CBSE Class 10 Science Sample Paper 2026-27: Solutions

In photosynthesis, what happens to carbon dioxide and water as glucose forms?

1. Photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.
2. Water is split; its hydrogen (with electrons) goes to carbon dioxide and its oxygen is released. Water loses electrons — it is oxidised.
3. Carbon dioxide gains hydrogen to become glucose — it is reduced.

Answer: Carbon dioxide gains hydrogen while water loses electrons — option (A).

Which statement about the experiment showing that sunlight is needed for photosynthesis is incorrect? (Solved for the version CBSE prints for visually impaired candidates, which needs no figure.)

1. The leaf is destarched first (kept in the dark), so any starch found later was made during the experiment.
2. The part exposed to sunlight makes starch and turns blue-black with iodine; starch is the stored end product.
3. The part kept in darkness cannot photosynthesise, so it shows no starch. The statement that it shows starch is wrong.

Answer: "The leaf kept in darkness will show the presence of starch" — option (B).

Which pair of substances is reabsorbed in the kidneys during urine formation?

1. As the filtrate passes along the tubule, useful substances such as glucose, amino acids, salts and much of the water are taken back into the blood.
2. Starch and glycogen are not present in the filtrate at all.

Answer: Glucose and salts — option (A).

Anthers of mustard flowers are removed at the bud stage and the bud is covered with a paper bag. What happens?

1. Removing the anthers removes the flower's own pollen, so self-pollination cannot happen.
2. The bag keeps out pollen from other flowers, so cross-pollination cannot happen either.
3. Without pollination there is no fertilisation, so no seeds or fruit form.

Answer: They will not produce any seeds or fruits — option (C).

Blocking the vas deferens (vasectomy) or the oviduct (tubectomy) prevents which common event?

1. The vas deferens carries sperm from the testes; the oviduct carries the egg from the ovary and is where fertilisation happens.
2. Gametes and hormones are still made; what is stopped is the sperm and egg reaching each other.

Answer: The meeting and fusion of the male and female gametes — option (C).

An apple plant's male gamete has 17 chromosomes. How many are in the female gamete and in the adult plant's cells?

1. Gametes are made by halving the chromosome number, so both gametes carry the same number: 17.
2. Fusion of two gametes restores the full number in the zygote and the body cells: 17 + 17 = 34.

Answer: 17 and 34 — option (B).

Algae and bacteria → insect larvae → fish. What is the fish's trophic level?

1. Algae and photosynthetic bacteria are producers; the larvae eating them are primary consumers.
2. The fish eats the primary consumers, so it is a secondary consumer.

Answer: Secondary consumer — option (C).

Assertion: aquatic organisms breathe much more slowly than land organisms. Reason: water holds less dissolved oxygen than air.

1. Water holds far less oxygen than air, so the reason is true.
2. Because of that, aquatic animals must breathe faster, not slower, to get enough oxygen. The assertion is false.

Answer: A is false but R is true — option (D).

Assertion: 1% of energy passes from one trophic level to the next. Reason: on average 10% of the food eaten is available to flow on as chemical energy.

1. The ten per cent law: about 10% of the energy at one level is available to the next, not 1%. The assertion is false.
2. The reason states the 10% figure correctly.

Answer: A is false but R is true — option (D).

(A) If the bile duct narrows and bile cannot reach the small intestine, how is digestion affected? (B) How does intestine length depend on an animal's food habit?

1. (A) Bile makes the food arriving from the stomach alkaline so pancreatic enzymes can work, and bile salts break fat into small droplets (emulsification) so lipase can act on more surface. Without bile, fats in particular are not digested properly.
2. (B) Plant food has cellulose, which takes longer to digest, so herbivores such as cows have a long small intestine. Meat is easier to digest, so carnivores such as tigers have a shorter one.

Answer: (A) Fat digestion suffers (no emulsification, food not made alkaline). (B) Herbivores: longer intestine; carnivores: shorter.

Which parts of the brain control actions we do not think about? What are these actions called? Give two examples.

1. Such actions are controlled by the mid-brain and hind-brain (the medulla in the hind-brain controls many of them).
2. They are called involuntary actions.

Answer: Mid-brain and hind-brain; involuntary actions — for example, blood pressure, salivation, vomiting.

