---
title: "CBSE Class 10 Science Question Paper 2026 (Set 31/1/1) with Solutions"
url: https://www.swavid.com/cbse/class-10/science/previous-year-papers/2026-31-1-1
dateModified: 2026-10-07T15:36:11+00:00
---

# CBSE Class 10 Science Question Paper 2026 (Set 31/1/1) with Solutions

Code 31/1/1 · 80 marks · 180 minutes

Solved CBSE Class 10 Science board paper from 2026, 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/2026/X/Science.zip. Solutions written by SwaVid.

Free PDF (26 pages): https://www.swavid.com/api/seo/pdf/papers/cbse/science/swavid-cbse-class-10-science-question-paper-2026-31-1-1-64016ad1a6.pdf

## SECTION - A

All questions are compulsory.

### Question 1

*1 mark · MCQ · Life Processes*

Which structure in a leaf is mainly responsible for gaseous exchange ?

- Xylem
- Stomata
- Phloem
- Cuticle

**Solution**

1. Stomata are tiny pores present on the surface of leaves that facilitate massive amounts of gaseous exchange for photosynthesis and respiration.
2. Xylem and phloem are transport tissues, while the cuticle is a protective waxy layer.

**Answer:** (b) Stomata

> Common mistake: Confusing stomata with cuticle or vascular tissues.

### Question 2

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

Identify the type of reproduction shown in the diagram given below :

- Budding
- Fragmentation
- Spore Formation
- Binary Fission

**Solution**

1. The diagram illustrates the splitting of a single-celled organism into two equal halves, which is characteristic of binary fission.
2. Budding involves a small outgrowth, while fragmentation involves breaking into pieces.

**Answer:** (d) Binary Fission

> Common mistake: Confusing binary fission with multiple fission or budding.

### Question 3

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

Identify the correct statement for spirogyra, leishmania and hydra :

- they reproduce sexually.
- they are unicellular.
- they are multicellular.
- they reproduce asexually.

**Solution**

1. Spirogyra (alga), Leishmania (protozoan), and Hydra (coelenterate) all reproduce primarily through asexual methods such as fragmentation, binary fission, and budding respectively.
2. Therefore, the common correct statement is that they reproduce asexually.

**Answer:** (d) they reproduce asexually.

> Common mistake: Assuming all three share the same cellular structure.

### Question 4

*1 mark · MCQ · Control and Coordination*

Human brain has various parts or regions that help in different actions, responses and coordination. From the following, identify the part responsible for precision of voluntary actions :

- Cerebrum
- Cerebellum
- Medulla
- Pons

**Solution**

1. The cerebellum is the part of the hindbrain responsible for maintaining posture, balance, and the precision of voluntary actions.
2. Cerebrum is for thinking, and medulla controls involuntary actions.

**Answer:** (b) Cerebellum

> Common mistake: Confusing cerebellum with cerebrum.

### Question 5

*1 mark · MCQ · Life Processes*

Pancreas secretes pancreatic juice which contain certain enzyme that helps in digestion of food. Choose the correct option from the following :

- Trypsin digests emulsified fats and lipase digests proteins.
- Trypsin digests proteins and lipase digests emulsified fats.
- Trypsin and lipase both digests fats.
- Trypsin digests proteins and lipase digests carbohydrates.

**Solution**

1. Pancreatic juice contains trypsin for digesting proteins and lipase for breaking down emulsified fats.
2. Therefore, the correct matching is trypsin digests proteins and lipase digests emulsified fats.

**Answer:** (b) Trypsin digests proteins and lipase digests emulsified fats.

> Common mistake: Mixing up the substrates for trypsin and lipase.

### Question 6

*1 mark · MCQ · Heredity*

Sex is determined by different factors in various species. However, in human beings, it is determined genetically. Which amongst the following option(s) is/are correct for human beings ?

- (ii) and (iii)
- (i) only
- (i) and (iii)
- (iii) only

**Solution**

1. Human females have two X chromosomes ($XX$) and produce all gametes (ova) carrying an X chromosome.
2. Human males have an X and a Y chromosome ($XY$) and produce two types of gametes in equal proportion: one carrying an X chromosome and the other carrying a Y chromosome.
3. Therefore, statements (ii) and (iii) correctly describe the gametes carrying sex chromosomes from the male parent that determine the sex of the child upon fusion with the female gamete.

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

> Common mistake: Confusing the contribution of male and female parents, or incorrectly selecting only one of the male gamete types.

### Question 7

*1 mark · MCQ · Our Environment*

Which of the following group is not 'biodegradable' ?

- Vegetable peels, dead leaves, paper
- Cow dung, leather bag, water
- Polythene bag, rubber band, ball pen
- Paper, fruits, bones

**Solution**

1. Substances that are broken down by biological processes are called biodegradable.
2. Polythene bags, rubber bands, and ball pens cannot be broken down easily by microorganisms and are non-biodegradable.

**Answer:** (c) Polythene bag, rubber band, ball pen

> Common mistake: Confusing rubber with natural rubber products that might degrade slowly, while synthetic polymers and plastics are strictly non-biodegradable.

### Question 8

*1 mark · Assertion and reason · Control and Coordination*

Assertion (A) : Reflex actions do not involve thinking.
Reason (R) : Most reflex actions are controlled by the spinal cord.

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

**Solution**

1. Reflex actions are sudden, involuntary responses to external stimuli that do not involve conscious thought.
2. Most reflex actions are mediated through the spinal cord for a quick response.

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

> Common mistake: Thinking that the reason does not explain the assertion because thinking happens in the brain.

### Question 9

*1 mark · Assertion and reason · Life Processes*

Assertion (A) : In human beings, the respiratory pigment is haemoglobin present in red blood cells.
Reason (R) : Haemoglobin has a very high affinity for carbon dioxide.

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

**Solution**

1. Haemoglobin is the respiratory pigment present in red blood cells that carries oxygen.
2. Haemoglobin has a very high affinity for oxygen, not carbon dioxide, although carbon dioxide is transported dissolved in blood or by haemoglobin to some extent.

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

> Common mistake: Assuming haemoglobin has a high affinity for carbon dioxide because it carries it as well.

