ScienceCBSE Class 10 · Board exam 2026 (session 2025-26)

CBSE Class 10 Science: 2026 board paper, set 31/4/1

The question paper set in the CBSE Class 10 board exam of 2026, 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/2026/X/Science.zip (a zip; this paper is the file 1190-1_31-4-1_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	1-16	Section A (Biology), 30 marks: MCQs 1-7, assertion-reason 8-9, VSA 10-12, SA 13-14, CBQ 15, LA 16
B	17-29	Section B (Chemistry), 25 marks: MCQs 17-24 (18 and 24 assertion-reason style as printed), VSA 25, SA 26-27, CBQ 28, LA 29
C	30-39	Section C (Physics), 25 marks: MCQs 30-31, assertion-reason 32, SA 33-37, CBQ 38, LA 39

Instructions in brief: All questions compulsory; no overall choice; internal choice (OR) given in Q12, Q15(c), Q16, Q27, Q29, Q33, Q38(c), Q39 - attempt only one alternative in each.

Section A

Section A (Biology), 30 marks: MCQs 1-7, assertion-reason 8-9, VSA 10-12, SA 13-14, CBQ 15, LA 16

Q1 · 1 mark · MCQCompetency-basedLife Processes
Observe the given figure and select the statement that correctly states how these fungi obtain nutrition.
1. (A)By eating the bread on which it is growing.
2. (B)By using nutrients from the bread to prepare their own food.
3. (C)By breaking down the nutrients of bread outside the body and then absorbing them.
4. (D)By allowing other organisms to grow on the bread and then consuming them.
Figure in the paper: Bread mould growing on a slice of bread. Labels: 'Bread mould' (pointing to dark spore-bearing heads on the stalks), 'Fungi body' (the network of thread-like hyphae), and 'Bread' (the slice on which the fungus grows). See it in the official paper.
Try it first, then show our solutionSolution
1. Bread mould (Rhizopus) is a saprophyte: its hyphae secrete digestive enzymes onto the bread, breaking complex food into simple soluble molecules outside the body, which are then absorbed. Hence option (C).
  1 mark
Answer: (C) By breaking down the nutrients of bread outside the body and then absorbing them
Where the marks are
- Identifies saprophytic nutrition, option (C) (1 mark)
Q2 · 1 mark · MCQHow do Organisms Reproduce?
If the part 'P' of the flower is removed, which of the following processes will not take place?
1. (A)Formation of pollens
2. (B)Formation of flower
3. (C)Formation of fruit
4. (D)Formation of seed
Figure in the paper: A flower showing sepals, petals, stamens with anthers, and a central pistil. Part 'P' is a label line pointing to the ovules inside the lower swollen part of the pistil (ovary). See it in the official paper.
Try it first, then show our solutionSolution
1. P points to the ovules inside the ovary, and ovules contain the female gametes. Without ovules there is no fertilisation, so no seed develops.
2. With no fertilisation, the ovary does not develop into a fruit. The scheme takes formation of fruit as the answer: option (C).
  1 mark
Answer: (C) Formation of fruit
Where the marks are
- Chooses formation of fruit, option (C) (1 mark)
Open Q2 on its own page
Q3 · 1 mark · MCQOur Environment
Given below are some of the statements regarding our environment:
(i) Change in our life-styles have resulted in greater amounts of waste material generation.
(ii) Changes in attitude have resulted in use of more disposable things.
(iii) Changes in packaging has resulted in much of our waste becoming non-biodegradable.
(iv) Organic farming.
Which amongst these is not leading to any damage of environment?
1. (A)(i), (ii), (iii)
2. (B)(iv)
3. (C)(i) and (iii)
4. (D)(ii) and (iv)
Try it first, then show our solutionSolution
1. Statements (i), (ii) and (iii) all increase waste or non-biodegradable waste. Organic farming avoids chemical fertilisers and pesticides, so it does not harm the environment. Answer: option (B).
  1 mark
Answer: (B) (iv)
Where the marks are
- Selects organic farming (iv), option (B) (1 mark)
Q4 · 1 mark · MCQHeredity
Identify the traits that cannot be transmitted from parents to the offsprings:
(i) Swimming technique
(ii) Blood group
(iii) Eye colour
(iv) Dancing skills
1. (A)(i) and (ii)
2. (B)(i) and (iv)
3. (C)(iii) and (iv)
4. (D)(ii) and (iii)
Try it first, then show our solutionSolution
1. Swimming technique and dancing skills are acquired traits learnt during life; they are not coded in DNA, so they are not inherited.
2. Blood group and eye colour are genetic, inherited traits. Answer: option (B).
  1 mark
Answer: (B) (i) and (iv)
Where the marks are
- Identifies acquired traits (i) and (iv), option (B) (1 mark)
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Q5 · 1 mark · MCQControl and Coordination
Following are the statements about the different parts of the brain. Select the correct statements.
(i) The thinking part of the brain is the forebrain.
(ii) Centres for smell, hearing, memory, sight are located in forebrain.
(iii) Salivation, vomiting, blood pressure are controlled by medulla in the hind brain.
(iv) Cerebellum does not control the posture and balance of the body.
1. (A)(i), (ii) and (iii)
2. (B)(ii), (iii) and (iv)
3. (C)(i), (ii) and (iv)
4. (D)(i), (iii) and (iv)
Try it first, then show our solutionSolution
1. (i), (ii) and (iii) are correct. Statement (iv) is wrong because the cerebellum does control posture and balance. Answer: option (A).
  1 mark
Answer: (A) (i), (ii) and (iii)
Where the marks are
- Rejects (iv) and chooses option (A) (1 mark)
Q6 · 1 mark · MCQHeredity
Two pea plants, one with round green seeds (RRyy) and another with wrinkled yellow (rrYY) seeds were crossed with each other which produced $F_{1}$ progeny that have only round yellow (RrYy) seeds. When $F_{1}$ plants are self-pollinated, the $F_{2}$ progeny will have which new combination of characters, as compared to the parents:
(i) Round, yellow
(ii) Round, green
(iii) Wrinkled, yellow
(iv) Wrinkled, green
Correct options are:
1. (A)(i) and (ii)
2. (B)(i) and (iv)
3. (C)(ii) and (iv)
4. (D)(i) and (iii)
Try it first, then show our solutionSolution
1. The parents were round green and wrinkled yellow.
2. Because the two traits sort independently, $F_{2}$ also shows round yellow and wrinkled green. These are the new combinations not seen in either parent. Answer: option (B).
  1 mark
Answer: (B) (i) and (iv)
Where the marks are
- Identifies round yellow and wrinkled green as new combinations, option (B) (1 mark)
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Q7 · 1 mark · MCQLife Processes
In human beings the oxygen rich blood is pumped out to the body by:
1. (A)contraction of right atrium
2. (B)relaxation of left ventricle
3. (C)contraction of right ventricle
4. (D)contraction of left ventricle
Try it first, then show our solutionSolution
1. Oxygenated blood from the lungs enters the left atrium and passes to the left ventricle. The left ventricle contracts to pump it into the aorta and on to the body. Answer: option (D).
  1 mark
Answer: (D) contraction of left ventricle
Where the marks are
- Chooses contraction of left ventricle, option (D) (1 mark)
Q8 · 1 mark · Assertion-reasonCompetency-basedHow do Organisms Reproduce?
Question Nos. 8 & 9 consist of two statements: Assertion (A) and Reason (R). Answer these questions selecting the appropriate option from (A), (B), (C) and (D) given below.
Assertion (A): Variations are not often seen in off-springs produced by asexual reproduction.
Reason (R): DNA molecule generated by replication is not similar to the original DNA.
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: offspring of asexual reproduction show very few variations.
2. R is false: DNA copying produces copies that are similar to the original DNA, with only minor errors. Answer: option (C).
