CBSE Class 10 Science Question Paper 2025 Solved Code 31-1-2

SECTION A

1. The formation of magnesium oxide is correctly shown in option: [1 Mark]

[Figure: The four options are electron-dot diagrams of Mg and O atoms, described below.]

(A) \( Mg \longrightarrow Mg^{2+}\,[O]^{2-} \) (one Mg atom transfers 2 electrons to one O atom)
(B) \( Mg \longrightarrow Mg^{+}\,[O]^{-} \) (Mg atom transfers only 1 electron to an O atom)
(C) \( Mg \longrightarrow Mg^{2+}\,[O]^{-}_{2} \) (Mg atom transfers 1 electron each to two O atoms)
(D) \( 2Mg \longrightarrow [Mg^{2+}]_{2}\,[O]^{2-} \) (two Mg atoms transfer electrons to a single O atom)

Answer: (A) One Mg atom transfers its 2 electrons to one O atom, forming \( Mg^{2+}\,[O]^{2-} \)

Teacher's Note:
a) MgO forms by complete transfer of 2 electrons from one Mg atom to one O atom, not by sharing.
b) The number of electrons lost by the metal must equal the number gained by the non-metal.

2. The metals obtained from their molten chlorides by the process of electrolytic reduction are: [1 Mark]
(A) Gold and silver
(B) Calcium and magnesium
(C) Aluminium and silver
(D) Sodium and iron

Answer: (B) Calcium and magnesium

Teacher's Note:
a) Highly reactive metals are extracted by electrolytic reduction of their molten (fused) chlorides.
b) Remember: metals at the top of the reactivity series cannot be reduced by carbon.

3. In one formula unit of salt 'X', seven molecules of water of crystallisation are present. The salt 'X' is: [1 Mark]
(A) \( CuSO_4 \)
(B) \( Na_2CO_3 \)
(C) \( FeSO_4 \)
(D) \( CaSO_4 \)

Answer: (C) \( FeSO_4 \) (as \( FeSO_4.7H_2O \))

Teacher's Note:
a) Remember the fixed water molecules for common salts: \( CuSO_4.5H_2O \), \( Na_2CO_3.10H_2O \).
b) Green vitriol (ferrous sulphate) always has 7 molecules of water of crystallisation.

4. Reaction between two elements A and B, forms a compound C. A loses electrons and B gains electrons. Which one of the following properties will not be shown by compound C? [1 Mark]
(A) It has high melting point.
(B) It is highly soluble in water.
(C) It has weak electrostatic forces of attraction between its oppositely charged ions.
(D) It conducts electricity in its molten state or aqueous solution.

Answer: (C) It has weak electrostatic forces of attraction between its oppositely charged ions.

Teacher's Note:
a) Compound C is ionic since electrons are transferred, so it actually has strong electrostatic forces.
b) Ionic compounds have high melting points due to these strong forces, so option (C) is false.

5. Electrolysis of water is a decomposition reaction. The mass ratio \( (M_H : M_O) \) of hydrogen and oxygen gases liberated at the electrodes during electrolysis of water is: [1 Mark]
(A) 8 : 1
(B) 2 : 1
(C) 1 : 2
(D) 1 : 8

Answer: (D) 1 : 8

Teacher's Note:
a) Do not confuse mass ratio with volume ratio; volume ratio of \( H_2:O_2 \) is 2:1.
b) Mass ratio is based on atomic masses: \( 2 \times 1 : 1 \times 16 = 2:16 = 1:8 \).

6. Consider the following reactions:
(i) Dilute hydrochloric acid reacts with sodium hydroxide.
(ii) Magnesium oxide reacts with dilute hydrochloric acid.
(iii) Carbon dioxide reacts with sodium hydroxide.
It is found that in each case: [1 Mark]
(A) Salt and water is formed.
(B) Neutral salts are formed.
(C) Hydrogen gas is formed.
(D) Acidic salts are formed.

Answer: (A) Salt and water is formed.

Teacher's Note:
a) All three are neutralisation-type reactions between an acid/acidic oxide and a base/basic oxide.
b) The salts formed are not all neutral (e.g., \( NaHCO_3 \) can form), so option (B) is incorrect.

7. The products formed when Aluminium and Magnesium are burnt in the presence of air respectively are: [1 Mark]
(A) \( Al_3O_4 \) and \( MgO_2 \)
(B) \( Al_2O_3 \) and \( MgO \)
(C) \( Al_3O_4 \) and \( MgO \)
(D) \( Al_2O_3 \) and \( MgO_2 \)

Answer: (B) \( Al_2O_3 \) and \( MgO \)

Teacher's Note:
a) Balance the valencies: Al is trivalent so its oxide is \( Al_2O_3 \); Mg is divalent so its oxide is \( MgO \).
b) Both metals burn with a bright white flame in air.