### Question 10

*2 marks · Short answer · Life Processes*

What is the function of diaphragm in human respiratory system ? Where is it present in human body ?

**Solution**

1. Function: The diaphragm flattens and contracts during inhalation to increase the thoracic cavity volume, and relaxes during exhalation to decrease the volume.
2. Location: It is a muscular partition present at the floor of the chest cavity (thorax), separating the thoracic cavity from the abdominal cavity.

**Answer:** The diaphragm helps in expanding and contracting the chest cavity during breathing and is located at the floor of the chest cavity.

> Common mistake: Forgetting to mention its exact location between the thoracic and abdominal cavities.

### Question 11

*2 marks · Short answer · Life Processes*

State two differences between the act of chewing food and salivation on sight of food.

**Solution**

1. Chewing food is a voluntary action controlled by the somatic nervous system, whereas salivation on sight of food is an involuntary reflex action.
2. Chewing involves skeletal muscles of the jaw, while salivation is mediated by the autonomic nervous system and salivary glands.

**Answer:** Chewing is a voluntary action controlled consciously, while salivation on seeing food is an involuntary reflex action.

> Common mistake: Confusing voluntary and involuntary actions in digestive processes.

### Question 11 (OR)

*2 marks · Short answer · How do Organisms Reproduce?*

State two differences between pollination and fertilization.

**Solution**

1. Pollination is the transfer of pollen grains from the anther to the stigma of a flower, whereas fertilization is the fusion of the male gamete with the female gamete.
2. Pollination is an external process aided by agents like wind, water, or insects, while fertilization is an internal cellular process occurring inside the ovule.

**Answer:** Pollination is the transfer of pollen to the stigma, while fertilization is the fusion of male and female gametes.

> Common mistake: Stating that pollination leads directly to seed formation without mentioning fertilization.

### Question 12

*2 marks · Short answer · How do Organisms Reproduce?*

Draw a neat diagram to show germination of pollen on the female reproductive part of the flower. Name and label only the following parts :
(a) The part that receives the pollen grain.
(b) The structure that carries the male germ cell to reach the female germ cell.

**Part (a)**

1. The part that receives the pollen grain is the stigma.

Answer (a): Stigma

**Part (b)**

1. The structure that carries the male germ cell to reach the female germ cell is the pollen tube.

Answer (b): Pollen tube

**Answer:** Diagram: Draw a longitudinal section of a carpel showing the stigma, style, and ovary. Show pollen grains on the stigma with pollen tubes growing down through the style into the ovary. Label (a) Stigma as the part receiving pollen and (b) Pollen tube as the structure carrying the male germ cell.

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

### Question 13

*3 marks · Short answer · Our Environment*

Given below is a pyramid showing various trophic levels in an ecosystem :
(iv) Tertiary consumers
(iii) Secondary consumers
(ii) Primary consumers
(i) Producers
(a) From the organisms listed below, identify which one is to be placed at which trophic level ?
Deer, Grass, Lion, Snake, Rabbit
(b) Discuss the reason why primary consumers will have more energy as compared to secondary consumers ?
(c) Why is the base of the pyramid broad ?

**Part (a)**

1. Producers: Grass
2. Primary consumers: Deer, Rabbit
3. Secondary consumers: Snake
4. Tertiary consumers: Lion

Answer (a): Producers: Grass; Primary consumers: Deer, Rabbit; Secondary consumers: Snake; Tertiary consumers: Lion.

**Part (b)**

1. According to the 10% law, only 10% of the energy is transferred to the next trophic level.
2. The rest of the energy is lost as heat to the environment during metabolic activities.

Answer (b): Primary consumers have more energy because energy is lost at each successive trophic level.

**Part (c)**

1. The base is broad because the number of producers is the highest in an ecosystem to support all other trophic levels.

Answer (c): The base is broad because producers are the most abundant organisms.

**Answer:** The trophic levels are organized based on energy flow and population size.

> Common mistake: Forgetting that energy decreases as we move up the trophic levels.

### Question 14

*3 marks · Short answer · Life Processes*

Give differences between the following :
(a) Nephron and neuron
(b) Sensory nerve and motor nerve
(c) Consumers and decomposers

**Part (a)**

1. Nephron is the structural and functional unit of the kidney involved in filtration.
2. Neuron is the structural and functional unit of the nervous system involved in signal transmission.

Answer (a): Nephron is for excretion; Neuron is for nerve impulse conduction.

**Part (b)**

1. Sensory nerves carry impulses from receptors to the central nervous system.
2. Motor nerves carry impulses from the central nervous system to effectors like muscles or glands.

Answer (b): Sensory nerves carry signals to the CNS; Motor nerves carry signals from the CNS.

**Part (c)**

1. Consumers are organisms that depend on other organisms for food.
2. Decomposers are organisms that break down dead and decaying organic matter.

Answer (c): Consumers are heterotrophs; Decomposers are saprotrophs.

**Answer:** Differences between biological structures and functions.

> Common mistake: Confusing the direction of nerve impulses in sensory and motor nerves.

### Question 15

*4 marks · Long answer · Heredity*

Mendel took garden pea plants with different characteristics, such as height to study the inheritance pattern of factors (genes). He crossed tall pea plant with short pea plant and obtained all the tall plants in the $F_1$ generation.
Answer the following questions :
(a) Why only tall pea plants were observed in $F_1$ progeny ?
(b) By which method did Mendel obtain $F_2$ progeny ?
(c) (i) Write one difference between dominant and recessive trait.

**Part (a)**

1. Only tall plants were observed because the trait for tallness is dominant over the trait for shortness.

Answer (a): Tallness is the dominant trait.

**Part (b)**

1. Mendel obtained the F2 progeny by self-pollinating the F1 generation plants.

Answer (b): Self-pollination of F1 plants.

**Part (c)(i)**

1. A dominant trait is expressed in the presence of a contrasting allele, while a recessive trait is expressed only in the presence of identical alleles.

Answer (c)(i): Dominant traits express in F1; recessive traits do not.

**Answer:** Mendel's experiments demonstrate the laws of inheritance.

> Common mistake: Stating that the short trait disappears; it is only masked.

