CBSE Class 10 Science: 2024 board paper, set 31/3/1

1. When calcium carbonate (limestone) is heated strongly, it breaks down into calcium oxide and carbon dioxide: $\mathrm{CaC}{\mathrm{O}}_3\stackrel{\mathrm{\Delta }}{\to }\mathrm{CaO}+\mathrm{C}{\mathrm{O}}_2$. This is thermal decomposition.

2. So 'X' is calcium oxide, $\mathrm{CaO}$, whose common name is quick lime. Option (A).

   1 mark

Answer: Quick lime (CaO), option (A)

1. Natural indicators come from plants or lichens. Litmus is extracted from lichen, and turmeric comes from the turmeric plant.

2. Methyl orange and phenolphthalein are synthetic indicators. Only option (B), turmeric and litmus, has both indicators natural.

   1 mark

Answer: Turmeric and litmus, option (B)

1. Washing soda, $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3\cdot 10{\mathrm{H}}_2\mathrm{O}$, is used in the glass, soap and paper industries. It is also used to make borax and to remove permanent hardness of water.

2. Baking soda is used in cooking and antacids, bleaching powder in bleaching and disinfecting, and common salt as a raw material and in food. The answer is (A).

   1 mark

Answer: Washing soda, option (A)

1. Cyclohexane, ${\mathrm{C}}_6{\mathrm{H}}_{12}$, is a saturated cyclic hydrocarbon. It is a ring of 6 carbon atoms joined only by single bonds.

2. Each carbon makes two C–C bonds, so it carries two H atoms. Diagram (B) shows a saturated ring with $\mathrm{C}{\mathrm{H}}_2$ at every corner. Diagrams with double or triple bonds are unsaturated (benzene-type and others).

   1 mark

Answer: Diagram (B), option (B)

1. Balance Al: 2 Al on the left, so $c=2$ (2 $\mathrm{AlC}{\mathrm{l}}_3$). Balance Cl: $2\times 3=6$ Cl, so $b=6$ HCl.

2. Balance O and H: 3 O come from $\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3$, so $d=3$ ${\mathrm{H}}_2\mathrm{O}$. This gives 6 H, which matches the 6 HCl.

3. Balanced equation: $\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3+6HCl\to 2AlC{\mathrm{l}}_3+3{\mathrm{H}}_2\mathrm{O}$. So a, b, c, d = 1, 6, 2, 3, option (A).

   1 mark

Answer: 1, 6, 2 and 3, option (A)

1. ${\mathrm{C}}_4{\mathrm{H}}_{10}$, ${\mathrm{C}}_5{\mathrm{H}}_{12}$ and ${\mathrm{C}}_2{\mathrm{H}}_6$ all fit ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2n+2}$, so they are alkanes (saturated).

2. ${\mathrm{C}}_7{\mathrm{H}}_{14}$ fits ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2n}$, so it is an alkene (or a cycloalkane). It is the odd one out, option (B).

   1 mark

Answer: ${\mathrm{C}}_7{\mathrm{H}}_{14}$, option (B)

1. In (A), (C) and (D), two compounds exchange ions to form two new compounds. These are double displacement reactions.

2. In (B), $\mathrm{CaO}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{Ca}(\mathrm{OH}{)}_2$, two substances form a single product. This is a combination reaction, so (B) is different.

   1 mark

Answer: $\mathrm{CaO}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{Ca}(\mathrm{OH}{)}_2$, option (B)

1. Auxins promote cell elongation, gibberellins promote stem growth, and abscisic acid inhibits growth (it causes wilting of leaves).

2. Cytokinins promote cell division and are found in areas of rapid cell division such as fruits and seeds. Option (D).

   1 mark

Answer: Cytokinins, option (D)

1. Petals are usually brightly coloured and often scented. This attracts insects for pollination.

2. Sepals are usually green and protect the bud. Anther and stigma are reproductive parts. So the answer is petals only, (C).

   1 mark

Answer: Petals only, option (C)

1. In a dihybrid cross (round yellow × wrinkled green), the ${\mathrm{F}}_1$ plants are all round yellow (RrYy).

2. Selfing ${\mathrm{F}}_1$ gives the ${\mathrm{F}}_2$ phenotypes round yellow : round green : wrinkled yellow : wrinkled green = 9 : 3 : 3 : 1. This shows independent inheritance. Option (C).

   1 mark

Answer: 9 : 3 : 3 : 1, option (C)

1. (a) is false. The left atrium (not the right) receives oxygenated blood from the lungs, through the pulmonary vein. (d) is false. The aorta carries blood from the left ventricle, not from the left atrium.

