Class 10 Science Test Paper Chapter 1-5-9| CBSE Board | Detailed Solutions & Marking Scheme

VEDANT SKILL ASSESSMENT SERIES

Academic Year 2026–27

Class: X

Subject: Science

Chapters: 1, 5 & 9 CBSE Pattern)

Maximum Marks: 30
Time Allowed: 1 Hour


GENERAL INSTRUCTIONS

  1. Read all questions carefully before answering.

  2. Show proper steps wherever required.

  3. Write neatly and clearly.

  4. Manage your time wisely.

  5. If you do not know the answer, you may cry silently. Loud crying, emotional speeches, and negotiations for hints are strictly prohibited.


SECTION A

Multiple Choice & Assertion–Reasoning Questions

(7 Marks)

Q1. [1]

When a magnesium ribbon is burnt in air, a dazzling white flame is observed along with the formation of a white powder X. If this powder X is dissolved in water, what will be the nature of the resulting solution and its effect on litmus?

A. Acidic, turns blue litmus red

B. Basic, turns red litmus blue

C. Amphoteric, no change in litmus

D. Neutral, no change in litmus


Q2. [1]

During an intense physical activity like sprinting, an athlete often experiences painful muscle cramps. This acute condition occurs primarily due to the non-availability of oxygen, leading to the conversion of pyruvate into:

A. Ethanol + Carbon dioxide in cytoplasm

B. Lactic acid in mitochondria

C. Lactic acid in the cytoplasm of muscle cells

D. Carbon dioxide + Water in mitochondria


Q3. [1]

A student performs an experiment with a convex lens and places an object at a distance of 2F from the optical centre. She observes a sharp image on a screen on the other side. If she now covers the entire lower half of the lens with a black opaque paper, how will it affect the final image?

A. Only the top half of the image will be formed.

B. Only the bottom half of the image will be formed.

C. A complete image is formed but its intensity is reduced.

D. The image will disappear completely from the screen.


Q4. [1]

Consider the following chemical reaction occurring in a closed container:

MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

Identify the substance that undergoes oxidation and the oxidizing agent respectively.

A. MnO₂ and HCl

B. HCl and MnO₂

C. MnCl₂ and H₂O

D. Cl₂ and MnO₂


Q5. [1]

Which of the following is the correct path sequence of deoxygenated blood through the human circulatory system starting from the body tissues?

A. Vena Cava → Right Atrium → Right Ventricle → Pulmonary Artery

B. Pulmonary Vein → Left Atrium → Left Ventricle → Aorta

C. Vena Cava → Left Atrium → Left Ventricle → Pulmonary Artery

D. Pulmonary Artery → Right Atrium → Right Ventricle → Vena Cava


Directions for Q6 and Q7

Select the correct option:

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.


Q6. [1]

Assertion (A): White silver chloride turns grey when exposed to sunlight.

Reason (R): Silver chloride decomposes in the presence of sunlight to form silver metal and chlorine gas.


Q7. [1]

Assertion (A): The medium inside the human stomach is highly acidic with a pH of about 1.5–3.0.

Reason (R): Hydrochloric acid provides an optimum acidic medium for the activation and proper functioning of the enzyme pepsin.


SECTION B

Short Answer Type–I Questions

(8 Marks)

Q8. [2]

A student added a small piece of shiny zinc metal to a test tube containing dilute sulphuric acid.

a) Write the balanced chemical equation for the reaction, mentioning the physical states of all reactants and products.

b) State the test used to identify the gas evolved.


Q9. [2]

In the mammalian respiratory system, how are the alveoli structurally adapted to achieve maximum efficiency during gaseous exchange?

Mention any two structural adaptations.


Q10. [2]

An object is placed at a distance of 15 cm in front of a concave mirror of focal length 10 cm.

Using the mirror formula,

a) Calculate the position of the image.

b) State whether the image formed is real or virtual.


Q11. [2]

The absolute refractive indices of water and glass are 4/3 and 3/2 respectively.

The speed of light in vacuum is 3 × 10⁸ m/s.

