CBSE Class 10 Science: sample paper 2023-24

1. The reaction is a double displacement reaction: $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})+\mathrm{BaC}{\mathrm{l}}_2(\mathrm{aq})\to \mathrm{BaS}{\mathrm{O}}_4(\mathrm{s})+2NaCl(\mathrm{aq})$

2. Barium sulphate does not dissolve in water. It separates out as a white precipitate, which is the solid state. Sodium chloride stays dissolved. So the answer is option (B) Barium sulphate.

   1 mark

Answer: Barium sulphate, option (B)

1. Zinc is more reactive than copper, so it displaces copper: $\mathrm{Zn}(\mathrm{s})+\mathrm{CuS}{\mathrm{O}}_4(\mathrm{aq})\to \mathrm{ZnS}{\mathrm{O}}_4(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$

2. The blue $\mathrm{CuS}{\mathrm{O}}_4$ is used up and colourless $\mathrm{ZnS}{\mathrm{O}}_4$ forms. The solution becomes colourless, so the answer is option (B).

   1 mark

Answer: Colourless, option (B)

1. $\mathrm{Ca}(\mathrm{OH}{)}_2$ and $\mathrm{NaOH}$ are bases, not salts, and $\mathrm{NaOH}$ is corrosive. $\mathrm{NaCl}$ is a neutral salt.

2. Sodium hydrogencarbonate, $\mathrm{NaHC}{\mathrm{O}}_3$ (baking soda), is a mild, non-corrosive basic salt. It is used in antacids. So the answer is option (D).

   1 mark

Answer: $\mathrm{NaHC}{\mathrm{O}}_3$, option (D)

1. A colourless gas that burns with a pop near a burning match stick is hydrogen. Hydrogen forms when a metal above hydrogen in the reactivity series reacts with a dilute acid.

2. Sulphur is a non-metal. Copper and silver lie below hydrogen and do not give ${\mathrm{H}}_2$. Sodium does: $2Na+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to \mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4+{\mathrm{H}}_2$. So the answer is option (A).

   1 mark

Answer: Sodium, option (A)

1. Na (2,8,1) loses 1 electron to form $\mathrm{N}{\mathrm{a}}^{+}$. O (2,6) gains 2 electrons to form ${\mathrm{O}}^{2-}$ with an octet.

2. To balance the charge, two $\mathrm{N}{\mathrm{a}}^{+}$ ions combine with one ${\mathrm{O}}^{2-}$ ion, giving $\mathrm{N}{\mathrm{a}}_2\mathrm{O}$. This is written $2N{\mathrm{a}}^{+}[\mathrm{O}{]}^{2-}$, which is option (B).

   1 mark

Answer: $2N{\mathrm{a}}^{+}[\mathrm{O}{]}^{2-}$, option (B)

1. Metals form basic oxides. A metal has few valence electrons (1 to 3) and loses them easily.

2. The configuration 2,8,1 (atomic number 11) is sodium, a metal, which forms the basic oxide $\mathrm{N}{\mathrm{a}}_2\mathrm{O}$. The other options are non-metals (N, Cl, Si). So the answer is option (D).

   1 mark

Answer: 11 (2,8,1), option (D)

1. The 2nd shell holds 8 electrons. 50% of 8 is 4, so the 3rd shell has 4 electrons.

2. The configuration is 2,8,4, so the atomic number is $2+8+4=14$. The answer is option (C).

   1 mark

Answer: 14, option (C)

1. Saprophytes such as fungi release digestive enzymes onto their food. The food is broken down outside the body and then absorbed.

2. Mushroom is a saprophytic fungus. Amoeba and Paramoecium digest food inside food vacuoles, and lice are parasites that suck blood. So the answer is option (B).

   1 mark

Answer: Mushroom, option (B)

1. Gustatory receptors detect taste.

2. Taste is sensed by the tongue. The nose has olfactory receptors for smell. So the answer is option (A).

   1 mark

Answer: tongue, option (A)

1. Plants raised by vegetative propagation come from parts of one parent plant. They are therefore genetically similar to the parent.

2. Banana is commonly propagated vegetatively, because cultivated bananas have no viable seeds. So the answer is option (C) Vegetative propagation.

   1 mark

Answer: Vegetative propagation, option (C)

1. A gene codes for a protein, often an enzyme.

2. That enzyme controls how much growth hormone is made, and the amount of hormone decides the height of the plant. So the answer is option (C).