2. (b) is true: left atrium → left ventricle → body. (c) is true: the right atrium receives deoxygenated blood through the vena cava. The answer is (b) and (c), option (C).

   1 mark

Answer: (b) and (c), option (C)

1. Dwarf plants are tt, so each parent must give a t allele. Both parents are tall, so each must carry T as well. Each parent is therefore Tt.

2. Tt × Tt gives TT : Tt : tt = 1 : 2 : 1, so about ¼ of the offspring are dwarf ('a few'). Option (A).

   1 mark

Answer: Tt and Tt, option (A)

1. At the raindrop, sunlight is refracted and dispersed into its colours. It is then reflected internally at the back of the drop, and refracted again as it comes out.

2. NCERT describes this as dispersion, refraction and internal reflection. Option (D).

   1 mark

Answer: Dispersion, refraction and internal reflection, option (D)

1. $R=\rho \frac{l}{A}=\rho \frac{4l}{\pi d^2}$, so $R\propto \frac{l}{d^2}$.

2. Compare $l/d^2$ for each option: (A) $\frac{L/4}{D^2/4}=\frac{L}{D^2}$; (B) $\frac{4L}{D^2/16}=\frac{64L}{D^2}$; (C) $\frac{L}{4D^2}$; (D) $\frac{2L}{16D^2}=\frac{L}{8D^2}$.

3. (D) has the smallest value, so its resistance is minimum. Option (D).

   1 mark

Answer: 4D and 2L, option (D)

1. By the 10% law, only about 10% of the energy passes to the next trophic level. The rest is lost at every step.

2. The fewer the trophic levels, the more energy reaches the top. So a chain with 2 trophic levels is most advantageous. Option (A).

   1 mark

Answer: 2 trophic levels, option (A)

1. (a) True: ozone is a deadly poison. (c) True: UV radiation splits ${\mathrm{O}}_2$ into O atoms, which combine with ${\mathrm{O}}_2$ to form ${\mathrm{O}}_3$.

2. (b) False: ozone absorbs ultraviolet radiation, not infrared. (d) False: ozone's essential function is in the upper atmosphere. The answer is (a) and (c), option (B).

   1 mark

Answer: (a) and (c), option (B)

1. Assertion is true. Zinc displaces copper: $\mathrm{Zn}+\mathrm{CuS}{\mathrm{O}}_4\to \mathrm{ZnS}{\mathrm{O}}_4+\mathrm{Cu}$, and the copper deposits as a reddish brown coating.

2. Reason is false. Zinc is more reactive than copper, which is why it displaces copper. So A is true and R is false, option (C).

   1 mark

Answer: Option (C)

1. Asexual reproduction involves only one parent, with no fusion of gametes. Both statements are true.

2. Because DNA comes from a single parent, the offspring are genetically similar to it. R explains A, option (A).

   1 mark

Answer: Option (A)

1. Assertion is true. Red light is used for stop signals.

2. Reason is false. Red light has the longest wavelength, so it is scattered the least by fog and smoke. That is why it is seen from far away. Option (C).

   1 mark

Answer: Option (C)

1. Both statements are true. Waste is of two kinds, biodegradable and non-biodegradable, and improper disposal pollutes the environment.

2. R is about the effects of improper disposal. It does not explain why waste falls into these two types. Option (B).

   1 mark

Answer: Option (B)

1. Balanced chemical equation: $2Mg(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})\to 2MgO(\mathrm{s})$

   ½ mark

2. The white ash is magnesium oxide, $\mathrm{MgO}$.

   ½ mark

3. Type of reaction: combination reaction.

   ½ mark

4. Justification: two reactants, magnesium and oxygen, combine to form a single product, magnesium oxide.

   ½ mark

Answer: $2Mg+{\mathrm{O}}_2\to 2MgO$; ash = magnesium oxide; combination reaction (two substances form a single product)

1. Auxins are synthesised at the shoot tip (and root tip).

   ½ mark

2. When light falls on one side of the shoot, auxin diffuses towards the shady side.

   ½ mark

3. The higher auxin concentration on the shady side makes those cells grow longer than the cells on the lit side.

   ½ mark

4. The unequal growth makes the shoot bend towards light. This is phototropism.

   ½ mark

Answer: Synthesised at the shoot tip; auxin moves to the shaded side, making cells there elongate more, so the shoot bends towards light.