  1 mark
Answer: (C)
Where the marks are
- A true, R false, option (C) (1 mark)
Open Q8 on its own page
Q9 · 1 mark · Assertion-reasonCompetency-basedOur Environment
Assertion (A): Highest percentage of pesticides is found in tertiary consumers in a food chain.
Reason (R): Chemicals / pesticides get accumulated progressively at each trophic level.
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. Through biological magnification, pesticides accumulate progressively at each trophic level. The highest trophic level (tertiary consumers) therefore has the maximum concentration.
2. R explains A, so both are true and R is the correct explanation. Answer: option (A).
  1 mark
Answer: (A)
Where the marks are
- Both true and R explains A via biomagnification, option (A) (1 mark)
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Q10 · 2 marks · Very short answerLife Processes
"Transpiration in plants is a necessary evil." Comment on the statement giving two reasons.
Try it first, then show our solutionSolution
1. Why transpiration is necessary: Evaporation of water from the leaves creates a suction pull (transpiration pull). This pull helps move water and dissolved minerals up through the xylem from the roots to the leaves. It also cools the leaf surface and so regulates the plant's temperature.
  1 mark
2. Why it is an evil: Most of the water absorbed by the roots is lost to the air as vapour through the stomata. In hot or dry conditions this loss can cause wilting and water stress. Transpiration is therefore useful but wasteful, a 'necessary evil'.
  1 mark
Answer: Necessary: it creates the pull that carries water and minerals from roots to leaves and cools the plant. Evil: it causes a large loss of water from the plant.
Where the marks are
- Gives a benefit: upward transport of water and minerals, or temperature regulation (1 mark)
- Gives the harm: loss of water from the plant (1 mark)
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Q11 · 2 marks · Very short answerOur Environment
Explain why food chains generally consist of three or four steps.
Try it first, then show our solutionSolution
1. By the 10% law, only about 10% of the energy at one trophic level is passed on to the next level. The remaining energy is used for life processes or lost as heat.
  1 mark
2. So the usable energy falls sharply at each step. For example: Grass (1000 kJ) → Grasshopper (100 kJ) → Frog (10 kJ) → Snake (1 kJ). After three or four steps too little energy is left to support another level, so food chains usually have only 3–4 steps.
  1 mark
Answer: Only about 10% of energy passes to the next trophic level, so after 3–4 levels too little energy remains to support further organisms.
Where the marks are
- States the 10% law, with most energy lost as heat (1 mark)
- Concludes that very little energy remains after 3–4 levels, or shows this with an energy-flow example (1 mark)
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Answer one of these 2 alternatives.
Q12(A) · 2 marks · Very short answerCompetency-basedControl and Coordination
Varun is diabetic. His latest ultrasound reports showed a blockage in his pancreatic duct. How will this affect the dual functioning of the pancreas?
Try it first, then show our solutionSolution
1. Exocrine (digestive) function: Pancreatic juice contains trypsin, lipase and pancreatic amylase, and it reaches the small intestine through the pancreatic duct. With the duct blocked, these enzymes cannot reach the intestine. Digestion of proteins, fats and carbohydrates will therefore be impaired.
  1 mark
2. Endocrine (hormonal) function: Insulin is made by the islets of Langerhans and passes directly into the blood without using the duct, so the blockage does not affect this route. However, Varun is diabetic, so his insulin secretion is already deficient and his blood sugar control remains affected.
  1 mark
Answer: Digestive function is affected because pancreatic enzymes cannot reach the small intestine. The hormonal function (insulin) is not affected by the blockage but is already impaired because he is diabetic.
Where the marks are
- Explains that pancreatic enzymes cannot reach the small intestine, so digestion of proteins, fats and carbohydrates suffers (1 mark)
- States that insulin secretion is already affected because of diabetes (1 mark)
Open Q12(A) on its own page
OR
Q12(B) · 2 marks · Very short answerCompetency-basedControl and Coordination
Sunil unconsciously touched the hot iron rod and immediately withdrew his hand. Which type of action is involved? Also define this action. Draw a flow chart to show the path followed for this action.
Try it first, then show our solutionSolution
1. The action involved is a reflex action.
  ½ mark
2. Definition: A reflex action is a sudden, automatic and quick response to a stimulus that happens without conscious thinking. It is controlled by the spinal cord.
  ½ mark
3. Flow chart (reflex arc): Stimulus (heat of the rod) → Receptor (in the skin) → Sensory neuron → Relay neuron (in the spinal cord) → Motor neuron → Effector (arm muscle) → Response (hand withdrawn).
  1 mark
Answer: Reflex action: a quick, automatic response to a stimulus without thinking. Path: Stimulus → Receptor → Sensory neuron → Relay neuron (spinal cord) → Motor neuron → Effector (muscle) → Response.
Where the marks are
- Names the action as a reflex action (½ mark)
- Defines it as an automatic, quick response without thinking (½ mark)
- Draws the correct flow chart of the reflex arc (1 mark)
Open Q12(B) on its own page
Q13 · 3 marks · Short answerControl and Coordination
Give reasons for the following:
1. (a)
  Adrenal gland is known as the stress gland of the body.
  0 marks
2. (b)
  Prenatal sex determination is banned in India.
  0 marks
3. (c)
  Vegetative propagation is advantageous in some plants.
  0 marks
Try it first, then show our solutionSolution
1. (a) In stress or danger, the adrenal gland secretes adrenaline into the blood. Adrenaline increases the heart rate and breathing rate and sends more blood to the muscles. This prepares the body to face or escape the emergency, so the gland is called the stress gland.
  1 mark
2. (b) Prenatal sex determination is banned to stop female foeticide, the killing of female foetuses. This practice causes a sharp decline in the female-to-male sex ratio.
  1 mark
3. (c) Vegetative propagation is advantageous because:
  plants raised this way flower and fruit earlier than plants grown from seeds;
  plants that have lost the ability to produce seeds, such as banana, orange, rose and jasmine, can still be propagated;
  the new plants are genetically similar to the parent, so desirable characters are kept.
  (Any one point earns the mark.)
  1 mark
Answer: (a) It secretes adrenaline, which prepares the body for emergencies. (b) To prevent female foeticide. (c) Earlier flowering and fruiting, propagation of seedless plants, and offspring identical to the parent.
Where the marks are
- Links the adrenal gland to adrenaline, which prepares the body for stress (1 mark)
- Gives prevention of female foeticide as the reason for the ban (1 mark)
- Gives any one valid advantage of vegetative propagation (1 mark)
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Q14 · 3 marks · Short answerCompetency-basedHeredity

Study the given table and answer the following questions:

Parents	$F_{1}$	$F_{2}$
A pea plant with red flowers was crossed with a pea plant with white flowers.	30 seeds were sown and all bore plants with red flowers.	44 seeds were sown, 33 plants bore red flowers and 11 plants bore white flowers.

