Free practice paper · 2027 format

Debrecen Entrance Exam Sample Test 1, with Answers

Biology, which everyone sits, then Chemistry and Physics, of which you take one on the day. Each subject has 20 five-option questions and the structured written tasks Debrecen's own sample tests set, with every answer explained and a model answer for every task. Written by IvyTest in the format of the University of Debrecen's official sample tests; it is not an official paper.

Start at question 1 The full Debrecen Entrance Exam guide

Before you start

How to sit this paper

Questions: 40: Biology 20 + Chemistry / Physics 20
Time: 90 minutes
Options per question: 5 (A to E)

Subject	Questions	Who sits it
Biology	20	Everyone
Chemistry	20	Your choice of one
Physics	20	Your choice of one

Sit Biology and then the second subject you will choose, each against its own clock: 45 minutes a subject on Debrecen's entrance-exam page (its FAQ says 50). Answer on paper before opening any explanation, and mark the written tasks yourself against the model answers. Debrecen publishes no marking scheme, so the tally at the end counts one mark per correct answer and deducts nothing.

Section 1 of 3

Biology

Compulsory for everyone · 20 questions

Question 1

Biology · Cell Structure and Function · Foundation

Proteins that a cell will secrete, such as digestive enzymes, are synthesised by ribosomes bound to the

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Answer: B

Ribosomes attached to the rough endoplasmic reticulum translate secretory proteins and feed them into the ER lumen as they are made. The Golgi apparatus is the tempting choice, but it modifies, sorts and packages proteins that arrive from the ER; it has no ribosomes and synthesises no protein. The smooth ER also lacks ribosomes and mainly makes lipids.

Question 2

Biology · Cell Structure and Function · Standard

Which statement about mitochondria is correct?

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Answer: A

Mitochondria carry their own small circular DNA and their own ribosomes, which is evidence for their origin from engulfed bacteria (endosymbiosis). The cristae are folds of the inner membrane, not the outer one, and that inner membrane is where the electron transport chain and ATP synthase sit. Mitochondria have two membranes and are found only in eukaryotic cells.

Question 3

Biology · Metabolism and Enzymes · Standard

A competitive inhibitor of an enzyme

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Answer: C

A competitive inhibitor resembles the substrate and occupies the active site, so at a high enough substrate concentration the substrate out-competes it and the maximum rate is still reached. Binding to a separate (allosteric) site and changing the shape of the active site describes non-competitive or allosteric inhibition, which extra substrate cannot overcome. Inhibitors do not increase the maximum rate.

Question 4

Biology · Metabolism and Enzymes · Challenging

During intense exercise, human skeletal muscle cells short of oxygen convert pyruvate into lactate. What is the main benefit of this reaction to the cell?

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Answer: D

Reducing pyruvate to lactate oxidises NADH back to NAD⁺; without that NAD⁺, glycolysis would stop and with it the cell's only oxygen-independent ATP supply. The reaction itself makes no ATP, which is why the extra-ATP option is wrong. Lactic acid fermentation releases no CO₂ at all.

Question 5

Biology · DNA, Replication and Protein Synthesis · Foundation

During translation, the anticodon that base-pairs with a codon is part of

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Answer: E

Each transfer RNA carries a three-base anticodon at one end and its specific amino acid at the other, so the anticodon matches it to the right codon. The codon itself is on the messenger RNA, which is the tempting confusion. Ribosomal RNA forms the ribosome and catalyses peptide bond formation, and DNA takes no direct part in translation.

Question 6

Biology · DNA, Replication and Protein Synthesis · Challenging

A section of the template strand of a gene reads 3'-TAC GGA CTT-5'. Which mRNA sequence is transcribed from it?

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Answer: A

The mRNA is complementary and antiparallel to the template: opposite 3'-TAC-5' it has 5'-AUG-3', and so on, giving 5'-AUG CCU GAA-3' with uracil in place of thymine. 5'-AAG UCC GUA-3' is the same letters written in the wrong direction, so it would be read as different codons. The ATG version contains thymine, which RNA does not have, and the two UAC versions copy the template instead of pairing with it.

Question 7

Biology · Cell Division · Foundation

In which stage of mitosis do sister chromatids separate and move towards opposite poles of the cell?

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Answer: C

In anaphase the cohesion between sister chromatids is broken and spindle fibres pull each chromatid, now a separate chromosome, to opposite poles. In metaphase the chromosomes are lined up at the equator but their chromatids are still joined, which makes it the tempting wrong answer. The S phase is when DNA is replicated, before mitosis begins.

Question 8

Biology · Cell Division · Challenging

A human cell is in metaphase of the second meiotic division. What are the numbers of chromosomes and of chromatids in it?

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Answer: B

The first meiotic division separates homologous chromosomes, so each daughter cell has 23 chromosomes, but each of them still consists of two sister chromatids: 46 chromatids. The sister chromatids separate only in anaphase II, so 23 chromosomes with 23 chromatids describes a cell after meiosis II, not one in metaphase II. 46 chromosomes and 92 chromatids is a cell in mitotic metaphase.

Question 9

Biology · Genetics and Inheritance · Standard

Haemophilia A is an X-linked recessive disorder. An unaffected woman whose father had haemophilia A has children with an unaffected man. What is the probability that any one of their sons will have haemophilia A?

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Answer: C

The woman received her father's only X chromosome, which carries the haemophilia allele, and she is unaffected, so she is a carrier (XᴴXʰ). A son gets his X chromosome from his mother, so each son has a 1 in 2 (50%) chance of receiving Xʰ and being affected. 25% is the chance that a child of unknown sex is an affected son, not the chance for a son.

Question 10

Biology · Genetics and Inheritance · Standard

In the ABO blood group system the alleles Iᴬ and Iᴮ are co-dominant, and both are dominant to i. A man with blood group AB and a woman with blood group O have a child. Which blood groups are possible for the child?

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Answer: A

The man (IᴬIᴮ) passes on either Iᴬ or Iᴮ and the woman (ii) always passes on i, so the child is either Iᴬi (group A) or Iᴮi (group B). Group AB is impossible because the mother has no Iᴬ or Iᴮ allele to give, and group O is impossible because the father has no i allele.

Question 11

Biology · Respiration and Circulation · Foundation

Which blood vessel carries oxygen-rich blood from the lungs to the heart?

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Answer: B

The pulmonary veins return oxygenated blood from the lungs to the left atrium. The pulmonary arteries are the tempting wrong answer: they carry deoxygenated blood the other way, from the right ventricle to the lungs. Arteries and veins are named by the direction of flow relative to the heart, not by their oxygen content.

Question 12

Biology · Respiration and Circulation · Standard

In which form is most of the carbon dioxide carried in the blood from the tissues to the lungs?

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Answer: C

About 70% of CO₂ travels as hydrogencarbonate: carbonic anhydrase in red blood cells rapidly converts CO₂ and water into carbonic acid, which dissociates into H⁺ and HCO₃⁻, and most of the HCO₃⁻ then moves into the plasma. Only a small fraction is dissolved. Haemoglobin does carry some CO₂, but bound to the amino groups of its globin chains, not to the haem iron, which is where oxygen binds.

Question 13

Biology · Digestion and Excretion · Standard

Which statement about bile is correct?

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Answer: D

Bile salts break large fat droplets into many small ones (emulsification), which gives pancreatic lipase a far larger surface to work on. Bile itself contains no digestive enzymes, so the lipase option is wrong even though bile helps fat digestion. Bile is made by the liver, only stored and concentrated in the gallbladder, and released into the duodenum; it is alkaline, not acidic.

