ScienceCBSE Class 10 · Sample paper, session 2025-26

CBSE Class 10 Science: sample paper 2025-26

The sample question paper CBSE published for the 2025-26 session, 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 sample question paper from cbseacademic.nic.in: https://cbseacademic.nic.in/web_material/SQP/ClassX_2025_26/Science-SQP.pdf. Questions are transcribed from that PDF; figures are described in words, so check the official paper for them. Questions printed only for visually impaired candidates are left out (1).

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. MCQ 1-7 (1 mark), assertion-reason 8-9 (1 mark), Q10 (2), Q11 (2, internal choice), Q12 (2), Q13 (3), Q14 (3), Q15 (4, internal choice), Q16 (5, internal choice)
B	17-29	Chemistry. MCQ 17-23 (1 mark), assertion-reason 24 (1), Q25 (2), Q26 (3, internal choice), Q27 (3), Q28 (4), Q29 (5, internal choice)
C	30-39	Physics. MCQ 30-31 (1 mark), assertion-reason 32 (1), Q33 (2), Q34 (2, internal choice), Q35 (3), Q36 (3), Q37 (3), Q38 (4, internal choice), Q39 (5, internal choice)

Instructions in brief: All questions are compulsory; an internal choice is provided in some questions and the student must attempt only one of the alternatives. Section A is Biology, Section B is Chemistry and Section C is Physics.

Section A

Biology. MCQ 1-7 (1 mark), assertion-reason 8-9 (1 mark), Q10 (2), Q11 (2, internal choice), Q12 (2), Q13 (3), Q14 (3), Q15 (4, internal choice), Q16 (5, internal choice)

