--- title: "CBSE Class 10 Science Question Paper 2023 (Set 31/1/1) with Solutions" url: https://www.swavid.com/cbse/class-10/science/previous-year-papers/2023-31-1-1 dateModified: 2026-10-07T15:14:33+00:00 --- # CBSE Class 10 Science Question Paper 2023 (Set 31/1/1) with Solutions Code 31/1/1 · 80 marks · 180 minutes Solved CBSE Class 10 Science board paper from 2023, set 31/1/1. Every question has step-by-step working written to the CBSE marking scheme, with the marks for each answer. Official question paper: https://www.cbse.gov.in/cbsenew/question-paper/2023/X/SCIENCE.zip. Solutions written by SwaVid. Free PDF (30 pages): https://www.swavid.com/api/seo/pdf/papers/cbse/science/swavid-cbse-class-10-science-question-paper-2023-31-1-1-b7011dec87.pdf ## SECTION - A Select and write one most appropriate option out of the four options given for each of the questions 1 - 20 ### Question 1 *1 mark · MCQ · Acids, Bases and Salts* In the experimental setup given below, it is observed that on passing the gas produced in the reaction in the solution 'X' the solution 'X' first turns milky and then colourless. [Diagram showing dilute hydrochloric acid reacting with sodium carbonate, passing gas into test tube 'X'] The option that justifies the above stated observation is that 'X' is aqueous calcium hydroxide and - it turns milky due to carbon dioxide gas liberated in the reaction and after sometime it becomes colourless due to formation of calcium carbonate. - it turns milky due to formation of calcium carbonate and on passing excess of carbon dioxide it becomes colourless due to formation of calcium hydrogen carbonate which is soluble in water. - it turns milky due to passing of carbon dioxide through it. It turns colourless as on further passing carbon dioxide, sodium hydrogen carbonate is formed which is soluble in water. - the carbon dioxide liberated during the reaction turns lime water milky due to formation of calcium hydrogen carbonate and after some time it turns colourless due to formation of calcium carbonate which is soluble in water. **Solution** 1. Dilute hydrochloric acid reacts with sodium carbonate to evolve carbon dioxide gas, which turns lime water (aqueous calcium hydroxide, X) milky due to the formation of insoluble calcium carbonate. 2. On passing excess carbon dioxide, the milky solution becomes colourless due to the formation of soluble calcium hydrogen carbonate. **Answer:** (b) it turns milky due to formation of calcium carbonate and on passing excess of carbon dioxide it becomes colourless due to formation of calcium hydrogen carbonate which is soluble in water. > Common mistake: Confusing calcium carbonate with calcium hydrogen carbonate regarding solubility in water. ### Question 2 *1 mark · MCQ · Chemical Reactions and Equations* The emission of brown fumes in the given experimental set-up is due to [Diagram showing thermal decomposition of lead nitrate producing brown fumes] - thermal decomposition of lead nitrate which produces brown fumes of nitrogen dioxide. - thermal decomposition of lead nitrate which produces brown fumes of lead oxide. - oxidation of lead nitrate forming lead oxide and nitrogen dioxide. - oxidation of lead nitrate forming lead oxide and oxygen. **Solution** 1. On heating lead nitrate, it undergoes thermal decomposition to form lead oxide, oxygen gas, and brown fumes of nitrogen dioxide. **Answer:** (a) thermal decomposition of lead nitrate which produces brown fumes of nitrogen dioxide. > Common mistake: Mistaking thermal decomposition for simple oxidation or confusing nitrogen dioxide fumes with lead oxide. ### Question 3 *1 mark · MCQ · Chemical Reactions and Equations* $\text{MnO}_2 + x\text{HCl} \rightarrow \text{MnCl}_2 + y\text{H}_2\text{O} + z\text{Cl}_2$ In order to balance the above chemical equation, the values of $x, y$ and $z$ respectively are : - 6, 2, 2 - 4, 1, 2 - 4, 2, 1 - 2, 2, 1 **Solution** 1. The given equation is $\text{MnO}_2 + x\text{HCl} \rightarrow \text{MnCl}_2 + y\text{H}_2\text{O} + z\text{Cl}_2$. 2. Balancing oxygen and hydrogen gives 4 $\text{HCl}$ and 2 $\text{H}_2\text{O}$, and balancing chlorine gives 1 $\text{Cl}_2$, so $x=4, y=2, z=1$. **Answer:** (c) 4, 2, 1 > Common mistake: Incorrectly balancing chlorine atoms on both sides of the reaction. ### Question 4 *1 mark · MCQ · Acids, Bases and Salts* The table below has information regarding $\text{pH}$ and the nature (acidic/basic) of four different solutions. Which one of the options in the table is correct ? - Lemon juice | Orange | 3 | Basic - Milk of magnesia | Blue | 10 | Basic - Gastric juice | Red | 6 | Acidic - Pure water | Yellow | 7 | Neutral **Solution** 1. Pure water is neutral with a pH of 7. 2. Milk of magnesia is basic with a pH of around 10, making option (b) the correct pairing from standard NCERT data. **Answer:** (b) Milk of magnesia | Blue | 10 | Basic > Common mistake: Memorizing incorrect pH values for common laboratory solutions. ### Question 5 *1 mark · MCQ · Metals and Non-metals* A metal 'X' is used in thermite process. When X is burnt in air it gives an amphoteric oxide 'Y'. 'X' and 'Y' are respectively : - $\text{Fe}$ and $\text{Fe}_2\text{O}_3$ - $\text{Al}$ and $\text{Al}_2\text{O}_3$ - $\text{Fe}$ and $\text{Fe}_3\text{O}_4$ - $\text{Al}$ and $\text{Al}_3\text{O}_4$ **Solution** 1. Aluminium ($\text{Al}$) is used in the thermite process to reduce metal oxides like iron oxide. 2. When burnt in air, aluminium forms aluminium oxide ($\text{Al}_2\text{O}_3$), which is an amphoteric oxide. **Answer:** (b) $\text{Al}$ and $\text{Al}_2\text{O}_3$ > Common mistake: Confusing iron used in thermite reaction mixture with the metal used as the reducing agent. ### Question 6 *1 mark · MCQ · Acids, Bases and Salts* Select washing soda from the following : - $\text{NaHCO}_3$ - $\text{Na}_2\text{CO}_3 \cdot 5\text{H}_2\text{O}$ - $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$ - $\text{NaOH}$ **Solution** 1. Washing soda is sodium carbonate decahydrate, which has the chemical formula $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$. **Answer:** (c) $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$ > Common mistake: Confusing washing soda formula with baking soda ($\text{NaHCO}_3$) or other hydrates. ### Question 7 *1 mark · MCQ · Carbon and its Compounds* Consider the structures of the three cyclic carbon compounds $\text{A}, \text{B}$ and $\text{C}$ given below and select the correct option from the following : [Structures of A, B, and C shown] - $\text{A}$ and $\text{C}$ are isomers of hexane and $\text{B}$ is benzene. - $\text{A}$ is an isomer of hexene, $\text{B}$ is benzene and $\text{C}$ is an isomer of hexene. - $\text{A}$ is a saturated cyclic hydrocarbon and $\text{B}$ and $\text{C}$ are unsaturated cyclic hydrocarbons. - $\text{A}$ is cyclohexane and $\text{B}$ and $\text{C}$ are the isomers of benzene. **Solution** 1. Compound A is cyclohexane ($\text{C}_6\text{H}_{12}$), which is a saturated cyclic hydrocarbon. 