CBSE Class 10 Science Question Paper 2024 PDF (Set 1 - 31/1/1) is available for download here. CBSE conducted the Science exam on March 2, 2024, from 10:30 AM to 1:30 PM. The total marks for the theory paper are 80. The question paper contains 20% MCQ-based questions, 40% competency-based questions, and 40% short and long answer type questions.

CBSE Class 10 Science Question Paper 2024 (Set 1 - 31/1/1) with Solution PDF

CBSE Class 10 Science Question Paper 2024 (Set 1 - 31/1/1) with Answer Key download iconDownload Check Solutions

CBSE Science Question Paper (Set 1 – 31/1/1) 2024 Solution

Question  Answer Detailed Solution
1. When 2 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:
(A) NaOH + Zn → Na2ZnO2 + H2O
(B) 2NaOH + Zn → Na2ZnO2 + H2
(C) 2NaOH + Zn → Na2ZnO2 + H2
(D) 2NaOH + Zn → Na2ZnO2 + H2O		 (B) 2NaOH + Zn → Na2ZnO2 + H2 Sodium hydroxide reacts with zinc to form sodium zincate (Na2ZnO2) and hydrogen gas (H2). The balanced reaction is:
2NaOH + Zn → Na2ZnO2 + H2
This reaction demonstrates the amphoteric nature of zinc, which reacts with alkali to release hydrogen gas.
2. Select from the following a decomposition reaction in which the source of energy for decomposition is light:
(A) 2FeSO4 → Fe2O3 + SO2 + SO3
(B) 2H2O → 2H2 + O2
(C) 2AgBr → 2Ag + Br2
(D) CaCO3 → CaO + CO2		 (C) 2AgBr → 2Ag + Br2 Photolysis refers to a decomposition reaction driven by light energy. Silver bromide (AgBr) decomposes under light to form silver (Ag) and bromine (Br2):
2AgBr → 2Ag + Br2
This reaction is extensively used in photographic films, where light exposure causes the breakdown of silver halides.
3. A metal and a non-metal that exist in liquid state at room temperature are, respectively:
(A) Bromine and Mercury
(B) Mercury and Iodine
(C) Mercury and Bromine
(D) Iodine and Mercury		 (C) Mercury and Bromine Mercury is the only metal that exists in a liquid state at room temperature. Bromine is the only non-metal in liquid state at room temperature. Hence, the correct combination is 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)		 (C) (ii) and (iv) Carbon compounds are generally poor conductors of electricity because they lack free electrons or ions. Additionally, these compounds exhibit weak intermolecular forces, leading to lower boiling and melting points.
 
5. Consider the following compounds: FeSO4, CuSO4, CaSO4, Na2CO3.
The compound having the maximum number of water of crystallization in its crystalline form in one molecule is:
(A) FeSO4
(B) CuSO4
(C) CaSO4
(D) Na2CO3		 (D) Na2CO3 The compound having the maximum number of water of crystallization in its crystalline form in one molecule is Na2CO3.
6. Oxides of aluminium and zinc are:
(A) Acidic
(B) Basic
(C) Amphoteric
(D) Neutral		 (C) Amphoteric Amphoteric oxides react with both acids and bases to form salts. Aluminium oxide (Al2O3) and zinc oxide (ZnO) exhibit this dual behavior.
7. The reaction given below is a redox reaction because:
MnO2 + 4HCl → MnCl2 + 2H2O + Cl2

(A) MnO2 is oxidized and HCl is reduced.
(B) HCl is oxidized.
(C) MnO2 is reduced.
(D) MnO2 is reduced and HCl is oxidized.		 (D) MnO2 is reduced and HCl is oxidized. In the reaction, MnO2 is reduced as the oxidation state of Mn decreases from +4 in MnO2 to +2 in MnCl2. Concurrently, HCl is oxidized as the oxidation state of Cl increases from -1 in HCl to 0 in Cl2.
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)		 (B) (ii) and (iv) Statement (ii): The sex of a child is determined by the sperm from the father, which carries either an X or Y chromosome. If the sperm carries an X chromosome, the child will be female (XX), and if it carries a Y chromosome, the child will be male (XY).