### Question 15 (OR)

*2 marks · Short answer · Heredity*

Mendel took garden pea plants with different characteristics, such as height to study the inheritance pattern of factors (genes). He crossed tall pea plant with short pea plant and obtained all the tall plants in the $F_1$ generation.
Answer the following questions :
(c) (ii) Write two observations made by Mendel about $F_1$ progeny.

**Part (c)(ii)**

1. All plants in the F1 generation were tall.
2. No plants of intermediate height were observed.

Answer (c)(ii): All F1 plants were tall and no medium-height plants were produced.

**Answer:** Mendel's observations on F1 progeny.

> Common mistake: Failing to mention the absence of intermediate height.

### Question 16

*5 marks · Case-based · How do Organisms Reproduce?*

Given below are certain situations. Analyse each and describe its possible impact :
(i) A population of bacteria living in temperate waters whose temperature increased by global warming.
(ii) The sperm encounters the egg when it reaches the oviduct in human females.
(iii) Self pollination does not occur in a flower that contains only pistil.
(iv) Egg does not get fertilised in a human female.
(v) When the seed is placed under appropriate condition of water and air in the soil ?

**Part (i)**

1. The bacteria will likely die as they are adapted to temperate conditions and cannot survive the heat.

Answer (i): The population will decline or die out.

**Part (ii)**

1. Fertilisation will occur, leading to the formation of a zygote.

Answer (ii): Fertilisation occurs.

**Part (iii)**

1. Cross-pollination must occur for reproduction to take place.

Answer (iii): Cross-pollination is required.

**Part (iv)**

1. The thickened uterine lining will break down, resulting in menstruation.

Answer (iv): Menstruation occurs.

**Part (v)**

1. The seed will germinate and develop into a seedling.

Answer (v): Germination occurs.

**Answer:** Analysis of biological situations.

> Common mistake: Confusing fertilisation with implantation.

### Question 16 (OR)

*5 marks · Case-based · How do Organisms Reproduce?*

Given below are certain situations. Analyse and describe what would happen when :
(i) Spores are liberated from blob-like structures of the bread mould ?
(ii) Leaves of bryophyllum fall on wet soil ?
(iii) A pollen from different species land on the stigma of totally unrelated species ?
(iv) Copper-T is placed in the uterus of a human female ?
(v) Spirogyra breaks into smaller fragments upon maturation ?

**Part (i)**

1. Spores are covered by thick walls that protect them until they come into contact with another moist surface and begin to grow.
2. Under favourable conditions, they germinate and develop into new Rhizopus individuals.

Answer (i): Spores germinate under favourable moist conditions to develop into new bread mould individuals.

**Part (ii)**

1. Leaves of Bryophyllum have buds in the notches along the leaf margin.
2. When these leaves fall on wet soil, the buds develop into new plants.

Answer (ii): Buds present in the leaf notches develop into new Bryophyllum plants.

**Part (iii)**

1. Pollen grain from a different species is not compatible with the stigma of an unrelated species.
2. No pollen germination or pollen tube growth takes place, preventing fertilization.

Answer (iii): Pollen fails to germinate and no fertilization takes place.

**Part (iv)**

1. Copper-T is a contraceptive device placed in the uterus.
2. It prevents implantation of the fertilized egg in the uterus.

Answer (iv): It prevents implantation in the uterus, acting as a method of contraception.

**Part (v)**

1. Spirogyra is a simple multicellular organism that undergoes fragmentation upon maturation.
2. Each fragment grows into a new individual.

Answer (v): Each fragment grows into a new Spirogyra individual.

**Answer:** Described the outcomes of the five biological situations based on NCERT Class 10 Science.

> Common mistake: Confusing the function of Copper-T with barrier methods that prevent fertilization instead of implantation.

## SECTION - B

All questions are compulsory.

### Question 17

*1 mark · MCQ · Chemical Reactions and Equations*

(i) $\text{AgNO}_3 + \text{NaCl} \longrightarrow \text{NaNO}_3 + \text{AgCl}$
(ii) $\text{K}_2\text{SO}_4 + \text{BaCl}_2 \longrightarrow \text{BaSO}_4 + 2\text{KCl}$
Which of the following options clearly describes both the reactions ?

- (i) is double displacement, (ii) is displacement reaction.
- Both, (i) and (ii) are displacement reactions and precipitation reactions.
- Both, (i) and (ii) are double displacement reactions and precipitation reactions.
- (i) is displacement, (ii) is double displacement reaction.

**Solution**

1. In reaction (i), $\text{AgNO}_3$ and $\text{NaCl}$ exchange ions to form $\text{AgCl}$ precipitate and $\text{NaNO}_3$, making it a double displacement and precipitation reaction.
2. In reaction (ii), $\text{K}_2\text{SO}_4$ and $\text{BaCl}_2$ exchange ions to form a white precipitate of $\text{BaSO}_4$, making it also a double displacement and precipitation reaction.

**Answer:** (c) Both, (i) and (ii) are double displacement reactions and precipitation reactions.

> Common mistake: Confusing double displacement reactions with single displacement reactions.

### Question 18

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

Which one of the following can be used as an acid-base indicator by a visually impaired (blind) student ?

- Turmeric
- Vanilla essence
- Methyl orange
- Litmus

**Solution**

1. Olfactory indicators are substances whose odour changes in acidic or basic media.
2. Vanilla essence is an olfactory indicator and can be used by visually impaired students, whereas turmeric, methyl orange, and litmus are visual indicators.

**Answer:** (b) Vanilla essence

> Common mistake: Choosing colour-based indicators like litmus or methyl orange.

### Question 19

*1 mark · MCQ · Chemical Reactions and Equations*

The gases evolved on heating lead (II) nitrate crystals are :

- NO and $\text{O}_2$
- $\text{N}_2$ and $\text{NO}_2$
- $\text{NO}_2$ and $\text{H}_2$
- $\text{NO}_2$ and $\text{O}_2$

**Solution**

1. When lead (II) nitrate crystals are heated, they undergo thermal decomposition to give lead oxide, nitrogen dioxide, and oxygen gas.
2. The brown fumes evolved are of nitrogen dioxide ($\text{NO}_2$) along with oxygen ($\text{O}_2$) gas.

