ScienceCBSE Class 10 · Board exam 2026 (session 2025-26)

CBSE Class 10 Science: 2026 board paper, set 31/2/1

The question paper set in the CBSE Class 10 board exam of 2026, as printed: 39 questions for 80 marks, in sections, with each OR choice shown as an alternative. Try each question first, then open our solution.

Every question here has our step-by-step solution, split into the steps that earn marks. Each one was checked by an independent re-solve from the question paper.

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2026/X/Science.zip (a zip; this paper is the file 31-2-1.pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

Practise a paper like this, timedMy board paper sets a full paper in this exact format from real board questions, with a timer.Get your written answers markedThe Answer Coach marks a photo of your handwritten answer step by step, like the examiner.

How the paper is set

Section	Questions	What it has
A	1-16	Biology (30 marks): MCQ 1-7, assertion-reason 8-9, VSA 10-12, SA 13-14, case based 15, LA 16
B	17-29	Chemistry (25 marks): MCQ 17-23, assertion-reason 24, case/short 25-27, case based 28, LA 29
C	30-39	Physics (25 marks): MCQ 30-31, assertion-reason 32, VSA 33-34, SA 35-37, case based 38, LA 39

Instructions in brief: All questions compulsory; no overall choice, internal choice in a few questions; three sections A Biology, B Chemistry, C Physics to be answered in separate parts of the answer sheet; CBQs carry 4 marks each.

Section A

Biology (30 marks): MCQ 1-7, assertion-reason 8-9, VSA 10-12, SA 13-14, case based 15, LA 16