Statement (iv): The sex of the child is indeed determined at the time of fertilization when the male (sperm) and female (egg) gametes fuse to form a zygote.

Explanation: - Statement (i) is incorrect because the mother always contributes an X chromosome, but the father's sperm determines the sex. - Statement (iii) is generally incorrect as the probability of having male or female children is approximately equal, with a slight natural variation.
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)
(C) (i), (ii), and (iv)
(D) (i) and (iv)		 (D) (i) and (iv) Statement (i): Chromosomes carry hereditary information from parents to the next generation.

Statement (iv): Chromosomes are actively involved in the process of cell division.

Explanation:
  • Statement (ii) is partially correct as chromosomes are thread-like structures inside the nucleus, but specifying "animal cell" excludes plant cells, making it not entirely accurate.
  • Statement (iii) is incorrect because human reproductive cells (gametes) are haploid and contain a single set of chromosomes, not pairs.
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		 (C) Neuromuscular junction The neuromuscular junction is the specific site where an electrical impulse (action potential) in a nerve cell is converted into a chemical signal. When the electrical impulse reaches the end of the motor neuron (axon terminal), it triggers the release of neurotransmitters (chemical signals) into the synaptic cleft. These neurotransmitters then bind to receptors on the muscle cell membrane, initiating muscle contraction.

Explanation: - Axon is responsible for transmitting electrical impulses away from the cell body. - Dendrites receive electrical signals from other neurons. - Cell body contains the nucleus and organelles of the neuron. - Only the Neuromuscular junction facilitates the conversion of electrical impulses into chemical signals necessary for muscle action.
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 statements are:
(A) (i) only
(B) (i) and (iii)
(C) (ii) and (iii)
(D) (ii) and (iv)		 (C) (ii) and (iii) Statement (ii): A stomata closes when it does not need carbon dioxide for photosynthesis, reducing water loss through transpiration.

Statement (iii): Water flows out of the guard cells, causing them to lose turgidity, which results in the stomata closing.

Explanation: - Statement (i) is incorrect because stomata open to allow carbon dioxide intake when needed for photosynthesis. - Statement (iv) is incorrect because water flowing into the guard cells causes stomata to open, not close.
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		 (C) At twice the focal length of the lens When an object is placed at twice the focal length (2F) of a convex lens, the image formed is real, inverted, and of the same size as the object.

Explanation: - Placing the object at 2F ensures that the image is also formed at 2F on the other side of the lens. - This positioning results in the image being the same size as the object, adhering to the principles of geometric optics.
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		 (D) Retina The retina is the light-sensitive screen at the back of the eye where the lens system focuses the image. It contains photoreceptors that detect light and send signals to the brain.

Explanation: - Cornea is the transparent front layer of the eye that covers the iris and pupil. - Ciliary muscles control the shape of the lens for focusing. - Optic nerves transmit visual information from the retina to the brain. - Only the Retina serves as the light-sensitive screen where images are formed.
14. The pattern of the magnetic field produced inside a current-carrying solenoid is:
(A) Option A
(B) Option B
(C) Option C
(D) Option D		 (A) Option A The magnetic field inside a current-carrying solenoid is uniform and straight, represented by field lines running parallel to each other along the length of the solenoid. This pattern aligns with diagram (A), showing uniform field lines within the solenoid.
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		 (C) Tiger, grass, snake, frog Explanation:
In option (C), the food chain is Tiger → Grass → Snake → Frog. The second trophic level, which should consist of primary consumers (herbivores that eat producers like grass), is missing. Typically, a grass-eating herbivore (e.g., goat or grasshopper) would occupy the second trophic level before the snake or tiger.
16. In which of the following organisms, multiple fission is a means of asexual reproduction?
(A) Yeast
(B) Leishmania
(C) Paramecium
(D) Plasmodium		 (D) Plasmodium Multiple fission occurs when a single organism divides into multiple offspring simultaneously. This process is commonly observed in Plasmodium, a protozoan parasite.

 
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.

The correct explanation is:
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.		 (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). Explanation:
Assertion (A) is true because when zinc reacts with nitric acid, hydrogen gas that might be produced is immediately oxidised by nitric acid, preventing its evolution.