**Answer:** (d) $\text{NO}_2$ and $\text{O}_2$

> Common mistake: Confusing the gases with nitric oxide or nitrogen gas.

### Question 20

*1 mark · MCQ · Carbon and its Compounds*

The formula of functional group for aldehyde is :

- $-\text{COOH}$
- $-\text{CHO}$
- $-\text{C}=\text{O}$
- $-\text{OH}$

**Solution**

1. The functional group for aldehydes is represented by $-\text{CHO}$.
2. $-\text{COOH}$ is for carboxylic acids, $-\text{C}=\text{O}$ is for ketones, and $-\text{OH}$ is for alcohols.

**Answer:** (b) $-\text{CHO}$

> Common mistake: Confusing the aldehyde group $-\text{CHO}$ with the ketone carbonyl group $-\text{C}=\text{O}$.

### Question 21

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

Which of the following is a poor conductor of electricity ?

- Pb
- Cu
- Ag
- Al

**Solution**

1. Silver (Ag), copper (Cu), and aluminium (Al) are very good conductors of electricity.
2. Lead (Pb) is a comparatively poor conductor of electricity among metals.

**Answer:** (a) Pb

> Common mistake: Choosing copper or aluminium thinking of common electrical wires while ignoring lead's lower conductivity.

### Question 22

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

The natural sources of oxalic acid, lactic acid and methanoic acid respectively are :

- tomato, curd, ant-sting
- tomato, orange, nettle-sting
- orange, milk, ant-sting
- orange, sour milk, nettle-sting

**Solution**

1. Oxalic acid is naturally present in tomato.
2. Lactic acid is present in curd (sour milk), and methanoic (formic) acid is present in ant-sting.

**Answer:** (a) tomato, curd, ant-sting

> Common mistake: Confusing lactic acid source with orange or citrus fruits.

### Question 23

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

When an element 'X' reacts with water, it starts floating. Identify the element 'X' :

- Potassium
- Calcium
- Sodium
- Iron

**Solution**

1. Calcium reacts with water to form calcium hydroxide and hydrogen gas.
2. The bubbles of hydrogen gas formed stick to the surface of the metal, making it float.

**Answer:** (b) Calcium

> Common mistake: Confusing calcium with sodium or potassium, which float due to hydrogen gas bubbles but react violently and catch fire.

### Question 24

*1 mark · Assertion and reason · Carbon and its Compounds*

Assertion (A) : Carbon shares its valence electrons with other atoms of carbon or with atoms of other elements.
Reason (R) : The shared electrons belong to the outermost shells of both the atoms and lead to both atoms attaining the noble gas configuration.

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

**Solution**

1. Carbon has 4 valence electrons and achieves noble gas configuration by sharing electrons.
2. Both Assertion and Reason are true, and Reason correctly explains the Assertion.

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

> Common mistake: Thinking carbon forms ionic bonds by losing or gaining four electrons.

### Question 25

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

(a) What are amphoteric oxides ?
(b) Categorise the following based on their nature :
$\text{ZnO}, \text{Na}_2\text{O}, \text{CO}_2$

**Part (a)**

1. Metal oxides that show both acidic as well as basic behaviour are known as amphoteric oxides.
2. They react with both acids and bases to form salt and water.

Answer (a): Metal oxides which react with both acids and bases to form salts and water are amphoteric oxides.

**Part (b)**

1. ZnO is an amphoteric oxide.
2. $\text{Na}_2\text{O}$ is a basic oxide because it is a metal oxide.
3. $\text{CO}_2$ is an acidic oxide because it is a non-metal oxide.

Answer (b): ZnO: Amphoteric, $\text{Na}_2\text{O}$: Basic, $\text{CO}_2$: Acidic

**Answer:** ZnO is amphoteric, Na2O is basic, and CO2 is acidic.

> Common mistake: Classifying non-metal oxides as basic or metal oxides as acidic.

### Question 26

*3 marks · Short answer · Chemical Reactions and Equations*

(a) Name the substance oxidised and reduced in the following reaction :
$\text{ZnO} + \text{C} \longrightarrow \text{Zn} + \text{CO}$
(b) Balance the following chemical reaction :
$\text{Pb}(\text{NO}_3)_2 + \text{KI} \longrightarrow \text{PbI}_2 + \text{KNO}_3$
(c) Give one example each of electrolytic decomposition and decomposition by sunlight.

**Part (a)**

1. Carbon (C) is oxidised to carbon monoxide (CO) as it gains oxygen.
2. Zinc oxide (ZnO) is reduced to zinc (Zn) as it loses oxygen.

Answer (a): Substance oxidised: Carbon (C); Substance reduced: Zinc oxide (ZnO).

**Part (b)**

1. Write the unbalanced equation: $\text{Pb}(\text{NO}_3)_2 + \text{KI} \longrightarrow \text{PbI}_2 + \text{KNO}_3$.
2. Balance potassium and nitrate ions by placing coefficient 2 before $\text{KI}$ and $\text{KNO}_3$: $\text{Pb}(\text{NO}_3)_2 + 2\text{KI} \longrightarrow \text{PbI}_2 + 2\text{KNO}_3$.

Answer (b): $\text{Pb}(\text{NO}_3)_2 + 2\text{KI} \longrightarrow \text{PbI}_2 + 2\text{KNO}_3$

**Part (c)**

1. Electrolytic decomposition: $2\text{H}_2\text{O} \xrightarrow{\text{Electricity}} 2\text{H}_2 + \text{O}_2$.
2. Decomposition by sunlight: $2\text{AgCl} \xrightarrow{\text{Sunlight}} 2\text{Ag} + \text{Cl}_2$.

Answer (c): Electrolysis of acidified water and photolytic decomposition of silver chloride.

**Answer:** Substance oxidised is C, reduced is ZnO. Balanced equation: $\text{Pb}(\text{NO}_3)_2 + 2\text{KI} \longrightarrow \text{PbI}_2 + 2\text{KNO}_3$.

> Common mistake: Confusing the substance oxidised with the oxidising agent.

### Question 27

*3 marks · Short answer · Acids, Bases and Salts*

Explain chlor-alkali process with chemical equation. Name the products formed at anode and cathode.

