CBSE Class 10 Science: sample paper 2026-27

1. In photosynthesis, water is split by light energy: $2H_{2}O\to 4H^{+}+4e^{-}+O_2$. Water loses electrons and hydrogen, so it is oxidised and oxygen is released.

2. The hydrogen (with the electrons) is added to carbon dioxide to build glucose: $6C{O}_2+12H_{2}O\to C_6{H}_{12}{O}_6+6O_2+6H_{2}O$. Gaining hydrogen means $C{O}_2$ is reduced.

3. So carbon dioxide gains hydrogen while water loses electrons. The correct option is (A).

   1 mark

Answer: Option (A): Carbon dioxide gains hydrogen, while water loses electrons.

1. Lime water turns milky because of $C{O}_2$, which living organisms give out in respiration. Green plants in sunlight use up $C{O}_2$ in photosynthesis.

2. The setup that gives the most $C{O}_2$ is the one where respiration happens and no photosynthesis removes the $C{O}_2$. For example, an organism that only respires, or a plant kept in the dark.

3. According to the setups in the figure, this is setup R, so lime water is cloudiest in R. The correct option is (C).

   1 mark

Answer: Option (C): R

1. As the filtrate passes along the tubule, useful substances are taken back into the blood. These include glucose, amino acids, salts and much of the water.

2. Starch and glycogen are large polysaccharides. They are not present in the blood plasma, so they are never filtered or reabsorbed.

3. Therefore the correct pair is glucose and salts, option (A).

   1 mark

Answer: Option (A): Glucose and salts

1. Removing the anthers in the bud stage removes the flower's own pollen, so self-pollination cannot happen.

2. Covering the bud with a paper bag stops pollen from other flowers reaching the stigma, so cross-pollination cannot happen either.

3. Without pollination there is no fertilisation. Without fertilisation, ovules do not become seeds and the ovary does not become a fruit. The correct option is (C).

   1 mark

Answer: Option (C): They will not produce any seeds or fruits.

1. The vas deferens carries sperms from the testes. The oviduct carries the egg from the ovary and is the site of fertilisation.

2. Blocking these tubes does not stop gamete or hormone production. It only stops sperm and egg from reaching each other.

3. So the event prevented is the meeting and fusion of male and female gametes (fertilisation). The correct option is (C).

   1 mark

Answer: Option (C): The successful meeting and fusion of the male and the female gametes.

1. Gametes are formed by meiosis and carry half the chromosome number. Male and female gametes of the same species have the same number, so the female gamete has 17.

2. On fertilisation, the zygote gets $17+17=34$ chromosomes. All body cells of the adult plant come from the zygote, so they have 34.

3. The answer is 17, 34, option (B).

   1 mark

Answer: Option (B): 17, 34

1. The food chain is: green algae / photosynthetic bacteria (producers) $\to$ insect larvae (primary consumers) $\to$ fish.

2. The fish eats primary consumers, so it is at the third trophic level, the secondary consumer. The correct option is (C).

   1 mark

Answer: Option (C): Secondary Consumer

1. Reason: the amount of oxygen dissolved in water is much less than the amount of oxygen in air. R is true.

2. Assertion: because water holds so little oxygen, aquatic organisms breathe much faster than terrestrial organisms to get enough oxygen. A says the rate is lower, so A is false.

3. A is false but R is true, option (D).

   1 mark

Answer: Option (D): A is false but R is true.

1. By the 10 percent law, about 10% (not 1%) of the energy at one trophic level is passed on to the next. A is false.

2. On average, about 10% of the food eaten is turned into the organism's body and is available to the next level as chemical energy. R is true.

3. A is false but R is true, option (D).

   1 mark

Answer: Option (D): A is false but R is true.

1. (A) Food reaching the small intestine from the stomach is acidic. Bile from the liver makes it alkaline, so that pancreatic enzymes can act. Bile salts also emulsify fats, breaking large fat drops into small globules so enzymes act on them better.

   ½ mark

2. If the bile duct is blocked, the food stays acidic and fats are not emulsified. So digestion of fats will be improper or incomplete.

   ½ mark

3. (B) Herbivores eating grass need a longer small intestine so that the cellulose in their food has enough time to be digested. Carnivores such as tigers eat meat, which is easier to digest, so they have a shorter small intestine.