   1 mark

Answer: Option (C)

1. During heavy exercise, the oxygen supply to muscles falls short of demand.

2. The muscles then respire anaerobically, and pyruvate is changed into lactic acid. Lactic acid builds up and causes cramps. So the answer is option (C).

   1 mark

Answer: Option (C)

1. A convex mirror always forms a virtual, erect and diminished image behind the mirror, between the pole P and the focus F.

2. As the object moves closer, the image moves from F towards P. Its position therefore depends on the object distance, so the answer is option (D).

   1 mark

Answer: Option (D)

1. When very fine particles in the atmosphere send light off in all directions, the effect is called scattering.

2. This is the reason for the blue colour of the sky and the Tyndall effect. So the answer is option (C).

   1 mark

Answer: Scattering of light, option (C)

1. In 1987, UNEP agreed to freeze CFC production at 1986 levels.

2. CFCs deplete the ozone layer, so X = CFCs and Y = Ozone. The answer is option (C).

   1 mark

Answer: CFCs; Ozone, option (C)

1. Biodegradable substances are broken down by microorganisms such as bacteria and fungi.

2. Remaining inert, persisting for a long time and harming the ecosystem are features of non-biodegradable substances. So the answer is option (A).

   1 mark

Answer: Broken down by biological processes, option (A)

1. Assertion: Rusting of iron is an oxidation that releases heat. It is exothermic, so the Assertion is false.

2. Reason: Rusting is very slow, so the heat given out is barely noticeable. The Reason is true. A is false and R is true, so the answer is option (D).

   1 mark

Answer: Option (D)

1. Sexual reproduction produces variations, so the offspring have a better chance of surviving in a changing environment. The Assertion is true.

2. Variations help some individuals survive, which is exactly why survival is higher. Both are true and R explains A, so the answer is option (A).

   1 mark

Answer: Option (A)

1. Reason: The magnetic field near a conductor gets stronger when the current increases. The Reason is true.

2. A stronger field deflects the needle more, so the deflection increases. The Assertion is false, and the answer is option (D).

   1 mark

Answer: Option (D)

1. Decomposers break down biodegradable waste into compost and return nutrients to the soil. The Assertion is true.

2. Decomposers break complex organic substances into simple inorganic substances, not the other way round. The Reason is false, so the answer is option (C).

   1 mark

Answer: Option (C)

1. Reaction: $\mathrm{Zn}(\mathrm{s})+2HCl(\mathrm{aq})\to \mathrm{ZnC}{\mathrm{l}}_2(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})$. A new substance forms, so a chemical change takes place. Two observations prove this:

2. Argument 1: Bubbles of a gas appear (evolution of gas). The gas is hydrogen, which burns with a pop sound.

   1 mark

3. Argument 2: The test tube becomes warm, showing a change in temperature (heat is released). The zinc granules may also turn from silvery grey to dark.

   1 mark

Answer: Evolution of hydrogen gas (bubbles) and change in temperature (or change in colour of Zn)

1. After fertilisation, the zygote divides many times and develops into an embryo inside the ovule.

   ½ mark

2. The ovule forms a tough seed coat and slowly turns into a seed.

   ½ mark

3. The ovary enlarges and ripens to become the fruit.

   ½ mark

4. The sepals, petals, stamens, style and stigma usually wither and fall off.

   ½ mark

Answer: Zygote → embryo; ovule → seed; ovary → fruit; other floral parts shrivel and fall

1. Purpose: Urine is made to remove nitrogenous wastes such as urea and uric acid,

   ½ mark

2. which are filtered out of the blood by the kidneys.

   ½ mark

3. Organ that stores urine: urinary bladder.

   ½ mark

4. Organ through which urine is released: urethra.

   ½ mark

Answer: To filter nitrogenous wastes (urea, uric acid) from blood; stored in urinary bladder; released through urethra

1. Arteries carry blood away from the heart, and this blood is under high pressure. They need thick, elastic walls to withstand and adjust to this pressure.

   1 mark

2. Veins carry blood at low pressure back to the heart. They have valves that stop the blood flowing backwards, so it moves in one direction only.

   1 mark

Answer: Arteries: thick elastic walls to bear high blood pressure; veins: valves to keep low-pressure blood flowing one way

1. (a) A ray bends towards the normal when it goes from an optically rarer medium (lower refractive index) into an optically denser medium (higher refractive index).

