CBSE Class 10 Science Sample Paper 2026-27 with Solutions (Free Online Mock Test)
Pandey Sir’s BEST Classes
CBSE Class 10 Science Sample Paper 2026-27 — Full Mock Test with Solutions
Original questions written to the official CBSE Sample Question Paper blueprint released for session 2026-27 (board exam February 2027).
80 Marks
3 Hours
39 Questions
Official 30 / 25 / 25 blueprint
Auto-scored objectives
By Pandey Sir, BEST Classes, Dammaiguda, Hyderabad · Updated 2 October 2026
Exam blueprint used for this paper
The official CBSE Class 10 Science paper for 2026-27 is organised by subject, not by question type. This is the structure your mock follows exactly.
| Section | Subject | Questions | Marks | Suggested time |
|---|---|---|---|---|
| A | Biology | Q1 – Q16 | 30 | 60 min |
| B | Chemistry | Q17 – Q29 | 25 | 55 min |
| C | Physics | Q30 – Q39 | 25 | 55 min |
| Total | 80 | 180 min + 10 min review | ||
General instructions (as per the official pattern): This paper has 39 questions in 3 sections. All questions are compulsory, but an internal choice is given in some questions — attempt only one alternative in those. The question paper design and assessment pattern are unchanged for session 2026-27.
Competency weighting in the official design: 50% knowledge and understanding, 30% application of concepts, 20% formulate, analyse, evaluate and create. Each question below is tagged with its type so you can check your own balance after scoring.
Section A — BiologyQ1–Q16 · 30 marks
During photosynthesis, carbon dioxide and water are converted into glucose and oxygen. Which statement correctly describes this change?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
Correct option(A). In CO2, carbon goes from oxidation state +4 to 0 in glucose — it gains hydrogen, so it is reduced. In water, oxygen goes from −2 to 0 in O2 — it loses hydrogen, so it is oxidised.
✅ Answer: (A)
A student wants to prove that germinating seeds release carbon dioxide by showing that lime water turns milky. Which setup is the best control for this experiment?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningA control must differ from the test setup in only one variable — here, whether respiration is happening. Boiled seeds are dead, so respiration has stopped; everything else (container, lime water, conditions) stays the same. If only the live seeds turn the lime water milky, respiration is proven to be the cause. Option (C) removes the seeds entirely, which changes more than one thing and makes a weaker control.
✅ Answer: (B)
During urine formation, which set of substances is completely reabsorbed by the kidney tubules?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningGlucose and amino acids are useful to the body, so they are reabsorbed completely; their appearance in urine indicates disease such as diabetes. Urea is a waste product and is not reabsorbed — it is concentrated and excreted. Salts and water are reabsorbed only partly, depending on the body’s needs.
✅ Answer: (A)
A gardener removes the anthers from a bisexual flower in the bud stage and covers the flower with a paper bag. What is the most likely outcome?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningRemoving the anthers removes the source of pollen, so self-pollination becomes impossible. The paper bag stops insects and wind from bringing in pollen from another flower, so cross-pollination is also blocked. With no pollination there is no fertilisation, and therefore no seeds or fruits.
✅ Answer: (C)
Vasectomy blocks the vas deferens and tubectomy blocks the oviduct. Which single biological event do both prevent?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningThese are blocking methods — they cut the transport route, not the production. Gametes are still produced; they simply cannot meet. Blocking the vas deferens stops sperm from reaching the egg, and blocking the oviduct stops the egg from meeting sperm. In both cases the gametes cannot fuse, so fertilisation — and hence pregnancy — is prevented.
✅ Answer: (B)
The cells in the root tip of a plant contain 24 chromosomes. How many chromosomes are present in the egg cell and in the zygote of this plant respectively?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningRoot-tip cells are body cells, so they are diploid (2n) = 24, giving n = 12. Gametes are haploid, so the egg cell has 12 chromosomes. The zygote forms by fusion of two haploid gametes, so it is diploid again: 12 + 12 = 24.
✅ Answer: (B) 12 and 24
In the food chain grass → grasshopper → frog → snake, what is the trophic level of the frog?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningGrass is the producer (first trophic level). The grasshopper eats grass, so it is the primary consumer (second level). The frog eats the grasshopper, so it is the secondary consumer (third level). The snake, eating the frog, is the tertiary consumer.
✅ Answer: (C)
The following consist of two statements — Assertion (A) and Reason (R).
A: The rate of breathing in aquatic organisms is much faster than in terrestrial organisms.
R: Water contains a much lower concentration of dissolved oxygen than air.
