CBSE Class 10 Science 2026 — Paper 31/8/1/SB
80 marks · 3 hours · 39 questions (45 with the OR choices)
28 of its 45 questions also appear in 31/8/2/SB, 31/8/3/SB and other papers; 17 are only in this paper.
Take this paper as a timed mock
Free with an account: the board's time limit, marking at the end, and the answer to every question.
Questions 1–9 · 1 mark each
- Q1 1 mark
Also in 31/8/2/SB (Q6), 31/8/3/SB (Q3)
Select the correct option that shows the path of flow of only the deoxygenated blood through the human circulatory system.- (a)Body left atrium left ventricle lungs right atrium right ventricle body.
- (b)Right atrium right ventricle lungs left atrium left ventricle body
- (c)Body left atrium left ventricle lungs
- (d)Body right atrium right ventricle lungs
Step-by-step solution →Show answer
(d) Body right atrium right ventricle lungs - Q2 1 mark
Also in 31/8/2/SB (Q4), 31/8/3/SB (Q5)
Which of the following groups consists entirely of substances that are broken down by an enzyme, lipase ?- (a)Carbohydrates, proteins and vitamins
- (b)Fats, oils and ghee
- (c)Minerals, starch and vitamins
- (d)Proteins, sugars and esters
Step-by-step solution →Show answer
(b) Fats, oils and ghee - Q3 1 markIdentify the gland that enables the human beings to deal with an emergency situation.
- (a)Pineal gland
- (b)Pituitary gland
- (c)Adrenal gland
- (d)Pancreatic gland
Step-by-step solution →Show answer
(c) Adrenal gland - Q4 1 mark
Also in 31/8/2/SB (Q2), 31/8/3/SB (Q7)
Examine the given food chains to establish the correct one in an ecosystem :- (a)Seed mouse snake deer
- (b)Algae small fish large fish hawk
- (c)Grasshopper frog snake elephant
- (d)Plants rabbit deer wolf
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(b) Algae small fish large fish hawk - Q5 1 mark
Also in 31/8/2/SB (Q7), 31/8/3/SB (Q1)
A man underwent MRI of brain which shows some defects in the cerebellum of his brain. Which activity of that person will be affected ?- (a)Heart beat
- (b)Feeling of hunger and thirst
- (c)Reflexes
- (d)Walking in straight line
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(d) Walking in straight line - Q6 1 markDuring the process of exhalation, air follows a definite path from lungs to outside. Choose the correct option showing the path of exhaled air.
- (a)Alveoli Trachea Larynx Pharynx Nostril
- (b)Nasal cavity Pharynx Larynx Trachea Bronchi Bronchioles
- (c)Body tissue Blood Lungs Nose
- (d)Alveoli Trachea Pharynx Larynx Nostril
Step-by-step solution →Show answer
(a) Alveoli Trachea Larynx Pharynx Nostril - Q7 1 mark
Also in 31/8/2/SB (Q1), 31/8/3/SB (Q4)
Two pea plants one with round green seeds (RRyy) and another with wrinkled yellow seeds (rrYY) were crossed. All the seeds of progeny have round, yellow seeds (RrYy). When plants are self-pollinated, the progeny will have new combination of characters.- (i)Round, yellow
- (ii)Round, green
- (iii)Wrinkled, yellow
- (iv)Wrinkled, green Choose the option that shows new combination of characters in progeny.
- (a)(i) and (ii)
- (b)(i) and (iv)
- (c)(iii) and (iv)
- (d)(i) and (iii)
Step-by-step solution →Show answer
(b) (i) and (iv) - Q8 1 markQuestion No. 8 & 9 consists of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option from (a), (b), (c) and (d) given below :
Assertion (A): In regeneration, organisms have the ability to give rise to new individual organisms from their body parts.
Reason (R): Simple animals like Hydra and Planaria show regeneration.
- (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.
Step-by-step solution →Show answer
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). - Q9 1 mark
Also in 31/8/2/SB (Q8), 31/8/3/SB (Q8)
Question No. 8 & 9 consists of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option from (a), (b), (c) and (d) given below :Assertion (A): The energy transferred to a lion when it eats a deer is only a small fraction of the deer’s total energy content.
Reason (R): Energy transfer is efficient at all trophic levels with nearly 90% of energy passed on to the next level.
- (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.
Step-by-step solution →Show answer
(c) Assertion (A) is true, but Reason (R) is false.
Questions 10–12 · 2 marks each
- Q10(a) 2 marksHow does translocation of food and other substances take place in plants ?Step-by-step solution →
Show answer
Phloem loads food into sieve tubes using ATP; water enters by osmosis and the pressure pushes it to sinks, upwards or downwards.
