CBSE Class 10 Science Sample Paper 2026 27 with Solutions PDF Download

Section - A

1. During photosynthesis, what happens to carbon dioxide and water as glucose is formed? [1 Mark]
(A) Carbon dioxide gains hydrogen, while water loses electrons.
(B) Carbon dioxide loses oxygen, while water gains oxygen.
(C) Carbon dioxide is oxidised, while water is reduced.
(D) Both carbon dioxide and water undergo reduction.

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

Teacher's Note:
a) In photosynthesis, carbon dioxide is reduced to glucose by adding hydrogen.
b) Water is split (oxidised) and releases oxygen gas, hydrogen and electrons.

2. Lime water turns cloudy in the presence of a gas, which is a by-product of respiration. Shown below are four setups kept in sunlight for 24 hours. In which setup is lime water expected to be the cloudiest? [1 Mark]
(A) P
(B) Q
(C) R
(D) S

[Figure: Four closed tanks, each with a dish of limewater and a pot. Set P: glass tank with limewater and an empty pot (no plant). Set Q: glass tank with a plant in a pot and limewater. Set R: tank coated with black paint, with a plant in a pot and limewater. Set S: tank coated with black paint, with an empty pot and limewater.]

Answer: (C) R

Teacher's Note:
a) In Set R the black paint blocks light, so the plant only respires and releases carbon dioxide.
b) In Set Q the plant in light uses carbon dioxide for photosynthesis, so the limewater stays clearer.

For Visually Impaired Candidates (in lieu of Q. 2)

Which of the following statement is incorrect concerning the experiment to show that sunlight is necessary for photosynthesis? [1 Mark]
(A) The leaf should be destarched before conducting the experiment.
(B) The leaf kept in darkness will show the presence of starch.
(C) The leaf exposed to sunlight will show the presence of starch.
(D) Starch is the end product of photosynthesis.

Answer: (B) The leaf kept in darkness will show the presence of starch.

Teacher's Note:
a) A leaf kept in darkness cannot photosynthesise, so it shows no starch with iodine.
b) Read the question carefully: it asks for the incorrect statement.

3. During urine formation, tubular reabsorption and secretion take place in the kidneys. Which of the following sets correctly indicates both the substances that are reabsorbed in the kidneys? [1 Mark]
(A) Glucose and salts
(B) Glucose and starch
(C) Glycogen and salts
(D) Glycogen and starch

Answer: (A) Glucose and salts

Teacher's Note:
a) Useful substances like glucose, amino acids, salts and water are reabsorbed in the tubules.
b) Starch and glycogen are not present in the filtrate, so they cannot be reabsorbed.

4. In a controlled experiment, the anthers of a mustard plant's flowers were carefully removed in the bud stage before they matured. The bud was covered with a paper bag.
What will be the most likely result for these flowers? [1 Mark]
(A) They will still produce seeds and fruit through self-pollination.
(B) They will attract more insects and produce larger flowers but no fruit.
(C) They will not produce any seeds or fruits.
(D) They will produce seeds and fruits that are genetically identical to the parent.

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

Teacher's Note:
a) Without anthers there is no pollen, and the paper bag stops pollen from other flowers.
b) No pollination means no fertilisation, so no seeds or fruits are formed.

5. The Vas deferens and the Oviduct are both targets for surgical contraception (Vasectomy and Tubectomy respectively). Analysing the process of reproduction, what common biological event is prevented by blocking both these structures? [1 Mark]
(A) The production of gametes (sperm and egg).
(B) The implantation of the fertilized egg into the uterine wall.
(C) The successful meeting and fusion of the male and the female gametes.
(D) The release of hormones - testosterone and estrogen.

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

Teacher's Note:
a) Blocking the vas deferens stops sperm transfer; blocking the oviduct stops the egg reaching the sperm.
b) Gametes and hormones are still produced, so options (A) and (D) are wrong.

6. The male gamete of an apple plant has seventeen chromosomes. What is the number of chromosomes in the female gamete and in the cells of the adult plant respectively? [1 Mark]
(A) 17,17
(B) 17,34
(C) 34,34
(D) 34,51

Answer: (B) 17,34

Teacher's Note:
a) Gametes have half the number of chromosomes, so the female gamete also has 17.
b) Fusion of two gametes restores the full number: 17 + 17 = 34 in body cells.

7. A fish living in a pond ecosystem feeds on small aquatic insects (larvae). These insect larvae, in turn, feed on green algae and small photosynthetic bacteria. In this specific food chain, what is the trophic level classification of fish? [1 Mark]
(A) Producer
(B) Primary Consumer
(C) Secondary Consumer
(D) Tertiary Consumer

Answer: (C) Secondary Consumer

Teacher's Note:
a) Chain: algae and bacteria (producers) - insect larvae (primary consumers) - fish (secondary consumer).
b) Count the levels from the producer to avoid mistakes.

The following two questions consist of two statements - Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below:
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

8. Assertion (A): The rate of breathing in aquatic organisms is much lower than in the terrestrial organisms.
Reason (R): The amount of oxygen dissolved in water is less as compared to the amount of oxygen in air. [1 Mark]
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

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

Teacher's Note:
a) Water has less dissolved oxygen than air, so aquatic organisms breathe much faster, not slower.
b) So the Assertion is false while the Reason is a true fact.

9. Assertion (A): In an ecosystem, 1% of energy is available for transfer from one trophic level to the next.
Reason (R): An average of 10% of the food eaten by organisms is available for flow in the form of chemical energy. [1 Mark]
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

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

Teacher's Note:
a) According to the 10 percent law, about 10% of energy passes to the next trophic level, not 1%.
b) The 1% figure refers to the sunlight captured by producers, a common point of confusion.

10. A. Due to narrowing of the bile duct, the bile secretion from the liver to the small intestine was obstructed. How will the process of digestion be affected?
B. How does the length of the intestine depend on the nature of food habit in different animals? [2 Marks]

Answer:
A. Bile from the liver makes the food alkaline so that pancreatic enzymes can act. Bile salts also break fats into smaller globules, which increases the efficiency of enzyme action. If the bile duct is blocked, fats will not be digested properly.
B. Herbivores eating grass need a longer small intestine so that cellulose can be digested. Meat is easier to digest, so carnivores like tigers have a shorter small intestine.

Teacher's Note:
a) For A, write both roles of bile: making the medium alkaline and emulsifying fats.
b) For B, link the length of the intestine with cellulose digestion in herbivores.

11. Students to attempt either option A or B. [2 Marks]

A. Which parts of the brain perform certain actions for which we do not have any thinking control? What are such actions called, give two examples. [2 Marks]

Answer:
1. Actions over which we have no thinking control are controlled by the mid-brain and hind-brain.
2. Such actions are called involuntary actions. Examples: blood pressure and salivation.

Teacher's Note:
a) Name both parts, mid-brain and hind-brain, for full credit.
b) Vomiting is another accepted example of an involuntary action.

