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

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

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.

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/2024/X/SCIENCE.zip (a zip; this paper is the file 31_1_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; internal choice in 23 and 25
C	27-33	short answer, 50-80 words; internal choice in 33
D	34-36	long answer, 80-120 words; internal choice in each question
E	37-39	source/case-based with sub-parts; internal choice in sub-part (iii)

Instructions in brief: All 39 questions are compulsory with no overall choice; internal choice (OR) is given in some questions, and only one alternative is to be attempted.

Section A

MCQ 1-16, assertion-reason 17-20

Q1 · 1 mark · MCQChemical Reactions and Equations
When 2 mL of sodium hydroxide solution is added to few pieces of granulated zinc in a test tube and then warmed, the reaction that occurs can be written in the form of a balanced chemical equation as :
1. (A) $N a O H + Z n \to N a Z n O_{2} + H_{2} O$
2. (B) $2 N a O H + Z n \to N a_{2} Z n O_{2} + H_{2}$
3. (C) $2 N a O H + Z n \to N a Z n O_{2} + H_{2}$
4. (D) $2 N a O H + Z n \to N a_{2} Z n O_{2} + H_{2} O$
Try it first, then show our solutionSolution
1. Zinc reacts with hot NaOH to give sodium zincate ( $N a_{2} Z n O_{2}$ ) and hydrogen gas. Balanced: $2 N a O H + Z n \to N a_{2} Z n O_{2} + H_{2}$ , so option (B).
  1 mark
Answer: (B) $2 N a O H + Z n \to N a_{2} Z n O_{2} + H_{2}$
Where the marks are
- Chooses the equation with sodium zincate and $H_{2}$ (1 mark)
Q2 · 1 mark · MCQChemical Reactions and Equations
Select from the following a decomposition reaction in which source of energy for decomposition is light :
1. (A) $2 F e S O_{4} \to F e_{2} O_{3} + S O_{2} + S O_{3}$
2. (B) $2 H_{2} O \to 2 H_{2} + O_{2}$
3. (C) $2 A g B r \to 2 A g + B r_{2}$
4. (D) $C a C O_{3} \to C a O + C O_{2}$
Try it first, then show our solutionSolution
1. Silver bromide decomposes in sunlight into silver and bromine (photolytic decomposition, used in black-and-white photography): $2 A g B r l i g h t 2 A g + B r_{2}$ . The others need heat or electricity.
  1 mark
Answer: (C) $2 A g B r \to 2 A g + B r_{2}$
Where the marks are
- Identifies decomposition of AgBr by light (1 mark)
Q3 · 1 mark · MCQMetals and Non-metals
A metal and a non-metal that exists in liquid state at the room temperature are respectively :
1. (A)Bromine and Mercury
2. (B)Mercury and Iodine
3. (C)Mercury and Bromine
4. (D)Iodine and Mercury
Try it first, then show our solutionSolution
1. Mercury is the only metal and bromine the only non-metal liquid at room temperature; order asked is metal then non-metal.
  1 mark
Answer: (C) Mercury and Bromine
Where the marks are
- Correct pair in correct order (1 mark)
Q4 · 1 mark · MCQCarbon and its Compounds
Carbon compounds :
(i) are good conductors of electricity.
(ii) are bad conductors of electricity.
(iii) have strong forces of attraction between their molecules.
(iv) have weak forces of attraction between their molecules.
The correct statements are :
1. (A)(i) and (ii)
2. (B)(ii) and (iii)
3. (C)(ii) and (iv)
4. (D)(i) and (iii)
Try it first, then show our solutionSolution
1. Carbon compounds are covalent, form no ions, so they are bad conductors; intermolecular forces are weak, hence low melting/boiling points. Statements (ii) and (iv).
  1 mark
Answer: (C) (ii) and (iv)
Where the marks are
- Selects (ii) and (iv) (1 mark)
Q5 · 1 mark · MCQAcids, Bases and Salts
Consider the following compounds :
$F e S O_{4}$ ; $C u S O_{4}$ ; $C a S O_{4}$ ; $N a_{2} C O_{3}$
The compound having maximum number of water of crystallisation in its crystalline form in one molecule is :
1. (A) $F e S O_{4}$
2. (B) $C u S O_{4}$
3. (C) $C a S O_{4}$
4. (D) $N a_{2} C O_{3}$
Try it first, then show our solutionSolution
1. Water of crystallisation: $F e S O_{4} \cdot 7 H_{2} O$ , $C u S O_{4} \cdot 5 H_{2} O$ , $C a S O_{4} \cdot 2 H_{2} O$ , $N a_{2} C O_{3} \cdot 10 H_{2} O$ . Maximum is 10 in washing soda.
  1 mark
Answer: (D) $N a_{2} C O_{3}$
Where the marks are
- Identifies $N a_{2} C O_{3} \cdot 10 H_{2} O$ (1 mark)
Q6 · 1 mark · MCQMetals and Non-metals
Oxides of aluminium and zinc are :
1. (A)acidic
2. (B)basic
3. (C)amphoteric
4. (D)neutral
Try it first, then show our solutionSolution
1. $A l_{2} O_{3}$ and $Z n O$ react with both acids and bases to form salt and water, so they are amphoteric.
  1 mark
Answer: (C) amphoteric
Where the marks are
- States amphoteric (1 mark)
Q7 · 1 mark · MCQChemical Reactions and Equations
$M n O_{2} + 4 H C l \to M n C l_{2} + 2 H_{2} O + C l_{2}$
The reaction given above is a redox reaction because in this case :
1. (A) $M n O_{2}$ is oxidised and HCl is reduced.
2. (B)HCl is oxidised.
3. (C) $M n O_{2}$ is reduced.
4. (D) $M n O_{2}$ is reduced and HCl is oxidised.
Try it first, then show our solutionSolution
1. $M n O_{2}$ loses oxygen → reduced; HCl loses hydrogen (gives $C l_{2}$ ) → oxidised. Both occur together, so redox: option (D).
  1 mark
Answer: (D) $M n O_{2}$ is reduced and HCl is oxidised
Where the marks are
- Both oxidation and reduction identified correctly (1 mark)
Q8 · 1 mark · MCQHeredity
Consider the following statements :
(i) The sex of a child is determined by what it inherits from the mother.
(ii) The sex of a child is determined by what it inherits from the father.
(iii) The probability of having a male child is more than that of a female child.
(iv) The sex of a child is determined at the time of fertilisation when male and female gametes fuse to form a zygote.
The correct statements are :
1. (A)(i) and (iii)
2. (B)(ii) and (iv)
3. (C)(iii) and (iv)
4. (D)(i), (iii) and (iv)
Try it first, then show our solutionSolution
1. Mother always gives X; father gives X or Y, so the father's contribution decides sex, at fertilisation. Probability of boy or girl is equal (50%). Correct: (ii) and (iv).
  1 mark
Answer: (B) (ii) and (iv)
Where the marks are
- Selects (ii) and (iv) (1 mark)
Q9 · 1 mark · MCQHeredity
Chromosomes :
(i) carry hereditary information from parents to the next generation.
(ii) are thread like structures located inside the nucleus of an animal cell.
(iii) always exist in pairs in human reproductive cells.
(iv) are involved in the process of cell division.
The correct statements are :
1. (A)(i) and (ii)
2. (B)(iii) and (iv)
3. (C)(i), (ii) and (iv)
4. (D)(i) and (iv)
Solution being checked. Our step-by-step solution is written and is waiting for an independent check before we show it.
Q10 · 1 mark · MCQControl and Coordination
In a nerve cell, the site where the electrical impulse is converted into a chemical signal is known as :
1. (A)Axon
2. (B)Dendrites
3. (C)Neuromuscular junction
4. (D)Cell body
Try it first, then show our solutionSolution