Q1 · 1 mark · MCQHeredity
The genotype of green stemmed tomato plant is denoted by ‘GG’ and that of purple stemmed tomato plant as ‘gg’. Hybrid of a cross between these two were self-pollinated to obtain $F_{2}$ progeny. What will be the ratio of GG, Gg and gg in this $F_{2}$ progeny ?
1. (A)2 : 1 : 1
2. (B)3 : 1 : 0
3. (C)1 : 1 : 2
4. (D)1 : 2 : 1
Try it first, then show our solutionSolution
1. F1 hybrid is Gg. Selfing Gg × Gg gives gametes G and g from each parent.
2. Checkerboard: GG, Gg, Gg, gg, so GG : Gg : gg = 1 : 2 : 1. Option (D).
  1 mark
Answer: 1 : 2 : 1, option (D)
Where the marks are
- Identifies genotypic ratio 1 : 2 : 1 (1 mark)
Q2 · 1 mark · MCQControl and Coordination
Identify the type of tropic movement that is responsible for the movement of the pollen tube towards an ovule from the following options.
1. (A)Hydrotropism
2. (B)Geotropism
3. (C)Chemotropism
4. (D)Phototropism
Try it first, then show our solutionSolution
1. The pollen tube grows towards the ovule in response to chemicals secreted by the ovule, which is a chemical stimulus, so the movement is chemotropism. Option (C).
  1 mark
Answer: Chemotropism, option (C)
Where the marks are
- Chemotropism (1 mark)
Q3 · 1 mark · MCQHow do Organisms Reproduce?
Spirogyra, despite being a multi-cellular organism, still shows a simple reproductive method. Select the correct option with regard to reproduction in Spirogyra :
1. (A)Asexual reproduction by division of cells into two cells
2. (B)Sexual reproduction by division of cells into many cells
3. (C)Asexual reproduction by breaking up of Spirogyra filaments into smaller parts
4. (D)Sexual reproduction by the formation of gametes after meiosis of cells of filaments followed by fusion of male and female gametes
Try it first, then show our solutionSolution
1. A Spirogyra filament simply breaks into small pieces, and each piece grows into a new individual. This is fragmentation, a form of asexual reproduction. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Fragmentation – breaking of filaments (1 mark)
Q4 · 1 mark · MCQLife Processes
The carbohydrates, which are not used immediately in plants, are stored in the form of ‘a’, which serves as an internal energy reserve. What does ‘a’ stand for ?
1. (A)Starch
2. (B)Sugar
3. (C)Fat
4. (D)Protein
Try it first, then show our solutionSolution
1. Glucose made in photosynthesis that is not used at once is converted to starch, an insoluble storage carbohydrate that acts as the plant's internal energy reserve. So 'a' is starch, option (A).
  1 mark
Answer: Starch, option (A)
Where the marks are
- Identifies starch as the stored form of carbohydrate (1 mark)
Q5 · 1 mark · MCQLife Processes
The following events occur during the process of photosynthesis. Choose the option that indicates the correct events.
(i) Oxidation of carbon dioxide to carbohydrates
(ii) Conversion of light energy into chemical energy
(iii) Absorption of light energy by chlorophyll
(iv) Conversion of hydrogen and oxygen into water
1. (A)(i) and (ii)
2. (B)(ii) and (iii)
3. (C)(iii) and (iv)
4. (D)(i) and (iv)
Try it first, then show our solutionSolution
1. In photosynthesis, chlorophyll absorbs light energy and converts it into chemical energy, so (ii) and (iii) are correct.
2. Statement (i) is wrong because $C O_{2}$ is reduced, not oxidised, to carbohydrate. Statement (iv) is wrong because water is split, not formed. Option (B).
  1 mark
Answer: (ii) and (iii), option (B)
Where the marks are
- Selects (ii) and (iii) (1 mark)
Open Q5 on its own page
Q6 · 1 mark · MCQCompetency-basedOur Environment
In the given figure, identify the trophic level having maximum concentration of harmful chemicals.
1. (A)T1
2. (B)T5
3. (C)T4
4. (D)T3
Figure in the paper: A pyramid of five stacked levels labelled from bottom to top T1 (widest), T2, T3, T4, T5 (narrowest, at the top). See it in the official paper.
Try it first, then show our solutionSolution
1. Harmful non-biodegradable chemicals accumulate at each step of a food chain (biological magnification). They are therefore maximum at the top level, T5. Option (B).
  1 mark
Answer: T5, option (B)
Where the marks are
- T5, due to biomagnification (1 mark)
Q7 · 1 mark · MCQOur Environment
Which of the following limits the number of trophic levels in a food chain ?
1. (A)Availability of water
2. (B)Deficient food supply
3. (C)Decrease in energy at higher levels
4. (D)Pollution in air
Try it first, then show our solutionSolution
1. Only about 10% of the energy passes on to each next trophic level (10% law). After 3 or 4 levels too little energy is left to support another level, so the chain stops. Option (C).
  1 mark
Answer: Decrease in energy at higher levels, option (C)
Where the marks are
- Links the limit on trophic levels to energy loss at each level (1 mark)
Q8 · 1 mark · Assertion-reasonCompetency-basedLife Processes
Assertion (A) : Blood plasma transports carbon dioxide in dissolved form, while oxygen is transported by respiratory pigments.
Reason (R) : Carbon dioxide is more soluble in water than in oxygen.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. Assertion: CO₂ is carried mainly dissolved in plasma, while O₂ is carried by haemoglobin, a respiratory pigment. True.
2. Reason: CO₂ is more soluble in water than O₂ is. This is why plasma can carry CO₂ in dissolved form but not enough O₂, so R explains A. Option (A).
  1 mark
Answer: Option (A) (scheme also accepts (C), because of the wording of R)
Where the marks are
- Selects (A); (C) also accepted (1 mark)
Open Q8 on its own page
Q9 · 1 mark · Assertion-reasonCompetency-basedOur Environment
Assertion (A) : There are generally a greater number of individuals at the lower trophic levels of an ecosystem.
Reason (R) : The greatest number is of the producers.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. Both statements are true. Lower trophic levels hold more individuals, and producers are the most numerous.
2. R only restates a fact about one level. It does not explain why numbers are greater at lower levels; the explanation is the energy loss at each transfer. So R is not the correct explanation. Option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Recognises both true but R does not explain A (1 mark)
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Answer one of these 2 alternatives.
Q10(a) · 2 marks · Very short answerControl and Coordination
“The timing and the amount of hormone secreted by a gland are regulated in the human body.” Justify this statement with the help of an example.
Try it first, then show our solutionSolution
1. The timing and amount of a hormone released are controlled by a feedback mechanism: the body detects the level of a substance and adjusts hormone release to match.
  1 mark
2. Example (insulin): when blood sugar rises, cells of the pancreas detect it and secrete more insulin. As the blood sugar level falls, insulin secretion is reduced. Insulin is therefore released only when needed and in the right amount.
  1 mark
Answer: Hormone secretion is regulated by feedback mechanisms, e.g. the pancreas secretes more insulin when blood sugar rises and less when it falls.
Where the marks are
- States that timing and amount are regulated by a feedback mechanism (1 mark)
- Gives a correct example, e.g. insulin release by the pancreas changing with blood sugar level (1 mark)
Open Q10(a) on its own page
OR
Q10(b) · 2 marks · Very short answerControl and Coordination
Electrical impulse is an excellent means to transmit information in the animal body, but there are some limitations to the use of electrical impulses. State any two limitations.
Try it first, then show our solutionSolution
1. Limitation 1: Electrical impulses reach only those cells that are connected by nervous tissue. They cannot reach every cell of the animal body.
  1 mark
2. Limitation 2: After a cell generates and transmits an impulse, it needs some time to reset its mechanism before it can send a new one. So electrical impulses cannot be transmitted continuously.
  1 mark
Answer: (1) Impulses reach only cells connected by nervous tissue; (2) cells need time to reset after each impulse, so transmission is not continuous.
Where the marks are
- Impulses reach only cells connected to nervous tissue (1 mark)
- Cells need time to reset before the next impulse, so transmission is not continuous (1 mark)
Open Q10(b) on its own page
Q11 · 2 marks · Very short answerHeredity
“Variation is useful for the survival of species over time.” Explain this statement with the help of an example.
Try it first, then show our solutionSolution
1. Variation means individuals of a species are not all identical. If the environment changes, some variants may be better suited to the new conditions. They survive and reproduce, so the species does not die out.
  1 mark
2. Example: Suppose a population of bacteria lives in temperate water and the water temperature rises because of global warming. Most bacteria would die, but the few variants that resist heat would survive and multiply. Variation thus allows the species to survive over time.
  1 mark
