ScienceCBSE Class 10 · Board exam 2025 (session 2024-25)

CBSE Class 10 Science: 2025 board paper, set 31/6/1

The question paper set in the CBSE Class 10 board exam of 2025, as printed: 39 questions for 80 marks, in sections, with each OR choice shown as an alternative. Try each question first, then open our solution.

44 of 45 questions (counting each OR alternative) have our step-by-step solution so far. We show a solution only after an independent re-solve agrees with it; the rest say "Solution being checked".

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2025/X/086_Science.zip (a zip; this paper is the file 31-6-1_Science.pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

Practise a paper like this, timedMy board paper sets a full paper in this exact format from real board questions, with a timer.Get your written answers markedThe Answer Coach marks a photo of your handwritten answer step by step, like the examiner.

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-16, assertion-reason 17-20
B	21-26	Very short answer, 30-50 words
C	27-33	Short answer, 50-80 words
D	34-36	Long answer, 80-120 words
E	37-39	Source/case-based with subparts

Instructions in brief: No overall choice; internal choice (OR) in some questions of Sections B, C, D and E; attempt only one alternative in each.

Section A

MCQ 1-16, assertion-reason 17-20

Q1 · 1 mark · MCQChemical Reactions and Equations
In the following experimental setup of electrolysis of water, if P and Q are the gases collected in the test tubes enclosing the electrodes R and S, then select the option/options in which the matching is correct :
1. (A)(i) and (ii)
2. (B)(iii) and (iv)
3. (C)(i) and (iii)
4. (D)(ii) and (iv)
Figure in the paper: Electrolysis of water apparatus: a beaker of water with two inverted test tubes over inverted electrodes. Test tube labelled P (left) and Q (right) at the top; electrodes labelled R (left) and S (right) connected by wires to a battery. Statements listed: (i) P – Oxygen gas, R – Anode; (ii) P – Hydrogen gas, R – Cathode; (iii) Q – Hydrogen gas, S – Cathode; (iv) Q – Oxygen gas, S – Anode. See it in the official paper.
Try it first, then show our solutionSolution
1. In electrolysis of water, oxygen collects at the anode (+) and hydrogen at the cathode (−), with hydrogen volume twice that of oxygen. So the tube over the anode R holds oxygen (P) and the tube over the cathode S holds hydrogen (Q). Statements (i) and (iii) match, option (C).
  1 mark
Answer: (i) and (iii), option (C)
Where the marks are
- Chooses option (C) (1 mark)
Q2 · 1 mark · MCQAcids, Bases and Salts
You have three aqueous solutions A, B and C as given below :
A - Potassium nitrate
B - Ammonium chloride
C - Sodium carbonate
The ascending order of the pH of these solutions is :
1. (A) $A < B < C$
2. (B) $B < C < A$
3. (C) $C < A < B$
4. (D) $B < A < C$
Try it first, then show our solutionSolution
1. NH₄Cl is a salt of a strong acid and a weak base, so it is acidic (pH < 7). KNO₃ is a salt of a strong acid and a strong base, so it is neutral (pH = 7). Na₂CO₃ is a salt of a weak acid and a strong base, so it is basic (pH > 7). Ascending order: B < A < C, option (D).
  1 mark
Answer: B < A < C, option (D)
Where the marks are
- Chooses option (D) (1 mark)
Q3 · 1 mark · MCQChemical Reactions and Equations
Select from the following a statement which is not true about burning of magnesium ribbon in air :
1. (A)It burns with a dazzling white flame.
2. (B)A white powder is formed on burning.
3. (C)It is an endothermic reaction.
4. (D)It is an example of a combination reaction.
Try it first, then show our solutionSolution
1. Burning Mg gives a dazzling white flame and forms white MgO: $2 M g + O_{2} \to 2 M g O$ . This is a combination reaction that releases heat, so it is exothermic. Calling it endothermic is false, option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Chooses option (C) (1 mark)
Q4 · 1 mark · MCQCarbon and its Compounds
A Hydrocarbon which does not belong to the same homologous series of carbon compounds is
1. (A) $C_{4} H_{10}$
2. (B) $C_{6} H_{14}$
3. (C) $C_{7} H_{14}$
4. (D) $C_{10} H_{22}$
Try it first, then show our solutionSolution
1. C₄H₁₀, C₆H₁₄ and C₁₀H₂₂ all fit $C_{n} H_{2 n + 2}$ , so they are alkanes. C₇H₁₄ fits $C_{n} H_{2 n}$ , so it is an alkene and belongs to a different series, option (C).
  1 mark
Answer: C₇H₁₄, option (C)
Where the marks are
- Chooses option (C) (1 mark)
Q5 · 1 mark · MCQChemical Reactions and Equations
The colour of the solution observed after about 1 hour of placing iron nails in copper sulphate solution is
1. (A)Blue
2. (B)Pale green
3. (C)Yellow
4. (D)Reddish brown
Try it first, then show our solutionSolution
1. Iron is more reactive than copper, so it displaces copper: $F e + C u S O_{4} \to F e S O_{4} + C u$ . The FeSO₄ formed makes the solution pale green, option (B).
  1 mark
Answer: Pale green, option (B)
Where the marks are
- Chooses option (B) (1 mark)
Q6 · 1 mark · MCQAcids, Bases and Salts
Juice of tamarind turns blue litmus to red. It is because of the presence of a chemical compound called
1. (A)Acetic acid
2. (B)Methanoic acid
3. (C)Oxalic acid
4. (D)Tartaric acid
Try it first, then show our solutionSolution
1. Tamarind contains tartaric acid, a natural acid that turns blue litmus red, option (D).
  1 mark
Answer: Tartaric acid, option (D)
Where the marks are
- Chooses option (D) (1 mark)
Q7 · 1 mark · MCQAcids, Bases and Salts
The water of crystallization is present in
1. (A)(ii) and (iv)
2. (B)(ii) and (iii)
3. (C)(i) and (iii)
4. (D)(i) and (iv)
Try it first, then show our solutionSolution
1. Plaster of Paris is $C a S O_{4} \cdot \frac{1}{2} H_{2} O$ and washing soda is $N a_{2} C O_{3} \cdot 10 H_{2} O$ , so both contain water of crystallisation. Bleaching powder (CaOCl₂) and baking soda (NaHCO₃) do not. The answer is (ii) and (iii), option (B).
  1 mark
Answer: (ii) and (iii), option (B)
Where the marks are
- Chooses option (B) (1 mark)
Q8 · 1 mark · MCQHeredity
A tall pea plant with round seeds (TTRR) is crossed with a short pea plant with wrinkled seeds (ttrr). The $F_{1}$ generation will be
1. (A)25% tall with round seeds
2. (B)50% tall with wrinkled seeds
3. (C)75% tall with wrinkled seeds
4. (D)100% tall with round seeds
Try it first, then show our solutionSolution
1. The TTRR parent gives only TR gametes and the ttrr parent gives only tr gametes. Every F₁ plant is therefore TtRr.
2. Tall and round are dominant, so all F₁ plants are tall with round seeds (100%), option (D).
  1 mark
Answer: 100% tall with round seeds, option (D)
Where the marks are
- Chooses option (D) (1 mark)
Open Q8 on its own page
Q9 · 1 mark · MCQControl and Coordination
A pair of endocrine glands located in the human brain is
1. (A)Parathyroid and Pituitary
2. (B)Pineal and Thymus
3. (C)Hypothalamus and Thymus
4. (D)Hypothalamus and Pineal
Try it first, then show our solutionSolution
1. The hypothalamus and the pineal gland both lie in the brain. The thymus, parathyroid and thyroid lie in the neck or chest, so the pair is Hypothalamus and Pineal, option (D).
