CBSE Class 10 Science: 2024 board paper, set 31/5/1

1. Balanced equation: $2\mathrm{Zn}(\mathrm{N}{\mathrm{O}}_3{)}_2\stackrel{\Delta }{\to }2\mathrm{ZnO}+4\mathrm{N}{\mathrm{O}}_2+{\mathrm{O}}_2$.

2. Check: Zn 2 = 2; N 4 = 4; O: $12=2+8+2$. So $x,y,z=2,2,4$, option (C).

   1 mark

Answer: 2, 2, 4, option (C)

1. (A), (B) and (C) each form a single product from two reactants, so they are combination reactions.

2. In (D), Fe₂O₃ is reduced to Fe and CO is oxidised to CO₂. It gives two products, so it is redox but not combination. Option (D).

   1 mark

Answer: Option (D)

1. Tooth enamel is made of calcium phosphate, the hardest substance in the body. Option (A).

   1 mark

Answer: Calcium phosphate, option (A)

1. NaCl is the salt of a strong acid (HCl) and a strong base (NaOH), so its solution is neutral with pH = 7. Option (C).

   1 mark

Answer: 7, option (C)

1. The thermit process uses aluminium. Al burns in oxygen to give Al₂O₃, which reacts with both acids and bases, so it is amphoteric. Option (B).

   1 mark

Answer: Al, Al₂O₃, option (B)

1. The transport of soluble products of photosynthesis through phloem is called translocation. Option (D).

   1 mark

Answer: Translocation, option (D)

1. In a reflex arc, the receptor sends the impulse along the sensory neuron to the relay neuron in the spinal cord. The motor neuron then carries it to the effector muscle. Option (C).

   1 mark

Answer: Option (C)

1. Olfactory receptors detect smell and are found in the nose. Option (A).

   1 mark

Answer: Nose, option (A)

1. The villi of the placenta give it a very LARGE surface area for exchange. So statement (C), which says the area is small, is incorrect. Option (C).

   1 mark

Answer: Option (C)

1. Rapid spread needs a large number of spores (i) and moisture and nutrients in the bread (ii). A low temperature slows growth, so (iii) is ruled out. Option (A).

   1 mark

Answer: (i) and (ii), option (A)

1. Every part of the lens forms the complete image. Covering half the lens lets fewer rays through, so the full image is still formed but its brightness is reduced. Option (C).

   1 mark

Answer: Option (C)

1. In a raindrop, sunlight is refracted and dispersed on entering, internally reflected at the back, and refracted again on leaving. Option (B).

   1 mark

Answer: Option (B)

1. Red light bends the least in glass and travels fastest in it, so the refractive index of glass is minimum for red. Option (A).

   1 mark

Answer: Red, option (A)

1. The field lines of a current-carrying solenoid resemble those of a bar magnet, with N and S poles at its ends. Option (C).

   1 mark

Answer: A solenoid, option (C)

1. An electron moving down is equivalent to a conventional current pointing up, which is the same as the positron moving up.

2. For an upward current in a field to the right, Fleming's left-hand rule gives a force into the paper for both particles. Option (A).

   1 mark

Answer: Option (A)

1. A cropland is created and maintained by humans, so it is an artificial ecosystem. Option (D).

   1 mark

Answer: Cropland ecosystem, option (D)

1. Vegetable oils contain long unsaturated carbon chains, which makes them healthier than saturated animal fats. Both statements are true, and R explains A. Option (A).

   1 mark

Answer: Option (A)

1. The mother always passes on an X chromosome, so sex depends on whether the father gives X or Y. The Assertion is false, but women do have XX, so the Reason is true. Option (D).

   1 mark

Answer: Option (D)

1. Electrons are negatively charged, so they move from lower to higher potential; the Assertion is true. A cell does maintain a potential difference; the Reason is true.

2. However, the direction in which electrons move is explained by their negative charge, not by the cell. So R does not explain A. Option (B).

   1 mark

Answer: Option (B)

1. Ozone does absorb harmful UV radiation, and CFCs do deplete the ozone layer, so both statements are true.

2. Ozone depletion by CFCs does not explain why ozone protects the Earth. Option (B).

   1 mark

Answer: Option (B)

1. On heating, the surface of the copper powder combines with oxygen of the air. The product formed is copper(II) oxide (CuO).