Select the correct option:
A: The rate of breathing in aquatic organisms is much faster than in terrestrial organisms.
R: Water contains a much lower concentration of dissolved oxygen than air.
Select the correct option:
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
Truth of A and RA is true — aquatic animals breathe faster. R is true — dissolved oxygen in water is far lower than in air. A follows directly from R: because each breath yields less oxygen, the animal must breathe more often. So R correctly explains A.
✅ Answer: (A)
A: Reflex actions are fast and involuntary.
R: Reflex arcs are formed in the spinal cord and do not involve the brain at all.
Select the correct option:
R: Reflex arcs are formed in the spinal cord and do not involve the brain at all.
Select the correct option:
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
Truth of A and RA is true. R is false — this is the classic trap. The decision to act is made at the spinal cord without waiting for the brain, which is why the response is fast, but the impulse does travel to the brain afterwards so the animal becomes aware of the stimulus. Saying the brain is not involved “at all” overstates it.
✅ Answer: (C)
Why is the small intestine in humans very long, and how is its inner lining adapted for absorption? Give two points.
Answer & marking scheme
Marking scheme: Large surface area / time for digestion 1 · villi and blood supply 1. Total 2 marks.
Point 1 — why it is longThe small intestine is long because it provides a large surface area and allows enough time for the complete digestion and absorption of food. The food stays in contact with the digesting juices for a longer period.
Point 2 — adaptation of the liningIts inner lining has millions of finger-like projections called villi. Villi greatly increase the surface area for absorption, and each villus has a rich network of blood capillaries that carry the absorbed food to every cell of the body.
✅ Full marks require both points: surface area / time, and villi with blood supply.
State two differences between nervous control and hormonal control in animals.
Answer & marking scheme
Marking scheme: Each correct difference 1 mark. Total 2 marks.
Difference 1 — speed and pathwayNervous control works through electrical impulses carried by neurons and is very fast. Hormonal control works through chemical messengers carried in the blood and is slower.
Difference 2 — duration and reachThe effect of nervous control is short-lived and acts at a specific site. The effect of hormonal control is long-lasting and can reach all cells of the body through the blood.
✅ Any two valid contrasting differences earn full marks.
State two functions of the placenta in a human pregnancy.
Answer & marking scheme
Marking scheme: Each correct function 1 mark. Total 2 marks.
Function 1 — supplyIt provides the developing embryo with nutrition (glucose) and oxygen taken from the mother’s blood.
Function 2 — removalIt carries the embryo’s waste products such as carbon dioxide and urea away into the mother’s blood for excretion. (It also acts as an endocrine tissue, secreting hormones such as oestrogen and progesterone.)
✅ Supply of nutrition / oxygen, and removal of wastes.
(a) Name the enzyme present in saliva and state its function.
(b) What is the role of hydrochloric acid in the stomach?
(c) Why does the stomach not digest itself?
(b) What is the role of hydrochloric acid in the stomach?
(c) Why does the stomach not digest itself?
Answer & marking scheme
Marking scheme: (a) 1 · (b) 1 · (c) 1. Total 3 marks.
(a) Salivary enzymeThe enzyme is salivary amylase. It breaks down starch into simpler sugars (maltose), beginning the digestion of carbohydrates in the mouth.
(b) Role of hydrochloric acidHydrochloric acid makes the medium in the stomach acidic, which is necessary for the enzyme pepsin to act. It also kills bacteria that enter with food and converts inactive pepsinogen into active pepsin.
(c) Self-protectionThe inner lining of the stomach secretes a layer of mucus. This mucus protects the stomach wall from the corrosive action of its own hydrochloric acid and pepsin.
✅ (a) salivary amylase / starch to sugar · (b) acidic medium + kills bacteria · (c) mucus layer.
In a monohybrid cross, a pure tall pea plant (TT) is crossed with a pure dwarf pea plant (tt).
(a) What is the phenotype of the F1 generation?
(b) What is the phenotypic ratio in the F2 generation?
(c) Why does the dwarf trait reappear in F2 but not in F1?
(a) What is the phenotype of the F1 generation?
(b) What is the phenotypic ratio in the F2 generation?
(c) Why does the dwarf trait reappear in F2 but not in F1?
Answer & marking scheme
Marking scheme: (a) 1 · (b) 1 · (c) 1. Total 3 marks.
(a) F1 phenotypeAll plants in F1 are tall. Their genotype is Tt.
(b) F2 phenotypic ratioCrossing Tt × Tt gives 3 tall : 1 dwarf (genotypes 1 TT : 2 Tt : 1 tt).
