CBSE Class 10 Science Board Paper 2023 (31/1/1): Solutions

Dilute HCl reacts with sodium carbonate and the gas is bubbled through solution X (aqueous calcium hydroxide). X first turns milky, then colourless. Which explanation is right?

Uses a figure printed in CBSE's paper — what the figure shows: Set-up: dilute HCl poured on sodium carbonate in a boiling tube; delivery tube leads the gas into test tube X.

1. Dilute HCl reacts with sodium carbonate to produce carbon dioxide gas, which is bubbled into the calcium hydroxide solution.
2. The first portion of CO2 reacts with calcium hydroxide to form insoluble calcium carbonate, causing the solution to turn milky.
3. Excess CO2 reacts with the calcium carbonate to form soluble calcium hydrogen carbonate, which dissolves and makes the solution colourless again, confirming option (b).

Answer: CO₂ forms insoluble calcium carbonate (milky); with excess CO₂ it changes to calcium hydrogen carbonate, which dissolves, so the liquid clears — option (b).

Lead nitrate is heated in a test tube over a burner and brown fumes come out. What causes the brown fumes?

1. Lead nitrate decomposes on heating to form lead oxide, nitrogen dioxide, and oxygen.
2. The brown fumes are caused by the nitrogen dioxide gas produced in this reaction.

Answer: Lead nitrate decomposes on heating and gives brown nitrogen dioxide gas — option (a).

Balance MnO₂ + x HCl → MnCl₂ + y H₂O + z Cl₂. Find x, y, z.

1. Count Cl atoms: x HCl gives x Cl on the left; MnCl₂ and z Cl₂ give 2 + 2z on the right, so x = 2 + 2z.
2. Count O atoms: MnO₂ has 2 O and only y H₂O carries O on the right, so y = 2.
3. Count H atoms: x = 2y = 2 × 2 = 4.
4. Then 4 = 2 + 2z, so z = 1. The equation is MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂, option (c).

Answer: MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂, so x = 4, y = 2, z = 1 — option (c).

A table gives solution, pH-paper colour, approximate pH and nature for lemon juice, milk of magnesia, gastric juice and pure water. Which row is fully correct?

1. Milk of magnesia is a base, so its pH is above 7 (about 10) and pH paper turns blue in it.
2. Lemon juice and gastric juice are acidic (pH below 7) and pure water is neutral (pH 7); the other rows mismatch these, so option (b) is the fully correct row.

Answer: Milk of magnesia: blue colour, pH about 10, basic — option (b).

Metal X is used in the thermite process and burns in air to give amphoteric oxide Y. Name X and Y.

1. Metal X is identified as aluminium because it is the primary metal used in the thermite process for welding railway tracks.
2. Aluminium burns in air to form aluminium oxide (Al₂O₃), which is amphoteric as it reacts with both acids and bases, confirming Y is aluminium oxide.

Answer: X is aluminium (Al) and Y is aluminium oxide (Al₂O₃) — option (b).

Pick the formula of washing soda.

1. Identify the chemical name of washing soda as sodium carbonate decahydrate.
2. Write the formula as Na₂CO₃·10H₂O, which corresponds to option (c).

Answer: Washing soda is Na₂CO₃·10H₂O — option (c).

Three six-carbon rings are drawn: A has only single bonds (each C has 2 H), B has three alternate double bonds (benzene), C has two double bonds. Choose the correct statement.

Uses a figure printed in CBSE's paper — what the figure shows: A: C₆H₁₂ ring, all single bonds. B: C₆H₆ ring with 3 alternate double bonds. C: C₆H₈ ring with 2 double bonds and two CH₂ groups.

1. A is cyclohexane (C₆H₁₂) with only single bonds, making it a saturated cyclic hydrocarbon.
2. B (benzene) and C contain C=C double bonds, classifying them as unsaturated cyclic hydrocarbons.
3. Thus, the correct statement is that A is saturated while B and C are unsaturated, corresponding to option (c).

Answer: A (cyclohexane) is a saturated cyclic hydrocarbon; B and C contain C=C bonds, so they are unsaturated cyclic hydrocarbons — option (c).

Which organism digests food outside its body and then absorbs it?

1. Fungi such as Rhizopus (bread mould) secrete enzymes to digest food outside their body.
2. They then absorb the digested nutrients, making the fungus the correct answer.

Answer: A fungus such as Rhizopus (bread mould) — option (d).

Which statement about the small intestine is NOT correct?

1. The small intestine is the primary site for the absorption of digested nutrients, such as sugars, amino acids, and fatty acids.
2. Water absorption from the remaining undigested food primarily occurs in the large intestine, not the small intestine.
3. Therefore, the statement claiming villi in the small intestine absorb water before egestion is incorrect.