How is the information of touch controlled and coordinated in the 'sensitive plant' (touch-me-not)?

1. The plant has no nerves or muscles. The touch is passed from cell to cell by electrical-chemical signals.
2. Cells at the base of the leaflets then change shape by losing or gaining water — they shrink or swell — and the leaflets fold.

Answer: Electrical-chemical signals from cell to cell; cells change shape by changing their water content.

What changes happen in the uterus after the embryo is implanted?

1. The lining of the uterus thickens and gets a rich blood supply to nourish the embryo.
2. A placenta develops: a disc-shaped tissue joining the embryo and the uterine wall, through which food and oxygen pass in and wastes pass out.

Answer: Thickened, blood-rich uterine lining; development of the placenta.

How do muscles move at the cellular level when they receive messages from the brain?

1. The brain's message reaches the muscle as a nerve impulse.
2. Muscle cells contain special proteins. In response to the impulse, these proteins change their shape and their arrangement in the cell.
3. The new arrangement makes the muscle cells shorter — the muscle contracts, and this pulls on the body part to move it.

Answer: Nerve impulse → special proteins in muscle cells change shape and arrangement → cells shorten → movement.

For CBSE's four-level food chain: (A) 10 kJ is available at the fourth level; how much is at the producer level? (B) Could three more levels be added? (C) Which level would hold the most pesticide washed into the water, and why?

Uses a figure printed in CBSE's paper — open the paper above to see it.

1. (A) About 10% passes on at each step. Going back three steps from level 4 to level 1 multiplies by 10 three times: 10 × 10 × 10 × 10 kJ = 10,000 kJ.
2. (B) No. So much energy is lost at each step that after a few levels too little usable energy is left to support more organisms.
3. (C) The highest level — the last organism in the chain printed in CBSE's paper. Pesticides are not broken down; they build up at each step (biological magnification), so the top consumer has the highest concentration.

Answer: (A) 10,000 kJ. (B) No — too little energy would remain. (C) The top consumer, because of biological magnification.

Neha crosses white (vv) with violet (VV) pea plants. What percentage of F1 and F2 plants have violet flowers? Why?

1. F1: VV × vv gives only Vv. Violet is dominant, so 100% of F1 are violet.
2. F2: Vv × Vv gives VV : Vv : vv = 1 : 2 : 1. VV and Vv are violet: 3 of 4 = 75%.

Answer: F1: 100% violet; F2: 75% violet — violet is the dominant trait.

Second cross: violet tall (VVTT) × white short (vvtt). What percentage of F2 plants are tall with white flowers?

1. F1 are all VvTt. In F2 the two traits sort independently: P(white, vv) = 1/4 and P(tall, TT or Tt) = 3/4.
2. P(tall and white) = 1/4 × 3/4 = 3/16 = 18.75%.

Answer: 3/16 = 18.75% (vvTT or vvTt).

What percentage of F1 and F2 plants in the second cross would be of medium height with light violet flowers? Why?

1. In Mendel's pea traits one copy of the dominant allele (T or V) gives the full dominant trait — Tt is as tall as TT, Vv as violet as VV.
2. So no 'in-between' plants appear in either generation.

Answer: 0% in F1 and in F2 — pea traits show no blending.

What percentage of F1 and F2 plants in the second cross are short with white flowers?

1. F1 are all VvTt, showing both dominant traits: 0% short and white.
2. F2: short needs tt (1/4) and white needs vv (1/4); together 1/4 × 1/4 = 1/16 = 6.25% (vvtt).

Answer: F1: 0%; F2: 1/16 = 6.25%.

(I) How many chambers does the heart of birds and mammals have, and how does this help them keep their body temperature? (II) How many chambers does a fish heart have, and how does its blood circulate? (Solved for the version CBSE prints for visually impaired candidates, which needs no figure.)

1. (I) Four chambers. The right and left sides are completely separated, so oxygenated and deoxygenated blood never mix.
2. This gives the body a very efficient oxygen supply — needed by birds and mammals, which have high energy needs because they constantly use energy to keep their body temperature constant.
3. (II) Two chambers. The heart pumps blood to the gills, where it picks up oxygen, and from there it goes straight on to the rest of the body.
4. So in one round of the body the blood passes through the heart only once (single circulation).

