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CBSE Class 10 Science 2026 — Paper 31/4/2

80 marks · 3 hours · 39 questions (45 with the OR choices)

32 of its 45 questions also appear in 31/4/1 or 31/4/3; 13 are only in this paper.

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Questions 1–9 · 1 mark each

  1. Q1 1 mark

    Also in 31/4/1 (Q2), 31/4/3 (Q3)

    If the part 'P' of the flower is removed, which of the following processes will not take place ?
    P
    1. (a)Formation of pollens
    2. (b)Formation of flower
    3. (c)Formation of fruit
    4. (d)Formation of seed
    Show answer
    (c) Formation of fruit
    Step-by-step solution →
  2. Q2 1 mark

    Also in 31/4/1 (Q3), 31/4/3 (Q1)

    Given below are some of the statements regarding our environment :
    1. (i)Change in our life-styles have resulted in greater amounts of waste material generation.
    2. (ii)Changes in attitude have resulted in use of more disposable things.
    3. (iii)Changes in packaging has resulted in much of our waste becoming non-biodegradable.
    4. (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)
    Show answer
    (b) (iv)
    Step-by-step solution →
  3. Q3 1 mark

    Also in 31/4/1 (Q5)

    Following are the statements about the different parts of the brain. Select the correct statements.
    1. (i)The thinking part of the brain is the forebrain.
    2. (ii)Centres for smell, hearing, memory, sight are located in forebrain.
    3. (iii)Salivation, vomiting, blood pressure are controlled by medulla in the hind brain.
    4. (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)
    Show answer
    (a) (i), (ii) and (iii)
    Step-by-step solution →
  4. Q4 1 mark

    Also in 31/4/1 (Q6), 31/4/3 (Q7)

    Two pea plants, one with round green seeds (RRyy) and another with wrinkled yellow (rrYY) seeds were crossed with each other which produced progeny that have only round yellow (RrYy) seeds. When plants are self-pollinated, the progeny will have which new combination of characters, as compared to the parents :
    1. (i)Round, yellow
    2. (ii)Round, green
    3. (iii)Wrinkled, yellow
    4. (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)
    Show answer
    (b) (i) and (iv)
    Step-by-step solution →
  5. Q5 1 mark

    Also in 31/4/1 (Q7), 31/4/3 (Q6)

    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
    Show answer
    (d) contraction of left ventricle
    Step-by-step solution →
  6. Q6 1 mark
    In terms of various trophic levels of a food chain, which of the following organisms belong to the same level ?
    1. (a)Snake and Grasshopper
    2. (b)Grass and tree
    3. (c)Lizard and elephant
    4. (d)Bacteria and grass
    Show answer
    (b) Grass and tree
    Step-by-step solution →
  7. Q7 1 mark
    Opening of stomata is regulated by :
    1. (a)Entry of water into guard cells
    2. (b)Exit of water from guard cells
    3. (c)Entry of air into the stomatal pore
    4. (d)Entry of water into the stomatal pore
    Show answer
    (a) Entry of water into guard cells
    Step-by-step solution →
  8. Q8 1 mark

    Also in 31/4/1 (Q9), 31/4/3 (Q8)

    This question consists of two statements — Assertion (A) and Reason (R). Answer the question by selecting the appropriate option from (a), (b), (c) and (d) given below :

    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.
    Show answer
    (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
    Step-by-step solution →
  9. Q9 1 mark
    This question consists of two statements — Assertion (A) and Reason (R). Answer the question by selecting the appropriate option from (a), (b), (c) and (d) given below :

    Assertion (A): Reproduction, like any other life process, is essential to maintain the life of an individual organism.

    Reason (R): Reproduction involves creation of a DNA copy and additional cellular apparatus by the cell involved in the process.

    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.
    Show answer
    (d) Assertion (A) is false, but Reason (R) is true.
    Step-by-step solution →

Questions 10–12 · 2 marks each

  1. Q10(a) 2 marks

    Also in 31/4/1 (Q12(a)), 31/4/3 (Q12(a))

    Varun is diabetic. His latest ultrasound reports showed a blockage in his pancreatic duct. How will this affect the dual functioning of the pancreas?
    Show answer

    The blocked duct affects digestion, as pancreatic juice cannot reach the small intestine, but insulin reaches the blood without the duct.

