CBSE Sample Papers for Class 10 Science Set-6

CBSE Sample Papers for Class 10 Science Set 6

Time: 3 Hours
Maximum Marks: 80

Instructions

• This question paper consists of 39 questions in 5 sections.
• All questions are compulsory. However, an internal choice is provided in some questions. A student is expected to attempt only one of these questions.
• Section A consists of 20 objective-type questions carrying 1 mark each.
• Section B consists of 6 Very Short questions carrying 2 marks each. Answers to these questions should be in the range of 30 to 50 words.
• Section C consists of 7 Short Answer type questions carrying 3 marks each. Answers to these questions should be in the range of 50 to 80 words.
• Section D consists of 3 Long Answer type questions carrying 5 marks each. Answers to these questions should be in the range of 80 to 120 words.
• Section E consists of 3 source-based/case-based assessment units of 4 marks each with sub-parts.

Section A

Select and write the most appropriate option out of the four options given for each of the questions 1-20.

Question 1.
Identify gas A in the following experiment.

(a) Nitrogen
(b) Hydrogen
(c) Oxygen
(d) Carbon dioxide
Answer:
(b) Hydrogen
In the given experiment, the gas evolved (A) is hydrogen as zinc reacts with dil.H₂SO₄ to form zinc chloride and hydrogen gas.

The presence of hydrogen is tested by bringing a burning candle near the soap bubbles filled with gas A (hydrogen). In doing so, hydrogen gas burns with a pop sound due to the reaction between hydrogen and oxygen present in the air.

Question 2.
The colour of the solution observed after 30 minutes of placing zinc metal to copper sulphate solution is [1]
(a) Blue
(b) Colourless
(c) Dirty green
(d) Reddish Brown
Solution:
(b) Colourless
Explanation: The colour of the solution observed after 30 minutes of placing zinc metal to copper sulphate solution is colourless because zinc being more reactive than copper displaces it from copper sulphate solution and forms a colourless solution of zinc sulphate.

Question 3.
A sample of soil is mixed with water and allowed to settle. The clear supernatant solution turns the pH paper yellowish-orange. Which of the following would change the colour of this pH paper to greenish-blue?
(a) Lemon juice
(b) Vinegar
(c) Common salt
(d) An antacid
Answer:
(d) An antacid
As pH paper turns greenish blue for weakly basic compounds and antacids contain weak bases like Mg(OH)₂. So, an antacid would change the colour of this pH paper to greenish-blue. Other options (a) and (b) contain acids and option (c) is a neutral salt.

Question 4.
On adding dilute sulphuric acid to a test tube containing a metal ‘X’, a colourless gas is produced when a burning match stick is brought near it. Which of the following correctly represents metal ‘X’? [1]
(a) Sodium
(b) Zinc
(c) Copper
(d) Silver
Solution:
(a) Sodium
Explanation: When dilute sulphuric acid is taken in a test tube containing sodium metal, sodium sulphate is formed along with the liberation of hydrogen gas which is colourless gas. Thus, metal ‘X’ is sodium metal. The balanced chemical reaction can be represented as:

Question 5.
__________ will displace hydrogen from dilute acids.
(a) Copper
(b) Gold
(c) Zinc
(d) Silver
Answer:
(c) Zinc
Zinc will displace hydrogen from dilute acids as it has a higher reactivity than hydrogen.

Question 6.
An element with atomic number ……………… will form a basic oxide. [1]
(a) 7 (2, 5)
(b) 17 (2, 8, 7)
(c) 14 (2, 8, 4)
(d) 11 (2., 8, 1)
Solution:
(d) 11(2, 8, 1)
Explanation: Generally, metals form basic oxide while non-metals form acidic oxide. Metals have 1 to 3 electrons in the outermost shell whereas non-metals have 4 to 8 electrons in the outermost shell. The electronic configuration of given elements are as follows:

Atomic Number	Electronic Configuration
7 (Nitrogen)	2, 5
17 (Chlorine)	2, 8, 7
14 (Silicon)	2, 8, 4
11 (Sodium)	2, 8, 1

Thus, element with atomic number 11 (sodium) having electronic configuration 2, 8, 1 is metal and it will form a basic oxide. Whereas, the other three elements having atomic numbers 7 (Nitrogen), 17 ; (Chlorine) and 14 (Silicon) are non-metals and hence they will form acidic oxide.

