Detailed Solutions

Detailed Solutions: CBSE Class 10 Science Sample Paper 2027 – Set 2

Every written question from the paper (2, 3, 4 and 5 marks) is solved here in order, with the marks given for each step, so you can check your answer sheet the way an examiner would. Where a question has an internal choice (OR), both options are solved. The 1-mark questions are answered, with explanations, in the quiz on the Question Paper page.

Section A - Biology (30 marks)

Question 10 (2 marks)

Answer the following:

(a) Some pepsin was added to two test tubes, each containing small pieces of boiled egg white in water. A few drops of dilute hydrochloric acid were also added to the first tube only. After an hour at body temperature, the egg white had been digested only in the first tube. What does this show about the role of hydrochloric acid in the stomach? (1 mark)

Answer.

1. Pepsin, the protein-digesting enzyme of the stomach, acts only in an acidic medium — 0.5 marks
2. Hydrochloric acid in the stomach creates this acidic medium, so that pepsin can digest proteins — 0.5 marks

(b) A colourless fluid that has leaked out of the blood capillaries bathes the cells of the body. It later gathers into small vessels of its own and finally drains back into the blood. Name this fluid and give one of its functions. (1 mark)

Answer.

1. Lymph (tissue fluid) — 0.5 marks
2. Any one: it carries digested and absorbed fat from the intestine; it drains the excess fluid from the spaces between cells back into the blood — 0.5 marks

Question 11 (2 marks)

In a mango orchard near Ratnagiri, a gardener removed a complete ring of bark, including the phloem, from one branch of a tree, leaving the wood (xylem) in place. The leaves on that branch stayed fresh and green for many weeks, but the bark just above the ring slowly became swollen. Explain these two observations, bringing out two differences between transport in the xylem and transport in the phloem.

Answer.

1. Leaves stay fresh: the xylem was not cut, so water and minerals still reach them from the roots (0.5); the bark swells above the ring: food made in the leaves moves down in the phloem, cannot cross the ring, and collects just above it (0.5) — 1 mark
2. Differences (0.5 each): the xylem carries water and minerals, only upwards from the roots, while the phloem carries food (sucrose) upwards or downwards to where it is needed; transport in the xylem is mainly by transpiration pull, without the plant spending energy, while translocation in the phloem uses energy from ATP; also accept: the xylem conducts through vessels and tracheids, the phloem through sieve tubes and companion cells — 1 mark

Question 12 (2 marks)

Attempt either option (A) or (B).

(A) One summer, the water of a lake became much warmer than usual. Most fish of one kind died, but a few survived and went on to breed. Explain how variation among the individuals of a species helps it to survive such a change. Why is a way of reproduction that produces more variation useful to a species in the long run?

Answer.

1. Individuals of a species are not all alike; some of them may, by chance, have variations that let them tolerate the warmer water, so they survive and breed when the others die — 1 mark
2. If a species produces more variation, it is more likely that some individuals will survive whenever the conditions change, so the species is not wiped out — 1 mark

OR

(B) A variety of rose raised only from stem cuttings for many years is badly hit when a new disease spreads, because every plant is affected. A plant breeder suggests raising new plants from seeds produced by crossing two different varieties. Give two advantages of sexual reproduction that support this suggestion.

Answer.

1. Sexual reproduction combines DNA from two different parents, so the offspring have new combinations of characters (more variation) — 1 mark
2. Any one: with more variation, some of the new plants are likely to resist the disease or cope with changed conditions, so the variety is not wiped out; useful traits of the two parent varieties can be brought together in one plant — 1 mark

Question 13 (3 marks)

Nandini peeled the thin lower skin of a leaf picked at 10 am on a mild morning and looked at it under a microscope. Most of the stomatal pores were open. When she did the same with a leaf picked from the same plant at 2 pm on a very hot, dry afternoon, most of the pores were closed.

(a) Draw a neat diagram of a stomatal pore as Nandini saw it at 2 pm, together with the cells next to it. Label any four of these parts: guard cell, stomatal pore, chloroplast, nucleus, epidermal cell. Use arrows to show the direction in which water moved in the guard cells as the pore reached this state. (2 marks)

Answer.

