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AMIT SIR · CBSE CLASS 10 · BOARD EXAM 2027
Class 10 Science: 15 Sample Papers
Turn every mock into better preparation. Attempt an original board-style paper, check the method behind each answer, and use your mistakes to guide your next revision.
80marks per paper3 hoursfull-paper practice39questions in Paper 01
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PRACTICE THAT LEADS SOMEWHERE
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Understand the method, not just the answer.
Open the worked solutions after your attempt. Compare your reasoning, calculations, diagrams and presentation with the suggested marking guidance.
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Practise weak Science chapters ↗What your ₹399 purchase includes
Start with Science Paper 01 today
- 39 original questions across Biology, Chemistry and Physics
- 80 marks · 3-hour full-paper attempt · all 13 Science chapters represented
- Worked solutions, balanced equations, numericals and labelled diagrams
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Practise the full Science paper
- Biology: 30 marks; Chemistry: 25 marks; Physics: 25 marks
- Objective questions, assertion–reason, written answers and case studies
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- CBSE Class 10 Science (086) · 2026–27 session · Board Exam 2027
- Paper 01 and its solutions are a free complete preview
- Online digital book; no printed copy is shipped
Original Science practice questions by Amit Sir, using the 39-question, three-subject-section structure of the CBSE 2026–27 Science sample paper. The worked solutions give suggested marking guidance for self-review. This is an original practice paper, not an official CBSE paper.
See the official CBSE paper pattern ↗PAPER 01 · FREE COMPLETE PREVIEW · 2026–27 SESSION
Original Science Practice Paper 01
Science (086) · Class 10
Maximum marks: 80Time: 3 hours
Before you begin
- Attempt all 39 numbered questions. Section A is Biology (30 marks), Section B Chemistry (25 marks), and Section C Physics (25 marks).
- Use a separate answer sheet and allow 3 hours. Keep solutions closed until you finish your attempt.
- Internal choices are provided in Q11, Q16, Q26, Q29, Q34 and Q39, and in part (iii) of Q15, Q28 and Q38. Attempt one alternative wherever a choice is given.
- For assertion–reason questions, choose from the four options printed below that question.
- Show formulae, substitutions, units and reasoning in calculations. Write balanced chemical equations with required conditions. Draw neat labelled diagrams where asked.
- Use the Cartesian sign convention for mirrors and lenses. Calculators are not allowed. Figures are schematic and are not to scale.
For a blank paper, leave solutions closed before printing. Solutions show a suggested marking breakdown for self-review.
Section A · Biology
Questions 1–16 · 30 marks. Q1–7 are MCQs and Q8–9 are assertion–reason. Internal choices appear in Q11, Q15(iii) and Q16.
Three tubes contain equal starch solution. P receives fresh saliva; Q receives boiled and cooled saliva; R receives water. All are kept near 37°C for the same period. Which tube is most likely to lose the blue-black iodine-test response first?
Answer & suggested solution
A. P
- Salivary amylase in fresh saliva breaks down starch into simpler sugars.
- Boiling inactivates the enzyme; water alone does not supply amylase. P therefore loses starch first.
A cut leafy shoot stands in coloured water. Later, the colour appears in the stem tissue that transports water towards the leaves. Which tissue is stained?
Answer & suggested solution
B. Xylem
- Xylem conducts water and dissolved minerals upwards through the shoot.
- The coloured water follows that pathway, staining the xylem.
Which change would most directly reduce efficient gas exchange at the alveoli?
Answer & suggested solution
C. Marked thickening of the alveolar walls
- Gas exchange depends on diffusion across a thin barrier.
- Thicker walls increase the diffusion distance, slowing exchange.
In a simple complete-dominance model, round seed shape R is dominant to wrinkled r. What is the expected F1 from RR × rr?
Answer & suggested solution
D. All Rr and round
- The RR parent produces R gametes and the rr parent produces r gametes.
- All F1 offspring are Rr and express the dominant round phenotype.
After a carbohydrate-rich meal, which hormone normally helps bring increased blood glucose towards its usual range?