   ½ mark

2. For A → B: $n_A=1.6$ and $n_B=1.8$, so $n_B>n_A$. B is denser than A. The ray going from A to B bends towards the normal. (For B → C, $n_C<n_B$, so that ray bends away from the normal.)

   ½ mark

3. (b) Speed of light in a medium: $v=\frac{c}{n}$. So the smaller the refractive index, the greater the speed of light.

   ½ mark

4. For B → C: $n_C=1.5<n_B=1.8$, so $v_C>v_B$. The speed of light increases when the ray goes from B to C.

   ½ mark

Answer: (a) Ray from A to B bends towards the normal, since $n_B>n_A$. (b) Speed increases for the ray from B to C, since $n_C<n_B$ and $v=c/n$.

1. Resistance is directly proportional to length ($R\propto l$). Each of the three equal parts therefore has resistance $\frac{R}{3}$.

   ½ mark

2. For three resistors in parallel: $\frac{1}{R_1}=\frac{1}{R/3}+\frac{1}{R/3}+\frac{1}{R/3}=\frac{3}{R}+\frac{3}{R}+\frac{3}{R}=\frac{9}{R}$

   1 mark

3. So $R_1=\frac{R}{9}$, which gives $\frac{R_1}{R}=\frac{1}{9}$, i.e. $R_1:R=1:9$.

   ½ mark

Answer: $R_1:R=1:9$

1. Where the magnetic field lines are closer together (crowded), the magnetic field is stronger.

   ½ mark

2. Outside the solenoid, the lines are most crowded near its ends (the poles). So the field is strongest near the ends, and it becomes weaker further away, where the lines spread out.

   ½ mark

3. When the current in the circuit is reversed, the direction of the magnetic field also reverses. The two ends of the solenoid swap their N and S polarities, and the field lines point the opposite way.

   1 mark

Answer: The field is stronger where the lines are crowded, i.e. near the ends (poles) of the solenoid. Reversing the current reverses the direction of the magnetic field (poles interchange).

1. (a) The trophic levels are: Leaf (producer, 1st), Caterpillar (2nd), Chameleon (3rd), Snake (4th), Mongoose (5th). By the 10% law, only about 10% of the energy at one level passes to the next level.

2. So the energy at the 2nd level is $100\times 10=1000$ J, and at the producer level it is $1000\times 10=10000$ J. Energy at producer level = 10000 J, because only 10% of the energy is transferred to each next trophic level.

   1 mark

3. (b) No. At each step about 90% of the energy is lost, mostly as heat. If two more levels were added before the fourth level, so little usable energy would remain that the higher levels could not be supported. That is why food chains usually have only 3–4 (rarely 5) trophic levels.

   1 mark

Answer: (a) 10000 J (10% law); (b) No, so much energy is lost at each step that too little usable energy would remain for the extra levels.

1. Balanced equation: $3Mn{\mathrm{O}}_2(\mathrm{s})+4Al(\mathrm{s})\to 3Mn(\mathrm{l})+2A{\mathrm{l}}_2{\mathrm{O}}_3(\mathrm{s})+\mathrm{Heat}$

2. (a) The reaction is highly exothermic. So much heat is given out that the metal (Mn/Fe) is produced in the molten (liquid) state. That is why it is called a thermite reaction.

   1 mark

3. (b) Al gains oxygen to form $\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3$, so the substance oxidised is Al(s).

   ½ mark

4. $\mathrm{Mn}{\mathrm{O}}_2$ loses oxygen to form Mn, so the **substance reduced is $\mathrm{Mn}{\mathrm{O}}_2$(s)**.

   ½ mark

5. (c) Aluminium is more reactive than Fe and Mn (it is placed above them in the reactivity series). It can therefore displace them from their oxides, acting as a strong reducing agent and releasing a lot of heat.

   1 mark

Answer: (a) Highly exothermic reaction; the metal is obtained in molten state. (b) Oxidised: Al; Reduced: $\mathrm{Mn}{\mathrm{O}}_2$. (c) Al is more reactive than Fe/Mn.

1. M (2, 8, 3) has 3 valence electrons, so its valency is 3 and it forms ${\mathrm{M}}^{3+}$. Formulae of the compounds: with $\mathrm{C}{\mathrm{l}}^{-}$ it forms $\mathrm{MC}{\mathrm{l}}_3$; with $\mathrm{S}{\mathrm{O}}_4^{2-}$ it forms ${\mathrm{M}}_2(\mathrm{S}{\mathrm{O}}_4{)}_3$.

