---
title: "CBSE Class 10 Science Question Paper 2025 (Set 31/1/1) with Solutions"
url: https://www.swavid.com/cbse/class-10/science/previous-year-papers/2025-31-1-1
dateModified: 2026-10-07T15:30:11+00:00
---

# CBSE Class 10 Science Question Paper 2025 (Set 31/1/1) with Solutions

Code 31/1/1 · 80 marks · 180 minutes

Solved CBSE Class 10 Science board paper from 2025, set 31/1/1. Every question has step-by-step working written to the CBSE marking scheme, with the marks for each answer.

Official question paper: https://www.cbse.gov.in/cbsenew/question-paper/2025/X/086_Science.zip. Solutions written by SwaVid.

Free PDF (28 pages): https://www.swavid.com/api/seo/pdf/papers/cbse/science/swavid-cbse-class-10-science-question-paper-2025-31-1-1-4de03f7093.pdf

## SECTION A

Select and write the most appropriate option out of the four options given for each of the questions no. 1 to 20.

### Question 1

*1 mark · MCQ · Chemical Reactions and Equations*

Electrolysis of water is a decomposition reaction. The mass ratio ($M_H : M_O$) of hydrogen and oxygen gases liberated at the electrodes during electrolysis of water is :

- 8 : 1
- 2 : 1
- 1 : 2
- 1 : 8

**Solution**

1. Water ($H_2O$) decomposes into hydrogen and oxygen gases in the ratio of $2:1$ by volume.
2. Since the molecular formula is $H_2O$, the mass ratio of hydrogen to oxygen is $M_H : M_O = 2 \times 2 : 16 = 4 : 16 = 1 : 8$.
3. Students often confuse volume ratio ($2:1$) with mass ratio ($1:8$), making option (b) a common wrong guess.

**Answer:** (d) 1 : 8

> Common mistake: Confusing the volumetric ratio of gases with their mass ratio.

### Question 2

*1 mark · MCQ · Metals and Non-metals*

The products formed when Aluminium and Magnesium are burnt in the presence of air respectively are :

- Al3O4 and MgO2
- Al2O3 and MgO
- Al3O4 and MgO
- Al2O3 and MgO2

**Solution**

1. Aluminium reacts with oxygen in the air to form aluminium oxide, $Al_2O_3$.
2. Magnesium burns in air to form magnesium oxide, $MgO$.

**Answer:** (b) Al2O3 and MgO

> Common mistake: Writing incorrect chemical formulae for metal oxides based on wrong valencies.

### Question 3

*1 mark · MCQ · Acids, Bases and Salts*

The following table shows the pH values of four solutions A, B, C and D on a pH scale:
The solutions A, B, C and D respectively are of a

- 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

**Solution**

1. A pH value less than 7 indicates an acidic solution (lower pH means stronger acid).
2. A pH value of 7 is neutral, and values greater than 7 indicate basic solutions (higher pH means stronger base).

**Answer:** (b) Weak acid, neutral, weak base, strong base

> Common mistake: Confusing the strength of acids and bases with respect to the numerical value of pH.

### Question 4

*1 mark · MCQ · Acids, Bases and Salts*

Consider the following reactions :
(i) Dilute hydrochloric acid reacts with sodium hydroxide.
(ii) Magnesium oxide reacts with dilute hydrochloric acid.
(iii) Carbon dioxide reacts with sodium hydroxide.
It is found that in each case :

- Salt and water is formed.
- Neutral salts are formed.
- Hydrogen gas is formed.
- Acidic salts are formed.

**Solution**

1. Acid-base neutralisation reactions (such as HCl + NaOH) produce salt and water.
2. Metal oxides reacting with acids (such as MgO + HCl) form salt and water.
3. Non-metal oxides reacting with bases (such as $CO_2$ + NaOH) form salt and water.

**Answer:** (a) Salt and water is formed.

> Common mistake: Assuming non-metal oxides with bases do not form water.

### Question 5

*1 mark · MCQ · Metals and Non-metals*

Reaction between two elements A and B, forms a compound C. A loses electrons and B gains electrons. Which one of the following properties will not be shown by compound C ?

- 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.

**Solution**

1. Element A loses electrons to form a cation, and B gains electrons to form an anion, forming an ionic compound C.
2. Ionic compounds possess strong electrostatic forces of attraction between oppositely charged ions, not weak forces.

**Answer:** (c) It has weak electrostatic forces of attraction between its oppositely charged ions.

> Common mistake: Assuming ionic bonds are weak because they involve charged ions.

### Question 6

*1 mark · MCQ · Metals and Non-metals*

The metals obtained from their molten chlorides by the process of electrolytic reduction are :

- Gold and silver
- Calcium and magnesium
- Aluminium and silver
- Sodium and iron

**Solution**

1. Highly reactive metals like sodium, calcium, and magnesium are obtained by electrolytic reduction of their molten chlorides.
2. Option (b) lists calcium and magnesium, which fit this extraction method.

**Answer:** (b) Calcium and magnesium

> Common mistake: Confusing electrolytic reduction of molten salts with carbon reduction.

### Question 7

*1 mark · MCQ · Metals and Non-metals*

The formation of magnesium oxide is correctly shown in option :

- Mg + O -> Mg2+ [O2-]
- Mg -> O -> Mg+ [O-]
- Mg + O -> Mg2+ [O-]2
- 2Mg + O -> [Mg2+]2 [O2-]2

**Solution**

1. Magnesium atom loses two electrons to form $\text{Mg}^{2+}$, and oxygen atom gains two electrons to form $\text{O}^{2-}$.
2. The correct representation for the formation of magnesium oxide involves balanced atoms and resulting ions.

**Answer:** (d) 2Mg + O -> [Mg2+]2 [O2-]2

> Common mistake: Incorrectly balancing the number of electrons transferred or omitting brackets for ions.

### Question 8

*1 mark · MCQ · Life Processes*

Secretion of less saliva in mouth will effect the conversion of :

- proteins into amino acids
- fats into fatty acids and glycerol
- starch into simple sugars
- sugars into alcohol

**Solution**

1. Saliva contains the enzyme salivary amylase that breaks down starch into simple sugars.
2. Less saliva secretion will therefore affect the conversion of starch into simple sugars.

**Answer:** (c) starch into simple sugars

> Common mistake: Thinking saliva digests proteins or fats.

