ScienceCBSE Class 10 · Board exam 2023 (session 2022-23)

CBSE Class 10 Science: 2023 board paper, set 31/2/1

The question paper set in the CBSE Class 10 board exam of 2023, 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/2023/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, 30-50 words
C	27-33	short answer, 50-80 words
D	34-36	long answer, 80-120 words
E	37-39	source/case-based with sub-parts

Instructions in brief: No overall choice; internal choice (OR) is provided in some questions in Sections B, C, D and E, and only one alternative is to be attempted.

Section A

MCQ 1-16, assertion-reason 17-20

Q1 · 1 mark · MCQChemical Reactions and Equations
Which of the following is an example of endothermic process ?
1. (A)Formation of slaked lime
2. (B)Decomposition of vegetable matter into compost
3. (C)Dissolution of ammonium chloride in water
4. (D)Digestion of food in our body
Try it first, then show our solutionSolution
1. An endothermic process absorbs heat. When ammonium chloride dissolves in water, the solution becomes cold because heat is absorbed from the surroundings.
  1 mark
2. Slaking of lime, composting and digestion all release heat, so they are exothermic.
Answer: (C) Dissolution of ammonium chloride in water
Where the marks are
- Chooses dissolution of NH $_{4}$ Cl as the endothermic process (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQAcids, Bases and Salts
In the given experimental set-up, if the experiment is carried out separately with each of the following solutions the cases in which the bulb will glow is/are :
(i) Dilute hydrochloric acid
(ii) Dilute sulphuric acid
(iii) Glucose solution
(iv) Alcohol
1. (A)(i) only
2. (B)(ii) only
3. (C)(i) and (ii)
4. (D)(ii), (iii) and (iv)
Figure in the paper: A battery (marked + and -) in series with a bulb and a switch, connected by wires to two electrodes dipped into a beaker containing a solution. The bulb is shown glowing. The label 'Solution' points to the liquid in the beaker. See it in the official paper.
Try it first, then show our solutionSolution
1. Dilute HCl and dilute H $_{2}$ SO $_{4}$ produce H $^{+}$ ions in water, so they conduct electricity and the bulb glows.
  1 mark
2. Glucose and alcohol do not ionise in water, so they do not conduct and the bulb does not glow.
Answer: (C) (i) and (ii)
Where the marks are
- Identifies the two acids as conductors (1 mark)
Open Q2 on its own page
Q3 · 1 mark · MCQChemical Reactions and Equations
In order to balance the following chemical equation, the values of the coefficients x and y respectively are :
$x Pb (N O_{3})_{2} Heat 2 PbO + y N O_{2} + O_{2}$
1. (A)2, 4
2. (B)2, 2
3. (C)2, 3
4. (D)4, 2
Try it first, then show our solutionSolution
1. Balancing Pb: $x = 2$ . Balancing N: $2 \times 2 = 4$ N atoms on the left, so $y = 4$ .
  1 mark
2. Checking O: left side has $12$ ; right side has $2 + 8 + 2 = 12$ . The equation $2 Pb (N O_{3})_{2} \to 2 PbO + 4 N O_{2} + O_{2}$ is balanced.
Answer: (A) x = 2, y = 4
Where the marks are
- Obtains x = 2 and y = 4 (1 mark)
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Q4 · 1 mark · MCQMetals and Non-metals
When zinc reacts with sodium hydroxide, the product formed is :
1. (A)Sodium oxide
2. (B)Sodium zincate
3. (C)Zinc hydroxide
4. (D)Zinc oxide
Try it first, then show our solutionSolution
1. $Zn + 2NaOH \to N a_{2} Zn O_{2} + H_{2}$ . The salt formed is sodium zincate.
  1 mark
Answer: (B) Sodium zincate
Where the marks are
- Identifies sodium zincate (1 mark)
Q5 · 1 mark · MCQMetals and Non-metals
Among the following, the metal with lowest density is :
1. (A)Lithium
2. (B)Lead
3. (C)Magnesium
4. (D)Aluminium
Try it first, then show our solutionSolution
1. Lithium has a density of about 0.53 g cm $^{- 3}$ . It is the lightest metal and floats even on water.
  1 mark
Answer: (A) Lithium
Where the marks are
- Chooses lithium (1 mark)
Q6 · 1 mark · MCQMetals and Non-metals
The number of electrons in the outermost shell of the atom of a non-metal can be :
1. (A)1, 2 or 3
2. (B)3, 4 or 5
3. (C)5, 6 or 7
4. (D)5, 6 or 8
Try it first, then show our solutionSolution
1. Non-metals usually gain or share electrons to complete an octet, so their atoms have 5, 6 or 7 valence electrons (e.g. N, O, Cl).
  1 mark
Answer: (C) 5, 6 or 7
Where the marks are
- Selects 5, 6 or 7 (1 mark)
Q7 · 1 mark · MCQCarbon and its Compounds
The total number of electrons shared in the formation of an ethyne molecule is :
1. (A)6
2. (B)3
3. (C)10
4. (D)4
Try it first, then show our solutionSolution
1. Ethyne is H–C≡C–H: two C–H bonds and one triple bond make 2 + 3 = 5 shared pairs.
2. 5 pairs × 2 = 10 electrons are shared.
  1 mark
Answer: (C) 10
Where the marks are
- Counts 5 bond pairs, i.e. 10 electrons (1 mark)
Q8 · 1 mark · MCQLife Processes
Sphincter muscles are present at the exit of :
1. (A)Stomach and small intestine
2. (B)Stomach and anus
3. (C)Small intestine and large intestine
4. (D)Oesophagus and stomach
Try it first, then show our solutionSolution
1. A sphincter muscle at the exit of the stomach regulates the passage of food into the small intestine. The anal sphincter controls the removal of faeces.
  1 mark
Answer: (B) Stomach and anus
Where the marks are
- Identifies stomach and anus (1 mark)
Q9 · 1 mark · MCQLife Processes
In the following diagram, identify the cells through which massive amounts of gaseous exchange takes place for photosynthesis :
1. (A)I
2. (B)IV
3. (C)III
4. (D)II
Figure in the paper: A cross-section of a leaf showing layers of cells. Four cell regions are labelled: I points to the top layer of flattened cells (upper epidermis); II points to the tall, closely packed cells below it (palisade mesophyll); III points to the loosely arranged, irregular cells with intercellular spaces (spongy mesophyll); IV points to the bottom layer of cells (lower epidermis). See it in the official paper.
Try it first, then show our solutionSolution
1. Massive gas exchange for photosynthesis occurs through stomata, which are mostly present in the lower epidermis (IV).
  1 mark
Answer: (B) IV
Where the marks are
- Identifies the lower epidermis with stomata, IV (1 mark)
Q10 · 1 mark · MCQLife Processes
During vigorous exercise, the occurrence of cramps in the outer muscles of an athlete is due to the conversion of pyruvate to :
1. (A)Glucose
2. (B)Ethanol
3. (C)Lactic acid
4. (D)Lactose
Try it first, then show our solutionSolution
1. When oxygen is short, muscle cells convert pyruvate to lactic acid. Lactic acid builds up and causes cramps.
  1 mark
Answer: (C) Lactic acid
Where the marks are
- Lactic acid (1 mark)
Q11 · 1 mark · MCQHow do Organisms Reproduce?
Plants which bear unisexual flowers are :
1. (A)Mustard and Papaya
2. (B)Hibiscus and Watermelon
3. (C)Mustard and Hibiscus
4. (D)Watermelon and Papaya