Q1 · 1 mark · MCQLife Processes
Select the group in which all organisms have the same mode of nutrition.
1. (A)Cuscuta, yeast, legumes, leeches and tapeworm
2. (B)Cactus, ticks, lice, leeches and cow
3. (C)Cuscuta, ticks, lice, leeches and tapeworm
4. (D)Cactus, grass, lice, lion and tapeworm
Try it first, then show our solutionSolution
1. Cuscuta is a plant parasite. Ticks, lice and leeches are ectoparasites, and the tapeworm is an endoparasite. Each one draws nutrition from a living host without killing it, so all of them show parasitic nutrition.
2. None of the other groups share one mode. (A) contains yeast, which is saprophytic, and legumes, which are autotrophic. (B) and (D) contain autotrophs (cactus, grass) mixed with holozoic animals (cow, lion) and parasites. So the answer is option (C).
  1 mark
Answer: Option (C): Cuscuta, ticks, lice, leeches and tapeworm (all parasites)
Where the marks are
- Chooses (C), recognising that all five are parasites (1 mark)
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Q2 · 1 mark · MCQLife Processes
Which of the following options indicates the products formed after breakdown of the glucose in our muscle cells when there is lack of oxygen?
1. (A)Ethanol + carbon dioxide + Energy
2. (B)Lactic acid + Energy
3. (C)Lactic acid + carbon monoxide + Energy
4. (D)Carbon dioxide + Water + Energy
Try it first, then show our solutionSolution
1. In muscle cells, glucose is first broken down in the cytoplasm into pyruvate, which is a 3-carbon molecule.
2. When oxygen is short, for example during hard exercise, pyruvate is converted into lactic acid. Only a small amount of energy is released. No CO $_{2}$ is produced, so the products are lactic acid + energy: option (B). Ethanol + CO $_{2}$ is what yeast produces.
  1 mark
Answer: Option (B): Lactic acid + Energy
Where the marks are
- Identifies anaerobic respiration in muscle as giving lactic acid + energy, option (B) (1 mark)
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Q3 · 1 mark · MCQControl and Coordination
Which of the following is a correct combination of function and part of the brain?
1. (A)Posture and balance: Cerebrum
2. (B)Salivation: Medulla in midbrain
3. (C)Hunger: Pons in hindbrain
4. (D)Blood pressure: Medulla in hindbrain
Try it first, then show our solutionSolution
1. The medulla is part of the hindbrain. It controls involuntary actions such as blood pressure, salivation and vomiting.
2. Each of the other options has an error. Posture and balance are controlled by the cerebellum, not the cerebrum. The medulla lies in the hindbrain, not the midbrain. Hunger is controlled by the hypothalamus in the forebrain. So the correct pair is (D) Blood pressure: Medulla in hindbrain.
  1 mark
Answer: Option (D): Blood pressure – Medulla in hindbrain
Where the marks are
- Selects (D), medulla (hindbrain) controlling blood pressure (1 mark)
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Q4 · 1 mark · MCQControl and Coordination
The blood glucose level in a patient was very high. It may be due to inadequate secretion of:
1. (A)growth hormone from pituitary gland
2. (B)oestrogen from ovary
3. (C)insulin from pituitary gland
4. (D)insulin from pancreas
Try it first, then show our solutionSolution
1. Insulin lowers blood sugar. It is secreted by the pancreas, not by the pituitary gland.
2. If too little insulin is secreted, blood glucose rises. This condition is diabetes. Answer: option (D).
  1 mark
Answer: Option (D): insulin from pancreas
Where the marks are
- Links high blood sugar to too little insulin from the pancreas, option (D) (1 mark)
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Q5 · 1 mark · MCQHeredity
In a cross between black furred rabbit (B) and white furred rabbit (b), all offspring were found to have black fur. What can be inferred about the genetic makeup of the parent rabbits?
1. (A) $B B \times bb$
2. (B) $B b \times B b$
3. (C) $B b \times bb$
4. (D) $bb \times bb$
Try it first, then show our solutionSolution
1. Black (B) is dominant over white (b). If even one parent were Bb, then crossing it with bb, or Bb with Bb, would give some white (bb) offspring.
2. All offspring are black, so the black parent must be pure (BB) and the white parent is bb. Every offspring is then Bb, which is black. Answer: **(A) $B B \times bb$ **.
  1 mark
Answer: Option (A): BB × bb
Where the marks are
- Reasons that 100% black offspring means a homozygous dominant parent crossed with bb, option (A) (1 mark)
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Q6 · 1 mark · MCQOur Environment
Which are the correct statements related to ozone?
(i) Ozone layer helps in increasing the UV radiations reaching earth.
(ii) Ozone is a deadly poison.
(iii) Ozone layer shields the earth from UV radiations.
(iv) Ozone layer prevents UV rays which cause skin cancer.
(v) Ozone is formed with the help of Chloroflurocarbons.
1. (A)(i), (ii), (iii)
2. (B)(ii), (iii), (iv)
3. (C)(iii), (iv), (v)
4. (D)(i), (iv), (v)
Try it first, then show our solutionSolution
1. (i) is wrong, because the ozone layer reduces the UV radiation reaching the earth. (v) is wrong, because CFCs deplete ozone. Ozone is formed when UV light acts on O $_{2}$ .
2. (ii) is correct: at ground level ozone is a deadly poison. (iii) and (iv) are correct: the layer shields the earth from UV rays, which cause skin cancer. So the answer is (B) (ii), (iii), (iv).
  1 mark
Answer: Option (B): (ii), (iii), (iv)
Where the marks are
- Rejects (i) and (v) and selects (B) (1 mark)
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Q7 · 1 mark · MCQOur Environment
Which of the following human activities has resulted in an increase of non-biodegradable substances?
1. (A)Organic farming
2. (B)Increase in tree plantation
3. (C)Use of plastic as packaging material
4. (D)Composting of kitchen waste
Try it first, then show our solutionSolution
1. Plastic cannot be broken down by microorganisms, so it is non-biodegradable. Using it widely for packaging adds more non-biodegradable waste.
2. Organic farming, planting trees and composting all reduce such waste or involve biodegradable materials. Answer: option (C).
  1 mark
Answer: Option (C): Use of plastic as packaging material
Where the marks are
- Identifies plastic packaging as the source of non-biodegradable waste, option (C) (1 mark)
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Q8 · 1 mark · Assertion-reasonCompetency-basedHeredity
The following two questions consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below:
A. Both A and R are true, and R is the correct explanation of A.
B. Both A and R are true, and R is not the correct explanation of A.
C. A is true but R is false.
D. A is false but R is true.
Assertion (A): Tallness of a pea plant is controlled by an enzyme.
Reason (R): The gene for that enzyme makes proteins which help the plant to be tall.
1. (A)Both A and R are true, and R is the correct explanation of A.
2. (B)Both A and R are true, and R is not the correct explanation of A.
3. (C)A is true but R is false.
4. (D)A is false but R is true.
Try it first, then show our solutionSolution
1. In Mendel's pea plants, tallness depends on how much growth hormone is made. That amount depends on how efficiently a particular enzyme works. So (A) is true.
2. The gene for that enzyme carries the information to make the enzyme protein. An efficient enzyme gives more hormone, and the plant grows tall. So (R) is true and it explains (A): option (A).
  1 mark
Answer: Option (A): Both A and R are true, and R is the correct explanation of A
Where the marks are
- Recognises the gene → protein (enzyme) → hormone → tallness link and chooses (A) (1 mark)
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Q9 · 1 mark · Assertion-reasonCompetency-basedOur Environment
Assertion (A): Vulture will always have the least amount of pesticides in a food chain.
Reason (R): Vulture occupies the last trophic level and it gets only 10% of energy of the previous trophic level.
1. (A)Both A and R are true, and R is the correct explanation of A.
2. (B)Both A and R are true, and R is not the correct explanation of A.
3. (C)A is true but R is false.
4. (D)A is false but R is true.
Try it first, then show our solutionSolution
1. Pesticides are not broken down, so they build up at each trophic level. This is biological magnification. The vulture is at the top, so it has the maximum amount of pesticides. (A) is false.
2. The vulture is at the last trophic level and receives only about 10% of the energy of the level before it (10% law). (R) is true. Answer: option (D).
  1 mark
Answer: Option (D): A is false but R is true
Where the marks are
- States that pesticides accumulate most at the top (biomagnification), so A is false and R is true, option (D) (1 mark)
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Q10 · 2 marks · Very short answerLife Processes
Unlike animals, plants do not have any excretory products as they do not eat food. Comment upon the statement with justification.
Try it first, then show our solutionSolution
1. The statement is incorrect. Plants also carry out metabolic activities such as photosynthesis and respiration, and these produce waste products. Plants simply get rid of their wastes in ways different from animals.
  1 mark
2. Way 1: Oxygen, a by-product of photosynthesis, and CO $_{2}$ from respiration are removed by diffusion through the stomata. Excess water is lost by transpiration.
  ½ mark
3. Way 2: Other wastes are stored in vacuoles, or in old leaves and bark that fall off. Some are stored as resins and gums in old xylem, and some are released into the soil around the roots.
  ½ mark
Answer: The statement is wrong; plants do produce wastes (O₂, CO₂, water, resins, gums, etc.) and remove them through stomata and transpiration, storage in vacuoles, resins and gums, and shedding of leaves.
Where the marks are
- States that the statement is wrong because plants produce metabolic wastes (1 mark)
- Gives any two valid ways plants get rid of wastes (½ mark each)
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Answer one of these 2 alternatives.
Q11(A) · 2 marks · Very short answerLife Processes
How many chambers are there in the heart of the following organisms? How is mixing of oxygenated and deoxygenated blood prevented in their body?
(i) Fishes
(ii) Humans
Try it first, then show our solutionSolution
1. (i) Fishes: 2 chambers (one atrium, one ventricle). The heart receives only deoxygenated blood and pumps it to the gills. There it is oxygenated and goes straight to the body tissues, without returning to the heart. Oxygenated blood never enters the heart, so no mixing occurs. This is single circulation.
  1 mark
2. (ii) Humans: 4 chambers (2 atria, 2 ventricles). A wall called the septum separates the right side, which carries deoxygenated blood, from the left side, which carries oxygenated blood. This keeps the two kinds of blood apart. This is double circulation.
  1 mark
Answer: (i) Fish: 2 chambers; blood goes heart → gills → body, so the heart holds only deoxygenated blood. (ii) Humans: 4 chambers; a septum separates the right and left sides.
Where the marks are
- Fish: 2 chambers and the reason there is no mixing (blood oxygenated in gills goes directly to the body) (1 mark)
- Human: 4 chambers and the septum separating oxygenated and deoxygenated blood (1 mark)
Open Q11(A) on its own page
OR
Q11(B) · 2 marks · Very short answerLife Processes
Explain the mechanism by which the water is transported in plants?
Try it first, then show our solutionSolution
1. Water and minerals move upward in one direction through the xylem (tracheids and vessels), which forms a continuous channel from roots to leaves.
2. Root pressure: Root cells actively take up ions from the soil. This makes the ion concentration in the root higher than in the soil. Water therefore moves into the root, and a steady column of water is pushed up the root xylem. This matters most at night.
  1 mark
3. Transpiration pull: Water evaporates from the leaves through the stomata. This creates a suction that pulls water up through the xylem from the roots. It is the main force in the daytime.
  1 mark
Answer: Water moves up through the xylem, driven by root pressure (active ion uptake by roots) and mainly by transpiration pull from the leaves.
Where the marks are
- Explains root pressure from active ion uptake causing water to enter the root xylem (1 mark)
- Explains transpiration pull through xylem vessels and tracheids (1 mark)