8. The plant hormone whose concentration stimulates the cells to grow longer on the side of the shoot which is away from light is: [1 Mark]
(A) Cytokinins
(B) Gibberellins
(C) Adrenaline
(D) Auxins

Answer: (D) Auxins

Teacher's Note:
a) Auxin diffuses to the shaded side, promoting more cell elongation there, causing the shoot to bend towards light (phototropism).
b) Adrenaline is an animal hormone, not a plant hormone; it can be eliminated immediately.

9. Secretion of less saliva in mouth will effect the conversion of: [1 Mark]
(A) proteins into amino acids
(B) fats into fatty acids and glycerol
(C) starch into simple sugars
(D) sugars into alcohol

Answer: (C) starch into simple sugars

Teacher's Note:
a) Saliva contains salivary amylase which breaks down starch into simpler sugars in the mouth.
b) Protein and fat digestion occur later in the stomach and small intestine, not in the mouth.

10. If pea plants with round and green seeds (RRyy) are crossed with pea plants having wrinkled and yellow seeds (rrYY), the seeds developed by the plants of \( F_1 \) generation will be: [1 Mark]
(A) 50% round and green
(B) 75% wrinkled and green
(C) 100% round and yellow
(D) 75% wrinkled and yellow

Answer: (C) 100% round and yellow

Teacher's Note:
a) Cross RRyy \(\times\) rrYY gives all offspring RrYy in \( F_1 \).
b) Since Round (R) and Yellow (Y) are dominant, all \( F_1 \) seeds appear round and yellow.

11. The correct/true statement(s) for a bisexual flower is/are:
(i) They possess both stamen and pistil.
(ii) They possess either stamen or pistil.
(iii) They exhibit either self-pollination or cross-pollination.
(iv) They cannot produce fruits on their own. [1 Mark]
(A) (i) only
(B) (iv) only
(C) (i) and (iii)
(D) (i) and (iv)

Answer: (C) (i) and (iii)

Teacher's Note:
a) A bisexual flower has both male (stamen) and female (pistil) parts, so it can undergo self- or cross-pollination.
b) Statement (iv) is false because a bisexual flower can produce fruit through self-pollination.

12. The breakdown of glucose has taken the following pathway:
Glucose \(\xrightarrow{(a)}\) Pyruvate + Energy \(\xrightarrow{(b)}\) Lactic acid + Energy
The sites 'a' and 'b' respectively are: [1 Mark]
(A) Mitochondria and Oxygen deficient muscle cells
(B) Cytoplasm and Oxygen rich muscle cells
(C) Cytoplasm and Yeast cells
(D) Cytoplasm and Oxygen deficient muscle cells

Answer: (D) Cytoplasm and Oxygen deficient muscle cells

Teacher's Note:
a) Glycolysis (glucose to pyruvate) always occurs in the cytoplasm.
b) Lactic acid is formed in muscle cells only under lack of oxygen, as in yeast it forms ethanol instead.

13. The white light entering a glass prism, gets split into its constituent colours. It is observed that: [1 Mark]
(A) Red light deviates the most.
(B) Violet light deviates the least.
(C) Yellow light deviates more than the blue light.
(D) Green light deviates more than the orange light.

Answer: (D) Green light deviates more than the orange light.

Teacher's Note:
a) The order of deviation (least to most) follows the VIBGYOR sequence: red deviates least, violet deviates most.
b) Since green comes after orange towards violet in this sequence, it deviates more than orange.

14. Mirror 'X' is used to concentrate sunlight in solar furnace and Mirror 'Y' is fitted on the side of the vehicle to see the traffic behind the driver. Which of the following statements are true for the two mirrors?
(i) The image formed by mirror 'X' is real, diminished and at its focus.
(ii) The image formed by mirror 'Y' is virtual, diminished and erect.
(iii) The image formed by mirror 'X' is virtual, diminished and erect.
(iv) The image formed by mirror 'Y' is real, diminished and at its focus. [1 Mark]
(A) (i) and (ii)
(B) (ii) and (iii)
(C) (iii) and (iv)
(D) (i) and (iv)

Answer: (A) (i) and (ii)

Teacher's Note:
a) Mirror X is a concave mirror; parallel rays (sunlight) converge to form a real image at its focus.
b) Mirror Y is a convex mirror, used in vehicles for its wide field of view giving a virtual, diminished, erect image.

15. The percentage of solar energy which is not converted into food energy by the leaves of green plants in a terrestrial ecosystem is about: [1 Mark]
(A) 1%
(B) 10%
(C) 90%
(D) 99%

Answer: (D) 99%

Teacher's Note:
a) Only about 1% of incident solar energy is actually captured and converted into food (chemical energy) by plants.
b) This is why the amount of energy available at successive trophic levels keeps decreasing sharply.

16. Which of the following groups do not constitute a food chain?
(i) Wolf, rabbit, grass, lion
(ii) Plankton, man, grasshopper, fish
(iii) Hawk, grass, snake, grasshopper, frog
(iv) Grass, snake, wolf, tiger [1 Mark]
(A) (i) and (iv)
(B) (i) and (iii)
(C) (ii) and (iii)
(D) (ii) and (iv)

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

Teacher's Note:
a) A valid food chain must show a correct feeding sequence: producer, then herbivore, then carnivore(s), in order.
b) In group (ii) and (iv), the organisms are not arranged in a logical feeding sequence, so they do not form a food chain.

