CBSE Class 10 Science Question Paper 2025 Solved Code 31-1-3

SECTION A

1. 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 electrolysis of their molten chlorides.
b) Remember that Na, Mg, Ca and Al are obtained by electrolytic reduction, not carbon reduction.

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

[Figure: Four electron dot diagrams. (A) shows Mg transferring 2 electrons to one O atom to form Mg\(^{2+}\) and \([O]^{2-}\). (B) shows Mg transferring only 1 electron to O forming Mg\(^{+}\) and \([O]^{-}\). (C) shows one Mg transferring electrons to two separate O atoms forming Mg\(^{2+}\) and two \([O]^{-}\) ions. (D) shows 2 Mg atoms combining with an O2 molecule to form \([Mg^{2+}]_2[O^{2-}]\).]

Answer: (A) - one Mg atom transfers 2 electrons to one oxygen atom, forming Mg\(^{2+}\) and \(O^{2-}\) ions, each with a stable octet.

Teacher's Note:
a) Magnesium has 2 valence electrons and oxygen needs 2 electrons to complete its octet, so a 1:1 ratio is correct.
b) Always check that both ions achieve a stable noble gas configuration.

3. 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 an ionic compound, which has strong (not weak) electrostatic forces between ions.
b) Ionic compounds have high melting points and conduct electricity when molten or dissolved.

4. 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) Acid-base and acid-oxide reactions always produce salt and water.
b) Note that (ii) and (iii) do not give neutral salts always, so option (B) is wrong.

5. Tooth enamel is made up of calcium hydroxyapatite (a crystalline form of calcium phosphate). This chemical starts corroding in the mouth when the pH is : [1 Mark]
(A) 7
(B) 5
(C) 10
(D) 14

Answer: (B) 5

Teacher's Note:
a) Tooth decay starts when the mouth pH falls below 5.5.
b) Bacteria act on sugar left in the mouth to produce acids that lower the pH.

6. The products formed when Aluminium and Magnesium are burnt in the presence of air respectively are : [1 Mark]
(A) Al₃O₄ and MgO₂
(B) Al₂O₃ and MgO
(C) Al₃O₄ and MgO
(D) Al₂O₃ and MgO₂

Answer: (B) Al₂O₃ and MgO

Teacher's Note:
a) Balance the valency of the metal with oxygen to get the correct oxide formula.
b) Aluminium is trivalent so it forms Al2O3, magnesium is divalent so it forms MgO.

7. 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) Water is H2O, so by mass ratio, hydrogen : oxygen = (2×1) : 16 = 1 : 8.
b) Do not confuse mass ratio with volume ratio, which is 2:1.

8. The breakdown of glucose has taken the following pathway :
Glucose (a) → Pyruvate + Energy (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) Glucose breaks down to pyruvate in the cytoplasm during glycolysis.
b) Pyruvate converts to lactic acid in muscle cells when oxygen supply is insufficient.

9. 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 F₁ 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) RRyy x rrYY gives all F1 offspring as RrYy.
b) Round (R) and yellow (Y) are dominant traits, so all F1 seeds are round and yellow.

10. The correct/true statement(s) for a bisexual flower is/are : [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 stamen and pistil, so statement (i) is correct.
b) Such flowers can show either self or cross pollination, hence statement (iii) is correct.

11. 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 accumulates on the shaded side of the shoot causing more elongation there.
b) This unequal growth causes the shoot to bend towards light (phototropism).

12. 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 simple sugars.
b) Less saliva means less amylase enzyme, reducing this conversion.

13. 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 the sunlight falling on leaves is converted into food energy.
b) The remaining 99% is reflected, transmitted or not absorbed.

14. Which of the following groups do not constitute a food chain ? [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 food chain must show organisms in order from producer to consumers correctly.
b) Check each group for correct sequential feeding order before answering.