(a)
Depict the information given in the table in the form of a cross.
0 marks
(b)
Deduce the $F_{2}$ phenotypic ratio/ $F_{2}$ ratio showing external look as well as genotypic/genetic makeup, ratio.
0 marks

Try it first, then show our solutionSolution

(a) Let R = dominant allele for red flowers and r = recessive allele for white flowers.
Parents: RR (red) × rr (white)
½ mark
Gametes: R and r.
F $_{1}$ : Rr, all plants red. This agrees with the table: all 30 F $_{1}$ plants bore red flowers.
½ mark
Selfing of F $_{1}$ : Rr × Rr.
Gametes from each parent: R and r.
½ mark
F $_{2}$ (checkerboard):
♂ \ ♀ R r
R RR (red) Rr (red)
r Rr (red) rr (white)
½ mark
(b) Phenotypic ratio: Red : White = 33 : 11 = 3 : 1.
½ mark
Genotypic ratio: RR : Rr : rr = 1 : 2 : 1.
½ mark

♂ \ ♀	R	r
R	RR (red)	Rr (red)
r	Rr (red)	rr (white)

Answer: Cross: RR × rr → F $_{1}$ Rr (all red); Rr × Rr → F $_{2}$ RR, Rr, Rr, rr. Phenotypic ratio 3 red : 1 white (33 : 11); genotypic ratio 1 RR : 2 Rr : 1 rr.

Where the marks are

Writes the parents as RR × rr (½ mark)
Shows the gametes and F $_{1}$ as Rr, all red (½ mark)
Shows F $_{1}$ selfing, Rr × Rr, with its gametes (½ mark)
Gives the F $_{2}$ checkerboard or offspring RR, Rr, Rr, rr (½ mark)
States the phenotypic ratio 3 : 1 (½ mark)
States the genotypic ratio 1 : 2 : 1 (½ mark)

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Q15 · 4 marks · Case-basedCompetency-basedControl and Coordination

Plants exhibit movements in response to various environmental stimuli to adopt and survive in their surroundings. These movements, influenced by external factors such as light, gravity, water and chemicals are regulated by specific plant hormones. For example, the differential growth of plant parts allows them to bend, grow or orient themselves in particular directions. Roots and shoots often show contrasting response to the same stimulus, highlighting the complex mechanisms involved in plant growth and development.

(a)
Why is hydrotropism essential for plants?
0 marks
(b)
What role does auxin play in the growth of plants?
0 marks
(c)
Differentiate between movement shown by the sensitive plant and tropic movement by plants (Two points).
0 marks
OR
(c)
OR: How does movement of roots differ from that of shoots in response to gravity and light?
0 marks

Try it first, then show our solutionSolution

(a) Hydrotropism makes roots grow towards a source of water. This lets the plant reach and absorb water and dissolved minerals, which it needs for survival.
1 mark
(b) Auxin promotes cell elongation. When light falls from one side, auxin moves to the shaded side of the shoot, and those cells grow longer. The shoot therefore bends towards light, so auxin controls tropic (directional) growth.
1 mark

Movement in sensitive plant (Mimosa)	Tropic movement
Not directional; does not depend on the direction of the stimulus	Directional; occurs towards or away from the stimulus
Not growth-related; caused by change in the water content of cells	Growth-related; caused by unequal growth
Fast	Slow

(Any two points.)

2 marks

Stimulus	Roots	Shoots
Gravity	Grow towards gravity (positively geotropic)	Grow away from gravity (negatively geotropic)
Light	Grow away from light (negatively phototropic)	Grow towards light (positively phototropic)

Answer: (a) It lets roots reach and absorb water and minerals. (b) Auxin promotes cell elongation and causes bending towards light. (c) Sensitive-plant movement is non-directional, fast and not growth-related; tropic movement is directional, slow and growth-related. OR: Roots show positive geotropism and negative phototropism; shoots show negative geotropism and positive phototropism.

Where the marks are

States that hydrotropism helps roots reach and absorb water and minerals (1 mark)
States that auxin promotes cell elongation or regulates bending towards light (1 mark)
Gives two correct differences between sensitive-plant and tropic movements (1 + 1 marks)
OR: Contrasts roots and shoots correctly for gravity and for light (1 + 1 marks)

Open Q15 on its own page

Answer one of these 2 alternatives.
Q16(A) · 5 marks · Long answerLife Processes
(a) Draw a labelled diagram showing open and closed stomatal pore.
(b) List the main events that occur during the process of photosynthesis.
(c) Write a balanced chemical equation showing photo-synthesis in plants.
Figure in the paper: Labelled diagram required showing a stomatal pore in open and closed states, with guard cells marked. See it in the official paper.
Try it first, then show our solutionSolution
1. (a) Open stomatal pore (diagram described): Two bean- or kidney-shaped guard cells are swollen (turgid) and curved outwards, leaving a wide gap between them. Labels: guard cells, stomatal pore (open), chloroplasts, nucleus, epidermal cells. Guard cells take in water and swell, so the pore opens.
  1 mark
2. Closed stomatal pore (diagram described): Two guard cells are shrunken (flaccid) and lie close together, so the pore is a thin closed slit. Labels: guard cells, stomatal pore (closed), chloroplasts, nucleus, epidermal cells. Guard cells lose water and shrink, so the pore closes.
  1 mark
3. (b) Main events of photosynthesis:
  1. Chlorophyll absorbs light energy.
  2. Light energy is converted into chemical energy, and water molecules split into hydrogen and oxygen.
  3. Carbon dioxide is reduced to carbohydrates.
  2 marks
4. (c) Balanced equation:
  $6 C O_{2} + 12 H_{2} O sunlight chlorophyll C_{6} H_{12} O_{6} + 6 O_{2} + 6 H_{2} O$
  1 mark
Answer: (a) Diagrams of open (turgid guard cells) and closed (flaccid guard cells) stomata. (b) Light absorbed by chlorophyll; light energy converted to chemical energy and water split; CO $_{2}$ reduced to carbohydrate. (c) $6 C O_{2} + 12 H_{2} O \to C_{6} H_{12} O_{6} + 6 O_{2} + 6 H_{2} O$ (in sunlight, with chlorophyll).
Where the marks are
- Draws the open stomatal pore with guard cells labelled (1 mark)
- Draws the closed stomatal pore with guard cells labelled (1 mark)
- Lists the three events of photosynthesis (2 marks)
- Writes the balanced equation with sunlight and chlorophyll shown (1 mark)
Open Q16(A) on its own page
OR
Q16(B) · 5 marks · Long answerLife Processes
(a) Draw a well-labelled diagram of nephron and label atleast 4 of its parts.
(b) Write the role of capillary cluster inside Bowman's capsule.
(c) What is meant by selective reabsorption?
Figure in the paper: Labelled diagram of a nephron required, labelling at least 4 parts. See it in the official paper.
Try it first, then show our solutionSolution
1. (a) Diagram of a nephron (described):
  A cup-shaped Bowman's capsule encloses a knot of capillaries, the glomerulus.
  The afferent arteriole enters the glomerulus and the efferent arteriole leaves it.
  From the capsule runs a coiled tubule, which leads into the collecting duct.
  The tubule is surrounded by a network of blood capillaries.
  1 mark
2. Labels (any four): Bowman's capsule, glomerulus, tubule (with its coiled part), collecting duct, capillary network, and branch of the renal artery or renal vein.
  2 marks
3. (b) The glomerulus (the capillary cluster inside Bowman's capsule) filters blood under pressure. Water, glucose, amino acids, salts and nitrogenous wastes such as urea pass into the capsule as the filtrate.
  1 mark
4. (c) Selective reabsorption is the process in the tubule by which useful substances in the filtrate are taken back into the blood capillaries around the tubule. These include glucose, amino acids, vitamins, salts and most of the water. The wastes are left behind to form urine.
  1 mark
Answer: (a) Labelled nephron: Bowman's capsule, glomerulus, tubule, collecting duct. (b) It filters blood to remove wastes. (c) Useful substances (glucose, amino acids, salts, water) are taken back from the filtrate into the blood.
Where the marks are
- Draws a correct diagram of the nephron (1 mark)
- Labels any four parts correctly (½ × 4 = 2 marks)
- States that the glomerulus filters blood to remove wastes (1 mark)
- Explains selective reabsorption of useful substances into the blood (1 mark)
Open Q16(B) on its own page