    The pancreas is a dual gland. Exocrine function: it secretes pancreatic juice, containing trypsin, pancreatic amylase and lipase, which reaches the small intestine (duodenum) through the pancreatic duct. With the duct blocked, these digestive enzymes cannot be transported to the small intestine, so the digestion of proteins, carbohydrates and fats will be affected. Endocrine function: its islets of Langerhans secrete insulin directly into the blood, without using the duct. Varun is diabetic, so his insulin secretion is already affected; the blockage does not change it further, because insulin does not travel through the duct.
    Step-by-step solution →
    OR
    Q10(b) 2 marks

    Also in 31/4/1 (Q12(b)), 31/4/3 (Q12(b))

    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.
    Show answer

    The action is a reflex action: a rapid, automatic, involuntary response to a stimulus, without conscious thinking.

    The action involved is a **reflex action**. A reflex action is a sudden, rapid, automatic and involuntary response of the body to a stimulus, which occurs without conscious thinking; the withdrawal of the hand is controlled by the spinal cord. The path followed is the reflex arc: Stimulus (heat of the hot iron rod) Receptor (heat and pain receptors in the skin of the hand) Sensory neuron Spinal cord (relay neuron) Motor neuron Effector (muscle of the arm) Response (the hand is withdrawn).
    Step-by-step solution →
  2. Q11 2 marks

    Also in 31/4/1 (Q10)

    "Transpiration in plants is a necessary evil." Comment on the statement giving two reasons.
    Show answer

    Transpiration is necessary (it pulls water up and cools leaves) but wasteful, as most absorbed water is lost and the plant may wilt.

    The statement is justified. Transpiration is **necessary** because (i) evaporation of water from the leaves creates the transpiration pull, the suction that draws water and dissolved minerals up the xylem from the roots to the top of the plant, and (ii) the water that evaporates carries heat away with it, cooling the leaves and protecting them from the sun. It is an **evil** because over of the water absorbed by the roots is lost as vapour, and when the loss outruns absorption the plant wilts and its growth suffers. The loss cannot be avoided, because the stomata that must open in daylight to admit carbon dioxide for photosynthesis are the same pores through which the water vapour escapes. A large, risky loss that the plant cannot do without is exactly a necessary evil.
    Step-by-step solution →
  3. Q12 2 marks
    1. (a)Explain the role of UV radiation in the formation of ozone layer.
    2. (b)Why is UV radiation considered as highly damaging to organisms ?
    Show answer

    UV splits oxygen into atoms that combine with oxygen to make ozone; its high energy breaks bonds in DNA and proteins, damaging organisms.

    (a) Ultraviolet radiation supplies the energy that makes ozone. High-energy UV radiation from the Sun splits some molecules of oxygen high in the atmosphere into free oxygen atoms, , and each free atom then combines with another oxygen molecule to give ozone, . Repeated throughout the upper atmosphere, these two steps build up the ozone layer. (b) Ultraviolet radiation is of much shorter wavelength and higher energy than visible light, so it carries enough energy to break chemical bonds. It is absorbed by the DNA and proteins of living cells and damages them, causing mutations; in human beings it is known to cause skin cancer, as well as cataract and a weakened immune system, and in plants it damages chlorophyll and lowers the rate of photosynthesis.
    Step-by-step solution →

Questions 13–14 · 3 marks each

  1. Q13 3 marks

    Also in 31/4/1 (Q14)

    Study the given table and answer the following questions :
    1. (a)Depict the information given in the table in the form of a cross.
    2. (b)Deduce the phenotypic ratio/ ratio showing external look as well as genotypic /genetic makeup, ratio.
    Cross between a red-flowered and a white-flowered pea plant
    Parents
    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.
    Show answer

    The cross gives all red in ; shows red : white and .

    (a) The cross is (the red-flowered parent) (the white-flowered parent). Their gametes are and , so the whole is and red, which is the 30 red plants recorded. Self-pollinating the gives , whose offspring are , , and . (b) phenotypic ratio (external look) red white, which is the printed ; genotypic ratio (genetic make-up) , that is of the 44 plants.
    Step-by-step solution →
  2. Q14 3 marks
    Give reasons for the following :
    1. (a)Plants do not have an excretory system so how they excrete their wastes ?
    2. (b)Why do mammals and birds have double circulation ?
    Show answer

    Plants excrete through stomata, lenticels, transpiration and by storing or shedding wastes; warm-blooded mammals and birds need a highly efficient oxygen supply.