Translocation is the transport of food - the soluble products of photosynthesis - and of other substances such as amino acids, and it is carried out by the phloem, whose sieve tubes and companion cells run through the plant. At a source such as a leaf, the food is loaded into the sieve tubes using energy from ATP; this raises the concentration inside the tube, so water enters it by osmosis and the pressure inside rises. The pressure pushes the contents along the sieve tubes to a sink - a root, a fruit or a growing bud - where the food is unloaded and the pressure falls. Because the pressure is created wherever food is loaded, the phloem can carry material both upwards and downwards, according to the needs of the plant.ORQ10(b) 2 marksExplain mechanism of opening and closing of stomata in plants.Step-by-step solution →Show answer
Water entering the guard cells makes them turgid and the pore opens; water leaving makes them flaccid and the pore closes.
A stoma is a tiny pore in the leaf epidermis bordered by a pair of bean-shaped guard cells, and the opening and closing of the pore is a function of these cells. When water flows into the guard cells by osmosis they swell and become turgid; because the wall facing the pore is thick and stiff while the outer wall is thin and elastic, each cell curves outwards, its inner wall is drawn away from its partner and the pore opens. When water passes out of the guard cells they shrink and become flaccid, straighten and come together again, so the pore closes. This closing is what saves the plant from losing water through the stomata when carbon dioxide is not needed. - Q11 2 marksDescribe how the sex of a new born individual is determined in humans.Step-by-step solution →
Show answer
Every egg carries X; an X sperm gives a girl and a Y sperm a boy, so the father's sperm decides the sex.
The sex of a new born child in human beings is decided by the sex chromosome it receives from its father. A human body cell has 23 pairs of chromosomes; 22 pairs are alike in both sexes and the remaining pair is the pair of sex chromosomes. A woman has a matched pair, XX, and a man has a mismatched pair, XY. When germ cells are formed each gamete gets only one chromosome of a pair, so every egg produced by the mother carries an X chromosome, while the father produces two kinds of sperm in equal numbers, half carrying X and half carrying Y. If a sperm carrying X fertilises the egg the zygote is XX and the child is a girl; if a sperm carrying Y fertilises it the zygote is XY and the child is a boy. Since the mother's contribution is the same either way, the sex of the child depends on the father, and a girl and a boy are equally likely. - Q12 2 marks
Also in 31/8/2/SB (Q11), 31/8/3/SB (Q10)
The sensitive plant shows movement of leaves in response to touch. Explain such movement briefly.Step-by-step solution →Show answer
On touch, cells at the leaflet base lose water and shrink, so the leaflets fold; this nastic movement does not depend on growth.
The folding of the leaves of the sensitive plant on being touched is a movement that is independent of growth, that is, a nastic movement. The stimulus is touch. The plant has no nervous tissue, so the information that a leaf has been touched is passed on from one cell to the next as an electrical and chemical signal until it reaches the cells at the base of the leaflets. Plants have no muscle cells; instead, their cells change shape by changing the amount of water in them. The cells in the swelling at the base of each leaflet lose water, become flaccid and shrink, so the leaflets fold up in pairs and the leaf droops. When these cells take up water again they swell, and the leaflets spread out as before, so the movement is reversible.
Questions 13–14 · 3 marks each
- Q13 3 marksWhat role does UV radiation play in the formation of the ozone layer ? Explain the process with relevant chemical reactions. Why is the excessive use of CFCs (Chlorofluoro Carbons) a cause of concern for organisms ?Step-by-step solution →
Show answer
UV splits into atoms that form ; CFCs release chlorine that destroys ozone, letting in UV that harms organisms.
Role of UV radiation and the process : Ozone at the higher levels of the atmosphere is a product of ultraviolet radiation acting on oxygen. The high-energy ultraviolet radiation of the Sun splits molecules of oxygen into free oxygen atoms, and each free atom then combines with a molecule of oxygen to form ozone. The reactions are and . Ultraviolet radiation is thus the source of the energy that breaks the oxygen molecule, without which the ozone layer could not form; the ozone so formed then absorbs the harmful ultraviolet radiation and shields the surface of the Earth. Why CFCs are a cause of concern : Chlorofluorocarbons, used as refrigerants and in fire extinguishers, are very stable and are not destroyed in the lower atmosphere, so they drift up to the ozone layer. There ultraviolet radiation frees chlorine atoms from them, and each chlorine atom destroys many molecules of ozone, so the layer becomes thinner. More ultraviolet radiation then reaches the Earth, and it is highly damaging to organisms, causing skin cancer and cataract and weakening the immune system in human beings, damaging plants and lowering crop yields, and killing plankton on which aquatic food chains depend. - Q14 3 marks
Also in 31/8/2/SB (Q13), 31/8/3/SB (Q13)
“Each organism has two sets of all genes, one inherited from each parent i.e. both parents contribute equally to the DNA of the progeny during sexual reproduction”. If such is the case, then how is the stability of DNA of the species ensured ? Discuss.Step-by-step solution →Show answer
Gametes get half the chromosomes (one set), and fertilisation restores the full number (), so the DNA of the species stays stable.
