CBSE Class 10 Science: 2023 board paper, set 31/1/1

1. Sodium carbonate reacts with dilute HCl and releases carbon dioxide: $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3+2HCl\to 2NaCl+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2$

2. $\mathrm{C}{\mathrm{O}}_2$ turns lime water milky because insoluble calcium carbonate forms: $\mathrm{Ca}(\mathrm{OH}{)}_2+\mathrm{C}{\mathrm{O}}_2\to \mathrm{CaC}{\mathrm{O}}_3\downarrow +{\mathrm{H}}_2\mathrm{O}$

3. With excess $\mathrm{C}{\mathrm{O}}_2$, soluble calcium hydrogen carbonate forms and the milkiness disappears: $\mathrm{CaC}{\mathrm{O}}_3+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2\to \mathrm{Ca}(\mathrm{HC}{\mathrm{O}}_3{)}_2$. This matches option (B).

   1 mark

Answer: Option (B)

1. On heating, lead nitrate breaks down into simpler substances. This is thermal decomposition, not oxidation: $2Pb(\mathrm{N}{\mathrm{O}}_3{)}_2\stackrel{\mathrm{heat}}{\to }2PbO+4N{\mathrm{O}}_2+{\mathrm{O}}_2$

2. The brown fumes are nitrogen dioxide ($\mathrm{N}{\mathrm{O}}_2$). Lead oxide is a yellow solid and oxygen is colourless. So the answer is (A).

   1 mark

Answer: Option (A)

1. Balance O first. There are 2 O on the left, so $y=2$ ($2{\mathrm{H}}_2\mathrm{O}$). Then there are 4 H on the right, so $x=4$ ($4HCl$).

2. Cl: there are 4 on the left. On the right, $\mathrm{MnC}{\mathrm{l}}_2$ has 2, so $\mathrm{C}{\mathrm{l}}_2$ must supply the other 2, giving $z=1$.

3. Balanced equation: $\mathrm{Mn}{\mathrm{O}}_2+4HCl\to \mathrm{MnC}{\mathrm{l}}_2+2{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{l}}_2$, so $x,y,z=4,2,1$, which is option (C).

   1 mark

Answer: x = 4, y = 2, z = 1; option (C)

1. Check each row:

   • (a) Lemon juice (pH about 2–3) is acidic, not basic.
   • (c) Gastric juice has pH about 1–2, not 6.
   • (d) Pure water (pH 7) turns universal indicator green, not yellow.

2. Milk of magnesia has pH about 10. It is basic and gives a blue colour on pH paper. So row (b) is correct.

   1 mark

Answer: Option (B)

1. The thermite process uses aluminium as the reducing agent: $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3+2Al\to 2Fe+\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3+\mathrm{heat}$. So X = Al.

2. Aluminium burns to form $\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3$, an amphoteric oxide. ($\mathrm{A}{\mathrm{l}}_3{\mathrm{O}}_4$ is not a real formula.) So the answer is (B).

   1 mark

Answer: X = Al, Y = Al₂O₃; option (B)

1. Washing soda is sodium carbonate decahydrate, which has 10 molecules of water of crystallisation.

2. Its formula is $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3.10{\mathrm{H}}_2\mathrm{O}$, which is option (C). $\mathrm{NaHC}{\mathrm{O}}_3$ is baking soda and NaOH is caustic soda.

   1 mark

Answer: Option (C) Na₂CO₃·10H₂O

1. A has only C–C single bonds. It is cyclohexane (${\mathrm{C}}_6{\mathrm{H}}_{12}$), a saturated cyclic hydrocarbon.

2. B (benzene) and C (cyclohexadiene) both contain C=C double bonds, so both are unsaturated. Their formulas are ${\mathrm{C}}_6{\mathrm{H}}_6$ and ${\mathrm{C}}_6{\mathrm{H}}_8$, so they are not isomers of each other.

3. So the correct statement is (C).

   1 mark

Answer: Option (C)

1. Saprophytes such as fungi release enzymes onto the food, digest it outside the body, and then absorb the simple products.

2. Rhizopus (bread mould) is a saprophytic fungus. So the answer is (D).

   1 mark

Answer: Option (D) Rhizopus

1. Statements (A), (B) and (C) are true. Herbivores have a longer small intestine than carnivores. Digestion is completed in the small intestine. It receives bile from the liver and pancreatic juice from the pancreas.

2. Water is absorbed from the unabsorbed food in the large intestine, not by the villi of the small intestine. So (D) is the incorrect statement.

   1 mark

Answer: Option (D)

1. A gene is a section of DNA that carries the information for making a protein. It occupies a fixed position on a particular chromosome in a species.

2. Options (B), (C) and (D) contradict this: genes do code for proteins, a chromosome carries many genes, and inherited traits are controlled by genes. So the answer is (A).

