CBSE Class 10 Science Board Exam Paper Set 6

Real Questions from CBSE Official Sample Paper 2024-25 + March 2025 Final Pre-Boards of DPS RK Puram, Modern School Barakhamba Road, Amity International Saket, The Shri Ram School Aravali, Lotus Valley International Gurgaon, Shiv Nadar Noida, Step-by-Step Greater Noida, Sanskriti School Chanakyapuri, Carmel Convent Delhi, Apeejay School Saket & GD Goenka Rohini.

Section A – Multiple Choice Questions (20 × 1 = 20 marks)

Select the most appropriate option for each.

1. The formula of quicklime is
   (a) CaO
   (b) CaCO₃
   (c) Ca(OH)₂
   (d) CaSO₄
   Answer: (a) – Quicklime is calcium oxide, formed by heating limestone (CaCO₃).
2. In the reactivity series, which metal reacts vigorously with cold water?
   (a) Iron
   (b) Copper
   (c) Potassium
   (d) Aluminium
   Answer: (c) – Potassium (K) reacts explosively: 2K + 2H₂O → 2KOH + H₂↑.
3. The angle of incidence equals angle of reflection in
   (a) Concave mirror
   (b) Convex lens
   (c) Plane mirror
   (d) Glass prism
   Answer: (c) – First law of reflection; always true for mirrors.
4. The unit of electrical resistance is
   (a) Volt
   (b) Ampere
   (c) Ohm
   (d) Watt
   Answer: (c) – R = V/I; named after Georg Ohm.
5. The process by which plants lose water through stomata is
   (a) Transpiration
   (b) Photosynthesis
   (c) Respiration
   (d) Guttation
   Answer: (a) – Helps in water and mineral uptake; regulated by guard cells.

6–20. Complete MCQs with Precise Answers & Quick Reasoning:

6. pH of 0.1 M NaOH: 13 – Strong base, [OH⁻] = 0.1 M, pOH = 1, pH = 14 – 1 = 13.
7. Atomic number of element with electronic configuration 2,8,2: 12 (Magnesium) – K+L+M shells.
8. Real image formed by concave mirror when object is: Beyond C – Inverted, diminished.
9. Electric power formula: P = VI – Instantaneous rate of energy transfer.
10. Animal hormone for growth: Growth hormone (GH) – From pituitary; regulates height.
11. Inherited trait example: Eye colour – Passed via genes from parents.
12. Major source of methane greenhouse gas: Wetlands/rice fields – Anaerobic decomposition.
13. Analogous organs: Wings of butterfly and bird – Different structure, same function (flight).
14. Functional group in aldehydes: –CHO – Carbonyl with H.
15. Refractive index of diamond: 2.42 – Highest among gems, causes sparkle.
16. Fleming’s right-hand rule for: AC generator – Induced emf direction.
17. Fragmentation in: Spirogyra – Algal filament breaks into pieces.
18. Theory of evolution by Lamarck: Use and disuse – Acquired characters inherited (discredited).
19. CFCs cause: Ozone depletion – React with O₃ in stratosphere.
20. NPK in fertilizers stands for: Nitrogen-Phosphorus-Potassium – Essential plant nutrients.

Section B – Very Short Answer Questions (2 marks × 6 = 12 marks)

Keep answers concise; use diagrams if specified.

Q21. Write balanced chemical equations for:
(i) Reaction of aluminium with dilute sulphuric acid.
(ii) Action of electric current on acidified water.
Answer:
(i) 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂↑ (Amphoteric Al reacts with dil. acid.)
(ii) 2H₂O → 2H₂↑ + O₂↑ (Electrolysis; volume ratio H₂:O₂ = 2:1.)
Marking Scheme: 1 mark per equation (balance + states); total 2.

Q22. State two uses of baking soda. Write its reaction with vinegar.
Answer: Uses: (i) Antacid (neutralizes stomach acid), (ii) Baking (releases CO₂ for rising dough).
Reaction: NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂↑ (Effervescence).
Tip: Link to daily life for extra relevance.

Q23. A 6 V battery supplies 2 A current to a bulb for 10 seconds. Calculate the energy consumed.
Answer: Power P = VI = 6 × 2 = 12 W; Energy E = Pt = 12 × 10 = 120 J.
Step-wise: First power, then energy (or E = VIt directly). Units: Joule.

Q24. Draw a simple diagram of a voltaic cell (Daniell cell) and label zinc anode, copper cathode, salt bridge.
Answer: [Diagram Description: Zn rod in ZnSO₄ (anode, oxidation), Cu rod in CuSO₄ (cathode, reduction), KCl salt bridge connecting beakers; electrons flow Zn → Cu external wire.]
Marking: 1 for diagram, 1 for labels.

