CBSE Class 10 Science Answer Key 2024 (All Sets)

CBSE Class 10 Science Answer Key 2024:The exam paper has been reviewed to be well-balanced and also was moderately easy. Now, students have been eagerly awaiting the answers or solutions to the CBSE Class 10 Science exam, as they seek the review as well as assessment of their performance.

In this article, we will discuss the answer key for today’s CBSE Class 10 Science Paper 2024. This would include the answers to all questions as well as answers, that can be checked for analysis of the exam.

CBSE Class 10 Science Paper Set Code 31/1/1 Answer Key 2024

Section – A

1. When 2 mL of sodium hydroxide solution is added to a 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:

(a) NaOH + Zn → NaZnO₂ + H₂O

(b) 2NaOH + Zn → Na₂ZnO₂ + H₂

(c) 2NaOH + Zn → NaZnO₂ + H₂

(d) 2NaOH + Zn → Na₂ZnO2 + H₂O

Answer:

(b) 2NaOH + Zn → Na₂ZnO₂ + H₂

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

(a) 2FeSO₄ → Fe₂O₃+ SO₂+ SO₃

(b) 2H₂O → 2H₂+ O₂

(c) 2AgBr → 2Ag + Br₂

(d) CaCO₃ → CaO + CO₂

Answer:

(c) 2AgBr → 2Ag + Br2

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

(a) Bromine and Mercury

(b) Mercury and lodine

(c) Mercury and Bromine

(d) Iodine and Mercury

Answer:

(c) Mercury and Bromine

4. 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:

(a) (i) and (ii)

(b) (ii) and (iii)

(c) (ii) and (iv)

(d) (i) and (iii)

Answer:

(d) (i) and (iii)

5. Consider the following compounds :

FeSO₄; CuSO₄; CaSO₄; Na₂CO₃

The compound having maximum number of water of crystallisation in its crystalline form in one molecule is :

(a) FeSO₄

(b) CuSO₄

(c) CaSO₄

(d) Na₂CO₃

Answer:

(d) Na₂CO₃

6. Oxides of aluminium and zine are :

(a) acidic

(b) basic

(c) amphoteric

(d) neutral

Answer:

(c) amphoteric; behave as both acidic and basic oxides

7. MnO₂ + 4HCI → MnCl₂ + 2H₂O + Cl₂

The reaction given above is a redox reaction because in this case

(a) MnO₂ is oxidised and HCI is reduced.

(b) HCI is oxidised.

(c) MnO₂ is reduced.

(d) MnO₂ is reduced and HCI is oxidised.

Answer:

(d) MnO₂ is reduced and HCI is oxidised

8. 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 :

(a) (i) and (iii)

(b) (ii) and (iv)

(c) (iii) and (iv)

(d) i), (iii) and (iv)

Answer:

(b) (ii) and (iv)

9. 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 :

(a) (i) and (ii)

(b) (iii) and (iv)

(a) (i), (ii) and (iv)

(d) (i) and (iv)

Answer:

(a) (i) and (ii)

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

(a) Axon

(b) Dendrites

(c) Neuromuscular junction

(d) Cell body

Answer:

(c) Neuromuscular junction

11. 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 :

(a) (i) only

(b) (i) and (iii)

(c) (ii) and (iii)

(d) (ii) and (iv)

Answer:

(c) (ii) and (iii);

12. 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?

(a) Beyond twice the focal length of the lens.

(b) At the principal focus of the lens.

(c) At twice the focal length of the lens.

(d) Between the optical centre of the lens and its principal focus.

Answer:

(c) At twice the focal length of the lens;​

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

(a) Cornea

(b) Ciliary muscles

(c) Optic nerves

(d) Retina

Answer:

(d) Retina

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

Answer:

(a)

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

(a) Grass, goat, lion

(b) Zooplankton , Phytoplankton, small fish, large fish

(c) Tiger, grass , snake , frog

(d) Grasshopper, grass , snake , frog, eagle

Answer:

(c) Tiger, grass, snake, frog

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

(a) Yeast

(b) Leishmania

(c) Paramoecium

(d) Plasmodium

Answer:

(d) Plasmodium

Section A Part B: Assertion-Based Questions

For Q. Nos. 17 to 20, two statements are given – One labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below :

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

17. 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.

Answer:

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

18. 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

Answer:

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

19. 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.

Answer:

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

20. 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.

Answer:

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

Section- B

21. 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.

Answer:

The chemical reaction in which calcium oxide reacts with water is known as a combination reaction. In a combination reaction, two or more substances combine to form a single new substance.