1. At the axon ending/junction the electrical impulse releases chemicals that cross the gap; among the options this is the neuromuscular junction (synapse with muscle).
  1 mark
Answer: (C) Neuromuscular junction
Where the marks are
- Chooses neuromuscular junction (1 mark)
Q11 · 1 mark · MCQLife Processes
A stomata closes when :
(i) it needs carbon dioxide for photosynthesis.
(ii) it does not need carbon dioxide for photosynthesis.
(iii) water flows out of the guard cells.
(iv) water flows into the guard cells.
The correct reason(s) in this process is/are :
1. (A)(i) only
2. (B)(i) and (iii)
3. (C)(ii) and (iii)
4. (D)(ii) and (iv)
Try it first, then show our solutionSolution
1. Stomata close when CO2 is not needed, and closing occurs as guard cells lose water (become flaccid). So (ii) and (iii).
  1 mark
Answer: (C) (ii) and (iii)
Where the marks are
- Selects (ii) and (iii) (1 mark)
Q12 · 1 mark · MCQLight – Reflection and Refraction
At what distance from a convex lens should an object be placed to get an image of the same size as that of the object on a screen ?
1. (A)Beyond twice the focal length of the lens.
2. (B)At the principal focus of the lens.
3. (C)At twice the focal length of the lens.
4. (D)Between the optical centre of the lens and its principal focus.
Try it first, then show our solutionSolution
1. For a convex lens, an object at $2 F_{1}$ gives a real, inverted, same-size image at $2 F_{2}$ , so object at twice the focal length.
  1 mark
Answer: (C) At twice the focal length
Where the marks are
- Object at 2F (1 mark)
Q13 · 1 mark · MCQThe Human Eye and the Colourful World
The lens system of human eye forms an image on a light sensitive screen, which is called as :
1. (A)Cornea
2. (B)Ciliary muscles
3. (C)Optic nerves
4. (D)Retina
Try it first, then show our solutionSolution
1. The eye lens focuses the image on the retina, a delicate light-sensitive membrane with rods and cones.
  1 mark
Answer: (D) Retina
Where the marks are
- Retina (1 mark)
Q14 · 1 mark · MCQMagnetic Effects of Electric Current
The pattern of the magnetic field produced inside a current carrying solenoid is :
1. (A)Diagram (a): several parallel, evenly spaced straight field lines with arrows pointing in one direction.
2. (B)Diagram (b): concentric circles with an arrow, centred on a dot.
3. (C)Diagram (c): parallel straight field lines inside, closed loops outside with arrows at the ends.
4. (D)Diagram (d): curved lines bending around a central straight line, with circled dot and cross symbols at the top and bottom.
Figure in the paper: Four labelled diagrams (a), (b), (c), (d) of magnetic field patterns. (a) Several parallel straight lines with arrows pointing right. (b) Concentric circles with an arrow on one circle, around a central dot. (c) Parallel straight lines inside with curved return lines outside forming closed loops, arrows on the lines. (d) Curved lines symmetric about a central horizontal straight line, with arrows pointing towards the centre, and circled dot and cross symbols at top and bottom. See it in the official paper.
Try it first, then show our solutionSolution
1. Inside a solenoid the field is uniform, shown by parallel, equally spaced straight lines in one direction — diagram (a).
  1 mark
Answer: (A) Diagram (a)
Where the marks are
- Parallel equidistant straight lines (1 mark)
Q15 · 1 mark · MCQOur Environment
Identify the food chain in which the organisms of the second trophic level are missing :
1. (A)Grass, goat, lion
2. (B)Zooplankton, Phytoplankton, small fish, large fish
3. (C)Tiger, grass, snake, frog
4. (D)Grasshopper, grass, snake, frog, eagle
Try it first, then show our solutionSolution
1. Arrange: grass → (herbivore missing) → frog → snake → tiger. The herbivore (e.g. grasshopper), the second trophic level, is absent in (C).
  1 mark
Answer: (C) Tiger, grass, snake, frog
Where the marks are
- Identifies chain lacking herbivore (1 mark)
Q16 · 1 mark · MCQHow do Organisms Reproduce?
In which of the following organisms, multiple fission is a means of asexual reproduction ?
1. (A)Yeast
2. (B)Leishmania
3. (C)Paramoecium
4. (D)Plasmodium
Try it first, then show our solutionSolution
1. Plasmodium divides into many daughter cells at once (multiple fission); Leishmania and Paramoecium show binary fission, yeast buds.
  1 mark
Answer: (D) Plasmodium
Where the marks are
- Plasmodium (1 mark)
Q17 · 1 mark · Assertion-reasonCompetency-basedMetals and Non-metals
Assertion (A): Hydrogen gas is not evolved when zinc reacts with nitric acid.
Reason (R): Nitric acid oxidises the hydrogen gas produced to water and itself gets reduced.
For Q. Nos. 17 to 20, select the correct answer from the codes below.
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, but 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. $H N O_{3}$ is a strong oxidising agent; it oxidises $H_{2}$ to water and itself is reduced to nitrogen oxides, so $H_{2}$ is not evolved. Both true and R explains A.
  1 mark
Answer: (A)
Where the marks are
- Both true, R explains A (1 mark)
Q18 · 1 mark · Assertion-reasonCompetency-basedOur Environment
Assertion (A): Accumulation of harmful chemicals is maximum in the organisms at the highest trophic level of a food chain.
Reason (R): Harmful chemicals are sprayed on the crops to protect them from diseases and pests.
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, but 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 due to biological magnification (non-biodegradable chemicals accumulate up the chain). R is true but only says why chemicals are sprayed, not why they concentrate at top levels: option (B).
  1 mark
Answer: (B)
Where the marks are
- Both true, R not explanation (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedLife Processes
Assertion (A): The rate of breathing in aquatic organisms is much faster than in terrestrial organisms.
Reason (R): The amount of oxygen dissolved in water is very high as compared to the amount of oxygen in air.
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, but 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. Aquatic organisms breathe faster because dissolved oxygen in water is much lower than oxygen in air. So A is true, R is false.
  1 mark
Answer: (C)
Where the marks are
- A true, R false (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedThe Human Eye and the Colourful World
Assertion (A): The rainbow is a natural spectrum of sunlight in the sky.
Reason (R): Rainbow is formed in the sky when the sun is overhead and water droplets are also present in air.
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, but 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. Rainbow is a natural spectrum (A true). It is seen opposite the sun, when the sun is behind the observer (low in sky), not overhead, so R is false.
  1 mark
Answer: (C)
Where the marks are
- A true, R false (1 mark)