### Question 9

*1 mark · MCQ · Control and Coordination*

The plant hormone whose concentration stimulates the cells to grow longer on the side of the shoot which is away from light is :

- Cytokinins
- Gibberellins
- Adrenaline
- Auxins

**Solution**

1. Auxin is the plant hormone that synthesizes at the shoot tip and helps the cells to grow longer.
2. When light comes from one side, auxin diffuses towards the shady side of the shoot, stimulating greater cell elongation on that side.

**Answer:** (d) Auxins

> Common mistake: Confusing auxins with cytokinins or gibberellins.

### Question 10

*1 mark · MCQ · How do Organisms Reproduce?*

The correct/true statement(s) for a bisexual flower is/are :
(i) They possess both stamen and pistil.
(ii) They possess either stamen or pistil.
(iii) They exhibit either self-pollination or cross-pollination.
(iv) They cannot produce fruits on their own.

- (i) only
- (iv) only
- (i) and (iii)
- (i) and (iv)

**Solution**

1. A bisexual flower possesses both male (stamen) and female (pistil) reproductive organs.
2. Due to the presence of both organs, they can exhibit self-pollination or cross-pollination.

**Answer:** (c) (i) and (iii)

> Common mistake: Confusing bisexual flowers with unisexual flowers which lack one of the reproductive whorls.

### Question 11

*1 mark · MCQ · Heredity*

If pea plants with round and green seeds ($RRyy$) are crossed with pea plants having wrinkled and yellow seeds ($rrYY$), the seeds developed by the plants of $F_1$ generation will be :

- 50% round and green
- 75% wrinkled and green
- 100% round and yellow
- 75% wrinkled and yellow

**Solution**

1. When a cross is made between pea plants with round green seeds ($RRyy$) and wrinkled yellow seeds ($rrYY$), the $F_1$ generation inherits one allele for each trait from each parent.
2. The resulting genotype is $RrYy$, which expresses the dominant traits, resulting in 100% round and yellow seeds.

**Answer:** (c) 100% round and yellow

> Common mistake: Confusing dominant and recessive traits in the F1 generation.

### Question 12

*1 mark · MCQ · Life Processes*

The breakdown of glucose has taken the following pathway :
Glucose -(a)-> Pyruvate + Energy -(b)-> Lactic acid + Energy
The sites 'a' and 'b' respectively are :

- Mitochondria and Oxygen deficient muscle cells
- Cytoplasm and Oxygen rich muscle cells
- Cytoplasm and Yeast cells
- Cytoplasm and Oxygen deficient muscle cells

**Solution**

1. The first step of glucose breakdown (glycolysis) takes place in the cytoplasm where glucose is converted into pyruvate.
2. The breakdown of pyruvate into lactic acid takes place in muscle cells during oxygen deficiency (lack of oxygen).

**Answer:** (d) Cytoplasm and Oxygen deficient muscle cells

> Common mistake: Confusing cytoplasm with mitochondria as the site for the first step of glucose breakdown.

### Question 13

*1 mark · MCQ · Light – Reflection and Refraction*

Mirror 'X' is used to concentrate sunlight in solar furnace and Mirror 'Y' is fitted on the side of the vehicle to see the traffic behind the driver. Which of the following statements are true for the two mirrors ?
(i) The image formed by mirror 'X' is real, diminished and at its focus.
(ii) The image formed by mirror 'Y' is virtual, diminished and erect.
(iii) The image formed by mirror 'X' is virtual, diminished and erect.
(iv) The image formed by mirror 'Y' is real, diminished and at its focus.

- (i) and (ii)
- (ii) and (iii)
- (iii) and (iv)
- (i) and (iv)

**Solution**

1. Mirror 'X' is a concave mirror used in solar furnaces to concentrate sunlight at its focus, forming a real and diminished image.
2. Mirror 'Y' is a convex mirror fitted as a rear-view mirror in vehicles, which always forms a virtual, erect, and diminished image.

**Answer:** (b) (ii) and (iii)

> Common mistake: Confusing the properties of concave mirrors used for convergence with convex mirrors used for wider field of view.

### Question 14

*1 mark · MCQ · The Human Eye and the Colourful World*

An old person is suffering from an eye defect caused by weakening of ciliary muscles and diminishing flexibility of the eye lens. If the defect of vision is 'a' which can be corrected by lens 'b', then 'a' and 'b' respectively are :

- hypermetropia and convex lens
- presbyopia and bifocal lens
- myopia and concave lens
- myopia and bifocal lens

**Solution**

1. The weakening of ciliary muscles and diminishing flexibility of the eye lens in old age leads to the defect of vision known as presbyopia.
2. Presbyopia is corrected using a bifocal lens consisting of both concave and convex lenses.

**Answer:** (b) presbyopia and bifocal lens

> Common mistake: Confusing presbyopia with hypermetropia or myopia.

### Question 15

*1 mark · MCQ · Our Environment*

Which of the following groups do not constitute a food chain ?
(i) Wolf, rabbit, grass, lion
(ii) Plankton, man, grasshopper, fish
(iii) Hawk, grass, snake, grasshopper, frog
(iv) Grass, snake, wolf, tiger

- (i) and (iv)
- (i) and (iii)
- (ii) and (iii)
- (ii) and (iv)

**Solution**

1. In group (ii), plankton, man, grasshopper, and fish do not form a linear trophic sequence because man does not eat plankton or grasshoppers directly in a standard chain.
2. In group (iv), grass, snake, wolf, and tiger do not form a correct food chain as snakes are not herbivores feeding on grass.

**Answer:** (d) (ii) and (iv)

> Common mistake: Failing to trace the correct trophic levels of organisms in the options.

### Question 16

*1 mark · MCQ · Our Environment*

The percentage of solar energy which is not converted into food energy by the leaves of green plants in a terrestrial ecosystem is about :

- 1%
- 10%
- 90%
- 99%

**Solution**

1. Green plants capture about $1\%$ of the solar energy that falls on their leaves and convert it into food energy.
2. Therefore, the percentage of solar energy which is not converted into food energy is $100\% - 1\% = 99\%$.

**Answer:** (d) 99%

> Common mistake: Confusing the $1\%$ energy captured by plants with the $10\%$ law of energy transfer between trophic levels.

### Question 17

*1 mark · Assertion and reason · Chemical Reactions and Equations*

Assertion (A) : Decomposition reactions are generally endothermic reactions.
Reason (R) : Decomposition of organic matter into compost is an exothermic process.

- 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.

**Solution**

1. Decomposition reactions require energy in the form of heat, light or electricity for breaking down the reactants, making them generally endothermic.
2. The decomposition of organic matter into compost is an exothermic process because it releases heat.
3. Both statements are correct facts from NCERT, but the fact that organic decomposition is exothermic does not explain why general decomposition reactions are endothermic.