The balanced chemical equation for the reaction between calcium oxide (CaO) and water (H₂O) to produce calcium hydroxide (Ca(OH)₂) is as follows

CaO + H₂O → Ca(OH)₂

In this reaction, calcium oxide and water combine to form calcium hydroxide, illustrating the characteristics of a combination reaction.

22.State one role of each of the following in human digestive system :
(i) Hydrochloric acid

Answer:

Hydrochloric Acid (HCl): Produced by the stomach, hydrochloric acid serves a critical role in digestion. It creates an acidic environment in the stomach, which helps in the denaturation of proteins, making them easier for digestive enzymes to break down. Additionally, the acidic environment kills most of the bacteria in food, protecting the body from infection.

(ii) Villi

Answer:

Villi are small, finger-like projections found in the lining of the small intestine. They play a crucial role in the absorption of nutrients from digested food into the bloodstream. Each villus contains blood vessels and a lymph vessel (lacteal), which help in the transportation of absorbed nutrients. The large surface area provided by the villi maximizes the absorption of nutrients and water.

(iii) Anal Sphincter

Answer:

Anal Sphincter is a muscle at the end of the digestive tract, specifically at the anus. It plays a key role in controlling the expulsion of feces from the body. The sphincter remains tightly closed to hold in the contents of the rectum and opens during a bowel movement to allow the passage of feces out of the body, thus playing a critical role in bowel continence.

(iv) Lipase

Answer:

Lipase: Lipase is an enzyme produced primarily by the pancreas and secreted into the small intestine. It is essential for the digestion of fats (lipids). Lipase breaks down dietary fats into smaller molecules called fatty acids and glycerol, which can then be easily absorbed by the villi in the small intestine. This enzymatic action is crucial for the body to utilize fats from the diet as a source of energy and for other metabolic processes.

23. (A) How is the movement of leaves of a sensitive plant different from the downward movement of the roots?

Answer:

The movement of leaves in a sensitive plant, like Mimosa pudica, is a rapid response to touch or other stimuli, resulting in the leaves folding or drooping quickly. This movement, known as nastic movement, is non-directional and does not depend on the direction of the stimulus. In contrast, the downward movement of roots, known as positive gravitropism, is a growth movement directed by gravity, causing roots to grow downwards into the soil.

OR

(B) 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?

Answer:

The hormone that regulates carbohydrate, protein, and fat metabolism in our body is insulin, which is secreted by the pancreas. However, the context of needing iodised salt in the diet relates to the thyroid gland and its hormone. So, the relevant hormone for this context would be thyroxine, which is secreted by the thyroid gland. Thyroxine plays a crucial role in metabolism, growth, and development.

Iodised salt is important in our diet because iodine is a critical component of thyroxine. Without adequate iodine, the thyroid gland cannot synthesize enough thyroxine, leading to hypothyroidism and related disorders, such as goiter (enlargement of the thyroid gland). Consuming iodised salt ensures that the body gets enough iodine to produce the necessary levels of thyroxine for normal metabolic functions.

24. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position of the image formed by the mirror.

Answer:

To find the image position for an object placed 10 cm from a convex mirror with a 15 cm focal length, we use the mirror formula:

1/f=1/v + 1/u .

Given f =15 cm and u = −10, solving for v yields the image position approximately 6 cm behind the mirror, indicating a virtual image.

25. (A) Show how you would connect three resistors each of resistance 6 Ω, so that the combination has a resistance of 9 Ω. Also justify your answer.

Answer:

To achieve a total resistance of 9 Ω using three 6 Ω resistors,

First connect two resistors in parallel, yielding 3 Ω due to the formula

1/R_parallel = 1/R₁ + 1/R₂

1/R_parallel = 1/6 + 1/6 , which simplify 3 Ω.

Then, connect this parallel arrangement in series with the third resistor. The series connection adds resistances directly, due to the formula:

R_series = R₁ + R₂

resulting in 3Ω + 6Ω = 9Ω , achieving the desired resistance.

25. (B) In the given circuit calculate the power consumed in watts in the resistor of 2 Ω:

Answer:

According to the figure given we have:

Potential Difference (V) = 6V

R₁ = 1Ω , R₂ = 2Ω

The resistors are connected in series, Hence total resistance is given by:

R_series = R₁ + R₂

R_series = 1 + 2 = 3Ω

Now, according to ohm’s law

V = IR

I = V/R

Putting the values given we get

I = 6/3

I = 2 A

As there is no current division in the circuit, equal amount of current will pass through each component , hence the 2Ω resistor will have a current of 2A.