If both parents really passed on the two sets of genes each of them carries, the progeny would have four sets, the next generation eight, and the amount of DNA would double in every generation. This does not happen, because sexual reproduction includes a halving step. Each body cell has two copies of each chromosome, one from the male parent and the other from the female parent. When germ cells are formed, each germ cell (gamete) takes only one chromosome from each pair, which may be of either maternal or paternal origin, so a gamete carries one set of genes instead of two and half the number of chromosomes; in human beings a body cell has 46 chromosomes (23 pairs) while a sperm or an egg has 23. When two germ cells unite at fertilisation, the zygote once again has two copies of each chromosome, 23 plus 23 giving 46, so the normal number of chromosomes is restored in the progeny. This re-establishment of the number of chromosomes, and so of the amount of DNA, in every generation ensures the stability of the DNA of the species.
Question 15 · 4 marks
- Q15 4 marks
Also in 31/8/2/SB (Q15), 31/8/3/SB (Q15)
The given diagram shows the human kidney and its associated structures. Study the Table I and Table II that shows the concentration of certain substances found within the structure P and Q.Table I (structure P) and Table II (structure Q) - concentration of components, in per cent Components Table I - Structure P, concentration % Table II - Structure Q, concentration % Urea 0.05 1.00 Amino acids 0.06 0.00 Salts 1.72 1.60 Glucose 0.10 0.00 - (A)Where are kidneys located in the human body ? Name the filtration unit of them.[1]
- (B)List two factors on which reabsorption of water from the filtrate depends.[1]
- (C)Why is there variation in the concentration of given substance in structure P and Q listed in table I and II respectively ?[2]
- (C)OR : If a person is suffering from diabetes, concentration of which substance present in structure Q (listed in table II) will be modified and why ?[2]
Step-by-step solution →Show answer
Kidneys lie in the abdomen beside the backbone, with the nephron as filtration unit; filtration then selective reabsorption changes blood into urine.
(A) The kidneys are a pair of bean-shaped organs lying in the abdomen, one on either side of the backbone, towards the back of the abdominal cavity. Their filtration unit is the nephron. (B) The amount of water reabsorbed from the filtrate depends on (i) how much excess water there is in the body, and (ii) how much dissolved waste there is to be excreted. (C) P is the blood vessel carrying blood into the kidney, so Table I is the composition of blood, and Q is the ureter carrying urine away, so Table II is the composition of urine. In the nephron the blood is first filtered at the glomerulus, so water and all the small dissolved substances enter the tubule, and the filtrate is then selectively reabsorbed as it flows along the tubule. All the glucose and all the amino acids are useful and are taken back completely, so they fall from 0.10% and 0.06% in P to 0.00% in Q. Most of the salts are taken back along with most of the water, so their concentration hardly changes, 1.72% to 1.60%. Urea is a nitrogenous waste and is not reabsorbed, while the water around it is, so it is left behind and becomes concentrated, rising from 0.05% in P to 1.00% in Q. OR (C) The concentration of glucose in structure Q will be modified: there will be more glucose in Q. In a healthy person it is 0.00% because all the filtered glucose is reabsorbed by the tubule. A person suffering from diabetes does not have sufficient insulin, so the glucose level of the blood rises; more glucose is then filtered than the tubule is able to reabsorb, the excess is not taken back and passes out in the urine, so glucose now appears in structure Q at a value above 0.00%.
Question 16 · 5 marks
- Q16(a) 5 marks
Also in 31/8/2/SB (Q16(b)), 31/8/3/SB (Q16(a))
In flowering plants, the process of pollination leads to fertilization.- (i)Explain with the help of a well-labelled diagram how the male germ cell from the pollen grain reaches the female germ cell located in the ovule.
- (ii)Differentiate between self-pollination and cross-pollination.
- (iii)Identify and name the parts ‘R’ and ‘S’ in the given diagram of seed.
Step-by-step solution →Show answer
Pollen tube carries the male germ cell to the egg; self-pollination is within a flower, cross-pollination between plants; R is plumule, S radicle.