   ½ mark

2. The brown copper powder becomes coated with a black substance; copper oxide is black.

   ½ mark

3. Balanced equation: $2\mathrm{Cu}(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})\stackrel{\text{Heat}}{\to }2\mathrm{CuO}(\mathrm{s})$ (black)

   1 mark

Answer: Copper(II) oxide, black in colour; $2Cu+O_2\stackrel{Heat}{\to }2CuO$

1. This is a double displacement (precipitation) reaction. A white precipitate of barium sulphate forms:
   $\mathrm{BaC}{\mathrm{l}}_2(\mathrm{aq})+\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})\to \mathrm{BaS}{\mathrm{O}}_4(\mathrm{s})+2NaCl(\mathrm{aq})$

   1 mark

2. The precipitate $\mathrm{BaS}{\mathrm{O}}_4$ contains barium ions, $\mathrm{B}{\mathrm{a}}^{2+}$,

   ½ mark

3. and sulphate ions, $\mathrm{S}{\mathrm{O}}_4^{2-}$.

   ½ mark

Answer: $BaC{l}_2(aq)+N{a}_{2}S{O}_4(aq)\to BaS{O}_4(s)+2NaCl(aq)$; ions $B{a}^{2+}$ and $S{O}_4^{2-}$

1. Low melting and boiling points: the forces of attraction between the molecules of carbon compounds are weak, so little energy is needed to separate them.

   1 mark

2. Non-conductors of electricity: the atoms in carbon compounds are held by covalent bonds (sharing of electrons), so no ions or charged particles are formed to carry current.

   1 mark

Answer: (1) Intermolecular forces in carbon compounds are weak. (2) Their bonding is covalent, so no ions are formed.

1. During vigorous running, oxygen supply to the muscle cells falls short. Pyruvate is then broken down anaerobically into lactic acid, and the build-up of lactic acid in the muscles causes cramps.

   1 mark

2. Difference: aerobic respiration occurs in the presence of oxygen and gives $\mathrm{C}{\mathrm{O}}_2$ + water + more energy. Here respiration occurs in the absence (lack) of oxygen and gives lactic acid + less energy.
   Glucose $\to$ Pyruvate $\stackrel{\text{lack of }{O}_2}{\to }$ Lactic acid + Energy

   1 mark

Answer: Cramps are due to lactic acid build-up from anaerobic respiration. This respiration occurs without enough oxygen and gives lactic acid and less energy; aerobic respiration uses oxygen and gives $C{O}_2$, water and more energy.

1. Lymph is also called tissue fluid (extracellular fluid).

   1 mark

2. Function 1: lymph carries digested and absorbed fats from the small intestine (through the lacteals).

   ½ mark

3. Function 2: it drains excess fluid from the intercellular spaces back into the blood. (It also contains lymphocytes that help fight infection.)

   ½ mark

Answer: Tissue fluid; it carries absorbed fats from the intestine and returns excess tissue fluid to the blood (also fights infection).

1. Plasmodium – multiple fission.

   ½ mark

2. The nucleus divides many times and the single cell then splits into many daughter cells at the same time.

   ½ mark

3. Leishmania – binary fission.

   ½ mark

4. The single cell splits into two daughter cells in a definite plane (orientation) with respect to its whip-like flagellum at one end.

   ½ mark

Answer: Plasmodium: multiple fission (one cell gives many daughter cells at once); Leishmania: binary fission in a definite plane (one cell gives two).

1. (a) A convex lens is a converging lens. It brings the parallel rays of sunlight together at one point, so the light energy is concentrated there. The heat produced raises the temperature of the paper at that point and it catches fire.

   1 mark

2. (b) The point is the principal focus of the convex lens.

   ½ mark

3. The bright spot is the real image of the Sun formed by the lens.

   ½ mark

Answer: (a) The lens converges sunlight to a point, concentrating heat that burns the paper. (b) Principal focus; the bright spot is the real image of the Sun.

1. Given $Q=500\mathrm{C}$ and $I=25\mathrm{mA}$. Use $Q=I\times t$, so $t=\frac{Q}{I}$.

   ½ mark

2. Convert the current to amperes: $I=25\mathrm{mA}=\frac{25}{1000}\mathrm{A}=0.025\mathrm{A}$

   ½ mark

3. $t=\frac{500\mathrm{C}}{0.025\mathrm{A}}=20000\mathrm{s}$

   The source is completely discharged in $20000$ s (about 5.56 h).