Answer: Variation gives some individuals traits that let them survive changed conditions, e.g. heat-resistant bacterial variants surviving when water warms.
Where the marks are
- Explains that variation helps a species survive changing or unfavourable conditions (1 mark)
- Gives a suitable example such as heat-resistant bacteria surviving a rise in water temperature (1 mark)
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Q12 · 2 marks · Very short answerCompetency-basedOur Environment
Suggest one best way to dispose of the following wastes :
(i) Kitchen waste
(ii) Metallic waste
(iii) Plastic bottles
(iv) Sanitary waste
Try it first, then show our solutionSolution
1. (i) Kitchen waste: composting (converting it into manure).
  ½ mark
2. (ii) Metallic waste: recycling or reuse.
  ½ mark
3. (iii) Plastic bottles: recycling or reuse.
  ½ mark
4. (iv) Sanitary waste: incineration (burning at high temperature).
  ½ mark
Answer: (i) Composting; (ii) Recycle/reuse; (iii) Recycle/reuse; (iv) Incineration
Where the marks are
- Kitchen waste – composting (½ mark)
- Metallic waste – recycling/reuse (½ mark)
- Plastic bottles – recycling/reuse (½ mark)
- Sanitary waste – incineration (½ mark)
Open Q12 on its own page
Q13 · 3 marks · Short answerLife Processes
Complete the pathways given below, showing the breakdown of glucose.
Identify ‘A’, ‘B’, ‘C’, ‘D’, ‘E’ and ‘F’.
Figure in the paper: Flow chart: Glucose --‘A’--> Pyruvate + Energy. From Pyruvate three branches: in yeast --> ‘B’ + ‘C’ + Energy; in muscle cells --> ‘D’ + Energy; in mitochondria --> ‘E’ + ‘F’ + Energy. See it in the official paper.
Try it first, then show our solutionSolution
1. A: Cytoplasm. Glucose (6-carbon) is broken down to pyruvate (3-carbon) in the cytoplasm.
  ½ mark
2. B: Ethanol (yeast, anaerobic respiration)
  ½ mark
3. C: Carbon dioxide. B and C may be written in either order.
  ½ mark
4. D: Lactic acid. This forms in muscle cells when oxygen is in short supply.
  ½ mark
5. E: Carbon dioxide (aerobic respiration in mitochondria)
  ½ mark
6. F: Water. E and F may be written in either order.
  ½ mark
Answer: A – Cytoplasm; B – Ethanol; C – Carbon dioxide; D – Lactic acid; E – Carbon dioxide; F – Water (B/C and E/F interchangeable)
Where the marks are
- A = cytoplasm (½ mark)
- B = ethanol (½ mark)
- C = carbon dioxide (½ mark)
- D = lactic acid (½ mark)
- E = carbon dioxide (½ mark)
- F = water (½ mark)
Open Q13 on its own page
Q14 · 3 marks · Short answerHeredity
Answer the following.
1. (a)
  Giving two examples, justify the statement that the sex of the offspring in all animals is not always genetically determined.
  0 marks
2. (b)
  Differentiate between the sex chromosomes of male and female human beings.
  0 marks
Try it first, then show our solutionSolution
1. (a) Example 1: In a few reptiles, the temperature at which the fertilised eggs are kept decides whether the animal developing from the egg will be male or female.
  1 mark
2. (a) Example 2: In snails, individuals can change sex. So sex is not always fixed by genes.
  1 mark
3. (b) Males have a mismatched pair of sex chromosomes, XY, in which Y is smaller than X. Females have a perfect pair, XX, with both X chromosomes of the same size.
  1 mark
Answer: (a) Temperature-dependent sex determination in some reptiles; sex change in snails. (b) Male: XY (mismatched, Y smaller); Female: XX (matched pair).
Where the marks are
- Example of reptiles where egg temperature decides sex (1 mark)
- Example of snails changing sex (1 mark)
- Difference: male XY mismatched pair vs female XX perfect pair (1 mark)
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Q15 · 4 marks · Case-basedCompetency-basedControl and Coordination
Read the following passage and answer the questions that follow :
A middle-aged person is facing some cognitive changes in himself. His early symptoms included mild forgetfulness. For example, forgetting where he left his mobile or failing to find words to complete his sentences. He was facing problems in driving a car, climbing up and down the stairs, inserting a plug into an outlet. He slowly developed a sad mood and anxiety.
1. (a)
  What are voluntary actions ?
  1 mark
2. (b)
  Which part of the brain is responsible for precision of voluntary actions ?
  1 mark
3. (c)(i)
  Explain the role of the medulla.
  2 marks
4. OR
  (c)(ii)
  How do the animal muscles move ?
  2 marks
Try it first, then show our solutionSolution
1. (a) Voluntary actions are actions that are under our control and done according to our will, such as writing or walking. They are controlled by the forebrain.
  1 mark
2. (b) The cerebellum (in the hindbrain) is responsible for the precision of voluntary actions and for maintaining posture and balance.
  1 mark
3. (c)(i) The medulla, in the hindbrain, controls involuntary actions such as heart rate, blood pressure and breathing. It also controls reflexes like sneezing and vomiting.
  2 marks
4. (c)(ii) Animal muscle cells contain special proteins. In response to nerve electrical impulses, these proteins change both their shape and their arrangement in the cell. The new arrangement gives the muscle cells a shorter form, so the muscles contract and the animal moves.
  2 marks
Answer: (a) Actions under our will, controlled by the forebrain; (b) Cerebellum; (c)(i) Medulla controls involuntary actions like heartbeat, blood pressure, breathing, sneezing and vomiting; (c)(ii) Muscle proteins change shape and arrangement on nerve impulses, shortening the cells.
Where the marks are
- Defines voluntary actions as those under our will or controlled by the forebrain (1 mark)
- Names cerebellum (1 mark)
- (c)(i) Medulla regulates involuntary functions with examples (2 marks)
- (c)(ii) Muscle proteins change shape and arrangement in response to impulses, shortening the cells (2 marks)
Open Q15 on its own page
Answer one of these 2 alternatives.
Q16(a) · 5 marks · Long answerHow do Organisms Reproduce?
Answer the following.
1. (i)
  Name an organ that helps in the formation of male germ cells in humans and also mention the condition that favours the formation of male germ cells.
  2 marks
2. (ii)
  In male reproductive system, write the pathway followed to deliver male germ cells to the site of fertilisation.
  2 marks
3. (iii)
  Mention the main physical feature of the male germ cell that helps it move towards the female germ cell.
  1 mark
Try it first, then show our solutionSolution
1. (i) Organ: Testes. They form the male germ cells (sperms).
  1 mark
2. (i) Condition: sperm formation needs a temperature lower than normal body temperature, which is why the testes lie in the scrotum outside the abdominal cavity. Testosterone secreted by the testes also regulates sperm formation.
  1 mark
3. (ii) Sperms formed in the testes are carried by the vas deferens. The vas deferens joins the urethra, the common passage for sperms and urine. Pathway: Testes → Vas deferens → Urethra → out of the body. Secretions of the seminal vesicles and prostate gland are added on the way.
  2 marks
4. (iii) The sperm has a long tail, which helps it swim towards the female germ cell.
  1 mark
Answer: (i) Testes; temperature lower than body temperature; (ii) Testes → Vas deferens → Urethra; (iii) Long tail
Where the marks are
- Names testes (1 mark)
- States lower than body temperature (or testosterone) as the condition (1 mark)
- Pathway testes → vas deferens → urethra (2 marks)
- Long tail of sperm (1 mark)
Open Q16(a) on its own page
OR
Q16(b) · 5 marks · Long answerCompetency-basedHow do Organisms Reproduce?
Radha has three children. She does not want to have more children. Radha visits a nearby hospital, to know the methods to prevent pregnancy.
1. (i)
  Mention the name of any two ways other than the surgical methods, following which Radha can avoid unwanted pregnancies.
  2 marks
2. (ii)
  Name the surgical methods to prevent pregnancy.
  2 marks
3. (iii)
  Give one example each of sexually transmitted diseases caused by bacteria and virus.
  1 mark
Try it first, then show our solutionSolution
1. (i) Any two non-surgical methods: (1) Mechanical barrier such as a condom, or a diaphragm or similar covering worn in the vagina; (2) Oral contraceptive pills, which change the hormonal balance so eggs are not released; (3) Copper-T or loop placed in the uterus.
  2 marks
2. (ii) Surgical methods: Vasectomy, in which the vas deferens of the male is blocked or cut so sperm cannot pass.
  1 mark
3. (ii) Tubectomy, in which the fallopian tube of the female is blocked or cut so the egg cannot reach the uterus.
  1 mark
4. (iii) STD caused by bacteria: gonorrhoea or syphilis.
  ½ mark
5. (iii) STD caused by virus: warts or HIV-AIDS.
  ½ mark
Answer: (i) Any two: condom/diaphragm, oral pills, Copper-T; (ii) Vasectomy and tubectomy; (iii) Bacterial: gonorrhoea/syphilis; Viral: warts/AIDS
Where the marks are
- Any two non-surgical contraceptive methods (1 + 1 marks)
- Vasectomy (vas deferens blocked) (1 mark)
- Tubectomy (fallopian tube blocked) (1 mark)
- Bacterial STD example (½ mark)
- Viral STD example (½ mark)
Open Q16(b) on its own page