OR

B. How is the information of touch in the 'sensitive plant' controlled and coordinated? [2 Marks]

Answer:
1. The information of touch is passed from cell to cell by electrical-chemical means.
2. Plant cells change their shape by changing the amount of water in them. They swell or shrink, and this change in shape causes the leaves to move.

Teacher's Note:
a) Plants have no nerves or muscles; stress the electrical-chemical signal and the change in water content.
b) Keywords: cell to cell communication, swelling and shrinking of cells.

12. What changes are observed in the uterus subsequent to implantation of the young embryo in a human female? [2 Marks]

Answer:
1. The lining of the uterus thickens and is richly supplied with blood vessels to nourish the embryo.
2. A placental connection develops between the uterus and the embryo.

Teacher's Note:
a) Two value points: thick, blood-rich uterine lining and formation of the placenta.
b) The placenta passes nutrients and oxygen from the mother's blood to the embryo.

13. How do muscles move at the cellular level and respond when processed messages are received from the brain? [3 Marks]

Answer:
1. When a nerve impulse reaches the muscle, the muscle cells move by changing their shape so that they shorten.
2. Muscle cells have special proteins that change both their shape and their arrangement in the cell in response to nervous electrical impulses.
3. The new arrangement of these proteins gives the muscle cells a shorter form, and this causes movement.

Teacher's Note:
a) Keywords examiners look for: nerve impulse, special proteins, change in shape and arrangement, shortening.
b) Do not write only "muscles contract"; explain what happens inside the cells.

14. Study the food chain given below:
A. In the given food chain, the amount of energy available at the fourth trophic level is 10 kJ, what will be the energy available at the producer level?
B. Is it possible to have three more trophic levels in the above food chain? Justify.
C. Which of these trophic levels will have the maximum amount of pesticide, if it had washed down in the water body from agricultural fields. Why? [3 Marks]

[Figure: Food chain: Phytoplankton - zooplankton - small fishes - large fish, joined by arrows]

Answer:
A. Only 10% of energy passes to the next level, so energy increases 10 times at each step going back:
Large fish (4th level) = 10 kJ
Small fishes (3rd level) = \( 10 \times 10 = 100 \) kJ
Zooplankton (2nd level) = \( 100 \times 10 = 1000 \) kJ
Phytoplankton (producer) = \( 1000 \times 10 = 10000 \) kJ
B. No. The loss of energy at each step is so large that very little usable energy will be left for three more levels.
C. Large fish, because of biological magnification of the pesticide.

Teacher's Note:
a) Multiply by 10 for every step you go down the food chain towards the producer.
b) Biological magnification means the pesticide concentration is highest at the top trophic level.

15. In order to study patterns of inheritance, Neha had performed typical Mendelian experiments with pea plants in the field. In experiment one, she crossed a pea plant having white flowers (vv) with another pea plant bearing violet flowers (VV). In another experiment, in addition to the colour of the flower she also focussed on the height of the pea plant. This time she crossed a pea plant having violet flowers and tall height (VVTT) with another pea plant having white flowers and short height (vvtt).
What percentage is expected for the following traits as a result of these experiments? Give a reason. [4 Marks]

A. Violet flowers in F1 and F2 generation in experiment one? [1 Mark]

Answer: In F₁, all plants (100%) will have violet flowers (Vv). In F₂, 75% (\( \frac{3}{4} \)) plants will have violet flowers (\( \frac{1}{4} \) VV and \( \frac{2}{4} \) Vv), because violet flower colour is the dominant trait in pea plants.

Teacher's Note:
a) Give both values, 100% in F₁ and 75% in F₂, with the reason (dominance).
b) In F₂ the ratio is VV : Vv : vv = 1 : 2 : 1.

B. Tall height plants with white flowers in F2 generation in experiment two? [1 Mark]

Answer: \( \frac{3}{16} \) or 18.75% of the plants will be tall with white flowers (vvTT or vvTt). This is a combination of one dominant trait (tall) and one recessive trait (white).

Teacher's Note:
a) The dihybrid F₂ ratio is 9 : 3 : 3 : 1; tall-white is one of the "3" groups.
b) \( \frac{3}{16} \times 100 = 18.75\% \).

Attempt either sub-part C or D.

C. Percentage of medium height plants with light violet flowers in F1 and F2 generations in experiment two? Give a reason. [2 Marks]

Answer:
1. Zero percent in both F₁ and F₂.
2. In pea plants, no half-way characteristics are produced. Even a single copy of T or V gives the same result as two copies of T or V. So no plants with medium height and light violet flowers are produced.

Teacher's Note:
a) Mendel's traits show complete dominance, so there is no blending of traits.
b) Write "0%" clearly for both generations.

OR

D. Percentage of short height plants with white flowers in F1 and F2 generations in experiment two? [2 Marks]

Answer:
1. F₁: zero percent. All F₁ plants (VvTt) show only dominant traits, so no plant is short with white flowers.
2. F₂: \( \frac{1}{16} \) or 6.25% plants will be short with white flowers (vvtt), as both traits are recessive.

Teacher's Note:
a) Recessive traits show only when both alleles are recessive (vvtt).
b) \( \frac{1}{16} \times 100 = 6.25\% \).

16. Attempt either option A or B [5 Marks]

A. Observe the structure of hearts A, B and C shown above and answer the following questions
I. Which of the hearts depicted above is present in both birds and Mammals and how are they able to maintain their body temperature? Give reason for your answer.
II. Which of the hearts depicted above is present in fishes, how does circulation of blood take place in them? [5 Marks]

[Figure: Three hearts labelled A, B and C. A: a two-chambered heart shown fully in blue (deoxygenated blood). B: a three-chambered heart in which red and blue blood mix in the single ventricle. C: a four-chambered heart with the right side blue and the left side red, kept fully separate.]

Answer:
I. C - Four chambered heart.
The right side and the left side of the heart are separated. This keeps oxygenated and deoxygenated blood from mixing. Such separation gives a highly efficient supply of oxygen to the body. This is useful in animals with high energy needs, such as birds and mammals, which constantly use energy to maintain their body temperature.
II. A - Two chambered heart.
The blood is pumped to the gills, gets oxygenated there and passes directly to the rest of the body. So blood goes only once through the heart in a fish during one cycle of passage through the body (single circulation).

Teacher's Note:
a) Link "no mixing of blood" with "efficient oxygen supply" and "high energy for body temperature".
b) For fishes, the key phrase is: blood passes through the heart only once in one cycle.

OR

B.
I. Observe both the diagrams A and B. Which diagram indicates inhalation? Describe how it occurs.
II. If diffusion were to move oxygen in our body, it is estimated that it would take 3 years for a molecule of oxygen to get to our toes from our lungs. Then, how is oxygen delivered to all parts of our body? [5 Marks]

[Figure: Two diagrams of the human chest labelled A and B, with the diaphragm labelled in both. In A, arrows point upwards and outwards from the ribs and the diaphragm is flat. In B, arrows point inwards and the diaphragm is dome-shaped (arched upwards).]