Reason (R) is true as nitric acid acts as an oxidizing agent, oxidizing any hydrogen gas produced to water and itself gets reduced in the process.

Since Reason (R) correctly explains Assertion (A), the correct option is (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.

The correct explanation is:
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.		 (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). Explanation:
Assertion (A) is true because harmful chemicals like pesticides and pollutants tend to accumulate in organisms at higher trophic levels through the process of biomagnification.

Reason (R) is also true as harmful chemicals are indeed sprayed on crops to protect them from diseases and pests.

However, while Reason (R) explains a source of harmful chemicals, it does not directly explain why their accumulation is maximum at the highest trophic level. The primary reason for accumulation at higher trophic levels is biomagnification, not just the application of chemicals on crops.
 

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.

The correct explanation is:
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

 (C) Assertion (A) is true, but Reason (R) is false. Explanation:
Assertion (A) is true because aquatic organisms often have a faster rate of breathing to compensate for the lower availability of oxygen dissolved in water compared to air.

Reason (R) is false because the amount of oxygen dissolved in water is actually much lower than the amount of oxygen available in air. Water can hold only a limited amount of dissolved oxygen, which is why aquatic organisms need to breathe faster to meet their oxygen requirements.
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.

The correct explanation is:
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.		 (C) Assertion (A) is true, but Reason (R) is false. Explanation:
Assertion (A) is true because a rainbow is indeed a natural spectrum of sunlight that appears in the sky.

Reason (R) is false because rainbows are typically formed when the sun is at a low angle in the sky (morning or late afternoon) and not when it is overhead. This positioning allows sunlight to enter water droplets at an angle, undergo refraction, dispersion, and reflection, and then exit at an angle that makes the spectrum visible to the observer.
21. Name the type of chemical reaction in which calcium oxide reacts with water.
Justify your answer by giving a balanced chemical equation for the chemical reaction.		 Exothermic combination reaction The type of chemical reaction is an exothermic combination reaction. When calcium oxide (quicklime) reacts with water, calcium hydroxide (slaked lime) is formed, and heat is released:

CaO + H2O → Ca(OH)2 + Heat.
22. State one role of each of the following in the human digestive system:
(i) Hydrochloric acid
(ii) Villi
(iii) Anal sphincter
(iv) Lipase		 (i) Hydrochloric acid: Creates an acidic medium in the stomach, activating pepsin for protein digestion.
(ii) Villi: Increase the surface area in the small intestine for nutrient absorption.
(iii) Anal sphincter: Controls the expulsion of undigested food and waste during defecation.
(iv) Lipase: Breaks down fats into fatty acids and glycerol.		 Explanation:
- Hydrochloric acid creates an acidic environment in the stomach, which is necessary for activating the enzyme pepsin that digests proteins.
- Villi are small, finger-like projections in the small intestine that significantly increase the surface area, facilitating the efficient absorption of digested nutrients.
- Anal sphincter muscles control the passage of waste material out of the body during defecation, preventing the involuntary loss of feces.
- Lipase is an enzyme that catalyzes the breakdown of lipids (fats) into fatty acids and glycerol, aiding in fat digestion.
 
23(A). How is the movement of leaves of a sensitive plant different from the downward movement of the roots? (A) Nastic movement vs. Tropic movement (Geotropism) The movement of leaves in a sensitive plant is a nastic movement caused by changes in turgor pressure and is independent of the direction of the stimulus. In contrast, the downward movement of roots is a tropic movement (specifically geotropism), which occurs in response to gravity.
23(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? (B) Thyroxine; Thyroid gland The hormone is Thyroxine, which is secreted by the thyroid gland. Iodine is important in our diet because it is essential for the synthesis of thyroxine. A deficiency in iodine leads to goitre, which is the swelling of the thyroid gland.
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. 6 cm behind the mirror To determine the position of the image formed by a convex mirror, we use the mirror formula:
1/f = 1/v + 1/u
Given:
f = +15 cm (Positive for convex mirrors)
u = -10 cm (Object distance is taken as negative in mirror conventions)
Calculations:
1/15 = 1/v + 1/(-10)
Simplifying the equation:
1/v = 1/15 + 1/10 = (2 + 3)/30 = 5/30 = 1/6
Therefore, v = +6 cm
The positive sign of the image distance indicates that the image is formed behind the mirror, which is characteristic of virtual images produced by convex mirrors.
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. Connect two resistors in parallel and one in series To achieve a total resistance of 9 Ω, connect two resistors in parallel, and then connect the third resistor in series with this parallel combination.

