CBSE Class 10 Science Question Paper 2026 Solved Code 31-1-3

SECTION A

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

Answer: (D) they reproduce asexually

Teacher's Note:
a) Spirogyra, leishmania and hydra reproduce mainly by asexual methods like fragmentation, binary fission and budding.
b) Note that spirogyra is multicellular while leishmania is unicellular, so option (B) and (C) cannot be common to all.

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

Answer: (B) Stomata

Teacher's Note:
a) Stomata are tiny pores on leaves that open and close to allow \(CO_2\) and \(O_2\) exchange.
b) Xylem and phloem are transport tissues, not gas exchange structures.

3. Pancreas secretes pancreatic juice which contain certain enzyme that helps in digestion of food. [1 Mark]
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 digests proteins and lipase digests emulsified fats.

Teacher's Note:
a) Remember: trypsin acts on proteins, lipase acts on fats, amylase acts on starch.
b) Do not confuse pancreatic lipase (fats) with bile (which only emulsifies fats, does not digest them).

4. From the given situations, identify 'Chemotropic' and 'Geotropic' movements in parts of plants, respectively : [1 Mark]
(i) Growth of pollen tube towards ovule.
(ii) Movement of sunflower towards sunlight.
(iii) Movement of root towards Earth/Gravity.
(iv) Movement of leaves due to breeze.
Choose the correct option :
(A) (i) and (iii) respectively
(B) (iii) and (i) respectively
(C) (i), (ii) and (iii), (iv) respectively
(D) (i), (iii) and (ii), (iv) respectively

Answer: (A) (i) and (iii) respectively

Teacher's Note:
a) Chemotropic movement is growth in response to a chemical stimulus, seen in the pollen tube moving towards the ovule.
b) Geotropic movement is growth in response to gravity, seen in root growth downward.

5. Which of the following group is not 'biodegradable' ? [1 Mark]
(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) Polythene bag, rubber band, ball pen

Teacher's Note:
a) Biodegradable substances are broken down by microorganisms; man-made plastics and rubber are not.
b) A quick way to check: if it is natural (plant or animal origin), it is usually biodegradable.

6. Study the given diagram of the heart, with blood vessels marked '1' and '2'. From the following statements, establish the relationship between heart and/or the two blood vessels : [1 Mark]
(i) Blood vessel 1 - It carries carbon dioxide rich blood to the lungs.
(ii) Blood vessel 2 - It carries oxygen rich blood from the lungs.
(iii) Blood vessel 2 - Left atrium relaxes as it receives blood from this blood vessel.
(iv) Blood vessel 1 - Right atrium has thick wall as it has to pump blood to this vessel.
The option with correct statements is :
(A) (i) and (ii)
(B) (ii) and (iii)
(C) (ii), (iii) and (iv)
(D) (i), (ii) and (iii)

[Figure: Diagram of the heart showing two blood vessels at the top of the heart labelled '1' and '2', with arrows showing the direction of blood flow into the heart chambers.]

Answer: (D) (i), (ii) and (iii)

Teacher's Note:
a) Statement (iv) is wrong because the right atrium wall is thin, not thick, since it only receives blood and does not pump it far.
b) Blood vessel 1 is the pulmonary artery (carries deoxygenated blood to lungs) and vessel 2 is the pulmonary vein (carries oxygenated blood from lungs).

7. Plants use variety of techniques to get rid of their waste materials. Some are mentioned below. Identify the incorrect one : [1 Mark]
(A) Excess water is given out by transpiration.
(B) Gums and Resins are wastes that are stored.
(C) Roots secrete some wastes into the soil.
(D) Flowers can store some waste products.

Answer: (D) Flowers can store some waste products.

Teacher's Note:
a) Plants store wastes mainly in old leaves, bark, gums, resins and vacuoles, not specifically in flowers.
b) Remember the four ways plants excrete waste: transpiration, storage in leaves/bark, excretion into soil by roots, and shedding of leaves.

Directions : Question numbers 8 and 9 are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below :
(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.

8. Assertion (A) : Reflex actions do not involve thinking.
Reason (R) : Most reflex actions are controlled by the spinal cord. [1 Mark]
(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: (A) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Teacher's Note:
a) Reflex actions are quick, involuntary responses controlled by the spinal cord, bypassing conscious thought in the brain.
b) The spinal cord acts through a reflex arc, which is faster than sending signals to the brain first.