1. Correct diagram of a closed stomatal pore between two shrunken guard cells — 0.5 marks
2. Arrows showing water moving out of the guard cells into the neighbouring cells — 0.5 marks
3. Any four parts correctly labelled, 0.25 each (in the figure: 1 guard cell, 2 stomatal pore (closed), 3 chloroplast, 4 nucleus, 5 epidermal cell; the arrows show water leaving the guard cells) — 1 mark

(b) How does this movement of water close the pore? How did closing the pores help the plant at 2 pm? (1 mark)

Answer.

1. As the guard cells lose water they shrink and straighten, so the pore between them closes (when they take in water they swell and curve, and the pore opens) — 0.5 marks
2. On the hot, dry afternoon the plant was losing a lot of water by transpiration; closing the pores cut down this loss of water — 0.5 marks

Question 14 (3 marks)

The figure shows a food chain in a lake near Udaipur. The letters stand for: A - phytoplankton, B - zooplankton, C - small fish, D - large fish, E - kingfisher. Each arrow points from the organism that is eaten to the organism that eats it.

(a) The phytoplankton trap 1,00,000 J of energy from sunlight. Using the 10 per cent law, find the energy available to D and to E. (1 mark)

Answer.

1. D: 1,00,000 → 10,000 → 1,000 → 100 J — 0.5 marks
2. E: 100 → 10 J — 0.5 marks

(b) A pesticide washed into the lake from nearby fields. Tests gave these amounts of the pesticide, in parts per million (ppm):

Name the process shown by these values. Why do the phytoplankton (A) already hold twice as much pesticide as the lake water in which they float? (1 mark)

Answer.

1. Biological magnification — 0.5 marks
2. The phytoplankton keep taking in the pesticide from the water along with water and minerals, but cannot break it down or get rid of it, so it goes on collecting in their bodies — 0.5 marks

(c) Only a small part of the energy in D reaches E. What happens to most of the rest? (1 mark)

Answer.

1. Most of it is used by D for its own life processes such as respiration, movement and growth, and is lost as heat — 0.5 marks
2. Part of it is in parts of D that are not eaten or digested, and in its dead remains, and passes to the decomposers — 0.5 marks

Question 15 (4 marks)

Kiran, a Class 10 student in Dehradun, crossed pea plants from a pure line with yellow seeds with plants from a pure line with green seeds in her school garden. All the seeds of the F1 generation were yellow. She grew F1 plants from these seeds and let them self-pollinate. The seeds that formed in the pods of these F1 plants make up the F2 generation. She counted them:

(a) Which seed colour is dominant? How does the F1 generation show this? (1 mark)

Answer.

1. Yellow is dominant: every F1 seed was yellow, although each F1 plant had received the factor for green from one parent — 1 mark

(b) Work out the ratio of yellow to green seeds in the F2 generation from Kiran's counts. What does the reappearance of green seeds tell us? (1 mark)

Answer.

1. 615 : 201 is about 3 : 1 — 0.5 marks
2. The factor for green was present in the F1 plants without showing; it was passed on unchanged and shows up again in F2 — 0.5 marks

(c) Using Y for yellow and y for green, show the cross between two F1 plants. In what ratio are the two kinds of yellow seeds expected in F2, and about how many of Kiran's 615 yellow seeds are expected to be pure-breeding? (2 marks)

Answer.

1. Yy × Yy: gametes Y and y from each parent, giving YY : Yy : yy = 1 : 2 : 1; so the yellow seeds are YY and Yy in the ratio 1 : 2 — 1 mark
2. Pure-breeding yellow (YY) $= \frac{1}{3} \times 615 = 205$ seeds (about) — 1 mark

OR

(c) Traits such as seed colour and plant height are controlled by genes. Explain, with the example of the height of a plant, how a gene controls a trait. (2 marks)

Answer.

1. A gene is a section of DNA that carries the information for making a particular protein, often an enzyme — 1 mark
2. If the gene makes an efficient enzyme, the plant makes a lot of the growth hormone and grows tall; if the gene is altered and the enzyme works less well, less hormone is made and the plant stays short — 1 mark

Question 16 (5 marks)

Attempt either option (A) or (B).

(A) The brain is the main coordinating centre of the body.

(i) Draw a neat diagram of the human brain and label any four of these parts: forebrain, midbrain, pons, cerebellum, medulla. (2 marks)

Answer.