Answer: The claim that villi of the small intestine absorb water from undigested food before egestion is wrong; water is absorbed in the large intestine — option (d).

Which statement about a gene is correct?

1. A gene is a specific segment of DNA that occupies a fixed position, or locus, on a particular chromosome within a species.
2. Therefore, the statement that a particular gene sits on a particular chromosome is correct, making option (a) the right answer.

Answer: In a species, a particular gene sits on a particular chromosome — option (a).

Which statement about tropic movement in plants is correct?

1. Tropic movements are growth responses to directional stimuli like light or gravity, distinguishing them from nastic movements which are non-directional.
2. Therefore, the statement that tropic movement is a growth-related movement is correct, making option (d) the right answer.

Answer: Tropic movement is a growth-related movement — option (d).

Find the wrong pair of plant and vegetative part used: Bryophyllum–leaf, potato–stem, money-plant–stem, rose–root.

1. Bryophyllum grows new plants from buds on its leaf margins, potato from buds on its stem tuber, and money-plant from stem cuttings, so these pairs are correct.
2. Rose is grown from stem cuttings, not from roots, so the pair rose–root is wrong — option (d).

Answer: Rose–root is wrong; rose is grown from stem cuttings — option (d).

Four 8 Ω resistors give Rs in series and Rp in parallel. Find Rs/Rp.

1. For four 8 Ω resistors in series, Rs = 8 + 8 + 8 + 8 = 32 Ω.
2. For four 8 Ω resistors in parallel, 1/Rp = 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2, so Rp = 2 Ω.
3. The ratio Rs/Rp = 32/2 = 16, which corresponds to option (d).

Answer: Rs = 32 Ω, Rp = 2 Ω, so Rs/Rp = 16 — option (d).

Domestic wiring rated 15 A is meant for which appliances?

1. Domestic wiring rated at 15 A is designed to handle high current loads, which corresponds to high-power appliances.
2. Therefore, it is meant for appliances such as a geyser, making option (a) the correct answer.

Answer: High-power appliances such as a geyser — option (a).

A voltmeter scale runs 0 to 3 V with 1.5 V at the top; between 0 and 1.5 there are 10 equal divisions. The needle is 2 small divisions past 1.5. Find the least count and the reading.

Uses a figure printed in CBSE's paper — what the figure shows: Semicircular scale 0–3 V, 1.5 V at the top; 10 small divisions between 0 and 1.5 (a longer tick at the 5th); needle points at the 2nd small tick beyond 1.5.

1. Least count = value between 0 and 1.5 V ÷ number of divisions = 1.5/10 = 0.15 V.
2. The needle is 2 divisions past 1.5 V, so reading = 1.5 + 2 × 0.15 = 1.5 + 0.3 = 1.8 V. Answer: 0.15 V and 1.8 V, option (c).

Answer: Least count = 1.5/10 = 0.15 V; reading = 1.5 + 2 × 0.15 = 1.8 V. Least count and reading: 0.15 V and 1.8 V — option (c).

Two horizontal parallel wires AB (top) and CD (bottom) carry equal current I in the same direction. What is the net magnetic field at P, midway between them?

Uses a figure printed in CBSE's paper — what the figure shows: AB above, CD below, both currents left to right; P is halfway between the wires.

1. Apply the right-hand thumb rule to wire AB (current left to right): at point P below it, the magnetic field B1 is directed into the page.
2. Apply the right-hand thumb rule to wire CD (current left to right): at point P above it, the magnetic field B2 is directed out of the page.
3. Since P is midway between the wires and currents are equal, the magnitudes are equal (B1 = B2), but directions are opposite, so the net magnetic field is zero.

Answer: The two fields at P are equal and opposite, so the net field is zero — option (c).

A: Blue copper sulphate solution loses its colour when lead is put in it. R: Lead is more reactive than copper and displaces it from its salt.

1. Lead is more reactive than copper, so it displaces copper from copper sulphate: Pb + CuSO₄ → PbSO₄ + Cu.
2. Copper ions, which give the blue colour, are removed from the solution, so the blue colour fades. Hence A is true.
3. R is true, and this displacement is exactly why the colour goes, so R explains A — option (a).

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

A: Genes from the parents decide a child's sex. R: A boy gets his X chromosome from his father.

1. Assertion A is true because the sex of a child is determined by the sex chromosomes inherited from the parents.
2. Reason R is false because a boy inherits the Y chromosome from his father and the X chromosome from his mother, not the X from the father.
3. Therefore, A is true but R is false, which corresponds to option (c).