Question 14

Biology · Digestion and Excretion · Standard

In a healthy kidney, where is almost all of the filtered glucose reabsorbed?

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Answer: E

Glucose is reabsorbed by carrier proteins (sodium-glucose co-transport) in the proximal convoluted tubule, so normally none reaches the urine. The glomerulus and Bowman's capsule are where glucose is filtered out of the blood, not taken back. The collecting duct mainly adjusts water reabsorption under the control of ADH.

Question 15

Biology · Nervous and Endocrine Regulation · Standard

The rapid depolarisation (rising phase) of an action potential in a neuron is caused mainly by

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Answer: B

When the membrane reaches threshold, voltage-gated sodium channels open and Na⁺ rushes in down its concentration and electrical gradients, making the inside of the membrane positive. Outflow of K⁺ through voltage-gated potassium channels is real but causes repolarisation, the falling phase. The Na⁺/K⁺-ATPase maintains the ion gradients over time rather than producing the rapid rise.

Question 16

Biology · Nervous and Endocrine Regulation · Challenging

A benign tumour of the thyroid gland secretes large amounts of thyroxine regardless of the level of TSH. Which pattern would you expect in the patient's blood?

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Answer: D

Thyroxine normally inhibits the release of TRH from the hypothalamus and of TSH from the anterior pituitary (negative feedback). Here the tumour keeps thyroxine high whatever TSH does, so the high thyroxine suppresses TSH. High thyroxine with high TSH would point to a problem in the pituitary instead, and low thyroxine with high TSH is the pattern of an underactive thyroid.

Question 17

Biology · Muscle and Locomotion · Foundation

In skeletal muscle, which ion is released from the sarcoplasmic reticulum and binds to troponin, allowing myosin to bind to actin?

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Answer: A

An action potential in the muscle fibre makes the sarcoplasmic reticulum release Ca²⁺, which binds to troponin; this shifts tropomyosin away from the myosin-binding sites on actin so that cross-bridges can form. Na⁺ is tempting because its inflow depolarises the muscle fibre membrane, but it does not bind troponin and is not stored in the sarcoplasmic reticulum.

Question 18

Biology · Muscle and Locomotion · Standard

When a skeletal muscle fibre contracts, which part of the sarcomere stays the same length?

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Answer: D

The A band is the length of the thick (myosin) filaments, and in the sliding filament model neither kind of filament changes length, so the A band stays constant. The thin filaments slide towards the centre, so the I band and H zone shorten, the sarcomere shortens, and the zone of overlap gets longer.

Question 19

Biology · Immunity and Defence · Foundation

Which cells secrete antibodies?

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Answer: E

Plasma cells are activated B lymphocytes that have differentiated into antibody factories, with an extensive rough endoplasmic reticulum for making this secreted protein. Helper T cells are tempting because they are needed to activate B cells, but they secrete cytokines, not antibodies. Cytotoxic T cells kill infected cells directly, and neutrophils and macrophages are phagocytes.

Question 20

Biology · Immunity and Defence · Standard

Which of the following is an example of passive immunity?

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Answer: E

In passive immunity the person receives ready-made antibodies and makes none of their own, as when maternal IgG crosses the placenta; the protection fades within months because no memory cells are formed. Vaccination is tempting because it is given from outside, but the vaccine is an antigen that makes the body produce its own antibodies and memory cells, so it is active (artificial) immunity. Recovery from infection and memory B cells are also active immunity.

Written tasks

Answer on paper, then mark your work against the model answer.

1. Definitions

5 points · Nervous and Endocrine Regulation

Define the following five terms in one or two sentences each (1 point each).

1. Homeostasis

2. Hormone

3. Synapse

4. Reflex arc

5. Myelin sheath

Show the model answer

1. Homeostasis (1 point): keeping the internal environment of the body (for example body temperature, blood glucose, pH and water content) relatively constant, within narrow limits, despite changes outside or inside the body; it is achieved by regulatory mechanisms, mostly negative feedback. The point needs both the idea of a constant internal environment and that it is actively regulated.

2. Hormone (1 point): a chemical messenger secreted by an endocrine (ductless) gland or cell directly into the blood, which carries it to target cells that have specific receptors for it, where it changes their activity. The point needs secretion into the blood and action on distant target cells.

3. Synapse (1 point): the junction between a neuron and another cell (another neuron, a muscle fibre or a gland cell) across which a signal is passed; at a chemical synapse the signal crosses the synaptic cleft as a neurotransmitter released from the presynaptic ending and binding to receptors on the postsynaptic membrane. The point needs the junction between two cells and the transfer of the signal across it.

4. Reflex arc (1 point): the nerve pathway of a reflex, from receptor to sensory (afferent) neuron to the integrating centre in the brain or spinal cord (often through an interneuron) to motor (efferent) neuron to effector; it produces a rapid, automatic (involuntary) response to a stimulus. The point needs the chain receptor, sensory neuron, centre, motor neuron, effector (or an equivalent description).

5. Myelin sheath (1 point): an insulating, lipid-rich layer of membrane wrapped around an axon, formed by Schwann cells in the peripheral nervous system and by oligodendrocytes in the central nervous system; it is interrupted at the nodes of Ranvier, so the impulse jumps from node to node (saltatory conduction) and travels much faster. The point needs the insulating layer around the axon and its effect of speeding up conduction.

2. Calculation

5 points · Genetics and Inheritance

In guinea pigs, assume that black coat colour (B) is dominant to brown (b), and short hair (S) is dominant to long hair (s). The two genes are on different chromosomes.

Two guinea pigs that are heterozygous for both genes are crossed, and over several litters they have 320 offspring in total.

(a) Give the genotype of the parents. (1 point)

(b) List the types of gamete each parent can produce, with their proportions. (1 point)

(d) How many of the 320 offspring are expected to be black with long hair? Show your working. (1 point)

(e) How many of these black, long-haired offspring are expected to be homozygous for both genes? Show your working. (1 point)

Show the model answer

(a) Both parents are BbSs (black, short-haired). The cross is BbSs × BbSs. (1 point)

(b) Because the genes are on different chromosomes they assort independently (Mendel's second law), so each parent produces four kinds of gamete in equal proportions: BS, Bs, bS and bs, each 1/4. (1 point)

(c) The 4 × 4 combinations give 16 equally likely outcomes, which fall into the ratio 9 black short-haired : 3 black long-haired : 3 brown short-haired : 1 brown long-haired. (1 point)

(d) Black and long-haired means B_ss, which is 3/16 of the offspring: probability of black (B_) = 3/4, probability of long hair (ss) = 1/4, and 3/4 × 1/4 = 3/16. Expected number = 3/16 × 320 = 60. (1 point for 60 with this working)

(e) Of the 3/16 black long-haired offspring, 1/16 are BBss and 2/16 are Bbss, so 1/3 of them are homozygous for both genes: 1/3 × 60 = 20 (equivalently 1/16 × 320 = 20). (1 point for 20 with this working)

3. Describe

10 points · Respiration and Circulation

Describe how air is drawn into the lungs during quiet (resting) inspiration and how it leaves them during quiet expiration. Name the muscles involved and explain the changes in volume and pressure that make the air move. (10 points)

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One point is awarded for each of the following ten points; any equivalent wording earns the point.