1. Pair 1, plant height: Tall is dominant.

   ½ mark

2. Dwarf (short) is recessive.

   ½ mark

3. Pair 2, seed colour: Yellow seed is dominant.

   ½ mark

4. Green seed is recessive. (Other valid pairs: round/wrinkled seeds, violet/white flowers.)

   ½ mark

Answer: Tall (dominant) / dwarf (recessive); yellow seed (dominant) / green seed (recessive)

1. Parents: Father XY × Mother XX.

   ½ mark

2. Gametes: sperms are X or Y (two types, 50% each); eggs are all X.

   ½ mark

3. Fertilisation: egg X + sperm X → XX (girl); egg X + sperm Y → XY (boy).

   ½ mark

4. Offspring: XX : XY = 1 : 1, so the chance of a female child is 50% and of a male child is 50%. The sex of the child depends on which sperm fertilises the egg.

   ½ mark

Answer: XY × XX → gametes X, Y and X → XX (female, 50%) and XY (male, 50%)

1. When: A person has hypermetropia (far-sightedness) when he can see distant objects clearly but cannot see nearby objects distinctly. The near point of the eye moves farther away than the normal 25 cm, because light from a nearby object focuses behind the retina.

   ½ mark

2. Cause 1: The focal length of the eye lens is too long, so the eye lens is not converging enough.

   ½ mark

3. Cause 2: The eyeball has become too small, so the distance from the lens to the retina is too short.

   ½ mark

4. Correction: A convex (converging) lens of suitable power is used. It brings the image back onto the retina.

   ½ mark

Answer: Hypermetropia: nearby objects are not seen clearly while far objects are seen clearly. Causes: focal length of eye lens too long; eyeball too small. Correction: convex (converging) lens.

1. Diagram (describe/draw): Draw a straight vertical wire XY passing through a horizontal cardboard sheet. Connect it in a circuit with a battery, key K, a variable resistance and an ammeter. On the cardboard, draw concentric circles centred on the wire. These are the magnetic field lines. They lie closer together near the wire and farther apart away from it. Label: conductor, cardboard, magnetic field lines, battery, key.

   1 mark

2. Direction of current: Mark an arrow on the wire. Take the current flowing vertically upward (from Y to X), i.e. from the + terminal of the battery round the circuit.

   ½ mark

3. Direction of field: By the right-hand thumb rule, with the thumb pointing along the upward current, the curled fingers give the field direction. The field lines are therefore anticlockwise when viewed from above. Mark arrows on the circles accordingly. (If the current is downward, the circles become clockwise.)

   ½ mark

Answer: Concentric circular field lines centred on the wire. For an upward current they are anticlockwise when seen from above (right-hand thumb rule).

1. Device: A current-carrying solenoid is used. When a soft iron piece is placed inside it, it forms an electromagnet.

   ½ mark

2. Diagram: Draw a soft iron cylinder (rod). Wind a coil of many turns of insulated copper wire closely around it, forming a solenoid. Connect the ends of the coil in series with a battery and a plug key K. Show the direction of current with arrows.

   1 mark

3. Labels: soft iron rod, current-carrying solenoid (coil), battery, key. When the key is closed, the strong magnetic field inside the solenoid magnetises the soft iron.

   ½ mark

Answer: Current-carrying solenoid (electromagnet). The soft iron cylinder is placed inside a coil connected to a battery and key.

1. Decomposers are micro-organisms such as bacteria and fungi. They break down the complex organic substances of dead plants, dead animals and wastes into simple inorganic substances, which return to the soil.

   1 mark

2. Consequence 1: Dead bodies and wastes would pile up in the environment, causing foul smell and the breeding of flies.

   ½ mark

3. Consequence 2: Nutrients would not be recycled, so the soil would not be replenished and plants would lack minerals.

   ½ mark

Answer: Decomposers are microbes that break dead organic matter into simple inorganic substances. Without them, dead matter would accumulate and nutrients would not be recycled or returned to the soil.

1. (a) Zinc oxide reacts with both acids and bases to give salt and water, so it is amphoteric. For example: $\mathrm{ZnO}+2HCl\to \mathrm{ZnC}{\mathrm{l}}_2+{\mathrm{H}}_2\mathrm{O}$ and $\mathrm{ZnO}+2NaOH\to \mathrm{N}{\mathrm{a}}_2\mathrm{Zn}{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}$.

   1 mark

2. (b) Sodium is very reactive. In open air it reacts vigorously with oxygen and moisture and may catch fire. Keeping it under kerosene cuts off contact with air and water and prevents accidental fire.

   1 mark

3. (c) Nitric acid is a strong oxidising agent. It oxidises the hydrogen produced to water, while itself being reduced to oxides of nitrogen. So ${\mathrm{H}}_2$ gas is generally not evolved. (Mg and Mn with very dilute $\mathrm{HN}{\mathrm{O}}_3$ are exceptions.)

   1 mark

Answer: (a) ZnO reacts with both acids and bases to give salt and water. (b) Na reacts vigorously with air/moisture and catches fire; kerosene prevents contact. (c) HNO₃ is a strong oxidising agent and oxidises H₂ to water.