9. Assertion (A) : Ozone at higher levels of atmosphere is a product of UV radiation acting on oxygen molecule.
Reason (R) : The higher energy of UV splits apart some molecular O₂. [1 Mark]
(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: (A) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Teacher's Note:
a) UV radiation splits \(O_2\) molecules into free oxygen atoms which combine with \(O_2\) to form ozone \((O_3)\).
b) This ozone layer protects the Earth by absorbing harmful UV radiation.

10. (a) Why is bile juice considered to be very important for digestion, even though it doesn't contain any digestive enzymes ? [1 Mark]

Answer: Bile emulsifies fats by breaking them into tiny droplets. This increases the surface area for pancreatic enzymes and also makes the medium alkaline for enzyme action.

Teacher's Note:
a) Key term to write is "emulsification of fats".
b) Also mention that bile makes the acidic food from the stomach alkaline for intestinal enzymes.

(b) Name some substances present in initial filtrate and subsequently selectively reabsorbed in the tubules of nephron. [1 Mark]

Answer: Glucose, water, amino acids and salts are present in the initial filtrate and are selectively reabsorbed by the tubules.

Teacher's Note:
a) Only useful substances like glucose, water, salts and amino acids are reabsorbed, not urea.
b) Reabsorption happens as the filtrate flows along the tubule of the nephron.

11. Draw a neat diagram to show germination of pollen on the female reproductive part of the flower. Name and label only the following parts : [2 Marks]
(a) The part that receives the pollen grain.
(b) The structure that carries the male germ cell to reach the female germ cell.

[Figure: Diagram of the pistil of a flower showing pollen grains landing on the stigma, with a pollen tube growing down through the style into the ovary and reaching the ovule inside.]

Answer:
(a) Stigma - the part that receives the pollen grain.
(b) Pollen tube - the structure that carries the male germ cell down to the female germ cell.

Teacher's Note:
a) Label the stigma at the top of the pistil and the pollen tube growing through the style.
b) Marks are given separately for the diagram and for correct labelling of both parts.

12. (a) State two differences between the act of chewing food and salivation on sight of food. [2 Marks]

Answer:
1. Chewing of food is a voluntary action controlled by the forebrain.
2. Salivation on sight of food is a reflex/involuntary action controlled by the medulla in the hindbrain.

Teacher's Note:
a) Voluntary actions are consciously controlled while reflex actions are automatic.
b) Remember the forebrain controls voluntary actions and the medulla controls many involuntary reflexes.

OR

(b) State two differences between pollination and fertilization. [2 Marks]

Answer:
1. Pollination is the transfer of pollen grains from the anther to a suitable stigma and occurs only in plants.
2. Fertilization is the fusion of the male gamete with the female gamete and occurs in both plants and animals.

Teacher's Note:
a) Pollination always happens before fertilization.
b) Do not confuse pollination (transfer) with fertilization (fusion of gametes).

13. Give differences between the following : [3 Marks]
(a) Nephron and neuron
(b) Sensory nerve and motor nerve
(c) Consumers and decomposers

Answer:
1. Nephron is the structural and functional unit of the kidney which filters nitrogenous wastes from blood, while neuron is the structural and functional unit of the nervous system which transmits information from one part of the body to another.
2. Sensory nerve carries impulses from receptors to the central nervous system, while motor nerve carries impulses from the CNS to the effector organ.
3. Consumers are organisms that feed on producers and other consumers and transfer energy through the food chain, while decomposers break down dead organic matter into simpler substances and recycle nutrients into the environment.

Teacher's Note:
a) Give at least one clear structural or functional point for each pair.
b) Keywords like "filtration unit", "impulse conduction", "recycling nutrients" fetch marks.

14. Given below is a pyramid showing various trophic levels in an ecosystem : [3 Marks]
Class Interval: (iv) Tertiary consumers | (iii) Secondary consumers | (ii) Primary consumers | (i) Producers

[Figure: A pyramid shape divided into four horizontal levels from bottom to top, labelled (i) Producers at the base, (ii) Primary consumers, (iii) Secondary consumers, and (iv) Tertiary consumers at the top.]

(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:
1. (i) Grass, (ii) Deer and Rabbit, (iii) Snake and Lion, (iv) Lion.
2. Primary consumers feed directly on green plants (producers) which have the largest amount of energy; only about 10% of this energy is transferred to the next trophic level (secondary consumers).
3. The base of the pyramid is broad because the number, energy and mass of producers is usually the highest among all trophic levels.

Teacher's Note:
a) Remember the 10% law of energy transfer between trophic levels.
b) Lion can be both a secondary and tertiary consumer depending on its prey.