Section B

Chemistry (25 marks): MCQ 17-23, assertion-reason 24, case/short 25-27, case based 28, LA 29

Q17 · 1 mark · MCQChemical Reactions and Equations
Which of the following are combination reactions ?
(i) $2 K C l O_{3} \to 2 K C l + 3 O_{2}$
(ii) $M g O + H_{2} O \to M g (O H)_{2}$
(iii) $4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$
(iv) $Z n + F e S O_{4} \to Z n S O_{4} + F e$
1. (A)(i) and (iii)
2. (B)(iii) and (iv)
3. (C)(ii) and (iv)
4. (D)(ii) and (iii)
Try it first, then show our solutionSolution
1. A combination reaction joins two or more reactants into a single product.
2. (ii) MgO + H₂O → Mg(OH)₂ and (iii) 4Al + 3O₂ → 2Al₂O₃ each give a single product. Reaction (i) is decomposition and (iv) is displacement. Option (D).
  1 mark
Answer: (ii) and (iii), option (D)
Where the marks are
- Identifies (ii) and (iii) as combination reactions (1 mark)
Open Q17 on its own page
Q18 · 1 mark · MCQAcids, Bases and Salts
A metal carbonate reacts with a solution X which forms a salt, water and a gas Y. What are X and Y ?
1. (A)X – Sodium hydroxide, Y – Carbon dioxide
2. (B)X – Sodium hydroxide, Y – Hydrogen
3. (C)X – Hydrochloric acid, Y – Carbon dioxide
4. (D)X – Hydrochloric acid, Y – Hydrogen
Try it first, then show our solutionSolution
1. Metal carbonate + acid → salt + water + CO₂. For example, $N a_{2} C O_{3} + 2 H C l \to 2 N a C l + H_{2} O + C O_{2}$ . So X is hydrochloric acid and Y is carbon dioxide, option (C).
  1 mark
Answer: X – HCl, Y – CO₂, option (C)
Where the marks are
- X = hydrochloric acid, Y = carbon dioxide (1 mark)
Q19 · 1 mark · MCQChemical Reactions and Equations
When ferrous sulphate crystals are heated in a test tube, we observe that :
1. (A)A colourless gas with no smell is evolved.
2. (B)A brown gas is evolved.
3. (C)The green colour of the salt fades and a gas with the smell of burning sulphur is evolved.
4. (D)The green colour of the salt fades and no gas is evolved.
Try it first, then show our solutionSolution
1. On heating, the green crystals lose their water of crystallisation, so the colour fades. The salt then decomposes: $2 F e S O_{4} \to F e_{2} O_{3} + S O_{2} + S O_{3}$ . The SO₂ released smells of burning sulphur. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Colour fades and SO₂ with burning-sulphur smell is evolved (1 mark)
Q20 · 1 mark · MCQMetals and Non-metals
When Mg metal is treated with hot water, it starts floating on the surface of the water. The reason for its different behaviour is :
1. (A)Mg is a lighter element than water.
2. (B)Mg reacts with water to produce $H_{2}$ gas which helps in floating.
3. (C)Mg reacts with water to produce $N_{2}$ gas which helps in floating.
4. (D)Mg reacts with water to form $C O_{2}$ gas which helps in floating.
Try it first, then show our solutionSolution
1. $M g + 2 H_{2} O \to M g (O H)_{2} + H_{2}$ . Bubbles of hydrogen stick to the surface of the metal and make it float. Option (B).
  1 mark
Answer: Option (B)
Where the marks are
- H₂ bubbles cause floating (1 mark)
Q21 · 1 mark · MCQAcids, Bases and Salts
The pH of the gastric juices released during digestion is :
1. (A)Less than 7
2. (B)More than 7
3. (C)Equal to 7
4. (D)Equal to 0
Try it first, then show our solutionSolution
1. The stomach secretes hydrochloric acid, so gastric juice is acidic and its pH is less than 7. Option (A).
  1 mark
Answer: Less than 7, option (A)
Where the marks are
- Acidic, pH < 7 (1 mark)
Q22 · 1 mark · MCQAcids, Bases and Salts
Baking powder is a mixture of :
1. (A)Sodium carbonate + Acetic acid
2. (B)Sodium carbonate + Tartaric acid
3. (C)Sodium hydrogen carbonate + Tartaric acid
4. (D)Sodium hydrogen carbonate + Acetic acid
Try it first, then show our solutionSolution
1. Baking powder is baking soda (NaHCO₃) mixed with a mild edible acid, tartaric acid. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- NaHCO₃ + tartaric acid (1 mark)
Q23 · 1 mark · MCQCompetency-basedMetals and Non-metals
A student notices that her silver jewellery turned dull and had a grey-black coating over it after wearing for a few months. What results in the change in colour of the silver metal ?
1. (A)The polish over the jewellery was removed after wearing for a few months.
2. (B)The jewellery comes in contact with air, moisture and acids and corrodes.
3. (C)Dust particles over the jewellery change its colour.
4. (D)Its colour changes due to rusting.
Try it first, then show our solutionSolution
1. Silver corrodes when it reacts with sulphur compounds in the air and moisture. This forms a black coating of silver sulphide (tarnishing). Rusting applies only to iron, so option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Identifies corrosion of silver (1 mark)
Q24 · 1 mark · Assertion-reasonCompetency-basedCarbon and its Compounds
Assertion (A) : Carbon compounds can form chains, branched and ring structures.
Reason (R) : Carbon exhibits the property of catenation.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. Catenation is carbon's ability to bond with other carbon atoms. It lets carbon form long chains, branched chains and rings, so R correctly explains A. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Both true, R explains A (1 mark)
Q25 · 2 marks · Very short answerCompetency-basedMetals and Non-metals
Read the following passage and answer the questions that follow :
One student was comparing the reactivity of different metals for his science project. He added iron filings in four test tubes A, B, C and D containing aqueous solutions of $Z n S O_{4}$ , $C u S O_{4}$ , $F e S O_{4}$ and $A l_{2} (S O_{4})_{3}$ .
1. (i)
  In which of the test tubes will he observe the reaction to be the most vigorous and why ?
  1 mark
2. (ii)
  Write a balanced equation for the reaction involved.
  1 mark
Try it first, then show our solutionSolution
1. (i) The reaction occurs in test tube B ( $C u S O_{4}$ solution).
  ½ mark