1. (i) Mercury – Process: Roasting followed by heating alone. Cinnabar is roasted: $2HgS+3{\mathrm{O}}_2\stackrel{\mathrm{heat}}{\to }2HgO+2S{\mathrm{O}}_2$, and then $2HgO\stackrel{\mathrm{heat}}{\to }2Hg+{\mathrm{O}}_2$.

   ½ mark

2. Reason: Mercury is low in the activity series (low reactivity), so heating alone is enough to reduce its oxide.

   ½ mark

3. (ii) Copper – Process: Roasting with self-reduction. $2C{\mathrm{u}}_2\mathrm{S}+3{\mathrm{O}}_2\to 2C{\mathrm{u}}_2\mathrm{O}+2S{\mathrm{O}}_2$, then $2C{\mathrm{u}}_2\mathrm{O}+\mathrm{C}{\mathrm{u}}_2\mathrm{S}\to 6Cu+\mathrm{S}{\mathrm{O}}_2$.

   ½ mark

4. Reason: Copper also has low reactivity, so heating in air is enough to obtain the metal.

   ½ mark

5. (iii) Sodium – Process: Electrolytic reduction of molten sodium chloride. Sodium is deposited at the cathode: $\mathrm{N}{\mathrm{a}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Na}$.

   ½ mark

6. Reason: Sodium is highly reactive, near the top of the activity series. Carbon cannot reduce its compounds, so electrolysis is needed.

   ½ mark

Answer: (i) Mercury: roasting/heating alone, because it has low reactivity. (ii) Copper: roasting/heating alone (self-reduction), because it has low reactivity. (iii) Sodium: electrolytic reduction of molten NaCl, because it is highly reactive.

1. (i) Silver: Shiny white silver turns black (tarnishes).

   ½ mark

2. Reason: It reacts with hydrogen sulphide in air to form a coating of silver sulphide, $\mathrm{A}{\mathrm{g}}_2\mathrm{S}$.

   ½ mark

3. (ii) Copper: Reddish-brown copper gains a green coating.

   ½ mark

4. Reason: It reacts with moist carbon dioxide in air to form basic copper carbonate, $\mathrm{CuC}{\mathrm{O}}_3\cdot \mathrm{Cu}(\mathrm{OH}{)}_2$.

   ½ mark

5. (iii) Iron: Grey iron becomes covered with a reddish-brown flaky layer.

   ½ mark

6. Reason: In the presence of moist air (oxygen and water), iron rusts, forming hydrated iron(III) oxide, $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3\cdot \mathrm{x}{\mathrm{H}}_2\mathrm{O}$.

   ½ mark

Answer: (i) Silver turns black (Ag₂S formed). (ii) Copper turns green (basic copper carbonate formed). (iii) Iron gets a reddish-brown coating (rust, hydrated iron oxide).

1. Absorption: Root cells take in water and minerals from the soil. Water enters the root xylem, and the vessels and tracheids of roots, stems and leaves form a continuous channel.

   1 mark

2. Transpiration: Water evaporates from the leaf cells and is lost through the stomata. This loss of water vapour from the aerial parts is called transpiration.

   1 mark

3. Transpiration pull: The loss of water creates a suction force (transpiration pull). This pulls the continuous column of water upward through the xylem from the roots to the topmost leaves. Root pressure also helps, especially at night.

   1 mark

Answer: Water lost from leaves by transpiration through the stomata creates a suction (transpiration pull), which draws water up through the continuous xylem from the roots to the leaves.

1. (a) Constituents: The CNS consists of the brain and the spinal cord.

   1 mark

2. (a) Protection of brain: The brain lies inside a bony box, the skull (cranium). It is cushioned by a fluid-filled balloon, the cerebrospinal fluid, which acts as a shock absorber.

   ½ mark

3. (a) Protection of spinal cord: The spinal cord is protected by the vertebral column (backbone).

   ½ mark

4. (b) Limitation 1: Electrical impulses reach only those cells that are connected by nervous tissue, not every cell of the body.

   ½ mark

5. (b) Limitation 2: After a cell generates and transmits an impulse, it needs time to reset before it can send a new one. So cells cannot generate impulses continuously.

   ½ mark

Answer: (a) Brain and spinal cord. The brain is protected by the skull (cranium) and fluid; the spinal cord by the vertebral column. (b) Impulses reach only cells connected to nerves; a cell needs time to reset before sending another impulse.

1. Phenomenon: Scattering of light, seen here as the Tyndall effect.

   1 mark

2. Explanation: The smoke contains fine colloidal particles. When the beam strikes them, they scatter (diffusely reflect) light in all directions. Some of this scattered light reaches our eyes, so the path of the beam becomes visible.

