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Class 10 Science competency-based questions 2026-27: what CBSE asks, with 8 of our own

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"Competency-based question" is the name CBSE uses for a question that gives you a situation, a table, a graph or an experiment and asks you to use what you learnt, rather than repeat a definition. This guide explains what CBSE says about these questions for 2026-27, shows where they sit in the official Class 10 Science sample paper, and then gives 8 practice questions of our own, two each from four chapters, with worked answers and the NCERT page each answer rests on.

The 8 questions are our own, drafted with AI help. Every number was re-computed by a script and every statement was checked against the NCERT chapter, but no practising teacher has reviewed them yet. The data in the questions is made up for practice unless a page number says it comes from the book. If you find a mistake, please tell us.

What CBSE means by competency-based

CBSE's secondary curriculum for 2026-27 says board papers are designed to test application, analysis and evaluation, and that at present about 50% of the questions are competency-focused. It lists the kinds it means: questions built on a case, a source or a data table, situation and application questions, and integrated questions. The rest are multiple-choice and short or long answers (Secondary Curriculum 2026-27, Part 1, section 3.1, PDF page 24). The same section, on PDF page 25, says the papers use real-life contexts, diagrams and passages.

The Class 10 Science curriculum adds two things (Science curriculum 2026-27, PDF page 9):

The same page lists the question types as very short answers (including objective and assertion-reason questions), short answers, long answers, and source-based, case-based, passage-based or integrated questions, with an internal choice in about 33% of the paper.

So "competency-based" is not a separate chapter or a separate paper. It is a way of asking about the same syllabus. A question on electric power can ask you to state the formula (knowledge) or to work out which of two fans is cheaper to run (application).

Where they sit in the 2026-27 sample paper

The 2026-27 Science sample paper has one 4-mark case-based question in each section: Biology, Chemistry and Physics (sample paper, PDF pages 3-4, 9 and 13-14). Each one starts with a short situation, then asks two 1-mark parts and lets you choose one of two 2-mark parts. The marking scheme gives marks part by part in the same way (marking scheme, PDF pages 2, 5 and 9).

QuestionSituation (in our words)What it asks you to do
Q15, BiologyA student crosses pea plants for flower colour, then for colour and height togetherWork out the percentage of plants with given traits in F1 and F2, and explain why
Q28, ChemistryRods of five metals are heated at one end and wax-fixed candles drop off at different times, shown on a graphRead the graph, place a new metal (gold) on it, predict what a longer or thicker rod would change
Q38, PhysicsA school door lock uses a solenoid that pulls an iron boltExplain the iron core, find the poles, then either predict what more current does or say how to make the field stronger

Competency questions are not only the 4-mark cases. Several 1-mark questions in the same paper are set in a situation too: a chameleon changing colour (is it a physical or chemical change?) and a student choosing between soap and detergent for hard water (sample paper, PDF page 5).

How to answer them

Acids, Bases and Salts (Chapter 2)

Question 1. A dentist's poster shows how the pH at the surface of a child's teeth changed after the child ate a sweet (made-up numbers for this question): before eating 7.0; after 5 minutes 5.0; after 20 minutes 5.4; after 40 minutes 6.2; after 60 minutes 6.8. (a) At which of these readings were the teeth at risk of decay? (b) Why does the pH drop after eating a sweet? (c) The child wants to shorten the risky time. What should they do, and what does it do chemically?

Answer: (a) at 5 and 20 minutes; (b) acids made from the sugar left in the mouth; (c) clean the teeth with toothpaste soon after eating, because the basic toothpaste neutralises the acid. (a) The chapter puts the danger line at pH 5.5: below it, the enamel of the teeth starts to decay (page 27). Only the 5-minute reading (5.0) and the 20-minute reading (5.4) are below 5.5; at 40 minutes (6.2) the mouth has come back above it. The 5-minute reading is the more acidic of the two, because a lower pH means more hydronium ions (page 25). (b) Bacteria in the mouth break down sugar and food particles left after eating, and this makes acids (page 27). (c) Cleaning the mouth after eating is the best way to prevent this, and toothpastes, which are generally basic, neutralise the excess acid (page 27). That is a neutralisation of an acid by a base.

Question 2. Two batches of dhokla batter are steamed. Batch A has only baking soda. Batch B has baking powder instead. Both rise. (a) Name the gas that makes both rise, and describe a test you could do with it in the lab. (b) Write the equation for what happens to baking soda when it is heated in Batch A, and name the solid it leaves behind. (c) Baking powder also contains tartaric acid. What does the acid do?

