CBSE Class 10 Science: 2023 board paper, set 31/4/1

1. $\mathrm{NaHC}{\mathrm{O}}_3+\mathrm{HCl}\to \mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2\uparrow$. The gas given off is carbon dioxide.

2. $\mathrm{C}{\mathrm{O}}_2$ turns lime water milky because insoluble $\mathrm{CaC}{\mathrm{O}}_3$ forms. This makes option (C) correct.

   1 mark

Answer: Carbon dioxide; turns lime water milky, option (C)

1. Lead is +2, so lead nitrate is $\mathrm{Pb}(\mathrm{N}{\mathrm{O}}_3{)}_2$ and lead iodide is $\mathrm{Pb}{\mathrm{I}}_2$.

2. The balanced equation is $2KI+\mathrm{Pb}(\mathrm{N}{\mathrm{O}}_3{)}_2\to \mathrm{Pb}{\mathrm{I}}_2\downarrow +2KN{\mathrm{O}}_3$, so option (B) is correct. Yellow $\mathrm{Pb}{\mathrm{I}}_2$ precipitates.

   1 mark

Answer: Option (B)

1. Magnesium ribbon burns with a dazzling white flame and forms white MgO: $2Mg+{\mathrm{O}}_2\to 2MgO$.

2. Two reactants form one product, so this is a combination reaction. Option (B) is correct.

   1 mark

Answer: X = Mg, Y = MgO, combination, option (B)

1. Tomato contains oxalic acid. (Vinegar has acetic acid, curd has lactic acid, and ant sting has methanoic acid.) Option (D) is correct.

   1 mark

Answer: Oxalic acid, option (D)

1. An alkali is a base that is soluble in water.

2. NaOH dissolves in water but $\mathrm{Fe}(\mathrm{OH}{)}_3$ does not, so only NaOH is an alkali. Option (B) is correct.

   1 mark

Answer: Option (B)

1. Washing soda, $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3.10{\mathrm{H}}_2\mathrm{O}$, precipitates the $\mathrm{C}{\mathrm{a}}^{2+}$ and $\mathrm{M}{\mathrm{g}}^{2+}$ ions as carbonates. This removes permanent hardness, so option (C) is correct.

   1 mark

Answer: Sodium carbonate decahydrate, option (C)

1. Each Cl atom (2,8,7) shares one electron, giving a single shared pair between the atoms. Each Cl is then left with 3 lone pairs, so both have 8 electrons.

2. According to the marking scheme, structure (c) shows this correctly, so the answer is option (C).

   1 mark

Answer: Option (C)

1. The figure shows water vapour leaving the leaves, which is transpiration.

2. Transpiration creates a suction pull that draws water up from the roots through the xylem. Option (B) is correct.

   1 mark

Answer: Transpiration, option (B)

1. When guard cells take in water they swell and the pore opens. When they lose water they shrink and the pore closes. Option (B) is correct.

   1 mark

Answer: Option (B)

1. $F_1$ is all Vv. Selfing Vv × Vv gives VV : Vv : vv = 1 : 2 : 1.

2. So VV : vv = 1 : 1, and option (A) is correct.

   1 mark

Answer: 1 : 1, option (A)

1. Cytokinins promote cell division and are present in regions of rapid cell division, such as fruits and seeds. Option (A) is correct.

   1 mark

Answer: Promote cell division, option (A)

1. Gametes are formed by meiosis and carry half the chromosome number. Fusion of two gametes then restores the original number in the zygote. Option (D) is correct.

   1 mark

Answer: Option (D)

1. In a series circuit the same current flows through every component, so the current through the 6 W bulb is also 0.06 A. Option (B) is correct.

   1 mark

Answer: 0.06 A, option (B)

1. The field lines form concentric circles around each point where the wire crosses the plane. Near the centre of the loop they become almost straight and all point in the same direction.

2. According to the scheme, pattern (c) shows this correctly, so the answer is option (C).

   1 mark

Answer: Option (C)

1. With the potential difference V and time t unchanged, the heat produced is $H=\frac{V^{2}t}{R}$.

2. If $R\to R/2$, then $H\to 2H$. The heat becomes two times, so option (B) is correct.

   1 mark

Answer: Two times, option (B)

1. An alpha particle is positive, so its motion (downward) acts like a current in that direction.

2. By Fleming's left-hand rule, with the field to the right and the current downward, the force is out of the page. Option (D) is correct.