    (a) A plant has no excretory organ, so each waste leaves by the route that suits it. Gaseous wastes - the oxygen made in photosynthesis and the carbon dioxide made in respiration - diffuse out through the stomata of the leaves and the lenticels of the stem. Excess water is lost as water vapour by transpiration. Other wastes are stored in the vacuoles of cells, deposited as resins and gums in old xylem, which is already dead tissue, or moved into leaves and bark that are then shed; some are passed out into the soil around the roots. (b) Because they are warm-blooded. A mammal or a bird holds its body temperature constant whatever the surroundings do, and it can only do that by oxidising food continuously, so its demand for energy - and therefore for oxygen - is high and constant. Double circulation, with a completely four-chambered heart, keeps the oxygenated and the deoxygenated blood from mixing, so the blood delivered to the tissues carries its full load of oxygen; and because the blood returns to the heart after the lungs, it is pumped out to the body a second time at high pressure. Together these give the efficient, high-rate oxygen supply that a warm-blooded body needs.
    Step-by-step solution →

Question 15 · 4 marks

  1. Q15 4 marks

    Also in 31/4/1 (Q15), 31/4/3 (Q15)

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

    Roots grow towards water; auxin lengthens cells so shoots bend to light; sensitive plants move without growth; roots and shoots respond oppositely.

    (a) A plant is fixed in one place and cannot move to find water, so its roots must grow to reach it. Roots are positively hydrotropic: they bend and grow towards the area of water availability in the soil, so that they can absorb water and the minerals dissolved in it, even when the water does not lie directly below the seed. That water is needed for photosynthesis, for carrying minerals up the plant, for keeping cells turgid and for transpiration, and the spreading root system it produces also anchors the plant. (b) Auxin is a hormone made at the tip of the shoot. It makes cells grow longer, so it controls the rate and the direction of growth. When light comes from one side, auxin diffuses to the shaded side; those cells elongate more than the ones in the light, so that side becomes longer and the shoot bends towards the light. (c) Two differences. First, the sensitive plant's movement is non-directional - the leaflets fold the same way whichever side is touched - whereas a tropic movement is directional, its direction set by the direction of the stimulus. Second, the sensitive plant's movement involves no growth: cells at the base of the leaflets rapidly lose water, so it takes seconds and is reversible, whereas a tropic movement is a slow, permanent growth movement caused by unequal cell elongation. OR (c) Roots and shoots respond in opposite directions. To gravity, roots grow downwards in the direction of gravity (positive geotropism) while shoots grow upwards against it (negative geotropism). To light, shoots grow towards the light (positive phototropism) while roots grow away from it (negative phototropism).
    Step-by-step solution →

Question 16 · 5 marks

  1. Q16(a) 5 marks

    Also in 31/4/1 (Q16(a)), 31/4/3 (Q16(a))

    1. (a)Draw a labelled diagram showing open and closed stomatal pore.
    2. (b)List the main events that occur during the process of photo-synthesis.
    3. (c)Write a balanced chemical equation showing photo-synthesis in plants.
    Show answer

    Turgid guard cells open the pore and flaccid ones close it; photosynthesis absorbs light, splits water and reduces carbon dioxide to carbohydrate.

    (a) Two guard cells face each other around the pore. In the open stoma they are turgid and bow apart, leaving a wide pore between them; in the closed stoma they are flaccid and their inner walls press together, so the pore is shut. Label the guard cells, the chloroplasts in them and the stomatal pore. (b) The main events are: (i) Absorption of light energy by chlorophyll; (ii) Conversion of light energy into chemical energy and splitting of water molecules into hydrogen and oxygen; (iii) Reduction of carbon dioxide to carbohydrates. These events need not follow one another immediately. (c) , where is glucose. Cancelling the six molecules of water that appear on both sides gives the equivalent net equation , which is also balanced.
    Step-by-step solution →
    OR
    Q16(b) 5 marks

    Also in 31/4/1 (Q16(b)), 31/4/3 (Q16(b))

    1. (a)Draw a well-labelled diagram of nephron and label atleast 4 of its parts.
    2. (b)Write the role of capillary cluster inside Bowman's capsule.
    3. (c)What is meant by selective reabsorption ?
    Show answer

    The glomerulus filters blood under high pressure into Bowman's capsule; the tubule then selectively reabsorbs useful substances, leaving wastes to form urine.