Section B

very short answer, 30-50 words; internal choice in 23 and 25

Q21 · 2 marks · Very short answerChemical Reactions and Equations
Name the type of chemical reaction in which calcium oxide reacts with water. Justify your answer by giving balanced chemical equation for the chemical reaction.
Try it first, then show our solutionSolution
1. Type of reaction: Combination reaction (it is also exothermic).
  ½ mark
2. Justification: two reactants, calcium oxide and water, combine to form a single product, calcium hydroxide.
  ½ mark
3. Balanced equation: $CaO (s) + H_{2} O (l) \to Ca (OH)_{2} (aq) + Heat$ . Here quick lime gives slaked lime.
  1 mark
Answer: Combination reaction: $CaO (s) + H_{2} O (l) \to Ca (OH)_{2} (aq)$ + heat
Where the marks are
- Names the reaction as a combination reaction (½ mark)
- Justifies it: a single product is formed from two reactants (½ mark)
- Writes the balanced equation forming $Ca (OH)_{2}$ (1 mark)
Open Q21 on its own page
Q22 · 2 marks · Very short answerLife Processes
State one role of each of the following in human digestive system :
(i) Hydrochloric acid
(ii) Villi
(iii) Anal Sphincter
(iv) Lipase
Try it first, then show our solutionSolution
1. (i) Hydrochloric acid: makes the medium in the stomach acidic so that the enzyme pepsin can act. It also kills germs that come in with food.
  ½ mark
2. (ii) Villi: these finger-like projections increase the surface area of the small intestine for absorption of digested food.
  ½ mark
3. (iii) Anal sphincter: controls (regulates) the exit of waste material (faeces) through the anus.
  ½ mark
4. (iv) Lipase: breaks down emulsified fats into fatty acids and glycerol.
  ½ mark
Answer: HCl: acidic medium for pepsin / kills germs; Villi: increase surface area for absorption; Anal sphincter: regulates egestion; Lipase: digests emulsified fats
Where the marks are
- Role of HCl: acidic medium for enzyme action or kills microbes (½ mark)
- Role of villi: increase surface area for absorption (½ mark)
- Role of anal sphincter: regulates exit of waste (½ mark)
- Role of lipase: digests emulsified fats (½ mark)
Open Q22 on its own page
Answer one of these 2 alternatives.
Q23(A) · 2 marks · Very short answerLife Processes
How is the movement of leaves of a sensitive plant different from the downward movement of the roots ?
Try it first, then show our solutionSolution
1. Difference 1: In the sensitive plant the stimulus is touch. The roots move downward in response to gravity.
  1 mark
2. Difference 2: The leaf movement of the sensitive plant involves no growth; it happens through changes in water content of cells. It is also non-directional (nastic). The downward root movement is a growth movement (geotropism). It is directional, depending on the direction of the stimulus.
  1 mark
Answer: Sensitive plant: touch stimulus, no growth, non-directional; Roots: gravity stimulus, growth involved, directional
Where the marks are
- Correct difference in stimulus (touch vs gravity) (1 mark)
- Correct difference in growth or directionality (1 mark)
Open Q23(A) on its own page
OR
Q23(B) · 2 marks · Very short answerControl and Coordination
There is a hormone which regulates carbohydrate, protein and fat metabolism in our body. Name the hormone and the gland which secretes it. Why is it important for us to have iodised salt in our diet?
Try it first, then show our solutionSolution
1. Hormone: Thyroxine.
  ½ mark
2. Gland: Thyroid gland.
  ½ mark
3. Iodised salt is needed because the thyroid gland requires iodine to make thyroxine. Lack of iodine in the diet causes goitre, a swollen neck.
  1 mark
Answer: Thyroxine; thyroid gland; iodine is needed to make thyroxine and its deficiency causes goitre
Where the marks are
- Names thyroxine (½ mark)
- Names thyroid gland (½ mark)
- Iodine is needed for thyroxine synthesis; deficiency causes goitre (1 mark)
Open Q23(B) on its own page
Q24 · 2 marks · Very short answerLight – Reflection and Refraction
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position of the image formed by the mirror.
Try it first, then show our solutionSolution
1. Given $u = - 10$ cm and $f = + 15$ cm (convex mirror). Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ .
  ½ mark
2. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{- 10} = \frac{2 + 3}{30} = \frac{1}{6}$
  ½ mark
3. $v = + 6$ cm. The image is formed 6 cm behind the mirror; it is virtual and erect.
  1 mark
Answer: v = +6 cm, i.e. 6 cm behind the mirror (virtual, erect)
Where the marks are
- Correct sign convention and mirror formula (½ mark)
- Correct substitution (½ mark)
- v = +6 cm, behind the mirror (1 mark)
Open Q24 on its own page
Answer one of these 2 alternatives.
Q25(A) · 2 marks · Very short answerElectricity
Show how you would connect three resistors each of resistance $6 Ω$ , so that the combination has a resistance of $9 Ω$ . Also justify your answer.
Try it first, then show our solutionSolution
1. Arrangement: connect two $6 Ω$ resistors in parallel and connect the third $6 Ω$ resistor in series with this parallel pair. Diagram: two 6 Ω resistors side by side between points A and B, then a third 6 Ω resistor from B to C.
  1 mark
2. Justification: $\frac{1}{R _{P}} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6}$ , so $R_{P} = 3 Ω$ . Then $R = R_{P} + 6 = 3 + 6 = 9 Ω$ .
  1 mark
Answer: Two 6 Ω in parallel (3 Ω) in series with the third 6 Ω gives 9 Ω
Where the marks are
- Correct arrangement stated or drawn (1 mark)
- Calculation showing $R_{P} = 3 Ω$ and total $9 Ω$ (1 mark)
Open Q25(A) on its own page
OR
Q25(B) · 2 marks · Very short answerElectricity
In the given circuit calculate the power consumed in watts in the resistor of $2 Ω$ .
Figure in the paper: Circuit: a rectangular loop containing two resistors in series along the top side, labelled 1 Ω (left) and 2 Ω (right). The bottom side contains a 6 V battery (cell symbols with + terminal on the left). See it in the official paper.
Try it first, then show our solutionSolution
1. The resistors are in series: $R = R_{1} + R_{2} = 1 Ω + 2 Ω = 3 Ω$ .
  ½ mark
2. Current: $I = \frac{V}{R} = \frac{6 V}{3 Ω} = 2$ A. The same current flows through both resistors.
  ½ mark
3. Power in the $2 Ω$ resistor: $P = I^{2} R$ .
  ½ mark
4. $P = (2 A)^{2} \times 2 Ω = 8$ W
  ½ mark
Answer: 8 W
Where the marks are
- Equivalent resistance 3 Ω (½ mark)
- Current 2 A (½ mark)
- Uses $P = I^{2} R$ (½ mark)
- P = 8 W (½ mark)
Open Q25(B) on its own page
Q26 · 2 marks · Very short answerMagnetic Effects of Electric Current
(i) Two magnetic field lines do not intersect each other. Why?
(ii) How is a uniform magnetic field in a given region represented? Draw a diagram in support of your answer.
Try it first, then show our solutionSolution
1. (i) If two field lines intersected, the point of intersection would have two directions of magnetic field. A compass needle placed there would have to point in two directions at once, which is not possible.
  1 mark
2. (ii) A uniform magnetic field is represented by parallel, equally spaced straight lines, all with arrows in the same direction.
  ½ mark
3. Diagram: draw 4–5 parallel straight lines at equal spacing, each with an arrow pointing the same way. An example is the field inside a long current-carrying solenoid.
  ½ mark
Answer: (i) There would be two directions of field at one point, which is impossible; (ii) equidistant parallel straight lines
Where the marks are
- Reason: two field directions at one point is impossible (1 mark)
- Uniform field shown by equidistant parallel straight lines (½ mark)
- Correct diagram with arrows (½ mark)
Open Q26 on its own page