(i) Pollination brings the pollen grain to the stigma of the carpel, while the female germ cell lies inside an ovule in the ovary. The pollen grain that lands on a suitable stigma germinates and puts out a pollen tube, which grows through the tissue of the stigma and downward through the style until it reaches the ovary and enters the ovule. The male germ cell travels down this pollen tube and is released inside the ovule, where it fuses with the female germ cell, that is, with the egg. This fusion is fertilisation and it forms the zygote, which divides to form the embryo; the ovule then becomes the seed and the ovary becomes the fruit. The diagram should show the stigma with a germinating pollen grain, the pollen tube growing down the style with the male germ cell in it, the ovary, the ovule and the female germ cell inside it, all labelled. (ii) Self-pollination: pollen grains are transferred from the anther to the stigma of the same flower; cross-pollination: pollen grains are transferred from the anther of one flower to the stigma of a different flower, usually a flower on another plant of the same species. Self-pollination usually needs no outside agent; cross-pollination needs an agent such as wind, water, insects or birds. The progeny of self-pollination show little variation; cross-pollination between different plants brings together genes from two individuals, so its progeny show more variation. (iii) R is the plumule, the part of the embryo above the attachment of the cotyledons, which grows into the shoot; S is the radicle, the part below it, which grows into the root.ORQ16(b) 5 marksAlso in 31/8/2/SB (Q16(a)), 31/8/3/SB (Q16(b))
- (i)Name the special tissue inside the uterus that provides nutrition to the developing embryo. Describe its structure. State its two functions besides nutrition.
- (ii)Write any two sexually transmitted diseases in humans and their causative microbial agents.
Step-by-step solution →Show answer
The placenta nourishes the embryo, exchanges gases and removes wastes; gonorrhoea (bacterium) and AIDS (virus) are two sexually transmitted diseases.
(i) The tissue is the placenta. It is a disc-shaped tissue embedded in the wall of the uterus; villi grow out from the embryo's side of the disc into blood spaces (sinuses) filled with the mother's blood, which gives a very large surface for exchange, and the embryo is joined to the disc by the umbilical cord. Two functions besides nutrition: it passes oxygen from the mother's blood to the embryo and carries carbon dioxide back; and it removes the embryo's nitrogenous wastes, chiefly urea, into the mother's blood. (The placenta also secretes hormones that help to maintain the pregnancy.) (ii) Gonorrhoea is caused by the bacterium Neisseria gonorrhoeae, and AIDS is caused by a virus, the Human Immunodeficiency Virus (HIV). Syphilis (bacterium Treponema pallidum) and warts (human papilloma virus) are equally acceptable pairs.
Questions 17–24 · 1 mark each
- Q17 1 markHow many bonds can one carbon atom form in a compound ?
- (a)
- (b)
- (c)
- (d)
Step-by-step solution →Show answer
(c) - Q18 1 markWhich of the following salts do not contain water of crystallization ?
- (i)Bleaching powder
- (ii)Baking soda
- (iii)Gypsum
- (iv)Washing soda
- (a)(ii) and (iv)
- (b)(i) and (iii)
- (c)(ii) and (iii)
- (d)(i) and (ii)
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(d) (i) and (ii) - Q19 1 markIdentify the unsaturated compounds from the following :
- (i)Propane
- (ii)Propene
- (iii)Propyne
- (iv)Chloropropane
- (a)(i) and (ii)
- (b)(ii) and (iv)
- (c)(iii) and (iv)
- (d)(ii) and (iii)
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(d) (ii) and (iii) - Q20 1 mark
Also in 31/8/2/SB (Q23), 31/8/3/SB (Q22)
During the preparation of hydrogen chloride gas on a humid day, the gas is usually passed through a guard tube containing calcium chloride. The role of taken in guard tube is to –- (a)Absorb the evolved gas
- (b)Moisten the gas
- (c)Absorb moisture from the gas
- (d)Absorb chloride ions from the evolved gas
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(c) Absorb moisture from the gas - Q21 1 markIn the compound which of the following functional group is present ?
- (a)Alcohol
- (b)Aldehyde
- (c)Carboxylic acid
- (d)Ketone
Step-by-step solution →Show answer
(d) Ketone - Q22 1 markWhich of the following statements about the given reaction are correct ?
- (i) is getting reduced.
- (ii) is getting oxidised.
- (iii) acts as an oxidising agent.
- (iv) acts as a reducing agent.
- (a)(i), (ii) and (iii)
- (b)(iii) and (iv)
- (c)(i), (ii) and (iv)
- (d)(ii) and (iv)
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(a) (i), (ii) and (iii) - Q23 1 mark
Also in 31/8/2/SB (Q21), 31/8/3/SB (Q21)
Which of the following acid-base indicators will be used by a visually challenged student to detect the presence of acid in a given solution ?- (a)Blue litmus
- (b)Clove oil
- (c)Red cabbage extract
- (d)Methyl orange
Step-by-step solution →Show answer
(b) Clove oil - Q24 1 mark
Also in 31/8/2/SB (Q24), 31/8/3/SB (Q24)
For question number 24, two statements are given-one labelled as Assertion (A) and the other labelled as Reason (R). Answer this question selecting the appropriate option from (a), (b), (c) and (d) given below :Assertion (A): Sodium acetate is a basic salt.
Reason (R): It is because it is a salt of weak acid and strong base.
- (a)Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
- (b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- (c)Assertion (A) is true, but Reason (R) is false.
- (d)Assertion (A) is false, but Reason (R) is true.