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Q12 · 2 marks · Very short answerCompetency-basedOur Environment
About 100 acres of forest land was declared as Natural reserve park. The following organisms were predominant in the Natural reserve park: rabbit, frog, grass, fish, fox, water insects, zebra, peacock, snake, trees, bird, owl, insects, tiger, vulture, duck.
Create a food web comprising two separate food chains with different producers by using the above data.
Try it first, then show our solutionSolution
1. Food chain 1 (producer: trees): Trees → Insects → Bird → Owl
  ½ mark
2. Food chain 2 (producer: grass): Grass → Insects → Frog → Snake → Owl
  ½ mark
3. Food web: draw the two chains with arrows pointing from food to feeder, joined at the organisms they share (insects and owl). The insects feed on both trees and grass. The insects are eaten by both the bird and the frog. The owl eats both the bird and the snake. This interlinking forms the web. Another valid example is Trees (leaves) → Zebra → Tiger and Grass → Zebra → Tiger, joined at the zebra.
  1 mark
Answer: Chain 1: Trees → Insects → Bird → Owl; Chain 2: Grass → Insects → Frog → Snake → Owl; joined at insects/owl to form a web (any correct pair with different producers is acceptable).
Where the marks are
- Correct food chain starting with one producer, e.g. trees (½ mark)
- Correct food chain starting with a different producer, e.g. grass (½ mark)
- Joins the chains at a common organism with correct arrows to form a web (1 mark)
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Q13 · 3 marks · Short answerControl and Coordination
Draw and explain how the nerve cells help in transmission of impulses?
Try it first, then show our solutionSolution
1. Diagram (structure of a neuron): Draw a star-shaped cell body containing a nucleus, with branched dendrites around it. A long axon runs from the cell body and ends in branched nerve endings. Next to these, show the synapse, the gap between the nerve ending and the dendrite of the next neuron. Label all of these parts.
  1 mark
2. Information from the environment is picked up at the dendrite tips of a nerve cell. This sets off a chemical reaction that creates an electrical impulse.
  1 mark
3. The impulse travels from the dendrite to the cell body, then along the axon to its end. There it causes certain chemicals to be released.
  ½ mark
4. These chemicals cross the synapse and start a similar electrical impulse in a dendrite of the next neuron, or in a muscle cell at a neuromuscular junction. This is how impulses are passed along in the body.
  ½ mark
Answer: Labelled neuron diagram; impulse path: dendrite → cell body → axon → nerve ending → chemicals released → cross synapse → impulse in next neuron.
Where the marks are
- Correct labelled neuron diagram showing dendrite, cell body, nucleus, axon and nerve ending (1 mark)
- Information at the dendrite converted into an electrical impulse by a chemical reaction (1 mark)
- Impulse travels through cell body and axon, and chemicals are released at the axon end (½ mark)
- Chemicals cross the synapse and start an impulse in the next neuron (½ mark)
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Q14 · 3 marks · Short answerHeredity
In a genetic experiment, plants with pure round green seeds ( $R R y y$ ) were crossed with plants with wrinkled yellow seeds ( $r r Y Y$ ).
1. (i)
  Show the gametes formed when F1 was self-pollinated.
  1 mark
2. (ii)
  A total of 144 seeds were produced which developed into saplings. Show the ratio in which these traits are independently inherited in these 144 saplings.
  2 marks
Try it first, then show our solutionSolution
1. Cross: RRyy (round green) × rrYY (wrinkled yellow). The parents' gametes are Ry and rY, so the F1 is RrYy (round yellow).
2. (i) When F1 (RrYy) is self-pollinated, each plant forms four kinds of gametes in equal numbers: RY, Ry, rY, ry.
  1 mark
3. (ii) A 4 × 4 checkerboard gives 16 combinations. Since the two traits are inherited independently, the phenotypes occur in the ratio Round yellow : Round green : Wrinkled yellow : Wrinkled green = 9 : 3 : 3 : 1.
  1 mark
4. For 144 saplings, one part = $144 \div 16 = 9$ . Round yellow = $9 \times 9 = 81$ ; Round green = $3 \times 9 = 27$ ; Wrinkled yellow = $3 \times 9 = 27$ ; Wrinkled green = $1 \times 9 = 9$ .
  1 mark
Answer: (i) Gametes: RY, Ry, rY, ry; (ii) 9:3:3:1, i.e. round yellow 81, round green 27, wrinkled yellow 27, wrinkled green 9
Where the marks are
- Writes the four F1 gametes RY, Ry, rY, ry (1 mark)
- Gives the 9:3:3:1 dihybrid ratio with correct phenotypes (1 mark)
- Converts the ratio to 81 : 27 : 27 : 9 out of 144 (1 mark)
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Q15 · 4 marks · Case-basedCompetency-basedLife Processes
Neha consumed boiled sweet potatoes and boiled eggs for breakfast. Help her to understand some steps in the process of digestion of the food taken by her by answering the questions given below.
1. A
  Which of these food items is rich in proteins? In which part of the alimentary canal is the digestion of this component initiated? Name the enzymes, conditions required and the glands associated with the digestion here.
  0 marks
2. OR
  B
  Which of these food items contains fats? How is it digested?
  0 marks
3. C
  Which of these food items is rich in starch? How is its digestion initiated?
  0 marks
4. D
  The figure given below represents parts of the human alimentary canal. Which of these parts will have the maximum amount of digested food as soon as the process of digestion is completed?
  0 marks
Figure in the paper: Figure of the human alimentary canal (outline of human body with digestive organs). Labels: L points to the mouth/oral cavity region at the top; M points to the oesophagus (food pipe) running down the neck and chest; N points to the stomach region; O points to the small intestine (coiled region); P points to the large intestine/rectum at the lower end. See it in the official paper.
Try it first, then show our solutionSolution
1. A: Boiled eggs are rich in proteins. Protein digestion begins in the stomach.
  1 mark
2. A (contd.): Gastric glands in the stomach wall release hydrochloric acid, the enzyme pepsin and mucus. The HCl makes the medium acidic, which pepsin needs in order to digest proteins. The mucus protects the stomach lining from the acid.
  1 mark
3. B (alternative to A; carries the same 2 marks): Eggs (the yolk) contain fats. In the small intestine, bile from the liver makes the medium alkaline and breaks large fat globules into smaller ones (emulsification). The enzyme lipase from the pancreas, together with intestinal lipase, then digests the emulsified fats into fatty acids and glycerol.
4. C: Sweet potatoes are rich in starch. Digestion of starch starts in the mouth, where the salivary glands secrete saliva containing salivary amylase. This enzyme breaks starch down into sugar.
  1 mark
5. D: The small intestine (part O) will hold the maximum amount of digested food, because the digestion of carbohydrates, proteins and fats is completed there.
  1 mark
6. D (for visually impaired students, alternative to D): The inner lining of the small intestine has many finger-like projections called villi. These increase the surface area, and the digested food is absorbed through them into the blood.
Answer: A: Eggs; stomach; gastric glands release HCl (acidic medium), pepsin and mucus. B: Eggs; bile emulsifies fats, pancreatic lipase digests them. C: Sweet potatoes; salivary amylase in the mouth begins starch digestion. D: Small intestine (O). [VI: absorbed by villi]
Where the marks are
- A: eggs as the protein source and stomach as the site where digestion begins (1 mark)
- A: gastric glands, pepsin and HCl giving the acidic medium (1 mark)
- B (alternative): emulsification by bile in alkaline medium and digestion by pancreatic lipase (2 marks)
- C: sweet potato starch digested by salivary amylase in the mouth (1 mark)
- D: small intestine (O), where digestion is completed (1 mark)
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Answer one of these 2 alternatives.
Q16(A) · 5 marks · Long answerCompetency-basedHow do Organisms Reproduce?
Puneet wanted to grow banana plants.
1. (i)
  Based on your knowledge on plant reproduction should he opt for seeds or any alternate method of reproduction. Justify your answer.
  2 marks
2. (ii)
  Offsprings of a banana plant usually show very little variation. What causes variation and are variations good or bad? Justify.
  3 marks
Try it first, then show our solutionSolution
1. (i) Puneet should not use seeds. Cultivated banana plants have lost the capacity to produce seeds.
  1 mark
2. (i) He should use vegetative propagation, growing new plants from parts of the parent such as the underground stem, suckers or stem cuttings. The new plants are also genetically like the parent and grow quickly.
  1 mark
3. (ii) Cause of variation: DNA copying during reproduction is never completely accurate. Small errors or changes in DNA copying create variation. Sexual reproduction mixes DNA from two parents and so gives more variation. Banana offspring come from vegetative propagation, a form of asexual reproduction, so they show very little variation.
  1 mark
4. (ii) Good: Variation helps a population survive. If the environment changes (heat, drought, disease), some variants may suit the new conditions and survive. The species is then not wiped out.
  1 mark
5. (ii) Bad: Some variations are harmful. Desirable characters of the parent may be lost, a variant may not survive in its conditions, or drastic DNA changes may stop the cell's machinery from working properly. So variation is useful over long periods, but not every variation is beneficial.
  1 mark
Answer: (i) Not seeds (banana has lost the capacity to produce seeds); use vegetative propagation. (ii) Errors in DNA copying cause variation; it is good because it helps populations survive changing conditions, and bad because some variants lose desirable traits or cannot survive.
Where the marks are
- Says seeds should not be used because banana does not produce seeds (1 mark)
- Suggests vegetative propagation (1 mark)
- Gives errors in DNA copying as the cause of variation (1 mark)
- Justifies why variation is good: survival of the species in changed conditions (1 mark)
- Justifies why variation can be bad: loss of desirable traits or variants unable to survive (1 mark)
Open Q16(A) on its own page
OR
Q16(B) · 5 marks · Long answerCompetency-basedHow do Organisms Reproduce?
Annie was conducting research on the number of fruits produced by watermelon under different conditions. She grew 25 watermelon plants each in both glass house A and B. She introduced pollinators in glass house A only.
1. (i)
  What difference will she observe in the number of fruits produced in the two glass houses? Explain with reason.
  3 marks
2. (ii)
  List 3 changes that will occur in a flower once it gets fertilized.
  2 marks
Try it first, then show our solutionSolution
1. (i) Glass house A, which has pollinators, will produce more fruits than glass house B.
  1 mark
2. (i) Reason: watermelon has unisexual flowers, with male and female flowers separate. Pollen must be carried from a male flower to the stigma of a female flower.
  1 mark
3. (i) Pollinators such as bees carry out this transfer (cross-pollination), so more female flowers are fertilised and more fruits form. In B there are no pollinators, and pollen rarely reaches a stigma, especially when the flowers are far apart. Fewer flowers are fertilised, so fewer fruits are produced.
  1 mark
4. (ii) Change 1: the zygote develops into an embryo, and the ovule develops a tough coat and becomes a seed.
  ½ mark
5. (ii) Change 2: the ovary grows and ripens to form the fruit.
  1 mark
6. (ii) Change 3: the petals, sepals, stamens, style and stigma shrivel and fall off.
  ½ mark
Answer: (i) Glass house A gives more fruits; watermelon flowers are unisexual and need pollinators for pollen transfer, so B gives few fruits. (ii) Ovule → seed; ovary → fruit; petals, sepals, stamens, style and stigma shrivel and fall off.
Where the marks are
- States that glass house A has more fruits than B (1 mark)
- Notes watermelon flowers are unisexual, needing pollen transfer between flowers (1 mark)
- Explains pollinators increase cross-pollination and fertilisation, while B has less (1 mark)
- Lists three post-fertilisation changes: ovule to seed, ovary to fruit, other floral parts fall (2 marks)
Open Q16(B) on its own page