Answer: (a) carbon dioxide, which turns lime water milky; (b) 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂, leaving sodium carbonate; (c) it reacts with the baking soda to give carbon dioxide. (a) Bubbled through lime water (calcium hydroxide solution), carbon dioxide forms a white precipitate of calcium carbonate, which makes the lime water look milky (page 20). The carbon dioxide is what makes a cake or bread rise and turn soft and spongy (page 31). (b) When baking soda (sodium hydrogencarbonate) is heated during cooking, it gives sodium carbonate, water and carbon dioxide (page 31). So Batch A rises, but sodium carbonate stays in the food. (c) Baking powder is baking soda mixed with tartaric acid, a weak edible acid. When it is heated or mixed with water, the hydrogen ions from the acid react with the sodium hydrogencarbonate to give carbon dioxide, water and the sodium salt of the acid (page 31). The gas then comes from an acid-base reaction, not only from heating.

Life Processes (Chapter 5)

Question 3. A student tests two digestive enzymes, X and Y, at different pH values and records how fast each works (made-up numbers; 10 = fastest). Enzyme X: pH 2, 10; pH 4, 6; pH 7, 1; pH 8, 0. Enzyme Y: pH 2, 0; pH 4, 1; pH 7, 7; pH 8, 10. (a) Which enzyme behaves like pepsin, and in which organ does pepsin work? (b) Which enzyme behaves like the pancreatic enzyme trypsin? (c) The food that leaves the stomach is acidic. What changes it so that Y-type enzymes can work, and where does that come from? (d) Why is the stomach's own lining not damaged by the acid?

Answer: (a) X, in the stomach; (b) Y; (c) bile juice from the liver makes it alkaline; (d) mucus protects the lining. (a) X works best at a very low pH, like pepsin. The gastric glands in the stomach wall release hydrochloric acid, which creates the acidic medium that pepsin needs (page 85). (b) Y works best in a slightly alkaline medium, like the pancreatic enzymes. The pancreas secretes trypsin for proteins and lipase for fats into the small intestine (page 86). (c) What leaves the stomach is still acidic, and pancreatic enzymes cannot work in an acid. The liver's bile juice turns it alkaline, and it also works on fats (page 86). (d) The gastric glands also release mucus, which coats the stomach wall from inside and shields it from the acid (page 85).

Question 4. Aarav sprints 200 m as fast as he can and gets a painful cramp in his leg. The next day he jogs slowly for 20 minutes and has no cramp. (a) What happened to the pyruvate in his leg muscles during the sprint, and why did it cause the cramp? (b) In both runs, where in the muscle cell is glucose first broken down, and into what? (c) During the slow jog, where is the pyruvate broken down, and what are the products? (d) Which run made more energy available per glucose molecule? Explain in one line.

Answer: (a) it turned into lactic acid, whose build-up causes cramps; (b) in the cytoplasm, into pyruvate; (c) in the mitochondria, into carbon dioxide and water; (d) the jog, because aerobic respiration makes more energy available. (a) In a sudden burst of activity the muscle cells run short of oxygen, so pyruvate takes another pathway and turns into lactic acid, a three-carbon molecule. The build-up of lactic acid in the muscles causes cramps (page 88). (b) Whichever pathway follows, glucose (six carbon atoms) is first split into pyruvate (three carbon atoms), and that split happens in the cytoplasm (page 87). (c) With enough oxygen, pyruvate is broken down in the mitochondria into carbon dioxide and water; this is aerobic respiration (pages 87 and 88). (d) The chapter summary says the aerobic route, which uses oxygen, gives the organism more energy than the routes without oxygen (page 99).

Light: Reflection and Refraction (Chapter 9)

Question 5. A student uses one convex lens and records where a sharp image forms for four object positions, with distances in cm measured from the optical centre and signed by the textbook's convention. Reading 1: u = −30, v = +60. Reading 2: u = −40, v = +40. Reading 3: u = −60, v = +30. Reading 4: u = −25, v = +60. (a) Use reading 1 to find the focal length. (b) One reading was copied wrongly. Which one, and what should v have been? (c) Find the power of the lens. (d) Describe the image in reading 1.

Answer: (a) f = +20 cm; (b) reading 4, where v should be +100 cm; (c) +5 D; (d) real, inverted and twice the size of the object. (a) The lens formula is 1/v − 1/u = 1/f (page 155). For reading 1: 1/60 − 1/(−30) = 1/60 + 2/60 = 3/60 = 1/20, so f = +20 cm. A positive focal length fits a convex lens (page 155). (b) Readings 2 and 3 also give 1/20 (1/40 + 1/40 = 2/40 and 1/30 + 1/60 = 3/60). Reading 4 gives 1/60 + 1/25 = 17/300, which would mean f ≈ 17.6 cm, so it does not fit. For u = −25 cm and f = +20 cm: 1/v = 1/20 − 1/25 = 1/100, so v = +100 cm. (c) Power is the reciprocal of the focal length in metres: P = 1/0.20 m = +5 D, positive for a convex lens (page 158). (d) Magnification m = v/u = 60/(−30) = −2 (page 156). The negative sign means the image is real and inverted, and 2 means it is twice the height of the object; v is positive, so it forms on the other side of the lens (pages 156 and 157).

