CBSE Class 10 Science (Code 086) - 2025 Board Solved Examination Paper

8 : 1

2 : 1

1 : 2

1 : 8

Official Verified Academic Outcome:

D: 1 : 8

VidyaTrail Step-by-Step Pedagogical Working:

Electrolysis of pure water occurs according to the fundamental stoichiometric equation:
2H₂O(l) ──(Electric Energy)──> 2H₂(g)↑ + O₂(g)↑
• Volume ratio liberated: 2 volumes of H₂ to 1 volume of O₂ (2:1).
• Mass calculation:
Hydrogen mass: 2 mol × 2.016 g/mol = 4.032 g
Oxygen mass: 1 mol × 32.00 g/mol = 32.00 g
Mass Ratio (MH : MO) = 4 / 32 = 1 / 8.

💡 VidyaTrail Examiner Insight: Students frequently confuse the volume ratio (2:1) with the mass ratio (1:8). Check whether the prompt asks for volume or mass!
Number of moles of H2 produced = 2, molar mass = 2 g/mol => Mass of H2 = 2 * 2 = 4 g.
Number of moles of O2 produced = 1, molar mass = 32 g/mol => Mass of O2 = 32 g.
Mass ratio MH : MO = 4 : 32 = 1 : 8.

CBSE Marking Guidelines Breakdown:

Step 1: Write balanced decomposition of water (½ mark)
Step 2: Calculate stoichiometric mass ratio MH : MO = 4 : 32 = 1 : 8 (½ mark)

Al3O4 and MgO2

Al2O3 and MgO

Al3O4 and MgO

Al2O3 and MgO2

Official Verified Academic Outcome:

B: Al2O3 and MgO

VidyaTrail Step-by-Step Pedagogical Working:

Oxidation of electropositive metals upon thermal ignition:
1. Aluminium (valency +3) combines with oxygen (valency -2) to generate amphoteric aluminium oxide:
4Al(s) + 3O₂(g) ──> 2Al₂O₃(s)
2. Magnesium (valency +2) burns with a brilliant white flame to yield basic magnesium oxide:
2Mg(s) + O₂(g) ──> 2MgO(s)

💡 VidyaTrail Examiner Insight: Al₂O₃ is an amphoteric oxide that dissolves in both strong acids and strong bases (forming aluminate salts), whereas MgO is strictly a basic oxide.
Magnesium burns with a dazzling white flame: 2Mg + O2 -> 2MgO (Magnesium oxide).

CBSE Marking Guidelines Breakdown:

Step 1: Identify stable oxides for group 13 (Al³⁺ -> Al2O3) and group 2 (Mg²⁺ -> MgO) (1 mark)

Strong acid, weak acid, neutral, strong base

Weak acid, neutral, weak base, strong base

Weak acid, neutral, strong base, weak base

Weak acid, neutral, strong base, strong acid

Official Verified Academic Outcome:

D: Weak acid, neutral, strong base, strong acid

VidyaTrail Step-by-Step Pedagogical Working:

On the pH scale:
• pH 2-3 is a weak/moderate acid
• pH 7 is strictly neutral
• pH 10 is a strong base
• pH 14 corresponds to maximum alkalinity / strong base per official key D.

CBSE Marking Guidelines Breakdown:

Step 1: Map pH ranges according to CBSE QP key (1 mark)

Salt and water is formed.

Neutral salts are formed.

Hydrogen gas is formed.

Acidic salts are formed.

Official Verified Academic Outcome:

A: Salt and water is formed.

VidyaTrail Step-by-Step Pedagogical Working:

(i) HCl + NaOH -> NaCl + H2O (Acid + Base -> Salt + Water)
(ii) MgO + 2HCl -> MgCl2 + H2O (Basic oxide + Acid -> Salt + Water)
(iii) CO2 + 2NaOH -> Na2CO3 + H2O (Acidic oxide + Base -> Salt + Water)
In all three cases, neutralization produces salt and water.

CBSE Marking Guidelines Breakdown:

Step 1: Recognize neutralization characteristic in all three equations (1 mark)

It has high melting point.

It is highly soluble in water.

It has weak electrostatic forces of attraction between its oppositely charged ions.

It conducts electricity in its molten state or aqueous solution.

Official Verified Academic Outcome:

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

VidyaTrail Step-by-Step Pedagogical Working:

Ionic Compound Fundamentals:
When electropositive atom A transfers valence electrons to electronegative atom B, positive cations and negative anions are generated, crystallizing into a dense giant ionic lattice.
• Because ions are bonded by intense non-directional electrostatic Coulomb forces, ionic solids exhibit high melting points, rigidity, and water solubility.
• Therefore, claiming that compound C has 'weak electrostatic forces' is completely contradictory.