   1 mark

Answer: Option (A)

1. Tropic movements are directional responses (towards or away from a stimulus such as light, gravity, water or chemicals). They occur in both roots and stems.

2. These movements happen through unequal growth of the plant part, so they are growth-related. The answer is (D).

   1 mark

Answer: Option (D)

1. Bryophyllum produces buds on the margins of its leaves. Potato tubers are stems. Money-plant grows from stem cuttings. All three matches are correct.

2. Rose is propagated by stem cuttings, not roots. So 'Rose, root' is the incorrect match, option (D).

   1 mark

Answer: Option (D)

1. In series: $R_s=8+8+8+8=32\Omega$

2. In parallel: $\frac{1}{R_p}=4\times \frac{1}{8}=\frac{1}{2}$, so $R_p=2\Omega$

3. $\frac{R_s}{R_p}=\frac{32}{2}=16$, which is option (D).

   1 mark

Answer: 16; option (D)

1. Domestic wiring has two circuits: 15 A for high-power appliances and 5 A for low-power ones such as fans and bulbs.

2. A geyser draws a large current, so it is connected to the 15 A line. The answer is (A).

   1 mark

Answer: Option (A)

1. Least count = (value of one main interval) ÷ (number of small divisions in it). The scale has 10 divisions between 0 and 1.5 V, so LC $=\frac{1.5}{10}=0.15$ V.

2. The needle stands 2 divisions beyond the 1.5 V mark. Reading $=1.5+2\times 0.15=1.8$ V. So the answer is (C).

   1 mark

Answer: LC = 0.15 V, reading = 1.8 V; option (C)

1. By the right-hand thumb rule, current from left to right in the top wire AB gives a field at P (below AB) pointing into the page. The same current in the bottom wire CD gives a field at P (above CD) pointing out of the page.

2. P is equidistant from the two wires and the currents are equal, so the two fields are equal in size and opposite in direction. The resultant field is zero, option (C).

   1 mark

Answer: Zero; option (C)

1. Lead is above copper in the reactivity series, so it displaces copper: $\mathrm{Pb}+\mathrm{CuS}{\mathrm{O}}_4\to \mathrm{PbS}{\mathrm{O}}_4+\mathrm{Cu}$. Reason (R) is true.

2. Because copper ions are removed from the solution, the blue colour fades. So (A) is true and (R) explains it, option (A).

   1 mark

Answer: Option (A)

1. Sex is decided by the sex chromosome inherited from the parents: XX gives a girl, XY gives a boy. So (A) is true.

2. A boy (XY) gets his Y chromosome from his father and his X chromosome from his mother. So (R) is false, and the answer is (C).

   1 mark

Answer: Option (C)

1. Clotting seals a leak in a blood vessel and prevents excessive blood loss. So (A) is true.

2. Clotting is brought about by platelets, not by plasma and white blood cells. WBCs fight infection. So (R) is false, and the answer is (C).

   1 mark

Answer: Option (C)

1. For a coil of n turns, the field at the centre is n times the field of a single turn. So (A) is true.

2. This happens because the current flows in the same direction in every turn, so the fields add up. (R) is true and explains (A), option (A).

   1 mark

Answer: Option (A)

1. (i) Gypsum heated to 373 K loses part of its water and forms Plaster of Paris (calcium sulphate hemihydrate). It sets into a hard mass when mixed with water. So X = Plaster of Paris.

   ½ mark

2. (i) Formula of X: $\mathrm{CaS}{\mathrm{O}}_4.\frac{1}{2}{\mathrm{H}}_2\mathrm{O}$

   ½ mark

3. (ii) Baking soda is a single compound, sodium hydrogen carbonate, $\mathrm{NaHC}{\mathrm{O}}_3$.

   ½ mark

4. (ii) Baking powder is a mixture of baking soda ($\mathrm{NaHC}{\mathrm{O}}_3$) and a mild edible acid such as tartaric acid.

   ½ mark

Answer: (i) X = Plaster of Paris, CaSO₄·½H₂O; (ii) baking soda is NaHCO₃ only, while baking powder is NaHCO₃ + tartaric acid (mild edible acid)

1. (i) On heating, the blue crystals lose their water of crystallisation and turn into white anhydrous copper sulphate:
   $\mathrm{CuS}{\mathrm{O}}_4.5{\mathrm{H}}_2\mathrm{O}\stackrel{\mathrm{heat}}{\to }\mathrm{CuS}{\mathrm{O}}_4+5{\mathrm{H}}_2\mathrm{O}$

   1 mark

2. (ii) On heating, baking soda decomposes to sodium carbonate, water and carbon dioxide:
   $2NaHC{\mathrm{O}}_3\stackrel{\mathrm{heat}}{\to }\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2$

   1 mark

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

1. (a) Insulin, secreted by the pancreas, lowers the blood sugar level by helping body cells take up and use glucose.

