ADEPT YOURSELF
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CBSE CLASS X – SCIENCE SAMPLE QUESTION PAPER
SET – 5 | ACADEMIC SESSION 2026–2027
General Instructions:
- This question paper consists of 39 questions divided into Sections A, B and C.
- All questions are compulsory.
- Internal choices are provided in some questions. Attempt only one alternative wherever applicable.
- Questions include MCQs, Assertion–Reason, Very Short Answer, Short Answer, Case-Based and Long Answer questions.
- Marks are indicated against each question.
- Write all answers clearly and show necessary calculations wherever required.
- Draw neat and properly labelled diagrams wherever asked.
SECTION A – BIOLOGY
Q1.
The opening and closing of the stomatal pore depends upon:[1]
Q2.
The filtration units present in the human kidneys are called:[1]
Q3.
Involuntary actions like salivation, blood pressure, and vomiting are controlled by which part of the brain?[1]
Q4.
Vegetative propagation by leaves is observed in which of the following organisms?[1]
Q5.
According to the 10% law of energy transfer in an ecosystem, if 10,000 J of solar energy falls on green plants, the amount of energy trapped and available to the primary consumers will be:[1]
Q6.
Which of the following represents the correct phenotypic ratio in the F₂ generation of a Mendelian dihybrid cross?[1]
Q7.Assertion (A):
Translocation of sucrose in the phloem tissue requires energy in the form of ATP.
Reason (R): Active transport creates osmotic pressure that moves water into the phloem and helps transport food to tissues with less pressure.[1]
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Reason (R): Active transport creates osmotic pressure that moves water into the phloem and helps transport food to tissues with less pressure.[1]
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Q8.Assertion (A):
The depletion of the ozone layer in the stratosphere causes severe health hazards including skin cancer in humans.
Reason (R): Ozone absorbs harmful ultraviolet (UV) radiations emitted by the Sun.[1]
Reason (R): Ozone absorbs harmful ultraviolet (UV) radiations emitted by the Sun.[1]
Q9.
Bile juice contains no digestive enzymes, yet it plays a crucial role in digestion. Justify this statement with two functions.[2]
Q10.
Differentiate between biodegradable and non-biodegradable substances. Give one example of each.[2]
Q11.(a) Name the structural and functional unit of the nervous system.
(b) How does an electrical impulse travel across a synapse between two successive neurons?
(c) State one difference between nervous and chemical (hormonal) control in animals.[3]
(b) How does an electrical impulse travel across a synapse between two successive neurons?
(c) State one difference between nervous and chemical (hormonal) control in animals.[3]
Q12.
A pure tall pea plant with round seeds (TTRR) is crossed with a pure dwarf pea plant with wrinkled seeds (ttrr).
(a) What will be the phenotype and genotype of the plants in the F₁ generation?
(b) State the four distinct combinations of traits obtained in the F₂ generation.
(c) Name the law of inheritance illustrated by this cross.[3]
(a) What will be the phenotype and genotype of the plants in the F₁ generation?
(b) State the four distinct combinations of traits obtained in the F₂ generation.
(c) Name the law of inheritance illustrated by this cross.[3]
Q13.
Explain the three major steps that take place during the process of photosynthesis in green plants. Why do desert plants have a different sequence of these steps?[3]
Q14. Read the following passage and answer the questions:[4]
(b) In a food chain comprising Grass → Grasshopper → Frog → Snake → Hawk, which organism will accumulate the highest concentration of toxic chemicals?
(c) Why is energy flow in an ecosystem always unidirectional?
OR
If 20,000 J of energy is available to producers (plants), calculate the amount of energy that reaches the tertiary consumer according to Lindeman’s 10% law.
In an ecosystem, energy enters through producers that convert solar energy into chemical energy during photosynthesis. This energy is then transferred across various trophic levels along a food chain. However, harmful non-biodegradable chemicals such as pesticides also enter the food chain through soil or water. Since these chemicals cannot be metabolised or excreted, their concentration progressively increases at each successive trophic level.
(a) Name the biological phenomenon in which the concentration of harmful non-biodegradable chemicals increases at successive trophic levels.(b) In a food chain comprising Grass → Grasshopper → Frog → Snake → Hawk, which organism will accumulate the highest concentration of toxic chemicals?
(c) Why is energy flow in an ecosystem always unidirectional?
OR
If 20,000 J of energy is available to producers (plants), calculate the amount of energy that reaches the tertiary consumer according to Lindeman’s 10% law.
Q15.(a) Draw a neat labelled diagram of the human male reproductive system.
