CBSE Class 10 Science Question Paper 2026 Solved Code 31-1-2

SECTION A (Biology)

1. [1 mark] Choose the correct statements with reference to chromosomes:
(i) carry hereditary information from parents to next generation.
(ii) are thread-like structures located inside the nucleus of an animal cell.
(iii) always exist in pairs in human gametes.
(iv) are involved in the process of cell division.
Options:
(A) (i) and (ii)
(B) (iii) and (iv)
(C) (i), (ii) and (iv)
(D) (ii), (iii) and (iv)

Answer : (C) statement (iii) is wrong, gametes carry a single set, not pairs, of chromosomes.

Teacher's Note:
Chromosomes carry genes and control heredity.
Gametes are haploid, so their chromosomes are never in pairs; only body cells have pairs.
Chromosomes become active and visible during cell division.

2. [1 mark] Which structure in a leaf is mainly responsible for gaseous exchange?
(A) Xylem
(B) Stomata
(C) Phloem
(D) Cuticle

Answer : (B) stomata are tiny pores that let gases move in and out.

Teacher's Note:
Stomata open and close with the help of guard cells.
Xylem carries water, phloem carries food, not gases.
Cuticle is a waxy layer that prevents water and gas loss.

3. [1 mark] A farmer wants to grow banana plants genetically similar to the plants already available in the fields. Which of the following method would you suggest for this purpose:
(A) Regeneration
(B) Budding
(C) Vegetative propagation
(D) Sexual reproduction

Answer : (C) it produces genetically identical (clone) plants from a parent plant part.

Teacher's Note:
Vegetative propagation gives clones, useful for banana, sugarcane, rose.
Sexual reproduction always produces variation, not identical copies.
Budding and regeneration are common in simple organisms, not banana crops.

4. [1 mark] Which of the following group is not 'biodegradable'?
(A) Vegetable peels, dead leaves, paper
(B) Cow dung, leather bag, water
(C) Polythene bag, rubber band, ball pen
(D) Paper, fruits, bones

Answer : (C) these are synthetic materials that microbes cannot break down.

Teacher's Note:
Biodegradable substances are broken down naturally by microorganisms.
Plastics, rubber and similar synthetic items stay in the environment for a very long time.
Non-biodegradable waste causes pollution as it does not decompose.

5. [1 mark] Human brain has various parts or regions that help in different actions, responses and coordination. From the following, identify the part responsible for precision of voluntary actions:
(A) Cerebrum
(B) Cerebellum
(C) Medulla
(D) Pons

Answer : (B) cerebellum controls balance and precision of voluntary movements.

Teacher's Note:
Cerebrum handles thinking, memory and initiating voluntary actions.
Cerebellum fine-tunes and balances those movements.
Medulla controls involuntary actions like heartbeat and breathing.

6. [1 mark] Identify the correct statement for spirogyra, leishmania and hydra:
(A) they reproduce sexually.
(B) they are unicellular.
(C) they are multicellular.
(D) they reproduce asexually.

Answer : (D) all three commonly reproduce by asexual methods.

Teacher's Note:
Spirogyra reproduces by fragmentation, leishmania by binary fission, hydra by budding.
Leishmania is unicellular while spirogyra and hydra can be multicellular, so options B and C are not true for all three.

7. [1 mark] Pancreas secretes pancreatic juice which contains certain enzymes that help in digestion of food. Choose the correct option from the following:
(A) Trypsin digests emulsified fats and lipase digests proteins.
(B) Trypsin digests proteins and lipase digests emulsified fats.
(C) Trypsin and lipase both digests fats.
(D) Trypsin digests proteins and lipase digests carbohydrates.

Answer : (B) trypsin acts on proteins, lipase acts on emulsified fats.

Teacher's Note:
Bile from the liver emulsifies fat before lipase can act on it.
Pancreatic juice also has amylase for carbohydrates.

8. [1 mark] Assertion (A): In human beings, the respiratory pigment is haemoglobin present in red blood cells.
Reason (R): Haemoglobin has a very high affinity for carbon dioxide.
(A) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer : (C) haemoglobin has a very high affinity for oxygen, not carbon dioxide.

Teacher's Note:
Haemoglobin binds oxygen strongly to carry it to body tissues.
Most CO2 is carried dissolved as bicarbonate ions in blood plasma.

9. [1 mark] Assertion (A): Plants have hormones that do not control directional growth.
Reason (R): Abscisic acid inhibits growth.

Answer : (B) both statements are true, but R does not correctly explain A.

Teacher's Note:
Abscisic acid is a growth-inhibiting hormone; it does not control directional (tropic) growth.
Auxin, not abscisic acid, is the hormone that causes directional growth such as bending towards light.