15. Mendel took garden pea plants with different characteristics, such as height to study the inheritance pattern of factors (genes). He crossed tall pea plant with short pea plant and obtained all the tall plants in the F₁ generation.
Answer the following questions :

(a) Why only tall pea plants were observed in F₁ progeny ? [1 Mark]

Answer: The F1 plants have genotype Tt, where 'T' (tall) is dominant over 't' (short). Since tallness is the dominant trait, all F1 plants were tall.

Teacher's Note:
a) Use the term "dominant trait" to explain the result.
b) Genotype Tt always shows the dominant phenotype.

(b) By which method did Mendel obtain F₂ progeny ? [1 Mark]

Answer: Mendel obtained the F2 progeny by self-pollinating the F1 plants.

Teacher's Note:
a) Self-pollination allows recessive traits hidden in F1 to reappear in F2.
b) F2 progeny showed both tall and short plants in a 3:1 ratio.

(c) (i) Write one difference between dominant and recessive trait. [2 Marks]

Answer: A dominant trait expresses itself even in the presence of a recessive trait (in both TT and Tt conditions), while a recessive trait can express itself only in the pure or homozygous condition (tt).

Teacher's Note:
a) Dominant traits mask recessive traits when both are present together.
b) Recessive traits show up only when both alleles are recessive.

OR

(c) (ii) Write two observations made by Mendel for F₁ progeny. [2 Marks]

Answer:
1. All plants of the F1 progeny were tall.
2. No medium or short height plants were observed in the F1 progeny; F1 progeny resembled only one parent.

Teacher's Note:
a) F1 always resembles only the dominant parent.
b) The recessive trait is hidden but not lost, it reappears in F2.

16. (a) Given below are certain situations. Analyse each and describe its possible impact : [5 Marks]
(i) A population of bacteria living in temperate waters whose temperature increased by 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:
1. Most of the bacteria would die, but a few heat-resistant variants would survive and grow further.
2. Fertilization occurs, forming a zygote.
3. Cross pollination may occur leading to fruit formation, otherwise no fertilization and no fruit is formed.
4. If the egg is not fertilised, the thick spongy uterine lining breaks down and is released as blood and mucus through the vagina, known as menstruation.
5. The seed will germinate and develop into a seedling.

Teacher's Note:
a) Each situation is worth 1 mark, so write a short, direct answer for each.
b) Link situation (i) to variation and survival of the fittest.

OR

(b) Given below are certain situations. Analyse and describe what would happen when : [5 Marks]
(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 land on the stigma of totally unrelated species ?
(iv) Copper-T is placed in the uterus of a human female ?
(v) Spirogyra breaks into smaller fragments upon maturation ?

Answer:
1. The spores will land on a suitable substance and, when they get adequate moisture and temperature, develop into a new bread mould (Rhizopus).
2. New plants will grow from the notches (buds) present on the margins of the fallen bryophyllum leaves.
3. The pollen tube will not form and no fertilisation will take place because the pollen is of an unrelated species.
4. Copper-T will act as a contraceptive device and prevent fertilization/pregnancy.
5. Each fragment of Spirogyra will grow into a new individual organism.

Teacher's Note:
a) These are all examples of asexual reproduction except (iii) and (iv) which relate to sexual reproduction and contraception.
b) Use precise biological terms like "spore germination", "vegetative propagation" and "incompatibility".

SECTION B

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

Answer: (D) NO₂ and O₂

Teacher's Note:
a) This is a thermal decomposition reaction: \(2Pb(NO_3)_2 \rightarrow 2PbO + 4NO_2 + O_2\).
b) Remember the brown fumes of \(NO_2\) gas are a characteristic observation of this reaction.

18. (i) AgNO₃ + NaCl → NaNO₃ + AgCl
(ii) K₂SO₄ + BaCl₂ → BaSO₄ + 2KCl
Which of the following options clearly describes both the reactions ? [1 Mark]
(A) (i) is double displacement, (ii) is 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) is displacement, (ii) is double displacement reaction.

Answer: (C) Both, (i) and (ii) are double displacement reactions and precipitation reactions.

Teacher's Note:
a) In both reactions, ions are exchanged between two compounds, forming an insoluble precipitate (AgCl and BaSO4).
b) A reaction that forms an insoluble solid from two solutions is called a precipitation reaction.

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

Answer: (B) Vanilla essence

Teacher's Note:
a) An olfactory indicator changes smell in acidic or basic medium, useful for visually impaired students.
b) Vanilla essence and onion extract are common examples of olfactory indicators.