Answer & suggested solution
A. Insulin
- Insulin is secreted by the pancreas.
- It helps glucose uptake and storage, reducing an elevated blood-glucose concentration.
A grower wants many potato plants that retain the useful characteristics of one selected parent. Which is the most suitable starting material?
Answer & suggested solution
B. Tuber pieces containing healthy buds
- Potato tuber buds can grow into new plants by vegetative propagation.
- This asexual method usually preserves the selected parent’s genetic characteristics, apart from possible mutations.
Which statement correctly explains the protective importance of stratospheric ozone?
Answer & suggested solution
C. It absorbs much harmful ultraviolet radiation.
- The ozone layer absorbs a substantial fraction of harmful ultraviolet radiation.
- This limits exposure at Earth’s surface; it does not eliminate all radiation or pollution.
Assertion (A): Sugar made in a leaf can be transported to a growing root through phloem. Reason (R): Phloem translocates soluble products of photosynthesis from source regions to sinks.
Answer & suggested solution
A. Both A and R are true, and R correctly explains A.
- A growing root can act as a sink for food made in leaves.
- Translocation through phloem explains this movement of sugars.
Assertion (A): Pollination and fertilisation are the same event in a flowering plant. Reason (R): Fertilisation involves fusion of a male gamete with the egg to form a zygote.
Answer & suggested solution
D. A is false, but R is true.
- Pollination is transfer of pollen to a stigma; fertilisation is gamete fusion.
- The assertion is false, but the reason gives the correct meaning of fertilisation.
Distinguish breathing from cellular respiration, giving one role of each in supplying energy to an active muscle.
Answer & suggested solution
Breathing exchanges gases; cellular respiration releases usable energy from food.
- [1 mark] Breathing moves air into and out of the lungs, helping supply oxygen and remove carbon dioxide.
- [1 mark] Cellular respiration is a set of reactions inside cells that releases energy from food; muscle cells use that energy for activity.
A hand is withdrawn quickly after touching a hot surface. Trace the signal from a skin receptor to the responding muscle, and state where the reflex is coordinated.
Answer & suggested solution
Receptor → sensory neuron → spinal cord → motor neuron → muscle; the spinal cord coordinates the reflex.
- [1 mark] A sensory neuron carries the signal from the receptor to the spinal cord, where the circuit connects to a motor neuron, usually through an interneuron.
- [1 mark] The motor neuron activates the effector muscle. Information also reaches the brain, but conscious interpretation need not precede withdrawal.
OR · attempt one alternative
A young shoot receives light mainly from one side. Explain how unequal auxin distribution can make the shoot bend towards the light.
Answer & suggested solution
More elongation on the shaded side bends the shoot towards light.
- [1 mark] In the standard shoot-phototropism model, auxin accumulates more on the shaded side.
- [1 mark] Auxin promotes greater cell elongation there, producing curvature towards the light.
A pure round-seeded RR pea plant is crossed with a pure wrinkled-seeded rr plant. State the gametes of the parents and the genotype and phenotype of F1, assuming complete dominance.
Answer & suggested solution
R and r gametes; all F1 are Rr and round.
- [1 mark] The RR parent contributes R; the rr parent contributes r.
- [1 mark] Every F1 combination is Rr. R is dominant, so the seeds are round.
In a healthy nephron, explain (i) why blood cells do not normally enter the filtrate, (ii) why filtered glucose is normally absent or very low in urine, and (iii) how water reabsorption relates to the body’s needs.
Answer & suggested solution
The filtration barrier retains blood cells; glucose is reabsorbed; water reabsorption varies with the body’s requirements.
- [1 mark] The glomerular filtration barrier allows water and small solutes through while normally retaining blood cells and most large proteins.
- [1 mark] Useful glucose is selectively reabsorbed from the tubule into blood.
- [1 mark] The extent of water reabsorption is adjusted according to water availability and the amount of dissolved waste to be removed.