Answer: A is true but R is false (a boy gets X from the mother and Y from the father) — option (c).

A: Clotting stops heavy blood loss. R: Clotting is caused by plasma and white blood cells.

1. Assertion A is true because blood clotting prevents excessive blood loss from wounds.
2. Reason R is false because clotting is caused by platelets, not plasma and white blood cells.
3. Therefore, the correct option is (c) where A is true but R is false.

Answer: A is true but R is false (platelets bring about clotting) — option (c).

A: More turns in a circular coil give a stronger field at its centre. R: Current in every turn flows the same way, so the fields of the turns add.

1. Assertion A is true because increasing the number of turns in a circular coil increases the magnetic field strength at its centre.
2. Reason R is true because the current flows in the same direction in all turns, causing their individual magnetic fields to add up.
3. Since R provides the correct explanation for why A is true, the correct option is (a).

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

(i) X is made from gypsum and sets hard when mixed with water. Name X and give its formula. (ii) How do baking soda and baking powder differ in composition?

1. (i) X is Plaster of Paris, calcium sulphate hemihydrate, CaSO₄·½H₂O.
2. (ii) Baking soda is only sodium hydrogen carbonate (NaHCO₃); baking powder is NaHCO₃ mixed with a mild edible acid such as tartaric acid.

Answer: (i) X is Plaster of Paris, calcium sulphate hemihydrate, CaSO₄·½H₂O. (ii) Baking soda is only sodium hydrogen carbonate (NaHCO₃); baking powder is NaHCO₃ mixed with a mild edible acid such as tartaric acid.

Write balanced equations for heating (i) blue copper sulphate crystals and (ii) sodium hydrogen carbonate during cooking.

1. For blue copper sulphate crystals, write the decomposition equation: CuSO₄·5H₂O → (heat) CuSO₄ + 5H₂O.
2. For sodium hydrogen carbonate, write the thermal decomposition equation: 2NaHCO₃ → (heat) Na₂CO₃ + H₂O + CO₂.

Answer: (i) CuSO₄·5H₂O → (heat) CuSO₄ + 5H₂O. (ii) 2NaHCO₃ → (heat) Na₂CO₃ + H₂O + CO₂.

(a) What does insulin do to blood sugar, and which disease results from its shortage? (b) What controls when and how much insulin is released?

1. Insulin acts to lower the level of sugar in the blood, and a shortage of this hormone leads to the disease diabetes.
2. The release of insulin is controlled by a feedback mechanism where a rise in blood sugar triggers the pancreas to secrete more insulin.
3. As the blood sugar level falls back to normal, the secretion of insulin is reduced to maintain balance.

Answer: (a) Insulin lowers blood sugar; its deficiency causes diabetes. (b) A feedback mechanism: a rise in blood sugar makes the pancreas release more insulin, and as sugar falls, secretion is reduced.

State the type of blood (oxygenated/deoxygenated) and its path for (i) vena cava, (ii) pulmonary artery.

1. The vena cava carries deoxygenated blood from the body to the heart.
2. The pulmonary artery carries deoxygenated blood from the heart to the lungs.

Answer: (i) Vena cava: deoxygenated blood from the body to the heart. (ii) Pulmonary artery: deoxygenated blood from the heart to the lungs.

Show by flow chart how glucose is broken down in a cell for energy (i) with oxygen, (ii) with too little oxygen.

1. In both cases, glucose is first broken down in the cytoplasm to form pyruvate (a 3-carbon compound) and energy.
2. (i) With oxygen: Pyruvate moves to the mitochondria, where it is completely broken down into CO₂, H₂O, and a large amount of energy.
3. (ii) With too little oxygen: Pyruvate is converted into lactic acid in the cytoplasm, releasing a small amount of energy (this occurs in muscle cells).

Answer: Glucose → (cytoplasm) pyruvate (3-carbon) + energy. (i) With oxygen: pyruvate → (mitochondria) CO₂ + H₂O + energy. (ii) Lack of oxygen (e.g. in muscle cells): pyruvate → lactic acid + energy.

Note: In yeast, lack of oxygen gives ethanol + CO₂ instead of lactic acid; that route is also accepted.

Where in the excretory system are nephrons found? Describe their structure and function.

1. Nephrons are the functional units of the excretory system and are located in the kidneys.
2. Structurally, a nephron consists of a knot of thin-walled capillaries called the glomerulus, which is enclosed by a cup-shaped structure called Bowman's capsule.
3. The Bowman's capsule continues as a long, coiled tubule that eventually opens into a collecting duct.
4. The function of the nephron is to filter nitrogenous wastes from the blood to form urine, while reabsorbing useful substances back into the blood.