1. Correct diagram of the human brain — 1 mark
2. Any four parts correctly labelled, 0.25 each (in the figure: 1 forebrain, 2 midbrain, 3 pons, 4 cerebellum, 5 medulla) — 1 mark

(ii) Three patients have had head injuries. For each, name the part of the brain most likely to be damaged, and state the normal function of that part.

(I) Patient P has strong muscles, but cannot walk in a straight line or pick up a pencil smoothly.
(II) In patient Q, the blood pressure and the flow of saliva are no longer regulated properly.
(III) Patient R can see and hear, but cannot think through a simple problem or remember what she learnt that morning. (3 marks)

Answer.

Model answer:

(I) Patient P's cerebellum is probably damaged. The cerebellum, in the hind-brain, keeps the posture and balance of the body and makes voluntary actions precise, such as walking in a straight line or picking up a pencil. (II) Patient Q's medulla is probably damaged. The medulla, also in the hind-brain, controls involuntary actions such as blood pressure, salivation and vomiting. (III) Patient R's forebrain is probably damaged. The forebrain is the main thinking part of the brain. It has areas for thinking, memory and voluntary actions, and areas that make sense of the messages coming from the eyes and ears.

Marking scheme:

1. (I) Cerebellum (0.5): it maintains posture and balance and is responsible for the precision of voluntary actions (0.5) — 1 mark
2. (II) Medulla (0.5): it controls involuntary actions such as blood pressure, salivation and vomiting (0.5) — 1 mark
3. (III) Forebrain (cerebrum) (0.5): it is the main thinking part of the brain, with areas for memory, reasoning and voluntary actions and for making sense of what we see and hear (0.5) — 1 mark

OR

(B) Anusha, a student in Hubballi, did two things in her school garden. She touched a leaflet of a sensitive plant (touch-me-not): within two seconds the leaflets folded up, and about twenty minutes later they had opened again. She also pushed a thin stick into the soil beside a young bitter-gourd plant. Over the next three days, a tendril that touched the stick coiled tightly round it.

(i) Anusha's friend says, "The leaflets folded because the cells on one side grew faster, just as in the tendril." Is the friend right? Explain how the leaflets fold after being touched, how the message of the touch is passed on, and why the leaflets can open again within minutes. (3 marks)

Answer.

Model answer:

The friend is not right, because the folding of the leaflets does not involve any growth. When Anusha touched a leaflet, the information of the touch was passed on from cell to cell by electrical-chemical means, since plants have no nerves. The cells in the affected region then changed their shape by changing the amount of water in them: they lost water and shrank, and the leaflets folded. As no new growth had taken place, the cells could take in water again after some time, swell back to their earlier shape, and the leaflets opened.

Marking scheme:

1. No: the folding is not a growth movement — 0.5 marks
2. Plants have no nervous tissue; the plant cells pass the information of the touch from cell to cell by electrical-chemical means — 1 mark
3. The cells in the affected region change shape by changing the amount of water in them (they swell or shrink), and this folds the leaflets — 1 mark
4. No growth takes place, so when water moves back into these cells, the leaflets open again — 0.5 marks

(ii) If the stick is pulled out after a week, will the tendril uncoil? Give a reason. State one more way in which the movement of the tendril differs from the folding of the leaflets. (2 marks)

Answer.

Model answer:

No, the tendril will not uncoil. It coiled because the part of the tendril away from the stick grew faster than the part in contact with it, and a growth movement is permanent. The tendril's movement is also slow, taking hours or days, and it is directional: the tendril bends towards the support. The folding of the leaflets is quick, taking only seconds, and does not depend on the direction from which the plant is touched.

Marking scheme:

1. No: the tendril coiled by growth, the part away from the stick growing faster than the part touching it, and growth cannot be undone — 1 mark
2. Any one: the tendril moves slowly (over hours or days), the leaflets in seconds; the tendril's movement depends on the direction of the stimulus (it coils towards the support), while the folding of the leaflets does not — 1 mark

Section B - Chemistry (25 marks)

Question 25 (2 marks)

Rehana dropped a small piece of a silvery-white metal M into a beaker of cold water. The metal reacted steadily but not violently. Bubbles of a colourless gas formed on its surface, and after a while the piece began to float.

(a) Identify M. Give one reason why M cannot be sodium, and one reason why it cannot be magnesium. (1 mark)

Answer.