Try it first, then show our solutionSolution
1. Papaya and watermelon flowers contain either stamens or a pistil, so they are unisexual. Hibiscus and mustard flowers are bisexual.
  1 mark
Answer: (D) Watermelon and Papaya
Where the marks are
- Selects watermelon and papaya (1 mark)
Q12 · 1 mark · MCQHeredity
When an 'X' bearing sperm fertilises the egg, the resulting zygote has the following combination of chromosomes :
1. (A)44 + XX
2. (B)44 + XY
3. (C)22 + XX
4. (D)22 + XY
Try it first, then show our solutionSolution
1. The egg (22 + X) fuses with an X-bearing sperm (22 + X). The zygote is 44 + XX, a girl.
  1 mark
Answer: (A) 44 + XX
Where the marks are
- 44 + XX (1 mark)
Q13 · 1 mark · MCQLight – Reflection and Refraction
An object is placed in front of a convex mirror at infinity. According to the New Cartesian Sign Convention, the sign of the focal length and the sign of the image distance in this case are respectively :
1. (A)+ , −
2. (B)− , +
3. (C)− , −
4. (D)+ , +
Try it first, then show our solutionSolution
1. For a convex mirror, the focus lies behind the mirror, so $f$ is positive.
2. For an object at infinity, the virtual image forms at F behind the mirror, so $v$ is also positive.
  1 mark
Answer: (D) +, +
Where the marks are
- Both signs positive (1 mark)
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Q14 · 1 mark · MCQElectricity
For verifying Ohm's law, we design an electric circuit diagram in which we show the arrangement of different circuit components. We find that with respect to the resistor, the :
1. (A)ammeter is connected in parallel and the voltmeter in series.
2. (B)ammeter is connected in series and the voltmeter in parallel.
3. (C)ammeter and voltmeter are both connected in series.
4. (D)ammeter and voltmeter are both connected in parallel.
Try it first, then show our solutionSolution
1. The ammeter is connected in series so that the same current flows through it as through the resistor. The voltmeter is connected in parallel so that it measures the potential difference across the resistor.
  1 mark
Answer: (B)
Where the marks are
- Ammeter in series, voltmeter in parallel (1 mark)
Q15 · 1 mark · MCQElectricity
In a resistive circuit if the current is increased to two times, the percentage change in the amount of heat dissipated in the circuit would be :
1. (A)400%
2. (B)300%
3. (C)200%
4. (D)100%
Try it first, then show our solutionSolution
1. $H = I^{2} R t$ . If the current doubles, $H^{'} = 4 H$ .
2. Percentage change $= \frac{4 H - H}{H} \times 100 = 300%$ .
  1 mark
Answer: (B) 300%
Where the marks are
- Uses H ∝ I² to get 300% (1 mark)
Q16 · 1 mark · MCQMagnetic Effects of Electric Current
A positron enters a uniform magnetic field at right angles to it as shown. The direction of force experienced by the positron will be :
1. (A)to the right
2. (B)to the left
3. (C)into the page
4. (D)out of the page
Figure in the paper: A vertical arrow labelled 'Positron' pointing upward, and four horizontal parallel arrows pointing to the right, labelled 'Magnetic field'. The positron moves upward across the horizontal field lines at right angles to them. See it in the official paper.
Try it first, then show our solutionSolution
1. A positron carries positive charge, so the current is in its direction of motion (upward). The field points to the right.
2. By Fleming's left-hand rule (field to the right, current upward), the thumb points into the page.
  1 mark
Answer: (C) into the page
Where the marks are
- Applies Fleming's left-hand rule to get into the page (1 mark)
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Q17 · 1 mark · Assertion-reasonCompetency-basedAcids, Bases and Salts
Assertion (A) : Sodium oxide is an amphoteric oxide.
Reason (R) : Metal oxides which react with both acids as well as bases are known as amphoteric oxides.
Codes: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (c) Assertion (A) is true, but Reason (R) is false. (d) Assertion (A) is false, but Reason (R) is true.
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. Na $_{2}$ O is a basic oxide; it reacts only with acids. So A is false.
2. R correctly defines amphoteric oxides (e.g. Al $_{2}$ O $_{3}$ , ZnO). So R is true.
  1 mark
Answer: (D)
Where the marks are
- A false, R true (1 mark)
Q18 · 1 mark · Assertion-reasonCompetency-basedHow do Organisms Reproduce?
Assertion (A) : When a bacterium divides into two, and the resultant two bacteria divide again, the four bacteria produced would be almost similar.
Reason (R) : DNA copying involves small inaccuracies in the reproduction process.
Codes: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (c) Assertion (A) is true, but Reason (R) is false. (d) Assertion (A) is false, but Reason (R) is true.
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. Asexual division copies DNA almost exactly. Small copying inaccuracies cause only minor variations, so the four bacteria are almost (not exactly) similar.
  1 mark
2. Both statements are true, and R explains why they are only 'almost' similar.
Answer: (A)
Where the marks are
- Both true and R explains A (1 mark)
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Q19 · 1 mark · Assertion-reasonCompetency-basedThe Human Eye and the Colourful World
Assertion (A) : A person suffering from myopia cannot see the distant objects clearly.
Reason (R) : A converging lens is used for the correction of myopic eye as it can form real as well as virtual images of the objects placed in front of it.
Codes: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (c) Assertion (A) is true, but Reason (R) is false. (d) Assertion (A) is false, but Reason (R) is true.
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 myopic eye cannot see distant objects clearly, so A is true.
2. Myopia is corrected by a diverging (concave) lens, not a converging lens, so R is false.
  1 mark
Answer: (C)
Where the marks are
- A true, R false (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedMagnetic Effects of Electric Current
Assertion (A) : Magnetic field lines do not intersect each other.
Reason (R) : Magnetic field lines are imaginary lines, the tangent to which at any point gives the direction of the field at that point.
Codes: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (c) Assertion (A) is true, but Reason (R) is false. (d) Assertion (A) is false, but Reason (R) is true.
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. Both statements are true. Field lines do not intersect because the field at a point cannot have two directions.
  1 mark
2. R only defines field lines and does not directly state this reason, so per the scheme R is not the correct explanation.
Answer: (B)
Where the marks are
- Both true, R not the correct explanation (1 mark)
Open Q20 on its own page

