---
title: "CBSE Class 10 Science Question Paper 2024 (Set 31/1/1) with Solutions"
url: https://www.swavid.com/cbse/class-10/science/previous-year-papers/2024-31-1-1
dateModified: 2026-10-07T15:28:45+00:00
---

# CBSE Class 10 Science Question Paper 2024 (Set 31/1/1) with Solutions

Code 31/1/1 · 80 marks · 180 minutes

Solved CBSE Class 10 Science board paper from 2024, 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/2024/X/SCIENCE.zip. Solutions written by SwaVid.

Free PDF (27 pages): https://www.swavid.com/api/seo/pdf/papers/cbse/science/swavid-cbse-class-10-science-question-paper-2024-31-1-1-0df6a223ce.pdf

## SECTION - A

Select and write the most appropriate option out of the four options given for each of the questions 1-20. There is no negative mark for the incorrect response.

### Question 1

*1 mark · MCQ · Chemical Reactions and Equations*

When $2\text{ mL}$ of sodium hydroxide solution is added to few pieces of granulated zinc in a test tube and then warmed, the reaction that occurs can be written in the form of a balanced chemical equation as :

- $\text{NaOH} + \text{Zn} \rightarrow \text{NaZnO}_2 + \text{H}_2\text{O}$
- $2\text{NaOH} + \text{Zn} \rightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2$
- $2\text{NaOH} + \text{Zn} \rightarrow \text{NaZnO}_2 + \text{H}_2$
- $2\text{NaOH} + \text{Zn} \rightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2\text{O}$

**Solution**

1. When sodium hydroxide reacts with zinc metal, it produces sodium zincate and hydrogen gas.
2. The balanced chemical equation is $2\text{NaOH} + \text{Zn} \rightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2$.

**Answer:** (b) $2\text{NaOH} + \text{Zn} \rightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2$

> Common mistake: Confusing sodium zincate formula as $\text{NaZnO}_2$ instead of $\text{Na}_2\text{ZnO}_2$.

### Question 2

*1 mark · MCQ · Chemical Reactions and Equations*

Select from the following a decomposition reaction in which source of energy for decomposition is light :

- $2\text{FeSO}_4 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2 + \text{SO}_3$
- $2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$
- $2\text{AgBr} \rightarrow 2\text{Ag} + \text{Br}_2$
- $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$

**Solution**

1. Decomposition reactions that require light energy are called photolytic decomposition reactions.
2. Silver bromide decomposes into silver and bromine in the presence of sunlight: $2\text{AgBr} \rightarrow 2\text{Ag} + \text{Br}_2$.

**Answer:** (c) $2\text{AgBr} \rightarrow 2\text{Ag} + \text{Br}_2$

> Common mistake: Choosing water electrolysis which uses electrical energy instead of light.

### Question 3

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

A metal and a non-metal that exists in liquid state at the room temperature are respectively :

- Bromine and Mercury
- Mercury and Iodine
- Mercury and Bromine
- Iodine and Mercury

**Solution**

1. Mercury is the only metal that exists in liquid state at room temperature.
2. Bromine is the only non-metal that exists in liquid state at room temperature.

**Answer:** (c) Mercury and Bromine

> Common mistake: Reversing the order of metal and non-metal as given in the question.

### Question 4

*1 mark · MCQ · Carbon and its Compounds*

Carbon compounds :
(i) are good conductors of electricity.
(ii) are bad conductors of electricity.
(iii) have strong forces of attraction between their molecules.
(iv) have weak forces of attraction between their molecules.
The correct statements are :

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

**Solution**

1. Carbon compounds form covalent bonds and do not contain ions, making them bad conductors of electricity.
2. Due to weak intermolecular forces between their molecules, they have low melting and boiling points.

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

> Common mistake: Assuming carbon compounds conduct electricity because graphite is a good conductor.

### Question 5

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

Consider the following compounds :
$\text{FeSO}_4$; $\text{CuSO}_4$; $\text{CaSO}_4$; $\text{Na}_2\text{CO}_3$
The compound having maximum number of water of crystallisation in its crystalline form in one molecule is :

- $\text{FeSO}_4$
- $\text{CuSO}_4$
- $\text{CaSO}_4$
- $\text{Na}_2\text{CO}_3$

**Solution**

1. $\text{FeSO}_4$ has $7\text{H}_2\text{O}$ (ferrous sulphate heptahydrate).
2. $\text{CuSO}_4$ has $5\text{H}_2\text{O}$, $\text{CaSO}_4$ has $2\text{H}_2\text{O}$, and $\text{Na}_2\text{CO}_3$ has $10\text{H}_2\text{O}$ (sodium carbonate decahydrate), which is the maximum.

**Answer:** (d) $\text{Na}_2\text{CO}_3$

> Common mistake: Confusing with copper sulphate pentahydrate which has only 5 molecules of water.

### Question 6

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

Oxides of aluminium and zinc are :

- acidic
- basic
- amphoteric
- neutral

**Solution**

1. Metal oxides that show both acidic and basic behavior are known as amphoteric oxides.
2. Oxides of aluminium and zinc react with both acids as well as bases to produce salt and water.

**Answer:** (c) amphoteric

> Common mistake: Classifying metal oxides strictly as basic without considering specific exceptions like aluminium and zinc.

### Question 7

*1 mark · MCQ · Chemical Reactions and Equations*

$\text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2$
The reaction given above is a redox reaction because in this case :

- $\text{MnO}_2$ is oxidised and $\text{HCl}$ is reduced.
- $\text{HCl}$ is oxidised.
- $\text{MnO}_2$ is reduced.
- $\text{MnO}_2$ is reduced and $\text{HCl}$ is oxidised.

**Solution**

1. In the given reaction, $\text{MnO}_2$ loses oxygen to form $\text{MnCl}_2$, so it is reduced.
2. $\text{HCl}$ loses hydrogen to form $\text{Cl}_2$ (or is oxidized), so $\text{HCl}$ is oxidized.

**Answer:** (d) $\text{MnO}_2$ is reduced and $\text{HCl}$ is oxidised.

> Common mistake: Confusing oxidation and reduction in terms of addition/removal of oxygen and hydrogen.