2. Compounds B (benzene, $\text{C}_6\text{H}_6$) and C are unsaturated cyclic hydrocarbons with alternating double bonds. **Answer:** (c) $\text{A}$ is a saturated cyclic hydrocarbon and $\text{B}$ and $\text{C}$ are unsaturated cyclic hydrocarbons. > Common mistake: Confusing the degree of saturation in ring structures. ### Question 8 *1 mark · MCQ · Life Processes* An organism which breaks down the food material outside the body and then absorbs it is - a plant parasite, Cuscuta - an animal parasite, Tapeworm - a bacteria, Rhizobium - a fungi, Rhizopus **Solution** 1. Organisms like fungi break down food material outside the body and then absorb it. 2. Rhizopus (bread mould) is an example of a saprophytic fungus that exhibits this mode of nutrition. **Answer:** (d) a fungi, Rhizopus > Common mistake: Confusing saprophytic nutrition in fungi with parasitic nutrition. ### Question 9 *1 mark · MCQ · Life Processes* Consider the following statements about small intestine and select the one which is NOT correct : - The length of the small intestine in animals differs as it depends on the type of food they eat. - The small intestine is the site of complete digestion of food. - The small intestine receives secretions from liver and pancreas. - The villi of the small intestine absorb water from the unabsorbed food before it gets removed from the body via the anus. **Solution** 1. Water absorption from unabsorbed food mainly takes place in the large intestine, not by the villi in the small intestine. 2. Therefore, the statement regarding villi absorbing water before removal via anus is incorrect. **Answer:** (d) The villi of the small intestine absorb water from the unabsorbed food before it gets removed from the body via the anus. > Common mistake: Confusing the functions of the small and large intestines regarding water absorption. ### Question 10 *1 mark · MCQ · Heredity* The statement that correctly describes the characteristic(s) of a gene is : - In individuals of a given species, a specific gene is located on a particular chromosome. - A gene is not the information source for making proteins in the cell. - Each chromosome has only one gene located all along its length. - All the inherited traits in human beings are not controlled by genes. **Solution** 1. A specific gene is located on a particular chromosome in individuals of a given species. 2. Chromosomes contain multiple genes, genes code for proteins, and inherited traits are controlled by genes. **Answer:** (a) In individuals of a given species, a specific gene is located on a particular chromosome. > Common mistake: Assuming each chromosome carries only one gene. ### Question 11 *1 mark · MCQ · Control and Coordination* Select from the following the correct statement about tropic movement in plants : - It is due to stimulus of touch and temperature. - It does not depend upon the direction of stimulus received. - It is observed only in roots and not in stems. - It is a growth related movement. **Solution** 1. Tropic movements are directional growth movements that occur in response to an external stimulus. 2. Since they involve cell elongation or growth, they are growth-related movements dependent on the direction of the stimulus. **Answer:** (d) It is a growth related movement. > Common mistake: Confusing tropic movements with nastic movements which are non-directional. ### Question 12 *1 mark · MCQ · How do Organisms Reproduce?* Select the INCORRECT match (between the plant and its vegetative part) from the following : - Bryophyllum, leaf - Potato, stem - Money-plant, stem - Rose, root **Solution** 1. Rose plants are artificially propagated primarily through stem cuttings, not roots. 2. Bryophyllum propagates through leaves, potato through tubers (stems), and money-plant through stems, making the match for Rose incorrect. **Answer:** (d) Rose, root > Common mistake: Assuming all plants reproduce vegetatively through roots. ### Question 13 *1 mark · MCQ · Electricity* If four identical resistors, of resistance $8\ \Omega$, are first connected in series so as to give an effective resistance $R_s$, and then connected in parallel so as to give an effective resistance $R_p$, then the ratio $\frac{R_s}{R_p}$ is - 32 - 2 - 0.5 - 16 **Solution** 1. For four identical resistors of resistance $R = 8\ \Omega$ connected in series, the equivalent resistance is $R_s = 4R = 4 \times 8\ \Omega = 32\ \Omega$. 2. For the same resistors connected in parallel, the equivalent resistance is $R_p = \frac{R}{4} = \frac{8}{4} = 2\ \Omega$. 3. The ratio of $R_s$ to $R_p$ is $\frac{R_s}{R_p} = \frac{32}{2} = 16$. **Answer:** (d) 16 > Common mistake: Students often invert the ratio and calculate $R_p/R_s$ instead of $R_s/R_p$, leading to the incorrect option 0.5. ### Question 14 *1 mark · MCQ · Electricity* In domestic electric circuits the wiring with $15\text{ A}$ current rating is for the electric devices which have - higher power ratings such as geyser. - lower power ratings such as fan. - metallic bodies and low power ratings. - non-metallic bodies and low power ratings. **Solution** 1. Two separate circuits are used in domestic wiring: a $5\text{ A}$ current rating for lower power appliances like fans and bulbs, and a $15\text{ A}$ current rating for higher power appliances like geysers, air conditioners, and electric irons. **Answer:** "(a) higher power ratings such as geyser." > Common mistake: Confusing the $5\text{ A}$ circuit with the $15\text{ A}$ circuit. ### Question 15 *1 mark · MCQ · Electricity* In the following diagram, the position of the needle is shown on the scale of a voltmeter. The least count of the voltmeter and the reading shown by it respectively are : [Diagram showing voltmeter scale with needle between 1.5 and 2] - $0.15\text{ V}$ and $1.6\text{ V}$ - $0.05\text{ V}$ and $1.6\text{ V}$ - $0.15\text{ V}$ and $1.8\text{ V}$ - $0.05\text{ V}$ and $1.8\text{ V}$ **Solution** 1. The voltmeter scale has $10$ divisions between $0$ and $1.5\text{ V}$ or between $1.5$ and $2.0\text{ V}$. Between $1.5$ and $2.0$, there are $10$ subdivisions, so the least count is $\frac{2.0 - 1.5}{10} = \frac{0.5}{10} = 0.05\text{ V}$. 2. The needle is at $2$ divisions after $1.5\text{ V}$, which gives a reading of $1.5 + 2 \times 0.05 = 1.6\text{ V}$. **Answer:** "(b) $0.05\text{ V}$ and $1.6\text{ V}$" > Common mistake: Dividing by the wrong number of divisions while calculating the least count. ### Question 16 *1 mark · MCQ · Magnetic Effects of Electric Current* The resultant magnetic field at point 'P' situated midway between two parallel wires (placed horizontally) each carrying a steady current $I$ is [Diagram showing parallel wires AB and CD with point P in between] - in the same direction as the current in the wires. - in the vertically upward direction. - zero - in the vertically downward direction. **Solution** 1. According to the right-hand thumb rule, the magnetic field produced by each parallel wire carrying current in the same direction points in opposite directions at the midway point P. 2. Since the magnitudes of the magnetic fields are equal and their directions are opposite, they cancel each other out, making the resultant magnetic field zero. **Answer:** "(c) zero" > Common mistake: Assuming the magnetic fields add up instead of opposing each other at the midpoint. ### Question 17 *1 mark · Assertion and reason · Metals and Non-metals* Assertion (A) : The colour of aqueous solution of copper sulphate turns colourless when a piece of lead is added to it. Reason (R) : Lead is more reactive than copper, and hence displaces copper from its salt solution. - Both (A) and (R) are true and (R) is the correct explanation of (A). - Both (A) and (R) are true, but (R) is not the correct explanation of (A). - (A) is true, but (R) is false. - (A) is false, but (R) is true. **Solution** 1. Lead is more reactive than copper and displaces copper from copper sulphate solution, forming lead sulphate and copper metal. 