Section B

very short answer, 30-50 words

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerChemical Reactions and Equations
What is observed when aqueous solutions of potassium iodide and lead nitrate are mixed together ? Name the type of reaction and write the chemical equation for the reaction that occurs.
Try it first, then show our solutionSolution
1. Observation: A yellow precipitate forms as soon as the two colourless solutions are mixed. This precipitate is lead iodide, $PbI_{2}$ .
  ½ mark
2. Type of reaction: This is a double displacement reaction, and because an insoluble solid forms it is also a precipitation reaction. The two compounds exchange their ions.
  ½ mark
3. Balanced equation: $Pb(NO_{3})_{2} (a q) + 2 KI (a q) \to PbI_{2} (s) ↓ + 2 KNO_{3} (a q)$
  1 mark
Answer: Yellow precipitate of PbI₂; double displacement (precipitation) reaction; Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃
Where the marks are
- States that a yellow precipitate of lead iodide forms (½ mark)
- Names the reaction as double displacement or precipitation (½ mark)
- Writes the correct balanced equation (1 mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerChemical Reactions and Equations
When copper powder is heated in a watch glass, a black substance is formed.
1. (i)
  Why is this black substance formed ? Name it.
  1 mark
2. (ii)
  How can this black substance be reversed to its original form ?
  1 mark
Try it first, then show our solutionSolution
1. (i) When copper is heated in air, it combines with oxygen, so copper is oxidised: $2 Cu (s) + O_{2} (g) Δ 2 CuO (s)$
  ½ mark
2. (i) The black substance is copper(II) oxide, $CuO$ .
  ½ mark
3. (ii) Pass hydrogen gas over the heated black copper oxide. It is reduced back to brown copper: $CuO (s) + H_{2} (g) Δ Cu (s) + H_{2} O (l)$
  1 mark
Answer: (i) Copper is oxidised by oxygen of air; black substance is copper oxide (CuO). (ii) By passing hydrogen gas over heated CuO: CuO + H₂ → Cu + H₂O
Where the marks are
- Reason: copper combines with oxygen, i.e. it is oxidised (½ mark)
- Names the substance as copper oxide, CuO (½ mark)
- Reverses it by passing hydrogen over heated CuO, with or without the equation (1 mark)
Open Q21(b) on its own page
Q22 · 2 marks · Very short answerLife Processes
(a) Why is it important to prevent oxygenated and deoxygenated blood from mixing in birds and mammals ?
(b) Which animals can tolerate some mixing of the oxygenated and deoxygenated blood streams ? On what factor does the body temperature of these animals depend ?
Try it first, then show our solutionSolution
1. (a) Birds and mammals are warm-blooded and use a lot of energy to keep their body temperature constant. Keeping oxygenated and deoxygenated blood separate gives a highly efficient supply of oxygen to the body, which meets this high energy demand.
  1 mark
2. (b) Amphibians (e.g. frogs) and many reptiles can tolerate some mixing of the two blood streams.
  ½ mark
3. (b) Their body temperature depends on the temperature of the surrounding environment, since they are cold-blooded.
  ½ mark
Answer: (a) Separation gives an efficient oxygen supply for the high energy needed to maintain body temperature. (b) Amphibians/reptiles; body temperature depends on environmental temperature.
Where the marks are
- Links separation of blood to an efficient oxygen supply for the high energy need of maintaining body temperature (1 mark)
- Names amphibians or reptiles (½ mark)
- States that their body temperature depends on the temperature of the environment (½ mark)
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Q23 · 2 marks · Very short answerHeredity
Mendel crossed a pure tall pea plant (TT) with a pure short pea plant (tt) and obtained all tall plants in $F_{1}$ generation.
1. (a)
  What is the gene combination present in the plants of $F_{1}$ generation ?
  1 mark
2. (b)
  Give reason why only tall plants are observed in $F_{1}$ progeny.
  1 mark
3. (c)
  What will be the ratio of the plants obtained in the $F_{2}$ generation when $F_{1}$ plants are self-pollinated ?
  1 mark
Try it first, then show our solutionSolution
1. (a) The cross is TT × tt. The tall parent gives gamete T and the short parent gives gamete t, so every $F_{1}$ plant is Tt (heterozygous).
  ½ mark
2. (b) The trait for tallness (T) is dominant over the trait for shortness (t). In Tt plants only the dominant trait is expressed, and the recessive trait stays hidden, so all $F_{1}$ plants are tall.
  1 mark
3. (c) Selfing Tt × Tt gives TT : Tt : tt = 1 : 2 : 1. So in $F_{2}$ , Tall : Short = 3 : 1.
  ½ mark
Answer: (a) Tt; (b) tallness is dominant over shortness, so only it is expressed; (c) Tall : Short = 3 : 1
Where the marks are
- Gene combination Tt (½ mark)
- Reason: tall trait is dominant, so only it is expressed in F₁ (1 mark)
- F₂ ratio Tall : Short = 3 : 1 (½ mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerThe Human Eye and the Colourful World
When and where does a rainbow appear in the sky ? Draw a labelled ray diagram to show its formation.
Try it first, then show our solutionSolution
1. When: A rainbow appears just after a rain shower, when tiny water droplets are still suspended in the air and the Sun is shining.
  ½ mark
2. Where: It always appears in the part of the sky opposite to the Sun, so the observer must stand with the Sun behind them.
  ½ mark
3. Labelled ray diagram (described):
  Draw a spherical raindrop with a ray of white sunlight entering it.
  The light refracts and disperses at the first surface.
  It undergoes internal reflection at the back of the drop.
  It refracts again as it leaves the drop, spreading into a spectrum.
  Label the incoming ray 'Sunlight', and label the emerging rays 'Red' and 'Violet'.
  Mark the steps 'Refraction and dispersion', 'Internal reflection' and 'Refraction'.
  Show arrows on all rays.
  1 mark
Answer: After a rain shower, in the direction opposite to the Sun; formed by refraction and dispersion, internal reflection and refraction in raindrops.
Where the marks are
- When: after a rain shower (½ mark)
- Where: opposite to the Sun (½ mark)
- Correct ray diagram through a raindrop showing refraction, dispersion, internal reflection and labelled red and violet rays (1 mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerThe Human Eye and the Colourful World
What is scattering of light ? Why does the clear sky appear blue ?
Try it first, then show our solutionSolution
1. Scattering of light: When light strikes very small particles, such as molecules of air or fine dust and colloidal particles, its direction of travel changes and it spreads out in different directions. This change of direction is called scattering of light.
  1 mark
2. Blue sky: Air molecules and other fine particles in the atmosphere are smaller than the wavelength of visible light. They scatter light of shorter wavelength (blue) much more strongly than light of longer wavelength (red). This scattered blue light enters our eyes from all directions, so the clear sky looks blue.
  1 mark
Answer: Scattering is the change in direction of light by small particles; the sky looks blue because fine air particles scatter blue (short wavelength) light more than red.
Where the marks are
- Correct definition of scattering of light (1 mark)
- Explains that fine particles scatter shorter-wavelength blue light more than red (1 mark)
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Q25 · 2 marks · Very short answerCompetency-basedElectricity
V – I graph for two conducting wires A and B are as shown. If both wires are of the same length and same diameter, which of the two is made of a material of high resistivity ? Give reasons to justify your answer.
Figure in the paper: V–I graph with V on the vertical axis and I on the horizontal axis. Two straight lines start from the origin: line B is steeper, and line A has a smaller slope (lies lower, closer to the I-axis). Both are labelled. See it in the official paper.
Try it first, then show our solutionSolution
1. With V on the y-axis and I on the x-axis, the slope of each line gives $\frac{V}{I} = R$ . Line B is steeper, so $R_{B} > R_{A}$ . Wire B is made of the material of higher resistivity.
  1 mark
2. Resistance is $R = ρ \frac{l}{A}$ . Both wires have the same length $l$ and the same diameter, so the same area $A$ . Hence $R \propto ρ$ . The wire with greater resistance (B) must have the greater resistivity.
  1 mark
Answer: Wire B
Where the marks are
- Identifies wire B (greater slope means greater resistance) (1 mark)
- Justifies using R = ρl/A: for the same dimensions, greater resistance means greater resistivity (1 mark)
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Q26 · 2 marks · Very short answerCompetency-basedOur Environment
We do not clean ponds or lakes but an aquarium needs to be cleaned regularly. Why ?
Try it first, then show our solutionSolution
1. An aquarium is an artificial, man-made ecosystem. It does not have enough natural decomposers (bacteria and fungi) to break down dead remains and wastes. These wastes build up, so the aquarium must be cleaned regularly.
  1 mark
2. Ponds and lakes are natural ecosystems. Decomposers in them break down dead organisms and wastes into simple substances that are recycled. These natural cleansing agents keep the water clean, so ponds and lakes need no cleaning.
  1 mark
Answer: An aquarium is an artificial ecosystem without natural decomposers; ponds and lakes are natural ecosystems where decomposers clean them.
Where the marks are
- Aquarium is man-made and lacks decomposers or natural cleansing agents, so it needs cleaning (1 mark)
- Ponds and lakes are natural ecosystems with decomposers that clean them (1 mark)
Open Q26 on its own page