### Question 8

*1 mark · MCQ · Heredity*

Consider the following statements :
(i) The sex of a child is determined by what it inherits from the mother.
(ii) The sex of a child is determined by what it inherits from the father.
(iii) The probability of having a male child is more than that of a female child.
(iv) The sex of a child is determined at the time of fertilisation when male and female gametes fuse to form a zygote.
The correct statements are :

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

**Solution**

1. Statement (ii) is correct because the sex of a child depends on whether the sperm carries an X or Y chromosome.
2. Statement (iv) is correct as sex is determined at the time of fertilisation when gametes fuse.

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

> Common mistake: Assuming human reproductive cells (gametes) have pairs of chromosomes instead of a single set.

### Question 9

*1 mark · MCQ · Heredity*

Chromosomes :
(i) carry hereditary information from parents to the next generation.
(ii) are thread like structures located inside the nucleus of an animal cell.
(iii) always exist in pairs in human reproductive cells.
(iv) are involved in the process of cell division.
The correct statements are :

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

**Solution**

1. Chromosomes carry hereditary information (DNA) and are thread-like structures in the nucleus involved in cell division.
2. Human reproductive cells contain only half the number of chromosomes (unpaired single set), making statement (iii) incorrect.

**Answer:** (c) (i), (ii) and (iv)

> Common mistake: Believing that reproductive cells contain paired chromosomes.

### Question 10

*1 mark · MCQ · Control and Coordination*

In a nerve cell, the site where the electrical impulse is converted into a chemical signal is known as :

- Axon
- Dendrites
- Neuromuscular junction
- Cell body

**Solution**

1. At the end of the axon, the electrical impulse sets off the release of some chemicals at the nerve ending.
2. These chemicals cross the synapse (neuromuscular junction or synapse between neurons) and start a similar electrical impulse in a dendrite of the next neurone.

**Answer:** (c) Neuromuscular junction

> Common mistake: Confusing the axon terminal/synapse with dendrites.

### Question 11

*1 mark · MCQ · Life Processes*

A stomata closes when :
(i) it needs carbon dioxide for photosynthesis.
(ii) it does not need carbon dioxide for photosynthesis.
(iii) water flows out of the guard cells.
(iv) water flows into the guard cells.
The correct reason(s) in this process is/are :

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

**Solution**

1. Stomata close when water flows out of the guard cells, causing them to shrink.
2. When the plant does not need carbon dioxide for photosynthesis, stomata remain closed.

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

> Common mistake: Reversing the movement of water into and out of guard cells during stomatal closing.

### Question 12

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

At what distance from a convex lens should an object be placed to get an image of the same size as that of the object on a screen ?

- Beyond twice the focal length of the lens.
- At the principal focus of the lens.
- At twice the focal length of the lens.
- Between the optical centre of the lens and its principal focus.

**Solution**

1. A convex lens produces a real image of the same size as the object only when the object is placed at $2F$ (twice the focal length).
2. The image is also formed at $2F$ on the other side of the lens.

**Answer:** (c) At twice the focal length of the lens.

> Common mistake: Confusing the condition for same-sized image with the focus or optical centre positions.

### Question 13

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

The lens system of human eye forms an image on a light sensitive screen, which is called as :

- Cornea
- Ciliary muscles
- Optic nerves
- Retina

**Solution**

1. The human eye uses a lens system to form an image on a light-sensitive screen called the retina.

**Answer:** (d) Retina

> Common mistake: Confusing the cornea with the retina as the screen for image formation.

### Question 14

*1 mark · MCQ · Magnetic Effects of Electric Current*

The pattern of the magnetic field produced inside a current carrying solenoid is :

- (a)
- (b)
- (c)
- (d)

**Solution**

1. The magnetic field inside a current-carrying solenoid is uniform, represented by parallel straight field lines.

**Answer:** (b)

> Common mistake: Confusing the uniform parallel field lines inside the solenoid with the concentric circles around a straight current-carrying conductor.

### Question 15

*1 mark · MCQ · Our Environment*

Identify the food chain in which the organisms of the second trophic level are missing :

- Grass, goat, lion
- Zooplankton, Phytoplankton, small fish, large fish
- Tiger, grass, snake, frog
- Grasshopper, grass, snake, frog, eagle

**Solution**

1. In the food chain 'Zooplankton, Phytoplankton, small fish, large fish', phytoplankton are producers (first trophic level) and zooplankton are herbivores (second trophic level), but they are written in reverse order, making the second trophic level missing or improperly placed in sequence.
2. Alternatively, considering the options, the food chain 'Zooplankton, Phytoplankton, small fish, large fish' lists zooplankton before phytoplankton, disrupting the producers-first rule of the second trophic level.

**Answer:** (b) Zooplankton, Phytoplankton, small fish, large fish

> Common mistake: Confusing the order of producers and primary consumers in aquatic food chains.

### Question 16

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

In which of the following organisms, multiple fission is a means of asexual reproduction ?

- Yeast
- Leishmania
- Paramoecium
- Plasmodium

**Solution**

1. Plasmodium, the malaria parasite, divides into many daughter cells simultaneously by multiple fission.
2. Yeast reproduces by budding, Leishmania by binary fission, and Paramoecium also by binary fission.

**Answer:** (d) Plasmodium

> Common mistake: Confusing multiple fission in Plasmodium with binary fission in Amoeba or Leishmania.

### Question 17

*1 mark · Assertion and reason · Metals and Non-metals*

Assertion (A) : Hydrogen gas is not evolved when zinc reacts with nitric acid.
Reason (R) : Nitric acid oxidises the hydrogen gas produced to water and itself gets reduced.

- 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. Nitric acid is a strong oxidising agent which oxidises the hydrogen gas produced during the reaction to water and gets reduced itself to nitrogen oxides.

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

> Common mistake: Thinking that nitric acid behaves like other dilute acids with all metals.

### Question 18

*1 mark · Assertion and reason · Our Environment*

Assertion (A) : Accumulation of harmful chemicals is maximum in the organisms at the highest trophic level of a food chain.
Reason (R) : Harmful chemicals are sprayed on the crops to protect them from diseases and pests.

- 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. Both statements are true and explain biological magnification, but Reason (R) is the source of chemicals, not the correct chemical explanation of why accumulation increases at higher trophic levels.

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

> Common mistake: Selecting option (a) by assuming the presence of chemicals on crops explains the progressive accumulation process.