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

> Common mistake: Confusing general inorganic decomposition reactions with biochemical exothermic processes like composting.

### Question 18

*1 mark · Assertion and reason · Heredity*

Assertion (A) : A human child bears all the basic features of human beings.
Reason (R) : It looks exactly like its parents, showing very little variations.

- 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.

**Solution**

1. A human child exhibits all basic human features due to heredity, making Assertion (A) true.
2. Reason (R) is false because offspring do not look exactly like their parents; sexual reproduction introduces significant variations.

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

> Common mistake: Assuming children are exact biological clones of their parents.

### Question 19

*1 mark · Assertion and reason · Magnetic Effects of Electric Current*

Assertion (A) : No two magnetic field lines are found to cross each other.
Reason (R) : The compass needle cannot point towards two directions at the point of intersection of two magnetic field lines.

- 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.

**Solution**

1. Magnetic field lines never intersect because if they did, the compass needle would point in two directions at the same time, which is not possible.

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

> Common mistake: Confusing the direction of magnetic field lines with electric field lines.

### Question 20

*1 mark · Assertion and reason · Our Environment*

Assertion (A) : The amount of ozone in the atmosphere began to drop sharply in the 1980s.
Reason (R) : The oxygen atoms combine with molecular oxygen to form ozone.

- 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.

**Solution**

1. The amount of ozone in the atmosphere began to drop sharply in the 1980s mainly due to synthetic chemicals like chlorofluorocarbons (CFCs).
2. Higher up in the atmosphere, UV radiations split molecular oxygen into free oxygen atoms, which then combine with molecular oxygen to form ozone.
3. Both statements are true independently according to NCERT, but the formation of ozone does not explain why its amount dropped sharply in the 1980s.

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

> Common mistake: Assuming the natural formation process of ozone explains its anthropogenic depletion.

## SECTION B

Questions no. 21 to 26 are Very Short Answer Type questions.

### Question 21

*2 marks · Very short answer · Chemical Reactions and Equations*

A student performs the following experiment in his school laboratory (dilute sulphuric acid with Zn granules). List two observations to justify that in this experiment a chemical change has taken place.

**Solution**

1. Evolution of hydrogen gas is observed as bubbles formed around zinc granules.
2. Change in temperature is observed as the conical flask becomes warm due to an exothermic reaction.

**Answer:** Evolution of hydrogen gas and change in temperature (or evolution of heat).

> Common mistake: Writing physical changes instead of chemical change characteristics.

### Question 22

*2 marks · Very short answer · How do Organisms Reproduce?*

Draw labelled diagrams to show different stages of budding in Hydra.

**Solution**

1. Draw a diagram showing a parent Hydra with a small outgrowth called a bud.
2. Draw subsequent stages showing the growth of the bud, development of tentacles, and finally its detachment from the parent body.

**Answer:** Labeled diagram showing parent Hydra, developing bud, tentacles, and new independent Hydra.

> Common mistake: Not labeling the bud and tentacles clearly.

### Question 23

*2 marks · Very short answer · Life Processes*

(a) Besides minimising the loss of blood, why is it essential to plug any leak in a blood vessel ? Name the component of blood which helps in this process and state how this component perform this function.

**Solution**

1. Plugging leaks is essential to maintain the pressure of the circulatory system which would otherwise result in loss of efficiency and internal bleeding.
2. Blood platelets circulate around the body and plug these leaks by clotting the blood at the site of injury.

**Answer:** To maintain blood pressure and circulatory efficiency; platelets perform this function by blood clotting.

> Common mistake: Forgetting to mention the role of blood pressure or platelets.

### Question 23 (OR)

*2 marks · Very short answer · Life Processes*

(b) (i) The transport system in plants is relatively slower than in animals. Give reasons.
(ii) State the role of phloem in the transport of materials in plants.

**Part (i)**

1. Plants do not move and have many dead cells in their plant body, so their energy needs are low.
2. Hence, the plant transport systems can be slower compared to animals.

Answer (i): Plants have lower energy needs due to stationary habit and dead cells.

**Part (ii)**

1. Phloem transports soluble products of photosynthesis from leaves to other parts of the plant.

Answer (ii): Phloem transports food and other substances from leaves to storage organs and growing regions.

**Answer:** Plants are stationary and have a high proportion of dead cells in many tissues, requiring less energy and slower transport systems.

> Common mistake: Confusing xylem function with phloem function.

### Question 24

*2 marks · Numerical · Light – Reflection and Refraction*

An object is placed at a distance of 60 cm from a concave lens of focal length 30 cm. Use lens formula to find the position of the image formed in this case.

**Solution**

1. Given: Object distance $u = -60\text{ cm}$, Focal length of concave lens $f = -30\text{ cm}$.
2. Formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
3. Substitution: $\frac{1}{v} - \frac{1}{-60} = \frac{1}{-30}$
4. Working: $\frac{1}{v} = \frac{1}{-30} - \frac{1}{60} = \frac{-2 - 1}{60} = \frac{-3}{60} = \frac{-1}{20}$
5. Result: $v = -20\text{ cm}$

**Answer:** The image is formed at a distance of 20 cm in front of the lens.

> Common mistake: Sign convention errors for concave lens focal length and object distance.

### Question 25

*2 marks · Numerical · Electricity*

(a) A wire of resistance $R$ is cut into three equal parts. If these three parts are then joined in parallel, calculate the total resistance of the combination so formed.

**Solution**

1. Given: Resistance of original wire is $R$. It is cut into three equal parts, so resistance of each part is $R' = \frac{R}{3}$.
2. Formula: For parallel combination, $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
3. Substitution: $\frac{1}{R_p} = \frac{1}{R/3} + \frac{1}{R/3} + \frac{1}{R/3} = \frac{3}{R} + \frac{3}{R} + \frac{3}{R}$
4. Working: $\frac{1}{R_p} = \frac{9}{R}$
5. Result: $R_p = \frac{R}{9}$

**Answer:** Total resistance of the combination is $\frac{R}{9}$.

> Common mistake: Taking resistance of each part as $3R$ instead of $R/3$.

### Question 25 (OR)

*2 marks · Very short answer · Electricity*

(b) Define electric power. When do we say that the power consumed in an electric circuit is 1 watt ?

**Solution**

1. Electric power is defined as the rate at which electric energy is dissipated or consumed in an electric circuit, given by $P = VI$.
2. The power consumed is 1 watt when a potential difference of 1 volt causes a current of 1 ampere to flow through the circuit.