Inspiration:

1. Quiet inspiration is an active process: it needs muscle contraction. (1 point)

2. The diaphragm contracts and flattens, moving downwards. (1 point)

3. The external intercostal muscles contract, raising the ribs and sternum upwards and outwards. (1 point)

4. As a result the volume of the thoracic cavity increases. (1 point)

5. The lungs follow the chest wall because the pleural membranes are held together by the thin layer of pleural fluid between them (the pressure in the pleural space is below atmospheric), so lung volume increases too. (1 point)

6. The increase in volume lowers the pressure in the alveoli below atmospheric pressure (Boyle's law: at constant temperature, pressure and volume of a gas are inversely related). (1 point)

7. Air flows into the lungs, down the pressure gradient, through the airways until the alveolar pressure equals atmospheric pressure. (1 point)

Expiration:

8. Quiet expiration is passive: the diaphragm and the external intercostal muscles relax. (1 point)

9. The elastic recoil of the stretched lungs and chest wall (and the surface tension in the alveoli) returns the thorax and lungs to their smaller resting volume. (1 point)

10. The smaller volume raises the alveolar pressure above atmospheric pressure, so air flows out of the lungs until the pressures are equal again. (1 point)

Note: forced expiration (for example during exercise) is active, using the internal intercostal muscles and the abdominal muscles; mentioning this is correct but earns no extra point, and it does not replace point 8.

4. Problem solving

10 points · Metabolism and Enzymes

A student investigated the effect of temperature on human salivary amylase. In each tube, 2 cm³ of 1% starch solution and 1 cm³ of the same amylase solution, both at pH 7, were first brought separately to the test temperature and then mixed. Every 30 seconds a drop of the mixture was added to iodine solution. The student recorded the time at which the drop no longer turned the iodine blue-black.

Results: 10 °C: 600 s; 20 °C: 300 s; 30 °C: 150 s; 40 °C: 120 s; 50 °C: 240 s; 60 °C: the iodine still turned blue-black after 900 s, when the test was stopped.

(a) The rate of reaction can be expressed as 1000 / time (in arbitrary units). Calculate the rate at 20 °C and at 40 °C, and state how many times faster the reaction was at 20 °C than at 10 °C. (2 points)

(b) What is the optimum temperature of the enzyme according to these results? (1 point)

(d) Explain the result at 60 °C. (2 points)

(e) The student then took the tube kept at 10 °C and the tube kept at 60 °C, warmed or cooled both to 40 °C, and added fresh starch to each. Predict what happened in each tube, and explain. (2 points)

(f) Describe a suitable control tube for this experiment and say what it shows. (1 point)

(g) Name the main product of the reaction catalysed by amylase. (1 point)

Show the model answer

(a) Rate at 20 °C = 1000 / 300 = 3.33 and rate at 40 °C = 1000 / 120 = 8.33 arbitrary units (1 point for both values). At 10 °C the rate is 1000 / 600 = 1.67, so the reaction at 20 °C was 3.33 / 1.67 = 2 times faster: doubling the rate for a 10 °C rise (equivalently, 600 s / 300 s = 2) (1 point).

(b) About 40 °C: it gave the shortest time (fastest rate) of the temperatures tested. Because only 10 °C steps were tested, the true optimum lies somewhere between 30 °C and 50 °C. (1 point for 40 °C; the note on its precision is good practice but not required)

(c) At low temperature the enzyme and substrate molecules have less kinetic energy, so they collide less often and fewer enzyme–substrate complexes form per unit time; the enzyme is not damaged, only slower. (1 point)

(d) At 60 °C the enzyme is denatured (1 point): the heat breaks the weak bonds (hydrogen bonds and ionic bonds) that hold its tertiary structure, so the shape of the active site changes and starch no longer fits into it, no enzyme–substrate complexes form and the starch is not broken down (1 point).

(e) In the tube from 10 °C the starch was digested at about the normal rate for 40 °C (the iodine soon stops turning blue-black), because cold only slows an enzyme and does not change its structure, so the inactivation is reversible (1 point). In the tube from 60 °C the starch was still not digested (iodine stays blue-black), because heat denaturation of the enzyme is irreversible: the active site does not regain its shape on cooling (1 point).

(f) A tube with starch solution and 1 cm³ of water (or of boiled, denatured amylase) instead of active amylase, kept at the same temperature: the iodine keeps turning blue-black, which shows that the starch disappears only because of the active enzyme and not by itself or because of the temperature alone. (1 point)

(g) Maltose (a disaccharide). (1 point; 'glucose' does not earn the point, as amylase produces mainly maltose)

Section 2 of 3

Chemistry

Your choice: this or Physics · 20 questions

Question 21

Chemistry · Measurement and Classification of Matter · Foundation

A blood plasma sample contains glucose at a concentration of 0.90 g/dm³. What is this concentration expressed in mg/cm³?

Show answer and explanation

Answer: D

0.90 g is 900 mg, and 1 dm³ is 1000 cm³, so the concentration is 900 mg / 1000 cm³ = 0.90 mg/cm³. The tempting 900 mg/cm³ comes from converting grams to milligrams but forgetting that the same mass is spread over 1000 times as many cubic centimetres.

Question 22

Chemistry · Atomic Structure and the Periodic Table · Foundation

An ion contains 26 protons, 30 neutrons and 23 electrons. Which ion is it? (Atomic numbers: Mn 25, Fe 26, Co 27, Zn 30.)

Show answer and explanation

Answer: A

The atomic number (26 protons) identifies the element as iron; the mass number is protons + neutrons = 26 + 30 = 56; the charge is protons − electrons = 26 − 23 = +3. Fe3− is tempting but wrong: an atom that has lost electrons has more protons than electrons, so its charge is positive.

Question 23

Chemistry · Atomic Structure and the Periodic Table · Challenging

Which order of first ionisation energy is correct for the period 3 elements sodium, magnesium and aluminium?

Show answer and explanation

Answer: B

Ionisation energy generally rises across a period, so sodium is lowest, but aluminium is an exception: its outermost electron is in a 3p orbital, which is higher in energy and partly shielded by the filled 3s subshell, so it is easier to remove than magnesium's 3s electron. The values are about 496 (Na), 578 (Al) and 738 (Mg) kJ/mol. Na < Mg < Al is tempting because it simply follows the general trend.

Question 24

Chemistry · Chemical Bonding and Intermolecular Forces · Standard

Which molecule contains polar covalent bonds but is non-polar overall?

Show answer and explanation

Answer: C

CO2 is linear (O=C=O), so its two equal C=O bond dipoles point in opposite directions and cancel. H2O is tempting because it is also a triatomic molecule with two polar bonds, but it is bent, so its bond dipoles do not cancel and the molecule is polar.

Question 25

Chemistry · Chemical Bonding and Intermolecular Forces · Foundation

Ethanol (CH3–CH2–OH) and dimethyl ether (CH3–O–CH3) have the same molar mass, 46 g/mol, yet ethanol boils at 78 °C and dimethyl ether at −24 °C. What is the main reason for the difference?

Show answer and explanation

Answer: A

A hydrogen bond needs a hydrogen atom bonded directly to O, N or F; ethanol has an O–H group, whereas every hydrogen in dimethyl ether is bonded to carbon. Dimethyl ether is not non-polar (its C–O–C unit is bent, so it has a dipole), and boiling separates molecules without breaking covalent bonds inside them.