Calculations:
1. Parallel combination of two resistors:
\[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} \] \[ R_p = \frac{6}{2} = 3 \, \Omega \]
2. Series combination with the third resistor:
\[ R_{\text{total}} = R_p + R_3 = 3 + 6 = 9 \, \Omega \]
25. (B) In the given circuit, calculate the power consumed in watts in the resistor of 2 Ω: 8 W The circuit has a 6V source with two resistors R1 = 1 Ω and R2 = 2 Ω connected in series.

Calculate the equivalent resistance:
Req = R1 + R2 = 1 Ω + 2 Ω = 3 Ω.

Calculate the current using Ohm’s Law:
I = V / Req = 6 V / 3 Ω = 2 A.

Calculate the electric power:
Power consumed in the R2 = 2 Ω resistor is:
P = I2 × R = (2 A)2 × 2 Ω = 4 × 2 W = 8 W.
The power consumed in the 2 Ω resistor is 8 watt.
26.(i) Two magnetic field lines do not intersect each other. Why? Because magnetic field lines never intersect as it would imply two different directions of the magnetic field at the intersection point, which is impossible. Magnetic field lines never intersect because at the point of intersection, the magnetic field would have two different directions, which is not possible.
26.(ii) How is a uniform magnetic field in a given region represented? Draw a diagram in support of your answer. By parallel and equally spaced straight lines. A uniform magnetic field is represented by parallel and equally spaced straight lines. This indicates that the field strength and direction are the same at every point.
27.(i) Write one chemical equation for the chemical reaction in which a change in colour has taken place. CuCO3 → CuO + CO2 A chemical reaction that causes a colour change is the heating of copper carbonate:

CuCO3 → CuO + CO2

Observation: Copper carbonate (green) changes to copper oxide (black).
27.(ii) Write one chemical equation for the chemical reaction in which a change in temperature has taken place. CaO + H2O → Ca(OH)2 + Heat An example of a chemical reaction that causes a change in temperature is the exothermic reaction of quicklime with water:

CaO + H2O → Ca(OH)2 + Heat

Observation: The reaction releases heat, causing a rise in temperature.
27.(iii) Write one chemical equation for the chemical reaction in which the formation of precipitate has taken place. Na2SO4 + BaCl2 → BaSO4↓ + 2NaCl An example of a chemical reaction that results in the formation of a precipitate is the reaction between sodium sulfate and barium chloride:

Na2SO4 + BaCl2 → BaSO4↓ + 2NaCl

Observation: A white precipitate of barium sulfate (BaSO4) is formed.
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. Slightly sour The taste of tomato juice will be slightly sour. The pH value of 4.6 indicates that tomato juice is acidic, as substances with a pH less than 7 are acids. Acids are typically sour in taste.
28.(ii) How do we differentiate between a strong acid and a weak base in terms of ion-formation in aqueous solutions? Strong acids completely ionize in water, releasing a large number of H⁺ ions, whereas weak bases partially ionize in water, releasing fewer OH⁻ ions. A strong acid ionizes completely in water, releasing a large number of H⁺ ions.
Example:
HCl → H⁺ + Cl⁻.

