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Class 10 Science chapter-wise test 2026-27: 13 practice tests with answers

CMDLearn · 5 November 2026

This page has 13 short tests for Class 10 Science, one for each chapter of the NCERT Science textbook (Reprint 2026-27). Every test is worth 10 marks and takes 25 minutes, and it is built like a small slice of the board paper:

Every question has a worked answer, a marking guide that says where the step marks go, and the textbook page the idea comes from.

These questions are our own, drafted with AI help. Every answer was checked against the textbook page named under it, and every number was re-computed by a test script, but no practising teacher has reviewed them yet. None of them copies a textbook question. Several practise a reaction, an example or a method that the textbook explains, in a new situation or with new numbers, and a few use the same method as a question in our chapter guides with different numbers, so each test is fresh practice after you finish the chapter. If you find a mistake, please tell us.

How to take a test

1. Finish the chapter and its NCERT exercises first.
2. Set a timer for 25 minutes. Write the answers on paper, with equations, units and the steps of any calculation.
3. Mark your work with the marking guide under each answer. Give yourself a step mark only if that step is written down.
4. Score 8 or more: move on. Score 5 to 7: redo the questions you lost marks on in a week. Below 5: go back to the chapter, then try the longer set of practice questions linked under the test.

How the 80 marks are split by unit

CBSE's Science curriculum for Class X, 2026-27, gives the theory paper 80 marks across five units (course structure, page 4 of the curriculum PDF). The other 20 marks are internal assessment in school.

CBSE gives marks per unit, not per chapter, so plan your tests by unit: the four Chemistry tests together cover 25 marks, and so do the four Biology tests.

The curriculum also says which topics are taught but assessed only in school, not in the board paper: periodic classification of elements, evolution, and the electric motor, electromagnetic induction and the electric generator (pages 4 to 6). Why the Sun looks reddish when it rises and sets is left out of the light unit, and information printed in boxes in the textbook is not assessed (page 6). These tests stay inside those limits. One line needs care: the curriculum's "Note for Teachers" (page 6) also lists heredity and the effects of electric current as not assessed in the year-end exam, while the unit lists on pages 5 and 6 include them. We have kept Tests 8 and 12, and you should check with your teacher which reading your school follows.

The 13 tests

Test 1: Chemical Reactions and Equations (Chapter 1, printed pages 1 to 16)

Q1 (1 mark). Burning magnesium keeps burning in carbon dioxide: 2Mg + CO₂ → 2MgO + C. The substance that is oxidised is: (A) Mg (B) CO₂ (C) MgO (D) C

Answer: (A) Mg. Magnesium gains oxygen and becomes magnesium oxide, so it is oxidised; carbon dioxide loses oxygen and becomes carbon, so it is reduced. (Oxidation and reduction, page 14.)

Q2 (2 marks). Aluminium burns in oxygen to form aluminium oxide. Write the balanced equation and name the type of reaction.

Answer. 4Al + 3O₂ → 2Al₂O₃. Two substances join to form a single product, so it is a combination reaction. Check: 4 Al and 6 O on each side.

Marking: 1 mark for the balanced equation, 1 mark for "combination".

Q3 (3 marks). Name the type of each reaction: (a) 2FeSO₄(s) → Fe₂O₃(s) + SO₂(g) + SO₃(g), on heating; (b) Zn(s) + 2AgNO₃(aq) → Zn(NO₃)₂(aq) + 2Ag(s); (c) BaCl₂(aq) + H₂SO₄(aq) → BaSO₄(s) + 2HCl(aq).

Answer. (a) Decomposition (thermal decomposition): a single reactant breaks down into simpler products. (b) Displacement: zinc, being more reactive, pushes silver out of its compound. (c) Double displacement: the two compounds exchange ions, and the insoluble white barium sulphate separates out, so it is also a precipitation reaction.

Marking: 1 mark each. (Decomposition, page 8; displacement and double displacement, pages 11 and 12.)

Q4 (4 marks, case study). Old black-and-white photographic film was coated with silver bromide. When light falls on it, this reaction takes place: 2AgBr(s) → 2Ag(s) + Br₂(g). The grey silver that forms makes the picture. Unused film was always kept in a light-proof box.