Question 26

Chemistry · Stoichiometry and Mass Relationships · Foundation

How many moles of atoms in total are present in 9.0 g of water? (Relative atomic masses: H = 1, O = 16.)

Show answer and explanation

Answer: C

The molar mass of H2O is 2 × 1 + 16 = 18 g/mol, so 9.0 g is 9.0 / 18 = 0.50 mol of molecules. Each molecule contains 3 atoms, so there are 3 × 0.50 = 1.5 mol of atoms. 0.50 mol is tempting but counts molecules, not atoms.

Question 27

Chemistry · Stoichiometry and Mass Relationships · Challenging

4.8 g of magnesium is added to a solution containing 0.30 mol of HCl. The reaction is Mg + 2HCl → MgCl2 + H2. What volume of hydrogen is produced, measured at STP (0 °C, 101.3 kPa, molar volume 22.41 dm³/mol)? (Relative atomic mass: Mg = 24.)

Show answer and explanation

Answer: B

Mg: 4.8 / 24 = 0.20 mol. 0.20 mol of Mg would need 0.40 mol of HCl, but only 0.30 mol is present, so HCl is the limiting reactant. 0.30 mol HCl gives 0.30 / 2 = 0.15 mol H2, and 0.15 × 22.41 = 3.36 dm³. 4.48 dm³ is tempting but assumes all the magnesium reacts.

Question 28

Chemistry · States of Matter and Gas Laws · Standard

A sealed, rigid steel container holds a gas at 27 °C and 150 kPa. The container is heated to 127 °C. Assuming ideal behaviour, what is the new pressure? (T/K = t/°C + 273.)

Show answer and explanation

Answer: C

At constant volume and amount, p is proportional to absolute temperature: p2 = p1 × T2 / T1 = 150 kPa × 400 K / 300 K = 200 kPa. About 706 kPa is tempting but comes from using Celsius temperatures (150 × 127 / 27), which is wrong because the gas laws need kelvin.

Question 29

Chemistry · Solutions and Chemical Equilibrium · Foundation

What mass of sodium hydroxide is needed to make 250 cm³ of a 0.20 mol/dm³ NaOH solution? (Relative atomic masses: Na = 23, O = 16, H = 1.)

Show answer and explanation

Answer: B

n = c × V = 0.20 mol/dm³ × 0.250 dm³ = 0.050 mol, and the molar mass of NaOH is 23 + 16 + 1 = 40 g/mol, so m = 0.050 × 40 = 2.0 g. 8.0 g is tempting but is the mass for a full 1 dm³ (0.20 × 40), ignoring that only 250 cm³ is made.

Question 30

Chemistry · Solutions and Chemical Equilibrium · Standard

Ammonia is made by the equilibrium N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ/mol. Which change, made to the mixture at equilibrium, increases the equilibrium yield of ammonia?

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Answer: E

The forward reaction turns 4 mol of gas into 2 mol of gas, so raising the pressure shifts the equilibrium towards ammonia (Le Chatelier's principle). A catalyst is tempting, but it speeds up the forward and reverse reactions equally, so equilibrium is reached sooner without changing the yield; raising the temperature favours the endothermic reverse reaction.

Question 31

Chemistry · Acids and Bases · Standard

What is the pH of a 0.010 mol/dm³ solution of barium hydroxide, Ba(OH)2, at 25 °C? Assume complete dissociation. (Kw = 1.0 × 10⁻¹⁴; log 2 = 0.30.)

Show answer and explanation

Answer: E

Each Ba(OH)2 releases two OH− ions, so [OH−] = 2 × 0.010 = 0.020 mol/dm³. pOH = −log(2 × 10⁻²) = 2 − 0.30 = 1.70, and pH = 14.00 − 1.70 = 12.30. 12.00 is tempting but treats Ba(OH)2 as if it released only one OH− ion.

Question 32

Chemistry · Acids and Bases · Standard

A weak monoprotic acid HA has Ka = 1.0 × 10⁻⁵. What is the pH of a 0.10 mol/dm³ solution of HA? (Assume the fraction of HA that dissociates is small.)

Show answer and explanation

Answer: E

Ka = [H+][A−] / [HA] ≈ [H+]² / 0.10, so [H+]² = 1.0 × 10⁻⁵ × 0.10 = 1.0 × 10⁻⁶ and [H+] = 1.0 × 10⁻³ mol/dm³, giving pH = 3.0. pH 1.0 is tempting but treats HA as a strong acid that dissociates completely.

Question 33

Chemistry · Thermochemistry · Standard

Standard enthalpies of formation: CH4(g) −75 kJ/mol, CO2(g) −394 kJ/mol, H2O(l) −286 kJ/mol. What is the standard enthalpy change for CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)?

Show answer and explanation

Answer: A

ΔH = ΣΔfH(products) − ΣΔfH(reactants) = [−394 + 2 × (−286)] − [−75 + 0] = −966 + 75 = −891 kJ/mol; O2 is an element in its standard state, so its ΔfH is zero. −1041 kJ/mol is tempting but comes from adding the enthalpy of formation of methane instead of subtracting it.

Question 34

Chemistry · Oxidation and Reduction · Standard

In acidic solution, permanganate ions oxidise iron(II) ions: MnO4− + Fe2+ + H+ → Mn2+ + Fe3+ + H2O (unbalanced). How many moles of Fe2+ are oxidised by 1 mol of MnO4−?

Show answer and explanation

Answer: D

Manganese goes from +7 in MnO4− to +2 in Mn2+, gaining 5 electrons, while each Fe2+ loses only 1 electron to become Fe3+, so 5 Fe2+ are needed. The balanced equation is MnO4− + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O (charge: +17 on each side). 8 is tempting but is the coefficient of H+, not of Fe2+.

Question 35

Chemistry · Inorganic Chemistry · Standard

Chlorine water is added to a colourless solution of potassium iodide, and the solution turns brown. Which statement describes what happens?

Show answer and explanation

Answer: B

Chlorine is the stronger oxidising agent, so Cl2 + 2I− → 2Cl− + I2: iodide loses electrons (oxidised, giving brown iodine) and chlorine gains them (reduced to chloride). The option with the roles reversed is tempting, but turning I− into I2 is a loss of electrons, which is oxidation, not reduction.

Question 36

Chemistry · Inorganic Chemistry · Standard

Which oxide reacts with water to form a strong acid?

Show answer and explanation

Answer: D

Sulfur trioxide reacts with water to give sulfuric acid, SO3 + H2O → H2SO4, which is a strong acid. CO2 is tempting because it is an acidic oxide, but it gives carbonic acid, H2CO3, which is a weak acid; Na2O gives the base NaOH, CO is a neutral oxide and SiO2 does not dissolve in water.

Question 37

Chemistry · Hydrocarbons and Aromatics · Foundation

Which compound reacts rapidly with bromine at room temperature, in the dark and without a catalyst, so that the red-brown colour of the bromine disappears?

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Answer: D

Cyclohexene has a C=C double bond, which adds bromine at once to give 1,2-dibromocyclohexane. Benzene is tempting because it is unsaturated, but its delocalised ring is very stable: it does not add bromine and reacts with it only by substitution with a catalyst such as FeBr3. Alkanes such as hexane and cyclohexane react with bromine only in light (UV).

Question 38

Chemistry · Functional Groups and Derivatives · Standard

Propan-2-ol, CH3–CH(OH)–CH3, is heated with acidified potassium dichromate solution. What is the organic product?