### Question 19

*1 mark · MCQ · Life Processes*

Assertion (A) : The rate of breathing in aquatic organisms is much faster than in terrestrial organisms.
Reason (R) : The amount of oxygen dissolved in water is very high as compared to the amount of oxygen in air.

- 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. Aquatic organisms take up oxygen dissolved in water. Since the amount of dissolved oxygen is fairly low compared to the amount of oxygen in the air, the rate of breathing in aquatic organisms is much faster than that in terrestrial organisms.
2. Therefore, Assertion (A) is true, but Reason (R) is false because the amount of dissolved oxygen is very low, not high.

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

> Common mistake: Assuming oxygen dissolved in water is high because water contains oxygen ($H_2O$).

### Question 20

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

Assertion (A) : The rainbow is a natural spectrum of sunlight in the sky.
Reason (R) : Rainbow is formed in the sky when the sun is overhead and water droplets are also present in air.

- 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. A rainbow is a natural spectrum of sunlight appearing in the sky, caused by dispersion of sunlight by tiny water droplets present in the atmosphere.
2. The Assertion (A) is true.
3. Rainbows are observed when the sun is behind the observer and water droplets are in front, not when the sun is overhead.
4. Therefore, the Reason (R) is false.

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

> Common mistake: Students often assume both statements are true because both mention rainbows and water droplets.

## SECTION - B

Question Nos. 21 to 26 are very short answer type questions.

### Question 21

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

Name the type of chemical reaction in which calcium oxide reacts with water. Justify your answer by giving balanced chemical equation for the chemical reaction.

**Solution**

1. The reaction between calcium oxide and water is a combination reaction because two or more reactants combine to form a single product.
2. The balanced chemical equation is: $\text{CaO(s)} + \text{H}_2\text{O(l)} \rightarrow \text{Ca(OH)}_2\text{(aq)} + \text{Heat}$

**Answer:** Combination reaction with $\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2$

> Common mistake: Forgetting to mention the release of heat or failing to balance the chemical equation.

### Question 22

*2 marks · Very short answer · Life Processes*

State one role of each of the following in human digestive system :
(i) Hydrochloric acid
(ii) Villi
(iii) Anal Sphincter
(iv) Lipase

**Part (i)**

1. Hydrochloric acid creates an acidic medium which facilitates the action of the enzyme pepsin.

Answer (i): Creates an acidic medium for pepsin.

**Part (ii)**

1. Villi increase the surface area for the absorption of digested food.

Answer (ii): Increase surface area for absorption.

**Part (iii)**

1. Anal sphincter regulates the exit of waste material from the body.

Answer (iii): Regulates the exit of waste material.

**Part (iv)**

1. Lipase breaks down emulsified fats into fatty acids and glycerol.

Answer (iv): Emulsifies and breaks down fats.

**Answer:** Roles of the components in the human digestive system.

> Common mistake: Confusing the function of lipase with pepsin or amylase.

### Question 23

*2 marks · Very short answer · Control and Coordination*

How is the movement of leaves of a sensitive plant different from the downward movement of the roots ?

**Solution**

1. The movement of leaves of a sensitive plant (such as Mimosa pudica) is a nastic movement, which is non-directional, rapid, and caused by changes in water content (turgor pressure).
2. The downward movement of roots is a tropic movement (gravitropism), which is a slow, directional growth movement in response to gravity.

**Answer:** Sensitive plant movement is rapid, non-directional and turgor-based, whereas root movement is slow, directional and growth-based.

> Common mistake: Not mentioning that root movement is a growth movement while touch-me-not plant movement is not a growth movement.

### Question 23 (OR)

*2 marks · Very short answer · Control and Coordination*

There is a hormone which regulates carbohydrate, protein and fat metabolism in our body. Name the hormone and the gland which secretes it. Why is it important for us to have iodised salt in our diet?

**Solution**

1. The hormone is thyroxine and it is secreted by the thyroid gland.
2. Iodine is essential for the synthesis of thyroxine hormone by the thyroid gland.
3. Iodised salt is important to prevent deficiency diseases like goitre.

**Answer:** Thyroxine, secreted by the thyroid gland; iodine is needed for its synthesis.

> Common mistake: Forgetting to mention the thyroid gland or the role of iodine in thyroxine synthesis.

### Question 24

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

An object is placed at a distance of $10\text{ cm}$ from a convex mirror of focal length $15\text{ cm}$. Find the position of the image formed by the mirror.

**Solution**

1. Given: Object distance $u = -10\text{ cm}$, Focal length $f = +15\text{ cm}$
2. Formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
3. Substitution: $\frac{1}{15} = \frac{1}{v} + \frac{1}{-10}$
4. Working: $\frac{1}{v} = \frac{1}{15} + \frac{1}{10} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$
5. Result: $v = +6\text{ cm}$

**Answer:** $+6\text{ cm}$

> Common mistake: Incorrect sign convention for $u$ and $f$ in a convex mirror.

### Question 25

*2 marks · Numerical · Electricity*

(A) Show how you would connect three resistors each of resistance $6\text{ }\Omega$, so that the combination has a resistance of $9\text{ }\Omega$. Also justify your answer.

**Solution**

1. To get an equivalent resistance of $9\text{ }\Omega$ using three $6\text{ }\Omega$ resistors, connect two $6\text{ }\Omega$ resistors in parallel and the third $6\text{ }\Omega$ resistor in series with this parallel combination.
2. For the two resistors in parallel: $\frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$, so $R_p = 3\text{ }\Omega$.
3. For the series combination with the third resistor: $R_{eq} = R_p + 6 = 3 + 6 = 9\text{ }\Omega$.
4. Justification: The calculated equivalent resistance matches the required $9\text{ }\Omega$.

**Answer:** Two resistors in parallel connected in series with the third resistor.

> Common mistake: Connecting all three in series or all three in parallel.

### Question 25 (OR)

*2 marks · Numerical · Electricity*

(B) In the given circuit calculate the power consumed in watts in the resistor of $2\text{ }\Omega$ :

**Solution**

1. Given: Resistance $R_1 = 1\text{ }\Omega$, $R_2 = 2\text{ }\Omega$, Voltage $V = 6\text{ V}$ connected in series.
2. Formula: Total resistance $R_s = R_1 + R_2$, current $I = \frac{V}{R_s}$, and power $P = I^2 R_2$.
3. Substitution: $R_s = 1 + 2 = 3\text{ }\Omega$, $I = \frac{6}{3} = 2\text{ A}$, and $P = (2)^2 \times 2$.
4. Result: $P = 4 \times 2 = 8\text{ W}$.