Also Power = I²R

Putting the values we get

P = 2² × 2 = 8 W

Hence, the power consumed in the resistor of 2 Ω is 8 Watts.

26. (i) Two magnetic field lines do not intersect each other. Why ?

Answer:

Two magnetic field lines do not intersect each other because if they did, it would imply that at the point of intersection, the compass needle would point in two different directions simultaneously, which is impossible. Magnetic field lines represent the direction of the magnetic field at any point in space, and at any given point, the direction of the magnetic field is unique. Therefore, magnetic field lines are drawn so that they never cross each other, ensuring that the direction of the field is clearly defined everywhere.

26. (ii) How is a uniform magnetic field in a given region represented? Draw a diagram in support of your answer.

Answer:

A uniform magnetic field in a given region is represented by parallel, equidistant lines. This indicates that the strength (magnitude) and the direction of the magnetic field are the same at all points in the region. Uniform magnetic fields are typically created between the poles of two large, flat magnets when they are close to each other.

Section – C

27. 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.

Answer:

(i) Change in colour

FeCl₃​(aq) + NaOH(aq) → Fe(OH)₃​(s) + NaCl(aq)

In this reaction, FeCl₃ reacts with sodium hydroxide (NaOH) to form iron(III) hydroxide (Fe(OH)₃​). The change in colour from transparent solutions of FeCl₃ and sodium hydroxide to a brown precipitate indicates the occurrence of the reaction.

(ii) Change in temperature (rise)

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

In this ₄reaction, methane (CH₄) reacts with oxygen gas (O₂) to form carbon dioxide (CO₂​) and water (H₂O). This reaction is highly exothermic, meaning it releases heat energy. As a result, there is a rise in temperature during the reaction.

(iii) Formation of precipitate

AgNO₃​(aq) + NaCl(aq) → AgCl(s) + NaNO₃​(aq)

In this reaction, silver nitrate (AgNO₃) reacts with sodium chloride (NaCl) to form silver chloride (AgCl), which is a white precipitate.

28. (i) The 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.

Answer:

Tomato juice with a pH of 4.6 is considered acidic. In terms of taste, the tomato juice with pH 4.6 taste tart or sour.

The reason for this is that acidic substances, with a low pH, tend to taste sour or tart. This is because the acidic pH stimulates taste receptors on the tongue that are sensitive to sourness.

28. (ii) How do we differentiate between a strong acid and a weak base in terms of ion-formation in aqueous solutions?

Answer:

Strong acids completely ionize in aqueous solutions to form ions. For example, hydrochloric acid (HCl), sulfuric acid (H₂SO₄​), and nitric acid (HNO₃​) are strong acids, whereas, Weak bases only partially ionize in aqueous solutions to form ions. For example, ammonia (NH₃​) is a weak base which partially ionize when dissolved in any aqueous solution.

28. (iii) The acid rain can make the survival of aquatic animals difficult. How ?

Answer:

Acid rain lowers the pH of water bodies, making them more acidic. This increased acidity can directly harm aquatic animals by damaging their gills, skin, and other tissues. Also, acidification of water bodies can lead to reduced oxygen levels, as acidic conditions can inhibit the ability of water to hold dissolved oxygen. This decrease in oxygen availability can suffocate aquatic organisms.

29. (i) Why is respiratory pigment needed in multicellular organisms with large body size?

Answer:

Respiratory pigments play a crucial role in facilitating oxygen transport in multicellular organisms. Multicellular organisms with large body sizes have a greater number of cells, which results in higher oxygen demands. Respiratory pigments increase the oxygen-carrying capacity of the circulatory system, allowing for efficient delivery of oxygen to tissues throughout the body.

(ii) Give reasons for the following:

(a) Rings of cartilage are present in the throat.

Answer:

Cartilage rings provide structural support to the trachea, preventing it from collapsing or narrowing during breathing. The rings of cartilage also help to maintain the flow of the air, ensuring air passage doesnot collapse.

(b) Lungs always contain a residual volume of air.

Answer:

The presence of a residual volume of air in the lungs is essential for maintaining efficient gas exchange which provides sufficient time for oxygen to be absorbed and carbon dioxide to be released.

(c) The diaphragm flattens and ribs are lifted up when we breathe in.

