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

SECTION - A

1. 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 small pores mostly on the lower surface of leaves that control the entry and exit of gases.
b) Do not confuse stomata with xylem and phloem, which are transport tissues, not gas exchange structures.

2. Identify the type of reproduction shown in the diagram given below: [1 Mark]
(A) Budding
(B) Fragmentation
(C) Spore Formation
(D) Binary Fission

[Figure: A single-celled organism shown dividing repeatedly - one cell splits into two equal daughter cells, which further split into smaller identical cells, indicating repeated equal division of one cell into two.]

Answer: (D) Binary Fission

Teacher's Note:
a) Binary fission is asexual reproduction in which a parent cell divides into two equal daughter cells, seen in Amoeba and bacteria.
b) Look for equal-sized daughter cells in the diagram to identify binary fission and not budding or fragmentation.

3. 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 reproduces by fragmentation, Leishmania by binary fission, and Hydra by budding - all asexual methods.
b) Note that Hydra is multicellular while Spirogyra and Leishmania are not always unicellular in the same way, so option (D) is the only common correct statement.

4. 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: [1 Mark]
(A) Cerebrum
(B) Cerebellum
(C) Medulla
(D) Pons

Answer: (B) Cerebellum

Teacher's Note:
a) The cerebellum maintains posture, balance and precision of voluntary movements.
b) Do not confuse it with the cerebrum, which controls thinking and voluntary actions but not their precision.

5. Pancreas secretes pancreatic juice which contain certain enzyme that helps in digestion of food.
Choose the correct option from the following: [1 Mark]
(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 and lipase acts on fats, both secreted in pancreatic juice.
b) A common mistake is swapping the substrates of trypsin and lipase.

6. Sex is determined by different factors in various species. However, in human beings, it is determined genetically.
Which amongst the following option(s) is/are correct for human beings?
(i) Gamete carrying X chromosome from female parent.
(ii) Gamete carrying X chromosome from male parent.
(iii) Gamete carrying Y chromosome from male parent.
Options: [1 Mark]
(A) (ii) and (iii)
(B) (i) only
(C) (i) and (iii)
(D) (iii) only

Answer: (A) (ii) and (iii)

Teacher's Note:
a) The male parent is heterogametic (XY) and produces both X and Y bearing sperms, while the female parent is homogametic (XX) and produces only X bearing eggs.
b) The sex of the child depends on whether the sperm carries X or Y chromosome.

7. 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; synthetic materials like plastics and rubber are non-biodegradable.
b) Remember: natural, organic materials degrade easily, man-made polymers do not.

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 without involving the brain's thinking process.
b) This is why reflex actions are faster than actions requiring conscious thought.

9. 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. [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) Haemoglobin has a very high affinity for oxygen, not carbon dioxide, which makes the Reason false.
b) The Assertion about haemoglobin being the respiratory pigment in red blood cells is correct.

10. What is the function of diaphragm in human respiratory system? Where is it present in human body? [2 Marks]

Answer:
1. The diaphragm flattens during inhalation, which expands the chest cavity and helps draw air into the lungs.
2. It is a muscular sheet located at the base of the chest cavity, separating it from the abdominal cavity.

Teacher's Note:
a) Remember the diaphragm's movement changes the volume of the thoracic cavity, driving breathing.
b) Students often forget to mention its location; always state it clearly for full marks.

11. (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, whereas salivation on sight of food is a reflex action controlled by the medulla in the hindbrain.
2. Chewing requires conscious will, while salivation on sight of food happens automatically without conscious control.

Teacher's Note:
a) Chewing is voluntary; salivation on sight is an involuntary reflex.
b) Link each action to the correct brain region for full marks.

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, whereas fertilization is the fusion of a male gamete with a female gamete.
2. Pollination occurs only in plants, while fertilization can occur in both plants and animals.

Teacher's Note:
a) Pollination always precedes fertilization in flowering plants.
b) Do not confuse pollination (transfer of pollen) with fertilization (fusion of gametes).