 A weak base partially ionizes in water, releasing fewer OH⁻ ions.
Example:
NH₄OH ⇌ NH₄⁺ + OH⁻.
28.(iii) The acid rain can make the survival of aquatic animals difficult. How? It lowers the pH of water bodies, damaging the gills of fish and reducing essential minerals by dissolving toxic metals like aluminum. Acid rain lowers the pH of water bodies, making the water acidic. This affects aquatic life because:
It damages the gills of fish and other aquatic organisms.
It reduces the availability of essential minerals by dissolving toxic metals like aluminum.
29.(i) Why is respiratory pigment needed in multicellular organisms with large body size? To transport oxygen efficiently Respiratory pigments like hemoglobin are needed to transport oxygen efficiently in multicellular organisms with large body sizes because:

- Simple diffusion is insufficient to meet their oxygen requirements.
- Respiratory pigments bind to oxygen, ensuring its transport to all parts of the body.
29.(ii)(a) Give reasons for the following:
Rings of cartilage are present in the throat.		 They keep the airway open and prevent its collapse during breathing. The rings of cartilage in the throat (trachea) keep the airway open and prevent its collapse during breathing.
29.(ii)(b) Give reasons for the following:
Lungs always contain a residual volume of air.		 It prevents the lungs from collapsing and allows continuous exchange of gases even during exhalation. The residual volume of air in the lungs prevents them from collapsing and allows continuous exchange of gases even during exhalation.
29.(ii)(c) Give reasons for the following:
The diaphragm flattens and ribs are lifted up when we breathe in.		 It increases the chest cavity’s volume and reduces lung pressure, causing air to flow in. During inhalation:

- The diaphragm contracts and flattens, increasing the chest cavity’s volume.
- The ribs lift up and outward, reducing air pressure in the lungs, causing air to flow in.
29.(ii)(d) Give reasons for the following:
Walls of alveoli contain an extensive network of blood vessels.		 They ensure efficient exchange of oxygen and carbon dioxide between the lungs and blood. The extensive network of blood vessels in alveoli ensures efficient exchange of oxygen and carbon dioxide between the lungs and blood.
30.Define reflex action. With the help of a flow chart, show the path of a reflex action such as sneezing. An automatic, involuntary response to a stimulus that does not involve the conscious part of the brain. A reflex action is an automatic, involuntary response to a stimulus that does not involve the conscious part of the brain. Reflex actions are designed to protect the body from harm and enable quick responses. For example, sneezing helps expel irritants from the nasal passages.
31.(i) Name the defect of vision represented in the diagram. Give reason for your answer. Hypermetropia (Far-sightedness) The defect of vision shown is Hypermetropia (Far-sightedness). This occurs because:
The image is formed behind the retina.
The near point for the person is farther away than the normal near point (25 cm), making it difficult to see nearby objects clearly.
31.(ii) List two causes of Hypermetropia (Far-sightedness). The focal length of the eye lens is too long.
The eyeball has become too small, causing the image to form behind the retina.		 The two causes of Hypermetropia (Far-sightedness) are:
The focal length of the eye lens is too long.
The eyeball has become too small, causing the image to form behind the retina.
31.(iii) With the help of a diagram, show how Hypermetropia (Far-sightedness) is corrected. Using a convex lens to converge incoming light rays before they enter the eye Hypermetropia is corrected using a convex lens. The convex lens converges incoming light rays before they enter the eye, so the image is formed on the retina.
Explanation:
- N: Near point of a hypermetropic eye.
- N′: Near point of a normal eye.
The convex lens brings the light rays together to form a clear image at the retina, thus correcting hypermetropia.
32.(i) Name and state the rule to determine the direction of the magnetic field produced around a current-carrying straight conductor. Right-Hand Thumb Rule The rule is the Right-Hand Thumb Rule.
According to this rule: If the thumb of the right hand points in the direction of the current, the curled fingers show the direction of the magnetic field around the conductor.
32.(ii) Force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it. Name and state the rule to determine its direction. Fleming’s Left-Hand Rule The rule is Fleming’s Left-Hand Rule.
According to this rule: Stretch the forefinger, middle finger, and thumb of the left hand mutually perpendicular to each other.
Forefinger: Direction of magnetic field.
Middle finger: Direction of current.
Thumb: Direction of force experienced by the conductor.
33. (A) Plants – Deer – Lion
In the given food chain, what will be the impact of removing all the organisms of the second trophic level on the first and third trophic levels? Will the impact be the same for the organisms of the third trophic level in a food web? Justify.		 On the first trophic level (plants): Population increases.
On the third trophic level (lion): Population declines.
In a food web: Impact on the third trophic level may be less severe.		 Removing the second trophic level (deer) reduces grazing pressure on plants, causing plant populations to increase.
The absence of deer leads to a decline in the lion population due to lack of food.
In a food web, lions may have alternative prey, so the impact on their population may not be as severe as in a simple food chain.
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? Gas: Ozone
Function: Absorbs harmful ultraviolet (UV) radiation
Chemical causing depletion: Chlorofluorocarbons (CFCs)
Measures taken: Montreal Protocol		 Gas ‘X’: Ozone.
Function: Ozone protects life on Earth by absorbing harmful ultraviolet (UV) radiation.
Chemical causing depletion: Chlorofluorocarbons (CFCs).
Measures taken: The Montreal Protocol was adopted to phase out the production and use of ozone-depleting substances like CFCs.
34.(A)(i) Define a homologous series of carbon compounds. A group of organic compounds with the same general formula, similar chemical properties, and successive members differing by a -CH₂ group. A homologous series is a group of organic compounds that share the same general formula and exhibit similar chemical properties. Each successive member of the series differs by a -CH₂- group, which adds to the molecular structure without altering the functional group.
Alkanes: CₙH₂ₙ₊₂ (e.g., Methane, Ethane, Propane)
Alkenes: CₙH₂ₙ (e.g., Ethene, Propene)
Alcohols: CₙH₂ₙ₊₁OH (e.g., Methanol, Ethanol)
34.(A)(ii) Why do we NOT see any gradation in chemical properties of homologous series compounds? Because the chemical properties are determined by the functional group, which remains constant in a homologous series. The chemical properties of compounds in a homologous series do not show gradation because they all contain the same functional group. This functional group dictates the chemical reactivity and behavior of the compounds, ensuring that despite differences in molecular size or the number of -CH₂- groups, their chemical properties remain similar.
34.(A)(iii) Write the name and structures of
(i) aldehyde and
(ii) ketone with molecular formula C₃H₆O.		 (i) Aldehyde: Propanal
(ii) Ketone: Propanone (Acetone)		 (i) Aldehyde: Propanal
Structure: CH₃-CH₂-CHO
(ii) Ketone: Propanone (commonly known as Acetone)
Structure: CH₃-CO-CH₃
Both propanal and propanone have the same molecular formula (C₃H₆O) but differ in the placement of the carbonyl group, which defines their functional groups and properties.
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’. Ethanol The compound is ethanol.
Structure: CH3-CH2-OH
34. (B)(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. Ethanol undergoes dehydration to form ethene and water. When ethanol (CH3-CH2-OH) is heated with concentrated H2SO4 at 443 K, it undergoes dehydration to form ethene (C2H4) and water.
Reaction:
CH3-CH2-OH → C2H4 + H2O
Conditions:
Temperature: 443 K
Reagents: Excess concentrated H2SO4
Role of H2SO4: Acts as a dehydrating agent by removing water from ethanol, facilitating the formation of ethene.
34. (B)(iii) Name and draw the electron dot structure of the hydrocarbon produced in the above reaction. Ethene (C2H4) The hydrocarbon produced is ethene (C2H4).
Electron Dot Structure:
H2C=CH2
35. (A)(i) Name three techniques/devices used by human females to avoid pregnancy. Mention the side effects caused by each. Oral Contraceptive Pills
Intrauterine Devices (IUDs)
Barrier Methods (e.g., condoms)		 Oral Contraceptive Pills: Prevent ovulation.
Side Effects: Nausea, weight gain, mood swings.
Intrauterine Devices (IUDs): Prevent implantation of the embryo.
Side Effects: Cramps, irregular bleeding.
Barrier Methods (e.g., condoms): Prevent sperm from reaching the egg.
Side Effects: None significant; rare latex allergies.
35. (A)(ii)(a) What will happen if in a human female fertilisation takes place? The zygote is formed, implants in the uterus, and develops into an embryo. If fertilisation takes place in a human female, the sperm and egg unite to form a zygote. This zygote then implants itself into the lining of the uterus, where it begins to develop into an embryo.
35. (A)(ii)(b) What will happen if in a human female an egg is not fertilised? The egg disintegrates, and the uterine lining sheds during menstruation. If an egg is not fertilised in a human female, it disintegrates. Consequently, the uterine lining, which was prepared to support the potential embryo, sheds off during menstruation.
35. (B)(i) Draw a diagram showing spore formation in Rhizopus and label the:
(a) Reproductive parts,
(b) Non-reproductive parts.

