Class 10 Science: CBSE Sample Question Paper (2020-21) - 5 | CBSE Sample Papers For Class 10 PDF Download

Class-X
Science Theory
TIME: 3 Hrs.
M.M: 80
General Instructions:
Read the following instructions very carefully and strictly follow them :
1. The question paper comprises four sections A, B, C and D. There are 36 questions in the question paper.
All questions are compulsory.
2. Section-A - Question no. 1 to 20 - all questions and parts thereof are of one mark each. These questions contain multiple-choice questions (MCQs), very short answer questions and assertion - reason type questions. Answers to these should be given in one word or one sentence.
3. Section-B - Question no. 21 to 26 are short answer type questions, carrying 2 marks each. Answers to these questions should be in the range of 30 to 50 words.
4. Section-C - Question no. 27 to 33 are short answer type questions, carrying 3 marks each. Answers to these questions should be in the range of 50 to 80 words.
5. Section-D - Question no. - 34 to 36 are long answer type questions carrying 5 marks each. Answers to these questions should be in the range of 80 to 120 words.
6. There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
7. Wherever necessary, neat and properly labelled diagrams should be drawn.

SECTION A

Q.1. What is the role of acid in our stomach?    (1 Mark)
Ans. HCl of gastric juice disinfects the food and acidifies it for the proper functioning of the proteolytic enzyme pepsin.

Q.2. Write the number of covalent bonds in the molecules of butane C₄H₁₀?    (1 Mark)
Ans. Thirteen covalent bonds.

Q.3. State one reason for placing Mg and Ca in the same group of the periodic table?    (1 Mark)
Ans. Due to the presence of 2 electrons in the valence shell and similar chemical properties.

Q.4. The human body is made up of five important components, of which water is the main component. Food as well as potable water are essential for every human being. The food is obtained from plants through agriculture, Pesticides are being used extensively for a high yield in the fields. These pesticides are absorbed by the plants from the soil along with water and minerals and from the water bodies, these pesticides are taken up by the aquatic animals and plants. As these chemicals are not biodegradable, they get accumulated progressively at each trophic level. The maximum concentration of these chemicals gets accumulated in our bodies and greatly affects the health of our mind and body.
(i) Why is the maximum concentration of pesticides found in human beings?     (1 Mark)
Ans. It is because humans are at the top of the food chain and due to biomagnification, the concentration of pesticides increases as one goes up the trophic levels.

(ii) Give one method which could be applied to reduce our intake of pesticides through food to some extent.      (1 Mark)
Ans. Organic farming should be done or more biopesticides should be used.

(iii) Various steps in a food chain represent:     (1 Mark)
(a) Food web 
(b) Trophic level
(c) Ecosystem 
(d) Biomagnification
Ans. b

(iv) With regard to various food chains operating in an ecosystem, man is a:      (1 Mark)
(a) Consumer
(b) Producer
(c) Producer and consumer
(d) Producer and decomposer
Ans. a

Q.5. Two bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1 A. The current through the 40 W bulb will be:      (1 Mark)
(a) 0.4 A 
(b) 0.6 A
(c) 0.8 A 
(d) 1 A
Ans. d
Solution:
1 A (because in series, current will remain the same.)

Q.6. Which one of the following is responsible for the sustenance of underground water?      (1 Mark)
(a) Loss of vegetation cover 
(b) Diversion for high water demanding crops
(c) Pollution from urban wastes 
(d) Afforestation
Ans. d
Solution:
Sustenance of the groundwater means the maintenance of groundwater, which can be achieved by afforestation.
Afforestation is the process of planting trees. Trees roots and trunks will form a barrier for run-off water, it will help seepage of water thus, increasing the groundwater level.
Thus, 'Afforestation' is responsible for the sustenance of groundwater.

Q.7. For question number 7, two statements are given - one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
Assertion (A): In a homologous series of alcohols, the formula for the second member is C₂H₅OH and the third member is C₃H₇OH.
Reason (R): The difference between the molecular masses of the two consecutive members of a homologous series is 144.       (1 Mark)
Options~
(a) Both A and R are true and R is the correct explanation of the Assertion.
(b) Both A and R are true but R is not the correct explanation of the Assertion.
(c) A is true but R is false.
(d) A is false but R is true.
Ans. c
Solution:
In homologous series of alcohols, the formula for the second member is C₂H₅OH and the third member is C₃H₇OH. The difference between the molecular masses of the two consecutive members of a homologous series is 14.