(c) Why the dwarf reappearsT (tall) is dominant over t (dwarf), so in F1 the single t allele is hidden and the plant is tall. The recessive allele is not lost — in F2 two t alleles come together to form tt, so the dwarf plant reappears.
✅ F1 all tall · F2 3:1 · recessive allele hidden in F1, pairs up in F2.
Case: In humans, blood passes through the heart twice in one complete circuit of the body. This is called double circulation.
(a) Name the two circuits involved in double circulation.
(b) Why is double circulation necessary in humans?
(c) State one difference between the blood carried by the pulmonary artery and the blood carried by the pulmonary vein.
(a) Name the two circuits involved in double circulation.
(b) Why is double circulation necessary in humans?
(c) State one difference between the blood carried by the pulmonary artery and the blood carried by the pulmonary vein.
Answer & marking scheme
Marking scheme: (a) 1 · (b) 1.5 · (c) 1.5. Total 4 marks.
(a) The two circuitsPulmonary circulation — blood is pumped from the heart to the lungs and back. Systemic circulation — blood is pumped from the heart to the rest of the body and back.
(b) Why it is necessaryDouble circulation keeps oxygenated and deoxygenated blood completely separate, so the body receives blood that is rich in oxygen. This efficient oxygen supply is essential because humans are warm-blooded and need a large amount of energy to maintain body temperature and carry out life processes.
(c) Difference in the two vesselsThe pulmonary artery carries deoxygenated blood (from the heart to the lungs), while the pulmonary vein carries oxygenated blood (from the lungs back to the heart). This is the exception to the usual rule that arteries carry oxygenated blood.
✅ (a) pulmonary + systemic · (b) separation of oxygenated and deoxygenated blood for efficient supply · (c) pulmonary artery deoxygenated, pulmonary vein oxygenated.
(a) Name the basic filtration unit of the kidney and state its two main parts.
(b) Explain the three processes involved in the formation of urine.
(c) Why does the volume of urine produced decrease in summer?
(b) Explain the three processes involved in the formation of urine.
(c) Why does the volume of urine produced decrease in summer?
Answer & marking scheme
Marking scheme: (a) 1 · (b) three processes 3 (1 each) · (c) 1. Total 5 marks.
(a) The filtration unitThe basic filtration unit is the nephron. Its two main parts are the glomerulus with Bowman’s capsule and the tubule (with the loop of Henle and collecting duct).
(b) Process 1 — FiltrationBlood enters the glomerulus under high pressure. Water, glucose, amino acids, salts and urea are filtered out into Bowman’s capsule, forming the filtrate. Blood cells and large proteins are not filtered.
(b) Process 2 — Selective reabsorptionAs the filtrate flows along the tubule, useful substances — glucose, amino acids, most salts and much of the water — are reabsorbed back into the blood.
(b) Process 3 — Tubular secretionAdditional waste ions such as hydrogen and potassium ions are secreted from the blood into the tubule, and the remaining fluid is concentrated into urine.
(c) Why urine volume falls in summerIn summer the body loses a large amount of water as sweat, used for cooling. To conserve water, the kidneys reabsorb more water, so less water is left to be excreted as urine.
✅ (a) nephron · (b) filtration, selective reabsorption, tubular secretion · (c) more water lost as sweat, so more is reabsorbed.
Section B — ChemistryQ17–Q29 · 25 marks
Which of the following is a physical change?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningMelting of wax is a physical change: only the state changes from solid to liquid, and the wax can be changed back on cooling. No new substance is formed. Rusting, burning and curdling all produce new substances, so they are chemical changes.
✅ Answer: (C)
In the chlor-alkali process, which product is formed at the cathode and what is it mainly used for?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningWhen electricity is passed through brine (concentrated sodium chloride solution), hydrogen is released at the cathode and chlorine at the anode, while sodium hydroxide forms near the cathode. Hydrogen is used as a fuel. (Chlorine is used for water treatment and to make bleaching powder; sodium hydroxide is used to make soap and detergents.)
✅ Answer: (B)
Why do detergents clean well in hard water while soaps do not?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningHard water contains Ca2+ and Mg2+ ions. Soap reacts with these to form an insoluble precipitate called scum, which wastes soap and leaves deposits on cloth. Detergents are generally ammonium or sulphonate salts whose calcium and magnesium salts remain soluble, so they lather and clean well even in hard water.
✅ Answer: (B)
Which of the following does not contain water of crystallisation?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningBlue vitriol has 5, gypsum has 2 and washing soda has 10 molecules of water of crystallisation in their formulae. Baking soda (NaHCO3) has none — its formula carries no water of crystallisation.