**Answer:** Electric power is the rate of consumption of electric energy ($P = VI$). Power is 1 watt when 1 A current flows through a potential difference of 1 V.

> Common mistake: Forgetting to mention the units (volt and ampere) while defining 1 watt.

### Question 26

*2 marks · Very short answer · Our Environment*

"Excessive use of chemicals and pesticides in agriculture adversely effect the environment." Justify this statement.

**Solution**

1. Excessive use of chemicals and pesticides leads to soil degradation by killing useful microorganisms and reducing natural fertility.
2. These non-biodegradable chemicals enter food chains and undergo biological magnification, harming higher trophic levels including humans.

**Answer:** Excessive use of chemicals degrades soil fertility by killing beneficial microbes and causes biological magnification of toxic substances in food chains.

> Common mistake: Writing general pollution effects instead of specific agricultural impacts like soil degradation and biological magnification.

## SECTION C

Questions no. 27 to 33 are Short Answer Type questions.

### Question 27

*3 marks · Short answer · Metals and Non-metals*

(a) "Displacement reactions also play a key role in extracting metals in the middle of the reactivity series." Justify this statement with two examples.
(b) Why can metals high up in the reactivity series not be obtained by reduction of their oxides by carbon ?

**Part a**

1. Highly reactive metals like sodium, calcium, and aluminium are used as reducing agents to displace metals of lower reactivity from their oxides.
2. Example 1: Heating manganese dioxide with aluminium powder: $3MnO_2(s) + 4Al(s) \rightarrow 3Mn(l) + 2Al_2O_3(s) + \text{Heat}$.
3. Example 2: Reaction of iron(III) oxide with aluminium used in thermit welding: $Fe_2O_3(s) + 2Al(s) \rightarrow 2Fe(l) + Al_2O_3(s) + \text{Heat}$.

Answer a: Displacement reactions reduce metal oxides using aluminium powder to obtain metals like manganese and iron.

**Part b**

1. Metals high up in the reactivity series (such as sodium, magnesium, calcium, and aluminium) have a much greater affinity for oxygen than carbon does.
2. Therefore, carbon cannot reduce the oxides of these metals.

Answer b: Carbon cannot reduce oxides of highly reactive metals because these metals have a higher affinity for oxygen than carbon.

**Answer:** Displacement reactions involving more reactive metals like aluminium are used to reduce metal oxides to molten metals, whereas carbon cannot reduce oxides of highly reactive metals because these metals have a higher affinity for oxygen than carbon.

> Common mistake: Writing incorrect chemical formulae or balancing coefficients for the thermit reaction.

### Question 28

*3 marks · Short answer · Metals and Non-metals*

(a) With the help of an activity, explain the conditions under which iron articles get rusted.

**Solution**

1. Take three test tubes and label them A, B, and C. Place clean iron nails in each.
2. In test tube A, add some tap water, cork it, and leave it so that nails are exposed to both air and water.
3. In test tube B, add boiled distilled water and about 1 mL of oil to prevent air from dissolving in the water, ensuring the nails are exposed to water only.
4. In test tube C, put anhydrous calcium chloride (a drying agent) to absorb moisture from the air, ensuring the nails are exposed to dry air only.
5. Observe the nails after a few days; rusting occurs only in test tube A where both air and moisture are present, while nails in B and C do not rust.

**Answer:** Iron articles get rusted in the presence of both oxygen and moisture (water vapor).

> Common mistake: Forgetting to mention that both air and moisture are required simultaneously for rusting.

### Question 28 (OR)

*3 marks · Short answer · Metals and Non-metals*

(b) (i) Name two metals which react violently with cold water. List any three observations which a student notes when these metal are dropped in a beaker containing water.
(ii) Write a test to identify the gas evolved (if any) during the reaction of these metals with water.

**Part i**

1. Potassium and sodium are the two metals that react violently with cold water.
2. Observation 1: The reaction is highly exothermic and violent.
3. Observation 2: Hydrogen gas evolved immediately catches fire on the surface of water.
4. Observation 3: The metal starts floating because bubbles of hydrogen gas formed stick to its surface.

Answer i: Potassium and sodium react with cold water, releasing hydrogen gas with catching fire and floating due to gas bubbles.

**Part ii**

1. Bring a burning matchstick near the evolved gas.
2. The gas burns with a pop sound, which confirms that the evolved gas is hydrogen.

Answer ii: Hydrogen gas burns with a pop sound when a burning matchstick is brought near it.

**Answer:** Sodium and potassium react violently with cold water, releasing hydrogen gas which catches fire, and the metal floats as it forms bubbles.

> Common mistake: Failing to state all three distinct observations for the metal-water reaction.

### Question 29

*3 marks · Short answer · Control and Coordination*

Plants have neither a nervous system nor muscles, even then they respond to stimuli. For example, the leaves of chhui-mui (touch-me-not) plant when touched begin to fold up and droop.
(a) How is the information communicated in "touch-me-not" plants ?
(b) What enables the plant cells to bring out the observable response ?
(c) Differentiate the movement mentioned above from the movement of tendrils in a pea plant.

**Part a**

1. Plants use electrical-chemical means to transmit information from one cell to another.

Answer a: Information is communicated through electrical-chemical signals across plant cells.

**Part b**

1. Plant cells change their shape by changing the amount of water in them, resulting in swelling or shrinking.

Answer b: Changes in cellular water content enable plant cells to alter their shape and bring about an observable response.

**Part c**

1. The movement of touch-me-not leaves is a nastic movement independent of growth and direction, whereas the movement of tendrils in a pea plant is a directional growth movement (thigmotropism).

Answer c: Leaf folding is non-directional and independent of growth, while tendril movement is directional and growth-dependent.

**Answer:** Plants communicate via electrical-chemical signals, bring about response by changing water content in cells, and show nastic movements unlike directional growth movements in tendrils.

> Common mistake: Confusing nastic movements with tropic movements.

### Question 30

*3 marks · Short answer · Heredity*

(a) What are chromosomes ?
(b) Explain in brief how stability of DNA content of a species is ensured in sexually reproducing organisms ?

**Part a**

1. Chromosomes are highly condensed thread-like structures composed of DNA and proteins found in the nucleus of a cell, which carry hereditary information in the form of genes.

Answer a: Chromosomes are nuclear structures made of DNA and protein that carry hereditary information.