2. The blue colour of copper sulphate fades or turns colourless as copper ions are removed from the solution, making both Assertion and Reason true with the Reason being the correct explanation. **Answer:** "Both (A) and (R) are true and (R) is the correct explanation of (A)." > Common mistake: Thinking lead is less reactive than copper. ### Question 18 *1 mark · Assertion and reason · Heredity* Assertion (A) : Genes inherited from the parents decide the sex of a child. Reason (R) : X chromosome in a male child is inherited from his father. - Both (A) and (R) are true and (R) is the correct explanation of (A). - Both (A) and (R) are true, but (R) is not the correct explanation of (A). - (A) is true, but (R) is false. - (A) is false, but (R) is true. **Solution** 1. The sex of a child depends on whether the fertilizing sperm carries an X chromosome or a Y chromosome, making the Assertion true. 2. A male child inherits an X chromosome from his mother and a Y chromosome from his father, making the Reason false. **Answer:** "(A) is true, but (R) is false." > Common mistake: Assuming a male child inherits the X chromosome from his father. ### Question 19 *1 mark · Assertion and reason · Life Processes* Assertion (A) : Blood clotting prevents excessive loss of blood. Reason (R) : Blood clotting is due to blood plasma and white blood cells present in the blood. - Both (A) and (R) are true and (R) is the correct explanation of (A). - Both (A) and (R) are true, but (R) is not the correct explanation of (A). - (A) is true, but (R) is false. - (A) is false, but (R) is true. **Solution** 1. Blood clotting prevents excessive loss of blood from the body during an injury, making Assertion (A) true. 2. Blood clotting is actually due to platelets and fibrinogen present in the blood plasma, not white blood cells, making Reason (R) false. **Answer:** (A) is true, but (R) is false. > Common mistake: Confusing the function of white blood cells with platelets in blood clotting. ### Question 20 *1 mark · Assertion and reason · Magnetic Effects of Electric Current* Assertion (A) : The strength of the magnetic field produced at the centre of a current carrying circular coil increases on increasing the number of turns in it. Reason (R) : The current in each circular turn has the same direction and the magnetic field due to each turn then just adds up. - Both (A) and (R) are true and (R) is the correct explanation of (A). - Both (A) and (R) are true, but (R) is not the correct explanation of (A). - (A) is true, but (R) is false. - (A) is false, but (R) is true. **Solution** 1. The strength of the magnetic field produced by a circular coil is directly proportional to the number of turns in it, so increasing the turns increases the magnetic field strength. 2. Since the current in each circular turn flows in the same direction, the magnetic field produced by each individual turn adds up. **Answer:** Both (A) and (R) are true and (R) is the correct explanation of (A). > Common mistake: Failing to recognize that magnetic fields of individual turns in the same direction reinforce each other. ## SECTION - B Q. No. 21 to 26 are very short answer questions. ### Question 21 *2 marks · Very short answer · Acids, Bases and Salts* (a) (i) A compound 'X' which is prepared from gypsum has the property of hardening when mixed with proper quantity of water. Identify 'X' and write its chemical formula. (ii) State the difference in chemical composition between baking soda and baking powder. **Part (i)** 1. Compound X is Plaster of Paris. 2. Its chemical formula is $\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}$. Answer (i): Plaster of Paris, $\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}$ **Part (ii)** 1. Baking soda is chemically sodium hydrogen carbonate ($\text{NaHCO}_3$). 2. Baking powder is a mixture of baking soda and a mild edible acid such as tartaric acid. Answer (ii): Baking soda is pure $\text{NaHCO}_3$, while baking powder is $\text{NaHCO}_3$ mixed with a mild edible acid. **Answer:** Compound X is Plaster of Paris with formula $\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}$. Baking soda is sodium hydrogen carbonate, whereas baking powder is a mixture of baking soda and a mild edible acid like tartaric acid. > Common mistake: Confusing Plaster of Paris formula with gypsum, or stating baking powder is just baking soda. ### Question 21 (OR) *2 marks · Very short answer · Acids, Bases and Salts* (b) Write balanced chemical equation for the reaction that occurs when : (i) blue coloured copper sulphate crystals are heated and (ii) Sodium hydrogen carbonate is heated during cooking. **Part (i)** 1. Blue copper sulphate crystals contain water of crystallisation ($\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$). 2. On heating, they lose water to become white anhydrous copper sulphate. Answer (i): $$\text{CuSO}_4 \cdot 5\text{H}_2\text{O}_{(s)} \xrightarrow{\text{Heat}} \text{CuSO}_4_{(s)} + 5\text{H}_2\text{O}_{(g)}$$ **Part (ii)** 1. Sodium hydrogen carbonate on heating decomposes to give sodium carbonate, water and carbon dioxide. Answer (ii): $$2\text{NaHCO}_3_{(s)} \xrightarrow{\text{Heat}} \text{Na}_2\text{CO}_3_{(s)} + \text{H}_2\text{O}_{(l)} + \text{CO}_2_{(g)}$$ **Answer:** Balanced equations: (i) $\text{CuSO}_4 \cdot 5\text{H}_2\text{O} \xrightarrow{\text{Heat}} \text{CuSO}_4 + 5\text{H}_2\text{O}$, (ii) $2\text{NaHCO}_3 \xrightarrow{\text{Heat}} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$. > Common mistake: Forgetting to balance the equation for the thermal decomposition of sodium hydrogen carbonate. ### Question 22 *2 marks · Very short answer · Control and Coordination* (a) Write the role of insulin in regulating blood sugar levels in human body. Mention the disease caused due to it. (b) How is the timing and the amount of release of insulin in the blood regulated? **Part (a)** 1. Insulin helps in regulating blood sugar levels by lowering excess glucose in the blood. 2. Deficiency of insulin causes a disease known as diabetes. Answer (a): Regulates blood sugar levels; deficiency causes diabetes. **Part (b)** 1. The timing and amount of insulin released are regulated by feedback mechanisms. 2. When blood sugar rises, insulin is secreted; as sugar levels fall, secretion reduces. Answer (b): Regulated by feedback mechanisms based on blood sugar levels. **Answer:** Insulin regulates blood sugar by lowering it, and its deficiency causes diabetes. The timing and amount of release are controlled by feedback mechanisms. > Common mistake: Writing diabetes mellitus as just diabetes without mentioning blood sugar regulation clearly. ### Question 23 *2 marks · Very short answer · Life Processes* (a) Name the type of blood (oxygenated / deoxygenated) transported by each of the following mentioning the path (i.e. from one organ (which place) to another (which place)). (i) Vena cava (ii) Pulmonary artery **Part (i)** 1. Vena cava transports deoxygenated blood. 2. It carries blood from various body organs to the right atrium of the heart. Answer (i): Deoxygenated blood from body organs to the right atrium. **Part (ii)** 1. Pulmonary artery transports deoxygenated blood. 