Step-by-step solution →Show answer
(a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Question 25 · 2 marks
- Q25 2 marksWrite chemical equations for the reactions taking place when :
- (i)Manganese dioxide is heated with aluminium powder.
- (ii)Steam is passed over red hot iron.
Step-by-step solution →Show answer
Aluminium takes oxygen from manganese dioxide, setting manganese free with much heat; red hot iron and steam give magnetic iron oxide and hydrogen.
(i) . Aluminium is more reactive than manganese, so it takes the oxygen and sets manganese free in a strongly exothermic displacement. (ii) . Red hot iron reacts with steam to give the magnetic oxide of iron and hydrogen gas.
Questions 26–27 · 3 marks each
- Q26(a) 3 marks
Also in 31/8/2/SB (Q27(a)), 31/8/3/SB (Q26(a))
- (i)State any two reasons for carbon forming a large number of compounds. Why does carbon form compounds mainly by covalent bonds ?
- (ii)Identify the compound which contains aldehyde as its functional group : , , ,
Step-by-step solution →Show answer
Catenation and tetravalency let carbon form many compounds; carbon shares its four valence electrons, forming covalent bonds; has the aldehyde group.
(i) Two reasons: (1) Catenation — carbon atoms link to one another by strong covalent bonds and build long straight chains, branched chains and closed rings; (2) Tetravalency — carbon has four valence electrons, so one carbon atom can bond with four other atoms of carbon or of other monovalent elements, and because the carbon atom is small these bonds are strong and stable. Carbon forms compounds mainly by covalent bonds because it has four electrons in its outermost shell: it can neither gain four electrons (a nucleus of six protons cannot hold ten electrons) nor lose four electrons (that would need a very large amount of energy), so it attains a noble-gas arrangement by **sharing** its four valence electrons, and a shared pair of electrons is a covalent bond. (ii) (propanal) — it carries the aldehyde group .ORQ26(b) 3 marksAlso in 31/8/2/SB (Q27(b)), 31/8/3/SB (Q26(b))
An organic compound ‘X’ is a constituent of wine. ‘X’ on reacting with acidified forms another compound ‘Y’ with molecular formula . ‘Y’ produces brisk effervescence with sodium carbonate.- (i)Identify ‘X’ and ‘Y’.
- (ii)Name the gas evolved when ‘Y’ reacts with sodium carbonate.
- (iii)Write the chemical equations involved for the chemical reactions taking place.
Step-by-step solution →Show answer
X is ethanol and Y is ethanoic acid; the gas evolved is carbon dioxide.
(i) ‘X’ is ethanol, — the alcohol present in wine; ‘Y’ is ethanoic acid (acetic acid), , whose molecular formula is . (ii) The gas evolved is carbon dioxide, . (iii) and - Q27 3 marks
Also in 31/8/2/SB (Q26), 31/8/3/SB (Q27)
(A) Name the acid present in ant sting and give its chemical formula. Give the common method to get relief from the pain and irritation caused by ant sting. (B) Two solutions ‘X’ and ‘Y’ have pH values and respectively. Which of these will turn blue litmus to red and which will turn phenolphthalein from colourless to pink ?Step-by-step solution →Show answer
Ant-sting acid is methanoic acid, relieved by baking soda; X (pH 3.0) turns blue litmus red; Y (pH 9.5) turns phenolphthalein pink.
(A) The acid present in an ant sting is methanoic acid (formic acid), . Relief is obtained by rubbing the stung area with a mild base such as moist baking soda (sodium hydrogencarbonate, ), which neutralises the acid; calamine solution, containing zinc carbonate, may be used in the same way. (B) ‘X’, of pH , is acidic and will turn blue litmus red. ‘Y’, of pH , is basic and will turn phenolphthalein from colourless to pink.
Question 28 · 4 marks
- Q28 4 marks
Also in 31/8/2/SB (Q28), 31/8/3/SB (Q28)
The Statue of Liberty is one of the seven wonders. It is present on Liberty island in New York harbour in U.S.A. It is made up of a metal. Over a period of time, the metal’s colour has changed and statue gives greenish look. Answer the following questions based on the above information :- (A)What is corrosion ?[1]
- (B)Why do silver articles turn black after some time ?[1]
- (C)Copper reacts with moist gas ‘A’ present in the air and slowly loses its shiny surface. It turns green. Identify ‘A’ and the substance of green coating formed on the surface of copper.[2]
- (C)OR : What is an alloy ? How is an alloy made ?[2]
Step-by-step solution →Show answer
Corrosion eats away a metal's surface; silver turns black due to silver sulphide; gas A is carbon dioxide, the green coating basic copper carbonate.