15. The phenomenon responsible for making the smoke particles visible when a beam of sunlight enters a smoke filled room through a narrow hole is : [1 Mark]
(A) scattering of light
(B) dispersion of light
(C) reflection of light
(D) internal reflection of light

Answer: (A) Scattering of light

Teacher's Note:
a) This is the Tyndall effect, caused by scattering of light by suspended particles.
b) Do not confuse scattering with dispersion, which separates white light into colours.

16. 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 from the sun converge at its focus.
b) Mirror Y is a convex mirror, always giving a virtual, erect, diminished image.

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.

17. 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) Reason (R) explains how ozone is formed, not why it began depleting in the 1980s.
b) The actual cause of ozone depletion is CFCs, which is not mentioned in R.

18. 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 magnetic field has only one definite direction shown by the compass needle.
b) If two field lines crossed, there would be two directions at that point, which is impossible.

19. 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 child does not look exactly like its parents; small variations always exist.
b) These variations arise due to sexual reproduction and DNA copying errors.

20. 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, so they are endothermic.
b) Composting is an exception which is exothermic, so R does not explain A.

SECTION B

21. An object is placed at a distance of 10 cm in front of a concave mirror of focal length 15 cm. Use mirror formula to determine the position of the image formed by this mirror. [2 Marks]

Answer:
Given: \( u = -10 \) cm, \( f = -15 \) cm
Using \( \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} \)
\( \dfrac{1}{-15} = \dfrac{1}{v} + \dfrac{1}{-10} \)
\( \dfrac{1}{v} = \dfrac{-1}{15} + \dfrac{1}{10} = \dfrac{1}{30} \)
\( v = +30 \) cm, so the image is formed 30 cm behind the mirror (virtual image).

Teacher's Note:
a) Always apply sign convention: distances measured against incident light are negative.
b) A positive value of v for a concave mirror means the image is virtual.

22. (a) Consider two lamps A and B of rating 50 W; 220 V and 25 W; 220 V respectively. Find the ratio of the resistances of the two lamps (i.e. R_A : R_B). [2 Marks]

Answer:
1. Using \( R = \dfrac{V^2}{P} \), \( R_A = \dfrac{(220)^2}{50} \) and \( R_B = \dfrac{(220)^2}{25} \).
2. Since voltage is same, \( R \) is inversely proportional to \( P \); so \( R_A : R_B = 1 : 2 \).

Teacher's Note:
a) Same voltage rating means power and resistance are inversely related.
b) The lamp with higher wattage has lower resistance.

OR

(b) Heat produced per second due to a current in a resistor of 4 Ω is 400 joules. Calculate the potential difference across the resistor. [2 Marks]

Answer:
1. Using \( H = \dfrac{V^2}{R} \times t \): \( 400 = \dfrac{V^2}{4} \times 1 \)
2. \( V^2 = 1600 \Rightarrow V = 40 \) V.

Teacher's Note:
a) Heat produced per second is the power dissipated in the resistor.
b) Substitute given values carefully to avoid calculation errors.

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

[Figure: A series of Hydra bodies showing a small bud developing as an outgrowth on the parent body, gradually growing tentacles, and finally detaching as a new independent individual.]

Answer:
1. A bud develops as an outgrowth due to repeated cell division at one specific site on the body of Hydra.
2. The bud grows, develops tentacles, and detaches from the parent body to become a new independent organism.

Teacher's Note:
a) Labelling of the bud and tentacles is essential to score full marks.
b) Budding is a form of asexual reproduction seen in simple organisms like Hydra.

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 drop 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.

Teacher's Note:
a) Platelets are the key word examiners look for in this answer.
b) Do not confuse platelets with red or white blood cells.

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 a large proportion of their tissues are made of dead cells, and they do not move.
2. Phloem translocates soluble products of photosynthesis from leaves to storage organs like roots, fruits, seeds and other growing parts.

Teacher's Note:
a) Translocation is the key term to use for phloem transport.
b) Phloem transport works in both directions, unlike xylem.