(b) State the functions of the following organs:
(i) Testes (ii) Vas deferens (iii) Prostate gland and Seminal vesicles
(c) Name one bacterial and one viral sexually transmitted disease (STD).[5]
(i) Hydrochloric acid in the stomach
(ii) Mucus in gastric juice
(iii) Trypsin and Lipase secreted by the pancreas
(b) How are the small intestine walls designed to maximise the absorption of digested food?
(b) State the functions of the following organs:
(i) Testes (ii) Vas deferens (iii) Prostate gland and Seminal vesicles
(c) Name one bacterial and one viral sexually transmitted disease (STD).[5]
OR
(a) Describe the role of the following secretions in human digestion:
(i) Hydrochloric acid in the stomach
(ii) Mucus in gastric juice
(iii) Trypsin and Lipase secreted by the pancreas
(b) How are the small intestine walls designed to maximise the absorption of digested food?
SECTION B – CHEMISTRY
Q16.
During the electrolysis of water, the ratio of the volume of hydrogen gas to oxygen gas collected at the respective electrodes is:[1]
Q17.
Which of the following compounds is known as Plaster of Paris?[1]
Q18.
Which of the following pairs of oxides are amphoteric in nature?[1]
Q19.
Two consecutive members of a homologous series differ in their molecular formula by:[1]
Q20.
When green crystals of ferrous sulphate are heated strongly in a dry test tube, the observation recorded is:[1]
Q21.Assertion (A):
Highly reactive metals like sodium and aluminium cannot be obtained from their oxides by heating with carbon.
Reason (R): These metals have a greater affinity for oxygen than carbon has.[1]
Reason (R): These metals have a greater affinity for oxygen than carbon has.[1]
Q22.
Lead nitrate crystals are heated in a boiling tube.
(a) Write a balanced chemical equation for the reaction.
(b) Name the brown gas evolved and state the type of reaction taking place.[2]
(a) Write a balanced chemical equation for the reaction.
(b) Name the brown gas evolved and state the type of reaction taking place.[2]
Q23.
Differentiate between roasting and calcination with one suitable chemical equation for each process during the extraction of zinc.[3]
(b) State two methods commonly used to prevent the rusting of iron.
OR
(a) What is corrosion? Name two necessary conditions for rusting of iron.(b) State two methods commonly used to prevent the rusting of iron.
Q24.(a) Define a homologous series of carbon compounds.
(b) Write the general formula for the homologous series of alkenes.
(c) Write the molecular formulas and names of the first two members of the alkene series.[3]
(b) Write the general formula for the homologous series of alkenes.
(c) Write the molecular formulas and names of the first two members of the alkene series.[3]
Q25. Read the following passage and answer the questions:[4]
(b) Name the gas released at the cathode.
(c) Write the balanced chemical equation for the chlor-alkali process.
OR
State one important industrial use of each of the three products formed in the chlor-alkali process (NaOH, Cl₂, H₂).
When electricity is passed through an aqueous solution of sodium chloride (called brine), it decomposes to form sodium hydroxide, chlorine gas, and hydrogen gas. This commercial method is known as the chlor-alkali process because of the products formed: chlor for chlorine and alkali for sodium hydroxide. All three products obtained in this process are valuable raw materials for various chemical industries.
(a) Name the gas released at the anode during the chlor-alkali process.(b) Name the gas released at the cathode.
(c) Write the balanced chemical equation for the chlor-alkali process.
OR
State one important industrial use of each of the three products formed in the chlor-alkali process (NaOH, Cl₂, H₂).
Q26.(a) What is an addition reaction? Explain the process of hydrogenation of vegetable oils with a chemical equation.
(b) Explain the substitution reaction of methane with chlorine in the presence of sunlight with a balanced chemical equation.
(c) Why are unsaturated hydrocarbons generally more reactive than saturated hydrocarbons?[5]
(b) How is baking soda prepared industrially? Write the balanced chemical equation involved.
(c) Explain what happens when baking soda is heated during cooking. Give the chemical equation.
(d) State two important uses of washing soda (Na₂CO₃ · 10H₂O).
(b) Explain the substitution reaction of methane with chlorine in the presence of sunlight with a balanced chemical equation.
(c) Why are unsaturated hydrocarbons generally more reactive than saturated hydrocarbons?[5]
OR
(a) Write the chemical formula and chemical name of Baking Soda.(b) How is baking soda prepared industrially? Write the balanced chemical equation involved.
(c) Explain what happens when baking soda is heated during cooking. Give the chemical equation.