Q.8. On which factor does the colour of the scattered white light depend?    (1 Mark)
Ans. Size of the particles of the medium through which it is passing.

Q.9. A prism ABC (with BC as a base) is placed in different orientations. A narrow beam of white light is incident on the prism as shown in Figure. In which of the following cases, after dispersion, the third colour from the top corresponds to the colour of the sky ?    (1 Mark)
(i)

(ii)

(iii)

(iv)

Options~
(a) (i) 
(b) (ii)
(c) (iii)
(d) (iv)
Ans. (b)
Solution:
If the prism is kept with base BC at the bottom, then the emergent band of colour would show violet at the bottom. If the prism is kept with base BC at the top, then violet would be at the top; followed by indigo and blue.

Q.10. What is the role of acid in our stomach?    (1 Mark)
Ans. HCl of gastric juice disinfects the food and acidifies it for the proper functioning of the proteolytic enzyme pepsin.

OR

Name the raw material required for photosynthesis.
Ans. CO₂ and water.
Q.11. List two functions of the ovary of the human female reproductive system.    (1 Mark)
Ans. Two functions of Ovary :
(i) To produce female gamete/ovum.
(ii) To secrete female hormones: estrogen and progesterone.

Q.12. What is the magnification of the images formed by the plane mirror and why?    (1 Mark)
Ans. Its magnification is +1 because the plane mirror always forms an image equal to the object's size.

Q.13. A real image is formed by the light rays after reflection or refraction when they :
(A) actually meet or intersect with each other.
(B) actually, converge at a point.
(C) appear to meet when they are produced in the backward direction.
(D) appear to diverge from a point.
Which of the above statements are correct?      (1 Mark)
Options~
(a) (A) and (D) 
(b) (B) and (D)
(c) (A) and (B) 
(d) (B) and (C)
Ans. c

Q.14 Consider the following properties of virtual images :
(A) cannot be projected on the screen
(B) are formed by both concave and convex lens
(C) are always erect
(D) are always inverted
The correct properties are :      (1 Mark)
(a) (A) and (D) 
(b) (A) and (B)
(c) (A), (B) and (C) 
(d) (A), (B) and (D)
Ans. c

Instructions: Q. No 15 contain five sub-parts each. You are expected to answer any four sub-parts in these questions.
Q.15. Read the following and answer any four questions from 15 (i) to 15 (v).    (1 Mark)
Metallic Character
The ability of an atom to donate electrons and form a positive ion (cation) is known as electropositivity or metallic character. Down the group, metallic character increases due to an increase in atomic size and across the period, from left to right electropositivity decreases due to a decrease in atomic size.
Non-Metallic Character
The ability of an atom to accept electrons to form a negative ion (anion) is called a non-metallic character or electronegativity. The elements having high electro-negativity have a higher tendency to gain electrons and form an anion. Down the group, electronegativity decreases due to an increase in atomic size and across the period, from left to right electronegativity increases due to a decrease in atomic size.
(i) Which of the following correctly represents the decreasing order of metallic character of Alkali metals plotted in the graph?
(a) Cs > Rb > Li > Na > K
(b) K > Rb > Li > Na > Cs
(c) Cs > Rb > K > Na > Li
(d) Cs > K > Rb > Na > Li
Ans. (i) (c) Cs > Rb > K > Na > Li

(ii) Hydrogen is placed along with Alkali metals in the modern periodic table though it shows nonmetallic character
(a) as Hydrogen has one electron & readily loses an electron to form a negative ion
(b) as Hydrogen can easily lose one electron like alkali metals to form a positive ion
(c) as Hydrogen can gain one electron easily like Halogens to form a negative ion
(d) as Hydrogen shows the properties of non-metals
Ans. (b) As Hydrogen can easily lose one electron like alkali metals to form a positive ion.