Question 7.
Which of the given options correctly represents the parent acid and base of calcium carbonate?

	Parent Acid	Parent Base
(a)	HCl	NaOH
(b)	H2CO3	Ca(OH)2
(c)	H3PO3	CaSO4
(d)	H2SO4	CaSO4

Answer:
(b) Parent Acid – H₂CO₃, Parent Base – Ca(OH)₂
The terms ‘parent acid’ and ‘parent base’ are used to describe the original compounds that combine to make the salt. So, calcium carbonate salt can be obtained by the reaction between H₂CO₃ (parent acid) and Ca(OH)₂ (parent base) to form calcium carbonate salt and water.

Question 8.
Generally, food is broken and absorbed within the body of organisms. In which of the following organisms is it done outside the body? [1]
(a) Amoeba
(b) Mushroom
(c) Paramoecium
(d) Lice
Solution:
(b) Mushroom
Explanation: The process of animals feeding on dead and decaying substances or organisms for : getting energy, food and nutrition is called as saprophytic nutrition and the organisms that follow saprophytic nutrition are called saprotrophs. They are the recycler of nutrients as they release specific enzymes that act on complex organic matter and break them into smaller and simpler particles that are easily consumable by the organism. Organisms like – Mushrooms, Yeast, and Bread moulds are saprophytic. They break down food outside the body and absorb the simpler digested particles. Hence, from the given options fungi, mushroom break down food outside the body and absorb the simpler digested particles.

Question 9.
Iodine is necessary for the synthesis of which hormone?
(a) Adrenaline
(b) Thyroxine
(c) Auxin
(d) Insulin
Answer:
(b) Thyroxine
Iodine is necessary for the synthesis of thyroxine hormone.

Question 10.
A farmer wants to grow banana plants genetically similar enough to the plants already available in his field. Which one of the following methods would you suggest for this purpose? [1]
(a) Regeneration
(c) Vegetative propagation
(b) Budding
(d) Sexual reproduction
Solution:
(c) Vegetative propagation
Explanation: Plants that are genetically similar enough can be grown by means of vegetative propagation. This method of asexual reproduction is a method of rapid propagation where the sex cells are not involved. The new plant formed is also genetically similar to the parent plant. In this type of asexual reproduction, parts of the plant like stem, roots or leaves can be used to grow new plants. Banana propagation is a type of vegetative reproduction.

Question 11.
The figure given below shows three stages in the cardiac cycle.

Which of the following sequences is correct regarding this
(a) 1, 2, 3
(b) 2, 1, 3
(c) 2, 3, 1
(d) 3, 1, 2
Answer:
(b) 2, 1, 3
In Figure (2), blood is entering into the right auricle through the superior and inferior vena cava and blood is entering into the left auricle through the pulmonary vein. Figure (1) shows the movement of blood from the auricles into the ventricles. Figure (3) shows the movement of blood from the right ventricle into the pulmonary artery and from the left ventricle into the aorta.

Question 12.
A sportsman, after a long break of his routine exercise, suffered muscular cramps during a heavy exercise session. This happened due to: [1]
(a) lack of carbon dioxide and formation of pyruvate.
(b) presence of oxygen and formation of ethanol.
(c) lack of oxygen and formation of lactic acid.
(d) lack of oxygen and formation of carbon dioxide.
Solution:
(c) lack of oxygen and formation of lactic acid.
Explanation: During heavy exercise, the energy demand is high but the supply of oxygen to produce energy is limited. Therefore, anaerobic respiration takes places in the muscles cells to fulfil the energy demand. This anaerobic breakdown of glucose leads to the formation of lactic acid in muscles. The accumulation of lactic acid in muscles leads to muscle cramps. Hence, due to lack of oxygen and formation of lactic acid, a sportsman suffered muscular cramps during a heavy exercise session.