✅ Answer: (D)
Which of the following pairs are isomers?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningIsomers have the same molecular formula but different structures. Butane and isobutane both have the formula C4H10, but a straight chain versus a branched chain. The other pairs have different molecular formulae, so they are not isomers.
✅ Answer: (A)
A manufacturer flushes a packet of potato chips with nitrogen gas before sealing it. Why?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningFats and oils in chips react with oxygen and become rancid, giving an unpleasant smell and taste. Nitrogen is an unreactive (inert) gas; filling the packet with it displaces oxygen, so oxidation is prevented and the chips stay fresh longer.
✅ Answer: (B)
Aluminium is a reactive metal, yet it does not corrode easily and is used for making cooking utensils and electrical wires. Why?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningAluminium reacts with oxygen in the air to form a thin layer of aluminium oxide (Al2O3). This layer is non-porous, hard and stable, and it seals the metal underneath from further attack. That is why aluminium does not corrode easily despite being reactive. (A porous, flaky oxide layer, as in iron, is exactly what allows continued rusting.)
✅ Answer: (A)
A: Onion and vanilla extract are called olfactory indicators.
R: Their smell changes in acidic and basic media, so the change can be detected by smell.
Select the correct option:
R: Their smell changes in acidic and basic media, so the change can be detected by smell.
Select the correct option:
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
Truth of A and RBoth statements are true. Olfactory indicators are substances whose smell tells us whether a medium is acidic or basic. Onion loses its characteristic smell in a basic medium; vanilla loses its smell in a basic medium. Since the definition is the smell change, R correctly explains A.
✅ Answer: (A)
Carbon forms millions of compounds whereas silicon, which lies just below it, forms far fewer. Give two reasons for this difference.
Answer & marking scheme
Marking scheme: Each correct reason 1 mark. Total 2 marks.
Reason 1 — CatenationCarbon shows strong catenation — it forms strong, stable covalent bonds with other carbon atoms, giving long chains, branched chains and rings. The Si–Si bond is much weaker, so silicon cannot build long stable chains.
Reason 2 — Small size and multiple bondsCarbon atoms are small, so they form stable double and triple bonds with themselves, oxygen, nitrogen and sulphur. Silicon atoms are larger and cannot form such stable multiple bonds.
✅ Catenation with strong C–C bonds, and small size enabling stable multiple bonds.
Attempt either option A or option B.
Option A
Compound X has the molecular formula C2H6O and is used as a solvent in tincture iodine and cough syrups.
(i) Identify X and write its structural formula.
(ii) Name the product formed when X is heated with ethanoic acid in the presence of a few drops of concentrated H2SO4, and write the equation.
(iii) Name the functional group present in X.
(i) Identify X and write its structural formula.
(ii) Name the product formed when X is heated with ethanoic acid in the presence of a few drops of concentrated H2SO4, and write the equation.
(iii) Name the functional group present in X.
— OR — (Attempt any ONE)
Option B
Compound Y is a carboxylic acid containing two carbon atoms.
(i) Identify Y and write its structural formula.
(ii) What is observed when Y reacts with sodium hydrogencarbonate? Write the equation.
(iii) Name the functional group present in Y.
(i) Identify Y and write its structural formula.
(ii) What is observed when Y reacts with sodium hydrogencarbonate? Write the equation.
(iii) Name the functional group present in Y.
Answer & marking scheme (both options)
Marking scheme: (i) 1 · (ii) 1 · (iii) 1. Total 3 marks. Either option is marked out of 3.
Option A — answers
(i) Identify X
X is ethanol. Structural formula: CH3–CH2–OH.
(ii) Reaction with ethanoic acidAn ester is formed — ethyl ethanoate (ethyl acetate), which has a sweet smell (esterification). The reaction needs a few drops of concentrated H2SO4.
CH3COOH + C2H5OH → CH3COOC2H5 + H2O
(iii) Functional groupCH3COOH + C2H5OH → CH3COOC2H5 + H2O
The alcohol group, –OH (hydroxyl).
✅ Option A: ethanol / ethyl ethanoate / –OH.
Option B — answers
(i) Identify Y
Y is ethanoic acid (acetic acid). Structural formula: CH3–COOH.
(ii) Reaction with NaHCO3Brisk effervescence of carbon dioxide gas is observed (the gas turns lime water milky).
CH3COOH + NaHCO3 → CH3COONa + H2O + CO2
(iii) Functional groupCH3COOH + NaHCO3 → CH3COONa + H2O + CO2
The carboxylic acid group, –COOH.
