CBSE Class 10 Science Question Paper 2024 PDF (Set 1 - 31/3/1) 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 - 31/3/1) with Answer Key
| CBSE Class 10 Science Question Paper 2024 (Set 1 - 31/3/1) with Answer Key |  Download |
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CBSE Science Question Paper (Set 1 – 31/3/1) 2024 Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 1. Identify the product 'X' obtained in the following chemical reaction: CaCO3 → X + CO2 (A) Quick lime (B) Gypsum (C) Lime Stone (D) Plaster of Paris |
(A) Quick lime | When calcium carbonate (CaCO3) is heated, it decomposes into quick lime (CaO) and carbon dioxide (CO2): CaCO3 → CaO + CO2 Quick lime is used in various industrial applications such as construction and steel manufacturing. |
| 2. Select a pair of natural indicators from the following: (A) Litmus and methyl orange (B) Turmeric and Litmus (C) Phenolphthalein and methyl orange (D) Methyl orange and Turmeric |
(B) Turmeric and Litmus | Turmeric and litmus are natural indicators. Turmeric turns red in a basic solution, while litmus changes color in acidic and basic solutions. |
| 3. A chemical compound used in glass, soap, and paper industries is: (A) Washing Soda (B) Baking Soda (C) Bleaching Powder (D) Common Salt |
(A) Washing Soda | Washing soda (Na2CO3) is commonly used in the glass, soap, and paper industries due to its alkaline nature. It is used in the manufacture of glass and as a cleansing agent in soaps. |
| 4. The structural formula of Cyclohexane is: (A) Option A (B) Option B (C) Option C (D) Option D |
(C) Option C | Cyclohexane is a cyclic alkane with the formula C6H12. It forms a six-membered ring with single bonds between carbon atoms. Each carbon is bonded to two hydrogen atoms. |
| 5. Consider the following chemical equation: aAl2O3 + bHCl → cAlCl3 + dH2O In order to balance this chemical equation, the values of a, b, c, and d must be: (A) 1, 6, 2, and 3 (B) 1, 6, 3, and 2 (C) 2, 6, 2, and 3 (D) 2, 6, 3, and 2 |
(A) 1, 6, 2, and 3 | The balanced equation is: Al2O3 + 6HCl → 2AlCl3 + 3H2O Steps: Balance Al, then Cl, then O and H. This ensures stoichiometric consistency. |
| 6. Which one of the following hydrocarbons is different from the others? (A) C4H10 (B) C7H14 (C) C5H12 (D) C2H6 |
(B) C7H14 | C7H14 is an alkene (unsaturated hydrocarbon), while the others are alkanes (saturated hydrocarbons). |
| 7. Which one of the following reactions is different from the remaining three? (A) NaCl + AgNO3 → AgCl + NaNO3 (B) CaO + H2O → Ca(OH)2 (C) KNO3 + H2SO4 → KHSO4 + HNO3 (D) ZnCl2 + H2S → ZnS + 2HCl |
(B) CaO + H2O → Ca(OH)2 | Option (B) is a combination reaction where calcium oxide reacts with water to form calcium hydroxide. The others involve displacement or exchange reactions. |
| 8. Select from the following a plant hormone which promotes cell division: (A) Gibberellins (B) Auxins (C) Abscisic Acid (D) Cytokinins |
(D) Cytokinins | Cytokinins are plant hormones that promote cell division and differentiation in plant tissues. |
| 9. Part(s) of a flower which attracts insects for pollination is (are): (A) Petals and Sepals (B) Anther and Stigma (C) Petals only (D) Sepals only |
(C) Petals only | Petals are brightly colored and fragrant, attracting insects for pollination. |
| 10. In an experiment to study independent inheritance of two separate traits: shape and color of seeds, the ratio of the different combinations in F2 progeny would be: (A) 1:3 (B) 1:2:1 (C) 9:3:3:1 (D) 9:1:1:3 |
(C) 9:3:3:1 | According to Mendel’s law of independent assortment, the F2 progeny exhibits a phenotypic ratio of 9:3:3:1 for two traits inherited independently. |
| 11. Which of the following statement(s) is (are) true about the human heart? (a) Right atrium receives oxygenated blood from lungs through pulmonary artery. (b) Left atrium transfers oxygenated blood to left ventricle which sends it to various parts of the body. (c) Right atrium receives deoxygenated blood from different parts of the body through vena cava. (d) Left atrium transfers oxygenated blood to aorta which sends it to different parts of the body. (A) (b) only (B) (a) and (d) (C) (b) and (c) (D) (b) and (d) |