25. A student performs the following experiment in his school laboratory.
Class Interval: Test tube contains dilute sulphuric acid | Zinc granules
List two observations to justify that in this experiment a chemical change has taken place. [2 Marks]

[Figure: A conical flask containing dilute sulphuric acid with zinc granules, closed with a cork through which a delivery tube passes.]

Answer:
1. Evolution of a gas (hydrogen) is observed as bubbles in the solution.
2. There is a change (rise) in the temperature of the reaction mixture.

Teacher's Note:
a) Gas evolution and temperature change are the standard indicators of a chemical change.
b) The gas evolved can be confirmed as hydrogen using a burning splint test (pop sound).

26. Translate the following statements into chemical equations and then balance them : [2 Marks]
(a) Nitric acid reacts with calcium hydroxide to form calcium nitrate and water.
(b) Sodium chloride reacts with silver nitrate to form silver chloride and sodium nitrate.

Answer:
(a) \( 2HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + 2H_2O \)
(b) \( NaCl + AgNO_3 \rightarrow AgCl + NaNO_3 \)

Teacher's Note:
a) Always balance the number of atoms of each element on both sides.
b) Reaction (b) is a double displacement (precipitation) reaction.

SECTION C

27. (a) Define one volt potential difference between two points in an electric field.
(b) Draw a schematic diagram of an electric circuit of a cell of 1.5 V, 5 Ω and 10 Ω resistor and a plug key, all connected in series. Calculate the current drawn from the cell when the key is closed. [3 Marks]

[Figure: A series circuit with a 1.5 V cell, a 5 Ω resistor, a 10 Ω resistor and a plug key, all connected in a single loop.]

Answer:
1. One volt potential difference is the work done (1 joule) in moving 1 coulomb of charge from one point to the other in an electric field: \( 1V = \dfrac{1J}{1C} \).
2. In series circuit, total resistance = \( 5 + 10 = 15 \, \Omega \).
3. Current = \( \dfrac{V}{R} = \dfrac{1.5}{15} = 0.1 \) A.

Teacher's Note:
a) In series circuits, resistances simply add up.
b) Always show the circuit symbols correctly - cell, resistor and key. c) Substitute values with correct units to avoid loss of marks.

28. Consider the following electric circuit :
15 V source with a plug key K in series with a parallel combination of 60 Ω and 40 Ω resistors, in series with an ammeter and a parallel combination of 10 Ω and 15 Ω resistors.
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: A circuit with a 15 V battery, a plug key K, an ammeter, a parallel combination of 10 Ω and 15 Ω resistors, and a parallel combination of 60 Ω and 40 Ω resistors, all connected in series in a single loop.]

Answer:
1. Parallel combination of 10 and 15: \( \dfrac{1}{R_1} = \dfrac{1}{10} + \dfrac{1}{15} = \dfrac{1}{6} \Rightarrow R_1 = 6\,\Omega \)
2. Parallel combination of 60 and 40: \( \dfrac{1}{R_2} = \dfrac{1}{60} + \dfrac{1}{40} = \dfrac{1}{24} \Rightarrow R_2 = 24\,\Omega \)
3. Total resistance = \( R_1 + R_2 = 6 + 24 = 30\,\Omega \)
4. Total current: \( I = \dfrac{V}{R} = \dfrac{15}{30} = 0.5 \) A
5. PD across 10&15 parallel combination = \( I \times R_1 = 0.5 \times 6 = 3.0 \) V

Teacher's Note:
a) Simplify parallel combinations first before adding in series.
b) Use \( V = IR \) for the last part, using current common to that branch. c) Keep track of units throughout the calculation.

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 (i): Parallel rays from an object at infinity strike a convex mirror and diverge after reflection, appearing to come from the focus F behind the mirror, forming a point-sized virtual image at F. Figure (ii): An object AB placed between infinity and pole P in front of a convex mirror, with rays diverging after reflection to form a small, virtual, erect image between P and F behind the mirror.]