10. [2 marks] (a) State two differences between the act of chewing food and salivation on sight of food.
OR
(b) State two differences between pollination and fertilization.

Answer (a) :

Chewing food is a voluntary action, controlled by the forebrain, we decide when to chew. Salivation on seeing food is a reflex (involuntary) action, controlled by the medulla in the hindbrain, and it happens automatically.

Answer (b) :

Pollination is the transfer of pollen grains from the anther to the stigma, and it occurs only in plants. Fertilization is the fusion of the male gamete with the female gamete to form a zygote, and it occurs in both plants and animals.

Teacher's Note:
Voluntary actions are controlled by the forebrain; reflex actions by lower brain regions.
Pollination always happens before fertilization in flowering plants.
Fertilization involves fusion of gametes; pollination does not.

11. [2 marks] A squirrel in a scary situation requires its body to prepare for either 'fight or flight' to save itself. State the immediate changes that will take place in its body so that it can face the situation.

Answer :

The squirrel's adrenal glands release the hormone adrenaline into the blood. This makes the heart beat faster, increases the breathing rate through faster contraction of the ribs and diaphragm, and sends more oxygen and blood to the limb muscles. At the same time, blood supply to the digestive system and skin is reduced so the muscles get extra energy to fight or run away quickly.

Teacher's Note:
Adrenaline is released from the adrenal glands during stress.
It raises heart rate and breathing rate and redirects blood to muscles.
This whole response is called the "fight or flight" response.

12. [2 marks] Give differences between the following:
(a) Sensory nerve and motor nerve
(b) Consumers and decomposers

Answer :

(a) A sensory nerve carries impulses from receptors (sense organs) towards the brain and spinal cord (CNS). A motor nerve carries impulses from the CNS towards an effector organ, like a muscle, to bring about a response.

(b) Consumers are organisms that feed on producers or other consumers to obtain food. Decomposers, like fungi and bacteria, break down dead and waste organic matter into simple substances and recycle nutrients back into the environment.

Teacher's Note:
Sensory nerve carries messages towards the brain; motor nerve carries them away, towards muscles.
Decomposers complete the food chain by recycling nutrients back to the soil.

13. [3 marks] (a) A couple are parents to 4 daughters in a sequence, and do not have any son. Does this indicate that the husband does not produce Y-chromosome bearing sperms? Explain.
(b) What are the chances of this couple bearing yet another daughter? Show with the help of a cross.

Answer :

(a) No, this does not mean the husband cannot produce Y-chromosome bearing sperms. It is purely a matter of chance. A man produces two types of sperm in equal numbers, 50% carrying an X chromosome and 50% carrying a Y chromosome, and either type can fertilise the egg randomly each time.

(b) The chance of having another daughter is 50%. This can be shown with a cross:
Parents: XX (mother) x XY (father)
Gametes from mother: X, X; Gametes from father: X, Y
Offspring: XX (girl), XY (boy), XX (girl), XY (boy), giving a ratio of 2 girls : 2 boys.
So there is a 50% chance the next child is again a daughter.

Teacher's Note:
The sex of the child is decided by the father's sperm (X or Y), not by the mother.
Each pregnancy is an independent 50:50 chance event.
Draw a simple cross showing gamete combinations to explain the ratio.

14. [3 marks] Given below is a pyramid showing various trophic levels in an ecosystem: (iv) Tertiary consumers, (iii) Secondary consumers, (ii) Primary consumers, (i) Producers.
(a) From the organisms listed below, identify which one is to be placed at which trophic level: Deer, Grass, Lion, Snake, Rabbit.
(b) Discuss the reason why primary consumers will have more energy as compared to secondary consumers.
(c) Why is the base of the pyramid broad?

Answer :

(a) Grass is the producer (i). Deer and Rabbit are primary consumers (ii), since they feed directly on grass. Snake and Lion are secondary consumers (iii), since they feed on primary consumers, and Lion also acts as a tertiary consumer (iv), since it can also feed on other carnivores like the snake.

(b) Primary consumers feed directly on green plants (producers), which have a large amount of energy stored in them. Only about 10% of this energy is passed on to the next level, the secondary consumers, so primary consumers have more energy available to them.

(c) The base of the pyramid is broad because producers are usually the most numerous and have the highest amount of energy or biomass, compared to the consumer levels above them.

Teacher's Note:
Only about 10% of energy passes from one trophic level to the next (the 10% law).
Producers (green plants) sit at the base and are always the most abundant.
The same organism (like a lion) can occupy more than one trophic level depending on what it eats.