Answer: Nephrons are in the kidneys. Structure: a knot of thin-walled capillaries (glomerulus) inside the cup-shaped Bowman's capsule, which continues as a long tubule ending in a collecting duct. Function: filters nitrogenous wastes from blood to make urine (and reabsorbs useful substances).

A narrow beam of white light XY enters glass prism ABC; a screen PQ is placed beyond the prism. Show the emergent beam on the screen, name the phenomenon and give its cause.

Uses a figure printed in CBSE's paper — what the figure shows: Ray XY hits face AB of an upright triangular prism ABC (base BC at the bottom); vertical screen PQ stands to the right of the prism.

1. Draw the emergent beam on screen PQ as a fan of seven distinct colours, with violet at the bottom and red at the top.
2. Name the phenomenon as dispersion of white light, caused by different wavelengths bending through different angles in the glass prism.

Answer: The light splits into a spectrum on PQ, violet bent most (lowest on screen) and red least (highest). This is dispersion of white light; it happens because different colours bend through different angles in glass (different wavelengths).

With age, the eye's power to see both near and far objects falls. (i) Why? (ii) Name the defect. (iii) Draw the corrective lens arrangement.

1. (i) With age, the ciliary muscles weaken and the eye lens loses flexibility, reducing its ability to adjust focus.
2. (ii) This defect is called presbyopia.
3. (iii) It is corrected using a bifocal lens, which has a concave upper part for distant vision and a convex lower part for near vision.

Answer: (i) The ciliary muscles weaken and the eye lens loses flexibility. (ii) Presbyopia. (iii) A bifocal lens: concave upper part for distant vision and convex lower part for near vision.

How do harmful chemicals build up more and more at each trophic level of a food chain?

1. Harmful chemicals like pesticides enter the soil and water, where they are absorbed by plants at the first trophic level.
2. These chemicals are non-biodegradable, so they cannot be broken down by the body and remain in the tissues of the organisms.
3. As organisms at higher trophic levels consume many individuals from the lower level, the chemicals accumulate in their bodies in increasing concentrations.
4. This progressive increase in the concentration of harmful chemicals at each successive trophic level is known as biological magnification.

Answer: Pesticides washed into soil and water are taken up by plants and then eaten by animals. They are not degradable, so they collect in higher amounts at each higher trophic level — biological magnification.

(a) Name the reducing agent in: (i) 4NH₃ + 5O₂ → 4NO + 6H₂O, (ii) H₂O + F₂ → HF + HOF, (iii) Fe₂O₃ + 3CO → 2Fe + 3CO₂, (iv) 2H₂ + O₂ → 2H₂O. (b) Define redox reaction in terms of oxygen gain/loss.

1. (a)(i) In 4NH₃ + 5O₂ → 4NO + 6H₂O, NH₃ gains oxygen (becomes NO), so NH₃ is oxidised and is the reducing agent.
2. (a)(ii) In H₂O + F₂ → HF + HOF, H₂O loses hydrogen to fluorine (forming HF), so H₂O is oxidised and is the reducing agent.
3. (a)(iii) In Fe₂O₃ + 3CO → 2Fe + 3CO₂, CO gains oxygen from Fe₂O₃ to become CO₂, so CO is the reducing agent.
4. (a)(iv) In 2H₂ + O₂ → 2H₂O, H₂ gains oxygen to become H₂O, so H₂ is the reducing agent.
5. (b) A redox reaction is one in which one substance gains oxygen (is oxidised) while another loses oxygen (is reduced) at the same time.

Answer: (a) (i) NH₃ (ii) H₂O (iii) CO (iv) H₂. (b) A redox reaction is one in which one substance gains oxygen (oxidation) while another loses oxygen (reduction) at the same time.

Note: CBSE's marking scheme gives full marks for (a)(ii) to every student who attempted it.

(a) Give a remedy for (i) excess stomach acid in indigestion, (ii) a bee sting or nettle sting. (b) Milk has pH 6. Does pH go up or down when it turns into curd? Why?

1. (a)(i) Excess stomach acid is neutralized by taking an antacid, which is a mild base like milk of magnesia.
2. (a)(ii) For a bee or nettle sting, apply a mild base such as baking soda solution to neutralize the acid, or rub a dock-plant leaf for nettle stings.
3. (b) The pH decreases when milk turns into curd because the process produces lactic acid, making the substance more acidic than the original milk.

Answer: (a) (i) Take an antacid (a mild base such as milk of magnesia). (ii) Apply baking soda solution (a mild base), or rub dock-plant leaf on a nettle sting. (b) pH decreases, because curd contains lactic acid and is more acidic than milk.