   ½ mark

2. (a) Too little insulin raises the blood sugar level and causes diabetes.

   ½ mark

3. (b) Insulin release is controlled by a feedback mechanism. When the blood sugar level rises, the pancreas detects it and secretes more insulin. As the sugar level falls, insulin secretion is reduced.

   1 mark

Answer: (a) Insulin lowers blood sugar; its deficiency causes diabetes; (b) a feedback mechanism: rising sugar increases insulin release and falling sugar reduces it

1. (i) Vena cava carries deoxygenated blood.

   ½ mark

2. (i) Path: from the body organs to the heart (right atrium).

   ½ mark

3. (ii) Pulmonary artery carries deoxygenated blood.

   ½ mark

4. (ii) Path: from the heart (right ventricle) to the lungs, where the blood is oxygenated.

   ½ mark

Answer: (i) Vena cava: deoxygenated, from body to heart (right atrium); (ii) Pulmonary artery: deoxygenated, from heart (right ventricle) to lungs

1. In both cases the first step is the same. In the cytoplasm, glucose (a 6-carbon molecule) breaks down into pyruvate (a 3-carbon molecule) and releases a small amount of energy. This step is called glycolysis.

2. (i) In the presence of oxygen (aerobic respiration):

   Glucose $\stackrel{\text{in cytoplasm}}{\to }$ Pyruvate + Energy $\underset{\text{in mitochondria}}{\overset{\text{presence of oxygen}}{\to }}$ Carbon dioxide + Water + Energy

   The breakdown is complete, so a large amount of energy is released.

   1 mark

3. (ii) In lack of oxygen (in human muscle cells):

   Glucose $\stackrel{\text{in cytoplasm}}{\to }$ Pyruvate + Energy $\underset{\text{in muscle cells}}{\overset{\text{lack of oxygen}}{\to }}$ Lactic acid + Energy

   Lactic acid builds up in the muscles and causes cramps. Yeast, under anaerobic conditions, gives ethanol + $\mathrm{C}{\mathrm{O}}_2$ + energy instead.

   1 mark

Answer: (i) Glucose → (cytoplasm) Pyruvate → (oxygen, mitochondria) CO₂ + H₂O + energy; (ii) Glucose → (cytoplasm) Pyruvate → (lack of oxygen, muscle cells) Lactic acid + energy

1. Nephrons are found in the kidneys. Each kidney contains a very large number of them.

   ½ mark

2. Structure of a nephron: It begins as a cup-shaped structure called Bowman's capsule. This capsule holds a cluster of thin-walled blood capillaries called the glomerulus. The capsule continues as a long coiled tubular part, which is surrounded by capillaries. The tubule opens into a collecting duct. (A labelled diagram showing these parts may be drawn instead.)

   1 mark

3. Function (any one): It filters nitrogenous wastes such as urea out of the blood to form urine. It also reabsorbs useful substances such as glucose, amino acids, salts and water from the filtrate, and helps maintain the water balance of the body (osmoregulation).

   ½ mark

Answer: Kidneys; a nephron is Bowman's capsule with glomerulus + tubule leading to the collecting duct; it filters wastes from blood to form urine (and reabsorbs useful substances).

1. Ray diagram (described in words):

   • Draw the prism ABC and the incident white ray XY striking face AB at Y.
   • Inside the prism, the ray splits into a fan of seven coloured rays.
   • These rays bend towards the base BC. Red bends the least and violet bends the most.
   • The rays leave through face AC, bending further towards the base.
   • On the screen PQ they form a band of colours. From top (P side) to bottom (Q side) the order is R, O, Y, G, B, I, V: red at the top and violet at the bottom.
   • Arrows should be shown on all rays.
   1 mark

2. Phenomenon: This is the dispersion of white light, that is, the splitting of white light into its seven component colours (the spectrum).

   ½ mark

3. Cause: Light of different colours (different wavelengths) travels at different speeds in glass. So each colour bends through a different angle with respect to the incident ray. Red is deviated the least and violet the most.

   ½ mark

Answer: The beam splits into a spectrum, with red at the top and violet at the bottom on PQ (VIBGYOR read from the bottom up). The phenomenon is dispersion, caused by different colours bending through different angles.

1. (i) Reason: With age, the ciliary muscles gradually become weaker. The eye lens also loses its flexibility. So the eye can no longer change the focal length of its lens properly (its accommodation falls), and both near and far vision suffer.

   1 mark

2. (ii) Defect: Presbyopia. When the person suffers from both near and far vision problems, this is presbyopia along with myopia.