   1 mark

Answer: Out of the page, option (D)

1. $\mathrm{CaO}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{Ca}(\mathrm{OH}{)}_2$ releases a large amount of heat. This release of heat is exactly what makes the reaction exothermic, so R correctly explains A. Option (A) is correct.

   1 mark

Answer: Option (A)

1. B is dominant, so BB and Bb give black eyes, but bb gives brown eyes. Therefore A is false.

2. R is true, because black is the dominant trait. Option (D) is correct.

   1 mark

Answer: Option (D)

1. Villi are finger-like projections that are richly supplied with blood vessels, so A is true.

2. Their large surface area is for the absorption of digested food, not for completing digestion. So R is false, and option (C) is correct.

   1 mark

Answer: Option (C)

1. A conductor carrying current at right angles to a magnetic field feels the maximum force, so A is true. The net charge on a conductor is zero, so R is also true.

2. The force arises from moving charges, not from net charge, so R does not explain A. Option (B) is correct.

   1 mark

Answer: Option (B)

1. (i) The compound formed is **copper(II) chloride, $\mathrm{CuC}{\mathrm{l}}_2$** (cupric chloride).

   ½ mark

2. The solution becomes blue-green. The black copper oxide dissolves to give this coloured salt solution.

   ½ mark

3. (ii) A metal oxide reacts with an acid to give a salt and water:
   $\mathrm{CuO}(\mathrm{s})+2HCl(\mathrm{aq})\to \mathrm{CuC}{\mathrm{l}}_2(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

   1 mark

Answer: (i) Copper(II) chloride, blue-green; (ii) $\mathrm{CuO}+2HCl\to \mathrm{CuC}{\mathrm{l}}_2+{\mathrm{H}}_2\mathrm{O}$

1. Caustic soda (NaOH) is made by the chlor-alkali process, the electrolysis of brine. So **X = sodium chloride, $\mathrm{NaCl}$**.

   ½ mark

2. The gas liberated at the cathode is **Y = hydrogen, ${\mathrm{H}}_2$**.

   ½ mark

3. The gas liberated at the anode is **Z = chlorine, $\mathrm{C}{\mathrm{l}}_2$**.
   $2NaCl(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2NaOH(\mathrm{aq})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})$

   ½ mark

4. Chlorine reacts with dry slaked lime to form **B = bleaching powder, $\mathrm{CaOC}{\mathrm{l}}_2$**:
   $\mathrm{Ca}(\mathrm{OH}{)}_2+\mathrm{C}{\mathrm{l}}_2\to \mathrm{CaOC}{\mathrm{l}}_2+{\mathrm{H}}_2\mathrm{O}$

   ½ mark

Answer: X = NaCl, Y = ${\mathrm{H}}_2$, Z = $\mathrm{C}{\mathrm{l}}_2$, B = bleaching powder $\mathrm{CaOC}{\mathrm{l}}_2$

1. (i) Maintaining posture and balance: cerebellum (hindbrain).

   ½ mark

2. (ii) Beating of heart: medulla oblongata (hindbrain).

   ½ mark

3. (iii) Thinking: cerebrum (forebrain).

   ½ mark

4. (iv) Blood pressure: medulla oblongata (hindbrain).

   ½ mark

Answer: (i) Cerebellum; (ii) Medulla; (iii) Cerebrum; (iv) Medulla

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

   ½ mark

2. (i) Positive phototropism (growth towards light): shoot / stem.

   ½ mark

3. (ii) Negative geotropism (growth away from gravity): shoot / stem.

   ½ mark

4. (iii) Positive hydrotropism (growth towards water): roots.

   ½ mark

Answer: Auxins: at shoot tip; (i) shoot; (ii) shoot; (iii) root

1. (i) Renal artery: brings blood containing nitrogenous wastes (urea) into the kidney for filtration.

   ½ mark

2. (ii) Urethra: carries urine out of the body from the urinary bladder.

   ½ mark

3. (iii) Glomerulus: filters blood. Water, urea, glucose and salts pass into Bowman's capsule.

   ½ mark

4. (iv) Tubular part of nephron: selectively reabsorbs useful substances such as glucose, amino acids, salts and water back into the blood.