20. The hydrocarbons with general formula C_nH_2n represents : [1 Mark]
(A) alkane
(B) alkene
(C) alkyne
(D) cyclic compounds

Answer: (B) alkene

Teacher's Note:
a) Remember: alkanes are \(C_nH_{2n+2}\), alkenes are \(C_nH_{2n}\), alkynes are \(C_nH_{2n-2}\).
b) Alkenes contain one double bond between carbon atoms.

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

Answer: (B) Calcium

Teacher's Note:
a) Calcium reacts with water moderately, and the bubbles of hydrogen gas formed stick to its surface, making it float.
b) Potassium and sodium react so vigorously that they catch fire, which is a different observation from simple floating.

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

Answer: (A) Pb

Teacher's Note:
a) Silver and copper are excellent conductors, aluminium is a good conductor, while lead is comparatively a poor conductor among metals.
b) Conductivity order to remember: Ag > Cu > Al > ... > Pb.

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

Answer: (A) tomato, curd, ant-sting

Teacher's Note:
a) Oxalic acid is found in tomatoes, lactic acid is found in curd, and methanoic acid is found in ant and bee stings.
b) Methanoic acid is also called formic acid and causes the burning sensation from stings.

Directions : For question number 24, two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to this question from the codes (A), (B), (C) and (D) as given below :
(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.

24. 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 lead to both atoms attaining the noble gas configuration. [1 Mark]
(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: (A) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Teacher's Note:
a) Carbon forms covalent bonds by sharing electrons rather than gaining or losing them.
b) Sharing of electrons helps both atoms achieve a stable noble gas electronic configuration.

25. Why does one feel pain and irritation when stung by honey-bee ? Rubbing of baking soda on the stung area gives relief. How ? [2 Marks]

Answer:
1. A honey-bee sting injects methanoic acid (an acid) into the skin, which causes pain and irritation.
2. Baking soda is a mild base (alkaline); rubbing it on the stung area neutralises the acid, giving relief.

Teacher's Note:
a) The key concept is neutralisation of acid by a base.
b) Remember baking soda (sodium bicarbonate) is a common household remedy for acid stings.

26. What happens when : [3 Marks]
(a) An iron nail is dipped in copper (II) sulphate solution ?
(b) Potassium iodide solution is mixed with lead nitrate solution ?
(c) Silver chloride is exposed to sunlight ?
Write balanced chemical equations to support your answer.

Answer:
1. The blue colour of copper sulphate solution changes to pale green as iron displaces copper: \(Fe(s) + CuSO_4(aq) \rightarrow FeSO_4(aq) + Cu(s)\).
2. A yellow precipitate of lead iodide is formed: \(Pb(NO_3)_2(aq) + 2KI(aq) \rightarrow PbI_2(s) + 2KNO_3(aq)\).
3. The white colour of silver chloride changes to grey on exposure to sunlight: \(2AgCl \xrightarrow{sunlight} 2Ag + Cl_2\).

Teacher's Note:
a) (a) is a displacement reaction, (b) is a double displacement/precipitation reaction, (c) is a photochemical decomposition reaction.
b) Colour change observations are important value points, do not skip them.

27. (a) Explain chlor-alkali process with chemical equation. Name the products formed at anode and cathode. [3 Marks]

Answer:
1. When electricity is passed through a concentrated solution of sodium chloride (brine), it decomposes to form sodium hydroxide, chlorine and hydrogen gas; this is called the chlor-alkali process: \(2NaCl(aq) + 2H_2O(l) \xrightarrow{electricity} 2NaOH(aq) + H_2(g) + Cl_2(g)\).
2. At the anode, chlorine gas \((Cl_2)\) is formed.
3. At the cathode, hydrogen gas \((H_2)\) is formed.

Teacher's Note:
a) NaOH remains in the solution near the cathode.
b) Remember the process name comes from "chlor" (chlorine) and "alkali" (NaOH).

OR

(b) Write the preparation of following compounds with balanced chemical equation : [3 Marks]
(i) Baking soda
(ii) Bleaching powder
(iii) Plaster of Paris

Answer:
1. Baking soda: \(NaCl + H_2O + NH_3 + CO_2 \rightarrow NaHCO_3 + NH_4Cl\).
2. Bleaching powder: \(Ca(OH)_2 + Cl_2 \rightarrow CaOCl_2 + H_2O\).
3. Plaster of Paris: \(CaSO_4 . 2H_2O \xrightarrow{373K} CaSO_4 . \frac{1}{2}H_2O + 1\frac{1}{2}H_2O\).

Teacher's Note:
a) Plaster of Paris is made by heating gypsum at 373 K.
b) Balance all equations correctly since marks are deducted for incorrect balancing.