(C) (b) and (c) | The right atrium receives deoxygenated blood via vena cava, and the left atrium sends oxygenated blood to the left ventricle for distribution. |
| 12. A cross between two tall pea plants resulted in offspring having a few dwarf plants. The gene-combination of the parental plants must be: (A) Tt and Tt (B) Tt and tt (C) TT and tt (D) TT and Tt |
(A) Tt and Tt | Crossing two heterozygous plants (Tt) results in a 3:1 phenotypic ratio of tall to dwarf plants. |
| 13. The phenomena of light involved in the formation of a rainbow in the sky are: (A) Refraction, dispersion and reflection (B) Refraction, dispersion and total internal reflection (C) Dispersion, scattering and reflection (D) Dispersion, refraction and internal reflection |
(D) Dispersion, refraction and internal reflection | A rainbow is formed by dispersion, refraction, and internal reflection of light through water droplets. |
| 14. In case of four wires of the same material, the resistance will be minimum if the diameter and length of the wire respectively are: (A) D/2 and L/4 (B) D/4 and 4L (C) 2D and L (D) 4D and 2L |
(D) 4D and 2L | Resistance decreases with larger diameter (increased cross-sectional area) and shorter length. |
| 15. A food chain will be more advantageous in terms of energy if it has: (A) 2 trophic levels (B) 3 trophic levels (C) 4 trophic levels (D) 5 trophic levels |
(A) 2 trophic levels | Shorter food chains are more energy-efficient, as less energy is lost at each level. |
| 16. Consider the following statements about ozone: (a) Ozone is poisonous gas. (b) Ozone shields the earth's surface from the infrared radiation from the sun. (c) Ozone is a product of UV radiations acting on oxygen molecule. (d) At the lower level of the earth's atmosphere, ozone performs most essential function. (A) (a) and (b) (B) (a) and (c) (C) (b) and (c) (D) (b) and (d) |
(B) (a) and (c) | Ozone is poisonous and formed by UV acting on oxygen, but it protects against UV radiation, not infrared. |
| 17. Assertion (A): A piece of Zinc metal gets reddish brown coating when kept in copper sulfate solution for some time. Reason (R): Copper is more reactive metal than Zinc. (A) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct explanation for Assertion (A). (B) Both Assertion (A) and Reason (R) are correct, but Reason (R) is not the correct explanation for Assertion (A). (C) Assertion (A) is correct, but Reason (R) is incorrect. (D) Assertion (A) is incorrect, but Reason (R) is correct. |
(C) Assertion (A) is correct, but Reason (R) is incorrect. | Zinc displaces copper due to its higher reactivity, forming a reddish-brown coating of copper. |
| 18. Assertion (A): Offsprings produced by asexual reproduction are genetically similar to the parents. Reason (R): Asexual reproduction involves a single parent. (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). | Asexual reproduction involves a single parent and results in genetically identical offspring. |
| 19. Assertion (A): Red light signals are used to stop the vehicles on the road. Reason (R): Red coloured light is scattered the most so as to be visible from a large distance. (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. | Red light has the longest wavelength and is scattered the least, making it visible over long distances. |
| 20. Assertion (A): The waste we generate daily may be biodegradable or non-biodegradable. Reason (R): The waste generated, if not disposed of properly, may cause serious environmental problems. (A) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct explanation for Assertion (A). (B) Both Assertion (A) and Reason (R) are correct, but Reason (R) is not the correct explanation for Assertion (A). (C) Assertion (A) is correct, but Reason (R) is incorrect. (D) Assertion (A) is incorrect, but Reason (R) is correct. |