   1 mark

Answer: t = 20000 s

1. Iron displaces copper from copper sulphate solution. The blue solution turns light green and a brown coating of copper forms on the nail:
   $\mathrm{Fe}(\mathrm{s})+\mathrm{CuS}{\mathrm{O}}_4(\mathrm{aq})\to \mathrm{FeS}{\mathrm{O}}_4(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$

   1 mark

2. Type of reaction: displacement reaction.

   ½ mark

3. Definition: a reaction in which a more reactive metal displaces a less reactive metal from a solution of its salt.

   ½ mark

4. Two other metals that can replace iron (they are more reactive than copper): zinc

   ½ mark

5. and aluminium (magnesium also works).

   ½ mark

Answer: $Fe(s)+CuS{O}_4(aq)\to FeS{O}_4(aq)+Cu(s)$; displacement reaction (a more reactive metal displaces a less reactive metal from its salt solution); zinc and aluminium.

1. The ore of mercury is **cinnabar ($\mathrm{HgS}$)**.

   ½ mark

2. It occurs in nature as a sulphide ore (mercury(II) sulphide).

   ½ mark

3. Roasting: cinnabar is heated strongly in air, which converts it into mercury oxide:
   $2HgS(\mathrm{s})+3{\mathrm{O}}_2(\mathrm{g})\stackrel{\text{Heat}}{\to }2HgO(\mathrm{s})+2S{\mathrm{O}}_2(\mathrm{g})$

   1 mark

4. Mercury is low in the reactivity series, so on further heating the oxide is reduced to the metal:
   $2HgO(\mathrm{s})\stackrel{\text{Heat}}{\to }2Hg(\mathrm{l})+{\mathrm{O}}_2(\mathrm{g})$

   1 mark

Answer: Cinnabar (HgS), a sulphide ore; $2HgS+3O_2\stackrel{Heat}{\to }2HgO+2S{O}_2$; $2HgO\stackrel{Heat}{\to }2Hg+O_2$

1. (i) Growth and development: Growth hormone,

   ½ mark

2. secreted by the pituitary gland.

   ½ mark

3. It regulates growth and development of the body, stimulating growth in all organs. Its deficiency in childhood causes dwarfism.

   ½ mark

4. (ii) Regulation of metabolism: Thyroxin,

   ½ mark

5. secreted by the thyroid gland.

   ½ mark

6. It regulates carbohydrate, protein and fat metabolism in the body so as to provide the best balance for growth. Iodine is needed to make it.

   ½ mark

Answer: (i) Growth hormone, pituitary gland: stimulates growth of all organs; (ii) Thyroxin, thyroid gland: regulates carbohydrate, protein and fat metabolism.

1. (a) All the $F_1$ plants were tall;

   ½ mark

2. their gene combination is Tt.

   ½ mark

3. (b) The gene for shortness (t) is recessive. In the presence of the dominant gene T it cannot express itself, so Tt plants look tall.

   1 mark

4. (c) $F_2$ ratio: Tall : Short = 3 : 1 (genotypes TT : Tt : tt = 1 : 2 : 1).

   ½ mark

5. Conclusion: the tall trait is dominant and the short trait is recessive. The recessive trait, hidden in $F_1$, is not lost and reappears in $F_2$.

   ½ mark

Answer: (a) All tall, Tt; (b) shortness is recessive and is masked by dominant T; (c) Tall : Short = 3 : 1, so tallness is dominant and shortness recessive.

1. (i) The defect is hypermetropia (long-sightedness): a nearby object, at the near point N, is not focused on the retina.

   ½ mark

2. The part responsible is the eye lens, whose converging power the ciliary muscles cannot increase enough.

   ½ mark

3. (ii) Cause 1: the focal length of the eye lens is too long.

   ½ mark

4. Cause 2: the eyeball has become too small (too short).

   ½ mark

5. (iii) It is corrected with a convex (converging) lens of suitable power.

   ½ mark

6. Role: the convex lens provides the additional focusing (converging) power needed to form the image of a nearby object on the retina.

   ½ mark

Answer: (i) Hypermetropia, eye lens/ciliary muscles; (ii) focal length of eye lens too long, eyeball too small; (iii) convex lens, which gives the extra converging power to focus the image on the retina.

1. Dispersion: the splitting of white light into its seven constituent colours (VIBGYOR) on passing through a transparent medium such as a glass prism.