12. 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 germinating on the stigma at the top, a pollen tube growing down through the style, and reaching the ovule inside the ovary at the base.]

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

Teacher's Note:
a) Always label the stigma at the top and pollen tube growing through the style in your diagram.
b) Marks are given for a correctly labelled diagram, not just the names.

13. Given below is a pyramid showing various trophic levels in an ecosystem: [3 Marks]
(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?

[Figure: A four-tier pyramid diagram with the widest tier at the bottom labelled (i) Producers, followed upward by (ii) Primary consumers, (iii) Secondary consumers, and the narrowest top tier (iv) Tertiary consumers.]

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

Teacher's Note:
a) Lion can appear at more than one trophic level since it may eat herbivores (secondary role) or eat carnivores like snakes (tertiary role).
b) Remember the 10% law of energy transfer while explaining energy loss between trophic levels.
c) A broad base always represents the maximum biomass/energy at the producer level.

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

Teacher's Note:
a) Always link nephron to kidney and neuron to nervous system.
b) Remember direction of impulse flow: sensory - towards CNS, motor - away from CNS.
c) Decomposers recycle matter; consumers only transfer energy forward.

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: In the F₁ progeny, the pea plants have the genotype 'Tt', where 'T' (tall) is dominant over 't' (short), so all F₁ plants show the tall trait.

Teacher's Note:
a) Dominant trait always masks the recessive trait in a heterozygous condition.
b) Use genotype 'Tt' to justify why only the dominant phenotype appears.

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

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

Teacher's Note:
a) Self-pollination allows recessive traits hidden in F₁ to reappear in F₂.
b) Remember the classic 3:1 ratio obtained by Mendel in F₂ generation.

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

Answer: A dominant trait expresses itself even in the heterozygous condition (Tt) as well as homozygous condition (TT), whereas a recessive trait can express itself only in the homozygous condition (tt), being masked in the heterozygous state.

Teacher's Note:
a) Dominant traits need only one copy of the allele to show; recessive traits need two copies.
b) Use examples like tall (T) versus short (t) to explain clearly.

OR

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

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

Teacher's Note:
a) F₁ progeny shows uniformity, resembling only the dominant parent.
b) This observation led Mendel to conclude the concept of dominance.

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 due to the rise in temperature, but a few heat-resistant variants would survive and continue to grow, showing variation aids survival.
2. Fertilisation occurs, forming a zygote when the sperm meets the egg in the oviduct.
3. Since the flower has only a pistil (no stamens), cross pollination may occur to allow fertilisation and fruit formation; without pollen from another source, no fertilisation or fruit will form.
4. If the egg is not fertilised, the thick, spongy lining of the uterus breaks down and is shed through the vagina as blood and mucus, called menstruation.
5. The seed will germinate and develop into a seedling under suitable conditions of water and air.

Teacher's Note:
a) Link each situation to the correct biological process: variation, fertilisation, pollination, menstrual cycle, or germination.
b) Keep each answer to one clear line, as this is a 5-part, 5-mark question.
c) Common mistake: confusing pollination with fertilisation in part (iii).

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. When spores land on a suitable substance and get adequate moisture and temperature, they develop into a new bread mould (Rhizopus).
2. New plants grow from the buds located in the notches of the bryophyllum leaf when it falls on wet soil.
3. The pollen tube will not form because it is from an unrelated species, so no fertilisation will take place.
4. Copper-T prevents fertilisation/pregnancy by acting as a contraceptive device.
5. Each fragment of Spirogyra grows into a new independent individual organism.

Teacher's Note:
a) These are all examples of asexual reproduction except the pollination case.
b) Remember Copper-T is a mechanical/chemical contraceptive method.
c) Keep each point short and specific to the situation described.

SECTION - B

17. (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) Both reactions form an insoluble precipitate (AgCl and BaSO₄) and involve exchange of ions between two compounds.
b) Precipitation reactions are always a type of double displacement reaction.