   1 mark

Answer: (A) Food is not made alkaline and fats are not emulsified, so fat digestion is improper; (B) herbivores (cellulose diet) have a longer small intestine, carnivores (meat diet) a shorter one.

1. Actions over which we have no thinking control are controlled by the mid-brain and hind-brain (for example, the medulla in the hind-brain).

   ½ mark

2. Such actions are called involuntary actions.

   ½ mark

3. Examples (any two): regulation of blood pressure, salivation, vomiting.

   1 mark

Answer: Mid-brain and hind-brain; involuntary actions, e.g. blood pressure regulation, salivation (also vomiting).

1. Plants have no nervous tissue. When the sensitive plant (touch-me-not) is touched, the information is passed from cell to cell by electrical-chemical means.

   1 mark

2. Plant cells respond by changing the amount of water in them. They swell or shrink, which changes their shape, and this makes the leaves fold and droop.

   1 mark

Answer: Touch is communicated from cell to cell by electrical-chemical signals; cells change shape by gaining or losing water (swelling or shrinking), so the leaves fold.

1. After implantation, the lining of the uterus thickens and gets a rich supply of blood vessels to nourish the growing embryo.

   1 mark

2. A special tissue, the placenta, forms a connection between the uterine wall and the embryo. It allows glucose and oxygen to pass to the embryo and wastes to be removed.

   1 mark

Answer: The uterine lining thickens and becomes richly supplied with blood; a placental connection develops between the uterus and the embryo.

1. When a nerve impulse from the brain reaches a muscle, the muscle cells move by changing their shape so that they shorten (contract).

   1 mark

2. Muscle cells contain special proteins that change both their shape and their arrangement in the cell in response to the electrical nerve impulse.

   1 mark

3. The new arrangement of these proteins gives the muscle cells a shorter form, and this shortening produces movement.

   1 mark

Answer: On receiving a nerve impulse, special proteins in muscle cells change their shape and arrangement, making the cells shorter (contract), which causes movement.

1. A. By the 10% law, each level gets only 10% of the energy of the level below. Going backwards, multiply by 10 at each step: large fish (4th) = 10 kJ $\to$ small fishes (3rd) = 100 kJ $\to$ zooplankton (2nd) = 1000 kJ $\to$ phytoplankton (1st) = 10000 kJ.

   1 mark

2. B. No. About 90% of the energy is lost at each step. After three more levels the energy would be $10\times \frac{1}{1000}=0.01$ kJ, which is far too little usable energy to support organisms. So food chains usually have only 3–4 trophic levels.

   1 mark

3. C. The large fish (top level) will have the maximum pesticide. Pesticides are non-biodegradable and accumulate at each trophic level. This is called biological magnification, so the highest level holds the greatest concentration.

   1 mark

Answer: A. 10000 kJ; B. No, very little usable energy would remain after 90% loss at each step; C. Large fish, due to biological magnification.

1. A. Cross: $VV\times vv$. All gametes are $V$ and $v$, so all F1 plants are $Vv$. Violet is dominant, so 100% violet in F1.

   ½ mark

2. F1 selfed: $Vv\times Vv\to 1VV:2Vv:1vv$. Violet plants $=\frac{3}{4}$, so 75% violet in F2, because a single dominant $V$ gives violet colour.

   ½ mark

3. B. Cross: $VVTT\times vvtt\to$ F1 all $VvTt$. Selfing F1 gives the dihybrid ratio 9 : 3 : 3 : 1. Tall white plants have genotype $vvTT$ or $vvTt$: $\frac{1}{4}(vv)\times \frac{3}{4}(T\_)=\frac{3}{16}$ = 18.75%. This is a combination of one recessive and one dominant trait.

   1 mark

4. C. (alternative to D, carries 2 marks if attempted instead): 0% in both F1 and F2. In pea plants these traits show complete dominance. A single $T$ or $V$ gives the same effect as two copies, so no blending or intermediate (medium height, light violet) plants are produced.

5. D. F1: all plants are $VvTt$, showing only dominant traits (violet, tall). So short plants with white flowers = 0%.

   1 mark

6. F2: short white plants must be double recessive $vvtt$: $\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$ = 6.25%.