   ½ mark

Answer: (i) Brings waste-laden blood to kidney; (ii) Passes urine out; (iii) Filtration of blood; (iv) Selective reabsorption

1. The plant kept in the dark cannot photosynthesise, so no oxygen is produced inside the oxygen-free container. Without oxygen it cannot carry out aerobic respiration to release energy, so it dies sooner.

   1 mark

2. The plant kept in sunlight photosynthesises. It uses $\mathrm{C}{\mathrm{O}}_2$ and water to produce glucose and oxygen. It uses this oxygen for respiration and gets energy, so it survives longer.

   1 mark

Answer: In the dark there is no photosynthesis, so no oxygen is available for respiration. In light, photosynthesis supplies oxygen for respiration.

1. (i) The parallel rays from a distant object meet in front of the retina, so the defect is myopia (short-sightedness).

   ½ mark

2. (ii) Causes:
   1. Excessive curvature of the eye lens.
   2. Elongation of the eyeball.

   1 mark

3. (iii) It is corrected with a concave (diverging) lens of suitable power.

   ½ mark

Answer: (i) Myopia; (ii) Excessive curvature of eye lens, elongated eyeball; (iii) Concave lens

1. The molecules of air and fine particles in the atmosphere are smaller than the wavelength of visible light. They scatter shorter wavelengths (blue) much more than longer ones (red). This scattered blue light enters our eyes, so the sky looks blue from the earth.

   1 mark

2. In space there is no atmosphere, so there are no particles to scatter sunlight. No scattered light reaches the eye, so the sky appears black (dark).

   1 mark

Answer: Blue from earth due to scattering of blue light by fine atmospheric particles; black from space since there is no atmosphere to scatter light.

1. Many pesticides are non-biodegradable. They are not broken down and get washed into rivers and ponds, where aquatic plants and animals absorb them.

   1 mark

2. These chemicals pass along the food chain and become more concentrated at each higher trophic level. This is biological magnification. Top consumers, including humans who eat fish, receive the highest dose, which harms their health. Hence it is a matter of deep concern.

   1 mark

Answer: Non-biodegradable pesticides enter food chains and concentrate at each trophic level (biomagnification), harming top consumers including humans.

1. (i) A few drops of acid are added to increase the electrical conductivity of water. Pure water is a poor conductor because it has very few ions.

   ½ mark

2. Gas at cathode: **hydrogen (${\mathrm{H}}_2$). Gas at anode: oxygen (${\mathrm{O}}_2$)**.

   1 mark

3. $2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2{\mathrm{H}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})$
   Volume of gas at anode : volume of gas at cathode = 1 : 2. The volume of hydrogen is twice that of oxygen.

   ½ mark

4. (ii) White silver chloride turns grey in sunlight because silver metal is formed:
   $2AgCl(\mathrm{s})\stackrel{\mathrm{sunlight}}{\to }2Ag(\mathrm{s})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})$

   ½ mark

5. This is a decomposition reaction (photolytic decomposition).

   ½ mark

Answer: (i) To increase conductivity; H₂ at cathode, O₂ at anode; anode : cathode volume = 1 : 2. (ii) White AgCl turns grey; photolytic decomposition.

1. (i) Always add the concentrated acid slowly to water, never water to acid, with constant stirring. The process is highly exothermic. Adding water to acid can make the mixture splash and the glass break.

   1 mark

2. (ii) ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4+2NaOH\to \mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4+2{\mathrm{H}}_2\mathrm{O}$
   The salt formed is **sodium sulphate, $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4$**.

   ½ mark

3. It is a salt of a strong acid and a strong base, so its solution is neutral with pH = 7.

   ½ mark

4. (iii) Dry HCl gas does not ionise in the absence of water, so no ${\mathrm{H}}^{+}$ (hydronium) ions are formed. Acidic behaviour needs ${\mathrm{H}}^{+}$ ions, so dry blue litmus does not turn red.

   1 mark

Answer: (i) Add acid slowly to water with stirring; (ii) Sodium sulphate, pH 7; (iii) No H⁺ ions form without water

1. (i) Paramecium is unicellular with a definite shape. Food is taken in at a specific spot (the oral groove/cytostome), where the beating of cilia moves it in. The food is then digested in a food vacuole.

   1 mark

2. (ii)(a) Hydrochloric acid: creates an acidic medium that activates pepsin, and kills germs that enter with food.

   ½ mark

3. (b) Trypsin: digests proteins into smaller peptides in the small intestine.