For Questions number 17 to 20, two statements are given, one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

17. Assertion (A): A human child bears all the basic features of human beings.
Reason (R): It looks exactly like its parents, showing very little variations. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) A human child never looks exactly like its parents; small variations always occur due to sexual reproduction.
b) The basic body plan is inherited (A is true) but "exactly like" makes R factually false.

18. Assertion (A): The amount of ozone in the atmosphere began to drop sharply in the 1980s.
Reason (R): The oxygen atoms combine with molecular oxygen to form ozone. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

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

Teacher's Note:
a) Ozone depletion in the 1980s was due to CFCs, not simply the formation mechanism of ozone.
b) The Reason correctly describes how ozone forms, but it does not explain why ozone was depleting.

19. Assertion (A): Decomposition reactions are generally endothermic reactions.
Reason (R): Decomposition of organic matter into compost is an exothermic process. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

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

Teacher's Note:
a) Most decomposition reactions need heat, light or electricity input, making them endothermic.
b) Composting is an exception that is exothermic, so it does not explain the general endothermic nature stated in A.

20. Assertion (A): No two magnetic field lines are found to cross each other.
Reason (R): The compass needle cannot point towards two directions at the point of intersection of two magnetic field lines. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

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

Teacher's Note:
a) At any point, the field has only one direction, shown by the tangent to the field line at that point.
b) If two lines crossed, the compass would need to point in two directions at once, which is impossible.

SECTION B

21. "Excessive use of chemicals and pesticides in agriculture adversely effect the environment." Justify this statement. [2 Marks]

Answer:
1. Chemicals and pesticides used by farmers get washed into soil and water bodies, harming both living and non-living components of the ecosystem.
2. Being mostly non-biodegradable, they accumulate progressively at each trophic level (biological magnification), harming organisms at all levels.

Teacher's Note:
a) The keyword "biological magnification" is essential for full marks.
b) Mention that these chemicals are non-biodegradable, which is the reason for their accumulation.

22. (a) Consider the following circuits:
In which circuit will the power dissipated in the circuit be (I) minimum (II) maximum? Justify your answer. [2 Marks]

[Figure: Three circuit diagrams each with a 12V battery: (i) one 5Ω resistor alone in the loop, (ii) two 5Ω resistors joined end to end (series) in the loop, (iii) two 5Ω resistors joined between the same two points (parallel) with the battery.]

Answer:
1. Power dissipated is minimum in circuit (ii) and maximum in circuit (iii).
2. Since \( P = \dfrac{V^2}{R} \) and voltage is the same (12 V) in all three cases, power is inversely proportional to the total resistance of the circuit; series combination gives the highest resistance (minimum power) and parallel combination gives the lowest resistance (maximum power).

Teacher's Note:
a) Use \( P = V^2/R \) since voltage is constant across all circuits, not \( P = I^2R \).
b) Remember: series resistance is always higher, parallel resistance is always lower than any individual resistor.

OR

(b) Two lamps, rated 100 W; 220 V and 60 W; 220 V are connected in parallel to electric main supply of 220 V. Find the current drawn by the two lamps from the supply. [2 Marks]

Answer:
1. Current through 100 W lamp: \( I_1 = \dfrac{P_1}{V} = \dfrac{100}{220} = 0.45 \, A \)
2. Current through 60 W lamp: \( I_2 = \dfrac{P_2}{V} = \dfrac{60}{220} = 0.27 \, A \); Total current drawn \( I = I_1 + I_2 = 0.72 \, A \)

Teacher's Note:
a) In a parallel connection, total current is the sum of individual branch currents.
b) Use \( P = VI \) directly for each lamp since voltage across both is the same (220 V).

23. An object is placed at a distance of 30 cm in front of a concave mirror of focal length 20 cm. Use mirror formula to determine the position of the image formed in this case. [2 Marks]

Answer:
1. Given: \( f = -20 \, cm \), \( u = -30 \, cm \); using \( \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} \), we get \( \dfrac{1}{v} = \dfrac{1}{-20} - \dfrac{1}{-30} = -\dfrac{1}{60} \)
2. So \( v = -60 \, cm \); the image is formed 60 cm in front of the mirror.

Teacher's Note:
a) Always apply the sign convention correctly: distances measured against incident light direction are negative.
b) A negative value of v confirms the image is real and formed in front of the mirror.

24. (a) Besides minimising the loss of blood, why is it essential to plug any leak in a blood vessel? Name the component of blood which helps in this process and state how this component perform this function. [2 Marks]

Answer:
1. Plugging the leak prevents a fall in blood pressure and maintains the efficiency of the pumping (circulatory) system.
2. Platelets help in this process by clotting the blood at the site of injury, sealing the leak.