Answer:
I. Diagram A indicates inhalation. It occurs as follows:
1. The ribs get lifted.
2. The diaphragm is flattened.
3. The chest cavity becomes larger.
4. Air is sucked into the lungs.
II. In human beings, the respiratory pigment haemoglobin has a very high affinity for oxygen. This pigment is present in the red blood corpuscles. It picks up oxygen in the lungs, and this oxygen-rich blood is pumped by the heart to all parts of the body.

Teacher's Note:
a) Each of the four steps of inhalation carries half a mark; write all four.
b) For II, the keywords are haemoglobin, high affinity for oxygen, RBCs and pumping by the heart.

For Visually Impaired Candidates (in lieu of Q. 16)

A.
I. How many chambers are present in the hearts of birds and mammals and how are they able to maintain their body temperature? Give reason for your answer
II. How many chambers are present in the heart of fishes, how does circulation of blood take place in them? [5 Marks]

Answer:
I. Four chambers are present.
The right side and the left side of the heart are separated. This keeps oxygenated and deoxygenated blood from mixing and gives a highly efficient supply of oxygen to the body. This is useful in animals with high energy needs, such as birds and mammals, which constantly use energy to maintain their body temperature.
II. Two chambers are present.
The blood is pumped to the gills, gets oxygenated there and passes directly to the rest of the body. So blood goes only once through the heart in a fish during one cycle of passage through the body.

Teacher's Note:
a) State the number of chambers first, then give the reason.
b) Separation of oxygenated and deoxygenated blood is the main value point for birds and mammals.

OR

B.
I. Describe the process of inhalation and also mention how are the lungs designed in human beings to maximise the area for exchange of gases?
II. If diffusion were to move oxygen in our body, it is estimated that it would take 3 years for a molecule of oxygen to get to our toes from our lungs then, how is oxygen delivered to all parts of our body? [5 Marks]

Answer:
I. Process of inhalation:
1. The ribs get lifted.
2. The diaphragm is flattened.
3. The chest cavity becomes larger.
4. Air is sucked into the lungs.
Design of lungs: Exchange of gases takes place on the surface provided by the alveoli. The walls of the alveoli are supplied with an extensive network of blood vessels. So the lungs have a large number of alveoli, richly supplied with blood, to maximise the area for gaseous exchange.
II. In human beings, the respiratory pigment haemoglobin has a very high affinity for oxygen. This pigment is present in the red blood corpuscles. This oxygen-rich blood is pumped by the heart to all parts of the body.

Teacher's Note:
a) Alveoli and a rich network of blood vessels are the two keywords for surface area.
b) Haemoglobin in RBCs carries oxygen much faster than diffusion could.

Section - B

17. Chameleons contain crystals that reflect light and produce colour. Adjusting the spacing between these crystals, they can alter the wavelengths of light reflected, thus changing their skin colour. Based on the above information, we can infer that the change in colour by chameleons is a: [1 Mark]
(A) chemical change because change in colour takes place.
(B) chemical change because wavelength of reflected light changes.
(C) physical change because the space between crystals is adjusted.
(D) physical change because light is reflected by crystals.

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

Teacher's Note:
a) No new substance is formed; only the spacing between crystals changes.
b) This changes the wavelength of reflected light, so the colour change is physical.

18. Rishita summarizes the chlor-alkali process in the following table.
Option | Product | Formed | Use
I | Hydrogen | At cathode | Used as a fuel
II | Chlorine | At anode | Used to make margarines
III | Sodium Hydroxide | Near cathode | To make soap
IV | Sodium Hydroxide | Near anode | To make bleaching powder
Identify the options that are correctly matched. [1 Mark]
(A) I, II and III
(B) I and III
(C) II and IV
(D) I, II, III and IV

Answer: (B) I and III

Teacher's Note:
a) II is wrong: chlorine is produced at the anode, but it is not used to make margarine.
b) IV is wrong: sodium hydroxide is formed near the cathode, not the anode.

19. Manjeet wants to wash clothes, but the water available is hard. He compares the action of soap and detergent in hard water, to decide which one to use for washing his clothes. He observes:
Soap forms scum in hard water.
Detergent produces lather easily in hard water.
Which of the following inferences is correct? [1 Mark]
(A) Ca²⁺ and Mg²⁺ ions react with soap, forming insoluble compounds, whereas detergents do not form such insoluble salts.
(B) Soap is acidic, and detergents are basic.
(C) Soap react with Na⁺ ions in hard water to form scum while detergents do not react with Na⁺ ions.
(D) Detergents form lather easily because they are ionic compounds and react with Ca²⁺ and Mg²⁺ ions while soaps do not react with hard water.

Answer: (A) Ca²⁺ and Mg²⁺ ions react with soap, forming insoluble compounds, whereas detergents do not form such insoluble salts.

Teacher's Note:
a) Hardness of water is due to calcium and magnesium salts, not sodium ions.
b) The insoluble substance formed by soap with these ions is called scum.

20. Which of the following salts contain water of crystallisation:
(i) Plaster of Paris, (ii) Washing soda,
(iii) Baking soda and (iv) Gypsum. [1 Mark]
(A) (i), (ii) and (iii)
(B) (i), (iii) and (iv)
(C) (ii), (iii) and (iv)
(D) (i), (ii) and (iv)

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

Teacher's Note:
a) Plaster of Paris is CaSO₄.\( \frac{1}{2} \)H₂O, washing soda is Na₂CO₃.10H₂O and gypsum is CaSO₄.2H₂O.
b) Baking soda (NaHCO₃) has no water of crystallisation.

21. Which of the following pair of compounds are isomers? [1 Mark]

[Figure: Each option shows two structural formulas of hydrocarbons. (A) A straight chain of 4 carbon atoms with 10 hydrogen atoms, and a branched chain of 3 carbon atoms with a CH₃ group attached to the middle carbon (4 carbon atoms, 10 hydrogen atoms). (B) A straight chain of 5 carbon atoms with 12 hydrogen atoms, and a branched chain of 3 carbon atoms with a CH₃ group attached to the middle carbon (4 carbon atoms). (C) A straight chain of 5 carbon atoms with only single bonds, and a chain of 5 carbon atoms with a double bond between the third and fourth carbon atoms. (D) A straight chain of 5 carbon atoms and a straight chain of 4 carbon atoms, both with single bonds only.]

(A) Straight-chain butane and branched-chain isobutane (both C₄H₁₀)
(B) Straight-chain pentane (C₅H₁₂) and branched-chain isobutane (C₄H₁₀)
(C) Pentane (C₅H₁₂) and pentene with one double bond (C₅H₁₀)
(D) Pentane (C₅H₁₂) and butane (C₄H₁₀)

Answer: (A) Butane and isobutane - same molecular formula but different structural formula

Teacher's Note:
a) Isomers have the same molecular formula but different structures.
b) Count the carbon and hydrogen atoms in both structures before choosing.