Section C

short answer, 50-80 words

Q27 · 3 marks · Short answerChemical Reactions and Equations
Silver chloride kept in a china dish turns grey in sunlight.
1. (a)
  Write the colour of silver chloride when it was kept in the china dish.
  1 mark
2. (b)
  Name the type of chemical reaction taking place and write the chemical equation for the reaction.
  1 mark
3. (c)
  State one use of the reaction. Name one more chemical which can be used for the same purpose.
  1 mark
Try it first, then show our solutionSolution
1. (a) Silver chloride is white in colour.
  ½ mark
2. (b) This is a decomposition reaction brought about by light, called photolytic decomposition.
  ½ mark
3. (b) $2 AgCl (s) Sunlight 2 Ag (s) + Cl_{2} (g)$ . The grey colour is due to silver metal.
  1 mark
4. (c) Use: This reaction is used in black-and-white photography.
  ½ mark
5. (c) Another chemical: Silver bromide, $AgBr$ , can also be used for the same purpose.
  ½ mark
Answer: (a) White; (b) photolytic decomposition, 2AgCl → 2Ag + Cl₂; (c) black-and-white photography; silver bromide (AgBr)
Where the marks are
- Colour white (½ mark)
- Decomposition (photolytic) reaction (½ mark)
- Balanced equation 2AgCl → 2Ag + Cl₂ (1 mark)
- Use in black-and-white photography (½ mark)
- Names silver bromide (½ mark)
Open Q27 on its own page
Q28 · 3 marks · Short answerMetals and Non-metals
With the help of suitable chemical equations, list the two main differences between roasting and calcination. How is metal reduced from the product obtained after roasting/calcination of the ore ? Write the chemical equation for the reaction involved.
Try it first, then show our solutionSolution
1. Difference 1: Roasting is done for sulphide ores, while calcination is done for carbonate ores.
  ½ mark
2. Difference 2: In roasting, the ore is heated strongly in excess of air. In calcination, the ore is heated strongly in the absence or limited supply of air.
  ½ mark
3. Equations:
  Roasting: $2 ZnS (s) + 3 O_{2} (g) Heat 2 ZnO (s) + 2 SO_{2} (g)$
  Calcination: $ZnCO_{3} (s) Heat ZnO (s) + CO_{2} (g)$
  1 mark
4. Reduction: The metal oxide obtained is reduced to the metal by heating it with a reducing agent such as carbon (coke).
  ½ mark
5. $ZnO (s) + C (s) Heat Zn (s) + CO (g)$
  ½ mark
Answer: Roasting: sulphide ores, excess air (2ZnS + 3O₂ → 2ZnO + 2SO₂). Calcination: carbonate ores, limited/no air (ZnCO₃ → ZnO + CO₂). The oxide is reduced by carbon: ZnO + C → Zn + CO.
Where the marks are
- Difference based on type of ore (½ mark)
- Difference based on supply of air (½ mark)
- Correct roasting and calcination equations (½ + ½ mark)
- Reduction of the oxide using carbon (½ mark)
- Correct reduction equation (½ mark)
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Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerLife Processes
(i) What is the first step in the breakdown of glucose during aerobic and anaerobic respiration ? Where does it take place ?
(ii) ATP is called the energy currency of the cell. Why ?
(iii) What is meant by "residual volume of air" in a breathing cycle ?
Try it first, then show our solutionSolution
1. (i) The first step in both kinds of respiration is the breakdown of glucose, a six-carbon molecule, into pyruvate, a three-carbon molecule: $Glucose \to Pyruvate + Energy$
  ½ mark
2. (i) This step takes place in the cytoplasm of the cell.
  ½ mark
3. (ii) The energy released during respiration is stored in ATP. When ATP is broken down to ADP, a fixed amount of energy is released. This energy drives the energy-requiring (endothermic) reactions and all other activities of the cell. Because ATP is used like money for energy everywhere in the cell, it is called the energy currency.
  1 mark
4. (iii) Residual volume is the volume of air that always stays in the lungs even after forceful exhalation. Because of it, there is enough time for oxygen to be absorbed into the blood and carbon dioxide to be released.
  1 mark
Answer: (i) Glucose → pyruvate, in the cytoplasm; (ii) ATP releases a fixed amount of energy that drives all cell activities; (iii) air left in the lungs after exhalation, allowing continuous gas exchange
Where the marks are
- Glucose broken into pyruvate (½ mark)
- Takes place in the cytoplasm (½ mark)
- ATP acts as fuel for all cell activities, releasing a fixed amount of energy on breakdown (1 mark)
- Residual volume is the air remaining in the lungs after exhalation, allowing time for gas exchange (1 mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerLife Processes
Write in sequence the steps for experimental verification of the fact that "sunlight is essential for photosynthesis".
Try it first, then show our solutionSolution
1. 1. Keep a healthy potted plant in the dark for about 24 hours so that its leaves are destarched.
  ½ mark
2. 2. Cover part of one leaf on both sides with a strip of black paper, so that this part gets no sunlight.
  ½ mark
3. 3. Place the plant in bright sunlight for several hours (about a day).
  ½ mark
4. 4. Pluck the leaf and remove the black paper. Boil the leaf in water and then in alcohol over a water bath to remove chlorophyll. Wash it and dip it in dilute iodine solution.
  ½ mark
5. 5. Observation: the covered part does not turn blue-black, showing that no starch was formed there in the absence of sunlight.
  ½ mark
6. 6. The uncovered part turns blue-black, showing that starch was made there. Conclusion: sunlight is essential for photosynthesis.
  ½ mark
Answer: Destarch the plant → cover part of a leaf with black paper → keep in sunlight → boil leaf in alcohol and test with iodine → covered part shows no starch, uncovered part turns blue-black; so sunlight is essential.
Where the marks are
- Destarching the plant in the dark (½ mark)
- Covering part of the leaf with black paper (½ mark)
- Exposing the plant to sunlight (½ mark)
- Removing chlorophyll with alcohol and testing with iodine (½ mark)
- Covered part shows no blue-black colour (½ mark)
- Uncovered part turns blue-black, so sunlight is essential (½ mark)
Open Q29(b) on its own page
Q30 · 3 marks · Short answerControl and Coordination
Name the hormone released and the gland which secretes it in human beings during scary situations. How does the body respond to enable it to deal with the situation ?
Try it first, then show our solutionSolution
1. The hormone released in scary situations is adrenaline.
  ½ mark