  1 mark
Answer: Hypothalamus and Pineal, option (D)
Where the marks are
- Chooses option (D) (1 mark)
Q10 · 1 mark · MCQHow do Organisms Reproduce?
Select the option having correct matching of the organism given in Column I with the mode of reproduction in Column II :
Column I Column II
P Leishmania 1. Regeneration
Q Spirogyra 2. Multiple Fission
R Planaria 3. Binary Fission
S Plasmodium 4. Fragmentation
5. Budding
1. (A)P-4, Q-2, R-1, S-3
2. (B)P-3, Q-4, R-5, S-2
3. (C)P-3, Q-4, R-1, S-2
4. (D)P-4, Q-3, R-2, S-1
Try it first, then show our solutionSolution
1. Leishmania reproduces by binary fission (3), Spirogyra by fragmentation (4), Planaria by regeneration (1) and Plasmodium by multiple fission (2). The matching is P-3, Q-4, R-1, S-2, option (C).
  1 mark
Answer: P-3, Q-4, R-1, S-2, option (C)
Where the marks are
- Chooses option (C) (1 mark)
Q11 · 1 mark · MCQLife Processes
The basic filtration unit of the excretory system in human beings is :
1. (A)Nephron
2. (B)Urethra
3. (C)Neuron
4. (D)Urinary bladder
Try it first, then show our solutionSolution
1. The nephron is the basic filtering unit of the kidney, option (A).
  1 mark
Answer: Nephron, option (A)
Where the marks are
- Chooses option (A) (1 mark)
Q12 · 1 mark · MCQLife Processes
In human alimentary canal, the digestive juice secreted by the gastric glands are
1. (A)Bile, Trypsin, Pepsin
2. (B)Hydrochloric acid, Pepsin, Mucus
3. (C)Lipase, Bile, Mucus
4. (D)Salivary amylase, Pepsin, Bile
Try it first, then show our solutionSolution
1. The gastric glands in the stomach wall secrete hydrochloric acid, pepsin and mucus, option (B). Bile comes from the liver and trypsin and lipase from the pancreas, so the other options are wrong.
  1 mark
Answer: Hydrochloric acid, Pepsin, Mucus, option (B)
Where the marks are
- Chooses option (B) (1 mark)
Q13 · 1 mark · MCQLight – Reflection and Refraction
Identify from the following the ray diagram which shows the correct path of the reflected ray for the ray incident on a concave mirror as shown :
1. (A)Option A ray diagram
2. (B)Option B ray diagram
3. (C)Option C ray diagram
4. (D)Option D ray diagram
Figure in the paper: Top figure: concave mirror with pole P, centre of curvature C and focus F on the principal axis; a ray parallel to the principal axis is incident. Options (A)-(D) show four ray diagrams with the same mirror and C, F, P marked; reflected rays are drawn as: (A) reflected ray going downward from the mirror at a point above the axis; (B) reflected ray through C after incident ray at the mirror; (C) reflected ray passing through F; (D) reflected ray directed away from the mirror, passing below C-side along a line that is not through F. See it in the official paper.
Try it first, then show our solutionSolution
1. A ray parallel to the principal axis, after reflection from a concave mirror, passes through the principal focus F. The diagram with the reflected ray through F is option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Chooses option (C) (1 mark)
Q14 · 1 mark · MCQThe Human Eye and the Colourful World
The part of human eye which controls the amount of light entering into it.
1. (A)Iris
2. (B)Cornea
3. (C)Ciliary muscles
4. (D)Pupil
Solution being checked. Our step-by-step solution is written and is waiting for an independent check before we show it.
Q15 · 1 mark · MCQOur Environment
Consider the following food chain :
Grass $\to$ Grasshopper $\to$ Frog $\to$ Snake $\to$ Eagle
If the amount of energy available at third trophic level is 50 kJ, the available energy at the producer level was :
1. (A)0.5 kJ
2. (B)5 kJ
3. (C)500 kJ
4. (D)5000 kJ
Try it first, then show our solutionSolution
1. By the 10% law, only 10% of the energy passes to the next level. Grass is trophic level 1 and the frog is level 3. Working back, level 2 has $50 \times 10 = 500$ kJ and level 1 has $500 \times 10 = 5000$ kJ, option (D).
  1 mark
Answer: 5000 kJ, option (D)
Where the marks are
- Chooses option (D) (1 mark)
Q16 · 1 mark · MCQOur Environment
The incorrect statement about ozone is
1. (A)It is a deadly poisonous gas.
2. (B)It shields the surface of the earth from UV radiation from sun.
3. (C)It is used as a refrigerant and in fire-extinguishers.
4. (D)It is formed by combining oxygen molecule with free oxygen atom.
Try it first, then show our solutionSolution
1. CFCs, not ozone, are used as refrigerants and in fire extinguishers, so statement (C) is incorrect. Ozone is poisonous, shields the earth from UV radiation and forms by $O_{2} + O \to O_{3}$ , so the other statements are true.
  1 mark
Answer: Option (C)
Where the marks are
- Chooses option (C) (1 mark)
Q17 · 1 mark · Assertion-reasonCompetency-basedCarbon and its Compounds
Assertion (A): Carbon and its compounds are our major sources of fuels.
Reason (R): Most of the carbon compounds on burning release a large amount of heat and light.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. Both statements are true. Carbon fuels are useful precisely because burning them releases a large amount of heat and light, so R correctly explains A, option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Chooses option (A) (1 mark)
Q18 · 1 mark · Assertion-reasonCompetency-basedLife Processes
Assertion (A): Xylem tissue moves water and minerals obtained from the soil by the roots.
Reason (R): Xylem tissue is found only in the roots of a plant.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. A is true, because xylem carries water and minerals from the roots upward. R is false, because xylem runs through the roots, stem and leaves. The answer is option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Chooses option (C) (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedElectricity
Assertion (A): In the common domestic circuits the earth wire is connected to a metallic plate buried deep inside the earth.
Reason (R): Earth wire ensures that any leakage of current to the metallic body of the appliance keeps its potential to that of the earth, so the user may not get a severe electric shock.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. Both statements are true. The earth wire is connected to a buried metal plate so that any leaked current flows to the earth and the appliance body stays at earth potential. R therefore explains A, option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Chooses option (A) (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedOur Environment
Assertion (A): Food web is a network of several food chains operating in an ecosystem.
Reason (R): Food web decreases the stability of an ecosystem.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. A is true, because a food web is a network of interconnected food chains. R is false, because a food web increases the stability of an ecosystem by providing alternative food paths. The answer is option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Chooses option (C) (1 mark)

Column I		Column II
P	Leishmania	1.	Regeneration
Q	Spirogyra	2.	Multiple Fission
R	Planaria	3.	Binary Fission
S	Plasmodium	4.	Fragmentation
		5.	Budding

