CBSE Class 10 Science Sample Paper 2025 Standard Set 1 with Solutions
Practicing the official CBSE Class 10 Science Sample Paper is the single most effective strategy for students aiming to score 95+ in the board examinations. This comprehensive CBSE Science sample paper with solutions is curated strictly in alignment with the latest CBSE syllabus, NCERT guidelines, and rationalized evaluation criteria. Designed to test foundational recall, analytical thinking, and experimental competencies, this full-length CBSE sample paper 2025 Science standard set provides complete step-by-step marking scheme answers so you can benchmark your preparation and eliminate common exam errors.
Sample Paper Format and BluePrint
The Class 10 Science theory paper carries a maximum weightage of 80 marks to be completed within a duration of 3 hours (180 minutes). An additional 15 minutes of reading time is allocated prior to writing. The question paper comprises 39 compulsory questions divided into five structured sections:
| Section | Question Numbers | Question Type | Marks per Question | Total Marks |
|---|---|---|---|---|
| Section A | Q1 – Q20 | Multiple Choice Questions (including 4 Assertion-Reasoning) | 1 Mark | 20 Marks |
| Section B | Q21 – Q26 | Very Short Answer (VSA) Questions | 2 Marks | 12 Marks |
| Section C | Q27 – Q33 | Short Answer (SA) Questions | 3 Marks | 21 Marks |
| Section D | Q34 – Q36 | Long Answer (LA) Questions | 5 Marks | 15 Marks |
| Section E | Q37 – Q39 | Case-Based / Integrated Competency Units | 4 Marks | 12 Marks |
| Total | 39 Questions | 80 Marks | ||
Internal choices are provided in 2 questions of Section B, 2 questions of Section C, and all 3 questions of Section D, as well as in the 2-mark sub-questions of Section E.
Complete CBSE Class 10 Science Sample Paper with Step-by-Step Solutions
Section A (20 Marks — 1 Mark Each)
Q1. When manganese dioxide is heated with concentrated hydrochloric acid, the reaction is represented by:
MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)
In this chemical reaction, the oxidising agent and the substance oxidised are respectively:
(a) MnO2 and HCl
(b) HCl and MnO2
(c) MnO2 and MnCl2
(d) HCl and Cl2
Answer: (a) MnO2 and HCl
Explanation: MnO2 loses oxygen to form MnCl2 (reduction); therefore, MnO2 acts as the oxidising agent. HCl loses hydrogen and its chloride ions are oxidised to form Cl2 (oxidation); hence, HCl is the substance oxidised.
Q2. A plant shoot bends towards unilateral light. Which phytohormone is responsible for this phototropic response, and where does it accumulate?
(a) Gibberellin, accumulates on the illuminated side
(b) Auxin, accumulates on the shaded side
(c) Cytokinin, accumulates on the illuminated side
(d) Abscisic acid, accumulates on the root tip
Answer: (b) Auxin, accumulates on the shaded side
Explanation: Auxin synthesised at the shoot tip diffuses toward the shaded side of the stem, stimulating faster cell elongation on that side compared to the illuminated side, causing the shoot to bend towards light.
Q3. A ray of light travels from medium A into medium B. If the speed of light in medium A is 2.25 × 108 m/s and in medium B is 1.8 × 108 m/s, the refractive index of medium B with respect to medium A (AnB) is:
(a) 0.80
(b) 1.25
(c) 1.50
(d) 1.33
Answer: (b) 1.25
Explanation: Relative refractive index AnB = vA / vB = (2.25 × 108) / (1.8 × 108) = 1.25.
Q4. Which of the following functional groups undergoes an addition reaction with hydrogen in the presence of a nickel catalyst?
(a) –COOH
(b) –OH
(c) –C≡C–
(d) –CHO
Answer: (c) –C≡C–
Explanation: Unsaturated hydrocarbons containing double (–C=C–) or triple (–C≡C–) carbon bonds undergo catalytic hydrogenation to form saturated alkanes.