15. [4 marks] Mendel took garden pea plants with different characteristics, such as height, to study the inheritance pattern of factors (genes). He crossed tall pea plants with short pea plants and obtained all the tall plants in the F1 generation. Answer the following questions:
(a) [1 mark] Why only tall pea plants were observed in F1 progeny?
(b) [1 mark] By which method did Mendel obtain F2 progeny?
(c)(i) [2 marks] Write one difference between dominant and recessive trait.
OR
(c)(ii) [2 marks] Write two observations made by Mendel about F1 progeny.

Answer :

(a) Tallness (T) is dominant over shortness (t). Mendel crossed a pure tall plant (TT) with a pure short plant (tt), so every F1 plant received one T and one t allele (Tt). Since the dominant T allele expresses itself, all F1 plants were tall.

(b) Mendel obtained the F2 progeny by self-pollinating the F1 plants.

Answer (c)(i) :

A dominant trait expresses itself even in the presence of a recessive trait, and shows up in both the pure (TT) and hybrid (Tt) condition. A recessive trait can express itself only when no dominant allele is present, that is, only in the pure recessive condition (tt).

Answer (c)(ii) :

Two observations Mendel made about the F1 progeny: (1) All the plants of the F1 progeny were tall, no plant of medium or short height was seen. (2) The F1 progeny resembled only one parent, the tall parent, and the trait of the short parent did not appear at all.

Teacher's Note:
Dominant allele expresses in both homozygous and heterozygous state; recessive shows only in homozygous state.
F1 is obtained by cross-pollinating two different pure-breeding parents; F2 by self-pollinating F1.
This is Mendel's Law of Dominance.

16. [5 marks] (a) Given below are certain situations. Analyse and describe what would happen when:
(i) Spores are liberated from blob-like structures of the bread mould?
(ii) Leaves of bryophyllum fall on wet soil?
(iii) A pollen from different species lands on the stigma of a totally unrelated species?
(iv) Copper-T is placed in the uterus of a human female?
(v) Spirogyra breaks into smaller fragments upon maturation?
OR
(b) Given below are certain situations. Write the analysis of each and describe its possible impact:
(i) A population of bacteria living in temperate waters whose temperature increased due to global warming.
(ii) The sperm encounters the egg when it reaches the oviduct in human females.
(iii) Self pollination does not occur in a flower that contains only pistil.
(iv) Egg does not get fertilised in a human female.
(v) When the seed is placed under appropriate condition of water and air in the soil?

Answer (a) :

(i) When spores land on a suitable food substance and get enough moisture and a favourable temperature, they germinate and grow into a new bread mould (Rhizopus) plant.
(ii) New buds present in the notches of the bryophyllum leaf develop into new independent plants when the leaf falls on wet soil.
(iii) The pollen tube usually does not form properly, so fertilisation does not take place, and no seed or fruit is formed.
(iv) The Copper-T prevents the fertilised egg from implanting in the uterus, so it works as a contraceptive and prevents pregnancy.
(v) Each fragment of the mature spirogyra grows into a new independent organism; this is called fragmentation.

Answer (b) :

(i) Most of the bacteria would die due to the rising temperature, but a few heat-resistant variants would survive and continue to grow.
(ii) When the sperm meets and fuses with the egg in the oviduct, fertilization takes place and a zygote is formed.
(iii) A flower with only a pistil cannot self-pollinate; cross-pollination from another flower would be needed for fertilization and fruit formation.
(iv) If the egg is not fertilised, the thick, spongy lining of the uterus breaks down and is shed as blood and mucus through the vagina; this is called menstruation.
(v) The seed absorbs water and air, and germinates into a seedling.

Teacher's Note:
Spore formation, budding and fragmentation are asexual methods of reproduction.
Fertilization needs fusion of a male and a female gamete to form a zygote.
Germination needs water, oxygen and a suitable temperature.
Copper-T and menstruation are important NCERT keywords in the reproduction chapter.

SECTION B (Chemistry)

17. [1 mark] When a certain element 'X' reacts with water, it starts floating. Identify the element 'X':
(A) Potassium
(B) Calcium
(C) Sodium
(D) Iron

Answer : (B) calcium floats due to hydrogen gas bubbles sticking to its surface.

Teacher's Note:
Calcium's reaction with water is not violent, so it floats.
Potassium and sodium react so vigorously with water that they can even catch fire.
Iron does not react with cold water.

18. [1 mark] The natural sources of oxalic acid, lactic acid and methanoic acid respectively are:
(A) tomato, curd, ant-sting
(B) tomato, orange, nettle-sting
(C) orange, milk, ant-sting
(D) orange, sour milk, nettle-sting

Answer : (A) tomato gives oxalic acid, curd gives lactic acid, ant-sting gives methanoic acid.

Teacher's Note:
These are common weak organic acids found in everyday sources.
Nettle sting also contains methanoic acid, like an ant's sting.