For Visually Impaired Candidates (in lieu of Q. 21)

The unsaturated hydrocarbon with three carbon atoms is: [1 Mark]
(A) propene
(B) propane
(C) pentane
(D) pentene

Answer: (A) propene

Teacher's Note:
a) "Prop-" means three carbon atoms and "-ene" means a double bond (unsaturated).
b) Propane is saturated; pentane and pentene have five carbon atoms.

22. A food label is given below. E304 and E307b are added to the flavoured biscuits.
Which of the following is added to food packets for the same reason? [1 Mark]
(A) salt
(B) vinegar
(C) nitrogen
(D) vanilla

[Figure: Food label reading: "FLAVOURED BISCUITS. Wheat Flour, Vegetable Oil, Sugar, Salt, Malt Extract (From Barley), Baking Powder, Flavour Enhancers (E621, E635), Onion Powder, Natural Flavour, Maltodextrin (Maize), Chicken, Spices, Antioxidants (E304, E307b From Soy), Emulsifier (Soy Lecithin). CONTAINS SOY AND GLUTEN CONTAINING CEREALS. MAY CONTAIN TRACES OF EGG, MILK, PEANUT, SESAME AND TREE NUT." Some ingredients, including the antioxidants, are highlighted.]

Answer: (C) nitrogen

Teacher's Note:
a) The label shows E304 and E307b are antioxidants; they prevent oxidation (rancidity) of fats.
b) Chip packets are flushed with nitrogen for the same reason, to prevent oxidation.

For Visually Impaired Candidates (in lieu of Q. 22)

In the food industry nitrogen is used as a/an [1 Mark]
(A) preservative.
(B) oxidising agent.
(C) antioxidant.
(D) colouring agent.

Answer: (C) antioxidant.

Teacher's Note:
a) Nitrogen is an unreactive gas that keeps oxygen away from the food.
b) This prevents rancidity of oily and fatty foods.

23. Electrical wires made of aluminum [1 Mark]
(A) corrode easily as aluminium is a very reactive metal.
(B) corrode easily because porous flaky aluminium oxide is formed.
(C) do not corrode easily as aluminium oxide forms a protective layer.
(D) do not corrode easily as aluminium is less reactive than metals like sodium, magnesium and zinc.

Answer: (C) do not corrode easily as aluminium oxide forms a protective layer.

Teacher's Note:
a) A thin, tough layer of aluminium oxide stops further reaction with air and moisture.
b) Aluminium is actually more reactive than zinc, so option (D) is wrong.

The following question consists of two statements - Assertion (A) and Reason (R). Answer this question by selecting the appropriate option given below:
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

24. Assertion (A): Onion extract loses its smell, when it is added to baking soda solution whereas it retains its smell when added to lemon juice.
Reason (R): Onion acts as an olfactory indicator. [1 Mark]
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

Answer: (B) Both A and R are true, and R is not the correct explanation of A.

Teacher's Note:
a) Onion smell is lost in a basic solution (baking soda) and stays in an acidic solution (lemon juice).
b) The marking scheme treats R as a true general statement that does not explain the observation in A.

25. Carbon forms millions of compounds whereas silicon, which is just below it in the Periodic Table, forms a few compounds. Explain the reason for the difference in their ability to form compounds. [2 Marks]

Answer:
1. Carbon has a small atomic size, so it forms strong and stable C-C bonds and shows extensive catenation. This lets it form millions of compounds.
2. Silicon is larger in size, so it forms weaker Si-Si bonds. It cannot show catenation to the same extent, so it forms fewer compounds.

Teacher's Note:
a) Keywords: small atomic size, strong C-C bonds, catenation.
b) Always compare both elements to answer a "difference" question.

26. Attempt either option A or B. [3 Marks]

A. An organic compound "X" with formula C₂H₆O is used to blend with petrol in order to reduce the dependence on fossil fuels and lower down the harmful emissions from the cars. It is also used as a solvent in medicines such as tincture iodine, cough syrups and many tonics.
(i) Draw the electron dot structure of "X".
(ii) Write the name and formula of the next homologue of "X".
(iii) Write the reaction of "X" with acetic acid to form a sweet smelling substance. [3 Marks]

Answer:
X is ethanol, C₂H₅OH.
(i) Electron dot structure of ethanol (CH₃CH₂OH): the two carbon atoms share one pair of electrons with each other. The first carbon shares one pair each with three hydrogen atoms. The second carbon shares one pair each with two hydrogen atoms and one pair with the oxygen atom. The oxygen atom shares one pair with a hydrogen atom and has two lone pairs.
(ii) Propanol, C₃H₇OH
(iii) C₂H₅OH + CH₃COOH \( \xrightarrow{\text{conc. } H_2SO_4} \) CH₃COOC₂H₅ + H₂O

Teacher's Note:
a) In the dot structure, show every shared pair between the atoms and the two lone pairs on oxygen.
b) The sweet smelling product is an ester (ethyl ethanoate); write conc. H₂SO₄ over the arrow.

OR

B. Ramya performs an experiment in the following two steps:
Step 1: Heats ethanol with conc. H₂SO₄ at \( 170^{\circ}C \) to get product A.
Step 2: Passes hydrogen gas over product A, using Ni as a catalyst and gets product B.
(i) Name the products A and B.
(ii) Write the reactions involved in both the steps. [3 Marks]

Answer:
(i) Product A - Ethene, Product B - Ethane
(ii) Step 1: CH₃CH₂OH \( \xrightarrow{\text{conc. } H_2SO_4} \) CH₂=CH₂ + H₂O
Step 2: CH₂=CH₂ + H₂ \( \xrightarrow{Ni} \) CH₃CH₃

Teacher's Note:
a) Step 1 is dehydration of ethanol; step 2 is hydrogenation (addition reaction).
b) Each equation carries 1 mark; write the catalyst or reagent over the arrow.

For Visually Impaired Candidates (in lieu of Q. 26)

Attempt either option A or B. [3 Marks]

A. An organic compound "X" with formula C₂H₆O is used to blend with petrol in order to reduce the dependence on fossil fuels and lower down the harmful emissions from the cars. It is also used as a solvent in medicines such as tincture iodine, cough syrups and many tonics.
(i) Name the compound "X".
(ii) Write the formula of the next homologue of "X".
(iii) Write the reaction of "X" with acetic acid to form a sweet smelling substance. [3 Marks]

Answer:
(i) Ethanol
(ii) C₃H₇OH
(iii) C₂H₅OH + CH₃COOH \( \xrightarrow{\text{conc. } H_2SO_4} \) CH₃COOC₂H₅ + H₂O

Teacher's Note:
a) Next homologue differs by one CH₂ unit: C₂H₅OH to C₃H₇OH.
b) The reaction of an alcohol with an acid to form an ester is called esterification.