Q5. The inner lining of the small intestine possesses millions of microscopic finger-like projections called villi. Their primary function is to:
(a) Secrete digestive enzymes and hydrochloric acid
(b) Increase the surface area for rapid absorption of digested nutrients
(c) Prevent backflow of partially digested chyme
(d) Synthesise bile salts and store glycogen
Answer: (b) Increase the surface area for rapid absorption of digested nutrients
Explanation: Villi are richly supplied with blood vessels and lacteals, significantly increasing the effective absorptive surface area of the ileum.
Q6. The magnetic field lines inside a current-carrying long straight solenoid are:
(a) Circular and non-uniform
(b) Converging at the geometric centre
(c) Parallel straight lines indicating a uniform magnetic field
(d) Completely absent
Answer: (c) Parallel straight lines indicating a uniform magnetic field
Explanation: Inside a solenoid, magnetic field lines run parallel to the axis, demonstrating that the field strength is identical at all interior points.
Q7. (Assertion-Reasoning)
Assertion (A): When zinc metal is dipped in a blue copper sulphate solution, the blue colour gradually fades and a reddish-brown deposit forms.
Reason (R): Zinc is chemically more reactive than copper and displaces copper from copper sulphate solution to form colourless zinc sulphate.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation: Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s). Zinc displaces Cu2+ ions because of its higher position in the electrochemical reactivity series.
Q8. (Assertion-Reasoning)
Assertion (A): The separation of the right and left sides of the human heart prevents the mixing of oxygenated and deoxygenated blood.
Reason (R): Such separation allows a highly efficient supply of oxygen to body cells, which is essential for warm-blooded animals with high energy needs to maintain body temperature.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation: Complete four-chambered double circulation ensures unmixed oxygen-rich systemic distribution for thermoregulation in homeothermic mammals and birds.
Section B (12 Marks — 2 Marks Each)
Q9. Solid lead(II) nitrate is heated strongly in a dry test tube:
(i) State the observable colour change and name the gas evolved as brown fumes.
(ii) Write the balanced chemical equation for this thermal decomposition reaction including physical states.
Solution:
(i) Observations: The white crystalline lead nitrate decomposes with crackling to leave a yellow residue of Lead(II) oxide (PbO), releasing pungent, reddish-brown fumes of Nitrogen dioxide (NO2) gas alongside colourless oxygen gas. [1 Mark]
(ii) Balanced Chemical Equation:
2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g) [1 Mark]
Q10. Differentiate between binary fission in Amoeba and binary fission in Leishmania with respect to the plane of cell division.
Solution:
| Feature | Binary Fission in Amoeba | Binary Fission in Leishmania |
|---|---|---|
| Plane of Division | Can take place along any arbitrary plane due to lack of fixed cell polarity. | Occurs strictly in a definite longitudinal orientation relative to the whip-like flagellum at one end. |
| Structural Specialization | Simple irregular body morphology. | Complex unicellular protozoan containing an anterior flagellum and basal body. |
[1 Mark for each point of distinction]
Q11. Two electric bulbs rated 100 W at 220 V and 60 W at 220 V are connected in parallel to an electric mains supply of 220 V. Calculate the total electric current drawn by the circuit from the line.
Solution:
Using the electric power relation P = V × I ⇒ I = P / V:
Current drawn by Bulb 1: I1 = P1 / V = 100 / 220 = 5 / 11 A ≈ 0.455 A [0.5 Mark]
Current drawn by Bulb 2: I2 = P2 / V = 60 / 220 = 3 / 11 A ≈ 0.273 A [0.5 Mark]
In a parallel circuit, total line current is the sum of branch currents:
Itotal = I1 + I2 = (5 / 11) + (3 / 11) = 8 / 11 A [0.5 Mark]
Final Answer: Itotal = 0.727 A (or 8/11 A) [0.5 Mark]
Section C (21 Marks — 3 Marks Each)
Q12. Zinc is an intermediate reactivity metal extracted from its carbonate and sulphide ores:
(a) Differentiate between Roasting and Calcination processes.
(b) Write balanced chemical equations for converting Zinc Blende (ZnS) and Calamine (ZnCO3) into Zinc Oxide (ZnO).