Teacher's Note:
a) The key term "platelets" must be mentioned by name to score full marks.
b) Do not confuse platelets with red blood cells, which only carry oxygen.

OR

(b) (i) The transport system in plants is relatively slower than in animals. Give reasons.
(ii) State the role of phloem in the transport of materials in plants. [2 Marks]

Answer:
1. Plants have low energy needs since they do not move and have a large proportion of dead cells in many tissues, so a slow transport system is sufficient.
2. Phloem translocates soluble products of photosynthesis (like sugars) and other substances from leaves to storage organs such as roots, fruits, seeds and growing parts of the plant.

Teacher's Note:
a) Do not mix up xylem (water transport) with phloem (food transport) in your answer.
b) Mention that phloem transport can occur in both upward and downward directions, unlike xylem.

25. Draw labelled diagrams to show different stages of budding in Hydra. [2 Marks]

[Figure: A sequence of three Hydra figures showing budding: a small bud emerging from the body wall of the parent Hydra, the bud growing larger and developing tentacles while attached to the parent, and the fully formed young Hydra detaching from the parent as an independent organism.]

Answer: The diagram should show a small outgrowth (bud) appearing on the body of the parent Hydra, developing tentacles as it grows, and finally detaching to become a new independent Hydra, with each stage clearly labelled.

Teacher's Note:
a) Marks are given for both the correct diagram and for proper labelling of each stage.
b) Budding is a form of asexual reproduction; ensure the bud is shown attached to the parent initially.

26. What happens when: (write balanced chemical equation)
(a) Lead nitrate is thermally decomposed,
(b) Natural gas burns in oxygen (or air)? [2 Marks]

Answer:
1. \( 2Pb(NO_3)_2 \xrightarrow{heat} 2PbO + 4NO_2 + O_2 \)
2. \( CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + \text{heat} + \text{light} \)

Teacher's Note:
a) Lead nitrate decomposition is a good example of thermal decomposition producing a brown gas (\( NO_2 \)).
b) Combustion of methane is exothermic, releasing heat and light; balance the equation carefully.

SECTION C

27. (a) Define one ampere.
(b) The resistance of a wire of 0.01 cm radius is 14 Ω. If the resistivity of the material of the wire is \( 44 \times 10^{-8} \, \Omega m \), find the length of the wire. (Given \( \pi = \dfrac{22}{7} \)) [3 Marks]

Answer:
1. One ampere is the current flowing through a conductor when one coulomb of charge passes through any cross-section of it in one second.
2. Given: radius \( r = 0.01 \times 10^{-2} \, m \), \( R = 14 \, \Omega \), \( \rho = 44 \times 10^{-8} \, \Omega m \); using \( R = \rho \dfrac{l}{A} \), \( l = \dfrac{RA}{\rho} = \dfrac{14 \times 22 \times (0.01 \times 10^{-2})^2}{7 \times 44 \times 10^{-8}} \)
3. On solving, \( l = 1.0 \, m \).

Teacher's Note:
a) Convert radius from cm to m carefully before calculating area.
b) Use \( A = \pi r^2 \) with the given value of \( \pi \) for accurate cancellation.

28. Consider the following electric circuit:
Calculate the values of the following: [3 Marks]
(a) The total resistance of the circuit
(b) The total current drawn from the source
(c) Potential difference across the parallel combination of 10 Ω and 15 Ω resistors

[Figure: Circuit with a 15V battery, an ammeter, a switch K, a parallel combination of 10Ω and 15Ω resistors, and a parallel combination of 60Ω and 40Ω resistors, all connected in a single series loop.]

Answer:
1. Parallel combination of 10 Ω and 15 Ω: \( \dfrac{1}{R_1} = \dfrac{1}{10} + \dfrac{1}{15} = \dfrac{1}{6} \Rightarrow R_1 = 6 \, \Omega \); Parallel combination of 60 Ω and 40 Ω: \( R_2 = 24 \, \Omega \); since these are in series, total resistance \( = 6 + 24 = 30 \, \Omega \)
2. Total current \( I = \dfrac{V}{R} = \dfrac{15}{30} = 0.5 \, A \)
3. Potential difference across \( R_1 \): \( V = IR_1 = 0.5 \times 6 = 3.0 \, V \)

Teacher's Note:
a) Simplify each parallel group into a single equivalent resistance before adding them in series.
b) The same current flows through both series groups, so use it directly to find voltage across each group.

29. Draw ray diagrams to show the nature, position and relative size of the image formed by a convex mirror when the object is placed (i) at infinity and (ii) between infinity and pole P of the mirror. [3 Marks]

[Figure: Two ray diagrams for a convex mirror. In the first, parallel rays from an object at infinity appear to diverge from a point-sized virtual image at the focus F behind the mirror. In the second, an object AB is placed between infinity and the pole P; rays reflected from the mirror appear to diverge from a virtual, erect, diminished image located between the pole and focus behind the mirror.]

