ScienceCBSE Class 10 · Board exam 2024 (session 2023-24)

CBSE Class 10 Science: 2024 board paper, set 31/2/1

The question paper set in the CBSE Class 10 board exam of 2024, 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.

Every question here has our step-by-step solution, split into the steps that earn marks. Each one was checked by an independent re-solve from the question paper.

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2024/X/SCIENCE.zip (a zip; this paper is the file 31_2_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; answer in 30-50 words
C	27-33	Short answer; answer in 50-80 words
D	34-36	Long answer; answer in 80-120 words
E	37-39	Source/case-based with sub-parts

Instructions in brief: All questions are compulsory; there is no overall choice, but internal choice (OR) is given in some questions/sub-parts, and only one alternative is to be attempted.

Section A

MCQ 1-16, assertion-reason 17-20

Q1 · 1 mark · MCQChemical Reactions and Equations
Solid Calcium oxide reacts vigorously with water to form Calcium hydroxide accompanied by the liberation of heat. From the information given above it may be concluded that this reaction
1. (A)is endothermic and pH of the solution formed is more than 7.
2. (B)is exothermic and pH of the solution formed is 7.
3. (C)is endothermic and pH of the solution formed is 7.
4. (D)is exothermic and pH of the solution formed is more than 7.
Try it first, then show our solutionSolution
1. The reaction is $CaO (s) + H_{2} O (l) \to Ca (OH)_{2} (aq)$ + heat. Heat is given out, so the reaction is exothermic.
  ½ mark
2. $Ca (OH)_{2}$ (slaked lime) is a base that dissolves in water as an alkali. So the solution is basic and its pH is more than 7. Option (D) is correct.
  ½ mark
Answer: Exothermic, pH more than 7: option (D)
Where the marks are
- Chooses option (D): exothermic reaction, pH > 7 (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQAcids, Bases and Salts
Juice of tamarind turns blue litmus to red. It is because of the presence of an acid called :
1. (A)methanoic acid
2. (B)acetic acid
3. (C)tartaric acid
4. (D)oxalic acid
Try it first, then show our solutionSolution
1. Blue litmus turning red shows that tamarind juice is acidic.
2. The natural acid found in tamarind is tartaric acid. Methanoic acid is in ant stings, acetic acid in vinegar and oxalic acid in tomato. Option (C) is correct.
  1 mark
Answer: Tartaric acid: option (C)
Where the marks are
- Identifies tartaric acid, option (C) (1 mark)
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Q3 · 1 mark · MCQChemical Reactions and Equations
Select from the following a process in which a combination reaction is involved :
1. (A)Black and White photography
2. (B)Burning of coal
3. (C)Burning of methane
4. (D)Digestion of food
Try it first, then show our solutionSolution
1. In a combination reaction, two or more reactants join to form a single product.
2. When coal burns, $C (s) + O_{2} (g) \to C O_{2} (g)$ . Two reactants give one product, so this is a combination reaction.
  ½ mark
3. Black and white photography is decomposition of AgBr. Burning methane gives two products, $C O_{2}$ and $H_{2} O$ . Digestion is breakdown of food. So the answer is option (B).
  ½ mark
Answer: Burning of coal: option (B)
Where the marks are
- Selects burning of coal, option (B) (1 mark)
Open Q3 on its own page
Q4 · 1 mark · MCQAcids, Bases and Salts
The oxide which can react with $HCl$ as well as $KOH$ to give corresponding salt and water is
1. (A) $CuO$
2. (B) $A l_{2} O_{3}$
3. (C) $N a_{2} O$
4. (D) $K_{2} O$
Try it first, then show our solutionSolution
1. An oxide that reacts with both an acid (HCl) and a base (KOH) to give salt and water is an amphoteric oxide.
  ½ mark
2. $A l_{2} O_{3} + 6HCl \to 2AlC l_{3} + 3 H_{2} O$ and $A l_{2} O_{3} + 2KOH \to 2KAl O_{2} + H_{2} O$ . CuO, $N a_{2} O$ and $K_{2} O$ are basic oxides. So the answer is option (B).
  ½ mark
Answer: $A l_{2} O_{3}$ : option (B)
Where the marks are
- Identifies amphoteric $A l_{2} O_{3}$ , option (B) (1 mark)
Open Q4 on its own page
Q5 · 1 mark · MCQMetals and Non-metals
Consider the following cases :
(a) $CaS O_{4} + Al \to$
(b) $CuS O_{4} + Ca \to$
(c) $FeS O_{4} + Cu \to$
(d) $ZnS O_{4} + Mg \to$
The cases in which new products will form are –
1. (A)(a) and (b)
2. (B)(b) and (c)
3. (C)(c) and (d)
4. (D)(b) and (d)
Try it first, then show our solutionSolution
1. A metal displaces another metal from its salt solution only if it is more reactive. Use the reactivity series: K > Ca > Na > Mg > Al > Zn > Fe > Pb > H > Cu.
2. (a) Al is less reactive than Ca, so there is no reaction. (b) Ca is more reactive than Cu, so a reaction occurs. (c) Cu is less reactive than Fe, so there is no reaction. (d) Mg is more reactive than Zn, so a reaction occurs.
  ½ mark
3. New products form in (b) and (d). Option (D) is correct.
  ½ mark
Answer: (b) and (d): option (D)
Where the marks are
- Chooses option (D), cases (b) and (d) (1 mark)
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Q6 · 1 mark · MCQChemical Reactions and Equations
Identify the correct statement about the following reaction :
$2 H_{2} S + S O_{2} \to 2 H_{2} O + S$
1. (A) $H_{2} S$ is oxidising agent and $S O_{2}$ is reducing agent.
2. (B) $H_{2} S$ is reduced to sulphur.
3. (C) $S O_{2}$ is oxidising agent and $H_{2} S$ is reducing agent.
4. (D) $S O_{2}$ is oxidised to sulphur.
Try it first, then show our solutionSolution
1. $H_{2} S$ loses hydrogen to become S, so it is oxidised. It is therefore the reducing agent.
  ½ mark
2. $S O_{2}$ loses oxygen to become S, so it is reduced. It is therefore the oxidising agent. Option (C) is correct.
  ½ mark
Answer: $S O_{2}$ is the oxidising agent and $H_{2} S$ is the reducing agent: option (C)
Where the marks are
- Selects option (C) (1 mark)
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Q7 · 1 mark · MCQCarbon and its Compounds
Consider the following statements about homologous series of carbon compounds :
(a) All succeeding members differ by $- C H_{2}$ unit.
(b) Melting point and boiling point increases with increasing molecular mass.
(c) The difference in molecular masses between two successive members is 16 u.
(d) $C_{2} H_{2}$ and $C_{3} H_{4}$ are NOT the successive members of alkyne series.
The correct statements are –
1. (A)(a) and (b)
2. (B)(b) and (c)
3. (C)(a) and (c)
4. (D)(c) and (d)
Try it first, then show our solutionSolution
1. (a) is true: successive members differ by a $- C H_{2} -$ unit. (b) is true: melting and boiling points rise as molecular mass increases.
  ½ mark
2. (c) is false: the mass difference is $12 + 2 = 14$ u, not 16 u. (d) is false: $C_{2} H_{2}$ and $C_{3} H_{4}$ are successive alkynes. So the correct statements are (a) and (b), option (A).
  ½ mark
Answer: (a) and (b): option (A)
Where the marks are
- Selects option (A) (1 mark)
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Q8 · 1 mark · MCQLife Processes
Which of the following statement(s) is (are) true about human heart ?
(a) Right atrium receives oxygenated blood from lungs through pulmonary artery.
(b) Left atrium transfers oxygenated blood to left ventricle which sends it to various parts of the body.
(c) Right atrium receives deoxygenated blood through vena cava from upper and lower body.
(d) Left atrium transfers oxygenated blood to aorta which sends it to different parts of the body.
1. (A)(a)
2. (B)(a) and (d)
3. (C)(b) and (c)
4. (D)(b) and (d)
Try it first, then show our solutionSolution
1. (a) is false: oxygenated blood from the lungs enters the left atrium through the pulmonary vein. (d) is false: the aorta leaves the left ventricle, not the atrium.
  ½ mark
2. (b) and (c) are true: the left atrium passes blood to the left ventricle, which pumps it to the body, and the right atrium receives deoxygenated blood from the vena cava. Option (C) is correct.
  ½ mark
Answer: (b) and (c): option (C)
Where the marks are
- Selects option (C) (1 mark)
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Q9 · 1 mark · MCQControl and Coordination
Select out of the following a gland which does NOT occur as a pair in the human body :
1. (A)Pituitary
2. (B)Ovary
3. (C)Testis
4. (D)Adrenal
Try it first, then show our solutionSolution
1. Ovaries, testes and adrenal glands each occur as a pair.
  ½ mark
2. The pituitary is a single gland at the base of the brain. Option (A) is correct.
  ½ mark
Answer: Pituitary: option (A)
Where the marks are
- Selects pituitary, option (A) (1 mark)
Q10 · 1 mark · MCQLife Processes
In human respiratory system, when a person breathes in, the position of ribs and diaphragm will be :
1. (A)lifted ribs and curve/dome shaped diaphragm.
2. (B)lifted ribs and flattened diaphragm.
3. (C)relaxed ribs and flattened diaphragm.
4. (D)relaxed ribs and curve/dome shaped diaphragm.