(c) State the reducing agent used to reduce Zinc Oxide to pure Zinc metal.
Solution:
(a) Roasting involves heating sulphide ores strongly in the presence of excess air, whereas Calcination involves heating carbonate/hydrated ores strongly in a limited supply or absence of air. [1 Mark]
(b) Chemical Equations:
• Roasting of Zinc Blende: 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g) [0.5 Mark]
• Calcination of Calamine: ZnCO3(s) → ZnO(s) + CO2(g) [0.5 Mark]
(c) Reduction: Zinc oxide is reduced to metallic zinc using Carbon (Coke) as the reducing agent: ZnO(s) + C(s) → Zn(s) + CO(g). [1 Mark]
Q13. Trace the sequence of events that occur during a reflex action when you accidentally touch a hot object. Why did reflex arcs evolve in animals even though the complex forebrain is present?
Solution:
Sequence of Reflex Arc Pathway:
1. Stimulus & Receptor: Heat is detected by thermo-receptors in the skin of the hand.
2. Sensory Neuron: Converts the stimulus into an electrochemical nerve impulse and transmits it toward the spinal cord.
3. Relay Neuron (in Spinal Cord): Processes the signal instantly in the grey matter and shifts transmission directly to the motor pathway without waiting for conscious cortical processing.
4. Motor Neuron: Conducts the response impulse from the central nervous system to the effector.
5. Effector (Muscle): Arm muscles contract rapidly to withdraw the hand immediately. [2 Marks]
Evolutionary Significance: Reflex arcs evolved as an emergency survival mechanism because the thinking process of the complex brain network takes longer to evaluate signals. Reflex arcs offer an instantaneous, hardwired protective pathway to prevent severe tissue injury. [1 Mark]
Q14. A concave mirror produces a real, inverted image that is 3 times magnified when an object is placed 10 cm in front of the mirror.
(a) Calculate the image distance (v) from the pole of the mirror.
(b) Determine the focal length (f) and radius of curvature (R) of the concave mirror.
(c) State the nature of the image formed if the object is shifted to 5 cm from the pole.
Solution:
Given: Object distance u = -10 cm. Magnification for real/inverted image m = -3.
(a) Using magnification formula m = -v / u:
-3 = -v / (-10) ⇒ v = -30 cm.
The image is formed 30 cm in front of the mirror (on the same side as the object). [1 Mark]
(b) Using the mirror formula 1/f = 1/v + 1/u:
1/f = 1/(-30) + 1/(-10) = (-1 - 3) / 30 = -4 / 30 = -2 / 15
f = -15 / 2 = -7.5 cm
Radius of curvature R = 2f = 2 × (-7.5) = -15.0 cm. [1 Mark]
(c) When u = -5 cm, the object lies between the Focus (f = -7.5 cm) and the Pole (P). Hence, the image formed will be virtual, erect, and magnified behind the mirror. [1 Mark]
Section D (15 Marks — 5 Marks Each)
Q15. (Chemistry: Carbon Compounds & Soaps)
(a) Define esterification. Write a balanced chemical equation for the reaction between ethanoic acid and ethanol in the presence of concentrated sulphuric acid.
(b) Explain the chemical process of saponification with its equation.
(c) Describe the mechanism of the cleansing action of soap with the help of a labeled micelle structure. Explain why soaps fail to lather effectively in hard water.
Solution:
(a) Esterification: The condensation reaction between a carboxylic acid and an alcohol in the presence of an acid catalyst to produce a sweet-smelling ester and water. [1 Mark]
CH3COOH(l) + C2H5OH(l) → (conc. H2SO4) → CH3COOC2H5(l) + H2O(l)
(b) Saponification: Alkaline hydrolysis of an ester (fat/oil) with a strong base (NaOH) to yield the sodium salt of a carboxylic acid (soap) and alcohol: [1 Mark]
CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
(c) Cleansing Mechanism & Micelle Formation:
• Soap molecules are sodium or potassium salts of long-chain fatty acids (e.g., sodium stearate, C17H35COO-Na+).