Two heterozygous round-seeded Rr pea plants are crossed. (i) List the four equally likely genotype combinations. (ii) Predict the expected number of wrinkled offspring among 80 plants. (iii) Explain why the actual count may differ.
Answer & suggested solution
RR, Rr, Rr, rr; expected wrinkled count 20; random fertilisation can cause deviations.
- [1 mark] Each parent produces R and r gametes. The combinations are RR, Rr, Rr and rr.
- [1 mark] Wrinkled plants are rr, with probability 1/4; 80 × 1/4 = 20.
- [1 mark] The ratio describes probabilities over many offspring. A finite sample can differ through chance.
A pond food chain is algae → insect larvae → small fish → fish-eating bird. Algae store 25,000 J of energy. Use the approximate 10% transfer model for parts (i) and (ii). A persistent non-biodegradable pesticide enters the pond.
(i) [1] Estimate the energy available to insect larvae.
Answer & suggested solution
2,500 J.
- [1 mark] 25,000 × 0.10 = 2,500 J.
(ii) [1] Estimate the energy available to small fish.
Answer & suggested solution
250 J.
- [1 mark] 2,500 × 0.10 = 250 J.
(iii) [2] Which organism is most likely to have the greatest pesticide concentration? Name and explain the process.
Answer & suggested solution
The fish-eating bird; biomagnification.
- [1 mark] The top consumer, the fish-eating bird, is expected to have the greatest concentration.
- [1 mark] A persistent chemical can accumulate in organisms and increase in concentration as predators consume contaminated prey.
OR · attempt one alternative for this part
Decomposers act on dead organisms in this pond. Explain their role in nutrient recycling and why this does not recycle all the energy back to algae.
Answer & suggested solution
They return nutrients to the environment; much energy has been dissipated as heat.
- [1 mark] Decomposition releases mineral nutrients that producers can use again.
- [1 mark] Energy is used and dissipated as heat during life processes, so it requires a continuing external supply rather than being fully recycled.
The diagram shows a simplified longitudinal view of a flower’s female reproductive part. (i) Identify A and C. [2] (ii) Describe how a compatible pollen grain at A delivers a male gamete to the egg. [2] (iii) State what the ovule becomes after fertilisation. [1]
Answer & suggested solution
A is stigma; C is ovary. A pollen tube grows through the style to an ovule, allowing a male gamete to fuse with the egg. The ovule becomes a seed.
- [2 marks] A is the receptive stigma; C is the ovary containing ovules.
- [2 marks] The pollen grain germinates; a pollen tube grows through the style towards an ovule and carries male gametes. A male gamete fuses with the egg to form a zygote.
- [1 mark] After fertilisation, the ovule develops into a seed containing the developing embryo.
OR · attempt one alternative
Explain (i) the usual site of human fertilisation, (ii) implantation, (iii) two exchange functions of the placenta, and (iv) why a surgical method preventing sperm transport does not itself protect against sexually transmitted infections. [1+1+2+1]
Answer & suggested solution
Fertilisation normally occurs in an oviduct; the embryo implants in the uterus; the placenta exchanges nutrients, gases and wastes; blocking sperm transport does not block pathogens.
- [1 mark] Egg and sperm normally fuse in an oviduct to form a zygote.
- [1 mark] The developing embryo embeds in the prepared uterine lining.
- [2 marks] The placenta supplies oxygen and nutrients to the embryo and enables transfer of carbon dioxide and other wastes towards the mother. Their blood need not mix directly.
- [1 mark] Blocking sperm transport is contraception, but it is not a physical barrier against transmission of pathogens.
Section B · Chemistry
Questions 17–29 · 25 marks. Q17–23 are MCQs and Q24 is assertion–reason. Internal choices appear in Q26, Q28(iii) and Q29.
A magnesium ribbon burns in oxygen to form a white solid. Which balanced equation represents the overall change?
Answer & suggested solution
B. 2Mg + O₂ → 2MgO
- Magnesium oxide has formula MgO.
- 2Mg + O₂ → 2MgO conserves two magnesium atoms and two oxygen atoms on each side.