(i) Role of ATP in respiration? (ii) What allows enough gas exchange in plants? (iii) On what does the direction of gas diffusion in plants depend?

1. ATP acts as the cell's energy currency, and its breakdown releases a fixed amount of energy to drive cellular reactions.
2. Gas exchange in plants is facilitated by stomata in leaves and diffusion through the surfaces of stems and roots.
3. The direction of gas diffusion depends on environmental conditions and the plant's requirements, such as photosynthesis during the day and respiration at night.

Answer: (i) ATP is the cell's energy currency; its breakdown releases a fixed amount of energy to drive cell reactions. (ii) Stomata in leaves plus diffusion through surfaces of stems and roots. (iii) The environmental conditions and the plant's own requirements (e.g. photosynthesis by day, respiration at night).

(i) What is the stored energy reserve in plants and in animals? (ii) How do desert plants photosynthesise when stomata stay shut by day?

1. Plants store their energy reserve in the form of starch.
2. Animals store their energy reserve in the form of glycogen.
3. Desert plants open their stomata at night to take in carbon dioxide.
4. They store the carbon dioxide as an intermediate compound.
5. During the day, energy absorbed by chlorophyll is used to convert this intermediate into food.

Answer: (i) Plants store starch; animals store glycogen. (ii) They open stomata at night, take in CO₂ and store it as an intermediate compound; in the day, energy absorbed by chlorophyll is used to turn this into food.

Object AB stands in front of a convex mirror (pole P; F marked R and centre C behind it). Ray 1 goes parallel to the axis; ray 2 heads towards C. (a) Complete the ray diagram. (b) Give nature, position and size of the image. (c) Sign of image distance?

Uses a figure printed in CBSE's paper — what the figure shows: Convex mirror facing left; object AB on the left of pole P; points R (focus) and C lie behind the mirror on the right; ray 1 parallel to axis, ray 2 directed towards C.

1. (a) Ray 1, parallel to the axis, reflects so that it appears to come from the focus behind the mirror; draw the reflected ray going away and extend it backwards as a dotted line to the focus.
2. Ray 2, aimed at the centre of curvature C, meets the mirror along the normal and reflects back along its own path; extend it as a dotted line towards C.
3. The two dotted lines meet behind the mirror; that point is the top of the image A′, and B′ is on the axis below it.
4. (b) The image is virtual and erect, behind the mirror between P and F, and diminished.
5. (c) The image is behind the mirror, on the side where distances are positive, so the image distance is positive.

Answer: (a) Ray 1 reflects as if coming from F behind the mirror; ray 2 retraces its path. Extended backwards they meet behind the mirror, giving image A′B′. (b) Virtual and erect, behind the mirror between P and F, diminished. (c) Positive.

Why: (a) danger lights at airports and on tall buildings are red; (b) the sky looks dark to people flying very high; (c) a light beam is visible through a colloid?

1. (a) Red light has the longest wavelength among visible colors, so it is scattered the least by fog or smoke.
2. Because it is scattered least, red danger lights can be seen from a greater distance in poor visibility.
3. (b) At very high altitudes, the air is extremely thin and contains very few particles to scatter sunlight.
4. Without sufficient scattering of blue light by air molecules, the sky appears dark to people flying at those heights.
5. (c) The particles in a colloid are large enough to scatter light effectively, a phenomenon known as the Tyndall effect.
6. This scattering makes the path of the light beam visible when it passes through the colloidal solution.

Answer: (a) Red has the longest wavelength and is scattered least by fog or smoke, so it is seen from far. (b) At great heights there is almost no air to scatter sunlight. (c) Colloid particles are large enough to scatter light (Tyndall effect), making the path visible.

(i) Which rule gives the force on a current-carrying conductor in a magnetic field? State it. (ii) An electron enters a field (lines left to right) (i) at right angles, (ii) at a slant, (iii) along the lines. Where is the force maximum and minimum? Why?

Uses a figure printed in CBSE's paper — what the figure shows: Three sketches with horizontal field lines: (i) electron moving vertically up, (ii) electron moving at an angle, (iii) electron moving along the field lines.

1. (i) The rule is Fleming's left-hand rule.
2. State: Stretch the thumb, forefinger, and middle finger of the left hand mutually at right angles.
3. Forefinger points in the direction of the magnetic field, middle finger in the direction of current, and thumb in the direction of force or motion.
4. (ii) The force is maximum in case (i) because the electron moves perpendicular to the magnetic field lines.
5. The force is minimum (zero) in case (iii) because the electron moves parallel to the magnetic field lines.
6. The force on a moving charge is given by F = qvB sin θ, which is maximum when θ = 90° and zero when θ = 0°.