(A) Corrosion is the process in which the surface of a metal is slowly eaten away by the action of air, moisture, acids or other chemicals around it, the metal being converted into a surface compound such as an oxide, a carbonate or a sulphide. (B) Silver articles turn black because silver reacts with the sulphur present in the air, as hydrogen sulphide, to form a black coating of silver sulphide, : . (C) ‘A’ is carbon dioxide, , and the green coating formed on copper is basic copper carbonate, : . (C) OR : An alloy is a homogeneous mixture of two or more metals, or of a metal and a non-metal. It is made by first melting the main metal, then dissolving the other metals or non-metals in it in a fixed proportion, and then cooling the mixture to room temperature.
Question 29 · 5 marks
- Q29(a) 5 marks
Also in 31/8/2/SB (Q29(a)), 31/8/3/SB (Q29(a))
Write chemical equations for the following when- (a)Silver bromide is exposed to sunlight.
- (b)A piece of lead metal is dropped into copper (II) chloride solution.
- (c)Hydrogen gas is passed over heated copper oxide.
- (d)Methane gas is burnt in air.
- (e)Quick lime is mixed with water.
Step-by-step solution →Show answer
Silver bromide decomposes in sunlight, lead displaces copper, hydrogen reduces copper oxide, methane burns, and quick lime combines with water.
(a) — photolytic decomposition. (b) — displacement. (c) — reduction of copper oxide. (d) — combustion. (e) — combination.ORQ29(b) 5 marksAlso in 31/8/2/SB (Q29(b)), 31/8/3/SB (Q29(b))
- (a)Study the experimental setup shown in the diagram and write chemical equation for the same. Name and define the type of reaction. Mention the colour of the products formed.
- (b)What type of reaction takes place when aqueous solution of lead (II) nitrate and potassium iodide are mixed together ? Write chemical equation also.
Step-by-step solution →Show answer
Heated lead nitrate decomposes into yellow lead oxide, brown nitrogen dioxide and oxygen; lead nitrate and potassium iodide solutions give a double displacement precipitate.
(a) . It is a thermal decomposition reaction - a reaction in which a single compound breaks down on heating into two or more simpler substances. Colours of the products: lead oxide is a yellow solid, nitrogen dioxide is a reddish brown gas and oxygen is colourless. (b) A double displacement reaction, and since an insoluble solid separates out it is a precipitation reaction: , the precipitate being yellow lead iodide.
Questions 30–32 · 1 mark each
- Q30 1 mark
Also in 31/8/2/SB (Q31)
When a convergent beam of light is incident on a convex mirror, the reflected rays appear to be coming from a point 20 cm behind the mirror and retrace their path. The focal length of the mirror is- (a)20 cm
- (b)10 cm
- (c)40 cm
- (d)30 cm
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(b) 10 cm - Q31 1 mark
Also in 31/8/2/SB (Q30), 31/8/3/SB (Q31)
In the given ray diagram the values of the angle of incidence and the angle of reflection are :- (a),
- (b),
- (c),
- (d),
Step-by-step solution →Show answer
(d) , - Q32 1 mark
Also in 31/8/2/SB (Q32), 31/8/3/SB (Q32)
For question number 32, two statements are given – one labelled as Assertion (A) and the other labelled as Reason (R). Answer this question selecting the appropriate option from (a), (b), (c) and (d) given below :Assertion (A): The bending of a straight wire in the form of current carrying circular coil will change the pattern of magnetic field around it.
Reason (R): Strength of magnetic field depends upon number of turns of coil and current flowing through the coil.
- (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.
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(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Questions 33–34 · 2 marks each
- Q33(a) 2 marks
Also in 31/8/2/SB (Q34(a)), 31/8/3/SB (Q34(a))
What is meant by scattering of light ? How do our observations differ when white light (sun light) is scattered from- (i)very fine particles of the medium and
- (ii)very large particles of the medium ?
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Scattering sends light off in different directions; very fine particles give blue scattered light, very large particles give white.
Scattering of light is the phenomenon in which a part of the light falling on the particles of a medium - molecules, dust, smoke or water droplets - is redirected (sent out again) in different directions, away from its original straight path. (i) Very fine particles: they scatter mainly the shorter wavelengths, so the scattered light appears **blue**. This is why the clear sky looks blue. (ii) Very large particles: they scatter all wavelengths almost equally, so the scattered light appears **white**. This is why clouds, fog and mist look white.ORQ33(b) 2 marksAlso in 31/3/1 (Q34(b)), 31/3/2 (Q33(b)), 31/3/3 (Q34(b)), 31/8/2/SB (Q34(b)), 31/8/3/SB (Q34(b))
With the help of a ray diagram, explain any phenomenon of formation of spectrum in nature.Step-by-step solution →Show answer
A rainbow forms when raindrops disperse sunlight by refraction, internal reflection and refraction, giving a spectrum with red outside and violet inside.