(iii) Which of the following has the highest electronegativity?
(a) F
(b) Cl
(c) Br
(d) I
Ans. (a) F

(iv) Identify the reason for the gradual change in electronegativity in halogens down the group.
(a) Electronegativity increases down the group due to a decrease in atomic size
(b) Electronegativity decreases down the group due to a decrease in the tendency to lose electrons
(c) Electronegativity decreases down the group due to an increase in the atomic radius/ tendency to gain electron decreases
(d) Electronegativity increases down the group due to an increase in forces of attractions between nucleus & valence electrons
Ans. (c) Electronegativity decreases down the group due to an increase in the atomic radius/ tendency to gain electron decreases.

(v) Which of the following reason correctly justifies that “Fluorine (72 pm) has a smaller atomic radius than Lithium (152 pm)”?
(a) F and Li are in the same group. Atomic size increases down the group
(b) F and Li are in the same period. Atomic size increases across the period due to the increase in the number of shells.
(c) F and Li are in the same group. Atomic size decreases down the group
(d) F and Li are in the same period and across the period atomic size/radius decreases from left to right.
Ans. (d) F and Li are in the same period and across the period atomic size/radius decreases from left to right.

Q.16. Two solutions X and Y are tested with universal indicator. Solution X turns orange whereas solution Y turns red. Which of the solutions is a stronger acid?    (1 Mark)
Ans. Solution Y is a stronger acid.

Q.17. Make a distinction between metals and non-metals with respect to the nature of their oxide.    (1 Mark)
Ans. Metallic oxides are basic, few are amphoteric.
Non-metallic oxides are acidic, few are neutral.

Q.18. Write the molecular formula of the 2nd and the 3rd member of the homologous series whose first member is methane.    (1 Mark)
Ans. Ethane (C₂H₆)
Propane (C₃H₈)

Q.19. Directions: Q. No 19 contain five sub-parts each. You are expected to answer any four subparts in this question.    (1 Mark)
Study the given flow chart and answer any four questions.
(a) Identify X, Y and Z?
(i) X-Glycolysis, Y-Anaerobic, Z - Aerobic
(ii) X - Krebs's cycle, Y - Aerobic, Z-Anaerobic
(iii) X - Glycolysis, Y - Aerobic, Z - Anaerobic
(iv) X - Glycolysis, Y - Aerobic, Z - Krebs's cycle
Ans. (iii) X - Glycolysis, Y - Aerobic, Z – Anaerobic

(b) The process X occurs in and Y occurs in the part of the cell.
(i) Mitochondria and cytoplasm respectively
(ii) Cytoplasm and mitochondria respectively
(iii) Both takes place in the cytoplasm
(iv) Both takes place in mitochondria
Ans. (ii) Cytoplasm, mitochondria respectively.

(c) In which of these organisms the process Z takes place?
(i) Bacteria
(ii) Humans
(iii) Yeast
(iv) Spirogyra
Ans. (iii) Yeast

(d) In which part of the human body do the process Z takes place?
(i) In muscle cells
(ii) In kidneys
(iii) In liver cells
(iv) In Leydig’s cell
Ans. (i) In muscle cells
(e) Where does aerobic respiration occur in a cell?
(i) Mitochondria
(ii) Cytoplasm
(iii) Nucleus
(iv) Plastid
Ans. (i) Aerobic respiration occurs in the mitochondria of the cell.

Q.20. Directions: Q. No 20 contain five sub-parts each. You are expected to answer any four subparts in this question.    (1 Mark)
From the following part of the periodic table, answer any of the four questions :
(a) Which is the most reactive metal?
(i) P
(ii) Q
(iii) Y
(iv) Z
Ans. (iv) Z

(b) The family of fluorine Q, R, T is known as :
(i) Halogens
(ii) Alkali metal
(iii) Transition metal
(iv) Noble gases
Ans. (i) Halogens.

(c) Which of these represent the correct element for each of group 2 and 15?
(i) Group 2- Magnesium, Group 15- Nitrogen
(ii) Group 2- Nitrogen, Group 15- Magnesium
(iii) Group 2- Carbon, Group 15- Nitrogen
(iv) Group 2- Magnesium, Group 15- Bromine
Ans. (i) Group 2- Magnesium, Group 15- Nitrogen.