Show answer and explanation

Answer: E

Propan-2-ol is a secondary alcohol, and secondary alcohols are oxidised to ketones: here propanone, CH3–CO–CH3, which resists further oxidation. Propanal (and then propanoic acid) is tempting, but aldehydes and carboxylic acids come from oxidising a primary alcohol such as propan-1-ol.

Question 39

Chemistry · Functional Groups and Derivatives · Challenging

Which statement correctly compares the acidity of ethanol, phenol and ethanoic acid?

Show answer and explanation

Answer: C

The order of acid strength is ethanol < phenol < ethanoic acid: the phenoxide ion is stabilised by delocalisation of its charge into the ring, and the ethanoate ion even more by sharing it between two oxygen atoms. Phenol is acidic enough to react with NaOH but weaker than carbonic acid, so it does not release CO2 from hydrogencarbonate, while ethanoic acid does. The claim that phenol does not react with NaOH is tempting but wrong: phenol's OH is attached to a benzene ring, which makes it far more acidic than an alcohol.

Question 40

Chemistry · Isomerism · Challenging

Consider all the constitutional (structural) isomers with the molecular formula C4H9Cl. How many are there, and how many of them contain a chiral (asymmetric) carbon atom? (Count each constitutional isomer once; do not count enantiomers separately.)

Show answer and explanation

Answer: A

The four isomers are 1-chlorobutane (CH3–CH2–CH2–CH2–Cl), 2-chlorobutane (CH3–CHCl–CH2–CH3), 1-chloro-2-methylpropane ((CH3)2CH–CH2–Cl) and 2-chloro-2-methylpropane ((CH3)3C–Cl). Only C-2 of 2-chlorobutane carries four different groups (H, Cl, CH3 and C2H5). 1-chloro-2-methylpropane is tempting, but its C-2 carries two identical CH3 groups, so it is not chiral.

Written tasks

Answer on paper, then mark your work against the model answer.

1. Complete the statements

5 points · Atomic Structure and the Periodic Table

Fill in the gap in each statement with the correct word or number. (1 point each)

1. The mass number of an atom is the total number of __________ in its nucleus.

2. Atoms of the same element that contain different numbers of neutrons are called __________.

3. The energy needed to remove one mole of electrons from one mole of gaseous atoms, forming one mole of gaseous 1+ ions, is called the __________.

4. According to the Pauli exclusion principle, one atomic orbital can hold at most __________ electrons, and these must have opposite spins.

5. Elements in the same group of the periodic table have the same number of __________, which is why their chemical properties are similar.

Show the model answer

1. protons and neutrons (nucleons). 1 point. "Neutrons" alone or "particles" earns no point.

2. isotopes. 1 point.

3. first ionisation energy. 1 point. "Ionisation energy" without "first" is accepted; "electron affinity" earns no point.

4. two (2). 1 point.

5. valence (outer-shell) electrons. 1 point. "Electrons" alone earns no point, since the total number of electrons differs within a group.

Total: 5 points.

2. Calculation

5 points · Acids and Bases

40.0 cm³ of 0.100 mol/dm³ hydrochloric acid is mixed with 60.0 cm³ of 0.0500 mol/dm³ sodium hydroxide solution at 25 °C. Assume that the volumes are additive and that both HCl and NaOH are fully dissociated.

a) Calculate the amount (in mol) of HCl and of NaOH in the two solutions before mixing. (1 point)

b) Which reactant is in excess, and by how many moles? (1 point)

c) Calculate the concentration of the excess ion in the final mixture. (1 point)

d) Calculate the pH of the final mixture. (1 point)

e) What further volume of the 0.0500 mol/dm³ NaOH solution would be needed to neutralise the mixture exactly? (1 point)

Show the model answer

a) n(HCl) = c × V = 0.100 mol/dm³ × 0.0400 dm³ = 4.00 × 10⁻³ mol. n(NaOH) = 0.0500 mol/dm³ × 0.0600 dm³ = 3.00 × 10⁻³ mol. 1 point (both values needed).

b) HCl + NaOH → NaCl + H2O reacts 1 : 1, so 3.00 × 10⁻³ mol of each reacts and HCl is in excess by 4.00 × 10⁻³ − 3.00 × 10⁻³ = 1.00 × 10⁻³ mol (of H+). 1 point.

c) Total volume = 40.0 + 60.0 = 100.0 cm³ = 0.1000 dm³. [H+] = 1.00 × 10⁻³ mol / 0.1000 dm³ = 1.00 × 10⁻² mol/dm³. 1 point.

d) pH = −log[H+] = −log(1.00 × 10⁻²) = 2.00. 1 point (follow-through from an incorrect [H+] is accepted).

e) The excess 1.00 × 10⁻³ mol of H+ needs 1.00 × 10⁻³ mol of OH−: V = n / c = 1.00 × 10⁻³ mol / 0.0500 mol/dm³ = 0.0200 dm³ = 20.0 cm³. 1 point.

Total: 5 points.

3. Write the equations

10 points · Functional Groups and Derivatives

Write a balanced equation for each reaction, giving organic compounds as condensed structural formulae (for example CH3–CH2–OH), and name the product(s) asked for. (2 points each: 1 point for the correct balanced equation, 1 point for the correct name(s).)

a) Ethanoic acid is heated with ethanol in the presence of a little concentrated sulfuric acid as catalyst. Name the ester formed.

b) The ester from part (a) is boiled with sodium hydroxide solution. Name both organic products.

c) Ethanal is oxidised (use [O] to represent the oxidising agent). Name the organic product.

d) Ethanoic acid reacts with sodium hydrogencarbonate solution. Name the organic product.

e) Ethylamine reacts with hydrochloric acid. Name the product.

Show the model answer

a) CH3–COOH + CH3–CH2–OH ⇌ CH3–COO–CH2–CH3 + H2O (H2SO4 catalyst). 1 point for the equation (water must be shown; an ordinary arrow is accepted). The ester is ethyl ethanoate (ethyl acetate). 1 point.

b) CH3–COO–CH2–CH3 + NaOH → CH3–COONa + CH3–CH2–OH. 1 point for the equation (CH3–COO− Na+ for the salt is accepted). The products are sodium ethanoate (sodium acetate) and ethanol. 1 point; both names are needed. This is alkaline hydrolysis (saponification) of the ester.

c) CH3–CHO + [O] → CH3–COOH. 1 point for the equation. The product is ethanoic acid (acetic acid). 1 point.

d) CH3–COOH + NaHCO3 → CH3–COONa + H2O + CO2. 1 point for the equation (CO2 and H2O must both be shown). The organic product is sodium ethanoate (sodium acetate). 1 point.

e) CH3–CH2–NH2 + HCl → CH3–CH2–NH3+ Cl−. 1 point for the equation (CH3–CH2–NH3Cl is accepted). The product is ethylammonium chloride; the amine acts as a Brønsted–Lowry base, accepting a proton. 1 point.

Total: 10 points.

4. Problem solving

10 points · Stoichiometry and Mass Relationships

An organic compound X contains only carbon, hydrogen and oxygen. Complete combustion of 1.48 g of X gives 3.52 g of carbon dioxide and 1.80 g of water.

In a separate experiment, 0.370 g of X is completely vaporised; the vapour occupies 200 cm³ at 127 °C and 83.14 kPa.