💡 VidyaTrail Examiner Insight: Remember that solid ionic crystals do NOT conduct electricity because ions are locked in place; electrical conductivity only occurs in molten or aqueous states where ions are free to migrate. Ionic compounds possess very strong electrostatic forces of attraction between oppositely charged ions, giving them high melting points and water solubility. Therefore, having weak electrostatic attraction is false.

CBSE Marking Guidelines Breakdown:

Step 1: Classify C as an ionic compound (½ mark)
Step 2: Identify false statement regarding electrostatic forces (½ mark)

Gold and silver

Calcium and magnesium

Aluminium and silver

Sodium and iron

Official Verified Academic Outcome:

B: Calcium and magnesium

VidyaTrail Step-by-Step Pedagogical Working:

Highly reactive metals located at the top of the reactivity series (such as Na, Ca, Mg) have very high affinity for oxygen and cannot be reduced by carbon. They are extracted via electrolysis of their molten chlorides.

CBSE Marking Guidelines Breakdown:

Step 1: Identify highly electropositive metals requiring electrolytic reduction (1 mark)

proteins into amino acids

fats into fatty acids and glycerol

starch into simple sugars

sugars into alcohol

Official Verified Academic Outcome:

C: starch into simple sugars

VidyaTrail Step-by-Step Pedagogical Working:

Saliva contains salivary amylase (ptyalin) which initiates the breakdown of complex carbohydrates (starch) into maltose (simple sugars).

CBSE Marking Guidelines Breakdown:

Step 1: Identify enzyme salivary amylase substrate and product (1 mark)

Cytokinins

Gibberellins

Adrenaline

Auxins

Official Verified Academic Outcome:

D: Auxins

VidyaTrail Step-by-Step Pedagogical Working:

When growing plant shoots detect light, auxin synthesized at the shoot tip diffuses towards the shady side. The higher concentration of auxin stimulates cells on the shady side to elongate faster, bending the stem towards light (phototropism).

CBSE Marking Guidelines Breakdown:

Step 1: Identify hormone responsible for cell elongation in phototropism (1 mark)

50% round and green

75% wrinkled and green

100% round and yellow

75% wrinkled and yellow

Official Verified Academic Outcome:

C: 100% round and yellow

VidyaTrail Step-by-Step Pedagogical Working:

Parental genotypes: RRyy (Round, Green) x rrYY (Wrinkled, Yellow).
Gametes: Ry and rY.
F1 generation genotype: RrYy.
Since Round (R) is dominant over wrinkled (r) and Yellow (Y) is dominant over green (y), 100% of F1 plants have round and yellow seeds.

CBSE Marking Guidelines Breakdown:

Step 1: Determine F1 genotype RrYy (½ mark)
Step 2: Apply law of dominance for phenotype (½ mark)

hypermetropia and convex lens

presbyopia and bifocal lens

myopia and concave lens

myopia and bifocal lens

Official Verified Academic Outcome:

B: presbyopia and bifocal lens

VidyaTrail Step-by-Step Pedagogical Working:

The progressive loss of accommodation power with old age due to ciliary muscle weakening and hardening of the eye lens is called presbyopia. It is corrected using bifocal lenses with upper concave portion for distant vision and lower convex portion for near vision.

CBSE Marking Guidelines Breakdown:

Step 1: Identify presbyopia from clinical description (½ mark)
Step 2: State corrective bifocal lens (½ mark)

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

Both Assertion (A) and Reason (R) are true, but Reason (R) is NOT the correct explanation of Assertion (A).

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

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

Official Verified Academic Outcome:

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

VidyaTrail Step-by-Step Pedagogical Working:

Assertion is TRUE: Most decomposition reactions require energy input in form of heat, light or electricity to break bonds.
Reason is TRUE: Decomposition of vegetable matter into compost by microbes releases thermal energy and is exothermic.
However, Reason does not explain why thermal decomposition is generally endothermic.

CBSE Marking Guidelines Breakdown:

Step 1: Verify truth value of Assertion and Reason (½ mark)
Step 2: Check causal explanation relationship (½ mark)

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

Both Assertion (A) and Reason (R) are true, but Reason (R) is NOT the correct explanation of Assertion (A).

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

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

Official Verified Academic Outcome:

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

VidyaTrail Step-by-Step Pedagogical Working:

If two magnetic field lines intersected, a compass needle placed at the point of intersection would have to point in two different directions simultaneously, which is physically impossible. Hence Reason correctly explains Assertion.

CBSE Marking Guidelines Breakdown:

Step 1: State compass behavior at intersection point (½ mark)
Step 2: Conclude Reason is correct causal explanation (½ mark)

Official Verified Academic Outcome:

Chemical Change Observations

VidyaTrail Step-by-Step Pedagogical Working:

1. Evolution of gas: Effervescence and gas bubbles (Hydrogen gas, H2) are released with rapid bubbling.
2. Change in temperature: The test tube becomes warm/hot to touch indicating that heat is released (exothermic reaction).
Balanced Chemical Equation: Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g) + Heat.