✅ Option B: ethanoic acid / brisk effervescence of CO2 / –COOH.
Give a reason for each of the following.
(a) A solution of common salt conducts electricity, but a solution of sugar does not.
(b) The milkiness of lime water disappears when excess carbon dioxide is passed through it.
(c) An iron gate near the sea rusts faster than an iron gate in a dry hilly region.
(a) A solution of common salt conducts electricity, but a solution of sugar does not.
(b) The milkiness of lime water disappears when excess carbon dioxide is passed through it.
(c) An iron gate near the sea rusts faster than an iron gate in a dry hilly region.
Answer & marking scheme
Marking scheme: (a) 1 · (b) 1 · (c) 1. Total 3 marks.
(a) Conduction of salt versus sugar solutionSalt is an ionic compound. In water it dissociates into free ions (Na+ and Cl−) which carry electric current. Sugar is a covalent compound; it dissolves as neutral molecules and does not form ions, so it does not conduct.
(b) Disappearance of milkinessCarbon dioxide first turns lime water milky by forming insoluble calcium carbonate. When excess CO2 is passed, this is converted into soluble calcium hydrogencarbonate, so the milkiness disappears.
(c) Rusting near the seaSea air contains a high amount of moisture and dissolved salts. Both moisture and salt speed up the electrochemical process of rusting. A dry hilly region lacks this moisture, so rusting there is much slower.
✅ (a) ions versus molecules · (b) soluble hydrogencarbonate formed · (c) moisture and salts speed up rusting.
Case: A student dips a clean iron nail into a copper sulphate solution and leaves it for 30 minutes. She observes that a reddish-brown deposit forms on the nail and the blue colour of the solution fades.
(a) Name the type of reaction taking place.
(b) Write the balanced chemical equation for the reaction.
(c) Explain the observation using the reactivity series.
(d) Which substance is oxidised and which is reduced?
(a) Name the type of reaction taking place.
(b) Write the balanced chemical equation for the reaction.
(c) Explain the observation using the reactivity series.
(d) Which substance is oxidised and which is reduced?
Answer & marking scheme
Marking scheme: (a) 1 · (b) 1 · (c) 1 · (d) 1. Total 4 marks.
(a) Type of reactionIt is a displacement reaction (a more reactive metal displaces a less reactive metal from its salt solution).
(b) Balanced equationFe + CuSO4 → FeSO4 + Cu
(c) Explanation using the reactivity seriesIn the reactivity series, iron is more reactive than copper. Iron therefore displaces copper from copper sulphate. The displaced copper deposits as a reddish-brown layer on the nail, and as blue CuSO4 changes to pale-green FeSO4, the blue colour fades.
(d) Oxidation and reductionIron is oxidised (it loses electrons and goes into solution as Fe2+). Copper is reduced (Cu2+ gains electrons to form copper metal).
✅ (a) displacement · (b) Fe + CuSO4 → FeSO4 + Cu · (c) iron more reactive than copper · (d) Fe oxidised, Cu2+ reduced.
Attempt either option A or option B.
Option A
(i) In the reaction ZnO + C → Zn + CO, identify the substance that is oxidised and the substance that is reduced. Name the oxidising and the reducing agent.
(ii) Why is respiration considered an exothermic reaction?
(iii) Balance the equation: Fe + H2O → Fe3O4 + H2.
(ii) Why is respiration considered an exothermic reaction?
(iii) Balance the equation: Fe + H2O → Fe3O4 + H2.
— OR — (Attempt any ONE)
Option B
(i) Define a redox reaction and give one example.
(ii) Why is the reaction between silver nitrate and sodium chloride called a precipitation reaction? Write the balanced equation.
(iii) Classify the reaction: CaCO3 (on heating) → CaO + CO2.
(ii) Why is the reaction between silver nitrate and sodium chloride called a precipitation reaction? Write the balanced equation.
(iii) Classify the reaction: CaCO3 (on heating) → CaO + CO2.
Answer & marking scheme (both options)
Marking scheme: (i) 2 · (ii) 1.5 · (iii) 1.5. Total 5 marks. Either option is marked out of 5.
Option A — answers
(i) Oxidation, reduction and the agents
Carbon (C) gains oxygen to form CO, so carbon is oxidised and acts as the reducing agent. ZnO loses oxygen to form Zn, so zinc oxide is reduced and acts as the oxidising agent.
(ii) Respiration is exothermicDuring respiration, glucose is broken down and energy is released (as heat and ATP). Since energy is given out, it is an exothermic reaction.