Section C

short answer, 50-80 words; internal choice in 33

Q27 · 3 marks · Short answerChemical Reactions and Equations
Write one chemical equation each for the chemical reaction in which the following have taken place :
(i) Change in colour
(ii) Change in temperature
(iii) Formation of precipitate
Mention colour change/temperature change (rise/fall)/compound precipitated along with equation.
Try it first, then show our solutionSolution
1. (i) Change in colour: an iron nail placed in copper sulphate solution turns the blue solution pale green.
  ½ mark
2. $Fe (s) + CuS O_{4} (aq) \to FeS O_{4} (aq) + Cu (s)$
  ½ mark
3. (ii) Change in temperature: neutralisation of sodium hydroxide with hydrochloric acid releases heat, so the temperature rises.
  ½ mark
4. $NaOH (aq) + HCl (aq) \to NaCl (aq) + H_{2} O (l) + Heat$
  ½ mark
5. (iii) Formation of precipitate: a **yellow precipitate of lead iodide ( $Pb I_{2}$ )** forms.
  ½ mark
6. $Pb (N O_{3})_{2} (aq) + 2KI (aq) \to Pb I_{2} (s) + 2KN O_{3} (aq)$
  ½ mark
Answer: (i) Fe + CuSO₄, blue to green; (ii) NaOH + HCl, temperature rises; (iii) Pb(NO₃)₂ + 2KI, yellow PbI₂ precipitate
Where the marks are
- Colour change stated (½ mark) with balanced equation (½ mark)
- Temperature rise stated (½ mark) with balanced equation (½ mark)
- Precipitate named (½ mark) with balanced equation (½ mark)
Open Q27 on its own page
Q28 · 3 marks · Short answerCompetency-basedAcids, Bases and Salts
(i) The pH of a sample of tomato juice is 4.6. How is this juice likely to be in taste? Give reason to justify your answer.
(ii) How do we differentiate between a strong acid and a weak base in terms of ion-formation in aqueous solutions?
(iii) The acid rain can make the survival of aquatic animals difficult. How?
Try it first, then show our solutionSolution
1. (i) The tomato juice will taste sour (slightly).
  ½ mark
2. Reason: a pH of 4.6 is less than 7, so the juice is acidic, and acids taste sour.
  ½ mark
3. (ii) A strong acid ionises almost completely in water and gives a large concentration of $H^{+}$ ( $H_{3} O^{+}$ ) ions.
  ½ mark
4. A weak base ionises only partially in water and gives a small concentration of $O H^{-}$ ions.
  ½ mark
5. (iii) Aquatic animals survive only in a narrow pH range of about 7.0 to 7.8. Acid rain (pH below 5.6) lowers the pH of river water outside this range, so their survival becomes difficult.
  1 mark
Answer: (i) Sour, since pH < 7 means acidic; (ii) strong acid gives many H⁺ ions, weak base gives few OH⁻ ions; (iii) acid rain lowers river pH below the 7.0–7.8 range animals need
Where the marks are
- Sour taste (½ mark) with reason that pH < 7 means acidic (½ mark)
- Strong acid gives more H⁺ ions (½ mark)
- Weak base gives fewer OH⁻ ions (½ mark)
- Acid rain lowers pH below the survivable range of 7.0–7.8 (1 mark)
Open Q28 on its own page
Q29 · 3 marks · Short answerLife Processes
(i) Why is respiratory pigment needed in multicellular organisms with large body size?
(ii) Give reasons for the following :
(a) Rings of cartilage are present in the throat.
(b) Lungs always contain a residual volume of air.
(c) The diaphragm flattens and ribs are lifted up when we breathe in.
(d) Walls of alveoli contain an extensive network of blood vessels.
Try it first, then show our solutionSolution
1. (i) In large organisms, diffusion alone is too slow to supply oxygen to all body cells. A respiratory pigment such as haemoglobin, which has high affinity for oxygen, picks up oxygen in the lungs and carries it to the tissues.
  1 mark
2. (ii)(a) Rings of cartilage support the throat (trachea) so that the air passage does not collapse when there is less air in it.
  ½ mark
3. (b) A residual volume of air stays in the lungs so that there is enough time for oxygen to be absorbed and carbon dioxide to be released.
  ½ mark
4. (c) When the diaphragm flattens and the ribs are lifted, the chest cavity becomes larger, so air is sucked into the lungs.
  ½ mark
5. (d) Exchange of gases takes place in the alveoli. The dense network of blood vessels lets oxygen pass into the blood and carbon dioxide pass out efficiently.
  ½ mark
Answer: (i) Diffusion alone cannot meet O₂ needs, so the pigment carries O₂; (ii)(a) prevents collapse of air passage; (b) time for gas exchange; (c) chest cavity enlarges; (d) gas exchange occurs in alveoli
Where the marks are
- Diffusion is insufficient to deliver oxygen in large bodies (1 mark)
- Cartilage rings prevent collapse of the air passage (½ mark)
- Residual air allows time for gas exchange (½ mark)
- Chest cavity enlarges during inhalation (½ mark)
- Blood vessels at alveoli for exchange of gases (½ mark)
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Q30 · 3 marks · Short answerControl and Coordination
Define reflex action. With the help of a flow chart show the path of a reflex action such as sneezing.
Try it first, then show our solutionSolution
1. Reflex action is a sudden, automatic (involuntary) response to a stimulus, made without conscious thinking. Example: sneezing.
  1 mark
2. Flow chart for sneezing: Stimulus (irritant/dust in nose) → Receptors in the nasal lining → Sensory neuron → Relay neuron in the CNS (spinal cord / brainstem) → Motor neuron → Effector (muscles of the chest/diaphragm) → Response (sneeze, forcing air out).
  2 marks
Answer: Reflex action is a sudden involuntary response to a stimulus; stimulus → receptor (nose) → sensory neuron → relay neuron (spinal cord) → motor neuron → effector (muscles) → response
Where the marks are
- Correct definition of reflex action (1 mark)