(a) What type of reaction is this? (1) (b) What supplies the energy for it? (1) (c) Is the reaction exothermic or endothermic? Give a reason, and say why the film is kept in a light-proof box. (2)

Answer. (a) Decomposition: one compound breaks into two substances. (b) Light (sunlight). (c) Endothermic: it takes in energy (light) to break the silver bromide down, and reactions that absorb energy are endothermic. Any light would start the reaction and spoil the film before it is used, so it is kept in the dark.

Marking: (a) 1. (b) 1. (c) 1 for endothermic with the reason, 1 for the light-proof box. (Decomposition by light and endothermic reactions, page 10.)

More practice: Chemical Reactions and Equations questions with answers.

Test 2: Acids, Bases and Salts (Chapter 2, printed pages 17 to 36)

Q1 (1 mark). Dilute hydrochloric acid is added to each of these. Which one gives off a gas that turns lime water milky? (A) zinc granules (B) marble chips (C) copper turnings (D) common salt

Answer: (B) marble chips. Marble is calcium carbonate, and an acid reacts with a metal carbonate to give a salt, water and carbon dioxide, which turns lime water milky. Zinc gives hydrogen instead. (Acids with metal carbonates, pages 20 and 21.)

Q2 (2 marks). How is bleaching powder made, and what are two of its uses?

Answer. Chlorine gas is passed over dry slaked lime, Ca(OH)₂, which takes it up to form bleaching powder. Uses (any two): bleaching cotton and linen, wood pulp or washed clothes; as an oxidising agent in the chemical industry; to make drinking water free from germs.

Marking: 1 mark for chlorine on dry slaked lime, 1 mark for two uses. (Bleaching powder, pages 30 and 31.)

Q3 (3 marks). A person has a burning pain in the stomach after a heavy, spicy meal, and takes a spoon of milk of magnesia. (a) What causes the pain? (1) (b) What kind of substance is milk of magnesia, and what does it do in the stomach? (1) (c) Name the type of reaction, and give its general word equation. (1)

Answer. (a) The stomach makes too much hydrochloric acid (indigestion). (b) It is magnesium hydroxide, a mild base (an antacid); it neutralises the extra acid. (c) Neutralisation: acid + base → salt + water.

Marking: 1 mark each. (Antacids, page 27; neutralisation, page 21.)

Q4 (4 marks, case study). A chemical plant passes electricity through brine (a strong solution of common salt in water). Three products come out, and each is sold.

(a) What is this process called? (1) (b) Which gas is given off at the anode? (1) (c) Write the balanced equation for the process, and name the product formed near the cathode in solution. (2)

Answer. (a) The chlor-alkali process ("chlor" for chlorine, "alkali" for sodium hydroxide). (b) Chlorine. (c) 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g). Sodium hydroxide solution forms near the cathode, where hydrogen gas is also given off.

Marking: (a) 1. (b) 1. (c) 1 for the balanced equation, 1 for sodium hydroxide. (Chlor-alkali process, page 30.)

More practice: Acids, Bases and Salts questions with answers.

Test 3: Metals and Non-metals (Chapter 3, printed pages 37 to 57)

Q1 (1 mark). Which of these oxides reacts with both acids and bases to give a salt and water? (A) Na₂O (B) MgO (C) ZnO (D) CO₂

Answer: (C) ZnO. Zinc oxide, like aluminium oxide, is an amphoteric oxide. (Amphoteric oxides, page 41.)

Q2 (2 marks). Potassium (2, 8, 8, 1) reacts with chlorine (2, 8, 7) to form potassium chloride. Show how the ions form, and name the type of bond.

Answer. Potassium loses its one outer electron and becomes K⁺ (2, 8, 8); chlorine gains that electron and becomes Cl⁻ (2, 8, 8). K → K⁺ + e⁻ and Cl + e⁻ → Cl⁻, so K⁺Cl⁻ forms. The oppositely charged ions are held by an ionic (electrovalent) bond.

Marking: 1 mark for the electron transfer with both ions, 1 mark for "ionic bond". (Formation of ionic compounds, pages 46 to 48.)

Q3 (3 marks). A white solid X melts only at a very high temperature. It does not conduct electricity as a solid, but it does when melted or dissolved in water. (a) Is X an ionic or a covalent compound? (1) (b) Why is the solid hard, and what happens to a crystal of it when strong pressure is applied? (1) (c) Why does it conduct only when molten or dissolved? (1)

Answer. (a) Ionic. (b) The strong attraction between the positive and negative ions makes it hard; but it is brittle, so a crystal breaks into pieces under pressure. (c) In the solid the ions cannot move; when the solid melts or dissolves, the ions are free to move and carry the current.