**Answer:** 8 W

> Common mistake: Using the total voltage across the circuit instead of finding the current first, or taking total resistance incorrectly.

### Question 26

*2 marks · Very short answer · Magnetic Effects of Electric Current*

(i) Two magnetic field lines do not intersect each other. Why ?
(ii) How is a uniform magnetic field in a given region represented ? Draw a diagram in support of your answer.

**Part (i)**

1. Two magnetic field lines never intersect each other because if they did, it would mean that at the point of intersection, the compass needle points in two different directions, which is not possible.

Answer (i): Intersection would imply two directions for the magnetic field at a single point.

**Part (ii)**

1. A uniform magnetic field is represented by parallel and equidistant field lines.
2. Diagram: Draw a set of straight, parallel, equally spaced horizontal lines with arrows pointing in the same direction.

Answer (ii): Represented by parallel, equidistant field lines.

**Answer:** Magnetic field lines never intersect and uniform fields are shown by parallel equidistant lines.

> Common mistake: Writing that the magnetic field has two north poles instead of two directions at the point of intersection.

## SECTION - C

Question Nos. 27 to 33 are short answer type questions.

### Question 27

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

Write one chemical equation each for the chemical reaction in which the following have taken place :
(i) Change in colour
(ii) Change in temperature
(iii) Formation of precipitate
Mention colour change/temperature change (rise/fall)/compound precipitated along with equation.

**Part (i)**

1. Reaction: $\text{Fe} (s) + \text{CuSO}_4 (aq) \to \text{FeSO}_4 (aq) + \text{Cu} (s)$
2. Colour change: The blue colour of copper sulphate solution fades and a reddish-brown coating of copper is deposited on iron.

Answer (i): $\text{Fe} + \text{CuSO}_4 \to \text{FeSO}_4 + \text{Cu}$, blue colour changes to light green with reddish-brown deposit.

**Part (ii)**

1. Reaction: $\text{Zn} (s) + \text{H}_2\text{SO}_4 (aq) \to \text{ZnSO}_4 (aq) + \text{H}_2 (g)$
2. Temperature change: There is a rise in temperature as it is an exothermic reaction.

Answer (ii): $\text{Zn} + \text{H}_2\text{SO}_4 \to \text{ZnSO}_4 + \text{H}_2$, rise in temperature.

**Part (iii)**

1. Reaction: $\text{BaCl}_2 (aq) + \text{Na}_2\text{SO}_4 (aq) \to \text{BaSO}_4 (s) + 2\text{NaCl} (aq)$
2. Precipitate formed: A white precipitate of barium sulphate is formed.

Answer (iii): $\text{BaCl}_2 + \text{Na}_2\text{SO}_4 \to \text{BaSO}_4 \downarrow + 2\text{NaCl}$, white precipitate of barium sulphate.

**Answer:** Chemical equations representing change in colour, change in temperature, and formation of precipitate are given in the respective parts.

> Common mistake: Forgetting to write physical states or failing to mention the specific observation such as colour change or temperature rise.

### Question 28

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

(i) The $\text{pH}$ of a sample of tomato juice is $4.6$. How is this juice likely to be in taste ? Give reason to justify your answer.
(ii) How do we differentiate between a strong acid and a weak base in terms of ion-formation in aqueous solutions ?
(iii) The acid rain can make the survival of aquatic animals difficult. How ?

**Part (i)**

1. The pH of tomato juice is $4.6$, which is less than $7$.
2. Therefore, it is acidic in nature and will taste sour.

Answer (i): Sour in taste because a pH of $4.6$ indicates an acidic nature.

**Part (ii)**

1. A strong acid dissociates completely in an aqueous solution to yield a high concentration of hydronium ions ($\text{H}_3\text{O}^+$).
2. A weak base dissociates only partially in an aqueous solution to yield a low concentration of hydroxide ions ($\text{OH}^-$).

Answer (ii): Strong acids dissociate completely into ions, whereas weak bases dissociate only partially in aqueous solutions.

**Part (iii)**

1. When acid rain flows into rivers, it lowers the pH of the river water.
2. This acidic water makes the survival of aquatic animals difficult.

Answer (iii): Acid rain lowers the pH of aquatic bodies, making it difficult for aquatic life to survive.

**Answer:** Tomato juice is sour in taste due to its acidic pH, strong acids dissociate completely while weak bases partially dissociate in water, and acid rain lowers water pH making aquatic survival difficult.

> Common mistake: Confusing the ion types released by acids and bases.

### Question 29

*3 marks · Short answer · Life Processes*

(i) Why is respiratory pigment needed in multicellular organisms with large body size ?
(ii) Give reasons for the following :
(a) Rings of cartilage are present in the throat.
(b) Lungs always contain a residual volume of air.
(c) The diaphragm flattens and ribs are lifted up when we breathe in.
(d) Walls of alveoli contain an extensive network of blood vessels.

**Part (i)**

1. In large multicellular organisms, the body size is large and diffusion pressure alone cannot take oxygen to all parts of the body efficiently.
2. Respiratory pigments like haemoglobin have high affinity for oxygen and transport it from the lungs to all body tissues.

Answer (i): Diffusion is insufficient to deliver oxygen to all cells in large bodies, so respiratory pigments are needed for efficient transport.

**Part (ii)**

1. (a) Rings of cartilage are present in the throat to ensure that the air-passage does not collapse.
2. (b) Lungs always contain a residual volume of air so that there is sufficient time for oxygen to be absorbed and for carbon dioxide to be released.
3. (c) During inhalation, the diaphragm flattens and ribs are lifted up to increase the chest cavity volume and reduce air pressure, drawing air into the lungs.
4. (d) The walls of alveoli contain an extensive network of blood vessels to provide a large surface area for gaseous exchange.

Answer (ii): (a) Prevent collapse of trachea (b) Provide sufficient time for gas exchange (c) Increase chest cavity volume (d) Provide large surface area for gas exchange.