- Correct reflex arc flow chart in the right sequence (2 marks)
Open Q30 on its own page
Q31 · 3 marks · Short answerCompetency-basedThe Human Eye and the Colourful World
Study the diagram given below and answer the questions that follow :
(i) Name the defect of vision represented in the diagram. Give reason for your answer.
(ii) List two causes of this defect.
(iii) With the help of a diagram show how this defect of vision is corrected.
Figure in the paper: Eye diagram: a point object N on the left on the principal axis, a point N′ closer to the eye. Light rays from N pass through the lens and converge to a point on the far side that lies behind the retina (shown by dashed lines extending beyond the retina), with arrows on the rays showing direction of travel toward the retina. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Defect: Hypermetropia (far-sightedness).
  ½ mark
2. Reason: light from a nearby object (N′) is focused behind the retina, so the near point (N) lies farther than 25 cm.
  ½ mark
3. (ii) Cause 1: the focal length of the eye lens is too long, so the lens is less converging.
  ½ mark
4. Cause 2: the eyeball has become too small (shortened).
  ½ mark
5. (iii) Correction diagram: place a convex lens of suitable power in front of the eye. Rays from the object at the normal near point N′ (25 cm) are made less divergent by the lens, so they appear to come from N. The eye lens then focuses them on the retina. Label N, N′, the convex lens, the retina and the arrows on the rays.
  1 mark
Answer: (i) Hypermetropia, since the image forms behind the retina; (ii) focal length too long, eyeball too small; (iii) corrected with a convex lens
Where the marks are
- Names hypermetropia (½ mark)
- Reason: image formed behind retina / near point farther than 25 cm (½ mark)
- Two causes (½ + ½ marks)
- Labelled diagram of correction with a convex lens (1 mark)
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Q32 · 3 marks · Short answerMagnetic Effects of Electric Current
Name and state the rule to determine the direction of a :
(i) magnetic field produced around a current carrying straight conductor.
(ii) force experienced by a current carrying straight conductor placed in a magnetic field which is perpendicular to it.
Try it first, then show our solutionSolution
1. (i) Rule: Right-hand thumb rule.
  ½ mark
2. Hold the current-carrying straight wire in the right hand with the thumb pointing in the direction of the current. The curled fingers then show the direction of the magnetic field lines around the wire.
  1 mark
3. (ii) Rule: Fleming's left-hand rule.
  ½ mark
4. Stretch the thumb, forefinger and middle finger of the left hand so that they are mutually perpendicular. Point the forefinger in the direction of the magnetic field and the middle finger in the direction of the current. The thumb then gives the direction of the force (motion) on the conductor.
  1 mark
Answer: (i) Right-hand thumb rule; (ii) Fleming's left-hand rule
Where the marks are
- Names right-hand thumb rule (½ mark)
- States it correctly (1 mark)
- Names Fleming's left-hand rule (½ mark)
- States it correctly with finger roles (1 mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(A) · 3 marks · Short answerCompetency-basedOur Environment
Plants $\to$ Deer $\to$ Lion
In the given food chain, what will be the impact of removing all the organisms of second trophic level on the first and third trophic level? Will the impact be the same for the organisms of the third trophic level in the above food chain if they were present in a food web? Justify.
Try it first, then show our solutionSolution
1. First trophic level: with no deer to eat them, the number of plants will increase.
  1 mark
2. Third trophic level: the lions will decrease in number (they may die) because they lose their only food.
  1 mark
3. No, the impact would not be the same in a food web.
  ½ mark
4. Justification: in a food web the lions have alternative prey (other herbivores) to feed on, so they would not starve. Food webs are therefore more stable.
  ½ mark
Answer: Plants increase; lions decrease; no, in a food web lions find alternative food
Where the marks are
- Plants (first trophic level) increase (1 mark)
- Lions (third trophic level) decrease (1 mark)
- Impact not the same (½ mark)
- Alternative food sources available in a food web (½ mark)
Open Q33(A) on its own page
OR
Q33(B) · 3 marks · Short answerOur Environment
A gas 'X' which is a deadly poison is found at the higher levels of atmosphere and performs an essential function.
Name the gas and write the function performed by this gas in the atmosphere. Which chemical is linked to the decrease in the level of this gas? What measures have been taken by an international organization to check the depletion of the layer containing this gas?
Try it first, then show our solutionSolution
1. Gas X is **ozone ( $O_{3}$ )**.
  1 mark
2. Function: the ozone layer shields the Earth's surface from harmful ultraviolet (UV) radiation from the Sun.
  1 mark
3. Chemical linked to its depletion: chlorofluorocarbons (CFCs), used in refrigerants and fire extinguishers.
  ½ mark
4. Measure: in 1987 UNEP brought about an agreement to freeze CFC production at 1986 levels. CFC-free refrigerators are now made.
  ½ mark
Answer: Ozone; absorbs/shields UV radiation; CFCs; UNEP agreement to freeze CFC production at 1986 levels
Where the marks are
- Names ozone (1 mark)
- Protects from UV radiation (1 mark)
- Names CFCs (½ mark)
- UNEP agreement to freeze CFC production at 1986 levels (½ mark)
Open Q33(B) on its own page