    (a) The nephron runs from Bowman's capsule, which cups the glomerulus, into the proximal convoluted tubule, then down and up the loop of Henle, into the distal convoluted tubule and finally into the collecting duct; blood enters the glomerulus by the afferent arteriole and leaves by the narrower efferent arteriole. Any four of these parts satisfy the question, and the diagram labels eight. (b) The capillary cluster is the glomerulus. Because the afferent arteriole bringing blood in is wider than the efferent arteriole taking it away, the blood pressure inside the tuft is high, and under that pressure the glomerulus filters the blood: water, glucose, amino acids, salts, urea and other small molecules are forced through the thin capillary walls into Bowman's capsule, while blood cells and large plasma proteins are too big to pass and remain in the blood. The fluid collected is the glomerular filtrate, and the process is ultrafiltration. (c) Selective reabsorption is the taking back into the blood, from the filtrate flowing along the tubule, of the substances the body still needs, while the wastes are left behind. Glucose, amino acids, most salts and a major part of the water are reabsorbed, chiefly in the proximal convoluted tubule, whereas the wastes such as urea, together with the excess salts and the remaining water, pass on to form urine. It is selective because the useful substances are recovered in preference to the wastes, and the amount of water taken back depends on how much excess water the body has and how much dissolved waste it must excrete.
    Step-by-step solution →

Questions 17–24 · 1 mark each

  1. Q17 1 mark
    When lead nitrate is heated in a boiling tube, we observe :
    1. (a)brown fumes of gas
    2. (b)white fumes of gas
    3. (c)yellow fumes of lead oxide
    4. (d)brown fumes of gas
    Show answer
    (d) brown fumes of gas
    Step-by-step solution →
  2. Q18 1 mark

    Also in 31/4/1 (Q20), 31/4/3 (Q18)

    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.
    Show answer
    (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.
    Step-by-step solution →
  3. Q19 1 mark

    Also in 31/4/1 (Q18), 31/4/3 (Q20)

    The above reaction is an example of -
    1. (a)Displacement Reaction
    2. (b)Endothermic Reaction
    3. (c)Exothermic Reaction
    4. (d)Neutralisation Reaction
    Show answer
    (c) Exothermic Reaction
    Step-by-step solution →
  4. Q20 1 mark

    Also in 31/4/1 (Q19), 31/4/3 (Q17)

    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
    Show answer
    (b) Hydrochloric acid
    Step-by-step solution →
  5. Q21 1 mark
    Which one of the following metals neither reacts with cold water nor reacts with hot water but reacts with steam ?
    1. (a)Sodium
    2. (b)Calcium
    3. (c)Iron
    4. (d)Magnesium
    Show answer
    (c) Iron
    Step-by-step solution →
  6. Q22 1 mark

    Also in 31/4/1 (Q23)

    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.
    Show answer
    (a) By adding acid to water with constant stirring.
    Step-by-step solution →
  7. Q23 1 mark

    Also in 31/4/1 (Q22)

    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
    Show answer
    (d) Steel
    Step-by-step solution →
  8. Q24 1 mark

    Also in 31/4/1 (Q24), 31/4/3 (Q24)

    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.
    Show answer
    (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
    Step-by-step solution →

Question 25 · 2 marks

  1. Q25 2 marks
    Write the balanced chemical equations involved in the preparation of
    1. (a)Bleaching powder
    2. (b)Baking soda
    Show answer

    Chlorine on dry slaked lime gives bleaching powder; carbon dioxide in ammonia-saturated brine gives baking soda.

    (a) Bleaching powder is prepared by the action of chlorine on dry slaked lime: It may also be written . (b) Baking soda, sodium hydrogencarbonate, is prepared from brine by the ammonia-soda process, also called the Solvay process. Carbon dioxide is passed into brine saturated with ammonia: Both equations are balanced as written.
    Step-by-step solution →

Questions 26–27 · 3 marks each

  1. Q26 3 marks
    Answer the following :
    1. (a)Identify the substance oxidized and the substance reduced in the following reaction :
    2. (b)What type of reaction occurs when Lead (II) nitrate solution is mixed with potassium iodide solution ?
    3. (c)Why is silver chloride stored in dark coloured bottles ?
    Show answer

    Sodium is oxidized and oxygen reduced; the lead nitrate and potassium iodide reaction is double displacement; dark bottles stop light decomposing silver chloride.