CBSE Marking Guidelines Breakdown:

• Observation 1: Evolution of hydrogen gas bubbles (1 mark)
• Observation 2: Rise in temperature / exothermic reaction (1 mark)

Official Verified Academic Outcome:

Blood Vessel Repair & Platelets

VidyaTrail Step-by-Step Pedagogical Working:

1. Why plugging is essential: Leaks lead to a loss of pressure, which reduces the pumping efficiency of the heart across the circulatory system.
2. Blood component: Platelets (thrombocytes).
3. Mechanism: Platelets circulate around the body and clot the blood at the site of injury by forming a mesh of fibrin threads.

CBSE Marking Guidelines Breakdown:

• Maintaining blood pressure / heart pumping efficiency (1 mark)
• Naming Platelets (½ mark)
• Clotting mechanism at injury site (½ mark)

Official Verified Academic Outcome:

Image position v = -20 cm

VidyaTrail Step-by-Step Pedagogical Working:

Optical Analysis for Diverging (Concave) Lens:
According to the Cartesian sign convention:
• Object distance, u = -60 cm (measured against incident light)
• Focal length, f = -30 cm (concave lens focus is virtual)
• Image distance, v = ?

Applying the Gaussian Lens Equation:
1/f = 1/v - 1/u ==> 1/v = 1/f + 1/u
1/v = (1 / -30) + (1 / -60) = (-2 - 1) / 60 = -3 / 60 = -1 / 20
v = -20 cm.

Conclusion: A virtual, erect, and diminished image is formed at 20 cm on the same side as the object.

💡 VidyaTrail Examiner Insight: For all concave lenses, image distance 'v' must always come out negative. If your calculated 'v' is positive, you applied incorrect sign conventions.
Focal length f = -30 cm
Lens formula: 1/f = 1/v - 1/u
=> 1/v = 1/f + 1/u = 1/(-30) + 1/(-60)
=> 1/v = (-2 - 1) / 60 = -3 / 60 = -1 / 20
=> v = -20 cm.
Negative sign indicates that the image is virtual, erect, and formed on the same side as the object at a distance of 20 cm from the optical centre.

CBSE Marking Guidelines Breakdown:

• Step 1: Write lens formula with proper sign convention (½ mark)
• Step 2: Substitute values: 1/v = -1/30 - 1/60 (½ mark)
• Step 3: Solve for v = -20 cm and conclude position 20 cm in front of lens (1 mark)

Official Verified Academic Outcome:

Total Resistance Rp = R / 9

VidyaTrail Step-by-Step Pedagogical Working:

Resistance is directly proportional to length (R = ρl/A).
When wire of resistance R is cut into 3 equal pieces, the resistance of each piece becomes: R1 = R2 = R3 = R / 3.
When connected in parallel:
1 / Rp = 1 / R1 + 1 / R2 + 1 / R3
1 / Rp = 3/R + 3/R + 3/R = 9 / R
=> Rp = R / 9.
The equivalent resistance decreases to one-ninth of the original wire.

CBSE Marking Guidelines Breakdown:

• Step 1: Resistance of each segment = R/3 (½ mark)
• Step 2: Parallel combination formula 1/Rp = 3/R + 3/R + 3/R (1 mark)
• Step 3: Result Rp = R / 9 (½ mark)

Official Verified Academic Outcome:

Thermite & Reduction Reactions

VidyaTrail Step-by-Step Pedagogical Working:

(a) Aluminothermic Reduction of Intermediate Metal Oxides:
Moderately active transition metals (such as Manganese and Iron) can be reduced using reactive aluminium powder as a strong reducing agent in exothermic thermite reactions:
1. 3MnO₂(s) + 4Al(s) ──> 3Mn(l) + 2Al₂O₃(s) + Heat
2. Fe₂O₃(s) + 2Al(s) ──> 2Fe(l) + Al₂O₃(s) + Heat (Used for railway track welding because molten iron fills gaps seamlessly).

(b) Carbon Affinity Limitation:
Metals at the summit of the reactivity series (Na, Mg, Ca) possess far higher chemical affinity for oxygen than carbon does. Carbon cannot outcompete these electropositive metals for oxygen; hence, they require electrolytic extraction.

💡 VidyaTrail Examiner Insight: Always specify that the metals are produced in the MOLTEN (l) phase due to huge heat evolution. as reducing agents to displace moderately reactive metals from their oxides:
1. Thermite Reaction for Manganese:
3MnO2(s) + 4Al(s) -> 3Mn(l) + 2Al2O3(s) + Heat
2. Thermite Welding of Railway Tracks:
Fe2O3(s) + 2Al(s) -> 2Fe(l) + Al2O3(s) + Heat (Molten iron is produced due to huge exothermic energy).

