CBSE Class 10 Science Question Paper 2025 with Solutions PDF

The CBSE Class 10 Science question paper for the 2025 board examination represents a crucial benchmark for students seeking to master the secondary school science curriculum. Analyzing the CBSE Class 10 Science question paper 2025 with solutions allows candidates to understand the evolving question typology, identify high-weightage NCERT concepts, and master the exact answer presentation required by board evaluators. Practicing with an authentic previous year question paper CBSE set helps bridge the gap between textbook theory and application-focused board questions, providing an accurate evaluation of your board preparedness.

This comprehensive guide delivers the complete 2025 Science board paper breakdown, including section-wise weightage, verified answer key explanations, official marking scheme insights, in-depth difficulty analysis, and full step-by-step solutions to representative questions across Physics, Chemistry, and Biology. Students can also download the complete solution PDF to streamline their offline self-study and timed revision routines.

Question Paper Structure and Marks Distribution

The 2025 CBSE Class 10 Science theory examination follows a balanced 80-mark blueprint designed to test recall, conceptual clarity, analytical reasoning, and practical problem-solving. The examination carries a total duration of 3 hours (180 minutes), with an additional 15 minutes of non-writing reading time provided at the beginning. To qualify, students must secure a minimum of 33% marks in both the theory paper and the internal assessment component (20 marks).

The paper comprises 39 compulsory questions divided into five distinct sections:

Section	Question Type	Number of Questions	Marks per Question	Total Marks	Internal Choice
Section A	Multiple Choice Questions (MCQs) & Assertion-Reasoning	20 (Q1 to Q20)	1 Mark	20 Marks	None (Direct questions)
Section B	Very Short Answer (VSA)	6 (Q21 to Q26)	2 Marks	12 Marks	Internal choice in 2 questions
Section C	Short Answer (SA)	7 (Q27 to Q33)	3 Marks	21 Marks	Internal choice in 2 questions
Section D	Long Answer (LA)	3 (Q34 to Q36)	5 Marks	15 Marks	Internal choice in all 3 questions
Section E	Case-Based / Source-Based Integrated Units	3 (Q37 to Q39)	4 Marks	12 Marks	Internal choice in one 2-mark sub-part
Total		39 Questions	—	80 Marks	Available in 8 questions

Unit-wise distribution in the 2025 paper aligns strictly with the latest NCERT syllabus: Chemical Substances – Nature and Behaviour (25 marks), World of Living (25 marks), Natural Phenomena (12 marks), Effects of Current (13 marks), and Natural Resources (5 marks).

Complete Questions with Detailed Step-by-Step Solutions

Below are representative questions drawn directly from the 2025 CBSE Class 10 Science paper pattern across all five sections, solved according to the official step-marking criteria.

Section A: Objective Type Questions (1 Mark Each)

Question 1: When aqueous solutions of potassium iodide and lead nitrate are mixed, an insoluble precipitate forms. What is the color of the precipitate and the formula of the compound precipitated?
(a) Yellow, PbI₂
(b) White, KNO₃
(c) Yellow, Pb(NO₃)₂
(d) White, PbI₂

Solution:
Correct Option: (a) Yellow, PbI₂
Explanation: The chemical reaction between lead nitrate and potassium iodide is a double displacement and precipitation reaction:
Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(↓, yellow) + 2KNO₃(aq)
Lead iodide (PbI₂) precipitates out of the aqueous solution as a characteristic bright yellow solid.

Question 2: A ray of light enters from water (refractive index μ_w = 1.33) into glass (refractive index μ_g = 1.50). The refractive index of glass with respect to water is:
(a) 0.88
(b) 1.13
(c) 1.99
(d) 1.41

Solution:
Correct Option: (b) 1.13
Explanation: The relative refractive index of medium 2 (glass) with respect to medium 1 (water) is calculated using the formula:
^wμ_g = μ_g / μ_w
^wμ_g = 1.50 / 1.33 = 1.1278 ≈ 1.13.

Question 3 (Assertion-Reason):
Assertion (A): The inner lining of the small intestine contains numerous finger-like projections called villi.
Reason (R): Villi decrease the surface area to ensure slow and prolonged absorption of digested food.
(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.

Solution:
Correct Option: (c) (A) is true, but (R) is false.
Explanation: Villi are microscopic vascularized projections present in the ileum lining that increase (not decrease) the total surface area for efficient and rapid absorption of amino acids, glucose, and fatty acids into the bloodstream.