2. Reason: iron is more reactive than copper, so it displaces copper from copper sulphate solution. Iron is less reactive than Zn and Al and cannot displace itself from $F e S O_{4}$ , so no reaction occurs in A, C and D.
  ½ mark
3. (ii) $F e (s) + C u S O_{4} (a q) \to F e S O_{4} (a q) + C u (s)$
  The blue solution turns pale green and reddish-brown copper is deposited.
  1 mark
Answer: (i) Test tube B, as iron is more reactive than copper and displaces it; (ii) $F e (s) + C u S O_{4} (a q) \to F e S O_{4} (a q) + C u (s)$
Where the marks are
- Identifies test tube B (½ mark)
- Reason: Fe is more reactive than Cu and displaces it (½ mark)
- Correct balanced equation (1 mark)
Open Q25 on its own page
Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerMetals and Non-metals
Answer the following.
1. (i)
  Write one example each of one metal that is so soft that it can be cut with a knife, and one non-metal that is a good conductor of electricity.
  1 mark
2. (ii)
  Using the electronic configuration, explain how Mg atom combines with oxygen to form magnesium oxide by transfer of electrons. [Atomic number : Mg = 12, O = 8]
  2 marks
Try it first, then show our solutionSolution
1. (i) Soft metal that can be cut with a knife: sodium (or lithium or potassium).
  ½ mark
2. (i) Non-metal that is a good conductor of electricity: graphite (a form of carbon).
  ½ mark
3. (ii) Mg (Z = 12) has configuration 2, 8, 2. It loses its 2 valence electrons to form $M g^{2 +}$ (2, 8): $M g \to M g^{2 +} + 2 e^{-}$
  ½ mark
4. O (Z = 8) has configuration 2, 6. It gains these 2 electrons to form $O^{2 -}$ (2, 8): $O + 2 e^{-} \to O^{2 -}$
  ½ mark
5. Electron-dot representation: $M g :$ (2 dots) $+ \ddot{.. O}$ (6 dots) $\to [M g]^{2 +} [: \ddot{.. O} :]^{2 -}$ . Arrows show the two electrons moving from Mg to O.
  ½ mark
6. The oppositely charged ions $M g^{2 +}$ and $O^{2 -}$ are held by strong electrostatic attraction, forming ionic magnesium oxide $M g O$ . Both ions now have a stable octet.
  ½ mark
Answer: (i) Sodium (or Li/K); graphite. (ii) Mg (2,8,2) loses 2e⁻ → Mg²⁺; O (2,6) gains 2e⁻ → O²⁻; Mg²⁺O²⁻ forms ionic MgO.
Where the marks are
- Soft metal example: Na/K/Li (½ mark)
- Graphite (½ mark)
- Electronic configurations and formation of Mg²⁺ (½ mark)
- Formation of O²⁻ (½ mark)
- Electron transfer shown by electron-dot diagram (½ mark)
- Ionic bond in Mg²⁺O²⁻ (½ mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerMetals and Non-metals
Give reasons :
1. (i)
  Metal sulphides and carbonates should be converted to their metal oxides in the process of extraction of metals.
  1 mark
2. (ii)
  Aluminium oxide is considered as an amphoteric oxide.
  1 mark
3. (iii)
  Metals like Na, K, Ca and Mg are never found in their free state in nature.
  1 mark
Try it first, then show our solutionSolution
1. (i) It is easier to obtain a metal from its oxide than from its sulphide or carbonate, because the oxide can be readily reduced (e.g. by carbon). So sulphide ores are roasted and carbonate ores are calcined to convert them to oxides first.
  1 mark
2. (ii) Aluminium oxide reacts with both acids and bases to form salt and water, so it is amphoteric:
  $A l_{2} O_{3} + 6 H C l \to 2 A l C l_{3} + 3 H_{2} O$
  $A l_{2} O_{3} + 2 N a O H \to 2 N a A l O_{2} + H_{2} O$
  1 mark
3. (iii) Na, K, Ca and Mg are highly reactive metals at the top of the reactivity series. They readily combine with oxygen, water and other substances, so they are always found in the combined state, never free.
  1 mark
Answer: (i) Oxides are easier to reduce to metal; (ii) Al₂O₃ reacts with both acids and bases giving salt and water; (iii) They are highly reactive, so exist only in combined state.
Where the marks are
- Metal oxides are easier to reduce to metals (1 mark)
- Al₂O₃ reacts with both acids and bases forming salt and water (1 mark)
- These metals are highly reactive, so found combined (1 mark)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerChemical Reactions and Equations
Seema took a small amount of copper powder in a china dish and heated it. What changes will she observe on heating ? When $H_{2}$ gas is passed over this heated substance, what visible changes will be seen in it ? Give the chemical equations of the reactions involved, along with the names and colours of the products in each case.
Figure in the paper: Apparatus: a china dish containing copper powder placed on a wire gauze over a tripod stand, heated by a burner below. Labels: China dish containing copper powder, Wire gauze, Tripod stand, Burner. See it in the official paper.
Try it first, then show our solutionSolution
1. On heating, the brown copper powder turns black as it gets coated with copper(II) oxide.
  ½ mark
2. When $H_{2}$ gas is passed over the heated black substance, the black coating turns brown again as copper is regenerated.
  ½ mark
3. Reaction 1 (oxidation): $2 C u + O_{2} h e a t 2 C u O$
  ½ mark
4. Product: copper(II) oxide, black in colour.
  ½ mark
5. Reaction 2 (reduction of CuO): $C u O + H_{2} h e a t C u + H_{2} O$
  ½ mark
6. Products: copper, brown in colour, and water (as vapour).
  ½ mark
Answer: Brown Cu turns black (2Cu + O₂ → 2CuO, copper oxide, black); with H₂ the black turns brown (CuO + H₂ → Cu + H₂O, copper, brown).
Where the marks are
- Observation: brown turns black on heating (½ mark)
- Observation: black turns brown on passing H₂ (½ mark)
- Equation 2Cu + O₂ → 2CuO (½ mark)
- Name and colour: copper oxide, black (½ mark)
- Equation CuO + H₂ → Cu + H₂O (½ mark)
- Name and colour: copper, brown (½ mark)
Open Q27 on its own page
Q28 · 4 marks · Case-basedCompetency-basedAcids, Bases and Salts
Read the following passage and answer the questions that follow :
In the given diagram, when electricity is passed through an aqueous solution of a common salt, a substance ‘Z’ is produced, along with the evolution of gases ‘X’ and ‘Y’. When a burning matchstick is brought near the gas ‘Y’, it burns with a pop sound, whereas ‘X’ is used for disinfecting drinking water. When ‘X’ is passed through slaked lime, an insoluble substance ‘A’ is produced.
1. (a)