1. M is calcium — 0.5 marks
2. Sodium would react violently and the hydrogen would catch fire; magnesium does not react with cold water — 0.5 marks

(b) Write the balanced chemical equation for the reaction, and describe a test for the gas. Why does the metal float? (1 mark)

Answer.

1. $\mathrm{Ca + 2H_2O \rightarrow Ca(OH)_2 + H_2}$ — 0.5 marks
2. The gas burns with a pop sound when a burning splinter is brought near it (hydrogen); the metal floats because bubbles of hydrogen stick to its surface — 0.5 marks

Question 26 (3 marks)

Pieces of four metals A, B, C and D were placed in solutions of salts of the other metals. The results are shown below.

(a) Arrange the four metals in decreasing order of reactivity. Explain how you used the table. (2 marks)

Answer.

1. A more reactive metal displaces a less reactive metal from its salt solution, so a deposit forms only when the metal is more reactive than the metal in the salt — 1 mark
2. B displaces all three, D displaces A and C, A displaces only C, and C displaces none; so the order is B > D > A > C — 1 mark

(b) Which of the other metals could be used to make a container for storing the salt solution of A? Give a reason. (1 mark)

Answer.

1. Only C — 0.5 marks
2. C is less reactive than A, so it cannot displace A from its salt; B and D would react with the solution — 0.5 marks

Question 27 (3 marks)

Attempt either option (A) or (B).

(A) Write the balanced chemical equation, with the physical states, for each of the following, and name the type of reaction.

(i) Hydrogen gas and chlorine gas combine in sunlight to give hydrogen chloride gas. (1 mark)

Answer.

1. $\mathrm{H_2(g) + Cl_2(g) \rightarrow 2HCl(g)}$ — 0.5 marks
2. Combination reaction — 0.5 marks

(ii) A strip of magnesium is dipped into copper sulphate solution; the blue colour fades and a reddish-brown coating forms on the strip. (1 mark)

Answer.

1. $\mathrm{Mg(s) + CuSO_4(aq) \rightarrow MgSO_4(aq) + Cu(s)}$ — 0.5 marks
2. Displacement reaction — 0.5 marks

(iii) Sodium carbonate solution is mixed with calcium chloride solution, and a white precipitate forms. (1 mark)

Answer.

1. $\mathrm{Na_2CO_3(aq) + CaCl_2(aq) \rightarrow CaCO_3(s) + 2NaCl(aq)}$ — 0.5 marks
2. Double displacement (precipitation) reaction — 0.5 marks

OR

(B) A family keeps a pair of silver anklets in a drawer, and after some months they have turned black. A packet of namkeen left open in the kitchen for many days begins to smell and taste bad.

(i) Name the process that has taken place in the anklets, and the black substance formed on them. (1 mark)

Answer.

1. Corrosion (tarnishing) of silver — 0.5 marks
2. Silver sulphide, formed by the reaction of silver with sulphur (compounds) in the air — 0.5 marks

(ii) Name the process that has taken place in the namkeen, and say why it happens. (1 mark)

Answer.

1. Rancidity — 0.5 marks
2. The fats and oils in the namkeen are oxidised by the oxygen of the air — 0.5 marks

(iii) Give two ways to prevent each of these changes. (1 mark)

Answer.

1. Silver: any two, 0.25 each, e.g. keep it wrapped in an airtight pouch or box; keep it away from air and moisture; clean and polish it regularly — 0.5 marks
2. Namkeen: any two, 0.25 each, e.g. keep it in an airtight container; add antioxidants; fill the packet with nitrogen; keep it in a cool place — 0.5 marks

Question 28 (4 marks)

In a school laboratory in Guwahati, Mrinal's group made some indicators of their own. They rubbed turmeric paste on strips of filter paper, boiled red cabbage leaves in water to get a purple extract, and kept strips of cloth with finely chopped onions in a closed plastic bag overnight. They also had a small bottle of clove oil. They tested two colourless solutions, X and Y, from the teacher's table and noted:

(a) Which of the two solutions is acidic? Give the observation from the table that shows it. (1 mark)

Answer.

1. Y is acidic (0.5), because it turns the purple red cabbage extract red (0.5) — 1 mark

(b) Why are onion and clove oil called olfactory indicators? What would happen to the smell of clove oil in solution X? (1 mark)

Answer.