2. It carries blood from the right ventricle of the heart to the lungs for oxygenation. Answer (ii): Deoxygenated blood from the right ventricle to the lungs. **Answer:** (i) Vena cava transports deoxygenated blood from body organs to the right atrium. (ii) Pulmonary artery transports deoxygenated blood from the right ventricle to the lungs. > Common mistake: Confusing pulmonary artery with pulmonary vein regarding the type of blood carried. ### Question 23 (OR) *2 marks · Very short answer · Life Processes* (b) With the help of a schematic flow chart, show the breakdown of glucose in a cell to provide energy - (i) in the presence of oxygen (ii) in lack of oxygen **Part (i)** 1. Glucose (6-carbon) breaks down into pyruvate (3-carbon) in the cytoplasm. 2. In the presence of oxygen (mitochondria), pyruvate breaks down to give carbon dioxide, water and energy. Answer (i): Glucose $\rightarrow$ Pyruvate (in cytoplasm) $\rightarrow$ $\text{CO}_2 + \text{H}_2\text{O} + \text{Energy}$ (in mitochondria). **Part (ii)** 1. In the lack of oxygen (such as in muscle cells during vigorous exercise), pyruvate is converted into lactic acid and energy. Answer (ii): Glucose $\rightarrow$ Pyruvate $\rightarrow$ Lactic acid (3-carbon molecule) + Energy. **Answer:** Glucose breakdown: (i) Presence of oxygen yields $\text{CO}_2$, $\text{H}_2\text{O}$ and energy in mitochondria. (ii) Lack of oxygen (in muscle cells) yields lactic acid and energy. > Common mistake: Confusing lack of oxygen in muscles (produces lactic acid) with absence of oxygen in yeast (produces ethanol and $\text{CO}_2$). ### Question 24 *2 marks · Very short answer · Life Processes* Name the part of the human excretory system where nephrons are found. Write the structure and function of nephrons. **Solution** 1. Nephrons are found in the kidneys, which are the basic filtering units of the human excretory system. 2. A nephron consists of a cup-shaped end called Bowman's capsule containing a cluster of capillaries called glomerulus, connected to a tubular part. 3. Its main function is to filter nitrogenous wastes from the blood and form urine. **Answer:** Nephrons are found in the kidneys. Each consists of a glomerulus, Bowman's capsule and a renal tubule, functioning to filter blood and form urine. > Common mistake: Stating nephrons are found in the urinary bladder instead of the kidneys. ### Question 25 *2 marks · Very short answer · The Human Eye and the Colourful World* (a) A narrow beam XY of white light is passing through a glass prism ABC as shown in the diagram : [Diagram of white light passing through prism] Trace it on your answer sheet and show the path of the emergent beam as observed on the screen PQ. Name the phenomenon observed and state its cause. **Solution** 1. Draw the prism ABC and the incident white beam XY refracting towards the base, splitting into a band of seven colours (VIBGYOR) with violet bending the most and red the least, and striking the screen PQ. 2. The phenomenon observed is dispersion of white light, caused by the bending of light rays of different wavelengths through different angles by the glass prism. **Answer:** Dispersion of white light; caused by different angles of bending for different colours. > Common mistake: Reversing the bending order of red and violet colours. ### Question 25 (OR) *2 marks · Very short answer · The Human Eye and the Colourful World* (b) It is observed that the power of an eye to see nearby objects as well as far off objects diminishes with age. (i) Give reason for the above statement. (ii) Name the defect that is likely to arise in the eyes in such a condition. (iii) Draw a labelled ray diagram to show the type of corrective lens used for restoring the vision of such an eye. **Part (i)** 1. It is caused by the gradual weakening of the ciliary muscles and diminishing flexibility of the eye lens with age. Answer (i): Weakening of ciliary muscles and loss of flexibility of the eye lens. **Part (ii)** 1. The defect of vision is called presbyopia. Answer (ii): Presbyopia **Part (iii)** 1. Diagram: Draw a bifocal lens showing both the upper concave part for distant vision and the lower convex part for near vision. Answer (iii): Corrective lens: Bifocal lens. **Answer:** Presbyopia is caused by the gradual weakening of ciliary muscles and diminishing flexibility of the eye lens, corrected using a bifocal lens. > Common mistake: Confusing presbyopia with simple myopia or hypermetropia. ### Question 26 *2 marks · Very short answer · Our Environment* How do harmful chemicals get accumulated progressively at each trophic level in a food chain? **Solution** 1. Harmful chemicals like pesticides enter the food chain through producers and get absorbed by them. 2. As these chemicals cannot be degraded, their concentration increases progressively at each higher trophic level, a phenomenon known as biological magnification. **Answer:** Progressive accumulation of non-biodegradable chemicals at each higher trophic level is called biological magnification. > Common mistake: Failing to mention that the chemicals are non-biodegradable. ## SECTION - C Q. No. 27 to 33 are short answer questions. ### Question 27 *3 marks · Short answer · Chemical Reactions and Equations* (a) Identify the reducing agent in the following reactions : (i) $4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}$ (ii) $\text{H}_2\text{O} + \text{F}_2 \rightarrow \text{HF} + \text{HOF}$ (iii) $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ (iv) $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ (b) Define a redox reaction in terms of gain or loss of oxygen. **Part (a)(i)** 1. Ammonia ($\text{NH}_3$) loses hydrogen and gets oxidized to NO, so it acts as the reducing agent. Answer (a)(i): $\text{NH}_3$ **Part (a)(ii)** 1. Water ($\text{H}_2\text{O}$) loses hydrogen to form HOF, so it acts as the reducing agent. Answer (a)(ii): $\text{H}_2\text{O}$ **Part (a)(iii)** 1. Carbon monoxide ($\text{CO}$) gains oxygen to form $\text{CO}_2$, so it acts as the reducing agent. Answer (a)(iii): $\text{CO}$ **Part (a)(iv)** 1. Hydrogen ($\text{H}_2$) gains oxygen to form $\text{H}_2\text{O}$, so it acts as the reducing agent. Answer (a)(iv): $\text{H}_2$ **Part (b)** 1. A reaction in which one reactant gains oxygen (gets oxidized) while another reactant loses oxygen (gets reduced) is called a redox reaction. Answer (b): A reaction involving gain of oxygen by one substance and loss of oxygen by another is called a redox reaction. **Answer:** (a) Identifying reducing agents in the given reactions, and (b) defining redox reaction in terms of gain or loss of oxygen. > Common mistake: Confusing oxidizing agent with reducing agent. ### Question 28 *3 marks · Short answer · Acids, Bases and Salts* (a) Suggest one remedial measure each to counteract the change in $\text{pH}$ in human beings in following cases : (i) Production of too much acid in stomach during indigestion (ii) Stung by a honey bee / nettle leaves (b) Fresh milk has a $\text{pH}$ of 6. When it changes into curd will its $\text{pH}$ increase or decrease? Why? **Part (a)(i)** 1. Use antacids such as milk of magnesia which are basic in nature to neutralize excess stomach acid. Answer (a)(i): Antacid (like Milk of Magnesia) **Part (a)(ii)** 1. Apply a mild base like baking soda on the affected area to neutralize the acid injected by the bee or nettle sting. Answer (a)(ii): Baking soda solution **Part (b)** 1. When fresh milk changes into curd, lactic acid is produced. 