Answer:

When we breathe in, diaphragm flattens and ribs are lifted up to increase the space in chest cavity and it helps the lungs to expand into it. This helps the outside air to rush into the lungs

(d) Walls of alveoli contain an extensive network of blood vessels.

Answer:

The walls of the alveoli contain an extensive network of blood vessels, primarily capillaries because it is essential for efficient gas exchange between the air in the alveoli and the bloodstream.

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

Answer:

A reflex action is an involuntary and automatic response to a stimulus, typically to protect the body from harm or maintain homeostasis. These actions are rapid and do not involve conscious thought or decision-making by the brain. Instead, they are mediated by reflex arcs, which are neural pathways that bypass the brain and involve a direct connection between sensory neurons and motor neurons in the spinal cord.

Here’s a flowchart showing the path of a reflex action such as sneezing:

Stimulus (e.g., irritant in nasal passages) → Sensory Receptor (e.g., sensory nerve endings in nasal mucosa) → Sensory Neuron (afferent neuron) → Spinal Cord (integration center) → Interneuron → Motor Neuron (efferent neuron) → Effector Organ (e.g., muscles of the diaphragm, chest, and throat) → Response (e.g., forceful expulsion of air through the nose and mouth)

In the case of sneezing, the stimulus could be an irritant, such as dust, pollen, or other particles, that enters the nasal passages. Sensory receptors in the nasal mucosa detect the presence of the irritant and send signals via sensory neurons (afferent neurons) to the spinal cord. In the spinal cord, the sensory information is integrated, and a reflex response is generated via an interneuron. This response activates motor neurons (efferent neurons), which innervate the muscles involved in sneezing, including the muscles of the diaphragm, chest, and throat. Finally, the coordinated contraction of these muscles leads to the forceful expulsion of air through the nose and mouth, resulting in a sneeze.

31. Study the diagram given below and answer the questions that follow:

Convex Lens

(i) Name the defect of vision represented in the diagram. Give reason for your answer.

Hypermetropic eye is the defect of vision represented in the diagram because the light rays are focused behind the retina rather than directly on it.

(ii) List two causes of this defect.

Two common causes of hypermetropic eye are:

1. Shorter Eyeball Length: In many cases of hypermetropia, the eyeball is shorter than normal.
2. Flat Cornea: Another cause of hypermetropia can be a cornea that is flatter than normal.

(iii) With the help of a diagram show how this defect of vision is corrected

32. Name and state the rule to determine the direction of a:

(i) Magnetic field produced around a current carrying straight conductor.

The direction of the magnetic field produced around a current-carrying straight conductor can be determined using the right-hand grip rule (or the right-hand thumb rule). The rule states:

If you wrap your right hand around the conductor with your thumb pointing in the direction of the current (the direction of the flow of positive charge), the direction in which your fingers curl around the conductor represents the direction of the magnetic field lines around the conductor.

(ii) Force experienced by a current carrying straight conductor placed in a magnetic field which is perpendicular to it.

The direction of the force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it can be determined using Fleming’s left-hand rule. The rule states:

• Extend your left hand, keeping the thumb, forefinger, and middle finger perpendicular to each other.
• Point your forefinger in the direction of the magnetic field lines (B).
• Point your middle finger in the direction of the current (I) flowing through the conductor.
• Your thumb will then point in the direction of the force experienced by the conductor (F).

33. (A) Plants → Deer → 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.

Answer:

In the given food chain “Plants → Deer → Lion,” the second trophic level consists of the deer.

Impact on the First Trophic Level (Plants):

With the removal of all organisms from the second trophic level (deer), there would likely be an increase in the population of plants. This is because deer are herbivores that consume plants. As a result, plant biomass may increase, and certain plant species may dominate the ecosystem.

Impact on the Third Trophic Level (Lion):

With the removal of all organisms from the second trophic level (deer), the lion, which is a predator at the third trophic level, would experience a decrease in its food supply. This is because without deer to prey upon, lions would face food scarcity, leading to a decline in their population size.

In a food web, organisms typically have multiple prey species and are not solely dependent on one specific species for food. Therefore, if lions were part of a food web where they had alternative prey species besides deer, the impact of removing all organisms from the second trophic level (deer) would not be as severe. Lions could switch to feeding on other available prey species, thereby mitigating the effects of the removal of deer on their population size and survival.

OR

33. (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?

Answer:

The gas ‘X’ referred to in the question is ozone (O₃). Despite being a deadly poison at ground level, ozone performs an essential function in the atmosphere, particularly in the stratosphere, where it forms the ozone layer.