Answer:
1. When the object is at infinity, the image is formed at the focus F, is highly diminished (point-sized), virtual and erect.
2. When the object is between infinity and pole P, the image is formed between P and F, is diminished, virtual and erect.
3. In both cases the image is virtual, erect and smaller than the object.

Teacher's Note:
a) Convex mirrors always give virtual, erect and diminished images for real objects.
b) Show at least two incident and reflected rays with arrows in each diagram.

30. (a) How many chromosomes are present in human beings ? Out of these how many are sex chromosomes ?
(b) Explain how, in sexually reproducing organisms, the number of chromosomes in the progeny is maintained. [3 Marks]

Answer:
1. Humans have 23 pairs (46) chromosomes, out of which 1 pair (2 chromosomes) are sex chromosomes.
2. In sexually reproducing organisms, the chromosome number is halved during gamete (germ cell) formation.
3. When two germ cells fuse to form a zygote, the normal chromosome number is restored, maintaining the same number as the parents in the progeny.

Teacher's Note:
a) 23 pairs and 1 pair sex chromosomes are the key numerical facts to remember.
b) Meiosis during gamete formation is responsible for halving the chromosome number.

31. A hormone 'X' is secreted in blood when a person is under scary situation.
(a) Identify the hormone 'X' and the gland that secretes it.
(b) Explain its role in dealing with scary or emergency situations. [3 Marks]

Answer:
1. Hormone 'X' is adrenaline, secreted by the adrenal gland.
2. It acts on the heart, making it beat faster, thereby supplying more oxygen to the skeletal muscles.
3. It reduces blood supply to the digestive system and skin, and increases breathing rate by acting on the diaphragm and rib muscles, preparing the body to deal with the emergency.

Teacher's Note:
a) Adrenaline is also called the "fight or flight" hormone.
b) Mention its effect on heart rate and breathing rate for full marks.

32. (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 containing a clean iron nail.
2. In test tube A, add ordinary water and cork it; in test tube B, add boiled distilled water, add oil on top, and cork it (prevents air dissolving); in test tube C, add anhydrous calcium chloride (absorbs moisture) and cork it.
3. The nail in test tube A rusts, but nails in B and C do not, showing that rusting occurs only when both air and water are present together.

Teacher's Note:
a) Both air (oxygen) and moisture are needed together for rusting.
b) A neat labelled diagram of the three test tubes can also be given for full marks.

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 (any two of Na, K, Li) react violently with cold water.
2. Observations: the reaction is very violent, a large amount of heat is evolved, and the evolved gas may catch fire.
3. Test: Bring a burning splint near the gas; it burns with a pop sound, confirming it is hydrogen gas.

Teacher's Note:
a) Sodium and potassium float and melt into a ball due to the heat produced.
b) The pop sound test confirms hydrogen gas.

33. (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 and Fe, are extracted from their oxides using a more reactive metal such as aluminium in a displacement reaction.
2. Example: \( 3MnO_2 + 4Al \rightarrow 3Mn + 2Al_2O_3 + heat \)
3. Example: \( Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3 + heat \)
4. Metals high up in the reactivity series (like Na, Mg, Ca) have a greater affinity for oxygen than carbon, so carbon cannot reduce their oxides.

Teacher's Note:
a) This displacement process is called the thermite reaction and is highly exothermic.
b) Affinity for oxygen is the key reason why carbon reduction fails for reactive metals.

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:
1. Since power is negative, lens X is a concave lens; \( f = \dfrac{100}{P} = \dfrac{100}{-2.5} = -40 \) cm. This type of lens is prescribed for myopia (short-sightedness).
2. Since m is negative, the image is real and inverted.
3. Since \(|m| = 2\), the image is magnified, twice the size of the object.
4. The image is formed beyond 2F, on the other side of the lens from the object.
5. The sign of the height of image is negative (since it is inverted); and the lens with focal length 10 cm has a higher degree of convergence/divergence, since power is inversely proportional to focal length.