Section B

Very short answer, 30-50 words

Q21 · 2 marks · Very short answerMetals and Non-metals
A light green coloured solution of sulphate salt of metal 'P' is taken in a beaker, a rod of another metal 'Q' is put in this solution as shown the following figures :
Identify the metals 'P' and 'Q' and write its chemical equation for the reaction that occurs. State the conclusion of this reaction in terms of reactivity series of metals.
Figure in the paper: Two beakers with a rod labelled Q dipping into the solution. Left beaker: solution labelled PSO4 (Light Green Solution). An arrow labelled 'After about 1 hour' points to the right beaker: solution labelled QSO4 (Colourless Solution), with rod Q still in the solution. See it in the official paper.
Try it first, then show our solutionSolution
1. A light green solution of a sulphate is iron(II) sulphate, so metal P = Iron (Fe).
  ½ mark
2. The solution turns colourless, so a colourless sulphate forms. Metal Q = Zinc (Zn) (magnesium is also acceptable).
  ½ mark
3. Reaction: $FeS O_{4} (aq) + Zn (s) \to ZnS O_{4} (aq) + Fe (s)$ . This is a displacement reaction.
  ½ mark
4. Conclusion: a more reactive metal displaces a less reactive metal from its salt solution. Zinc is above iron in the reactivity series, so it is more reactive than iron.
  ½ mark
Answer: P = Fe, Q = Zn; $FeS O_{4} + Zn \to ZnS O_{4} + Fe$ ; Zn is more reactive than Fe, so it displaces Fe.
Where the marks are
- Identifies P as iron (½ mark)
- Identifies Q as zinc or magnesium (½ mark)
- Writes the correct displacement equation (½ mark)
- States that the more reactive metal displaces the less reactive one, i.e. Zn is above Fe (½ mark)
Open Q21 on its own page
Q22 · 2 marks · Very short answerControl and Coordination
Answer the following.
1. (a)
  How is brain protected in our body ?
  0 marks
2. (b)
  A doctor finds in one of his patients that he is not maintaining a proper posture and balance of his body. State the region of brain and also the part of brain which is responsible for it.
  0 marks
Try it first, then show our solutionSolution
1. (a) The brain sits inside a bony box called the skull (cranium). Inside the skull it is surrounded by a fluid-filled balloon (cerebrospinal fluid), which absorbs shocks.
  1 mark
2. (b) Region of the brain: hind-brain.
  ½ mark
3. (b) Part of the brain: cerebellum, which maintains posture and balance of the body.
  ½ mark
Answer: (a) Bony skull (cranium) with a fluid-filled cushion; (b) Hind-brain, cerebellum
Where the marks are
- States that the brain is protected by the skull/cranium and fluid (1 mark)
- Names the hind-brain (½ mark)
- Names the cerebellum (½ mark)
Open Q22 on its own page
Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerHeredity
"Proteins control the expression of various characters." Explain this statement by taking an example of "tallness" as a characteristic in plants.
Try it first, then show our solutionSolution
1. The height of a plant depends on the amount of a particular plant growth hormone it makes. This hormone is produced with the help of an enzyme, which is a protein. The enzyme is made according to the information in a gene, which is a section of DNA.
  1 mark
2. If the gene gives an efficient enzyme, plenty of hormone is made and the plant grows tall. If the gene is altered and the enzyme works less efficiently, less hormone is made and the plant stays short. So proteins (enzymes) control the expression of the character.
  1 mark
Answer: Gene → efficient enzyme (protein) → more growth hormone → tall plant; a less efficient enzyme gives less hormone and a short plant.
Where the marks are
- Links plant height to the amount of growth hormone, made with the help of an enzyme (protein) coded by a gene (1 mark)
- Explains that an efficient enzyme gives more hormone and a tall plant, and a less efficient one gives a short plant (1 mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerHeredity
Explain the mechanism of inheritance used by sexually reproducing organisms to ensure the stability of DNA of the species.
Try it first, then show our solutionSolution
1. Every body cell has two copies of each chromosome, one from the male parent and one from the female parent. During gamete formation (meiosis), each germ cell receives only one chromosome from each pair, so a gamete has half the number of chromosomes (a single set).
  1 mark
2. At fertilisation, a male gamete and a female gamete fuse. The zygote then gets the normal (paired) number of chromosomes again. In this way the chromosome number and DNA content stay constant from generation to generation, keeping the DNA of the species stable.
  1 mark
Answer: Gametes carry a single set of chromosomes (half the number); their fusion at fertilisation restores the full number, so DNA content stays constant.
Where the marks are
- States that gametes get one chromosome of each pair, i.e. half the chromosome number (1 mark)
- Explains that fusion of gametes restores the normal number, keeping DNA stable (1 mark)
Open Q23(b) on its own page
Q24 · 2 marks · Very short answerLight – Reflection and Refraction
Study the figure in which the path of a ray of light going from Medium 1 to Medium 2 is shown.
1. (a)
  Out of the two Media – Medium 1 and Medium 2, in which is the speed of light more ?
  0 marks
2. (b)
  State reason of bending of the refracted ray away from the normal.
  0 marks
3. (c)
  Express refractive index of Medium 2 with respect to Medium 1 in terms of speed of light in two media.
  0 marks
Figure in the paper: A rectangular box showing a horizontal boundary between Medium 1 (above) and Medium 2 (below), with a dashed normal perpendicular to the boundary. Incident Ray comes in from the upper left in Medium 1 and the Refracted Ray goes to the lower right in Medium 2, bending away from the normal. See it in the official paper.
Try it first, then show our solutionSolution
1. (a) The speed of light is more in Medium 2.
  ½ mark
2. (b) The ray travels from an optically denser medium (Medium 1) to an optically rarer medium (Medium 2). Light speeds up on entering Medium 2, so it bends away from the normal.
  ½ mark
3. (c) $n_{21} = \frac{speed of light in Medium 1}{speed of light in Medium 2} = \frac{v _{1}}{v _{2}}$
  1 mark
Answer: (a) Medium 2; (b) light goes from a denser to a rarer medium (it speeds up); (c) $n_{21} = v_{1} / v_{2}$
Where the marks are
- Medium 2 (½ mark)
- Reason: denser to rarer medium / speed increases (½ mark)
- $n_{21} = v_{1} / v_{2}$ (1 mark)
Open Q24 on its own page
Answer one of these 2 alternatives.
Q25(a) · 2 marks · Very short answerCompetency-basedThe Human Eye and the Colourful World
Give reasons :
1. (i)
  The sky appears dark to passengers flying at very high altitude.
  1 mark
2. (ii)
  'Danger' signal lights are red in colour.
  1 mark
Try it first, then show our solutionSolution
1. (i) At very high altitudes there is almost no atmosphere, so there are hardly any particles to scatter sunlight. Since scattering is not prominent, no scattered light reaches the eye and the sky looks dark.
  1 mark
2. (ii) Red light has the longest wavelength in the visible spectrum, so it is scattered least by fog, smoke and dust. It therefore travels farthest and can be seen from a long distance.
  1 mark
Answer: (i) Negligible scattering at high altitude due to very few particles; (ii) red light is scattered least because of its longest wavelength.
Where the marks are
- Scattering of light is negligible at such heights (1 mark)
- Red light is scattered least by fog/smoke because it has the longest wavelength (1 mark)
Open Q25(a) on its own page
OR
Q25(b) · 2 marks · Very short answerThe Human Eye and the Colourful World
What is a rainbow ? "We see a rainbow in the sky only after the rainfall." Why ?
Try it first, then show our solutionSolution
1. A rainbow is a natural spectrum of seven colours that appears in the sky after a rain shower, on the side opposite to the Sun.
  1 mark
2. After rainfall, tiny water droplets remain suspended in the air. Each droplet acts like a small prism: it refracts and disperses the incoming sunlight, reflects it internally, and refracts it again as it comes out. The colours then reach the eye separately. Without these droplets no rainbow can form, so it is seen only after rain.
  1 mark
Answer: A rainbow is a natural spectrum in the sky; it forms only after rain because suspended water droplets act as tiny prisms that refract and disperse sunlight.
Where the marks are
- Defines a rainbow as a natural spectrum seen after a rain shower (1 mark)
- Water droplets act as tiny prisms that refract, disperse and internally reflect sunlight (1 mark)
Open Q25(b) on its own page
Q26 · 2 marks · Very short answerCompetency-basedOur Environment
We do not clean natural ponds or lakes whereas an aquarium or a swimming pool needs to be cleaned regularly. Why ?
Try it first, then show our solutionSolution
1. Ponds and lakes are natural, self-sustaining ecosystems. Decomposers such as bacteria and fungi break down dead matter and wastes there, and nutrients are recycled, so they clean themselves naturally.
  1 mark
2. An aquarium or a swimming pool is an artificial (man-made) ecosystem with few or no decomposers. Wastes keep building up and become harmful to fish, plants or swimmers, so these must be cleaned regularly.
  1 mark
Answer: Natural ponds have decomposers that break down wastes; aquaria and pools are artificial ecosystems lacking decomposers, so wastes accumulate.
Where the marks are
- Ponds/lakes are natural ecosystems where decomposers break down wastes (1 mark)
- Aquarium/pool is an artificial ecosystem without decomposers, so wastes build up and must be cleaned (1 mark)
Open Q26 on its own page