**Answer:** Multicellular organisms need respiratory pigments like haemoglobin because diffusion alone is too slow to transport oxygen to all body cells.

> Common mistake: Missing out on mentioning the increase in surface area or pressure changes during breathing.

### Question 30

*3 marks · Short answer · Control and Coordination*

Define reflex action. With the help of a flow chart show the path of a reflex action such as sneezing.

**Solution**

1. A reflex action is a rapid, automatic response to a stimulus which is controlled by the spinal cord without the conscious will of the brain.
2. Stimulus (Irritant in nasal cavity) $\to$ Sensory nerve / Receptor $\to$ Spinal cord (CNS) $\to$ Motor neuron $\to$ Effector (Respiratory muscles / Diaphragm) $\to$ Response (Sneezing).

**Answer:** Reflex action is a rapid, automatic response to a stimulus. Flow chart: Stimulus $\to$ Receptor $\to$ Sensory Neuron $\to$ Spinal Cord $\to$ Motor Neuron $\to$ Effector $\to$ Response.

> Common mistake: Forgetting to include the spinal cord or mixing up the order of sensory and motor neurons in the reflex arc.

### Question 31

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

Study the diagram given below and answer the questions that follow :
(i) Name the defect of vision represented in the diagram. Give reason for your answer.
(ii) List two causes of this defect.
(iii) With the help of a diagram show how this defect of vision is corrected.

**Part (i)**

1. The defect shown is myopia (near-sightedness).
2. Reason: The rays are converging in front of the retina instead of on the retina.

Answer (i): Myopia; rays converge in front of the retina.

**Part (ii)**

1. Excessive curvature of the eye lens.
2. Elongation of the eyeball.

Answer (ii): 1. Excessive curvature of eye lens. 2. Elongation of the eyeball.

**Part (iii)**

1. This defect is corrected by using a concave lens of suitable power.
2. Diagram: A concave lens placed in front of the eye diverges the incoming rays so that they focus on the retina.

Answer (iii): Corrected using a suitable concave lens.

**Answer:** The defect is myopia, caused by excessive curvature of the eye lens or elongation of the eyeball, and is corrected using a concave lens.

> Common mistake: Confusing myopia with hypermetropia and recommending a convex lens for correction.

### Question 32

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

Name and state the rule to determine the direction of a :
(i) magnetic field produced around a current carrying straight conductor.
(ii) force experienced by a current carrying straight conductor placed in a magnetic field which is perpendicular to it.

**Part (i)**

1. Rule name: Right-hand thumb rule.
2. Statement: If you hold a current-carrying straight conductor in your right hand such that the thumb points in the direction of current, then your fingers wrap around the conductor in the direction of the field lines of the magnetic field.

Answer (i): Right-hand thumb rule: Thumb points along current, wrapped fingers show magnetic field direction.

**Part (ii)**

1. Rule name: Fleming's left-hand rule.
2. Statement: Stretch the thumb, forefinger, and middle finger of your left hand such that they are mutually perpendicular. If the forefinger points in the direction of magnetic field and the middle finger in the direction of current, then the thumb will point in the direction of motion or force acting on the conductor.

Answer (ii): Fleming's left-hand rule: Forefinger = magnetic field, middle finger = current, thumb = force.

**Answer:** Right-hand thumb rule determines the magnetic field around a straight conductor, and Fleming's left-hand rule determines the force on a current-carrying conductor.

> Common mistake: Mixing up Fleming's left-hand rule (used for force on current-carrying conductor) with Fleming's right-hand rule (used for electromagnetic induction).

### Question 33

*3 marks · Short answer · Our Environment*

(A) $\text{Plants} \rightarrow \text{Deer} \rightarrow \text{Lion}$
In the given food chain, what will be the impact of removing all the organisms of second trophic level on the first and third trophic level ? Will the impact be the same for the organisms of the third trophic level in the above food chain if they were present in a food web ? Justify.

**Solution**

1. In the given food chain, plants form the first trophic level, deer the second trophic level, and lion the third trophic level.
2. If all organisms of the second trophic level (deer) are removed, the population of the first trophic level (plants) will increase due to lack of grazing.
3. The third trophic level (lions) will starve and their population will decrease due to the shortage of food.
4. No, the impact will not be the same if they were present in a food web because lions would have alternative food sources such as other herbivores available in the food web.

**Answer:** Removal of second trophic level increases first trophic level and decreases third in a food chain, but a food web provides alternative food sources.

> Common mistake: Students often confuse the flow of energy and state that the first trophic level will decrease.

### Question 33 (OR)

*3 marks · Short answer · Our Environment*

(B) A gas 'X' which is a deadly poison is found at the higher levels of atmosphere and performs an essential function.
Name the gas and write the function performed by this gas in the atmosphere. Which chemical is linked to the decrease in the level of this gas? What measures have been taken by an international organization to check the depletion of the layer containing this gas?

**Solution**

1. The gas 'X' is ozone ($O_3$), which shields the surface of the earth from harmful ultraviolet (UV) radiations from the Sun.
2. Chlorofluorocarbons (CFCs) are the synthetic chemicals linked to the decrease in the ozone level.
3. The United Nations Environment Programme (UNEP) forged an agreement to freeze CFC production at 1986 levels (Montreal Protocol) worldwide.

**Answer:** Ozone gas protects from UV rays, CFCs cause its depletion, and UNEP's Montreal Protocol regulates CFC production.

> Common mistake: Students write carbon dioxide instead of CFCs as the chemical responsible for ozone depletion.

## SECTION - D

Question Nos. 34 to 36 are long answer type questions.

### Question 34

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

(i) Define a homologous series of carbon compounds.
(ii) Why is the melting and boiling points of $\text{C}_4\text{H}_8$ higher than that of $\text{C}_3\text{H}_6$ or $\text{C}_2\text{H}_4$ ?
(iii) Why do we NOT see any gradation in chemical properties of a homologous series compounds ?
(iv) Write the name and structures of (i) aldehyde and (ii) ketone with molecular form $\text{C}_3\text{H}_6\text{O}$.

**Part (i)**

1. A homologous series is a series of carbon compounds in which the same functional group substitutes for hydrogen in a carbon chain.
2. Successive members differ by a $-\text{CH}_2-$ unit and a molecular mass difference of $14\text{ u}$.