28. Read the following passage and answer the questions given below :

Most of metals occur in combined state in form of ores. Carbonate ores are converted into oxides by calcination and sulphide ores by roasting. Oxides are reduced with suitable reducing agent like carbon to get free metal. Highly reactive metals like Al, Mg are also used as reducing agents to obtain metal from their oxides. Most reactive metals are obtained by electrolytic reduction of their molten ores. Alloying is a very good method of improving the properties of a metal. We can get desired properties by this method. The electrical conductivity and melting point of an alloy is less than that of pure metals.

(a) Why carbonate or sulphide ores are converted to oxides before extraction of metal from it ? [1 Mark]

Answer: Because it is easier to reduce metal oxides to free metal compared to carbonates or sulphides.

Teacher's Note:
a) Metal oxides are more reactive towards common reducing agents like carbon.
b) Calcination is used for carbonate ores, roasting for sulphide ores.

(b) Write a reaction in which Aluminium is used as a reducing agent to obtain metal from its oxide. [1 Mark]

Answer: \(Fe_2O_3(s) + 2Al(s) \rightarrow 2Fe(l) + Al_2O_3(s) + Heat\)

Teacher's Note:
a) This is known as the thermite reaction, used to join railway tracks.
b) A large amount of heat is released, which melts the iron formed.

(c) (i) How is copper obtained from its ore (Cu₂S) ? Give equations of the reactions. [2 Marks]

Answer:
1. The sulphide ore is first roasted in air to partially convert it to copper(I) oxide: \(2Cu_2S + 3O_2(g) \xrightarrow{Heat} 2Cu_2O(s) + 2SO_2(g)\).
2. The remaining copper(I) sulphide then reacts with copper(I) oxide to give copper metal: \(2Cu_2O + Cu_2S \xrightarrow{Heat} 6Cu(s) + SO_2(g)\).

Teacher's Note:
a) This is a self-reduction process, no external reducing agent is needed.
b) Both equations must be written to get full marks.

OR

(c) (ii) (I) Why highly reactive metals cannot be obtained from their oxides by using carbon as a reducing agent ? [1 Mark]

Answer: Because highly reactive metals have a greater affinity for oxygen than carbon does, so carbon cannot remove oxygen from their oxides.

Teacher's Note:
a) Such metals are extracted by electrolytic reduction instead.
b) Compare the activity series to justify this reasoning.

(II) Why solder, an alloy of lead and tin, is used for welding electrical wires together ? [1 Mark]

Answer: Solder has a low melting point, which allows it to melt easily and join electrical wires without damaging them.

Teacher's Note:
a) Low melting point is the key property tested here.
b) Solder also does not conduct as much heat as pure metals, protecting nearby components.

29. (a) (i) Give reasons for the following : [3 Marks]
(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.

Answer:
1. Covalent compounds do not give rise to charged particles/ions, so they cannot conduct electricity.
2. Soap reacts with calcium and magnesium salts present in hard water to form an insoluble substance called scum, so no lather is formed.
3. Carbon-carbon bonds are strong and stable while silicon-silicon bonds are relatively weak, so silicon does not show catenation like carbon.

Teacher's Note:
a) Absence of ions is the key reason for poor conductivity of covalent compounds.
b) Bond strength explains why catenation is unique to carbon.

(ii) Write chemical equations for the following : [2 Marks]
(I) Oxidation of ethanol by acidified K₂Cr₂O₇.
(II) Hydrogenation of ethene.

Answer:
1. \(C_2H_5OH \xrightarrow{acidified\ K_2Cr_2O_7 + heat} CH_3COOH\)
2. \(CH_2=CH_2 + H_2 \xrightarrow{Ni} CH_3-CH_3\)

Teacher's Note:
a) Ethanol oxidises to ethanoic acid (acetic acid) using acidified potassium dichromate as oxidising agent.
b) Hydrogenation of ethene needs a nickel catalyst and gives ethane.

OR

(b) Mohan heated ethanol with a compound 'X' in the presence of a few drops of conc. H₂SO₄ and observed a sweet smelling compound 'Y' is formed. When 'Y' is treated with sodium hydroxide it gives back ethanol and a compound 'Z'. [5 Marks]
(i) Identify 'X', 'Y' and 'Z'.
(ii) Write the role of conc. H₂SO₄ in the reaction.
(iii) Write the chemical equations involved and name the reactions.