Teacher's Note:
a) Power and focal length are related by \(P = 1/f(m)\); smaller focal length means higher power.
b) Negative magnification always indicates a real, inverted image (as formed by convex lens or mirror). c) Concave lens for myopia and convex lens for hypermetropia is a key fact to remember.

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: A light ray travelling in air strikes the first surface of a rectangular glass slab obliquely, bending towards the normal as it enters glass, travelling straight through the slab, then bending away from the normal as it emerges into air, parallel to the original incident ray but laterally displaced.]

Answer:
1. The ray diagram shows the incident ray bending towards the normal on entering the glass (denser medium) and bending away from the normal on emerging into air, with the emergent ray parallel to the incident ray but laterally displaced.
2. Snell's law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media and a given colour of light: \( \dfrac{\sin i}{\sin r} = constant \).
3. For a convex lens, the object must be placed between the optical centre O and the focus F to get a virtual image, while for a concave lens, the object can be placed anywhere in front of the lens to always get a virtual image.
4. A convex lens gives a magnified virtual image, while a concave lens always gives a diminished virtual image.

Teacher's Note:
a) Show clear direction arrows on the ray diagram; marks are deducted otherwise.
b) Concave lens forms only virtual, erect, diminished images for all object positions. c) Convex lens forms a virtual image only when the object is within its focal length.

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

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

Teacher's Note:
a) Learn the specific function of each organ, not just its general role.
b) Condom and vasectomy are the two accepted answers for male contraception.

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 flower's pistil showing two stigmas labelled A at the top with pollen grains on them, a pollen tube labelled B growing down through the style, and an ovule/egg cell labelled C inside the ovary at the base.]

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

Teacher's Note:
a) Self-pollination occurs within the same flower; cross-pollination occurs between flowers of the same species.
b) The pollen tube is the pathway for the male gamete to reach the ovule.

36. (a) (i) Draw electron dot structure of chlorine molecule.
(Atomic Number of Chlorine = 17)
(ii) What happens when chlorine reacts with methane in the presence of sunlight ? Write the name of the reaction.
(iii) Name the two oxidising agents used for the conversion of alcohols to acids.
(iv) List four differences in properties between covalent compounds and ionic compounds. [5 Marks]

[Figure: Electron dot structure showing two chlorine atoms, each with three lone pairs of dots, sharing one pair of electrons (shown by a cross and dot) between them to form a single covalent bond, Cl-Cl.]

Answer:
1. Two chlorine atoms (each with 7 valence electrons) share one pair of electrons to complete their octets, forming the Cl2 molecule with a single covalent bond.
2. Chlorine reacts with methane in the presence of sunlight to form chloromethane and hydrogen chloride: \( CH_4 + Cl_2 \xrightarrow{sunlight} CH_3Cl + HCl \). This is called a substitution reaction.
3. Alkaline KMnO4 and acidified K2Cr2O7 are used as oxidising agents to convert alcohols to acids.
4. Covalent compounds have low melting/boiling points, are poor conductors of electricity, are generally soft solids and dissolve in non-polar solvents; ionic compounds have high melting/boiling points, conduct electricity in molten/aqueous state, are hard crystalline solids and dissolve in polar solvents like water.

Teacher's Note:
a) The substitution reaction is a key term examiners look for regarding chlorine-methane reaction.
b) Remember any two accepted oxidising agents (alkaline/acidified KMnO4 or acidified K2Cr2O7). c) Give differences in pairs (property of covalent vs property of ionic) for clarity.

OR

(b) (i) Give reason why carbon forms compounds mainly by covalent bonding.
(ii) Why do covalent compounds have low melting and boiling points ?
(iii) Give reason for the following :
I. Covalent compounds are bad conductors of electricity.
II. Carbon shows catenation. [5 Marks]

Answer:
1. Carbon has 4 valence electrons; it would require too much energy either to gain 4 electrons (forming C4-) or lose 4 electrons (forming C4+), so it shares its electrons to form covalent bonds instead.
2. Covalent compounds have low melting and boiling points because there are weak forces of attraction between their molecules.
3. Covalent compounds do not form ions, so they cannot conduct electricity.
4. Carbon shows catenation because the carbon-carbon single bond is very strong and stable, allowing carbon atoms to link with each other to form long chains, branched chains and rings.