At room temperature, solutions P, Q, R and S have pH 5, 9, 7 and 2 respectively. Which is most acidic?
Answer & suggested solution
D. S
- A smaller pH corresponds to a more acidic aqueous solution.
- S has the lowest pH, 2.
An ionic compound does not conduct electricity as a solid but conducts when molten. What explains this?
Answer & suggested solution
A. Its ions are mobile in the molten state.
- Solid lattice ions are fixed in place.
- Molten ions can move and carry electric charge.
Which hydrocarbon can undergo addition of hydrogen across a carbon–carbon double bond?
Answer & suggested solution
C. Ethene, C₂H₄
- Ethene contains a C=C double bond.
- Hydrogen can add across it, using a suitable catalyst such as nickel, to form ethane.
Which pair belongs to the same homologous series?
Answer & suggested solution
B. Methanol and ethanol
- Methanol and ethanol both contain the alcohol functional group.
- Successive homologues differ by CH₂ and have similar chemical properties.
A zinc coating can help protect a small scratch in an iron sheet because:
Answer & suggested solution
D. Zinc oxidises preferentially because it is more reactive.
- Zinc is more reactive than iron.
- It can provide sacrificial protection by oxidising preferentially while sufficient zinc remains.
Plaster of Paris hardens after mixing with water mainly because it:
Answer & suggested solution
A. Forms gypsum by hydration.
- Plaster of Paris is calcium sulphate hemihydrate.
- It takes up water to form gypsum, CaSO₄·2H₂O, setting into a hard mass.
Assertion (A): Copper does not normally liberate hydrogen from dilute hydrochloric acid. Reason (R): Copper lies above hydrogen in the reactivity series.
Answer & suggested solution
C. A is true, but R is false.
- Copper lies below hydrogen in the reactivity series.
- It therefore does not normally displace hydrogen from this dilute non-oxidising acid. The assertion is true but the reason is false.
Magnesium has electron arrangement 2,8,2 and chlorine 2,8,7. Explain the ions formed and deduce the simplest formula of their neutral ionic compound.
Answer & suggested solution
Mg²⁺ and Cl⁻; MgCl₂.
- [1 mark] Magnesium loses two electrons to form Mg²⁺; each chlorine atom gains one to form Cl⁻.
- [1 mark] Two chloride ions balance one magnesium ion, so the formula is MgCl₂.
Ethanol is oxidised under suitable conditions to ethanoic acid. (i) Identify the functional groups before and after oxidation. (ii) Write the oxidation equation using [O]. (iii) State the gas released when the acid reacts with sodium hydrogen carbonate.
Answer & suggested solution
Alcohol → carboxylic acid; CH₃CH₂OH + 2[O] → CH₃COOH + H₂O; carbon dioxide.
- [1 mark] Ethanol contains the alcohol group –OH; ethanoic acid contains –COOH.
- [1 mark] CH₃CH₂OH + 2[O] → CH₃COOH + H₂O.
- [1 mark] Ethanoic acid reacts with NaHCO₃ to release CO₂.
OR · attempt one alternative
Write balanced equations for (i) complete combustion of methane and (ii) hydrogenation of ethene. (iii) Name the catalyst commonly used for the second reaction.
Answer & suggested solution
CH₄ + 2O₂ → CO₂ + 2H₂O; C₂H₄ + H₂ → C₂H₆; nickel.
- [1 mark] CH₄ + 2O₂ → CO₂ + 2H₂O.
- [1 mark] C₂H₄ + H₂ → C₂H₆.
- [1 mark] Nickel is a suitable hydrogenation catalyst.
Write a balanced equation and identify the reaction type for each: (i) quicklime reacts with water; (ii) silver chloride is exposed to sunlight; (iii) barium chloride solution is mixed with sodium sulphate solution.
Answer & suggested solution
Combination; photochemical decomposition; double displacement with precipitation.
- [1 mark] CaO + H₂O → Ca(OH)₂: combination, also exothermic.