1. Their smell changes (is lost) in a basic medium, so we can tell an acid from a base by smell — 0.5 marks
2. Solution X is basic, so the smell of clove oil would disappear — 0.5 marks

(c) Mrinal says, "Turmeric paper alone is enough to show whether a solution is acidic." Is he right? Explain using the table. Which of the group's indicators would suit a visually impaired student, and why? (2 marks)

Answer.

1. No: turmeric changes colour only in a base; it stays yellow in an acid and also in a neutral solution, so 'no change' does not prove that a solution is acidic — 1 mark
2. An olfactory indicator such as onion or clove oil, because the change can be detected by smell — 1 mark

OR

(c) The teacher tells them that X is sodium hydroxide solution and Y is dilute sulphuric acid. To some of X coloured with red cabbage extract, they add Y drop by drop. The colour changes from green to purple and then to red. Explain these colour changes, and write the balanced chemical equation for the reaction. (2 marks)

Answer.

1. Green: the solution is basic; purple: the acid has just neutralised the base, so the solution is neutral; red: extra acid makes the solution acidic — 1 mark
2. $\mathrm{2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O}$ — 1 mark

Question 29 (5 marks)

Attempt either option (A) or (B).

(A) Answer the following about naming carbon compounds and their isomers.

(i) Priya named five compounds as shown. Check each name. Where it is wrong, write the correct name. Also name the functional group in each compound.

(I) $\mathrm{CH_3CH_2Br}$: bromoethane
(II) $\mathrm{CH_3CH_2CH_2OH}$: propanone
(III) $\mathrm{HCHO}$: methanal
(IV) $\mathrm{CH_3COCH_3}$: propanal
(V) $\mathrm{CH_3CH_2COOH}$: butanoic acid (3 marks)

Answer.

Model answer:

(I) $\mathrm{CH_3CH_2Br}$ has two carbon atoms and a bromine atom, so bromoethane is correct; it has the halo (bromo) group. (II) $\mathrm{CH_3CH_2CH_2OH}$ has three carbon atoms and an -OH group, so it is propanol, not propanone; it has the alcohol group. (III) $\mathrm{HCHO}$ has one carbon atom and a -CHO group, so methanal is correct; it has the aldehyde group. (IV) $\mathrm{CH_3COCH_3}$ has the C=O group on the middle carbon, so it is a ketone, propanone, not propanal. (V) $\mathrm{CH_3CH_2COOH}$ has only three carbon atoms, counting the carbon of the -COOH group, so it is propanoic acid, not butanoic acid; it has the carboxylic acid group.

Marking scheme:

1. (I) Correct; halo (bromo) group — 0.5 marks
2. (II) Wrong: propanol; alcohol group — 0.5 marks
3. (III) Correct; aldehyde group — 0.5 marks
4. (IV) Wrong: propanone; ketone group — 0.5 marks
5. (V) Wrong: propanoic acid, because the carbon atom of the -COOH group is counted in the chain of three carbon atoms (0.5); carboxylic acid group (0.5) — 1 mark

(ii) Pentane, $\mathrm{C_5H_{12}}$, has three structural isomers. Draw their structures, showing all the atoms. What do the three have in common, and how do they differ? (2 marks)

Answer.

1. Three correct structures: a straight chain of five carbon atoms, a chain of four with one branch, and one carbon atom joined to four others (0.5 each) — 1.5 marks
2. All three have the same molecular formula $\mathrm{C_5H_{12}}$ but a different arrangement of the carbon atoms — 0.5 marks

OR

(B) Answer the following about soaps and the oxidation of ethanol.

(i) Draw a diagram of a soap micelle formed round a drop of oil in water, and label the oil drop, the hydrocarbon tail, the ionic end and the water. Why does soap give very little lather and form a scum in hard water? (3 marks)

Answer.

Model answer:

In water, soap molecules gather round a drop of oil with their long hydrocarbon tails inside the oil and their ionic ends outside, facing the water. This cluster is a micelle, and it can be rinsed away with the oil inside it. Hard water contains calcium and magnesium salts. These react with soap to form an insoluble substance called scum. So a large amount of soap is used up in forming scum, and very little lather is formed.