Calculate the speed of light in:

a) Water

b) Glass


SECTION C

Short Answer Type–II Questions

(6 Marks)

Q12. [3]

Lead nitrate powder is heated strongly in a dry test tube.

a) State any two observable changes during the reaction.

b) Write the balanced chemical equation.

c) Classify the reaction based on:

  • Process type

  • Energy change


Q13. [3]

Differentiate between Xylem and Phloem on the basis of:

a) Nature of constituent cells (living/dead)

b) Material transported

c) Direction of transport


SECTION D

Long Answer Type Question

(5 Marks)

Q14. [5]

State the two laws of refraction of light.

A convex lens has a focal length of 20 cm.

An object of height 5 cm is placed 30 cm from the optical centre.

Calculate:

a) Position of the image using the lens formula.

b) Magnification and height of the image.

c) State the characteristics of the image (real/virtual, erect/inverted, magnified/diminished).


SECTION E

Case-Based / Competency Question

(4 Marks)

Q15.

Read the given passage carefully and answer the following questions:

The green leaves of terrestrial plants are highly specialized micro-factories responsible for synthesizing complex organic carbohydrates from simple inorganic raw materials like carbon dioxide and water. This light-driven physiological phenomenon is called photosynthesis. The raw materials are gathered via structural adaptations: water is drawn up from the soil through roots, while atmospheric carbon dioxide enters through minute cellular pores called stomata scattered across the epidermal layer. The opening and closing of these stomatal pores is an active mechanism strictly regulated by specialized kidney-shaped guard cells. This cellular control depends on the turgidity changes within the guard cells, ensuring that optimal gaseous exchange takes place while minimizing excessive water loss through transpiration.

i) Name the cell organelle where the light reaction and dark reaction of photosynthesis take place. [1]

ii) Explain how guard cells regulate the opening of stomata when water enters them. [1]

iii) Write the balanced chemical equation of photosynthesis, clearly mentioning the essential conditions above the reaction arrow. [2]


End of Question Paper

Best of Luck!

VEDANT SKILL ASSESSMENT SERIES

SOLVED ANSWER KEY WITH DETAILED MARKING SCHEME

Class: X
Subject: Science
Chapters: 1, 6 & 10 (CBSE Pattern)
Maximum Marks: 30


SECTION A

Multiple Choice & Assertion-Reasoning Questions (7 Marks)

Q1. [1 Mark]

Correct Answer: B

Explanation:
When magnesium burns in air, it forms magnesium oxide (MgO).

Chemical equation:

2Mg + O₂ → 2MgO

MgO is a basic oxide. When dissolved in water, it forms magnesium hydroxide.

MgO + H₂O → Mg(OH)₂

Magnesium hydroxide is basic and turns red litmus blue.

Marking Scheme

  • Correct option (B) – 1 Mark

Q2. [1 Mark]

Correct Answer: C

Explanation:

During vigorous exercise, muscles do not receive sufficient oxygen.

Therefore, pyruvate is converted into lactic acid in the cytoplasm of muscle cells.

Accumulation of lactic acid causes muscle cramps.

Marking Scheme

  • Correct option (C) – 1 Mark

Q3. [1 Mark]

Correct Answer: C

Explanation:

Every part of a convex lens contributes to the formation of the complete image.

Covering half the lens blocks some light rays but does not prevent image formation.

Therefore,

  • Complete image is formed.
  • Brightness (intensity) decreases.

Marking Scheme

  • Correct option (C) – 1 Mark

Q4. [1 Mark]

Reaction:

MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

Correct Answer: B

Explanation:

Chloride ions from HCl lose electrons to form chlorine gas.

Hence,

  • HCl undergoes oxidation.
  • MnO₂ acts as the oxidizing agent.

Marking Scheme

  • Correct option (B) – 1 Mark

Q5. [1 Mark]

Correct Answer: A

Explanation:

Correct pathway:

Body tissues
→ Vena Cava
→ Right Atrium
→ Right Ventricle
→ Pulmonary Artery
→ Lungs

Marking Scheme

  • Correct option (A) – 1 Mark

Q6. [1 Mark]

Correct Answer: A

Explanation:

Assertion is true.