Answer: (i) Fleming's left-hand rule: stretch thumb, forefinger and middle finger of the left hand at right angles; forefinger = field, middle finger = current, thumb = force/motion. (ii) Maximum in case (i), as motion is perpendicular to the field; minimum (zero) in case (iii), as the electron moves parallel to the field.

(i) Draw field lines of (1) a current-carrying solenoid and (2) a bar magnet. (ii) Give two differences between these fields.

1. Draw the field lines for the solenoid and bar magnet, showing lines leaving the North pole, curving around, and entering the South pole.
2. Indicate that inside the solenoid, the field lines are parallel and straight, representing a uniform magnetic field.
3. State the first difference: A solenoid's field strength can be changed by altering the current, and its direction is reversed by reversing the current.
4. State the second difference: A solenoid's field is temporary and exists only while current flows, whereas a bar magnet's field is permanent and cannot be changed in strength or direction.

Answer: (i) Both: lines leave the N end, curve round and enter the S end; inside a solenoid the lines are parallel and straight (uniform field). (ii) A solenoid's field strength can be changed by changing the current, and its direction reversed by reversing the current; it exists only while current flows (temporary). A bar magnet's field is permanent and cannot be changed in strength or direction.

(i) Why is a kitchen garden an artificial ecosystem but a forest natural? (ii) Name two things to keep in mind to make a home ecosystem self-sustaining, with reasons.

1. A kitchen garden is an artificial ecosystem because it is created and maintained by human intervention, whereas a forest is a natural ecosystem maintained by nature itself.
2. To make a home ecosystem self-sustaining, include green aquatic plants to produce food and oxygen through photosynthesis.
3. Include animals or other organisms to ensure the cycling of nutrients and waste decomposition within the system.

Answer: (i) A kitchen garden is made and maintained by people (not self-sustaining); a forest is maintained by nature by itself. (ii) E.g. in an aquarium jar: keep green aquatic plants (they make food and oxygen) and supply oxygen/aeration with a pump; include animals so that nutrients cycle.

(i) Make a four-level food chain from hawk, snake, plants, rat. (ii) Producers pass 20,000 J to the second trophic level. How much energy goes from the third to the fourth level?

1. The food chain starts with the producer, which is the plant, followed by the primary consumer, the rat.
2. The secondary consumer is the snake, and the tertiary consumer is the hawk, forming the chain: Plants → rat → snake → hawk.
3. According to the 10% law, only 10% of energy is transferred from the second trophic level to the third level.
4. Energy at the third level = 10% of 20,000 J = 2,000 J.
5. Energy transferred from the third level to the fourth level = 10% of 2,000 J = 200 J.

Answer: (i) Plants → rat → snake → hawk. (ii) By the 10% law: third level gets 2000 J, and it passes 200 J to the fourth level.

Alcohol A, heated with concentrated H₂SO₄ at 443 K, gives unsaturated B (molecular mass 28 u). B adds one mole of H₂ over nickel to give saturated C. (i) Identify A, B, C. (ii) Equation for A → B. (iii) What happens when C burns? (iv) One industrial use of hydrogenation. (v) Products when A reacts with sodium.

1. B has molecular mass 28 u and is unsaturated, so B is ethene (CH₂=CH₂).
2. A is an alcohol that dehydrates to ethene, so A is ethanol (CH₃CH₂OH).
3. C is formed by adding H₂ to ethene, so C is ethane (CH₃–CH₃).
4. The equation for A to B is CH₃CH₂OH → (hot conc. H₂SO₄, 443 K) CH₂=CH₂ + H₂O.
5. When C burns, it produces CO₂ and water with heat: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O + heat.
6. One industrial use of hydrogenation is converting vegetable oils into vegetable ghee.
7. When A reacts with sodium, the products are sodium ethoxide and hydrogen gas.

Answer: (i) A = ethanol (CH₃CH₂OH), B = ethene (CH₂=CH₂), C = ethane (CH₃–CH₃). (ii) CH₃CH₂OH → (hot conc. H₂SO₄, 443 K) CH₂=CH₂ + H₂O. (iii) It burns to CO₂ and water with a lot of heat: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O + heat. (iv) Turning vegetable oils into vegetable ghee (fats). (v) Sodium ethoxide and hydrogen gas.

(i) Draw how micelles form when soap acts on oily dirt. (ii) Two tubes with 10 mL hard water each: soap added to X, detergent to Y, both shaken equally. (1) Which foams more and why? (2) Which forms a curdy solid and why?