Question 13.
What is the maximum resistance that can be made using five resistors each of \(\frac{1}{5}\) Ω?
(a) 1 Ω
(b) 5 Ω
(c) 2 Ω
(d) 2.5 Ω
Answer:
(a) 1 Ω
The maximum resistance is obtained when resistors are connected in a series combination. Thus, equivalent resistance,
R_s = n × R = 5 × \(\frac{1}{5}\) = 1 Ω
where, R_s = equivalent resistance for series combination. (1)

Question 14.
When light enters the atmosphere it strikes extremely fine particles, which deflect the rays of light in all possible directions, This is due to – [1]
(a) reflection of light
(c) scattering of light
(b) atmospheric refraction
(d) dispersion of light
Solution:
(c) scattering of light
Explanation: The earth’s atmosphere is a heterogeneous mixture of minute particles like smoke, tiny water droplets, suspended particles of dust, and molecules of air. When a beam of light enters the atmosphere it strike on such extremely fine particles, which deflect the rays of light in all possible direction due to scattering of light. As a result of this phenomenon, the path of the beam becomes visible.

Question 15.
During a Mendelian experiment, cross-breeding is done between tall pea plants bearing violet flowers and dwarf pea plants with white flowers. The F₁ generation produced all violet flowers but half of them were short. What can be the genotype of the tall parent?
(a) TTww
(b) TTWW
(c) TtWW
(d) TtWw
Answer:

Since all the flowers are purple so tall parent is homozygous dominant for this character. Half of the plants are tall and half are dwarf, so a tall parent is heterozygous dominant for this character. Thus, the genotype of tall parent-TtWW.

Question 16.
Which of the following features relates to biodegradable substances? [1]
(a) Broken down by biological processes
(b) Remain inert
(c) Persist in the environment for a long time
(d) May harm the ecosystem
Solution:
(a) Broken down by biological processes
Explanation: The substances that can be easily broken down by biological processes are called biodegradable wastes. These substances are decomposed through the actions of fungi, bacteria, and other living organisms. For example Food waste, paper, wood, cloth, cow-dung, human and animal excreta, etc.

On the other hand, the substances that cannot be broken down by biological processes are called non-biodegradable wastes. These substances may be in solid, liquid or gaseous form. These substances are inert and simply persist in the environment for a long time or may harm the various members of the ecosystem. For example, DDT, insecticides, pesticides, mercury, plastics, polythene bags, glass, and radioactive wastes. These non-biodegradable wastes are major pollutants of the environment.

Directions (Q.Nos. 17-20) consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below.

(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 17.
Assertion (A): Generally the candle flame is yellow.
Reason (R): The flame of a candle is yellow due to the presence of unburnt carbon particles.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
A candle flame is generally yellow due to the presence of unburnt carbon particles when light falls on these particles, they scatter a yellow colour. This shows, that the combustion of hydrocarbons in wax (candle) is not complete.

Question 18.
Assertion: The probability of survival of an organism produced through sexual reproduction is more than that of organism produced through asexual mode.
Reason: Variations provide advantages to individuals for survival. [1]
Solution:
(a) Both A and R are true, and R is the correct explanation of A.
Explanation: The Probability of survival of an organism produced through sexual reproduction is more than that of organism produced through asexual mode. This is because an organism produced by sexual reproduction has a greater survival rate as variations provide advantages to individuals for survival and they can adapt to various environments. Thus, both assertion and reason are true and reason is the correct explanation of assertion.

Question 19.
Assertion (A): Ureter forms the common passage for both the sperm and urine.
Reason (R): It never carries sperm.
Answer:
(d) A is false, but R is true.
In males, the urethra forms the common passage for both the sperms and urine, whereas ureters are tubes – that propel urine from the kidneys to the urinary bladder. It never carries sperm.