**Solution**

1. When electricity is passed through an aqueous solution of sodium chloride (brine), it decomposes to form sodium hydroxide, chlorine gas, and hydrogen gas.
2. Chemical equation: $2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \longrightarrow 2\text{NaOH}(aq) + \text{Cl}_2(g) + \text{H}_2(g)$.
3. Product formed at anode: Chlorine gas ($\text{Cl}_2$).
4. Product formed at cathode: Hydrogen gas ($\text{H}_2$).

**Answer:** Chlorine at anode, hydrogen at cathode, and sodium hydroxide near cathode.

> Common mistake: Interchanging the products formed at anode and cathode.

### Question 27 (OR)

*3 marks · Short answer · Acids, Bases and Salts*

Write the preparation of following compounds with balanced chemical equation :
(i) Baking soda
(ii) Bleaching powder
(iii) Plaster of Paris

**Part (i)**

1. Baking soda (Sodium hydrogen carbonate) is prepared by reacting cold and concentrated sodium chloride solution with ammonia and carbon dioxide.
2. Equation: $\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 + \text{NH}_3 \longrightarrow \text{NH}_4\text{Cl} + \text{NaHCO}_3$.

Answer (i): $\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 + \text{NH}_3 \longrightarrow \text{NH}_4\text{Cl} + \text{NaHCO}_3$

**Part (ii)**

1. Bleaching powder is prepared by the action of chlorine on dry slaked lime.
2. Equation: $\text{Ca}(\text{OH})_2 + \text{Cl}_2 \longrightarrow \text{CaOCl}_2 + \text{H}_2\text{O}$.

Answer (ii): $\text{Ca}(\text{OH})_2 + \text{Cl}_2 \longrightarrow \text{CaOCl}_2 + \text{H}_2\text{O}$

**Part (iii)**

1. Plaster of Paris is prepared by heating gypsum at $373\text{ K}$.
2. Equation: $\text{CaSO}_4 \cdot 2\text{H}_2\text{O} \xrightarrow{373\text{ K}} \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + 1\frac{1}{2}\text{H}_2\text{O}$.

Answer (iii): $\text{CaSO}_4 \cdot 2\text{H}_2\text{O} \xrightarrow{373\text{ K}} \text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O} + 1\frac{1}{2}\text{H}_2\text{O}$

**Answer:** Prepared using NaCl, slaked lime, and gypsum respectively.

> Common mistake: Writing incorrect water of crystallisation coefficients for gypsum and Plaster of Paris.

### Question 28

*4 marks · Case-based · Metals and Non-metals*

Read the following passage and answer the questions given below :
Most of metals occur in combined state in form of ores. Carbonate ores are converted into oxides by calcination and sulphide ores by roasting. Oxides are reduced with suitable reducing agent like carbon to get free metal. Highly reactive metals like Al, Mg are also used as reducing agents to obtain metal from their oxides. Most reactive metals are obtained by electrolytic reduction of their molten ores. Alloying is a very good method of improving the properties of a metal. We can get desired properties by this method. The electrical conductivity and melting point of an alloy is less than that of pure metals.
(a) Why carbonate or sulphide ores are converted into oxides before extraction of metal from it ?
(b) Write a reaction in which Aluminium is used as a reducing agent to obtain metal from its oxide.
(c) (i) How is copper obtained from its ore ($\text{Cu}_2\text{S}$) ? Give equations of the reactions.

**Part (a)**

1. Metal oxides are much easier to reduce to their respective metals than carbonates or sulphides.
2. Therefore, ores are first converted into metal oxides by calcination or roasting before reduction.

Answer (a): Metal oxides are easier to reduce than carbonates or sulphides.

**Part (b)**

1. Aluminium is a highly reactive metal and can displace less reactive metals from their oxides.
2. A well-known example is the thermite reaction where manganese dioxide is reduced by aluminium powder.
3. $3\text{MnO}_2(s) + 4\text{Al}(s) \rightarrow 3\text{Mn}(l) + 2\text{Al}_2\text{O}_3(s) + \text{Heat}$

Answer (b): $3\text{MnO}_2(s) + 4\text{Al}(s) \rightarrow 3\text{Mn}(l) + 2\text{Al}_2\text{O}_3(s) + \text{Heat}$

**Part (c)**

1. Copper(I) sulphide ($\text{Cu}_2\text{S}$) is first heated in air to get some copper(I) oxide.
2. $2\text{Cu}_2\text{S} + 3\text{O}_2(g) \rightarrow 2\text{Cu}_2\text{O}(s) + 2\text{SO}_2(g)$
3. When the supply of air is stopped and the temperature is raised, the remaining copper sulphide reacts with copper oxide to give molten copper.
4. $2\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \rightarrow 6\text{Cu}(s) + \text{SO}_2(g)$

Answer (c): Copper is obtained by roasting partial $\text{Cu}_2\text{S}$ to oxide, followed by self-reduction with remaining $\text{Cu}_2\text{S}$.

**Answer:** Carbonate and sulphide ores are converted to oxides because it is easier to reduce oxide ores than carbonate or sulphide ores to obtain the free metal.

> Common mistake: Writing equations without balancing them or omitting physical states in metallurgical reactions.

### Question 28 (OR)

*2 marks · Case-based · Metals and Non-metals*

Read the following passage and answer the questions given below :
Most of metals occur in combined state in form of ores. Carbonate ores are converted into oxides by calcination and sulphide ores by roasting...
(c) (ii) Why highly reactive metals cannot be obtained from their oxides by using carbon as a reducing agent ?
(iii) Why solder, an alloy of lead and tin, is used for welding electrical wires together ?

**Part (c) (ii)**

1. Highly reactive metals like sodium, magnesium, and aluminium have a greater affinity for oxygen than carbon does.
2. Hence, carbon cannot reduce their oxides to free metals.

Answer (c) (ii): Highly reactive metals have more affinity for oxygen than carbon.

**Part (c) (iii)**

1. Solder is an alloy of lead and tin which has a very low melting point.
2. This property makes it suitable for welding electrical wires together.

Answer (c) (iii): Solder has a low melting point, making it ideal for electrical soldering.