- [1 mark] 2AgCl → 2Ag + Cl₂ in sunlight: photochemical decomposition.
- [1 mark] BaCl₂ + Na₂SO₄ → BaSO₄↓ + 2NaCl: double displacement producing insoluble barium sulphate.
A laboratory compares three aqueous samples at room temperature: P has pH 3, Q has pH 8 and R has pH 7. Equal portions of P are used in separate dilution and neutralisation trials.
(i) [1] Which sample is neutral?
Answer & suggested solution
R.
- [1 mark] A neutral aqueous sample has pH 7 at ordinary room temperature.
(ii) [1] Which of P and Q turns red litmus blue?
Answer & suggested solution
Q.
- [1 mark] Q is basic at pH 8 and turns red litmus blue.
(iii) [2] P is diluted with pure water. Predict the direction of pH change and explain whether dilution alone makes it strongly basic.
Answer & suggested solution
The pH rises towards 7; dilution alone does not make it strongly basic.
- [1 mark] Dilution lowers hydrogen-ion concentration, increasing the pH towards neutral.
- [1 mark] Water does not add an excess of base, so dilution alone does not turn this acid into a strongly basic solution.
OR · attempt one alternative for this part
Dilute hydrochloric acid is mixed with a suitable quantity of sodium hydroxide. Write the balanced neutralisation equation and name both products.
Answer & suggested solution
HCl + NaOH → NaCl + H₂O; sodium chloride and water.
- [1 mark] HCl + NaOH → NaCl + H₂O.
- [1 mark] The products are sodium chloride and water.
Answer the following: (i) Write the reaction when hydrogen reduces hot copper(II) oxide. (ii) Identify the substance reduced and the substance oxidised. (iii) Explain why copper does not displace iron from iron(II) sulphate. (iv) State the two essential conditions for rusting. (v) Explain how intact paint protects iron. [1+1+1+1+1]
Answer & suggested solution
CuO + H₂ → Cu + H₂O; CuO is reduced and H₂ oxidised; copper is less reactive than iron; oxygen and moisture cause rusting; paint forms a barrier.
- [1 mark] CuO + H₂ → Cu + H₂O.
- [1 mark] CuO loses oxygen and is reduced; H₂ gains oxygen and is oxidised.
- [1 mark] Copper is below iron in the reactivity series, so it does not displace iron from this salt.
- [1 mark] Rusting requires both oxygen and moisture.
- [1 mark] Intact paint reduces contact between the iron surface and those substances.
OR · attempt one alternative
A metal X displaces Y from Y sulphate, while Y displaces Z from Z sulphate. (i) Rank X, Y and Z. (ii) Predict whether Z displaces X from X sulphate and justify. (iii) Explain why an ionic salt conducts when molten but not as a solid. (iv) Explain why a highly reactive metal such as aluminium is extracted electrolytically rather than by heating its oxide with carbon. [1+1+2+1]
Answer & suggested solution
X > Y > Z; Z does not displace X; mobile molten ions carry current; aluminium oxide is too stable for ordinary carbon reduction.
- [1 mark] Displacement establishes the order X > Y > Z.
- [1 mark] Z is less reactive than X, so it cannot displace X under these conditions.
- [2 marks] Solid ions are fixed in a lattice; molten ions are free to move and carry charge.
- [1 mark] Aluminium is more reactive than carbon and binds oxygen strongly, so electrolysis is used to extract it from a suitable molten mixture.
Section C · Physics
Questions 30–39 · 25 marks. Q30–31 are MCQs and Q32 is assertion–reason. Internal choices appear in Q34, Q38(iii) and Q39.
An object is placed between the focus and optical centre of a convex lens. The image is:
Answer & suggested solution
C. Virtual, erect and enlarged
- An object inside the focal length of a convex lens gives diverging refracted rays.
- Their backward extensions form an erect, enlarged virtual image on the object side.
An ideal lamp operates at its rating of 48 W on a 12 V supply. What current does it draw?