Section D

long answer, 80-120 words; internal choice in each question

Answer one of these 2 alternatives.
Q34(A) · 5 marks · Long answerCarbon and its Compounds
(i) Define a homologous series of carbon compounds.
(ii) Why is the melting and boiling points of $C_{4} H_{8}$ higher than that of $C_{3} H_{6}$ or $C_{2} H_{4}$ ?
(iii) Why do we NOT see any gradation in chemical properties of a homologous series compounds ?
(iv) Write the name and structures of (i) aldehyde and (ii) ketone with molecular form $C_{3} H_{6} O$ .
Try it first, then show our solutionSolution
1. (i) A homologous series is a series of carbon compounds that have the same functional group and similar chemical properties. Successive members differ by a $- C H_{2} -$ unit (14 u in mass).
  1 mark
2. (ii) $C_{4} H_{8}$ has a higher molecular mass than $C_{3} H_{6}$ and $C_{2} H_{4}$ . In a homologous series, melting and boiling points increase as molecular mass increases.
  1 mark
3. (iii) Chemical properties depend on the functional group. All members of a homologous series have the same functional group, so their chemical properties remain similar and show no gradation.
  1 mark
4. (iv)(i) Aldehyde: Propanal, $C H_{3} - C H_{2} - CHO$ .
  ½ mark
5. Structure: $H_{3} C - C H_{2} - C (= O) - H$ , i.e. the C=O is at the end of the chain bonded to an H.
  ½ mark
6. (iv)(ii) Ketone: Propanone (acetone), $C H_{3} COC H_{3}$ .
  ½ mark
7. Structure: $H_{3} C - C (= O) - C H_{3}$ , i.e. the C=O is on the middle carbon.
  ½ mark
Answer: (i) Same functional group, similar properties, differ by CH₂; (ii) higher molecular mass; (iii) same functional group decides properties; (iv) propanal CH₃CH₂CHO, propanone CH₃COCH₃
Where the marks are
- Definition of homologous series (1 mark)
- Melting/boiling point rises with molecular mass (1 mark)
- Same functional group, so similar chemical properties (1 mark)
- Propanal name and structure (½ + ½ marks)
- Propanone name and structure (½ + ½ marks)
Open Q34(A) on its own page
OR
Q34(B) · 5 marks · Long answerCarbon and its Compounds
(i) Write the name and structure of an organic compound 'X' having two carbon atoms in its molecule and its name is suffixed with '-ol'.
(ii) What happens when 'X' is heated with excess concentrated sulphuric acid at 443 K? Write chemical equation for the reaction stating the conditions for the reaction. Also state the role played by concentrated sulphuric acid in the reaction.
(iii) Name and draw the electron dot structure of hydrocarbon produced in the above reaction.
Try it first, then show our solutionSolution
1. (i) X is ethanol, $C_{2} H_{5} OH$ .
  ½ mark
2. Structure: $H - H C H - H C H - O - H$ , i.e. $C H_{3} - C H_{2} - OH$ .
  ½ mark
3. (ii) Ethanol is dehydrated to ethene.
  ½ mark
4. $C H_{3} C H_{2} OH hot conc. H_{2} S O_{4} 443 K C H_{2} = C H_{2} + H_{2} O$
  1 mark
5. Role of concentrated $H_{2} S O_{4}$ : it acts as a dehydrating agent and removes water from ethanol.
  1 mark
6. (iii) Hydrocarbon formed: ethene, $C_{2} H_{4}$ .
  ½ mark
7. Electron dot structure: the two C atoms share two electron pairs (a double bond, written $C :: C$ ). Each C also shares one electron pair with each of two H atoms. In summary: $H : C .. H :: C .. H : H$ , with every C having an octet.
  1 mark
Answer: (i) Ethanol, CH₃CH₂OH; (ii) ethene forms: C₂H₅OH → (conc. H₂SO₄, 443 K) C₂H₄ + H₂O, with H₂SO₄ as dehydrating agent; (iii) ethene, H₂C::CH₂ electron dot structure
Where the marks are
- Names ethanol (½ mark) and gives its structure (½ mark)
- States ethene is formed (½ mark)
- Equation with conditions conc. H₂SO₄ and 443 K (1 mark; ½ deducted if conditions missing)
- Conc. H₂SO₄ acts as dehydrating agent (1 mark)
- Names ethene (½ mark) and draws correct electron dot structure (1 mark)
Open Q34(B) on its own page
Answer one of these 2 alternatives.
Q35(A) · 5 marks · Long answerHow do Organisms Reproduce?
(i) Name three techniques/devices used by human females to avoid pregnancy. Mention the side effects caused by each.
(ii) What will happen if in a human female (a) fertilisation takes place, (b) an egg is not fertilised ?
Try it first, then show our solutionSolution
1. (i) Technique 1: Chemical method, i.e. oral contraceptive pills.
  ½ mark
2. Side effect: they change the hormonal balance of the body.
  ½ mark
3. Technique 2: Intrauterine device such as the Copper-T or loop.
  ½ mark
4. Side effect: it can cause irritation of the uterus.
  ½ mark
5. Technique 3: Surgical method, in which the fallopian tubes are blocked (tubectomy).
  ½ mark
6. Side effect: if not done properly, it may cause infections and other problems.
  ½ mark
7. (ii)(a) If fertilisation takes place, the zygote starts dividing and gets implanted in the thick, spongy lining of the uterus. It develops into an embryo and pregnancy begins.
  1 mark
8. (b) If the egg is not fertilised, the thickened uterine lining is not needed. It breaks down and comes out through the vagina as blood and mucus; this is menstruation.
  1 mark
Answer: (i) Oral pills (hormonal imbalance), Copper-T (uterine irritation), surgical blocking of fallopian tubes (infection); (ii)(a) zygote implants in uterus; (b) lining sheds as menstruation
Where the marks are
- Three methods named (½ mark each)
- Side effect of each (½ mark each)