**Answer:** Highly reactive metals have a higher affinity for oxygen than carbon, and solder has a low melting point.

> Common mistake: Confusing the reactivity of carbon with that of alkali and alkaline earth metals.

### Question 29

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

(a) (i) Give reasons for the following :
(I) Covalent compounds are poor conductor of electricity.
(II) Soap does not form lather in hard water.
(III) Carbon shows catenation but silicon does not.
(ii) Write chemical equations for the following :
(I) Oxidation of ethanol by acidified $\text{K}_2\text{Cr}_2\text{O}_7$.
(II) Hydrogenation of ethene.

**Part (a)(I)**

1. Covalent compounds do not form ions when dissolved in water.
2. They lack charged particles (ions or free electrons) responsible for conducting electricity.

Answer (a)(I): They do not form ions in solution and lack free electrons.

**Part (a)(II)**

1. Hard water contains calcium and magnesium salts.
2. Soap reacts with these calcium and magnesium ions to form an insoluble precipitate called scum.
3. This consumes the soap and prevents lather formation.

Answer (a)(II): Soap reacts with calcium and magnesium ions in hard water to form insoluble scum.

**Part (a)(III)**

1. Carbon has a very small size, which enables its nucleus to hold onto the shared pair of electrons strongly.
2. Silicon has a larger atomic size, making its $\text{Si}-\text{Si}$ bonds much weaker and less stable than $\text{C}-\text{C}$ bonds.

Answer (a)(III): Carbon has a smaller size and strong inter-atomic $\text{C}-\text{C}$ bonds compared to silicon.

**Part (b)(I)**

1. Ethanol is oxidized to ethanoic acid when treated with acidified potassium dichromate.
2. $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Acidified } \text{K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH}$

Answer (b)(I): $\text{CH}_3\text{CH}_2\text{OH} + 2[\text{O}] \xrightarrow{\text{Acidified } \text{K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH} + \text{H}_2\text{O}$

**Part (b)(II)**

1. Ethene undergoes addition reaction with hydrogen in the presence of nickel or palladium catalyst to form ethane.
2. $\text{CH}_2=\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni catalyst}} \text{CH}_3-\text{CH}_3$

Answer (b)(II): $\text{CH}_2=\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni}} \text{CH}_3-\text{CH}_3$

**Answer:** Carbon compounds show characteristic bonding, soap action, and chemical reactions as explained in the parts.

> Common mistake: Confusing catenation property between carbon and silicon due to atomic size differences.

### Question 29 (OR)

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

(b) Mohan heated ethanol with a compound 'X' in the presence of a few drops of conc. $\text{H}_2\text{SO}_4$ and observed a sweet smelling compound 'Y' is formed. When 'Y' is treated with sodium hydroxide it gives back ethanol and a compound 'Z'.
(i) Identify 'X', 'Y' and 'Z'.
(ii) Write the role of conc. $\text{H}_2\text{SO}_4$ in the reaction.
(iii) Write the chemical equations involved and name the reactions.

**Part (i)**

1. Compound 'X' is ethanoic acid ($\text{CH}_3\text{COOH}$), which reacts with ethanol to form a sweet-smelling ester.
2. Compound 'Y' is ethyl ethanoate ($\text{CH}_3\text{COOCH}_2\text{CH}_3$), an ester.
3. Compound 'Z' is sodium ethanoate ($\text{CH}_3\text{COONa}$), formed during alkaline hydrolysis of the ester.

Answer (i): X: Ethanoic acid, Y: Ethyl ethanoate, Z: Sodium ethanoate.

**Part (ii)**

1. Conc. $\text{H}_2\text{SO}_4$ acts as a dehydrating agent in the esterification reaction.
2. It removes the water molecule formed during the reaction, shifting the equilibrium forward to increase the yield of the ester.

Answer (ii): It acts as a dehydrating agent to remove water and drive the reaction forward.

**Part (iii)**

1. Esterification reaction: $\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Conc. } \text{H}_2\text{SO}_4} \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}$
2. Saponification reaction: $\text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{CH}_3\text{CH}_2\text{OH}$

Answer (iii): Esterification and Saponification equations with proper names.

**Answer:** X is ethanoic acid, Y is ethyl ethanoate, and Z is sodium ethanoate along with ethanol.

> Common mistake: Forgetting to name the reactions (esterification and saponification) alongside writing the chemical equations.

## SECTION - C

All questions are compulsory.

### Question 30

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

A convex lens of focal length $15\text{ cm}$, is forming a real image. If the size of image is same as the size of object, then position of object and position of image will be, respectively :

- $-15\text{ cm}$ and $-15\text{ cm}$ from lens
- $-15\text{ cm}$ and $+15\text{ cm}$ from lens
- $-30\text{ cm}$ and $+30\text{ cm}$ from lens
- $-30\text{ cm}$ and $-30\text{ cm}$ from lens

**Solution**

1. For a convex lens, the size of the image is equal to the size of the object when the object is placed at $2F$, i.e., at a distance of $2f$ from the optical centre.
2. Given focal length $f = 15\text{ cm}$, the object is at $-30\text{ cm}$ and the real image is formed at $+30\text{ cm}$.
3. Therefore, the correct option is (c).

**Answer:** (c) $-30\text{ cm}$ and $+30\text{ cm}$ from lens

> Common mistake: Confusing object and image distances with sign conventions or choosing the focal point instead of $2F$.

### Question 31

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

When you look at an object very close to your eyes, the :

- Ciliary muscles of your eye contract and the eye lens becomes thick.
- Ciliary muscles of your eye get relaxed and the eye lens becomes thick.
- Ciliary muscles of your eye contract and the eye lens becomes thin.
- Ciliary muscles of your eye get relaxed and the eye lens becomes thin.

**Solution**

1. When an object is placed very close to the eyes, the ciliary muscles contract to increase the curvature of the eye lens.
2. This makes the eye lens thicker so that its focal length decreases to focus the image on the retina.
3. Therefore, the correct option is (a).

**Answer:** (a) Ciliary muscles of your eye contract and the eye lens becomes thick.

> Common mistake: Confusing the action of ciliary muscles for near and distant vision.