Section B

Chemistry. MCQ 17-23 (1 mark), assertion-reason 24 (1), Q25 (2), Q26 (3, internal choice), Q27 (3), Q28 (4), Q29 (5, internal choice)

Q17 · 1 mark · MCQChemical Reactions and Equations
Which of the following equations represent redox reactions and what are the values for 'p' and 'q' in these equations?
Equation 1: $F e_{2} O_{3} (s) + 2Al (s) \to A l_{2} O_{3} (s) + p Fe (l) + heat$
Equation 2: $2 C_{4} H_{10} (g) + 13 O_{2} (g) △ 8C O_{2} (g) + q H_{2} O (g)$
1. (A)Only equation 1 is a redox reaction, p = 1 and q = 3
2. (B)Both equations 1 and 2 are redox reactions, p = 2 and q = 4
3. (C)Only equation 2 is a redox reaction, p = 2 and q = 10
4. (D)Both equations 1 and 2 are redox reactions, p = 2 and q = 10
Try it first, then show our solutionSolution
1. Equation 1: the left side has 2 Fe in Fe $_{2}$ O $_{3}$ , so the right side needs 2Fe, giving p = 2. Fe $_{2}$ O $_{3}$ loses oxygen (reduced) and Al gains oxygen (oxidised), so it is redox.
2. Equation 2: the left side has $2 \times 10 = 20$ H atoms, so 10H $_{2}$ O is needed, giving q = 10. Check O: $13 \times 2 = 26$ on the left and $16 + 10 = 26$ on the right. Butane is oxidised and O $_{2}$ is reduced, so it is redox.
3. Both are redox reactions, with p = 2 and q = 10: option (D).
  1 mark
Answer: Option (D): both redox, p = 2, q = 10
Where the marks are
- Balances to get p = 2 and q = 10 and identifies both as redox, option (D) (1 mark)
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Q18 · 1 mark · MCQAcids, Bases and Salts
Four statements about the reactions of oxides with dilute hydrochloric acid and aqueous sodium hydroxide are listed.
I. Aluminium oxide reacts with both dilute hydrochloric acid and aqueous sodium hydroxide.
II. Calcium oxide reacts with dilute hydrochloric acid and aqueous sodium hydroxide.
III. Zinc oxide reacts with both dilute hydrochloric acid and aqueous sodium hydroxide.
IV. Sulphur dioxide does not react with either dilute hydrochloric acid or aqueous sodium hydroxide.
Which statements are correct?
1. (A)I and II
2. (B)I and III
3. (C)II and IV
4. (D)III and IV
Try it first, then show our solutionSolution
1. Al $_{2}$ O $_{3}$ and ZnO are amphoteric oxides, so they react with both acids and bases. Statements I and III are correct.
2. CaO is a basic oxide and does not react with NaOH, so II is wrong. SO $_{2}$ is an acidic oxide and reacts with NaOH, so IV is wrong. Answer: (B) I and III.
  1 mark
Answer: Option (B): I and III
Where the marks are
- Recognises Al₂O₃ and ZnO as amphoteric and rejects II and IV, option (B) (1 mark)
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Q19 · 1 mark · MCQChemical Reactions and Equations
An iron nail is added to each of the two test tubes 'P' and 'Q' containing aqueous copper (II) sulphate, and aqueous silver nitrate respectively. Which of the following observation is correct?
1. (A)In test tube 'P' iron nail is coated with a blue coating and in test tube 'Q' there is no reaction.
2. (B)Iron nail is coated with a brown coating in test tube 'P' and silver coating in test tube 'Q'.
3. (C)There is no reaction in either of the test tubes 'P' or 'Q'.
4. (D)There is no reaction in test tube 'P' but a silver coating on iron nail is seen in test tube 'Q'.
Try it first, then show our solutionSolution
1. Iron is more reactive than both copper and silver, so it displaces them from their salt solutions.
2. In P: Fe + CuSO $_{4}$ → FeSO $_{4}$ + Cu, which deposits a reddish-brown coating of copper on the nail. In Q: Fe + 2AgNO $_{3}$ → Fe(NO $_{3}$ ) $_{2}$ + 2Ag, which deposits a silvery coating. Answer: option (B).
  1 mark
Answer: Option (B): brown coating in P, silver coating in Q
Where the marks are
- Uses the reactivity series (Fe above Cu and Ag) to predict copper and silver deposits, option (B) (1 mark)
Open Q19 on its own page
Q20 · 1 mark · MCQAcids, Bases and Salts
Methyl orange is added to dilute hydrochloric acid and to aqueous sodium hydroxide. What is the colour of the methyl orange in each solution?
Sample colour in dilute hydrochloric acid colour in aqueous sodium hydroxide
A Orange Red
B Red Yellow
C Red Orange
D Yellow Red
Try it first, then show our solutionSolution
1. Methyl orange is an acid–base indicator. It is red in acidic solutions and yellow in basic solutions.
2. Dilute HCl is acidic, so methyl orange turns red. Aqueous NaOH is basic, so methyl orange turns yellow.
3. The row Red / Yellow is option (B).
  1 mark
Answer: Red in HCl, yellow in NaOH; option (B)
Where the marks are
- Chooses option (B): red in acid, yellow in base (1 mark)
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Q21 · 1 mark · MCQAcids, Bases and Salts
Which of the following substances when dissolved in equal volume of water, will have the highest pH value?
1. (A)Sulphuric acid
2. (B)Acetic acid
3. (C)Magnesium hydroxide
4. (D)Sodium hydroxide
Try it first, then show our solutionSolution
1. Acids have pH < 7, so sulphuric acid and acetic acid are ruled out. Bases have pH > 7.
2. Magnesium hydroxide is only sparingly soluble, so it is a weak base and gives few $O H^{-}$ ions. Sodium hydroxide is a highly soluble strong alkali and gives the most $O H^{-}$ ions.
3. More $O H^{-}$ ions means a higher pH. So NaOH has the highest pH, option (D).
  1 mark
Answer: Sodium hydroxide, option (D)
Where the marks are
- Selects sodium hydroxide, the strong soluble base (1 mark)
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Q22 · 1 mark · MCQChemical Reactions and Equations
When excess of carbon dioxide is passed through lime water, the milkiness disappears because
1. (A)water soluble calcium carbonate converts to water soluble calcium bicarbonate.
2. (B)insoluble calcium carbonate converts to water soluble calcium bicarbonate.
3. (C)water soluble calcium carbonate converts to insoluble calcium bicarbonate.
4. (D)insoluble calcium carbonate converts to insoluble calcium bicarbonate.
Try it first, then show our solutionSolution
1. Lime water turns milky with $C O_{2}$ because insoluble calcium carbonate forms: $C a (O H)_{2} + C O_{2} \to C a C O_{3} ↓ + H_{2} O$ .
2. With excess $C O_{2}$ , this carbonate changes into soluble calcium hydrogencarbonate: $C a C O_{3} + H_{2} O + C O_{2} \to C a (H C O_{3})_{2}$ (soluble). The milkiness therefore disappears.
3. The correct description is insoluble carbonate → soluble bicarbonate, option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Identifies conversion of insoluble CaCO₃ to soluble Ca(HCO₃)₂, option (B) (1 mark)
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Q23 · 1 mark · MCQChemical Reactions and Equations
In the reaction of aqueous solution of barium chloride with aqueous solution of sodium sulphate, the aqueous solution formed will be:
1. (A) $B a C l_{2}$
2. (B) $B a S O_{4}$
3. (C) $N a_{2} S O_{4}$
4. (D) $N a C l$
Try it first, then show our solutionSolution
1. This is a double displacement (precipitation) reaction: $B a C l_{2} (a q) + N a_{2} S O_{4} (a q) \to B a S O_{4} (s) ↓ + 2 N a C l (a q)$ .
2. $B a S O_{4}$ is a white insoluble precipitate. The product that stays dissolved is $N a C l$ , option (D).
  1 mark
Answer: NaCl, option (D)
Where the marks are
- Writes the products and identifies NaCl as the aqueous product (1 mark)
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Q24 · 1 mark · Assertion-reasonCompetency-basedCarbon and its Compounds
The following question consists of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below:
A. Both A and R are true, and R is the correct explanation of A.
B. Both A and R are true, and R is not the correct explanation of A.
C. A is true but R is false.
D. A is false but R is true.
Assertion (A): $C_{4} H_{8}$ , $C_{4} H_{6}$ and $C_{4} H_{10}$ are members of the same homologous series.
Reason (R): $C_{4} H_{8}$ , $C_{4} H_{6}$ , $C_{3} H_{4}$ , $C_{3} H_{6}$ , $C_{2} H_{4}$ , $C_{2} H_{2}$ are unsaturated hydrocarbons.
1. (A)Both A and R are true, and R is the correct explanation of A.
2. (B)Both A and R are true, and R is not the correct explanation of A.
3. (C)A is true but R is false.
4. (D)A is false but R is true.
Try it first, then show our solutionSolution