(B) Both Assertion (A) and Reason (R) are correct, but Reason (R) is not the correct explanation for Assertion (A). | Waste can be biodegradable or non-biodegradable, and improper disposal leads to environmental problems, but the reason doesn't directly explain the assertion. |
| 21. When magnesium ribbon is burnt in air, an ash of white colour is produced. Write the chemical equation for the reaction giving the chemical name of the ash produced. State the type of chemical reaction, giving justification for your answer. | 2Mg + O2 → 2MgO | When magnesium is heated in air, it reacts with oxygen to form magnesium oxide (MgO), a white ash: 2Mg + O2 → 2MgO This is a combustion reaction as magnesium reacts with oxygen, releasing heat and light to form a new compound. |
| 22. Where are auxins synthesized? How do they promote phototropism? | Apical meristem; by differential cell elongation | Auxins are synthesized in the apical meristem (tip of the stem and roots). They promote phototropism by accumulating more on the shaded side of the plant, causing cells on that side to elongate more, which results in the plant bending towards the light source. |
| 23 (a). List any two pairs of visible contrasting characters of garden pea plants used by Mendel for his experiments, stating the dominant and recessive characters in each pair. | 1. Round (Dominant) vs Wrinkled (Recessive) 2. Yellow (Dominant) vs Green (Recessive) |
Mendel used several pairs of contrasting characters in his experiments. Two examples are: 1. Seed Shape: Dominant - Round; Recessive - Wrinkled 2. Seed Color: Dominant - Yellow; Recessive - Green |
| 23 (b). In human beings, the probability of getting a male or a female child is 50 percent. Explain with the help of a flow diagram only. | 50% Male; 50% Female | The probability of having a male or female child depends on the combination of sex chromosomes: Father (XY) → Sperm carries X or Y Mother (XX) → Egg carries X Possible Combinations: XY (Male) and XX (Female) Hence, there is a 50% probability for both male and female offspring. |
| 24. When do we say that a particular person is suffering from hypermetropia? List two causes of this defect. Name the type of lens used to correct this defect. | Convex lens | A person suffers from hypermetropia (farsightedness) when they can see distant objects clearly but have difficulty focusing on nearby objects. This occurs due to: 1. Shortened eyeball. 2. Weakened eye lens. A convex lens is used to correct this defect as it converges light rays onto the retina. |
| 25 (a). Draw a labelled diagram to show the pattern of magnetic field lines produced due to a current carrying straight conductor. Mark on it the direction of current in the conductor and the direction of magnetic field lines. | Right-hand thumb rule | Magnetic field lines form concentric circles around a current-carrying conductor. The direction of the magnetic field is determined by the right-hand thumb rule: if the thumb points in the direction of the current, the curled fingers indicate the direction of the magnetic field lines. |
| 25 (b). Name the device used to magnetize a piece of magnetic material. Draw a labelled diagram to show the arrangement used for the magnetization of a cylinder made of soft iron. | Electromagnet | An electromagnet is used to magnetize a piece of magnetic material. A solenoid (a coil of wire) carrying current is used to magnetize a soft iron cylinder placed inside it. |
| 26. What are decomposers? List two consequences of their absence in an ecosystem. | Accumulation of waste; Nutrient deficiency | Decomposers are organisms like bacteria and fungi that break down dead organic matter into simpler substances, recycling nutrients in the ecosystem. Consequences of their absence: 1. Accumulation of dead matter. 2. Disruption of the nutrient cycle, leading to nutrient deficiency in the environment. |