- Fertilisation: zygote implants in uterine lining (1 mark)
- No fertilisation: lining breaks down, menstruation (1 mark)
Open Q35(A) on its own page
OR
Q35(B) · 5 marks · Long answerHow do Organisms Reproduce?
(i) Draw a diagram showing spore formation in Rhizopus and label the (a) reproductive and (b) non-reproductive parts. Why does Rhizopus not multiply on a dry slice of bread ?
(ii) Name and explain the process by which reproduction takes place in Hydra.
Figure in the paper: Students are asked to draw a diagram of spore formation in Rhizopus (sporangiophore with sporangium and related structures) and label reproductive and non-reproductive parts. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Diagram: draw Rhizopus on bread showing thread-like hyphae spread on the surface, with upright sporangiophores bearing round sporangia full of spores at their tips.
  1 mark
2. (a) Reproductive part: sporangium (containing spores).
  ½ mark
3. (b) Non-reproductive part: hyphae.
  ½ mark
4. Rhizopus does not grow on dry bread because its spores need moisture and nutrients to germinate and multiply, and dry bread lacks moisture.
  1 mark
5. (ii) Process: Budding (asexual reproduction).
  1 mark
6. Explanation: in Hydra, regenerative cells divide repeatedly at one specific site to form an outgrowth called a bud. The bud develops into a tiny individual and, when mature, detaches from the parent to live as a new Hydra. (Hydra can also regenerate: each cut piece grows into a whole new organism.)
  1 mark
Answer: (i) Reproductive: sporangium; non-reproductive: hyphae; dry bread lacks the moisture needed for spores to germinate; (ii) budding: a bud forms by repeated cell division, grows and detaches
Where the marks are
- Neat diagram of Rhizopus (1 mark)
- Labels sporangium (½ mark) and hyphae (½ mark)
- Reason: dry bread lacks moisture/nutrients for germination (1 mark)
- Names budding or regeneration (1 mark)
- Correct explanation of the process (1 mark)
Open Q35(B) on its own page
Answer one of these 2 alternatives.
Q36(A) · 5 marks · Long answerElectricity
(i) Define electric power. Express it in terms of potential difference (V) and resistance (R).
(ii) An electric oven is designed to work on the mains voltage of 220 V. This oven consumes 11 units of electrical energy in 5 hours. Calculate :
(a) power rating of the oven
(b) current drawn by the oven
(c) resistance of the oven when it is red hot
Try it first, then show our solutionSolution
1. (i) Electric power is the rate at which electrical energy is consumed or dissipated in a circuit, i.e. $P = \frac{energy}{time}$ .
  1 mark
2. Since $P = V I$ and, by Ohm's law, $I = \frac{V}{R}$ , we get $P = V \cdot \frac{V}{R} = \frac{V ^{2}}{R}$ .
  1 mark
3. (ii)(a) Energy consumed $E = 11$ units $= 11$ kWh and time $t = 5$ h.
  $P = \frac{E}{t} = \frac{11 kWh}{5 h} = 2.2 kW = 2200 W$ .
  1 mark
4. (b) $I = \frac{P}{V} = \frac{2200 W}{220 V} = 10 A$ .
  1 mark
5. (c) $R = \frac{V ^{2}}{P} = \frac{( 220 ) ^{2}}{2200} = \frac{48400}{2200} = 22 Ω$ . (Check: $R = V / I = 220/10 = 22 Ω$ .)
  1 mark
Answer: (i) Rate of consumption of electrical energy; $P = V^{2} / R$ . (ii)(a) 2200 W (2.2 kW); (b) 10 A; (c) 22 Ω
Where the marks are
- Correct definition of electric power (1 mark)
- Expression $P = V^{2} / R$ (1 mark)
- Power = 11 kWh / 5 h = 2.2 kW = 2200 W (1 mark)
- Current $I = P / V = 10$ A (1 mark)
- Resistance $R = V^{2} / P = 22 Ω$ (1 mark)
Open Q36(A) on its own page
OR
Q36(B) · 5 marks · Long answerElectricity
(i) Write the relation between resistance R and electrical resistivity $ρ$ of the material of a conductor in the shape of cylinder of length $l$ and area of cross-section A. Hence derive the SI unit of electrical resistivity.
(ii) The resistance of a metal wire of length 3 m is $60 Ω$ . If the area of cross-section of the wire is $4 \times 1 0^{- 7} m^{2}$ , calculate the electrical resistivity of the wire.
(iii) State how would electrical resistivity be affected if the wire (of part 'ii') is stretched so that its length is doubled. Justify your answer.
Try it first, then show our solutionSolution
1. (i) Resistance of a cylindrical conductor: $R = ρ \frac{l}{A}$ , where $ρ$ is the resistivity of the material.
  1 mark
2. Rearranging, $ρ = \frac{R A}{l}$ . SI unit $= \frac{Ω \times m ^{2}}{m} = Ω m$ (ohm metre).
  1 mark
3. (ii) Given $l = 3$ m, $R = 60 Ω$ , $A = 4 \times 1 0^{- 7} m^{2}$ .
  $ρ = \frac{R A}{l} = \frac{60 \times 4 \times 1 0 ^{- 7}}{3} = 80 \times 1 0^{- 7} = 8 \times 1 0^{- 6} Ω m$ .
  1 mark
4. (iii) The resistivity will not change when the wire is stretched to double its length.
  1 mark
5. Justification: resistivity depends only on the nature of the material (and temperature), not on the length or area of cross-section of the conductor. (Only the resistance changes on stretching.)
  1 mark
Answer: (i) $R = ρl / A$ ; SI unit of ρ = Ω m. (ii) $ρ = 8 \times 1 0^{- 6} Ω$ m. (iii) No change, as resistivity depends only on the material.
Where the marks are
- Writes $R = ρl / A$ (1 mark)
- Derives $ρ = R A / l$ and unit Ω m (1 mark)
- Calculates $ρ = 80 \times 1 0^{- 7} Ω$ m (1 mark)
- States resistivity remains unchanged (1 mark)
- Reason: depends only on nature of material, not dimensions (1 mark)
Open Q36(B) on its own page

