking: formula 2 marks, value and unit 2 marks)
4	(d) A horizontal force of 25 N moves the block 4.0 m in 5.0 s. Find the work done and the average power.	Answer: work = 25 x 4.0 = 100 J; average power = 100 / 5.0 = 20 W. (Marking: 2 marks each)

Section B: Circuits and Electrical Power (16 marks)

Data 2: a 12 V battery is connected to a 4.0 ohm resistor. Separately, a 6.0 ohm and a 3.0 ohm resistor are joined in parallel and then in series with a 4.0 ohm resistor across the same 12 V battery. A 60 W lamp is used for 5.0 hours a day and electricity costs 1.20 dollars per kWh.

#	Question	Answer and marking notes
5	(a) Find the current through the 4.0 ohm resistor and the power it dissipates.	Answer: current = 12 / 4.0 = 3.0 A; power = 12 x 3.0 = 36 W. (Marking: 2 marks each)
6	(b) Find the equivalent resistance of the parallel part, the total resistance and the current from the battery.	Answer: parallel part = (6.0 x 3.0) / (6.0 + 3.0) = 2.0 ohm; total = 2.0 + 4.0 = 6.0 ohm; current = 12 / 6.0 = 2.0 A. (Marking: about 1.5 marks each)
7	(c) Find the electrical energy the lamp uses each day in kWh and the daily cost.	Answer: energy = 0.060 kW x 5.0 h = 0.30 kWh; cost = 0.30 x 1.20 = 0.36 dollars. (Marking: 2 marks each)
8	(d) If a current of 2.0 A flows for 300 s, find the charge that passes a cross section of the wire.	Answer: charge = 2.0 x 300 = 600 C. (Marking: formula 2 marks, value and unit 2 marks)

Section C: Neutralisation and Salts (16 marks)

Data 3: 25.0 cm3 of 0.10 M sodium hydroxide solution is neutralised by 0.10 M hydrochloric acid. Take the molar mass of sodium chloride as 58.5 g/mol. A separate bottle holds 0.0010 M hydrochloric acid.

#	Question	Answer and marking notes
9	(a) Find the volume of acid required and the number of moles of sodium hydroxide.	Answer: moles = 0.10 x 25.0 / 1000 = 0.0025 mol; with equal concentrations the acid volume is 25.0 cm3. (Marking: 2 marks each)
10	(b) What mass of sodium chloride is formed?	Answer: mass = 0.0025 x 58.5 = 0.146 g to three significant figures. (Marking: mole ratio 2 marks, mass 2 marks)
11	(c) Find the pH of 0.0010 M hydrochloric acid and the pH after tenfold dilution.	Answer: pH = minus log of 0.0010 = 3.0; after tenfold dilution the concentration is 0.00010 M and pH = 4.0. (Marking: 2 marks each)
12	(d) State two observations of this neutralisation and name a suitable indicator with a reason.	Answer: the mixture warms up because the reaction is exothermic; the pH changes sharply at the end point. Phenolphthalein suits because it is pink in alkali and colourless in acid, changing between pH 8 and 10, close to the equivalence point. (Marking: 1 mark per observation, 1 for indicator, 1 for reason)

Section D: Density and the Particle Model (16 marks)

Data 4: a piece of metal has mass 54.0 g and volume 20.0 cm3. An empty bottle has mass 150 g and capacity 500 cm3. A piece of rubber has mass 40.0 g and density 0.80 g/cm3; take the density of water as 1.0 g/cm3.

#	Question	Answer and marking notes
13	(a) Find the density of the metal and suggest which metal it may be.	Answer: density = 54.0 / 20.0 = 2.7 g/cm3, which matches aluminium. (Marking: density 3 marks, identification 1 mark)
14	(b) Fill the bottle with water and find the mass of the water and the total mass.	Answer: mass of water = 250 x 1.0 = 250 g; total = 150 + 250 = 400 g. (Marking: 2 marks each)
15	(c) Predict whether the rubber floats or sinks in water and find the volume of water it displaces.	Answer: 0.80 g/cm3 is less than 1.0 g/cm3 so it floats; a floating body displaces its own mass of water, 40.0 / 1.0 = 40 cm3, while its own volume is 50 cm3. (Marking: 2 marks each)
16	(d) Use the particle model to explain why a gas is easily compressed but a liquid is almost incompressible.	Answer: gas particles are far apart with large empty spaces, so pressure pushes them closer and the volume drops; liquid particles are closely packed and attract each other, leaving little empty space, so the volume hardly changes. (Marking: 2 marks each)

Section E: Inheritance and Ecology (18 marks)

Data 5: a cross Tt x Tt of pea plants gives 400 offspring, where T is the dominant allele for tall. Parents of blood groups AO and BO have 400 children, A and B being codominant. In a food chain the producers fix 5000 kJ and 10 per cent of energy passes to the next level.

#	Question	Answer and marking notes
17	(a) Find the expected numbers of tall and short plants among 400 offspring and give the genotype ratio.	Answer: genotype ratio TT to Tt to tt is 1 to 2 to 1; phenotype ratio tall to short is 3 to 1; so tall = 400 x 3 / 4 = 300 and short = 100. (Marking: ratio 3 marks, 1 mark per number)
18	(b) Find the expected numbers of the four blood groups among 400 children.	Answer: the cross AO x BO gives AB, AO, BO and OO in the ratio 1 to 1 to 1 to 1, so each group has 400 / 4 = 100 children. (Marking: genotypes 2 marks, ratio 1 mark, numbers 2 marks)
19	(c) Find the maximum energy available to the primary and the secondary consumers.	Answer: primary consumers 5000 x 10 per cent = 500 kJ; secondary consumers 500 x 10 per cent = 50 kJ. (Marking: 2 marks each)
20	(d) Explain why a food chain rarely has more than five trophic levels and give one breeding method that raises crop yield.	Answer: only about one tenth of the energy passes on, the rest is lost in respiration, excretion and heat, so higher levels receive too little energy to sustain a population. One breeding method is selective breeding of high yielding varieties, or hybridisation combining disease resistance with yield. (Marking: 2 marks each)

Section F: Experimental Data and Errors (16 marks)

Data 6: a student times a trolley rolling down a slope four times and gets 4.8 s, 5.0 s, 4.9 s and 5.3 s. Another group measures a length as 25.0 cm with an uncertainty of plus or minus 0.5 cm from the smallest scale division.

#	Question	Answer and marking notes
21	(a) Find the mean of the four readings, to a suitable number of significant figures.	Answer: mean = (4.8 + 5.0 + 4.9 + 5.3) / 4 = 5.0 s; the data are given to 0.1 s so the answer stays at 5.0 s. (Marking: 2 marks each)
22	(b) Find the percentage error of that length measurement.	Answer: percentage error = 0.5 / 25.0 x 100 per cent = 2.0 per cent. (Marking: 2 marks each)
23	(c) State one practical way to reduce the random error in the timing.	Answer: repeat the timing several times and average, or use light gates instead of a hand held stopwatch so a sensor records start and stop, removing reaction time bias. (Marking: 2 marks each)
24	(d) A classmate suggests reporting only the fastest time, 4.8 s. Explain why this is inappropriate.	Answer: keeping only the fastest run selects data in a favourable direction, which is bias rather than random error; the other runs are equally valid and dropping them makes the result too low. All readings should be kept and averaged, with any outlier noted and checked. (Marking: 2 marks each)

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