Section B

Section B (Chemistry), 25 marks: MCQs 17-24 (18 and 24 assertion-reason style as printed), VSA 25, SA 26-27, CBQ 28, LA 29

Q17 · 1 mark · MCQChemical Reactions and Equations
The volume ratio of hydrogen and oxygen gases liberated during electrolysis of water is
1. (A)1 : 1
2. (B)2 : 1
3. (C)4 : 1
4. (D)1 : 2
Try it first, then show our solutionSolution
1. $2 H_{2} O \to 2 H_{2} + O_{2}$ gives 2 volumes of hydrogen for every 1 volume of oxygen. Answer: option (B), 2 : 1.
  1 mark
Answer: (B) 2 : 1
Where the marks are
- States 2 : 1 from the balanced equation, option (B) (1 mark)
Q18 · 1 mark · MCQChemical Reactions and Equations
$C_{6} H_{12} O_{6 (aq)} + 6 O_{2 (g)} \to 6C O_{2 (g)} + 6 H_{2} O_{(l)}$
The above reaction is an example of -
1. (A)Displacement Reaction
2. (B)Endothermic Reaction
3. (C)Exothermic Reaction
4. (D)Neutralisation Reaction
Try it first, then show our solutionSolution
1. This is respiration (oxidation of glucose), which releases energy, so it is an exothermic reaction. Answer: option (C).
  1 mark
Answer: (C) Exothermic Reaction
Where the marks are
- Identifies respiration as exothermic, option (C) (1 mark)
Q19 · 1 mark · MCQAcids, Bases and Salts
An aqueous solution turns red litmus blue. Excess addition of which of the following solution would reverse the change?
1. (A)Sodium hydroxide
2. (B)Hydrochloric acid
3. (C)Baking soda
4. (D)Slaked lime
Try it first, then show our solutionSolution
1. Turning red litmus blue shows the solution is basic. Adding excess acid (HCl) makes it acidic, which turns the litmus red again. The other three options are bases. Answer: option (B).
  1 mark
Answer: (B) Hydrochloric acid
Where the marks are
- Chooses the only acid, HCl, option (B) (1 mark)
Q20 · 1 mark · MCQCarbon and its Compounds
While washing dirty clothes by soap, micelles are formed. Select the correct statement about micelles.
1. (A)The hydrophobic tail of the soap molecule is in the interior of the cluster, whereas hydrophilic end is on the surface of the cluster.
2. (B)The hydrophobic tail is on the surface of the cluster, whereas hydrophilic end is in the interior of cluster.
3. (C)Both hydrophobic tail and hydrophilic end are in the interior of the cluster.
4. (D)Both hydrophobic tail and hydrophilic end are on the surface of the cluster.
Try it first, then show our solutionSolution
1. In water, the hydrocarbon (hydrophobic) tails turn inward towards the oily dirt. The ionic (hydrophilic) heads face outward towards the water, forming a micelle. Answer: option (A).
  1 mark
Answer: (A)
Where the marks are
- Tail inside and head outside, option (A) (1 mark)
Q21 · 1 mark · MCQMetals and Non-metals
Which one of the following metals does not react with cold water?
1. (A)Sodium
2. (B)Potassium
3. (C)Calcium
4. (D)Magnesium
Try it first, then show our solutionSolution
1. Na, K and Ca react with cold water. Magnesium reacts only with hot water. Answer: option (D).
  1 mark
Answer: (D) Magnesium
Where the marks are
- Selects magnesium, option (D) (1 mark)
Q22 · 1 mark · MCQMetals and Non-metals
Alloys are homogeneous mixture of metal with a metal or non-metal. Which among the following alloys contain non-metal as one of its constituents?
1. (A)Brass
2. (B)Bronze
3. (C)Amalgam
4. (D)Steel
Try it first, then show our solutionSolution
1. Steel is iron with carbon, and carbon is a non-metal. Brass (Cu, Zn), bronze (Cu, Sn) and amalgam (Hg with a metal) contain only metals. Answer: option (D).
  1 mark
Answer: (D) Steel
Where the marks are
- Identifies steel (Fe + C), option (D) (1 mark)
Q23 · 1 mark · MCQAcids, Bases and Salts
How will you protect yourself from the heat generated while diluting a concentrated acid?
1. (A)By adding acid to water with constant stirring.
2. (B)By adding water to acid with constant stirring.
3. (C)By adding water to acid followed by base.
4. (D)By adding base to acid with constant stirring.
Try it first, then show our solutionSolution
1. Dilution is highly exothermic. Adding acid slowly to water with stirring lets the large volume of water absorb the heat, which prevents splashing and burns. Answer: option (A).
  1 mark
Answer: (A) By adding acid to water with constant stirring
Where the marks are
- Acid added to water slowly with stirring, option (A) (1 mark)
Q24 · 1 mark · Assertion-reasonCompetency-basedCarbon and its Compounds
In the question, two statements are given - One labelled as Assertion (A) and other labelled as Reason (R). Write the correct answer selecting the appropriate option from (A), (B), (C) & (D) given below.
Assertion (A): Pure ethanoic acid is also known as glacial acetic acid.
Reason (R): Pure ethanoic acid often freezes during winter due to its low melting point of 290 K.
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. The melting point of pure ethanoic acid is 290 K, so it freezes in cold weather into an ice-like solid. That is why it is called glacial acetic acid.
2. Both statements are true and R explains A. Answer: option (A).
  1 mark
Answer: (A)
Where the marks are
- Both true, R explains A, option (A) (1 mark)
Open Q24 on its own page
Q25 · 2 marks · Very short answerAcids, Bases and Salts
Answer the following.
1. (a)
  A compound 'A' is used in soda-lime fire-extinguisher and decomposes on heating to form compound 'B'. Compound 'B' is used for removing permanent hardness of water. Identify 'A' and 'B'.
  1 mark
2. (b)
  Write chemical formula of Plaster of Paris. At what temperature gypsum is heated to obtain Plaster of Paris?
  1 mark
Try it first, then show our solutionSolution