(iii) Balanced equation3Fe + 4H2O → Fe3O4 + 4H2
✅ Option A: C oxidised (reducing agent), ZnO reduced (oxidising agent) · energy released · 3Fe + 4H2O → Fe3O4 + 4H2.
Option B — answers
(i) Redox reaction
A reaction in which oxidation and reduction occur simultaneously is called a redox reaction. Example: ZnO + C → Zn + CO.
(ii) Precipitation reactionSilver nitrate reacts with sodium chloride to form an insoluble white precipitate of silver chloride. Any reaction that produces an insoluble solid (precipitate) is called a precipitation reaction.
AgNO3 + NaCl → AgCl + NaNO3
AgCl is the insoluble white precipitate.
(iii) ClassificationAgNO3 + NaCl → AgCl + NaNO3
AgCl is the insoluble white precipitate.
CaCO3 breaks down into two simpler substances on heating, so it is a thermal decomposition reaction.
✅ Option B: simultaneous oxidation and reduction · insoluble AgCl precipitate · thermal decomposition.
Section C — PhysicsQ30–Q39 · 25 marks
A convex lens is used to obtain a sharp image of a distant tree on a screen. The distance between the lens and the screen gives:
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
ReasoningA very distant object sends parallel rays to the lens. A convex lens brings parallel rays to a focus at the principal focus. The image of a distant object therefore forms at the focus, and the lens-to-screen distance equals the focal length.
✅ Answer: (B)
An electric heater is rated 1500 W. What current does it draw when connected to a 250 V supply?
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
WorkingP = V × I
1500 = 250 × I
I = 1500 / 250
I = 6 A
1500 = 250 × I
I = 1500 / 250
I = 6 A
✅ Answer: (C) 6 A
A: The magnetic field inside a long current-carrying solenoid is uniform.
R: The magnetic field lines inside a long solenoid are parallel straight lines spaced equally.
Select the correct option:
R: The magnetic field lines inside a long solenoid are parallel straight lines spaced equally.
Select the correct option:
Answer & marking scheme
Marking scheme: Correct option 1 mark (no partial).
Truth of A and RBoth are true, and R explains A. A field is uniform when its lines are parallel, equally spaced and in the same direction. Inside a long solenoid the field lines have exactly this pattern, so the field strength is the same at every point inside.
✅ Answer: (A)
An object is placed 20 cm in front of a concave mirror of focal length 10 cm. Find the position of the image.
Answer & marking scheme
Marking scheme: Correct formula and signs 1 · substitution and answer 1. Total 2 marks.
Sign conventionObject is on the left, so u = −20 cm. Concave mirror, so f = −10 cm. Using the mirror formula:
Mirror formula1/v + 1/u = 1/f
1/v = 1/f − 1/u
1/v = (1/−10) − (1/−20)
1/v = −1/10 + 1/20
1/v = (−2 + 1)/20
1/v = −1/20
v = −20 cm
Interpretation1/v = 1/f − 1/u
1/v = (1/−10) − (1/−20)
1/v = −1/10 + 1/20
1/v = (−2 + 1)/20
1/v = −1/20
v = −20 cm
v is negative, so the image forms 20 cm in front of the mirror (on the same side as the object). The object is at the centre of curvature, so the image is real, inverted and of the same size.
✅ Image at 20 cm from the mirror, on the same side as the object; real, inverted, same size.
Two resistors of 6 Ω and 3 Ω are connected in parallel. Calculate the equivalent resistance of the combination.
Answer & marking scheme
Marking scheme: Formula 1 · answer with unit 1. Total 2 marks.
Formula for parallel resistors1/Rp = 1/R1 + 1/R2
Substitution1/Rp = 1/6 + 1/3
1/Rp = 1/6 + 2/6
1/Rp = 3/6
1/Rp = 1/2
Rp = 2 Ω
Check1/Rp = 1/6 + 2/6
1/Rp = 3/6
1/Rp = 1/2
Rp = 2 Ω
The equivalent resistance (2 Ω) is less than the smallest resistance (3 Ω), which is always true for a parallel combination.
✅ Equivalent resistance = 2 Ω
A 4 Ω resistor is connected across a 12 V battery. Calculate:
(i) the current through the resistor,
(ii) the power dissipated in the resistor,
(iii) the electrical energy consumed in 5 minutes.
(i) the current through the resistor,
(ii) the power dissipated in the resistor,
(iii) the electrical energy consumed in 5 minutes.