2. Due to the presence of acid, the pH of the solution decreases. Answer (b): pH decreases because lactic acid is produced during curd formation. **Answer:** (a) Antacids and mild bases, (b) pH decreases due to formation of lactic acid. > Common mistake: Writing that pH increases when acid is formed. ### Question 29 *3 marks · Short answer · Life Processes* (a) (i) State the role of ATP in cellular respiration. (ii) What ensures sufficient exchange of gases in plants? (iii) State the conditions on which the direction of diffusion of gases in plant depend upon. **Part (i)** 1. ATP acts as the energy currency for cellular processes, releasing energy when broken down to drive endergonic reactions in the cell. Answer (i): ATP provides energy for various cellular activities. **Part (ii)** 1. Large intercellular spaces ensure that all cells are in contact with air, and stomata facilitate the exchange of gases with the atmosphere. Answer (ii): Stomata and large intercellular spaces ensure sufficient exchange of gases. **Part (iii)** 1. The direction of diffusion depends upon the environmental conditions and the requirement of the plant at the time. Answer (iii): Environmental conditions and the physiological requirements of the plant. **Answer:** (a)(i) ATP provides energy, (ii) stomata and intercellular spaces ensure exchange, (iii) environmental conditions and plant requirements determine diffusion direction. > Common mistake: Writing incomplete roles of ATP. ### Question 29 (OR) *3 marks · Short answer · Life Processes* (b) (i) What is the internal energy reserve in plants and animals? (ii) How do desert plants perform photosynthesis if their stomata remain closed during the day? **Part (i)** 1. The internal energy reserve in plants is starch, and in animals, it is glycogen. Answer (i): Starch in plants and glycogen in animals. **Part (ii)** 1. Desert plants take up carbon dioxide at night and prepare an intermediate compound. 2. This intermediate is acted upon by the energy absorbed by chlorophyll during the day when their stomata are open or using the stored $\text{CO}_2$. Answer (ii): They take up $\text{CO}_2$ at night to form an intermediate which is used for photosynthesis during the day. **Answer:** (a)(i) Starch and glycogen, (ii) desert plants take up $\text{CO}_2$ at night and form an intermediate. > Common mistake: Confusing plant and animal energy reserves. ### Question 30 *3 marks · Short answer · Light – Reflection and Refraction* (a) Complete the following ray diagram to show the formation of image : [Diagram showing concave mirror with object AB and rays] (b) Mention the nature, position and size of the image formed in this case. (c) State the sign of the image distance in this case using the Cartesian sign convention. **Solution** 1. A ray parallel to the principal axis passes through the principal focus after reflection. 2. A ray passing through the focus reflects parallel to the principal axis, forming a real, inverted and enlarged image beyond the centre of curvature when the object is placed between F and C. **Answer:** Nature: Real and inverted; Position: Beyond C; Size: Enlarged; Sign of image distance: Negative. > Common mistake: Forgetting that image distance for a real image formed by a concave mirror is negative. ### Question 31 *3 marks · Short answer · The Human Eye and the Colourful World* Give reasons for the following : (a) Danger signals installed at airports and at the top of tall buildings are of red colour. (b) The sky appears dark to the passengers flying at very high altitudes. (c) The path of a beam of light passing through a colloidal solution is visible. **Part (a)** 1. Red colour has the longest wavelength among visible colours and is least scattered by fog or smoke, so it can be seen over long distances. Answer (a): Red light is least scattered by fog or smoke. **Part (b)** 1. At very high altitudes, there is no atmosphere containing particles to scatter sunlight, hence the sky appears dark. Answer (b): Due to the absence of atmosphere and scattering at high altitudes. **Part (c)** 1. Colloidal particles are large enough to scatter a beam of light passing through them, making the path of light visible (Tyndall effect). Answer (c): Due to scattering of light by colloidal particles (Tyndall effect). **Answer:** (a) Red light is least scattered, (b) absence of atmosphere for scattering, (c) scattering of light by colloidal particles (Tyndall effect). > Common mistake: Writing that red light is scattered most instead of least. ### Question 32 *3 marks · Short answer · Magnetic Effects of Electric Current* (a) (i) State the rule used to find the force acting on a current carrying conductor placed in a magnetic field. (ii) Given below are three diagrams showing entry of an electron in a magnetic field. Identify the case in which the force will be (1) maximum and (2) minimum respectively. Give reason for your answer. [Diagrams (i), (ii), (iii) showing electron entering magnetic field] **Part (a) (i)** 1. Stretch the thumb, forefinger and middle finger of your left hand such that they are mutually perpendicular. 2. If the first finger points in the direction of magnetic field and the second finger in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor. Answer (a) (i): Fleming's Left-Hand Rule is used to find the direction of force. **Part (a) (ii)** 1. The magnetic force on a moving charge is given by $F = Bqv \sin\theta$, where $\theta$ is the angle between the velocity of the charge and the magnetic field. 2. The force is maximum when $\theta = 90^\circ$ (perpendicular entry) and minimum when $\theta = 0^\circ$ or $180^\circ$ (parallel or anti-parallel entry). Answer (a) (ii): Maximum force occurs when the electron enters perpendicular to the magnetic field, and minimum (zero) force occurs when it enters parallel to the magnetic field. **Answer:** Fleming's Left-Hand Rule is used; force is maximum when the direction of motion is perpendicular to the magnetic field and minimum when parallel. > Common mistake: Confusing Fleming's Left-Hand Rule (used for motor/force) with Fleming's Right-Hand Rule (used for generator/induced current). ### Question 32 (OR) *3 marks · Short answer · Magnetic Effects of Electric Current* (b) (i) Draw the pattern of magnetic field lines of (1) a current carrying solenoid (2) a bar magnet (ii) List two distinguishing features between the two fields. **Part (b) (i)** 1. Draw a solenoid with closed loops of magnetic field lines passing through it, emerging from the North pole and entering the South pole outside. 2. Draw a bar magnet showing closed continuous magnetic field lines directed from North pole to South pole outside the magnet and South to North inside. Answer (b) (i): The field line patterns resemble each other, with the solenoid having uniform parallel lines inside. **Part (b) (ii)** 1. The magnetic field inside a solenoid is uniform and parallel, whereas inside a bar magnet it is non-uniform. 2. The strength of the magnetic field can be changed in a current-carrying solenoid by altering the current or number of turns, but for a permanent bar magnet, it is fixed. Answer (b) (ii): 1. Field inside a solenoid is uniform while inside a bar magnet it varies. 