1. Draw a cluster where soap molecules surround an oily drop.
2. Show the hydrocarbon tails of the soap molecules pointing inward into the oil.
3. Show the ionic heads of the soap molecules pointing outward toward the water.
4. Explain that this structure forms a micelle which traps the oil and allows it to be rinsed away.
5. Identify that Tube Y foams more because detergents do not form scum in hard water.
6. Identify that Tube X forms a curdy solid because soap reacts with calcium and magnesium salts to form insoluble scum.

Answer: (i) Soap molecules arrange with hydrocarbon tails dissolved in the oil drop and ionic heads facing the water, forming a cluster (micelle) that traps the oil and is rinsed away. (ii) (1) Tube Y: detergents work well in hard water. (2) Tube X: soap reacts with calcium and magnesium salts of hard water to form insoluble scum.

(a) Which parts of a bisexual flower take no direct part in reproduction? (b) Differentiate self- and cross-pollination; give two uses of pollination. (c) What do ovules and ovary become after fertilisation?

1. (a) The parts of a bisexual flower that do not take direct part in reproduction are the sepals (collectively called calyx) and the petals (collectively called corolla).
2. (b) Self-pollination is the transfer of pollen from the anther to the stigma of the same flower or another flower on the same plant.
3. (b) Cross-pollination is the transfer of pollen from the anther of one flower to the stigma of a flower on a different plant.
4. (b) Two uses of pollination are that it is essential for fertilisation and seed formation, and cross-pollination introduces genetic variation in offspring.
5. (c) After fertilisation, the ovules develop into seeds.
6. (c) After fertilisation, the ovary develops into the fruit.

Answer: (a) Sepals (calyx) and petals (corolla). (b) Self-pollination: pollen moves from anther to stigma of the same flower or another flower on the same plant. Cross-pollination: pollen moves to a flower on a different plant. Uses: needed for fertilisation and seed formation; cross-pollination brings variation (also stimulates fruit growth). (c) Ovules become seeds; the ovary becomes the fruit.

(a) An iron uses 880 W at maximum heating and 330 W at minimum, on 220 V. Find current and resistance in each case. (b) What is the heating effect of current? (c) Derive the heat produced in a resistor in time t.

1. For maximum heating, calculate current using I = P/V, so I = 880/220 = 4 A.
2. For maximum heating, calculate resistance using R = V/I, so R = 220/4 = 55 Ω.
3. For minimum heating, calculate current using I = P/V, so I = 330/220 = 1.5 A.
4. For minimum heating, calculate resistance using R = V/I, so R = 220/1.5 = 146.67 Ω.
5. The heating effect of current is the conversion of electrical energy into heat energy when current flows through a conductor.
6. Work done moving charge Q through p.d. V is W = VQ, and since Q = It, W = VIt.
7. Substituting V = IR into W = VIt gives W = I²Rt, which is the heat H produced in time t.

Answer: (a) I = P/V, R = V/I. Maximum: 4 A, 55 Ω; minimum: 1.5 A, 146.67 Ω. (b) When current flows through a resistor, electrical energy is converted into heat. (c) Work done moving charge Q through p.d. V is VQ = VIt; with V = IR, heat H = I²Rt.

Note: CBSE's marking scheme prints 146.6 Ω; 220/1.5 = 146.67 Ω.

What happens when copper is heated in air? Give the equation.

1. When copper is heated in air, it reacts with oxygen to form a black layer of copper(II) oxide.
2. The balanced chemical equation for this reaction is 2Cu + O₂ → 2CuO.

Answer: Copper turns into black copper(II) oxide: 2Cu + O₂ → 2CuO.

Why are some metal oxides called amphoteric? Give an example.

1. Metal oxides are called amphoteric because they react with both acids and bases to form salt and water.
2. An example is aluminium oxide (Al₂O₃) or zinc oxide (ZnO).

Answer: They react with both acids and bases to give salt and water, e.g. Al₂O₃ (or ZnO).

Complete: (i) Na₂O(s) + H₂O(l) → (ii) Al₂O₃ + 2NaOH →

1. (i) Sodium oxide reacts with water to form sodium hydroxide, so the product is 2NaOH(aq).
2. (ii) Aluminium oxide reacts with sodium hydroxide to form sodium aluminate and water, so the products are 2NaAlO₂ and H₂O.

Answer: (i) Na₂O + H₂O → 2NaOH(aq). (ii) Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O.

Sulphur burns in oxygen to give a colourless gas. (i) Equation? (ii) Name the gas. (iii) Its nature? (iv) Effect on dry litmus paper?