19. [1 mark] Which of the following is a poor conductor of electricity?
(A) Pb
(B) Cu
(C) Ag
(D) Al

Answer : (A) lead (Pb) conducts electricity poorly compared to the other metals listed.

Teacher's Note:
Silver is the best conductor among common metals, followed by copper.
Lead conducts electricity but much less efficiently.

20. [1 mark] Which of the following functional groups is for carboxylic acid?
(A) -OH
(B) -C-OH with double bond O
(C) -C- with double bond O
(D) -C-H with double bond O

Answer : (B) the -COOH group, carbon double-bonded to O and singly bonded to -OH.

Teacher's Note:
-OH alone is the alcohol functional group, not carboxylic acid.
-CHO (option D type) is an aldehyde group.

21. [1 mark] The gases evolved on heating lead (II) nitrate crystals are:
(A) NO and O2
(B) N2 and NO2
(C) NO2 and H2
(D) NO2 and O2

Answer : (D) 2Pb(NO₃)₂(s) → 2PbO(s) + 4NO₂(g) + O₂(g)

Teacher's Note:
This is a thermal decomposition reaction.
Brown fumes seen on heating lead nitrate are due to NO2 gas.

22. [1 mark] Which one of the following can be used as an acid-base indicator by a visually impaired (blind) student?
(A) Turmeric
(B) Vanilla essence
(C) Methyl orange
(D) Litmus

Answer : (B) its smell disappears in basic solution, so it can be identified without seeing colour.

Teacher's Note:
This is called an olfactory indicator, an indicator identified by smell, not colour.
Litmus, turmeric and methyl orange are all colour-change indicators, not usable by a blind student.

23. [1 mark] (i) AgNO₃ + NaCl → NaNO₃ + AgCl
(ii) K₂SO₄ + BaCl₂ → BaSO₄ + 2KCl
Which of the following options clearly describes both the reactions?
(A) (i) double displacement, (ii) displacement reaction.
(B) Both, (i) and (ii) are displacement reactions and precipitation reactions.
(C) Both, (i) and (ii) are double displacement reactions and precipitation reactions.
(D) (i) displacement, (ii) double displacement reaction.

Answer : (C) ions exchange partners in both, and an insoluble precipitate is formed in each case.

Teacher's Note:
AgCl and BaSO4 are both insoluble white precipitates.
Any reaction forming a precipitate through ion exchange is a double displacement + precipitation reaction.

24. [1 mark] Assertion (A): Carbon shares its valence electrons with other atoms of carbon or with atoms of other elements.
Reason (R): The shared electrons belong to the outermost shells of both the atoms, and as a result, both atoms attain the noble gas configuration.

Answer : (A) both statements are true, and R correctly explains why carbon forms covalent bonds.

Teacher's Note:
Carbon forms covalent bonds by sharing electrons, not by losing or gaining them.
Sharing electrons lets both atoms complete their outer shell (octet), like noble gases.

25. [2 marks] How is tooth decay related to pH? How can it be prevented?

Answer :

Tooth decay starts when the pH inside the mouth falls below 5.5. At this pH, the acid produced (when bacteria act on sugar left on the teeth) begins to corrode the tooth enamel. It can be prevented by cleaning the mouth after eating and by using toothpaste, which is basic in nature and neutralises the acid formed in the mouth.

Teacher's Note:
Tooth enamel is the hardest substance in the body, but it is attacked by acid below pH 5.5.
Bacteria break down leftover sugar in the mouth to produce acids.
Basic toothpaste neutralises this acid and protects the enamel.

26. [3 marks] (a) Explain chlor-alkali process with chemical equation. Name the products formed at anode and cathode.
OR
(b) Write the preparation of the following compounds with balanced chemical equation:
(i) Baking soda
(ii) Bleaching powder
(iii) Plaster of Paris

Answer (a) :

When electricity is passed through a concentrated aqueous solution of sodium chloride (brine), it decomposes to form sodium hydroxide (an alkali), chlorine gas and hydrogen gas. This process is called the chlor-alkali process.

2NaCl(aq) + 2H₂O(l) --Electricity--> 2NaOH(aq) + H₂(g) + Cl₂(g)

At the anode, chlorine gas (Cl2) is produced. At the cathode, hydrogen gas (H2) is produced, while sodium hydroxide forms in the solution.

Answer (b) :

(i) Baking soda: NaCl + H₂O + NH₃ + CO₂ → NaHCO₃ + NH₄Cl
(ii) Bleaching powder: Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O
(iii) Plaster of Paris: CaSO₄.2H₂O --373K--> CaSO₄.½H₂O + 1½H₂O

Teacher's Note:
Chlor-alkali process: Cl2 forms at the anode, H2 at the cathode, NaOH stays in solution.
Bleaching powder's formula is CaOCl2.
Plaster of Paris is calcium sulphate hemihydrate; heating it further removes all its water.