• They possess two distinct ends: a hydrophobic hydrocarbon tail (water-repelling, oil-soluble) and a hydrophilic ionic head –COO-Na+ (water-attracting). [1 Mark]
• In water containing oily dirt, the hydrophobic tails dissolve into the central oil droplet while the polar hydrophilic heads project outward into the surrounding water. This spherical aggregate is called a micelle. Mechanical agitation pulls the oily dirt into suspension as an emulsion, which rinses away with water. [1 Mark]
• Action in Hard Water: Hard water contains dissolved calcium (Ca2+) and magnesium (Mg2+) ions. These ions react with soluble soap molecules to form an insoluble white curd-like precipitate called scum: 2C17H35COONa + Ca2+ → (C17H35COO)2Ca↓ + 2Na+, which wastes soap and hinders cleansing action. [1 Mark]
Q16. (Biology: Human Excretory System & Nephron Function)
(a) Draw a conceptual flow diagram detailing the three basic physiological steps in urine formation by a functional nephron.
(b) List four initial constituents present in the glomerular filtrate that are selectively reabsorbed by the tubular epithelial cells.
(c) Explain two physiological factors that regulate the volume and concentration of urine produced in humans.
Solution:
(a) Physiological Steps of Urine Formation:
1. Ultrafiltration (Glomerular Filtration): Blood enters the glomerulus under high hydrostatic pressure via the afferent arteriole. Water, glucose, amino acids, urea, and salts pass across the fenestrated podocyte membranes into Bowman’s capsule as primary filtrate.
2. Selective Reabsorption: As filtrate passes through the Proximal Convoluted Tubule (PCT) and Loop of Henle, essential solutes and variable quantities of water are reabsorbed back into the peritubular capillaries.
3. Tubular Secretion: Excess ions (K+, H+, NH4+) and drug metabolites are actively secreted from peritubular capillaries into the distal tubule and collecting duct to maintain electrolyte and acid-base homeostasis. [2 Marks]
(b) Selectively Reabsorbed Substances: Glucose, amino acids, inorganic salts (Na+, Cl-), and major proportions of water. [1 Mark]
(c) Regulation of Urine Volume:
1. Quantity of excess water present in the body: When body hydration is high, less water is reabsorbed, resulting in dilute, high-volume urine.
2. Amount of dissolved metabolic waste (urea and salts) to be excreted: Higher solute waste requires a proportional volume of solvent water for safe excretion, modulated by the Antidiuretic Hormone (ADH / Vasopressin). [2 Marks]
Q17. (Physics: Electricity and Circuit Analysis)
(a) State Joule’s Law of Heating and express it mathematically.
(b) Derive the equivalent resistance expression for three resistors R1, R2, and R3 connected in series.
(c) An electric circuit contains a 12 V battery connected across a parallel combination of two resistors (6 Ω and 12 Ω), which is further connected in series with a 4 Ω resistor and an ammeter. Calculate:
(i) The total equivalent resistance of the entire circuit.
(ii) The reading of the ammeter (total circuit current).
(iii) The heat produced in the 4 Ω resistor in 10 seconds.
Solution:
(a) Joule’s Law of Heating: The heat (H) produced in a resistor is directly proportional to (i) the square of current (I2), (ii) the resistance (R), and (iii) the time duration (t) for which current flows. H = I2Rt. [1 Mark]
(b) Series Combination Derivation:
In series, the same current I passes through all resistors. Total potential drop V = V1 + V2 + V3.
By Ohm’s Law: V1 = I R1, V2 = I R2, V3 = I R3.
Substituting: I Req = I R1 + I R2 + I R3 ⇒ Req = R1 + R2 + R3. [1.5 Marks]
(c) Numerical Calculation:
(i) Parallel branch resistance Rp:
1 / Rp = 1/6 + 1/12 = (2 + 1) / 12 = 3 / 12 = 1 / 4 ⇒ Rp = 4 Ω.