2. Adrenaline is secreted by the adrenal glands, which lie on top of the kidneys.
  ½ mark
3. Response 1: The heart beats faster, so more oxygen is supplied to the muscles. The breathing rate also increases because of contraction of the diaphragm and rib muscles.
  1 mark
4. Response 2: Blood supply to the digestive system and skin is reduced, and blood is diverted to the skeletal muscles. This prepares the body to fight or run away.
  1 mark
Answer: Adrenaline, from the adrenal glands; faster heartbeat and breathing, and blood diverted from the digestive system and skin to the skeletal muscles
Where the marks are
- Names adrenaline (½ mark)
- Names the adrenal gland (½ mark)
- Any two responses, e.g. faster heartbeat, increased breathing rate, blood diverted to skeletal muscles (1 + 1 marks)
Open Q30 on its own page
Answer one of these 2 alternatives.
Q31(a) · 3 marks · Short answerLight – Reflection and Refraction
A student has focussed the image of an object of height 3 cm on a white screen using a concave mirror of focal length 12 cm. If the distance of the object from the mirror is 18 cm, find the values of the following :
1. (i)
  Distance of the image from the mirror
  1 mark
2. (ii)
  Height of the image
  1 mark
Try it first, then show our solutionSolution
1. Given (New Cartesian sign convention): $h = + 3$ cm, $f = - 12$ cm (concave mirror), $u = - 18$ cm.
  ½ mark
2. (i) Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ , so $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$
  ½ mark
3. $\frac{1}{v} = \frac{1}{- 12} - \frac{1}{- 18} = \frac{- 3 + 2}{36} = - \frac{1}{36}$
  ½ mark
4. $v = - 36$ cm. The image is 36 cm in front of the mirror, which is where the screen is.
  ½ mark
5. (ii) Magnification: $m = \frac{h ^{'}}{h} = - \frac{v}{u}$ , so $h^{'} = - \frac{v}{u} \times h$
  ½ mark
6. $h^{'} = - \frac{( - 36 )}{( - 18 )} \times 3 = - 2 \times 3 = - 6$ cm. The image is 6 cm tall, real and inverted.
  ½ mark
Answer: (i) v = −36 cm (36 cm in front of the mirror); (ii) h′ = −6 cm (6 cm, inverted)
Where the marks are
- Correct data with signs (½ mark)
- Mirror formula (½ mark)
- Correct substitution (½ mark)
- v = −36 cm (½ mark)
- Magnification formula (½ mark)
- h′ = −6 cm (½ mark)
Open Q31(a) on its own page
OR
Q31(b) · 3 marks · Short answerLight – Reflection and Refraction
Define power of a lens. The focal length of a lens is −10 cm. Write the nature of the lens and find its power. If an object is placed at a distance of 20 cm from the optical centre of this lens, according to the New Cartesian Sign Convention, what will be the sign of magnification in this case ?
Try it first, then show our solutionSolution
1. Power of a lens: It is the degree of convergence or divergence of light rays achieved by a lens. It equals the reciprocal of the focal length in metres, $P = \frac{1}{f ( m )}$ . Its SI unit is the dioptre (D).
  1 mark
2. The focal length is negative, so the lens is a concave (diverging) lens.
  ½ mark
3. $P = \frac{1}{f ( m )} = \frac{100}{f ( cm )}$
  ½ mark
4. $P = \frac{100}{- 10} = - 10$ D
  ½ mark
5. A concave lens always forms a virtual, erect, diminished image on the same side as the object. Hence the magnification is positive (+).
  ½ mark
Answer: Concave (diverging) lens; P = −10 D; magnification is positive
Where the marks are
- Correct definition of power (1 mark)
- Identifies concave/diverging lens (½ mark)
- Formula P = 1/f(m) = 100/f(cm) (½ mark)
- P = −10 D (½ mark)
- Sign of magnification positive (½ mark)
Open Q31(b) on its own page
Q32 · 3 marks · Short answerElectricity
Explain the meaning of overloading of an electrical circuit. List two possible causes due to which overloading may occur in household circuits. Write one preventive measure that should be taken to avoid overloading of domestic circuits.
Try it first, then show our solutionSolution
1. Overloading: It is the condition in which an excessively large current, more than the rated value, flows through the wires of a circuit. This heats the wires and may cause a fire.
  1 mark
2. Cause 1: Connecting too many appliances to a single socket at the same time.
  ½ mark
3. Cause 2: Accidental hike in the supply voltage, or the live and neutral wires coming into direct contact (short-circuiting).
  ½ mark
4. Preventive measure: Use a fuse or MCB of proper rating in the circuit, which melts or trips when the current is too large. Also avoid connecting too many appliances to one socket.
  1 mark
Answer: Overloading is an excessive current flowing through the circuit; causes: too many appliances in one socket, supply voltage hike or short circuit; prevention: use a fuse/MCB of proper rating
Where the marks are
- Meaning of overloading: excessive current through the wires (1 mark)
- Any two causes (½ + ½ mark)
- Any one preventive measure, e.g. fuse or not overloading sockets (1 mark)
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Q33 · 3 marks · Short answerOur Environment
Explain how some harmful chemicals enter our bodies through the food chain. Why is the concentration of these harmful chemicals found to be maximum in human beings ?
Try it first, then show our solutionSolution
1. Farmers use chemicals such as pesticides and fertilisers to protect and grow crops. Rain washes these chemicals into the soil and water bodies. Plants absorb them along with water and minerals, and aquatic animals take them in from the water.
  1 mark
2. These chemicals are non-biodegradable, so they are not broken down in organisms. When animals eat these plants, and we eat the plants or animals (for example grains, vegetables, fish or milk), the chemicals pass from one trophic level to the next. In this way they enter our bodies.
  1 mark
3. Human beings occupy the topmost trophic level in most food chains. The chemicals keep accumulating at each successive level (biological magnification). Hence their concentration is maximum in human beings.
  1 mark
Answer: Pesticides reach soil and water, are absorbed by plants and animals, and pass up the food chain into humans; humans are at the top level, so biomagnification makes the concentration maximum in them.
Where the marks are
- Explains how pesticides enter soil and water and are absorbed by plants and animals, then reach us through food (2 marks)
- Humans occupy the top trophic level, so the chemicals accumulate most in them (biomagnification) (1 mark)
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Section D