The rainbow is a natural spectrum formed after rain by the dispersion of sunlight by tiny water droplets in the atmosphere. Each droplet acts like a small prism: sunlight is refracted and dispersed as it enters the drop, reflected internally at the back surface, and refracted again as it comes out (see the ray diagram). The colours emerge at slightly different angles - red at about , violet at about to the direction opposite the Sun - so an observer standing with the Sun behind sees a band of colours in the sky, red on the outside and violet on the inside, always on the side opposite the Sun. - Q34 2 marksA bulb consumed 24 J of energy per second, when connected to a 12 V supply. Calculate the power consumed when it is connected to a 6 V supply.Step-by-step solution →
Show answer
W: the bulb consumes 6 W on the 6 V supply.
W. The bulb consumes J per second at V, so W and its resistance is . The same resistance on a V supply gives W.
Questions 35–37 · 3 marks each
- Q35 3 marks
Also in 31/8/2/SB (Q36), 31/8/3/SB (Q37)
(A) A vertical wire is carrying a current in the upward direction. It is placed in a magnetic field, pointing towards the East direction. Find out the direction of force on the wire. (B) Out of the two current carrying circular coils, coil ‘A’ has 10 turns while coil B has 15 turns. Which of the two coils will produce a stronger magnetic field at its centre if same current is flowing through both the coils ? Explain with reason. (C) In the given figure, AB and CD are two magnetic field lines intersecting each other at point O. If a compass is placed at point O, what will happen ? State your conclusions.Step-by-step solution →Show answer
Force is towards North; coil B (15 turns) gives the stronger field; the compass cannot point two ways, so field lines never intersect.
(A) The force on the wire is horizontal and directed towards the **north**. By Fleming's left-hand rule, with the forefinger along the field (East) and the middle finger along the current (vertically up), the thumb points North. (B) Coil B, with turns, produces the stronger field. At the centre of a circular coil , so with the same current and the same radius the field is proportional to the number of turns - each turn adds its own field in the same direction - and , that is times as strong. (C) The figure shows the field at O in two directions at once, so a compass needle placed at O would have to point in two directions (along OB and along OD) at the same time. That is not possible, since a needle can settle in only one direction - the situation shown is unreal. Conclusions: the magnetic field at any point has a single definite direction, and therefore two magnetic field lines never intersect each other. - Q36 3 marks(A) The elderly people, may have both near sightedness and far sightedness. Write the possible solutions, can be proposed for correction of their eye defect. (B) How does the eye control the amount of light entering into it ?Step-by-step solution →
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Bi-focal lenses or two pairs of spectacles correct both defects; the iris changes the pupil size to control the light entering the eye.
(A) An elderly person who has both near sightedness (myopia) and far sightedness (hypermetropia) needs a concave lens for distant vision and a convex lens for near vision. The possible solutions are: (i) two separate sets of spectacles - one (concave lenses) for correcting the near sightedness and the other (convex lenses) for correcting the far sightedness; (ii) bi-focal spectacles, in which each lens has two parts: the upper portion is a **concave** lens, looked through for distant vision, and the lower portion is a **convex** lens, looked down through for reading - or progressive (varifocal) lenses, which give the two powers with a gradual change between them. Contact lenses or corrective surgery are further options. (B) The amount of light entering the eye is controlled by the **iris**, the dark muscular diaphragm behind the cornea, which changes the size of the **pupil** - the opening at its centre through which light enters. In bright light the iris contracts the pupil so that less light enters and the retina is protected; in dim light the iris widens the pupil so that more light enters. The pupil therefore acts as a variable aperture. - Q37 3 marks
Also in 31/8/2/SB (Q35), 31/8/3/SB (Q36)
(A) Write any two precautions to be taken to avoid electrical accidents. (B) State how these precautions prevent possible damage to the electric circuit.Step-by-step solution →Show answer
Fit a fuse in the live wire and earth metal-bodied appliances; the fuse breaks the circuit, earthing gives a low-resistance path.
(A) Two precautions: (i) Fit an electric fuse (or an MCB) of the proper rating in series in the **live** wire of every circuit and of every high-power appliance. (ii) Earth every metal-bodied appliance - connect its metal casing through the green earth wire and the third pin of a three-pin plug to a metal plate buried deep in the ground. (B) How they prevent damage to the circuit: The fuse is a short piece of wire of high resistivity and low melting point. If the current rises above the rated value because of overloading or a short circuit, the heat produced in it, , melts the fuse wire. The circuit is broken at once and the current stops, so the household wiring, the switches and the appliances are never heated to the point where their insulation melts, they are damaged or a fire starts. Earthing provides the metal body of the appliance with a path of very low resistance to the ground. If the live wire touches the casing, the body is held at the potential of the earth (about V), so nobody gets a shock; and because that path has a very low resistance the leakage current is large, so it melts the fuse and disconnects the faulty appliance before its damaged wiring can overheat. (Other acceptable precautions, with the same kind of explanation: do not overload one socket through a multi-plug; use only ISI-marked, properly insulated wires of adequate thickness; switch off the mains before repairs and never touch switches with wet hands.)