1. Assertion: $C_{4} H_{8}$ fits $C_{n} H_{2 n}$ (alkene), $C_{4} H_{6}$ fits $C_{n} H_{2 n - 2}$ (alkyne) and $C_{4} H_{10}$ fits $C_{n} H_{2 n + 2}$ (alkane). Members of one homologous series share the same general formula, so these do not form one series. A is false.
2. Reason: $C_{4} H_{8}$ , $C_{3} H_{6}$ and $C_{2} H_{4}$ are alkenes, and $C_{4} H_{6}$ , $C_{3} H_{4}$ and $C_{2} H_{2}$ are alkynes. All contain double or triple bonds, so all are unsaturated. R is true.
3. A is false but R is true, option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Recognises A false (different general formulae) and R true (all unsaturated), option (D) (1 mark)
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Q25 · 2 marks · Very short answerCompetency-basedMetals and Non-metals
The following activity is set-up in the science lab by the teacher. He clamped an aluminium wire on a stand and fixed a pin to the free end of the wire using wax. Then he heated the wire with a burner from the end where the wire is clamped. Students observed the pin fall off.
A. If the teacher replaces aluminium wire by silver wire, will the students' observation change? Justify your answer.
B. Will the aluminium wire melt? Give reason for your answer.
Figure in the paper: A stand with a clamp holds a horizontal metal wire at one end. A pin is attached with wax to the free end of the wire. A spirit lamp with flame is placed below the wire near the clamped end. Labels: Stand, Clamp, Metal wire, Free end of wire, Pin, Wax, spirit lamp. See it in the official paper.
Try it first, then show our solutionSolution
1. A. The pin falls because heat is conducted along the wire and melts the wax at the free end.
2. With a silver wire the pin will still fall, but sooner. Silver is a better conductor of heat than aluminium, so the wax melts faster.
  1 mark
3. B. No, the aluminium wire will not melt. Metals such as aluminium have high melting points (about 660 °C for Al). The flame only heats the wire enough to melt the wax, which has a low melting point.
  1 mark
Answer: A. Pin still falls but in less time, since silver conducts heat better. B. No, aluminium has a high melting point.
Where the marks are
- States the pin falls sooner with silver because silver is a better heat conductor (1 mark)
- States aluminium will not melt because of its high melting point (1 mark)
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Answer one of these 2 alternatives.
Q26(A) · 3 marks · Short answerMetals and Non-metals
An element 'X' is stored in kerosene, and cannot be extracted from its ore using a reducing agent. 'X' forms an ionic compound on reaction with chlorine.
1. (i)
  Can we store 'X' in water? Give reason to support your answer.
  1 mark
2. (ii)
  Identify element 'X'. Name the process used and write the equation for extraction of 'X' from its ore.
  2 marks
Try it first, then show our solutionSolution
1. (i) No. X is stored in kerosene because it is very reactive. It reacts vigorously and exothermically with water, and the hydrogen produced catches fire.
  1 mark
2. (ii) X is a highly reactive metal at the top of the activity series that forms an ionic chloride: X = sodium (Na). Potassium is also acceptable.
  ½ mark
3. Such metals cannot be reduced by carbon, so they are extracted by electrolytic reduction (electrolysis of molten NaCl).
  ½ mark
4. At cathode: $N a^{+} + e^{-} \to N a$
  At anode: $2 C l^{-} \to C l_{2} + 2 e^{-}$
  1 mark
Answer: (i) No, it reacts violently with water and catches fire. (ii) Sodium; electrolytic reduction of molten NaCl; cathode $N a^{+} + e^{-} \to N a$ , anode $2 C l^{-} \to C l_{2} + 2 e^{-}$
Where the marks are
- No, with reason: violent reaction with water / catches fire (1 mark)
- Identifies sodium (or potassium) (½ mark)
- Names electrolytic reduction (½ mark)
- Correct electrode equations (1 mark)
Open Q26(A) on its own page
OR
Q26(B) · 3 marks · Short answerCompetency-basedMetals and Non-metals
The domes of many building in Europe are made of copper. These domes now appear greenish in colour.
1. (i)
  Why do the domes appear greenish though copper is orange-red in colour?
  1 mark
2. (ii)
  In your opinion, should the copper domes be replaced by iron domes to overcome the problem of change of colour of copper domes?
  1 mark
3. (iii)
  Domes used to be made from thin sheets of metals. Why did the ancient architects use copper to make domes?
  1 mark
Try it first, then show our solutionSolution
1. (i) Copper slowly reacts with moist carbon dioxide and oxygen in air. It forms a green layer of basic copper carbonate, $C u C O_{3} \cdot C u (O H)_{2}$ , so the domes look greenish.
  1 mark
2. (ii) No. Iron would rust in moist air, and the reddish-brown rust flakes off. This keeps exposing fresh iron, so corrosion continues and the dome weakens. The green layer on copper sticks to it and protects the metal below from further corrosion.
  1 mark
3. (iii) Copper is highly malleable, so it can be beaten into thin sheets and shaped into curved domes. It also resists corrosion, so it lasts long.
  1 mark
Answer: (i) Green basic copper carbonate layer forms. (ii) No, iron rusts and flakes; copper's coating is protective. (iii) Copper is highly malleable (and corrosion resistant).
Where the marks are
- Green basic copper carbonate due to corrosion (1 mark)
- No, with reasoning about flaking rust versus protective copper layer (1 mark)
- Malleability of copper (1 mark)
Open Q26(B) on its own page
Q27 · 3 marks · Case-basedCompetency-basedChemical Reactions and Equations
Amrita electrolysed distilled water using the set-up shown in figure 1. She was expecting two gases to be evolved at the anode and cathode respectively. Suddenly, she realised that the bulb in the circuit did not glow when she used distilled water (figure 2). After this realization, she added a substance to the distilled water for electrolysis to take place. Answer the following questions based on the information given above:
1. A
  Which gas was she expecting to be formed at the anode and which one at the cathode respectively?
  1 mark
2. B
  Why did the bulb not glow when Amrita passed electricity through distilled water?
  1 mark
3. C
  Which substance was added by Amrita to distilled water to get the expected result?
  1 mark
Figure in the paper: Figure 1: A plastic mug containing water, with two test tubes inverted over two graphite rods (electrodes) dipped in the water. The rod under the left tube is the anode and the rod under the right tube is the cathode. Both rods are connected by wires through a switch to a 6 V battery. Figure 2: A circuit with a battery, a bulb and a conducting wire, the circuit completed by dipping the two ends of the wire into a beaker of distilled water. See it in the official paper.
Try it first, then show our solutionSolution
1. A. In the electrolysis of water, oxygen is expected at the anode and hydrogen at the cathode. The volume of hydrogen is double that of oxygen: $2 H_{2} O \to 2 H_{2} + O_{2}$ .
  1 mark
2. B. Distilled water has almost no ions, so it is a very poor conductor of electricity. Too little current flows for the bulb to glow.
  1 mark
3. C. She added a few drops of dilute sulphuric acid (or some common salt / any strong electrolyte). This provides ions so that current flows and electrolysis takes place.
  1 mark
4. For visually impaired students (in place of A–C, 1 mark each): A. $Z n O + C \to Z n + C O$ is a redox reaction (ZnO is reduced, C is oxidised). B. $Z n C O_{3} \to Z n O + C O_{2}$ is a decomposition reaction and is endothermic (needs heat). C. $2 M g + O_{2} \to 2 M g O$ + heat is a combination reaction and is exothermic.
Answer: A. Oxygen at anode, hydrogen at cathode. B. Distilled water is a poor conductor (no ions). C. Few drops of dilute H₂SO₄ (or NaCl). VI: redox; decomposition (endothermic); combination (exothermic).
Where the marks are
- O₂ at anode and H₂ at cathode (1 mark)
- Distilled water is a poor conductor / lacks ions (1 mark)
- Adds dilute H₂SO₄ or NaCl or another electrolyte (1 mark)
- VI: redox; decomposition/endothermic; combination/exothermic (1 mark each)
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Q28 · 4 marks · Case-basedCompetency-basedAcids, Bases and Salts