### Question 32

*1 mark · Assertion and reason · The Human Eye and the Colourful World*

Assertion (A) : When rays of white light pass through a prism, on emerging they give spectrum of seven colours.
Reason (R) : It is due to the scattering of light that red light bends minimum and violet light bends the maximum.

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

**Solution**

1. Assertion (A) is true because white light splits into its seven constituent colours when passing through a glass prism.
2. Reason (R) is false because the spectrum is due to refraction and dispersion (different colours bending at different angles), not scattering of light.
3. Therefore, Assertion (A) is true, but Reason (R) is false.

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

> Common mistake: Mistaking dispersion and refraction through a prism for atmospheric scattering of light.

### Question 33

*2 marks · Case-based · Light – Reflection and Refraction*

The given figure shows image formation by a lens. Analyse the figure and answer the following questions :
(a) What is the type of lens used for image formation in the given ray diagram ?
(b) If the real image is formed at a distance of $30\text{ cm}$ from the lens and the size of image is twice the size of the object, then where was the object placed ?

**Part (a)**

1. Observe the ray diagram where a real and enlarged image is formed by the lens.
2. A convex lens can form a real and magnified image when the object is placed between $F$ and $2F$.

Answer (a): Convex lens

**Part (b)**

1. Given: Image distance $v = +30\text{ cm}$ (real image), magnification $m = \frac{h_i}{h_o} = -2$ (real image is inverted).
2. Formula for magnification of a lens: $m = \frac{v}{u}$.
3. Substitute the values: $-2 = \frac{30}{u}$, which gives $u = \frac{30}{-2} = -15\text{ cm}$.
4. Result: The object was placed at a distance of $15\text{ cm}$ in front of the lens.

Answer (b): $15\text{ cm}$ in front of the lens

**Answer:** Analyse the lens ray diagram and calculate object position.

> Common mistake: Forgetting the negative sign for magnification in the case of a real, inverted image.

### Question 34

*2 marks · Numerical · Electricity*

The resistance of a wire of $0.01\text{ cm}$ radius and $1.0\text{ cm}$ length is $7\ \Omega$. Calculate its resistivity.

**Solution**

1. Given: radius $r = 0.01\text{ cm} = 10^{-4}\text{ m}$, length $l = 1.0\text{ cm} = 10^{-2}\text{ m}$, resistance $R = 7\ \Omega$
2. Formula: $R = \rho \frac{l}{A}$, where $A = \pi r^2$, so resistivity $\rho = \frac{R \cdot \pi r^2}{l}$
3. Substitution: $\rho = \frac{7 \times \frac{22}{7} \times (10^{-4}\text{ m})^2}{10^{-2}\text{ m}}$
4. Result: $\rho = 2.2 \times 10^{-6}\ \Omega\text{ m}$

**Answer:** $2.2 \times 10^{-6}\ \Omega\text{ m}$

> Common mistake: Forgetting to convert centimeters to meters before substituting into the formula.

### Question 34 (OR)

*2 marks · Numerical · Electricity*

An electric heater is rated $220\text{ V}; 11\text{ A}$. Calculate the power consumed if the heater is operated at $200\text{ V}$.

**Solution**

1. Given: Rated voltage $V_1 = 220\text{ V}$, rated current $I = 11\text{ A}$.
2. Formula: Resistance of the heater $R = \frac{V_1}{I} = \frac{220}{20}= 20\ \Omega$.
3. When operated at $V_2 = 200\text{ V}$, the power consumed is $P = \frac{V_2^2}{R}$.
4. Substitution: $P = \frac{(200)^2}{20} = \frac{40000}{20} = 2000\text{ W}$.

**Answer:** $2000\text{ W}$

> Common mistake: Using the initial power directly or assuming resistance changes with applied voltage.

### Question 35

*3 marks · Short answer · The Human Eye and the Colourful World*

A person is unable to read a book placed closer than $1\text{ meter}$ from his eyes. Identify the defect of vision in his eyes. Draw the ray diagrams to show the defect of vision and its correction.

**Solution**

1. The defect of vision is Hypermetropia or farsightedness.
2. Diagram: Draw a convex lens of the eye with the image forming behind the retina.
3. Diagram: Draw a convex lens placed in front of the eye to converge light rays onto the retina.

**Answer:** The defect is Hypermetropia.

> Common mistake: Confusing Hypermetropia with Myopia and drawing the wrong lens for correction.

### Question 36

*3 marks · Short answer · Magnetic Effects of Electric Current*

(a) Describe an activity to show that a current carrying conductor, placed in an external magnetic field experiences a force.
(b) Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally towards the front wall from the back wall, is deflected by a strong magnetic field to your right side. Find the direction of the magnetic field.

**Solution**

1. Activity: Suspend an aluminum rod horizontally between the poles of a strong horseshoe magnet and connect it to a battery and switch.
2. Observation: When current flows through the rod, it experiences a force and gets displaced.
3. Direction: Using Fleming's Left-Hand Rule, since the current is towards the back wall and force is to the right, the magnetic field is vertically downwards.

**Answer:** The magnetic field is directed vertically downwards.

> Common mistake: Forgetting that the direction of conventional current is opposite to the direction of electron flow.

### Question 37

*3 marks · Short answer · Magnetic Effects of Electric Current*

(a) The pattern of magnetic field due to a current carrying wire depends upon the shape made by that wire. Justify.
(b) A current carrying straight wire AB is shown in the given diagram. Out of X, Y and Z on which point will the strength of magnetic field be maximum and why ?

**Solution**

1. The magnetic field pattern depends on the shape: a straight wire produces concentric circles, a loop produces circles near the wire, and a solenoid produces a pattern similar to a bar magnet.
2. The strength of the magnetic field is inversely proportional to the distance from the wire.
3. Point X is closest to the wire, so the magnetic field strength is maximum at X.

**Answer:** The strength is maximum at point X because it is closest to the wire.

> Common mistake: Stating that the field is uniform everywhere near the wire.