Try it first, then show our solutionSolution
1. During inhalation the rib muscles lift the ribs and the diaphragm contracts and flattens.
  ½ mark
2. This enlarges the chest cavity, so air rushes into the lungs. Option (B) is correct.
  ½ mark
Answer: Lifted ribs and flattened diaphragm: option (B)
Where the marks are
- Selects option (B) (1 mark)
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Q11 · 1 mark · MCQHow do Organisms Reproduce?
Identify the mode of asexual reproduction in the following organism :
1. (A)Fragmentation
2. (B)Multiple fission
3. (C)Budding
4. (D)Binary fission
Figure in the paper: A single elongated, flask-like organism (Hydra-type) with several slender tentacles at the top and a small projection (bud) on its side near the lower body. See it in the official paper.
Try it first, then show our solutionSolution
1. The figure shows Hydra with a small outgrowth (bud) on its side.
  ½ mark
2. The bud forms by repeated cell division at one site, grows into a tiny individual and detaches. This is budding, option (C).
  ½ mark
Answer: Budding: option (C)
Where the marks are
- Identifies budding, option (C) (1 mark)
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Q12 · 1 mark · MCQHeredity
A cross made between two pea plants produces 50% tall and 50% short pea plants. The gene combination of the parental pea plants must be
1. (A) $Tt$ and $Tt$
2. (B) $TT$ and $Tt$
3. (C) $Tt$ and $tt$
4. (D) $TT$ and $tt$
Try it first, then show our solutionSolution
1. Cross Tt × tt. The Tt parent gives gametes T and t; the tt parent gives only t.
  ½ mark
2. Offspring are Tt (tall) and tt (short) in a 1 : 1 ratio, which is 50% tall and 50% short. Option (C) is correct.
  ½ mark
Answer: Tt and tt: option (C)
Where the marks are
- Selects option (C) with 1 : 1 ratio reasoning (1 mark)
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Q13 · 1 mark · MCQThe Human Eye and the Colourful World
Consider the following statements in the context of human eye :
(a) The diameter of the eye ball is about 2.3 cm.
(b) Iris is a dark muscular diaphragm that controls the size of the pupil.
(c) Most of the refraction for the light rays entering the eye occurs at the crystalline lens.
(d) While focusing on the objects at different distances the distance between the crystalline lens and the retina is adjusted by ciliary muscles.
The correct statements are –
1. (A)(a) and (b)
2. (B)(a), (b) and (c)
3. (C)(b), (c) and (d)
4. (D)(a), (c) and (d)
Try it first, then show our solutionSolution
1. (a) is true: the eyeball is about 2.3 cm in diameter. (b) is true: the iris controls the size of the pupil.
  ½ mark
2. (c) is false: most refraction happens at the outer surface of the cornea. (d) is false: ciliary muscles change the focal length of the lens, not its distance from the retina. Option (A) is correct.
  ½ mark
Answer: (a) and (b): option (A)
Where the marks are
- Selects option (A) (1 mark)
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Q14 · 1 mark · MCQElectricity
The maximum resistance of a network of five identical resistors of $\frac{1}{5} Ω$ each can be –
1. (A) $1 Ω$
2. (B) $0.5 Ω$
3. (C) $0.25 Ω$
4. (D) $0.1 Ω$
Try it first, then show our solutionSolution
1. Resistance is maximum when all the resistors are joined in series.
  ½ mark
2. $R_{s} = 5 \times \frac{1}{5} Ω = 1 Ω$ . Option (A) is correct.
  ½ mark
Answer: $1 Ω$ : option (A)
Where the marks are
- Uses series combination and gets $1 Ω$ , option (A) (1 mark)
Q15 · 1 mark · MCQElectricity
Study the I-V graph for three resistors of resistances $R_{1}$ , $R_{2}$ and $R_{3}$ and select the correct statement from the following :
1. (A) $R_{1} = R_{2} = R_{3}$
2. (B) $R_{1} > R_{2} > R_{3}$
3. (C) $R_{3} > R_{2} > R_{1}$
4. (D) $R_{2} > R_{3} > R_{1}$
Figure in the paper: Graph of current I (ampere) on the vertical axis against potential difference V (volts) on the horizontal axis. Three straight lines pass through the origin, labelled R1 (steepest), R2 (intermediate) and R3 (least steep). See it in the official paper.
Try it first, then show our solutionSolution
1. From $I = \frac{V}{R}$ , the slope of an I–V graph is $\frac{1}{R}$ . A steeper line means a smaller resistance.
  ½ mark
2. R1 is steepest and R3 is least steep, so $R_{3} > R_{2} > R_{1}$ . Option (C) is correct.
  ½ mark
Answer: $R_{3} > R_{2} > R_{1}$ : option (C)
Where the marks are
- Relates slope to $1/ R$ and selects option (C) (1 mark)
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Q16 · 1 mark · MCQMagnetic Effects of Electric Current
Strength of magnetic field produced by a current carrying solenoid DOES NOT depend upon :
1. (A)number of turns in the solenoid
2. (B)direction of the current flowing through it
3. (C)radius of solenoid
4. (D)material of core of the solenoid
Try it first, then show our solutionSolution
1. The strength of the field depends on the number of turns, the current, the radius and the core material.
  ½ mark
2. Reversing the current only reverses the direction of the field, not its strength. Option (B) is correct.
  ½ mark
Answer: Direction of current: option (B)
Where the marks are
- Selects option (B) (1 mark)
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Q17 · 1 mark · Assertion-reasonCompetency-basedMetals and Non-metals
Assertion (A) : Different metals have different reactivities with water and dilute acids.
Reason (R) : Extraction of a metal from its ore depends on its position in the reactivity series.
1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A).
2. (B)Both (A) and (R) are true, but (R) is not the correct explanation of (A).
3. (C)(A) is true, but (R) is false.
4. (D)(A) is false, but (R) is true.
Try it first, then show our solutionSolution
1. (A) is true: for example, Na reacts violently with cold water while Cu does not react with water or dilute acids.
  ½ mark
2. (R) is true: metals at the top, middle and bottom of the reactivity series are extracted by different methods. But (R) is about extraction and does not explain why reactivities differ. Option (B) is correct.
  ½ mark
Answer: Option (B)
Where the marks are
- Both statements true, R not the explanation: option (B) (1 mark)
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Q18 · 1 mark · Assertion-reasonCompetency-basedHeredity
Assertion (A) : Human female has a perfect pair of sex chromosome.
Reason (R) : Sex chromosome contributed by the human male in the zygote decides the sex of a child.
1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A).
2. (B)Both (A) and (R) are true, but (R) is not the correct explanation of (A).
3. (C)(A) is true, but (R) is false.
4. (D)(A) is false, but (R) is true.
Try it first, then show our solutionSolution
1. (A) is true: females have a perfect pair of sex chromosomes, XX.
  ½ mark
2. (R) is true: the X or Y from the father decides the sex of the child. But this does not explain why the female has XX. Option (B) is correct.
  ½ mark
Answer: Option (B)
Where the marks are
- Selects option (B) (1 mark)
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Q19 · 1 mark · Assertion-reasonCompetency-basedThe Human Eye and the Colourful World
Assertion (A) : Myopic eye cannot see distant objects distinctly.
Reason (R) : For the correction of myopia converging lenses of appropriate power are prescribed by eye-surgeons.
1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A).
2. (B)Both (A) and (R) are true, but (R) is not the correct explanation of (A).
3. (C)(A) is true, but (R) is false.
4. (D)(A) is false, but (R) is true.
Try it first, then show our solutionSolution
1. (A) is true: a myopic eye sees near objects clearly but not distant ones.
  ½ mark
2. (R) is false: myopia is corrected with a concave (diverging) lens of suitable power. Option (C) is correct.
  ½ mark
Answer: Option (C)
Where the marks are
- Selects option (C) (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedMagnetic Effects of Electric Current
Assertion (A) : The deflection of a compass needle placed near a current carrying wire decreases when the magnitude of an electric current in the wire is increased.
Reason (R) : Strength of the magnetic field at a point due to a current carrying conductor increases on increasing the current in the conductor.
1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A).
2. (B)Both (A) and (R) are true, but (R) is not the correct explanation of (A).
3. (C)(A) is true, but (R) is false.
4. (D)(A) is false, but (R) is true.
Try it first, then show our solutionSolution
1. (R) is true: the magnetic field strength increases when the current increases.
  ½ mark
2. A stronger field deflects the compass needle more, so the deflection increases. (A) is false. Option (D) is correct.
  ½ mark
Answer: Option (D)
Where the marks are
- Selects option (D) (1 mark)
Open Q20 on its own page