   1 mark

2. Cause: the different colours of white light bend through different angles with respect to the incident ray. Red bends the least and violet bends the most, so the colours separate.

   1 mark

3. Diagram (described): draw a triangular glass prism. A narrow beam of white light falls obliquely on one refracting face. Inside the prism the beam spreads out, and it emerges from the second face as a band of colours on a screen. Label the colours from top to bottom as Red, Orange, Yellow, Green, Blue, Indigo, Violet (violet deviated most, at the bottom; red at the top). Show arrows on all rays and label 'white light' and 'spectrum'.

   1 mark

Answer: Dispersion is the splitting of white light into its seven colours. It occurs because different colours bend through different angles (red least, violet most). Diagram: prism forming a VIBGYOR spectrum with red at the top and violet at the bottom.

1. (a) The soft iron bundle gets strongly magnetised by the magnetic field of the current-carrying solenoid.

   ½ mark

2. The device obtained is an electromagnet.

   ½ mark

3. It is called so because it behaves as a magnet only as long as current flows through the solenoid; it loses its magnetism when the current is switched off.

   ½ mark

4. (b) Diagram (described): draw a solenoid connected to a cell and key. Inside the coil draw several straight, parallel, equally spaced field lines along the axis with arrows from the S end to the N end. Outside, the lines curve from the N end back round to the S end, like a bar magnet's field. Mark N and S and the direction of current.

   1 mark

5. The straight, parallel, equally spaced lines indicate that the magnetic field inside the solenoid is uniform (same strength and direction at all points).

   ½ mark

Answer: (a) It gets magnetised to form an electromagnet, so called because it is a magnet only while current flows. (b) Parallel straight lines inside the solenoid show a uniform magnetic field.

1. Food chain: a single linear series of organisms in which each organism feeds on the one before it, at successive trophic levels, e.g. grass → deer → tiger.

   1 mark

2. Food web: a network of many interconnected food chains (branching lines) that gives several alternative feeding connections among the organisms of an ecosystem.

   1 mark

3. In grass → deer → tiger, if deer decrease, fewer deer feed on grass, so the population of grass (first trophic level) will increase.

   ½ mark

4. Tigers get less food, so the population of tigers (third trophic level) will decrease.

   ½ mark

Answer: Food chain is a single linear feeding sequence; food web is a network of interconnected food chains. Grass (first level) increases and tiger (third level) decreases.

1. (i) The droplets come from the water of crystallisation in ferrous sulphate crystals ($\mathrm{FeS}{\mathrm{O}}_4\cdot 7{\mathrm{H}}_2\mathrm{O}$). On heating, this water is driven off as vapour and condenses on the cooler upper walls of the tube.

   1 mark

2. (ii) The light green crystals turn white (anhydrous $\mathrm{FeS}{\mathrm{O}}_4$) as they lose water. On strong heating they finally form brown $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3$.

   ½ mark

3. (iii) Seven molecules of water are attached per molecule: $\mathrm{FeS}{\mathrm{O}}_4\cdot 7{\mathrm{H}}_2\mathrm{O}$.

   ½ mark

4. (I) Crystalline copper sulphate: $\mathrm{CuS}{\mathrm{O}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}$

   ½ mark

5. (II) Crystalline sodium carbonate (washing soda): $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3\cdot 10{\mathrm{H}}_2\mathrm{O}$

   ½ mark

6. (iv) Gypsum is heated at 373 K, where it loses part of its water of crystallisation and forms Plaster of Paris:
   $\mathrm{CaS}{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}\stackrel{373\mathrm{K}}{\to }\mathrm{CaS}{\mathrm{O}}_4\cdot \frac{1}{2}{\mathrm{H}}_2\mathrm{O}+1\frac{1}{2}{\mathrm{H}}_2\mathrm{O}$

   1 mark

7. Uses of Plaster of Paris: (1) supporting fractured bones in the right position; (2) making toys and materials for decoration, and smoothing surfaces.

   1 mark

Answer: (i) From the water of crystallisation lost on heating; (ii) green to white; (iii) seven, $CuS{O}_4\cdot 5H_{2}O$, $N{a}_{2}C{O}_3\cdot 10H_{2}O$; (iv) heating gypsum at 373 K gives $CaS{O}_4\cdot \frac{1}{2}{H}_{2}O$; used for plastering fractured bones and making toys/decorative items.