(d) State two important uses of washing soda (Na₂CO₃ · 10H₂O).
Q27.
Explain why tooth decay starts when the pH of the mouth is lower than 5.5. How does the use of toothpaste prevent this condition?[2]
SECTION C – PHYSICS
Q28.
When a ray of light travels from an optically rarer medium to an optically denser medium, it:[1]
Q29.
The defect of vision in which a person cannot clearly see distant objects is called:[1]
Q30.
According to Joule’s law of heating, the heat produced in a resistor is directly proportional to:[1]
Q31.
The magnetic field lines inside a current-carrying long straight solenoid are:[1]
Q32.Assertion (A):
A ray of light passing through the optical centre of a thin spherical lens emerges without suffering any deviation.
Reason (R): The central portion of a thin spherical lens acts like a thin rectangular glass slab with parallel faces.[1]
Reason (R): The central portion of a thin spherical lens acts like a thin rectangular glass slab with parallel faces.[1]
Q33.Assertion (A):
Tungsten metal is almost exclusively used for making filaments of incandescent electric lamps.
Reason (R): Tungsten has an exceptionally high melting point and high resistivity.[1]
Reason (R): Tungsten has an exceptionally high melting point and high resistivity.[1]
Q34.
A concave mirror of focal length 15 cm forms a real image at a distance of 30 cm from the mirror. Calculate the object distance from the mirror and the magnification produced.[2]
Q35.
List the four factors on which the electrical resistance of a uniform metallic conductor depends.[2]
OR
Compute the amount of heat generated while transferring 96,000 coulombs of charge in 1 hour through a potential difference of 50 V.Q36.(a) What is dispersion of white light?
(b) Draw a labelled ray diagram showing the dispersion of white light through a triangular glass prism.
(c) Which colour of light deviates the most and which deviates the least? Give a reason.[3]
(b) Draw a labelled ray diagram showing the dispersion of white light through a triangular glass prism.
(c) Which colour of light deviates the most and which deviates the least? Give a reason.[3]
Q37.
State three major disadvantages of connecting electrical appliances in a series arrangement in household domestic circuits.[3]
Q38. Read the following passage and answer the questions:[4]
(b) Which type of corrective lens is used to restore normal vision for a person suffering from Hypermetropia?
(c) What is the near point and far point for a normal human adult eye?
OR
A person with a vision defect requires a corrective lens of power -2.5 D. Name the vision defect and calculate the focal length of the lens.
The human eye functions like an optical camera. The crystalline lens forms an inverted real image of an object on the light-sensitive retina. The ciliary muscles adjust the curvature and focal length of the eye lens to see nearby and distant objects clearly. This ability is called accommodation. When the eye fails to focus light properly on the retina due to changes in eyeball length or lens curvature, defects of vision occur.
(a) Name the defect of vision in which the eyeball becomes too long or the focal length of the eye lens becomes too short.(b) Which type of corrective lens is used to restore normal vision for a person suffering from Hypermetropia?
(c) What is the near point and far point for a normal human adult eye?
OR
A person with a vision defect requires a corrective lens of power -2.5 D. Name the vision defect and calculate the focal length of the lens.
Q39.(a) State Joule’s Law of Heating.
(b) Derive an expression for the equivalent resistance of three resistors R₁, R₂, and R₃ connected in parallel across a battery of voltage V.
(c) Three resistors of 5 Ω, 10 Ω, and 30 Ω are connected in parallel across a 12 V battery. Calculate:
(i) Total effective resistance of the circuit.
(ii) Total current drawn from the battery.[5]
(b) Why do two magnetic field lines never cross (intersect) each other?
(c) State three factors on which the strength of the magnetic field produced by a current-carrying circular coil depends.
(b) Derive an expression for the equivalent resistance of three resistors R₁, R₂, and R₃ connected in parallel across a battery of voltage V.
(c) Three resistors of 5 Ω, 10 Ω, and 30 Ω are connected in parallel across a 12 V battery. Calculate:
(i) Total effective resistance of the circuit.
(ii) Total current drawn from the battery.[5]
OR
(a) Draw the magnetic field lines around a straight current-carrying conductor. State the rule used to determine the direction of the magnetic field.(b) Why do two magnetic field lines never cross (intersect) each other?
(c) State three factors on which the strength of the magnetic field produced by a current-carrying circular coil depends.