| 27. State reasons for the following: (a) Zinc oxide is an amphoteric oxide. (b) Sodium metal is stored in a bottle filled with kerosene oil. (c) In the reactions of nitric acid with metals, generally hydrogen gas is not evolved. |
Amphoteric; Reactive metal; Oxidizing acid | (a) Zinc oxide reacts with both acids and bases, showing its amphoteric nature. (b) Sodium is highly reactive with air and water, so it is stored in kerosene to prevent contact. (c) Nitric acid is a strong oxidizing agent, and it oxidizes hydrogen gas to water during reactions with metals. |
| 28 (a). State the reduction process used to obtain the following metals from their compounds: (i) Mercury (ii) Copper (iii) Sodium |
Roasting; Carbon reduction; Electrolysis | (i) Mercury: Roasting cinnabar (HgS) in air forms mercury. (ii) Copper: Reduction of copper oxide using carbon. (iii) Sodium: Electrolysis of molten sodium chloride (NaCl). |
| 28 (b). State the changes observed when the following metals are exposed to air: (i) Silver (ii) Copper (iii) Iron |
Tarnish; Patina; Rust | (i) Silver: Forms silver sulfide (tarnish). (ii) Copper: Forms green patina (copper carbonate). (iii) Iron: Forms rust (iron oxide). |
| 29. We water the soil but it reaches the topmost leaves of the plants. Explain in brief the process involved. | Transpiration pull | Water reaches the topmost leaves through the process of transpiration pull, which involves: 1. Absorption of water by roots via osmosis. 2. Capillary action in xylem vessels that allows water to rise. 3. Evaporation of water from stomata creates a suction force pulling water upwards. |
| 30 (a). List two constituents of the Central Nervous System (CNS). How are these components protected from injuries? (b) Write two limitations of the use of electrical impulses. |
Brain; Spinal Cord Slow response; Limited distance |
(a) Constituents: 1. Brain: Protected by the skull and cerebrospinal fluid (CSF). 2. Spinal Cord: Protected by the vertebral column and CSF. (b) Limitations of electrical impulses: 1. They cannot efficiently travel long distances without losing strength. 2. Point-to-point transmission slows down coordination involving multiple body parts. |
| 31. Name and explain the phenomenon of light due to which the path of a beam of light becomes visible when it enters a smoke-filled room through a small hole. Also state the dependence of the color of the light we receive on the size of the particle of the medium through which the beam of light passes. | Tyndall Effect | Phenomenon: Tyndall Effect Explanation: The path of light becomes visible due to scattering by particles in smoke. The intensity and visibility depend on particle size. Smaller particles scatter shorter wavelengths (blue light), while larger particles scatter all wavelengths, making the light appear white. |
| 32. Explain in brief the function of an electric fuse in a domestic circuit. An electric heater of current rating 3 kW; 220 V is to be operated in an electric circuit of rating 5 A. What is likely to happen when the heater is switched 'ON'? Justify your answer with necessary calculation. | Fuse melts due to overcurrent | Function: A fuse prevents overloading and short circuits by melting when excessive current flows through it. Calculation: Power (P) = 3 kW = 3000 W; Voltage (V) = 220 V Current (I) = P / V = 3000 / 220 ≈ 13.64 A The heater's current requirement exceeds the circuit rating (5 A), causing the fuse to melt and break the circuit. |
| 33 (a). State Ohm’s law. Write the formula for the equivalent resistance (R) of the parallel combination of three resistors with values R1, R2, and R3. | V = IR; Rparallel = (1/R1 + 1/R2 + 1/R3)-1 | (a) Ohm's Law: The current (I) through a conductor is directly proportional to the potential difference (V) across its ends, provided temperature remains constant. Formula for parallel resistance: 1/R = 1/R1 + 1/R2 + 1/R3 |