1. (i) Sulphur reacts with oxygen to form sulphur dioxide, so the balanced chemical equation is S + O₂ → SO₂.
2. (ii) The colourless gas produced in this reaction is named sulphur dioxide.
3. (iii) Sulphur dioxide is a non-metal oxide, which makes its nature acidic.
4. (iv) Dry sulphur dioxide gas does not produce ions, so it causes no change on dry litmus paper; it only acts when the paper is moist.

Answer: (i) S + O₂ → SO₂. (ii) Sulphur dioxide. (iii) Acidic. (iv) No change on dry litmus (it acts only when moist).

Apart from seed shape and seed colour, give two pairs of contrasting pea characters Mendel used.

1. Mendel studied seven contrasting characters in pea plants, including plant height and flower colour.
2. Two pairs of contrasting characters are tall versus dwarf plant height and violet versus white flower colour.

Answer: Tall vs dwarf plant; violet (purple) vs white flowers.

Differentiate dominant and recessive traits.

1. A dominant trait is expressed even when only one copy of the gene is present in the organism.
2. A recessive trait is masked by the dominant trait and is expressed only when both copies of the gene are recessive.

Answer: A dominant trait shows even when only one copy of its gene is present; a recessive trait is masked by the dominant one and shows only when both copies are recessive.

Round yellow peas crossed with wrinkled green; F₁ selfed. Give the F₂ ratio of seed types and what it shows.

1. The F1 generation consists of dihybrid heterozygotes (RrYy) that produce four types of gametes (RY, Ry, rY, ry) in equal proportions.
2. Selfing these F1 plants results in the F2 ratio of Round yellow : Round green : Wrinkled yellow : Wrinkled green = 9 : 3 : 3 : 1.
3. This ratio demonstrates that the genes for seed shape and seed colour are inherited independently of each other.

Answer: Round yellow : round green : wrinkled yellow : wrinkled green = 9 : 3 : 3 : 1. Seed shape and seed colour are inherited independently.

Cross pure violet (VV) with pure white (vv) pea plants. Show the F₁ and F₂ progeny.

1. Cross pure violet (VV) with pure white (vv) to get F₁ generation.
2. All F₁ offspring are Vv and show violet phenotype.
3. Self-cross F₁ (Vv × Vv) to obtain F₂ generation.
4. F₂ genotypic ratio is VV : Vv : vv = 1 : 2 : 1, and phenotypic ratio is violet : white = 3 : 1.

Answer: F₁: all Vv, violet. F₁ selfed (Vv × Vv) gives VV : Vv : vv = 1 : 2 : 1, i.e. violet : white = 3 : 1.

Give two uses of concave mirrors.

1. Concave mirrors are used in torches, headlights, and searchlights to produce a powerful parallel beam of light.
2. They are also used as shaving mirrors or dentist's mirrors to obtain a magnified and upright image of the object.

Answer: Any two: torches/headlights/searchlights, shaving mirrors, dentist's mirrors, solar furnaces.

The focus of a concave mirror is 15 cm from it. Find the radius of curvature.

1. Given that the focal length (f) of the concave mirror is 15 cm.
2. Using the relation R = 2f, the radius of curvature is calculated as R = 2 × 15 = 30 cm.

Answer: R = 2f = 2 × 15 = 30 cm.

Draw the ray diagram for an object placed between the pole and focus of a concave mirror.

1. Place the object AB on the axis between the pole P and the focus F of the concave mirror.
2. Draw a ray from A parallel to the axis; after reflection it passes through F.
3. Draw a second ray from A along the line from C; it strikes the mirror normally and reflects back along itself.
4. The reflected rays spread apart, so extend them as dotted lines behind the mirror; they meet at A′, giving a virtual, erect, enlarged image A′B′ behind the mirror.

Answer: A ray parallel to the axis reflects through F; a ray from the top along the line from C reflects back on itself. The reflected rays diverge; extended behind the mirror they meet, giving a virtual, erect, enlarged image behind the mirror.

A 10 cm object is 100 cm in front of a concave mirror and its image forms at the object's own position. Find (i) focal length, (ii) magnification with sign.

1. Given object distance u = -100 cm and image forms at the object's position, so image distance v = -100 cm.
2. Using the mirror formula 1/f = 1/v + 1/u, substitute values to get 1/f = -1/100 - 1/100 = -2/100.
3. Solving for focal length gives f = -50 cm.
4. Calculate magnification using m = -v/u, which is -(-100)/(-100) = -1, indicating a real, inverted image of the same size.

Answer: u = v = -100 cm; 1/f = 1/v + 1/u = -2/100, so f = -50 cm and m = -v/u = -1 (real, inverted, same size).