Answer (i): A series of compounds with the same functional group and similar chemical properties, differing by a $-\text{CH}_2-$ group.

**Part (ii)**

1. Melting and boiling points increase with increasing molecular mass in a homologous series.
2. As the molecular size and surface area of $\text{C}_4\text{H}_8$ is greater than $\text{C}_3\text{H}_6$ or $\text{C}_2\text{H}_4$, intermolecular forces of attraction are stronger.

Answer (ii): Higher molecular mass leads to stronger intermolecular forces, requiring more energy to boil.

**Part (iii)**

1. Chemical properties of carbon compounds in a homologous series are determined by the functional group.
2. Since all members of a series possess the same functional group, their chemical properties remain identical or show no gradation.

Answer (iii): Chemical properties depend on the functional group, which is the same for all members of a homologous series.

**Part (iv)**

1. The aldehyde with molecular formula $\text{C}_3\text{H}_6\text{O}$ is Propanal ($\text{CH}_3-\text{CH}_2-\text{CHO}$).
2. The ketone with molecular formula $\text{C}_3\text{H}_6\text{O}$ is Propanone ($\text{CH}_3-\text{CO}-\text{CH}_3$).

Answer (iv): Propanal: $\text{CH}_3\text{CH}_2\text{CHO}$; Propanone: $\text{CH}_3\text{COCH}_3$.

**Answer:** Homologous series defined, boiling point explanation, chemical properties uniformity, and structures for propanal and propanone given.

> Common mistake: Confusing the structures of aldehyde and ketone having the same molecular formula.

### Question 34 (OR)

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

(B) (i) Write the name and structure of an organic compound 'X' having two carbon atoms in its molecule and its name is suffixed with '-ol'.
(ii) What happens when 'X' is heated with excess concentrated sulphuric acid at $443\text{ K}$? Write chemical equation for the reaction stating the conditions for the reaction. Also state the role played by concentrated sulphuric acid in the reaction.
(iii) Name and draw the electron dot structure of hydrocarbon produced in the above reaction.

**Part (i)**

1. The organic compound having two carbon atoms with '-ol' suffix is ethanol.
2. Its molecular formula is $\text{C}_2\text{H}_5\text{OH}$ and structure is $\text{CH}_3-\text{CH}_2-\text{OH}$.

Answer (i): Ethanol, $\text{CH}_3-\text{CH}_2-\text{OH}$.

**Part (ii)**

1. When ethanol is heated with excess concentrated sulphuric acid at $443\text{ K}$, it undergoes dehydration to form ethene gas.
2. Chemical equation: $\text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4, \, 443\text{ K}} \text{C}_2\text{H}_4 + \text{H}_2\text{O}$.
3. Concentrated sulphuric acid acts as a dehydrating agent which removes water molecules from ethanol.

Answer (ii): Ethene is formed and concentrated $\text{H}_2\text{SO}_4$ acts as a dehydrating agent.

**Part (iii)**

1. The hydrocarbon produced is ethene ($\text{C}_2\text{H}_4$).
2. Its electron dot structure shows a double covalent bond between the two carbon atoms and single bonds with four hydrogen atoms.

Answer (iii): Ethene, with a double bond between carbon atoms.

**Answer:** Compound X is ethanol, dehydration reaction produces ethene and acts as a dehydrating agent.

> Common mistake: Forgetting to mention the temperature $443\text{ K}$ or the role of concentrated sulphuric acid.

### Question 35

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

(A) (i) Name three techniques/devices used by human females to avoid pregnancy. Mention the side effects caused by each.
(ii) What will happen if in a human female (a) fertilisation takes place, (b) an egg is not fertilised ?

**Part (i)**

1. Three contraceptive techniques are: (1) Barrier methods (e.g., condoms), (2) Oral pills (hormonal pills), and (3) Intrauterine contraceptive devices (IUDs like Copper-T).
2. Side effects: Barrier methods have negligible side effects; oral pills can cause hormonal imbalance and related health issues; IUDs can cause side effects due to uterine irritation or infection if not monitored.

Answer (i): Barrier methods (low side effects), oral pills (hormonal imbalance), IUDs (uterine irritation/infection).

**Part (ii)**

1. (a) If fertilisation takes place, the zygote gets implanted in the lining of the uterus, leading to pregnancy, and menstruation stops.
2. (b) If the egg is not fertilised, the uterine lining thickens and then breaks down, leading to menstruation.

Answer (ii): (a) Pregnancy occurs; (b) Menstruation occurs.

**Answer:** Contraceptive methods with side effects and consequences of fertilisation or non-fertilisation explained.

> Common mistake: Listing methods of contraception without specifying their respective side effects.

### Question 35 (OR)

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

(B) (i) Draw a diagram showing spore formation in Rhizopus and label the (a) reproductive and (b) non-reproductive parts. Why does Rhizopus not multiply on a dry slice of bread ?
(ii) Name and explain the process by which reproduction takes place in Hydra.

**Part (i)**

1. Diagram: Draw sporangium showing tiny blob-like structures (sporangia) on thread-like structures (hyphae). Label (a) reproductive part as sporangium (containing spores) and (b) non-reproductive part as hyphae/stolon.
2. Rhizopus does not multiply on a dry slice of bread because spores require moisture and nutrients to germinate and grow, which are absent in dry conditions.

Answer (i): Sporangium is the reproductive part, hyphae are non-reproductive; lacks moisture on dry bread.

**Part (ii)**

1. Hydra reproduces by budding, which is an asexual mode of reproduction.
2. A bud develops as an outgrowth due to repeated cell division at a specific site, grows into a tiny individual, and detaches from the parent body to live independently.

Answer (ii): Budding: a bud grows and detaches from the parent Hydra.

**Answer:** Rhizopus spore formation details, lack of moisture on dry bread, and budding in Hydra explained.

> Common mistake: Failing to mention the requirement of moisture for spore germination in fungi.

### Question 36

*5 marks · Long answer · Electricity*

(A) (i) Define electric power. Express it in terms of potential difference (V) and resistance (R).
(ii) An electric oven is designed to work on the mains voltage of $220\text{ V}$. This oven consumes $11\text{ units}$ of electrical energy in $5\text{ hours}$. Calculate :
(a) power rating of the oven
(b) current drawn by the oven
(c) resistance of the oven when it is red hot

**Part (i)**

1. Electric power is defined as the rate at which electric energy is dissipated or consumed in an electric circuit: $P = \frac{W}{t} = VI$.
2. Using Ohm's law ($V = IR$), electric power in terms of potential difference ($V$) and resistance ($R$) is expressed as: $P = \frac{V^2}{R}$.