   ½ mark

3. (iii) Corrective lens: bifocal lens (diagram described in words):

   • Draw a spectacle lens divided into two halves.
   • Label the upper part concave lens (for distant vision).
   • Label the lower part convex lens (for near vision).
   ½ mark

Answer: (i) Weakening of the ciliary muscles and reduced flexibility of the eye lens; (ii) presbyopia; (iii) bifocal lens, with a concave upper part for distant vision and a convex lower part for near vision.

1. Entry into the food chain: Pesticides and other chemicals sprayed on crops are washed into the soil and water bodies. Plants absorb them from the soil along with water and minerals. Animals then eat these plants, so the chemicals enter the food chain.

   1 mark

2. Progressive accumulation: These chemicals are non-biodegradable, so they are neither broken down nor excreted. At every step, each organism eats many organisms of the lower trophic level. The chemical therefore keeps building up, and its concentration increases at each successive trophic level. It is highest in the top consumers, such as humans. This is called biological magnification.

   1 mark

Answer: Non-biodegradable chemicals from crops pass from the soil to plants and then to animals. Because they cannot be broken down, their concentration builds up at each trophic level. This is biological magnification.

1. (a) A reducing agent is the substance that gets oxidised: it gains oxygen or loses hydrogen.

2. (a)(i) In $4\mathrm{N}{\mathrm{H}}_3+5{\mathrm{O}}_2\to 4\mathrm{NO}+6{\mathrm{H}}_2\mathrm{O}$, $\mathrm{N}{\mathrm{H}}_3$ loses hydrogen and gains oxygen to form NO. **Reducing agent: $\mathrm{N}{\mathrm{H}}_3$**.

   ½ mark

3. (a)(ii) In ${\mathrm{H}}_2\mathrm{O}+{\mathrm{F}}_2\to \mathrm{HF}+\mathrm{HOF}$, ${\mathrm{F}}_2$ gains hydrogen (it is reduced), so ${\mathrm{H}}_2\mathrm{O}$ is the one giving up hydrogen. **Reducing agent: ${\mathrm{H}}_2\mathrm{O}$**.

   ½ mark

4. (a)(iii) In $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3+3\mathrm{CO}\to 2\mathrm{Fe}+3\mathrm{C}{\mathrm{O}}_2$, CO gains oxygen to become $\mathrm{C}{\mathrm{O}}_2$. Reducing agent: CO.

   ½ mark

5. (a)(iv) In $2{\mathrm{H}}_2+{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}$, ${\mathrm{H}}_2$ gains oxygen. **Reducing agent: ${\mathrm{H}}_2$**.

   ½ mark

6. (b) A redox reaction is one in which one substance gains oxygen (is oxidised) while another substance loses oxygen (is reduced), both at the same time.

   1 mark

Answer: (a) (i) NH₃, (ii) H₂O, (iii) CO, (iv) H₂; (b) a reaction in which gain of oxygen by one substance and loss of oxygen by another occur simultaneously.

1. (a)(i) Excess acid in the stomach lowers the pH. Taking an antacid, a mild base such as milk of magnesia (magnesium hydroxide), neutralises the excess acid.

   1 mark

2. (a)(ii) A bee sting or nettle sting injects methanoic acid into the skin. Rubbing on a mild base such as baking soda solution neutralises it and gives relief. For a nettle sting, the leaf of the dock plant can also be rubbed on.

   1 mark

3. (b) The pH will decrease.

   ½ mark

4. Reason: when milk sets into curd, bacteria convert lactose into lactic acid. Curd is therefore more acidic than milk, so its pH falls below 6.

   ½ mark

Answer: (a)(i) Take an antacid (e.g., milk of magnesia); (ii) apply baking soda solution or a mild base (dock leaf for nettle); (b) pH decreases because curd contains lactic acid and is more acidic.

1. (i) ATP is the energy currency of the cell. The energy released during cellular respiration is stored in ATP. When needed, ATP breaks down and releases a fixed amount of energy. This energy drives energy-requiring (endothermic) reactions in the cell, such as muscle contraction and protein synthesis.

   1 mark

2. (ii) Plants have stomata in their leaves, and their leaves, stems and roots have a large surface area. Gases diffuse in and out through the stomata and the surfaces of leaves, stems and roots. This ensures sufficient exchange of gases.

   1 mark

3. (iii) The direction of diffusion of gases depends on:
   1. The environmental conditions, for example, the presence or absence of light.

   ½ mark

4. 2. The requirements of the plant. In the daytime, photosynthesis uses $\mathrm{C}{\mathrm{O}}_2$ and releases ${\mathrm{O}}_2$. At night only respiration occurs, so ${\mathrm{O}}_2$ goes in and $\mathrm{C}{\mathrm{O}}_2$ comes out.

   ½ mark

Answer: (i) ATP is the energy currency that supplies energy for cellular processes; (ii) stomata and the large surface area of leaves, stems and roots; (iii) environmental conditions and the requirements of the plant.