   1 mark

2. Nature of bond: ionic. M is a metal that easily loses its 3 valence electrons to form the ${\mathrm{M}}^{3+}$ cation. It then attains the stable configuration of neon (2, 8), and the cation is held to the anions by electrostatic attraction.

   1 mark

3. Conductivity: M, a metal, conducts electricity in the solid state. Its ionic compounds do not conduct in the solid state, because the ions are fixed in the crystal. They do conduct in the molten state or in aqueous solution, where the ions are free to move.

   1 mark

Answer: $\mathrm{MC}{\mathrm{l}}_3$ and ${\mathrm{M}}_2(\mathrm{S}{\mathrm{O}}_4{)}_3$; ionic bond (M loses 3 electrons to attain the neon configuration); the compounds conduct only when molten/dissolved, unlike M, which conducts in the solid state.

1. X is a reddish-brown metal used in electrical wiring, so X = Copper (Cu). On heating in air: $2Cu+{\mathrm{O}}_2\to 2CuO$, so Y = Copper(II) oxide, CuO (black). With hydrogen: $\mathrm{CuO}+{\mathrm{H}}_2\to \mathrm{Cu}+{\mathrm{H}}_2\mathrm{O}$.

   1 mark

2. Diagram of electrolytic refining of copper (to be drawn and labelled): a tank containing acidified copper sulphate solution (electrolyte). A thick block of impure copper is the anode, connected to the positive terminal of the battery. A thin strip of pure copper is the cathode, connected to the negative terminal. A battery and a key complete the circuit. Show $\mathrm{C}{\mathrm{u}}^{2+}$ ions moving towards the cathode, and anode mud (impurities) collecting at the bottom below the anode.

   1½ marks

3. Working: at the anode, $\mathrm{Cu}\to \mathrm{C}{\mathrm{u}}^{2+}+2{\mathrm{e}}^{-}$. At the cathode, $\mathrm{C}{\mathrm{u}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$, so pure copper is deposited on the cathode. Insoluble impurities settle as anode mud.

   ½ mark

Answer: X = Copper (Cu), Y = Copper(II) oxide (CuO); refining by electrolysis with an impure Cu anode, a pure Cu cathode and acidified $\mathrm{CuS}{\mathrm{O}}_4$ electrolyte.

1. Iodine is needed by the thyroid gland to make the hormone thyroxin. Iodised salt is an easy way to get enough iodine in the daily diet.

   1 mark

2. Thyroxin regulates carbohydrate, protein and fat metabolism in the body.

   1 mark

3. Thyroxin provides the best balance for growth of the body. If iodine is lacking, too little thyroxin is made and the person may suffer from goitre (swollen neck). That is why doctors advise iodised salt.

   1 mark

Answer: Iodine is needed to make thyroxin, which regulates carbohydrate, protein and fat metabolism and balances growth; its deficiency causes goitre.

1. The probability of a girl or a boy being born is equal: 50% each (1 in 2).

   1 mark

2. Humans have 23 pairs of chromosomes: 22 pairs of autosomes and 1 pair of sex chromosomes. Women have a perfect pair of sex chromosomes: XX.

   ½ mark

3. Men have a mismatched pair: one normal-sized X and one short Y, i.e. XY. So half the sperms carry X and half carry Y.

   ½ mark

4. Every egg from the mother carries an X chromosome, so the child always gets X from the mother.

   ½ mark

5. If the child gets X from the father, it is XX, a girl. If it gets Y from the father, it is XY, a boy. Both kinds of sperm are equally likely to fertilise the egg, so the chance is 50 : 50.

   ½ mark

Answer: 50% chance of a girl and 50% chance of a boy; sex is decided by whether the father's sperm carries X or Y.

1. (i) Refractive index of a medium with respect to air: $n=\frac{\text{speed of light in air}}{\text{speed of light in the medium}}$

   ½ mark

2. Light travels slower in any material medium than in air, so the denominator is always smaller than the numerator. Hence $n>1$.

   ½ mark

3. (ii) The ray travels towards the centre of the semicircle, i.e. along a radius. A radius is normal to the curved surface, so the ray strikes the curved boundary along the normal (angle of incidence $=0^{\circ }$). A ray falling normally on a surface passes through undeviated.