  Write the names of the gases ‘X’ and ‘Y’.
  1 mark
2. (b)
  Write the balanced chemical equation for the formation of substance ‘Z’.
  1 mark
3. (c)(i)
  What happens :
  (I) If a drop of red litmus solution is added to the aqueous solution of substance ‘Z’ ?
  (II) If $C O_{2}$ gas is passed through ammoniacal solution of aqueous NaCl ?
  2 marks
4. OR
  (c)(ii)
  Identify the substance ‘A’ and write the balanced chemical equation of the reaction which takes place in its preparation.
  2 marks
Figure in the paper: Electrolytic cell for brine with a membrane (dotted vertical line) in the middle. Anode (+) electrode on the left with gas X leaving at top-left; Cathode (−) electrode on the right with gas Y leaving at top-right. Brine enters at the left; Z leaves at the right. Near the membrane arrows show Cl⁻ and OH⁻ moving left, and Na⁺ and H⁺ moving right. Membrane labelled at bottom. See it in the official paper.
Try it first, then show our solutionSolution
1. (a) X: chlorine gas ( $C l_{2}$ ), released at the anode and used to disinfect water.
  ½ mark
2. (a) Y: hydrogen gas ( $H_{2}$ ), released at the cathode; it burns with a pop sound.
  ½ mark
3. (b) Z is sodium hydroxide, formed in the chlor-alkali process:
  $2 N a C l (a q) + 2 H_{2} O (l) \to 2 N a O H (a q) + C l_{2} (g) + H_{2} (g)$
  1 mark
4. (c)(i)(I) Z (NaOH) is a base, so the red litmus solution turns blue.
  1 mark
5. (c)(i)(II) Sodium hydrogen carbonate ( $N a H C O_{3}$ ) and ammonium chloride ( $N H_{4} C l$ ) are formed:
  $N a C l + H_{2} O + C O_{2} + N H_{3} \to N H_{4} C l + N a H C O_{3}$
  1 mark
6. (c)(ii) A is bleaching powder ( $C a O C l_{2}$ ).
  1 mark
7. (c)(ii) Equation: $C a (O H)_{2} + C l_{2} \to C a O C l_{2} + H_{2} O$
  1 mark
Answer: (a) X = Cl₂, Y = H₂; (b) 2NaCl + 2H₂O → 2NaOH + Cl₂ + H₂; (c)(i) red litmus turns blue; NaHCO₃ and NH₄Cl formed; (c)(ii) A = bleaching powder, Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O
Where the marks are
- X = chlorine (½ mark)
- Y = hydrogen (½ mark)
- Balanced chlor-alkali equation (1 mark; ½ deducted if unbalanced)
- (c)(i)(I) Red litmus turns blue (1 mark)
- (c)(i)(II) NaHCO₃ and NH₄Cl formed or equation (1 mark)
- (c)(ii) Identifies bleaching powder (1 mark)
- (c)(ii) Balanced equation (1 mark)
Open Q28 on its own page
Answer one of these 2 alternatives.
Q29(a) · 5 marks · Long answerCarbon and its Compounds
A carbon compound ‘P’ of molecular formula $C_{2} H_{6} O$ is found to be neutral when tested with red and blue litmus. A gas that burns with a pop sound is produced when sodium metal reacts with compound ‘P’.
1. (i)
  Identify the compound ‘P’.
  1 mark
2. (ii)
  The carbon compound ‘P’ is heated with conc. $H_{2} S O_{4}$ at 443 K to produce compound ‘Q’.
  (I) Write the chemical equation for the reaction.
  (II) What is the role of conc. $H_{2} S O_{4}$ in this reaction ?
  2 marks
3. (iii)
  What is likely to be observed on heating compound ‘P’ with ethanoic acid with an acid as a catalyst ? Write the chemical equation for the reaction.
  2 marks
Try it first, then show our solutionSolution
1. (i) P is ethanol, $C_{2} H_{5} O H$ ( $C H_{3} C H_{2} O H$ ). It is neutral to litmus and reacts with sodium to give $H_{2}$ , which burns with a pop.
  1 mark
2. (ii)(I) $C H_{3} C H_{2} O H hot conc. H_{2} S O_{4} 443 K C H_{2} = C H_{2} + H_{2} O$
  Q is ethene.
  1 mark
3. (ii)(II) Concentrated $H_{2} S O_{4}$ acts as a dehydrating agent, removing water from ethanol.
  1 mark
4. (iii) A sweet-smelling substance, an ester (ethyl ethanoate), is formed. This reaction is esterification.
  1 mark
5. (iii) $C H_{3} C O O H + C H_{3} C H_{2} O H acid C H_{3} C O O C H_{2} C H_{3} + H_{2} O$
  1 mark
Answer: (i) Ethanol; (ii) C₂H₅OH → CH₂=CH₂ + H₂O (hot conc. H₂SO₄, 443 K), H₂SO₄ is dehydrating agent; (iii) Sweet-smelling ester formed: CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O
Where the marks are
- Identifies P as ethanol (1 mark)
- Equation for dehydration to ethene (1 mark)
- Role: dehydrating agent (1 mark)
- Observation: sweet-smelling ester (1 mark)
- Esterification equation (1 mark)
Open Q29(a) on its own page
OR
Q29(b) · 5 marks · Long answerCarbon and its Compounds
Answer the following.
1. (i)
  Write the structural formula of benzene.
  1 mark
2. (ii)
  State the reason why carbon can neither form $C^{4 +}$ cation nor $C^{4 -}$ anion.
  1 mark
3. (iii)
  What type of bond is formed between carbon and hydrogen ?
  1 mark
4. (iv)
  How is ethanol converted to ethanoic acid ?
  1 mark
5. (v)
  Complete the following chemical equation :
  $C H_{3} C O O C_{2} H_{5} + N a O H \to$
  1 mark
Try it first, then show our solutionSolution
1. (i) Benzene, $C_{6} H_{6}$ , is a hexagonal ring of six carbon atoms with alternate single and double bonds (three C=C). Each carbon carries one H atom. Either resonating structure is acceptable.
  1 mark
2. (ii) Forming $C^{4 +}$ needs removal of four electrons, which requires a very large amount of energy. Forming $C^{4 -}$ is not possible because a nucleus with six protons cannot hold ten electrons.
  1 mark
3. (iii) A covalent bond, formed by sharing a pair of electrons.
  1 mark
4. (iv) Ethanol is oxidised by alkaline $K M n O_{4}$ or acidified $K_{2} C r_{2} O_{7}$ on heating:
  $C H_{3} C H_{2} O H alk. K M n O_{4} + heat C H_{3} C O O H$
  1 mark
5. (v) $C H_{3} C O O C_{2} H_{5} + N a O H \to C H_{3} C O O N a + C_{2} H_{5} O H$
  This is saponification; the products are sodium ethanoate and ethanol.
  1 mark
Answer: (i) Six-carbon hexagonal ring with alternate single/double bonds; (ii) energy needed to lose 4e⁻ is too high, and 6 protons cannot hold 10 electrons; (iii) Covalent; (iv) Oxidation with alkaline KMnO₄/acidified K₂Cr₂O₇; (v) CH₃COONa + C₂H₅OH
Where the marks are
- Correct structure of benzene (1 mark)
- Reasons for no C⁴⁺ and no C⁴⁻ (1 mark)
- Covalent bond (1 mark)
- Oxidation using alkaline KMnO₄ or acidified K₂Cr₂O₇ (1 mark)
- Products sodium ethanoate and ethanol (1 mark)
Open Q29(b) on its own page