Answer:
1. X is ethanoic acid \((CH_3COOH)\), Y is ethyl ethanoate/ester \((CH_3COOC_2H_5)\), Z is sodium ethanoate \((CH_3COONa)\).
2. Concentrated \(H_2SO_4\) acts as a catalyst and a dehydrating agent in the reaction.
3. Esterification: \(CH_3COOH + C_2H_5OH \xrightarrow{acid} CH_3COOC_2H_5 + H_2O\).
4. Saponification/de-esterification: \(CH_3COOC_2H_5 + NaOH \rightarrow CH_3COONa + C_2H_5OH\).

Teacher's Note:
a) The sweet smell is a characteristic identification point for an ester.
b) Name both reactions correctly: esterification (forward) and saponification (reverse) for full marks.

SECTION C

30. 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 : [1 Mark]
(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) - 30 cm and + 30 cm from lens

Teacher's Note:
a) Equal size real image forms only when the object is placed at \(2F\), i.e. \(u = -2f = -30\) cm, and image also forms at \(2F\) on the other side, i.e. \(v = +30\) cm.
b) Remember sign convention: distances measured against incident light direction are negative.

31. When you look at an object very close to your eyes, the : [1 Mark]
(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 of your eye contract and the eye lens becomes thick.

Teacher's Note:
a) This is called accommodation, allowing the eye to focus on nearby objects.
b) For distant objects, the ciliary muscles relax and the lens becomes thin.

Directions : For question number 32, two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to this question from the codes (A), (B), (C) and (D) as given below :
(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.

32. Assertion (A) : When rays of white light pass through a prism, on emerging they give 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. [1 Mark]
(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) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) The splitting of white light into seven colours is due to dispersion, not scattering.
b) Different colours bend by different amounts because each colour has a different wavelength and refractive index in glass.

33. Draw the ray diagram for the image formation by a lens which shows a magnification of + 2. [2 Marks]

[Figure: Ray diagram of a convex lens with object AB placed between the focus \(F_1\) and the optical centre O. Two rays are drawn from the top of the object: one parallel to the principal axis refracting through \(F_2\), and one passing through the optical centre O undeviated. Both refracted rays are extended backward on the same side as the object to meet at A', forming a virtual, erect and magnified image A'B' larger than the object.]

Answer: A magnification of +2 means the image is virtual, erect and twice the size of the object. This happens for a convex lens when the object is placed between the optical centre and the principal focus \(F_1\), forming a virtual, magnified, erect image on the same side as the object.

Teacher's Note:
a) Positive magnification always means the image is virtual and erect.
b) Show at least two standard rays clearly and mark the virtual image with dotted lines.

34. (a) The resistance of a wire of 0.01 cm radius and 1.0 cm length is 7 Ω. Calculate its resistivity. [2 Marks]

Answer:
Given: \(r = 0.01\ cm = 1 \times 10^{-4}\ m\), \(l = 1\ cm = 0.01\ m\), \(R = 7\ \Omega\)
Formula: \(R = \rho \frac{l}{A}\), so \(\rho = \frac{RA}{l} = \frac{R \pi r^2}{l}\)
\(\rho = \dfrac{7 \times \frac{22}{7} \times (1\times 10^{-4})^2}{0.01}\)
\(\rho = \dfrac{22 \times 10^{-8}}{0.01} = 22 \times 10^{-6}\ \Omega m\)
\(\rho = 2.2 \times 10^{-5}\ \Omega m\)

Teacher's Note:
a) Always convert radius and length to metres before substitution.
b) Remember area of cross-section is \(A = \pi r^2\), not \(\pi d^2\).

OR

(b) An electric heater is rated 220 V ; 11 A. Calculate the power consumed if the heater is operated at 200 V. [2 Marks]

Answer:
Resistance of heater: \(R = \dfrac{V}{I} = \dfrac{220}{11} = 20\ \Omega\)
Power at 200 V: \(P = \dfrac{V^2}{R} = \dfrac{200 \times 200}{20} = 2000\ W = 2\ kW\)

Teacher's Note:
a) Resistance of the heater stays constant; only the operating voltage changes.
b) Use \(P = V^2/R\) since resistance is known and voltage is given.

35. (a) The pattern of magnetic field due to a current carrying wire depends upon the shape made by that wire. Justify. [3 Marks]

Answer:
1. A straight current-carrying wire produces a magnetic field in the form of concentric circles around it.
2. A current-carrying circular loop produces a magnetic field pattern of concentric circles that become straight and parallel near the centre of the loop.
3. A current-carrying solenoid produces a magnetic field pattern similar to that of a bar magnet, with field lines emerging from one end and entering the other.

Teacher's Note:
a) Give any two of the three cases with correct field line patterns for full marks.
b) This shows that the shape of the conductor changes the magnetic field pattern, justifying the statement.