CBSE CLASS X SCIENCE – COMPLETE SOLUTIONS & MARKING SCHEME (SET – 5)
SECTION A – BIOLOGY SOLUTIONS
Q1. (c) Water content in guard cells [1 Mark]
Explanation: The opening and closing of the stomatal pore is a function of the guard cells. When water flows into the guard cells, they swell, become turgid, and curve outward, opening the pore. When they lose water, they shrink and become flaccid, closing the pore.
Explanation: The opening and closing of the stomatal pore is a function of the guard cells. When water flows into the guard cells, they swell, become turgid, and curve outward, opening the pore. When they lose water, they shrink and become flaccid, closing the pore.
Q2. (b) Nephrons [1 Mark]
Explanation: Nephrons are the basic structural and functional filtration units of the kidney, consisting of a glomerulus, Bowman’s capsule, and tubular system.
Explanation: Nephrons are the basic structural and functional filtration units of the kidney, consisting of a glomerulus, Bowman’s capsule, and tubular system.
Q3. (c) Medulla in the hindbrain [1 Mark]
Explanation: All involuntary actions including blood pressure, salivation, peristalsis, and vomiting are controlled by the medulla located in the hindbrain.
Explanation: All involuntary actions including blood pressure, salivation, peristalsis, and vomiting are controlled by the medulla located in the hindbrain.
Q4. (a) Bryophyllum [1 Mark]
Explanation: In Bryophyllum, adventitious buds develop in the notches along the leaf margins. When these buds fall on moist soil, they grow into new independent plants.
Explanation: In Bryophyllum, adventitious buds develop in the notches along the leaf margins. When these buds fall on moist soil, they grow into new independent plants.
Q5. (b) 100 J [1 Mark]
Explanation: Green plants capture only about 1% of total incident solar energy: 1% of 10,000 J = 100 J. Then, primary consumers receive 10% of this trapped energy (10% of 100 J = 10 J). Note: The question asks for the energy trapped by plants (100 J) which is available to the primary consumers.
Explanation: Green plants capture only about 1% of total incident solar energy: 1% of 10,000 J = 100 J. Then, primary consumers receive 10% of this trapped energy (10% of 100 J = 10 J). Note: The question asks for the energy trapped by plants (100 J) which is available to the primary consumers.
Q6. (c) 9 : 3 : 3 : 1 [1 Mark]
Explanation: The classical dihybrid phenotypic ratio in the F₂ generation is 9 (Dominant-Dominant) : 3 (Dominant-Recessive) : 3 (Recessive-Dominant) : 1 (Recessive-Recessive).
Explanation: The classical dihybrid phenotypic ratio in the F₂ generation is 9 (Dominant-Dominant) : 3 (Dominant-Recessive) : 3 (Recessive-Dominant) : 1 (Recessive-Recessive).
Q7. (a) Both A and R are true and R is the correct explanation of A. [1 Mark]
Explanation: Translocation in phloem uses ATP energy to build osmotic pressure, which pulls water from xylem into phloem and drives food to lower pressure tissues.
Explanation: Translocation in phloem uses ATP energy to build osmotic pressure, which pulls water from xylem into phloem and drives food to lower pressure tissues.
Q8. (a) Both A and R are true and R is the correct explanation of A. [1 Mark]
Explanation: The stratospheric ozone layer blocks harmful solar UV rays. Its depletion allows UV rays to penetrate the atmosphere, causing mutations, skin cancer, and cataracts.
Explanation: The stratospheric ozone layer blocks harmful solar UV rays. Its depletion allows UV rays to penetrate the atmosphere, causing mutations, skin cancer, and cataracts.