Answer & marking scheme
Marking scheme: (i) 1 · (ii) 1 · (iii) 1. Total 3 marks.
(i) CurrentV = I × R
I = V / R
I = 12 / 4
I = 3 A
(ii) PowerI = V / R
I = 12 / 4
I = 3 A
P = V × I
P = 12 × 3
P = 36 W
Check: P = I²R = 9 × 4 = 36 W ✓
(iii) Energy in 5 minutesP = 12 × 3
P = 36 W
Check: P = I²R = 9 × 4 = 36 W ✓
Time t = 5 × 60 = 300 s
E = P × t
E = 36 × 300
E = 10800 J (or 1.08 × 104 J)
E = P × t
E = 36 × 300
E = 10800 J (or 1.08 × 104 J)
✅ (i) 3 A · (ii) 36 W · (iii) 10800 J
(a) State Snell’s law of refraction.
(b) The refractive index of glass is 1.5 and that of water is 1.33. In which medium does light travel faster, and why?
(c) Define the absolute refractive index of a medium.
(b) The refractive index of glass is 1.5 and that of water is 1.33. In which medium does light travel faster, and why?
(c) Define the absolute refractive index of a medium.
Answer & marking scheme
Marking scheme: (a) 1 · (b) 1 · (c) 1. Total 3 marks.
(a) Snell’s lawThe ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media and a given colour of light:
sin i / sin r = constant = n21
(b) Faster mediumsin i / sin r = constant = n21
Light travels faster in water. A higher refractive index means the medium is optically denser and slows light down more. Since glass (1.5) has a higher refractive index than water (1.33), light is slowed more in glass and travels faster in water.
(c) Absolute refractive indexThe absolute refractive index of a medium is the ratio of the speed of light in vacuum (or air) to the speed of light in that medium:
n = c / v
n = c / v
✅ (a) sin i / sin r constant · (b) water, because it has the lower refractive index · (c) n = c / v.
An electric current flows through a long straight vertical conductor.
(a) State the rule used to find the direction of the magnetic field produced around it.
(b) Describe the pattern of the magnetic field lines around the conductor.
(c) How does the strength of the field change as we move away from the conductor?
(a) State the rule used to find the direction of the magnetic field produced around it.
(b) Describe the pattern of the magnetic field lines around the conductor.
(c) How does the strength of the field change as we move away from the conductor?
Answer & marking scheme
Marking scheme: (a) 1 · (b) 1 · (c) 1. Total 3 marks.
(a) The ruleRight-hand thumb rule: if you hold the conductor in your right hand with the thumb pointing along the direction of the current, the direction in which your fingers curl gives the direction of the magnetic field lines.
(b) Pattern of field linesThe field lines form concentric circles around the conductor, in a plane perpendicular to it. The circles are stronger (closer together) near the conductor.
(c) Change with distanceThe strength of the magnetic field is inversely proportional to the distance from the conductor: as we move away, the field becomes weaker and the field lines become farther apart.
✅ (a) right-hand thumb rule · (b) concentric circles · (c) strength decreases with distance.
Case: A household operates three appliances at 220 V: a 100 W bulb, a 60 W fan and a 1000 W electric iron.
(a) Calculate the current drawn by the iron.
(b) Which appliance has the highest resistance? Justify with a calculation.
(c) If all three appliances run together for 2 hours, calculate the total electrical energy consumed in kWh.
(a) Calculate the current drawn by the iron.
(b) Which appliance has the highest resistance? Justify with a calculation.
(c) If all three appliances run together for 2 hours, calculate the total electrical energy consumed in kWh.
Answer & marking scheme
Marking scheme: (a) 1 · (b) 1.5 · (c) 1.5. Total 4 marks.
(a) Current drawn by the ironP = V × I
I = P / V
I = 1000 / 220
I = 4.55 A (approximately)
(b) Highest resistanceI = P / V
I = 1000 / 220
I = 4.55 A (approximately)
R = V² / P, so for the same voltage, the lowest power appliance has the highest resistance. That is the 60 W fan.
R = 220² / 60
R = 48400 / 60
R = 806.7 Ω
For comparison: bulb R = 48400/100 = 484 Ω; iron R = 48400/1000 = 48.4 Ω.
(c) Total energy for 2 hoursR = 220² / 60
R = 48400 / 60
R = 806.7 Ω
For comparison: bulb R = 48400/100 = 484 Ω; iron R = 48400/1000 = 48.4 Ω.