long answer, 80-120 words

Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerCarbon and its Compounds
An acid 'X' and an alcohol 'Y' react with each other in the presence of an acid catalyst to form a sweet smelling substance 'Z'. Identify 'X', 'Y' and 'Z'. Write the chemical equation for the reaction involved and name it. The substance 'Z' on treatment with sodium hydroxide produces back the alcohol 'Y' and sodium ethanoate. Write the chemical equation for the reaction involved and name it, giving justification for the name.
Try it first, then show our solutionSolution
1. X is ethanoic acid (acetic acid), $CH_{3} COOH$ .
  ½ mark
2. Y is ethanol, $C_{2} H_{5} OH$ .
  ½ mark
3. Z is ethyl ethanoate, $CH_{3} COOC_{2} H_{5}$ , a sweet-smelling ester.
  ½ mark
4. $CH_{3} COOH + C_{2} H_{5} OH Acid CH_{3} COOC_{2} H_{5} + H_{2} O$
  1 mark
5. This reaction is called esterification.
  ½ mark
6. $CH_{3} COOC_{2} H_{5} + NaOH \to C_{2} H_{5} OH + CH_{3} COONa$
  1 mark
7. This reaction is called saponification.
  ½ mark
8. Justification: hydrolysis of an ester by an alkali gives the alcohol and the sodium salt of the acid. The same reaction is used to make soap from fats and oils, hence the name saponification.
  ½ mark
Answer: X = CH₃COOH (ethanoic acid), Y = C₂H₅OH (ethanol), Z = CH₃COOC₂H₅ (ethyl ethanoate); esterification; saponification (used in making soap)
Where the marks are
- Identifies X, Y and Z (½ mark each)
- Equation for esterification (1 mark)
- Names esterification (½ mark)
- Equation of ester with NaOH (1 mark)
- Names saponification (½ mark)
- Justification: the reaction is used in preparing soap (½ mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerCarbon and its Compounds
(i) Name the simplest saturated hydrocarbon. Draw its electron dot structure. Which type of bonds exist in this compound ?
(ii) Name any two mixtures of the carbon compound used as a fuel in daily life, of which the above mentioned compound is an important component.
(iii) In which homologous series of carbon compounds can this compound be placed ? Write the general formula of the series.
(iv) Which type of flame is produced on burning it ?
Try it first, then show our solutionSolution
1. (i) The simplest saturated hydrocarbon is methane, $CH_{4}$ .
  ½ mark
2. (i) Electron dot structure (described):
  Write C at the centre with four H atoms placed above, below, left and right of it.
  Between C and each H, draw one shared pair of electrons: one electron from carbon (×) and one from hydrogen (•).
  This gives four shared pairs in total.
  Carbon then has an octet and each hydrogen has a duplet.
  1 mark
3. (i) Methane contains four single covalent bonds (C–H).
  ½ mark
4. (ii) Two fuels containing methane as a major component are biogas and CNG (compressed natural gas).
  1 mark
5. (iii) Methane belongs to the alkane homologous series.
  ½ mark
6. (iii) The general formula of the series is $C_{n} H_{2 n + 2}$ .
  ½ mark
7. (iv) Being saturated, methane burns completely with a clean blue flame that gives no soot.
  1 mark
Answer: (i) Methane, CH₄; four single covalent bonds; (ii) biogas and CNG; (iii) alkanes, CₙH₂ₙ₊₂; (iv) clean blue flame
Where the marks are
- Names methane (½ mark)
- Correct electron dot structure (1 mark)
- Single covalent bonds (½ mark)
- Biogas and CNG (½ + ½ mark)
- Alkanes (½ mark)
- General formula CₙH₂ₙ₊₂ (½ mark)
- Blue/clean flame (1 mark)
Open Q34(b) on its own page
Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerControl and Coordination
(i) List three points of difference between nervous and hormonal mechanisms for control and coordination in animals.
(ii) How are auxins related with the bending of plant shoot towards unidirectional light ? Explain.
Try it first, then show our solutionSolution
1. (i) Difference 1 (nature of message): In nervous control, information travels as electrical impulses. In hormonal control, it is carried as chemical messengers (hormones).
  1 mark
2. (i) Difference 2 (route): Nerve impulses travel along neurons. Hormones are carried by the blood to their target organs.
  1 mark
3. (i) Difference 3 (speed and duration): Nervous responses are rapid and short-lived. Hormonal responses are slow but long-lasting.
  1 mark
4. (ii) When a growing shoot receives light from one side, auxin, a growth hormone made at the shoot tip, moves towards the shaded side of the shoot.
  1 mark
5. (ii) The higher auxin concentration makes the cells on the shaded side elongate faster than those on the lit side. So the shoot bends towards the light (positive phototropism).
  1 mark