Marking scheme:

1. Correct micelle: the tails of the soap molecules pointing into the oil drop and the ionic ends facing the water all round — 1 mark
2. Four labels, 0.25 each (in the figure: 1 oil drop, 2 hydrocarbon tail, 3 ionic end, 4 water) — 1 mark
3. The calcium and magnesium salts in hard water react with soap to form an insoluble substance (scum), so much of the soap is wasted and little lather forms — 1 mark

(ii) Vikram added alkaline potassium permanganate solution drop by drop to warm ethanol in one test tube and to warm ethanoic acid in another. In the ethanol tube the purple colour disappeared as each of the first few drops went in; in the ethanoic acid tube the colour stayed from the very first drop. His friend says, "The colour vanished in the ethanol tube only because the permanganate got diluted." Is the friend right? Explain the two observations, and write the chemical equation for the change in the ethanol. (2 marks)

Answer.

Model answer:

The friend is not right. If the colour had vanished only because of dilution, it would also have faded in the ethanoic acid tube, where the same drops were added; but there the purple colour stayed. Alkaline potassium permanganate is an oxidising agent. In the first tube it adds oxygen to the ethanol and is used up, so its colour disappears: $\mathrm{CH_3CH_2OH \rightarrow CH_3COOH}$ (with alkaline $\mathrm{KMnO_4}$ and heat). Ethanoic acid is itself the product of this oxidation, so it does not react with the permanganate, and the purple colour remains from the first drop.

Marking scheme:

1. The friend is wrong: dilution would make the colour fade in the ethanoic acid tube too, but there it stays (0.5); the permanganate is an oxidising agent that is used up in oxidising ethanol, while ethanoic acid is already the oxidised product, so nothing uses up the permanganate there (0.5) — 1 mark
2. $\mathrm{CH_3CH_2OH \rightarrow CH_3COOH}$ (with alkaline $\mathrm{KMnO_4}$ and heat): ethanol is oxidised to ethanoic acid — 1 mark

Section C - Physics (25 marks)

Question 33 (2 marks)

Attempt either option (A) or (B).

(A) Mr Menon, who is seventy, has worn concave lenses for years to see distant objects clearly. Now he also has to hold his newspaper at arm's length to read it.

(I) Name the defect that has now developed, and state its cause. (1 mark)

Answer.

1. Presbyopia — 0.5 marks
2. With age, the ciliary muscles weaken and the eye lens loses its flexibility, so the near point moves away — 0.5 marks

(II) What kind of spectacle lens will his doctor now prescribe? Describe how it is made up. (1 mark)

Answer.

1. A bifocal lens — 0.5 marks
2. Its upper part is a concave lens for distant vision and its lower part is a convex lens for reading — 0.5 marks

OR

(B) Aman says, "The sky looks blue because the blue sea reflects its colour on to the sky."

(I) Give one observation which shows that Aman is wrong. (1 mark)

Answer.

1. Any one: a clear sky looks just as blue over a desert or a place hundreds of kilometres from any sea; the sea is often grey or green while the sky above is blue — 1 mark

(II) Explain the real reason why a clear sky looks blue. (1 mark)

Answer.

1. The molecules of air and other fine particles in the atmosphere are smaller than the wavelength of visible light — 0.5 marks
2. They scatter blue light (shorter wavelength) much more strongly than red light, and this scattered blue light reaches our eyes from all parts of the sky — 0.5 marks

Question 34 (2 marks)

In an activity, Sneha connected a nichrome wire in a circuit with a battery, an ammeter and a voltmeter across the wire. She changed the number of cells and recorded:

(a) Show from her readings that the wire obeys Ohm's law, and find its resistance. (1 mark)

Answer.

1. $\frac{V}{I} = \frac{1.0}{0.2} = \frac{2.0}{0.4} = \frac{3.0}{0.6} = \frac{4.0}{0.8} = 5$ for every reading, so V is proportional to I — 0.5 marks
2. Resistance R = 5 Ω — 0.5 marks

(b) What is the shape of the graph of V (along the y-axis) against I (along the x-axis)? Her friend's readings for another wire gave a similar graph with a smaller slope. Which wire has the lower resistance? Give one way in which the friend's wire may differ from Sneha's. (1 mark)

Answer.