   1 mark

4. (iii) Ray diagram: at the flat top face, the ray meets the surface at the centre with angle of incidence $40^{\circ }$ inside the plastic. It goes from the denser medium (plastic) into the rarer medium (air), so it bends away from the normal. Draw the emergent ray in air on the opposite side of the normal (upper right), with angle of refraction $r>40^{\circ }$.

   ½ mark

5. Mark arrows on the incident, internal and emergent rays to show the direction of travel, and label the normal, $\angle i=40^{\circ }$ and $\angle r$ ($>40^{\circ }$).

   ½ mark

Answer: (i) $n=c_{air}/v_{medium}$ and light is slower in any medium, so n > 1; (ii) the ray moves along a radius, so it strikes the curved surface normally and is undeviated; (iii) at the flat face the ray emerges into air bending away from the normal ($r>40^{\circ }$).

1. (i) The law is Joule's law of heating.

   ½ mark

2. It states that the heat H produced in a resistor is directly proportional to: (a) the square of the current through it, $H\propto I^2$; (b) its resistance, $H\propto R$; (c) the time for which the current flows, $H\propto t$.

   1 mark

3. Combining these: $H=I^{2}Rt$ (H in joules, I in amperes, R in ohms, t in seconds).

   ½ mark

4. (ii) The resistance of a conductor depends on: its length ($R\propto l$); its area of cross-section ($R\propto 1/A$); the nature of its material (resistivity); and its temperature.

   1 mark

Answer: (i) Joule's law of heating: $H=I^{2}Rt$, i.e. heat ∝ $I^2$, ∝ R, ∝ t. (ii) Length, area of cross-section, nature of material, temperature.

1. (i) Anannya's answer is incorrect. A three-pin connection does not reduce heating of the connecting wires.

   1 mark

2. The third pin connects the metallic body of the appliance to the earth wire. If current leaks to the metal body, the earth wire gives it a low-resistance path to the ground. This keeps the body at earth potential, so the user does not get a severe electric shock. An electric bulb has no metallic body that a user touches, so a two-pin plug is enough.

   1 mark

3. (ii) A fuse is a safety device connected in series in the circuit. It protects the circuit and appliances from overloading and short-circuiting. When excess current flows, its wire (of low melting point) melts and breaks the circuit.

   1 mark

Answer: (i) Incorrect: the third pin provides earthing, a low-resistance path for leakage current that protects the user from shock. (ii) A fuse melts and breaks the circuit when excess current flows, protecting the circuit and appliances.

1. (a) Rehmat is correct. In this reaction a hydrogen atom of methane is replaced by a chlorine atom. A reaction in which one atom or group is replaced by another is a substitution reaction. Saturated hydrocarbons like methane react this way in sunlight.

   1 mark

2. Balanced equation: $\mathrm{C}{\mathrm{H}}_4+\mathrm{C}{\mathrm{l}}_2\stackrel{\mathrm{sunlight}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}+\mathrm{HCl}$ (further substitution can give $\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{l}}_2$, $\mathrm{CHC}{\mathrm{l}}_3$ and $\mathrm{CC}{\mathrm{l}}_4$).

   1 mark

3. (b) Electrolysis of brine (chlor-alkali process): $2NaCl(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2NaOH(\mathrm{aq})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})$. Chlorine is released at the anode and hydrogen at the cathode.

   1 mark

4. Other products: sodium hydroxide (NaOH), formed near the cathode, and **hydrogen gas (${\mathrm{H}}_2$)**, released at the cathode.

   1 mark

5. Uses (any one each): NaOH is used to make soaps and detergents (also for degreasing metals, paper making, artificial fibres). ${\mathrm{H}}_2$ is used as a fuel (also to hydrogenate oils into margarine, and to make ammonia for fertilisers).

   1 mark

Answer: (a) Substitution: an H of methane is replaced by Cl; $\mathrm{C}{\mathrm{H}}_4+\mathrm{C}{\mathrm{l}}_2\stackrel{\mathrm{sunlight}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}+\mathrm{HCl}$. (b) $2NaCl+2{\mathrm{H}}_2\mathrm{O}\to 2NaOH+\mathrm{C}{\mathrm{l}}_2+{\mathrm{H}}_2$; other products NaOH (soaps/detergents) and ${\mathrm{H}}_2$ (fuel/margarine/ammonia).