Marking: 1 mark each. (Properties of ionic compounds, page 49.)

Q4 (4 marks, case study). A museum shows a copper bell that has turned green, a silver plate that has turned black, and a gold coin that still shines after centuries.

(a) Name the green coating on the copper. (1) (b) What has formed the black layer on the silver? (1) (c) Name the general process behind (a) and (b), and explain why the gold coin is not affected. (2)

Answer. (a) Basic copper carbonate, formed as copper slowly reacts with damp air that contains carbon dioxide. (b) Silver sulphide, from the reaction of silver with sulphur in the air. (c) Corrosion. Gold sits near the very bottom of the reactivity (activity) series, so it hardly reacts with the air.

Marking: (a) 1. (b) 1. (c) 1 for corrosion, 1 for low reactivity of gold. (Corrosion, page 53; activity series, pages 44 and 45.)

More practice: Metals and Non-metals questions with answers.

Test 4: Carbon and its Compounds (Chapter 4, printed pages 58 to 78)

Q1 (1 mark). Which of these has a triple bond between its two carbon atoms? (A) ethane, C₂H₆ (B) ethene, C₂H₄ (C) ethyne, C₂H₂ (D) methane, CH₄

Answer: (C) ethyne. Each carbon in C₂H₂ can satisfy its valency of four only with three bonds to the other carbon and one to a hydrogen. Compounds with double or triple bonds between carbon atoms are unsaturated. (Saturated and unsaturated compounds, page 64.)

Q2 (2 marks). Name these compounds: (a) CH₃CH₂CH₂CHO (b) CH₃CH₂CH₂COOH.

Answer. (a) Butanal: four carbons, aldehyde group. (b) Butanoic acid: four carbons, carboxylic acid group.

Marking: 1 mark each. (Nomenclature of carbon compounds, pages 67 and 68.)

Q3 (3 marks). Ethanoic acid is warmed with absolute ethanol and a few drops of a strong acid. (a) Name the reaction and the class of product. (1) (b) Write the equation. (1) (c) What happens when the product is heated with sodium hydroxide, and what is that reaction called? (1)

Answer. (a) Esterification; the product is an ester (ethyl ethanoate), a sweet-smelling substance. (b) CH₃COOH + CH₃CH₂OH → CH₃COOCH₂CH₃ + H₂O (acid catalyst). (c) The alkali splits the ester again, giving back the alcohol and a sodium salt of the acid; this is saponification.

Marking: 1 mark each. (Esterification and saponification, page 73.)

Q4 (4 marks, case study). A bottle in a school laboratory is labelled "glacial acetic acid". In a cold winter its contents had turned solid. A diluted sample turns blue litmus red and fizzes when baking soda is added.

(a) Give the chemical name of acetic acid. (1) (b) Why is it called glacial? (1) (c) Name the gas given off with baking soda and give a test for it; and what is the 5 to 8% solution of this acid in water called? (2)

Answer. (a) Ethanoic acid (CH₃COOH). (b) Pure ethanoic acid melts at 290 K, so it often freezes in cold weather. (c) Carbon dioxide, which turns lime water milky. The 5 to 8% solution in water is vinegar.

Marking: (a) 1. (b) 1. (c) 1 for CO₂ and its test, 1 for vinegar. (Ethanoic acid, page 73; acids with carbonates and hydrogencarbonates, page 74.)

More practice: Carbon and its Compounds questions with answers.

Test 5: Life Processes (Chapter 5, printed pages 79 to 99)

Q1 (1 mark). The small pores on a leaf are opened and closed by: (A) xylem cells (B) guard cells (C) phloem cells (D) root hairs

Answer: (B) guard cells. They swell when water flows into them and open the pore; the pore closes when they shrink. (Stomata, page 83.)

Q2 (2 marks). What is transpiration? Give two ways it helps the plant.

Answer. Transpiration is the evaporation of water from the parts of the plant above the ground, mostly through the stomata of the leaves. It helps (1) pull water, with the minerals dissolved in it, up from the roots to the leaves, and (2) regulate the plant's temperature.

Marking: 1 mark for the definition, 1 mark for two uses. (Transpiration, page 95.)