OR

B. Ramya performs an experiment in the following two steps:
Step 1: Heats ethanol with conc. H₂SO₄ at \( 170^{\circ}C \) to get product A.
Step 2: Passes hydrogen gas over product A, using Ni as a catalyst and gets product B.
(i) Name the products A and B.
(ii) Write the reactions involved in both the steps. [3 Marks]

Answer:
(i) Product A - Ethene, Product B - Ethane
(ii) Step 1: CH₃CH₂OH \( \xrightarrow{\text{conc. } H_2SO_4} \) CH₂=CH₂ + H₂O
Step 2: CH₂=CH₂ + H₂ \( \xrightarrow{Ni} \) CH₃CH₃

Teacher's Note:
a) Hot conc. H₂SO₄ acts as a dehydrating agent and removes water from ethanol.
b) Adding hydrogen to ethene in the presence of nickel gives the saturated compound ethane.

27. Give reason for the following :
A. A solution of sports electrolyte drink conducts electricity, but coconut oil does not.
B. Milkiness disappears when excess of carbon dioxide is passed through lime water.
C. An iron bridge near the sea rusts much faster than an iron bridge located in a dry hilly region. [3 Marks]

Answer:
A. A sports drink contains ions (electrolytes), which conduct electricity. Coconut oil does not contain ions, so it cannot conduct electricity.
B. Lime water turns milky due to the formation of insoluble calcium carbonate (CaCO₃). When excess CO₂ is passed, calcium carbonate reacts further to form soluble calcium hydrogen carbonate, so the milkiness disappears.
C. Air near the sea has high moisture (humidity), which speeds up rusting. In a dry hilly region the moisture is low, so rusting is much slower.

Teacher's Note:
a) For B, name both compounds: insoluble calcium carbonate and soluble calcium hydrogen carbonate.
b) For C, remember that rusting needs both air and moisture.

28. Aavya and Vihan are studying the properties of metals in school. They decide to design and conduct an experiment to test the relative conduction of heat by metals. They use the apparatus (Figure) with rods of five different metals aluminium, silver, copper, iron and zinc attached to a central point. This central point is heated using a burner. The metal rods are of equal length and thickness. They observe that the candles at the end of the rods fall at different times. (Study the graph) [4 Marks]

[Figure: Photograph of the apparatus: five metal rods joined at a central point placed over a burner, with a candle attached at the end of each rod. Below it, a bar graph titled "Time take for the candle to drop", with Time/s on the y-axis (0 to 70) and Metals on the x-axis: Silver 10 s, Copper 15 s, Aluminium 30 s, Zinc 50 s, Iron 60 s.]

A. On the basis of the graph obtained, which metal is the best conductor of heat? [1 Mark]

Answer: Silver, because its candle drops first (in the shortest time, 10 s).

Teacher's Note:
a) The shorter the time for the candle to drop, the better the conductor.
b) Read the smallest bar on the graph.

B. Gold is a better conductor of heat than aluminium but it is not as good as copper. The time taken for the candle to drop if one of the rods were made of gold would have been: [1 Mark]
(i) 15 s
(ii) 20 s
(iii) 35 s
(iv) 10 s

Answer: (ii) 20 s

Teacher's Note:
a) Gold's time must lie between copper (15 s) and aluminium (30 s).
b) Only 20 s lies in this range.

Attempt either sub-part C or D.

C. The length of all the metal rods is increased by 20% without any change in the thickness. Will the time required for the candles to drop be affected? How and why? [2 Marks]

Answer:
1. Yes. The time required for the candles to drop will increase for all the metals, but the order of falling of candles will remain the same.
2. In conduction, particles vibrate and pass heat to one another. Heat has to travel a longer distance from one end to the other, so it takes more time.

Teacher's Note:
a) Mention both points: time increases and the order stays the same.
b) Give the reason in terms of the longer distance for heat to travel.

OR

D. Mayank tries the experiment with an apparatus similar to figure (1). He ensures the length of all metal rods is the same. However, the thickness of rods is not the same in his apparatus. Will the inference he draws on the basis of his experiment be correct? Justify your answer. [2 Marks]

Answer:
1. No. If the thickness of the rods is changed, the inference will not be correct.
2. The dimensions of the rods have to be the same for a fair test, because a thicker rod will take more time to conduct heat.

Teacher's Note:
a) In a fair comparison, only one variable (the metal) should change.
b) Always justify a Yes/No answer with the reason.

For Visually Impaired Candidates (in lieu of Q. 28)

Aavya and Vihan are studying the properties of metals in school. They decide to design and conduct an experiment to test the relative conduction of heat by metals. They use rods of equal length and thickness made of five different metals - A, B, C, D and E. A heat sensor is attached at one end of each rod. The other end of the rod is heated and time required for the heat sensor to beep is noted for all metal rods. The heat sensor at the end of B rod beeps in 10 seconds, C rod in 15 seconds, A a 30 seconds, E in 50 seconds and D sensor beeps in 60 seconds. [4 Marks]

A. On the basis of the experimental result, which metal is the best conductor of heat? [1 Mark]

Answer: Metal B, because its sensor beeps first (in 10 seconds).

Teacher's Note:
a) The best conductor takes the least time to pass heat along the rod.
b) Arrange the times in order before answering.

B. Metal G is a better conductor of heat than metal A but it is not as good as metal C. The time taken for the sensor to beep if one of the rods were made of G would have been: [1 Mark]
(i) 15 s
(ii) 20 s
(iii) 35 s
(iv) 10 s

Answer: (ii) 20 s

Teacher's Note:
a) G's time must lie between metal C (15 s) and metal A (30 s).
b) Only 20 s fits in this range.

Attempt either sub-part C or D.

C. The length of all the metal rods is increased by 20% without any change in the thickness. Will the time required for the sensor to beep be affected? How and why? [2 Marks]

Answer:
1. Yes. The time required for the sensors to beep will increase for all the metals, but the order will remain the same.
2. In conduction, particles vibrate and pass heat to one another. Heat energy has to cover a longer distance from one end to the other, so it takes more time.

Teacher's Note:
a) Time increases, but the order of the metals does not change.
b) The reason is the longer distance heat has to travel.

OR

D. Mayank tries the experiment with little variation. He keeps the length of all the rods of the same length, but, the thickness of rods is not the same. Will the inference he draws on the basis of his experiment be correct? Justify your answer. [2 Marks]

Answer:
1. No. If the thickness of the rods is changed, the inference will not be correct.
2. The dimensions of the rods have to be the same, because a thicker rod will take more time to conduct heat.

Teacher's Note:
a) Only the type of metal should differ between the rods.
b) Keyword: the dimensions of the rods must be the same.