(d) What is the name of the element P placed below carbon?
(i) Silicon
(ii) Hydrogen
(iii) Nitrogen
(iv) Magnesium
Ans. (i) Silicon

(e) Compare X and P with respect to the size of atoms.
(i) X is bigger in size than P
(ii) X is smaller in size than P
(iii) X is the same as P
(iv) None of these
Ans. X has a bigger size than P because X has a less effective nuclear charge.

Section B

Q.21. Name one nitrogenous waste present in urine. What is the basic filtration unit of the kidney called?    (2 Mark)
Ans. Nitrogenous waste present in urine is uric acid or urea. The basic filtration unit of the kidney is the nephron.

Q.22. A man with blood group 'A' marries a woman with blood group 'O' and their daughter has blood group 'O'. Is this information enough to tell you which of the traits - blood group 'A' or 'O' is dominant? What criteria would you use to assess it?     (2 Mark)
Ans. The given information is not enough to decide which blood group is dominant.
But universally, blood group A is dominant whereas blood group O is recessive. Here, the father’s Blood group has genotype AA or genotype AO, whereas that of the mother can be OA or OO.
For a daughter to be born with blood group O, she must receive O type gene-one each from father and mother. For this father blood group must have genotype AO and the mother’s blood group must have genotype OO.

Q.23. (a) Why did Mendel choose garden pea for his experiments? Write two reasons.    (2 Mark)
(b) List two contrasting visible characters of garden pea Mendel used for his experiment.
Ans. (a) Reasons :
(i) Pea plant is small and easy to grow.
(ii) A large number of true-breeding varieties of the pea plant are available.
(iii) Short life cycle.
(iv) Both self and cross-pollination can be made possible.
(b) Contrasting characters :
(i) Round / Wrinkled seeds
(ii) Tall / Short plants
(iii) White / Purple flowers
(iv) Green / Yellow seeds

Q.24. Explain the processes of aerobic respiration in mitochondria of a cell and anaerobic respiration in yeast and muscle with the help of word equations ?    (2 Mark)
Ans.

Q.25. A student observes the above phenomenon in the lab as white light passes through a prism. Among many other colours, he observed the position of the two colours Red and Violet.
What is the phenomenon called? What is the reason for the violet light to bend more than the red light?    (2 Mark)
Ans. The phenomenon is called dispersion. The speed of violet light inside the prism is slowest and that of red is the highest. Hence, the deviation of violet light is maximum and that of red is minimum.

Q.26. Name the parts (a) to (e) in the following diagram.     (3 Mark)What is the term given to the sequence of events occurring in the diagram?
Ans. a- Receptor, b- Sensory neuron, c- Spinal cord, d- Relay neuron, e- Effector.
The term given to the sequence of events occurring in the diagram is the reflex arc.

OR

(a) What is tropism? How do auxins promote the growth of a tendril around a support?
Ans. (a) Tropism: It is the directional growth movement of a plant organ in response to an external stimulus. Auxins produced in the shoot tip move downwards in the plant. These auxins cause cell elongation in the growing tissues. In the tendrils, auxins move away from the point of contact with the supporting object. More growth occurs on the side away from the support. As a result of unequal growth on the two sides, the tendril coils around the support.

Section C

Q.27. Name the plant Mendel used for his experiment? What type of progeny was obtained by Mendel in F₁ and F₂ generations when he crossed the tall and short plants? Write the ratio he obtained in F₂ generation plants.    (3 Mark)
Ans. Pea Plant / Garden pea / Pisum sativum F₁ – All tall; 
F₂- Tall and short Phenotypic Ratio – Tall:Short

Detailed Answer: Mendel used Pisum sativum (Pea plant) for his experiment. Mendel took a tall pea (TT) plant and a short pea (tt) plant. When he crossed both, the first filial generation (F₁) obtained were tall. When F₁ progeny was self-pollinated, all plants obtained in F₂ generation were not tall. Instead, three tall peas (dominant) plants and one short pea (recessive) plant were obtained.

The phenotypic ratio and genotypic ratio should be in the same horizontal line. 

Q.28. 1 g of copper powder was taken in a China dish and heated. What change takes place on heating? When hydrogen is passed over this heated substance, a visible change is seen in it. Give the chemical equations of reactions, the name and the colour of the products formed in each case.    (3 Mark) 
Ans. The black colour substance is formed by the reaction of copper with oxygen is
Copper (II) oxide (CuO).
Chemical Reaction: 2Cu + O₂ → 2CuO
Hydrogen gas is passed over this heated material (CuO) the black coating on the surface turns brown as the reverse reaction takes place and copper is obtained.