Section C

Short answer, 50-80 words

Q27 · 3 marks · Short answerChemical Reactions and Equations
Write balanced chemical equation for the reactions that occur when
1. (a)
  steam is passed over red hot iron.
  0 marks
2. (b)
  natural gas is burnt in air.
  0 marks
3. (c)
  glucose reacts with oxygen in the cells of our body and provides energy.
  0 marks
Try it first, then show our solutionSolution
1. (a) $3Fe (s) + 4 H_{2} O (g) \to F e_{3} O_{4} (s) + 4 H_{2} (g)$
  1 mark
2. (b) Natural gas is mainly methane: $C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$ + heat and light
  1 mark
3. (c) $C_{6} H_{12} O_{6} (aq) + 6 O_{2} (g) \to 6C O_{2} (g) + 6 H_{2} O (l)$ + energy
  1 mark
Answer: (a) $3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}$ ; (b) $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ ; (c) $C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O$ + energy
Where the marks are
- Balanced equation for iron with steam (1 mark)
- Balanced equation for combustion of methane (1 mark)
- Balanced equation for respiration of glucose (1 mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerAcids, Bases and Salts
State the chemical property in each case on which the following uses of baking soda are based upon :
1. (i)
  as an anti-acids
  0 marks
2. (ii)
  as a constituent in making baking powder
  0 marks
3. (iii)
  in soda-acid fire-extinguishers
  0 marks
Try it first, then show our solutionSolution
1. (i) Baking soda ( $NaHC O_{3}$ ) is a mild, non-corrosive basic salt. It neutralises excess hydrochloric acid in the stomach, so it works as an antacid.
  1 mark
2. (ii) When heated, or when it reacts with a mild edible acid such as tartaric acid in baking powder, it releases carbon dioxide: $2NaHC O_{3} heat N a_{2} C O_{3} + H_{2} O + C O_{2}$ . The $C O_{2}$ makes the dough rise and become soft and spongy.
  1 mark
3. (iii) In a soda-acid fire extinguisher, baking soda reacts with an acid (sulphuric acid) and produces a large amount of $C O_{2}$ gas. This gas cuts off the air supply and puts out the fire.
  1 mark
Answer: (i) Mild base that neutralises excess acid; (ii) releases CO2 on heating or with an edible acid; (iii) reacts with acid to produce CO2, which extinguishes fire.
Where the marks are
- Mild basic nature that neutralises excess stomach acid (1 mark)
- Liberates CO2 on reaction with a mild edible acid or on heating (1 mark)
- Reacts with acid to give CO2 that extinguishes fire (1 mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerAcids, Bases and Salts
Write chemical equations to show what happens when an acid reacts with a
1. (i)
  metal
  0 marks
2. (ii)
  base and
  0 marks
3. (iii)
  carbonate
  0 marks
Try it first, then show our solutionSolution
1. (i) Acid + metal → salt + hydrogen: $Zn (s) + 2HCl (aq) \to ZnC l_{2} (aq) + H_{2} (g)$
  1 mark
2. (ii) Acid + base → salt + water (neutralisation): $2NaOH (aq) + H_{2} S O_{4} (aq) \to N a_{2} S O_{4} (aq) + 2 H_{2} O (l)$
  1 mark
3. (iii) Acid + carbonate → salt + water + carbon dioxide: $N a_{2} C O_{3} (s) + 2HCl (aq) \to 2NaCl (aq) + H_{2} O (l) + C O_{2} (g)$
  1 mark
Answer: (i) $Z n + 2 H C l \to Z n C l_{2} + H_{2}$ ; (ii) $2 N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O$ ; (iii) $N a_{2} C O_{3} + 2 H C l \to 2 N a C l + H_{2} O + C O_{2}$
Where the marks are
- Correct balanced equation for acid with a metal (1 mark)
- Correct balanced equation for acid with a base (1 mark)
- Correct balanced equation for acid with a carbonate (1 mark)
Open Q28(b) on its own page
Q29 · 3 marks · Short answerLife Processes
Name the blood vessel that brings (i) oxygenated blood (ii) deoxygenated blood, to the human heart. Also name that chamber of the heart which receives deoxygenated blood and state how deoxygenated blood from this chamber is sent to lungs for oxygenation.
Try it first, then show our solutionSolution
1. (i) Oxygenated blood is brought to the heart (left atrium) from the lungs by the pulmonary vein.
  ½ mark
2. (ii) Deoxygenated blood is brought to the heart from the body by the vena cava.
  ½ mark
3. The chamber that receives deoxygenated blood is the right atrium.
  ½ mark
4. After the right atrium fills with blood, it contracts.
  ½ mark
5. This pushes the blood into the right ventricle, which expands to receive it.
  ½ mark
6. The right ventricle then contracts and pumps the deoxygenated blood to the lungs through the pulmonary artery for oxygenation.
  ½ mark
Answer: (i) Pulmonary vein; (ii) vena cava; right atrium → contracts → right ventricle → contracts → pulmonary artery → lungs
Where the marks are
- Pulmonary vein (½ mark)
- Vena cava (½ mark)
- Right atrium (½ mark)
- Right atrium contracts, blood passes to the right ventricle, which contracts and sends blood to the lungs via the pulmonary artery (½ × 3 = 1½ marks)
Open Q29 on its own page
Q30 · 3 marks · Short answerHeredity
The gene combination of purple flowered pea plants is denoted as (WW) and that of white flowered pea plants as (ww), when these two plants are crossed $F_{1}$ generation is obtained.
1. (a)
  List two observations made by Mendel in $F_{1}$ generation plants.
  0 marks
2. (b)
  Give the (i) percentage of white flowered plants and (ii) ratio of the gene combinations WW, Ww and ww in $F_{2}$ generation.
  0 marks
3. (c)
  Write one difference between dominant and recessive trait.
  0 marks
Try it first, then show our solutionSolution
1. (a) Observation 1: all the $F_{1}$ plants were purple flowered.
  ½ mark
2. (a) Observation 2: no white or intermediate (mixed) coloured flowers appeared. Only the dominant parental trait was expressed.
  ½ mark
3. (b)(i) On selfing $F_{1}$ (Ww × Ww), the offspring are WW, Ww, Ww, ww. Only ww is white, so white flowered plants = 1 out of 4 = 25%.