Answer: (I) Four chambers; no mixing of blood gives the efficient oxygen supply warm-blooded animals need. (II) Two chambers; single circulation through the gills.

(I) Describe inhalation, and how human lungs are built to give a large area for gas exchange. (II) Diffusion alone would take years to carry oxygen to our toes — so how is oxygen delivered? (Solved for the version CBSE prints for visually impaired candidates, which needs no figure.)

1. (I) Breathing in: the ribs are lifted and the diaphragm flattens, so the chest cavity becomes larger; air is sucked into the lungs.
2. Inside, the air passages end in millions of tiny balloon-like alveoli. Together they give a very large surface, and their walls have a dense network of blood vessels, so gases are exchanged quickly.
3. (II) Oxygen is carried by haemoglobin, a respiratory pigment in red blood cells with a very high affinity for oxygen.
4. The heart pumps this oxygen-rich blood through the blood vessels to every part of the body, far faster than diffusion could.

Answer: (I) Ribs up, diaphragm flat, chest bigger, air in; many alveoli rich in blood vessels. (II) Haemoglobin in red blood cells, pumped by the heart.

A chameleon changes colour by adjusting the spacing between reflecting crystals in its skin. Is this a physical or chemical change?

1. No new substance is formed; only the spacing of the crystals changes, which changes which wavelengths are reflected.
2. A change with no new substance is a physical change.

Answer: Physical change, because the space between the crystals is adjusted — option (C).

Chlor-alkali process table: which rows are correct? I: hydrogen at cathode, a fuel. II: chlorine at anode, for margarine. III: sodium hydroxide near cathode, for soap. IV: sodium hydroxide near anode, for bleaching powder.

1. Electrolysis of brine: chlorine forms at the anode, hydrogen at the cathode, and sodium hydroxide near the cathode.
2. I is right (hydrogen is used as a fuel). III is right (NaOH is used for soap).
3. II is wrong: hydrogen, not chlorine, is used to make margarine (hydrogenation of oils). IV is wrong: NaOH forms near the cathode.

Answer: I and III — option (B).

Soap forms scum in hard water but detergent lathers. Which inference is correct?

1. Hard water contains Ca²⁺ and Mg²⁺ ions. Soap molecules react with them to form an insoluble precipitate — scum.
2. Detergents do not form insoluble salts with these ions, so they still lather.

Answer: Ca²⁺ and Mg²⁺ ions react with soap to form insoluble compounds; detergents do not — option (A).

Which contain water of crystallisation: plaster of Paris, washing soda, baking soda, gypsum?

1. Plaster of Paris: CaSO₄·½H₂O — yes. Washing soda: Na₂CO₃·10H₂O — yes. Gypsum: CaSO₄·2H₂O — yes.
2. Baking soda: NaHCO₃ — no water of crystallisation.

Answer: (i), (ii) and (iv) — option (D).

Which is the unsaturated hydrocarbon with three carbon atoms: propene, propane, pentane or pentene? (Solved for the version CBSE prints for visually impaired candidates, which needs no figure.)

1. 'Prop-' means three carbon atoms; 'pent-' means five, so pentane and pentene are out.
2. Propane (C₃H₈) has only single bonds — saturated. Propene (C₃H₆) has a C=C double bond — unsaturated.

Answer: Propene — option (A).

In the food industry, nitrogen is used as a: preservative, oxidising agent, antioxidant or colouring agent? (Solved for the version CBSE prints for visually impaired candidates, which needs no figure.)

1. Fats and oils in packed food go rancid when they are oxidised by oxygen in the air.
2. Flushing packets with nitrogen keeps oxygen out and so prevents this oxidation — the job an antioxidant does.

Answer: Antioxidant — option (C).

Why do aluminium electrical wires not corrode easily (or do they)?

1. Aluminium is quite reactive, but in air it quickly forms a thin, tight layer of aluminium oxide.
2. This layer protects the metal underneath from further corrosion.

Answer: They do not corrode easily, as aluminium oxide forms a protective layer — option (C).

Assertion: onion extract loses its smell in baking soda solution but keeps it in lemon juice. Reason: onion is an olfactory indicator.

1. Onion's smell disappears in a basic solution (baking soda is basic) and stays in an acidic one (lemon juice) — the assertion is true.
2. Onion is indeed an olfactory indicator — the reason is true.
3. But the reason only names what onion is; it does not explain why the smell goes in a base. So R is not the explanation of A.