1. Identifying X: X has a vinegar-like smell. Vinegar is a dilute (5–8%) solution of ethanoic acid, so X is **ethanoic acid (acetic acid), $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$**.

   1 mark

2. Identifying Y: Y is used as an industrial solvent and reacts with an acid to give an ester, so Y is an alcohol: **ethanol (ethyl alcohol), ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$**.

   1 mark

3. Identifying Z: A sweet-smelling liquid formed from an acid and an alcohol is an ester. With formula ${\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2$, Z is **ethyl ethanoate, $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5$**. Check: carbons $2+2=4$, hydrogens $3+5=8$, oxygens $=2$.

   1 mark

4. Formation of Z (esterification):
   $$\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\underset{\text{warm (water bath)}}{\overset{\text{conc. }{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }}\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$$
   The acid (X) and the alcohol (Y) give the ester (Z). Conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ acts as the catalyst.

   1 mark

5. Action of caustic soda on Z (saponification):
   $$\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{NaOH}\stackrel{\text{heat}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$$
   The ester gives back sodium ethanoate (the sodium salt of X) and ethanol (Y).

   1 mark

Answer: X = ethanoic acid ($\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$); Y = ethanol (${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$); Z = ethyl ethanoate ($\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5$). Esterification: $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$ (conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$). Saponification: $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{NaOH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$.

1. (a) Testes do not descend into the scrotum: Sperm formation needs a temperature slightly lower than normal body temperature. The scrotum keeps the testes outside the body cavity for this reason. If the testes stay inside the abdomen, they are at body temperature, so sperm formation is badly affected. The person may be infertile.

   1 mark

2. (b) Vas deferens plugged: The vas deferens carries sperms from the testis towards the urethra. If it is blocked, sperms cannot be carried out and will not be present in the semen. The man cannot fertilise an egg. This is the basis of vasectomy, a surgical method of contraception.

   1 mark

3. (c) Prostate and seminal vesicles not functional: These glands add their secretions to the sperms. The secretions provide nourishment and a fluid medium in which sperms can move and be transported. Without them, sperms will be poorly nourished and will not be transported easily, so fertility is reduced.

   1 mark

4. (d) Egg not fertilised: The unfertilised egg survives for about one day. The thick, blood-rich lining of the uterus is then not needed. It breaks down and comes out through the vagina as blood and mucus along with the egg. This is menstruation, which occurs roughly every month.

   1 mark

5. (e) Placenta not attached to the uterus properly: The placenta passes glucose and oxygen from the mother's blood to the embryo and removes the embryo's wastes. Poor attachment means the embryo gets too little nutrition and oxygen. Its growth and development are affected, which may even cause serious complications such as loss of the pregnancy.

   1 mark

Answer: (a) Sperm formation affected because the testes are at body temperature. (b) Sperms cannot be transported, so no sperms in semen. (c) No secretions for nourishment and transport of sperms. (d) Uterine lining breaks down and menstruation occurs. (e) Embryo gets less nutrition and oxygen, so its growth is affected.

1. (a) The doctor asked Sameer to cut down sugar and exercise. This means his blood sugar level is high, so Sameer is suffering from diabetes (diabetes mellitus).

   1 mark

2. (a) The hormone involved is insulin. Insulin lowers blood sugar. When it is secreted in too little amount, the blood sugar level rises.

   1 mark

3. (a) Insulin is produced by the pancreas.

   1 mark

4. (b) Cytokinins are present in greater concentration in areas of rapid cell division, such as fruits and seeds, and they promote cell division.

   1 mark

5. (b) Abscisic acid inhibits growth. It also causes effects such as wilting of leaves.

   1 mark

Answer: (a) Diabetes; hormone: insulin; organ: pancreas. (b) Cytokinins in areas of rapid cell division; abscisic acid inhibits growth.

1. (i) The lens in the figure is thicker at the middle than at the edges. It has focal points on both sides of the principal axis. It is a convex (converging) lens.

   1 mark

2. (ii) Use the lens formula $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ with the New Cartesian sign convention.

   For a convex lens $f=+5\text{ m}$. A real image forms on the other side of the lens, so $v=+7\text{ m}$.

   ½ mark

3. Substituting: $\frac{1}{5}=\frac{1}{7}-\frac{1}{u}$

   ½ mark

4. $\frac{1}{u}=\frac{1}{7}-\frac{1}{5}=\frac{5-7}{35}=-\frac{2}{35}$

   ½ mark

5. $u=-\frac{35}{2}=-17.5\text{ m}$

   The object must be placed 17.5 m from the optical centre, on the left side (the side opposite to the image). Since $2f=10$ m, the object is beyond $2F_2$.