Answer:
1. When the object is at infinity, the image formed by a convex mirror is virtual, highly diminished (point-sized) and located exactly at the focus F, behind the mirror.
2. When the object is placed between infinity and the pole P, the image formed is virtual, erect, diminished in size, and located between the pole P and the focus F, behind the mirror.

Teacher's Note:
a) A convex mirror always forms a virtual, erect and diminished image, regardless of object position.
b) Marks are lost if arrowheads showing direction of light rays are missing on the diagram.

30. (a) With the help of an activity, explain the conditions under which iron articles get rusted. [3 Marks]

Answer:
1. Take three test tubes A, B and C, each with a clean iron nail; in tube A, add ordinary water and leave it open to air.
2. In tube B, add boiled distilled water and cover it with a layer of oil (to prevent air from dissolving in water); in tube C, add anhydrous calcium chloride (to absorb any moisture) and cork it tightly.
3. After a few days, the nail in tube A rusts, but nails in tubes B and C do not, showing that both air and water together are necessary for rusting.

Teacher's Note:
a) Full marks are also given if the activity is explained with a properly labelled diagram instead of text.
b) The key conclusion is that rusting needs both oxygen (air) and moisture simultaneously.

OR

(b) (i) Name two metals which react violently with cold water. List any three observations which a student notes when these metal are dropped in a beaker containing water.
(ii) Write a test to identify the gas evolved (if any) during the reaction of these metals with water. [3 Marks]

Answer:
1. Sodium and potassium react violently with cold water (any two, e.g., sodium, potassium, lithium).
2. Observations: the reaction is very vigorous; a large amount of heat is evolved; the evolved gas may catch fire.
3. Test: the gas evolved (hydrogen) burns with a pop sound when brought near a burning matchstick.

Teacher's Note:
a) Sodium and potassium float and melt into a ball due to the heat released during the reaction.
b) The "pop sound" test is the standard confirmatory test for hydrogen gas.

31. (a) "Displacement reactions also play a key role in extracting metals in the middle of the reactivity series." Justify this statement with two examples.
(b) Why can metals high up in the reactivity series not be obtained by reduction of their oxides by carbon? [3 Marks]

Answer:
1. Metals in the middle of the reactivity series (like Mn, Fe) can displace less reactive metals from their oxides using a more reactive metal such as aluminium, as in the thermite reaction: \( 3MnO_2 + 4Al \rightarrow 3Mn + 2Al_2O_3 + heat \)
2. Similarly, \( Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3 + heat \), both reactions being highly exothermic displacement reactions used industrially.
3. Metals high up in the reactivity series (like Na, Mg, Ca) have a greater affinity for oxygen than carbon does, so carbon cannot remove oxygen from their oxides; they must be extracted by electrolytic reduction instead.

Teacher's Note:
a) The thermite reaction (with Al and Fe2O3) is a very commonly asked example; remember it well.
b) The core reason for part (b) is the comparison of affinity for oxygen between carbon and the reactive metal.

32. In one of Mendalian experiments, when \( F_1 \) generation pea plants with round yellow seeds were self-pollinated, pea seeds with the following combinations were obtained in \( F_2 \) generation:
Seeds: Round yellow | Round green | Wrinkled yellow | Wrinkled green
Number: 800 | 275 | 268 | 90
Analyse the result and describe the mechanism of inheritance of traits which explains the above results. [3 Marks]

Answer:
1. The four combinations occur approximately in the ratio 9:3:3:1 (round yellow : round green : wrinkled yellow : wrinkled green), including new combinations (round green and wrinkled yellow) not seen in the parents.
2. The parents were RRYY (round yellow) and rryy (wrinkled green); their \( F_1 \) was RrYy, which on self-pollination gave the 9:3:3:1 ratio in \( F_2 \).
3. This shows that the two traits (seed shape and seed colour) are inherited independently of each other, which is Mendel's Law of Independent Assortment.

Teacher's Note:
a) Recognising the 9:3:3:1 ratio pattern is the key value point examiners look for.
b) Clearly name the law as "Law of Independent Assortment" for full marks.

33. (a) Name the glands that secrete:
(i) Adrenaline
(ii) Thyroxin
(b) Explain with example how the timing and amount of hormone released are regulated in the human body. [3 Marks]

Answer:
1. (i) Adrenaline is secreted by the adrenal gland. (ii) Thyroxin is secreted by the thyroid gland.
2. If the blood sugar level rises, this is detected by cells of the pancreas, which respond by producing more insulin to lower it.
3. As the blood sugar level falls back to normal, this is again detected by the pancreas and insulin secretion is correspondingly reduced, showing a feedback mechanism.

Teacher's Note:
a) The example of insulin and blood sugar regulation is the standard example for this feedback mechanism.
b) Emphasise that hormone release is controlled by feedback based on the level of the substance being regulated.