Answer: Both true, but R is not the correct explanation of A — option (B).

Carbon forms millions of compounds but silicon, just below it in the Periodic Table, forms only a few. Why?

1. Carbon atoms are small, so carbon-carbon bonds are short and strong. Carbon can therefore join to itself in long chains, branches and rings (catenation).
2. Silicon atoms are larger, so Si-Si bonds are weaker and less stable; silicon shows catenation only to a small extent (chains of up to about seven or eight atoms), so it forms far fewer compounds.

Answer: Carbon's small size gives strong C-C bonds and extensive catenation; silicon's larger size gives weak Si-Si bonds.

Compound X (C₂H₆O) is blended with petrol and used as a solvent in tincture iodine and cough syrups. (i) Electron dot structure of X. (ii) Name and formula of its next homologue. (iii) Its reaction with acetic acid giving a sweet-smelling substance.

1. X is ethanol, CH₃CH₂OH.
2. (i) Electron dot structure in words: two carbon atoms joined by one shared pair; the first carbon shares one pair with each of three hydrogens; the second shares one pair with each of two hydrogens and one pair with the oxygen; the oxygen shares one pair with a hydrogen and keeps two lone pairs.
3. (ii) The next homologue has one more CH₂: propanol, C₃H₇OH.
4. (iii) Esterification (with a little concentrated sulphuric acid): C₂H₅OH + CH₃COOH → CH₃COOC₂H₅ + H₂O. The product, ethyl ethanoate, is a sweet-smelling ester.

Answer: X = ethanol; next homologue propanol (C₃H₇OH); ethanol + ethanoic acid → ethyl ethanoate + water.

Ethanol is heated with concentrated H₂SO₄ at 170 °C to give A; hydrogen is passed over A with nickel to give B. Name A and B and write both reactions.

1. Step 1: hot concentrated sulphuric acid removes water from ethanol (dehydration): CH₃CH₂OH → CH₂=CH₂ + H₂O. A is ethene.
2. Step 2: hydrogen adds across the double bond in the presence of nickel (hydrogenation): CH₂=CH₂ + H₂ → CH₃CH₃. B is ethane.

Answer: A = ethene, B = ethane.

Give reasons: (A) a sports electrolyte drink conducts electricity but coconut oil does not; (B) the milkiness of lime water disappears with excess CO₂; (C) an iron bridge near the sea rusts faster than one in a dry hilly area.

1. (A) The drink has dissolved salts that give free ions, and ions carry current. Coconut oil has no ions.
2. (B) CO₂ first makes insoluble calcium carbonate (milky): Ca(OH)₂ + CO₂ → CaCO₃ + H₂O. Excess CO₂ turns it into soluble calcium hydrogen carbonate: CaCO₃ + H₂O + CO₂ → Ca(HCO₃)₂, so the milkiness goes.
3. (C) Rusting needs both oxygen and moisture. Sea air is humid (and its salt speeds rusting further); dry hill air has little moisture.

Answer: (A) ions vs no ions; (B) soluble Ca(HCO₃)₂ forms; (C) more moisture near the sea.

Rods of metals A to E of equal length and thickness are heated at one end; sensors beep after B 10 s, C 15 s, A 30 s, E 50 s, D 60 s. (A) Best conductor? (B) Metal G conducts better than A but worse than C: its time (15, 20, 35 or 10 s)? (C) If all rods are 20% longer, does the time change? (D) If rod thickness differs, is the inference still valid? (Solved for the version CBSE prints for visually impaired candidates, which needs no figure.)

1. (A) The fastest beep means the fastest heat flow: metal B (10 s).
2. (B) Better than A means less than 30 s; worse than C means more than 15 s. Of the choices only 20 s lies between them.
3. (C) Yes — every rod takes longer, because heat passed on from particle to particle has a longer path. The order of the metals stays the same.
4. (D) No. A fair test changes only the metal; a thicker or thinner rod changes how fast heat arrives, so the comparison would not be valid.

Answer: (A) Metal B. (B) 20 s. (C) Yes, all times increase; order unchanged. (D) No — rod size must be the same.