Section B

Very short answer; answer in 30-50 words

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerChemical Reactions and Equations
"No precipitation reaction can occur without exchange of ions between the two reactants." Justify this statement giving a balanced chemical equation for the reaction.
Try it first, then show our solutionSolution
1. A precipitate is an insoluble salt formed when ions from two reactants swap partners. This happens in a double displacement reaction, so a precipitate cannot form unless ions are exchanged.
  ½ mark
2. Example: sodium sulphate solution is mixed with barium chloride solution.
3. $N a_{2} S O_{4} (aq) + BaC l_{2} (aq) \to BaS O_{4} (s) + 2NaCl (aq)$
  1 mark
4. $B a^{2 +}$ and $S O_{4}^{2 -}$ ions combine to give the white precipitate of $BaS O_{4}$ , while $N a^{+}$ and $C l^{-}$ stay in solution. This exchange of ions is what produces the precipitate.
  ½ mark
Answer: A precipitate forms only by exchange of ions (double displacement), e.g. $N a_{2} S O_{4} + BaC l_{2} \to BaS O_{4} ↓ + 2NaCl$
Where the marks are
- Explains that a precipitate forms by exchange of ions in a double displacement reaction (½ mark)
- Gives a correct balanced equation with the precipitate shown (1½ marks)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerChemical Reactions and Equations
Giving one example of each, differentiate between a displacement reaction and a double displacement reaction.
Try it first, then show our solutionSolution
1. Displacement reaction: a more reactive metal displaces a less reactive metal from its salt solution. Example: $Fe (s) + CuS O_{4} (aq) \to FeS O_{4} (aq) + Cu (s)$
  1 mark
2. Double displacement reaction: two compounds exchange their ions. Example: $N a_{2} S O_{4} (aq) + BaC l_{2} (aq) \to BaS O_{4} (s) + 2NaCl (aq)$
  1 mark
Answer: Displacement: a more reactive element replaces a less reactive one (Fe + CuSO₄). Double displacement: two compounds exchange ions (Na₂SO₄ + BaCl₂).
Where the marks are
- Defines displacement reaction with a correct example (1 mark)
- Defines double displacement reaction with a correct example (1 mark)
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Q22 · 2 marks · Very short answerLife Processes
Photosynthesis takes place in the leaves and the food prepared by it reaches other parts of the plants. Name the process involved and explain it.
Try it first, then show our solutionSolution
1. The process is translocation.
  ½ mark
2. Soluble products of photosynthesis, such as sucrose, are carried through the phloem, in the sieve tubes with the help of companion cells.
  ½ mark
3. Transport happens in both upward and downward directions, to roots, fruits, seeds and growing parts.
  ½ mark
4. Energy from ATP is used to load sucrose into the phloem. This raises the osmotic pressure, so water enters and pushes the food to the parts where it is needed.
  ½ mark
Answer: Translocation: transport of food through phloem (sieve tubes and companion cells) in both directions, using energy (ATP)
Where the marks are
- Names translocation (½ mark)
- Food carried through phloem sieve tubes with companion cells (½ mark)
- Movement both upward and downward (½ mark)
- Uses energy (ATP) and osmotic pressure (½ mark)
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Q23 · 2 marks · Very short answerHeredity
"Stability of DNA in a species is ensured during sexual reproduction." Justify the statement.
Try it first, then show our solutionSolution
1. Germ cells are made by a special division, so each germ cell gets only one chromosome from each pair. That chromosome may be of maternal or paternal origin, so a gamete has half the number of chromosomes and half the DNA.
  1 mark
2. When the male and female gametes fuse at fertilisation, the zygote gets back the full original number of chromosomes. So the chromosome number and DNA content stay the same from generation to generation, which keeps the DNA of the species stable.
  1 mark
Answer: Gametes carry half the chromosomes (one of each pair); their fusion restores the original number, keeping DNA content constant in the species.
Where the marks are
- Gametes receive one chromosome from each pair, i.e. half the number (1 mark)
- Fusion at fertilisation restores the original number, keeping DNA stable (1 mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerLight – Reflection and Refraction
State two laws of refraction of light.
Try it first, then show our solutionSolution
1. First law of refraction: The incident ray, the refracted ray and the normal drawn to the interface of the two transparent media at the point of incidence all lie in the same plane.
  1 mark
2. Second law of refraction (Snell's law): For light of a given colour and a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. That is, $\frac{sin i}{sin r} = constant$ . This constant is the refractive index of the second medium with respect to the first.
  1 mark
Answer: (1) Incident ray, refracted ray and normal at the point of incidence lie in one plane. (2) $sin i / sin r$ is constant for a given colour and a given pair of media (Snell's law).
Where the marks are
- States that the incident ray, the refracted ray and the normal lie in the same plane (1 mark)
- States Snell's law in words, mentioning a given colour of light and a given pair of media (1 mark); only writing $sin i / sin r$ = constant without the statement earns ½ mark
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerLight – Reflection and Refraction
Define the term absolute refractive index of a medium. A ray of light enters from vacuum to glass of absolute refractive index 1.5. Find the speed of light in glass. The speed of light in vacuum is $3 \times 1 0^{8}$ m/s.
Try it first, then show our solutionSolution
1. Definition: The absolute refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium: $n_{m} = \frac{c}{v}$ .
  1 mark
2. Given: $c = 3 \times 1 0^{8}$ m/s and $n_{g} = 1.5$ . We need $v$ in glass. Using $n_{g} = \frac{c}{v}$ , we get $v = \frac{c}{n _{g}}$ .
  ½ mark
3. $v = \frac{3 \times 1 0 ^{8}}{1.5} = 2 \times 1 0^{8}$ m/s. So the speed of light in glass is $2 \times 1 0^{8}$ m/s.
  ½ mark
Answer: Absolute refractive index = speed of light in vacuum ÷ speed of light in the medium; speed of light in glass = $2 \times 1 0^{8}$ m/s
Where the marks are
- Correct definition of absolute refractive index (1 mark)
- Writes $n = c / v$ and rearranges it to $v = c / n$ (½ mark)
- Correct answer $2 \times 1 0^{8}$ m/s with unit (½ mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerElectricity
Use Ohm's law to determine the potential difference across the $3 Ω$ resistor in the circuit shown in the following diagram when key is closed :
Figure in the paper: Circuit diagram: a rectangular loop containing a key K, a 2 V cell (positive terminal marked +), a $1 Ω$ resistor on the upper right side, a $2 Ω$ resistor on the lower left side and a $3 Ω$ resistor on the lower right side, all in series in the loop. See it in the official paper.
Try it first, then show our solutionSolution
1. All three resistors are in series with the 2 V cell, so the total resistance is $R_{s} = R_{1} + R_{2} + R_{3} = 1 + 2 + 3 = 6 Ω$ .
  ½ mark
2. By Ohm's law, the current in the circuit is $I = \frac{V}{R _{s}}$ .
  ½ mark
3. $I = \frac{2 V}{6 Ω} = \frac{1}{3}$ A. The same current flows through every resistor because they are in series.
  ½ mark
4. Potential difference across the $3 Ω$ resistor: $V = I R = \frac{1}{3} A \times 3 Ω = 1$ V.
  ½ mark
Answer: 1 V
Where the marks are
- Equivalent series resistance $6 Ω$ (½ mark)
- Writes Ohm's law $I = V / R$ (½ mark)
- Current $= 1/3$ A (½ mark)
- $V = I R = 1$ V across the $3 Ω$ resistor (½ mark)
Open Q25 on its own page
Q26 · 2 marks · Very short answerOur Environment
Name the term used for the materials which cannot be broken down by biological processes. Give two ways by which they harm various components of an ecosystem.
Try it first, then show our solutionSolution
1. Materials that cannot be broken down by biological processes (by decomposers such as bacteria and fungi) are called non-biodegradable substances, for example plastics and DDT.
  1 mark
2. Harm 1: They are inert, so they stay in the environment for a long time and pollute the soil, water and air.
  ½ mark
3. Harm 2: Harmful non-biodegradable chemicals enter food chains and build up at each trophic level (biological magnification). Another harm is that they lower soil fertility. Any two of these are enough.
  ½ mark
Answer: Non-biodegradable substances; they persist and pollute the environment, and they cause biological magnification (or reduce soil fertility).
Where the marks are
- Names non-biodegradable substances (1 mark)
- Any two correct harms, e.g. persistence and pollution, biomagnification, loss of soil fertility (½ + ½ mark)
Open Q26 on its own page