27. [3 marks] Translate the following statements into chemical equations and then balance them:
(a) Water is added to quicklime
(b) Burning of natural gas
(c) Thermal decomposition of ferrous sulphate

Answer :

(a) CaO(s) + H₂O(l) → Ca(OH)₂(aq) + Heat
(b) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
(c) 2FeSO₄(s) --Heat--> Fe₂O₃(s) + SO₂(g) + SO₃(g)

Teacher's Note:
Adding water to quicklime is called slaking of lime, a highly exothermic combination reaction.
Burning natural gas (mainly methane) is a combustion reaction giving CO2 and water.
Thermal decomposition of ferrous sulphate gives ferric oxide and two gases, SO2 and SO3.

28. [4 marks] Read the following passage and answer the questions given below: Most metals occur in nature in the combined state, as ores. Carbonate ores are converted into oxides by calcination, and sulphide ores by roasting. Oxides are then reduced with a suitable reducing agent, such as carbon, to obtain the free metal. Highly reactive metals, like Al and Mg, are also used as reducing agents to obtain metal from oxides. Very reactive metals are obtained by electrolytic reduction of their molten ores. Alloying is a good method of improving the properties of a metal. We can get the desired properties of metals by this method. The electrical conductivity and melting point of an alloy is lower than that of pure metals.
(a) [1 mark] Why are carbonate or sulphide ores converted into oxides before extraction of metal from them?
(b) [1 mark] Write a reaction in which aluminium is used as a reducing agent to obtain metal from a metal oxide.
(c)(i) [2 marks] How is copper obtained from its ore (Cu₂S)? Give equations for the reactions.
OR
(c)(ii) [1+1 marks] (I) Why can highly reactive metals not be obtained from their oxides by using carbon as a reducing agent? (II) Why is solder, an alloy of lead and tin, used for welding electrical wires together?

Answer :

(a) Carbonate or sulphide ores are converted into oxides because it is much easier to obtain (reduce) a metal from its oxide than directly from its carbonate or sulphide.

(b) Fe₂O₃(s) + 2Al(s) → 2Fe(l) + Al₂O₃(s) (this is the thermite reaction, where aluminium, being more reactive, reduces iron oxide to iron.)

Answer (c)(i) :

Copper is obtained from its sulphide ore Cu2S by roasting it in air, which partly converts it to Cu2O:
2Cu₂S + 3O₂(g) --Heat--> 2Cu₂O(s) + 2SO₂(g)
The Cu2O then reacts with the remaining Cu2S to give copper metal:
2Cu₂O + Cu₂S --Heat--> 6Cu(s) + SO₂(g)

Answer (c)(ii) :

(I) Highly reactive metals have a greater affinity for oxygen than carbon does, so carbon cannot displace them from their oxides.
(II) Solder has a low melting point, so it melts easily and can join (weld) electrical wires together without damaging them.

Teacher's Note:
Calcination is used for carbonate ores, roasting for sulphide ores.
More reactive metals like Al or Mg can reduce oxides of less reactive metals (thermite reaction).
Alloys generally have a lower melting point and lower conductivity than the pure metal.

29. [5 marks] (a)(i) [3 marks] Give reasons for the following:
(I) Covalent compounds are poor conductor of electricity.
(II) Soap does not form lather in hard water.
(III) Carbon shows catenation but silicon does not.
(a)(ii) [2 marks] Write chemical equations for the following:
(I) Oxidation of ethanol by acidified K₂Cr₂O₇.
(II) Hydrogenation of ethene.
OR
(b) [5 marks] Mohan heated ethanol with a compound 'X' in the presence of a few drops of conc. H₂SO₄ and observed that a sweet smelling compound 'Y' is formed. When 'Y' is treated with sodium hydroxide it gives back ethanol and a compound 'Z'.
(i) Identify 'X', 'Y' and 'Z'.
(ii) Write the role of conc. H₂SO₄ in this reaction.
(iii) Write the chemical equations involved and name the reactions.

Answer (a) :

(I) Covalent compounds are poor conductors of electricity because they do not have charged particles (ions) to carry the current.
(II) Soap reacts with the calcium and magnesium salts present in hard water to form an insoluble substance called scum, so it does not form lather easily.
(III) Carbon forms strong and stable C-C bonds, so it can link into long chains, a property called catenation. Si-Si bonds are comparatively weak, so silicon does not show catenation to the same extent.