Q3 (3 marks). How does Amoeba take in and digest its food? How is Paramoecium different?

Answer. Amoeba pushes out temporary finger-like extensions of its cell surface, which fuse around a food particle to form a food vacuole. Inside it, complex substances are broken into simpler ones that diffuse into the cytoplasm, and undigested material is thrown out at the cell surface. Paramoecium has a definite shape and takes in food only at a specific spot.

Marking: 1 mark for the food vacuole forming, 1 mark for digestion and removal of waste, 1 mark for the Paramoecium difference. (Nutrition in single-celled organisms, page 84.)

Q4 (4 marks, case study). A student counts the stomata on a leaf peel of one plant (made-up counts): about 30 per square millimetre on the upper surface and 250 on the lower surface. On a very hot afternoon the plant's leaves droop.

(a) Through which surface would this leaf lose more water as vapour? (1) (b) What happens to the guard cells and the stomata when the plant is short of water? (1) (c) Why does the plant close its stomata at such times, and what does that cost it? (2)

Answer. (a) The lower surface, which has many more stomata. (b) The guard cells lose water and shrink, so the stomata close. (c) Large amounts of water are lost through open stomata, so closing them saves water. The cost is that carbon dioxide for photosynthesis also enters through the stomata, so less food is made while they are shut.

Marking: (a) 1. (b) 1. (c) 1 for saving water, 1 for less CO₂ and photosynthesis. (Stomata, page 83.)

More practice: Life Processes questions with answers.

Test 6: Control and Coordination (Chapter 6, printed pages 100 to 112)

Q1 (1 mark). Which plant hormone, made at the shoot tip, makes a shoot bend towards light? (A) auxin (B) cytokinin (C) abscisic acid (D) thyroxin

Answer: (A) auxin. It diffuses to the shady side and makes the cells there grow longer, so the shoot bends towards the light. Thyroxin is an animal hormone. (Plant hormones, page 108.)

Q2 (2 marks). The changes of puberty come from hormones. Name the hormone that brings them about (a) in boys and (b) in girls, and the organ that makes each.

Answer. (a) Testosterone, made by the testes. (b) Oestrogen, made by the ovaries.

Marking: 1 mark each (hormone and organ both needed). (Hormones in animals, pages 110 and 111.)

Q3 (3 marks). When a nerve impulse reaches a muscle, the muscle shortens. (a) How does a muscle cell manage to shorten? (b) What brings about this change inside the cell? (c) Which tissue does the final job of an action, and which tissue carries the decision to it?

Answer. (a) The muscle cell changes its shape so that it becomes shorter. (b) Muscle cells contain special proteins. When a nerve impulse arrives, these proteins alter their shape and the way they are arranged, and the cell shortens. (c) Muscle tissue does the final job; nervous tissue collects the information, makes the decision and conveys it to the muscles.

Marking: 1 mark each. (How nervous tissue causes action, page 105.)

Q4 (4 marks, case study). At a railway station, Ravi hears an announcement that his train is coming in on platform 3. He checks the display board, decides the announcement is right, and walks to the platform.

(a) Which part of the brain first receives what he hears and what he sees? (1) (b) Where in that part is the sound put together with the sight and with what he already knows? (1) (c) Which areas pass on his decision, and to which kind of muscles? Is walking to the platform a voluntary or an involuntary action? (2)

Answer. (a) The fore-brain, which has separate areas for hearing and for sight. (b) In its association areas, where sensory information is interpreted together with information from other receptors and information already stored in the brain. (c) The motor areas pass the decision on to the voluntary muscles, here his leg muscles. Walking there is a voluntary action: he decides to do it.

Marking: (a) 1. (b) 1. (c) 1 for motor areas and voluntary (leg) muscles, 1 for voluntary. (The human brain, pages 103 and 104.)

More practice: Control and Coordination questions with answers.

Test 7: How do Organisms Reproduce? (Chapter 7, printed pages 113 to 127)

Q1 (1 mark). Which of these divides into many daughter cells at the same time (multiple fission)? (A) Amoeba (B) Plasmodium (C) Hydra (D) yeast

Answer: (B) Plasmodium, the malarial parasite. Amoeba splits into two, and Hydra and yeast form buds. (Fission, pages 115 and 116.)

Q2 (2 marks). Spirogyra and Planaria can both form new individuals from pieces of the body. How do the two methods differ?