Section C

Short answer; answer in 50-80 words

Q27 · 3 marks · Short answerChemical Reactions and Equations
It is observed that Calcium on reaction with water floats on its surface. Explain why it happens. Also write a balanced chemical equation for the reaction that occurs. What happens when the aqueous solution of the product of this reaction reacts with Carbon dioxide gas ? Write a balanced chemical equation for the reaction.
Try it first, then show our solutionSolution
1. Calcium reacts with water less vigorously than sodium. The bubbles of hydrogen gas formed stick to the surface of the calcium metal. This makes it lighter than water, so it floats.
  ½ mark
2. $Ca (s) + 2 H_{2} O (l) \to Ca (OH)_{2} (aq) + H_{2} (g)$
  1 mark
3. When carbon dioxide is passed through the aqueous calcium hydroxide (lime water), the solution turns milky because insoluble calcium carbonate is formed.
  ½ mark
4. $Ca (OH)_{2} (aq) + C O_{2} (g) \to CaC O_{3} (s) + H_{2} O (l)$
  1 mark
Answer: Hydrogen bubbles stick to calcium, so it floats: $Ca + 2 H_{2} O \to Ca (OH)_{2} + H_{2}$ . With $C O_{2}$ the lime water turns milky: $Ca (OH)_{2} + C O_{2} \to CaC O_{3} + H_{2} O$ .
Where the marks are
- Reason: hydrogen bubbles stick to the calcium surface (½ mark)
- Balanced equation for calcium with water (1 mark)
- States that the solution turns milky (½ mark)
- Balanced equation forming $CaC O_{3}$ (1 mark)
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Q28 · 3 marks · DiagramMetals and Non-metals
Draw a labelled diagram to show electrolytic refining of copper. State what happens when electric current is passed through the electrolyte taken in this case.
Figure in the paper: Labelled diagram required of an electrolytic refining cell for copper (impure copper anode, pure copper cathode, copper sulphate electrolyte, battery), to be drawn by the candidate. See it in the official paper.
Try it first, then show our solutionSolution
1. Diagram (described): Draw a tank containing acidified copper sulphate solution. Dip two copper rods into it. Join a thick block of impure copper to the positive terminal of a battery (through a key); this is the anode. Join a thin strip of pure copper to the negative terminal; this is the cathode. Show anode mud (impurities) settling at the bottom below the anode.
  1 mark
2. Labels: Anode (impure copper), cathode (pure copper), electrolyte (acidified copper sulphate solution). Anode mud may also be labelled.
  1 mark
3. What happens: When current is passed, copper from the impure anode dissolves into the electrolyte as $C u^{2 +}$ ions. An equal amount of pure copper from the solution is deposited on the cathode. Soluble impurities stay in the solution and insoluble impurities settle as anode mud.
  At anode: $Cu \to C u^{2 +} + 2 e^{-}$
  At cathode: $C u^{2 +} + 2 e^{-} \to Cu$
  1 mark
Answer: Impure Cu anode, pure Cu cathode, acidified $CuS O_{4}$ electrolyte. Cu dissolves from the anode and an equal amount of pure Cu is deposited on the cathode; impurities form anode mud.
Where the marks are
- Correct diagram of the refining cell (1 mark)
- Labels for the electrodes and the electrolyte (1 mark)
- Explains that Cu dissolves at the anode and pure Cu deposits at the cathode, or writes the two electrode equations (1 mark)
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Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerLife Processes
Give reasons for the following :
1. (i)
  Alveoli in lungs are richly supplied with blood capillaries.
  0 marks
2. (ii)
  Respiratory pigment in the blood takes up oxygen and not carbon dioxide.
  0 marks
3. (iii)
  During anaerobic respiration, a 3-carbon molecule is formed as an end product instead of $C O_{2}$ in human beings.
  0 marks
Try it first, then show our solutionSolution
1. (i) Alveoli are surrounded by a dense network of blood capillaries so that oxygen and carbon dioxide can be exchanged efficiently between the air in the alveoli and the blood. Oxygen diffuses into the blood and $C O_{2}$ diffuses out over a very large area.
  1 mark
2. (ii) The respiratory pigment haemoglobin has a very high affinity for oxygen. So it picks up oxygen in the lungs. Carbon dioxide is more soluble in water and is carried mostly dissolved in plasma.
  1 mark
3. (iii) In human muscles, when oxygen is in short supply (for example during vigorous exercise), glucose is not oxidised completely. Pyruvate (a 3-carbon molecule) is converted into lactic acid, another 3-carbon molecule, instead of being broken down to $C O_{2}$ and water.
  1 mark
Answer: (i) For efficient exchange of gases; (ii) haemoglobin has a high affinity for oxygen; (iii) lack of oxygen means incomplete oxidation of glucose, forming 3-carbon lactic acid.
Where the marks are
- (i) Efficient gas exchange (1 mark)
- (ii) High affinity of haemoglobin for oxygen (1 mark)
- (iii) Lack of oxygen leads to incomplete oxidation of glucose, forming lactic acid, a 3-carbon molecule (1 mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerLife Processes
Answer the following.
1. (i)
  Name the movements that occur all along the gut in human digestive system. How do they help in digestion ?
  0 marks
2. (ii)
  Where is bile juice stored in human body ? List two roles of bile juice.
  0 marks
Try it first, then show our solutionSolution
1. (i) The movements are called peristaltic movements (peristalsis).
  ½ mark
2. The muscles of the gut wall contract and relax rhythmically. This pushes the food forward in a regulated way so that each part of the gut gets enough time to digest and absorb it properly.
  1 mark
3. (ii) Bile juice is made in the liver and stored in the gall bladder.
  ½ mark
4. Role 1: It makes the acidic food coming from the stomach alkaline, so that pancreatic enzymes can act.
  ½ mark
5. Role 2: Bile salts emulsify fats, breaking large fat globules into smaller ones and increasing the surface area for lipase to act.
  ½ mark
Answer: (i) Peristalsis, which pushes food along at a regulated pace for proper digestion; (ii) gall bladder; bile makes the medium alkaline and emulsifies fats.
Where the marks are
- Names peristaltic movements (½ mark)
- Explains rhythmic contraction moving food forward in a regulated way (1 mark)
- Gall bladder (½ mark)
- Two roles: emulsification of fats and making the medium alkaline (½ + ½ mark)
Open Q29(b) on its own page
Q30 · 3 marks · Short answerHow do Organisms Reproduce?
Explain the events that take place once a sperm reaches the oviduct till it becomes a foetus. Write the role of placenta in pregnancy.
Try it first, then show our solutionSolution
1. In the oviduct the sperm meets the egg (ovum), and fertilisation takes place to form a zygote.
  ½ mark
2. The zygote starts dividing repeatedly and forms a ball of cells, the embryo.
  ½ mark
3. The embryo moves to the uterus and is implanted in the thick, blood-rich lining of the uterus.
  ½ mark
4. There it keeps growing and develops organs, and it becomes a foetus.
  ½ mark
5. Role of placenta (1): The placenta is a disc-like tissue embedded in the uterine wall. Its villi give a large surface area for glucose and oxygen to pass from the mother's blood to the embryo.
  ½ mark
6. Role of placenta (2): Wastes produced by the embryo are passed into the mother's blood through the placenta for removal.
  ½ mark