   1 mark

Answer: A. F1 100%, F2 75% violet; B. 3/16 = 18.75%; C. 0% in F1 and F2 (no intermediate traits, complete dominance); D. F1 0%, F2 1/16 = 6.25%.

1. I. Birds and mammals have heart C, the four-chambered heart. Its right and left sides are completely separated. (In heart B the ventricle shows mixed blue-red blood, so B is not fully separated.)

   1 mark

2. The separation of the right and left sides keeps oxygenated and deoxygenated blood from mixing. This allows a highly efficient supply of oxygen to the body.

   1 mark

3. Birds and mammals have high energy needs, as they constantly use energy to maintain their body temperature. The efficient oxygen supply supports the high rate of respiration that releases this energy.

   1 mark

4. II. Fishes have heart A, the two-chambered heart.

   1 mark

5. The heart pumps blood to the gills, where it is oxygenated. It then passes directly to the rest of the body and returns to the heart. Thus blood goes only once through the heart in one cycle (single circulation).

   1 mark

Answer: I. Heart C (four-chambered): separation of oxygenated and deoxygenated blood gives an efficient O₂ supply for the high energy used to maintain body temperature. II. Heart A (two-chambered): heart → gills → body → heart; blood passes through the heart once per cycle.

1. I. Diagram A shows inhalation (ribs moving up and out, diaphragm moving down).

   1 mark

2. During inhalation the ribs are lifted (moved upwards and outwards) by the rib muscles.

   ½ mark

3. The diaphragm contracts and flattens.

   ½ mark

4. As a result the chest cavity becomes larger and the pressure in the lungs falls.

   ½ mark

5. Air is sucked into the lungs and fills the alveoli.

   ½ mark

6. II. Diffusion alone is far too slow. Humans use the respiratory pigment haemoglobin, present in red blood corpuscles, which has a very high affinity for oxygen and picks it up in the lungs. The heart then pumps this oxygen-rich blood quickly through the blood vessels to all parts of the body, including the toes.

   2 marks

Answer: I. Diagram A (per the scheme): ribs lift, diaphragm flattens, chest cavity enlarges, air is sucked in; II. Haemoglobin in RBCs binds oxygen strongly, and the heart pumps this oxygenated blood to all body parts.

1. The chameleon only changes the spacing between the crystals. The crystals themselves are not converted into any new substance.

2. A change in which no new substance is formed is a physical change. The colour changes only because the altered spacing changes which wavelength of light is reflected. The correct option is (C).

   1 mark

Answer: Option (C): physical change because the space between crystals is adjusted.

1. In the chlor-alkali process: $2NaCl(aq)+2H_{2}O(l)\to 2NaOH(aq)+C{l}_2(g)+H_2(g)$. Chlorine forms at the anode, hydrogen at the cathode, and sodium hydroxide near the cathode.

2. I: hydrogen at the cathode, used as fuel, is correct. III: NaOH near the cathode, used for making soap, is correct.

3. II is wrong: chlorine is formed at the anode, but margarine is made using hydrogen, not chlorine. IV is wrong: NaOH forms near the cathode, not the anode, and bleaching powder is made from chlorine. So only I and III are correct, option (B).

   1 mark

Answer: Option (B): I and III

1. Hard water contains $C{a}^{2+}$ and $M{g}^{2+}$ ions. These react with soap (sodium salts of long-chain carboxylic acids) to form insoluble calcium and magnesium salts, called scum, so lather does not form easily.

2. Detergents (sulphonate or ammonium salts) do not form insoluble salts with $C{a}^{2+}$ and $M{g}^{2+}$, so they lather well in hard water.

3. So the correct inference is option (A). Options C and D are wrong because $N{a}^{+}$ does not cause hardness and detergents do not react with the hardness ions. Option B is wrong because soap solutions are basic.

   1 mark

Answer: Option (A)

1. Formulae: Plaster of Paris $CaS{O}_4\cdot \frac{1}{2}{H}_{2}O$, washing soda $N{a}_{2}C{O}_3\cdot 10H_{2}O$, gypsum $CaS{O}_4\cdot 2H_{2}O$. All three contain water of crystallisation.