Q9. Role of Bile Juice: [2 Marks]
- Emulsification of Fats: Bile salts break down large fat globules into smaller globules, significantly increasing the surface area for the action of the enzyme lipase. [1 Mark]
- Alkaline Medium: It neutralises the acidic food (chyme) arriving from the stomach and makes it alkaline, which is necessary for pancreatic enzymes to function. [1 Mark]
Q10. Biodegradable vs Non-Biodegradable: [2 Marks]
- Biodegradable substances: Substances that can be broken down into simpler, harmless substances by biological processes/microorganisms. Example: Vegetable peels, paper, cow dung. [1 Mark]
- Non-biodegradable substances: Substances that cannot be broken down by microorganisms and persist in the environment for a long time. Example: Plastics, DDT, glass. [1 Mark]
Q11. [3 Marks]
- (a) Neuron (nerve cell) is the structural and functional unit. [0.5 Mark]
- (b) Impulse transmission across a synapse: When an electrical impulse reaches the axonal end of a neuron, it triggers the release of neurotransmitter chemicals. These chemicals diffuse across the synaptic gap and bind to receptors on the dendrite of the next neuron, setting off a new electrical impulse. [1.5 Marks]
- (c) Difference: Nervous control is rapid and transmits electrical impulses through nerves with short-lived effects, whereas hormonal control is comparatively slower, transmitted through blood chemical messengers, and produces longer-lasting widespread effects. [1 Mark]
Q12. Dihybrid Cross: [3 Marks]
- (a) F₁ Generation: Genotype = TtRr; Phenotype = Tall plants with Round seeds. [1 Mark]
- (b) F₂ Traits Combinations: (i) Tall Round, (ii) Tall Wrinkled, (iii) Dwarf Round, (iv) Dwarf Wrinkled (in the ratio 9 : 3 : 3 : 1). [1 Mark]
- (c) Law Illustrated: Law of Independent Assortment (Mendel’s third law). [1 Mark]
Q13. Photosynthesis Steps: [3 Marks]
- Absorption of light energy by chlorophyll. [0.5 Mark]
- Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen. [0.5 Mark]
- Reduction of carbon dioxide to carbohydrates. [1 Mark]
Q14. Case-Based Solutions (Biology): [4 Marks]
- (a) Biomagnification (or Biological Magnification). [1 Mark]
- (b) Hawk (top consumer / highest trophic level). [1 Mark]
- (c) Energy captured by autotrophs does not revert back to the solar input and energy passed to herbivores does not return to autotrophs; as it moves progressively through trophic levels, energy is dissipated as metabolic heat and cannot be reused. [2 Marks]
- Producers = 20,000 J
Primary Consumers (Herbivores) = 10% of 20,000 = 2,000 J
Secondary Consumers (Carnivores) = 10% of 2,000 = 200 J
Tertiary Consumers (Top Carnivores) = 10% of 200 = 20 J [2 Marks]
Q15. Long Answer (Biology): [5 Marks]
- (a) Diagram: Neat diagram of the male reproductive system labelling Testis, Scrotum, Vas deferens, Prostate gland, Seminal vesicle, Urethra, and Penis. [2 Marks]
- (b) Functions:
- Testes: Produce male gametes (sperms) and secrete the hormone testosterone. [1 Mark]
- Vas deferens: Transports mature sperms from the epididymis/testes towards the urethra. [0.5 Mark]
- Prostate gland & Seminal vesicles: Secrete fluid that provides nourishment and a liquid medium to make sperm transport easier. [0.5 Mark]
- (c) STDs: Bacterial: Gonorrhoea / Syphilis (any one); Viral: HIV-AIDS / Genital Warts (any one). [1 Mark]
- (a) Digestive Secretions:
- Hydrochloric acid (HCl): Creates an acidic medium (pH ~1.5–2.0) necessary for the activation and action of the enzyme pepsin; kills ingested bacteria. [1 Mark]
- Mucus: Protects the inner lining of the stomach from the corrosive action of HCl under normal conditions. [1 Mark]
- Trypsin: Digests proteins into peptones/peptides. Lipase: Breaks down emulsified fats into fatty acids and glycerol. [1.5 Marks]
- (b) Adaptations of small intestine: The inner lining has millions of tiny finger-like projections called villi that dramatically increase the surface area for absorption. The villi are richly supplied with blood vessels that carry absorbed nutrients to every cell of the body. [1.5 Marks]
SECTION B – CHEMISTRY SOLUTIONS
Q16. (b) 2 : 1 [1 Mark]
Explanation: Water has the chemical formula H₂O. During electrolysis, 2 molecules of water dissociate into 2 molecules of hydrogen gas (at cathode) and 1 molecule of oxygen gas (at anode): 2H₂O(l) → 2H₂(g) + O₂(g).
Explanation: Water has the chemical formula H₂O. During electrolysis, 2 molecules of water dissociate into 2 molecules of hydrogen gas (at cathode) and 1 molecule of oxygen gas (at anode): 2H₂O(l) → 2H₂(g) + O₂(g).
Q17. (b) CaSO₄ · ½H₂O [1 Mark]
Explanation: Plaster of Paris is calcium sulphate hemihydrate, CaSO₄ · ½H₂O.
Explanation: Plaster of Paris is calcium sulphate hemihydrate, CaSO₄ · ½H₂O.
Q18. (c) Al₂O₃ and ZnO [1 Mark]
Explanation: Aluminium oxide and zinc oxide react with both acids and bases to produce salt and water, hence they are amphoteric oxides.
Explanation: Aluminium oxide and zinc oxide react with both acids and bases to produce salt and water, hence they are amphoteric oxides.