29. Attempt either option A or B. [5 Marks]

A.
I. Refer to the dialogue between Alisha, Drishti and Bhushan in the chemistry lab. They are discussing rusting of iron and formation of ammonia gas. Who all do you agree with and why?
Alisha: The formation of ammonia by reaction of nitrogen with hydrogen is a reduction reaction.
Dristhti: Both rusting of iron and formation of ammonia are redox reactions.
Bhushan: Rusting of iron is an oxidation reaction.
II. Assess the following chemical equations. Identify the error(s) and rewrite the equation after correction.
(i) MgCl₂ + AgNO₃ → AgCl₂ + MgNO₃
(ii) Al + H₂SO₄ → Al₂SO₄ + H₂O [5 Marks]

Answer:
I. I agree with Drishti. Alisha and Bhushan are only partially correct.
Oxidation and reduction go hand in hand. In the formation of ammonia, nitrogen is reduced and hydrogen is oxidised. In rusting of iron, iron is oxidised while oxygen is reduced. So both are redox reactions.
II. (i) The formulas AgCl₂ and MgNO₃ are wrong and the equation is not balanced. Correct equation:
MgCl₂ + 2AgNO₃ → 2AgCl + Mg(NO₃)₂
(ii) The formula Al₂SO₄ is wrong, the gas formed is hydrogen (not water), and the equation is not balanced. Correct equation:
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Teacher's Note:
a) Write formulas using valencies: Ag is +1, Mg is +2, Al is +3, NO₃ is -1, SO₄ is -2.
b) A metal reacting with a dilute acid gives a salt and hydrogen gas; balance the equation last.

OR

B.
I. Refer to the dialogue between Daljeet and Avik in the chemistry lab. They have learnt about decomposition reactions. Who has understood decomposition reactions correctly? Justify your answer.
Daljeet: During a decomposition reaction, a substance decomposes to give two simpler compounds. MgCO₃ decomposes on heating to magnesium oxide and carbon dioxide.
Avik: No, during a decomposition reaction, a substance decomposes to give two elements. Water on electrolysis gives hydrogen and oxygen, both are elements.
II. Analyse the observations given in following two cases. Are these observations correct? Give reasons to support your answer.
(i) Iron fillings are added to an aqueous solution of magnesium sulphate.
Observation: The colourless solution turns greenish due to formation of iron sulphate.
(ii) Aluminium metal is added to dilute sulphuric acid
Observation: Brisk effervescence are seen due to evolution of a gas. [5 Marks]

Answer:
I. Both have understood decomposition only partially.
In a decomposition reaction, a single substance breaks down into simpler substances. The substance that decomposes must be a compound, but the products can be either elements or compounds.
MgCO₃ decomposes on heating to magnesium oxide and carbon dioxide. Both products are compounds, and it is still a decomposition reaction.
Water on electrolysis gives hydrogen and oxygen, which are both elements. This is also a decomposition reaction.
II. (i) The observation is incorrect. Iron is less reactive than magnesium, so it cannot displace magnesium from magnesium sulphate. No iron sulphate is formed.
(ii) The observation is incorrect. Hydrogen gas is formed, which produces a few bubbles. Brisk effervescence is seen when carbon dioxide is formed.

Teacher's Note:
a) Products of decomposition may be elements or compounds, so neither student is fully correct.
b) Use the reactivity series: a less reactive metal cannot displace a more reactive metal from its salt solution.

Section - C

30. A student wants to determine the focal length of a convex lens using sunlight.
Which of the following practical procedures will give the most accurate result? [1 Mark]
(A) Move the lens close to a wall and measure the blurred circle.
(B) Hold the lens above a sheet of paper and adjust until the smallest, brightest spot is formed.
(C) Measure the thickness of the lens with a scale.
(D) Look through the lens at a distant object.

Answer: (B) Hold the lens above a sheet of paper and adjust until the smallest, brightest spot is formed.

Teacher's Note:
a) Sun rays are nearly parallel, so they meet at the principal focus of the convex lens.
b) The distance from the lens to the sharpest, brightest spot is the focal length.

31. An electrical appliance rated 1000 W is operated at 250 V. What current does it draw? [1 Mark]
(A) 0.25 A
(B) 2.5 A
(C) 4.0 A
(D) 10 A

Answer: (C) 4.0 A

Teacher's Note:
a) Use \( P = VI \), so \( I = \frac{P}{V} = \frac{1000}{250} = 4.0 \) A.
b) Check the units: watt divided by volt gives ampere.

The following one question consists of two statements - Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below:
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

32. Assertion (A): The apparent depth of a coin in water is less than its real depth.
Reason (R): Light bends away from the normal when it travels from water to air. [1 Mark]
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

Answer: (A) Both A and R are true, and R is the correct explanation of A.

Teacher's Note:
a) Light from the coin bends away from the normal as it leaves water, so the coin appears raised.
b) This bending is exactly why the apparent depth is less, so R explains A.

33. A. State Snell's Law of refraction.
B. Define the refractive index of a material with respect to another material giving the mathematical expression. [2 Marks]

Answer:
A. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for light of a given colour and for the given pair of media: \( \frac{\sin i}{\sin r} = \) constant.
B. Consider a ray of light travelling from medium 1 into medium 2. Let \( v_1 \) be the speed of light in medium 1 and \( v_2 \) be the speed of light in medium 2. The refractive index of medium 2 with respect to medium 1 is the ratio of these speeds:
\( n_{21} = \frac{v_1}{v_2} \)

Teacher's Note:
a) Snell's law needs the condition: given colour of light and given pair of media.
b) The marking scheme prints \( n_{21} = \frac{v_2}{v_1} \); the correct expression is \( n_{21} = \frac{v_1}{v_2} \) (speed in medium 1 divided by speed in medium 2).

34. Attempt either option A or B. [2 Marks]

A.
I. State Joule's Law of heating.
II. What happens to the heat produced in a conductor if the resistance is doubled while keeping other variables constant? [2 Marks]

Answer:
I. Joule's law of heating: The heat produced in a resistor is (i) directly proportional to the square of the current for a given resistance, (ii) directly proportional to the resistance for a given current, and (iii) directly proportional to the time for which the current flows through the resistor.
\( H = I^2Rt \), where I is current, R is resistance and t is time.
II. If the resistance is doubled for the same current I and time t:
\( H' = I^2(2R)t = 2I^2Rt = 2H \)
So the heat produced doubles.

Teacher's Note:
a) Write all three proportionalities along with the formula \( H = I^2Rt \).
b) Current and time stay constant here, so heat is directly proportional to resistance.

OR

B.
I. Define resistivity.
II. Justify with an example why resistivity is an intrinsic property of a material. [2 Marks]

Answer:
I. The resistivity (\( \rho \)) of a material is the resistance of a wire of that material having unit length and unit cross-sectional area. It comes from \( R = \frac{\rho L}{A} \), where R is the resistance of a wire of length L and cross-sectional area A.
II. Resistivity depends only on the material, not on its dimensions. Example: two copper wires of different lengths and thicknesses give the same value of \( \rho \) when calculated using \( \rho = \frac{RA}{L} \).

Teacher's Note:
a) Resistance changes with length and area, but resistivity does not.
b) SI unit of resistivity is ohm metre (\( \Omega \) m).