Relative atomic masses: C = 12, H = 1.0, O = 16. R = 8.314 J/(mol·K); T/K = t/°C + 273; molar volume at STP (0 °C, 101.3 kPa) = 22.41 dm³/mol.

a) Calculate the mass of carbon and the mass of hydrogen in 1.48 g of X. (2 points)

b) Calculate the mass of oxygen in 1.48 g of X and determine the empirical formula of X. (2 points)

c) Using the ideal gas equation, calculate the molar mass of X. (2 points)

d) Give the molecular formula of X. (1 point)

e) Write the balanced equation for the complete combustion of X, and calculate the volume of oxygen, measured at STP, used to burn 1.48 g of X. (2 points)

f) X reacts with sodium metal, releasing hydrogen, and is oxidised by acidified potassium dichromate to a ketone. Give the condensed structural formula and the name of X. (1 point)

Show the model answer

a) n(CO2) = 3.52 g / 44 g/mol = 0.0800 mol, so n(C) = 0.0800 mol and m(C) = 0.0800 × 12 = 0.960 g. 1 point. n(H2O) = 1.80 g / 18 g/mol = 0.100 mol, so n(H) = 2 × 0.100 = 0.200 mol and m(H) = 0.200 × 1.0 = 0.200 g. 1 point.

b) m(O) = 1.48 − 0.960 − 0.200 = 0.320 g, so n(O) = 0.320 / 16 = 0.0200 mol. 1 point. C : H : O = 0.0800 : 0.200 : 0.0200 = 4 : 10 : 1, so the empirical formula is C4H10O. 1 point.

c) pV = nRT, so n = pV / RT = (83 140 Pa × 2.00 × 10⁻⁴ m³) / (8.314 J/(mol·K) × 400 K) = 16.628 / 3325.6 = 5.00 × 10⁻³ mol. 1 point (correct use of SI units: Pa, m³, K). M = m / n = 0.370 g / 5.00 × 10⁻³ mol = 74.0 g/mol. 1 point.

d) The empirical formula mass of C4H10O is 4 × 12 + 10 × 1.0 + 16 = 74, equal to the molar mass, so the molecular formula is C4H10O. 1 point.

e) C4H10O + 6O2 → 4CO2 + 5H2O (oxygen atoms: 1 + 12 = 13 on the left, 8 + 5 = 13 on the right). 1 point. n(X) = 1.48 / 74 = 0.0200 mol, so n(O2) = 6 × 0.0200 = 0.120 mol and V(O2) = 0.120 × 22.41 = 2.69 dm³. 1 point.

f) Reaction with sodium shows an O–H group (an alcohol, not an ether), and oxidation to a ketone shows a secondary alcohol. The only secondary alcohol with formula C4H10O is butan-2-ol, CH3–CH(OH)–CH2–CH3. 1 point (formula and name both needed).

Total: 10 points.

Section 3 of 3

Physics

Your choice: this or Chemistry · 20 questions

Question 41

Physics · Kinematics and Vectors · Foundation

A stone is dropped from rest from a high bridge and falls freely. Taking g = 10 m/s² and neglecting air resistance, how far has it fallen after 3.0 s?

Show answer and explanation

Answer: C

For free fall from rest, s = ½gt² = ½ × 10 × 3.0² = ½ × 10 × 9.0 = 45 m. The tempting 90 m forgets the factor ½, and 30 m is the speed after 3.0 s (v = gt = 30 m/s), not the distance.

Question 42

Physics · Kinematics and Vectors · Standard

A swimmer swims at 0.80 m/s relative to the water. She crosses a river 60 m wide, with straight parallel banks, whose water flows at a uniform 0.60 m/s. She always heads straight across, perpendicular to the banks. How far downstream of her starting point does she reach the opposite bank?

Show answer and explanation

Answer: A

The two velocity components are independent: crossing takes t = 60 m ÷ 0.80 m/s = 75 s, and in that time the current carries her 0.60 m/s × 75 s = 45 m downstream. 75 m is the length of her actual path (resultant speed √(0.80² + 0.60²) = 1.0 m/s for 75 s), not the downstream distance.

Question 43

Physics · Newton's Laws and Forces · Foundation

A book rests on a horizontal table. The Earth pulls the book downwards with a gravitational force (the book's weight). According to Newton's third law, which force is the reaction (the other member of the pair) to this force?

Show answer and explanation

Answer: D

A Newton's third law pair comes from one interaction and acts on two different bodies: the Earth pulls the book down, so the book pulls the Earth up with a gravitational force of equal size. The table's normal force is equal and opposite to the weight here only because the book is in equilibrium (Newton's second law); it acts on the same body, the book, and is a contact force, so it is not the third-law partner.

Question 44

Physics · Newton's Laws and Forces · Standard

Two blocks, of mass 2.0 kg and 3.0 kg, sit side by side in contact on a frictionless horizontal surface. A horizontal force of 20 N is applied to the 2.0 kg block, pushing it against the 3.0 kg block so that both accelerate together. What is the magnitude of the force that the 2.0 kg block exerts on the 3.0 kg block?

Show answer and explanation

Answer: D

Both blocks share one acceleration: a = 20 N ÷ (2.0 + 3.0) kg = 4.0 m/s². The only horizontal force on the 3.0 kg block is the push from the 2.0 kg block, so it equals 3.0 kg × 4.0 m/s² = 12 N. The tempting 8.0 N (2.0 kg × 4.0 m/s²) is the net force on the 2.0 kg block, not the contact force.

Question 45

Physics · Work, Energy and Momentum · Standard

A 2.0 kg trolley moving at 6.0 m/s along a level, frictionless track collides with a stationary 1.0 kg trolley. The two trolleys stick together. How much kinetic energy is converted into other forms in the collision?

Show answer and explanation

Answer: C

Momentum is conserved: 2.0 × 6.0 = 12 kg·m/s = 3.0 kg × v, so v = 4.0 m/s. Kinetic energy falls from ½ × 2.0 × 6.0² = 36 J to ½ × 3.0 × 4.0² = 24 J, so 36 − 24 = 12 J is converted. 24 J is the kinetic energy remaining after the collision, not the amount lost; 0 J would apply only to an elastic collision.

Question 46

Physics · Work, Energy and Momentum · Foundation

An electric hoist lifts a 50 kg load vertically through 12 m at constant speed in 20 s. Taking g = 10 m/s², what is the useful power output of the hoist?

Show answer and explanation

Answer: B

At constant speed the hoist does work W = mgh = 50 × 10 × 12 = 6000 J, so its power is P = W ÷ t = 6000 J ÷ 20 s = 300 W. 6000 is the work done in joules, not the rate of doing work.

Question 47

Physics · Rotational Motion and Gravitation · Standard

A uniform plank 3.0 m long is balanced horizontally on a pivot at its midpoint. A child of mass 30 kg sits 0.60 m to the left of the pivot. How far to the right of the pivot must a child of mass 20 kg sit for the plank to stay balanced horizontally?

Show answer and explanation

Answer: E

The plank's own weight acts at the pivot, so it has no torque about it. Balancing the children's torques about the pivot: 30g × 0.60 = 20g × d, so d = 18 ÷ 20 = 0.90 m (g cancels). 0.40 m comes from inverting the mass ratio; the lighter child must sit farther out, not closer in.

Question 48

Physics · Solids and Fluids · Foundation

What is the gauge pressure (the pressure in excess of atmospheric pressure) at a depth of 5.0 m in a freshwater lake? Take the density of water as 1000 kg/m³ and g = 10 m/s².