Section E

source/case-based with sub-parts; internal choice in sub-part (iii)

Q37 · 4 marks · Case-basedCompetency-basedMetals and Non-metals
The metals produced by various reduction processes are not very pure. They contain impurities, which must be removed to obtain pure metals. The most widely used method for refining impure metals is electrolytic refining.
1. (i)
  What is the cathode and anode made of in the refining of copper by this process ?
  1 mark
2. (ii)
  Name the solution used in the above process and write its formula.
  1 mark
3. (iii)(A)
  How copper gets refined when electric current is passed in the electrolytic cell ?
  2 marks
4. OR
  (iii)(B)
  You have two beakers 'A' and 'B' containing copper sulphate solution. What would you observe after about 2 hours if you dip a strip of zinc in beaker 'A' and a strip of silver in beaker 'B'? Give reason for your observations in each case.
  2 marks
Try it first, then show our solutionSolution
1. (i) Cathode: a thin strip of pure copper; Anode: a thick block of impure copper.
  1 mark
2. (ii) Electrolyte: acidified copper sulphate solution, formula $CuSO_{4}$ .
  1 mark
3. (iii)(A) On passing current, copper from the impure anode dissolves into the electrolyte as $Cu^{2 +}$ ions and an equal amount of pure copper from the electrolyte is deposited on the cathode:
  Anode: $Cu \to Cu^{2 +} + 2 e^{-}$ ; Cathode: $Cu^{2 +} + 2 e^{-} \to Cu$ (pure).
  1 mark
4. The soluble impurities go into the solution, while the insoluble impurities settle at the bottom below the anode as anode mud. Thus pure copper collects on the cathode.
  1 mark
5. (iii)(B) Beaker A (zinc strip): the blue colour of the solution fades and becomes colourless; reddish-brown copper deposits on the zinc strip.
  ½ mark
6. Reason: zinc is more reactive than copper, so it displaces copper: $Zn(s) + CuSO_{4} (aq) \to ZnSO_{4} (aq) + Cu(s)$ .
  ½ mark
7. Beaker B (silver strip): no change is observed; the solution stays blue.
  ½ mark
8. Reason: silver is less reactive than copper, so it cannot displace copper from copper sulphate.
  ½ mark
Answer: (i) Cathode – pure Cu, anode – impure Cu; (ii) acidified copper sulphate, CuSO₄; (iii)(A) anode Cu dissolves, pure Cu deposits on cathode, impurities form anode mud; (iii)(B) A: blue colour fades (Zn more reactive); B: no change (Ag less reactive)
Where the marks are
- Cathode pure copper, anode impure copper (½ + ½ mark)
- Acidified copper sulphate and formula CuSO₄ (½ + ½ mark)
- (iii)(A) Anode copper dissolves and pure copper deposits on cathode, with electrode reactions (1 mark)
- (iii)(A) Soluble impurities in solution, insoluble ones settle as anode mud (1 mark)
- (iii)(B) Observation and reason for beaker A (½ + ½ mark)
- (iii)(B) Observation and reason for beaker B (½ + ½ mark)
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Q38 · 4 marks · Case-basedCompetency-basedHeredity
Mendel worked out the rules of heredity by working on garden pea using a number of visible contrasting characters. He conducted several experiments by making a cross with one or two pairs of contrasting characters of pea plant. On the basis of his observations he gave some interpretations which helped to study the mechanism of inheritance.
1. (i)
  When Mendel crossed pea plants with pure tall and pure short characteristics to produce $F_{1}$ progeny, which two observations were made by him in $F_{1}$ plants?
  1 mark
2. (ii)
  Write one difference between dominant and recessive trait.
  1 mark
3. (iii)(A)
  In a cross with two pairs of contrasting characters
  $R R Y Y \times r r y y$
  (Round Yellow) (Wrinkled Green)
  Mendel observed 4 types of combinations in $F_{2}$ generation. By which method did he obtain $F_{2}$ generation? Write the ratio of the parental combinations obtained and what conclusions were drawn from this experiment.
  2 marks
4. OR
  (iii)(B)
  Justify the statement: "It is possible that a trait is inherited but may not be expressed."
  2 marks
Try it first, then show our solutionSolution
1. (i) Observation 1: all the $F_{1}$ plants were tall; no short plants appeared.
  ½ mark
2. Observation 2: no plants of medium height appeared, so there was no blending; only the dominant parental trait (tallness) was seen.
  ½ mark
3. (ii) A dominant trait is expressed even when only a single copy of its factor is present (e.g. Tt is tall), whereas a recessive trait is expressed only when both copies are recessive (e.g. tt is short).
  1 mark
4. (iii)(A) Method: the $F_{2}$ generation was obtained by self-pollination of the $F_{1}$ plants (RrYy).
  ½ mark
5. $F_{2}$ ratio was Round Yellow : Round Green : Wrinkled Yellow : Wrinkled Green = 9 : 3 : 3 : 1. The parental combinations Round Yellow : Wrinkled Green = 9 : 1.
  ½ mark
6. Conclusion: new combinations (round green, wrinkled yellow) appeared, so the two traits (seed shape and seed colour) are inherited independently of each other.
  1 mark
7. (iii)(B) When pure tall (TT) pea plants are crossed with pure short (tt) plants, all $F_{1}$ plants (Tt) are tall, so shortness is not expressed although the factor 't' is present in them.
  1 mark
8. On self-pollinating $F_{1}$ , short plants (tt) reappear in $F_{2}$ in the ratio tall : short = 3 : 1. This shows the recessive trait was inherited by $F_{1}$ but stayed unexpressed, so a trait may be inherited without being expressed.
  1 mark
Answer: (i) All F₁ tall; no intermediate height. (ii) Dominant expressed with one copy; recessive only in pair. (iii)(A) Self-pollination of F₁; parental ratio 9 : 1; traits inherited independently. (iii)(B) Recessive trait hidden in F₁ reappears in F₂.
Where the marks are
- All F₁ plants tall (½ mark)
- No medium-height / blended plants (½ mark)
- Correct difference between dominant and recessive trait (1 mark)
- Self-pollination of F₁ (½ mark)
- Parental combination ratio 9 : 1 (½ mark)
- Conclusion: independent inheritance of traits (1 mark)
- (iii)(B) Explanation with F₁ showing only dominant trait and F₂ reappearance of recessive trait (2 marks)
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Q39 · 4 marks · Case-basedCompetency-basedLight – Reflection and Refraction
Study the data given below showing the focal length of three concave mirrors A, B and C and the respective distances of objects placed in front of the mirrors :
Case Mirror Focal Length (cm) Object Distance (cm)
1 A 20 45
2 B 15 30
3 C 30 20
1. (i)
  In which one of the above cases the mirror will form a diminished image of the object? Justify your answer.
  1 mark
2. (ii)
  List two properties of the image formed in case 2.
  1 mark
3. (iii)(A)
  What is the nature and size of the image formed by mirror C? Draw ray diagram to justify your answer.
  2 marks
4. OR
  (iii)(B)
  An object is placed at a distance of 18 cm from the pole of a concave mirror of focal length 12 cm. Find the position of the image formed in this case.
  2 marks
Try it first, then show our solutionSolution
1. (i) Case 1 (mirror A) forms a diminished image.
  ½ mark
2. Justification: for A, $f = 20$ cm, so $C = 2 f = 40$ cm; the object at 45 cm lies beyond C, and a concave mirror forms a diminished image between F and C.
  ½ mark
3. (ii) Case 2: $f = 15$ cm, object at 30 cm $= 2 f$ , i.e. at C. The image is formed at C and is real and inverted, and of the same size as the object (any two).
  1 mark
4. (iii)(A) Mirror C: $f = 30$ cm, object at 20 cm, i.e. between P and F. The image is virtual and erect,
  ½ mark
5. and magnified (larger than the object), formed behind the mirror.
  ½ mark
6. Ray diagram (described): draw the principal axis with P, F (30 cm) and C (60 cm); place object AB at 20 cm. Ray 1 from A parallel to the axis reflects passing through F. Ray 2 from A directed towards the pole reflects symmetrically about the axis (or a ray along the line from C through A reflects back on itself). The reflected rays diverge; extend them backward with dotted lines behind the mirror — they meet at A′, giving an enlarged, erect image A′B′ behind the mirror. Mark arrows on all rays.
  1 mark
7. (iii)(B) Using sign convention: $u = - 18$ cm, $f = - 12$ cm. Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ .
  ½ mark
8. $\frac{1}{v} = \frac{1}{- 12} - \frac{1}{- 18} = - \frac{1}{12} + \frac{1}{18} = \frac{- 3 + 2}{36} = - \frac{1}{36}$
  ½ mark
9. $v = - 36$ cm
  ½ mark
10. The image is formed 36 cm in front of the mirror (on the same side as the object), so it is real.
  ½ mark
Answer: (i) Case 1 (mirror A), object beyond C; (ii) real, inverted, same size; (iii)(A) virtual, erect, magnified; (iii)(B) v = −36 cm, 36 cm in front of the mirror
Where the marks are
- Identifies mirror A / case 1 (½ mark)
- Justifies: object beyond C (½ mark)
- Any two properties of image in case 2 (½ + ½ mark)
- Nature virtual and erect (½ mark)
- Size magnified (½ mark)
- Correct ray diagram with arrows (1 mark; ½ deducted if ray directions not marked)
- Mirror formula with correct signs (½ mark)
- Substitution and simplification (½ mark)
- v = −36 cm (½ mark)
- Image position stated: 36 cm in front of mirror (½ mark)
Open Q39 on its own page

Case	Mirror	Focal Length (cm)	Object Distance (cm)
1	A	20	45
2	B	15	30
3	C	30	20

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