Answer: Fertilisation in the oviduct gives a zygote, which divides to form an embryo; the embryo implants in the uterine lining and develops organs to become a foetus. The placenta supplies glucose and oxygen to the embryo and removes its wastes into the mother's blood.
Where the marks are
- Fertilisation in the oviduct (½ mark)
- Zygote divides to form a ball of cells, the embryo (½ mark)
- Implantation in the uterine lining (½ mark)
- Develops organs and becomes a foetus (½ mark)
- Placenta passes glucose and oxygen to the embryo over a large surface area (½ mark)
- Placenta removes the embryo's wastes into the mother's blood (½ mark)
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Answer one of these 2 alternatives.
Q31(a) · 3 marks · Short answerThe Human Eye and the Colourful World
Define the term power of accommodation of human eye. Write the name of the part of eye which plays a major role in the process of accommodation and explain what happens when human eye focuses (i) nearby objects and (ii) distant objects.
Try it first, then show our solutionSolution
1. Power of accommodation: the ability of the eye lens to adjust its focal length so that both near and distant objects can be seen clearly.
  1 mark
2. The ciliary muscles play the major role by changing the curvature of the eye lens.
  1 mark
3. (i) Nearby objects: the ciliary muscles contract, the lens becomes thicker (more curved) and its focal length decreases.
  ½ mark
4. (ii) Distant objects: the ciliary muscles relax, the lens becomes thinner and its focal length increases.
  ½ mark
Answer: Ability of the eye lens to adjust its focal length; ciliary muscles; near: muscles contract, lens thick, focal length decreases; distant: muscles relax, lens thin, focal length increases.
Where the marks are
- Definition of power of accommodation (1 mark)
- Ciliary muscles (1 mark)
- Near object: contract, thick lens, shorter focal length (½ mark)
- Distant object: relax, thin lens, longer focal length (½ mark)
Open Q31(a) on its own page
OR
Q31(b) · 3 marks · DiagramThe Human Eye and the Colourful World
Draw a ray diagram to show the formation of a rainbow in the sky. On this diagram mark A – where dispersion of light occurs, B – where internal reflection of light occurs and C – where refraction of light occurs. List two necessary conditions to observe a rainbow.
Figure in the paper: Ray diagram to be drawn showing sunlight entering a water droplet, dispersion, internal reflection and refraction on exit, with points A (dispersion), B (internal reflection) and C (refraction) marked. See it in the official paper.
Try it first, then show our solutionSolution
1. Ray diagram (described): Draw a circle for a raindrop. A ray of sunlight strikes the drop on the upper left at point A, refracts into the drop and splits into colours. These rays travel to the back of the drop at B, reflect inside, and travel to C on the lower left, where they refract out towards the observer as separate coloured rays.
  ½ mark
2. Marking: A, entry point where refraction and dispersion of white light occur; B, back surface where internal reflection occurs; C, exit point where refraction occurs as light leaves the drop.
  1½ marks
3. Condition 1: Tiny water droplets must be present in the atmosphere (just after a shower).
  ½ mark
4. Condition 2: The Sun must be behind the observer, with the observer facing the droplets.
  ½ mark
Answer: Diagram: raindrop with dispersion at A, internal reflection at B, refraction at C. Conditions: water droplets in the air; Sun behind the observer.
Where the marks are
- Correct ray diagram through a raindrop (½ mark)
- Correct marking of A, B and C (½ × 3 marks)
- Water droplets present in the atmosphere (½ mark)
- Sun behind the observer (½ mark)
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Q32 · 3 marks · DiagramMagnetic Effects of Electric Current
Draw a diagram to show the pattern of magnetic field lines on a horizontal sheet of paper due to a straight conductor passing through its centre and carrying current vertically upwards. Mark on it (i) the direction of current in the conductor and (ii) the corresponding magnetic field lines. State right hand thumb rule and check whether the directions marked by you are in accordance with this rule or not.
Figure in the paper: Diagram to be drawn: a straight vertical conductor passing perpendicularly through the centre of a horizontal sheet of paper, carrying current vertically upwards, with concentric circular magnetic field lines around it; the candidate marks current direction and field line direction. See it in the official paper.
Try it first, then show our solutionSolution
1. Diagram (described): Draw a horizontal sheet of paper with a straight vertical wire passing through its centre, joined to a battery and key. Around the wire on the sheet, draw several concentric circles, with spacing increasing outwards.
  1 mark
2. (i) Mark an arrow on the wire pointing vertically upwards to show the direction of current.
  ½ mark
3. (ii) Mark arrows on the circles showing the field lines going anticlockwise when the sheet is viewed from above.
  ½ mark
4. Right-hand thumb rule: Imagine holding the current-carrying straight conductor in your right hand so that the thumb points in the direction of the current. Then the direction in which your fingers curl around the conductor gives the direction of the magnetic field lines.
  Check: With the thumb pointing upwards, the fingers curl anticlockwise as seen from above. This matches the marked field lines, so the diagram agrees with the rule.
  1 mark
Answer: Concentric circular field lines around the wire, anticlockwise when viewed from above for upward current; this agrees with the right-hand thumb rule.
Where the marks are
- Diagram of concentric field lines around a vertical wire through the sheet (1 mark)
- Upward current direction marked (½ mark)
- Anticlockwise field-line direction (viewed from above) marked (½ mark)
- Correct statement of the right-hand thumb rule and the check (1 mark)
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Q33 · 3 marks · Short answerOur Environment
Use of pesticides to protect our crops affect organisms at various trophic levels especially human beings. Name the phenomenon involved and explain how does it happen.
Try it first, then show our solutionSolution
1. The phenomenon is biological magnification (biomagnification).
  1 mark
2. Pesticides sprayed on crops are washed down into the soil and water bodies.
  ½ mark
3. Plants absorb them from the soil along with water and minerals, so the pesticides enter the food chain. Aquatic organisms take them up from water.
  ½ mark
4. These chemicals are non-biodegradable, so they are not broken down or excreted. They get accumulated progressively at each higher trophic level.
  ½ mark
5. Human beings occupy the top level in most food chains, so the maximum concentration of these chemicals accumulates in our bodies.
  ½ mark
Answer: Biological magnification: pesticides reach soil and water, enter plants and the food chain, are non-biodegradable and accumulate at each trophic level, reaching the highest concentration in humans.
Where the marks are
- Names biological magnification (1 mark)
- Pesticides washed into soil and water (½ mark)
- Absorbed by plants and enter the food chain (½ mark)
- Non-biodegradable, so they accumulate at each trophic level (½ mark)
- Highest concentration in humans at the top of the food chain (½ mark)
Open Q33 on its own page