Answer. Spirogyra simply breaks up into fragments on maturing, and each piece grows into a new individual (fragmentation). In Planaria, a body cut into pieces can regrow the missing parts from each piece, using specialised cells (regeneration).

Marking: 1 mark for fragmentation, 1 mark for regeneration. (Fragmentation and regeneration, pages 116 and 117.)

Q3 (3 marks). Amoeba and Leishmania are both single cells that reproduce by binary fission. (a) What happens to the cell in binary fission? (b) How is the split different in the two organisms? (c) What part of the Leishmania cell is the split lined up with?

Answer. (a) The cell splits into two equal halves, each a new individual. (b) In Amoeba the split can happen in any plane; in Leishmania it happens in a definite orientation. (c) Leishmania's body is more organised: one end of the cell carries a whip-like part, and the split is lined up in relation to it.

Marking: 1 mark each. (Fission, page 115.)

Q4 (4 marks, case study). A farmer has one grape vine whose fruit sells well. Instead of sowing its seeds, she grows new vines from parts of its stems.

(a) What is this way of growing plants called? (1) (b) Name two methods of this kind that the chapter mentions for grapes, roses and sugarcane. (1) (c) The parent vine gives sweet grapes and resists a common disease. Will the new vines have these features, and why? Name two other plants the chapter says are grown this way because they have lost the capacity to produce seeds. (2)

Answer. (a) Vegetative propagation (an asexual method). (b) Layering and grafting. (c) Yes: plants raised by vegetative propagation are genetically similar enough to the parent to have all its characteristics. Plants grown this way because they no longer make seeds (any two): banana, orange, rose, jasmine.

Marking: (a) 1. (b) 1. (c) 1 for the new vines keeping the features, 1 for two plants. (Vegetative propagation, page 117.)

More practice: How do Organisms Reproduce? questions with answers.

Test 8: Heredity (Chapter 8, printed pages 128 to 133)

Q1 (1 mark). A pure-breeding tall pea plant (TT) is crossed with a pure-breeding short one (tt). The F1 plants are: (A) all short (B) all tall (C) half tall, half short (D) all of medium height

Answer: (B) all tall. Mendel found no halfway plants in F1: only the dominant trait shows. (Mendel's crosses, page 130.)

Q2 (2 marks). Pure round-yellow seeded peas (RRYY) are crossed with pure wrinkled-green ones (rryy), and the F1 plants are self-pollinated. In an F2 of 320 seeds, about how many would you expect to be wrinkled and yellow?

Answer: about 60. The F2 shows the four combinations in the ratio 9 round-yellow : 3 round-green : 3 wrinkled-yellow : 1 wrinkled-green, so wrinkled-yellow is 3/16 of 320 = 60.

Marking: 1 mark for the 9 : 3 : 3 : 1 ratio, 1 mark for 60. (Independent inheritance, page 131.)

Q3 (3 marks). A hospital records 1,200 births in a year: 610 boys and 590 girls. (a) About how many boys would the chapter's explanation of sex determination lead you to expect, and do the records fit? (1) (b) Of the 590 girls, how many got an X chromosome from their father? (1) (c) Of the 610 boys, how many got an X chromosome from their mother? (1)

Answer. (a) About 600, because half the children are expected to be boys; 610 is close, so the records fit. (b) All 590: a child who gets an X from the father is a girl. (c) All 610: every child, boy or girl, gets an X from the mother.

Marking: 1 mark each. (Sex determination, page 132.)

Q4 (4 marks, case study). A gardener crosses a violet-flowered pea plant with a white-flowered one. She gets 52 plants with violet flowers and 48 with white flowers. Violet is dominant over white (V for violet, v for white).

(a) What is the genotype of the white-flowered parent? (1) (b) What is the genotype of the violet-flowered parent? (1) (c) Two of the violet-flowered offspring are crossed with each other. What ratio of violet to white would you expect, and why? (2)

Answer. (a) vv: white shows only when there is no dominant V. (b) Vv: about half the offspring are white (vv), so the violet parent must have passed on a v to them. (c) The violet offspring are all Vv (they got v from the white parent). Vv × Vv gives VV, Vv, Vv and vv, so 3 violet : 1 white.

Marking: (a) 1. (b) 1. (c) 1 for Vv × Vv, 1 for 3 : 1. (Dominant and recessive traits, pages 130 and 131.)