   1 mark

3. Dependence on particle size: Very fine particles scatter mainly blue light (shorter wavelengths). Larger particles scatter light of longer wavelengths. If the particles are large enough, all colours are scattered and the light appears white.

   1 mark

Answer: Tyndall effect (scattering of light). Smoke particles scatter light so the beam's path is visible. Very fine particles scatter mainly blue (short wavelength) light; larger particles scatter longer wavelengths (white if very large).

1. Function of fuse: A fuse is a short wire of low melting point connected in series with the circuit. If excess current flows because of overloading or a short circuit, the wire heats up (Joule heating), melts and breaks the circuit. This protects the appliances and the wiring from damage.

   1 mark

2. Given: $P=3\text{ kW}=3000\text{ W}$, $V=220\text{ V}$. Formula used: $P=VI$.

   ½ mark

3. $I=\frac{P}{V}=\frac{3000\text{ W}}{220\text{ V}}\approx 13.6\text{ A}$

   1 mark

4. Conclusion: The heater draws about 13.6 A, which is more than the 5 A rating of the circuit. When the heater is switched ON, the fuse wire will melt and break the circuit, and the heater will not work in this circuit.

   ½ mark

Answer: The fuse protects against overloading and short circuits by melting. The heater draws I = 3000/220 ≈ 13.6 A > 5 A, so the fuse will blow and the circuit will break.

1. (a) Ohm's law: The potential difference $V$ across the ends of a metallic conductor is directly proportional to the current $I$ through it, provided its temperature stays the same. So $V\propto I$, i.e. $V=IR$.

   1 mark

2. (a) Parallel combination: $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

   1 mark

3. (b) Parallel pair: The two resistors $R$ in parallel give $\frac{1}{R^{\prime }}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}$, so $R^{\prime }=\frac{R}{2}$. This combination is in series with the first $R$. Hence $R_{XY}=R+\frac{R}{2}=\frac{3R}{2}$.

   1 mark

Answer: (a) V ∝ I at constant temperature; 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃. (b) Equivalent resistance = 3R/2.

1. (i)(1) Strongly acidic: Solution B (pH 1). pH far below 7 means strongly acidic.

   ½ mark

2. (i)(2) Strongly alkaline: Solution C (pH 13).

   ½ mark

3. (i)(3) Weakly acidic: Solution A (pH 4), which is below 7 but close to it.

   ½ mark

4. (i)(4) Neutral: Solution D (pH 7).

   ½ mark

5. (i)(5) Weakly alkaline: Solution E (pH 10).

   ½ mark

6. **Increasing order of ${\mathrm{H}}^{+}$ concentration:** The lower the pH, the higher the $[{\mathrm{H}}^{+}]$. So the order of decreasing pH gives $\mathrm{C}(13)<\mathrm{E}(10)<\mathrm{D}(7)<\mathrm{A}(4)<\mathrm{B}(1)$.

   ½ mark

7. (ii)(1) Acidic salt: Ammonium chloride, $\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$. It is formed from a strong acid and a weak base, so its solution has pH < 7.

   ½ mark

8. Parent acid: hydrochloric acid, $\mathrm{HCl}$. Parent base: ammonium hydroxide, $\mathrm{N}{\mathrm{H}}_4\mathrm{OH}$.

   ½ mark

9. (ii)(2) Basic salt: Sodium carbonate, $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3$. It is formed from a weak acid and a strong base, so its solution has pH > 7.

   ½ mark

10. Parent acid: carbonic acid, ${\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_3$. Parent base: sodium hydroxide, $\mathrm{NaOH}$.

    ½ mark

Answer: (i) Strongly acidic B; strongly alkaline C; weakly acidic A; neutral D; weakly alkaline E. Order of increasing [H⁺]: C < E < D < A < B. (ii) Acidic salt NH₄Cl (HCl + NH₄OH); basic salt Na₂CO₃ (H₂CO₃ + NaOH).

1. Name of the process: Sodium hydroxide is made from sodium chloride by the chlor-alkali process. It is called this because its products are chlorine (chlor) and sodium hydroxide (alkali).

   1 mark

2. The process in brief: Electricity is passed through brine, a concentrated aqueous solution of sodium chloride. The brine decomposes to give sodium hydroxide, chlorine and hydrogen:
   $$2\mathrm{NaCl}(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2\mathrm{NaOH}(\mathrm{aq})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})$$
   Sodium hydroxide solution forms near the cathode.