Q19. (b) –CH₂– and 14 u [1 Mark]
Explanation: Successive members of any homologous series differ by one methylene unit (–CH₂–), corresponding to a molecular mass difference of 12 + (2 × 1) = 14 u.
Explanation: Successive members of any homologous series differ by one methylene unit (–CH₂–), corresponding to a molecular mass difference of 12 + (2 × 1) = 14 u.
Q20. (a) Formation of a brown residue and evolution of pungent smelling gases [1 Mark]
Explanation: 2FeSO₄(s) → Fe₂O₃(s) (reddish-brown) + SO₂(g) + SO₃(g) (pungent gases like burning sulphur).
Explanation: 2FeSO₄(s) → Fe₂O₃(s) (reddish-brown) + SO₂(g) + SO₃(g) (pungent gases like burning sulphur).
Q21. (a) Both A and R are true and R is the correct explanation of A. [1 Mark]
Explanation: Highly electropositive metals have a greater chemical affinity for oxygen than carbon, so carbon cannot reduce their oxides.
Explanation: Highly electropositive metals have a greater chemical affinity for oxygen than carbon, so carbon cannot reduce their oxides.
Q22. Thermal Decomposition of Lead Nitrate: [2 Marks]
- (a) Equation: 2Pb(NO₃)₂(s) $\xrightarrow{\Delta}$ 2PbO(s) + 4NO₂(g) + O₂(g) [1 Mark]
- (b) Brown gas: Nitrogen dioxide (NO₂). Type: Thermal decomposition reaction. [1 Mark]
Q23. Roasting vs Calcination: [3 Marks]
- Roasting: The process of heating a sulphide ore strongly in the presence of excess air to convert it into a metal oxide.
Equation: 2ZnS(s) + 3O₂(g) $\xrightarrow{\Delta}$ 2ZnO(s) + 2SO₂(g) [1.5 Marks] - Calcination: The process of heating a carbonate ore strongly in limited or no air to convert it into a metal oxide.
Equation: ZnCO₃(s) $\xrightarrow{\Delta}$ ZnO(s) + CO₂(g) [1.5 Marks]
- (a) Corrosion: The slow eating away of a metal surface due to attack by atmospheric moisture, oxygen, and chemicals.
Conditions for rusting: (i) Presence of moisture (water), (ii) Presence of air (oxygen). [1.5 Marks] - (b) Prevention methods: (i) Galvanisation (coating with thin layer of zinc), (ii) Painting / Greasing / Alloying (making stainless steel). [1.5 Marks]
Q24. Homologous Series: [3 Marks]
- (a) Definition: A series of organic compounds having the same functional group and similar chemical properties, where successive members differ by a –CH₂– unit. [1 Mark]
- (b) General Formula of Alkenes: $C_n H_{2n}$ (where $n \ge 2$). [1 Mark]
- (c) First two members:
- $n=2$: Ethene ($C_2H_4$)
- $n=3$: Propene ($C_3H_6$) [1 Mark]
Q25. Case-Based Solutions (Chemistry): [4 Marks]
- (a) Chlorine gas (Cl₂) is released at the anode (+). [1 Mark]
- (b) Hydrogen gas (H₂) is released at the cathode (–). [1 Mark]
- (c) Balanced Equation: 2NaCl(aq) + 2H₂O(l) $\xrightarrow{\text{electricity}}$ 2NaOH(aq) + Cl₂(g) + H₂(g) [2 Marks]
- Sodium hydroxide (NaOH): Manufacturing of soaps and detergents / paper making. [0.5 Mark]
- Chlorine (Cl₂): Water purification / manufacturing of PVC and bleaching powder. [0.5 Mark]
- Hydrogen (H₂): Used as rocket fuel / hydrogenation of vegetable oils to form margarines / making ammonia for fertilisers. [1 Mark]
Q26. Long Answer (Chemistry): [5 Marks]
- (a) Addition Reaction: A reaction in which unsaturated hydrocarbons add hydrogen or other reagents across double/triple bonds in the presence of catalysts (like Ni/Pd) to become saturated hydrocarbons.
Hydrogenation: Vegetable Oil (unsaturated) + H₂ $\xrightarrow{\text{Ni, } 473\text{ K}}$ Vegetable Ghee (saturated fat). [2 Marks] - (b) Substitution Reaction: Methane reacts with chlorine in the presence of sunlight where chlorine replaces hydrogen atoms one by one.