Answer & suggested solution
B. 4 A
- P = VI, so I = P/V.
- I = 48/12 = 4 A.
Assertion (A): A ray entering glass obliquely from air bends towards the normal. Reason (R): For ordinary glass, the speed of light is lower than in air.
Answer & suggested solution
A. Both A and R are true, and R correctly explains A.
- Glass has a higher refractive index than air.
- Light slows and refracts towards the normal on entering glass obliquely, so the reason explains the assertion.
Light travels through a transparent material at 2.0 × 10⁸ m/s. Taking the vacuum speed as 3.0 × 10⁸ m/s, calculate its refractive index and predict how a ray entering it obliquely from air bends.
Answer & suggested solution
n = 1.5; towards the normal.
- [1 mark] n = c/v = (3.0 × 10⁸)/(2.0 × 10⁸) = 1.5.
- [1 mark] The material has a higher refractive index than air, so the ray bends towards the normal.
A 5 Ω resistor carries a steady 2 A for 30 s. Calculate the heat generated and state the effect of doubling the current while keeping resistance and time fixed.
Answer & suggested solution
600 J initially; doubling current gives four times as much heat, 2,400 J.
- [1 mark] H = I²Rt = 2² × 5 × 30 = 600 J.
- [1 mark] H is proportional to I², so doubling current multiplies heat by 4.
OR · attempt one alternative
Two wires of the same material and temperature have equal length. Wire Q has twice the cross-sectional area of wire P. Compare their resistances and state why the resistivity is unchanged.
Answer & suggested solution
RQ = RP/2; resistivity is the same for the same material at the same temperature.
- [1 mark] R = ρL/A, so doubling A halves resistance.
- [1 mark] Resistivity depends on material and temperature, rather than this change of wire dimensions.
An object is 24 cm to the left of a convex lens of focal length 16 cm. Use the Cartesian sign convention to find the image distance, magnification and image nature.
Answer & suggested solution
v = +48 cm; m = −2; real, inverted and enlarged.
- [1 mark] u = −24 cm and f = +16 cm. From 1/f = 1/v − 1/u, 1/v = 1/16 − 1/24 = 1/48, so v = +48 cm.
- [1 mark] m = v/u = 48/(−24) = −2.
- [1 mark] Positive image distance means the image is on the opposite side; negative magnification means inverted. It is real and twice the object size.
A straight wire carries conventional current vertically downwards. (i) State the right-hand thumb rule. (ii) Give the field direction when viewed from above. (iii) State how the field strength generally changes as distance from the wire increases.
Answer & suggested solution
Thumb along current, curled fingers along field; clockwise from above; field strength decreases with distance.
- [1 mark] Point the right thumb along conventional current; curled fingers show the circular magnetic-field direction.
- [1 mark] Downward current points away from an observer above the wire, giving clockwise circles.
- [1 mark] For the same current in a long straight wire, field strength decreases as distance from the wire increases.
In the illustrated ideal circuit, a 3 Ω resistor is in series with a parallel combination of 6 Ω and 3 Ω across an 18 V supply. Calculate (i) total resistance, (ii) supply current, and (iii) voltage across the parallel combination.
Answer & suggested solution
5 Ω; 3.6 A; 7.2 V.
- [1 mark] The parallel resistance is 1/Rp = 1/6 + 1/3 = 1/2, so Rp = 2 Ω. Total resistance is 3 + 2 = 5 Ω.
- [1 mark] I = V/R = 18/5 = 3.6 A.
- [1 mark] The parallel pair has voltage I × Rp = 3.6 × 2 = 7.2 V.
A laboratory relay uses a wire coil wound around a soft-iron core. When current flows, the core attracts a nearby iron armature. The current is then switched off and a spring returns the armature.
(i) [1] Why does the coil act as a magnet only when current flows in this model?
Answer & suggested solution
Electric current produces the magnetic field of the coil.
- [1 mark] Current in the coil creates a field that magnetises the core. With current off, this field is removed.
(ii) [1] Why is a soft-iron core useful here?