Answer: (i) Electrical impulse vs chemical hormone; via neurons vs via blood; rapid and short-lived vs slow and long-lasting. (ii) Auxin moves to the shaded side, causing faster growth there, so the shoot bends towards light.
Where the marks are
- Any three correct differences between nervous and hormonal control (1 mark each, 3 marks)
- Auxin shifts towards the shaded side under unidirectional light (1 mark)
- More growth on the shaded side bends the shoot towards light (1 mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerControl and Coordination
(i) Name the disorder which a person is likely to suffer from due to the following : (I) Over-secretion of growth hormone (II) Deficiency of oestrogen in females (III) Less secretion of thyroxine. Also name the gland that secretes each of the hormones mentioned above.
(ii) How is the timing and amount of hormone released regulated ? Explain with the help of an example.
Try it first, then show our solutionSolution
1. (i)(I) Over-secretion of growth hormone causes gigantism. The hormone is secreted by the pituitary gland.
  1 mark
2. (i)(II) Deficiency of oestrogen in females causes delayed puberty (secondary sexual characters do not develop in time). Oestrogen is secreted by the ovaries.
  1 mark
3. (i)(III) Less secretion of thyroxine causes goitre (swelling in the neck). Thyroxine is secreted by the thyroid gland.
  1 mark
4. (ii) The timing and amount of hormone released are controlled by a feedback mechanism. The level of the substance that the hormone controls signals the gland to increase or decrease its secretion.
  1 mark
5. (ii) Example: insulin from the pancreas. When the blood sugar level rises, the pancreas detects it and secretes more insulin. As the blood sugar level falls, insulin secretion is reduced.
  1 mark
Answer: (i) (I) Gigantism – pituitary; (II) Delayed puberty – ovary; (III) Goitre – thyroid. (ii) By a feedback mechanism, e.g. a rise in blood sugar increases insulin secretion and a fall in blood sugar reduces it.
Where the marks are
- Gigantism and pituitary gland (½ + ½ mark)
- Delayed puberty and ovary (½ + ½ mark)
- Goitre and thyroid gland (½ + ½ mark)
- States that a feedback mechanism regulates timing and amount (1 mark)
- Insulin–blood sugar example explained correctly (1 mark)
Open Q35(b) on its own page
Q36 · 5 marks · Long answerLight – Reflection and Refraction
(a) An object is placed in front of a convex lens of focal length f. If the distance of the object from the lens is 2f, draw a ray diagram to show the formation of the image. Write the value of magnification in this case.
(b) A student has focussed the image of a candle flame on a white screen using a convex lens. The situation is as given below: Length of the flame = 2 cm; Focal length of the lens = 12 cm; Distance of the flame from the lens = 16 cm. If the flame is perpendicular to the principal axis of the lens, calculate the values of the following :
(i) Distance of the image from the lens
(ii) Length of the image formed
Try it first, then show our solutionSolution
1. (a) Ray diagram in words: draw the principal axis and the convex lens, and mark F₁, 2F₁ on the left and F₂, 2F₂ on the right. Place object AB upright at 2F₁. Ray 1 from A runs parallel to the principal axis and, after refraction, passes through F₂. Ray 2 from A passes through the optical centre O without deviation. The two rays meet at A′ below the axis at 2F₂. Draw the image A′B′ at 2F₂, with arrows on all rays. The image is real, inverted and the same size as the object.
  1½ marks
2. Since the image is the same size and inverted, the magnification is $m = - 1$ .
  ½ mark
3. (b) Data: $h = + 2$ cm, $f = + 12$ cm, $u = - 16$ cm. Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
  ½ mark
4. (i) $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{12} + \frac{1}{- 16} = \frac{4 - 3}{48}$
  ½ mark
5. $\frac{1}{v} = \frac{1}{48}$ , so $v = + 48$ cm. The image is 48 cm from the lens, on the side opposite to the flame.
  ½ mark
6. (ii) Magnification: $m = \frac{h ^{'}}{h} = \frac{v}{u}$ , so $h^{'} = h \times \frac{v}{u}$
  ½ mark
7. $h^{'} = 2 cm \times \frac{48}{- 16}$
  ½ mark
8. $h^{'} = - 6$ cm. The image is 6 cm long, real and inverted.
  ½ mark
Answer: (a) Image at 2F₂: real, inverted, same size; m = −1. (b)(i) v = +48 cm; (ii) image length = 6 cm (h′ = −6 cm, inverted).
Where the marks are
- Correct ray diagram with object at 2F₁ and image at 2F₂, arrows shown (1½ marks; ½ deducted if arrows missing)
- Magnification = −1 (½ mark)
- Correct sign convention and lens formula (½ mark)
- Correct substitution (½ mark)
- v = +48 cm (½ mark)
- Magnification formula h′ = hv/u (½ mark)
- Correct substitution (½ mark)
- h′ = −6 cm, i.e. image 6 cm long (½ mark)
Open Q36 on its own page