   ½ mark

2. Baking soda is $NaHC{O}_3$ and has no water of crystallisation. So the answer is (i), (ii) and (iv), option (D).

   ½ mark

Answer: (i), (ii) and (iv), option (D)

1. Isomers have the same molecular formula but different structures. In pair A, both compounds are $C_5{H}_{12}$: n-pentane and a branched isomer (isopentane).

   ½ mark

2. Pair B has different carbon counts (6 C and 5+1 = 6 C is possible, but the description shows different formulae). Pair C differs in H atoms ($C_5{H}_{12}$ and $C_5{H}_{10}$). Pair D shows the same compound drawn twice. So the answer is (A).

   ½ mark

Answer: Option (A)

1. The label shows that E304 and E307b are antioxidants. They stop the fats in the biscuits from being oxidised and turning rancid.

   ½ mark

2. Food packets are flushed with nitrogen, an unreactive gas, for the same purpose: it keeps oxygen out and prevents rancidity. Answer: (C).

   ½ mark

Answer: Nitrogen, option (C)

1. Aluminium is reactive, but on exposure to air it forms a thin, tough layer of $A{l}_2{O}_3$.

   ½ mark

2. This oxide layer is non-porous and protects the metal underneath from further corrosion. Answer: (C).

   ½ mark

Answer: Option (C)

1. Assertion A is true: baking soda solution is basic, and onion loses its smell in a base. Lemon juice is acidic, so the smell remains.

   ½ mark

2. Reason R is also true, since onion is an olfactory indicator. However, R only names the type of indicator and does not explain why the smell is lost in a base. So the answer is (B).

   ½ mark

Answer: Option (B)

1. Carbon atoms are very small, so the nucleus holds the shared electron pair strongly. Carbon therefore forms very strong and stable C–C bonds and can link with other carbon atoms to form long chains, branched chains and rings. This property is called catenation. Together with its tetravalency, this lets carbon form millions of compounds.

   1 mark

2. Silicon is larger in size, so its nucleus holds the shared electrons less strongly. Si–Si bonds are weaker and less stable, and silicon shows catenation only to a limited extent (chains of about 7–8 atoms, which are reactive). Hence silicon forms only a few compounds.

   1 mark

Answer: Carbon's small size gives strong, stable C–C bonds and extensive catenation. Larger silicon forms weak Si–Si bonds and shows little catenation, so it forms few compounds.

1. (i) X = ethanol, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ (structure ${\mathrm{H}}_3\mathrm{C}-\mathrm{C}{\mathrm{H}}_2-\mathrm{O}-\mathrm{H}$). In the electron dot structure:

   • the two carbon atoms share one electron pair between them;
   • the first C shares three electron pairs with three H atoms;
   • the second C shares two electron pairs with two H atoms and one pair with O;
   • O shares one pair with H and carries two lone pairs.

   Each C and O has an octet and each H has a duplet.

   1 mark

2. (ii) The next homologue differs by one $-\mathrm{C}{\mathrm{H}}_2-$ unit. It is propanol, ${\mathrm{C}}_3{\mathrm{H}}_7\mathrm{OH}$ ($\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$).

   1 mark

3. (iii) Esterification gives the sweet-smelling ester ethyl ethanoate:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\stackrel{conc.{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: (i) Electron dot structure of ethanol ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$ (O with two lone pairs); (ii) Propanol, ${\mathrm{C}}_3{\mathrm{H}}_7\mathrm{OH}$; (iii) $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\stackrel{conc.{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$

1. (i) Product A = ethene ($\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2$); product B = ethane ($\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_3$).

   1 mark

2. (ii) Step 1: hot conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ acts as a dehydrating agent:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\underset{443\mathrm{K}}{\overset{\mathrm{hot}conc.{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }}\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$

   1 mark

3. Step 2: hydrogenation (addition of hydrogen) in the presence of a nickel catalyst:
   $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\stackrel{\mathrm{Ni}}{\to }\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_3$

   1 mark

Answer: (i) A = ethene, B = ethane; (ii) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{conc.{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4,170^{\circ }\mathrm{C}}{\to }\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$; $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\stackrel{\mathrm{Ni}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_3$

1. A. A sports electrolyte drink contains dissolved salts that exist as free ions (such as $\mathrm{N}{\mathrm{a}}^{+}$, ${\mathrm{K}}^{+}$ and $\mathrm{C}{\mathrm{l}}^{-}$). These ions carry current, so the drink conducts electricity. Coconut oil is made of covalent molecules and has no ions, so it does not conduct.