| 33 (b) Find the resistance of the following network of resistors. | R = 3R/2 | (b) For the given network, first find the equivalent resistance of parallel resistors, then add it to series resistors to calculate total resistance. |
| 34 (a) (i). Five solutions A, B, C, D, and E were tested with pH paper and showed pH as 4, 1, 13, 7, and 10, respectively. Which solution is: 1. Strongly acidic 2. Strongly alkaline 3. Weakly acidic 4. Neutral 5. Weakly alkaline |
Strongly acidic: B; Strongly alkaline: C | (i) pH classification: 1. Strongly acidic: Solution B (pH = 1). 2. Strongly alkaline: Solution C (pH = 13). 3. Weakly acidic: Solution A (pH = 4). 4. Neutral: Solution D (pH = 7). 5. Weakly alkaline: Solution E (pH = 10). |
| 34 (a) (ii) Write the name and formula of: 1. An acidic salt 2. A basic salt |
Acidic salt: NH4Cl; Basic salt: Na2CO3 | (ii) Examples: 1. Acidic salt: Ammonium chloride (NH4Cl). 2. Basic salt: Sodium carbonate (Na2CO3). |
| 34 (b). Name and state in brief the process which is used to prepare sodium hydroxide from sodium chloride. In this process, along with the main product, two gases 'X' and 'Y' are also given off at the two electrodes. Name 'X' and 'Y' specifying the name of their respective electrodes at which each gas is obtained. One of these gases, when reacts with dry calcium hydroxide, produces a compound 'Z' which is widely used in water treatment plants and textile industries. Name 'Z' and write the chemical equation for the reaction involved in its formation. | Chlor-alkali process X: Chlorine (Anode) Y: Hydrogen (Cathode) Z: Calcium hypochlorite |
The process used to prepare sodium hydroxide (NaOH) from sodium chloride (NaCl) is called the Chlor-alkali process. In this process: Gas 'X': Chlorine (Cl2) is produced at the anode. Gas 'Y': Hydrogen (H2) is produced at the cathode. When chlorine reacts with dry calcium hydroxide (Ca(OH)2), it forms calcium hypochlorite (Ca(OCl)2), which is used in water treatment and textile industries. Reaction: Ca(OH)2 + Cl2 → Ca(OCl)2 + H2O |
| 35 (a) (i). What are spores? On which structures are they formed? How do they overcome unfavorable conditions? Name the organism which multiplies with the help of these structures. | Spores: Asexual reproductive bodies | (i) Spores are asexual reproductive bodies formed in structures like sporangia. They help organisms survive in unfavorable conditions by remaining dormant. Example: Moss. |
| 35 (a) (ii) Give two reasons why some plants are grown by vegetative propagation. List two methods used to grow plants vegetatively. | Reasons: Cloning; Faster reproduction Methods: Cutting; Grafting |
(ii) Reasons for vegetative propagation: 1. Ensures cloning of plants with desirable traits. 2. Faster reproduction bypassing seed stage. Methods: 1. Cutting: Plant part (stem/leaf) is used for growth. 2. Grafting: Joining parts of two plants for combined traits. |
| 35 (b). (i) Study the diagram given below and name the parts marked as A, B, and C. What happens when B reaches C in the ovary? Mention its significance. | (i) A: Stigma; B: Pollen tube; C: Ovule | (i) Parts of the diagram: A: Stigma - Receives pollen. B: Pollen tube - Transfers male gametes. C: Ovule - Contains the egg cell. When B (pollen tube) reaches C (ovule), fertilization occurs. The male gamete from the pollen tube fuses with the egg cell in the ovule to form a zygote. This is significant as it leads to the formation of seeds for reproduction. |
| 35 (b).(ii) Write the post-fertilization changes that occur in a flower. | (ii) Formation of seed and fruit | (ii) Post-fertilization changes: 1. The ovule develops into a seed. 2. The ovary grows into a fruit. 3. Stigma, style, and anthers wither and fall off. 4. Petals fall as they are no longer needed for pollination. |
| 36 (a) (i). Draw a ray diagram to show the path of the refracted ray in each of the following cases: 1. A ray of light incident on a concave lens parallel to its principal axis. 2. A ray of light incident on a concave lens directed towards its principal focus. |