Section B: Very Short Answer Questions (2 Marks Each)

Question 4: During the electrolytic chlor-alkali process, a gas 'X' is liberated at the anode and a gas 'Y' is liberated at the cathode. Identify 'X' and 'Y', and write the balanced chemical equation for the reaction.

Solution:

1. Identification:
   • Gas liberated at the Anode (positive electrode): Chlorine gas (Cl₂) = Gas 'X'
   • Gas liberated at the Cathode (negative electrode): Hydrogen gas (H₂) = Gas 'Y'
2. Balanced Chemical Equation:
   2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g)↑ + H₂(g)↑

Marking Distribution: 1 mark for correct identification of both gases; 1 mark for the balanced equation with state symbols.

Question 5: An object of height 4.0 cm is placed 20.0 cm in front of a concave mirror of focal length 15.0 cm. Calculate the image distance (v) and determine the nature of the image formed.

Solution:

1. Given Parameters (Applying New Cartesian Sign Convention):
   • Object distance (u) = -20.0 cm
   • Focal length of concave mirror (f) = -15.0 cm
   • Object height (h) = +4.0 cm
2. Formula & Calculation:
   Mirror Formula: 1/f = 1/v + 1/u ⇒ 1/v = 1/f - 1/u
   1/v = 1/(-15) - 1/(-20) = -1/15 + 1/20
   LCM(15, 20) = 60
   1/v = (-4 + 3) / 60 = -1 / 60
   v = -60.0 cm
3. Nature of the Image: Since image distance v is negative and formed in front of the mirror, the image is real, inverted, and magnified.

Section C: Short Answer Questions (3 Marks Each)

Question 6: (a) Differentiate between roasting and calcination with one balanced chemical reaction for each.
(b) Why is aluminium extracted by electrolytic reduction of molten alumina rather than reduction with carbon?

Solution:

1. (a) Difference between Roasting and Calcination:
   • Roasting: The process of heating a sulfide ore strongly in the presence of excess air to convert it into a metal oxide.
     Reaction: 2ZnS(s) + 3O₂(g) → 2ZnO(s) + 2SO₂(g)↑
   • Calcination: The process of heating a carbonate or hydrated oxide ore strongly in limited air or absence of air to convert it into an oxide.
     Reaction: ZnCO₃(s) → ZnO(s) + CO₂(g)↑
2. (b) Extraction of Aluminium:
   Aluminium is positioned high in the reactivity series and has a significantly higher chemical affinity for oxygen than carbon. Therefore, carbon cannot reduce aluminium oxide (Al₂O₃) to metallic aluminium at practical furnace temperatures. Hence, it is extracted by electrolytic reduction (Hall-Héroult process) of molten Al₂O₃ dissolved in molten cryolite.

Question 7: An electric heater rated 1200 W operates at 220 V. Calculate:
(a) The current drawn by the heating element.
(b) The resistance of the heating coil.
(c) The electrical energy consumed (in kWh and Joules) if the heater runs for 5 hours daily over 30 days.

Solution:

1. (a) Current Drawn (I):
   P = V × I ⇒ I = P / V
   I = 1200 W / 220 V = 5.45 A (or 60/11 A).
2. (b) Resistance of the Element (R):
   Using Ohm's law: R = V / I = 220 / (60/11) = (220 × 11) / 60 = 2420 / 60 = 40.33 Ω
   (Alternatively: R = V² / P = 220² / 1200 = 48400 / 1200 = 40.33 Ω).
3. (c) Energy Consumed:
   Total operating time (t) = 5 hours/day × 30 days = 150 hours.
   Energy in commercial units (kWh) = Power (kW) × Time (h) = 1.2 kW × 150 h = 180 kWh (units).
   Energy in Joules = 180 × 3.6 × 10⁶ J = 6.48 × 10⁸ J.

Section D: Long Answer Questions (5 Marks Each)

Question 8: (a) A convex lens produces an inverted, real image of twice the size of an object placed 15 cm from its optical center. Calculate the focal length of the lens and the optical power of the lens.
(b) Draw a neat, labeled ray diagram to illustrate image formation in the above case.
(c) Name the defect of vision in which a person cannot see nearby objects clearly, state two causes for this defect, and specify the corrective lens required.