   ½ mark

3. Gas X: X is **chlorine gas ($\mathrm{C}{\mathrm{l}}_2$). It is given off at the anode** (positive electrode).

   1 mark

4. Gas Y: Y is **hydrogen gas (${\mathrm{H}}_2$). It is given off at the cathode** (negative electrode).

   1 mark

5. Compound Z: Chlorine (gas X) reacts with dry slaked lime, $\mathrm{Ca}(\mathrm{OH}{)}_2$, to form **bleaching powder (calcium oxychloride), $\mathrm{CaOC}{\mathrm{l}}_2$** = Z.

   Z is used to disinfect drinking water in treatment plants and to bleach cotton and linen in the textile industry.

   ½ mark

6. Equation for the formation of Z:
   $$\mathrm{Ca}(\mathrm{OH}{)}_2+\mathrm{C}{\mathrm{l}}_2\to \mathrm{CaOC}{\mathrm{l}}_2+{\mathrm{H}}_2\mathrm{O}$$

   1 mark

Answer: Chlor-alkali process ($2NaCl+2H_{2}O\to 2NaOH+C{l}_2+H_2$); X = $C{l}_2$ at anode, Y = $H_2$ at cathode; Z = bleaching powder $CaOC{l}_2$; $Ca(OH{)}_2+C{l}_2\to CaOC{l}_2+H_{2}O$

1. (i) Spores: Spores are tiny, single-celled reproductive bodies. They are released from the parent and each one can grow into a new individual when it lands on a moist, suitable surface.

   ½ mark

2. (i) Where they form: Spores are formed inside blob-like structures called sporangia (singular: sporangium), at the tips of thin hyphae.

   ½ mark

3. (i) Surviving unfavourable conditions: Each spore is covered by a thick protective wall. This wall protects the spore from drying out and from harsh conditions until it reaches a moist surface and can germinate.

   ½ mark

4. (i) Organism: Rhizopus (bread mould) reproduces by spore formation.

   ½ mark

5. (ii) Reason 1: Some plants have lost the ability to produce seeds, for example banana, orange, rose and jasmine. Such plants can only be multiplied vegetatively.

   1 mark

6. (ii) Reason 2: Plants raised vegetatively flower and fruit earlier than plants grown from seeds. Also, all the new plants are genetically similar to the parent, so useful characters are kept.

   1 mark

7. (ii) Two methods of vegetative propagation: (1) Layering and (2) Grafting. Stem cutting is also accepted.

   1 mark

Answer: (i) Spores are single-celled reproductive bodies; formed in sporangia; thick protective walls help them survive; Rhizopus. (ii) Seedless plants (banana, rose) can be grown, and plants flower/fruit earlier and are genetically similar to the parent; methods: layering and grafting

1. (i) Labels (as in NCERT's figure of pollen germinating on the stigma):

   • A – male germ cell (male gamete)
   • B – pollen tube
   • C – female germ cell (female gamete, the egg inside the ovule)
   1½ marks

2. (i) When B reaches C: The pollen tube (B) grows down through the style and enters the ovule. It carries the male germ cell (A) with it. The male germ cell fuses with the female germ cell (C). This fusion is fertilisation, and it forms a zygote.

   1 mark

3. (i) Significance: The zygote develops into an embryo, which grows into a new plant. Because genetic material comes from two parents, the offspring also show variation.

   ½ mark

4. (ii) Post-fertilisation change 1: The zygote divides many times to form an embryo inside the ovule.

   ½ mark

5. (ii) Post-fertilisation change 2: The ovule develops a tough coat and becomes the seed.

   ½ mark

6. (ii) Post-fertilisation change 3: The ovary grows rapidly and ripens into the fruit.

   ½ mark

7. (ii) Post-fertilisation change 4: The petals, sepals, stamens, style and stigma shrivel and fall off.

   ½ mark

Answer: (i) A – male germ cell, B – pollen tube, C – female germ cell; male germ cell fuses with female germ cell (fertilisation) to form a zygote, which grows into a new plant. (ii) Zygote forms embryo, ovule becomes seed, ovary becomes fruit, other floral parts shrivel and fall

1. (i)(1) Ray diagram: ray parallel to the principal axis of a concave lens.

   • Draw a concave lens with optical centre O, its principal axis, and foci $F_1$ (on the object side) and $F_2$.
   • Draw a ray, with an arrow, travelling parallel to the principal axis and striking the lens.
   • After refraction the ray diverges away from the axis.
   • Extend it backwards with a dotted line; the extension passes through $F_1$ on the same side as the incoming ray.
   • So the refracted ray appears to come from the principal focus $F_1$.
   1 mark

2. (i)(2) Ray diagram: ray directed towards the principal focus of a concave lens.