   ½ mark

6. (iii) Ray diagram (described): Draw the principal axis and the convex lens with $F_1$, $2F_1$ on the right and $F_2$, $2F_2$ on the left. Place object AB upright at 17.5 m on the left, beyond $2F_2$.

   • Ray 1 from A, parallel to the principal axis, refracts and passes through $F_1$.
   • Ray 2 from A passes through the optical centre O without deviation.
   1 mark

7. The two refracted rays meet at A′ below the axis, between $F_1$ and $2F_1$ (at 7 m). Draw A′B′ perpendicular to the axis. The image is real, inverted and diminished. Show arrows on all rays.

   1 mark

Answer: (i) Convex lens. (ii) $u=-17.5$ m: object 17.5 m on the left of the lens (beyond $2F$). (iii) Ray diagram with object beyond $2F_2$ and a real, inverted, diminished image between $F_1$ and $2F_1$.

1. (i) Given $R=40$ cm, so $f=\frac{R}{2}=20$ cm. For a concave mirror $f=-20$ cm. Object distance $u=-5$ cm and object height $h=10$ cm.

   ½ mark

2. Mirror formula: $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$, so $-\frac{1}{20}=\frac{1}{v}-\frac{1}{5}$

   ½ mark

3. $\frac{1}{v}=-\frac{1}{20}+\frac{1}{5}=\frac{-1+4}{20}=\frac{3}{20}$

   ½ mark

4. $v=\frac{20}{3}\approx +6.67$ cm. The positive sign means the image is 6.67 cm behind the mirror, so it is virtual and erect.

   ½ mark

5. (ii) Magnification $m=\frac{h^{\prime }}{h}=-\frac{v}{u}=-\frac{20/3}{-5}=\frac{4}{3}$.

   Image size $h^{\prime }=m\times h=\frac{4}{3}\times 10\approx 13.3$ cm. The image is erect and enlarged.

   1 mark

6. (iii) Ray diagram (described): Draw the concave mirror with pole P, focus F (20 cm) and centre of curvature C (40 cm) on the principal axis. Place the pencil AB upright at 5 cm, between P and F.

   • Ray 1 from A, parallel to the axis, reflects and passes through F.
   • Ray 2 from A, directed as if coming from C (along the normal), reflects back along itself.
   1 mark

7. The reflected rays diverge. Extend them backwards (dotted lines) behind the mirror; they meet at A′. Draw A′B′ behind the mirror. The image is virtual, erect and enlarged. Show arrows on all rays.

   1 mark

Answer: (i) $v=+6.67$ cm, i.e. 6.67 cm behind the mirror (virtual, erect). (ii) $m=4/3$, image size $\approx 13.3$ cm (enlarged, erect). (iii) Ray diagram with object between P and F and a virtual image behind the mirror.

1. Reasoning from the hints: C has two carbon atoms and is used as a preservative (in pickles), so C is ethanoic acid. A is oxidised to C, and A is denatured with methanol, benzene and pyridine to stop its misuse, so A is ethanol. Heating A with conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ dehydrates it to F, which takes up hydrogen, so F is an alkene, ethene.

2. (a) IUPAC names: A = ethanol ($\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$); F = ethene ($\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2$).

   1 mark

3. (b) $A\to C$ (oxidation):
   $$\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{alkaline }\mathrm{KMn}{\mathrm{O}}_4+\text{heat}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{O}}^{-}(\text{salt})\stackrel{{\mathrm{H}}^{+}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$$
   Ethanol is oxidised to ethanoate in the alkaline medium. Acidification gives ethanoic acid (C).

   1½ marks

4. (b) $A\to F$ (dehydration):
   $$\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{hot conc. }{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4,443\text{ K}}{\to }\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$$
   Conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ acts as a dehydrating agent.

   1½ marks

5. (b) Alternative, part 1: Step 2 (ethanol to ethanoic acid) is an oxidation reaction. Step 5 (ethene + ${\mathrm{H}}_2$ with Ni/Pd catalyst) is an addition reaction (hydrogenation).

   1 mark

6. (b) Alternative, part 2: The next homologue of ethanol is propanol, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$.