   ½ mark

4. (c) Muscular walls of stomach: churn and mix the food thoroughly with the digestive juices.

   ½ mark

5. (d) Salivary amylase: converts starch into sugar (maltose) in the mouth.

   ½ mark

Answer: (i) Food enters at a specific spot by ciliary movement; (ii) HCl: acidic medium / kills germs; trypsin: digests proteins; stomach muscles: churn food with juices; salivary amylase: starch to sugar

1. (i) Double circulation: blood passes through the heart twice in one complete cycle through the body. It goes once through the pulmonary circuit (heart → lungs → heart) and once through the systemic circuit (heart → body → heart).

   1 mark

2. (ii) Separating the right and left sides keeps oxygenated blood (left side) from mixing with deoxygenated blood (right side). This allows a highly efficient supply of oxygen to the body.

   1 mark

3. This helps birds and mammals because they have high energy needs. They constantly use energy to maintain a constant body temperature, so they need an efficient oxygen supply.

   1 mark

Answer: (i) Blood passes through the heart twice per cycle; (ii) Prevents mixing of oxygenated and deoxygenated blood, giving efficient oxygen supply needed by warm-blooded birds and mammals with high energy needs

1. (i) Principal focus of a diverging (convex) mirror: the point on the principal axis from which rays parallel to the principal axis appear to diverge after reflection from the mirror.

   1 mark

2. It lies behind the mirror and is virtual, because the reflected rays only appear to come from it.

   ½ mark

3. (ii) Focal length: the distance between the pole (P) and the principal focus (F) of the mirror.

   1 mark

4. For a spherical mirror it is half the radius of curvature, $f=\frac{R}{2}$. For a convex mirror it is taken as positive by the sign convention.

   ½ mark

Answer: (i) Point on principal axis from which reflected rays (incident parallel to axis) appear to diverge; (ii) Distance between pole and principal focus

1. Given: $h=+10$ cm, $u=-25$ cm, $f=+15$ cm (converging lens).
   Lens formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, so $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$

   ½ mark

2. $\frac{1}{v}=\frac{1}{15}+\frac{1}{-25}=\frac{5-3}{75}=\frac{2}{75}$

   ½ mark

3. $v=\frac{75}{2}=+37.5$ cm. The image is formed 37.5 cm on the other side of the lens.

   ½ mark

4. Magnification: $m=\frac{h^{\prime }}{h}=\frac{v}{u}$, so $h^{\prime }=h\times \frac{v}{u}$

   ½ mark

5. $h^{\prime }=10\times \frac{37.5}{-25}$

   ½ mark

6. $h^{\prime }=-15$ cm. The image is 15 cm tall; the negative sign shows it is real and inverted.

   ½ mark

Answer: v = +37.5 cm; h' = −15 cm (real, inverted, 15 cm tall)

1. $f=\frac{1}{P}$ (in metres) $=\frac{1}{+4\text{D}}$

   ½ mark

2. $f=+0.25$ m $=+25$ cm. It is a convex (converging) lens.

   ½ mark

3. Object distance is 50 cm $=2f$, so the object is at $2F_1$. The image forms at $2F_2$ (50 cm behind the lens) and is real and inverted, of the same size.

   ½ mark

4. Magnification $m=\frac{v}{u}=\frac{+50}{-50}=-1$.

   ½ mark

5. Ray diagram (described):

   • Draw a convex lens with the principal axis; mark $F_1$, $2F_1$ on the left and $F_2$, $2F_2$ on the right.
   • Place an upright object AB at $2F_1$.
   • Ray 1 from B runs parallel to the axis and refracts through $F_2$.
   • Ray 2 from B passes undeviated through the optical centre O.
   • The rays meet at B′ below the axis at $2F_2$. Image A′B′ is inverted and equal in size to AB.
   • Show arrows on all rays.
   1 mark

Answer: f = +25 cm; image real and inverted, same size, m = −1 (formed at 2F₂)

1. (i) AC can be transmitted over long distances at high voltage without much loss of electrical energy. Its voltage can easily be stepped up or down, which DC cannot do easily.

   1 mark

2. (ii) Household supply is alternating current (AC): its direction reverses periodically (50 Hz in India).

   ½ mark

3. A battery of dry cells gives direct current (DC), which always flows in one direction.

   ½ mark

4. (iii) A fuse is a wire of low melting point connected in series with the circuit. When the current exceeds its rating during a short circuit or overload, the fuse wire heats up, melts and breaks the circuit. This stops the current and protects the circuit and appliances.