Diagram | (i) Ray diagrams: 1. A ray parallel to the principal axis diverges after refraction and appears to originate from the focal point. 2. A ray directed towards the focal point refracts and travels parallel to the principal axis. |
| 36 (a) (ii) A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24 cm. The distance of the object from the lens is 16 cm. Find the position and size of the image formed. | v = -48 cm; hi = 12 cm | (ii) Using lens formula: 1/f = 1/v - 1/u Substituting values: f = 24 cm, u = -16 cm v = -48 cm Magnification (M): hi/ho = v/u hi = 12 cm The image is real, inverted, and magnified. |
| 36 (b) (i). Draw a ray diagram to show the path of the reflected ray in each of the following cases: 1. A ray of light incident on a convex mirror parallel to its principal axis. 2. A ray of light incident on a convex mirror directed towards its principal focus. |
Parallel ray diverges; Focused ray reflects parallel | 1. A ray parallel to the principal axis reflects and diverges away from the mirror. The reflected ray appears to originate from the focal point behind the mirror. 2. A ray directed towards the principal focus reflects and travels parallel to the principal axis. |
| 36 (b) (ii). A 1.5 cm tall candle flame is placed perpendicular to the principal axis of a concave mirror of focal length 12 cm. If the distance of the flame from the pole of the mirror is 18 cm, use the mirror formula to determine the position and size of the image formed. | v = -36 cm; hi = -3.0 cm | Using the mirror formula: 1/f = 1/v + 1/u Given: f = -12 cm (concave mirror), u = -18 cm Rearrange: 1/v = 1/f - 1/u 1/v = 1/(-12) - 1/(-18) = -1/12 + 1/18 = -3/36 + 2/36 = -1/36 v = -36 cm The image is formed 36 cm in front of the mirror. Magnification (M): M = hi/ho = -v/u M = -(-36)/(-18) = -2 Image height: hi = M × ho = -2 × 1.5 = -3.0 cm The image is real, inverted, and reduced. |
| 37. More than three million carbon compounds have been discovered in the field of chemistry. The diversity of these compounds is due to the capacity of carbon atoms for bonding with one another as well as with other atoms. Most of the carbon compounds are poor conductors of electricity and have low melting and boiling points. (a) Write the molecular formula of the first two members of the homologous series having the functional group -Br. |
(a) C2H5Br, C3H7Br | (a) First two members of homologous series with -Br group are: 1. Bromoethane: C2H5Br 2. Bromopropane: C3H7Br |
| 37. More than three million carbon compounds have been discovered in the field of chemistry. The diversity of these compounds is due to the capacity of carbon atoms for bonding with one another as well as with other atoms. Most of the carbon compounds are poor conductors of electricity and have low melting and boiling points. (b) Given below are the formulae of some functional groups: (i) −CHO (ii) −C=O Write the name of these functional groups. |
(b) Aldehyde, Carbonyl |
(b) Functional groups: (i) −CHO: Aldehyde (ii) −C=O: Carbonyl group |
| 37. More than three million carbon compounds have been discovered in the field of chemistry. The diversity of these compounds is due to the capacity of carbon atoms for bonding with one another as well as with other atoms. Most of the carbon compounds are poor conductors of electricity and have low melting and boiling points. (c) What would be observed on adding a 5% alkaline potassium permanganate drop by drop to some warm ethanol taken in a test tube? State the role of KMnO4 in the reaction and write the chemical equation for the reaction involved. |
(c) Pink color fades; Oxidizing agent | (c) On adding 5% alkaline KMnO4 to warm ethanol, the pink color of KMnO4 fades. KMnO4 acts as an oxidizing agent, converting ethanol to ethanoic acid. Reaction: CH3CH2OH + 2[O] → CH3COOH + H2O |