    (a) Sodium is the substance oxidized and oxygen is the substance reduced. Sodium goes from an oxidation number of in the free metal to in , a rise of one, losing one electron per atom, so it is oxidized and is the reducing agent. Oxygen goes from in to in , a fall of two, gaining two electrons per atom, so it is reduced and is the oxidising agent. (b) It is a double displacement reaction: the two solutions exchange their ions. Because the new pair is insoluble and separates as a yellow solid, the same reaction is also called a precipitation reaction. (c) Silver chloride is decomposed by light. Sunlight splits it into finely divided grey silver and chlorine gas, , so a sample kept in a clear bottle slowly turns grey and is spoilt. A dark coloured bottle keeps the light out and prevents this photolytic decomposition.
    Step-by-step solution →
  2. Q27(a) 3 marks

    Also in 31/4/1 (Q27(a)), 31/4/3 (Q26(a))

    (i) Which method will be used to convert each of them to their respective metal oxides ?
    1. (a)Carbonate ore
    2. (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 ?
    Show answer

    Carbonate ore: calcination; sulphide ore: roasting; the thermite reaction welds rails; impure copper is the anode and pure copper the cathode.

    (i) (a) A carbonate ore is converted to its oxide by calcination, that is by heating it strongly in a limited supply of air: . (b) A sulphide ore is converted to its oxide by roasting, that is by heating it strongly in excess air: . (ii) The thermite reaction is used: . It is so strongly exothermic that the iron is produced molten, and it runs into the crack between the rails and solidifies there, welding them. (iii) Anode: a thick block of impure copper. Cathode: a thin strip or sheet of pure copper. The electrolyte is copper sulphate solution acidified with sulphuric acid, and the insoluble impurities collect below the anode as anode mud.
    Step-by-step solution →
    OR
    Q27(b) 3 marks

    Also in 31/4/1 (Q27(b)), 31/4/3 (Q26(b))

    1. (i)Write all the reactions involved in the extraction of from .
    2. (ii)Hydrogen gas is not evolved when most of the metals react with nitric acid. Give reason.
    Show answer

    Zinc sulphide is roasted to zinc oxide, then reduced with carbon; nitric acid oxidises the hydrogen formed to water.

    (i) Zinc is extracted from zinc blende in two reactions. Roasting, in which the concentrated ore is heated in excess air: . Reduction, in which the oxide is heated with coke: . The zinc obtained is finally purified by electrolytic refining. (ii) Nitric acid is a strong oxidising agent. The hydrogen that the metal displaces from it is oxidised to water as soon as it is formed, while the nitric acid is itself reduced to an oxide of nitrogen (, or according to the concentration of the acid and the metal used). No hydrogen therefore escapes as gas. Magnesium and manganese with very dilute nitric acid are the exceptions and do evolve hydrogen.
    Step-by-step solution →

Question 28 · 4 marks

  1. Q28 4 marks

    Also in 31/4/1 (Q28), 31/4/3 (Q28)

    The pH plays an important role in our daily life. Strength of acids and bases depends on the number of ions and 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 ?[1]
    2. (b)Three solutions A, B and C showed pH as 3, 5 and 7. Which solution has the highest concentration of ions ?[1]
    3. (c)What are strong acids and weak acids ?[2]
    4. (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.[2]
    Show answer

    A universal indicator shows strength; solution A (pH 3) has the highest ; strong acids ionise completely, weak acids partially.

    (a) A universal indicator - a mixture of several indicators, used as a solution or as pH paper. It gives a different colour at each pH, so its colour reports how strong the acid or base is, which litmus or phenolphthalein cannot do. (b) Solution A. Since , the concentrations are mol/L in A, mol/L in B and mol/L in C, so the smallest pH, 3, carries the highest hydrogen ion concentration. (c) A strong acid ionises completely in water and so gives a large number of ions (for example , , ); a weak acid ionises only partially and so gives far fewer ions (for example , ). OR (c) (i) Rain water 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; nitric, hydrochloric and sulphuric acids are strong.
    Step-by-step solution →

Question 29 · 5 marks

  1. Q29(a) 5 marks

    Also in 31/4/1 (Q29(a)), 31/4/3 (Q29(a))

    1. (i)Carbon cannot form cation and anion. Explain.
    2. (ii)Define homologous series of carbon compounds. Why are the melting and boiling points of higher than or ?
    3. (iii)What happens when ethanol is heated with acidified potassium dichromate ?
    Show answer

    Carbon cannot lose or gain four electrons, so it shares them; heavier has stronger intermolecular forces; acidified dichromate oxidises ethanol to ethanoic acid.