1. (i) In plants, food is stored as the internal energy reserve starch.

   1 mark

2. In animals (including humans), the internal energy reserve is glycogen.

   1 mark

3. (ii) Desert plants keep their stomata open at night and take in $\mathrm{C}{\mathrm{O}}_2$. They convert it into an intermediate compound (an acid) and store it. During the day, the stomata stay closed to reduce water loss. Chlorophyll absorbs sunlight, and this energy is used to release $\mathrm{C}{\mathrm{O}}_2$ from the stored intermediate. This $\mathrm{C}{\mathrm{O}}_2$ is then used to make carbohydrates.

   1 mark

Answer: (i) Starch in plants, glycogen in animals; (ii) they take in CO₂ at night and store it as an intermediate, which is used in the day with light energy absorbed by chlorophyll.

1. (a) The mirror is a convex mirror: its reflecting surface faces the object, and F and C lie behind it. The point marked R on the axis is the principal focus F.

   • Ray 1 (parallel to the principal axis) is reflected so that it appears to come from F behind the mirror. Draw it diverging outward, with a dotted line back to F.
   • Ray 2 (directed towards C) strikes the mirror normally, so it is reflected back along the same path.
   • Extend both reflected rays backwards as dotted lines. They meet behind the mirror at A′.
   • Drop a perpendicular from A′ to the axis to get B′. A′B′ is the image. It lies between P and F.
   1 mark

2. (b) Nature: virtual and erect.

   ½ mark

3. Position: behind the mirror, between P and F.

   ½ mark

4. Size: diminished (smaller than the object).

   ½ mark

5. (c) The image is formed behind the mirror, on the side opposite the object. By the Cartesian sign convention, the image distance $v$ is positive.

   ½ mark

Answer: (a) Image A′B′ is formed behind the mirror between P and F; (b) virtual, erect, behind the mirror between P and F, diminished; (c) positive.

1. (a) Red light has the longest wavelength among visible colours, so it is scattered the least by fog, smoke and air particles. A red signal therefore keeps its colour and can be seen clearly from a long distance.

   1 mark

2. (b) At very high altitudes there is almost no atmosphere (air molecules and dust) to scatter sunlight. With no scattered light reaching the eyes, the sky appears dark instead of blue.

   1 mark

3. (c) The particles in a colloidal solution are large enough to scatter light. The beam is scattered sideways by these particles and reaches our eyes, so its path becomes visible. This is the Tyndall effect.

   1 mark

Answer: (a) Red has the longest wavelength and is scattered least, so it is visible from far; (b) there is no atmosphere to scatter light at high altitude; (c) colloid particles scatter light (Tyndall effect).

1. (i) Fleming's left-hand rule: Stretch the thumb, forefinger and middle (central) finger of the left hand so that they are mutually perpendicular.

   • If the forefinger points in the direction of the magnetic field,
   • and the middle finger points in the direction of the current,
   • then the thumb gives the direction of the force (motion) on the conductor.
   1 mark

2. (ii)(1) The force is maximum in Fig. (i).

   ½ mark

3. Reason: the electron moves at right angles (perpendicular) to the magnetic field, and the force is greatest when the motion is perpendicular to the field.

   ½ mark

4. (ii)(2) The force is minimum (zero) in Fig. (iii).

   ½ mark

5. Reason: the electron moves along (parallel to) the magnetic field lines, and no force acts when the motion is parallel to the field.

   ½ mark

Answer: (i) Fleming's left-hand rule; (ii) (1) maximum in Fig. (i), since the motion is perpendicular to the field; (2) minimum (zero) in Fig. (iii), since the motion is parallel to the field.

1. (i)(1) Solenoid (diagram described in words):

   • Draw a coil connected to a battery and a key.
   • Inside the coil, draw straight, parallel, equally spaced field lines, which show a uniform field.
   • Outside, draw closed loops going from one end (N) to the other (S), like those of a bar magnet.
   • Mark arrows going from S to N inside the coil and from N to S outside.
   1 mark

2. (i)(2) Bar magnet (diagram described in words):

   • Draw closed curves that come out of the N pole and enter the S pole outside the magnet, and run from S to N inside.
   • The lines are crowded near the poles and never cross each other.
   • Mark arrows on the lines.
   1 mark

3. (ii) Distinguishing features (any two):
   1. The strength of a solenoid's field can be changed by changing the current, but a bar magnet's field strength is fixed.

   ½ mark

4. 2. The direction of a solenoid's field can be reversed by reversing the current, but a bar magnet's field direction cannot be changed. (Also accepted: a solenoid's field is temporary and exists only while current flows, whereas a bar magnet's field is permanent.)