Section D

Long answer; answer in 80-120 words

Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerCarbon and its Compounds
Answer the following.
1. (i)
  Give reason why carbon can neither form $C^{4 +}$ cations nor $C^{4 -}$ anions but form covalent compounds.
  0 marks
2. (ii)
  What is homologous series of carbon compound ? Write the molecular formula of any two consecutive members of homologous series of aldehydes.
  0 marks
3. (iii)
  Draw the structure of the molecule of cyclohexane ( $C_{6} H_{12}$ ).
  0 marks
Try it first, then show our solutionSolution
1. (i) Carbon has 4 valence electrons. To form $C^{4 +}$ it would have to lose 4 electrons, which needs a very large amount of energy. To form $C^{4 -}$ it would have to gain 4 electrons, and a nucleus with only 6 protons would find it hard to hold 10 electrons.
  1 mark
2. So carbon completes its octet by sharing its valence electrons with other atoms, forming covalent bonds and covalent compounds.
  1 mark
3. (ii) A homologous series is a series of carbon compounds in which the same functional group replaces hydrogen in a carbon chain. Members have similar chemical properties, and successive members differ by a $- C H_{2} -$ unit (14 u in mass).
  1 mark
4. Two consecutive aldehydes: ethanal, $C H_{3} CHO$ ( $C_{2} H_{4} O$ ), and propanal, $C_{2} H_{5} CHO$ ( $C_{3} H_{6} O$ ). Methanal, HCHO, and ethanal, $C H_{3} CHO$ , are also correct.
  1 mark
5. (iii) Structure of cyclohexane (described): Six carbon atoms are joined by single bonds in a closed hexagonal ring. Each carbon is bonded to its two neighbouring carbons and to two hydrogen atoms, so each corner is a $C H_{2}$ group. In total there are 6 C–C single bonds and 12 C–H bonds.
  1 mark
Answer: (i) Losing or gaining 4 electrons is energetically too difficult, so carbon shares electrons and forms covalent bonds; (ii) homologous series: same functional group, similar properties, successive members differ by $C H_{2}$ ; e.g. $C H_{3} CHO$ and $C_{2} H_{5} CHO$ ; (iii) hexagonal ring of six $C H_{2}$ groups joined by single bonds.
Where the marks are
- Explains why $C^{4 +}$ and $C^{4 -}$ do not form (1 mark)
- States that carbon shares electrons and forms covalent compounds (1 mark)
- Definition of homologous series (1 mark)
- Two consecutive aldehyde formulae (½ + ½ mark)
- Correct structure of cyclohexane with all 12 H atoms (1 mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerCarbon and its Compounds
Answer the following.
1. (i)
  Name a commercially important carbon compound having functional group $- OH$ and write its molecular formula.
  0 marks
2. (ii)
  Write chemical equation to show its reaction with (1) Sodium metal (2) Excess conc. sulphuric acid (3) Ethanoic acid in the presence of an acid catalyst (4) Acidified potassium dichromate. Also write the name of the product formed in each case.
  0 marks
Try it first, then show our solutionSolution
1. (i) A commercially important carbon compound with the $- OH$ (alcohol) functional group is ethanol (ethyl alcohol).
  ½ mark
2. (i) Molecular formula of ethanol: $C_{2} H_{5} OH$ .
  ½ mark
3. (ii)(1) Reaction with sodium metal. Hydrogen gas is evolved: $2 C_{2} H_{5} OH + 2Na \to 2 C_{2} H_{5} ONa + H_{2} ↑$
  ½ mark
4. Product: sodium ethoxide ( $C_{2} H_{5} ONa$ ), along with hydrogen gas.
  ½ mark
5. (ii)(2) Heating with excess concentrated sulphuric acid at 443 K. The acid acts as a dehydrating agent and removes water: $C_{2} H_{5} OH Excess conc. H_{2} S O_{4} 443 K C H_{2} = C H_{2} + H_{2} O$
  ½ mark
6. Product: ethene ( $C H_{2} = C H_{2}$ ).
  ½ mark
7. (ii)(3) Reaction with ethanoic acid in the presence of an acid catalyst (esterification): $C_{2} H_{5} OH + C H_{3} COOH Acid catalyst C H_{3} COO C_{2} H_{5} + H_{2} O$
  ½ mark
8. Product: ethyl ethanoate, a sweet-smelling ester ( $C H_{3} COO C_{2} H_{5}$ ).
  ½ mark
9. (ii)(4) Oxidation by acidified potassium dichromate, which acts as an oxidising agent: $C H_{3} C H_{2} OH Acidified K_{2} C r_{2} O_{7} C H_{3} COOH$
  ½ mark
10. Product: ethanoic acid ( $C H_{3} COOH$ ).
  ½ mark
Answer: (i) Ethanol, C₂H₅OH; (ii) (1) sodium ethoxide + H₂; (2) ethene; (3) ethyl ethanoate (ester); (4) ethanoic acid
Where the marks are
- Names ethanol (½ mark)
- Writes its formula C₂H₅OH (½ mark)
- Equation with Na (½ mark) and names sodium ethoxide (½ mark)
- Equation of dehydration with excess conc. H₂SO₄ at 443 K (½ mark) and names ethene (½ mark)
- Esterification equation (½ mark) and names the ester, ethyl ethanoate (½ mark)
- Oxidation equation with acidified K₂Cr₂O₇ (½ mark) and names ethanoic acid (½ mark)
Open Q34(b) on its own page
Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerControl and Coordination
Answer the following.
1. (i)
  Distinguish between hormonal co-ordination in plants and animals.
  0 marks
2. (ii)
  Which part of the brain is responsible for – (1) intelligence (2) riding a bicycle (3) vomiting (4) controlling hunger
  0 marks
3. (iii)
  How is brain and spinal-cord protected against mechanical injuries ?
  0 marks
Try it first, then show our solutionSolution
1. (i) Difference 1: In plants, hormones reach the target cells by simple diffusion, from cell to cell. In animals, hormones are carried by the blood to distant target organs.
  1 mark
2. (i) Difference 2: In plants, there are no specialised glands; hormones are made in growing regions such as the root and shoot tips. In animals, hormones are secreted by specialised endocrine glands such as the thyroid and pituitary.
  1 mark
3. (ii)(1) Intelligence: cerebrum (forebrain).
  ½ mark
4. (ii)(2) Riding a bicycle, which needs balance, posture and precise voluntary movement: cerebellum (hindbrain).
  ½ mark
5. (ii)(3) Vomiting, an involuntary action: medulla oblongata (hindbrain).