    (i) Carbon has four valence electrons (configuration ). To form it would have to lose all four, which needs an enormous amount of energy and would leave a nucleus of six protons holding only two electrons; to form it would have to gain four, and a nucleus of only six protons cannot hold on to ten electrons. Carbon therefore shares its four valence electrons instead, forming four covalent bonds - it is tetravalent. (ii) A homologous series is a series of carbon compounds having the same general formula and the same functional group, in which successive members differ by one unit, that is by 14 u; the members have similar chemical properties and a gradual gradation in physical properties. (58 u) is a larger, heavier molecule than (44 u) or (30 u), so the intermolecular forces of attraction between its molecules are stronger and more energy is needed to separate them; hence its higher melting and boiling points. (iii) Acidified potassium dichromate acts as an oxidising agent and oxidises ethanol to ethanoic acid, the orange colour of the dichromate turning green:
    Step-by-step solution →
    OR
    Q29(b) 5 marks

    Also in 31/4/1 (Q29(b)), 31/4/3 (Q29(b))

    1. (i)How would you name the following : I. II.
    2. (ii)When ethanol reacts with ethanoic acid in the presence of conc. , 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.
    3. (iii)Why soaps do not produce foam with hard water ?
    I. CH₃ – CH₂ – C = O H II. CH₃ – C ≡ CH
    Show answer

    Propanal and propyne; the sweet-smelling compounds are esters; soap forms insoluble scum with the calcium and magnesium ions of hard water, so no foam.

    (i) I. Propanal (), an aldehyde. II. Propyne (), also written prop-1-yne. (ii) I. The sweet smelling compounds are esters, and the reaction that forms them is called esterification. II. , that is ethanoic acid and ethanol give ethyl ethanoate and water. (iii) Hard water contains and ions. These react with soap, which is the sodium salt of a long chain carboxylic acid, to form insoluble calcium and magnesium salts of that acid - a curdy white precipitate called scum. The soap is used up as scum instead of dissolving to form micelles, so it does not lather and a large amount of soap is wasted.
    Step-by-step solution →

Questions 30–32 · 1 mark each

  1. Q30 1 mark

    Also in 31/4/1 (Q31), 31/4/3 (Q31)

    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
    Show answer
    (c) Presbyopia
    Step-by-step solution →
  2. Q31 1 mark

    Also in 31/4/1 (Q30), 31/4/3 (Q30)

    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
    Show answer
    (a) Iris and pupil
    Step-by-step solution →
  3. Q32 1 mark

    Also in 31/4/1 (Q32), 31/4/3 (Q32)

    For this question, 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.
    Show answer
    (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
    Step-by-step solution →

Questions 33–34 · 2 marks each

  1. Q33(a) 2 marks

    Also in 31/4/1 (Q33(a)), 31/4/3 (Q33(a))

    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.
    Show answer

    No; the heater draws 10 A, more than the 4 A rating, so the fuse would melt.

    No, a A fuse cannot be used. The heater draws A in normal working. The current rating of a fuse is the largest current its wire can carry without melting, and A is two and a half times the fuse's A rating, so the fuse wire would melt and break the circuit the moment the heater is switched on. A fuse rated just above A must be used instead.
    Step-by-step solution →
    OR
    Q33(b) 2 marks

    Also in 31/4/1 (Q33(b)), 31/4/3 (Q33(b))

    How is an electromagnet made ? Write any two ways by which the strength of electromagnet can be increased.
    Show answer

    Wind insulated copper wire as a solenoid over a soft iron core and pass current; strength rises with more current or more turns.