2. Solenoid field strength is adjustable, bar magnet field strength is constant. **Answer:** Magnetic field lines of a solenoid are similar to those of a bar magnet; key difference is that field lines inside a solenoid are parallel uniform straight lines. > Common mistake: Drawing field lines intersecting each other or omitting arrowheads indicating the direction of field lines. ### Question 33 *3 marks · Short answer · Our Environment* (a) (i) Why does a kitchen garden called an artificial ecosystem while a forest is considered to be a natural ecosystem? (ii) While designing an artificial ecosystem at home, write any two things to be kept in mind to convert it into a self-sustaining system. Give reason to justify your answer. **Part (a) (i)** 1. A kitchen garden is created, managed, and modified by humans, lacking natural complexity and self-regulation, making it an artificial ecosystem. 2. A forest is a self-sustaining natural unit comprising diverse biotic and abiotic components interacting freely without human intervention, making it a natural ecosystem. Answer (a) (i): A kitchen garden is man-made and artificially maintained, whereas a forest is self-regulating and natural. **Part (a) (ii)** 1. Include producers (plants) and primary consumers along with decomposers to ensure nutrient recycling. 2. Provide a continuous energy source like sunlight and regular water supply to maintain biological processes. Answer (a) (ii): Provisions for a light source and balanced biotic components (producers, consumers, decomposers) must be included to ensure energy flow and nutrient recycling. **Answer:** Kitchen gardens are created and maintained by humans (artificial), whereas forests exist naturally. To make an artificial system self-sustaining, producers, consumers, and decomposers must be included with adequate light and water. > Common mistake: Omitting decomposers while listing components of a self-sustaining artificial ecosystem. ### Question 33 (OR) *3 marks · Short answer · Our Environment* (b) (i) Construct a food chain of four trophic levels comprising the following : Hawk, snake, plants, rat. (ii) $20,000\text{ J}$ of energy was transferred by the producers to the organism of second trophic level. Calculate the amount of energy that will be transferred by organisms of the third trophic level to the organisms of the fourth trophic level. **Part (b) (i)** 1. Identify the four organisms: Plants, rat, snake, hawk. 2. Arrange them in proper trophic sequence: Plants (Producer) $\to$ Rat (Herbivore) $\to$ Snake (Carnivore) $\to$ Hawk (Top Carnivore). Answer (b) (i): Plants $\to$ Rat $\to$ Snake $\to$ Hawk **Part (b) (ii)** 1. Energy at the second trophic level (Rat) = $20,000\text{ J}$. 2. According to the 10% law, only 10% of energy is transferred to the next trophic level. 3. Energy at third trophic level (Snake) = $10\% \text{ of } 20,000\text{ J} = 2,000\text{ J}$. 4. Energy at fourth trophic level (Hawk) = $10\% \text{ of } 2,000\text{ J} = 200\text{ J}$. Answer (b) (ii): $200\text{ J}$ **Answer:** Food chain: Plants $\to$ Rat $\to$ Snake $\to$ Hawk. Energy transferred to fourth trophic level is $2\text{ J}$. > Common mistake: Calculating 10% from the producer level directly for the fourth trophic level without stepping through the third trophic level, or making calculation errors with zeroes. ## SECTION - D Q. No. 34 to 36 are long answer questions. ### Question 34 *5 marks · Long answer · Carbon and its Compounds* (a) A saturated organic compound 'A' belongs to the homologous series of alcohols. On heating 'A' with concentrated sulphuric acid at $443\text{ K}$, it forms an unsaturated compound 'B' with molecular mass $28\text{ u}$. The compound 'B' on addition of one mole of hydrogen in the presence of Nickel, changes to a saturated hydrocarbon 'C'. (i) Identify A, B and C. (ii) Write the chemical equations showing the conversion of A into B. (iii) What happens when compound C undergoes combustion? (iv) State one industrial application of hydrogenation reaction. (v) Name the products formed when compound A reacts with sodium. **Part (i)** 1. Compound B is an alkene with molecular mass $28\text{ u}$, which corresponds to ethene ($\text{C}_2\text{H}_4$). 2. Compound A is an alcohol that forms ethene on heating with concentrated $\text{H}_2\text{SO}_4$, so A is ethanol ($\text{C}_2\text{H}_5\text{OH}$). 3. Compound C is formed by the hydrogenation of B, so C is ethane ($\text{C}_2\text{H}_6$). Answer (i): A: Ethanol ($\text{C}_2\text{H}_5\text{OH}$), B: Ethene ($\text{C}_2\text{H}_4$), C: Ethane ($\text{C}_2\text{H}_6$). **Part (ii)** 1. Ethanol is heated with excess concentrated sulphuric acid at $443\text{ K}$. 2. $\text{CH}_3-\text{CH}_2-\text{OH} \xrightarrow{\text{Conc. }\text{H}_2\text{SO}_4, 443\text{ K}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$ Answer (ii): $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Conc. }\text{H}_2\text{SO}_4, 443\text{ K}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$ **Part (iii)** 1. Compound C is ethane, which is a saturated hydrocarbon. 2. It undergoes combustion in the presence of air to give carbon dioxide, water, heat, and light. 3. $2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} + \text{Heat and light}$ Answer (iii): It burns with a clean flame to produce carbon dioxide, water, heat, and light. **Part (iv)** 1. Hydrogenation is used in the food industry for the hardening of vegetable oils. 2. Vegetable oils are converted into vegetable ghee using nickel catalyst. Answer (iv): Hydrogenation of vegetable oils to form vegetable ghee (vanaspati fat). **Part (v)** 1. Ethanol reacts with sodium to form sodium ethoxide and hydrogen gas. 2. $2\text{CH}_3\text{CH}_2\text{OH} + 2\text{Na} \rightarrow 2\text{CH}_3\text{CH}_2\text{O}^-\text{Na}^+ + \text{H}_2\uparrow$ Answer (v): Sodium ethoxide and hydrogen gas. **Answer:** Identified compounds A, B, C along with their chemical reactions, combustion behavior, industrial application, and reaction with sodium. > Common mistake: Confusing the molecular mass of ethene with ethane or writing incorrect balancing for combustion reactions. ### Question 34 (OR) *5 marks · Long answer · Carbon and its Compounds* (b) (i) With the help of diagram, show the formation of micelles, when soap is applied on oily dirt. (ii) Take two test tubes X and Y with $10\text{ mL}$ of hard water in each. In test tube 'X', add few drops of soap solution and in test tube 'Y' add a few drops of detergent solution. Shake both the test tubes for the same period. (1) In which test tube the formation of foam will be more? Why? (2) In which test tube is a curdy solid formed? Why? **Part (i)** 1. Diagram: Draw a spherical aggregate of soap molecules (micelle). Show the hydrophobic hydrocarbon tail pointing towards the oily dirt at the centre and the hydrophilic ionic head ($-\text{COO}^-\text{Na}^+$) facing outward towards water. 2. When soap is dissolved in water, its hydrophobic tails attach to the oily dirt and the hydrophilic heads dissolve in water, forming a structure called a micelle. Answer (i): Micelle formation diagram showing hydrophobic tails towards oil and hydrophilic heads towards water. **Part (ii)** 1. 1. Foam formation will be more in test tube 'Y' containing detergent solution because detergents do not form insoluble precipitates (scum) with calcium and magnesium ions present in hard water. 2. 2. A curdy solid (scum) is formed in test tube 'X' containing soap solution because soap reacts with calcium and magnesium salts in hard water to form insoluble salts. Answer (ii): 1. More foam in test tube Y (detergent). 