35. A concave lens forms an image that is 1/3rd the size of the object, as shown in the diagram. If the object is placed 24 cm in front of the lens, determine:
A. the type of lens.
B. the image distance
C. the focal length of the lens [3 Marks]

[Figure: Ray diagram of a concave lens with optical centre O on the principal axis. Points marked on the axis from left to right: 2F₁, A, F₁, A', O, F₂, 2F₂. The object AB stands at A. A ray from B parallel to the axis meets the lens at C and diverges; its backward extension (dashed) passes through F₁. A second ray from B passes straight through O. The image A'B' is upright, smaller, and lies between F₁ and O.]

Answer:
A. Type of lens: Concave (diverging) lens (given). Object distance, u = -24 cm.
B. Image distance:
Magnification \( m = \frac{h_i}{h_o} = \frac{v}{u} \)
For a concave lens the image is upright and diminished, so \( m = +\frac{1}{3} \)
\( v = m \times u = \frac{1}{3} \times (-24) = -8 \) cm
The image is 8 cm from the lens, on the same side as the object (virtual image).
C. Focal length, using the lens formula:
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
\( \frac{1}{f} = \frac{1}{-8} - \frac{1}{-24} = -\frac{3}{24} + \frac{1}{24} = -\frac{2}{24} = -\frac{1}{12} \)
f = -12 cm

Teacher's Note:
a) The negative focal length confirms a concave (diverging) lens.
b) The marking scheme writes the lens formula as \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \) and u as "m = -24 cm"; the correct formula is \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), and the final answer f = -12 cm is the same.

For Visually Impaired Candidates (in lieu of Q. 35)

A student places an object 24 cm from a convex lens of focal length 12 cm. Calculate the image position and determine whether the image is real or virtual. [3 Marks]

Answer:
Given: u = -24 cm, f = +12 cm
Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \), so \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \)
\( \frac{1}{v} = \frac{1}{12} + \frac{1}{-24} = \frac{2 - 1}{24} = \frac{1}{24} \)
v = +24 cm
The image is formed 24 cm from the lens on the opposite side of the object, so it is real (and inverted).

Teacher's Note:
a) The object is at 2F, so the image is also at 2F on the other side, same size.
b) A positive value of v for a lens means the image is real.

36. A. State the Right-Hand Thumb Rule and describe how it helps determine the direction of magnetic field around a current-carrying conductor.
B. Explain how the magnetic field changes with distance from the conductor. [3 Marks]

Answer:
A. Right-hand thumb rule: Imagine that you are holding a current-carrying straight conductor in your right hand such that the thumb points towards the direction of current. Then your fingers will wrap around the conductor in the direction of the field lines of the magnetic field.
So if the thumb points along the conventional current, the curled fingers show the direction of the magnetic field lines around the conductor. This is how the rule gives the direction of the magnetic field.
B. The magnitude of the magnetic field around a long straight conductor at a distance r is:
\( B \propto \frac{I}{r} \)
So the magnetic field decreases as the distance from the conductor increases (it is inversely proportional to the distance).

Teacher's Note:
a) Thumb shows current and curled fingers show the magnetic field; do not mix them up.
b) The field lines are concentric circles that get farther apart away from the wire.

37. The figure shows a circuit with a 15 V battery connected to several resistors. Using the given circuit diagram:
A. Find the total current supplied by the battery.
B. Calculate the potential difference between points A and B. [3 Marks]

[Figure: Circuit with a battery V = 15 V. Between points A and B, a 2 Ω resistor is connected directly, and in parallel with it two 1 Ω resistors are joined in series along the two sides of a triangle above. A 5 Ω resistor connects point A to the battery. From point B, two 3 Ω resistors connected in parallel (drawn as a V-shaped arrangement) lead to the other terminal of the battery.]

Answer:
A. Step 1: The two 1 Ω resistors are in series: 1 + 1 = 2 Ω. This 2 Ω combination is in parallel with the 2 Ω resistor.
\( \frac{1}{R_{AB}} = \frac{1}{2} + \frac{1}{2} = 1 \), so \( R_{AB} = 1\ \Omega \)
Step 2: The two 3 Ω resistors are in parallel.
\( \frac{1}{R} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \), so \( R = 1.5\ \Omega \)
Step 3: 5 Ω, 1 Ω and 1.5 Ω are in series.
\( R_s = 5 + 1 + 1.5 = 7.5\ \Omega \)
Step 4: Using V = IR:
\( I = \frac{V}{R_s} = \frac{15}{7.5} = 2 \) A
B. The full current of 2 A flows between A and B, and \( R_{AB} = 1\ \Omega \).
\( V_{AB} = I \times R_{AB} = 2 \times 1 = 2 \) V

Teacher's Note:
a) Simplify the circuit in parts: first the series pair, then each parallel group, then the total series.
b) Check: 10 V across 5 Ω + 2 V across AB + 3 V across the 3 Ω pair = 15 V.

For Visually Impaired Candidates (in lieu of Q. 37)

A 15 V battery is connected in series with three resistors of 2 Ω, 3 Ω, and 5 Ω. Calculate:
A. The total resistance in the circuit.
B. The total current supplied by the battery.
C. The voltage drop across the 3Ω resistor. [3 Marks]

Answer:
A. \( R_{total} = 2 + 3 + 5 = 10\ \Omega \)
B. \( I = \frac{V}{R_{total}} = \frac{15}{10} = 1.5 \) A
C. \( V_{3\Omega} = I \times 3 = 1.5 \times 3 = 4.5 \) V

Teacher's Note:
a) In series, resistances add up and the same current flows through each resistor.
b) Write the unit with every answer: \( \Omega \), A and V.

38. A school has installed an automatic electromagnetic door-locking system at the main entrance. The locking mechanism uses a solenoid controlled by a switch at the security desk. When electricity flows through the solenoid, it creates a magnetic field strong enough to pull an iron bolt, unlocking the door.
As soon as the switch is turned off, the magnetic field disappears, and a spring pushes the bolt back, locking the door. [4 Marks]

[Figure: Diagram of a solenoid lock: a coil wound around a soft iron core on a base, with a plunger at the top and an upward arrow above it. Labels: "Plunger", "Switch", "Electric current energizes a coil", "Soft iron core", and a second "Switch" on the circuit connected to the coil.]

A. Why is an iron core kept inside the solenoid in this locking mechanism? [1 Mark]

Answer: The iron core concentrates and strengthens the magnetic field, so the solenoid produces a stronger pulling force on the bolt.

Teacher's Note:
a) A solenoid with a soft iron core becomes a strong electromagnet.
b) Keywords: concentrates and amplifies the magnetic field.

B. How can we determine the magnetic poles in this solenoid? [1 Mark]

Answer:
1. Using the right-hand thumb rule: curl the fingers of the right hand in the direction of the current; the thumb then points towards the north pole of the solenoid.
2. Or, bring the north pole of a bar magnet near one end of the solenoid; repulsion shows that this end is the north pole.

Teacher's Note:
a) Either method earns the mark.
b) Repulsion, not attraction, is the sure test of a pole.

Attempt either subpart (C) or (D).