**Part b**

1. Sexually reproducing organisms produce specialized germ cells (gametes) that have half the number of chromosomes (haploid) found in normal body cells.
2. This reduction is brought about by a special type of cell division called meiosis.
3. When two gametes fuse during fertilization, the original number of chromosomes (diploid) is restored in the zygote, ensuring stability of DNA content of the species.

Answer b: Meiosis halves the chromosome number in gametes, and subsequent fertilization restores the original number in the offspring.

**Answer:** Chromosomes are thread-like structures of DNA and protein, and their constancy across generations is maintained by gamete formation involving meiosis.

> Common mistake: Omitting the roles of meiosis and fertilization in maintaining chromosome number.

### Question 31

*3 marks · Short answer · Light – Reflection and Refraction*

Draw ray diagrams to show the nature, position and relative size of the image formed by a convex mirror when the object is placed (i) at infinity and (ii) between infinity and pole P of the mirror.

**Solution**

1. Diagram for object at infinity: Draw a convex mirror with principal axis, focus F, and center of curvature C behind the mirror. Draw parallel incident rays reflecting away such that they appear to diverge from focus F. State nature: virtual, erect, and point-sized at F.
2. Diagram for object between infinity and pole P: Draw a convex mirror with object AB in front of it. Draw a ray parallel to the principal axis that gets reflected diverging from F, and a ray directed towards the center of curvature C that reflects back along the same path. Produce the reflected rays backward to intersect between P and F behind the mirror.
3. State nature for object between infinity and pole: virtual, erect, and diminished, formed behind the mirror between P and F.

**Answer:** A convex mirror always forms virtual, erect images behind the mirror: point-sized at focus F when at infinity, and diminished between P and F when placed anywhere in front.

> Common mistake: Drawing diverging rays incorrectly or failing to show virtual rays meeting behind the mirror with dotted lines.

### Question 32

*3 marks · Short answer · Electricity*

Consider the following electric circuit:
Calculate the values of the following :
(a) The total resistance of the circuit
(b) The total current drawn from the source
(c) Potential difference across the parallel combination of $10\ \Omega$ and $15\ \Omega$ resistors

**Part (a)**

1. The $10\ \Omega$ and $15\ \Omega$ resistors are connected in parallel, so their equivalent resistance $R_1$ is given by $\frac{1}{R_1} = \frac{1}{10} + \frac{1}{15} = \frac{3+2}{30} = \frac{5}{30}$, which gives $R_1 = 6\ \Omega$.
2. The $60\ \Omega$ and $40\ \Omega$ resistors are connected in parallel, so their equivalent resistance $R_2$ is given by $\frac{1}{R_2} = \frac{1}{60} + \frac{1}{40} = \frac{2+3}{120} = \frac{5}{120}$, which gives $R_2 = 24\ \Omega$.
3. The total resistance of the circuit is $R = R_1 + R_2 = 6\ \Omega + 24\ \Omega = 30\ \Omega$.

Answer (a): $30\ \Omega$

**Part (b)**

1. Using Ohm's law, the total current drawn from the source is $I = \frac{V}{R}$.
2. Substituting the values, $I = \frac{15\ \text{V}}{30\ \Omega} = 0.5\ \text{A}$.

Answer (b): $0.5\ \text{A}$

**Part (c)**

1. The potential difference across the parallel combination of $10\ \Omega$ and $15\ \Omega$ resistors is the voltage across $R_1$.
2. $V_1 = I \times R_1 = 0.5\ \text{A} \times 6\ \Omega = 3\ \text{V}$.

Answer (c): $3\ \text{V}$

**Answer:** Total resistance is $30\ \Omega$, total current is $0.5\ \text{A}$, and potential difference across the parallel combination is $3\ \text{V}$.

> Common mistake: Adding all resistor values directly without identifying series and parallel combinations correctly.

### Question 33

*3 marks · Short answer · Electricity*

(a) Write the relationship between resistivity and resistance of a cylindrical conductor of length $l$ and area of cross-section $A$. Hence derive the SI unit of resistivity.
(b) Why are alloys used in electrical heating devices ?

**Part (a)**

1. The resistance $R$ of a cylindrical conductor is directly proportional to its length $l$ and inversely proportional to its cross-sectional area $A$, written as $R = \rho \frac{l}{A}$, where $\rho$ is resistivity.
2. Rearranging the formula for resistivity gives $\rho = \frac{R \times A}{l}$.
3. Substituting the SI units, $\text{SI unit of } \rho = \frac{\Omega \times \text{m}^2}{\text{m}} = \Omega\text{m}$.

Answer (a): $\rho = \frac{RA}{l}$, SI unit is $\Omega\text{m}$

**Part (b)**

1. Alloys are used in electrical heating devices because their resistivity is much higher than that of pure metals, and they do not burn (oxidize) easily at high temperatures.

Answer (b): High resistivity and high melting point without oxidation

**Answer:** Resistivity $\rho = \frac{RA}{l}$ with SI unit $\Omega\text{m}$, and alloys are used because they have high resistivity and do not oxidize at high temperatures.

> Common mistake: Writing the unit of resistivity incorrectly as $\Omega/\text{m}$ instead of $\Omega\text{m}$.

## SECTION D

Questions no. 34 to 36 are Long Answer Type questions.

### Question 34

*5 marks · Long answer · Carbon and its Compounds*

(a) (i) Draw two isomeric structures of Butene ($C_4H_8$).
(ii) Name the following compounds :
(I) $CH_3-CH_2-CH_2-CH_2-Cl$
(II) $CH_3-CH_2-CH_2-CHO$
(iii) Write the chemical equations for the following reactions. Mention one essential condition each for these reactions to take place.
(I) Ethanol undergoes complete oxidation
(II) Propene undergoes hydrogenation
(III) Ethanoic acid reacts with ethanol

**Part (a) (i)**

1. Draw But-1-ene: $CH_3-CH_2-CH=CH_2$
2. Draw But-2-ene: $CH_3-CH=CH-CH_3$

Answer (a) (i): But-1-ene and But-2-ene

**Part (a) (ii)**

1. (I) $CH_3-CH_2-CH_2-CH_2-Cl$ is 1-chlorobutane.
2. (II) $CH_3-CH_2-CH_2-CHO$ is butanal.

Answer (a) (ii): (I) 1-Chlorobutane, (II) Butanal

**Part (a) (iii)**

1. (I) $CH_3CH_2OH + 3O_2 \rightarrow 2CO_2 + 3H_2O$ (Condition: Heat/Combustion).
2. (II) $CH_3-CH=CH_2 + H_2 \xrightarrow{Ni/Pd catalyst} CH_3-CH_2-CH_3$ (Condition: Presence of nickel or palladium catalyst and heat).
3. (III) $CH_3COOH + CH_3CH_2OH \xrightarrow{Conc. H_2SO_4} CH_3COOCH_2CH_3 + H_2O$ (Condition: Presence of acid catalyst).