  ½ mark
4. (b)(ii) WW : Ww : ww = 1 : 2 : 1
  ½ mark
5. (c) A dominant trait expresses itself even in the presence of its contrasting trait (e.g. in Ww). A recessive trait remains unexpressed in the presence of the dominant form and appears only when both copies are recessive (ww).
  1 mark
Answer: (a) All F1 plants purple; no white/mixed flowers; (b)(i) 25%; (ii) 1 : 2 : 1; (c) dominant expresses in presence of contrasting trait, recessive does not.
Where the marks are
- Any two correct F1 observations (½ × 2 = 1 mark)
- 25% white flowered plants (½ mark)
- Ratio 1 : 2 : 1 (½ mark)
- One correct difference between dominant and recessive trait (1 mark)
Open Q30 on its own page
Q31 · 3 marks · Short answerCompetency-basedLight – Reflection and Refraction
A student placed a candle flame at different distances from a convex lens and focused its image on a screen. He recorded his observation in tabular form as given below :
S.No. Distance of flame from the lens (cm) Distance of the image from the lens
1 −90 +18
2 −60 +20
3 −40 +24
4 −30 +30
5 −24 +40
6 −20 +60
7 −18 +90
8 −12 +120
Analyse the observation table and on the basis of your analysis only, answer the following questions (without doing any calculations) :
1. (a)
  What is the focal length of the convex lens used ? Give reason to justify your answer.
  0 marks
2. (b)
  Which one of the sets of observations is not correct and why ?
  0 marks
3. (c)
  Draw ray diagram to show image formation for any correct set of observation.
  0 marks
Try it first, then show our solutionSolution
1. (a) Focal length $f = + 15$ cm.
  ½ mark
2. (a) Reason: in observation 4, object distance = image distance = 30 cm. This happens only when the object is at $2 F_{1}$ and the image is at $2 F_{2}$ . So $2 f = 30$ cm, giving $f = 15$ cm.
  ½ mark
3. (b) Observation No. 8 ( $u = - 12$ cm, $v = + 120$ cm) is not correct.
  ½ mark
4. (b) Reason: 12 cm is less than $f = 15$ cm, so the object is between the optical centre and the focus. A convex lens then forms a virtual, erect image on the same side as the object, which cannot be caught on a screen. So $v$ cannot be $+ 120$ cm; it should be negative.
  ½ mark
5. (c) Ray diagram for observation 4 (object at $2 F_{1}$ , 30 cm):
  Draw the convex lens on the principal axis. Mark $F_{1}$ , $2 F_{1}$ on the left and $F_{2}$ , $2 F_{2}$ on the right, 15 cm and 30 cm from O.
  Place an upright object AB at $2 F_{1}$ .
  A ray from B parallel to the principal axis refracts through $F_{2}$ .
  A ray from B through the optical centre O goes on undeviated.
  The rays meet below the axis at B′, forming image A′B′ at $2 F_{2}$ .
  The image is real, inverted and the same size as the object. Arrows show the direction of the rays.
  1 mark
Answer: (a) f = +15 cm (u = v = 30 cm means object at 2F); (b) Observation 8, since object within f gives a virtual image on the same side; (c) ray diagram for object at 2F giving a real, inverted, same-size image at 2F′.
Where the marks are
- Focal length 15 cm (½ mark)
- Reason: u = v at 2F, so 2f = 30 cm (½ mark)
- Identifies observation 8 (½ mark)
- Reason: object lies between O and F, so the image is virtual and on the same side (½ mark)
- Correct labelled ray diagram for any correct observation (1 mark)
Open Q31 on its own page
Q32 · 3 marks · Short answerThe Human Eye and the Colourful World
A person uses lenses of +2.0 D power in his spectacles for the correction of his vision.
1. (a)
  Name the defect of vision the person is suffering from.
  0 marks
2. (b)
  List two causes of this defect.
  0 marks
3. (c)
  Determine the focal length of the lenses used in the spectacles.
  0 marks
Try it first, then show our solutionSolution
1. (a) The positive power shows a convex (converging) lens, so the defect is hypermetropia (far-sightedness).
  1 mark
2. (b) Cause 1: the eyeball has become too small (shortened).
  ½ mark
3. (b) Cause 2: the focal length of the eye lens is too long, i.e. the lens has become less converging.
  ½ mark
4. (c) $f = \frac{1}{P} = \frac{1}{+ 2.0 D}$
  ½ mark
5. $f = + 0.5$ m $= + 50$ cm (convex lens)
  ½ mark
Answer: (a) Hypermetropia; (b) eyeball too short, focal length of eye lens too long; (c) f = +0.5 m (+50 cm)
Where the marks are
- Names hypermetropia (1 mark)
- Two causes (½ × 2 = 1 mark)
- Uses f = 1/P (½ mark)
- f = +0.5 m (½ mark)
Open Q32 on its own page
Q33 · 3 marks · Short answerElectricity
Answer the following.
1. (a)
  Explain the statement "Potential difference between two points is 1 volt".
  0 marks
2. (b)
  What do the symbols given below represent in an electric circuit ? Write one function of each.
  0 marks
Figure in the paper: Two circuit symbols: (i) a circle with letter A (ammeter) with terminals marked + and −; (ii) a resistor (zig-zag) with an arrow passing diagonally through it, indicating a variable resistor (rheostat). See it in the official paper.
Try it first, then show our solutionSolution
1. (a) The potential difference between two points is 1 volt if 1 joule of work is done to move a charge of 1 coulomb from one point to the other: $1 V = \frac{1 J}{1 C}$ .
  1 mark
2. (b)(i) The symbol is an ammeter.
  ½ mark
3. Function: it measures the electric current in a circuit. It is connected in series.
  ½ mark
4. (b)(ii) The symbol is a rheostat (variable resistance).
  ½ mark
5. Function: it changes the resistance of the circuit, and so regulates the current.
  ½ mark
Answer: (a) 1 J of work is done to move 1 C of charge between the points; (b)(i) Ammeter, measures current; (ii) Rheostat, varies resistance/current.
Where the marks are
- Explains 1 V as 1 J of work per 1 C of charge (1 mark)
- Ammeter (½ mark) and its function (½ mark)
- Rheostat (½ mark) and its function (½ mark)
Open Q33 on its own page