  ½ mark
6. (ii)(4) Controlling hunger: hypothalamus (forebrain).
  ½ mark
7. (iii) The brain lies inside a bony box, the skull (cranium). It is also cushioned by a fluid-filled balloon (cerebrospinal fluid within the membranes), which absorbs shocks.
  ½ mark
8. (iii) The spinal cord is enclosed and protected by the vertebral column (backbone).
  ½ mark
Answer: (i) Plants: diffusion, no glands; animals: blood transport, endocrine glands. (ii) (1) cerebrum (2) cerebellum (3) medulla (4) hypothalamus. (iii) Brain: skull + fluid-filled balloon; spinal cord: vertebral column
Where the marks are
- Any two correct differences between plant and animal hormonal coordination (1 + 1 marks)
- Cerebrum, cerebellum, medulla, hypothalamus correctly matched (½ mark each)
- Brain protected by the skull/cranium with a fluid cushion (½ mark)
- Spinal cord protected by the vertebral column (½ mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerControl and Coordination
Answer the following.
1. (i)
  What are tropic movements ? Give an example of a plant hormone which (1) inhibits growth and (2) promotes cell division.
  0 marks
2. (ii)
  Explain directional movement of a tendril in pea plant in response to touch. Name the hormone responsible for this movement.
  0 marks
Try it first, then show our solutionSolution
1. (i) Tropic movements are directional growth movements of a plant part in response to an external stimulus such as light, gravity, water, chemicals or touch. The growth is towards the stimulus (positive) or away from it (negative). Example: a shoot bends towards light (phototropism).
  1 mark
2. (i)(1) A plant hormone that inhibits growth: abscisic acid (ABA). It also causes wilting of leaves.
  ½ mark
3. (i)(2) A plant hormone that promotes cell division: cytokinin.
  ½ mark
4. (ii) The tendril of a pea plant is sensitive to touch (thigmotropism). When it touches a support, auxin moves away from the point of contact towards the opposite side.
  1 mark
5. (ii) The side of the tendril away from the support therefore grows faster than the side in contact with it. This unequal growth on the two sides makes the tendril bend and coil around the support.
  1 mark
6. (ii) Hormone responsible: auxin.
  1 mark
Answer: (i) Directional growth movements in response to a stimulus; (1) abscisic acid, (2) cytokinin. (ii) Unequal growth on the two sides makes the tendril coil around the support; hormone: auxin
Where the marks are
- Defines tropic movement as directional growth movement in response to a stimulus (1 mark)
- Abscisic acid as the growth inhibitor (½ mark)
- Cytokinin as the promoter of cell division (½ mark)
- Explains auxin moving away from the contact point, faster growth on the far side and coiling (2 marks)
- Names auxin (1 mark)
Open Q35(b) on its own page
Q36 · 5 marks · Long answerLight – Reflection and Refraction
Answer the following.
1. (a)
  Upper half of a convex lens is covered with a black paper. Draw a ray diagram to show the formation of image of an object placed at a distance of 2F from such a lens. Mention the position and nature of the image formed. State the observable difference in the image obtained if the lens is uncovered. Give reason to justify your answer.
  0 marks
2. (b)
  An object is placed at a distance of 30 cm from the optical centre of a concave lens of focal length 15 cm. Use lens formula to determine the distance of the image from the optical centre of the lens.
  0 marks
Try it first, then show our solutionSolution
1. (a) Ray diagram:
  Draw the principal axis with a convex lens at O and the upper half shaded black. Mark $F_{1}$ , $2 F_{1}$ on the left and $F_{2}$ , $2 F_{2}$ on the right.
  Place an upright object AB at $2 F_{1}$ .
  Ray 1: from the top B, passing through $F_{1}$ and striking the uncovered lower half. It emerges parallel to the principal axis.
  Ray 2: from B, directed at the optical centre O, just below the black paper. It goes straight on undeviated.
  The two rays meet below the axis at $2 F_{2}$ . Draw the inverted image A′B′ there. Mark arrows on all rays.
  1 mark
2. (a) Position of image: at $2 F_{2}$ , on the other side of the lens.
  ½ mark
3. (a) Nature of image: real, inverted and of the same size as the object. Covering half the lens does not cut off half the image; the full image still forms.
  ½ mark
4. (a) Observable difference when the lens is uncovered: the image forms at the same place with the same size, but it is brighter.
  ½ mark
5. (a) Reason: with the whole lens uncovered, more rays from the object pass through the lens and meet at the image. More light reaches the image, so it is brighter.
  ½ mark
6. (b) Given (sign convention, light from left): object distance $u = - 30$ cm; focal length of the concave lens $f = - 15$ cm; image distance $v = ?$
  ½ mark
7. Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
  ½ mark
8. Substituting: $\frac{1}{v} = \frac{1}{- 15} + \frac{1}{- 30} = \frac{- 2 - 1}{30} = - \frac{1}{10}$ So $v = - 10$ cm. The image is 10 cm from the optical centre, on the same side as the object. It is virtual and erect.
  1 mark
Answer: (a) Image at 2F₂ on the other side, real, inverted, same size; it becomes brighter when the lens is uncovered because more rays pass through. (b) v = −10 cm (10 cm from the optical centre, same side as the object)
Where the marks are
- Correct ray diagram using the uncovered half, with arrows (1 mark)
- Image at 2F on the other side (½ mark)
- Real and inverted (½ mark)
- Image becomes brighter when uncovered (½ mark)
- Reason: more rays pass through the lens (½ mark)
- Writes the given values with signs (½ mark)
- Writes the lens formula (½ mark)
- Gets v = −10 cm (1 mark)
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Section E