   1 mark

2. B. Passing $\mathrm{C}{\mathrm{O}}_2$ forms insoluble calcium carbonate, which makes the lime water milky:
   $\mathrm{Ca}(\mathrm{OH}{)}_2(\mathrm{aq})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})\to \mathrm{CaC}{\mathrm{O}}_3(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

   With excess $\mathrm{C}{\mathrm{O}}_2$, the carbonate changes into soluble calcium hydrogen carbonate, so the milkiness disappears:
   $\mathrm{CaC}{\mathrm{O}}_3(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})\to \mathrm{Ca}(\mathrm{HC}{\mathrm{O}}_3{)}_2(\mathrm{aq})$

   1 mark

3. C. Rusting needs both oxygen and moisture. Air near the sea is very humid and carries salt, which speeds up rusting. Air in a dry hilly region has little moisture, so rusting there is much slower.

   1 mark

Answer: A. The drink has free ions; coconut oil has none. B. Excess $\mathrm{C}{\mathrm{O}}_2$ converts insoluble $\mathrm{CaC}{\mathrm{O}}_3$ into soluble $\mathrm{Ca}(\mathrm{HC}{\mathrm{O}}_3{)}_2$. C. High humidity (and salt) near the sea speeds up rusting; dry hill air has little moisture.

1. A. The candle on the silver rod drops first (about 10 s). Heat reached its end fastest, so silver is the best conductor of heat.

   1 mark

2. B. Gold conducts better than aluminium (30 s) but worse than copper (15 s). Its time must lie between 15 s and 30 s, so the answer is (ii) 20 s.

   1 mark

3. C. Yes. The time taken for every candle to drop will increase, but the order in which the candles fall stays the same (silver, copper, aluminium, zinc, iron).

   1 mark

4. C (reason). In conduction, heat passes from particle to particle by vibration. In a longer rod, heat must travel a greater distance to reach the candle, so it takes more time.

   1 mark

5. D. No, Mayank's inference will not be correct. For a fair test, only the material of the rod should change; length and thickness must be the same for all rods.

   1 mark

6. D (reason). The time also depends on thickness, because a thicker rod takes more time for its far end to heat up. The candle times would then reflect both the metal and the thickness, so the metals cannot be compared correctly.

   1 mark

Answer: A. Silver; B. (ii) 20 s; C. Yes, every candle takes longer (order unchanged) because heat travels a longer distance; D. No, thickness also affects the time, so the comparison is not fair.

1. I. I agree fully with Drishti. Alisha and Bhushan are only partially correct, because oxidation and reduction always occur together (redox).

   1 mark

2. In ${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2N{\mathrm{H}}_3$, nitrogen gains hydrogen and is reduced, while hydrogen is oxidised. So ammonia formation is redox, not just reduction (Alisha saw only half of it).

   In rusting, $4Fe+3{\mathrm{O}}_2+\mathrm{x}{\mathrm{H}}_2\mathrm{O}\to 2F{\mathrm{e}}_2{\mathrm{O}}_3\cdot \mathrm{x}{\mathrm{H}}_2\mathrm{O}$, iron gains oxygen and is oxidised, while oxygen is reduced. So rusting is also redox, not just oxidation (Bhushan saw only half of it).

   2 marks

3. II (i). Errors: silver has valency 1, so the product is AgCl, not $\mathrm{AgC}{\mathrm{l}}_2$. Magnesium has valency 2, so the nitrate is $\mathrm{Mg}(\mathrm{N}{\mathrm{O}}_3{)}_2$, not $\mathrm{MgN}{\mathrm{O}}_3$. The equation must then be balanced.