More practice: Heredity questions with answers.

Test 9: Light: Reflection and Refraction (Chapter 9, printed pages 134 to 160)

Q1 (1 mark). A ray travelling parallel to the principal axis strikes a concave mirror. After reflection it: (A) passes through the centre of curvature (B) passes through the principal focus (C) comes back along the same path (D) goes parallel to the axis

Answer: (B) passes through the principal focus. (Rules for rays in image formation, pages 138 and 139.)

Q2 (2 marks). A spherical mirror gives a magnification of −2 for an object. Describe the image, and say which kind of mirror it must be.

Answer. A negative magnification means the image is real and inverted; its size is 2, so it is twice as large as the object. Only a concave mirror gives a real, enlarged image (a convex mirror always gives a virtual, erect, smaller image).

Marking: 1 mark for real, inverted, enlarged; 1 mark for concave. (Images by spherical mirrors, page 142; magnification, pages 143 and 144.)

Q3 (3 marks). A concave mirror has a focal length of 10 cm, and an object stands 15 cm from it. Find the position of the image and the magnification, using the sign convention.

Answer. u = −15 cm, f = −10 cm. 1/v = 1/f − 1/u = −1/10 + 1/15 = −1/30, so v = −30 cm: the image is 30 cm in front of the mirror. m = −v/u = −(−30)/(−15) = −2: real, inverted and twice the size.

Marking: 1 mark for the signs and the mirror formula, 1 mark for v, 1 mark for m with its meaning. (Mirror formula and magnification, page 143.)

Q4 (4 marks, case study). In a school laboratory, a sharp image of a candle flame forms on a screen 36 cm behind a convex lens of focal length 12 cm.

(a) Is the image real or virtual? (1) (b) How far in front of the lens is the candle? (1) (c) Find the magnification and the power of the lens. (2)

Answer. (a) Real: it forms on a screen. (b) v = +36 cm, f = +12 cm. 1/u = 1/v − 1/f = 1/36 − 1/12 = −2/36, so u = −18 cm: the candle is 18 cm in front of the lens. (c) m = v/u = 36/(−18) = −2 (inverted, twice the size). P = 1/f (in metres) = 1/0.12 = +8.3 D (about).

Marking: (a) 1. (b) 1. (c) 1 for m, 1 for P. (Lens formula, magnification and power, pages 155 to 158.)

More practice: Light: Reflection and Refraction questions with answers.

Test 10: The Human Eye and Colours (Chapter 10, printed pages 161 to 170)

Q1 (1 mark). Most of the refraction of the light entering the eye happens at: (A) the eye lens (B) the outer surface of the cornea (C) the pupil (D) the retina

Answer: (B) the outer surface of the cornea. The eye lens only provides the finer adjustment of focal length. (The human eye, page 161.)

Q2 (2 marks). What happens at the retina when an image forms on it, and how does the information reach the brain?

Answer. The retina has many light-sensitive cells, which are activated when light falls on them and generate electrical signals. The signals are sent to the brain through the optic nerves, and the brain interprets them.

Marking: 1 mark for the light-sensitive cells and signals, 1 mark for the optic nerve and the brain. (The human eye, pages 161 and 162.)

Q3 (3 marks). A beam of sunlight coming through a gap in the roof of a dusty storeroom is clearly visible as a bright shaft. Name the effect, explain it, and say why the beam would not be seen in perfectly clean air.

Answer. It is the Tyndall effect. The air in the room carries fine particles of dust (and smoke or tiny droplets). When the beam strikes them, the light is scattered, and some of it reaches our eyes, so the path of the beam becomes visible. In clean air without such particles, very little light is scattered towards us, so the path is not seen.

Marking: 1 mark for the name, 1 mark for scattering by fine particles, 1 mark for why clean air shows no beam. (Tyndall effect, page 169.)

Q4 (4 marks, case study). A 68-year-old man can see the far hills clearly but cannot read a newspaper held at a comfortable distance. His optician says his eye can no longer focus well on near objects.

(a) Name this defect. (1) (b) What has happened to his near point as he has grown older? (1) (c) Give the two causes of this defect, and the kind of lens that corrects it. (2)

Answer. (a) Presbyopia. (b) It has receded, moving farther away from his eye, because the eye's ability to adjust its focal length (its power of accommodation) tends to weaken as people age. (c) It arises from the gradual weakening of the ciliary muscles and the diminishing flexibility of the eye lens with age. It is corrected with a convex lens (a bi-focal lens if he also has trouble with distant vision).