Question 20.
Assertion: Biodegradable substances result in the formation of compost and natural replenishment. Reason: It is due to the breakdown of complex inorganic substances into simple organic substances. [1]
Solution:
(c) A is true but R is false.
Explanation: Biodegradable substances such as food waste, humans and animal excreta, plant products, dried leaves, grass, fruits, flowers, food wastes, wood and other remains of the death of living creature result in the formation of compost and natural replenishment. This is because complex organic substances like carbohydrates, proteins, and lipids present in these substances are converted into simple inorganic substances like hydrogen, oxygen, calcium, iron, and sodium. These inorganic substances are released back into the soil and they serve as nutrients for the growth of plants, thereby balancing the ecosystem. Thus, assertion is true but reason is false.

Section B

Questions No. 21 to 26 are Very Short Answer Questions.

Question 21.
A metal forms two types of oxide and rust in moisture. Write the formulas of oxides and name the metal. Also, give the name of the metal used in the hot water apparatus.
Answer:
Iron forms two oxides iron (II) oxide and iron (III) oxide, i.e. ferrous and ferric oxide respectively.
Formula of ferrous oxide = FeO
Ferric oxide = Fe₂O₃
Copper is used in hot water apparatus since it is a good conductor of heat. (2)

Question 22.
State the post-fertilisation changes that lead to fruit formation in plants. [2]
Solution:
The post-fertilisation changes that lead to fruit formation in plants are:

1. Zygote divides several times to form an embryo within the ovule.
2. The ovule develops a tough coat and is gradually converted into a seed.
3. The ovary grows rapidly and ripens to form a fruit.
4. The petals, sepals, stamens, style and stigma may shrivel and fall off.

Question 23.
State the functions of the following plant hormones.
(a) Abscisic acid
(b) Cytokinin
Or
Mention the correct positions of the pancreas, thyroid gland, pituitary gland, and adrenal gland in the human body. Also, mention the hormones released by them.
Answer:
(a) Function of Abscisic acid:
(i) It inhibits growth.
(ii) It causes dormancy of seeds, wilting of leaves,
(iii) Abscisic acid leads to stomatal closure. (1)
(b) Function of Cytokinin:
(i) Cytokinin promotes cell division and reduces apical dominance.
(ii) It delays aging in leaves. (1)
Or
(a) The pancreas is found just below the stomach. It secretes insulin.
(b) The thyroid gland is found just below the neck and it secretes thyroxine.
(c) The pituitary gland is the master gland. It is present at the base of the brain. It growth hormone.
(d) Adrenal glands are present on top of kidneys it secrete steroid hormones like aldosterone. (2)

Question 24.
The refractive indices of three media arc given below:

Medium	Refractive Index
A	1.6
B	1.8
C	1.5

A ray of light is travelling from A to B and another ray is travelling from B to C.
(a) In which of the two cases the refracted ray bends towards the normal?
(b) In which case does the speed of light increase in the second medium?
Give reasons for your answer. [2]
Solution:
(a) When light travels from an optically rarer medium to an optically denser medium it moves towards the normal. Since n_B > n_A, hence the light ray will bend towards the normal on passing from medium A to B.
(b) The speed of the light will increase when the light travels from B to C, Since n_C < n_B and υ =(c/n), the speed of light ray will increase in the second medium i.e., medium C.

Question 25.
Beams of light are incident through holes A and B and emerge out of box through the holes C and D respectively, as shown in the figure.

What could be inside the box?
Or
Two lamps, one rated at 100W-220V and the other at 60W-220V are connected in parallel to the electric mains supply. What current is drawn from the line, if the supply voltage is 220V?
Answer:
Here, the emergent rays are parallel to the direction of the incident ray. Therefore, a rectangular glass slab could be inside the box as the extent of bending of the light ray at the opposite parallel faces AB (air-glass interface) and CD (glass-air interface) of the rectangular glass slab are equal and opposite. This is why the ray emerges parallel to the incident ray. (2)
Or