   Corrected: $\mathrm{MgC}{\mathrm{l}}_2+2AgN{\mathrm{O}}_3\to 2AgCl+\mathrm{Mg}(\mathrm{N}{\mathrm{O}}_3{)}_2$

   1 mark

4. II (ii). Errors: aluminium has valency 3 and sulphate has valency 2, so the salt is $\mathrm{A}{\mathrm{l}}_2(\mathrm{S}{\mathrm{O}}_4{)}_3$, not $\mathrm{A}{\mathrm{l}}_2\mathrm{S}{\mathrm{O}}_4$. A metal reacting with a dilute acid gives hydrogen gas, not water. The equation must also be balanced.

   Corrected: $2Al+3{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to \mathrm{A}{\mathrm{l}}_2(\mathrm{S}{\mathrm{O}}_4{)}_3+3{\mathrm{H}}_2$

   1 mark

Answer: I. Drishti (Alisha and Bhushan are partly correct), as both reactions are redox. II. (i) $\mathrm{MgC}{\mathrm{l}}_2+2AgN{\mathrm{O}}_3\to 2AgCl+\mathrm{Mg}(\mathrm{N}{\mathrm{O}}_3{)}_2$; (ii) $2Al+3{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to \mathrm{A}{\mathrm{l}}_2(\mathrm{S}{\mathrm{O}}_4{)}_3+3{\mathrm{H}}_2$

1. I. Neither has understood it completely; each is only partially correct. In a decomposition reaction, a single compound breaks down into two or more simpler substances. These products may be elements or compounds.

   1 mark

2. Daljeet's example is valid ($\mathrm{MgC}{\mathrm{O}}_3(\mathrm{s})\stackrel{\mathrm{heat}}{\to }\mathrm{MgO}(\mathrm{s})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$), but products need not always be compounds.

   Avik's example is also valid ($2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\stackrel{\mathrm{electricity}}{\to }2{\mathrm{H}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})$). However, products need not always be elements, as Daljeet's example shows. So each one's rule is too narrow.

   2 marks

3. II (i). The observation is incorrect. Iron is less reactive than magnesium, so it cannot displace Mg from $\mathrm{MgS}{\mathrm{O}}_4$. No reaction occurs, and the solution stays colourless.

   1 mark

4. II (ii). The observation is incorrect. The gas evolved is hydrogen:
   $2Al+3{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to \mathrm{A}{\mathrm{l}}_2(\mathrm{S}{\mathrm{O}}_4{)}_3+3{\mathrm{H}}_2$

   Only a few bubbles are seen, because the reaction is slow at first owing to the protective oxide layer on aluminium. Brisk effervescence is typical of $\mathrm{C}{\mathrm{O}}_2$ released from carbonates.

   1 mark

Answer: I. Both are only partially correct: in decomposition, a single compound breaks into simpler substances that may be elements or compounds. II. (i) Incorrect, as Fe is less reactive than Mg, so no reaction occurs; (ii) Incorrect, as ${\mathrm{H}}_2$ is evolved with only a few bubbles, not brisk effervescence.

1. Sunlight comes from a very distant object, so its parallel rays converge at the principal focus of the convex lens.

   ½ mark

2. The smallest, brightest spot on the paper marks the focus. The lens-to-paper distance is then the focal length. Answer: (B).

   ½ mark

Answer: Option (B)

1. Use $P=VI$, so $I=\frac{P}{V}$.

   ½ mark

2. $I=\frac{1000}{250}=4.0$ A. Answer: (C).

   ½ mark

Answer: 4.0 A, option (C)

1. Assertion A is true: a coin in water appears raised, so its apparent depth is less than its real depth.

   ½ mark

2. Reason R is true: light from the coin bends away from the normal as it passes from water (denser) into air (rarer). The refracted rays seem to come from a higher point, which is exactly why the coin looks raised. So R explains A. Answer: (A).

   ½ mark

Answer: Option (A)

1. A. Snell's law: for light of a given colour and a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.
   $\frac{\sin i}{\sin r}=\text{constant}$

   1 mark

2. B. The refractive index of medium 2 with respect to medium 1 is the ratio of the speed of light in medium 1 to the speed of light in medium 2:
   $n_{21}=\frac{v_1}{v_2}$
   Here $v_1$ and $v_2$ are the speeds of light in media 1 and 2.