(b) A current carrying straight wire AB is shown in the given diagram. X, Y and Z in which point will the strength of magnetic field be maximum and why ?

[Figure: A vertical straight wire AB carrying electric current i (direction marked), with three points X, Y and Z placed along a horizontal line to the right of the wire, with X closest to the wire, then Y, then Z farthest away.]

Answer: The magnetic field strength will be maximum at point X, because the strength of the magnetic field decreases as the distance from the current-carrying wire increases.

Teacher's Note:
a) Magnetic field strength around a straight wire is inversely proportional to the distance from the wire.
b) Remember X is closest to the wire among X, Y and Z, so it experiences the strongest field.

36. (a) Describe an activity to show that a current carrying conductor, placed in an external magnetic field experiences a force. [3 Marks]

Answer:
1. Take a small aluminium rod AB and suspend it horizontally using two connecting wires from a stand.
2. Place a strong horse-shoe magnet so that the rod lies between its two poles, perpendicular to the magnetic field.
3. Connect the rod in series with a battery and a key, then pass current through it from one end to the other; it is observed that the rod gets displaced, showing that it experiences a force due to the magnetic field.

Teacher's Note:
a) The displacement of the rod on passing current is the key observation to mention.
b) This activity demonstrates the motor effect of current, later used in Fleming's left hand rule.

(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: The magnetic field is directed vertically downwards.

Teacher's Note:
a) Use Fleming's left hand rule, remembering that current direction is opposite to the direction of electron motion.
b) Point fingers along conventional current (opposite to electron flow) and thumb along the force (to the right) to find the field direction.

37. (a) What is hypermetropia ? [1 Mark]

Answer: Hypermetropia is an eye defect in which a person can see distant objects clearly but cannot see nearby objects clearly.

Teacher's Note:
a) Also called far-sightedness or long-sightedness.
b) Do not confuse it with myopia, which is the opposite defect.

(b) Write any one cause of hypermetropia. [1 Mark]

Answer: The focal length of the eye lens becomes too long, or the eyeball becomes too small.

Teacher's Note:
a) Either cause is acceptable for full marks.
b) This results in the image forming behind the retina.

(c) With the help of a suitable ray diagram, explain how hypermetropia is corrected ? [1 Mark]

[Figure: Ray diagram showing a convex lens placed in front of the eye. Rays from a nearby object N converge after passing through the convex lens, appearing to come from a farther point N' before entering the eye lens, allowing the eye to focus the image correctly on the retina.]

Answer: Hypermetropia is corrected by using a convex lens, which converges the incoming rays slightly before they enter the eye, so that the image forms correctly on the retina.

Teacher's Note:
a) Convex lens is the key correcting lens for hypermetropia.
b) The diagram should show the object N appearing to shift to N' after passing through the corrective lens.

38. 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 lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.

(a) 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 image ? [1 Mark]

Answer:
Given: \(f = +20\ cm\), \(u = -40\ cm\)
Using lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{20} + \frac{1}{-40} = \frac{1}{40}\)
\(v = +40\ cm\)
Position: image forms at 40 cm on the other side of the lens (at \(2F\)).
Nature: Real and inverted, same size as the object.

Teacher's Note:
a) Object at \(2F\) always produces a real, inverted, same-size image at \(2F\) on the other side.
b) Show the correct sign convention (u negative, f positive for convex lens).

(b) Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens. [1 Mark]

[Figure: Ray diagram of a concave lens with object AB placed between the optical centre O and the principal focus \(F_1\). A ray parallel to the principal axis diverges after refraction as if coming from \(F_1\); a ray through the optical centre goes straight. Both refracted rays are extended backward to meet at A', giving a virtual, erect and diminished image A'B' on the same side as the object, closer to the lens.]

Answer: When the object is placed between the optical centre and the principal focus of a concave lens, a virtual, erect and diminished image is formed on the same side as the object, between the optical centre and the focus.

Teacher's Note:
a) A concave lens always forms virtual, erect and diminished images, regardless of object position.
b) Use dotted lines to show the virtual rays and the virtual image.

(c) (i) A lens combination consists of a convex lens of focal length 30 cm and a concave lens of focal length 15 cm placed together. Find the equivalent focal length and power of this lens combination. [2 Marks]

Answer:
Given: \(f_1 = +30\ cm = 0.3\ m\), \(f_2 = -15\ cm = -0.15\ m\)
\(P_1 = \frac{1}{0.3} = +3.33\ D\), \(P_2 = \frac{-1}{0.15} = -6.67\ D\)
Equivalent power: \(P = P_1 + P_2 = -3.33\ D\)
Equivalent focal length: \(f = \frac{1}{P} = \frac{-1}{3.33} = -0.3\ m = -30\ cm\)

Teacher's Note:
a) Convert focal lengths to metres before calculating power in dioptres.
b) A negative equivalent focal length means the combination behaves like a concave lens.