   • Step 2: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{alk. }\mathrm{KMn}{\mathrm{O}}_4,{\mathrm{H}}^{+}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}$ gives propanoic acid.
   1 mark
7. • Step 5: propanol dehydrates (step 4) to propene, $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$. Hydrogenation then gives propane:

   $$\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\stackrel{Ni/Pd}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$$

   1 mark

Answer: (a) A = ethanol, F = ethene. (b) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ (alk. $\mathrm{KMn}{\mathrm{O}}_4$, then ${\mathrm{H}}^{+}$); $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\to \mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$ (hot conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$). OR (b): step 2 oxidation, step 5 addition (hydrogenation). With propanol, step 2 gives propanoic acid; propanol gives propene, which is hydrogenated in step 5 to propane.

1. (a) Free ear lobe appears to be dominant. In the survey, $36+31=67$ of 100 students have free ear lobes and only $14+19=33$ have attached ones. The trait seen in the large majority is likely to be the dominant one. Also, Rahul has attached lobes although both his parents have free lobes, so attached is recessive.

   1 mark

2. (b) No, it is not linked with sex. Free ear lobes occur in both males (36 of 50) and females (31 of 50) in nearly equal proportion. In the family too, both the father and Nisha (female) have free lobes. A sex-linked trait would show a clear bias towards one sex.

   1 mark

3. (c) From the figures: Father has a free ear lobe, Mother has a free ear lobe, Rahul has an attached ear lobe and Nisha has a free ear lobe.

4. Rahul has attached lobes, so his genotype must be ff. He received one f from each parent. Both parents show the dominant free trait, so each must be Ff: Father Ff, Mother Ff.

   1 mark

5. Cross $\mathrm{Ff}\times \mathrm{Ff}$ gives FF : Ff : ff = 1 : 2 : 1. Nisha has free lobes, so she is Ff (or FF; one F is enough for the free trait). Rahul is ff.

   1 mark

6. (c) Alternative (Suresh): Both parents have attached ear lobes (recessive), so each is ff. They can pass on only the f allele. Cross: $\mathrm{ff}\times \mathrm{ff}$ gives only ff.

   1 mark

7. So Suresh is ff and Siya is ff. Both can only have attached ear lobes.

   1 mark

Answer: (a) Free ear lobe is dominant (found in the majority, 67 of 100). (b) No, it is found in both males and females in similar proportion. (c) Father Ff (free), Mother Ff (free), Rahul ff (attached), Nisha Ff (or FF) (free). OR (c): Father ff, Mother ff; Suresh ff and Siya ff, so both have attached ear lobes.

1. (i)(a) With switch A at Position 2, only the branch with the two 12 Ω lamps is connected to the battery. Only the two 12 Ω lamps light up; the 4 Ω lamps stay off.

   ½ mark

2. (i)(b) With switch A at Position 3, only the branch with the two 4 Ω lamps is connected. Only the two 4 Ω lamps light up; the 12 Ω lamps stay off.

   ½ mark

3. (ii) All the lamps are connected in parallel directly across the 12 V battery. In a parallel connection, every branch gets the full supply voltage. So the potential difference across each lamp = 12 V when lit.

   1 mark

4. (iii)(a) Each 12 Ω lamp has $V=12$ V across it. By Ohm's law, $I=\frac{V}{R}=\frac{12\text{ V}}{12\Omega }=1\text{ A}$.

   1 mark

5. (iii)(b) Each 4 Ω lamp has $V=12$ V across it. By Ohm's law, $I=\frac{V}{R}=\frac{12\text{ V}}{4\Omega }=3\text{ A}$.

   1 mark

6. (iv) (alternative to (iii)) Use $P=\frac{V^2}{R}$. The voltage is the same (12 V) for every lamp because they are in parallel.

   • 4 Ω lamp: $P=\frac{12\times 12}{4}=36\text{ W}$
   • 12 Ω lamp: $P=\frac{12\times 12}{12}=12\text{ W}$
   1½ marks

7. Since $36\text{ W}>12\text{ W}$, the 4.0 Ω lamps have the higher power.

   ½ mark

Answer: (i)(a) Only the 12 Ω lamps light; (i)(b) only the 4 Ω lamps light. (ii) 12 V across each lamp. (iii)(a) 1 A; (iii)(b) 3 A. OR (iv) 4 Ω lamp: 36 W, 12 Ω lamp: 12 W, so the 4 Ω lamp has higher power.