   • Draw a ray, with an arrow, aimed towards $F_2$ on the far side of the lens. Show its continuation beyond the lens as a dotted line meeting $F_2$.
   • After refraction the ray emerges parallel to the principal axis.

   Arrows must be shown on all rays; half a mark is lost if they are missing.

   1 mark

3. (ii) Given (using the sign convention): height of object $h=+4$ cm, object distance $u=-16$ cm, focal length of convex lens $f=+24$ cm.

   ½ mark

4. (ii) Lens formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

   ½ mark

5. (ii) Substituting:
   $$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{24}+\frac{1}{(-16)}=\frac{2-3}{48}=-\frac{1}{48}$$
   $$\therefore v=-48\text{ cm}$$
   The image is formed 48 cm from the lens, on the same side as the object. It is virtual and erect.

   1 mark

6. (ii) Magnification: $m=\frac{h^{\prime }}{h}=\frac{v}{u}$

   ½ mark

7. (ii) Size of image:
   $$\frac{h^{\prime }}{4}=\frac{-48}{-16}=3$$
   $$h^{\prime }=3\times 4=+12\text{ cm}$$
   The image is 12 cm tall, erect and magnified 3 times.

   ½ mark

Answer: (i) (1) refracted ray diverges and appears to come from $F_1$; (2) refracted ray emerges parallel to the principal axis. (ii) v = −48 cm (on the same side as the object, virtual, erect); image height = 12 cm

1. (i)(1) Ray diagram: ray parallel to the principal axis of a convex mirror.

   • Draw a convex mirror, its principal axis, and F and C behind the mirror.
   • Draw a ray, with an arrow, parallel to the principal axis striking the mirror.
   • The reflected ray goes back outwards, away from the axis.
   • Extend it behind the mirror with a dotted line; the extension passes through the principal focus F.
   • So the reflected ray appears to diverge from F.
   1 mark

2. (i)(2) Ray diagram: ray directed towards the principal focus of a convex mirror.

   • Draw a ray, with an arrow, aimed towards F behind the mirror. Show the part behind the mirror as a dotted line.
   • After reflection the ray travels back parallel to the principal axis.

   Arrows must be shown on all rays; half a mark is lost if they are missing.

   1 mark

3. (ii) Given (using the sign convention): $h=+1.5$ cm, $u=-18$ cm, focal length of concave mirror $f=-12$ cm. Unknowns: $v=?$, $h^{\prime }=?$

   ½ mark

4. (ii) Mirror formula: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

   ½ mark

5. (ii) Substituting:
   $$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-12}-\frac{1}{-18}=\frac{-3+2}{36}=-\frac{1}{36}$$
   $$\therefore v=-36\text{ cm}$$
   The image is formed 36 cm in front of the mirror, on the same side as the object. It is a real image.

   1 mark

6. (ii) Magnification: $m=\frac{h^{\prime }}{h}=-\frac{v}{u}$

   ½ mark

7. (ii) Size of image:
   $$\frac{h^{\prime }}{1.5}=-\frac{(-36)}{(-18)}=-2$$
   $$h^{\prime }=-2\times 1.5=-3.0\text{ cm}$$
   The image is 3 cm tall. It is real, inverted and twice the size of the flame.

   ½ mark

Answer: (i) (1) reflected ray appears to diverge from F; (2) reflected ray goes parallel to the principal axis. (ii) v = −36 cm (36 cm in front of the mirror, real); h' = −3.0 cm (3 cm tall, inverted, magnified)

1. (a) First member: In a homologous series, each member differs from the next by a $-\mathrm{C}{\mathrm{H}}_2-$ unit. The first member with the –Br group is bromomethane, $\mathrm{C}{\mathrm{H}}_3\mathrm{Br}$.

   ½ mark

2. (a) Second member: The second member is bromoethane, ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Br}$ ($\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$).

   ½ mark

3. (b)(i) $-\mathrm{CHO}$ (carbonyl carbon bonded to H): aldehyde group.

   ½ mark

4. (b)(ii) $-\mathrm{CO}-$ (carbonyl carbon bonded to two carbons): ketone group.

   ½ mark

5. (c)(a) Observation: The purple (pink) colour of alkaline $\mathrm{KMn}{\mathrm{O}}_4$ disappears as it is added to warm ethanol. Once excess is added, the colour no longer goes.

   ½ mark

6. (c)(a) **Role of $\mathrm{KMn}{\mathrm{O}}_4$: It acts as an oxidising agent**. It supplies oxygen to convert ethanol into ethanoic acid, and is used up itself in the process, so its colour is lost.