   1 mark

Answer: (i) Less energy loss over long distances; (ii) Household: AC, dry cells: DC; (iii) Fuse melts on excess current and breaks the circuit

1. Diagram (described): inside the solenoid, draw straight, parallel, closely spaced field lines along the axis. Outside, the lines emerge from one end (N) and curve around to re-enter the other end (S), like a bar magnet. Mark the direction of current in the turns and the direction of the field lines with arrows, using the right-hand thumb rule.

   1 mark

2. Field is maximum at A (inside the solenoid).

   ½ mark

3. Reason: inside, the field lines are crowded together and uniform, because the fields of all $n$ turns add up.

   ½ mark

4. Field is minimum at B (just outside, beside the middle of the solenoid).

   ½ mark

5. Reason: there the field lines are far apart. Point C near the end is closer to a pole, where the lines are more concentrated than at B.

   ½ mark

Answer: Maximum at A (crowded, uniform lines inside); minimum at B (lines far apart outside)

1. Difference: biodegradable wastes (e.g. vegetable peels, paper) can be broken down by biological processes such as the action of bacteria and fungi. Non-biodegradable wastes (e.g. plastics, DDT, glass) cannot be broken down by biological processes.

   1 mark

2. Impacts of accumulated biodegradable waste:
   (i) It decomposes and gives a foul smell.

   ½ mark

3. (ii) It becomes a breeding ground for flies, mosquitoes and other disease carriers.

   ½ mark

4. Impacts of accumulated non-biodegradable waste:
   (i) Harmful chemicals enter food chains and cause biological magnification.

   ½ mark

5. (ii) It reduces soil fertility and blocks drains and water bodies.

   ½ mark

Answer: Biodegradable: broken down by microbes; non-biodegradable: not. Impacts: foul smell and disease carriers; biomagnification and loss of soil fertility.

1. (i)(a) Butanoic acid (${\mathrm{C}}_3{\mathrm{H}}_7\mathrm{COOH}$): a chain of 4 carbons with the –COOH group on the end carbon.
   $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{COOH}$
   Expanded: H₃C–CH₂–CH₂–C(=O)–O–H. Each of the first three carbons carries its H atoms (3, 2, 2), and the last carbon has a double bond to O and a single bond to –OH.

   1 mark

2. (b) Chloropentane (1-chloropentane, ${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{Cl}$): a 5-carbon chain with Cl on the end carbon.
   $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{Cl}$
   Expanded: each carbon has four single bonds; carbons 2–4 carry 2 H each, C-5 carries 2 H and Cl, and C-1 carries 3 H.

   1 mark

3. (ii) Structure (i), $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CH}-\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3{)}_2$, and structure (ii), $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$, are structural isomers (chain isomers).

   ½ mark

4. Reason: both have the same molecular formula ${\mathrm{C}}_6{\mathrm{H}}_{14}$ but different arrangements of carbon atoms (different structures).

   ½ mark

5. Another possible isomer is n-hexane:
   $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_3$
   (or 2-methylpentane, $\mathrm{C}{\mathrm{H}}_3-\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_3$)

   1 mark

6. (iii) Saturated compounds (alkanes) have the general formula ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2n+2}$. Unsaturated compounds have ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2n}$ (alkenes) or ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2n-2}$ (alkynes). So in saturated compounds the number of H atoms is more than twice the number of C atoms; in unsaturated compounds it is equal to or less than twice.

   1 mark

Answer: (i) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}$; $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Cl}$; (ii) Isomers (same formula ${\mathrm{C}}_6{\mathrm{H}}_{14}$, different structures); e.g. n-hexane; (iii) Saturated ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2n+2}$, unsaturated ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2n}$ / ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2n-2}$

1. (i) Sodium reacts with ethanol, forming sodium ethoxide, and hydrogen gas is evolved with brisk effervescence.

   ½ mark

2. (i) Equation: $2\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+2\mathrm{Na}\to 2\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{ONa}+{\mathrm{H}}_2\uparrow$ (sodium ethoxide + hydrogen)

   1 mark

3. (ii) Pure ethanoic acid has a low melting point of 290 K (about 17 °C). In cold climates during winter it freezes into an ice-like solid, so it is called glacial acetic acid.