Answer (a) (iii): Equations and conditions for oxidation, hydrogenation, and esterification written.

**Answer:** Isomeric structures of butene, IUPAC names of given compounds, and chemical equations for oxidation, hydrogenation, and esterification.

> Common mistake: Forgetting to write the reaction conditions such as catalysts or temperatures in chemical equations.

### Question 34 (OR)

*5 marks · Long answer · Carbon and its Compounds*

(b) (i) A carbon compound $X$ is a good solvent. On reaction with sodium, $X$ forms two products $Y$ and $Z$. $Z$ is used to convert vegetable oil into vegetable ghee. Identify and name $X$, $Y$ and $Z$. Also write the equation of reaction of $X$ with sodium to justify your answer.
(ii) Write chemical equation to show what happens when ethanol :
(I) burns in oxygen/air.
(II) is heated at 443 K in excess conc. $H_2SO_4$.
(III) reacts with acidified potassium dichromate.

**Part (b) (i)**

1. Compound X is ethanol ($CH_3CH_2OH$), which is a good solvent.
2. Ethanol reacts with sodium to form sodium ethoxide (Y) and hydrogen gas (Z).
3. Hydrogen ($Z$) is used in the hydrogenation of vegetable oils to vegetable ghee.
4. Equation: $2CH_3CH_2OH + 2Na \rightarrow 2CH_3CH_2ONa + H_2$

Answer (b) (i): X: Ethanol, Y: Sodium ethoxide, Z: Hydrogen gas

**Part (b) (ii)**

1. (I) $CH_3CH_2OH + 3O_2 \rightarrow 2CO_2 + 3H_2O + \text{Heat and light}$
2. (II) $CH_3CH_2OH \xrightarrow[443\text{ K}]{\text{Conc. } H_2SO_4} CH_2=CH_2 + H_2O$
3. (III) $CH_3CH_2OH + \text{Acidified } K_2Cr_2O_7 + \text{Heat} \rightarrow CH_3COOH$

Answer (b) (ii): Chemical equations for burning, dehydration, and oxidation of ethanol.

**Answer:** Identification of X as ethanol, Y as sodium ethoxide, Z as hydrogen gas, along with chemical equations.

> Common mistake: Writing incorrect temperature or catalyst for the dehydration of ethanol.

### Question 35

*5 marks · Long answer · How do Organisms Reproduce?*

(a) (i) Write the functions of the following parts of human female reproductive system :
(I) Ovary
(II) Fallopian tube
(III) Uterus
(ii) State briefly two contraceptive methods used by human males.

**Part (a) (i)**

1. (I) Ovary: Produces female gametes (ova/eggs) and secretes female hormones (oestrogen and progesterone).
2. (II) Fallopian tube: Carries the egg from the ovary to the uterus and is the site of fertilisation.
3. (III) Uterus: Site of implantation of the fertilized egg, and nourishes the developing embryo.

Answer (a) (i): Functions of ovary, fallopian tube, and uterus stated.

**Part (a) (ii)**

1. Condoms: Used over the penis as a mechanical barrier to prevent sperm from reaching the vagina.
2. Vasectomy: Surgical method in males where the vas deferens is blocked to prevent sperm transfer.

Answer (a) (ii): Condoms and vasectomy.

**Answer:** Functions of ovary, fallopian tube, uterus, and two male contraceptive methods.

> Common mistake: Confusing the site of fertilization (fallopian tube) with the site of implantation (uterus).

### Question 35 (OR)

*5 marks · Long answer · How do Organisms Reproduce?*

(b) (i) Differentiate between self-pollination and cross-pollination.
(ii) Identify A, B and C in the diagram given below and write one function of each.

**Part (b) (i)**

1. Self-pollination: Transfer of pollen grains from the anther to the stigma of the same flower or another flower of the same plant.
2. Cross-pollination: Transfer of pollen grains from the anther of one flower to the stigma of another flower of a different plant of the same species.

Answer (b) (i): Definitions of self-pollination and cross-pollination.

**Part (b) (ii)**

1. A: Stigma / Pollen grain - Receives the pollen grain and provides surface for germination.
2. B: Pollen tube - Grows down through the style to carry male gametes to the ovary.
3. C: Female gamete / Ovule - Contains the egg cell which fuses with the male gamete during fertilization.

Answer (b) (ii): A: Stigma/Pollen grain, B: Pollen tube, C: Female gamete/Ovule with functions.

**Answer:** Difference between self and cross pollination, and identification of A, B, C with functions.

> Common mistake: Confusing pollen tube function with the style.

### Question 36

*5 marks · Long answer · Light – Reflection and Refraction*

(a) (i) The power of a lens 'X' is $-2.5\text{ D}$. Name the lens and determine its focal length in cm. For which eye defect of vision will an optician prescribe this type of lens as a corrective lens ?
(ii) "The value of magnification 'm' for a lens is $-2$." Using new Cartesian Sign Convention and considering that an object is placed at a distance of $20\text{ cm}$ from the optical centre of this lens, state :
(I) the nature of the image formed;
(II) size of the image compared to the size of the object;
(III) position of the image, and
(IV) sign of the height of the image.
(iii) The numerical values of the focal lengths of two lenses A and B are $10\text{ cm}$ and $20\text{ cm}$ respectively. Which one of the two will show higher degree of convergence/divergence ? Give reason to justify your answer.

**Part (a)(i)**

1. The lens has a negative power, so lens X is a concave lens.
2. The focal length $f = \frac{1}{P} = \frac{1}{-2.5\text{ D}} = -0.4\text{ m} = -40\text{ cm}$.
3. This type of lens is prescribed for the correction of myopia or nearsightedness.

Answer (a)(i): Concave lens, focal length $-40\text{ cm}$, prescribed for myopia.

**Part (a)(ii)**

1. A negative sign of magnification ($m = -2$) indicates that the image formed is real and inverted.
2. Since the magnitude of $m$ is $2$ (greater than $1$), the size of the image is enlarged or magnified compared to the size of the object.
3. Using magnification formula $m = \frac{v}{u} = -2$, with $u = -20\text{ cm}$, we get $v = -2 \times (-20\text{ cm}) = +40\text{ cm}$, so the image is formed at a distance of $40\text{ cm}$ on the other side of the lens.
4. Since the image is real and inverted, the sign of the height of the image is negative.