18. 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) Vanilla essence is an olfactory indicator - its smell disappears in basic medium, making it useful for visually impaired students.
b) Turmeric, litmus and methyl orange rely on colour change, which cannot be detected without sight.

19. 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) Heating lead nitrate is a classic example of a thermal decomposition reaction.
b) Remember the brown fumes of NO₂ gas evolved along with oxygen.

20. The formula of functional group for aldehyde is: [1 Mark]
(A) -COOH
(B) -CHO
(C) -C(=O)-
(D) -OH

[Figure: Option (C) is printed as a vertical structural formula showing a carbon atom double bonded to an oxygen atom below it, representing a carbonyl-type group.]

Answer: (B) -CHO

Teacher's Note:
a) -CHO is the aldehyde functional group; -COOH is for carboxylic acid and -OH is for alcohol.
b) Option (C) shows a simple carbonyl group without a hydrogen, which is not specific to aldehydes.

21. 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, while lead has comparatively higher resistivity.
b) Metals generally conduct electricity, but their conductivities differ significantly.

22. 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 in curd, and methanoic (formic) acid in ant and nettle stings.
b) Remember these natural sources as they are commonly asked in board exams.

23. 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, causing it to float.
b) Potassium and sodium react vigorously and may catch fire, unlike calcium's steady reaction.

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 to complete its octet, achieving a stable noble gas configuration.
b) This sharing of electrons is the basis of covalent bonding in carbon compounds.

25. (a) What are amphoteric oxides? [1 Mark]
(b) Categorise the following based on their nature: [1 Mark]
ZnO, Na₂O, CO₂

Answer:
(a) Amphoteric oxides are metal oxides that can react with both acids and bases to produce salt and water.
(b) ZnO - amphoteric oxide, Na₂O - basic oxide, CO₂ - acidic oxide.

Teacher's Note:
a) ZnO and Al₂O₃ are the two most common amphoteric oxides asked in exams.
b) Metal oxides are generally basic, non-metal oxides are generally acidic; amphoteric ones are exceptions.

26. (a) Identify the substances oxidised and reduced in the following reaction: [1 Mark]
ZnO + C → Zn + CO
(b) Balance the following chemical reaction: [1 Mark]
Pb(NO₃)₂ + KI → PbI₂ + KNO₃
(c) Give one example each of electrolytic decomposition and decomposition by sunlight. [1 Mark]

Answer:
(a) Substance oxidised - C (carbon); substance reduced - ZnO.
(b) Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃
(c) Electrolytic decomposition: \(2H_2O \xrightarrow{electricity} 2H_2 + O_2\); Decomposition by sunlight: \(2AgCl \xrightarrow{sunlight} 2Ag + Cl_2\)

Teacher's Note:
a) The substance that gains oxygen is oxidised; the one that loses oxygen is reduced.
b) Always check that atoms of every element balance on both sides of the equation.
c) Remember these two classic examples of decomposition reactions for quick recall.

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 an aqueous solution of sodium chloride (brine), it decomposes to form sodium hydroxide and chlorine gas, so the process is called the chlor-alkali process.
2. \(2NaCl(aq) + 2H_2O(l) \xrightarrow{electricity} 2NaOH(aq) + H_2(g) + Cl_2(g)\)
3. At the anode, chlorine gas (Cl₂) is produced; at the cathode, hydrogen gas (H₂) is produced.

Teacher's Note:
a) "Chlor" stands for chlorine and "alkali" stands for sodium hydroxide, both formed in this process.
b) Always mention that the equation must be balanced and state the products at each electrode separately.

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: \(CaSO_4.2H_2O \xrightarrow{373K} CaSO_4.\tfrac{1}{2}H_2O + 1\tfrac{1}{2}H_2O\)

Teacher's Note:
a) Baking soda is made using the Solvay process ingredients: NaCl, water, ammonia and carbon dioxide.
b) Bleaching powder is made by passing chlorine gas over dry slaked lime.
c) Plaster of Paris is made by heating gypsum at 373 K; heating beyond this gives dead-burnt plaster.