SECTION D

34. (a) (i) The power of a lens 'X' is - 2.5 D. Name the lens and determine its focal length in cm. For which eye defect of vision will an optician prescribe this type of lens as a corrective lens?
(ii) "The value of magnification 'm' for a lens is - 2." Using new Cartesian Sign Convention and considering that an object is placed at a distance of 20 cm from the optical centre of this lens, state:
(I) the nature of the image formed;
(II) size of the image compared to the size of the object;
(III) position of the image, and
(IV) sign of the height of the image.
(iii) The numerical values of the focal lengths of two lenses A and B are 10 cm and 20 cm respectively. Which one of the two will show higher degree of convergence/divergence? Give reason to justify your answer. [5 Marks]

Answer:
(i) Since the power is negative, lens X is a concave lens. Focal length \( f = \dfrac{1}{P} = \dfrac{1}{-2.5} = -0.4 \, m = -40 \, cm \). This lens is prescribed to correct Myopia (short-sightedness).
(ii) (I) Since \( m = -2 \) is negative, the image is real and inverted.
(II) The image is magnified: it is double the size of the object.
(III) The image is formed beyond 2F, on the other side of the lens (here \( v = m \times u = -2 \times -20 = +40 \, cm \)).
(IV) The height of the image is negative, because the image is inverted.
(iii) Lens A (focal length 10 cm) shows the higher degree of convergence/divergence, because power \( P = \dfrac{1}{f} \): the shorter the focal length, the greater the power.

Teacher's Note:
a) Negative power always indicates a concave/diverging lens, used to correct myopia.
b) Remember: smaller focal length always means higher power (more converging or diverging ability).
c) A negative magnification with magnitude greater than 1 always means a real, inverted, magnified image.

OR

(b) (i) Draw a ray diagram to show the refraction of a ray of light through a rectangular glass slab when it falls obliquely from air into glass.
(ii) State Snell's law of refraction of light.
(iii) Differentiate between the virtual images formed by a convex lens and a concave lens on the basis of:
(I) object distance, and
(II) magnification. [5 Marks]

[Figure: Ray diagram of a rectangular glass slab. A ray of light travelling in air strikes the top surface obliquely at point O, bending towards the normal as it enters the denser glass medium, travelling through the slab, and emerging from the bottom face bending away from the normal, parallel to the original incident ray but laterally displaced.]

Answer:
1. The ray diagram should show the incident ray bending towards the normal on entering the glass (denser medium) and then bending away from the normal on emerging into air, resulting in an emergent ray parallel to the incident ray but laterally shifted.
2. Snell's law states that 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 a given pair of media: \( \dfrac{\sin i}{\sin r} = constant \)
3. For a convex lens, a virtual image is formed only when the object is placed between the optical centre O and the focus F.
4. For a concave lens, a virtual image is formed no matter where the object is placed in front of the lens.
5. The virtual image formed by a convex lens is always magnified, while the virtual image formed by a concave lens is always diminished.

Teacher's Note:
a) Half a mark may be deducted if arrowheads showing the direction of light are missing from the ray diagram.
b) Snell's law formula with correct symbols (i and r) is a key value point.
c) Concave lenses always form virtual, erect, diminished images regardless of object position; this is a common exam distinction.

35. (a) A saturated organic compound 'A' with two carbon atoms belongs to the homologous series of alcohols. On oxidation, it forms an organic acid 'B' with molecular mass 60 u. On heating 'A' with excess concentrated sulphuric acid at 443 K, an unsaturated hydrocarbon 'C' is formed.
(i) Name A, B and C.
(ii) Calculate molecular mass of C.
(iii) What happens when a pinch of sodium carbonate is added to compound B? Write chemical equation for the reaction.
(iv) Draw electron dot structure of compound B. [5 Marks]

Answer:
1. A = Ethanol (\( C_2H_5OH \)); B = Ethanoic acid (\( CH_3COOH \)); C = Ethene (\( C_2H_4 \))
2. Molecular mass of ethene \( (C_2H_4) = 12 \times 2 + 1 \times 4 = 28 \, u \)
3. Brisk effervescence (bubbles of \( CO_2 \) gas) is observed; \( 2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2 \)
4. The electron dot structure of ethanoic acid shows a methyl carbon bonded to three hydrogen atoms and to a carboxyl carbon, which is double bonded to one oxygen and single bonded to an -OH group, with all shared electron pairs shown between bonded atoms.

Teacher's Note:
a) Ethanol to ethanoic acid to ethene is a classic reaction sequence tested every year; remember all three names.
b) Effervescence with sodium carbonate/bicarbonate is the standard test for carboxylic acids (evolves \( CO_2 \)).
c) In the electron dot diagram, remember the C=O double bond needs two shared pairs of electrons.