1. (i) X = tartaric acid (the acid in tamarind).

   ½ mark

2. Y = baking soda.

   ½ mark

3. Z = baking powder (baking soda + tartaric acid).

   ½ mark

4. Chemical formula of Y: $\mathrm{NaHC}{\mathrm{O}}_3$ (sodium hydrogencarbonate).

   ½ mark

5. (ii) Preparation: common salt (brine) reacts with water, carbon dioxide and ammonia:
   $\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2+\mathrm{N}{\mathrm{H}}_3\to \mathrm{N}{\mathrm{H}}_4\mathrm{Cl}+\mathrm{NaHC}{\mathrm{O}}_3$

   1 mark

6. In the dough, on heating/wetting, baking soda reacts with the acid of baking powder:
   $\mathrm{NaHC}{\mathrm{O}}_3+{\mathrm{H}}^{+}\to \mathrm{C}{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}+\text{Sodium salt of acid}$

   ½ mark

7. The $\mathrm{C}{\mathrm{O}}_2$ gas released forms bubbles in the dough, which make the cake rise and become soft and spongy. The tartaric acid neutralises the bitter sodium carbonate that would otherwise form.

   ½ mark

8. (iii) Mild base used as an antacid: magnesium hydroxide (milk of magnesia), $\mathrm{Mg}(\mathrm{OH}{)}_2$.

   1 mark

Answer: (i) X tartaric acid, Y baking soda ($NaHC{O}_3$), Z baking powder; (ii) $NaCl+H_{2}O+C{O}_2+N{H}_3\to N{H}_{4}Cl+NaHC{O}_3$; $NaHC{O}_3+H^{+}\to C{O}_2+H_{2}O$ + sodium salt, and the released $C{O}_2$ makes the cake spongy; (iii) Magnesium hydroxide, $Mg(OH{)}_2$.

1. Take two healthy potted plants, A and B, of nearly the same size.

   ½ mark

2. Keep both in darkness for about three days so that their leaves are destarched.

   ½ mark

3. Place a watch glass containing potassium hydroxide (KOH) beside plant A only; plant B gets no KOH. KOH absorbs carbon dioxide.

   ½ mark

4. Cover each plant with a separate bell jar on a glass plate, and seal the bottom of the jars with Vaseline so that the set-up is airtight.

   ½ mark

5. Keep both plants in sunlight for about two hours.

   ½ mark

6. Pluck a leaf from each plant, boil it in water and then in alcohol to remove chlorophyll, wash it and test it with iodine solution for starch.

   ½ mark

7. Observation: the leaf of plant A (with KOH) does not turn blue-black. The leaf of plant B turns blue-black, showing that starch is present.

   1 mark

8. Conclusion: KOH absorbed the $\mathrm{C}{\mathrm{O}}_2$ in jar A, so no photosynthesis (no starch) took place in plant A. Hence carbon dioxide is essential for photosynthesis.

   1 mark

Answer: Destarched plants A (with KOH) and B (without) are kept under sealed bell jars in sunlight and their leaves tested with iodine. Leaf A shows no blue-black colour and leaf B turns blue-black, so $C{O}_2$ is essential for photosynthesis.

1. (i) Lime water in set-up (I) takes more time to turn milky than in set-up (II).

   1 mark

2. Reason: exhaled air contains a much higher percentage of $\mathrm{C}{\mathrm{O}}_2$ (about 4%) than atmospheric air (about 0.03–0.04%). $\mathrm{C}{\mathrm{O}}_2$ turns lime water milky: $\mathrm{Ca}(\mathrm{OH}{)}_2+\mathrm{C}{\mathrm{O}}_2\to \mathrm{CaC}{\mathrm{O}}_3+{\mathrm{H}}_2\mathrm{O}$. So the exhaled air of (II) does it faster.

   1 mark

3. (ii) Diagram (described): draw an open stoma as two bean-shaped (kidney-shaped) guard cells facing each other with their concave sides enclosing an elliptical gap, the stomatal pore. Show a nucleus and small oval chloroplasts in each guard cell, with epidermal cells around them. Title it 'Open stomatal pore'.