Reason is also true.

Silver chloride decomposes in sunlight.

2AgCl → 2Ag + Cl₂

Grey colour appears because of metallic silver.

Reason correctly explains Assertion.

Marking Scheme

  • Correct option (A) – 1 Mark

Q7. [1 Mark]

Correct Answer: A

Explanation:

Hydrochloric acid creates an acidic medium inside the stomach.

This acidic medium converts pepsinogen into active pepsin and helps protein digestion.

Hence both statements are true and the reason correctly explains the assertion.

Marking Scheme

  • Correct option (A) – 1 Mark

SECTION B

Short Answer Type-I (8 Marks)

Q8. [2 Marks]

(a) Balanced Chemical Equation

Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

1 Mark


(b) Test for Hydrogen Gas

Bring a burning splint near the mouth of the test tube.

Hydrogen burns with a "pop" sound, confirming its presence.

1 Mark

Marking Scheme

Balanced equation with state symbols – 1 Mark

Correct identification test (Pop sound) – 1 Mark

Total = 2 Marks


Q9. [2 Marks]

Any two of the following:

• Alveoli provide a large surface area for diffusion.

• Walls are one cell thick, reducing diffusion distance.

• Rich network of blood capillaries maintains concentration gradient.

• Moist surface helps diffusion.

Marking Scheme

Any two correct points

1 Mark each

Total = 2 Marks


Q10. [2 Marks]

Given

Object distance (u) = –15 cm

Focal length (f) = –10 cm

Mirror Formula

1/f = 1/v + 1/u

Substituting values

1/–10 = 1/v + 1/–15

1/v

= –1/10 + 1/15

= (–3 + 2)/30

= –1/30

v = –30 cm

Negative sign indicates image forms in front of the mirror.

Therefore,

Image position = –30 cm

Nature = Real and Inverted

Marking Scheme

Correct substitution and calculation – 1 Mark

Correct nature – 1 Mark


Q11. [2 Marks]

Formula

Speed = c/n

where

c = 3 × 10⁸ m/s


(a) Water

n = 4/3

Speed

= (3 × 10⁸)/(4/3)

= (9/4) × 10⁸

= 2.25 × 10⁸ m/s

1 Mark


(b) Glass

n = 3/2

Speed

= (3 × 10⁸)/(3/2)

= (6/3) × 10⁸

= 2.0 × 10⁸ m/s

1 Mark

Marking Scheme

Correct speed in water – 1 Mark

Correct speed in glass – 1 Mark

Total = 2 Marks


SECTION C

Short Answer Type-II Questions (6 Marks)


Q12. [3 Marks]

Lead nitrate powder is heated strongly over a flame in a dry test tube.

(a) State two observable changes during the reaction. (1 Mark)

Answer:

Any two of the following:

  1. Brown fumes of nitrogen dioxide (NO₂) are evolved.
  2. A yellow residue of lead oxide (PbO) is formed (yellow when hot).
  3. Oxygen gas is released.

Marking Scheme

  • Brown fumes observed – ½ Mark
  • Yellow residue (PbO) observed – ½ Mark

Total = 1 Mark


(b) Write the balanced chemical equation. (1 Mark)

Answer:

2Pb(NO₃)₂(s) —Heat→ 2PbO(s) + 4NO₂(g) + O₂(g)

Marking Scheme

  • Correct balanced equation – 1 Mark

(c) Classify the reaction. (1 Mark)

Answer:

  • It is a Thermal Decomposition Reaction because heat breaks down lead nitrate.
  • It is an Endothermic Reaction because heat is absorbed.

Marking Scheme

  • Thermal decomposition – ½ Mark
  • Endothermic – ½ Mark

Total = 1 Mark


Q13. [3 Marks]

Differentiate between Xylem and Phloem.