Answer (i): Electric power $P = VI = \frac{V^2}{R}$.

**Part (ii)(a)**

1. Given: Energy $E = 11\text{ units} = 11\text{ kWh}$, Time $t = 5\text{ hours}$.
2. Power rating $P = \frac{E}{t} = \frac{11\text{ kWh}}{5\text{ h}} = 2.2\text{ kW} = 2200\text{ W}$.

Answer (ii)(a): $2.2\text{ kW}$ (or $2200\text{ W}$)

**Part (ii)(b)**

1. Given: Voltage $V = 220\text{ V}$, Power $P = 2200\text{ W}$.
2. Formula: $P = VI \implies I = \frac{P}{V} = \frac{2200\text{ W}}{220\text{ V}} = 10\text{ A}$.

Answer (ii)(b): $10\text{ A}$

**Part (ii)(c)**

1. Formula: $R = \frac{V}{I}$ or $R = \frac{V^2}{P}$.
2. Substitution: $R = \frac{220\text{ V}}{10\text{ A}} = 22\text{ }\Omega$.

Answer (ii)(c): $22\text{ }\Omega$

**Answer:** Power definitions, power rating = $2.2\text{ kW}$, current = $10\text{ A}$, resistance = $22\text{ }\Omega$.

> Common mistake: Forgetting to convert kilowatts to watts or confusing electrical units (kWh) with power.

### Question 36 (OR)

*5 marks · Long answer · Electricity*

(B) (i) Write the relation between resistance $R$ and electrical resistivity $\rho$ of the material of a conductor in the shape of cylinder of length $l$ and area of cross-section $A$. Hence derive the SI unit of electrical resistivity.
(ii) The resistance of a metal wire of length $3\text{ m}$ is $60\text{ }\Omega$. If the area of cross-section of the wire is $4 \times 10^{-7}\text{ m}^2$, calculate the electrical resistivity of the wire.
(iii) State how would electrical resistivity be affected if the wire (of part 'ii') is stretched so that its length is doubled. Justify your answer.

**Part (i)**

1. Relation: $R = \rho \frac{l}{A}$, where $\rho$ is electrical resistivity.
2. Rearranging for resistivity: $\rho = \frac{R A}{l}$.
3. Substituting SI units: $\rho = \frac{\Omega \cdot \text{m}^2}{\text{m}} = \Omega \cdot \text{m}$. Hence, SI unit is ohm-metre ($\Omega\text{m}$).

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

**Part (ii)**

1. Given: $l = 3\text{ m}$, $R = 60\text{ }\Omega$, $A = 4 \times 10^{-7}\text{ m}^2$.
2. Formula: $\rho = \frac{R \cdot A}{l}$.
3. Substitution: $\rho = \frac{60 \times 4 \times 10^{-7}}{3} = 20 \times 4 \times 10^{-7} = 8 \times 10^{-6}\text{ }\Omega\text{m}$.

Answer (ii): $8 \times 10^{-6}\text{ }\Omega\text{m}$

**Part (iii)**

1. Electrical resistivity does not change when the wire is stretched.
2. Justification: Resistivity is a characteristic property of the material of the conductor and depends only on the nature of the material and temperature, not on its dimensions.

Answer (iii): Remains unchanged as resistivity depends only on material and temperature.

**Answer:** Resistivity formula and SI unit derived, resistivity = $8 \times 10^{-6}\text{ }\Omega\text{m}$, resistivity remains unchanged on stretching.

> Common mistake: Stating that resistivity changes when a wire is stretched (confusing it with resistance).

## SECTION - E

Q. Nos. 37-39 are source-based/case-based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts :

### Question 37

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

The metals produced by various reduction processes are not very pure. They contain impurities, which must be removed to obtain pure metals. The most widely used method for refining impure metals is electrolytic refining.
(i) What is the cathode and anode made of in the refining of copper by this process ?
(ii) Name the solution used in the above process and write its formula.
(iii) (A) How copper gets refined when electric current is passed in the electrolytic cell ?

**Part (i)**

1. The cathode is made of a thin strip of pure copper.
2. The anode is made of impure copper.

Answer (i): Cathode: pure copper strip; Anode: impure copper block.

**Part (ii)**

1. The solution used is acidified copper sulphate solution.
2. Its chemical formula is $\text{CuSO}_4$.

Answer (ii): Acidified copper sulphate solution, $\text{CuSO}_4$.

**Part (iii)**

1. When electric current is passed, copper ions from the electrolyte go to the cathode and deposit as pure copper.
2. An equivalent amount of copper from the impure anode dissolves into the electrolyte.

Answer (iii): Pure copper from anode dissolves into the electrolyte and deposits on the cathode.

**Answer:** Refining of copper involves pure copper cathode, impure copper anode, and acidified copper sulphate solution.

> Common mistake: Confusing the roles of anode and cathode during electrolytic refining.

### Question 37 (OR)

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

(iii) (B) You have two beakers 'A' and 'B' containing copper sulphate solution. What would you observe after about $2\text{ hours}$ if you dip a strip of zinc in beaker 'A' and a strip of silver in beaker 'B'? Give reason for your observations in each case.

**Part (iii) (B)**

1. In beaker A, the blue color of copper sulphate fades and a reddish-brown coating appears on the zinc strip because zinc is more reactive than copper and displaces it.
2. In beaker B, no reaction takes place because silver is less reactive than copper and cannot displace it from copper sulphate solution.

Answer (iii) (B): Zinc displaces copper (reaction occurs); silver does not displace copper (no reaction).

**Answer:** Zinc displaces copper from its solution, while silver does not.

> Common mistake: Stating that silver will react with copper sulphate solution.