S.No.	Distance of flame from the lens (cm)	Distance of the image from the lens
1	−90	+18
2	−60	+20
3	−40	+24
4	−30	+30
5	−24	+40
6	−20	+60
7	−18	+90
8	−12	+120

Section D

Long answer, 80-120 words

Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerCarbon and its Compounds
Name an alcohol and a carboxylic acid having two carbon atoms in their structures. Draw their structures and state how this alcohol can be converted into a carboxylic acid. What happens when these two compounds react in the presence of an acid ? Write chemical equations for the reactions involved in the two cases mentioned above.
Try it first, then show our solutionSolution
1. Alcohol with two carbon atoms: ethanol, $C H_{3} C H_{2} OH$ . Carboxylic acid with two carbon atoms: ethanoic acid, $C H_{3} COOH$ .
  1 mark
2. Structures:
  Ethanol: H–C(H)(H)–C(H)(H)–O–H. The first carbon carries 3 H atoms, the second carbon carries 2 H atoms and is bonded to an –OH group.
  Ethanoic acid: H–C(H)(H)–C(=O)–O–H. A CH₃ group is joined to a carbon that is double bonded to O and single bonded to an –OH group.
  1 mark
3. Conversion: when ethanol is heated with an oxidising agent such as alkaline potassium permanganate (or acidified potassium dichromate), it is oxidised to ethanoic acid.
  ½ mark
4. $C H_{3} C H_{2} OH alkaline KMn O_{4} heat C H_{3} COOH$
  1 mark
5. When ethanol and ethanoic acid are warmed together with a little concentrated sulphuric acid (catalyst), a sweet-smelling ester, ethyl ethanoate, is formed. This reaction is called esterification.
  ½ mark
6. $C H_{3} COOH + C H_{3} C H_{2} OH conc. H_{2} S O_{4} C H_{3} COOC H_{2} C H_{3} + H_{2} O$
  1 mark
Answer: Ethanol (CH3CH2OH) and ethanoic acid (CH3COOH); oxidation with alkaline KMnO4/acidified K2Cr2O7 converts ethanol to ethanoic acid; with an acid catalyst they form the ester ethyl ethanoate (esterification).
Where the marks are
- Names ethanol and ethanoic acid (½ + ½ mark)
- Correct structures of both (½ + ½ mark)
- States oxidation by alkaline KMnO4 or acidified K2Cr2O7 (½ mark)
- Oxidation equation (1 mark)
- States that an ester is formed (½ mark)
- Esterification equation (1 mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerCarbon and its Compounds
What are soaps ? Write the structure of a soap molecule. Explain the cleansing action of a soap. Why are soaps not considered suitable for washing clothes in a region where water is hard ? How is this problem overcomed ?
Try it first, then show our solutionSolution
1. Soaps are sodium or potassium salts of long-chain carboxylic (fatty) acids, e.g. sodium stearate $C_{17} H_{35} CO O^{-} N a^{+}$ .
  1 mark
2. Structure of a soap molecule: a long hydrocarbon tail, $C H_{3} (C H_{2})_{16}$ –, attached to an ionic head, $- CO O^{-} N a^{+}$ . The ionic head is hydrophilic (water-loving). The hydrocarbon tail is hydrophobic (water-repelling) and dissolves in oil.
  1 mark
3. Cleansing action: most dirt is oily and does not dissolve in water. In soap solution, the hydrocarbon tails of soap molecules attach to the oily dirt, while the ionic heads face the water. The molecules form spherical clusters called micelles, with the oily dirt trapped in the centre. This forms an emulsion in water, and rinsing washes the micelles and the dirt away, leaving the clothes clean.
  1 mark
4. Hard water contains calcium and magnesium salts. These react with soap to form an insoluble, curdy precipitate called scum. Much soap is wasted, little lather forms and the clothes are not cleaned properly.
  1 mark
5. Solution: use detergents, which do not form insoluble precipitates with calcium and magnesium ions and work well in hard water. Alternatively, remove the hardness of the water before washing.
  1 mark
Answer: Soaps are Na/K salts of long-chain carboxylic acids with a hydrophilic ionic head and a hydrophobic tail; micelles trap oily dirt; hard water (Ca²⁺, Mg²⁺) forms scum with soap; use detergents or remove hardness.
Where the marks are
- Definition of soap (1 mark)
- Structure with hydrophilic head and hydrophobic tail (1 mark)
- Cleansing action through micelle formation (1 mark)
- Ca/Mg salts in hard water form scum with soap (1 mark)
- Use of detergents or removal of hardness (1 mark)
Open Q34(b) on its own page
Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerHow do Organisms Reproduce?
Answer the following.
1. (a)
  Define Puberty. List any two changes seen in boys at the time of puberty.
  0 marks
2. (b)
  Why are testes in human males located outside the abdominal cavity in scrotum ?
  0 marks
3. (c)
  List any three techniques of contraception used by humans. Which one of these is not meant for males ?
  0 marks
Try it first, then show our solutionSolution
1. (a) Puberty is the period of adolescence when the rate of general body growth begins to slow down and the reproductive tissues begin to mature.
  1 mark
2. (a) Change 1 in boys: thick hair grows on the face (beard and moustache), and in the armpits and genital area.
  ½ mark
3. (a) Change 2 in boys: the voice begins to crack (becomes deeper).
  ½ mark
4. (b) Sperm formation needs a temperature lower than normal body temperature. The scrotum keeps the testes outside the abdominal cavity, where the temperature is about 2–3 °C lower.
  1 mark
5. (c) Technique 1, mechanical barrier: a condom, which stops sperm from reaching the egg.
  ½ mark
6. (c) Technique 2, chemical/hormonal method: oral pills.
  ½ mark
7. (c) Technique 3, intrauterine device: a loop or Copper-T placed in the uterus. (Surgical methods such as vasectomy or tubectomy are also acceptable.)
  ½ mark
8. The methods not meant for males are oral pills and the loop or Copper-T, which are used by females.
  ½ mark
Answer: (a) Puberty: growth slows and reproductive tissues mature; e.g. facial hair, cracking of voice; (b) sperm formation needs a lower temperature; (c) condom, oral pills, Copper-T; oral pills / Copper-T not meant for males.
Where the marks are
- Definition of puberty (1 mark)
- Any two changes in boys (½ × 2 = 1 mark)
- Lower temperature needed for sperm formation (1 mark)
- Any three contraceptive techniques (½ × 3 = 1½ marks)
- Identifies oral pills or Copper-T as not meant for males (½ mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerHow do Organisms Reproduce?
Answer the following.
1. (a)
  Name the part performing following functions in human female reproductive system :
  0 marks
2. (i)
  production of eggs
  0 marks
3. OR
  (ii)
  site of fertilization
  0 marks
4. OR
  (iii)
  site of implantation
  0 marks
5. OR
  (iv)
  entry of the sperms
  0 marks
6. (b)