Marking: (a) 1. (b) 1. (c) 1 for the two causes, 1 for the convex lens. (Accommodation and presbyopia, pages 162 to 164.)

More practice: The Human Eye and the Colourful World questions with answers.

Test 11: Electricity (Chapter 11, printed pages 171 to 194)

Q1 (1 mark). The SI unit of resistivity is: (A) ohm (B) ohm per metre (C) ohm metre (D) volt per ampere

Answer: (C) ohm metre (Ω m). (Resistivity, page 178.)

Q2 (2 marks). A 9 V battery is connected to a 6 Ω resistor and a 12 Ω resistor in series. Find the current, and the potential difference across the 12 Ω resistor.

Answer. Total resistance = 6 + 12 = 18 Ω. I = V/R = 9/18 = 0.5 A. Across the 12 Ω resistor, V = IR = 0.5 × 12 = 6 V.

Marking: 1 mark for 18 Ω and the current, 1 mark for 6 V. (Resistors in series, pages 183 and 184.)

Q3 (3 marks). A wire has a resistance of 12 Ω. What will the resistance of a wire of the same material be if it is (a) one-third as long, with the same thickness; (b) the same length, with three times the area of cross-section; (c) the original wire, pulled out evenly until it is twice as long (its volume stays the same)?

Answer. Resistance grows in step with the length and falls as the area of cross-section grows (R = ρl/A). (a) 12 ÷ 3 = 4 Ω. (b) 12 ÷ 3 = 4 Ω. (c) Twice the length with the same volume means half the area, so the resistance becomes 12 × 2 × 2 = 48 Ω.

Marking: 1 mark each. (Factors on which resistance depends, page 178.)

Q4 (4 marks, case study). A family uses a 1,500 W geyser for 40 minutes every day, on a 220 V supply.

(a) How much electrical energy does the geyser use in a day, in kilowatt hours? (1) (b) How many units does it use in a 30-day month? (1) (c) Find the current it draws and its resistance while it is working. (2)

Answer. (a) 1.5 kW × (40/60) h = 1 kW h. (b) 1 × 30 = 30 units (1 unit = 1 kW h). (c) I = P/V = 1,500/220 ≈ 6.8 A. R = V²/P = 220 × 220 ÷ 1,500 ≈ 32.3 Ω.

Marking: (a) 1. (b) 1. (c) 1 for 6.8 A, 1 for 32.3 Ω. (Electric power and the kilowatt hour, pages 191 and 192.)

More practice: Electricity questions with answers.

Test 12: Magnetic Effects of Electric Current (Chapter 12, printed pages 195 to 207)

Q1 (1 mark). Outside a bar magnet, the magnetic field lines go: (A) from the south pole to the north pole (B) from the north pole to the south pole (C) round in circles centred on the magnet (D) straight out in all directions

Answer: (B) from the north pole to the south pole. (Field lines, pages 196 and 197.)

Q2 (2 marks). A straight horizontal wire carries a current from west to east, in a uniform magnetic field pointing from south to north. In which direction is the force on the wire? Name the rule you used.

Answer. Vertically upwards. By Fleming's left-hand rule: forefinger along the field (north), middle finger along the current (east), and the thumb then points up.

Marking: 1 mark for upwards, 1 mark for the rule. (Fleming's left-hand rule, page 203.)

Q3 (3 marks). (a) What is a solenoid? (b) What is an electromagnet? (c) Name two devices that the chapter says use current-carrying conductors and magnetic fields (other than the motor and the generator).

Answer. (a) Insulated copper wire wound closely in many circular turns, so that the coil has the shape of a cylinder. (b) A piece of soft iron placed inside the current-carrying coil becomes magnetised; it is an electromagnet. (c) Any two: loudspeakers, microphones, measuring instruments.

Marking: 1 mark each. (Solenoid and electromagnet, page 201; devices, page 203.)

Q4 (4 marks, case study). In a house, the supply comes in with three wires: one with red insulation, one with black, and one with green, which goes to a metal plate deep in the ground.