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: It is because it is easier to obtain metal by reducing metal oxides rather than carbonate or sulphide ores directly.

Teacher's Note:
a) Carbonate ores are converted to oxides by calcination; sulphide ores by roasting.
b) Oxides are more easily reduced using carbon than carbonates or sulphides.

(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 an example of the thermite reaction, which is highly exothermic.
b) Aluminium is used because it is more reactive than iron and manganese.

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

Answer:
1. \(2Cu_2S + 3O_2(g) \xrightarrow{Heat} 2Cu_2O(s) + 2SO_2(g)\)
2. \(2Cu_2O + Cu_2S \xrightarrow{Heat} 6Cu(s) + SO_2(g)\)

Teacher's Note:
a) Copper is obtained by roasting the sulphide ore to partly convert it to oxide, then self-reduction occurs.
b) This process is called auto-reduction as Cu₂S itself reduces Cu₂O.

OR

(c) (ii) (I) Why highly reactive metals cannot be obtained from their oxides by using carbon as a reducing agent? [1 Mark]
(II) Why solder, an alloy of lead and tin, is used for welding electrical wires together? [1 Mark]

Answer:
(I) Highly reactive metals have a greater affinity for oxygen than carbon, so carbon cannot displace them from their oxides.
(II) Solder is used for welding electrical wires because of its low melting point, which allows easy joining without damaging the wires.

Teacher's Note:
a) Highly reactive metals like Al and Mg are extracted by electrolytic reduction instead of using carbon.
b) A low melting alloy like solder melts quickly and solidifies to form strong electrical joints.

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

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 it does not form lather.
3. Carbon-carbon bonds are strong and stable, allowing long chains, whereas silicon-silicon bonds are relatively weak, so silicon does not show catenation as strongly.
4. \(CH_3CH_2OH \xrightarrow{acidified\ K_2Cr_2O_7 + heat} CH_3COOH\)
5. \(CH_2=CH_2 + H_2 \xrightarrow{Ni} CH_3-CH_3\)

Teacher's Note:
a) Absence of free ions explains poor conductivity of covalent compounds.
b) Scum formation, not lack of cleaning ability, is why soap fails in hard water.
c) Catenation depends on bond strength; carbon's bonds are stronger than silicon's.

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 (CH₃COOH); Y is ethyl ethanoate, an ester (CH₃COOC₂H₅); Z is sodium ethanoate (CH₃COONa).
2. Conc. H₂SO₄ acts as a catalyst and dehydrating agent in the reaction.
3. \(CH_3COOH + CH_3CH_2OH \underset{Acid}{\rightleftharpoons} CH_3COOC_2H_5 + H_2O\) - this is called esterification.
4. \(CH_3COOC_2H_5 + NaOH \rightarrow CH_3COONa + C_2H_5OH\) - this is called saponification (de-esterification).

Teacher's Note:
a) Esters are known for their sweet, fruity smell, a key clue to identify compound Y.
b) Remember conc. H₂SO₄ removes water to drive the esterification reaction forward.
c) Reaction with NaOH regenerating alcohol and salt is called saponification.

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) An image of the same size as the object is formed only when the object is placed at 2F (twice the focal length).
b) Since f = 15 cm, 2F = 30 cm, so object is at -30 cm and image at +30 cm.

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) For near objects, ciliary muscles contract, decreasing focal length by making the lens thicker (more curved).
b) This process of adjusting lens shape is called accommodation.

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) Red light bends the least and violet bends the most because of their different wavelengths and refractive indices.

33. The given figure shows image formation by a lens. Analyse the figure and answer the following questions: [2 Marks]
(a) What is the type of lens used for image formation in the given ray diagram?
(b) If the real image is formed at a distance of 30 cm from the lens and the size of image is twice the size of the object, then where was the object placed?

[Figure: Ray diagram showing object AB (A on the principal axis, B above it) in front of a lens, with rays from B passing through the lens and converging below the axis at B' to form an inverted, larger image A'B', with A' below the axis further from the lens than A.]