  What changes are observed in the uterus : (i) subsequent to implantation of zygote and (ii) if an egg does not get fertilized ?
  0 marks
Try it first, then show our solutionSolution
1. (a)(i) Production of eggs: ovary
  ½ mark
2. (a)(ii) Site of fertilisation: oviduct (fallopian tube)
  ½ mark
3. (a)(iii) Site of implantation: uterus
  ½ mark
4. (a)(iv) Entry of sperms: vagina
  ½ mark
5. (b)(i) After implantation, the zygote keeps dividing and forms the embryo.
  ½ mark
6. The thickened uterine lining, rich in blood vessels, nourishes the developing embryo.
  ½ mark
7. A special tissue called the placenta develops. Through it the embryo receives glucose and oxygen from the mother's blood, and its wastes are passed into the mother's blood for removal.
  ½ mark
8. (b)(ii) If the egg is not fertilised, the egg lives for about a day and the thick, spongy uterine lining is no longer needed. The lining slowly breaks down and comes out through the vagina as blood and mucus, along with the unfertilised egg. This is menstruation, which occurs roughly every month and lasts about 2–8 days.
  1½ marks
Answer: (a) (i) Ovary (ii) Oviduct/fallopian tube (iii) Uterus (iv) Vagina; (b)(i) lining thickens and nourishes the embryo, placenta forms; (ii) lining breaks down and is shed as blood and mucus (menstruation).
Where the marks are
- Four correct parts named (½ × 4 = 2 marks)
- Changes after implantation: zygote/embryo develops, thick vascular lining nourishes it, placenta forms (½ × 3 = 1½ marks)
- Uterine lining breaks down and comes out as blood and mucus – menstruation (1½ marks)
Open Q35(b) on its own page
Answer one of these 2 alternatives.
Q36(a) · 5 marks · Long answerMagnetic Effects of Electric Current
What are magnetic field lines ? How is the direction of magnetic field at a point determined ? Draw the pattern of magnetic field lines of the magnetic field produced by a current carrying circular loop. Mark on it the direction of (i) current and (ii) magnetic field lines. Name the two factors on which the magnitude of the magnetic field due to a current carrying coil depends.
Try it first, then show our solutionSolution
1. Magnetic field lines are imaginary closed curves drawn around a magnet (or a current-carrying conductor). The tangent at any point on a field line gives the direction of the magnetic field at that point.
  ½ mark
2. Direction of field at a point: place a small magnetic compass needle at that point. The north pole of the needle points along the direction of the magnetic field there.
  ½ mark
3. Figure (described in words): Draw a circular loop held vertically and cut by a horizontal cardboard sheet at two points, P and Q. Mark the current with arrows, going up through the sheet at P and down at Q.
  Around P, draw small concentric circles with anticlockwise arrows (seen from above). Around Q, draw concentric circles with clockwise arrows. Moving away from the wire, the circles grow larger and less curved.
  At the centre of the loop, the lines become almost straight and parallel, all pointing the same way through the loop.
  Label (i) the direction of current on the loop with arrows, as given by the right-hand thumb rule.
  1 mark
4. Label (ii) the direction of the magnetic field lines with arrows on the circles and on the straight lines at the centre, consistent with the current direction.
  1 mark
5. Factor 1: the magnitude of the field is directly proportional to the current flowing through the coil.
  1 mark
6. Factor 2: it is directly proportional to the number of turns of the coil, because the fields of the turns add up. (It also depends on the radius of the coil: field $\propto \frac{1}{r}$ .)
  1 mark
Answer: Field lines are imaginary curves whose tangent gives the field direction; direction is found with a compass needle (N-pole points along the field). The figure shows concentric circles near the wire and straight parallel lines at the centre, with current and field directions marked. Field depends on the current and the number of turns (also the radius).
Where the marks are
- Defines magnetic field lines (½ mark)
- States that a compass needle's north pole gives the field direction (½ mark)
- Correct field pattern with the direction of current marked (1 mark)
- Direction of magnetic field lines marked correctly (1 mark)
- Any two factors: current, number of turns, radius of coil (1 + 1 marks)
Open Q36(a) on its own page
OR
Q36(b) · 5 marks · Long answerMagnetic Effects of Electric Current
Why can't two magnetic field lines cross each other ? Draw magnetic field lines showing the direction of the magnetic field due to a current carrying long straight solenoid. State the conclusion which can be drawn from the pattern of magnetic field lines inside the solenoid. Name any two factors on which the magnitude of the magnetic field due to this solenoid depends.
Try it first, then show our solutionSolution
1. Two field lines cannot cross. At the point of intersection, the magnetic field would have two directions, and a compass needle placed there would point two ways at once. This is not possible.
  1 mark
2. Figure (described in words): Draw a long coil (solenoid) with the current shown by arrows entering at one end and leaving at the other.
  Inside the solenoid, draw straight, parallel, equally spaced lines running along the axis. Outside, draw closed curves from one end to the other, like those of a bar magnet. Mark the end where the lines emerge as N and the other end as S.
  Put arrows on the lines: inside the solenoid from S to N, outside from N to S.
  1 mark
3. Conclusion: the field lines inside the solenoid are parallel straight lines, so the magnetic field is uniform, i.e. the same at all points inside the solenoid.
  1 mark
4. Factor 1: the field is stronger when the number of turns in the solenoid is larger.
  1 mark
5. Factor 2: the field is stronger when the current through the solenoid is larger. (Another factor is the nature of the core material, e.g. a soft iron core increases the field.)
  1 mark
Answer: Crossing would give two directions of field at one point, which is impossible. Inside a solenoid the lines are straight, parallel and equally spaced, so the field is uniform. The field depends on the number of turns and the current (also the core material).
Where the marks are
- Reason: two directions of field at one point is impossible (1 mark)
- Correct diagram of solenoid field lines with arrows (1 mark)
- Conclusion: uniform field inside the solenoid (1 mark)
- Any two factors: number of turns, current, core material (1 + 1 marks)
Open Q36(b) on its own page