1. A straight line passing through the origin — 0.5 marks
2. The slope of this graph is V/I = R, so the friend's wire has the lower resistance; it may be shorter, or thicker, or made of a material of lower resistivity (any one) — 0.5 marks

Question 35 (3 marks)

On a sunny morning in Shillong, just after a shower, Daphira saw a rainbow in the sky.

(a) In which part of the sky, east or west, did she see the rainbow? Give a reason. (1 mark)

Answer.

1. In the western sky — 0.5 marks
2. A rainbow is always seen on the side opposite to the Sun; in the morning the Sun is in the east, behind her — 0.5 marks

(b) Daphira notices that red is always on the outer (upper) edge of the bow and violet on the inner edge. Draw a labelled diagram of the path of sunlight through one raindrop, showing the red and violet light that leaves it, and use your diagram to explain her observation. (2 marks)

Answer.

1. Correct diagram: sunlight refracted and split on entering the drop, reflected inside at the back, and refracted again on leaving, with red and violet marked (in the figure: 1 sunlight, 2 refraction and dispersion, 3 internal reflection, 4 red, 5 violet) — 1 mark
2. Red light leaves each drop at a larger angle to the incoming sunlight (about 42 degrees) than violet (about 40 degrees); so the drops that send red to her eye lie higher in the sky than those that send violet, and red forms the outer edge — 1 mark

Question 36 (3 marks)

Hritik passed a thick copper wire vertically through the centre of a horizontal sheet of cardboard and sprinkled iron filings evenly on the cardboard. He switched on a current flowing upwards through the wire and tapped the cardboard gently.

(a) Draw the pattern formed by the iron filings, as seen from above, and show the direction of the magnetic field lines. (1 mark)

Answer.

1. Concentric circles with the wire at their centre — 0.5 marks
2. Arrows showing an anticlockwise direction, as seen from above, for the upward current — 0.5 marks

(b) Name and state the rule you used to find the direction of the field lines. (1 mark)

Answer.

1. The right-hand thumb rule — 0.5 marks
2. Hold the wire in the right hand with the thumb pointing in the direction of the current; the fingers wrapped round the wire give the direction of the magnetic field lines — 0.5 marks

(c) P is a point on the cardboard 4 cm from the wire. How will the magnetic field at P change if (I) the current is doubled, (II) the point is taken 8 cm from the wire instead, with the same current? (1 mark)

Answer.

1. (I) It becomes stronger (about double) — 0.5 marks
2. (II) It becomes weaker, since the field decreases as the distance from the wire increases — 0.5 marks

Question 37 (3 marks)

The figure shows the magnetic field lines round a bar magnet whose ends are marked X and Y. P, Q and R are three points near the magnet.

(a) At which of the points P, Q and R is the magnetic field the strongest? How does the figure show this? (1 mark)

Answer.

1. At P — 0.5 marks
2. The field lines are closest together (most crowded) near P — 0.5 marks

(b) Which end of the magnet, X or Y, is its north pole? Explain using the figure. (1 mark)

Answer.

1. X is the north pole — 0.5 marks
2. Outside a magnet, field lines come out of the north pole and go into the south pole; in the figure they come out of X and go into Y — 0.5 marks

(c) A student draws two field lines of a magnet crossing each other at a point. Explain why this can never happen. (1 mark)

Answer.

1. The direction of the field at a point is given by the tangent to the field line there; if two lines crossed, a compass needle at that point would have to point in two directions at once, which is impossible — 1 mark

Question 38 (4 marks)

For its annual exhibition, the science club of a school in Kochi built a simple slide projector. A bright lamp lights up a small transparent slide, and a convex lens of focal length 20 cm forms an image of the slide on a white screen. Anna, the club secretary, puts the slide in upside down, places it 25 cm from the lens and moves the screen until the picture is sharp.

(a) State the nature of the image formed on the screen. Why does Anna put the slide in upside down? (1 mark)

Answer.

1. Real, inverted and enlarged (0.5); since the image is inverted (upside down and left-right reversed), a slide put in inverted gives an upright picture (0.5) — 1 mark

(b) Find the power of the lens. (1 mark)

Answer.

1. $P = \frac{1}{f}$ with f in metres $= \frac{1}{0.20} = +5$ D — 1 mark

(c) How far from the lens must the screen be? How many times larger than the slide is the picture? (2 marks)

Answer.