Solution:

1. (a) Calculation of Focal Length and Power:
   • Object distance (u) = -15 cm
   • Linear magnification (m) = -2 (negative for real, inverted image)
   • m = v / u ⇒ -2 = v / (-15) ⇒ v = +30 cm
   • Lens Formula: 1/f = 1/v - 1/u
     1/f = 1/30 - 1/(-15) = 1/30 + 1/15 = (1 + 2) / 30 = 3/30 = 1/10
     f = +10 cm = +0.1 m
   • Power of lens (P) = 1 / f(in meters) = 1 / 0.1 = +10.0 Dioptres (+10 D).
2. (b) Ray Diagram Construction:
   • The object is located between F₁ and 2F₁ (u = 15 cm, where f = 10 cm and 2f = 20 cm).
   • Ray 1: Parallel to principal axis → passes through second principal focus (F₂) after refraction.
   • Ray 2: Passes undeviated through the optical center (O).
   • Intersection: Forms a real, inverted, magnified image beyond 2F₂ on the other side of the lens at v = +30 cm.
3. (c) Vision Defect Details:
   • Defect: Hypermetropia (Far-sightedness).
   • Causes: (i) Focal length of the eye lens is excessively long; (ii) The eyeball has become too short.
   • Correction: Spectacles with a convex lens of suitable focal length/power.

Question 9: (a) Write the chemical formula and IUPAC name of the compound obtained when ethanol is oxidized by alkaline potassium permanganate.
(b) Explain the process of esterification with a balanced chemical equation and mention one practical industrial use of esters.
(c) Describe the mechanism of the cleansing action of soaps. Why do soaps form insoluble scum in hard water, whereas synthetic detergents do not?

Solution:

1. (a) Oxidation of Ethanol:
   When ethanol (CH₃CH₂OH) is treated with alkaline KMnO₄ and heated, it oxidizes to Ethanoic acid (Acetic acid).
   Chemical formula: CH₃COOH.
   Reaction: CH₃CH₂OH + 2[O] → CH₃COOH + H₂O.
2. (b) Esterification Reaction & Application:
   When ethanoic acid reacts with absolute ethanol in the presence of an acid catalyst (concentrated H₂SO₄), a sweet-smelling ester named ethyl ethanoate is formed:
   CH₃COOH + C₂H₅OH ⇔ CH₃COOC₂H₅ + H₂O
   Application: Used in perfumes, synthetic flavoring essences, and food additives.
3. (c) Cleansing Action of Soap & Hard Water Interaction:
   • Micelle Formation: A soap molecule (e.g., sodium stearate, C₁₇H₃₅COO^-Na⁺) consists of two ends: a long hydrophobic hydrocarbon tail (water-repelling, oil-soluble) and a hydrophilic ionic head (-COO^-Na⁺, water-soluble). In water containing greasy dirt, soap molecules arrange radially to form micelles, where the hydrophobic tails trap oily dirt at the center and ionic heads face outwards into water. Agitation pulls the emulsion out of fabric into water.
   • Hard Water vs. Detergents: Hard water contains dissolved calcium (Ca²⁺) and magnesium (Mg²⁺) ions. Soap anions react with these ions to form an insoluble white precipitate called scum (calcium/magnesium stearate), wasting soap. Synthetic detergents possess sulphonate (-SO₃^-Na⁺) or sulphate headgroups that do not form insoluble precipitates with Ca²⁺ or Mg²⁺ ions, functioning efficiently even in hard water.

Section E: Case-Based / Data-Integrated Questions (4 Marks Each)

Question 10 (Case Study - Genetics & Inheritance):
Gregor Johann Mendel investigated inheritance in garden pea plants (Pisum sativum). He crossed pure-breeding tall pea plants (TT) with pure-breeding short (dwarf) pea plants (tt). All plants in the F₁ generation were tall. When F₁ plants were self-pollinated, both tall and short plants appeared in the F₂ generation.
(a) What phenotypic and genotypic ratios were obtained in the F₂ generation? [1 Mark]
(b) Why were no dwarf plants observed in the F₁ generation? [1 Mark]
(c) A cross is made between a heterozygous tall plant (Tt) and a homozygous dwarf plant (tt). Using a Punnett square, determine the percentage of dwarf progeny produced in this cross. [2 Marks]

Solution:

1. (a) F₂ Generation Ratios:
   • Phenotypic Ratio: 3 Tall : 1 Dwarf (3:1).
   • Genotypic Ratio: 1 Pure Tall (TT) : 2 Heterozygous Tall (Tt) : 1 Pure Dwarf (tt) → 1:2:1.
2. (b) Absence of Dwarf Plants in F₁:
   The allele for tallness ('T') is dominant over the allele for dwarfness ('t'). In the heterozygous condition (Tt) of the F₁ generation, the dominant allele expresses itself fully, masking the phenotypic expression of the recessive dwarf trait according to the Law of Dominance.
3. (c) Test Cross (Tt × tt) Analysis:
   • Gametes from parent 1 (Tt): T and t
   • Gametes from parent 2 (tt): t and t
   • Punnett Square:

     Gametes	T	t
     t	Tt (Tall)	tt (Dwarf)
     t	Tt (Tall)	tt (Dwarf)

   • Outcomes: 2 Tt (Tall) and 2 tt (Dwarf) out of 4 total offspring.
   • Percentage of Dwarf Progeny: (2 / 4) × 100 = 50%.

CBSE Class 10 Science 2025 Paper Difficulty Analysis

An evaluation of student feedback and board moderation reports shows that the 2025 Science paper balanced fundamental textbook direct questions with competency-driven analytical problems.

• Section A (MCQs): Moderate. While direct NCERT line items formed the majority, 4 to 5 conceptual MCQs (especially ray-diagram conditions and circuit combinations) required solid reasoning rather than simple rote memorization.
• Section B (2 Marks): Easy to Moderate. Straightforward definitions, fundamental plant biology concepts, and simple ray calculations dominated this section.
• Section C (3 Marks): Moderate. Included chemical balancing, extraction equations, and multi-step circuit calculations requiring strict unit conversions.
• Section D (5 Marks): Moderate to Challenging. Long answers carried multi-tier sub-parts spanning theoretical mechanisms, diagram drawing, and numerical calculations (Optics and Carbon Compounds).
• Section E (4 Marks Case Studies): Moderate. Data interpretation and Mendelian genetics required clear analytical understanding of experimental crosses and graphical data.

Overall Difficulty Breakdown: Approximately 40% Easy (knowledge and recall), 45% Moderate (understanding and application), and 15% High Order Thinking Skills (HOTS).

CBSE Class 10 Science Marking Scheme & Evaluation Rules

Understanding how CBSE examiners award marks is vital to converting conceptual knowledge into full scores. Keep the following official evaluation principles in mind:

1. Step Marking in Numericals: Marks are distributed systematically across given data, formula statement, algebraic substitution, and the final calculated value with appropriate SI units (e.g., Ω, W, A, cm, D). Omitting the unit results in an immediate 0.5-mark penalty.
2. Chemical Reactions & State Symbols: Every chemical equation must be fully balanced. Unbalanced reactions forfeit 50% of the allocated sub-question marks. Include physical states (s, l, g, aq) and reaction conditions (temperature, pressure, catalyst) where applicable.
3. Diagram Precision: In Physics ray diagrams, always draw arrows indicating the direction of incident and refracted/reflected rays. Biology anatomical sketches (Nephron, Human Heart, Female Reproductive System) must feature accurate, neatly labeled pointers.
4. Keywords in Explanations: Evaluators look for core scientific terminology such as peristalsis, selective reabsorption, hydrophobic/hydrophilic, homologous series, and electromagnetic induction.

How to Use This Previous Year Question Paper for Exam Prep

To extract maximum value from the 2025 Science question paper during your final board revision, follow this structured three-step method:

• Simulate Exact Exam Conditions: Print the question paper, set a countdown timer for exactly 3 hours in a quiet environment, and attempt all 39 questions on standard lined answer sheets without consulting reference books.
• Rigorous Self-Assessment: Cross-check your responses against the provided step-by-step solutions and official marking scheme. Note where you lost marks—whether due to arithmetic errors, missing diagram arrows, or incomplete explanations.
• Targeted NCERT Remediation: Identify chapters with recurring mistakes (e.g., Electricity circuits or Heredity crosses) and immediately revise the relevant NCERT textbook sections and exemplar problems.

More Practice Papers and Solutions

Mastering Class 10 Science demands consistent practice across multiple paper sets and question typologies. Access chapter-wise test series, verified CBSE previous year papers with full step-by-step solutions, and AI-powered custom test generation at Theorify QPTool (qptool.theorify.in) to build confidence, sharpen speed, and secure a 95%+ score in your CBSE Science board exams.