Teacher's Note:
a) Energy consideration is the key reason carbon avoids ion formation.
b) Weak intermolecular forces (not weak bonds) explain low melting/boiling points. c) Catenation is unique to carbon due to the strength and stability of the C-C bond.

SECTION E

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.

[Figure: not applicable, this is a text-based case.]

(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) Also remember the earth wire is green in colour.
b) These colour codes are standard for domestic wiring in India.

(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 \( I = P/V = 1000/220 = 4.54 \) A, the current rating of the circuit should be 5 A.

Teacher's Note:
a) Always round up the current rating to the next standard fuse value.
b) Standard domestic circuit fuses are commonly rated 5A or 15A.

(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 leakage, keeping the metallic body of the appliance at the same potential as the earth.
2. This ensures that the user does not get an electric shock if there is current leakage in the appliance.

Teacher's Note:
a) Earth wire is a safety feature, not part of the normal current-carrying circuit.
b) It is usually connected to the metal casing of appliances.

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: it prevents damage to the circuit due to overloading or short-circuiting by melting and breaking the circuit.
2. Using an earth wire: it keeps the appliance body at earth potential, preventing severe electric shock in case of current leakage.

Teacher's Note:
a) Fuse protects the circuit; earth wire protects the user.
b) Both devices are essential safety components in domestic wiring.

38. 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: \( 2NaCl + 2H_2O \xrightarrow{electricity} 2NaOH + H_2 + Cl_2 \)

Teacher's Note:
a) This process is called the chlor-alkali process.
b) The three products formed are sodium hydroxide, hydrogen and chlorine.

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

Answer: Sodium hydroxide (NaOH) is used for degreasing metals and for making soaps and detergents.

Teacher's Note:
a) Other uses of NaOH include paper making and preparation of artificial fibres.
b) Uses of Cl2 (disinfectant, PVC) or H2 (as fuel) are equally acceptable.

(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, NaHCO3) and B is sodium carbonate (Na2CO3).
2. \( 2NaHCO_3 \xrightarrow{heat} Na_2CO_3 + H_2O + CO_2 \)

Teacher's Note:
a) Baking soda decomposes on heating to give washing soda, water and carbon dioxide.
b) Washing soda 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 (CuSO4.5H2O) and Gypsum, calcium sulphate dihydrate (CaSO4.2H2O).

Teacher's Note:
a) Washing soda (Na2CO3.10H2O) is another commonly used example.
b) This water can be removed by heating, which changes the colour or form of the salt.

39. 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 use sunlight, chlorophyll, water and carbon dioxide to prepare food (carbohydrates), converting light energy into chemical energy.

Teacher's Note:
a) The definition must include chlorophyll and sunlight as key words.
b) Photosynthesis occurs mainly in the leaves of green plants.

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

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

Teacher's Note:
a) Mention sunlight and chlorophyll as conditions above/below the arrow.
b) Glucose (C6H12O6) is the main product of photosynthesis.

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

Answer:
1. Desert plants take up carbon dioxide at night through open stomata and prepare an intermediate compound.
2. During the day, chlorophyll absorbs light energy, which converts this intermediate into the final carbohydrate product, and light energy is converted into chemical energy with reduction of carbon dioxide.

Teacher's Note:
a) This special adaptation reduces water loss through stomata in desert plants.
b) The sequence (night CO2 uptake, day light reaction) is the key point to mention.

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 CO2 and release of O2) is reduced.

Teacher's Note:
a) Light and CO2 availability are both essential raw materials for photosynthesis.
b) Blocking stomata affects gas exchange, not light absorption directly.

Please click the link below to download pdf file of CBSE Class 10 Science Question Paper 2025 Solved Code 31-1-3