Section E

source/case-based with sub-parts

Q37 · 4 marks · Case-basedCompetency-basedAcids, Bases and Salts
The teacher while conducting practicals in the laboratory divided the students into three groups and gave them various solutions to find out their pH and classify them into acidic, basic and neutral solutions.
Group A – Lemon juice, vinegar, colourless aerated drink
Group B – Tomato juice, coffee, ginger juice
Group C – Sodium hydroxide, sodium chloride, lime water
1. (a)
  For the solutions provided, which group is/are likely to have pH value (i) less than 7, and (ii) greater than 7 ?
  1 mark
2. (b)
  List two ways of determining pH of a solution.
  1 mark
3. (c)
  Explain, why the sour substances such as lemon juice are effective in cleaning the tarnished copper vessels.
  2 marks
4. OR
  (c)
  "pH has great importance in our daily life." Justify this statement by giving two examples.
  2 marks
Try it first, then show our solutionSolution
1. (a)(i) Groups A and B have pH less than 7. Lemon juice, vinegar, aerated drink, tomato juice, coffee and ginger juice are all acidic.
  ½ mark
2. (a)(ii) Group C has pH greater than 7, because sodium hydroxide and lime water are basic. (Sodium chloride is neutral, with pH 7.)
  ½ mark
3. (b) Two ways to find pH: (1) using pH paper; (2) using universal indicator, then matching the colour with the pH colour chart.
  1 mark
4. (c) Tarnished copper has a dull layer of basic copper oxide / basic copper carbonate on it, formed by reaction with air and moisture. This layer is basic in nature.
  1 mark
5. (c) Lemon juice contains acid (citric acid). The acid neutralises the basic layer and forms a salt that dissolves and is washed away. The shiny copper surface is then exposed again.
  1 mark
6. (c) OR – Example 1: Our stomach produces HCl so that the pH is right for digestion. Excess acid causes indigestion, which is treated with antacids such as milk of magnesia.
  1 mark
7. (c) OR – Example 2: Tooth decay starts when the pH in the mouth falls below 5.5. Bacteria produce acids from food particles, and these acids corrode the enamel. Using a basic toothpaste prevents this. (Other valid examples: survival of aquatic life becomes difficult when rain lowers the pH of river water; bee stings and nettle stings inject acid, which is relieved by a mild base.)
  1 mark
Answer: (a)(i) Groups A and B; (ii) Group C. (b) pH paper and universal indicator. (c) The acid in lemon juice neutralises the basic copper oxide/carbonate layer and removes it. OR: pH in digestion and in tooth decay (or aquatic life, stings).
Where the marks are
- Groups A and B: pH < 7 (½ mark)
- Group C: pH > 7 (½ mark)
- pH paper and universal indicator (½ + ½ mark)
- Tarnish is a basic copper oxide/carbonate layer (1 mark)
- Acidic lemon juice neutralises it to a salt that washes away (1 mark)
- OR: any two correct examples of the importance of pH in daily life (1 + 1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedHow do Organisms Reproduce?
All the reproductive methods of living organisms are broadly categorized into two types : 1. Asexual reproduction, and 2. Sexual reproduction.
Asexual reproduction involves the participation of a single parent without the formation of gametes, fertilisation and transfer of genetic material. This method is a common means of rapidly increasing offsprings under favourable conditions.
1. (a)
  Name the type of fission that occurs in Leishmania and Plasmodium.
  1 mark
2. (b)
  Write one advantage of sexual mode of reproduction over asexual reproduction.
  1 mark
3. (c)
  Give reasons why: (i) Colonies of yeast fail to multiply in water but multiply in sugar solution. (ii) Rhizopus individuals do not grow on a dry slice of bread.
  2 marks
4. OR
  (c)
  Name the filamentous structures a student could identify when he collected water from a pond that appeared dark green. How do these organisms multiply ? Explain.
  2 marks
Try it first, then show our solutionSolution
1. (a) Leishmania reproduces by binary fission (along a definite plane). Plasmodium reproduces by multiple fission.
  1 mark
2. (b) Sexual reproduction combines DNA from two parents, so it produces more variations. These variations help a species survive changing environmental conditions.
  1 mark
3. (c)(i) Water contains no food. Sugar solution gives yeast the nutrients and energy it needs to grow and to multiply by budding, so yeast colonies multiply only in sugar solution.
  1 mark
4. (c)(ii) Rhizopus needs moisture to germinate its spores and grow hyphae. A dry slice of bread has no moisture, so Rhizopus cannot grow on it.
  1 mark
5. (c) OR – The dark green filamentous structures are Spirogyra (a filamentous green alga).
  ½ mark
6. (c) OR – Spirogyra multiplies by fragmentation.
  ½ mark
7. (c) OR – On maturing, a Spirogyra filament simply breaks up into two or more small pieces (fragments). Each fragment grows into a new filament.
  1 mark
Answer: (a) Leishmania – binary fission; Plasmodium – multiple fission. (b) Produces more variations, which aid survival of the species. (c)(i) Sugar provides nutrition; plain water does not. (ii) Rhizopus needs moisture. OR: Spirogyra; multiplies by fragmentation.
Where the marks are
- Binary fission for Leishmania and multiple fission for Plasmodium (½ + ½ mark)
- More variations aiding survival of the species (1 mark)
- Sugar solution supplies nutrients for growth and budding (1 mark)
- Moisture is needed for Rhizopus growth (1 mark)
- OR: names Spirogyra (½ mark); fragmentation (½ mark); explains breaking into pieces, each growing into a new individual (1 mark)
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Q39 · 4 marks · Case-basedCompetency-basedElectricity
Consider the following electrical circuit diagram in which nine identical resistors of 3 Ω each are connected as shown. If the reading of the ammeter A $_{1}$ is 1 ampere, answer the following questions :
1. (a)
  What is the relationship between the readings of A $_{1}$ and A $_{3}$ ? Give reasons for your answer.
  1 mark
2. (b)
  What is the relationship between the readings of A $_{2}$ and A $_{3}$ ?
  1 mark
3. (c)
  Determine the reading of the voltmeter V $_{1}$ .
  2 marks
4. OR
  (c)
  Find the total resistance of the circuit.
  2 marks
Figure in the paper: A circuit with a battery at the bottom, ammeter A1 (marked +) in series on the left branch and ammeter A3 (marked +) in series on the right branch. Three parallel branches sit between: the first section has voltmeter V1 (marked +) across two 3 Ω resistors in parallel; the second section has voltmeter V2 (marked +) across three 3 Ω resistors in parallel; the third section has ammeter A2 (marked +) in a top branch in series with a 3 Ω resistor, in parallel with two more 3 Ω resistors and a third 3 Ω resistor, all between the same two junctions. See it in the official paper.
Try it first, then show our solutionSolution
1. (a) The readings are equal: $A_{1} = A_{3} = 1$ A. Both ammeters are in series in the main circuit, and the same current flows through every part of a series circuit.
  1 mark
2. (b) The main current of 1 A divides equally among the four identical 3 Ω resistors in parallel. So $A_{2} = \frac{1}{4} A_{3} = 0.25$ A, i.e. $A_{2} < A_{3}$ .
  1 mark
3. (c) For two 3 Ω resistors in parallel: $\frac{1}{R _{p}} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ , so $R_{p} = \frac{3}{2} = 1.5 Ω$
  1 mark
4. Using $V = I R$ : $V_{1} = 1 A \times 1.5 Ω = 1.5$ V
  1 mark
5. (c) OR – Section 1 (two in parallel): $\frac{1}{R _{p 1}} = \frac{1}{3} + \frac{1}{3}$ , so $R_{p 1} = \frac{3}{2} Ω$ . Section 2 (three in parallel): $\frac{1}{R _{p 2}} = \frac{3}{3}$ , so $R_{p 2} = 1 Ω$
  1 mark
6. Section 3 (four in parallel): $\frac{1}{R _{p 3}} = \frac{4}{3}$ , so $R_{p 3} = \frac{3}{4} Ω$ . The sections are in series: $R = \frac{3}{2} + 1 + \frac{3}{4} = \frac{13}{4} = 3.25 Ω$
  1 mark
Answer: (a) A₁ = A₃ (series); (b) A₂ = ¼ A₃ = 0.25 A; (c) V₁ = 1.5 V; OR total resistance = 13/4 Ω = 3.25 Ω.
Where the marks are
- A₁ = A₃ (½ mark), because both are in series (½ mark)
- A₂ = ¼ of A₃ = 0.25 A, as current divides equally in four identical resistors (1 mark)
- Parallel combination of two 3 Ω resistors = 1.5 Ω (1 mark)
- V₁ = IR = 1.5 V (1 mark)
- OR: R_p1 = 1.5 Ω, R_p2 = 1 Ω (1 mark); R_p3 = 0.75 Ω and total 3.25 Ω (1 mark)
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Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.