Q.29. List the important products of the Chlor-alkali process. Write one important use of each.    (3 Mark) 
Ans. Products of the Chlor-alkali process are: Sodium hydroxide (NaOH), Chlorine gas (Cl₂) and Hydrogen gas (H₂).
Uses of sodium hydroxide:
(i) In the manufacture of paper.
(ii) In making soaps and detergents.
Uses of chlorine gas:
(i) In the production of bleaching powder
(ii) To make plastics (PVC), pesticides, chloroform, paints etc.
Uses of hydrogen gas:
(i) As fuel for rockets
(ii) In the hydrogenation of oils to obtain vegetable ghee.

OR

How is washing soda prepared from sodium carbonate? Give its chemical equation. State the type of this salt. Name the type of hardness of water that can be removed by it?
Ans. Washing soda is prepared from sodium carbonate by recrystallisation.         
Na₂CO₃ + 10H2O → Na₂CO₃.10H₂O
Washing soda is a basic salt. It is used for removing the permanent hardness of the water.

Q.30. 3 mL of ethanol is taken in a test tube and warmed gently in a water bath. A 5% solution of alkaline potassium permanganate is added the first drop by drop to this solution, then in excess.    (3 Mark) 

(i) How is the 5% solution of KMnO4 prepared?
Ans. Preparation of 5 g of KMnO₄: By dissolving 5 g potassium permanganate in 100 mL of water.

(ii) State the role of alkaline potassium permanganate in this reaction. What happens on adding it in excess?
Ans. (ii) Alkaline KMnO₄ acts as an oxidizing agent as it adds oxygen to alcohol and converts it into an acid.
Initially, when we add potassium permanganate all potassium permanganate is used up in the reaction. After completion of the reaction, there is no more ethanol in the solution. Adding more potassium permanganate after this endpoint makes the solution red.

(iii) Write the chemical equation of this reaction.
Ans. 

Q.31. A squirrel is in a scary situation. Its body has to prepare for either fighting or running away. State the immediate changes that take place in its body so that the squirrel is able to either fight or run?    (3 Mark) 
Ans. Adrenaline hormone will be secreted in the body as the squirrel is in a scary
situation. This will result in :

• Speeding up of heartbeat
• Rise in blood pressure
• Release of more glucose in the blood.

OR

Why is chemical communication better than electrical impulses as a means of communication between cells in a multi-cellular organism?
Ans. Chemical communication is better than electrical impulses because chemical communication is mediated through hormones that can diffuse to different regions of the body, thereby allowing cells to communicate even without interacting with each other. Moreover, this type of communication can be maintained at a steady rate and is easy to regulate.

Q.32. Define the term pollination. Differentiate between self-pollination and cross-pollination. What is the significance of pollination?    (3 Mark) 
Ans. The transfer of pollen grains from the anther to the stigma of a flower is known as pollination. The two types of pollination:
(a) Self-pollination: When the pollen grains from the stamens of a flower fall on the stigma of the same flower, then it is called self-pollination.
(b) Cross-pollination: When pollen grains from the stamens of a flower fall on the stigma of another flower, it is called cross-pollination.
Significance of pollination:
(i) It is a significant event because it precedes fertilization.
(ii) It brings the male and female gametes closer to the process of fertilization.
(iii) Cross-pollination introduces variations in plants because of the mixing of different genes. These variations further increase the adaptability of plants towards the environment or surroundings.

Q.33. What are homologous structures? Give an example. Is it necessary that homologous structures always have a common ancestor? Justify your answer.    (3 Mark)
Ans. Homologous Structures: Structure that is similar in origin but performs different functions. For e.g. forelimbs of humans and the wings of birds perform different functions but their skeletal structures are similar.
Yes, homology indicates common ancestry. Homologous organs follow the same basic plan of the organization during their development but in the adult condition, these organs are modified to perform different functions as an adaptation to different environments.