   1 mark

4. Label (I) Guard cells: the two bean-shaped cells around the pore.

   ½ mark

5. Label (II) Chloroplast: the small green bodies inside the guard cells.

   ½ mark

6. Function 1: exchange of gases (${\mathrm{O}}_2$ and $\mathrm{C}{\mathrm{O}}_2$) for photosynthesis and respiration.

   ½ mark

7. Function 2: loss of water vapour by transpiration.

   ½ mark

Answer: (i) Exhaled air has more $C{O}_2$ than atmospheric air, so lime water in (II) turns milky faster and (I) takes longer. (ii) Diagram of an open stoma with guard cells and chloroplasts labelled; stomata carry out gaseous exchange and transpiration.

1. (i) By Ohm's law, $I=\frac{V}{R}$. With $R$ constant and $V^{\prime }=\frac{V}{3}$, we get $I^{\prime }=\frac{V/3}{R}=\frac{I}{3}$. So the current becomes one-third of its initial value.

   ½ mark

2. The law used is Ohm's law.

   ½ mark

3. Statement: The potential difference across the ends of a conductor is directly proportional to the current flowing through it, provided its temperature (and other physical conditions) remain the same. That is, $V\propto I$, or $V=IR$.

   1 mark

4. (ii) Circuit diagram (in words): Draw four cell symbols joined positive to negative to form a battery of $4\times 1.5\text{V}$. Connect it in one single loop with a plug key and the resistors $5\Omega$, $10\Omega$ and $15\Omega$, one after the other in series. Mark the direction of current from the + terminal round the loop.

   1 mark

5. (I) Total voltage: $V=4\times 1.5\text{V}=6\text{V}$.

   Series resistance: $R_s=R_1+R_2+R_3=5+10+15=30\Omega$.

   Current: $I=\frac{V}{R_s}=\frac{6\text{V}}{30\Omega }=0.2\text{A}$.

   1 mark

6. (II) The same current flows through every resistor in series. So the p.d. across the $10\Omega$ resistor is $V_{10}=IR=0.2\text{A}\times 10\Omega =2\text{V}$.

   1 mark

Answer: (i) Current becomes one-third; Ohm's law (V ∝ I at constant temperature). (ii) (I) 0.2 A; (II) 2 V

1. (i) The potential difference between two points is 1 volt when 1 joule of work is done to move a charge of 1 coulomb from one point to the other: $1\text{V}=\frac{1\text{J}}{1\text{C}}$.

   1 mark

2. (ii) Given: $d=0.2\text{mm}=2\times 10^{-4}\text{m}$, $\rho =1.6\times 10^{-8}\Omega \text{m}$, $R=14\Omega$.

   Area of cross-section: $A=\frac{\pi d^2}{4}$.

   ½ mark

3. From $R=\rho \frac{l}{A}$ we get $l=\frac{RA}{\rho }=\frac{R\pi d^2}{4\rho }$.

   ½ mark

4. $l=\frac{14\times \frac{22}{7}\times (2\times 10^{-4}{)}^2}{4\times 1.6\times 10^{-8}}=\frac{14\times \frac{22}{7}\times 4\times 10^{-8}}{6.4\times 10^{-8}}=\frac{44}{1.6}=27.5\text{m}$

   1 mark

5. Diameter doubled: $d^{\prime }=2d$, so $A^{\prime }=\frac{\pi (2d{)}^2}{4}=4A$. Since $R\propto \frac{1}{A}$ (same $\rho$ and $l$), $\frac{R^{\prime }}{R}=\frac{A}{A^{\prime }}=\frac{1}{4}$.

   ½ mark

6. $R^{\prime }=\frac{R}{4}=\frac{14}{4}=3.5\Omega$

   1 mark

7. Change in resistance $=14.0-3.5=10.5\Omega$. The resistance decreases by $10.5\Omega$, becoming one-fourth of its original value.

   ½ mark

Answer: (i) When 1 J of work is done to move 1 C of charge between the points. (ii) l = 27.5 m; on doubling the diameter, R becomes 3.5 Ω, a decrease of 10.5 Ω

1. (a) Hydrocarbons are compounds made up of only carbon and hydrogen, e.g. methane ($\mathrm{C}{\mathrm{H}}_4$) and ethane (${\mathrm{C}}_2{\mathrm{H}}_6$).