Given, potential, V = 220 V
Power, P₁ = 100 W
Power, P₂ = 60 W
∴ Current, I₁ = \(\frac{P_1}{V}\) = \(\frac{100}{220}\) = 0.45 A
Current, I₂ = \(\frac{P_2}{V}\) = \(\frac{60}{220}\) = 0.27 A
∴ Total current drawn, I = I₁ + I₂
= 0.45 + 0.27
= 0.72 A (2)

Question 26.
Study the food chain given below and answer the questions that follow:

(a) If the amount of energy available at the third trophic level is 100 joules, then how much energy will be available at the producer level? Justify your answer.
(b) Is it possible to have 2 more trophic levels in this food chain just before the fourth trophic level?
Justify your answer. [2]
Solution:
(a) According to the 10 percent law, only 10% of the energy is transferred to each trophic level starting from the first trophic level or producer level. If the amount of energy available at the third trophic level is 100 joules, then only 1000 joules of energy will be available at the second trophic level and 10000 Joules will be available at the producer level or first trophic level.

(b) No, since the loss of energy at each step is so great that very little usable energy will remain after 4 trophic levels.

Section C

Questions No. 27 to 33 are Short Answer Questions.

Question 27.
Answer the following questions.
(a) State the functional group present in alcohols.
(b) Give the general formula of alcohol.
(c) What is meant by denatured alcohol?
Answer:
(a) -OH (1)
(b) C_nH_2n+1OH (1)
(c) To prevent the misuse of alcohol produced for industrial purposes, it is made unfit for drinking by adding poisonous substances like methanol to it. This is called denatured alcohol. (1)

Question 28.
An element ‘M’ with electronic configuration 2 8 3 combines separately with Cl^–, \(\mathrm{SO}_4^{-2}\) anions. Write the chemical formulae of the compounds formed. Predict with the suitable reason the nature of the bond formed by element ‘M’ in general. How will the electrical conductivity of the compounds formed vary concerning ‘M’?
OR
A reddish-brown metal ‘X’, when heated in air, gives a black compound ‘Y, which when heated in presence of H2 gas gives ‘X’ back. ‘X’ is refined by the process of electrolysis; this refined form of ‘X’ is used in electrical wiring.
Identify ‘X’ and ‘Y. Draw a well-labeled diagram to represent the process of refining ‘X’. [3]
Solution:
The Chemical formulae of the compounds formed will be MCl₃ and M₂(SO₄ )₃. The nature of bond formed by element ‘M’ in general is Ionic bond because it can acquire a stable electronic configuration of neon (2, 8) by losing its three valence electrons to form M3+ cation. Compounds formed will conduct electricity in liquid or molten state but not in solid state in contrast to ‘M’.
OR
(a) ‘X’ – Copper! Cu and ‘Y’ – CuO

Question 29.
Why is the flow of signals in a synapse from the axonal end of one neuron to the dendritic end of another neuron, but not the reverse?
Answer:
The synapse acts, like a one-way valve because the chemical substance is present only on one side of the gap. This chemical diffuses towards the dendrite end of the next neuron where it generates an electrical signal. Since the chemicals are absent at the dendritic end of the neuron, the nerve impulse can go across only from one side (which contains the chemical substance). In this way, it is ensured that nerve impulses travel in only one direction (through a particular set of neurons). (3)

Question 30.
What is the probability of a girl or a boy being bom in a family? Justify your answer. [3]
Solution:
There are 50% chances that a girl may be born and 50% chances that a boy may be born. It can be explained as follows: Most human chromosomes have a maternal copy and a paternal copy. We have 22 such chromosomes. One pair of chromosomes called sex chromosomes is odd and is in not always being a perfect pair. Women have a perfect pair of sex chromosomes, both called X, (XX). But men have a mismatched pair of sex chromosomes in which one is normal-sized – X chromosome while the other is a short one called Y chromosome (XY). A child receives one chromosome from mother which is essentially X chromosome. A child who inherits an X chromosome from her father will be a girl, and one who inherits a Y chromosome from him will be a boy.