Sara took 2 mL of dilute NaOH solution in a test tube and added two drops of phenolphthalein solution to it. The solution turned pink in colour. She added dilute H $_{2}$ SO $_{4}$ to the above solution drop by drop until the solution in the test tube became colourless. 40 drops of dilute H $_{2}$ SO $_{4}$ were used for the change in colour from pink to colourless. When Sara added a drop of NaOH to the solution, the colour changed back to pink again. Sara now tried the activity with different volumes of NaOH and recorded her observation in the table given below:
S. No. Volume of dil. NaOH taken (mL) Drops of dil. H $_{2}$ SO $_{4}$ used
1 2 20
2 3 30
3 4 40
Answer the following questions based on the above information:
1. A
  If Sara used concentrated H $_{2}$ SO $_{4}$ in place of dilute H $_{2}$ SO $_{4}$ , how many drops will be required for the change in colour to be observed?
  (a) 40
  (b) < 40
  (c) > 40
  Justify your answer.
  1 mark
2. B
  Sara measured 20 drops of dil. H $_{2}$ SO $_{4}$ and found its volume to be 1 mL. If Sara observed a change in colour of NaOH solution by using 3 mL of H $_{2}$ SO $_{4}$ , how many mL of NaOH did she add to the test tube initially?
  1 mark
3. OR
  B (OR)
  Sara takes 10 drops of dilute H $_{2}$ SO $_{4}$ in the test tube and adds two drops of phenolphthalein solution to it. Then she adds NaOH dropwise. Sara observes a change in colour after adding 20 drops of NaOH. What change in colour would she observe and why?
  1 mark
4. C
  Write a balanced chemical equation for the reaction taking place in the above experiment. Which of the following is true and why? The reaction is a
  (a) neutralisation and double displacement reaction
  (b) neutralisation and precipitation reaction
  (c) precipitation and double displacement reaction
  (d) neutralisation, double displacement as well as precipitation reaction.
  2 marks
Try it first, then show our solutionSolution
1. A. Option (b) < 40. Concentrated $H_{2} S O_{4}$ supplies more $H^{+}$ ions per drop than the dilute acid, so fewer drops neutralise the same amount of NaOH.
  1 mark
2. B. From the table, 2 mL NaOH needs 20 drops = 1 mL of acid. So 1 mL of acid neutralises 2 mL of NaOH. Then 3 mL of acid (60 drops) neutralises $3 \times 2 =$ 6 mL of NaOH.
  1 mark
3. B (OR) (alternative to B, 1 mark): The colour changes from colourless to pink. Phenolphthalein is colourless in the acid. Once all the acid is neutralised, the next drop of NaOH makes the solution basic and the indicator turns pink.
4. C. $2 N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O$
  1 mark
5. Option (a) neutralisation and double displacement is true. A base reacts with an acid to give salt and water (neutralisation), and the ions are exchanged: $N a^{+}$ pairs with $S O_{4}^{2 -}$ and $H^{+}$ with $O H^{-}$ (double displacement). It is not a precipitation reaction because $N a_{2} S O_{4}$ is soluble in water.
  1 mark
Answer: A. (b) < 40; B. 6 mL NaOH (OR: colourless to pink); C. $2 N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O$ , option (a)
Where the marks are
- (b) < 40 with reason: more H⁺ per drop (1 mark)
- 6 mL NaOH using 1 mL acid : 2 mL NaOH, or colourless to pink with reason (1 mark)
- Balanced equation (1 mark)
- Option (a) with justification, including Na₂SO₄ being soluble (1 mark)
Open Q28 on its own page
Answer one of these 2 alternatives.
Q29(A) · 5 marks · Case-basedCompetency-basedCarbon and its Compounds
A hydrocarbon with the formula $C_{x} H_{y}$ undergoes complete combustion as shown in the following equation:
$2 C_{x} H_{y} + 9 O_{2} \to 6 C O_{2} + 6 H_{2} O$
1. (a)
  What are the values of 'x' and 'y'?
  1 mark
2. (b)
  Give the chemical (IUPAC) name of the hydrocarbon.
  1 mark
3. (c)
  Draw its electron dot structure.
  1 mark
4. (d)
  Name the alcohol which on heating with conc. H $_{2}$ SO $_{4}$ will produce the above hydrocarbon $C_{x} H_{y}$ .
  1 mark
5. (e)
  Write a balanced chemical equation for the reaction of $C_{x} H_{y}$ with hydrogen gas in presence of Nickel.
  1 mark
Try it first, then show our solutionSolution
1. (a) Balance carbon: $2 x = 6 \Rightarrow x = 3$ . Balance hydrogen: $2 y = 6 \times 2 = 12 \Rightarrow y = 6$ . Check oxygen: $18 = 12 + 6$ ✓. So x = 3, y = 6 and the hydrocarbon is $C_{3} H_{6}$ .
  1 mark
2. (b) $C_{3} H_{6}$ fits $C_{n} H_{2 n}$ , so it is an alkene: propene, $C H_{2} = C H - C H_{3}$ .
  1 mark
3. (c) Electron dot structure: C1 and C2 share two pairs of electrons (double bond), and C2 and C3 share one pair. C1 shares one pair each with 2 H atoms, C2 with 1 H atom and C3 with 3 H atoms. Each carbon thus has 8 electrons around it and each H has 2.
  1 mark
4. (d) Propanol (propan-1-ol) on heating with concentrated $H_{2} S O_{4}$ loses water (dehydration) to give propene: $C H_{3} C H_{2} C H_{2} O H co n c . H_{2} S O_{4}, h e a t C H_{3} C H = C H_{2} + H_{2} O$ .
  1 mark
5. (e) Hydrogenation: $C H_{2} = C H - C H_{3} + H_{2} N i C H_{3} - C H_{2} - C H_{3}$ , i.e. $C_{3} H_{6} + H_{2} N i C_{3} H_{8}$ (propane).
  1 mark
6. For visually impaired students: (a) x = 3, y = 6; (b) propene; (c) unsaturated hydrocarbon, as it contains a C=C double bond; (d) propanol; (e) $C_{3} H_{6} + H_{2} N i C_{3} H_{8}$ .
Answer: (a) x = 3, y = 6; (b) propene; (c) dot structure with C=C double bond; (d) propanol; (e) $C_{3} H_{6} + H_{2} N i C_{3} H_{8}$
Where the marks are
- x = 3, y = 6 (1 mark)
- Propene (1 mark)
- Correct electron dot structure showing the double bond (1 mark)
- Propanol (1 mark)
- Balanced hydrogenation equation with Ni (1 mark)
Open Q29(A) on its own page
OR
Q29(B) · 5 marks · Case-basedCompetency-basedCarbon and its Compounds
The electronic structures of atoms P and Q are shown below. Based on the information given above, answer the following questions:
1. (a)
  If P and Q combine to form a compound, what type of bond is formed between them?
  1 mark
2. (b)
  Give the chemical formula of the compound formed.
  1 mark
3. (c)
  The compound so formed is dissolved in water. Is the resultant solution acidic or basic in nature? Justify your answer.
  1 mark
4. (d)
  Write the chemical equation for the reaction between 'Q' and ethanol.
  1 mark
5. (e)
  What will be the formula of the compound formed when 'P' undergoes bonding with carbon?
  1 mark
Figure in the paper: Two Bohr-type atomic diagrams. Atom P: a nucleus with a shell of 2 electrons (dots at top) and an outer shell with 6 electrons... (electrons shown as dots on the orbits: two dots on the inner circle region, and the outer shell carries the remaining dots). Atom Q: nucleus with two shells, inner shell with 2 electrons and outer shell with 8 electrons arranged around the circle, and a third shell with 2 electrons at the bottom and top. Labels P and Q are written in the centres. See it in the official paper.
Try it first, then show our solutionSolution
1. From the diagrams: P has configuration 2, 6, so it is oxygen (non-metal, needs 2 electrons). Q has one electron in its outermost (third) shell, configuration 2, 8, 1, so it is sodium (metal, loses 1 electron).
2. (a) The metal Q transfers electrons to the non-metal P, so an ionic (electrovalent) bond is formed.
  1 mark
3. (b) Q has valency 1 and P has valency 2, so the formula is ** $Q_{2} P$ ** (i.e. $N a_{2} O$ ).
  1 mark
4. (c) Basic. $Q_{2} P$ is a metal oxide, and metallic oxides dissolve in water to form alkalis: $N a_{2} O + H_{2} O \to 2 N a O H$ .
  1 mark
5. (d) $2 C_{2} H_{5} O H + 2 Q \to 2 C_{2} H_{5} O Q + H_{2}$ (sodium ethoxide and hydrogen gas).
  1 mark
6. (e) Carbon has valency 4 and P (oxygen) has valency 2, so the compound is ** $C P_{2}$ ** ( $C O_{2}$ ).
  1 mark
7. For visually impaired students: (a) C–O: covalent bond; (b) Ca–O: ionic bond; (c) CaO conducts in the molten state because it has free ions; (d) CaO is a solid while $C O_{2}$ is a gas; (e) valency of carbon in $C O_{2}$ = 4.
Answer: (a) Ionic bond; (b) $Q_{2} P$ ; (c) Basic, metal oxide forms alkali; (d) $2 C_{2} H_{5} O H + 2 Q \to 2 C_{2} H_{5} O Q + H_{2}$ ; (e) $C P_{2}$
Where the marks are
- Ionic bond (1 mark)
- Formula $Q_{2} P$ (1 mark)
- Basic, with reason that metallic oxides are basic (1 mark)
- Balanced equation of Q with ethanol (1 mark)
- $C P_{2}$ (1 mark)
Open Q29(B) on its own page