Answer:
(a) Convex lens (converging lens).
(b) Using \(m = \dfrac{v}{u}\), with \(m = -2\) and \(v = 30\) cm: \(u = \dfrac{v}{m} = \dfrac{30}{-2} = -15\) cm. The object was placed at 15 cm in front of the lens.

Teacher's Note:
a) A real, inverted, magnified image is formed only by a convex lens.
b) Take magnification as negative for a real inverted image while substituting in the formula.

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

Teacher's Note:
a) Always convert radius and length to metres before substituting in the formula.
b) Resistivity depends only on the material, not on dimensions of the wire.

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:
\(R = \dfrac{V}{I} = \dfrac{220}{11} = 20\ \Omega\)
\(P = \dfrac{V^2}{R} = \dfrac{200 \times 200}{20} = 2000\ W = 2\ kW\)

Teacher's Note:
a) First find the resistance using rated values, since resistance stays constant.
b) Then use the new voltage to calculate the actual power consumed.

35. A person is unable to read a book placed closer than 1 meter from his eyes. Identify the defect of vision in his eyes. Draw the ray diagrams to show the defect of vision and its correction. [3 Marks]

Answer:
1. The defect is Hypermetropia (long-sightedness / far-sightedness).
2. In this defect, the image of a nearby object forms behind the retina because the eyeball is too short or the eye lens's focal length is too long; the ray diagram shows converging rays from a near object meeting beyond the retina.
3. Correction is done using a convex lens, which converges the light rays a little before they enter the eye, so the image forms exactly on the retina.

[Figure: Two eye diagrams - first shows rays from a near object converging behind the retina (uncorrected hypermetropic eye); second shows the same rays passing through a convex lens placed in front of the eye and converging exactly on the retina (corrected eye).]

Teacher's Note:
a) Hypermetropia is corrected using a convex lens of suitable power.
b) Always show the near point shift and direction of light rays in both diagrams.

36. (a) Describe an activity to show that a current carrying conductor, placed in an external magnetic field experiences a force. [3 Marks]
(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:
1. Take a small aluminium rod AB and suspend it horizontally between the poles of a strong horse-shoe magnet using connecting wires from a stand, so that the rod lies perpendicular to the magnetic field.
2. Connect the rod in series with a battery and a key, then pass current through it.
3. It is observed that the rod gets displaced when the current flows, showing that a current-carrying conductor experiences a force in a magnetic field.
4. For part (b): the magnetic field must be directed vertically downwards.

Teacher's Note:
a) The direction of force can be found using Fleming's Left Hand Rule.
b) In part (b), apply Fleming's Left Hand Rule with current direction (electron flow reversed) and given deflection to get the field direction.

37. (a) The pattern of magnetic field due to a current carrying wire depends upon the shape made by that wire. Justify. [3 Marks]
(b) A current carrying straight wire AB is shown in the given diagram. Out of X, Y and Z on which point will the strength of magnetic field be maximum and why?

[Figure: A vertical straight wire AB carrying electric current i flowing upward from B to A. Points X, Y and Z lie in a horizontal line to the right of the wire, with X nearest to the wire, Y in the middle, and Z farthest from the wire.]

Answer:
1. The magnetic field pattern for a current-carrying straight conductor consists of concentric circles around the wire.
2. The magnetic field pattern for a current-carrying circular loop is a pair of concentric circles, becoming nearly parallel straight lines at the centre.
3. The magnetic field pattern for a current-carrying solenoid resembles the field lines of a bar magnet, confirming that field pattern depends on the shape of the conductor.
4. The magnetic field 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) Give at least two shapes (straight wire, circular loop, or solenoid) with their field patterns for full marks.
b) Magnetic field strength is inversely related to the distance from the current-carrying conductor.

38. Read the following passage and answer the questions given:

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: Using the lens formula with \(f = +20\) cm and \(u = -40\) cm, the image is formed at \(v = +40\) cm on the other side of the lens; the image is real and inverted, of the same size as the object (since object is at 2F).