   ½ mark

Answer: (i) Field-line patterns of a solenoid (uniform parallel lines inside, bar-magnet-like outside) and of a bar magnet (N to S outside); (ii) a solenoid's field strength and direction can be changed by changing the current and its field is temporary, whereas a bar magnet's field is fixed and permanent.

1. (i) A kitchen garden is made and maintained by humans: they choose the plants and supply water, manure and care. It cannot sustain itself, so it is an artificial ecosystem. A forest develops and maintains itself naturally, without human help, so it is a self-sustaining natural ecosystem.

   1 mark

2. (ii) Example: an aquarium. Things to keep in mind:
   1. Supply oxygen, using an aerator or pump, or ensure enough light for the plants.
   2. Add aquatic plants (producers) along with animals (consumers) such as fish, and allow decomposers to be present.

   1 mark

3. Justification: The oxygen supply is replenished continuously for the organisms to respire. The aquatic plants make food by photosynthesis and serve as food for the animals, while decomposers recycle wastes into nutrients. So all the components needed for energy flow and nutrient cycling are present, and the system becomes self-sustaining.

   1 mark

Answer: (i) A kitchen garden is man-made and maintained by humans (non-sustainable), while a forest is maintained by nature (self-sustaining); (ii) e.g., an aquarium with a continuous oxygen supply and aquatic plants as producers, so oxygen is replenished and the plants provide food.

1. (i) Food chain: Plants → Rat → Snake → Hawk. Here plants are the producers, the rat is the primary consumer, the snake is the secondary consumer and the hawk is the tertiary consumer.

   1 mark

2. (ii) By the 10% law, only about 10% of the energy at one trophic level is passed on to the next level.

   Energy available at the second trophic level (rat) = 20,000 J.

   Energy transferred from the second to the third trophic level (snake) $=10\%$ of $20000=\frac{10}{100}\times 20000=2000$ J.

   1 mark

3. Energy transferred from the third to the fourth trophic level (hawk) $=10\%$ of $2000=\frac{10}{100}\times 2000=200$ J.

   So 200 J is transferred by the third trophic level to the fourth.

   1 mark

Answer: (i) Plants → Rat → Snake → Hawk; (ii) 200 J

1. (i) B is an unsaturated compound with molecular mass 28 u. For an alkene ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2n}$: $12n+2n=28\Rightarrow n=2$. So **B = ethene, $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2$**.

   ½ mark

2. Ethene is formed by dehydration of an alcohol with the same number of carbon atoms. So **A = ethanol (ethyl alcohol), $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$**.

   ½ mark

3. Adding one mole of ${\mathrm{H}}_2$ (Ni catalyst) to ethene gives a saturated hydrocarbon. So **C = ethane, $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_3$**: $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\stackrel{\mathrm{Ni}}{\to }\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_3$.

   ½ mark

4. (ii) Dehydration of ethanol with hot concentrated sulphuric acid. The acid acts as a dehydrating agent and removes water:
   $$\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\underset{443\mathrm{K}}{\overset{\text{hot conc. }{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }}\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$$

   1 mark

5. (iii) Ethane burns in air or oxygen to give carbon dioxide and water. A large amount of heat (and light) is released:
   $$2{\mathrm{C}}_2{\mathrm{H}}_6+7{\mathrm{O}}_2\to 4C{\mathrm{O}}_2+6{\mathrm{H}}_2\mathrm{O}+\mathrm{heat}\mathrm{and}\mathrm{light}$$

   1 mark

6. (iv) Hydrogenation is used in industry to convert vegetable oils into vegetable ghee (solid fats), using a nickel catalyst.

   ½ mark

7. (v) Ethanol reacts with sodium to give sodium ethoxide and hydrogen gas:
   $$2C{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+2Na\to 2C{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{ONa}+{\mathrm{H}}_2\uparrow$$

   1 mark

Answer: (i) A = ethanol (CH₃CH₂OH), B = ethene (CH₂=CH₂), C = ethane (C₂H₆); (ii) CH₃CH₂OH → CH₂=CH₂ + H₂O (hot conc. H₂SO₄, 443 K); (iii) CO₂ and H₂O are formed with release of a lot of heat; (iv) converting vegetable oils into vegetable ghee (fats); (v) sodium ethoxide and hydrogen

1. (i) A soap molecule is a sodium salt of a long-chain carboxylic acid. It has an ionic (hydrophilic) end, $-\mathrm{CO}{\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+}$, which dissolves in water, and a long hydrocarbon (hydrophobic) tail, which dissolves in oil.

   ½ mark

2. When soap is applied on oily dirt, the hydrocarbon tails dissolve in the oil droplet. The ionic heads stay outside, facing the water.