CH₄ + Cl₂ $\xrightarrow{\text{sunlight}}$ CH₃Cl + HCl [1.5 Marks] - (c) Unsaturated hydrocarbons have reactive double or triple bonds with weaker $\pi$-bonds that readily open up to add reagents, making them more reactive than saturated hydrocarbons with strong single $\sigma$-bonds. [1.5 Marks]
- (a) Baking Soda: Chemical Formula = NaHCO₃; Chemical Name = Sodium Hydrogen Carbonate. [1 Mark]
- (b) Industrial Preparation: NaCl + H₂O + CO₂ + NH₃ → NH₄Cl + NaHCO₃ [1.5 Marks]
- (c) Heating during cooking: 2NaHCO₃ $\xrightarrow{\Delta}$ Na₂CO₃ + H₂O + CO₂↑. The released CO₂ gas makes cakes and bread soft and spongy. [1.5 Marks]
- (d) Uses of Washing Soda: (i) Used in glass, soap, and paper industries. (ii) Used for removing permanent hardness of water. [1 Mark]
Q27. Tooth Decay and pH: [2 Marks]
- Bacteria present in the mouth produce acids by degradation of sugar and food particles remaining in the mouth after eating. When the pH drops below 5.5, the acid starts corroding the tooth enamel (made of calcium hydroxyapatite), causing tooth decay. [1 Mark]
- Toothpastes are generally basic (alkaline) in nature. They neutralise the excess acid produced in the mouth and prevent enamel erosion. [1 Mark]
SECTION C – PHYSICS SOLUTIONS
Q28. (b) Bends towards the normal and slows down [1 Mark]
Explanation: When light travels into an optically denser medium, its speed decreases and it refracts towards the normal line.
Explanation: When light travels into an optically denser medium, its speed decreases and it refracts towards the normal line.
Q29. (c) Myopia [1 Mark]
Explanation: Myopia (short-sightedness) is the defect where a person can clearly see nearby objects but cannot see distant objects distinctly.
Explanation: Myopia (short-sightedness) is the defect where a person can clearly see nearby objects but cannot see distant objects distinctly.
Q30. (a) The square of the current (I²) [1 Mark]
Explanation: According to Joule’s Law: $H = I^2 R t$, where heat is directly proportional to the square of current for a given resistance and time.
Explanation: According to Joule’s Law: $H = I^2 R t$, where heat is directly proportional to the square of current for a given resistance and time.
Q31. (c) Parallel straight lines [1 Mark]
Explanation: The magnetic field lines inside a current-carrying solenoid are uniform, parallel straight lines pointing from south to north inside.
Explanation: The magnetic field lines inside a current-carrying solenoid are uniform, parallel straight lines pointing from south to north inside.
Q32. (a) Both A and R are true and R is the correct explanation of A. [1 Mark]
Explanation: The central section of a thin lens has parallel opposite faces. Refraction produces an imperceptibly tiny lateral shift without changing the direction of propagation.
Explanation: The central section of a thin lens has parallel opposite faces. Refraction produces an imperceptibly tiny lateral shift without changing the direction of propagation.
Q33. (a) Both A and R are true and R is the correct explanation of A. [1 Mark]
Explanation: Tungsten has an extremely high melting point (~3422°C) and high resistivity, so it glows incandescently without melting.
Explanation: Tungsten has an extremely high melting point (~3422°C) and high resistivity, so it glows incandescently without melting.
Q34. Numerical on Concave Mirror: [2 Marks]
Given: $f = -15\text{ cm}$, Real image $\implies v = -30\text{ cm}$.
Using Mirror Formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\implies \frac{1}{u} = \frac{1}{f} – \frac{1}{v} = \frac{1}{-15} – \frac{1}{-30} = -\frac{1}{15} + \frac{1}{30} = \frac{-2 + 1}{30} = -\frac{1}{30}$
$\implies u = -30\text{ cm}$ (Object is at $30\text{ cm}$ in front of mirror, at Centre of Curvature $C$). [1 Mark]
Magnification: $m = -\frac{v}{u} = -\frac{-30}{-30} = -1$ (Real, inverted, same size). [1 Mark]
Given: $f = -15\text{ cm}$, Real image $\implies v = -30\text{ cm}$.
Using Mirror Formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\implies \frac{1}{u} = \frac{1}{f} – \frac{1}{v} = \frac{1}{-15} – \frac{1}{-30} = -\frac{1}{15} + \frac{1}{30} = \frac{-2 + 1}{30} = -\frac{1}{30}$
$\implies u = -30\text{ cm}$ (Object is at $30\text{ cm}$ in front of mirror, at Centre of Curvature $C$). [1 Mark]
Magnification: $m = -\frac{v}{u} = -\frac{-30}{-30} = -1$ (Real, inverted, same size). [1 Mark]
Q35. Factors Affecting Resistance ($R = \rho \frac{l}{A}$): [2 Marks]
Given: $Q = 96,000\text{ C}$, $t = 1\text{ hr} = 3600\text{ s}$, $V = 50\text{ V}$.