Source/case-based with sub-parts

Q37(a) · 4 marks · Case-basedCompetency-basedAcids, Bases and Salts
Salts play a very important role in our daily life. Sodium chloride which is known as common salt is used almost in every kitchen. Baking soda is also a salt used in faster cooking as well as in baking industry. The family of salts is classified on the basis of cations and anions present in them.
1. (a)
  Identify the acid and base from which Sodium chloride is formed.
  1 mark
2. (b)
  Find the cation and the anion present in Calcium sulphate.
  1 mark
3. (c)
  "Sodium chloride and washing soda both belong to the same family of salts." Justify this statement.
  2 marks
4. (c)
  Define the term pH scale. Name the salt obtained by the reaction of Potassium hydroxide and Sulphuric acid and give the pH value of its aqueous solution.
  2 marks
Try it first, then show our solutionSolution
1. (a) Sodium chloride forms when an acid neutralises a base: $HCl + NaOH \to NaCl + H_{2} O$ . Acid: hydrochloric acid (HCl).
  ½ mark
2. (a) Base: sodium hydroxide (NaOH).
  ½ mark
3. (b) Calcium sulphate is $CaS O_{4}$ . Cation: $C a^{2 +}$ .
  ½ mark
4. (b) Anion: $S O_{4}^{2 -}$ (sulphate ion).
  ½ mark
5. (c)(a) Salts that have the same cation (or the same anion) are grouped into one family.
  1 mark
6. (c)(a) Sodium chloride ( $NaCl$ ) and washing soda ( $N a_{2} C O_{3} \cdot 10 H_{2} O$ ) both contain the same cation, $N a^{+}$ . So both belong to the family of sodium salts, even though their anions ( $C l^{-}$ and $C O_{3}^{2 -}$ ) differ.
  1 mark
7. (c)(b) [OR] The pH scale is a scale, usually from 0 to 14, for measuring the hydrogen ion ( $H^{+}$ ) concentration of a solution. A pH below 7 is acidic, 7 is neutral and above 7 is basic.
  1 mark
8. (c)(b) $2KOH + H_{2} S O_{4} \to K_{2} S O_{4} + 2 H_{2} O$ . The salt formed is potassium sulphate ( $K_{2} S O_{4}$ ).
  ½ mark
9. (c)(b) It comes from a strong acid and a strong base, so its aqueous solution is neutral, with pH = 7.
  ½ mark
Answer: (a) HCl and NaOH; (b) Ca²⁺ and SO₄²⁻; (c)(a) both contain the Na⁺ cation, so both are sodium salts; (c)(b) pH scale measures H⁺ ion concentration (0–14); K₂SO₄; pH = 7
Where the marks are
- HCl as the acid and NaOH as the base (½ + ½ marks)
- Ca²⁺ cation and SO₄²⁻ anion (½ + ½ marks)
- Justifies that both contain the Na⁺ cation, so they are in the same (sodium) family of salts (2 marks)
- OR: defines the pH scale (1 mark), names potassium sulphate (½ mark), states pH 7 (½ mark)
Open Q37(a) on its own page
Q38(a) · 4 marks · Case-basedCompetency-basedHow do Organisms Reproduce?
Asexual reproduction involves a single parent to produce offsprings without the formation of gametes. It occurs by the following ways : Fission, Budding, Fragmentation, Spore formation and Regeneration. In one of the methods like regeneration, Planaria A is cut horizontally into three pieces – L, M and N and Planaria B is cut vertically into two equal halves – O and P.
1. (a)
  Which of the cut pieces of the two Planaria could regenerate to form a complete organism ?
  1 mark
2. (b)
  Give an example of another organism which follows the same mode of reproduction as Planaria.
  1 mark
3. (c)
  What is the meaning of 'development' in regeneration ?
  2 marks
4. (c)
  Differentiate between regeneration and fragmentation.
  2 marks
Figure in the paper: Planaria A (left) is shown being cut horizontally into three pieces labelled L (head end), M (middle) and N (tail end). Planaria B (right) is shown with a vertical line through its length, separated into two equal halves labelled O (one half with head) and P (other half with head), with arrows pointing from B to O and P. See it in the official paper.
Try it first, then show our solutionSolution
1. (a) Planaria has a very high power of regeneration. Its body contains specialised cells that can form all the missing parts. So all the cut pieces can grow into complete planarians: L, M and N of Planaria A, and O and P of Planaria B.
  1 mark
2. (b) Another organism that reproduces by regeneration (like Planaria): Hydra.
  1 mark
3. (c)(a) In regeneration, specialised cells in the cut piece first proliferate, that is, divide repeatedly to make a large mass of cells.
  1 mark
4. (c)(a) This mass of cells then changes (differentiates) into different cell types and tissues, which form the missing organs. These changes happen in an organised sequence. That organised sequence is called development.
  1 mark
5. (c)(b) [OR] Regeneration: specialised cells of a body part proliferate and differentiate to form the lost parts. The whole new individual arises from any cut piece through this organised development. Example: Planaria, Hydra.
  1 mark
6. (c)(b) Fragmentation: the body simply breaks into pieces on maturing. Each fragment grows by ordinary cell division into a new organism, without needing specialised cells. Example: Spirogyra.
  1 mark
Answer: (a) All pieces L, M, N, O and P; (b) Hydra; (c)(a) proliferation of specialised cells, then their organised change into different cell types and tissues; (c)(b) regeneration uses specialised cells to rebuild lost parts, while in fragmentation each broken piece grows by simple cell division
Where the marks are
- States that all the cut pieces (L, M, N, O, P) regenerate (1 mark)
- Names Hydra (1 mark)
- Proliferation of specialised cells into a cell mass (1 mark)
- Organised change of the cell mass into different cell types and tissues, called development (1 mark)
- OR: correct features of regeneration (1 mark) and of fragmentation (1 mark)
Open Q38(a) on its own page
Q39(a) · 4 marks · Case-basedCompetency-basedElectricity
When electric current flows in a purely resistive circuit electrical energy gets fully converted into heat energy. The amount of heat produced (H) in the circuit is found to be directly proportional to (i) the square of current ( $I^{2}$ ) (ii) the resistance (R) of the conductor and (iii) the time (t) for which current flows. In other words $H = I^{2} Rt$ . Electrical devices such an electric fuse, electric heater, electric iron etc. are all based on this effect called heating effect of electric current.
1. (a)
  List two properties of heating elements.
  1 mark
2. (b)
  List two properties of electric fuse.
  1 mark
3. (c)
  Name the principle on which an electric fuse works. Explain how a fuse wire is capable of saving electrical appliances from getting damaged due to accidently produced high currents.
  2 marks
4. (c)
  The power of an electric heater is 1100 W. If the potential difference between the two terminals of the heater is 220 V, find the current flowing in the circuit. What will happen to an electric fuse of rating 5 A connected in this circuit ?
  2 marks
Try it first, then show our solutionSolution
1. (a) Heating elements (for example, nichrome in heaters) should have: (1) high resistivity, so more heat is produced ( $H = I^{2} R t$ ).
  ½ mark
2. (a) (2) A high melting point, and they should not oxidise (burn) readily at high temperatures. For this reason alloys are used.
  ½ mark
3. (b) An electric fuse wire should have: (1) high resistivity, so it heats up quickly.
  ½ mark
4. (b) (2) A low melting point, so it melts as soon as the current goes above its rating.
  ½ mark
5. (c)(a) Principle: an electric fuse works on the heating effect of electric current ( $H = I^{2} R t$ ).
  1 mark
6. (c)(a) The fuse is connected in series with the appliance. If an unduly high current flows by accident, for example in a short circuit or overload, the heat produced in the fuse wire rises sharply. The wire melts and breaks the circuit, which stops the current and saves the appliance from damage.
  1 mark
7. (c)(b) [OR] Given: $P = 1100$ W, $V = 220$ V. Use $P = V I$ .
  ½ mark
8. (c)(b) $I = \frac{P}{V} = \frac{1100 W}{220 V} = 5 A$
  ½ mark
9. (c)(b) The current (5 A) does not exceed the fuse rating (5 A). So the fuse will not melt; it has no effect and the heater works normally.
  1 mark
Answer: (a) High resistivity; high melting point / does not oxidise at high temperature. (b) High resistivity; low melting point. (c)(a) Heating effect of current; the wire melts on high current and breaks the circuit. (c)(b) I = 5 A; the 5 A fuse is not affected (does not blow)
Where the marks are
- Two properties of heating elements: high resistivity, high melting point/no oxidation (½ + ½ marks)
- Two properties of fuse wire: high resistivity, low melting point (½ + ½ marks)
- Names the heating effect of current (1 mark)
- Explains that the fuse melts on high current and breaks the circuit (1 mark)
- OR: uses P = VI (½ mark), gets I = 5 A (½ mark), states no effect on the 5 A fuse (1 mark)
Open Q39(a) 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.