1. $u = -25$ cm, $f = +20$ cm; $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} - \frac{1}{25} = \frac{1}{100}$, so $v = 100$ cm — 1 mark
2. $m = \frac{v}{u} = \frac{100}{-25} = -4$: the picture is 4 times as large as the slide (and inverted) — 1 mark

OR

(c) Draw a ray diagram to show how the lens forms the image of the slide placed between F1 and 2F1. (2 marks)

Answer.

1. A ray parallel to the principal axis passing through F2 after refraction, and a ray through the optical centre going straight on (0.5 each) — 1 mark
2. The rays meeting beyond 2F2 to give a real, inverted and enlarged image — 1 mark

Question 39 (5 marks)

Attempt either option (A) or (B).

(A) Study the circuit shown. The battery gives 6 V, and the key K is closed.

(i) Work out the equivalent resistance of the whole circuit, and hence the reading of the ammeter. (1 mark)

Answer.

1. Parallel part: $R_p = \frac{20 \times 20}{20 + 20} = 10$ Ω; total $= 10 + 10 = 20$ Ω — 0.5 marks
2. $I = \frac{V}{R} = \frac{6}{20} = 0.3$ A — 0.5 marks

(ii) Find the reading of the voltmeter and the current through each 20 Ω resistor. (1 mark)

Answer.

1. $V_{MN} = I R_p = 0.3 \times 10 = 3$ V — 0.5 marks
2. Current in each 20 Ω resistor $= \frac{3}{20} = 0.15$ A — 0.5 marks

(iii) Find the power used in the 10 Ω resistor. (1 mark)

Answer.

1. $P = I^2 R = 0.3^2 \times 10 = 0.9$ W — 1 mark

(iv) One of the 20 Ω resistors burns out and breaks. Find the new readings of the ammeter and the voltmeter. Does the 10 Ω resistor now use more power or less? (2 marks)

Answer.

1. Total resistance $= 10 + 20 = 30$ Ω, so $I = \frac{6}{30} = 0.2$ A — 1 mark
2. $V_{MN} = 0.2 \times 20 = 4$ V (0.5); power in the 10 Ω resistor $= 0.2^2 \times 10 = 0.4$ W, i.e. less (0.5) — 1 mark

OR

(B) A room heater has two identical nichrome coils of 110 Ω each. A switch connects the two coils either in series or in parallel to the 220 V mains.

(i) Find the power of the heater when the coils are in series. (1 mark)

Answer.

1. $R = 110 + 110 = 220$ Ω; $P = \frac{V^2}{R} = \frac{220 \times 220}{220} = 220$ W — 1 mark

(ii) Find the power of the heater when the coils are in parallel. (1 mark)

Answer.

1. $R = \frac{110}{2} = 55$ Ω; $P = \frac{V^2}{R} = \frac{220 \times 220}{55} = 880$ W — 1 mark

(iii) Which setting heats the room faster? Find the current drawn from the mains in that setting. (1 mark)

Answer.

1. The parallel setting, as it gives more heat every second (0.5); $I = \frac{P}{V} = \frac{880}{220} = 4$ A (0.5) — 1 mark

(iv) In the series setting both coils are working, yet the heater gives only a quarter of the power it gives in the parallel setting. Explain why, by finding the potential difference across each coil and the power used in each coil in the two settings. (2 marks)

Answer.

Model answer:

In the series setting the total resistance is 220 Ω, so the current is $\frac{220}{220} = 1$ A. The supply is shared equally, so each coil has only 110 V across it and uses $110 \times 1 = 110$ W; the two together use 220 W. In the parallel setting each coil is connected directly across the 220 V mains, carries 2 A and uses $220 \times 2 = 440$ W, so the two use 880 W. Both coils are working in either case, but in series each coil gets half the potential difference and so half the current. Since $P = \frac{V^2}{R}$, each coil then gives only one-quarter of the power.

Marking scheme:

1. Series: the current is $\frac{220}{220} = 1$ A, so each coil has only 110 V across it and uses $110 \times 1 = 110$ W (0.5); parallel: each coil has the full 220 V across it, carries 2 A and uses $220 \times 2 = 440$ W (0.5) — 1 mark
2. Halving the potential difference across a coil also halves the current through it, so its power $P = \frac{V^2}{R}$ falls to one-quarter; hence 220 W in all instead of 880 W — 1 mark