Show answer and explanation

Answer: B

The pressure of a liquid column is p = ρgh = 1000 × 10 × 5.0 = 5.0 × 10⁴ Pa. 1.5 × 10⁵ Pa is the absolute pressure (adding atmospheric pressure of about 1.0 × 10⁵ Pa), but the question asks only for the excess over atmospheric.

Question 49

Physics · Solids and Fluids · Challenging

A solid metal object hangs from a spring balance. The balance reads 30 N when the object is in air and 20 N when the object is completely submerged in water of density 1000 kg/m³. Taking g = 10 m/s², what is the density of the metal?

Show answer and explanation

Answer: E

The buoyant force is 30 − 20 = 10 N, which by Archimedes' principle equals the weight of displaced water: V = 10 ÷ (1000 × 10) = 1.0 × 10⁻³ m³. The object's mass is 30 ÷ 10 = 3.0 kg, so its density is 3.0 ÷ (1.0 × 10⁻³) = 3.0 × 10³ kg/m³. Using the apparent weight of 20 N for the mass gives the wrong answer 2.0 × 10³ kg/m³.

Question 50

Physics · Thermodynamics · Standard

0.20 kg of water at 80 °C is mixed with 0.30 kg of water at 20 °C in an insulated container whose own heat capacity is negligible. What is the final temperature of the mixture?

Show answer and explanation

Answer: B

Heat lost by the hot water equals heat gained by the cold water, and the specific heat cancels: 0.20 × (80 − T) = 0.30 × (T − 20), so 16 − 0.20T = 0.30T − 6 and T = 22 ÷ 0.50 = 44 °C. 50 °C is the simple average, which would be right only for equal masses; the larger cold mass pulls the result below 50 °C.

Question 51

Physics · Thermodynamics · Standard

An ideal gas in a cylinder absorbs 1000 J of heat while it expands at a constant pressure of 2.0 × 10⁵ Pa from a volume of 3.0 L to 5.0 L. What is the change in the internal energy of the gas? (1 L = 1.0 × 10⁻³ m³)

Show answer and explanation

Answer: C

The gas does work on its surroundings W = pΔV = 2.0 × 10⁵ × 2.0 × 10⁻³ = 400 J. By the first law, ΔU = Q − W(by gas) = 1000 − 400 = +600 J. +1400 J adds the work instead of subtracting it, but an expanding gas gives energy away as work.

Question 52

Physics · Vibrations and Waves · Foundation

A tuning fork produces a sound of frequency 680 Hz in air, where the speed of sound is 340 m/s. What is the wavelength of the sound?

Show answer and explanation

Answer: A

From v = fλ, λ = v ÷ f = 340 ÷ 680 = 0.50 m. 2.0 m is f ÷ v (the ratio inverted), and 2.3 × 10⁵ m is the product of the two values.

Question 53

Physics · Vibrations and Waves · Standard

On a windless day, an ambulance with its siren sounding at a constant frequency drives at constant speed in a straight line directly towards a stationary listener who is standing in its path, well ahead of it. Which statement is true?

Show answer and explanation

Answer: A

A source moving towards the listener crowds its wavefronts together ahead of it, so the wavelength is shorter and the heard frequency is higher; with constant source speed along the line to the listener this shifted frequency stays constant. The sound gets louder as the ambulance approaches, but the pitch does not keep rising, and the speed of sound depends only on the air, not on the motion of the source.

Question 54

Physics · Electricity and Circuits · Standard

A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel with each other across the terminals of a 12 V battery of negligible internal resistance. What is the current drawn from the battery?

Show answer and explanation

Answer: D

For parallel resistors 1/R = 1/6.0 + 1/3.0 = 0.50 Ω⁻¹, so R = 2.0 Ω and I = V ÷ R = 12 ÷ 2.0 = 6.0 A (equivalently 2.0 A + 4.0 A through the two branches). 1.3 A treats the resistors as if they were in series (12 ÷ 9.0), and 24 A forgets to invert 1/R.

Question 55

Physics · Electricity and Circuits · Challenging

A 4.0 Ω resistor is connected in series with a pair of resistors, 6.0 Ω and 12 Ω, that are connected in parallel with each other. The whole combination is connected across a 24 V battery of negligible internal resistance. What is the current in the 12 Ω resistor?

Show answer and explanation

Answer: B

The parallel pair is equivalent to (6.0 × 12) ÷ (6.0 + 12) = 4.0 Ω, so the total resistance is 8.0 Ω and the battery current is 24 ÷ 8.0 = 3.0 A. The 4.0 Ω resistor takes 3.0 × 4.0 = 12 V, leaving 12 V across the pair, so the 12 Ω resistor carries 12 ÷ 12 = 1.0 A. 2.0 A is the current in the 6.0 Ω branch, and 1.5 A wrongly splits the current equally.

Question 56

Physics · Magnetism and Induction · Challenging

A proton moves in a circular path in a uniform magnetic field, with its velocity perpendicular to the field. If the proton's speed were doubled, still perpendicular to the same field (relativistic effects negligible), what would happen to the radius of its path and to the time it takes to complete one revolution?

Show answer and explanation

Answer: E

The magnetic force provides the centripetal force, qvB = mv²/r, so r = mv/(qB) is proportional to v and doubles. The period T = 2πr/v = 2πm/(qB) does not depend on the speed, so it stays the same. Saying the period doubles forgets that the proton covers the larger circle at twice the speed.

Question 57

Physics · Light and Optics · Challenging

An object is placed 30 cm in front of a thin converging lens of focal length 10 cm. Which statement correctly describes the image?

Show answer and explanation

Answer: D

From the thin-lens equation 1/p + 1/q = 1/f: 1/q = 1/10 − 1/30 = 2/30, so q = +15 cm (a real image on the far side of the lens). The magnification is M = −q/p = −15/30 = −0.5, so the image is inverted and half the height of the object. 7.5 cm comes from adding 1/10 and 1/30 instead of subtracting; an object outside the focal length of a converging lens always gives a real, inverted image.

Question 58

Physics · Light and Optics · Foundation

A ray of monochromatic light passes from air into glass, striking the surface at 30° to the normal. Which property of the light is the same in the glass as it was in the air?

Show answer and explanation

Answer: E

The frequency is set by the source and does not change at a boundary. In glass the speed drops to c/n, so the wavelength λ = v/f becomes shorter too, and because the ray arrives at an angle it bends towards the normal. The wavelength is the tempting choice, but it shrinks in proportion to the speed.

Question 59

Physics · Modern Physics · Standard

Both the fission of a heavy nucleus such as uranium-235 and the fusion of light nuclei such as isotopes of hydrogen release energy. Which statement explains why?

Show answer and explanation

Answer: A

Binding energy per nucleon is greatest for nuclei near iron, so splitting a very heavy nucleus or joining very light ones produces more tightly bound nuclei. Their total mass is therefore smaller, and the mass difference is released as energy, E = Δmc². The number of nucleons does not change in either reaction; it is the mass of the nuclei that decreases.

Question 60

Physics · Modern Physics · Standard

A radioactive sample used in a hospital has an activity of 800 Bq. Its half-life is 6.0 hours. What will its activity be 24 hours later?

Show answer and explanation

Answer: C

24 hours is 24 ÷ 6.0 = 4 half-lives, so the activity is halved four times: 800 → 400 → 200 → 100 → 50 Bq, that is 800 ÷ 2⁴ = 50 Bq. 200 Bq comes from dividing by the number of half-lives (800 ÷ 4) instead of halving four times.