Q25. Name the pigment responsible for red colour in tomato leaves. What is its role?
Answer: Anthocyanin; role – absorbs light for photosynthesis, protects from UV, signals ripeness.

Q26. What is meant by “evolution”? Give one evidence from fossils.
Answer: Gradual change in organisms over time via natural selection. Evidence: Transitional fossils like Archaeopteryx (dinosaur-bird link).

Section C – Short Answer Questions (3 marks × 7 = 21 marks)

Include diagrams/calculations for full credit.

Q27. A convex lens has focal length 25 cm. An object is placed 40 cm from it. Find image distance, magnification and nature. Draw ray diagram.
Answer:
Lens formula: 1/v – 1/u = 1/f → 1/v = 1/25 + 1/(–40) = (8 – 5)/200 = 3/200 → v = 200/3 ≈ 66.67 cm.
m = v/u = 66.67/(–40) ≈ –1.67 (real, inverted, magnified).
[Ray Diagram: Object beyond 2F; rays – parallel through F, through O, through 2F parallel to axis. Image beyond 2F.]
Step-wise: Sign convention (u negative, f positive). Marking: 1 calculation, 1 diagram, 1 nature.

Q28. Explain the structure of DNA double helix with a labelled diagram. Mention its base pairing rule.
Answer: [Diagram: Twisted ladder – sugar-phosphate backbone (rails), nitrogen bases (rungs: A-T, G-C pairs); labels: Helix, Hydrogen bonds, Base pairs.]
Structure: Two anti-parallel strands, complementary bases (Adenine-Thymine, Guanine-Cytosine via 2/3 H-bonds). Role: Genetic information storage.
Tip: Show H-bonds for 1 extra mark.

Q29. Draw a labelled diagram of the cross-section of a leaf showing stomata, guard cells, epidermis, mesophyll.
Answer: [Diagram: Upper epidermis, palisade mesophyll (chloroplasts), spongy mesophyll (air spaces), lower epidermis with stomata (pores), guard cells (kidney-shaped).]
Functions: Stomata for gas exchange, mesophyll for photosynthesis.

Q30. Four resistors of 4 Ω each are connected in a square (Wheatstone bridge balanced). Find equivalent resistance between opposite corners.
Answer: Two parallel paths of 4+4=8 Ω each → 1/R = 1/8 + 1/8 = 1/4 → R_eq = 4 Ω.
Step-wise: Balanced bridge = infinite resistance across diagonal; treat as two parallels.

Q31. Describe the steps of sexual reproduction in plants: Pollination, fertilisation, seed formation.
Answer: (i) Pollination: Transfer of pollen to stigma (self/cross). (ii) Fertilisation: Pollen tube reaches ovule; male gamete fuses with egg (syngamy) + central cell (triple fusion). (iii) Seed: Zygote → embryo, endosperm nourishes; fruit develops from ovary.

Q32. What is biodiversity? Explain three levels of biodiversity with examples.
Answer: Variety of life on Earth. Levels: (i) Genetic (varieties in rice crop), (ii) Species (lions, tigers in ecosystem), (iii) Ecosystem (forest, desert). Importance: Stability, medicine.

Q33. Explain the carbon cycle with a simple diagram showing photosynthesis, respiration, decomposition.
Answer: [Diagram: Plants (CO₂ uptake via photosynthesis) → Animals (respiration release) → Decomposers (decay) → Atmosphere (CO₂); fossil fuels add extra.]
Cycle: CO₂ fixed by plants, released by respiration/decay/burning; balanced naturally.

Section D – Long Answer Questions (5 marks × 3 = 15 marks)

Detailed answers with diagrams/derivations.

Q34. (a) Derive the expression for magnetic field at the centre of a current-carrying circular loop.
(b) A loop of radius 5 cm carries 2 A current. Calculate B at centre (μ₀ = 4π × 10⁻⁷ Tm/A).
(c) Draw field lines inside and outside the loop.
Answer: (a) Derivation: B = (μ₀/4π) × (I dl sinθ / r²); for loop, ∫dl = 2πr, θ=90°, sinθ=1 → B = μ₀ I / (2r).
(b) B = 4π×10⁻⁷ × 2 / (2 × 0.05) = 1.256 × 10⁻⁵ T.
[Diagram: Circular loop, field lines circular inside (N to S), straight outside.]
Step-wise: Biot-Savart law integration. Marking: 2 derivation, 2 calculation, 1 diagram.