Question 31.
The figure shows three identical bulbs A, B, and C which are connected to a battery of supply voltage V. When the switch S is closed, discuss the change in

(a) the illumination of the three bulbs.
(b) the power dissipated in the circuit.
Answer:
When the switch is open,
V_A = V_B = V_C = \(\frac{V}{3}\)
and P_A = P_B = P_C = \(\frac{(V / 3)^2}{R}=\frac{V^2}{9 R}\) = P (say)
(a) When the switch closed, then bulb C is short-circuited, and will be no current through C, P_C = 0
V_A = V_B = \(\frac{V}{2}\)
⇒ P_A = P_B = \(\frac{(N / 2)^2}{4 R}=\frac{9}{4} P\) (1\(\frac{1}{2}\))
(b) Power dissipated, P₁ = P_A + P_B + P_C = 3P
Thus, P_t = P_A + P_B + P_C
= \(\frac{9}{4} P+\frac{9}{4} P\) + 0
= \(\frac{9}{2}\)P
= \(\frac{3}{2}\)P1 (1\(\frac{1}{2}\))

Question 32.
(i) State the law that explains the heating effect of current concerning the measurable properties in an electrical circuit.
(ii) List the factors on which the resistance of a conductor depends. [2 + 1]
Solution:
(i) Joule’s law of heating states that the heat dissipated across a resistor is directly proportional to the square of the current flowing through it, the resistance of the conductor and duration of flow of current.
H = I²Rt
Where H = Heat dissipated across resistor
I = Current
R = Resistance of the conductor t = time or duration of flow of current

(ii) The resistance of a conductor depends on:
(a) Length of the conductor
(b) Area of the cross-section
(c) Nature of material
(d) Temperature of the conductor.

Question 33.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
Mendel crossed a pure tall pea plant (TT) with a pure dwarf pea plant (tt) and observed that all the progeny were hybrid tall (Tt), i.e. only one of the traits was able to express itself in the F₁ generation, which is the dominant trait. The other trait is called the recessive trait which remains suppressed.

However, when he self-crossed the plants of F₁ generation, he observed that one-fourth of the plants were dwarf and three-fourths were tall.

The expressed trait T for tallness is dominant, while the trait ‘t’ of dwarfness is recessive. Thus, Mendel’s experiments show that traits may be dominant or recessive. (2)

Section D

Questions No. 34 to 36 are Long Answer Questions.

Question 34.
(a) Rehmat classified the reaction between Methane and Chlorine in the presence of sunlight as a substitution reaction. Support Rehmat’s view with suitable justification and illustrate the reaction with the help of a balanced chemical equation.
(b) Chlorine gas was prepared using electrolysis of brine solution. Write the chemical equation to represent the change. Identify the other products formed in the process and give one application of each.
OR
Raina while doing certain reactions observed that heating of substance ‘X’ with vinegar like smell with a substance ‘Y (which is used as an industrial solvent) in presence of cone. Sulphuric acid on a water bath gives a sweet-smelling liquid ‘Z’ having molecular formula C₄H₈O₂. When heated with caustic soda (NaOH), ‘Z’ gives back the sodium salt of and the compound ‘Y.
Identify ‘X’, ‘Y, and ‘Z’. Illustrate the changes with the help of suitable chemical equations. [5]
Solution:
(a) Rehmat’s observation is correct as the hydrogen atoms are substituted by hetero atom i.e., Cl Reaction between methane and chlorine in presence of sunlight can be repersented as:

In this reaction, the chlorine free radial, abstracts hydrogen from methyl group which then riacts
with Cl₂ to form chioromethane, It is can be represented as

In the above reaction, the hydrogen atoms are substituted by chlorine atom and hence this reaction is considered as substitution reaction.