(I) CH₃CH₂OH --acidified K₂Cr₂O₇, heat--> CH₃COOH
(II) CH₂=CH₂ + H₂ --Ni--> CH₃-CH₃

Answer (b) :

(i) X = CH₃COOH (ethanoic acid/acetic acid); Y = CH₃COOC₂H₅ (ethyl ethanoate, an ester); Z = CH₃COONa (sodium ethanoate).

(ii) Concentrated H2SO4 acts as a catalyst and a dehydrating agent in this reaction.

(iii) CH₃COOH + CH₃CH₂OH --Acid--> CH₃COOC₂H₅ + H₂O. This is called esterification (acid reacting with alcohol to form a sweet-smelling ester).
CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH. This reverse reaction is called saponification (de-esterification).

Teacher's Note:
Covalent compounds lack free ions, so they never conduct electricity.
Esters smell sweet and form by esterification (acid + alcohol + conc. H2SO4).
Saponification is the reverse of esterification and is also how soap is made.
Catenation is carbon's unique ability to form long chains with itself.

SECTION C (Physics)

30. [1 mark] When you look at an object very close to your eyes, the:
(A) Ciliary muscles of your eye contract and the eye lens becomes thick.
(B) Ciliary muscles of your eye get relaxed and the eye lens becomes thick.
(C) Ciliary muscles of your eye contract and the eye lens becomes thin.
(D) Ciliary muscles of your eye get relaxed and the eye lens becomes thin.

Answer : (A) ciliary muscles contract, making the lens thicker and more powerful for near objects.

Teacher's Note:
For near objects, ciliary muscles contract and lens curvature (power) increases.
For far objects, ciliary muscles relax and the lens becomes thin.

31. [1 mark] A convex lens of focal length 15 cm is forming a real image. If the size of image is same as the size of object, then position of object and position of image will be, respectively:
(A) - 15 cm and - 15 cm from lens
(B) - 15 cm and + 15 cm from lens
(C) - 30 cm and + 30 cm from lens
(D) - 30 cm and - 30 cm from lens

Answer : (C) an equal-size real image forms only when the object is at 2F, that is 30 cm here.

Teacher's Note:
A convex lens gives a real, equal-size image only when the object is placed at 2F (twice the focal length).
Object distance is taken as negative, real image distance as positive.

32. [1 mark] Assertion (A): When rays of white light pass through a prism, on emerging they give a spectrum of seven colours.
Reason (R): It is due to the scattering of light that red light bends minimum and violet light bends the maximum.

Answer : (C) A is true, but R is false, this splitting is due to dispersion (refraction), not scattering.

Teacher's Note:
Splitting of white light into 7 colours by a prism is called dispersion, not scattering.
Violet bends the most and red the least because of their different wavelengths.

33. [2 marks] (a) The resistance of a wire of 0.01 cm radius and 1.0 cm length is 7 Ω. Calculate its resistivity.
OR
(b) An electric heater is rated 220 V ; 11 A. Calculate the power consumed if this heater is operated at 200 V.

Answer (a) :

r = 0.01 cm = 1×10⁻⁴ m
l = 1 cm = 0.01 m
Using R = ρl/A, so ρ = RA/l = R×πr²/l
ρ = (7 × 22/7 × (1×10⁻⁴)²) / 0.01 = (22 × 10⁻⁸) / 0.01
ρ = 22 × 10⁻⁶ Ω m = 2.2 × 10⁻⁵ Ω m

Answer (b) :

Resistance of the heater, R = V/I = 220/11 = 20 Ω. This resistance stays the same when operated at a different voltage.
Power consumed at 200 V, P = V²/R = (200×200)/20 = 2000 W = 2 kW

Teacher's Note:
Resistivity ρ = RA/l, where A = πr² is the cross-sectional area.
Always convert cm to metres before substituting values.
Resistance of an appliance stays fixed; power changes with the operating voltage using P = V²/R.

34. [2 marks] A mirror always forms a virtual, erect and diminished image. Identify the mirror and draw a labelled ray diagram to show image formation by this mirror.

Answer :

This is a convex mirror (a diverging mirror). It always forms a virtual, erect and smaller (diminished) image, no matter where the object is placed, because it spreads out (diverges) the reflected rays.

Ray diagram (described): For an object AB placed in front of a convex mirror with pole P and focus F behind the mirror, draw one ray from the top of the object parallel to the principal axis, which after reflection appears to come from F (extend it backward through F). Draw a second ray aimed towards the centre of curvature C behind the mirror, which reflects straight back along the same path. Extending both reflected rays backward behind the mirror, they appear to meet at a point, this gives the virtual, erect, diminished image A'B'.

Teacher's Note:
Only a convex mirror gives a virtual, erect, diminished image for every object position.
A concave mirror gives different image types depending on where the object is placed.
In the ray diagram, reflected rays are shown as dotted lines behind the mirror to locate the virtual image.