   ½ mark

7. (c)(a) Equation:
   $$\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{alkaline }\mathrm{KMn}{\mathrm{O}}_4+\text{heat}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$$

   1 mark

8. (c)(b) OR alternative (marked instead of (c)(a) if chosen) – Product: Ethene ($\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2$) is formed. [½ mark]

9. (c)(b) **Role of conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$: It is a dehydrating agent**; it removes a water molecule from ethanol. [½ mark]

10. (c)(b) Equation: [1 mark]
    $$\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\underset{443\mathrm{K}}{\overset{\text{hot conc. }{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }}\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$$

Answer: (a) $C{H}_{3}Br$, $C_2{H}_{5}Br$; (b) (i) aldehyde, (ii) ketone; (c)(a) purple colour of $KMn{O}_4$ disappears, $KMn{O}_4$ is an oxidising agent, $C{H}_{3}C{H}_{2}OH\to C{H}_{3}COOH$; (c)(b) ethene, conc. $H_{2}S{O}_4$ is a dehydrating agent, $C{H}_{3}C{H}_{2}OH\to C{H}_2=C{H}_2+H_{2}O$

1. (a) Glands: The glands of the buccal cavity are the salivary glands.

   Food acted on: Saliva contains the enzyme salivary amylase, which acts on starch (a carbohydrate) and breaks it down into sugar.

   1 mark

2. (b) Organs with sphincters at their exit:
   1. The stomach – the pyloric sphincter controls the release of food into the small intestine.
   2. The anus – the anal sphincter controls the removal of undigested waste.

   1 mark

3. (c)(a)(i) No mucus from the gastric glands: Mucus protects the inner lining of the stomach. Without it, the lining would be exposed to hydrochloric acid and pepsin. The acid would damage the lining and cause acidity and ulcers.

   1 mark

4. (c)(a)(ii) No villi in the small intestine: Villi greatly increase the surface area for absorption. Without them, absorption of digested food would be greatly reduced. Most of the food would pass out unabsorbed, and the body would lack nutrients.

   1 mark

5. (c)(b) OR alternative (marked instead of (c)(a) if chosen) – Emulsification: Bile salts break large fat globules into tiny droplets. This increases the surface area for lipase from pancreatic juice to act on. [1 mark]

6. (c)(b) Making the medium alkaline: Food coming from the stomach is acidic. Bile neutralises it and makes the medium alkaline, which pancreatic enzymes need in order to work. So bile has no enzyme but is still important for digestion. [1 mark]

Answer: (a) Salivary glands; act on starch. (b) Stomach and anus. (c)(a) (i) Acid would damage the stomach lining (ulcers); (ii) absorption of digested food would be greatly reduced. (c)(b) Bile emulsifies fats and makes the acidic food alkaline for pancreatic enzymes to act

1. (a)(i) **Key $K_1$ closed: The branch containing bulb A is completed, so bulb A glows**.

   ½ mark

2. (a)(ii) **Key $K_2$ closed: The branch containing the series combination of bulbs B, C, D and E is completed, so bulbs B, C, D and E glow**.

   ½ mark

3. (b) Formula: For bulb B, power $P=11$ W and voltage $V=55$ V. Use $P=VI$.

   ½ mark

4. (b) Current drawn by B:
   $$I=\frac{P}{V}=\frac{11\text{W}}{55\text{V}}=\frac{1}{5}\text{A}=0.2\text{A}$$

   ½ mark

5. (c)(a)(i) Resistance of bulb B:
   $$R=\frac{V}{I}=\frac{55\text{V}}{0.2\text{A}}=275\Omega$$
   Check: $R=\frac{V^2}{P}=\frac{55^2}{11}=\frac{3025}{11}=275\Omega$.

   1 mark

6. (c)(a)(ii) Total resistance of B, C, D, E: The four identical bulbs are in series, so their resistances add:
   $$R_s=4\times 275\Omega =1100\Omega$$
   Check: $220\text{V}/1100\Omega =0.2$ A, which is the rated current of each bulb.

   1 mark

7. (c)(b) OR alternative (marked instead of (c)(a) if chosen) – Bulb A: It keeps glowing with the same brightness. [½ mark]

8. (c)(b) Bulbs B, D and E: They stop glowing. [½ mark]

9. (c)(b) Reason: B, D and E are in series with C. When C fuses, that branch becomes an open circuit, so no current flows through B, D or E. Bulb A is on a separate parallel branch. It still gets the full 220 V and the same current, so it is not affected. [1 mark]

Answer: (a) (i) Bulb A glows; (ii) bulbs B, C, D, E glow. (b) 0.2 A. (c)(a) (i) 275 Ω, (ii) 1100 Ω. (c)(b) A glows with the same brightness; B, D, E stop glowing, because they are in series with fused C, while A is on a parallel branch