| OR 37 (c). Write the name of the compound formed when ethanol is heated at 443 K with excess of concentrated H2SO4. What is the role of concentrated H2SO4 in the reaction? Write the chemical equation for the reaction involved. | Ethene; Dehydrating agent | The compound formed is ethene (C2H4). Role of concentrated H2SO4: It acts as a dehydrating agent, removing a water molecule from ethanol. Reaction: CH3CH2OH → C2H4 + H2O Conditions: Concentrated H2SO4, 443 K |
| 38. Human digestive system is a tube running from mouth to anus. Its main function is to break down complex molecules present in the food which cannot be absorbed as such into smaller molecules. These molecules are absorbed across the walls of the tube and the absorbed food reaches each and every cell of the body where it is utilized for obtaining energy. (a) Name the glands present in the buccal cavity and write the components of food on which the secretion of these glands act upon. |
(a) Salivary glands | (a) The glands in the buccal cavity are salivary glands. They secrete saliva, which acts on carbohydrates using the enzyme salivary amylase, breaking starch into maltose. |
| 38.(b) Two organs have a sphincter muscle at their exit. Name them. | (b) Stomach, Anus | (b) Organs with sphincter muscles: 1. Stomach (pyloric sphincter). 2. Anus (anal sphincter). |
| 38. (c) What will happen if: (i) Mucus is not secreted by the gastric glands. (ii) Villi are absent in the small intestine. |
(c) Ulcers, Malnutrition | (c) Effects: (i) Without mucus, gastric acid (HCl) will damage the stomach lining, leading to ulcers. (ii) Without villi, the surface area for nutrient absorption decreases, leading to malnutrition. |
| OR 38 (c). “Bile juice does not contain any enzyme, yet it has important roles in digestion.” Justify the statement. | Emulsification and neutralization | Although bile juice does not contain enzymes, it plays a crucial role in digestion by: 1. Emulsification of fats: Bile salts break down large fat globules into smaller droplets, increasing their surface area for enzyme action. 2. Neutralizing acidity: Bile neutralizes the acidic chyme from the stomach, creating an alkaline medium for pancreatic enzyme activity. These functions ensure efficient digestion and absorption of fats. |
| 39. In a domestic circuit, five LED bulbs are arranged as shown. The source voltage is 220 V, and the power rating of each bulb is marked in the circuit diagram. Based on the circuit diagram, answer the following questions: (a) State what happens when: (i) Key K1 is closed. (ii) Key K2 is closed. (b) Calculate the current through the circuit when Key K1 is closed. (c) Calculate the voltage across each bulb when Key K2 is closed. |
(a) (i) One bulb glows (ii) Four bulbs glow (b) 0.1 A (c) 55 V |
(a) (i) When K1 is closed, the single 22 W bulb glows normally as the circuit completes for it. (ii) When K2 is closed, the four 11 W bulbs in series glow normally as the circuit completes for them. (b) Current when K1 is closed: I = P/V = 22 W / 220 V = 0.1 A (c) Voltage across each bulb when K2 is closed: In series, total voltage = 220 V. Each bulb shares equal voltage: Veach = 220 V / 4 = 55 V. |
| 39. In a domestic circuit, five LED bulbs are arranged as shown. The source voltage is 220 V, and the power rating of each bulb is marked in the circuit diagram. Based on the circuit diagram, answer the following questions: (a) State what happens when: (i) Key K1 is closed. (ii) Key K2 is closed. |
(a) (i) One bulb glows (ii) Four bulbs glow |
(a) (i) When K1 is closed, the single 22 W bulb glows normally as the circuit completes for it. (ii) When K2 is closed, the four 11 W bulbs in series glow normally as the circuit completes for them. |
| 39. (b) Calculate the current through the circuit when Key K1 is closed. | (b) 0.1 A |
(b) Current when K1 is closed: I = P/V = 22 W / 220 V = 0.1 A |
| 39.(c) Calculate the voltage across each bulb when Key K2 is closed. | (c) 55 V | (c) Voltage across each bulb when K2 is closed: In series, total voltage = 220 V. Each bulb shares equal voltage: Veach = 220 V / 4 = 55 V. |
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