(b) When electricity is passed through a concentrated solution of NaCl, also called as Brine, it decomposes and results in the formation of Sodium Hydroxide (NaOH), Chlorine gas (Cl₂), and Hydrogen gas (H₂). The chemical equation can be represented as:
2NaCl(aq) + 2H₂O(l) 2NaOH(aq) + Cl₂(g) + H₂(g)
Apart from chlorine gas (Cl₂), the other products formed in the reaction are sodium hydroxide or caustic soda (NaOH) and hydrogen gas (H₂).
Uses
Sodium hydroxide (NaOH) or Caustic soda is used in the preparation of soaps and detergents
Hydrogen gas (H₂) is used in the manufacture of ammonia for fertilizers.
OR
When ethanoic acid with vinegar like smell is heated with ethanol(which is used as an industrial solvent) in the presence of cone, sulphuric acid on a water bath, a sweet-smelling liquid called ethyl ethanoate is formed. The molecular formula of ethyl ethanoate is C₄H₈O₂. The chemical equation can be represented as:

Therefore, the substances,
X – Ethanoic acid/ acetic acid/ CH₃COOH
Y – Ethanol/ Ethyl alcohol/ C₂H₅OH
Z – Ethyl ethanoate/ Ester – CH₃COOC₂H₅
When ethyl ethanoate (Z) is heated with a dilute solution of caustic soda or sodium hydroxide (Na), it gives back the original ethyl alcohol and sodium salt of the original carboxylic acid. Such a reaction is known as a saponification reaction and it can be represented as:

Question 35.
(a) Explain the difference between androecium and gynoecium.
(b) How does reproduction help in providing stability to the population of a species?
Or
(a) What is cellular respiration? How many ATP molecules are obtained by the oxidation of one glucose atom?
(b) With the help of an experiment, show that chlorophyll is necessary for photosynthesis.
Answer:
(a) Differences between androecium and gynoecium are as follows:

Androecium	Gynoecium
This is the male reproductive organ of a plant.	This is the female reproductive organ of a plant.
Each unit of this is called a stamen.	Each unit of this is called a carpel/pistil.
The terminal bloated part of the stamen is called the anther, in which male gametes or pollen grains are produced.	The lower bloated part of the carpel/pistil is called an ovary, in which the ovule is present, A Female gamete or egg is produced in the ovule.

(b) A species occupies a well-defined niche in an ecosystem, using its ability to reproduce. During reproduction, copies of DNA pass from one generation to the next. This copying of DNA takes place with consistency in reproducing organisms and This is important for the maintenance of body design features (physiological as well as structural) which allows the organism to use that particular niche. Reproduction is, therefore, linked to the stability of the population of a species. (2\(\frac{1}{2}\))
Or
(a) In cells, the biochemical process in which glucose is oxidized in the presence of oxygen is known as aerobic or cellular respiration. During the oxidation of food, energy gets released which is stored in the form of ATP through the electron transport system. On complete oxidation, one molecule of glucose forms 38 ATP molecules. (2)

(b) Experiment to show that chlorophyll is necessary for photosynthesis: In this experiment plants having variegated leaves are selected. These plants are kept in the dark for 48 hours to make them starch-free. Then, the paint is researched by keeping it in the sun. After a few hours, the green and non-green areas are marked on a leaf. Following this, the leaf is tested for starch in which only green areas i.e. chlorophyll containing part turn blue-black due to the presence of starch. This shows that chlorophyll is essential for photosynthesis.

Question 36.
The above image shows a thin lens of focal length 5m. [1 + 2 + 2]

(i) What is the kind of lens shown in the above figure?
(ii) If a real inverted image is to be formed by this lens at a distance of 7m from the pole, then show with calculation where should the object be placed?
(iii) Draw a neatly labelled diagram of the image formation mentioned in (ii)
OR
A 10 cm long pencil is placed 5 cm in front of a concave mirror having a radius of curvature of 40 cm. [2 + 1 + 2]
(i) Determine the position of the image formed by this mirror.
(ii) What is the size of the image?
(iii) Draw a ray diagram to show the formation of the image as mentioned in the part (i).
Solution:
(i) Convex Lens
(ii) Given, υ = 7 m, f = 5 m
We know that, \(\frac{1}{f}=\frac{1}{υ}-\frac{1}{u}\)
Putting the values in the equation, we get,
\(\frac{1}{5}=\frac{1}{7}-\frac{1}{u}\)
\(\frac{1}{u}=\frac{1}{7}-\frac{1}{5}=\frac{5-7}{35}=\frac{-2}{35}\)
u = – \(\frac{35}{2}\) = -17.5 m
Hence, the object will be placed 17.5 m on the left of the convex lens.