   1 mark

Answer: A. $\frac{\sin i}{\sin r}$ = constant for a given colour and pair of media; B. $n_{21}=\frac{v_1}{v_2}$

1. I. Joule's law of heating: the heat produced in a resistor is directly proportional to

   • the square of the current for a given resistance,
   • the resistance for a given current, and
   • the time for which the current flows.

   So $H=I^{2}Rt$.

   1 mark

2. II. Keeping the current and time the same, the new heat is
   $H^{\prime }=I^2(2R)t=2I^{2}Rt=2H$
   So the heat produced doubles.

   1 mark

Answer: I. $H=I^{2}Rt$; II. The heat produced doubles ($H^{\prime }=2H$) for the same current and time.

1. I. Resistance of a wire is $R=\rho \frac{L}{A}$. The resistivity $\rho$ of a material is the resistance of a wire of that material having unit length and unit area of cross-section. So $\rho =\frac{RA}{L}$, and its SI unit is $\Omega \text{m}$.

   1 mark

2. II. Resistivity depends only on the nature of the material (and its temperature), not on its length or thickness.

   Example: two copper wires of different lengths and thicknesses have different resistances. But when $\rho =RA/L$ is calculated for each, both give the same value (about $1.6\times 10^{-8}\Omega \text{m}$). Hence resistivity is an intrinsic property of copper.

   1 mark

Answer: I. $\rho =RA/L$: the resistance of a wire of unit length and unit cross-sectional area; II. Copper wires of any size give the same $\rho$, so it depends only on the material.

1. A. Type of lens: The image is virtual, erect and diminished (1/3rd the size of the object), and it is formed between $F_1$ and $O$ on the same side as the object. Only a concave (diverging) lens forms such an image for every object position. Object distance $u=-24$ cm (the object is on the left of the lens, against the direction of incident light).

   1 mark

2. B. Image distance: Magnification of a lens is $m=\frac{h_i}{h_o}=\frac{v}{u}$. The image is erect and one-third the size of the object, so $m=+\frac{1}{3}$.

3. $v=m\times u=\frac{1}{3}\times (-24)=-8$ cm. The image is 8 cm from the lens, on the same side as the object. It is virtual.

   1 mark

4. C. Focal length: Lens formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$.

5. $\frac{1}{f}=\frac{1}{-8}-\frac{1}{-24}=-\frac{3}{24}+\frac{1}{24}=-\frac{2}{24}=-\frac{1}{12}$, so $f=-12$ cm. The negative sign confirms a diverging (concave) lens.

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Answer: A. Concave (diverging) lens, u = −24 cm; B. v = −8 cm (virtual, same side as object); C. f = −12 cm

1. A. Right-Hand Thumb Rule: Imagine holding a straight current-carrying conductor in your right hand so that the stretched thumb points in the direction of the (conventional) current. The fingers then wrap around the conductor in the direction of the magnetic field lines.

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2. How it helps: The field lines around a straight conductor are concentric circles. The rule tells us which way they circulate. Point the thumb along the current and read the direction of the curled fingers. For example, for current flowing upward the field is anticlockwise when viewed from above. If the current is reversed, the field direction also reverses.

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3. B. Variation with distance: The magnetic field due to a long straight conductor is strongest close to the wire and becomes weaker farther away. The concentric circles of field lines become more widely spaced. The magnitude varies as $B\propto \frac{I}{r}$, so for a given current the field is inversely proportional to the distance $r$ from the conductor.

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Answer: A. Thumb along current, curled fingers give the direction of the circular field lines; B. Field decreases with distance, B ∝ I/r

1. A. In the left triangle, the two 1 Ω resistors are in series: $1+1=2\Omega$. This 2 Ω path is in parallel with the 2 Ω resistor joining A and B: $\frac{1}{R_{AB}}=\frac{1}{2}+\frac{1}{2}=1$, so $R_{AB}=1\Omega$.

   ½ mark

2. In the right-hand portion, the two 3 Ω resistors are in parallel: $\frac{1}{R^{\prime }}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$, so $R^{\prime }=1.5\Omega$.

   ½ mark

3. The 5 Ω resistor, $R_{AB}$ and $R^{\prime }$ are in series: $R_s=5+1+1.5=7.5\Omega$.

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4. By Ohm's law, $I=\frac{V}{R_s}=\frac{15\text{V}}{7.5\Omega }=2$ A. The total current supplied by the battery is 2 A.