   1 mark

4. (iii) Ethanol is dehydrated to ethene: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\underset{443K}{\overset{\text{hot conc. }{H}_{2}S{O}_4}{\to }}\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$

   1 mark

5. (iii) Role of conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$: it acts as a dehydrating agent. It removes a water molecule from ethanol.

   ½ mark

6. (iv) Saponification is the alkaline hydrolysis of an ester: $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{NaOH}\stackrel{\Delta }{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$ (ethyl ethanoate gives sodium ethanoate and ethanol).

   1 mark

Answer: (i) H₂ gas evolves and sodium ethoxide forms: 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂; (ii) it freezes at 290 K in winter, looking like ice; (iii) ethene forms, and conc. H₂SO₄ acts as a dehydrating agent; (iv) CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH

1. (i) Mode 1: Budding.

   ½ mark

2. In budding, a bud develops as an outgrowth from the body of hydra because of repeated cell division at one specific site. The bud grows into a tiny individual. When fully mature, it detaches from the parent and becomes a new independent hydra.

   1 mark

3. (i) Mode 2: Regeneration.

   ½ mark

4. In regeneration, hydra can be cut into many pieces. Each piece grows into a complete organism through specialised cells that proliferate and differentiate into the different tissues.

   1 mark

5. (ii) Vegetative propagation is the method of growing new plants from a vegetative part of the parent plant, such as the root, stem or leaf, without seeds.

   1 mark

6. Advantage 1: plants grown this way bear flowers and fruits earlier than plants raised from seeds.

   ½ mark

7. Advantage 2: it allows propagation of plants that have lost the capacity to produce seeds, such as banana, orange, rose and jasmine. (Also accepted: the new plants are genetically similar to the parent, so they keep all its desirable characters.)

   ½ mark

Answer: (i) Budding and regeneration; (ii) vegetative propagation is raising new plants from roots, stems or leaves. Advantages: earlier flowering and fruiting; propagates seedless plants such as banana and rose (also retains parental characters).

1. (i) Ohm's law: at constant temperature, the current through a conductor is directly proportional to the potential difference across its ends. So $I\propto V$, i.e. $V=IR$.

   1 mark

2. (i) Circuit diagram (described): a battery, a plug key, an ammeter (A) and a resistor R (nichrome wire XY) are all connected in series. A voltmeter (V) is connected in parallel across the resistor. The +/– terminals of the ammeter and voltmeter are marked, and arrows show the direction of current. Varying the number of cells changes V, and each time V and I are noted; V/I comes out constant.

   1 mark

3. (ii) An ammeter is connected in series, so the whole current passes through it. If it has low resistance, it does not add to the total resistance of the circuit and does not change the current it is measuring.

   1 mark

4. (iii) In the I–V graph, slope $=\frac{I}{V}=\frac{1}{R}$. So a smaller slope means a larger resistance.

5. (a) Graph A shows the series combination: it has the smaller slope, so its resistance is larger. Series resistance $R_s=R_1+R_2$ is the greater value.

   1 mark

6. (b) Graph B shows the parallel combination: it has the larger slope, so its resistance is smaller. Parallel resistance $R_p$ is less than the smallest individual resistance.

   1 mark

Answer: (i) I ∝ V (Ohm's law), verified with the ammeter in series and the voltmeter across the resistor; (ii) so that it does not change the current in the circuit; (iii) (a) series – A (less slope, more R); (b) parallel – B (more slope, less R)

1. (i) Mg (2,8,2) loses its two valence electrons, one to each of two Cl atoms (2,8,7): $\mathrm{M}{\mathrm{g}}^{\times \times }+2\ddot{\underset{..}{\mathrm{Cl}}}{:}^{\cdot }\to \mathrm{M}{\mathrm{g}}^{2+}+2[:\ddot{\underset{..}{\mathrm{Cl}}}{:}^{-}]\Rightarrow \mathrm{MgC}{\mathrm{l}}_2$. This gives $\mathrm{M}{\mathrm{g}}^{2+}$ (2,8) and two $\mathrm{C}{\mathrm{l}}^{-}$ (2,8,8) ions.

   1 mark

2. (ii) Ionic compounds are hard, brittle solids, and they are generally soluble in water but insoluble in solvents like kerosene and petrol.

   ½ mark

3. They also conduct electricity in the molten state or in aqueous solution.

   ½ mark

4. (iii)(A) Sodium (atomic number 11) has the configuration 2,8,1, so it has only one electron in its outermost shell.