### Question 38

*4 marks · Case-based · Light – Reflection and Refraction*

Read the following passage and answer the questions given :
Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.
(a) A convex lens of focal length $20\text{ cm}$ is used to form an image. If an object is placed at $40\text{ cm}$ from the lens, what will be the position and nature of image ?
(b) Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens.
(c) (i) A lens combination consists of a convex lens of focal length $30\text{ cm}$ and a concave lens of focal length $15\text{ cm}$ placed together. Find the equivalent focal length and power of this lens combination.

**Part (a)**

1. Given: f = +20 cm, u = -40 cm.
2. Formula: 1/v - 1/u = 1/f.
3. Calculation: 1/v = 1/20 - 1/40 = 1/40.
4. Result: v = +40 cm, image is real, inverted, and same size as object.

Answer (a): v = 40 cm, real and inverted.

**Part (b)**

1. Diagram: Draw a concave lens with an object between optical centre and focus.
2. The rays diverge and appear to meet behind the object to form a virtual, erect, and diminished image.

Answer (b): Virtual, erect, and diminished image.

**Part (c)(i)**

1. P1 = 100/30 = +3.33 D, P2 = 100/-15 = -6.67 D.
2. P = P1 + P2 = 3.33 - 6.67 = -3.34 D.
3. f = 100/P = -29.94 cm (approx -30 cm).

Answer (c)(i): f = -30 cm, P = -3.34 D.

**Answer:** The image is real, inverted, and formed at 40 cm.

> Common mistake: Ignoring sign conventions for focal length and object distance.

### Question 38 (OR)

*2 marks · Case-based · Light – Reflection and Refraction*

Read the following passage and answer the questions given :
Lenses can form different types of images depending upon their focal length and position of object...
(c) (ii) Two lenses are placed in contact. One is a concave lens with focal length $2\text{ m}$ and the other is a convex lens with focal length $1.5\text{ m}$. What type of lens will the combination behave as (convex or concave) ? Give reason.

**Part (c)(ii)**

1. P1 = 1/-2 = -0.5 D, P2 = 1/1.5 = +0.67 D.
2. P = -0.5 + 0.67 = +0.17 D.
3. Since the net power is positive, the combination behaves as a convex lens.

Answer (c)(ii): Convex lens.

**Answer:** The combination behaves as a convex lens.

> Common mistake: Incorrectly calculating the sum of powers.

### Question 39

*5 marks · Long answer · Electricity*

(a) (i) Due to change in length and area of cross-section of a conductor, resistance of conductor changes while resistivity does not change. Why ?
(ii) Conductors of electric toasters and electric iron are made of an alloy rather than a pure metal. Why ?
(iii) Define the S.I. unit of electric current.

**Solution**

1. (i) Resistivity is a characteristic property of the material of the conductor and does not depend on its dimensions.
2. (ii) Alloys have higher resistivity than pure metals and do not oxidize (burn) easily at high temperatures.
3. (iii) 1 Ampere is defined as the flow of 1 Coulomb of electric charge through a cross-section of a conductor in 1 second.

**Answer:** Resistivity is a material property; alloys are used for high resistance and oxidation resistance; 1 A = 1 C/s.

> Common mistake: Confusing resistance with resistivity.

### Question 39 (OR)

*5 marks · Long answer · Electricity*

(b) (i) How many bulbs of resistance $8\ \Omega$ each should be connected in parallel combination to draw a current of $2\text{ A}$ from a battery of $4\text{ V}$ ?
(ii) Name the device used for measuring electric current. How is it connected in a circuit ?
(iii) State Joule's law of heating.

**Part (i)**

1. Given potential difference $V = 4\text{ V}$, current $I = 2\text{ A}$, and resistance of each bulb $R = 8\ \Omega$.
2. Let the total equivalent resistance of the parallel combination be $R_p$.
3. Using Ohm's law, $R_p = \frac{V}{I} = \frac{4\text{ V}}{2\text{ A}} = 2\ \Omega$.
4. Let $n$ bulbs of resistance $R$ be connected in parallel, so $\frac{R}{R_p} = n$ or $\frac{1}{R_p} = n \times \frac{1}{R}$.
5. Substituting the values, $\frac{1}{2} = n \times \frac{1}{8}$, which gives $n = \frac{8}{2} = 4$.

Answer (i): 4 bulbs

**Part (ii)**

1. The device used for measuring electric current is an ammeter.
2. It is always connected in series in the electric circuit through which the current is to be measured.

Answer (ii): Ammeter, connected in series

**Part (iii)**

1. Joule's law of heating states that the heat produced in a resistor is directly proportional to the square of current for a given resistance, directly proportional to resistance for a given current, and directly proportional to the time for which the current flows through the resistor.
2. Mathematically, $H = I^2Rt$.

Answer (iii): $H = I^2Rt$

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

> Common mistake: Connecting the ammeter in parallel instead of in series, or forgetting to write the mathematical expression for Joule's law of heating.

## Frequently asked questions

### What is the paper pattern and section-wise breakdown for the CBSE Class 10 Science Question Paper 2026 Set 31/1/1?

The 80-mark question paper is divided into three sections and is to be completed in 180 minutes. Section A contains 16 questions carrying 30 marks, Section B has 13 questions carrying 25 marks, and Section C includes 10 questions carrying 25 marks.

### Which chapters carry the most marks in the CBSE Class 10 Science Set 31/1/1 paper for 2026?

Life Processes carries the highest weightage with 10 marks, followed by How do Organisms Reproduce? with 9 marks and Metals and Non-metals with 8 marks. Other important chapters include Carbon and its Compounds, Light - Reflection and Refraction, and Electricity, which carry 7 marks each.

### How should students write answers in the Class 10 Science exam to secure full marks?

Students should write clear, step-by-step answers with proper scientific terms, labeled diagrams where necessary, and correct units for numerical problems. Following the exam pattern and structuring answers according to the marks allocated helps ensure maximum scores.

### Is the solutions PDF for the CBSE Class 10 Science Question Paper 2026 Set 31/1/1 free to download?

Yes, the complete solutions PDF for this set is available for free on the SwaVid page. Students can easily access and download it to check their answers and understand the step-by-step marking scheme.

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

The exam is conducted for a total of 80 marks with a duration of 180 minutes. Students must manage their time efficiently across all three sections to attempt every question within this timeframe.

## Related pages

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

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student