Sample	colour in dilute hydrochloric acid	colour in aqueous sodium hydroxide
A	Orange	Red
B	Red	Yellow
C	Red	Orange
D	Yellow	Red

S. No.	Volume of dil. NaOH taken (mL)	Drops of dil. H $_{2}$ SO $_{4}$ used
1	2	20
2	3	30
3	4	40

Section C

Physics. MCQ 30-31 (1 mark), assertion-reason 32 (1), Q33 (2), Q34 (2, internal choice), Q35 (3), Q36 (3), Q37 (3), Q38 (4, internal choice), Q39 (5, internal choice)

Q30 · 1 mark · MCQLight – Reflection and Refraction
Arnav was making notes and he wrote down the following statements from his understanding of reflection from curved surfaces.
I. Concave mirrors can produce both real and virtual images depending on the position of the object.
II. Convex mirrors always produce real, inverted images regardless of the object's position.
III. In both concave and convex mirrors, the image location can be determined using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ where f is the focal length, v is the image distance, and u is the object distance.
Choose from the following the correct option that lists the correct statements about reflection from curved surfaces.
1. (A)I and II
2. (B)I, II and III
3. (C)II and III
4. (D)I and III
Try it first, then show our solutionSolution
1. I is true. A concave mirror forms real images when the object is beyond F, and a virtual image when the object is between P and F.
2. II is false. A convex mirror always forms virtual, erect, diminished images.
3. III is true. The mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ applies to all spherical mirrors when the sign convention is used. Correct statements: I and III, option (D).
  1 mark
Answer: I and III, option (D)
Where the marks are
- Identifies I and III as correct, option (D) (1 mark)
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Q31 · 1 mark · MCQThe Human Eye and the Colourful World
Choose the correct option from the below which explains the reason for us to perceive the day sky as blue.
1. (A)As sunlight passes through the atmosphere, shorter wavelengths, such as blue are scattered more than other colors.
2. (B)The sky appears blue because all colors are scattered equally, but blue light is stronger and more visible to the human eye.
3. (C)The blue color of the sky is due to longer wavelengths like red and orange scattering more than shorter wavelengths, making blue stand out more.
4. (D)The atmosphere contains blue-colored particles that give the sky its blue appearance.
Try it first, then show our solutionSolution
1. Air molecules are much smaller than the wavelength of visible light. They scatter shorter wavelengths (blue end) much more strongly than longer ones (red end).
2. This scattered blue light reaches our eyes from all directions, so the sky looks blue. Our eyes are also more sensitive to blue than to violet. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Chooses scattering of shorter (blue) wavelengths, option (A) (1 mark)
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Q32 · 1 mark · Assertion-reasonCompetency-basedLight – Reflection and Refraction
The following question consists of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below:
A. Both A and R are true, and R is the correct explanation of A.
B. Both A and R are true, and R is not the correct explanation of A.
C. A is true but R is false.
D. A is false but R is true.
Assertion (A): A point object is placed at a distance of 26 cm from a convex mirror of focal length 26 cm. The image will not form at infinity.
Reason (R): For above given system the equation $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ gives $v = \infty$ .
1. (A)Both A and R are true, and R is the correct explanation of A.
2. (B)Both A and R are true, and R is not the correct explanation of A.
3. (C)A is true but R is false.
4. (D)A is false but R is true.
Try it first, then show our solutionSolution
1. For a convex mirror, $f = + 26$ cm and $u = - 26$ cm.
2. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{26} + \frac{1}{26} = \frac{2}{26} \Rightarrow v = + 13$ cm. A virtual image forms 13 cm behind the mirror, not at infinity.
3. So A is true and R (which claims $v = \infty$ ) is false. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Uses the sign convention to get v = +13 cm and concludes A true, R false (1 mark)
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Q33 · 2 marks · Case-basedCompetency-basedLight – Reflection and Refraction
The above image shows the formation of an image with an optical instrument.
1. A
  Identify the optical instrument (shown schematically as a rectangle) in the image.
  1 mark
2. B
  What type of image is formed in this case?
  1 mark
3. C
  Based on the measurements given in the image, calculate the focal length of the instrument.
  1 mark
Figure in the paper: A horizontal principal axis with a vertical rectangle (the optical element) in the middle. An upright object arrow stands on the left of the rectangle; a smaller upright arrow appears on the same left side, with rays drawn from the top of the object through the element and emerging diverging to the right, and dotted lines extended backwards to meet at the image top. Measurements: 10 cm marked from the rectangle to the foot of the smaller arrow, and 20 cm marked from the object foot to the rectangle. See it in the official paper.
Try it first, then show our solutionSolution
1. A. The rays diverge after passing through the element, and the diminished, erect image is on the same side as the object. So the instrument is a concave (diverging) lens.
  ½ mark
2. B. The image is virtual, erect and diminished. It is formed where the backward extensions of the rays meet.
  ½ mark
3. C. Lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ with $u = - 20$ cm and $v = - 10$ cm (image on the object side).
4. $\frac{1}{f} = \frac{1}{- 10} - \frac{1}{- 20} = - \frac{2}{20} + \frac{1}{20} = - \frac{1}{20} \Rightarrow f = - 20$ cm
  1 mark
5. For visually impaired students: A. A convex lens forms a virtual image when the object is placed between its optical centre and principal focus F. B. A convex lens converges the parallel rays of sunlight to its focus. The light energy concentrated at that small spot raises the temperature enough to ignite the paper.
Answer: A. Concave lens; B. Virtual (erect, diminished); C. f = −20 cm
Where the marks are
- Identifies concave lens (½ mark)
- Virtual image (½ mark)
- Correct substitution with signs and f = −20 cm (1 mark)
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Answer one of these 2 alternatives.
Q34(A) · 2 marks · Case-basedCompetency-basedElectricity
Find out the following in the electric circuit given in the figure:
1. (i)
  Effective resistance of two 8 ohm resistors in the combination.
  1 mark
2. (ii)
  Current flowing through the 4-ohm resistor.
  1 mark
Figure in the paper: Circuit: 8 V battery connected from point A through a 4 Ω resistor to point B, then ammeter A1 (+ at B side, − at C side) to junction C. Between C and D are two 8 Ω resistors in parallel. From D, ammeter A2 (− at D side, + at E side) returns to the battery at E, completing the loop. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The two 8 Ω resistors are in parallel: $R_{p} = \frac{R _{1} R _{2}}{R _{1} + R _{2}} = \frac{8 \times 8}{8 + 8} = 4 Ω$
  1 mark
2. (ii) The 4 Ω resistor is in series with $R_{p}$ : $R_{t o t a l} = 4 + 4 = 8 Ω$
3. $I = \frac{V}{R} = \frac{8 V}{8 Ω} = 1$ A. In series the same current flows through the 4 Ω resistor, so the current is 1 A.
  1 mark
Answer: (i) 4 Ω; (ii) 1 A
Where the marks are
- Parallel combination = 4 Ω (1 mark)
- Total 8 Ω and I = 1 A through the 4 Ω resistor (1 mark)
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OR
Q34(B) · 2 marks · Case-basedCompetency-basedElectricity
Study the circuit and find out-
1. (i)
  Current in 12 ohm resistor.
  1 mark
2. (ii)
  Difference in the readings of ammeter A1 and A2 if any.
  1 mark
Figure in the paper: Circuit: a 6 V battery with a 12 Ω resistor in series. Two 24 Ω resistors are connected in parallel between two junctions, with ammeter A1 in series on the right side of the parallel combination, and ammeter A2 in the lower wire in series with the battery. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Parallel part: $\frac{1}{R _{p}} = \frac{1}{24} + \frac{1}{24} = \frac{2}{24} \Rightarrow R_{p} = 12 Ω$
2. Total resistance $R_{T} = 12 + 12 = 24 Ω$ . Then $I = \frac{V}{R _{T}} = \frac{6}{24} = 0.25$ A. The 12 Ω resistor is in series with the battery, so the current in it is 0.25 A.
  1 mark
3. (ii) Both A₁ and A₂ are in the main (series) line, and current is the same everywhere in a series path. So they show the same reading (0.25 A) and there is no difference.
  1 mark
Answer: (i) 0.25 A; (ii) No difference, A₁ and A₂ read the same
Where the marks are
- R_p = 12 Ω, R_T = 24 Ω, I = 0.25 A (1 mark)
- Same readings of A₁ and A₂ (1 mark)
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Q35 · 3 marks · Short answerThe Human Eye and the Colourful World
The above image shows a corrective measure for a particular defect of vision.
(i) Identify the defect of vision and state what kind of lens is used to correct this deficiency.
(ii) Draw and label a ray diagram that shows the defect of vision in the above case before correction.
For visually impaired students:
(i) What is dispersion of light?
(ii) Explain the condition under which dispersion happens?
(iii) Give one reason that causes presbyopia.
Figure in the paper: An eye (large circle on the right with lens and retina) with a diverging (concave) lens placed in front of the eye. Rays from a point object on the left diverge and are shown as dashed lines extended backwards to a virtual image; after passing through the lens, the rays become parallel and enter the eye to focus on the retina. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In the figure the object sits at the eye's near point (about 25 cm). After passing through the correcting lens, its rays appear to come from a virtual image N′ that is farther away. The eye can focus rays from N′ onto the retina. This is how NCERT shows the correction of a near point that has moved too far away, which happens in long-sightedness.
2. Defect: hypermetropia (long-sightedness / far-sightedness). The person sees distant objects clearly but cannot see nearby objects clearly. The image of a near object would form behind the retina.
  ½ mark
3. Correcting lens: a convex (converging) lens of suitable power. It adds converging power so the image of a near object forms on the retina.
  ½ mark
4. (ii) Ray diagram of the defective eye, before correction. Draw these parts in order:
  An eye (circle) with the eye lens and the retina.
  Point N′, the near point of the defective eye, which lies farther than 25 cm.
  Point N, the normal near point at 25 cm, placed closer to the eye than N′.
  Rays from N′ passing through the eye lens and meeting on the retina.
  1 mark
5. Then show the defect:
  Rays from an object at N pass through the eye lens and would meet at a point behind the retina.
  So the image of N is blurred.
  Label N, N′, eye lens, retina and the image point behind the retina. Title the figure 'Hypermetropic eye'.
  1 mark
6. For visually impaired students (alternative, 1 + 1 + 1 marks).
  (i) Dispersion is the splitting of white light into its seven component colours (VIBGYOR) when it passes through a prism. Each colour bends through a different angle, so the colours spread out into a spectrum.
7. (ii) Condition for dispersion:
  Light must pass from one medium into another (for example air into glass) and be refracted.
  The medium must have a refractive index that is different for different colours (wavelengths), as glass and water do.
  Then the colours travel at different speeds in the medium and bend by different amounts. Violet bends the most and red the least.
8. (iii) Cause of presbyopia: with age, the ciliary muscles weaken and the eye lens loses its flexibility. The eye therefore loses its power of accommodation, the near point moves away, and nearby objects cannot be seen clearly.
Answer: (i) Hypermetropia; corrected by a convex (converging) lens. (ii) Ray diagram: rays from N (25 cm) meet behind the retina; the near point of the defective eye is N′, farther away. VI: (i) splitting of white light into its colours; (ii) refraction in a medium whose refractive index differs for different colours; (iii) weakening of ciliary muscles / loss of lens flexibility with age.
Where the marks are
- Names the defect as hypermetropia (½ mark)
- States that a convex (converging) lens corrects it (½ mark)
- Diagram shows the eye with rays from N′ (shifted near point) focusing on the retina (1 mark)
- Diagram shows rays from N (25 cm) meeting behind the retina, with correct labels (1 mark)