   1 mark

5. It loses this one electron to chlorine and forms $\mathrm{N}{\mathrm{a}}^{+}$ with configuration 2,8: $\mathrm{Na}\to \mathrm{N}{\mathrm{a}}^{+}+e^{-}$. This is the stable octet of its nearest noble gas, neon.

   1 mark

6. (iii)(B)(i) In the solid state the ions are held tightly in a rigid crystal structure and cannot move. Without mobile ions, no current can flow.

   1 mark

7. (iii)(B)(ii) Hydrogen gas is liberated at the cathode (sodium hydroxide forms near it, as in the chlor-alkali process).

   1 mark

Answer: (i) Mg → Mg²⁺ + 2e⁻, with each Cl gaining 1e⁻ to form Cl⁻, giving Mg²⁺(Cl⁻)₂; (ii) hard solids, soluble in water, conduct when molten/aqueous (any two); (iii)(A) Na (2,8,1) loses 1e⁻ to form Na⁺ (2,8), the neon configuration; (iii)(B) ions are immobile in the rigid solid; H₂ gas is evolved at the cathode

1. (i) Sexual reproduction involves the fusion of a male gamete and a female gamete, each carrying half the DNA, from two different parents. The offspring gets a new combination of characters from both parents, so variations arise and it is not a true copy of either parent.

   1 mark

2. (ii) The F₁ generation is the first set of offspring from crossing the two parent plants, and only the dominant trait is expressed in it. The F₂ generation is obtained by self-crossing F₁ plants, and both dominant and recessive traits are expressed in it.

   1 mark

3. (iii)(A) If the niche of a population changes drastically, for example through a rise in temperature, a population with no variation could be wiped out completely.

   1 mark

4. If variations are present, some individuals may suit the new conditions. For example, a few heat-resistant bacteria would survive global warming and multiply. So variations help the species survive over time.

   1 mark

5. (iii)(B) New combinations in F₂: Round green and Wrinkled yellow (besides the parental Round yellow and Wrinkled green, in the ratio 9:3:3:1).

   1 mark

6. Reason: when two or more traits are involved, their genes are inherited independently of each other (law of independent assortment). They can therefore recombine in ways not present in either parent.

   1 mark

Answer: (i) Fusion of gametes from two parents creates new combinations (variation); (ii) F₁ shows only dominant traits, while F₂ (from selfing F₁) shows both dominant and recessive traits; (iii)(A) variants survive drastic changes in the niche; (iii)(B) round green and wrinkled yellow, due to independent inheritance of traits

1. (i) Use $n=\frac{c}{v}$, so $v=\frac{c}{n}$.

   ½ mark

2. $v=\frac{3\times 10^8\text{m/s}}{2.42}\approx 1.24\times 10^8$ m/s

   ½ mark

3. (ii) For the same angle of incidence, a higher refractive index bends light more towards the normal, so the angle of refraction is smaller. Increasing order: $r_{\text{carbon disulphide}}<r_{\text{glass}}<r_{\text{water}}$.

   1 mark

4. (iii)(A)(a) Glass is optically denser.

   ½ mark

5. Light travels slower in glass ($2\times 10^8$ m/s) than in water ($2.25\times 10^8$ m/s), so glass has the higher refractive index.

   ½ mark

6. (iii)(A)(b) The ray goes straight into the glass without bending. For normal incidence $\angle i=0^{\circ }$, so $\angle r=0^{\circ }$ and the ray continues along the normal (only its speed decreases).

   1 mark

7. (iii)(B)(i) $n_{glass}=\frac{c}{v_{glass}}\Rightarrow c=n_{glass}{v}_{glass}=\frac{3}{2}\times 2\times 10^8=3\times 10^8$ m/s

   1 mark

8. (iii)(B)(ii) $v_{water}=\frac{c}{n_{water}}=\frac{3\times 10^8}{4/3}=\frac{9}{4}\times 10^8=2.25\times 10^8$ m/s

   1 mark

Answer: (i) ≈1.24 × 10⁸ m/s; (ii) r(CS₂) < r(glass) < r(water); (iii)(A) (a) glass (light is slower, higher n); (b) the ray passes undeviated since i = 0 gives r = 0; (iii)(B) vacuum: 3 × 10⁸ m/s; water: 2.25 × 10⁸ m/s