OR

(c) (ii) 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 will the combination behave as (convex or concave) ? Give reason. [2 Marks]

Answer:
Given: \(f_1 = -2\ m\) (concave), \(f_2 = +1.5\ m\) (convex)
\(P_1 = \frac{-1}{2} = -0.5\ D\), \(P_2 = \frac{1}{1.5} = +0.67\ D\)
Equivalent power: \(P = P_1 + P_2 = +\frac{1}{6}\ D\)
Equivalent focal length: \(f = 6\ m\) (positive)
Since the equivalent focal length and power are positive, the combination behaves as a convex lens.

Teacher's Note:
a) A positive equivalent power means the combination is convergent, like a convex lens.
b) Convex lens has a smaller focal length here, so it dominates the combination.

39. (a) Study the given electric circuit in which 2 A electric current is flowing between points X and Y.

[Figure: A circuit showing points X and Y connected by two resistors in the top path, 2Ω and 4Ω in series, with a voltmeter connected across them showing polarity - and +. An ammeter reading 2A is connected near Y. A battery and a key are connected in the main circuit, with 2A current flowing through the whole loop as marked by arrows.]

(i) Using the battery, key, voltmeter and ammeter in this given electric circuit, redraw a circuit diagram in which 2 Ω and 4 Ω resistors are connected between X and Y in parallel combination. [3 Marks]

Answer: The circuit is redrawn with the 2 Ω and 4 Ω resistors connected between the same two points X and Y as two separate parallel branches, each branch having one resistor, both branches joined at X and Y, connected to the battery and key in the main circuit, with an ammeter in series with the battery and a voltmeter connected across the parallel combination (across X and Y).

Teacher's Note:
a) In a parallel circuit, both resistors are connected between the same two points, unlike in series where they are one after another.
b) The voltmeter is always connected across the combination, and the ammeter in the main line.

(ii) Circuit drawn by you, in which resistors are connected in parallel combination, calculate the electric current flowing through 4 Ω resistor. [2 Marks]

Answer:
From the series circuit given, potential difference across the battery: \(V = I(R_1+R_2) = 2 \times (2+4) = 12\ V\)
This same voltage is applied across the parallel combination.
Current through the 4 Ω resistor: \(I = \frac{V}{R} = \frac{12}{4} = 3\ A\)

Teacher's Note:
a) First find the battery voltage from the series circuit data given in the diagram.
b) In a parallel combination, the same voltage appears across each resistor branch.

OR

(b) (i) Two lamps 'A' and 'B' of rating 50 W ; 220 V and 100 W, 220 V are connected in series combination. Find out the ratio of the resistances (R_A : R_B) of these lamps. [5 Marks]

Answer:
\(R_A = \dfrac{V^2}{P_A} = \dfrac{220 \times 220}{50} = 968\ \Omega\)
\(R_B = \dfrac{V^2}{P_B} = \dfrac{220 \times 220}{100} = 484\ \Omega\)
\(\dfrac{R_A}{R_B} = \dfrac{968}{484} = 2\)
\(R_A : R_B = 2 : 1\)

Teacher's Note:
a) Use \(R = V^2/P\) since voltage rating is the same for both lamps.
b) Lower wattage lamp has higher resistance for the same rated voltage.

(ii) Derive the expression for the equivalent resistance of three resistors R₁, R₂ and R₃ connected in parallel combination.

Answer:
1. In a parallel combination, the potential difference V across each resistor is the same, while the total current I splits into \(I_1, I_2, I_3\) through \(R_1, R_2, R_3\) respectively.
2. Currents through each resistor: \(I_1 = \frac{V}{R_1}\), \(I_2 = \frac{V}{R_2}\), \(I_3 = \frac{V}{R_3}\).
3. If \(R_P\) is the equivalent resistance, then total current drawn from the battery is \(I = \frac{V}{R_P}\).
4. Since \(I = I_1 + I_2 + I_3\): \(\frac{V}{R_P} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}\).
5. Dividing throughout by V: \(\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\).

Teacher's Note:
a) In parallel, voltage is common and current divides, unlike series where current is common.
b) The final formula shows equivalent resistance in parallel is always less than the smallest individual resistance.

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