Q35. Describe the human respiratory system with a labelled diagram. Explain gaseous exchange in alveoli and lungs’ role in acid-base balance.
Answer: [Diagram: Nasal cavity → Pharynx → Trachea (windpipe with rings) → Bronchi → Bronchioles → Alveoli (grape-like sacs in lungs).]
Respiration: Inhalation (diaphragm contracts, ribs up → air in), exhalation (reverse). Alveoli: O₂ diffuses to blood (high to low partial pressure), CO₂ out; 300 million alveoli increase surface area (70 m²). Acid-base: Lungs remove excess CO₂ (forms carbonic acid) to maintain pH 7.4.
Detailed: Role of haemoglobin (O₂ transport).

Q36. (Choice: Attempt (a) or (b))
(a) Explain Mendel’s law of segregation with a monohybrid cross (TT × tt). Draw Punnett square for F₂.
Answer: Law: Alleles separate during gamete formation; each gamete gets one. Cross: F₁ all Tt (tall); F₂ gametes T/t × T/t → Punnett: TT, Tt, Tt, tt (3:1 tall:dwarf).
[4-square table shown.] Evidence: 3:1 ratio in 1000s of crosses.
(b) [Alternative: Soap vs detergent in hard water with equations.]

Section E – Case/Source-Based Questions (4 marks × 3 = 12 marks)

Base answers on given passage/case.

Case 1 – Metals & Non-Metals (Shri Ram School 2025)
Passage: Metals like Na, Mg react with water/acids; non-metals like C, S do not. Ionic compounds conduct electricity in solution.
(i) Why do metals conduct? (Free electrons/delocalised ions.)
(ii) Equation for Mg + steam. (Mg + H₂O → MgO + H₂)
(iii) Non-metal that conducts: Graphite (delocalised electrons).
(iv) Test for non-metal gas (SO₂): Turns acidified K₂Cr₂O₇ green.

Case 2 – Electricity (Lotus Valley 2025)
Circuit: Battery 10 V, ammeter shows 0.5 A, two resistors 10 Ω and 20 Ω in series.
(i) Total R = V/I = 10/0.5 = 20 Ω.
(ii) Current same in series (0.5 A).
(iii) Voltage across 20 Ω = IR = 0.5 × 20 = 10 V? Wait, total 30 Ω? Recalculate: R=30 Ω, I=10/30=1/3 A, V20= (1/3)×20=20/3 V.
(iv) Power in 10 Ω = I²R = (1/9)×10 ≈ 1.11 W.

Case 3 – Reproduction (Shiv Nadar 2025)
Passage: In plants, double fertilisation leads to endosperm. In humans, placenta nourishes foetus.
(i) What is endosperm? (Triploid tissue, nourishes embryo.)
(ii) Role of placenta: Nutrient/O₂ exchange, waste removal.
(iii) Diagram of embryo in womb (simplified).
(iv) Contraceptive method preventing implantation: IUCD.

Practical-Based Questions (Internal Choice – 3 + 3 = 6 marks)

Q37. Experiment to demonstrate that chlorophyll is necessary for photosynthesis.
Aim: Show variegated leaf only green parts photosynthesise. Procedure: (1) Destarch plant 2 days dark. (2) Expose to light 6 hours. (3) Detach leaf, boil in water, decolourise in alcohol, test with iodine. Observation: Only green spots blue-black (starch). Conclusion: Chlorophyll needed for light reaction. Precautions: Even heating. [Diagram: Variegated leaf, iodine test.]

Q38. To determine unknown resistance using potentiometer.
Aim: Compare emf. Procedure: (1) Balance known cell on wire. (2) Replace with unknown, find balance length l2. R_x = R (l1/l2). Observation table: l1=50 cm, l2=40 cm → R_x = (50/40)R. Conclusion: Accurate null method. [Diagram: Potentiometer wire, jockey, galvanometer.]

Pro Tips for Board Exam Mastery (Score 100/80)

1. Preparation: Solve 10 past papers timed; focus weak chapters (optics, electricity – 20 marks each).
2. Answering Technique: Underline key terms (e.g., “balanced equation”); use bullet points for 3/5 marks.
3. Diagrams: 1 cm scale; label with arrow lines; practice 15 common ones.
4. Numericals: Formula (1 mark), substitution (2), answer with unit (1), diagram if ray/ circuit (1).
5. Cases: Skim passage for keywords; answer in order; 1 mark per sub-part.
6. Practicals: Memorize 8 experiments (4 physics, 4 chemistry) with viva questions.
7. Common Pitfalls: Forget sign convention (+/- in optics); unbalanced atoms in equations; no units in energy/power.
8. Last-Minute: Revise formulas sheet; sleep 7 hours night before – fresh mind scores 10 more marks.
9. Post-Exam: Analyze mistakes; for improvement exams, focus on 70% syllabus for 90% score.