Chlorofluorocarbons (CFCs), halons, and other ozone-depleting substances (ODS) are primarily responsible for the decrease in the level of ozone in the stratosphere.

Measures taken by an international organization to check the depletion of the ozone layer: The Montreal Protocol on Substances that Deplete the Ozone Layer is an international treaty aimed at phasing out the production and use of ozone-depleting substances. It was adopted in 1987 and has been ratified by almost all countries worldwide.

SECTION- D

34. (A) (i) Define a homologous series of carbon compounds.

Answer:

A homologous series of carbon compounds is a family of organic compounds that have similar chemical properties and structural features but differ by a repeating unit of -CH₂-. Members of a homologous series share a common functional group and exhibit a gradual increase in molecular size and mass as the number of carbon atoms increases. Examples of homologous series include alkanes, alkenes, alkynes, alcohols, carboxylic acids, and ethers.

(ii) Why is the melting and boiling points of C₄H₈ higher than that of C₃H₆ Or C₂H₄?

Answer:

The melting and boiling points of C₄H₈ (butene) are higher than those of C₃H₆ (propene) or C₂H₄ (ethylene) due to the increase in molecular size and mass. As the number of carbon atoms in the molecule increases, the van der Waals forces between molecules also increase. These intermolecular forces play a significant role in determining the melting and boiling points of organic compounds.

(iii) Why do we NOT see any gradation in chemical properties of a homologous series compounds ?

Answer:

The chemical properties of compounds in a homologous series remain largely consistent throughout the series due to the presence of the same functional group. Although the molecular size and mass increase gradually from one member to the next, the functional group remains unchanged, leading to similar reactivity patterns. Therefore, there is no gradation in chemical properties because the functional group dictates the behavior of the compounds, regardless of their molecular size.

(iv) Write the name and structures of (i) aldehyde and ii) ketone with molecular form C₃H₆O.

Answer:

Aldehyde: Propanal (CH₃CH₂CHO)

Ketone: Propanone (also known as acetone) (CH₃COCH₃)

OR

34. (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’.

Answer:

The compound ‘X’ with two carbon atoms in its molecule and suffixed with ‘-ol’ is called ethanol.

Structure of ethanol:

(ii) What happens when ‘X’ is heated with excess concentrated sulphuric acid at 443 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.

Answer:

When ethanol is heated with excess concentrated sulphuric acid at 443 K, it undergoes dehydration to form ethene (ethylene) and water. This reaction is known as dehydration of alcohols.

Chemical equation

C₂H₅OH → C₂H₄ + H₂O

Conditions for the reaction:

• Excess concentrated sulphuric acid (H₂SO₄) is used.
• The reaction is carried out at 443 K.

Role of concentrated sulphuric acid in the reaction: Concentrated sulphuric acid acts as a dehydrating agent in the reaction. It abstracts the elements of water (H and OH) from ethanol molecules to facilitate the elimination of a water molecule, resulting in the formation of ethene. Sulphuric acid has a strong affinity for water and can remove water molecules from alcohols, enabling the dehydration reaction to proceed efficiently.

(iii) Name and draw the electron dot structure of hydrocarbon produced in the above reaction

The hydrocarbon produced in the above reaction is ethene (ethylene), which has the molecular formula C₂H₄.

Electron dot structure of ethene:

35. (A) (i) Name three techniques/devices used by human females to avoid pregnancy. Mention the side effects caused by each.

Answer:

Contraceptive Pills:

• Side effects:
  • Nausea and vomiting: Some women may experience nausea or vomiting shortly after taking contraceptive pills, especially during the initial months of use.
  • Breast tenderness: Contraceptive pills may cause breast tenderness or enlargement in some women.

Intrauterine Device (IUD):

• Side effects:
  • Cramping and discomfort: Insertion of the IUD may cause temporary cramping and discomfort, which usually subsides within a few days.
  • Irregular bleeding: Some women may experience irregular or heavier menstrual bleeding, especially during the initial months after IUD insertion.

Barrier Methods (e.g., Condoms):

• Side effects:
  • Allergic reactions: Some individuals may be allergic to latex condoms, leading to symptoms such as itching, redness, or irritation in the genital area.
  • Reduced sensation: Condom use may reduce sensation during sexual intercourse for some individuals.

(ii) What will happen if in a human female (a) fertilisation takes place, (b) an egg is not fertilised ?

Answer:

(a) Fertilisation takes place:

• If fertilisation occurs, the fertilised egg (zygote) implants into the lining of the uterus and begins to develop into an embryo. The embryo continues to grow and develop, ultimately leading to the formation of a fetus and pregnancy.