   ½ mark

3. Diagram (described): draw an oil droplet at the centre. Draw many soap molecules arranged radially around it, with zig-zag tails pointing inwards into the droplet and circular heads on the outside. Mark $\mathrm{N}{\mathrm{a}}^{+}$ ions around the outside. Label: 'oil droplet', 'hydrophobic tail', 'ionic/hydrophilic head', 'micelle'. This spherical cluster is a micelle.

   ½ mark

4. The micelles stay in solution as a colloid and do not come together, because their charged surfaces repel each other. When the cloth is rinsed with water, the oily dirt trapped inside the micelles is washed away.

   ½ mark

5. (ii)(1) More foam forms in test tube Y (detergent).

   ½ mark

6. Detergents do not form insoluble salts with the calcium and magnesium ions of hard water. So they stay effective in hard water and lather well.

   1 mark

7. (ii)(2) A curdy solid forms in test tube X (soap).

   ½ mark

8. Soap reacts with the calcium and magnesium salts in hard water and forms an insoluble precipitate called scum (curdy white solid). So little lather forms.

   1 mark

Answer: (i) Micelle: hydrocarbon tails dissolve in the oil droplet at the centre and ionic heads face the water outside (diagram); (ii)(1) Test tube Y, because detergents work in hard water; (2) Test tube X, because soap forms insoluble scum with Ca²⁺/Mg²⁺ salts

1. (a) The parts of a bisexual flower not directly involved in reproduction are the sepals (calyx)

   ½ mark

2. and the petals (corolla). Stamens and pistil are the reproductive parts.

   ½ mark

3. (b) Self-pollination: pollen grains are transferred from the anther to the stigma of the same flower, or of another flower on the same plant.

   1 mark

4. Cross-pollination: pollen grains are transferred from the anther of a flower on one plant to the stigma of a flower on a different plant of the same species, by agents such as wind, water, insects or birds.

   1 mark

5. Significance 1: pollination brings the male gamete to the female part, which leads to fertilisation and seed formation.

   ½ mark

6. Significance 2: cross-pollination brings genetic variation into the offspring. Pollination also stimulates fruit development.

   ½ mark

7. (c) After fertilisation, the ovules develop into seeds

   ½ mark

8. and the ovary develops (ripens) into the fruit.

   ½ mark

Answer: (a) Sepals (calyx) and petals (corolla); (b) Self: pollen to stigma of the same flower or another flower of the same plant; Cross: pollen to stigma of a flower on a different plant; significance: leads to fertilisation and seed formation, brings variation, stimulates fruit formation; (c) Ovules become seeds, ovary becomes fruit

1. (a) Maximum heating: $P=880\text{W}$, $V=220\text{V}$. Using $P=VI$, $I=\frac{P}{V}$

   ½ mark

2. $I=\frac{880}{220}=4\text{A}$

   ½ mark

3. Using Ohm's law, $R=\frac{V}{I}$

   ½ mark

4. $R=\frac{220}{4}=55\Omega$

   ½ mark

5. Minimum heating: $P=330\text{W}$, $V=220\text{V}$, so $I=\frac{330}{220}=1.5\text{A}$

   ½ mark

6. $R=\frac{V}{I}=\frac{220}{1.5}\approx 146.7\Omega$

   ½ mark

7. (b) When an electric current flows through a resistor, part of the electrical energy is dissipated as heat. The resistor gets heated. This is called the heating effect of electric current (Joule heating).

   1 mark

8. (c) Let a current $I$ flow through a resistor $R$ for time $t$, with potential difference $V$ across it. Charge passed: $Q=It$. Work done in moving this charge: $W=VQ=VIt$. This work appears as heat, so $H=VIt$. Putting $V=IR$:
   $$H=I^{2}Rt$$
   (This is Joule's law of heating.)

   1 mark

Answer: (a) Maximum: I = 4 A, R = 55 Ω; Minimum: I = 1.5 A, R ≈ 146.7 Ω; (b) electrical energy dissipated as heat in a resistor; (c) H = I²Rt (= VIt)

1. (a) When copper is heated in air, it combines with oxygen and forms black copper(II) oxide. The surface turns black:
   $$2Cu+{\mathrm{O}}_2\stackrel{\mathrm{heat}}{\to }2CuO$$

   1 mark

2. (b) Some metal oxides are called amphoteric because they react with both acids and bases to give salt and water.

   ½ mark

3. Example: aluminium oxide, $\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3$ (or zinc oxide, ZnO).

   ½ mark

4. (c)(a)(i) Sodium oxide dissolves in water and forms sodium hydroxide, an alkali:
   $$\mathrm{N}{\mathrm{a}}_2\mathrm{O}(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2NaOH(\mathrm{aq})$$

   1 mark

5. (c)(a)(ii) Here aluminium oxide behaves as an acidic oxide and forms sodium aluminate and water:
   $$\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3+2NaOH\to 2NaAl{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}$$

   1 mark

6. (c)(b) [alternative choice, 2 marks if attempted instead of (c)(a)] (i) $\mathrm{S}+{\mathrm{O}}_2\stackrel{\mathrm{burn}}{\to }\mathrm{S}{\mathrm{O}}_2$ (½ mark)

7. (c)(b)(ii) The gas is sulphur dioxide (½ mark). (iii) It is acidic in nature, being an oxide of a non-metal (½ mark).