Answer (a)(ii): (I) Real and inverted, (II) Magnified, (III) At $40\text{ cm}$ on the other side of the lens, (IV) Negative.

**Part (a)(iii)**

1. Power of a lens is inversely proportional to its focal length, given by $P = \frac{1}{f}$.
2. Lens A has a smaller focal length ($10\text{ cm}$) compared to lens B ($20\text{ cm}$), which means lens A has a higher power.
3. Therefore, lens A will show a higher degree of convergence or divergence.

Answer (a)(iii): Lens A, because it has a shorter focal length and hence higher power.

**Answer:** Lens X is a concave lens with focal length $-40\text{ cm}$ used for myopia. For magnification $-2$, the image is real, inverted, magnified to twice the object size, formed at $40\text{ cm}$ on the other side, and has a negative height. Lens A has a higher degree of convergence/divergence.

> Common mistake: Confusing the sign of the focal length for a concave lens or misinterpreting the sign of magnification for real versus virtual images.

### Question 36 (OR)

*5 marks · Long answer · Light – Reflection and Refraction*

(b) (i) Draw a ray diagram to show the refraction of a ray of light through a rectangular glass slab when it falls obliquely from air into glass.
(ii) State Snell's law of refraction of light.
(iii) Differentiate between the virtual images formed by a convex lens and a concave lens on the basis of :
(I) object distance, and
(II) magnification.

**Part (b) (i)**

1. Draw a rectangular glass slab PQRS.
2. Incident ray strikes face PQ obliquely, bends towards the normal inside the slab as the refracted ray, and emerges out from face SR as the emergent ray parallel to the incident ray (lateral displacement).

Answer (b) (i): Ray diagram showing refraction through glass slab.

**Part (b) (ii)**

1. Snell's law of refraction: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for the light of a given colour and for the given pair of media.
2. Formula: $\frac{\sin i}{\sin r} = \text{constant} = n_{21}$.

Answer (b) (ii): $\frac{\sin i}{\sin r} = \text{constant}$

**Part (b) (iii)**

1. (I) Object distance: For a convex lens, a virtual image is formed when the object is placed between the optical centre and the principal focus ($u < f$). For a concave lens, a virtual image is formed for any position of the object ($u$ can be anywhere in front of the lens).
2. (II) Magnification: For a convex lens, the magnification for a virtual image is greater than 1 ($m > 1$, enlarged). For a concave lens, the magnification for a virtual image is always less than 1 ($m < 1$, diminished).

Answer (b) (iii): Differences based on object distance and magnification stated.

**Answer:** Ray diagram description, Snell's law, and difference between virtual images of convex and concave lenses.

> Common mistake: Writing that a concave lens can form a real image, or confusing the magnification values for virtual images of convex and concave lenses.

## SECTION E

The following questions are Source-based/Case-based questions. Read the case carefully and answer the questions that follow.

### Question 37

*4 marks · Case-based · Acids, Bases and Salts*

Seawater contains many salts dissolved in it. Common salt is separated from these salts. Deposits of solid salt are also found in several parts of the world. These large crystals are often brown due to impurities. This is called rock salt and is mined like coal. The common salt is an important raw material for chemicals of daily use.
(a) Write balanced chemical equations to show the products formed during electrolysis of brine.
(b) List two uses of any one product obtained during electrolysis of brine.
(c) (i) A mild non-corrosive basic salt 'A', used for faster cooking, is strongly heated to produce a compound 'B', that is used for removing permanent hardness of water. Identify A and B and also write the equation for the reaction that occurs when A is heated.

**Part (a)**

1. 1. Write the balanced chemical equation for the chlor-alkali process where electricity is passed through an aqueous solution of sodium chloride.
2. 2. $2\text{NaCl (aq)} + 2\text{H}_2\text{O (l)} \rightarrow 2\text{NaOH (aq)} + \text{Cl}_2\text{ (g)} + \text{H}_2\text{ (g)}$

Answer (a): $2\text{NaCl (aq)} + 2\text{H}_2\text{O (l)} \rightarrow 2\text{NaOH (aq)} + \text{Cl}_2\text{ (g)} + \text{H}_2\text{ (g)}$

**Part (b)**

1. 1. Choose any product such as chlorine gas.
2. 2. Two uses of chlorine gas are water treatment (swimming pools) and production of polyvinyl chloride (PVC) or bleaching powder.

Answer (b): Water treatment and manufacturing of PVC

**Part (c) (i)**

1. 1. Identify the mild non-corrosive basic salt A used for faster cooking as sodium hydrogen carbonate (baking soda, $\text{NaHCO}_3$).
2. 2. Identify compound B used for removing permanent hardness of water as sodium carbonate ($\text{Na}_2\text{CO}_3$).
3. 3. Write the thermal decomposition equation: $2\text{NaHCO}_3 \xrightarrow{\text{Heat}} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$

Answer (c) (i): A is $\text{NaHCO}_3$, B is $\text{Na}_2\text{CO}_3$, and the reaction is $2\text{NaHCO}_3 \xrightarrow{\text{Heat}} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$

**Answer:** Electrolysis of brine produces hydrogen, chlorine, and sodium hydroxide; A is sodium hydrogen carbonate and B is sodium carbonate.

> Common mistake: Confusing baking soda ($\text{NaHCO}_3$) with washing soda ($\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$).

### Question 37 (OR)

*2 marks · Case-based · Acids, Bases and Salts*

(c) (ii) Define water of crystallisation. Give two examples of salts that have water of crystallisation.

**Part (i)**

1. 1. Define water of crystallisation as the fixed number of water molecules present in one formula unit of a salt.
2. 2. Give two examples: Copper sulphate pentahydrate ($\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$) and Sodium carbonate decahydrate ($\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$).

Answer (i): Water of crystallisation is the fixed number of water molecules in one formula unit of a salt; examples are $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ and $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$.

**Answer:** Water of crystallisation is the fixed number of water molecules present in one formula unit of a salt.

> Common mistake: Forgetting to mention that the number of water molecules is fixed for a particular salt.

### Question 38

*4 marks · Case-based · Life Processes*

The maintenance functions of all living organisms must go on even when they are not doing anything particular. Even when we are just sitting in a class or even asleep, this maintenance job has to go on. These maintenance processes require energy to prevent damage and break-down of cells and tissues, which is obtained by the individual organism from the food prepared by the autotrophs, called producers.
(a) Name and define the process by which green plants prepare food.
(b) Write chemical equation involved in the above process.
(c) (i) State in proper sequence the events that occur in synthesis of food by desert plants.