OR

(b) (i) What is a homologous series of carbon compounds? Write the name and formula of three successive members of the homologous series of compounds having functional group - COOH.
(ii) Write the name of two carbon compounds in which carbon atoms are arranged in the form of a ring. Draw the structure of any one of the two. [5 Marks]

Answer:
1. A homologous series is a series of carbon compounds having the same functional group, showing similar chemical properties, with each successive member differing from the previous one by a \( CH_2 \) unit and a fixed difference in molecular mass.
2. Three successive members with the -COOH group: Methanoic acid (\( HCOOH \)), Ethanoic acid (\( CH_3COOH \)), Propanoic acid (\( C_2H_5COOH \)).
3. Two carbon compounds with carbon atoms arranged in a ring are Benzene and Cyclohexane.
4. Structure of Benzene: a six-membered ring of carbon atoms with alternating single and double bonds, each carbon atom also bonded to one hydrogen atom.

Teacher's Note:
a) Successive members of a homologous series must be given strictly in increasing order of carbon atoms.
b) Benzene and cyclohexane are the two most common examples of ring (cyclic) carbon compounds asked in exams.
c) Remember benzene has three alternating double bonds within its six-membered ring.

36. (a) (i) Write the functions of the following parts of human female reproductive system:
(I) Ovary
(II) Fallopian tube (oviduct)
(III) Uterus
(ii) State briefly two contraceptive methods used by human males. [5 Marks]

Answer:
1. Ovary: produces the female gamete (egg/ovum) and secretes female hormones such as oestrogen.
2. Fallopian tube: it is the site where fertilisation of the egg by the sperm takes place.
3. Uterus: it is the site of implantation of the embryo and its subsequent development.
4. Mechanical barrier method: use of condoms, which prevent sperm from reaching the egg.
5. Surgical method: vasectomy, in which the vas deferens is blocked/cut in males to prevent sperm from reaching the semen.

Teacher's Note:
a) Do not mix up the function of the fallopian tube (fertilisation) with the uterus (implantation).
b) Vasectomy and condom are the two standard male contraceptive methods expected in this answer.

OR

(b) (i) Differentiate between self-pollination and cross-pollination.
(ii) Identify A, B and C in the diagram given below and write one function of each. [5 Marks]

[Figure: Diagram of a pistil showing two pollen grains resting on the stigma at the top (labelled A), a long tube running down through the style into the ovary (labelled B), and an egg cell inside the ovary at the base (labelled C).]

Answer:
1. In self-pollination, pollen grains are transferred from the stamen to the stigma of the same flower.
2. In cross-pollination, pollen grains are transferred from the stamen of one flower to the stigma of another flower of the same species.
3. A is the Stigma: it receives the pollen grains and provides a suitable environment for their germination.
4. B is the Pollen tube: it carries the male germ cells (gametes) down to the female gamete located in the ovary.
5. C is the Egg cell (female germ cell): it fuses with the male gamete to form the zygote.

Teacher's Note:
a) The main difference to state is "same flower" (self) versus "different flower of the same species" (cross).
b) In the diagram, correctly identifying the pollen tube (B) growing down through the style is essential.
c) Remember fertilisation occurs when the male gamete carried by the pollen tube fuses with the egg cell (C).

SECTION E

The following questions are Source-based/Case-based questions. Read the case carefully and answer the questions that follow.

37. In our homes, we receive the supply of electric power through a main supply also called mains, either supported through overhead electric poles or by underground cables. In our country the potential difference between the two wires (live wire and neutral wire) of this supply is 220 V.

(a) Write the colours of the insulation covers of the line wires through which supply comes to our homes. [1 Mark]

Answer: The live wire has red insulation and the neutral wire has black insulation.

Teacher's Note:
a) Remember the standard colour code: red for live, black for neutral, and green for earth wire.
b) This is a factual recall question worth full marks for stating both colours correctly.

(b) What should be the current rating of the electric circuit (220 V) so that an electric iron of 1 kW power rating can be operated? [1 Mark]

Answer: Using \( P = VI \), \( I = \dfrac{1000}{220} = 4.55 \, A \); therefore the circuit should have a current rating of 5 A.

Teacher's Note:
a) Always round up the current rating to the next standard fuse value (5 A here), not down.
b) Convert power from kW to W before applying the formula \( P = VI \).

(c) (i) What is the function of the earth wire? State the advantage of the earth wire in domestic electric appliances such as electric iron. [2 Marks]

Answer:
1. The earth wire provides a low resistance path for current, so that any leakage of current to the metallic body of the appliance flows safely to the earth, keeping the body's potential equal to that of the earth.
2. This ensures that the user does not get an electric shock if there is a current leakage in the appliance.

Teacher's Note:
a) The key phrase to remember is "low resistance path to the earth" for current leakage.
b) State the direct benefit clearly: prevention of electric shock to the user.

OR

(c) (ii) List two precautions to be taken to avoid electrical accidents. State how these precautions prevent possible damage to the circuit/appliance. [2 Marks]

Answer:
1. Using a fuse wire of correct rating in the circuit prevents damage to the circuit due to overloading or short-circuiting.
2. Using an earth wire connected to the metallic body of appliances prevents electric shock due to leakage of current, by directing it safely to the ground.