2. Curdy solid in test tube X (soap due to scum formation). **Answer:** Explained micelle formation with diagram and compared the action of soap and detergent in hard water. > Common mistake: Failing to label the hydrophilic head and hydrophobic tail correctly in the micelle diagram. ### Question 35 *5 marks · Long answer · How do Organisms Reproduce?* (a) Name the parts of a bisexual flower that are not directly involved in reproduction. (b) Differentiate between self pollination and cross pollination. List any two significance of pollination. (c) What is the fate of ovules and ovary after fertilization in a flower? **Part (a)** 1. The non-essential whorls of a bisexual flower that are not directly involved in reproduction are sepals (calyx) and petals (corolla). Answer (a): Sepals and petals. **Part (b)** 1. Self-pollination is the transfer of pollen grains from the anther to the stigma of the same flower or another flower of the same plant, whereas cross-pollination is the transfer of pollen from the anther of one flower to the stigma of another flower of a different plant of the same species. 2. Significance of pollination: (1) It brings male and female gametes together for fertilization. (2) It leads to the formation of seeds and fruits, ensuring continuity of species. Answer (b): Self-pollination occurs within the same plant; cross-pollination occurs between different plants. Significance: fertilization and seed formation. **Part (c)** 1. After fertilization, the ovules develop into seeds and the ovary ripens and develops into a fruit. Answer (c): Ovules develop into seeds and the ovary develops into a fruit. **Answer:** Answered plant reproduction parts, differences between pollination types, and post-fertilization changes. > Common mistake: Confusing the fate of ovary and ovule after fertilization. ### Question 36 *5 marks · Long answer · Electricity* (a) An electric iron consumes energy at a rate of $880\text{ W}$ when heating is at the maximum rate and $330\text{ W}$ when the heating is at the minimum. If the source voltage is $220\text{ V}$, calculate the current and resistance in each case. (b) What is heating effect of electric current? (c) Find an expression for the amount of heat produced when a current passes through a resistor for some time. **Part (a)** 1. Given: Voltage $V = 220\text{ V}$. Case 1 power $P_1 = 880\text{ W}$. 2. Current $I_1 = \frac{P_1}{V} = \frac{880}{220} = 4\text{ A}$. 3. Resistance $R_1 = \frac{V}{I_1} = \frac{220}{4} = 55\text{ }\Omega$. 4. Case 2 power $P_2 = 330\text{ W}$. Current $I_2 = \frac{P_2}{V} = \frac{330}{220} = 1.5\text{ A}$. 5. Resistance $R_2 = \frac{V}{I_2} = \frac{220}{1.5} = 146.67\text{ }\Omega$. Answer (a): Case 1: $4\text{ A}$, $55\text{ }\Omega$; Case 2: $1.5\text{ A}$, $146.67\text{ }\Omega$. **Part (b)** 1. When an electric current is passed through a high resistance wire, the resistance wire becomes very hot and produces heat. This is called the heating effect of electric current. Answer (b): The generation of heat in a resistor due to the flow of electric current through it. **Part (c)** 1. Let a current $I$ flow through a resistor of resistance $R$ for time $t$, with potential difference $V$ across its ends. 2. Work done $W$ in moving charge $Q$ is $W = V \times Q = V \times I \times t$. 3. Since $V = I \times R$ according to Ohm's law, substituting this gives $H = (I \times R) \times I \times t = I^2Rt$. Answer (c): Expression: $H = I^2Rt$. **Answer:** Calculated current and resistance for both cases, defined heating effect, and derived Joule's law expression. > Common mistake: Using incorrect formulas for resistance calculation like $R = \frac{V^2}{P}$ without deriving or properly substituting values. ## SECTION - E Q. No. 37 to 39 are case based/data based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts. ### Question 37 *4 marks · Case-based · Metals and Non-metals* Almost all metals combine with oxygen to form metal oxides. Metal oxides are generally basic in nature. But some metal oxides show both basic as well as acidic behaviour. Different metals show different reactivities towards oxygen. Some react vigorously while some do not react at all. (a) What happens when copper is heated in air ? (Give the equation of the reaction involved). (b) Why are some metal oxides categorized as amphoteric ? Give one example. (c) Complete the following equations : (i) $\text{Na}_2\text{O}_{(s)} + \text{H}_2\text{O}_{(l)} \rightarrow$ (ii) $\text{Al}_2\text{O}_3 + 2\text{NaOH} \rightarrow$ **Part (a)** 1. When copper is heated in air, it combines with oxygen to form copper(II) oxide, which is a black-coloured coating. 2. The chemical equation for the reaction is: $2\text{Cu}_{(s)} + \text{O}_{2(g)} \xrightarrow{\text{Heat}} 2\text{CuO}_{(s)}$ Answer (a): Copper reacts with oxygen on heating to form black copper(II) oxide ($2\text{CuO}$). **Part (b)** 1. Metal oxides that react with both acids as well as bases to produce salt and water are known as amphoteric oxides. 2. Example: Aluminium oxide ($\text{Al}_2\text{O}_3$) or Zinc oxide ($\text{ZnO}$). Answer (b): Metal oxides showing both basic and acidic behaviour are called amphoteric oxides, for example, $\text{Al}_2\text{O}_3$. **Part (c)(i)** 1. Soluble metal oxides dissolve in water to form alkalis. 2. The completed equation is: $\text{Na}_2\text{O}_{(s)} + \text{H}_2\text{O}_{(l)} \rightarrow 2\text{NaOH}_{(aq)}$ Answer (c)(i): $\text{Na}_2\text{O}_{(s)} + \text{H}_2\text{O}_{(l)} \rightarrow 2\text{NaOH}_{(aq)}$ **Part (c)(ii)** 1. Aluminium oxide reacts with sodium hydroxide to form sodium aluminate and water. 2. The completed equation is: $\text{Al}_2\text{O}_{3(s)} + 2\text{NaOH}_{(aq)} \rightarrow 2\text{NaAlO}_{2(aq)} + \text{H}_2\text{O}_{(l)}$ Answer (c)(ii): $\text{Al}_2\text{O}_{3(s)} + 2\text{NaOH}_{(aq)} \rightarrow 2\text{NaAlO}_{2(aq)} + \text{H}_2\text{O}_{(l)}$ **Answer:** Copper forms a black copper(II) oxide on heating, amphoteric oxides react with both acids and bases, and the completed reactions yield $\text{NaOH}$ and $\text{NaAlO}_2$ respectively. > Common mistake: Forgetting to balance chemical equations or missing physical states. ### Question 37 (OR) *2 marks · Case-based · Metals and Non-metals* On burning Sulphur in oxygen a colourless gas is produced. (i) Write chemical equation for the reaction. (ii) Name the gas formed. (iii) State the nature of the gas. (iv) What will be the action of this on a dry litmus paper? **Part (i)** 1. Chemical equation: $\text{S}_{(s)} + \text{O}_{2(g)} \rightarrow \text{SO}_{2(g)}$ Answer (i): $\text{S} + \text{O}_2 \rightarrow \text{SO}_2$ **Part (ii)** 1. The colourless gas formed is Sulphur dioxide ($\text{SO}_2$). Answer (ii): Sulphur dioxide **Part (iii)** 1. Sulphur dioxide is acidic in nature. Answer (iii): Acidic **Part (iv)** 1. There will be no action on dry litmus paper because moisture is required to produce hydrogen ions for the acidic behaviour. Answer (iv): No action **Answer:** Sulphur dioxide gas is formed which is acidic in nature but shows no action on dry litmus paper. > Common mistake: Stating that dry litmus paper turns red, forgetting that water is needed to form acid from acidic gases. ### Question 38 *4 marks · Case-based · Heredity* In order to trace the inheritance of traits Mendel crossed pea plants having one contrasting character or a pair of contrasting characters. When he crossed pea plants having round and yellow seeds with pea plants having wrinkled and green seeds, he observed that no plants with wrinkled and green seeds were obtained in the $F_1$ generation. When the $F_1$ generation pea plants were cross-bred by self-pollination, the $F_2$ generation had seeds with different combinations of shape and colour also. (a) Write any two pairs of contrasting characteristics of pea plant used by Mendel other than those mentioned above. (b) Differentiate between dominant and recessive traits. (c) State the ratio of the combinations observed in the seeds of $F_2$ generation (in the above case). What do you interpret from this result? **Part (a)** 1. Pair 1: Tall / Dwarf stem height 2. Pair 2: Violet / White flower colour (or Inflated / Constricted pod shape) Answer (a): Tall/dwarf stem and violet/white flowers **Part (b)** 1. Dominant trait: The trait that expresses itself in the presence or absence of another contrasting trait in the $F_1$ generation. 