Total equivalent resistance Rtotal = Rp + Rseries = 4 Ω + 4 Ω = 8 Ω. [1 Mark]
(ii) Ammeter reading (Current I):
I = V / Rtotal = 12 V / 8 Ω = 1.5 A. [0.5 Mark]
(iii) Heat generated in 4 Ω series resistor in t = 10 s:
H = I2 × R × t = (1.5)2 × 4 × 10 = 2.25 × 40 = 90 Joules. [1 Mark]
Section E (12 Marks — 4 Marks Each Case Study)
Q18. Case Study 1 (Physics: Domestic Electric Circuits & Safety Devices)
In domestic wiring, electric power is supplied through mains at 220 V with a frequency of 50 Hz. The wiring comprises three insulated wires: Live wire (red/brown), Neutral wire (black/blue), and Earth wire (green/yellow). All appliances are connected in parallel to ensure independent operation and full supply voltage. To prevent damage from sudden surges or faults, safety devices like fuses and earthing are integrated into the distribution board.
Questions:
(i) Why are domestic electrical appliances connected in parallel rather than in series? State two distinct reasons. (2 Marks)
(ii) What is the function of the earth wire in heavy electrical appliances like refrigerators and electric irons? (1 Mark)
(iii) What happens during an electrical ‘short circuit’, and how does an electric fuse provide protection? (1 Mark)
Solution:
(i) Reasons for parallel connection: [2 Marks]
• Constant Potential: Every appliance receives the full rated voltage of 220 V.
• Independent Switching: If one appliance fails or is turned off, other appliances in parallel branches remain functional without interruption.
(ii) Function of Earth Wire: It connects the metallic casing of heavy appliances to a metal plate buried deep in the ground, providing a low-resistance path for leakage current. This trips the fuse/MCB and protects the user from severe electric shocks. [1 Mark]
(iii) Short Circuit & Fuse Action: A short circuit occurs when bare live and neutral wires come in direct contact, causing the circuit resistance to drop near zero and current to surge dangerously. The high current heats the low-melting-point fuse wire via Joule heating, melting it to break the circuit and prevent electrical fires. [1 Mark]
Q19. Case Study 2 (Biology: Genetics and Mendel’s Monohybrid Cross)
Gregor Johann Mendel performed monohybrid inheritance experiments on pea plants (Pisum sativum). He crossed pure-breeding tall pea plants (TT) with pure-breeding dwarf pea plants (tt). In the F1 generation, all offspring were tall. When the F1 plants were self-pollinated, both tall and dwarf phenotypes appeared in the F2 generation.
Questions:
(i) State the phenotype and genotype of the plants obtained in the F1 generation. (1 Mark)
(ii) Using a Punnett square, show the self-pollination cross of F1 progeny to determine the phenotypic and genotypic ratios of the F2 generation. (2 Marks)
(iii) State Mendel’s Law of Segregation demonstrated by this experiment. (1 Mark)
Solution:
(i) F1 Generation: Phenotype = 100% Tall plants; Genotype = Heterozygous tall (Tt). [1 Mark]
(ii) F2 Self-Pollination Cross (Tt × Tt): [2 Marks]
| Gametes | T | t |
|---|---|---|
| T | TT (Tall) | Tt (Tall) |
| t | Tt (Tall) | tt (Dwarf) |
• Phenotypic Ratio: 3 Tall : 1 Dwarf (3:1)
• Genotypic Ratio: 1 TT : 2 Tt : 1 tt (1:2:1)
(iii) Law of Segregation: Alleles of a gene separate during gamete formation so that each gamete carries only one allele for each gene with equal probability. [1 Mark]
Q20. Case Study 3 (Chemistry: Universal Indicator & pH in Daily Life)
The pH scale measures the hydrogen ion concentration [H+] in an aqueous solution. Living organisms function within narrow pH ranges. For example, human blood pH is maintained between 7.35 and 7.45. Soil pH directly controls plant nutrient bioavailability, and the pH of the human digestive tract varies from strongly acidic (pH 1.5–2.0 in the stomach) to slightly alkaline (pH 7.8 in the duodenum).