C. If the current in the solenoid is increased from 1 A to 2 A, explain how the magnetic effect changes and its effect on unlocking the door. [2 Marks]

Answer:
1. The magnetic field inside a solenoid is directly proportional to the current, so when the current increases from 1 A to 2 A the magnetic field also increases.
2. The stronger magnetic field gives a stronger pull on the bolt, so the door is unlocked more effectively.

Teacher's Note:
a) State the relation (field is directly proportional to current) and then its effect on the door.
b) Both the change and its effect are needed for 2 marks.

OR

D. If the source of the current is fixed, how can we increase the magnetic field in the solenoid? [2 Marks]

Answer:
1. Increase the number of turns (N) of the coil.
2. Insert a soft iron core inside the solenoid.

Teacher's Note:
a) Reducing the length of the coil is also accepted; give any two points.
b) Current cannot be changed here, so do not write "increase the current".

For Visually Impaired Candidates (in lieu of Q. 38)

A student places a small magnetic compass on a wooden table and aligns it so that the compass needle points naturally toward North-South. She then stretches a straight charging cable above the compass so that the wire is parallel to the compass needle. When the phone charger is switched on and current begins to flow through the wire, the compass needle deflects away from its North-South position. [4 Marks]

A. What causes the compass needle to deflect when current flows through the wire? [1 Mark]

Answer: The magnetic field produced by the current-carrying wire acts on the compass needle, so the needle deflects.

Teacher's Note:
a) A current-carrying conductor produces a magnetic field around it.
b) The compass needle is a small magnet, so it responds to this field.

B. What happens to the compass needle when the current in the wire is switched off? [1 Mark]

Answer: The needle comes back and aligns itself with the Earth's magnetic field in the north-south direction.

Teacher's Note:
a) No current means no magnetic field around the wire.
b) Only the Earth's magnetic field acts on the needle then.

C. Explain how the amount of current in the wire affects the magnitude of deflection of the compass needle? [2 Marks]

Answer:
1. The deflection of the needle increases when the current increases.
2. This is because the magnetic field of a current-carrying conductor is directly proportional to the current flowing through it.

Teacher's Note:
a) More current means a stronger field and a larger deflection.
b) Give the reason (direct proportionality) for the second mark.

OR

D. When the direction of current is reversed, the compass deflects in the opposite direction. Explain how the Right-Hand Thumb Rule accounts for this change. [2 Marks]

Answer:
1. By the right-hand thumb rule, when the current (thumb) is reversed, the fingers curl the opposite way, so the direction of the magnetic field reverses.
2. The compass needle, being a tiny magnet, aligns with this new field and deflects in the opposite direction. This shows that the magnetic field is now in the opposite direction.

Teacher's Note:
a) Direction of current decides the direction of the magnetic field.
b) Link the thumb (current) and the curled fingers (field) in your answer.

39. Attempt either option A or B. [5 Marks]

A.
I. What is meant by accommodation of the human eye?
II. What is the near point of a normal eye and what does it signify?
III. Name which type of lens is used to correct the defect in which a person cannot see distant objects clearly.
IV. Draw a diagram to show how the choice of your lens in (d) helps to see distant objects clearly? [5 Marks]

Answer:
I. Accommodation is the ability of the eye lens to change its focal length to focus objects at different distances on the retina.
II. The near point is the nearest point (about 25 cm for a normal eye) at which an object can be seen distinctly. It signifies the closest distance of comfortable focus.
III. A diverging (concave) lens is used to correct myopia (short-sightedness).
IV. Diagram: A concave lens is placed in front of the myopic eye. Parallel rays from a distant object (at infinity) fall on the concave lens and diverge. These diverged rays are then focused by the eye lens exactly on the retina, so the distant object is seen clearly.

Teacher's Note:
a) In the diagram, show the rays diverging after the concave lens and meeting on the retina.
b) Without correction, a myopic eye focuses distant objects in front of the retina.

OR

B.
I. What is dispersion of light?
II. Which colour of light deviates the least during dispersion through a prism?
III. State why the sky appears dark to an astronaut in outer space.
IV. Draw a simple labelled diagram to show how white light splits into different colours while passing through a glass prism. [5 Marks]

Answer:
I. Dispersion is the splitting of white light into its component colours due to wavelength-dependent refraction.
II. Red light deviates the least (it has the longest wavelength).
III. In outer space there is no atmosphere to scatter sunlight, so the sky appears dark.
IV. Diagram: A white light beam falls obliquely on one face of a glass prism. It refracts and splits into a band of colours inside the prism. On coming out of the other face, the colours spread out into a spectrum, with red (R) at the top (least deviated) and violet (V) at the bottom (most deviated). Labels: White light beam, Glass prism, White light spectrum, R, V.

Teacher's Note:
a) Label the prism, the incident white light and the spectrum with red at the top and violet at the bottom.
b) Remember VIBGYOR: violet bends the most and red the least.

For Visually Impaired Candidates (in lieu of Q. 39)

A.
I. What is meant by hypermetropia (long-sightedness)?
II. What is the far point of a normal human eye and what does it signify?
III. Name the type of lens used to correct hypermetropia.
IV. Explain why a hypermetropic eye is unable to focus nearby objects on the retina. [5 Marks]

Answer:
I. Hypermetropia is a defect of vision in which a person can see distant objects clearly but cannot see nearby objects clearly.
II. The far point of a normal human eye is at infinity. It signifies that a normal eye can see objects clearly even when they are extremely far away.
III. A convex (converging) lens is used to correct hypermetropia.
IV. A hypermetropic eye cannot focus nearby objects on the retina because (a) the eyeball is too short, or (b) the eye lens is not converging enough (its curvature is insufficient). So the image of a nearby object forms behind the retina instead of on it. A convex lens increases the convergence of light so that the image forms on the retina.

Teacher's Note:
a) Give both causes of hypermetropia for part IV.
b) Hypermetropia: image behind the retina, corrected by a convex lens.

OR

B.
I. Why does Earth's atmosphere exhibit Tyndall's effect?
II. Which colour of light has the shortest wavelength?
III. Why do danger/stop signs use red colour?
IV. State why the sky appears blue even though sunlight is white. [5 Marks]

Answer:
I. The Earth's atmosphere is a heterogeneous mixture of minute particles such as smoke, tiny water droplets, suspended dust particles and molecules of air. When a beam of light strikes such fine particles, the path of the beam becomes visible. This scattering of light by colloidal particles is the Tyndall effect.
II. Violet light has the shortest wavelength.
III. Red light is scattered the least by air because it has the longest wavelength. It stays visible over long distances even in fog, smoke or dust, so red is used for danger and stop signals.
IV. When sunlight enters the atmosphere, the tiny particles of air scatter shorter wavelengths (blue and violet) more strongly than others. Our eyes are more sensitive to blue than to violet, so the sky appears blue.

Teacher's Note:
a) Scattering is stronger for shorter wavelengths; this explains both the blue sky and red signals.
b) Part I carries 2 marks, so name the particles and the scattering of light.