2. Recessive trait: The trait that remains hidden or masked in the presence of a dominant trait and expresses only in homozygous condition. Answer (b): Dominant traits express in $F_1$ while recessive traits remain hidden. **Part (c)** 1. Ratio of combinations in $F_2$ generation: $9:3:3:1$ (Round yellow : Round green : Wrinkled yellow : Wrinkled green). 2. Interpretation: The two pairs of traits are inherited independently of each other (Law of Independent Assortment). Answer (c): Ratio is 9:3:3:1; traits assort independently. **Answer:** Contrasting traits include tall/dwarf stem, dominant traits express themselves while recessive remain hidden, and dihybrid ratio is 9:3:3:1 showing independent assortment. > Common mistake: Confusing monohybrid phenotypic ratio (3:1) with dihybrid phenotypic ratio (9:3:3:1). ### Question 38 (OR) *2 marks · Case-based · Heredity* Given below is a cross between a pure violet flowered pea plant (V) and a pure white flowered pea plant (v). Diagrammatically explain what type of progeny is obtained in $F_1$ generation and $F_2$ generation : Pure violet flowered plant $\times$ Pure white flowered plant ($V\,V$) $\qquad\qquad\qquad\qquad$ ($v\,v$) **Part (i)** 1. Parents: Pure violet ($V\,V$) $\times$ Pure white ($v\,v$) 2. $F_1$ generation: All plants are violet flowered with genotype $V\,v$ due to dominance of allele $V$. Answer (i): All violet-flowered plants ($V\,v$) **Part (ii)** 1. Selfing of $F_1$ ($V\,v \times V\,v$) yields $F_2$ generation. 2. $F_2$ progeny has phenotypic ratio $3:1$ (violet to white) and genotypic ratio $1:2:1$ ($V\,V : V\,v : v\,v$). Answer (ii): 3 violet to 1 white flowered plants **Answer:** F1 progeny are all violet-flowered (Vv) and F2 progeny show violet and white flowers in 3:1 ratio. > Common mistake: Not showing genotypes alongside phenotypes in crosses. ### Question 39 *4 marks · Case-based · Light – Reflection and Refraction* Hold a concave mirror in your hand and direct its reflecting surface towards the sun. Direct the light reflected by the mirror on to a white card-board held close to the mirror. Move the card-board back and forth gradually until you find a bright, sharp spot of light on the board. This spot of light is the image of the sun on the sheet of paper; which is also termed as "Principal Focus" of the concave mirror. [Diagram showing parallel rays converging at principal focus of concave mirror] (a) List two applications of concave mirror. (b) If the distance between the mirror and the principal focus is $15\text{ cm}$, find the radius of curvature of the mirror. (c) Draw a ray diagram to show the type of image formed when an object is placed between pole and focus of a concave mirror. **Part (a)** 1. Application 1: Used in torches, search-lights and vehicle headlights to get powerful parallel beams of light. 2. Application 2: Used as shaving mirrors to see a larger image of the face. Answer (a): Used in torches and as shaving mirrors **Part (b)** 1. Focal length ($f$) = $15\text{ cm}$ 2. Radius of curvature ($R$) = $2f = 2 \times 15\text{ cm} = 30\text{ cm}$ Answer (b): 30 cm **Part (c)** 1. Diagram: Draw a concave mirror with pole P, focus F, and centre of curvature C. 2. Place object between P and F. Draw one ray parallel to principal axis passing through F, and another ray through C. 3. Produce rays backward to intersect behind the mirror, forming an enlarged, virtual and erect image. Answer (c): Ray diagram showing virtual, erect, and magnified image behind the mirror. **Answer:** Concave mirrors are used in torches and shaving mirrors, radius of curvature is 30 cm, and ray diagram shows a virtual and erect image. > Common mistake: Forgetting arrows on light rays in ray diagrams. ### Question 39 (OR) *2 marks · Case-based · Light – Reflection and Refraction* An object $10\text{ cm}$ in size is placed at $100\text{ cm}$ in front of a concave mirror. If its image is formed at the same point where the object is located, find : (i) focal length of the mirror, and (ii) magnification of the image formed with sign as per Cartesian sign convention. **Part (i)** 1. Given object size $h = 10\text{ cm}$, object distance $u = -100\text{ cm}$. 2. Since the image is formed at the same point where the object is located, the object is placed at the centre of curvature C. 3. Therefore, image distance $v = -100\text{ cm}$ and radius of curvature $R = -100\text{ cm}$. 4. Focal length $f = \frac{R}{2} = \frac{-100\text{ cm}}{2} = -50\text{ cm}$. Answer (i): -50 cm **Part (ii)** 1. Magnification $m = -\frac{v}{u}$ 2. Substitute values: $m = -\frac{-100\text{ cm}}{-100\text{ cm}} = -1$ Answer (ii): -1 **Answer:** Focal length is -50 cm and magnification is -1. > Common mistake: Writing positive sign for magnification of a real and inverted image formed at the centre of curvature. ## Frequently asked questions ### What is the paper pattern and section breakdown for the CBSE Class 10 Science Question Paper 2023 Set 31/1/1? The paper carries a total of 80 marks and is to be completed in 180 minutes. It is divided into five sections from Section A to Section E, containing a total of 39 questions. ### What are the marks allotted to each section in this Science paper? Section A has 20 questions for 20 marks and Section B has 6 questions for 12 marks. Section C contains 7 questions for 21 marks, Section D has 3 questions for 15 marks, and Section E includes 3 questions for 12 marks. ### Which chapters carry the most marks in the CBSE Class 10 Science 2023 Set 31/1/1 paper? Life Processes carries the highest weightage with 10 marks. Acids, Bases and Salts and Electricity carry 8 marks each, followed by Light – Reflection and Refraction at 7 marks, and Metals and Non-metals and Carbon and its Compounds at 6 marks each. ### How should students write answers to score full marks in this examination? Students should write clear, step-by-step answers using proper scientific terms and labelled diagrams where necessary. For numerical problems, write the given values, formula, and final answer with correct SI units. ### Is the solutions PDF for this CBSE Class 10 Science question paper available for free? Yes, the complete solutions PDF for the CBSE Class 10 Science 2023 Set 31/1/1 paper is available for free download on SwaVid. You can access it easily to check your answers and understand the correct marking scheme. ## Related pages - [All CBSE Class 10 Science papers](https://www.swavid.com/cbse/class-10/science/previous-year-papers) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student