Questions:
(i) Explain why tooth decay begins when the pH inside the oral cavity drops below 5.5. (1 Mark)
(ii) A farmer finds that his agricultural field has a soil pH of 4.8. Name two chemical compounds he should add to neutralize the excess acidity. (1 Mark)
(iii) Name the acid secreted by the gastric glands. What antacid compound is commonly prescribed to relieve hyperacidity, and what is its chemical neutralization reaction? (2 Marks)
Solution:
(i) Tooth Decay Mechanism: Tooth enamel is composed of calcium hydroxyapatite (a crystalline form of calcium phosphate), which is the hardest substance in the human body. When oral bacteria ferment leftover sugars, they release acids. At a pH below 5.5, the enamel undergoes demineralization and corrosion. [1 Mark]
(ii) Soil Treatment: To neutralize acidic soil (pH 4.8), the farmer should add basic substances such as Quicklime (Calcium oxide, CaO) or Slaked lime (Calcium hydroxide, Ca(OH)2) or Chalk (CaCO3). [1 Mark]
(iii) Gastric Acid & Antacids:
• Gastric glands secrete Hydrochloric acid (HCl).
• Commonly used mild antacids include Magnesium hydroxide [Milk of Magnesia, Mg(OH)2] or Sodium hydrogen carbonate [Baking soda, NaHCO3].
• Neutralization Equation: Mg(OH)2(aq) + 2HCl(aq) → MgCl2(aq) + 2H2O(l). [2 Marks]
Section-Wise Difficulty and Chapter Weightage Analysis
Analyzing the balance of conceptual and numerical demands across Physics, Chemistry, and Biology helps optimize your revision schedule. The distribution of marks across key curriculum domains is summarized below:
| Curriculum Unit | Chapters Covered | Total Marks | Difficulty Level |
|---|---|---|---|
| Unit I: Chemical Substances – Nature and Behaviour | Chemical Reactions, Acids Bases Salts, Metals & Non-Metals, Carbon Compounds | 25 Marks | Moderate to High (Balancing, Mechanisms, Reaction Conditions) |
| Unit II: World of Living | Life Processes, Control & Coordination, Reproduction, Heredity | 25 Marks | Moderate (Pathways, Physiological Processes, Punnett Squares) |
| Unit III: Natural Phenomena | Light: Reflection & Refraction, Human Eye & Colourful World | 12 Marks | Moderate to High (Ray Diagrams, Sign Conventions, Formula Application) |
| Unit IV: Effects of Current | Electricity, Magnetic Effects of Electric Current | 13 Marks | Moderate (Circuit Numericals, Right-Hand Thumb Rule, Solenoid) |
| Unit V: Natural Resources | Our Environment (Trophic levels, 10% Energy Law, Ozone depletion) | 05 Marks | Easy to Moderate (Direct Ecological Concepts) |
How to Practice and Score 95+ with This Sample Paper
Practicing this CBSE sample paper standard set under timed conditions bridges the gap between conceptual understanding and exam-day execution. Follow these targeted preparation guidelines:
- Simulate 3-Hour Exam Conditions: Sit in an isolated environment without textbooks or digital devices. Allocate strictly 180 minutes: spend 25 minutes on Section A, 25 minutes on Section B, 45 minutes on Section C, 50 minutes on Section D, 25 minutes on Section E, and reserve the final 10 minutes exclusively for checking calculations and ray diagram arrows.
- Master NCERT Step-Marking Guidelines: In physics numericals, always write the given parameters with units, state the foundational formula explicitly (e.g.,
1/f = 1/v + 1/u), maintain negative/positive sign conventions strictly, and underline your final answer with its standard SI unit. - Include State Symbols in Chemical Equations: Full marks in chemistry descriptive questions require balanced equations featuring physical states such as
(s),(l),(g),(aq), and reaction catalysts over the reaction arrow. - Annotate Biological Flowcharts: For physiology questions (e.g., nephron filtration, double circulation, reflex arcs), support textual answers with structured directional arrow flowcharts to maximize evaluation speed and precision.
More CBSE Class 10 Science Practice Papers and Resources
To access an extensive repository of solved CBSE sample paper basic sets, chapter-wise competency question banks, and authentic board-style mock tests, visit Theorify QPTool. Generate customized, syllabus-aligned test papers and elevate your CBSE Class 10 board exam preparation today.