Section D

Q.34. (a) Explain with the help of the pattern of magnetic field lines the distribution of magnetic field due to a current-carrying a circular loop.
(b) Why is it that the magnetic field of a current-carrying coil having n turns, is V times as large as that produced by a single turn (loop)?     (5 Mark)
Ans. (a) Magnetic field due to current through a circular loop: It can be represented by a concentric circle at every point. Circles become larger and larger as we move away. Every point on a wire carrying current would rise to a magnetic field appearing as the straight line at the centre of the loop. The direction of the magnetic field inside the loop is the same.

Fig: Magnetic field lines due to a current through a circular loop
(b) Magnetic field is directly proportional to the number of turns in the coil. As the number of turns in the coil increase, the magnetic strength at the centre increases, because the current in each circular turn is having the same direction, thus the field due to each turn adds up.

Q.35. (A) A reaction between 'X' and 'Y' forms compound 'Z'. 'X' Loses an electron and 'Y' gains an electron. Show the formation of compound 'Z'. What type of compound is it?
Write two properties shown by compound 'Z'.
(B) Compare the properties of metals and non-metals on the basis of the following:
(i) Malleability and ductility
(ii) Conductivity
(iii) Nature of the oxide formed by them.     (5 Mark)
Ans.
The type of compound ‘Z‘ is ionic compound Properties of Ionic compound ‘Z‘.
Ionic compounds are solids and hard because of a strong force of attraction between the positive and negative ions.
They have high melting and boiling points as a considerable amount of energy is required to break the strong inter-ionic attraction.
They are generally soluble in water and insoluble in organic solvents like kerosene, petrol etc.
They conduct electricity in the molten state as the ions can move freely and conduct electricity. (Any two)
(B) (iii) Nature of the oxide formed by them.

OR

You are given a hammer, a battery, a bulb, wires and a switch.
(A) How would you use them to distinguish between samples of metals and non-metals?
(B) Assess the usefulness of these tests in distinguishing between metals and non-metals.
Ans. (A) Given: a hammer, a battery, a bulb, wires and a switch.
(1) Using a hammer: If the sample breaks, this means it is brittle and hence it is a non-metal. On the other hand, if the sample flattens to form a sheet, this means that the sample is malleable and hence is a metal.
One can also use a battery, wires, bulb and switch to make a circuit and use the given material to complete the circuit one by one, if the electric current flows and the bulb glow then it is metal and if not then it is non-metal.
(2) Though these tests give us a rough idea in distinguishing between metals and non-metals they are not reliable because there are exceptions both in metals as well as non-metals. For example Sodium (Na) metal is brittle and graphite can show conductivity.
(B) (1) The most reactive element is Y as it displaces both X and Z from their compounds.
(2) The least reactive element is Z as it is being replaced by both X and Y.

Q.36. (a) What is the thermit process? Where is this process used? Write the balanced chemical equation for the reaction involved.     
(b) Where does the metal aluminium, used in the process, occurs in the reactivity series of metals?
(c) Name the substances that are getting oxidised and reduced in the process.    (5 Mark)
Ans. (a) Thermite reaction: Reaction in which iron oxide reacts with aluminium to
produce molten iron. It is an exothermic process.
Fe₂O₃(s) + 2Al(s) → 2Fe(l) + Al₂O₃(s) + Heat
The thermite reaction is used to join railway tracks or cracked machine parts. This process is called thermite welding.
(b) As aluminium is more reactive than iron, so it is placed above iron in the reactivity series.
(c) Aluminium is getting oxidized to aluminium oxide and iron oxide is getting reduced to iron.

OR

(a) What is genetics?
(b) What are genes? Where are the genes located?
(c) State and define three factors responsible for the rise of a new species.    (5 Mark) 
Ans. (a) Genetics: Branch of biology that deals with the study of genes and heredity in organisms.
(b) Genes are the basic unit of heredity. They are linear segments of DNA that code for a gene product. Genes are located on chromosomes.
(c) Factors responsible for the rise of a new species (speciation) are :
(i) Geographic isolation: wherein geographic barrier prevents interaction between species. Over a period of time, the sub-populations become more and more diversified from one another and finally form two different species.
(ii) Genetic drift: The accidental change in the frequency of genes in a small population is called genetic drift.
(iii) Natural selection: It is the process that results in the increased survival and reproductive success of individuals that are well suited to their environment.