    An electromagnet is made by winding a large number of closely spaced turns of insulated copper wire, in the form of a solenoid, over a soft iron core, and passing an electric current through the coil from a battery through a plug key. While the current flows, the soft iron core is strongly magnetised and the whole arrangement behaves as a magnet with a north and a south pole; as soon as the current is switched off the core loses almost all of its magnetism. Two ways of increasing its strength: (i) increase the current flowing through the coil; (ii) increase the number of turns of the coil.
    Step-by-step solution →
  2. Q34 2 marks
    What do the symbols given below represent in an electric circuit ? Write one function of each.
    1. (i)[symbol printed - see figure]
    2. (ii)[symbol printed - see figure]
    (i) V + − (ii)
    Show answer

    Symbol (i) is a voltmeter, which measures potential difference; symbol (ii) is a variable resistance (rheostat), which changes the current.

    (i) The symbol represents a voltmeter. Function: it measures the potential difference, in volts, between two points of a circuit - in practice across a component. It is connected in parallel with that component, with its terminal towards the positive terminal of the cell, and its very high resistance means it draws almost no current from the circuit. (ii) The symbol represents a variable resistance, that is, a rheostat (the arrow is its sliding contact). Function: connected in series in the circuit, it lets the resistance be changed while the circuit is running, and so varies the current in the circuit without changing the cell - by Ohm's law , more resistance means less current.
    Step-by-step solution →

Questions 35–37 · 3 marks each

  1. Q35 3 marks

    Also in 31/4/1 (Q37), 31/4/3 (Q36)

    On the basis of the given circuit arrangement, answer the following questions :
    1. (a)What happens to the rod AB when the key K is closed and why ?
    2. (b)Name and state the rule that determines this behaviour of the rod AB in the above situation.
    A B S N − + K
    Show answer

    The rod carrying current B to A in the upward field is displaced sideways, to the left; Fleming's left-hand rule.

    (a) When K is closed a current flows through the rod AB, from end B to end A. The rod is then a current-carrying conductor lying across the magnetic field of the horseshoe magnet, which in the gap is directed vertically upwards because the N pole is below the rod and the S pole above it. A current-carrying conductor placed in a magnetic field experiences a mechanical force perpendicular both to the current and to the field, so the freely suspended rod is displaced: it swings horizontally, at right angles to its own length, towards the left. Taking the plug out of K stops the current, the force vanishes and the rod returns to rest. (b) The rule is Fleming's left-hand rule. Stretch the thumb, the forefinger and the central finger of the left hand so that all three are mutually perpendicular. If the forefinger points in the direction of the magnetic field and the central finger points in the direction of the current, then the thumb points in the direction of the force acting on the conductor, that is, in the direction in which the rod moves.
    Step-by-step solution →
  2. Q36 3 marks

    Also in 31/4/1 (Q35)

    (a) Observe the given ray diagram and compare :
    1. (i)the speed of light in medium A and the speed of light in medium B.
    2. (ii)the optical densities of medium B and medium C. (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’.
    N i r Medium ‘A’ Medium ‘B’ Medium ‘C’
    Show answer

    Light is faster in A than in B; B and C have equal optical density; in (b) the ray emerges parallel to the incident ray.

    (a) (i) At the A-B face the ray bends towards the normal, that is , so medium B is optically denser than medium A. Since , light travels faster in the optically rarer medium, and therefore the speed of light in medium A is greater than the speed of light in medium B (). (ii) At the B-C face the ray passes on without any change of direction, so : media B and C have the same optical density. (b) See the diagram. With the optical density of ‘A’ equal to that of ‘C’ and the optical density of ‘B’ greater than both, the ray bends towards the normal on entering B and away from the normal by the same amount on entering C, so : the emergent ray in medium C is parallel to the incident ray in medium A and is only laterally displaced.
    Step-by-step solution →
  3. Q37 3 marks
    A person is suffering from a defect of vision due to the elongation of his/her eye ball.
    1. (a)Write the name of the defect of vision and write the other possible cause of this defect of vision.
    2. (b)This defect can be corrected by using a suitable lens. Draw the ray diagram for correction of this defect of vision.
    Show answer

    The defect is myopia; its other cause is excessive curvature of the eye lens; a concave lens of suitable power corrects it.