(b) An egg is not fertilised:

• If an egg is not fertilised during the menstrual cycle, it passes through the fallopian tube and into the uterus, where it disintegrates and is shed along with the uterine lining during menstruation. This process marks the end of the menstrual cycle and prepares the uterus for the next ovulation

OR

35. (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 ?

Answer:

The Diagram of spore formation of Rhizopus is given below:

• (a) Reproductive part: The sporangium (spore case) is the reproductive structure of Rhizopus. It produces asexual spores called sporangiospores.
• (b) Non-reproductive part: Rhizoids are root-like structures that anchor the fungus to the substrate (such as bread) and absorb nutrients.

Rhizopus does not multiply on a dry slice of bread because it requires moisture to grow and reproduce. Without sufficient moisture, the fungal spores cannot germinate and develop into new fungal colonies. Additionally, Rhizopus relies on organic matter for nutrients, and a dry slice of bread may not provide the necessary nutrients for fungal growth and reproduction. Therefore, the absence of moisture and adequate nutrients prevents Rhizopus from multiplying on a dry slice of bread.

(ii) Name and explain the process by which reproduction takes place in Hydra.

Answer:

The reproduction process in Hydra is called budding.

Explanation of the process:

• Budding is an asexual reproductive process in which a small outgrowth, called a bud, develops from the body of the parent organism.
• The bud grows and develops into a miniature replica of the parent organism, forming tentacles and other body structures.
• Eventually, the bud detaches from the parent organism and becomes an independent individual.
• Hydra can also reproduce sexually by producing gametes (sperm and eggs) through a process called gametogenesis. Fertilization occurs when sperm released by one individual fertilizes eggs released by another individual, leading to the formation of a zygote. The zygote develops into a new individual through a process of embryonic development.

36. (A) i) Define electric power. Express it in terms of potential difference (V) and resistance (R).

Answer:

Electric power is the rate at which electrical energy is transferred by an electric circuit. It is the product of the potential difference (voltage) across the circuit and the current flowing through it. Mathematically, electric power (P) can be expressed as:

P = V × I

where:

• P = electric power (in watts, W)
• V = potential difference or voltage (in volts, V)
• I= current (in amperes, A)

Alternatively, electric power can also be expressed in terms of resistance (R) using Ohm’s law:

P = I² × R

or

P = V² / R

where:

• R= resistance (in ohms, Ω)

(ii) An electric oven is designed to work on the mains voltage of 220 V. This oven consumes 11 units of electrical energy in 5 hours. Calculate :

Given:

• Mains voltage (potential difference), V =220 V
• Electrical energy consumed, E = 11 units (1 unit = 1 kWh = 1000 watt-hours)
• Time, t = 5 hours

(a) Power rating of the oven

Power(P) = Energy consumed (in watt-hours)​/ Time (in hours)

P = (11 units×1000Watt-hours/unit​)/ 5 hours

P = 2200W

(b) Current drawn by the oven

Using the formula, P = V × I, rearrange it to find the current:

I = P/ V

I = 2200/220

I = 10A

(c) Resistance of the oven when it is red hot

Since P = V²/R ​, we can rearrange it to find the resistance:

R = V²/P

R =(220 V)²/2200 W

R = 48400 V²/2200 W

R = 22Ω

OR

(B) (i) Write the relation between resistance R and electrical resistivity ρ 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.

Answer:

The resistance R of a conductor with length l, cross-sectional area A, and resistivity ρ can be expressed as:

R=ρ⋅l/ A​

From this relation, we can derive the SI unit of electrical resistivity.

Derivation of the SI unit of electrical resistivity:

R=ρ⋅l/ A​

ρ = R⋅A/l

The SI unit of resistance (R) is ohms (Ω), length (l) is measured in meters (m), and area (A) is measured in square meters (m²). Therefore, the SI unit of electrical resistivity (ρ) is: ohm-meter

(ii) The resistance of a metal wire of length 3 m is 60Ω . If the area of cross-section of the wire is 4×10^-7 m², calculate the electrical resistivity of the wire.

Answer:

Given:

• Length of wire (l) = 3 m
• Resistance of wire (R) = 60 Ω
• Area of cross-section of wire (A) = 4×10⁻⁷m²

Using the formula:

ρ = R⋅A/l

ρ = 60Ω × (4×10⁻⁷m²)/3m

ρ = 8 × 10⁻⁷ Ωm

(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.