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5. B. The whole 2 A passes through the combination between A and B, whose resistance is $R_{AB}=1\Omega$. So $V_{AB}=I\times R_{AB}=2\times 1=2$ V.

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Answer: A. I = 2 A (R_total = 7.5 Ω); B. V_AB = 2 V

1. A. Soft iron is a magnetic material. Placed inside the solenoid, it becomes strongly magnetised and concentrates the field lines, so the magnetic field becomes much stronger. This gives a stronger pull on the iron bolt. Soft iron also loses its magnetism as soon as the current is switched off, so the spring can push the bolt back and lock the door.

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2. B. Use the right-hand rule for a solenoid. Curl the fingers of the right hand in the direction of the current in the turns, and the thumb points towards the North pole. Seen from an end, clockwise current marks a South pole and anticlockwise current a North pole. To check, bring the known North pole of a bar magnet near one end of the solenoid. Repulsion shows that end is a North pole.

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3. C. The magnetic field inside a solenoid is directly proportional to the current through it ($B\propto I$). Increasing the current from 1 A to 2 A therefore doubles the strength of the magnetic field, other factors being unchanged.

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4. C (contd.) A stronger field gives a stronger magnetic pull on the iron bolt. The bolt is pulled in more firmly and quickly against the spring, so the door unlocks more effectively and reliably.

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5. D (alternative to C). With the current fixed, the field can be increased by: (i) increasing the number of turns of the coil, (ii) winding the same turns over a shorter length so there are more turns per unit length, or (iii) inserting a soft iron core inside the coil. Any two of these.

   2 marks

Answer: A. Iron core concentrates and strengthens the field (and demagnetises when switched off); B. Right-hand rule for a solenoid / repulsion with a known N pole; C. Field doubles (B ∝ I), stronger pull, door unlocks more effectively; D. More turns, shorter coil (more turns per length), soft iron core (any two)

1. I. Accommodation: The ability of the eye lens to change its focal length, through the ciliary muscles changing its curvature, so that both near and distant objects are focused sharply on the retina.

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2. II. Near point: The nearest point at which an object can be seen clearly without strain. For a normal adult eye it is about 25 cm. It signifies the least distance of distinct vision, the limit up to which the eye lens can increase its converging power (minimum focal length).

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3. III. A person who cannot see distant objects clearly has myopia (near-sightedness). It is corrected with a concave (diverging) lens of suitable power.

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4. IV. Diagram (correction of myopia): Draw the eye with its lens and retina, and place a concave lens in front of the eye. Draw parallel rays from a distant object (at infinity) falling on the concave lens. Show them diverging after the lens so that, traced back as dashed lines, they appear to come from the far point O of the myopic eye.

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5. Show the eye lens bending these diverging rays so that they meet exactly on the retina. Label: concave lens, eye lens, retina, far point O, object at infinity.

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Answer: I. Ability of the eye lens to adjust its focal length; II. About 25 cm, the least distance of distinct vision; III. Concave (diverging) lens, for myopia; IV. Diagram: concave lens diverges parallel rays as if from the far point, so the image forms on the retina

1. I. Dispersion: The splitting of white light into its component colours (VIBGYOR) when it passes through a transparent medium such as a glass prism. It happens because light of different colours (wavelengths) bends through different angles.

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2. II. Red light deviates the least, because it has the longest wavelength and bends least in glass. Violet deviates the most.

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3. III. Outer space has no atmosphere, so there are no air molecules or fine particles to scatter sunlight. No scattered light reaches the astronaut's eyes from the surroundings, so the sky appears dark.

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4. IV. Diagram: Draw a triangular glass prism. Show a ray of white light striking one face, bending towards the base inside the prism, and emerging from the second face as a band of colours.

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5. Label the spectrum in order: R, O, Y, G, B, I, V, with red at the top (least deviated) and violet at the bottom (most deviated). Also label the glass prism, the white light beam and the spectrum (on a screen).

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Answer: I. Splitting of white light into its colours; II. Red; III. No atmosphere, so no scattering of sunlight; IV. Labelled prism diagram with red least and violet most deviated