Written tasks

Answer on paper, then mark your work against the model answer.

1. True or false, with the reason

5 points · Rotational Motion and Gravitation

Decide whether each statement is true or false, and give a short reason for each answer. (1 point each: ½ point for the correct verdict and ½ point for a correct reason.)

1) An astronaut floats inside a spacecraft orbiting the Earth because there is no gravity at that height.

2) An object moving in a circle at constant speed has zero acceleration.

3) If the distance between two point masses is doubled, the gravitational force between them becomes one quarter as large.

4) A net torque must act on a wheel to keep it spinning at constant angular velocity.

5) Of two objects with the same mass, the one whose mass lies farther from the rotation axis has the larger moment of inertia about that axis.

Show the model answer

1) False. Gravity at the height of a typical orbit (a few hundred kilometres) is still nearly 90% of its value at the surface. The astronaut and the spacecraft are both in free fall around the Earth with the same acceleration, so nothing presses her against the floor and she appears weightless. (½ + ½ point)

2) False. The velocity keeps changing direction, so there is a centripetal acceleration of magnitude v²/r, directed towards the centre of the circle. (½ + ½ point)

3) True. By Newton's law of gravitation F = Gm₁m₂/r², the force is inversely proportional to the square of the distance, so doubling r divides F by 2² = 4. (½ + ½ point)

4) False. By the rotational form of Newton's first (and second) law, Στ = Iα: a net torque is needed only to change the angular velocity. With no net torque (for example, no friction at the axle) the wheel keeps spinning at constant angular velocity. (½ + ½ point)

5) True. The moment of inertia is I = Σmr², so mass placed at a larger distance r from the axis contributes more (it grows with r²). (½ + ½ point)

Total: 5 points.

2. A block on a spring

5 points · Vibrations and Waves

A block of mass 0.50 kg is attached to a horizontal spring of spring constant 200 N/m and rests on a frictionless horizontal surface. The block is pulled 0.10 m from its equilibrium position and released from rest, so that it performs simple harmonic motion.

(a) Calculate the period of the oscillation. (2 points)

(b) Calculate the maximum speed of the block. (3 points)

Show the model answer

(a) For a mass on a spring T = 2π√(m/k) (1 point). T = 2π√(0.50 ÷ 200) = 2π√(0.0025) = 2π × 0.050 s = 0.10π s ≈ 0.31 s (1 point).

(b) Mechanical energy is conserved because the surface is frictionless. At release the block is at rest, so all the energy is elastic potential energy: ½kA² = ½ × 200 × 0.10² = 1.0 J (1 point).

At the equilibrium position the spring is unstretched, so all this energy is kinetic and the speed is greatest there: ½mv²max = 1.0 J (1 point).

v²max = 2 × 1.0 ÷ 0.50 = 4.0 m²/s², so vmax = 2.0 m/s (1 point). Check: vmax = Aω = A√(k/m) = 0.10 × √400 = 0.10 × 20 = 2.0 m/s.

Total: 5 points.

3. State the laws and apply them: electromagnetic induction

10 points · Magnetism and Induction

(a) State Faraday's law of electromagnetic induction. If you write an equation, explain every symbol in it. (3 points)

(b) State Lenz's law. (2 points)

(c) A flat coil of 200 turns, each of area 5.0 × 10⁻³ m², lies with its plane perpendicular to a uniform magnetic field. The field strength decreases uniformly from 0.40 T to 0.10 T in 0.050 s. The ends of the coil are joined so that the coil forms a closed circuit of total resistance 4.0 Ω.

Calculate the magnitude of the average induced emf (3 points) and the induced current (1 point). State, with a reason, whether the magnetic field produced by the induced current inside the coil points in the same direction as the external field or in the opposite direction (1 point).

Show the model answer

(a) The emf induced in a circuit is equal to the rate of change of the magnetic flux through it (1 point). In equation form ε = −N ΔΦ/Δt, where ε is the average induced emf (V), N is the number of turns, ΔΦ is the change in magnetic flux through one turn (Wb) and Δt is the time taken for the change (s); the minus sign expresses Lenz's law (1 point). The magnetic flux through a flat loop is Φ = BA cos θ, where B is the magnetic field strength, A the area of the loop and θ the angle between the field and the normal to the loop (1 point).

(b) The induced current flows in such a direction that its own magnetic field opposes the change in magnetic flux that produced it (1 point). It is a consequence of the conservation of energy: if the induced current reinforced the change, energy would be created from nothing (1 point).

(c) The plane of the coil is perpendicular to the field, so the flux through one turn is Φ = BA. The change in flux per turn is ΔΦ = ΔB × A = (0.40 − 0.10) T × 5.0 × 10⁻³ m² = 0.30 × 5.0 × 10⁻³ = 1.5 × 10⁻³ Wb (1 point).

|ε| = N ΔΦ/Δt (1 point) = 200 × 1.5 × 10⁻³ ÷ 0.050 = 0.30 ÷ 0.050 = 6.0 V (1 point).

Induced current: I = ε/R = 6.0 V ÷ 4.0 Ω = 1.5 A (1 point).

Direction: the flux through the coil is decreasing, so by Lenz's law the induced current opposes the decrease by producing a magnetic field inside the coil in the same direction as the external field (1 point).

Total: 10 points.

4. Problem solving: ramp, collision and friction

10 points · Work, Energy and Momentum

Block A, of mass 2.0 kg, is released from rest at a height of 0.80 m above the floor on a frictionless curved ramp. At the foot of the ramp it moves horizontally and collides with block B, also of mass 2.0 kg, which is at rest on the floor. The two blocks stick together. The floor beyond the foot of the ramp, where B rests, is rough: the coefficient of kinetic friction between the blocks and the floor is 0.20. Take g = 10 m/s² and treat the collision as instantaneous.

(a) Calculate the speed of block A just before the collision. (3 points)

(b) Calculate the speed of the two blocks just after the collision. (3 points)

(d) Calculate how far the two blocks slide along the rough floor before they stop. (3 points)

Show the model answer

(a) The ramp is frictionless, so mechanical energy is conserved: the gravitational potential energy lost equals the kinetic energy gained, mgh = ½mv² (1 point). So v = √(2gh) (1 point) = √(2 × 10 × 0.80) = √16 = 4.0 m/s (1 point).

(b) During the instantaneous collision external horizontal forces have no time to act, so the total momentum is conserved (1 point): mAvA = (mA + mB)v′ (1 point). 2.0 × 4.0 = 4.0 × v′, so v′ = 8.0 ÷ 4.0 = 2.0 m/s (1 point).

(c) Kinetic energy before: ½ × 2.0 × 4.0² = 16 J. Kinetic energy after: ½ × 4.0 × 2.0² = 8.0 J. Kinetic energy converted (mostly to internal energy and sound): 16 − 8.0 = 8.0 J (1 point).

(d) The friction force on the combined blocks is f = μ(mA + mB)g = 0.20 × 4.0 × 10 = 8.0 N (1 point). By the work–energy theorem, the work done against friction equals the kinetic energy after the collision: f × d = 8.0 J (1 point), so d = 8.0 J ÷ 8.0 N = 1.0 m (1 point). (Equivalently, the deceleration is a = μg = 2.0 m/s² and d = v′²/(2a) = 2.0² ÷ (2 × 2.0) = 1.0 m.)

Total: 10 points.

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