Answer:

If the wire is stretched so that its length is doubled, its resistivity (ρ) will not change. This is because electrical resistivity (ρ) is an intrinsic property of the material and is independent of the shape or size of the conductor. Stretching the wire only changes its length and does not alter the material’s resistivity. Therefore, doubling the length of the wire will not affect its electrical resistivity.

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:

37. 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 ?

Answer:

• Cathode: The cathode in the refining of copper is typically made of pure copper metal. As the impure copper dissolves at the anode and pure copper ions migrate towards the cathode, they are deposited onto the pure copper cathode, gradually building up a layer of pure copper.
• Anode: The anode in the refining of copper is made of impure copper. As electric current passes through the electrolytic cell, the impure copper at the anode dissolves into the electrolyte solution, releasing copper ions into the solution.

(ii) Name the solution used in the above process and write its formula.

Answer:

The solution used in the electrolytic refining of copper is typically an aqueous solution of copper sulfate (CuSO₄​)

(iii) (A) How copper gets refined when electric current is passed in the electrolytic cell?

Answer:

In the electrolytic cell used for refining copper, the impure copper anode and a pure copper cathode are immersed in an electrolyte solution of copper sulfate (CuSO₄​).

1. When electric current is passed through the cell, copper ions (Cu²⁺) from the copper sulfate solution migrate towards the cathode.
2. At the cathode (pure copper electrode), copper ions gain electrons and are reduced to form solid copper metal. This deposition of pure copper on the cathode gradually increases its mass.
3. At the anode (impure copper electrode), the impure copper metal dissolves into the electrolyte solution as copper ions (Cu²⁺). This process replenishes the copper ions in the solution and maintains the concentration of copper sulfate.
4. As a result of the electrolysis process, the impurities present in the impure copper anode settle down as anode mud or sludge at the bottom of the electrolytic cell, leaving behind pure copper at the cathode.
5. The pure copper deposited on the cathode is periodically removed, washed, and dried to obtain high-purity copper metal. This electrolytic refining process helps to purify the copper obtained from various reduction processes, ensuring its high quality and purity for industrial applications.

OR

(iii) (B) You have two beakers ‘A’ and ‘B’ containing copper sulphate solution. What would you observe after about 2 hours if you dip a strip of zine in beaker “A’ and a strip of silver in beaker ‘B”? Give reason for your observations in each case.

Answer:

If you dip a strip of zinc in beaker ‘A’ containing copper sulfate solution and a strip of silver in beaker ‘B’ containing copper sulfate solution, the following observations can be made:

Beaker A (Zinc strip):

Reason: This reddish-brown coating is due to the displacement reaction between zinc and copper ions present in the copper sulfate solution. Zinc is more reactive than copper, so it displaces copper from copper sulfate solution, forming solid copper metal. The reaction can be represented as:

Zn (s) + CuSO₄(aq) → ZnSO₄(aq) + Cu (s)

After about 2 hours, you would observe that the zinc strip appears to have a reddish-brown coating on its surface.

The solid copper formed deposits onto the surface of the zinc strip, giving it a reddish-brown appearance.

Beaker B (Silver strip):

Reason: Silver is less reactive than copper, so it cannot displace copper from copper sulfate solution. Therefore, no chemical reaction occurs between silver and copper sulfate solution, and the silver strip remains unchanged.

After about 2 hours, you would observe that there is no visible change in the appearance of the silver strip.

38. 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 F₁ progeny, which two observations were made by him in F₁ plants?

Answer:

When Mendel crossed pea plants with pure tall and pure short characteristics to produce F1 progeny, two observations were made:

1. All the F₁ progeny plants were tall.
2. The trait of shortness disappeared in the F1 generation, indicating that the tall trait was dominant over the short trait.

(ii) Write one difference between dominant and recessive trait.

Answer:

The difference between dominant and recessive trait is:

• Dominant trait: A dominant trait is expressed phenotypically when at least one copy of the dominant allele is present in the genotype. It masks the expression of the recessive allele in heterozygous individuals.
• Recessive trait: A recessive trait is expressed phenotypically only when two copies of the recessive allele are present in the genotype. It is masked by the dominant allele in heterozygous individuals.

(iii) (A) In a cross with two pairs of contrasting characters

RRYY X rryy

(round Yellow) (wrinkled green)

Mendel observed 4 types of combinations in F₂ generation. By which method did he obtain F₂ generation? Write the ratio of the parental combinations obtained and what conclusions were drawn from this experiment.