Why does Rhizopus not multiply on a dry slice of bread?		 Rhizopus does not multiply on a dry slice of bread because it requires moisture for spore germination and growth. Spore Formation in Rhizopus:
Rhizopus reproduces asexually through the formation of sporangia, which contain spores.
(a) Reproductive Parts: Sporangium and spores.
(b) Non-reproductive Parts: Mycelium (the vegetative part consisting of hyphae).
Reason for Lack of Multiplication on Dry Bread:
Rhizopus requires moisture for spore germination and hyphal growth. A dry slice of bread lacks the necessary moisture, inhibiting the spores from germinating and the mycelium from growing, thereby preventing multiplication.
35. (B)(ii) Name and explain the process by which reproduction takes place in Hydra. Process: Budding
Explanation: Asexual reproduction

Process: Reproduction in Hydra occurs through budding.
A small outgrowth, known as a bud, develops on the parent Hydra.
This bud grows by elongating and differentiating into various body parts.
Once fully developed, the bud detaches from the parent organism to become an independent individual.
Type of Reproduction: Budding is an asexual reproduction method, allowing Hydra to reproduce rapidly without the need for gametes or fertilization.
36. (A)(i) Define electric power. Express it in terms of potential difference (V) and resistance (R). Electric power is the rate at which electrical energy is consumed or converted into another form of energy.  Electric power is defined as the rate at which electrical energy is consumed or transformed into another form of energy, such as heat or light. The relationship between power, potential difference, and resistance can be derived using Ohm’s Law 
Therefore, electric power ( P ) can be expressed in terms of potential difference ( V ) and resistance ( R ) as:
P = V²/R
36. (A)(ii)(a) 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 the power rating of the oven. 2200 W Voltage (V) = 220 V
Energy consumed = 11 units (1 unit = 1 kWh) = 11 × 1000 = 11000 Wh
Time ( t ) = 5 hours.
The power rating of the oven is 2200 W.
36. (A)(ii)(b) Calculate the current drawn by the oven. 10 A The current drawn by the oven is 10 A.
To find the current drawn by an appliance, divide the power rating by the voltage supply.
36. (A)(ii)(c) Calculate the resistance of the oven when it is red hot. 22 Ω Voltage (V) = 220 V
Power ( P ) = 2200 W
The resistance of the oven when it is red hot is 22 Ω.
Resistance can be calculated using the formula (R = V^2/P) when voltage and power are known.
36.(B)
(i) Write the relation between resistance \( R \) and electrical resistivity \( \rho \) of the material of a conductor in the shape of a 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 m is 60 Ω. 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 electrical resistivity would be affected if the wire (of part ’ii’) is stretched so that its length is doubled. Justify your answer.		 (i) \( R = \rho \frac{l}{A} \); SI Unit: \( \Omega \cdot \text{m} \)
(ii) \( 8 \times 10^{-6} \, \Omega \cdot \text{m} \)
(iii) No change in resistivity		 (i) Relation between resistance and resistivity:
\[ R = \rho \frac{l}{A}, \quad \text{where } \rho = \frac{RA}{l} \]
SI Unit of Resistivity:
\[ \rho = \frac{\text{Resistance} \times \text{Area}}{\text{Length}} = \frac{\Omega \cdot \text{m}^2}{\text{m}} = \Omega \cdot \text{m}. \]
(ii) Calculation of electrical resistivity:
Given:
\( R = 60 \, \Omega, \, A = 4 \times 10^{-7} \, \text{m}^2, \, l = 3 \, \text{m} \)
\[ \rho = \frac{R \cdot A}{l} = \frac{60 \cdot 4 \times 10^{-7}}{3} = 8 \times 10^{-6} \, \Omega \cdot \text{m}. \]
(iii) Effect of stretching on resistivity:
Electrical resistivity is a material property and does not depend on the dimensions of the conductor. Stretching the wire will not change its resistivity as it depends only on the material, not its shape or size.
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?
(ii) Name the solution used in the above process and write its formula.
(iii) (A) How does copper get refined when electric current is passed in the electrolytic cell?		 (i) Cathode: Pure copper plate