(I) Alisha: forming ammonia from N₂ and H₂ is a reduction. Drishti: rusting and ammonia formation are both redox. Bhushan: rusting is an oxidation. Who is right? (II) Correct: MgCl₂ + AgNO₃ → AgCl₂ + MgNO₃ and Al + H₂SO₄ → Al₂SO₄ + H₂O.

1. (I) Oxidation and reduction always happen together. In N₂ + 3H₂ → 2NH₃ nitrogen gains hydrogen (reduced) and hydrogen is oxidised; in rusting iron is oxidised and oxygen is reduced.
2. So Drishti is fully right; Alisha and Bhushan each describe only one half of the redox reaction.
3. (II)(i) Silver chloride is AgCl and magnesium nitrate is Mg(NO₃)₂; balance: MgCl₂ + 2AgNO₃ → 2AgCl + Mg(NO₃)₂.
4. (II)(ii) Aluminium sulphate is Al₂(SO₄)₃ and a metal with dilute acid gives hydrogen gas, not water: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂.

Answer: (I) Drishti (the other two are partly right). (II) MgCl₂ + 2AgNO₃ → 2AgCl + Mg(NO₃)₂; 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂.

(I) Daljeet: decomposition gives two simpler compounds (MgCO₃ → MgO + CO₂). Avik: it gives two elements (water → H₂ + O₂). Who is right? (II) Check: (i) iron filings in MgSO₄ solution turn it green; (ii) aluminium in dilute H₂SO₄ gives brisk effervescence.

1. (I) Both are partly right. In decomposition one compound breaks into simpler substances, which can be compounds (MgO and CO₂) or elements (H₂ and O₂). Both their examples are decompositions; each rule on its own is too narrow.
2. (II)(i) Wrong: iron is less reactive than magnesium, so it cannot displace magnesium; no iron sulphate forms and the solution stays colourless.
3. (II)(ii) Not right as stated: the gas is hydrogen, which comes off as a steady stream of bubbles; 'brisk effervescence' is the term used for carbon dioxide from carbonates.

Answer: (I) Both partly. (II)(i) Incorrect — no reaction. (II)(ii) Incorrect description — hydrogen bubbles, not brisk effervescence.

The most accurate way to find a convex lens's focal length using sunlight?

1. Sun rays are nearly parallel, so they converge at the principal focus.
2. Adjust the lens above paper until the spot is smallest and brightest; the lens-to-paper distance is the focal length.

Answer: Hold the lens above paper and adjust until the smallest, brightest spot forms — option (B).

A 1000 W appliance runs at 250 V. What current does it draw?

1. P = VI, so I = P/V = 1000/250 = 4 A.

Answer: 4.0 A — option (C).

Assertion: a coin in water looks less deep than it is. Reason: light bends away from the normal going from water to air.

1. Light from the coin bends away from the normal as it leaves the water; traced back, the rays seem to come from a point higher up.
2. So the coin appears raised: both statements are true and the reason explains the assertion.

Answer: Both true, and R explains A — option (A).

(A) State Snell's law. (B) Define the refractive index of one medium with respect to another, with the expression.

1. (A) For light of a given colour passing between a given pair of media, sin i / sin r is a constant (i = angle of incidence, r = angle of refraction). The incident ray, refracted ray and normal lie in one plane.
2. (B) For light going from medium 1 into medium 2, the refractive index of medium 2 with respect to medium 1 is the ratio of the speed of light in medium 1 to that in medium 2: n₂₁ = v₁/v₂.

Answer: sin i / sin r = constant; n₂₁ = v₁ / v₂.

(I) State Joule's law of heating. (II) What happens to the heat produced if the resistance is doubled, other things the same?

1. (I) The heat produced in a resistor is proportional to the square of the current, to the resistance and to the time the current flows: H = I²Rt.
2. (II) With I and t unchanged, H′ = I²(2R)t = 2I²Rt.

Answer: H = I²Rt; doubling R doubles the heat.

(I) Define resistivity. (II) Show with an example that resistivity is a property of the material.

1. (I) R = ρL/A, so ρ = RA/L: the resistivity of a material is the resistance of a piece of it of unit length and unit cross-sectional area (SI unit Ω m).
2. (II) A long thin copper wire and a short thick copper wire have different resistances, but RA/L comes out the same for both — about 1.6 × 10⁻⁸ Ω m at 20 °C. It changes only if the material (or its temperature) changes.