Section C

Physics (25 marks): MCQ 30-31, assertion-reason 32, VSA 33-34, SA 35-37, case based 38, LA 39

Q30 · 1 mark · MCQLight – Reflection and Refraction
From the following options choose the one which gives negative magnification greater than one :
1. (A)Concave mirror; Nature of image – real
2. (B)Concave mirror; Nature of image – virtual
3. (C)Convex mirror; Nature of image – real
4. (D)Convex mirror; Nature of image – virtual
Try it first, then show our solutionSolution
1. Negative magnification means the image is real and inverted. Of the two mirrors, only a concave mirror can form a real image.
2. When the object lies between F and C, the image is formed beyond C, is enlarged, and has $m < - 1$ . Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Concave mirror, real image (1 mark)
Open Q30 on its own page
Q31 · 1 mark · MCQThe Human Eye and the Colourful World
A thin beam of white light falls on a face of a glass prism and a spectrum is obtained on a screen. Which of the following options is the correct sequence of colours from top to bottom on the screen ?
1. (A)Red, Green, Yellow, Violet, Blue
2. (B)Red, Yellow, Green, Blue, Violet
3. (C)Violet, Blue, Green, Yellow, Red
4. (D)Violet, Red, Green, Yellow, Blue
Try it first, then show our solutionSolution
1. Red deviates least and violet deviates most. With the prism's base downward, red appears at the top and violet at the bottom: Red, Yellow, Green, Blue, Violet (VIBGYOR reversed). Option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Correct order red to violet (1 mark)
Q32 · 1 mark · Assertion-reasonCompetency-basedMagnetic Effects of Electric Current
Assertion (A) : Direction of force exerted on a current-carrying conductor placed in a magnetic field remains same if the directions of both magnetic field and current flowing through the conductor are reversed.
Reason (R) : The direction of force exerted on a current-carrying conductor placed in an external magnetic field can be determined by using magnetic compass.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. Assertion: reversing both the field and the current reverses the force twice, so its direction stays the same. True.
2. Reason: the direction of the force is found using Fleming's left-hand rule, not a magnetic compass. False. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- A true, R false (1 mark)
Open Q32 on its own page
Q33 · 2 marks · Very short answerLight – Reflection and Refraction
With the help of the given ray diagram, calculate the refractive index of Medium 2 with respect to Medium 1.
(Given that $3$ = 1·732)
Figure in the paper: Horizontal interface separating Medium 1 (above) and Medium 2 (below), with a dashed normal N vertical through point O. Incident ray AO comes from upper left making an angle of 30° with the interface (the horizontal surface). Refracted ray OB goes into Medium 2 making an angle of 30° with the normal. See it in the official paper.
Try it first, then show our solutionSolution
1. By Snell's law, $n_{21} = \frac{sin i}{sin r}$
  ½ mark
2. The incident ray makes 30° with the surface, so $i = 90° - 30° = 60°$ . The refracted ray makes 30° with the normal, so $r = 30°$ .
  $n_{21} = \frac{sin 60°}{sin 30°} = \frac{3 /2}{1/2}$
  1 mark
3. $n_{21} = 3 = 1.732$
  ½ mark
Answer: Refractive index of Medium 2 w.r.t. Medium 1 = √3 = 1.732
Where the marks are
- Writes Snell's law sin i / sin r = n (½ mark)
- Finds i = 60°, r = 30° and substitutes (1 mark)
- n = √3 = 1.732 (½ mark)
Open Q33 on its own page
Answer one of these 2 alternatives.
Q34(a) · 2 marks · Very short answerThe Human Eye and the Colourful World
How does the focal length of eye lens change as the distance of the object from the eyes is altered ? Explain.
Try it first, then show our solutionSolution
1. When the object is moved farther from the eye, the ciliary muscles relax. The eye lens becomes thinner and its focal length increases, so a distant object is focused on the retina.
  1 mark
2. When the object is brought closer to the eye, the ciliary muscles contract. The eye lens becomes thicker and its focal length decreases, so a near object is focused on the retina. This ability is called accommodation.
  1 mark
Answer: Farther object: ciliary muscles relax, lens thins, focal length increases; nearer object: muscles contract, lens thickens, focal length decreases.
Where the marks are
- Distant object: ciliary muscles relax, lens thin, focal length increases (1 mark)
- Near object: ciliary muscles contract, lens thick, focal length decreases (1 mark)
Open Q34(a) on its own page
OR
Q34(b) · 2 marks · Very short answerThe Human Eye and the Colourful World
Why is the concave lens used as a corrective lens for a myopic eye ?
Try it first, then show our solutionSolution
1. In a myopic (near-sighted) eye, the image of a distant object is formed in front of the retina.
  1 mark
2. A concave (diverging) lens of suitable power diverges the incoming rays slightly before they enter the eye. This moves the image back onto the retina, so distant objects are seen clearly.
  1 mark
Answer: Myopic eye forms the image in front of the retina; a concave (diverging) lens of suitable power shifts it back onto the retina.
Where the marks are
- States the image forms in front of the retina in myopia (1 mark)
- A concave/diverging lens brings the image back onto the retina (1 mark)
Open Q34(b) on its own page
Q35 · 3 marks · Short answerLight – Reflection and Refraction
Answer the following.
1. (a)
  Write the expression for the magnification produced by a lens in terms of object distance and image distance.
  0 marks
2. (b)
  A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. Calculate the size of the image formed, if the distance of the object from the lens is 10 cm.
  0 marks
Try it first, then show our solutionSolution
1. (a) Magnification produced by a lens: $m = \frac{image distance}{object distance} = \frac{v}{u}$
  1 mark
2. (b) Given (sign convention): $h_{o} = + 4$ cm, $f = + 20$ cm (convex lens), $u = - 10$ cm.
  Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ , so $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
  ½ mark
3. $\frac{1}{v} = \frac{1}{20} + \frac{1}{- 10} = \frac{1 - 2}{20} = - \frac{1}{20}$ , so $v = - 20$ cm (virtual image on the same side as the object).
  ½ mark
4. $m = \frac{v}{u} = \frac{- 20}{- 10} = + 2$ .
  Since $m = \frac{h _{i}}{h _{o}}$ : $h_{i} = m \times h_{o} = 2 \times 4 = 8$ cm. The image is 8 cm tall, virtual and erect.
  1 mark
Answer: (a) $m = v / u$ ; (b) $v = - 20$ cm, $m = + 2$ , image height = 8 cm (virtual, erect)
Where the marks are
- Writes $m = v / u$ (1 mark)
- Uses lens formula with correct signs (½ mark)
- Finds $v = - 20$ cm (½ mark)
- Finds $m = 2$ and image size 8 cm (1 mark)
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Q36 · 3 marks · Short answerThe Human Eye and the Colourful World
What is scattering of light ? How is scattering of light different from reflection of light ? Explain.
Try it first, then show our solutionSolution
1. Scattering of light: When light strikes very small particles of a medium (dust, smoke, air molecules), it gets spread out in different directions. This is called scattering of light.
  1 mark
2. Difference from reflection: In reflection, light bounces back from a surface in one fixed direction, following the laws of reflection (angle of incidence = angle of reflection). In scattering, light is thrown in all directions.
  1 mark
3. Reflection does not depend on the size of the reflecting surface's particles, but in scattering, the colour of the scattered light depends on the size of the particles. Very fine particles scatter mainly blue light (blue sky), while larger particles scatter all colours and appear white.
  1 mark
Answer: Scattering is the spreading of light in many directions by small particles of a medium; reflection sends light back in one fixed direction, and it does not depend on particle size, whereas the colour of scattered light does.
Where the marks are
- Correct definition of scattering of light (1 mark)
- Difference: reflection is in a fixed direction, scattering is in all directions / scattered colour depends on particle size (2 marks)
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Q37 · 3 marks · Short answerElectricity
The values of current $I$ flowing in a given resistor for the corresponding values of potential difference $V$ applied across the ends of resistor are given below in the table :
$I$ (Amperes) 0·4 0·8 1·2 1·6 2·0
$V$ (Volts) 1·2 2·4 3·6 4·8 6·0
Plot a graph between $V$ and $I$ and calculate the resistance of that resistor.
Try it first, then show our solutionSolution