Heat generated: $H = V \times Q$ [1 Mark]
$H = 50\text{ V} \times 96,000\text{ C} = 4,800,000\text{ J} = 4.8 \times 10^6\text{ J}$ (or $4800\text{ kJ}$). [1 Mark]
- Length of conductor ($R \propto l$) [0.5 Mark]
- Area of cross-section of conductor ($R \propto \frac{1}{A}$) [0.5 Mark]
- Nature of material (resistivity $\rho$) [0.5 Mark]
- Temperature of conductor [0.5 Mark]
Given: $Q = 96,000\text{ C}$, $t = 1\text{ hr} = 3600\text{ s}$, $V = 50\text{ V}$.
Heat generated: $H = V \times Q$ [1 Mark]
$H = 50\text{ V} \times 96,000\text{ C} = 4,800,000\text{ J} = 4.8 \times 10^6\text{ J}$ (or $4800\text{ kJ}$). [1 Mark]
Q36. Dispersion of Light: [3 Marks]
- (a) Definition: The splitting of white light into its seven component colours (VIBGYOR) when passing through a transparent prism. [1 Mark]
- (b) Ray Diagram: Diagram showing a narrow beam of white light entering a glass prism, splitting inside into rays of 7 colours, and exiting to form the VIBGYOR spectrum on a screen. [1 Mark]
- (c) Violet deviates the most (shortest wavelength, lowest speed in glass), while Red deviates the least (longest wavelength, highest speed in glass). [1 Mark]
Q37. Disadvantages of Series Arrangement: [3 Marks]
- If one appliance fails or is switched off, the circuit is broken and all other appliances stop working. [1 Mark]
- All appliances receive different voltages depending on their individual resistances, so none operates at its rated voltage. [1 Mark]
- It is not possible to provide separate on/off switches for individual appliances. [1 Mark]
Q38. Case-Based Solutions (Physics): [4 Marks]
- (a) Myopia (Short-sightedness). [1 Mark]
- (b) Convex lens (converging lens). [1 Mark]
- (c) For a normal human adult eye, the near point is 25 cm and the far point is infinity ($\infty$). [2 Marks]
- Power $P = -2.5\text{ D} \implies$ The defect is Myopia (uses concave lens). [1 Mark]
Focal length $f = \frac{1}{P} = \frac{1}{-2.5\text{ D}} = -0.4\text{ m} = -40\text{ cm}$. [1 Mark]
Q39. Long Answer (Physics): [5 Marks]
- (a) Joule’s Law: Heat produced in a conductor is directly proportional to (i) square of current ($I^2$), (ii) resistance ($R$), and (iii) time ($t$) for which current flows: $H = I^2 R t$. [1 Mark]
- (b) Parallel Derivation:
In parallel connection, potential difference $V$ across each resistor is the same.
Total current $I = I_1 + I_2 + I_3$.
From Ohm’s law, $I_1 = \frac{V}{R_1}$, $I_2 = \frac{V}{R_2}$, $I_3 = \frac{V}{R_3}$, and $I = \frac{V}{R_p}$.
Substituting: $\frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} \implies \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$. [2 Marks] - (c) Calculations:
- (i) $\frac{1}{R_p} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30} = \frac{6 + 3 + 1}{30} = \frac{10}{30} = \frac{1}{3} \implies R_p = \mathbf{3\ \Omega}$. [1 Mark]
- (ii) Total current $I = \frac{V}{R_p} = \frac{12\text{ V}}{3\ \Omega} = \mathbf{4\text{ A}}$. [1 Mark]
- (a) Magnetic Field around Straight Conductor: Concentric circular loops centered on the wire.
Rule: Right-Hand Thumb Rule — If you hold a current-carrying conductor in your right hand with the thumb pointing in the direction of current, your curled fingers show the direction of magnetic field lines. [2 Marks] - (b) If two field lines crossed, at the point of intersection a compass needle would point in two different directions at the same instant, which is physically impossible. [1.5 Marks]
- (c) Factors: (i) Current magnitude ($B \propto I$), (ii) Number of turns ($B \propto N$), (iii) Radius of the coil ($B \propto \frac{1}{r}$). [1.5 Marks]