Answer:

Mendel obtained the F₂ generation by allowing the F₁ plants resulting from the cross between the parental plants with two pairs of contrasting characters to self-pollinate.

Ratio of the parental combinations obtained: In the F₂ generation, Mendel observed four types of combinations:

1. Round yellow (RRYY)
2. Round green (RRYy)
3. Wrinkled yellow (rrYY)
4. Wrinkled green (rrYy)

OR

(iii) (B) Justify the statement :

“It is possible that a trait is inherited but may not be expressed.”

Answer:

The statement “It is possible that a trait is inherited but may not be expressed” can be justified by the concept of dominant and recessive alleles and the phenomenon of incomplete dominance and codominance in genetics.

1. Recessive Alleles: In cases where a trait is controlled by a recessive allele, it may not be expressed phenotypically if the individual possesses at least one dominant allele for that trait. Recessive alleles are only expressed phenotypically when the individual is homozygous for the recessive allele. In heterozygous individuals, the dominant allele masks the expression of the recessive allele, resulting in the trait not being expressed.
2. Incomplete Dominance: In incomplete dominance, neither allele is completely dominant over the other. Instead, the heterozygous phenotype is an intermediate blend of the phenotypes associated with the two homozygous genotypes. In such cases, the trait may not be fully expressed in heterozygous individuals, resulting in a phenotype that differs from both the homozygous dominant and homozygous recessive phenotypes.
3. Codominance: In codominance, both alleles contribute equally to the phenotype of the heterozygous individual. As a result, both alleles are expressed fully, without one dominating over the other. However, if the trait associated with one of the alleles is not observable or detectable, it may seem as though the trait is not expressed, even though it is inherited.
4. Environmental Factors: In some cases, the expression of a trait may be influenced by environmental factors. Even though the trait is inherited genetically, its expression may be influenced or modified by external environmental conditions, leading to variability in the phenotypic expression of the trait.

39. 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:

Case	Mirror	Focal Length(cm)	Object Distance(cm)
1	A	20	45
2	B	15	30
3	C	30	20

(i) In which one of the above cases the mirror will form a diminished image of the object ? Justify your answer.

Answer:

For a concave mirror to form a diminished image of the object, the object distance must be greater than the focal length of the mirror. This situation occurs when the object is placed beyond the focal point of the mirror.

From the given data:

• Case 1: Mirror A has a focal length of 20 cm and an object distance of 45 cm. Object distance > Focal length.
• Case 2: Mirror B has a focal length of 15 cm and an object distance of 30 cm. Object distance > Focal length.
• Case 3: Mirror C has a focal length of 30 cm and an object distance of 20 cm. Object distance < Focal length.

Therefore, in Case 2 (Mirror B), the mirror will form a diminished image of the object because the object distance is greater than the focal length of the mirror.

(ii) List two properties of the image formed in case 2.

Answer:

Two properties of the image formed in Case 2 (Mirror B):

1. Real: Since the object distance is greater than the focal length, the image formed by Mirror B will be real.
2. Inverted: Concave mirrors always produce inverted images, irrespective of the position of the object. Therefore, the image formed by Mirror B will be inverted

(iii) (A) What is the nature and size of the image formed by mirror C ? Draw ray diagram to justify your answer.

Answer:

Nature: The nature of the image formed by Mirror C can be determined using the mirror formula:

1/f = 1/u + 1/v

Where:

• f = focal length of the mirror (30 cm)
• v = image distance
• u = object distance (-20 cm)

Given that the object distance u is negative (since it is measured on the same side as the incident light), the image distance v will also be negative, indicating that the image is formed on the same side as the object. This implies that the image formed is a virtual image.

Size: Since the object distance u is less than the focal length f, the image formed will be magnified. The magnification can be calculated using the magnification formula: Magnification (m) = −Magnification (m) = u − v​

OR

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

Answer:

To find the position of the image formed by a concave mirror when an object is placed at a distance of 18 cm from the pole and the focal length of the mirror is 12 cm, we can use the mirror formula:

1/f = 1/u + 1/v

Given:

f = −12 cm (negative because it’s a concave mirror)

u = −18 cm (negative because the object is placed on the same side as the incident light)

Substituting the given values into the mirror formula:

1/−12 =1/v + 1/−18

Solving for v, we get :

1/v = -1/36

v = −36

Hence, the position of the image formed by the concave mirror when the object is placed at a distance of 18 cm from the pole is 36 cm on the same side as the object. Therefore, the image is virtual and magnified.

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