Answer & suggested solution
It strengthens the field and can lose most of its induced magnetism when current stops.
- [1 mark] Soft iron is readily magnetised and demagnetised, making it suitable for an electromagnet.
(iii) [2] If only the current direction is reversed, what happens to the poles and will a nearby unmagnetised iron armature necessarily be repelled?
Answer & suggested solution
The poles interchange; the unmagnetised iron can still be attracted.
- [1 mark] Reversing current reverses the solenoid field and interchanges its poles.
- [1 mark] Soft iron is magnetised by induction and is attracted by either pole; pole reversal does not necessarily cause repulsion.
OR · attempt one alternative for this part
For the same coil geometry and core, predict the effect of increasing current within a safe range. Name one other change that can strengthen the field.
Answer & suggested solution
The field generally becomes stronger; increasing the turns per unit length is another way.
- [1 mark] For fixed geometry and core, increasing the current generally increases field strength.
- [1 mark] Increasing turns per unit length also strengthens the field, within the material and circuit limits.
A myopic student’s far point is 4 m from the eye. Neglect the distance between spectacles and eye. (i) Calculate the correcting-lens power for a very distant object. [2] (ii) Explain the position of the uncorrected distant image and how the lens helps. [2] (iii) Name the muscles that help change the eye-lens curvature when looking from far to near. [1]
Answer & suggested solution
A concave lens of −0.25 D; uncorrected image in front of the retina; the lens diverges rays before the eye; ciliary muscles.
- [2 marks] For an effectively infinite object, the lens must form a virtual image at v = −4 m. Thus f = −4 m and P = 1/f = −0.25 D.
- [2 marks] A myopic eye focuses distant rays in front of the retina. The concave lens diverges incident rays so the combined system can focus on the retina.
- [1 mark] Ciliary muscles alter the curvature of the eye lens during accommodation.
OR · attempt one alternative
A person has near point 50 cm and wants to read a book at 25 cm. Neglect lens-to-eye distance. (i) Calculate the correcting-lens power. [2] (ii) Explain why a convex lens is appropriate for this near-vision defect. [1] (iii) Explain dispersion through an ordinary glass prism and name the least-deviated colour. [2]
Answer & suggested solution
+2.0 D; a convex lens pre-converges rays; dispersion separates colours because their refractive indices differ, with red least deviated.
- [2 marks] u = −0.25 m and v = −0.50 m. 1/f = 1/v − 1/u = −2 + 4 = 2 m⁻¹, so P = +2.0 D.
- [1 mark] A convex lens makes the nearby object appear at the person’s near point, helping the eye focus on the retina.
- [2 marks] Different visible colours travel with different refractive indices in ordinary glass and deviate by different amounts. Red usually deviates the least.
Finished your attempt?
Review your work against the suggested solutions. Note the marks lost to concepts, calculation, presentation or time, then revise those topics before your next paper.
Get the 15-paper series · ₹399 ↑Before you buy
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Currently 1 paper is available: Paper 01. The remaining 14 papers are in preparation and will be added to this same digital book as they are published. Release dates have not been announced.
What does the ₹399 purchase include?
A one-time purchase gives you account-linked access to this 15-paper digital series, including future papers as they are published at no additional charge. Paper 01 and its solutions are available as a free complete preview.
Which Science subjects does this book cover?
This book is for CBSE Class 10 Science (086), covering Biology, Chemistry and Physics for the 2026–27 session and Board Exam 2027. Paper 01 represents all 13 Science chapters.
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This is an online digital practice book. No physical copy is shipped. You can use the browser print control to print the currently available paper, with solutions closed for a blank attempt.
Does Paper 01 include answers and solutions?
Yes. The complete paper has 39 questions across three sections, 80 marks and a suggested 3-hour attempt, with internal choices, labelled diagrams and worked solutions for self-review.
Is this an official CBSE question paper?
These are original practice questions by Amit Sir using the section structure of the CBSE Science sample paper. The solutions provide suggested marking guidance; they are not official CBSE marking schemes.