1. (a) A = sodium hydrogen carbonate (baking soda), $N a H C O_{3}$ . It is used in soda-acid fire extinguishers.
  ½ mark
2. On heating it gives B = sodium carbonate (washing soda), $N a_{2} C O_{3}$ , which removes permanent hardness of water:
  $2 N a H C O_{3} heat N a_{2} C O_{3} + H_{2} O + C O_{2}$
  ½ mark
3. (b) Formula of Plaster of Paris: $C a S O_{4} \cdot \frac{1}{2} H_{2} O$ (calcium sulphate hemihydrate).
  ½ mark
4. Gypsum ( $C a S O_{4} \cdot 2 H_{2} O$ ) is heated at 373 K (100 °C) to obtain Plaster of Paris.
  ½ mark
Answer: (a) A = $N a H C O_{3}$ (baking soda), B = $N a_{2} C O_{3}$ (washing soda). (b) $C a S O_{4} \cdot \frac{1}{2} H_{2} O$ ; 373 K (100 °C).
Where the marks are
- Identifies A as sodium hydrogen carbonate (½ mark)
- Identifies B as sodium carbonate (½ mark)
- Writes the formula $C a S O_{4} \cdot \frac{1}{2} H_{2} O$ (½ mark)
- States the temperature 373 K (½ mark)
Open Q25 on its own page
Q26 · 3 marks · Short answerChemical Reactions and Equations
Account for the following:
(a) White silver chloride turns grey when exposed to sunlight.
(b) Lead (II) nitrate on heating releases brown fumes.
(c) Bags of oil and fat containing food items are flushed with nitrogen gas.
Try it first, then show our solutionSolution
1. (a) Sunlight decomposes silver chloride into grey silver metal and chlorine gas (photolytic decomposition):
  $2 A g C l (s) sunlight 2 A g (s) + C l_{2} (g)$
  1 mark
2. (b) On heating, lead(II) nitrate decomposes and gives off nitrogen dioxide, which is seen as brown fumes:
  $2 P b (N O_{3})_{2} (s) heat 2 P b O (s) + 4 N O_{2} (g) + O_{2} (g)$
  1 mark
3. (c) Nitrogen is unreactive and displaces oxygen from the bag. This prevents oxidation of the oils and fats, so the food does not become rancid and its taste and smell are kept.
  1 mark
Answer: (a) AgCl decomposes in sunlight into grey Ag and Cl $_{2}$ . (b) Pb(NO $_{3}$ ) $_{2}$ decomposes to give brown NO $_{2}$ gas. (c) Nitrogen prevents oxidation (rancidity) of fats and oils.
Where the marks are
- Explains the decomposition of AgCl to form silver in sunlight (1 mark)
- Identifies the brown fumes as NO $_{2}$ from decomposition of lead nitrate (1 mark)
- States that nitrogen prevents oxidation or rancidity of fats (1 mark)
Open Q26 on its own page
Answer one of these 2 alternatives.
Q27(A) · 3 marks · Short answerMetals and Non-metals
(i) Which method will be used to convert each of them to their respective metal oxides?
(a) Carbonate ore
(b) Sulphide ore
(ii) Write a chemical reaction to illustrate the use of aluminium for joining cracked railway tracks.
(iii) During extraction of metals, electrolytic refining is used to obtain pure metals. Which material will be used as anode and cathode for the refining of copper metal by this process?
Try it first, then show our solutionSolution
1. (i)(a) A carbonate ore is converted to the oxide by calcination, i.e. heating strongly in a limited supply of air. Example: $Z n C O_{3} heat Z n O + C O_{2}$ .
  ½ mark
2. (i)(b) A sulphide ore is converted to the oxide by roasting, i.e. heating strongly in excess air. Example: $2 Z n S + 3 O_{2} heat 2 Z n O + 2 S O_{2}$ .
  ½ mark
3. (ii) In the thermit reaction, aluminium reduces iron(III) oxide. The reaction is so exothermic that the iron formed is molten, and it is used to join cracked railway tracks:
  $F e_{2} O_{3} (s) + 2 A l (s) \to 2 F e (l) + A l_{2} O_{3} (s) + heat$
  1 mark
4. (iii) Anode: a thick rod or block of impure copper.
  ½ mark
5. Cathode: a thin strip of pure copper. The electrolyte is acidified copper sulphate solution.
  ½ mark
Answer: (i) (a) Calcination, (b) Roasting. (ii) $F e_{2} O_{3} + 2 A l \to 2 F e + A l_{2} O_{3} + heat$ . (iii) Anode: impure copper; cathode: pure copper strip.
Where the marks are
- States calcination for carbonate ore (½ mark)
- States roasting for sulphide ore (½ mark)
- Writes the balanced thermit reaction (1 mark)
- Identifies the anode as impure copper (½ mark)
- Identifies the cathode as pure copper (½ mark)
Open Q27(A) on its own page
OR
Q27(B) · 3 marks · Short answerMetals and Non-metals
(i) Write all the reactions involved in the extraction of Zn from ZnS.
(ii) Hydrogen gas is not evolved when most of the metals react with nitric acid. Give reason.
Try it first, then show our solutionSolution
1. (i) Roasting: Zinc blende is heated strongly in excess air, giving zinc oxide:
  $2 Z n S (s) + 3 O_{2} (g) heat 2 Z n O (s) + 2 S O_{2} (g)$
  1 mark
2. Reduction: Zinc oxide is reduced with carbon to give zinc metal:
  $Z n O (s) + C (s) heat Z n (s) + C O (g)$
  1 mark
3. (ii) Nitric acid is a strong oxidising agent. It oxidises the hydrogen produced to water and is itself reduced to oxides of nitrogen ( $N_{2} O$ , $N O$ , $N O_{2}$ ). So hydrogen gas is not evolved (except with Mg and Mn and very dilute HNO $_{3}$ ).
  1 mark
Answer: (i) $2 Z n S + 3 O_{2} \to 2 Z n O + 2 S O_{2}$ ; $Z n O + C \to Z n + C O$ . (ii) HNO $_{3}$ is a strong oxidising agent and oxidises the H $_{2}$ formed to water.
Where the marks are
- Writes the balanced roasting equation (1 mark)
- Writes the reduction of ZnO with carbon (1 mark)
- States that HNO $_{3}$ is a strong oxidising agent that oxidises H $_{2}$ to water (1 mark)
Open Q27(B) on its own page
Q28 · 4 marks · Case-basedCompetency-basedAcids, Bases and Salts
The pH plays an important role in our daily life. Strength of acids and bases depends on the number of $H^{+}$ ions and $O H^{-}$ ions produced by these respectively. pH scale helps in predicting the strength of acids and bases by measuring hydrogen ion concentration in a solution.
Answer the following:
1. (a)
  What type of indicator is used to predict how strong an acid or base is?
  0 marks
2. (b)
  Three solutions A, B and C showed pH as 3, 5 and 7. Which solution has the highest concentration of $H^{+}$ ions?
  0 marks
3. (c)
  What are strong acids and weak acids?
  0 marks
4. OR
  (c)
  OR (i) Under what condition is rainwater called acid rain? (ii) In the following acids, identify weak acids: Nitric acid, hydrochloric acid, acetic acid, sulphuric acid, formic acid.
  0 marks
Try it first, then show our solutionSolution
1. (a) A universal indicator, a mixture of indicators that shows different colours at different pH values, is used to judge how strong an acid or base is.
  1 mark
2. (b) Solution A (pH = 3) has the highest H $^{+}$ ion concentration, because the lower the pH, the higher the H $^{+}$ concentration.
  1 mark
3. (c) Strong acids ionise almost completely in water and give a large number of H $^{+}$ ions, e.g. HCl and H $_{2}$ SO $_{4}$ (pH about 1–2).
  1 mark
4. Weak acids ionise only partly in water and give fewer H $^{+}$ ions, e.g. acetic acid (pH closer to 7).
  1 mark
5. (c) OR (i) Rainwater is called acid rain when its pH is less than 5.6.
  (ii) The weak acids in the list are acetic acid and formic acid.
Answer: (a) Universal indicator. (b) A (pH 3). (c) Strong acids give many H $^{+}$ ions; weak acids give few H $^{+}$ ions. OR: (i) pH < 5.6; (ii) acetic acid and formic acid.
Where the marks are
- Names the universal indicator (1 mark)
- Chooses solution A (1 mark)
- Explains that strong acids give more H $^{+}$ ions in water (1 mark)
- Explains that weak acids give fewer H $^{+}$ ions in water (1 mark)
- OR: States pH < 5.6 for acid rain (1 mark); names acetic and formic acid as weak acids (½ + ½ marks)
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Answer one of these 2 alternatives.
Q29(A) · 5 marks · Short answerCarbon and its Compounds
(i) Carbon cannot form $C^{4 +}$ cation and $C^{4 -}$ anion. Explain.
(ii) Define homologous series of carbon compounds. Why are the melting and boiling points of $C_{4} H_{10}$ higher than $C_{3} H_{8}$ or $C_{2} H_{6}$ ?
(iii) What happens when ethanol is heated with acidified potassium dichromate?
Try it first, then show our solutionSolution
1. (i) Carbon (2, 4) has 4 valence electrons.
  **C $^{4 +}$ :** To form this cation, carbon must lose 4 electrons. That needs a very large amount of energy, so it is energetically not possible.
  1 mark
2. **C $^{4 -}$ :** To form this anion, carbon must gain 4 electrons. A nucleus with only 6 protons cannot hold 10 electrons. Carbon therefore shares electrons and forms covalent bonds instead.
  1 mark
3. (ii) Homologous series: a series of carbon compounds that have the same functional group replacing hydrogen in the carbon chain. Successive members differ by a $- C H_{2} -$ unit (14 u) and have similar chemical properties. Example: $C H_{4}, C_{2} H_{6}, C_{3} H_{8}, C_{4} H_{10}$ .
  1 mark
4. In a homologous series, melting and boiling points increase as molecular mass increases. $C_{4} H_{10}$ (58 u) has a higher molecular mass than $C_{3} H_{8}$ (44 u) and $C_{2} H_{6}$ (30 u), so its melting and boiling points are higher.
  1 mark
5. (iii) Acidified potassium dichromate is an oxidising agent. It oxidises ethanol to ethanoic acid, and its orange colour turns green:
  $C H_{3} C H_{2} O H acidified K_{2} C r_{2} O_{7} heat C H_{3} C O O H$
  1 mark
Answer: (i) Losing 4 electrons needs too much energy; 6 protons cannot hold 10 electrons. (ii) Same functional group, successive members differ by CH $_{2}$ ; C $_{4}$ H $_{10}$ has a higher molecular mass. (iii) Ethanol is oxidised to ethanoic acid.
Where the marks are
- Explains why C $^{4 +}$ and C $^{4 -}$ cannot form (2 marks)
- Defines homologous series correctly (1 mark)
- Gives the higher molecular mass of C $_{4}$ H $_{10}$ as the reason (1 mark)
- States that ethanoic acid is formed by oxidation (1 mark)
Open Q29(A) on its own page
OR
Q29(B) · 5 marks · Short answerCarbon and its Compounds
(i) How would you name the following:
I. $C H_{3} - C H_{2} - C ∥ O - H$
II. $C H_{3} - C \equiv CH$
(ii) When ethanol reacts with ethanoic acid in the presence of conc. $H_{2} S O_{4}$ , a substance with sweet smell is produced. Answer the following:
I. State the class of compounds to which the sweet smelling compounds belong.
II. Write the chemical equation for the reaction.
(iii) Why soaps do not produce foam with hard water?
Try it first, then show our solutionSolution
1. (i) I. $C H_{3} - C H_{2} - C H O$ has three carbons and an aldehyde group: Propanal.
  1 mark
2. II. $C H_{3} - C \equiv C H$ has three carbons and a triple bond: Propyne.
  1 mark
3. (ii) I. The sweet-smelling compounds are esters (ester group, $- C O O -$ ).
  1 mark
4. II. Esterification:
  $C H_{3} C O O H + C H_{3} C H_{2} O H conc. H_{2} S O_{4} C H_{3} C O O C_{2} H_{5} + H_{2} O$
  The product is ethyl ethanoate.
  1 mark
5. (iii) Hard water contains calcium and magnesium salts. Soap reacts with them to form an insoluble precipitate called scum, so the soap is wasted and little foam forms.
  1 mark
Answer: (i) I. Propanal; II. Propyne. (ii) I. Esters; II. $C H_{3} C O O H + C_{2} H_{5} O H \to C H_{3} C O O C_{2} H_{5} + H_{2} O$ (conc. H $_{2}$ SO $_{4}$ ). (iii) Soap forms insoluble scum with Ca $^{2 +}$ /Mg $^{2 +}$ salts.
Where the marks are
- Names propanal (1 mark)
- Names propyne (1 mark)
- States that the products are esters (1 mark)
- Writes the esterification equation (1 mark)
- Explains scum formation with Ca and Mg salts (1 mark)
Open Q29(B) on its own page