Section E

Questions No. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts.

Question 37.
Four groups of students were assigned separately to the experiment on the interaction of iron nails with a solution of copper sulphate. Each group recorded the observations as given below in the table.

Group of Students	Initial Colour of Solution	Final Colour of Solution	Change in the Iron Nail
A	Blue	Colourless	Brown Coat
B	Green	Green	Brown Coat
C	Blue	Blue	Brown Coat
D	Blue	Light Green	Brown Coat

(a) Which type of reaction is observed in the given experiment?
(b) Which group of students recorded all the observations correctly? Also, write the chemical equation involved.
Or
Discuss the reason behind the brown coating on iron nails.
Answer:
(a) The change in colour of the initial and final solution and the deposition of a layer on an element indicates the occurrence of displacement reaction. (2)

(b) The blue colour of copper sulphate changes into a light green colour solution due to the formation of ferrous sulphate by a displacement reaction.

The displaced copper gets deposited on the iron nail and forms a brown coating on it. Hence, students of the D group recorded all the correct observations. (2)
Or
The brown coating on iron nails is due to copper deposition. According to the reactivity series, iron is more reactive than copper, so, it will displace copper from its sulphate solution, and then solid copper will get deposited on the iron nail. (2)

Question 38.
Figures (a) to (d) given below represent the type of ear lobes present in a family consisting of 2 children – Rahul, Nisha and their parents. [4]


Excited by his observation of different types of ear lobes present in his family, Rahul conducted a survey of the type of ear lobes found {Figure (e) and (f)} in his classmates. He found two types of ear lobes in his classmates as per the frequency given below:

Sex	Free	Attached
Male	36	14
Female	31	19

Based on the above data answer the following questions.
(a) Which of the two characteristics – ‘free ear lobe’ or ‘attached ear lobe’ appears to be dominant in this case? Why?
(b) Is the inheritance of the free ear lobe linked with sex of the individual? Give reason for your answer.
(c) What type of ear lobe is present in father, mother, Rahul and his sister Nisha? Write the genetic constitution of each of these family members which explains the inheritance of this character in this family?
(Gene for Free ear lobe is represented by F and gene for attached ear lobe is represented by f for writing the genetic constitution).
OR
Suresh’s parents have attached earl obes. What type of ear lobe can be seen in Suresh and his sister , ‘ Siya? Explain by giving the genetic composition of all.
Solution:
(a) Free ear lobe is dominant because it is found in a large majority of the population.

(b) No, it is not sex-linked. As per the data of the family as well as the class, it is indicated that free ear lobe is present in males as well as in females.

(c) Father – Ff (free ear lobe), Mother – Ff (free ear lobe), Rahul – ff (attached ear lobe) and Nisha – Ff (free ear lobe).
OR
Both Suresh’s father and mother have attached ear lobes (ff). If both parents have recessive character, then all the children will have recessive character only. Therefore, Suresh and her sister siya will also have an attached ear lobe (ff).

Suresh’s Father ff (attached ear lobe),
Suresh’s Mother ff (attached ear lobe),
Suresh – ff (attached ear lobe),
Siya – ff (attached ear lobe).

Question 39.

The above images show the position and height of an object in front of a convex lens. A 6 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 25 cm. The distance of the object from the lens is 40 cm.
(a) Based on the text and data in the above paragraph, determine the position of the image.
(b) The magnification produced by the convex lens.
(c) Size and nature of the image formed.
Or
Find the power of the lens.
Answer:
(a) Given, the height of the object, h₀ = 6 cm
Focal length, f = 25 cm
Distance of object, u = -40 cm
Using lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)