Question 6. A materials lab tests two clear layers for a phone screen (made-up data). It finds that light travels at 1.6 × 10⁸ m/s in layer P and at 2.4 × 10⁸ m/s in layer Q. Take the speed of light in air as 3 × 10⁸ m/s. (a) Find the refractive index of each layer. (b) Which layer is optically denser? (c) Find the refractive index of P with respect to Q. (d) Light from the display passes obliquely from P into Q on its way out. At the P–Q boundary, does the ray bend towards or away from the normal, and does the light speed up or slow down?

Answer: (a) P 1.875, Q 1.25; (b) P; (c) 1.5; (d) away from the normal, and it speeds up. (a) A medium's refractive index is the speed of light in air divided by the speed of light in that medium, n = c/v (page 149). P: 3 ÷ 1.6 = 1.875. Q: 3 ÷ 2.4 = 1.25. (b) Of two media, the one with the higher refractive index counts as optically denser (page 149), so P. (c) The refractive index of one medium with respect to another is the speed of light in the second medium divided by its speed in the first; the textbook writes this as n₂₁ = v₁/v₂ (page 148). For P with respect to Q: v(Q) ÷ v(P) = 2.4 ÷ 1.6 = 1.5. As a check, 1.875 ÷ 1.25 is also 1.5. (d) The light leaves the denser layer P for the rarer layer Q, so it gains speed (from 1.6 to 2.4 × 10⁸ m/s) and the ray turns away from the normal (page 149).

Electricity (Chapter 11)

Question 7. A family must replace a ceiling fan that runs about 10 hours a day. Fan A is rated 75 W and costs ₹1,800. Fan B is rated 30 W and costs ₹3,150 (made-up prices). Electricity costs ₹8 per unit (our example; real tariffs differ by state). (a) How many units does each fan use in a day? (b) How much does Fan B save per day? (c) After how many days has Fan B's saving paid back its higher price? (d) At 220 V, what current does each fan draw?

Answer: (a) Fan A 0.75 units, Fan B 0.3 units; (b) ₹3.60; (c) 375 days, about a year; (d) about 0.34 A and 0.14 A. (a) The commercial unit of electrical energy is the kilowatt hour: 1 kW used for 1 hour (page 191). Fan A: 0.075 kW × 10 h = 0.75 kW h. Fan B: 0.030 kW × 10 h = 0.3 kW h. (b) Fan B uses 0.45 units less a day, and 0.45 × ₹8 = ₹3.60. (c) Fan B costs ₹3,150 − ₹1,800 = ₹1,350 more. ₹1,350 ÷ ₹3.60 = 375 days. After that, it keeps saving about ₹1,314 a year (₹3.60 × 365). (d) P = VI (page 191), so I = P/V. Fan A: 75 ÷ 220 ≈ 0.34 A. Fan B: 30 ÷ 220 ≈ 0.14 A. The fan with the lower power rating draws less current from the same supply.

Question 8. For a science-fair model house, Riya has a 9 V battery and two identical bulbs, each of 18 Ω. (a) If she joins the bulbs in series, find the current and the power of each bulb. (b) If she joins them in parallel, find the current through each bulb, the total current and the power of each bulb. (c) In which wiring do the bulbs glow brighter, and what happens in each wiring if one bulb fuses? (d) How many times faster does the parallel wiring take energy from the battery?

Answer: (a) 0.25 A, 1.125 W each; (b) 0.5 A each, 1 A in total, 4.5 W each; (c) brighter in parallel; in series both go out, in parallel the other stays lit; (d) 4 times. (a) In series the resistances add: 18 + 18 = 36 Ω (page 184). I = V/R = 9/36 = 0.25 A, and each bulb uses P = I²R = 0.25² × 18 = 1.125 W (page 191). (b) In parallel each bulb gets the full 9 V, so each carries 9/18 = 0.5 A, and the currents add to 1 A (page 186). The equivalent resistance is 9 Ω (1/R = 1/18 + 1/18). Each bulb uses P = V²/R = 81/18 = 4.5 W. (c) 4.5 W is four times 1.125 W, so the parallel bulbs glow brighter. In a series circuit, when one part fails the circuit is broken and nothing works; a parallel circuit divides the current between the branches, so the other bulb keeps its own path (page 187). (d) Total power in series: 2 × 1.125 = 2.25 W. In parallel: 2 × 4.5 = 9 W. 9 ÷ 2.25 = 4, so the battery runs down about four times as fast. This is one reason homes are wired in parallel and each appliance gets a fixed voltage.

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Sources

All opened and read on 3 November 2026. CBSE page numbers are PDF pages; NCERT page numbers are the printed page numbers of the textbook.

The board-pattern facts are from CBSE's documents above, and the science in each answer is from the textbook pages named. The situations, data, questions and worked answers are our own, and every number was re-computed by script. This guide is general information; follow your teacher, the textbook and CBSE's own notices.

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