Section C

Section C (Physics), 25 marks: MCQs 30-31, assertion-reason 32, SA 33-37, CBQ 38, LA 39

Q30 · 1 mark · MCQThe Human Eye and the Colourful World
The pair of eye parts responsible for admitting different amount of light in to the eyes is
1. (A)Iris and pupil
2. (B)Ciliary muscles and pupil
3. (C)Retina and Iris
4. (D)Ciliary muscles and cornea
Try it first, then show our solutionSolution
1. The iris adjusts the size of the pupil, and the pupil is the opening that lets light in. Together they control how much light enters the eye. Answer: option (A).
  1 mark
Answer: (A) Iris and pupil
Where the marks are
- Iris controls pupil size, option (A) (1 mark)
Q31 · 1 mark · MCQThe Human Eye and the Colourful World
The vision defect which arises due to gradual weakening of the ciliary muscles and diminishing flexibility of eye lens is:
1. (A)Myopia
2. (B)Hypermetropia
3. (C)Presbyopia
4. (D)Myopia and hypermetropia both
Try it first, then show our solutionSolution
1. Presbyopia is the age-related loss of the eye's power of accommodation. It is caused by weakening of the ciliary muscles and reduced flexibility of the eye lens. Answer: option (C).
  1 mark
Answer: (C) Presbyopia
Where the marks are
- Identifies presbyopia, option (C) (1 mark)
Q32 · 1 mark · Assertion-reasonCompetency-basedMagnetic Effects of Electric Current
Direction: For question number 32, two statements are given - one labelled as Assertion (A) and other labelled as Reason (R). Select the correct answer to this question from the codes (A), (B), (C) & (D) as given below.
Assertion (A): With the help of compass, magnetic field lines can be traced outside the bar magnet.
Reason (R): The north pole of a compass needle always points towards the south pole of the magnet when brought close to it.
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: field lines outside a bar magnet are traced by marking the successive positions of a compass needle.
2. R is true: unlike poles attract, so the needle's north pole points towards the magnet's south pole. The scheme treats R as the explanation of A. Answer: option (A).
  1 mark
Answer: (A)
Where the marks are
- Both true and R explains A, option (A) (1 mark)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(A) · 2 marks · Short answerCompetency-basedElectricity
A fuse in electric circuit is rated 4 A. Can it be used with an electric heater of rating 2 kW, 200 V? Explain your answer.
Try it first, then show our solutionSolution
1. Current drawn by the heater: $I = \frac{P}{V} = \frac{2000 W}{200 V}$
  ½ mark
2. $I = 10 A$
  ½ mark
3. The heater needs 10 A, which is much more than the fuse rating of 4 A. The fuse wire would melt and break the circuit as soon as the heater is switched on. So this fuse cannot be used with the heater.
  1 mark
Answer: No. The heater draws 10 A, which is more than 4 A, so the fuse would melt.
Where the marks are
- Uses $I = P / V$ with the values substituted (½ mark)
- Obtains I = 10 A (½ mark)
- Concludes that the fuse cannot be used because 10 A > 4 A and it would melt (1 mark)
Open Q33(A) on its own page
OR
Q33(B) · 2 marks · Very short answerMagnetic Effects of Electric Current
How is an electromagnet made? Write any two ways by which the strength of electromagnet can be increased.
Try it first, then show our solutionSolution
1. An electromagnet is made by winding many turns of insulated copper wire as a coil (solenoid) around a soft iron core and passing an electric current through it. The soft iron becomes a strong magnet while current flows.
  1 mark
2. The strength can be increased by:
  (i) increasing the number of turns in the coil;
  ½ mark
3. (ii) increasing the current flowing through the coil.
  ½ mark
Answer: A current-carrying insulated copper coil is wound on a soft iron core. Strength increases with more turns and with more current.
Where the marks are
- Describes making an electromagnet with a coil on a soft iron core carrying current (1 mark)
- Gives increasing the number of turns (½ mark)
- Gives increasing the current (½ mark)
Open Q33(B) on its own page
Q34 · 2 marks · Very short answerElectricity
Answer the following.
1. (a)
  Define volt, the unit of potential difference.
  0 marks
2. (b)
  Calculate the work done required to move an electron between two points A and B located at a potential difference of 100 V in an electric field.
  0 marks
Try it first, then show our solutionSolution
1. (a) One volt is the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other. So $1 V = \frac{1 J}{1 C}$ .
  1 mark
2. (b) Using $V = \frac{W}{Q}$ , so $W = Q V$ . The charge on an electron is $Q = 1.6 \times 1 0^{- 19}$ C and $V = 100$ V.
  ½ mark
3. $W = 1.6 \times 1 0^{- 19} \times 100 = 1.6 \times 1 0^{- 17} J$
  ½ mark
Answer: (a) 1 V = 1 J/1 C. (b) W = $1.6 \times 1 0^{- 17}$ J.
Where the marks are
- Defines one volt correctly (1 mark)
- Writes $W = Q V$ with the electron's charge (½ mark)
- Obtains $1.6 \times 1 0^{- 17}$ J (½ mark)
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Q35 · 3 marks · Short answerCompetency-basedLight – Reflection and Refraction
Answer the following.
1. (a)
  Observe the given ray diagram and compare: (i) the speed of light in medium A and the speed of light in medium B. (ii) the optical densities of medium B and medium C.
  0 marks
2. (b)
  Draw a ray diagram for refraction where the optical density of medium 'A' becomes equal to the optical density of medium 'C' and optical density of medium 'B' is greater than the optical density of medium 'A' and medium 'C'.
  0 marks
Figure in the paper: A ray of light travels from Medium 'A' (top) into Medium 'B' (a rectangular block) and then into Medium 'C' (below). The normal N is drawn at the upper surface, with angle i marked between the incident ray and the normal and angle r marked between the refracted ray and the normal. The ray bends towards the normal on entering Medium B and bends away from the normal on leaving it into Medium C. See it in the official paper.
Try it first, then show our solutionSolution
1. (a)(i) At the A–B surface the ray bends towards the normal ( $r < i$ ). So medium B is optically denser than A, and light slows down in B. Hence speed of light in A > speed of light in B.
  ½ mark
2. (a)(ii) Inside the ray diagram the ray continues from B into C without changing its direction. So B and C have the same optical density.
  ½ mark
3. (b) Ray diagram (glass-slab type: A = C, B denser). Draw a rectangular block B with medium A above it and medium C below it. Draw a normal at the top surface. Draw an incident ray in A, with an arrow, striking the top surface at angle $i$ .
  ½ mark
4. Inside B, draw the refracted ray bending towards the normal, with angle $r < i$ . This happens because B is denser than A.
  ½ mark
5. At the lower surface draw a second normal. Draw the emergent ray in C bending away from the normal, with an arrow. Since C has the same optical density as A, the angle of emergence equals $i$ .
  ½ mark
6. Show the emergent ray parallel to the incident ray but shifted sideways (lateral displacement). Mark the direction of the rays with arrows throughout.
  ½ mark
Answer: (a)(i) Speed in A > speed in B; (ii) B and C have equal optical density. (b) Ray bends towards the normal entering B and away from it entering C; the emergent ray is parallel to the incident ray and laterally displaced.
Where the marks are
- Speed of light in A greater than in B (½ mark)
- B and C have the same optical density (½ mark)
- Correct ray diagram: bending towards the normal in B, away from it in C, emergent ray parallel to incident ray, arrows shown (2 marks; ½ deducted if ray directions are not shown)
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Q36 · 3 marks · Short answerThe Human Eye and the Colourful World
Explain the phenomenon of rainbow formation in the sky. Draw the labelled ray diagram in support of your explanation.
Try it first, then show our solutionSolution
1. A rainbow is a natural spectrum seen in the sky after a shower, always on the side opposite to the Sun. It is caused by dispersion of sunlight by tiny water droplets in the atmosphere, which act like small prisms.
  1 mark
2. In each droplet, sunlight is first refracted and dispersed on entering. The light is then internally reflected at the back surface of the drop. Finally it is refracted again as it comes out. The colours are separated, so red and violet emerge at different angles and reach the observer to form the bands of the rainbow.
  1 mark
3. Labelled diagram: a spherical raindrop with a ray of sunlight entering it. The ray splits into colours at the first refraction, is internally reflected at the back of the drop, and is refracted again on emerging. Label the incoming sunlight, the raindrop, internal reflection, and the emergent red and violet rays.
  1 mark
Answer: Rainbow is formed by dispersion of sunlight by raindrops: refraction with dispersion, then internal reflection, then refraction again on leaving the drop; seen opposite the Sun.
Where the marks are
- Rainbow caused by dispersion of sunlight by water droplets acting as prisms (1 mark)