**Part (a)**

1. The process by which green plants prepare their food is called photosynthesis.
2. Photosynthesis is the process by which autotrophs take in substances from outside and convert them into stored forms of energy in the presence of sunlight and chlorophyll.

Answer (a): Photosynthesis; it is the conversion of carbon dioxide and water into carbohydrates using sunlight.

**Part (b)**

1. Write the reactants: carbon dioxide and water in the presence of sunlight and chlorophyll.
2. Write the balanced chemical equation: $6\text{CO}_2 + 12\text{H}_2\text{O} \xrightarrow{\text{Sunlight}_{\text{Chlorophyll}}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 + 6\text{H}_2\text{O}$.

Answer (b): $6\text{CO}_2 + 12\text{H}_2\text{O} \xrightarrow{\text{Sunlight}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 + 6\text{H}_2\text{O}$

**Part (c)**

1. Desert plants take up carbon dioxide at night and prepare an intermediate compound.
2. This intermediate compound is acted upon by the energy absorbed by chlorophyll during the day to form food.

Answer (c): Desert plants take up carbon dioxide at night to form an intermediate compound which is absorbed by energy from sunlight during the day.

**Answer:** Refer to sub-parts for the detailed answers.

> Common mistake: Forgetting to write water as a product in the photosynthesis equation or missing the specific nocturnal absorption of CO2 in desert plants.

### Question 38 (OR)

*2 marks · Case-based · Life Processes*

(c) (ii) Explain giving reasons what happens to the rate at which the green plants will prepare food
(I) during cloudy weather, and
(II) when stomata get blocked due to dust.

**Part (i)**

1. 1. Explain that during cloudy weather, sunlight intensity decreases, which reduces the rate of photosynthesis.
2. 2. Explain that when stomata get blocked by dust, the intake of carbon dioxide is prevented, which also decreases the rate of photosynthesis.

Answer (i): (I) Rate decreases due to less sunlight. (II) Rate decreases due to restricted $\text{CO}_2$ intake.

**Answer:** Rate of photosynthesis decreases during cloudy weather and when stomata are blocked due to dust.

> Common mistake: Not relating stomatal block to $\text{CO}_2$ uptake.

### Question 39

*4 marks · Case-based · Electricity*

In our homes, we receive the supply of electric power through a main supply also called mains, either supported through overhead electric poles or by underground cables. In our country the potential difference between the two wires (live wire and neutral wire) of this supply is 220 V.
(a) Write the colours of the insulation covers of the line wires through which supply comes to our homes.
(b) What should be the current rating of the electric circuit (220 V) so that an electric iron of 1 kW power rating can be operated ?
(c) (i) What is the function of the earth wire ? State the advantage of the earth wire in domestic electric appliances such as electric iron.

**Part (a)**

1. 1. State the colour of the insulation cover of the live wire in domestic wiring.
2. 2. Red (or brown according to the new international convention).

Answer (a): Red (or brown)

**Part (b)**

1. 1. State the given values: Power $P = 1 \text{ kW} = 1000 \text{ W}$, Voltage $V = 220 \text{ V}$.
2. 2. Use the formula $P = V \times I$ to find current $I = \frac{P}{V}$.
3. 3. Substitute values: $I = \frac{1000}{220} = 4.55 \text{ A}$.

Answer (b): $4.55 \text{ A}$

**Part (c) (i)**

1. 1. State that the earth wire provides a low-resistance conducting path to the earth.
2. 2. State that it ensures any leakage of current to the metallic body of an appliance keeps its potential to that of the earth, thus preventing severe electric shocks to the user.

Answer (c) (i): Earth wire prevents electric shocks by sending any leakage current from the metallic body of the appliance to the earth.

**Answer:** Live wire is red, current rating is 4.55 A, and earth wire protects against electric shocks.

> Common mistake: Forgetting to convert kW to W when calculating current.

### Question 39 (OR)

*2 marks · Case-based · Electricity*

(c) (ii) List two precautions to be taken to avoid electrical accidents. State how these precautions prevent possible damage to the circuit/appliance.

**Part (i)**

1. 1. List two precautions such as using an appropriate fuse in every circuit and avoiding the connection of too many appliances to a single socket.
2. 2. State how they prevent damage: fuses melt and break the circuit when current exceeds the safe limit, preventing overheating and short-circuiting.

Answer (i): Precautions: use a fuse and avoid overloading; they break the circuit or prevent overheating during excessive current flow.

**Answer:** Precautions include using an electric fuse and not overloading circuits to prevent damage.

> Common mistake: Failing to explain the mechanism of how the precaution prevents damage.

## Frequently asked questions

### What is the paper pattern and section-wise breakdown for the CBSE Class 10 Science Question Paper 2025 (Set 31/1/1)?

The question paper carries a total of 80 marks and has to be completed in 180 minutes. It is divided into five sections: Section A has 20 questions for 20 marks, Section B has 6 questions for 12 marks, Section C has 7 questions for 21 marks, Section D has 3 questions for 15 marks, and Section E has 3 questions for 12 marks.

### Which chapters carry the most marks in the CBSE Class 10 Science Set 31/1/1 paper?

In this paper, the top scoring chapters include Electricity with 12 marks and Light – Reflection and Refraction with 11 marks. Other high-weightage chapters are Metals and Non-metals with 10 marks, Life Processes and How do Organisms Reproduce with 8 marks each, and Acids, Bases and Salts with 6 marks.

### How should students write answers to score full marks in this Science question paper?

Students should write step-by-step answers, include necessary ray diagrams in physics, balance chemical equations in chemistry, and label biology diagrams clearly. Writing precise scientific terms and showing intermediate steps in numerical problems ensures that you do not lose marks.

### Is the solutions PDF for CBSE Class 10 Science 2025 Set 31/1/1 free on SwaVid?

Yes, the complete solutions PDF for the CBSE Class 10 Science Question Paper 2025 (Set 31/1/1) is completely free to download on SwaVid. You can access step-by-step answers prepared by subject experts to verify your responses and understand the marking scheme.

### What are the total duration and maximum marks for the CBSE Class 10 Science 2025 Set 31/1/1 exam?

The exam is conducted for a maximum of 80 marks with a total duration of 180 minutes. The paper tests conceptual understanding across physics, chemistry, and biology through a total of 39 questions distributed across five sections.

## Related pages

- [All CBSE Class 10 Science papers](https://www.swavid.com/cbse/class-10/science/previous-year-papers)

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