Teacher's Note:
a) Fuse protects the circuit/wiring; earth wire protects the user - keep this distinction clear.
b) Any two valid precautions (fuse, earthing, using ELCB) with correct reasoning earn full marks.

38. The maintenance functions of all living organisms must go on even when they are not doing anything particular. Even when we are just sitting in a class or even asleep, this maintenance job has to go on. These maintenance processes require energy to prevent damage and break-down of cells and tissues, which is obtained by the individual organism from the food prepared by the autotrophs, called producers.

(a) Name and define the process by which green plants prepare food. [1 Mark]

Answer: The process is Photosynthesis, in which green plants capture sunlight and convert it into chemical energy, using chlorophyll to convert carbon dioxide and water into carbohydrates.

Teacher's Note:
a) Both naming the process and giving its correct definition are needed for the full 1 mark.
b) Mention chlorophyll, sunlight, carbon dioxide and water as key words in the definition.

(b) Write chemical equation involved in the above process. [1 Mark]

Answer: \( 6CO_2 + 12H_2O \xrightarrow[\text{Chlorophyll}]{\text{Sunlight}} C_6H_{12}O_6 + 6O_2 + 6H_2O \)

Teacher's Note:
a) The conditions "sunlight" and "chlorophyll" must be shown above/below the arrow.
b) Glucose (\( C_6H_{12}O_6 \)) is the main product to be remembered along with oxygen.

(c) (i) State in proper sequence the events that occur in synthesis of food by desert plants. [2 Marks]

Answer:
1. Light energy is first absorbed by chlorophyll and converted into chemical energy, and water molecules are split into hydrogen and oxygen.
2. Desert plants take up carbon dioxide at night and prepare an intermediate compound, which is then used along with the energy absorbed by chlorophyll during the day to reduce carbon dioxide into carbohydrates.

Teacher's Note:
a) The special adaptation of desert (CAM) plants taking in CO2 at night is the key point examiners look for.
b) Mention the correct sequence: light absorption, energy conversion, then CO2 reduction.

OR

(c) (ii) Explain giving reasons what happens to the rate at which the green plants will prepare food
(I) during cloudy weather, and
(II) when stomata get blocked due to dust. [2 Marks]

Answer:
1. During cloudy weather, the rate of photosynthesis decreases due to the low amount of sunlight available.
2. When stomata get blocked by dust, the rate of photosynthesis decreases because gaseous exchange (uptake of \( CO_2 \)) is reduced.

Teacher's Note:
a) Always link the decrease in rate to the specific limiting factor mentioned (light or gas exchange).
b) Stomata are essential for CO2 entry; their blockage directly affects the raw material supply for photosynthesis.

39. Seawater contains many salts dissolved in it. Common salt is separated from these salts. Deposits of solid salt are also found in several parts of the world. These large crystals are often brown due to impurities. This is called rock salt and is mined like coal. The common salt is an important raw material for chemicals of daily use.

(a) Write balanced chemical equations to show the products formed during electrolysis of brine. [1 Mark]

Answer: \( 2NaCl + 2H_2O \xrightarrow{\text{electricity}} 2NaOH + H_2 + Cl_2 \)

Teacher's Note:
a) This process is called the chlor-alkali process; remember all three products: NaOH, \( H_2 \), and \( Cl_2 \).
b) The equation must be balanced for full marks.

(b) List two uses of any one product obtained during electrolysis of brine. [1 Mark]

Answer: Uses of NaOH: it is used for degreasing metals, and in making soaps and detergents.

Teacher's Note:
a) Alternative valid answers include uses of \( H_2 \) (as fuel) or \( Cl_2 \) (as disinfectant/water treatment).
b) Only two uses of any one single product are required, not multiple products.

(c) (i) A mild non-corrosive basic salt 'A', used for faster cooking, is strongly heated to produce a compound 'B', that is used for removing permanent hardness of water. Identify A and B and also write the equation for the reaction that occurs when A is heated. [2 Marks]

Answer:
1. A is Sodium hydrogen carbonate/baking soda (\( NaHCO_3 \)) and B is Sodium carbonate (\( Na_2CO_3 \)).
2. \( 2NaHCO_3 \xrightarrow{heat} Na_2CO_3 + H_2O + CO_2 \)

Teacher's Note:
a) Baking soda decomposing to washing soda on heating is a frequently tested reaction.
b) Remember that washing soda (Na2CO3) is used to remove permanent hardness of water.

OR

(c) (ii) Define water of crystallisation. Give two examples of salts that have water of crystallisation. [2 Marks]

Answer:
1. Water of crystallisation is the fixed number of water molecules present in one formula unit of a salt.
2. Examples: Copper sulphate pentahydrate (\( CuSO_4.5H_2O \)) and Washing soda / Sodium carbonate decahydrate (\( Na_2CO_3.10H_2O \)).

Teacher's Note:
a) The word "fixed number" is essential in the definition to earn full marks.
b) Other valid examples include Gypsum (\( CaSO_4.2H_2O \)) and Green vitriol (\( FeSO_4.7H_2O \)).

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