35. [3 marks] (a) A current carrying conductor placed in an external magnetic field experiences a force, describe an activity to show this fact.
(b) Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally towards the front wall from the back wall, is deflected by a strong magnetic field to your right side. Find the direction of the magnetic field.

Answer (a) :

Take a small aluminium rod AB and, using two connecting wires, suspend it horizontally from a stand so it hangs freely. Place a strong horse-shoe magnet so that the rod lies horizontally between its two poles. Connect the rod in series with a battery and a key, and pass current through the rod.
Observation: as soon as current flows through the rod, it is seen to get displaced (it jumps sideways). This shows that a current-carrying conductor placed in a magnetic field experiences a force.

Answer (b) :

The magnetic field will be directed vertically downwards.

Teacher's Note:
This activity demonstrates the motor effect, force on a current-carrying conductor in a magnetic field.
The direction of the force or field can be found using Fleming's Left-Hand Rule.
Remember, electron flow is opposite to the direction of conventional current.

36. [3 marks] (a) The pattern of magnetic field due to a current carrying wire depends upon the shape made by that wire. Justify.
(b) A current carrying straight wire AB is shown in a given diagram, with points X, Y and Z marked at increasing distances from the wire along a line. Out of X, Y and Z, at which point will the strength of magnetic field be maximum and why?

Answer (a) :

The magnetic field pattern does depend on the shape of the current-carrying conductor. For a straight current-carrying wire, the field lines form concentric circles around the wire. For a current-carrying solenoid (coil), the field pattern resembles that of a bar magnet, nearly uniform inside the solenoid. For a current-carrying circular loop, the pattern is concentric circles that become nearly straight, parallel lines at the centre of the loop. This shows that the magnetic field pattern changes with the shape of the current-carrying conductor.

Answer (b) :

The magnetic field will be strongest at point X, because it is closest to the wire, and the strength of the magnetic field due to a straight current-carrying wire decreases as the distance from the wire increases.

Teacher's Note:
Around a straight wire, field lines are concentric circles; inside a solenoid, they are nearly parallel, like a bar magnet.
Magnetic field strength decreases as distance from the current-carrying wire increases.

37. [3 marks] (a) What is Tyndall effect?
(b) What happens when sunlight is scattered by particles of very large size?
(c) 'Danger' signals are always red in colour. Why?

Answer :

(a) The scattering of light by colloidal-sized particles, which makes the path of a beam of light visible, is called the Tyndall effect.

(b) When sunlight is scattered by particles of very large size, such as in fog or clouds, the scattered light appears white, because all colours of light are scattered almost equally by such large particles.

(c) Red light has the longest wavelength among visible colours and is scattered the least by fog, smoke or dust particles in the air, so it can be seen clearly from a large distance even in poor visibility. This is why danger signals are made red.

Teacher's Note:
Tyndall effect proves a mixture is a colloid, like a visible light beam in a dusty room.
Large particles scatter all wavelengths almost equally, giving white scattered light.
Red light (longest wavelength) scatters the least, so it travels the farthest.

38. [4 marks] Read the following passage and answer the questions given below: Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of a lens. By convention, a convex lens has a positive focal length while a concave lens has a negative focal length. When lenses are placed in contact, their combined power is determined by the sum of the individual powers of the lenses. Ray diagrams help to visualize how light converges or diverges through a lens to form an image.
(a) [1 mark] A convex lens of focal length 20 cm is used to form an image. If an object is placed at 40 cm from the lens, what will be the position and nature of the image?
(b) [1 mark] Illustrate the formation of image with the help of a ray diagram, when the object is placed between the optical centre and principal focus of a concave lens.
(c)(i) [2 marks] A lens combination consists of a convex lens of focal length 30 cm and a concave lens of focal length 15 cm placed together in contact. Find the equivalent focal length and power of this lens combination.
OR
(c)(ii) [2 marks] Two lenses are placed in contact. One is a concave lens with focal length 2 m and the other is a convex lens with focal length 1.5 m. What type of lens (convex or concave) will the combination behave as? Give reason.

Answer (a) :

Using the lens formula 1/v - 1/u = 1/f, with f = +20 cm and u = -40 cm:
1/v = 1/f + 1/u = 1/20 + 1/(-40) = 2/40 - 1/40 = 1/40, so v = +40 cm.
Since the object is at 2F (40 cm), the image also forms at 2F on the other side of the lens, at 40 cm. It is real, inverted, and of the same size as the object.