Answer: ρ = RA/L; same value for any size of a given material.

A concave lens forms an image one-third the size of the object, which is 24 cm in front of the lens. Find (A) the type of lens, (B) the image distance, (C) the focal length.

1. (A) A concave (diverging) lens — it always gives an upright, diminished, virtual image.
2. (B) Magnification m = v/u = +1/3 (upright). With u = −24 cm, v = (1/3)(−24) = −8 cm: the image is 8 cm from the lens on the object's side.
3. (C) Lens formula 1/f = 1/v − 1/u = 1/(−8) − 1/(−24) = −3/24 + 1/24 = −2/24 = −1/12, so f = −12 cm.

Answer: Concave lens; v = −8 cm; f = −12 cm.

(A) State the right-hand thumb rule and how it gives the field direction round a current-carrying wire. (B) How does the field change with distance from the wire?

1. (A) Hold the straight wire in your right hand with the thumb pointing along the current. Your curled fingers show the direction of the magnetic field lines, which are circles round the wire.
2. (B) The field is strongest near the wire and gets weaker with distance; for a long straight wire it is inversely proportional to the distance (B ∝ 1/r).

Answer: Thumb = current, fingers = field; B decreases as 1/r.

A 15 V battery is in series with resistors of 2 Ω, 3 Ω and 5 Ω. Find (A) the total resistance, (B) the current, (C) the voltage across the 3 Ω resistor. (Solved for the version CBSE prints for visually impaired candidates, which needs no figure.)

1. (A) In series, resistances add: R = 2 + 3 + 5 = 10 Ω.
2. (B) I = V/R = 15/10 = 1.5 A (the same current flows through each resistor).
3. (C) V₃ = IR = 1.5 × 3 = 4.5 V.

Answer: 10 Ω; 1.5 A; 4.5 V.

A door lock uses a solenoid that pulls an iron bolt when current flows. (A) Why is there an iron core? (B) How can the poles of the solenoid be found? (C) What changes if the current goes from 1 A to 2 A? (D) With a fixed current source, how else can the field be made stronger?

1. (A) Soft iron becomes strongly magnetised inside the coil, concentrating and strengthening the field, so the pull on the bolt is stronger.
2. (B) Curl the fingers of the right hand along the current in the turns; the thumb points to the north pole. Or bring the north pole of a known bar magnet near one end: repulsion means that end is north.
3. (C) The field inside a solenoid is proportional to the current, so doubling the current roughly doubles the field and the pull, unlocking the door more surely.
4. (D) Use more turns in the same length of coil, or put in (or improve) a soft iron core.

Answer: (A) Iron core strengthens the field. (B) Right-hand rule or a test magnet. (C) Field and pull increase. (D) More turns / soft iron core.

(I) What is accommodation? (II) What is the near point of a normal eye and what does it mean? (III) Which lens corrects the defect where distant objects look blurred? (IV) Diagram of how it helps.

1. (I) The eye lens's ability to change its focal length (the ciliary muscles change its curvature) so that both near and far objects are focused on the retina.
2. (II) About 25 cm for a normal adult eye: the closest distance at which an object can be seen clearly without strain.
3. (III) That defect is myopia (short-sightedness); it is corrected with a concave (diverging) lens of suitable power.
4. (IV) Diagram in words: parallel rays from a distant object pass through the concave lens and spread out as if coming from the eye's far point; the eye lens then focuses them on the retina instead of in front of it.

Answer: Accommodation = adjusting focal length; near point ≈ 25 cm; myopia → concave lens.

(I) What is dispersion? (II) Which colour deviates least in a prism? (III) Why does the sky look dark to an astronaut in space? (IV) Diagram of white light through a prism.

1. (I) The splitting of white light into its component colours, because each colour bends by a different amount in a medium such as glass.
2. (II) Red deviates least (it has the longest wavelength); violet deviates most.
3. (III) In space there is no atmosphere to scatter sunlight, so no scattered light reaches the eye from other directions.
4. (IV) Diagram in words: a narrow beam of white light enters one face of a triangular prism, bends towards the base, and leaves the other face as a spread band of colours — violet nearest the base, red farthest (VIBGYOR).

Answer: Dispersion = splitting into colours; red deviates least; no atmosphere to scatter light.