1. Graph: Take current $I$ (A) on the x-axis and potential difference $V$ (V) on the y-axis with a suitable scale (e.g. 1 cm = 0.2 A, 1 cm = 0.6 V). Plot the points (0.4, 1.2), (0.8, 2.4), (1.2, 3.6), (1.6, 4.8), (2.0, 6.0). Joining them gives a straight line passing through the origin, which shows $V \propto I$ (Ohm's law).
  2 marks
2. Resistance = slope of the $V$ – $I$ graph. Take points A(0.4 A, 1.2 V) and B(2.0 A, 6.0 V) on the line, with C at (2.0 A, 1.2 V):
  $R = \frac{B C}{A C} = \frac{6.0 - 1.2}{2.0 - 0.4}$
  ½ mark
3. $R = \frac{4.8}{1.6} = 3 Ω$
  ½ mark
Answer: Straight-line $V$ – $I$ graph through the origin; $R = 3 Ω$
Where the marks are
- Correctly plotted straight-line $V$ – $I$ graph with labelled axes (2 marks)
- Uses slope = $Δ V /Δ I$ with substitution (½ mark)
- $R = 3 Ω$ (½ mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedElectricity
Read the following passage and answer the questions that follow :
Swati, a class 10 student, observes that when she passes close to the refrigerator in her kitchen, she feels the heat, although the things kept inside the refrigerator are cool.
1. (a)
  Describe the cause of heating in the above-mentioned case.
  1 mark
2. (b)
  A current $I$ flows through a resistor of resistance $R$ when the potential difference across it is $V$ . Applying Ohm’s law, write the formula for amount of heat produced by the resistor in time $t$ .
  1 mark
3. (c)(i)
  Write any two practical applications of heating effect of electric current.
  2 marks
4. OR
  (c)(ii)
  Define the commercial unit of electric energy and express it in Joules (J).
  2 marks
Try it first, then show our solutionSolution
1. (a) The refrigerator's compressor and circuits carry electric current. Only part of the electrical energy does useful work (cooling); the rest is turned into heat because of the resistance of the wires and coils (heating effect of current). This heat is given out at the back and sides, so Swati feels it.
  1 mark
2. (b) Work done in time $t$ : $W = V Q = V I t$ . By Ohm's law $V = I R$ , so the heat produced is $H = I^{2} R t$ (equivalently $H = \frac{V ^{2}}{R} t$ ).
  1 mark
3. (c)(i) Applications of the heating effect of current: (1) electric heater / electric iron / toaster / oven; (2) electric fuse (melts on excess current) / filament of electric bulb.
  2 marks
4. (c)(ii) The commercial unit of electric energy is the kilowatt-hour (kWh): the energy used when an appliance of power 1 kW runs for 1 hour.
  $1 kWh = 1000 W \times 3600 s = 3.6 \times 1 0^{6}$ J
  2 marks
Answer: (a) Heating effect of current – part of the electrical energy is lost as heat; (b) $H = I^{2} R t$ ; (c)(i) electric heater, electric iron (also fuse, toaster); (c)(ii) 1 kWh = energy used by 1 kW in 1 h = $3.6 \times 1 0^{6}$ J
Where the marks are
- (a) Part of electric energy converted into heat due to resistance (1 mark)
- (b) $H = I^{2} R t$ derived using $V = I R$ (1 mark)
- (c)(i) Any two correct applications (1 + 1 marks)
- (c)(ii) Definition of kWh (1 mark) and $1 kWh = 3.6 \times 1 0^{6}$ J (1 mark)
Open Q38 on its own page
Answer one of these 2 alternatives.
Q39(a) · 5 marks · Long answerMagnetic Effects of Electric Current
Answer the following.
1. (i)
  State the rule which gives the direction of force acting on a current-carrying conductor placed in a magnetic field.
  0 marks
2. (ii)
  What is a fuse ? Why is it called a safety device ?
  0 marks
3. (iii)
  Why are electrical appliances with metallic bodies connected to the mains through a three-pin plug, whereas an electric bulb is connected with a two-pin plug ?
  0 marks
4. (iv)
  What will be the direction of force experienced by the conductor AB due to the magnetic field, when current flows in the direction shown ? Name the law you applied to find the direction.
  0 marks
Figure in the paper: Vertical conductor AB with A at top and B at bottom; electric current is shown flowing upward (from B to A). Four horizontal magnetic field lines with arrows pointing to the right cross the conductor. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Fleming's left-hand rule: Stretch the thumb, forefinger and middle finger of the left hand so that they are mutually perpendicular. If the forefinger points along the magnetic field and the middle finger along the current, the thumb points in the direction of the force (motion) on the conductor.
  1 mark
2. (ii) A fuse is a safety device, a short wire of low melting point connected in series with the circuit, which melts and breaks the circuit when excess current flows (overloading or short-circuit).
  ½ mark
3. It is called a safety device because by breaking the circuit it stops the excess current and so prevents damage to the wiring and appliances and avoids fire.
  ½ mark
4. (iii) A three-pin plug connects the metallic body of the appliance to the earth wire. The earth wire gives a low-resistance path for any leaked current to flow to the ground, so the body stays at earth potential and the user does not get an electric shock. A bulb has no metallic body that can be touched, so it needs no earthing and a two-pin plug is enough.
  1 mark
5. (iv) Current in AB is upward (B to A) and the field is towards the right. Applying the rule with forefinger to the right and middle finger upward, the thumb points into the page. So the force on AB is perpendicular to the plane of the paper, directed into it.
  1 mark
6. Law/rule applied: Fleming's left-hand rule.
  1 mark
Answer: (i) Fleming's left-hand rule; (ii) fuse – low-melting wire in series that melts on excess current, protecting circuit and appliances; (iii) three-pin plug earths the metallic body to prevent shock; bulb has no metallic body; (iv) into the page, by Fleming's left-hand rule
Where the marks are
- (i) Correct statement of Fleming's left-hand rule (1 mark)
- (ii) Definition of fuse (½ mark) and reason it is a safety device (½ mark)
- (iii) Earthing of metallic body gives low-resistance path and prevents shock; bulb needs no earthing (1 mark)
- (iv) Force into the page (1 mark)
- (iv) Names Fleming's left-hand rule (1 mark)
Open Q39(a) on its own page
OR
Q39(b) · 5 marks · Long answerMagnetic Effects of Electric Current
Answer the following.
1. (i)
  Draw the pattern of magnetic field lines through and around a current-carrying circular loop.
  0 marks
2. (ii)
  State the rule to find out the direction of the magnetic field inside and around the circular loop.
  0 marks
3. (iii)
  Give reasons :
  (I) Why should a burnt-out fuse be replaced by another fuse of identical rating only, in an electric circuit ?
  (II) Why is soft iron-core used in making an electromagnet ?
  0 marks
Try it first, then show our solutionSolution
1. (i) Diagram (described): Draw a circular loop standing perpendicular to the plane of the paper, cutting it at two points P and Q. Around each of P and Q draw concentric circles, small near the wire and larger farther away. Near the centre of the loop the field lines become nearly straight and parallel, passing through the loop perpendicular to its plane.
  1 mark
2. Mark directions: current coming out of the paper at P and going into the paper at Q; by the right-hand thumb rule the field lines are anticlockwise around P and clockwise around Q, so inside the loop all lines point in the same direction.
  1 mark
3. (ii) Right-hand thumb rule: Imagine holding the current-carrying wire (each part of the loop) in the right hand with the thumb pointing along the current; the direction in which the fingers curl gives the direction of the magnetic field lines.
  1 mark
4. (iii)(I) A fuse of higher rating would not melt when the current exceeds the safe limit, so it would not protect against overloading; one of lower rating would melt even at normal current and the appliance would not work. Hence a fuse of identical rating must be used.
  1 mark
5. (iii)(II) Soft iron becomes strongly magnetised as soon as current flows through the solenoid and loses its magnetism immediately when the current is switched off, so the electromagnet can be turned on and off as needed.
  1 mark
Answer: (i) Concentric circles around each point where the loop cuts the paper, becoming straight and parallel at the centre; (ii) right-hand thumb rule; (iii)(I) higher rating fails to protect, lower rating melts in normal use; (II) soft iron magnetises and demagnetises quickly
Where the marks are
- (i) Correct field pattern of a circular loop (1 mark)
- (i) Correct directions of current and field lines (1 mark)
- (ii) Statement of right-hand thumb rule (1 mark)
- (iii)(I) Reason for same fuse rating (1 mark)
- (iii)(II) Soft iron is easily magnetised and demagnetised (1 mark)
Open Q39(b) on its own page

$I$ (Amperes)	0·4	0·8	1·2	1·6	2·0
$V$ (Volts)	1·2	2·4	3·6	4·8	6·0

Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.