(b) Metals high up in the reactivity series (such as Na, Mg, Ca) have a much greater affinity for oxygen than carbon does. Therefore, carbon cannot reduce these metallic oxides to free metals.

CBSE Marking Guidelines Breakdown:

• Example 1 equation: MnO2 + Al (1 mark)
• Example 2 equation: Fe2O3 + Al (1 mark)
• Reason for high affinity for oxygen over carbon (1 mark)

Official Verified Academic Outcome:

R = 30 Ω, I = 0.5 A, V1 = 3.0 V

VidyaTrail Step-by-Step Pedagogical Working:

(a) First pair in parallel (R1):
1/R1 = 1/10 + 1/15 = (3 + 2)/30 = 5/30 => R1 = 6 Ω.
Second pair in parallel (R2):
1/R2 = 1/60 + 1/40 = (2 + 3)/120 = 5/120 => R2 = 24 Ω.
Since R1 and R2 are in series:
R_total = R1 + R2 = 6 Ω + 24 Ω = 30 Ω.

(b) By Ohm's Law (V = I * R):
I = V / R_total = 15 V / 30 Ω = 0.5 A.

(c) Potential difference across first pair:
V1 = I * R1 = 0.5 A * 6 Ω = 3.0 V.

CBSE Marking Guidelines Breakdown:

• Part (a): Equivalent resistance calculation (1 mark)
• Part (b): Ohm's law current I = 0.5 A (1 mark)
• Part (c): Voltage drop V1 = 3.0 V (1 mark)

Official Verified Academic Outcome:

Chlor-Alkali & Sodium Compounds

VidyaTrail Step-by-Step Pedagogical Working:

(a) Electrolysis of Brine:
2NaCl(aq) + 2H2O(l) --(electricity)--> 2NaOH(aq) + Cl2(g) + H2(g)

(b) Uses of Products:
• NaOH: Manufacture of soaps & detergents, paper making, artificial fibres.
• Cl2: Water treatment disinfectant, PVC manufacturing, bleaching agent.
• H2: Clean rocket fuel, hydrogenation of oils into margarine, ammonia synthesis.

(c) Identification:
• Salt A is Baking Soda: Sodium hydrogen carbonate (NaHCO3).
• Compound B is Washing Soda precursor: Sodium carbonate (Na2CO3).
Heating Reaction:
2NaHCO3(s) --(Heat)--> Na2CO3(s) + H2O(g) + CO2(g).

CBSE Marking Guidelines Breakdown:

• Part (a): Balanced Chlor-alkali reaction (1 mark)
• Part (b): Two valid commercial uses of product (1 mark)
• Part (c): Identifying A (NaHCO3) and B (Na2CO3) with balanced heating equation (2 marks)

Official Verified Academic Outcome:

Domestic Circuit Safety & Ratings

VidyaTrail Step-by-Step Pedagogical Working:

(a) Standard Indian Electrical Wiring Color Codes:
• Live (Phase) Wire: Red (or modern International Brown)
• Neutral Wire: Black (or modern International Light Blue)
• Protective Earth Wire: Green (or Yellow-Green stripe).

(b) Fuse Sizing Calculation:
Given: P = 1 kW = 1000 W, Supply V = 220 V
Operating Current drawn (I):
I = P / V = 1000 W / 220 V = 4.545 A
Standard Consumer Fuse Chosen: A 5 A fuse rating is selected to allow normal 4.55 A operation without nuisance tripping while providing immediate surge protection.

(c) Protective Function of Grounding:
The green earth wire bonds to the outer metal chassis. If an internal wire frays and contacts the metal frame, fault current safely shunts to earth through the low-resistance grounding wire, instantly melting the fuse before a person touching the frame can experience a lethal electric shock.

💡 VidyaTrail Examiner Insight: Remember that fuses and circuit switches must ALWAYS be placed in the LIVE wire, never in the neutral or earth wire!
• Neutral wire: Black (or light blue)
• Earth wire: Green (or yellow/green).

(b) Current calculation:
P = V * I => I = P / V = 1000 W / 220 V = 4.54 A.
Since operating current is 4.54 A, the standard fuse rating chosen must be 5 A.

(c) Earth Wire Function:
The earth wire provides a low-resistance path to the ground. If insulation fails and live wire touches metallic casing, heavy current rushes to ground instead of passing through human body, blowing the fuse and preventing fatal electric shocks.

CBSE Marking Guidelines Breakdown:

• Part (a): Identifying Red and Black wires (1 mark)
• Part (b): Current calculation I = 4.54 A and rating 5 A (1 mark)
• Part (c): Low resistance path & prevention of electrical shock (2 marks)