(a) What are the red and black wires called? (1) (b) What is the potential difference between them, and the frequency of the supply in India? (1) (c) Why are the appliances in each circuit connected in parallel, and name two ways in which a circuit can become overloaded. (2)

Answer. (a) Red is the live wire; black is the neutral wire. (b) 220 V, at 50 Hz (alternating current). (c) In parallel, each appliance gets the same potential difference and can be switched on or off by its own switch. A circuit can be overloaded (any two): when the live and neutral wires come into direct contact (short-circuiting); by an accidental hike in the supply voltage; by plugging too many appliances into one socket.

Marking: (a) 1. (b) 1. (c) 1 for equal potential difference, 1 for two causes of overloading. (Domestic electric circuits, pages 204 to 206.)

More practice: Magnetic Effects of Electric Current questions with answers.

Test 13: Our Environment (Chapter 13, printed pages 208 to 217)

Q1 (1 mark). Which of these is biodegradable? (A) a polythene bag (B) a glass bottle (C) an old cotton shirt (D) an aluminium can

Answer: (C) the cotton shirt: it is broken down by bacteria and other saprophytes. (Biodegradable and non-biodegradable substances, page 214.)

Q2 (2 marks). How is ozone formed in the upper atmosphere? Give the two steps.

Answer. High-energy ultraviolet radiation splits oxygen molecules into free oxygen atoms: O₂ → O + O. A free oxygen atom then combines with an oxygen molecule to form ozone: O + O₂ → O₃.

Marking: 1 mark each step. (Ozone layer, pages 212 and 213.)

Q3 (3 marks). Sort these into natural and human-made (artificial) ecosystems: a forest, a crop field, a lake, a garden, a pond. Then say what makes a garden an ecosystem.

Answer. Natural: forest, lake, pond. Human-made: crop field, garden. A garden is an ecosystem because its living parts, the biotic components (plants, animals and microorganisms), interact with one another, while non-living factors, the abiotic components (such as temperature, rainfall, wind and soil), affect how they grow and reproduce.

Marking: 1 mark for the natural ones, 1 mark for the human-made ones, 1 mark for the reason. (What is an ecosystem?, page 208.)

Q4 (4 marks, case study). From the 1980s, the ozone in the atmosphere fell steeply.

(a) Which synthetic chemicals have been linked to this? (1) (b) What were they used for? (1) (c) What did the 1987 agreement under UNEP decide, and what must refrigerator makers now do? (2)

Answer. (a) Chlorofluorocarbons (CFCs). (b) As refrigerants and in fire extinguishers. (c) In 1987, under the United Nations Environment Programme (UNEP), countries agreed to hold CFC production at its 1986 level. Manufacturers everywhere must now make CFC-free refrigerators.

Marking: (a) 1. (b) 1. (c) 1 for the 1987 freeze, 1 for CFC-free refrigerators. (Ozone depletion, page 213.)

More practice: Our Environment questions with answers.

Practise more

• All 13 chapters with their questions, notes and pages in one place: Class 10 Science question answer.
• Longer sets by marks: Class 10 Science important questions 2026-27, competency-based questions and past-paper questions chapter-wise.
• How the board paper is laid out: our Class 10 Science sample paper guide and syllabus guide.
• Free practice, marked instantly in your browser: practice by chapter, and printable worksheets with answer keys: Class 10 worksheets.

Where CMDLearn fits

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Sources

All opened and read on 5 November 2026. CBSE page numbers are PDF pages; NCERT page numbers are the printed page numbers of the textbook.

• CBSE, Science (Subject Code 086), Class X (2026-27) curriculum: page 4 (course structure and unit marks), pages 4 to 6 (unit contents, topics assessed only formatively, sunrise and sunset excluded, Note for Teachers, boxed text not assessed), page 9 (question paper design): cbseacademic.nic.in/web_material/CurriculumMain27/SecPart1/Science_SecP1_2026-27.pdf
• NCERT, Science, the textbook for Class X (Reprint 2026-27), Chapters 1 to 13, the pages named under each answer above: ncert.nic.in/textbook/pdf/jesc101.pdf (and so on to jesc113.pdf)

The unit marks and assessment rules are from CBSE's curriculum above, and the science in each answer is from the textbook pages named. The situations, data, questions and worked answers are our own. This guide is general information; follow your teacher, the textbook and CBSE's own notices.

Practise this class: Class 10 worksheets, chapter by chapter · free practice, marked instantly · timed revision papers