Teacher's Note:
a) When the object is at 2F, the image forms at 2F on the other side, real, inverted and of the same size.
b) Substitute values carefully with correct sign convention.

(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 showing an object AB placed between the optical centre O and principal focus F1 of a concave lens on the same side; one ray parallel to the principal axis appears to diverge from F1 after refraction, and another ray through the optical centre goes straight; both rays are extended backward to meet, forming a virtual, erect, diminished image A'B' between O and F1 on the same side as the object.]

Answer: The image formed by a concave lens is always virtual, erect and diminished, located between the optical centre and the focus, on the same side as the object.

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

(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:
\(f_1 = +0.3\ m\), \(f_2 = -0.15\ m\)
\(P_1 = \dfrac{1}{0.3} = +3.33\ D\), \(P_2 = \dfrac{1}{-0.15} = -6.67\ D\)
\(P = P_1 + P_2 = -3.33\ D\)
\(f = \dfrac{1}{P} = -0.3\ m = -30\ cm\)

Teacher's Note:
a) Convert focal lengths to metres before calculating power.
b) The equivalent power is the algebraic sum of individual lens powers.

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:
\(f_1 = -2\ m\) (concave), \(f_2 = +1.5\ m\) (convex)
\(P_1 = -0.5\ D\), \(P_2 = +0.667\ D\)
\(P = P_1 + P_2 = +\dfrac{1}{6}\ D\)
Since the equivalent power (and focal length) is positive, the combination will behave as a convex lens.

Teacher's Note:
a) A positive equivalent power always means the combination behaves like a convex lens.
b) Compare the magnitude of powers of both lenses to predict combination behaviour quickly.

39. (a) (i) Due to change in length and area of cross-section of a conductor, resistance of conductor changes while resistivity does not change. Why? [5 Marks]
(ii) Conductors of electric toasters and electric iron are made of an alloy rather than a pure metal. Why?
(iii) Define the S.I. unit of electric current.

Answer:
1. Resistance, \(R = \rho \dfrac{l}{A}\), depends on the length and area of cross-section of the conductor, so it changes when these dimensions change.
2. Resistivity is a characteristic property of the material only, so it does not change with a change in dimensions.
3. Alloys generally have higher resistivity than their constituent pure metals and do not oxidise (burn) easily at high temperatures, making them suitable for heating appliances.
4. The S.I. unit of electric current, the ampere, is defined as the current when 1 coulomb of charge flows through a conductor per second (\(1\ A = 1\ C/1\ s\)).

Teacher's Note:
a) Do not confuse resistance (depends on dimensions) with resistivity (a material property).
b) Alloys are preferred in heating elements due to higher resistivity and resistance to oxidation.
c) Always give the exact definition of ampere in terms of charge and time.

OR

(b) (i) How many bulbs of resistance 8 Ω each should be connected in parallel combination to draw a current of 2 A from a battery of 4 V? [5 Marks]
(ii) Name the device used for measuring electric current. How is it connected in a circuit?
(iii) State Joule's law of heating.

Answer:
1. Given \(V = 4\) V, \(I = 2\) A, so total resistance \(R = \dfrac{V}{I} = \dfrac{4}{2} = 2\ \Omega\).
2. For 'n' bulbs of 8 Ω in parallel: \(\dfrac{1}{R} = \dfrac{n}{8}\), so \(\dfrac{1}{2} = \dfrac{n}{8}\), giving \(n = 4\) bulbs.
3. The device used for measuring electric current is an ammeter, and it is always connected in series in a circuit.
4. Joule's law of heating states that the heat produced in a conductor is directly proportional to the square of the current, the resistance of the conductor, and the time for which the current flows: \(H = I^2 R t\).

Teacher's Note:
a) First find the total circuit resistance from V and I, then apply the parallel resistance formula.
b) Remember ammeter is connected in series and voltmeter in parallel - a common point of confusion.
c) The formula H = I²Rt should be quoted exactly for full marks in the Joule's law statement.

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