BasisXylemPhloem
Nature of cellsMostly dead cellsMostly living cells
Material transportedWater and mineralsFood (sucrose, amino acids)
Direction of transportUnidirectional (roots → leaves)Bidirectional

Marking Scheme

Each correct comparison = 1 Mark

  • Nature of cells – 1 Mark
  • Material transported – 1 Mark
  • Direction of transport – 1 Mark

Total = 3 Marks


SECTION D

Long Answer Type Question (5 Marks)

Q14.

(a) State the laws of refraction of light. (1 Mark)

Answer

First Law

The incident ray, refracted ray and the normal at the point of incidence lie in the same plane.

Second Law (Snell's Law)

For a given pair of media,

sin i / sin r = Constant

or

n = sin i / sin r

Marking Scheme

  • First law – ½ Mark
  • Second law – ½ Mark

Total = 1 Mark


Given

Focal length (f) = +20 cm

Object distance (u) = –30 cm

Object height (h₀) = 5 cm


(b) Calculate the image position. (1.5 Marks)

Lens Formula

1/f = 1/v – 1/u

Substituting values

1/20 = 1/v – (–1/30)

1/20 = 1/v + 1/30

1/v

= 1/20 – 1/30

= (3 – 2)/60

= 1/60

Therefore,

v = +60 cm

Hence, the image is formed 60 cm on the other side of the lens.

Marking Scheme

  • Correct formula – ½ Mark
  • Correct substitution – ½ Mark
  • Correct answer (v = +60 cm) – ½ Mark

Total = 1.5 Marks


(c) Calculate magnification and image height. (1.5 Marks)

Magnification

m = v/u

= 60/–30

= –2

Also,

m = hᵢ/h₀

–2 = hᵢ/5

hᵢ = –10 cm

Answer

Magnification = –2

Image height = –10 cm

Marking Scheme

  • Correct magnification – 1 Mark
  • Correct image height – ½ Mark

Total = 1.5 Marks


(d) State the characteristics of the image. (1 Mark)

Since

v is positive

and

m is negative,

the image is

  • Real
  • Inverted
  • Magnified (twice the size of the object)

Marking Scheme

Any two correct characteristics = ½ + ½ Mark

Total = 1 Mark


SECTION E

Case-Based / Competency Question (4 Marks)

Q15.

(i) Name the organelle where photosynthesis occurs. (1 Mark)

Answer

Chloroplast

  • Light reaction takes place in the grana (thylakoids).
  • Dark reaction takes place in the stroma.

Marking Scheme

  • Chloroplast – 1 Mark

(ii) Explain how guard cells open the stomata. (1 Mark)

Answer

When water enters the guard cells by osmosis, they become turgid and swell.

The outer wall stretches more than the thick inner wall, causing the guard cells to bend outward.

As a result, the stomatal pore opens.

Marking Scheme

  • Guard cells become turgid – ½ Mark
  • Pore opens due to bending of guard cells – ½ Mark

Total = 1 Mark


(iii) Write the balanced equation of photosynthesis. (2 Marks)

Answer

6CO₂ + 6H₂O —Sunlight, Chlorophyll→ C₆H₁₂O₆ + 6O₂

OR (Fully balanced form)

6CO₂ + 12H₂O —Sunlight, Chlorophyll→ C₆H₁₂O₆ + 6O₂ + 6H₂O

Marking Scheme

  • Correct balanced equation – 1 Mark
  • Sunlight and Chlorophyll mentioned over the arrow – 1 Mark

Total = 2 Marks


MARKS DISTRIBUTION

SectionMarks
Section A7
Section B8
Section C6
Section D5
Section E4
Total30

CBSE Examiner's Notes

  • Award full marks for scientifically correct alternative answers where applicable.
  • Ignore minor spelling mistakes that do not change the scientific meaning (e.g., chlorophyl for chlorophyll).
  • Accept both balanced forms of the photosynthesis equation.
  • State symbols are desirable in chemical equations; if omitted but the equation is otherwise correct, partial credit may be awarded as per CBSE practice.
  • In numerical questions, allow minor rounding differences if the method is correct.
  • Award step marks even if the final answer is incorrect due to an arithmetic error, provided the correct formula and procedure are used.

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