8. (c)(b)(iv) On dry litmus paper there is no change, because the gas forms an acid (sulphurous acid) only when water is present (½ mark).

Answer: (a) 2Cu + O₂ → 2CuO (black copper oxide); (b) they react with both acids and bases to form salt and water, e.g. Al₂O₃/ZnO; (c)(a) (i) 2NaOH(aq), (ii) 2NaAlO₂ + H₂O; (c)(b) (i) S + O₂ → SO₂, (ii) sulphur dioxide, (iii) acidic, (iv) no change on dry litmus

1. (a) Two other pairs of contrasting characters Mendel used: tall vs dwarf plants (height of stem)

   ½ mark

2. and violet (purple) vs white flowers (flower colour). Inflated vs constricted pods or green vs yellow pods are also accepted.

   ½ mark

3. (b) Dominant trait: a trait that is expressed even when only one copy of its gene (factor) is present, e.g. Tt shows tallness.

   ½ mark

4. Recessive trait: a trait that is suppressed when a dominant gene is present. It is expressed only when both copies are recessive, e.g. tt is dwarf.

   ½ mark

5. (c)(a) In the $F_2$ generation the seeds appear as round-yellow : round-green : wrinkled-yellow : wrinkled-green in the ratio 9 : 3 : 3 : 1.

   1 mark

6. Interpretation: new combinations (round-green and wrinkled-yellow) appear. This shows that the genes for seed shape and seed colour are inherited independently of each other (law of independent assortment).

   1 mark

7. (c)(b) [alternative choice, 2 marks if attempted instead of (c)(a)] Parents: VV (pure violet) × vv (pure white). Gametes: V and v (½ mark).

8. $F_1$: all Vv, all violet flowered, because V is dominant (½ mark).

9. Self-pollination of $F_1$: Vv × Vv. Gametes V, v from each parent (½ mark).

10. $F_2$ (Punnet square): VV, Vv, Vv, vv. 3 violet : 1 white phenotypically, and genotype ratio 1 VV : 2 Vv : 1 vv (½ mark).

Answer: (a) Tall/dwarf plants, violet/white flowers (or any other valid pairs); (b) dominant trait shows even with one copy, recessive trait shows only when both copies are recessive; (c)(a) 9 : 3 : 3 : 1, meaning the traits are inherited independently; (c)(b) F₁ all Vv violet; F₂ VV : Vv : vv = 1 : 2 : 1, i.e. 3 violet : 1 white

1. (a) Applications of a concave mirror: (1) in torches, searchlights and vehicle headlights, to get a strong parallel beam of light;

   ½ mark

2. (2) as shaving mirrors / dentist's mirrors, to see an enlarged image (also used in solar furnaces).

   ½ mark

3. (b) Focal length $f=15$ cm. For a spherical mirror, $R=2f$

   ½ mark

4. $R=2\times 15=30$ cm

   ½ mark

5. (c)(a) Ray diagram (described): draw a concave mirror with pole P, focus F and centre of curvature C on the principal axis. Place a small object AB between P and F. Ray 1: from A parallel to the principal axis; after reflection it passes through F. Ray 2: from A along the line joining C to A, striking the mirror normally; it reflects back along the same path. Mark arrows on all rays.

   1 mark

6. The two reflected rays diverge. Extend them backwards with dotted lines; they meet behind the mirror at A′. Draw A′B′ perpendicular to the axis. The image is behind the mirror, virtual, erect and enlarged.

   1 mark

7. (c)(b) [alternative choice, 2 marks if attempted instead of (c)(a)] (i) $h=+10$ cm, $u=-100$ cm. The image forms at the object's position, so $v=-100$ cm. Mirror formula: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ (½ mark)

8. $\frac{1}{f}=-\frac{1}{100}-\frac{1}{100}=-\frac{2}{100}$, so $f=-50$ cm. (Equivalently, the object is at C, so $f=R/2=-100/2=-50$ cm.) (½ mark)

9. (ii) $m=-\frac{v}{u}$ (½ mark) $=-\frac{(-100)}{(-100)}=-1$. The image is real, inverted and the same size (10 cm). (½ mark)

Answer: (a) e.g. torches/headlights/searchlights and shaving/dentist's mirrors; (b) R = 30 cm; (c)(a) image virtual, erect, enlarged, behind the mirror (ray diagram); (c)(b) (i) f = −50 cm, (ii) m = −1