Section E

Source/case-based with subparts

Q37 · 4 marks · Case-basedCompetency-basedMetals and Non-metals
Many pure metals like copper, iron and gold are very soft and as such are considered unsuitable for certain uses. Metallic objects around us such as cooking utensils, statues, ornaments, guns etc. are actually not made up of pure metals. Instead of pure metals, alloys are used in the design of most of the useful objects. Making alloys enhances the basic properties of a metal which is the primary constituent (metal) of an alloy.
1. (I)
  How does electrical conductivity and melting point of a metal change when it is converted to its alloy by mixing a small amount of an element in it ?
  1 mark
2. (II)
  Name an alloy used for welding two wires together in an electric circuit. Write its major constituents.
  1 mark
3. (III)(a)
  What are alloys ? How is 'Brass' (an alloy) prepared ?
  2 marks
4. OR
  (III)(b)
  What is stainless steel ? How is it prepared ? Write one important property which makes it more useful in making cooking utensils as compared to its primary metal.
  2 marks
Try it first, then show our solutionSolution
1. (I) When a pure metal is converted into its alloy, its electrical conductivity decreases and its melting point (generally) decreases, compared with the pure metal.
  1 mark
2. (II) The alloy used for welding (joining) electrical wires is solder.
  ½ mark
3. (II) Its major constituents are lead (Pb) and tin (Sn).
  ½ mark
4. (III)(a) An alloy is a homogeneous mixture of two or more metals, or of a metal and a non-metal.
  1 mark
5. (III)(a) Brass is prepared by melting copper and adding zinc to it in a definite proportion. The molten mixture is then cooled to room temperature.
  1 mark
6. (III)(b) Stainless steel is an alloy of iron with chromium and nickel (and a small amount of carbon).
  1 mark
7. (III)(b) Preparation: iron is first mixed with a very small amount of carbon (about 0.05%) to make it hard and strong. This steel is then melted and mixed with nickel and chromium in definite proportions, and cooled.
  ½ mark
8. (III)(b) Property: stainless steel does not rust (it resists corrosion), unlike pure iron. This makes it better for cooking utensils.
  ½ mark
Answer: (I) Conductivity and melting point both decrease. (II) Solder: lead and tin. (III)(a) An alloy is a homogeneous mixture of metals (or a metal and a non-metal); brass is made by melting copper with zinc. (III)(b) Stainless steel is iron + Cr + Ni (+ a little C), made by melting iron with carbon and then with Ni and Cr; it does not rust.
Where the marks are
- (I) Both conductivity and melting point decrease (1 mark)
- (II) Names solder (½ mark) and gives lead and tin as constituents (½ mark)
- (III)(a) Definition of alloy (1 mark); brass from copper and zinc in definite proportion (1 mark)
- (III)(b) Composition of stainless steel (1 mark); method of preparation (½ mark); does not rust (½ mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedControl and Coordination
The growth movements of plant parts in which the direction of the stimulus determines the direction of the response is known as tropic movements or tropism. Plants also have non-directional movements which may not be growth dependent.
1. (I)
  Name the movement which causes 'X' and 'Y' to grow downwards and upwards respectively. (Refer above figure)
  1 mark
2. (II)
  Write the name of a hormone that plays a major role in (i) falling of leaves (ii) rapid cell division
  1 mark
3. (III)(a)
  Leaves of the sensitive plant move very quickly in response to 'touch'. How is this stimulus of touch communicated and explain how the movement takes place.
  2 marks
4. OR
  (III)(b)
  Name the plant hormone which is synthesized at the shoot tip. How does this hormone helps the plant to bend towards light ?
  2 marks
Figure in the paper: A potted plant: a pot on the left with roots labelled X spreading downward-left out of the pot; a shoot labelled Y extending to the right and upward with leaves on branches. See it in the official paper.
Try it first, then show our solutionSolution
1. (I) The roots 'X' grow downwards because of positive geotropism (they also show negative phototropism).
  ½ mark
2. (I) The shoot 'Y' grows upwards because of negative geotropism (it also shows positive phototropism).
  ½ mark
3. (II)(i) Falling of leaves is promoted by abscisic acid.
  ½ mark
4. (II)(ii) Rapid cell division is promoted by cytokinin.
  ½ mark
5. (III)(a) The touch is sensed at one spot. The information is passed from cell to cell by electrical–chemical means, since plants have no nervous tissue.
  1 mark
6. (III)(a) The cells near the base of the leaflets then change shape by changing the amount of water in them. They swell or shrink, and this makes the leaves fold and droop quickly.
  1 mark
7. (III)(b) The hormone synthesised at the shoot tip is auxin.
  1 mark
8. (III)(b) When light falls on the shoot from one side, auxin diffuses towards the shady side. The higher auxin concentration there makes those cells grow longer, so the shady side grows faster. As a result, the shoot bends towards the light.
  1 mark
Answer: (I) X: positive geotropism; Y: negative geotropism. (II)(i) Abscisic acid; (ii) Cytokinin. (III)(a) Touch is conveyed by electrical–chemical signals; cells swell or shrink by changing their water content, so the leaves fold. (III)(b) Auxin; it moves to the shady side and makes those cells elongate, so the shoot bends towards light.
Where the marks are
- (I) X – positive geotropism (½ mark); Y – negative geotropism (½ mark)
- (II) Abscisic acid (½ mark); cytokinin (½ mark)
- (III)(a) Electrical–chemical communication from cell to cell (1 mark); change in water content makes cells swell or shrink (1 mark)
- (III)(b) Names auxin (1 mark); explains its diffusion to the shady side and cell elongation (1 mark)
Open Q38 on its own page
Q39 · 4 marks · Case-basedCompetency-basedElectricity
Study the circuit shown in which two resistors X and Y of resistances 3 $Ω$ and 6 $Ω$ respectively are joined in series with a battery of 2 V.
1. (I)
  Draw a circuit diagram showing the above two resistors X and Y joined in parallel with same battery and same ammeter and voltmeter.
  1 mark
2. (II)
  In which combination of resistors will the (i) potential difference across X and Y and (ii) current through X and Y, be the same ?
  1 mark
3. (III)(a)
  Find the current drawn from the battery by the series combination of the two resistors (X and Y).
  2 marks
4. OR
  (III)(b)
  Determine the equivalent resistance of the parallel combination of the two resistors (X and Y).
  2 marks
Figure in the paper: Circuit: resistor X (3 Ω) and resistor Y (6 Ω) in series along the top wire, with a voltmeter V connected across X and Y (− on X side, + on Y side). Right side has an ammeter A (− at top, + at bottom). Battery of 2 V at the bottom left with + terminal on the right, and a point-contact switch symbol in the bottom wire; arrow shows current direction. See it in the official paper.
Try it first, then show our solutionSolution
1. (I) Circuit diagram (described in words): Connect the 2 V battery, the plug key (switch) and the ammeter A in series in the main line. The ammeter's + terminal goes towards the battery's + terminal.
  Join X (3 Ω) and Y (6 Ω) side by side between two common junction points, so that each forms a separate branch.
  Connect the voltmeter V across these two junction points, i.e. in parallel with X and Y, with its + terminal on the side joined to the battery's + terminal. Show the direction of current with arrows.
  1 mark
2. (II)(i) The potential difference across X and Y is the same in the parallel combination.
  ½ mark
3. (II)(ii) The current through X and Y is the same in the series combination.
  ½ mark
4. (III)(a) Equivalent resistance in series: $R_{s} = R_{X} + R_{Y} = 3 Ω + 6 Ω = 9 Ω$
  1 mark
5. (III)(a) By Ohm's law, $I = \frac{V}{R _{s}} = \frac{2 V}{9 Ω} \approx 0.22 A$
  1 mark
6. (III)(b) For parallel combination: $\frac{1}{R _{p}} = \frac{1}{R _{X}} + \frac{1}{R _{Y}}$
  ½ mark
7. (III)(b) Substituting: $\frac{1}{R _{p}} = \frac{1}{3} + \frac{1}{6} = \frac{2 + 1}{6} = \frac{3}{6} = \frac{1}{2}$
  ½ mark
8. (III)(b) Therefore $R_{p} = 2 Ω$
  1 mark
Answer: (I) X and Y in parallel, with the ammeter in series with the battery and the voltmeter across the combination. (II)(i) Parallel; (ii) Series. (III)(a) I ≈ 0.22 A. (III)(b) $R_{p} = 2 Ω$
Where the marks are
- (I) Correct parallel circuit with ammeter in series and voltmeter across the combination (1 mark)
- (II) (i) parallel (½ mark); (ii) series (½ mark)
- (III)(a) Series resistance 9 Ω (1 mark); current 2/9 ≈ 0.22 A (1 mark)
- (III)(b) Formula for parallel resistance (½ mark); substitution (½ mark); $R_{p} = 2 Ω$ (1 mark)
Open Q39 on its own page

Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.