### Question 38

*4 marks · Case-based · Heredity*

Mendel worked out the rules of heredity by working on garden pea using a number of visible contrasting characters. He conducted several experiments by making a cross with one or two pairs of contrasting characters of pea plant. On the basis of his observations he gave some interpretations which helped to study the mechanism of inheritance.
(i) When Mendel crossed pea plants with pure tall and pure short characteristics to produce $F_1$ progeny, which two observations were made by him in $F_1$ plants?
(ii) Write one difference between dominant and recessive trait.
(iii) (A) In a cross with two pairs of contrasting characters
$\text{RRYY} \times \text{rryy}$
(Round Yellow) (Wrinkled Green)
Mendel observed 4 types of combinations in $\text{F}_2$ generation. By which method did he obtain $\text{F}_2$ generation ? Write the ratio of the parental combinations obtained and what conclusions were drawn from this experiment.

**Part (i)**

1. All $F_1$ plants were tall.
2. None of the $F_1$ plants were short (intermediate trait was not observed).

Answer (i): All plants were tall and no medium-sized plants were observed.

**Part (ii)**

1. A dominant trait expresses itself even in the presence of an alternative trait, whereas a recessive trait gets masked in the presence of a dominant trait.

Answer (ii): Dominant trait expresses in both homozygous and heterozygous conditions, while recessive trait expresses only in homozygous condition.

**Part (iii)**

1. He obtained the $\text{F}_2$ generation by self-pollinating the $\text{F}_1$ plants.
2. The ratio of parental combinations to new combinations in the dihybrid cross is $9:3:3:1$, where parental combinations are $9$ (round yellow) and $1$ (wrinkled green).
3. It was concluded that the two pairs of traits are inherited independently of each other (Law of Independent Assortment).

Answer (iii): Self-pollination; parental ratio is 9:1 (out of 9:3:3:1); traits are inherited independently.

**Answer:** Mendel observed tallness in F1 progeny, and obtained F2 generation through self-pollination.

> Common mistake: Forgetting that selfing of F1 progeny produces the F2 generation.

### Question 38 (OR)

*2 marks · Case-based · Heredity*

(iii) (B) Justify the statement :
"It is possible that a trait is inherited but may not be expressed."

**Part (iii) (B)**

1. Every inherited trait is influenced by paternal and maternal DNA.
2. In a heterozygous condition, the recessive allele is inherited but remains unexpressed because the dominant allele suppresses its expression.

Answer (iii) (B): In a heterozygous individual, the recessive trait is inherited from one parent but is masked by the dominant allele, hence not expressed.

**Answer:** Recessive traits are inherited from both parents but remain unexpressed in the presence of a dominant allele.

> Common mistake: Writing that traits can be lost instead of being masked.

### Question 39

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

Study the data given below showing the focal length of three concave mirrors A, B and C and the respective distances of objects placed in front of the mirrors :
(i) In which one of the above cases the mirror will form a diminished image of the object ? Justify your answer.
(ii) List two properties of the image formed in case 2.
(iii) (A) What is the nature and size of the image formed by mirror C ? Draw ray diagram to justify your answer.

**Part (i)**

1. In Case 1, the object distance ($u = -45\text{ cm}$) is greater than the radius of curvature ($2f = -40\text{ cm}$), so the object is placed beyond C.
2. When an object is placed beyond C in a concave mirror, the image formed is real, inverted, and diminished.

Answer (i): Case 1, because the object is placed beyond the center of curvature.

**Part (ii)**

1. In Case 2, the object is at $u = -30\text{ cm}$ and $f = -15\text{ cm}$, meaning the object is at the center of curvature C ($u = 2f$).
2. The image formed is real, inverted, and of the same size as the object.

Answer (ii): The image is real and inverted (or equal in size to the object).

**Part (iii) (A)**

1. In Case 3, the object distance $u = -20\text{ cm}$ is less than the focal length $f = -30\text{ cm}$, so the object lies between the pole and the principal focus.
2. The nature of the image formed is virtual and erect, and its size is enlarged.
3. Diagram: Draw a concave mirror with object between P and F, showing two rays (one parallel to principal axis passing through focus, another passing through center of curvature) meeting behind the mirror to form a virtual, erect, and enlarged image.

Answer (iii) (A): Virtual, erect, and enlarged image. (Refer to standard textbook ray diagram for object between P and F of a concave mirror).

**Answer:** Case 1 forms a diminished image, case 2 forms a real and inverted image, and case 3 forms an enlarged virtual image.

> Common mistake: Incorrectly identifying object positions with respect to focal length and radius of curvature.

### Question 39 (OR)

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

(iii) (B) An object is placed at a distance of $18\text{ cm}$ from the pole of a concave mirror of focal length $12\text{ cm}$. Find the position of the image formed in this case.

**Solution**

1. Given: Object distance $u = -18\text{ cm}$, Focal length $f = -12\text{ cm}$.
2. Formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
3. Substitution: $\frac{1}{-12} = \frac{1}{v} + \frac{1}{-18}$
4. Working: $\frac{1}{v} = -\frac{1}{12} + \frac{1}{18} = \frac{-3 + 2}{36} = -\frac{1}{36}$
5. Result: $v = -36\text{ cm}$

**Answer:** The image is formed at a distance of $36\text{ cm}$ in front of the concave mirror.

> Common mistake: Sign convention errors for object distance and focal length of a concave mirror.

## Frequently asked questions

### What is the paper pattern and section-wise breakdown for the CBSE Class 10 Science Question Paper 2024 Set 31/1/1?

The paper carries a total of 80 marks and is 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.

### How are the marks distributed across different sections in this question paper?

Section A contains 20 objective-type questions worth 20 marks in total. Section B includes 6 short answer questions of 12 marks, Section C consists of 7 short answer questions carrying 21 marks, Section D features 3 long answer questions for 15 marks, and Section E comprises 3 source-based or case-based units totaling 12 marks.

### Which chapters carry the most marks in the CBSE Class 10 Science 2024 Set 31/1/1 paper?

The highest weightage in this paper goes to Chemical Reactions and Equations with 8 marks. This is followed by Metals and Non-metals, Life Processes, Light – Reflection and Refraction, and Electricity, each carrying 7 marks, while Carbon and its Compounds accounts for 6 marks.

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

Students should write precise, step-by-step answers using proper scientific terms and chemical equations where necessary. For numerical problems in physics, clearly state the given values, write the formula, show the calculation steps, and include proper SI units in the final answer.

### Is the solutions PDF for the CBSE Class 10 Science Question Paper 2024 Set 31/1/1 available for free on SwaVid?

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

## 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