Anode: Impure copper block
(ii) Solution: Acidified copper sulphate solution
Formula: CuSO4
(iii) Copper ions are reduced at cathode; impurities settle at anode.		 (i) Cathode and Anode in Copper Refining:
Cathode: Pure copper plate.
Anode: Impure copper block.
(ii) Solution and Formula:
Solution: Acidified copper sulphate solution.
Formula: CuSO4
(iii) Refining Process:
When electric current is passed through the electrolytic cell:
- Copper ions (Cu2+) from the solution are reduced at the cathode, depositing pure copper.
- At the anode, impure copper oxidizes, releasing Cu2+ ions into the solution. Impurities settle as anode mud.
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?
(ii) Write one difference between dominant and recessive trait.
(iii) (A) In a cross with two pairs of contrasting characters.
RRYY (Round Yellow) × rryy (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.		 (i) All F1 plants were tall; short trait not expressed.
(ii) Dominant: Expressed in both homozygous and heterozygous.
Recessive: Expressed only in homozygous.
(iii) Method: Self-pollination of F1 progeny (RrYy).
Phenotypic Ratio: 9:3:3:1		 (i) Observations in F1 Plants:
• All F1 plants were tall, showing that tallness is a dominant trait.
• The short (recessive) trait was not expressed in the F1 generation.
(ii) Difference between Dominant and Recessive Traits:
• Dominant Trait: Expressed in the presence of both homozygous and heterozygous alleles.
• Recessive Trait: Expressed only in the homozygous condition.
(iii) Method and Results of the Experiment with Two Pairs of Contrasting Characters:
• Method: Self-pollination of F1 progeny (RrYy).
• F2 Generation Phenotypic Ratio: 9 (Round Yellow): 3 (Round Green): 3 (Wrinkled Yellow): 1 (Wrinkled Green).
• Conclusions: Traits segregate independently, demonstrating the Law of Independent Assortment.
38. (B) Justify the statement: "It is possible that a trait is inherited but may not be expressed." A recessive trait may be inherited but not expressed if the individual has a dominant allele. A trait can be inherited but not expressed if it is recessive and the individual possesses a dominant allele for that trait. In heterozygous conditions, the dominant trait masks the recessive trait, making it unexpressed but still inheritable.
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 (f) (cm) Object Distance (u) (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.
(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.		 (i) Case 1 (Mirror A)
(ii) Real, inverted, same size
(iii) Virtual, erect, magnified		 (i) Case 1 (Mirror A):
The object distance (45 cm) is beyond the center of curvature (\( C = 2f = 40 \, \text{cm} \)).
When the object is beyond \( C \), the image formed is diminished, inverted, and between \( f \) and \( C \).
(ii) Properties of the image in Case 2 (Mirror B):
• The image is real and inverted.
• The image is of the same size as the object (object distance is at the center of curvature).
(iii) Nature and size of the image formed by Mirror C:
In Case 3 (Mirror C): The object distance (20 cm) is less than the focal length (30 cm).
• Nature of Image: Virtual, erect, and magnified.
Ray Diagram:
Draw a ray originating from the object parallel to the principal axis and refracting through the focal point. Another ray passes through the center of curvature, and the virtual rays appear to meet behind the mirror.
39. (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. Image position: -36 cm Using the mirror formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Here, \( f = -12 \, \text{cm} \) (concave mirror), \( u = -18 \, \text{cm} \) (object distance).
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \] Substituting the values: \[ \frac{1}{v} = \frac{1}{-12} - \frac{1}{-18} \] To simplify, take the LCM of 12 and 18, which is 36: \[ \frac{1}{v} = \frac{-3}{36} + \frac{2}{36} \] \[ \frac{1}{v} = \frac{-1}{36} \] Therefore, \[ v = -36 \, \text{cm}. \]

The image is formed 36 cm in front of the mirror from the pole of the concave mirror. The image is real, inverted, and larger than the object.