    (a) The defect is myopia, also called short-sightedness or near-sightedness: the image of a distant object is formed in front of the retina, so distant objects are not seen distinctly. Besides elongation of the eye ball, the other possible cause of myopia is excessive curvature of the eye lens - a lens curved so strongly that its converging power is too great and its focal length too short. (b) The defect is corrected by placing a concave (diverging) lens of suitable power in front of the eye; its power is negative and its focal length is equal to the distance of the far point of the defective eye. In the ray diagram, rays from a distant object arrive parallel to the principal axis and fall on the concave lens; after the lens they diverge, and their dashed backward extensions meet at the far point of the defective eye, so the rays entering the eye appear to come from . The eye lens then converges them to a sharp point exactly on the retina.
    Step-by-step solution →

Question 38 · 4 marks

  1. Q38 4 marks

    Also in 31/4/1 (Q38), 31/4/3 (Q38)

    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.
    Parallel Light Rays from distant star Plane Mirror Curved Mirror Eye-piece
    1. (a)What kind of image of a star is seen by the observer in the eye-piece ?
    2. (b)What kind of mirror is used in this reflecting telescope ?
    3. (c)Explain with the reason, what kind of optical device (type of lens or mirror) is used at the eye-piece.
    4. (c)OR : What is the role of plane mirror in image formation by this telescope ?
    Show answer

    Virtual, magnified, inverted image; a concave mirror with a plane mirror; a convex lens as eye-piece; or the plane mirror redirects the light.

    (a) A virtual, magnified image, inverted with respect to the star. The curved mirror first forms a real, inverted image of the star at its focus, and the eye-piece, used as a magnifying glass on that image, gives the enlarged virtual image the observer actually sees. (b) A concave (converging) mirror - the large curved mirror that brings the parallel rays from a distant star to a real focus - together with a small plane mirror that turns the converging beam towards the eye-piece. (c) A convex (converging) lens is used at the eye-piece. It magnifies the image formed by the curved mirror: that small real image lies at or just within the focal length of the lens, so the lens acts as a magnifying glass and gives a much enlarged virtual image of it; a concave lens would only diminish the image, and a mirror there would send the light back into the tube instead of to the eye. OR (c) The plane mirror redirects (reflects) the light rays coming from the curved mirror towards the eye-piece. Held at in the path of the converging beam, it turns the beam through and brings the real image out through the side of the tube, so that the eye-piece and the observer's head do not block the starlight falling on the curved mirror; it forms no image of its own and does not change the size of the image.
    Step-by-step solution →

Question 39 · 5 marks

  1. Q39(a) 5 marks
    - graph for a nichrome wire is shown in the diagram.
    1. (i)Prove that the graph follows Ohm's law and calculate the resistance of nichrome wire from the given - graph.
    2. (ii)(I) Calculate the equivalent resistance of the combination of resistors connected in the given circuit. (II) Find the value of the total current drawn from the battery.
    0.1 0.4 0.2 0.8 0.3 1.2 0.4 1.6 0.5 2.0 0.6 2.2 Potential difference (V) Electric Current (A)
    3 Ω 7 Ω 10 Ω 5 Ω 5 Ω 5 V K
    Show answer

    at every reading, so the graph obeys Ohm's law; and the battery current is A.

    (i) The plotted points lie on a straight line through the origin and the ratio has the same value at the readings, , so , which is Ohm's law; the resistance of the nichrome wire is the slope, . (ii) (I) The and are in series through the apex and give , which is in parallel with the bridge and gives ; with the two arms in series the equivalent resistance is . (II) The total current drawn from the battery is A.
    Step-by-step solution →
    OR
    Q39(b) 5 marks
    1. (i)Define the SI unit of power. Express the electric power of an electric appliance in terms of potential difference and current flowing through it.
    2. (ii)In a house, 3 bulbs of 100 W 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.
    3. (iii)Convert the commercial unit of electric energy into Joule (J).
    Show answer

    Power is (unit watt); the house uses 60 kWh a month costing ₹ 216; J.

    (i) The SI unit of power is the watt (W): an appliance has a power of watt when it consumes energy at the rate of joule per second, that is when it carries a current of ampere at a potential difference of volt. Since the work done in carrying a charge through a potential difference is and , the electric power is . (ii) The three bulbs use kWh a day and the heater kWh a day, so the house uses kWh a day and kWh in the month; at ₹ 3.60 per kWh the cost is ₹216. (iii) The commercial unit of electrical energy is the kilowatt-hour, and kWh W s J.
    Step-by-step solution →

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