- VI: correct definition of dispersion (1 mark); condition, i.e. refraction with a refractive index that depends on colour (1 mark); valid cause of presbyopia (1 mark)
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Q36 · 3 marks · Short answerElectricity
A student needs to make a 0.12 Ω resistor. She has some copper wire of 0.80 mm diameter. Resistivity of copper is $1.8 \times 1 0^{- 8} Ω m$ .
(i) Determine the cross-sectional area of the wire.
(ii) Calculate the length of wire required for the 0.12 Ω resistor.
Try it first, then show our solutionSolution
1. (i) The wire has a circular cross-section, so $A = π r^{2} = π (\frac{d}{2})^{2}$ .
  Diameter $d = 0.80 mm = 0.80 \times 1 0^{- 3} m$ , so radius $r = 0.40 \times 1 0^{- 3} m$ .
  ½ mark
2. $A = 3.14 \times (0.40 \times 1 0^{- 3})^{2} = 3.14 \times 0.16 \times 1 0^{- 6} \approx 5.0 \times 1 0^{- 7} m^{2}$
  ½ mark
3. (ii) Use $R = \frac{ρl}{A}$ and rearrange for length:
  $l = \frac{R A}{ρ}$
  ½ mark
4. Substitute $R = 0.12 Ω$ , $A = 5 \times 1 0^{- 7} m^{2}$ and $ρ = 1.8 \times 1 0^{- 8} Ω m$ :
  $l = \frac{0.12 \times 5 \times 1 0 ^{- 7}}{1.8 \times 1 0 ^{- 8}} = \frac{6 \times 1 0 ^{- 8}}{1.8 \times 1 0 ^{- 8}}$
  1 mark
5. $l \approx 3.33 m$ (about 3.3 m of the copper wire is needed).
  ½ mark
Answer: (i) A ≈ 5 × 10⁻⁷ m²; (ii) l ≈ 3.33 m
Where the marks are
- Writes $A = π (d /2)^{2}$ with radius 0.40 × 10⁻³ m (½ mark)
- Obtains A ≈ 5 × 10⁻⁷ m² (½ mark)
- Uses $R = ρl / A$ and rearranges for l (½ mark)
- Correct substitution with SI units (1 mark)
- Final length ≈ 3.33 m (½ mark)
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Q37 · 3 marks · Short answerMagnetic Effects of Electric Current
Magnetic field lines are shown in the given diagram. A student makes a statement that the magnetic field at X is stronger than at Y.
(i) Explain with reason if the student's claim is correct.
(ii) Also redraw the diagram and mark the direction of magnetic field lines.
Figure in the paper: A bar magnet lies horizontally with N pole on the left and S pole on the right. Several closed magnetic field lines emerge from the N pole, arc over the top, and return to the S pole. Point X is on the left, close to the N pole at the lower end of the innermost loop; point Y is at the top of the outermost/innermost arc, further from the magnet. The lines are closer together near X than at Y. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The student's claim is correct: the magnetic field at X is stronger than at Y.
  1 mark
2. Reason: the strength of a magnetic field is shown by how close together the field lines are. Where the lines are crowded, the field is stronger.
  X is near the N pole, where the lines are close together. Y lies far from the magnet, where the lines are spread apart. So the field at X is stronger than at Y.
  1 mark
3. (ii) Redraw the bar magnet with N on the left and S on the right, and copy the same curved field lines. Put arrows on every line:
  Outside the magnet, the arrows point from the N pole towards the S pole.
  Inside the magnet, the arrows run from S to N, so each line forms a closed loop.
  1 mark
Answer: (i) Correct: field lines are closer together at X (near the pole) than at Y, so the field at X is stronger. (ii) Arrows go from N to S outside the magnet (and S to N inside).
Where the marks are
- States that the claim is correct (1 mark)
- Reason: field lines are more crowded at X, so the field is stronger there (1 mark)
- Redrawn diagram with arrows from N to S outside the magnet (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedLight – Reflection and Refraction
The above image is that of a Digital Single Lense Reflector (DSLR) Camera which are used to take high resolution photographs by professional photographers. The second image of the above two is a schematic diagram of how an image is formed on the sensor of the camera. Based on your understanding of the lenses, answer the following questions.
1. A
  What type of lens is used in the DSLR camera shown in the image?
  1 mark
2. B
  What type of image is formed on the sensor?
  1 mark
3. C
  A photographer is using a DSLR camera with a lens of focal length f = 50 mm to take a close-up photograph of a small object. The lens projects an image onto the camera sensor that is located 60 mm behind the lens. Calculate the object distance (i.e., the distance between the object and the lens).
  2 marks
4. OR
  D
  A photographer is using a DSLR camera to take a picture of a flower. The flower is positioned 150 mm away from the camera lens. The actual height of the flower is 80 mm, and the image height formed on the camera's sensor is measured to be 20 mm. Calculate the focal length of the camera lens.
  2 marks
Figure in the paper: Top: photograph of a DSLR camera with a large lens barrel on the left and the camera body with dials and viewfinder. Bottom: schematic ray diagram showing an object (a palm tree on the left) with a lens drawn as a rectangle with a converging (convex) lens symbol; rays from the top and bottom of the object pass through the lens and converge to form an inverted image of the tree on the camera sensor, labelled 'image', inside the camera body at right. See it in the official paper.
Try it first, then show our solutionSolution
1. A. The rays from the object converge to form an image on the sensor, so the camera uses a convex (converging) lens.
  1 mark
2. B. The image falls on a screen (the sensor) and is upside down, so it is real and inverted (and diminished).
  1 mark
3. C. Lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ , which gives $\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$ .
  With the sign convention: $f = + 50$ mm and $v = + 60$ mm (the image is real, behind the lens).
  ½ mark
4. $\frac{1}{u} = \frac{1}{60} - \frac{1}{50} = \frac{5 - 6}{300} = - \frac{1}{300}$
  1 mark
5. $u = - 300$ mm. The object is 300 mm (30 cm) in front of the lens.
  ½ mark
6. C/D are alternatives. D (OR, worth 2 marks).
  Given: $u = - 150$ mm, $h = + 80$ mm, and $h^{'} = - 20$ mm (the image is inverted).
  Magnification: $m = \frac{h ^{'}}{h} = \frac{- 20}{80} = - 0.25$ [½ mark]
7. D (continued): from $m = \frac{v}{u}$ , we get $v = m \times u = (- 0.25) \times (- 150) = + 37.5$ mm [½ mark]
8. D (continued): $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{37.5} + \frac{1}{150} = \frac{4 + 1}{150} = \frac{1}{30}$
  So $f = + 30$ mm [1 mark]
9. For visually impaired students (alternative).
  VI A: a concave lens is used in the flashlight and a convex lens in the solar-cooker concentrator [1 mark].
10. VI B: a concave lens diverges light, so the flashlight beam spreads over a wider area. A convex lens converges sunlight to a point, which raises the temperature there for cooking [1 mark].
11. VI C: the solar cooker uses a convex lens, and the image is real, inverted and twice the size, so $m = - 2$ . With $u = - 40$ cm, $v = m u = (- 2) (- 40) = + 80$ cm.
  $\frac{1}{f} = \frac{1}{80} - \frac{1}{- 40} = \frac{1 + 2}{80} = \frac{3}{80}$ , so $f = \frac{80}{3} \approx 26.67$ cm [2 marks].
  (If a virtual image with $m = + 2$ were taken instead, $v = - 80$ cm and $f = 80$ cm.)
12. VI D (OR): $u = - 20$ cm and $v = - 10$ cm (the image is on the same side as the object).
  $\frac{1}{f} = \frac{1}{- 10} - \frac{1}{- 20} = \frac{- 2 + 1}{20} = - \frac{1}{20}$
  So $f = - 20$ cm, which means a concave (diverging) lens [2 marks].
Answer: A: convex lens; B: real and inverted (diminished); C: u = −300 mm (object 300 mm in front of the lens); D: f = 30 mm. VI: A concave (flashlight), convex (solar cooker); C f ≈ 26.67 cm; D f = −20 cm
Where the marks are
- A: convex lens (1 mark)
- B: real and inverted image (1 mark)
- C: lens formula with correct signs, f = +50 mm and v = +60 mm (½ mark)
- C: correct substitution and working to 1/u = −1/300 (1 mark)
- C: u = −300 mm, i.e. 300 mm in front of the lens (½ mark)
- D (alternative): m = −0.25 (½ mark), v = 37.5 mm (½ mark), f = 30 mm via the lens formula (1 mark)
- VI: lens choices (1 mark), reasons (1 mark), C f ≈ 26.67 cm or D f = −20 cm (2 marks)
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Answer one of these 2 alternatives.
Q39(A) · 5 marks · Case-basedCompetency-basedElectricity
The arrangement of resistors shown in the above figure is connected to a battery. The power dissipation in the 100 Ω resistor is 81 W. Calculate
(i) the current in the circuit
(ii) the reading in the voltmeter V $_{2}$
(iii) the reading in the voltmeter V $_{1}$
Figure in the paper: A battery at the top with voltmeter V1 connected across its terminals. A 100 Ω resistor is in series on the left side. Two 25 Ω resistors are connected in parallel to each other in the right-hand branch. Voltmeter V2 is connected across the parallel 25 Ω combination (the parallel branch). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) All the current from the battery flows through the 100 Ω resistor, because it is in series. Use $P = I^{2} R$ :
  $81 = I^{2} \times 100$
  1 mark
2. $I^{2} = \frac{81}{100} = 0.81$ , so $I = 0.9$ A. This is the current in the circuit.
  1 mark
3. (ii) The two 25 Ω resistors are in parallel:
  $\frac{1}{R _{p}} = \frac{1}{25} + \frac{1}{25} = \frac{2}{25}$ , so $R_{p} = 12.5 Ω$
  1 mark
4. V₂ reads the voltage across this combination:
  $V_{2} = I R_{p} = 0.9 \times 12.5 = 11.25$ V
  1 mark
5. (iii) Voltage across the 100 Ω resistor: $0.9 \times 100 = 90$ V.
  V₁ is across the battery, so it reads the total voltage:
  $V_{1} = 90 + 11.25 = 101.25$ V
  (Check: $I \times R_{t o t a l} = 0.9 \times 112.5 = 101.25$ V.)
  1 mark
Answer: (i) 0.9 A; (ii) V₂ = 11.25 V; (iii) V₁ = 101.25 V
Where the marks are
- Uses P = I²R for the 100 Ω resistor (1 mark)
- Current I = 0.9 A (1 mark)
- Equivalent resistance of the parallel 25 Ω pair = 12.5 Ω (1 mark)
- V₂ = 11.25 V (1 mark)
- V₁ = 90 V + 11.25 V = 101.25 V (1 mark)
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OR
Q39(B) · 5 marks · Case-basedCompetency-basedElectricity
An electric heater consists of three similar heating elements A, B and C, connected as shown in the figure above. Each heating element is rated as 1.2 kW, 240 V and has constant resistance. S $_{1}$ , S $_{2}$ and S $_{3}$ are respective switches. The circuit is connected to a 240 V supply.
1. (i)
  Calculate the resistance of one heating element.
  1 mark
2. (ii)
  Calculate the current in each resistor when only S $_{1}$ and S $_{3}$ are closed.
  2 marks
3. (iii)
  Calculate the power dissipated across A when S $_{1}$ , S $_{2}$ and S $_{3}$ are closed.
  2 marks
Figure in the paper: A 240 V supply line at the top left with switch S1 on the top right branch. Heating element A is in the middle branch on the left, connected in series with element B to its right. Switch S2 connects the junction between A and B to the right-hand return wire. Element C is in the bottom branch, in series with switch S3. The branches are joined at both ends to form a parallel arrangement. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Use $P = \frac{V ^{2}}{R}$ , so $R = \frac{V ^{2}}{P} = \frac{240 \times 240}{1200} = 48 Ω$
  1 mark
2. (ii) With S₁ and S₃ closed and S₂ open, A and B are in series in one branch. C is in a parallel branch. Each branch has 240 V across it.
  Current in C: $I_{C} = \frac{V}{R} = \frac{240}{48} = 5$ A
  1 mark
3. A and B in series: $R = 48 + 48 = 96 Ω$ , so $I_{A} = I_{B} = \frac{240}{96} = 2.5$ A
  1 mark
4. (iii) With S₂ also closed, the junction between A and B is joined straight to the return wire. B is short-circuited, so A alone has 240 V across it.
  $I_{A} = \frac{240}{48} = 5$ A
  1 mark
5. $P_{A} = I^{2} R = 5^{2} \times 48 = 1200$ W $= 1.2$ kW
  1 mark
6. For visually impaired students (alternative, 5 marks).
  The resistors are in parallel, so each has 24 V across it.
  $I_{1} = \frac{24}{6} = 4$ A and $I_{2} = \frac{24}{12} = 2$ A [1 mark]
7. Time $t = 5 \times 60 = 300$ s. Using Joule's law $H = I^{2} R t$ :
  $H_{1} = 4^{2} \times 6 \times 300 = 28800$ J
  $H_{2} = 2^{2} \times 12 \times 300 = 14400$ J [1 mark]
8. Total heat $= 28800 + 14400 = 43200$ J [1 mark]
9. Power in each resistor:
  $P_{1} = V I_{1} = 24 \times 4 = 96$ W
  $P_{2} = 24 \times 2 = 48$ W
  Both are below the 100 W rating, so both resistors are safe to use [2 marks].
Answer: (i) 48 Ω; (ii) C: 5 A, A and B: 2.5 A each; (iii) 1200 W (1.2 kW). VI: 43200 J; both resistors safe (96 W and 48 W < 100 W)
Where the marks are
- R = V²/P = 48 Ω (1 mark)
- Current in C = 5 A (1 mark)
- Current in A and B = 240/96 = 2.5 A (1 mark)
- Recognises that B is shorted, so the current in A is 5 A (1 mark)
- P_A = 1200 W (1 mark)
- VI: currents 4 A and 2 A (1 mark); heats 28800 J and 14400 J (1 mark); total 43200 J (1 mark); powers 96 W and 48 W, both safe (2 marks)
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Source: official sample paper, cbseacademic.nic.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.