   1 mark

2. (b) Carbon forms a large number of compounds because of:

   • Catenation: it can link with other carbon atoms to form long chains, branched chains and rings.
   • Tetravalency: it has a valency of four, so it can bond with four other atoms of carbon or of other elements.
   1 mark

3. (c)(i) Functional groups:

   (1) Aldehyde: $-\mathrm{CHO}$, i.e. $-\stackrel{O}{\stackrel{\Vert }{C}}-H$

   (2) Ketone: $>\mathrm{C}=\mathrm{O}$, i.e. $-\stackrel{O}{\stackrel{\Vert }{C}}-$ (the carbonyl carbon is bonded to two carbon atoms)

   1 mark

4. Esterification (an acid acts as the catalyst):

   $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\stackrel{\text{Acid}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$

   The products are ethyl ethanoate (an ester) and water.

   1 mark

5. (c)(ii) Structural isomers are compounds that have the same molecular formula but different structures.

   1 mark

6. Two isomers of butane, ${\mathrm{C}}_4{\mathrm{H}}_{10}$:

   1. n-Butane (straight chain): $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_3$

   2. Isobutane, or 2-methylpropane (branched chain): $\mathrm{C}{\mathrm{H}}_3-\underset{\mathrm{C}{\mathrm{H}}_3}{\underset{|}{\mathrm{CH}}}-\mathrm{C}{\mathrm{H}}_3$

   1 mark

Answer: (a) Compounds of only C and H; (b) Catenation and tetravalency; (c)(i) Aldehyde –CHO, ketone >C=O; CH3COOH + C2H5OH → (acid) CH3COOC2H5 + H2O; (c)(ii) Same molecular formula, different structures; n-butane CH3CH2CH2CH3 and isobutane CH3CH(CH3)CH3

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

   Cross-pollination: pollen grains are transferred from the anther of one flower to the stigma of a flower on another plant of the same species.

   1 mark

2. (b) The petals, which are brightly coloured and often scented, attract insects. After fertilisation, the petals dry up (wither) and fall off.

   1 mark

3. (c)(i) Fertilisation is the fusion of the male gamete with the female gamete (egg) to form a zygote.

   1 mark

4. After fertilisation, the ovules develop into seeds and the ovary develops (ripens) into the fruit.

   1 mark

5. (c)(ii) In a germinating seed, the plumule is the future shoot and the radicle is the future root.

   1 mark

6. Cotyledons store food, which nourishes the growing embryo during germination.

   1 mark

Answer: (a) Self: pollen to stigma of the same flower; cross: pollen to stigma of a flower on another plant. (b) Petals; they wither and fall off. (c)(i) Fusion of male and female gametes to form a zygote; ovules → seeds, ovary → fruit. (c)(ii) Plumule = future shoot, radicle = future root; cotyledon stores food

1. (a) The principal axis of a concave mirror is the imaginary straight line passing through its pole (P) and its centre of curvature (C).

   1 mark

2. (b) A ray parallel to the principal axis passes through the principal focus after reflection, so $f=10\text{cm}$.

   Radius of curvature: $R=2f=2\times 10=20\text{cm}$.

   1 mark

3. (c)(i) For a convex mirror, using the sign convention: $u=-10\text{cm}$, $f=+15\text{cm}$.

   Mirror formula: $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

   ½ mark

4. $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}-\frac{1}{-10}=\frac{1}{15}+\frac{1}{10}$

   ½ mark

5. $\frac{1}{v}=\frac{2+3}{30}=\frac{1}{6}$, so $v=+6\text{cm}$.

   The image is formed 6 cm behind the mirror. It is virtual and erect.

   1 mark

6. (c)(ii) A mirror that always forms a virtual, erect and diminished image is a convex mirror (diverging mirror).

   ½ mark

7. Ray diagram (in words): Draw a convex mirror with pole P on the principal axis, and F and C behind the mirror. Place object AB upright in front of the mirror.

   • Ray 1 from A runs parallel to the principal axis. After reflection it diverges, appearing to come from F; extend it backwards as a dotted line.
   • Ray 2 from A is directed towards C. It reflects straight back along the same path; extend it as a dotted line towards C.
   • The dotted lines meet at A' behind the mirror, between P and F. Drop A'B' perpendicular to the axis.

   A'B' is virtual, erect and diminished. Arrows must show the direction of the rays.

   1½ marks

Answer: (a) Line through the pole and centre of curvature; (b) R = 20 cm; (c)(i) v = +6 cm (6 cm behind the mirror, virtual and erect); (c)(ii) Convex mirror, with the ray diagram showing the image between P and F behind the mirror