Total power = 100 + 60 + 1000 = 1160 W = 1.16 kW
Energy = Power (kW) × time (h)
E = 1.16 × 2
E = 2.32 kWh
Energy = Power (kW) × time (h)
E = 1.16 × 2
E = 2.32 kWh
✅ (a) 4.55 A · (b) 60 W fan, 806.7 Ω · (c) 2.32 kWh
(a) Draw ray diagrams to show the image formed by a concave mirror when the object is placed (i) between the pole and the focus, and (ii) beyond the centre of curvature.
(b) State the mirror formula and the sign convention used with it.
(c) An object 5 cm tall is placed 15 cm in front of a concave mirror of focal length 10 cm. Find the position, nature and size of the image.
(b) State the mirror formula and the sign convention used with it.
(c) An object 5 cm tall is placed 15 cm in front of a concave mirror of focal length 10 cm. Find the position, nature and size of the image.
Answer & marking scheme
Marking scheme: (a) two ray diagrams 2 (1 each) · (b) formula and sign convention 1 · (c) numerical 2. Total 5 marks.
(a) Ray diagrams — expected features
(i) Object between pole and focus: the image is formed behind the mirror, is virtual, erect and enlarged. The two reflected rays appear to diverge from a point behind the mirror, so the rays are shown extended backwards with dotted lines.
(ii) Object beyond the centre of curvature: the image forms between the focus and the centre of curvature, and is real, inverted and diminished. The two reflected rays actually meet in front of the mirror.
Draw the principal axis, mark P, F and C, and show at least two rays: one parallel to the axis reflecting through F, and one through C reflecting back along itself. Mark arrow directions on all rays.
(b) Mirror formula and sign convention(ii) Object beyond the centre of curvature: the image forms between the focus and the centre of curvature, and is real, inverted and diminished. The two reflected rays actually meet in front of the mirror.
Draw the principal axis, mark P, F and C, and show at least two rays: one parallel to the axis reflecting through F, and one through C reflecting back along itself. Mark arrow directions on all rays.
1/v + 1/u = 1/f
Sign convention (New Cartesian): all distances are measured from the pole; distances measured in the direction of the incident light (to the right) are positive, and those against it (to the left) are negative. Heights above the principal axis are positive, below are negative. For a concave mirror, f is negative.
(c) Numerical — given dataSign convention (New Cartesian): all distances are measured from the pole; distances measured in the direction of the incident light (to the right) are positive, and those against it (to the left) are negative. Heights above the principal axis are positive, below are negative. For a concave mirror, f is negative.
Object height h = +5 cm
Object distance u = −15 cm
Focal length f = −10 cm
(c) Position of the imageObject distance u = −15 cm
Focal length f = −10 cm
1/v = 1/f − 1/u
1/v = (1/−10) − (1/−15)
1/v = −1/10 + 1/15
1/v = (−3 + 2)/30
1/v = −1/30
v = −30 cm
(c) Magnification and size1/v = (1/−10) − (1/−15)
1/v = −1/10 + 1/15
1/v = (−3 + 2)/30
1/v = −1/30
v = −30 cm
m = −v / u
m = −(−30) / (−15)
m = 30 / (−15)
m = −2
Image height h′ = m × h = −2 × 5 = −10 cm
(c) Nature of the imagem = −(−30) / (−15)
m = 30 / (−15)
m = −2
Image height h′ = m × h = −2 × 5 = −10 cm
v is negative, so the image is real and forms 30 cm in front of the mirror. m is negative and greater than 1 in magnitude, so the image is inverted and magnified (10 cm tall).
✅ (c) Image 30 cm in front of the mirror; real, inverted and magnified 2× (10 cm tall).
Source and accuracy: this paper is built to the official CBSE Sample Question Paper and Marking Scheme for Class X, session 2026-27, published by the CBSE Academics Unit.
View the official CBSE SQP page.
Not affiliated with CBSE. Pandey Sir’s BEST Classes is an independent coaching channel. This is an original mock test written to the published blueprint. It is not published, endorsed or approved by the Central Board of Secondary Education. CBSE names and paper patterns are referenced for study purposes only.
Items to verify: the exact chapter-wise marks split within each unit is not published by CBSE; only unit-wise weightage is official. Time allocations per section are our suggestions, not CBSE’s.
Not affiliated with CBSE. Pandey Sir’s BEST Classes is an independent coaching channel. This is an original mock test written to the published blueprint. It is not published, endorsed or approved by the Central Board of Secondary Education. CBSE names and paper patterns are referenced for study purposes only.
Items to verify: the exact chapter-wise marks split within each unit is not published by CBSE; only unit-wise weightage is official. Time allocations per section are our suggestions, not CBSE’s.
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