Answer (b) :

Ray diagram (described): when the object is placed between the optical centre (O) and the principal focus (F1) of a concave lens, a ray parallel to the axis appears, after refraction, to diverge from the near-side focus, and a ray through the optical centre passes straight. Tracing both rays backward, they meet on the same side as the object, giving a virtual, erect image that is smaller than the object, located between O and F1.

Answer (c)(i) :

f1 = 30 cm = 0.3 m (convex, positive), f2 = -15 cm = -0.15 m (concave, negative)
P1 = 1/f1 = 1/0.3 = +3.33 D; P2 = 1/f2 = 1/(-0.15) = -6.67 D
Equivalent power, P = P1 + P2 = 3.33 + (-6.67) = -3.33 D
Equivalent focal length, f = 1/P = 1/(-3.33) = -0.3 m = -30 cm (the combination behaves as a diverging/concave lens overall)

Answer (c)(ii) :

f1 = -2 m (concave), f2 = +1.5 m (convex)
P1 = 1/f1 = 1/(-2) = -0.5 D; P2 = 1/f2 = 1/1.5 = +0.67 D
Equivalent power, P = P1 + P2 = -0.5 + 0.67 = +0.17 D (approximately)
Since the equivalent power comes out positive (equivalent focal length is about +6 m, positive), the combination behaves like a convex (converging) lens overall.

Teacher's Note:
Convex lens focal length is positive, concave lens focal length is negative.
Combined power of lenses in contact, P = P1 + P2, where P = 1/f (f in metres), unit is dioptre (D).
An object at 2F of a convex lens always gives a real, inverted, same-size image.
A concave lens always gives a virtual, erect, diminished image, wherever the object is placed.

39. [5 marks] (a)(i) Consider the following electric circuit: a 2 Ω resistor in series with another 2 Ω resistor forms one branch, connected in parallel with a 4 Ω resistor; this parallel combination is in series with a 3 Ω resistor, an ammeter, a key and a 10 V battery. Calculate the values of the following:
(I) Total resistance of the circuit.
(II) The total electric current drawn from the source.
(III) Potential difference across the 3 Ω resistor.
(a)(ii) Two bulbs, rated as 100 W ; 220 V and 60 W ; 220 V are connected in parallel to an electric main supply of 220 V. Calculate the electric current drawn from the mains.
OR
(b)(i) State Ohm's law and draw the V-I graph for a conductor which follows Ohm's law. Show that the slope of this graph gives resistance of the conductor.
(b)(ii) Derive an expression for the equivalent resistance of a series combination of three resistors having resistances R₁, R₂ and R₃.

Answer (a) :

(I) The two 2 Ω resistors in series give 2+2 = 4 Ω. This 4 Ω is in parallel with the other 4 Ω resistor:
1/R' = 1/4 + 1/4 = 2/4, so R' = 2 Ω
This R' is in series with the 3 Ω resistor: R'' = R' + 3 = 2 + 3 = 5 Ω
So the total resistance of the circuit is 5 Ω.

(II) I = V/R = 10/5 = 2 A

(III) V = IR = 2 × 3 = 6 V

(ii) Since the two bulbs are connected in parallel, the voltage across each bulb is the same, 220 V.
Current through the 100 W bulb, I1 = P/V = 100/220 = 5/11 A
Current through the 60 W bulb, I2 = P/V = 60/220 = 3/11 A
Total current drawn from the mains, I = I1 + I2 = 5/11 + 3/11 = 8/11 A ≈ 0.73 A

Answer (b) :

(i) Ohm's law states that the potential difference (V) across the ends of a metallic conductor is directly proportional to the current (I) flowing through it, provided its temperature remains constant, that is, V = IR, where R is the resistance of the conductor.

V-I graph (described): plotting V on the y-axis and I on the x-axis for a conductor obeying Ohm's law gives a straight line through the origin. Taking two points on this line, the slope (change in V over change in I) equals V/I, which is constant and equal to the resistance R, so the slope of the V-I graph gives the resistance of the conductor.

(ii) Consider three resistors R1, R2 and R3 connected in series, with a battery supplying potential difference V across the combination, and the same current I flowing through all three resistors.
Potential difference across each: V1 = IR1, V2 = IR2, V3 = IR3
Let Rs be the equivalent series resistance, so V = I·Rs
Also, the total potential difference equals the sum of the individual potential differences:
V = V1 + V2 + V3, so I·Rs = IR1 + IR2 + IR3
Dividing throughout by I: Rs = R1 + R2 + R3

Teacher's Note:
In series, current stays the same through every resistor; in parallel, voltage stays the same across every branch.
For parallel resistors use 1/R = 1/R1 + 1/R2; for series, simply add resistances.
Power formula P = VI = V²/R is useful when only power and voltage are given.
Slope of the V-I graph for an ohmic conductor gives its resistance, R = V/I.

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