- Sequence: refraction + dispersion, internal reflection, refraction on emergence (1 mark)
- Correct labelled ray diagram of a raindrop showing red and violet rays (1 mark)
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Q37 · 3 marks · Short answerCompetency-basedMagnetic Effects of Electric Current
On the basis of the given circuit arrangement, answer the following questions:
(a) What happens to the rod AB when the key K is closed and why?
(b) Name and state the rule that determines this behaviour of the rod AB in the above situation.
Figure in the paper: A horseshoe magnet with N and S poles, with a rod AB (a straight conductor suspended horizontally between the magnet's poles from a stand, connected via flexible leads) placed between the poles. A battery (marked + and -) in series with a key K completes the circuit through the rod AB. Current direction in the rod is marked by an arrow from A to B, and the magnetic field lines of the horseshoe magnet are shown between the poles. See it in the official paper.
Try it first, then show our solutionSolution
1. (a) When key K is closed, current flows through rod AB and the rod gets displaced (moves) from its original position.
  ½ mark
2. This happens because a current-carrying conductor placed in a magnetic field experiences a force. The force acts perpendicular to both the current and the field.
  ½ mark
3. (b) The direction of this force is given by Fleming's Left-Hand Rule.
  1 mark
4. Statement: stretch the thumb, forefinger and middle finger of the left hand so that they are mutually perpendicular. If the forefinger points along the magnetic field and the middle finger along the current, then the thumb points in the direction of the force (motion) on the conductor.
  1 mark
Answer: (a) Rod AB is displaced because a current-carrying conductor in a magnetic field experiences a force. (b) Fleming's Left-Hand Rule, as stated.
Where the marks are
- Rod AB gets displaced (½ mark)
- Reason: force on current-carrying conductor in magnetic field (½ mark)
- Names Fleming's Left-Hand Rule (1 mark)
- Correct statement of the rule (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedLight – Reflection and Refraction
The above ray diagram is to show image formation by a reflecting telescope. Reflecting telescope revolutionized our way of looking into the sky. Reflecting telescopes employ large mirrors to gather and focus light rather than relying solely on lenses as we do in the refracting telescopes. These telescopes utilize precisely shaped and polished mirrors to capture incoming light and reflect it to a focal point, where the telescope forms an image for observation.
1. (a)
  What kind of image of a star is seen by the observer in the eye-piece?
  0 marks
2. (b)
  What kind of mirror is used in this reflecting telescope?
  0 marks
3. (c)
  Explain with the reason, what kind of optical device (type of lens or mirror) is used at the eye-piece.
  0 marks
4. OR
  (c)
  OR: What is the role of plane mirror in image formation by this telescope?
  0 marks
Figure in the paper: Ray diagram of a reflecting telescope. Parallel light rays from a distant star enter from the left and strike a large curved (concave) mirror on the right. The reflected rays converge towards a small plane mirror placed at an angle in the tube, which deflects them downwards out of the side of the tube to an eye-piece (a convex lens) at the bottom, where the rays emerge converging towards the observer. Labels: 'Parallel Light Rays from distant star', 'Curved Mirror', 'Plane Mirror', 'Eye-piece'. See it in the official paper.
Try it first, then show our solutionSolution
1. (a) The concave mirror forms a real image of the star near its focus. The eye-piece magnifies it, so the observer sees a magnified image of the star.
  1 mark
2. (b) The main mirror is a concave (converging) mirror. A plane mirror is also used to deflect the rays.
  1 mark
3. (c) A convex lens is used at the eye-piece.
  1 mark
4. Reason: the real image formed by the concave mirror lies within the focal length of the convex eye-piece, which then acts as a magnifier. It produces an enlarged image that the eye can view clearly.
  1 mark
5. (c) OR: The plane mirror intercepts the converging rays reflected by the concave mirror. It reflects (redirects) them sideways out of the tube towards the eye-piece, so the observer can view the image without blocking the incoming starlight.
  2 marks
Answer: (a) Real, magnified image; (b) concave (converging) mirror, along with a plane mirror; (c) convex lens, as it magnifies the image formed by the curved mirror. OR (c) The plane mirror redirects the reflected rays towards the eye-piece.
Where the marks are
- Image real/magnified (1 mark)
- Concave (converging) mirror, plane mirror (1 mark)
- Convex lens at eye-piece (1 mark)
- Reason: magnifies image formed by curved mirror (1 mark)
- OR: plane mirror reflects/redirects rays from curved mirror to eye-piece (2 marks)
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Answer one of these 2 alternatives.
Q39(A) · 5 marks · Long answerElectricity
Answer the following.
1. (i)
  The given electric circuit is a part of an electrical device. Use the information given in the electric circuit diagram to calculate: (I) Potential difference across the ends of resistor $R_{2}$ . (II) Value of resistor $R_{2}$ . (III) Value of resistor $R_{1}$ .
  0 marks
2. (ii)
  Write the factors on which resistance of a conductor depends and derive the formula for resistance of a given conductor.
  0 marks
Figure in the paper: Circuit diagram: a 12 V battery with key K in the top branch. Left branch contains resistor R1 with current 2.0 A marked downwards. At the bottom, the current splits into two parallel branches: the upper branch has a 4.0 Ω resistor carrying 1.5 A, and the lower branch contains resistor R2. The branches rejoin, and the 2.0 A current flows up through a 2.0 Ω resistor on the right side back to the battery. See it in the official paper.
Try it first, then show our solutionSolution
1. (i)(I) $R_{2}$ is in parallel with the 4.0 Ω resistor, so both have the same potential difference.
  $V = I_{1} R = 1.5 \times 4 = 6 V$
  1 mark
2. (II) Current through $R_{2}$ : $I_{2} = I - I_{1} = 2.0 - 1.5 = 0.5 A$ .
  $R_{2} = \frac{V}{I _{2}} = \frac{6}{0.5} = 12 Ω$
  1 mark
3. (III) Potential difference across the 2.0 Ω resistor: $V = I R = 2 \times 2 = 4 V$ .
  Potential difference across $R_{1}$ : $12 - (6 + 4) = 2 V$ .
  $R_{1} = \frac{V}{I} = \frac{2}{2} = 1 Ω$
  1 mark
4. (ii) Resistance of a conductor depends on:
  its length ( $R \propto l$ );
  its area of cross-section ( $R \propto \frac{1}{A}$ );
  the nature of its material (and its temperature).
  1 mark
5. Combining the two proportionalities: $R \propto \frac{l}{A}$ , so $R = ρ \frac{l}{A}$ .
  Here $ρ$ is the proportionality constant called the resistivity of the material.
  1 mark
Answer: (I) 6 V; (II) R₂ = 12 Ω; (III) R₁ = 1 Ω; (ii) R depends on length, area of cross-section and material, giving R = ρl/A.
Where the marks are
- P.d. across R₂ = 1.5 × 4 = 6 V (1 mark)
- I₂ = 0.5 A and R₂ = 12 Ω (1 mark)
- P.d. across 2 Ω = 4 V, across R₁ = 2 V, R₁ = 1 Ω (1 mark)
- Factors: length, area of cross-section (material) (1 mark)
- Derivation R = ρl/A with ρ as resistivity (1 mark)
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OR
Q39(B) · 5 marks · Long answerCompetency-basedElectricity
Answer the following.
1. (i)
  How much electric current will an electric iron draw from 220 V source if the resistance of its heating element when hot, is 55 $Ω$ ? Calculate the power consumed by the electric iron when it is operated at 220 V.
  0 marks
2. (ii)
  In a house, 3 bulbs of 100 watt each, are lit for 5 hours daily and an electric heater of 1.0 kW is used for half an hour daily. Calculate the total energy consumed in a month of 30 days and its cost at the rate of ₹3.60 per kWh.
  0 marks
3. (iii)
  With reason explain, why are alloys commonly used to make elements of electrical heating devices.
  0 marks
Try it first, then show our solutionSolution
1. (i) $I = \frac{V}{R} = \frac{220}{55} = 4 A$
  1 mark
2. $P = V I = 220 \times 4 = 880 W$
  1 mark
3. (ii) Using $E = P \times t$ :
  Bulbs: $3 \times 100 \times 5 = 1500 Wh = 1.5 kWh$ per day
  Heater: $1.0 \times 0.5 = 0.5 kWh$ per day
  Total per day $= 2 kWh$
  1 mark
4. Energy in 30 days $= 30 \times 2 = 60 kWh = 60$ units.
  Cost $= 60 \times 3.60 = ₹ 216$
  1 mark
5. (iii) Alloys are used because their resistivity is higher than that of their constituent metals, so they produce more heat. They also do not oxidise (burn) readily at high temperatures.
  1 mark
Answer: (i) 4 A, 880 W; (ii) 60 kWh, ₹216; (iii) Alloys have high resistivity and do not oxidise readily at high temperature.
Where the marks are
- I = 220/55 = 4 A (1 mark)
- P = 220 × 4 = 880 W (1 mark)
- Daily energy 1.5 + 0.5 = 2 kWh (1 mark)
- Monthly 60 kWh and cost ₹216 (1 mark)
- Alloys: higher resistivity / do not oxidise at high temperature (1 mark)
Open Q39(B) on its own page

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