Question 38 · 4 marks
- Q38 4 marks
Also in 31/8/2/SB (Q38), 31/8/3/SB (Q38)
Four cells, a thin wire AB, a voltmeter, an ammeter, a plug key and some thick connecting wires are taken. The circuit is set up as shown below. The key is inserted into the plug and readings of ammeter (I) and voltmeter (V) are noted. Then the plug key is removed. Now two cells are connected in the circuit in series. Plugging in the key the new values of ‘V’ and ‘I’ are noted. The experiment is repeated by connecting three cells in series and then four cells. In each case ‘V’ and ‘I’ are noted.- (A)Why was the plug key removed after each measurement of ‘V’ and ‘I’ and inserted again for the next step ?
- (B)What change takes place in the reading of the voltmeter when a new cell is added in series in the circuit ? Give reason.
- (C)In which case will the readings of voltmeter and ammeter be maximum in the above experiment ? What conclusion can be drawn from the straight line graph between ‘V’ and ‘I’ ?
- (C)OR : Suppose this experiment is carried out using only one cell but wires of same length & thickness but made of different materials like nichrome wire, fuse wire, copper wire and iron wire which are replaced one by one. Out of V and I which will change and why ?
Step-by-step solution →Show answer
Key removed so wire AB stays cool; each added cell raises the voltmeter reading; four cells give maximum readings; the line shows Ohm's law.
(A) So that current flows through the circuit only while a reading is being taken. A current left flowing through the thin wire AB would heat it (); the resistance of a metal rises with temperature, so would drift and the V-I graph would not come out straight, since Ohm's law holds only at constant temperature. Removing the key lets AB cool to room temperature between readings and also saves the cells. (B) The voltmeter reading increases. Cells joined in series have their emfs added, so each extra cell applies a larger potential difference across AB; as the wire is unchanged its resistance is unchanged, so the current increases in the same proportion and stays constant. (C) Both readings are maximum when all four cells are connected in series, since that gives the greatest total emf. The straight line through the origin shows that , i.e. is constant for the conductor at constant temperature - this is Ohm's law - and that constant, the slope of the line, is the resistance of the wire AB. (C) OR: will change and will remain the same. The same single cell applies the same potential difference in every trial, while the wires have the same length and the same area of cross-section , so by their resistances differ only through the resistivity of the material; hence is different for each wire - largest for copper, which has the lowest resistivity of the four, and smallest for nichrome, which has the highest. This shows that resistance depends on the nature of the material.
Question 39 · 5 marks
- Q39(a) 5 marks
- (i)In which mirror we can always see the full image of a distant object ? Support your answer by a ray diagram for it.
- (ii)A convex mirror forms an erect image of an object of size, one-third of the size of the object. Find the position of the object if the radius of curvature of convex mirror is 30 cm.
Step-by-step solution →Show answer
A convex mirror always shows the full image (virtual, erect, diminished); the object is 30 cm in front of the mirror.
(i) A convex mirror. Whatever the position of the object, a convex mirror always forms a virtual, erect and diminished image lying between the pole P and the focus F, so the whole of even a large or distant object is compressed into the mirror and its full image can always be seen (this, with its wide field of view, is why a convex mirror is used as a rear-view mirror). The ray diagram is drawn with a ray from the tip of the object parallel to the principal axis, reflected as though it came from F, and a second ray aimed at C, reflected back along itself; their dashed backward extensions meet at the image tip behind the mirror. (ii) cm - the object is cm in front of the convex mirror. Working: cm; gives ; substituting in gives , so and cm. The image is then at cm, i.e. virtual and erect, cm behind the mirror.ORQ39(b) 5 marks- (i)Define the optical centre of a lens.
- (ii)The light from an object placed on a meter scale at 8 cm mark, was focused on a white screen placed at 92 cm mark, with the help of a converging lens placed on the scale at 50 cm mark : (I) Find the focal length of converging lens. (II) Find the position of image if the object is shifted towards the lens and kept at 29 cm mark. (III) What will be the nature of the image formed if the object is further shifted towards the lens ?
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cm; at the 29 cm mark the image forms at infinity; nearer the lens it is virtual, erect and magnified.
(i) The optical centre of a lens is the central point of the lens, lying on its principal axis midway between its two surfaces, such that a ray of light directed towards it passes through the lens without suffering any deviation. (ii) (I) Measuring from the lens at the cm mark, cm and cm, so , giving cm. (II) At the cm mark, cm , so the object is exactly at the principal focus. Then , so : the refracted rays emerge parallel to one another and the image is formed at infinity (real, inverted and highly magnified); it cannot be caught on the screen at any finite distance. (III) Brought still nearer, the object lies between the optical centre and the focus, and the image is then virtual, erect and magnified, formed on the same side of the lens as the object and not obtainable on a screen - the magnifying-glass case.