Electrical

CBSE Class 10 Science Question Paper 2020 (Series: JBB/1) with Solutions

Students can find that CBSE Previous Year Question Papers Class 10 Science with Solutions and CBSE Class 10 Science Question Paper 2020 (Series: JBB/1) effectively boost their confidence.

CBSE Class 10 Science Board Question Paper 2020 (Series: JBB/1) with Solutions

Time allowed: 3 hours
Maximum marks: 80

General Instructions

  • Question Paper comprises three sections – A, B and C.
    There are 30 questions in the question paper. All questions are compulsory.
  • Section A – questions no. 1 to 14- all questions or part thereof are of one mark each. These questions comprise multiple choice questions (MCQ), very short answer (VSA), and Assertion-Reason type questions. Answers to these questions should be given in one word or one sentence.
  • Section B – questions no. 15 to 24 are short answer type questions, carrying 3 marks each. Answers to these questions should not exceed 50 to 60 words.
  • Section C – questions no. 25 to 30 are long answer type questions, carrying 5 marks each. Answers to these questions should not exceed 80 to 90 words.
  • Answers should be brief and to the point. Also the above mentioned word limit be adhered to as far as possible.
  • There is no overall choice in the question paper. However, an internal choice has been provided in some questions in each Section. Only one of the choices in such questions has to be attempted.
  • In addition to this, separate instructions are given with each section and question, wherever necessary.

SET – I Code No. 31/1/1

Section – A

Question 1.
Name a cyclic unsaturated carbon compound.
Answer:
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 1

Question 2.
List two sources of magnetic fields.
Answer questions number 3(a) to 3(d) and 4(a) to 4(d) on the basis of your understanding of the following paragraphs and the related studied concepts.
Answer:
Two sources of magnetic fields. (i) A bar magnet and (ii) An electromagnet

Question 3.
The growing size of the human population is a cause of concern for all people. The rate of birth and death in a given population will determine its size. Reproduction is the process by which organisms increase their population. The process of sexual maturation for reproduction is gradual and takes place while general body growth is still going on.

Some degree of sexual maturation does not necessarily mean that the mind or body is ready for sexual acts or for having and bringing up children. Various contraceptive devices are being used by human beings to control the size of population.
(a) List two common signs of sexual maturation in boys and girls.
(b) What is the result of reckless female foeticide?
(c) Which contraceptive method changes the hormonal balance of the body?
(d) Write two factors that determine the size of a population.
Answer:
(a) Common signs of sexual maturation in boys and girls:
(i) On attaining puberty, the male gonads called testes start producing male gametes called sperms and the female gonads called ovaries start producing female gametes called ova.
(ii) Hair grow on body parts like armpits, pubic area and face.

(b) For a healthy society, the male-female sex ratio must be maintained at almost the same level. Due to reckless female foeticide, the male-female child sex ratio is declining at an alarming rate in some sections of our society.
(c) Chemical method of contraception changes the hormonal balance of the body.
(d) The rate of births and deaths in a given population will determine the size of a population.

CBSE Class 10 Science Question Paper 2020 (Series: JBB/1) with Solutions

Question 4.
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.
(a) Why is the maximum concentration of pesticides found in human beings?
(b) Give one method which could be applied to reduce our intake of pesticides through food to some extent.
(c) Various steps in a food chain represent:
(i) Food web
(ii) Trophic level
(iii) Ecosystem
(iv) Biomagnification
(d) With regard to various food chains operating in an ecosystem, man is a:
(i) Consumer
(ii) Producer
(iii) Producer and consumer
(iv) Producer and decomposer
Answer:
(a) Pesticides are non-biodegradable and get accumulated progressively at each trophic level. Since humans occupy the top level in any food chain, so the maximum amount of harmful chemical pesticides gets accumulated in human beings.
(b) Washing of foodgrains, vegetables, fruits etc. with warm water is useful to reduce the intake of pesticides through food to some extent.
(c) (ii) Trophic level
(d) (i) Consumer

Question 5.
Calcium oxide reacts vigorously with water to produce slaked lime.
CaO(s) + H2O(l) → Ca(OH)2(aq)
This reaction can be classified as:
(A) Combination reaction
(B) Exothermic reaction
(C) Endothermic reaction
(D) Oxidation reaction
Which of the following is a correct option?
(a) (A) and (C)
(b) (C) and (D)
(c) (A), (C) and (D)
(d) (A) and (B)
OR
When hydrogen sulphide gas is passed through a blue solution of copper sulphate, a black precipitate of copper sulphide is obtained and the sulphuric acid so formed remains in the solution. The reaction is an example of a:
(a) Combination reaction
(b) Displacement reaction
(c) Decomposition reaction
(d) Double displacement reaction
Answer:
(d) (A) and (B)
OR
(d) Double displacement reaction
H2S + CuSO4 → CuS + H2SO4

Question 6.
In a double displacement reaction such as the reaction between sodium sulphate solution and barium chloride solution:
(A) exchange of atoms takes place
(B) exchange of ions takes place
(C) a precipitate is produced
(D) an insoluble salt is produced
The correct option is:
(a) (B) and (D)
(b) (A) and (C)
(c) only (B)
(d) (B), (C) and (D)
Answer:
(d) (B), (C) and (D)
Na2SO4 + BaCl2 → 2NaCl + BaSO4

Question 7.
Baking soda is a mixture of:
(a) Sodium carbonate and acetic acid
(b) Sodium carbonate and tartaric acid
(c) Sodium hydrogen carbonate and tartaric acid
(d) Sodium hydrogen carbonate and acetic acid
Answer:
(c) Sodium hydrogen carbonate and tartaric acid

Question 8.
The chemical formula for Plaster of Paris is:
(a) CaSO4 . 2H2O
(b) CaSO4 . H2O
(c) CaSO4 . 1/2H2O
(d) 2CaSO4 . H2O
Answer:
(c) CaSO4 . 1/2H2O

Question 9.
The laws of reflection hold true for:
(a) plane mirrors only
(b) concave mirrors only
(c) convex mirrors only
(d) all reflecting surfaces
OR
When an object is kept within the focus of a concave mirror, an enlarged image is formed behind the mirror. This image is:
(a) real
(b) inverted
(c) virtual and inverted
(d) virtual and erect
Answer:
(d) all reflecting surfaces
OR
(d) virtual and erect

CBSE Class 10 Science Question Paper 2020 (Series: JBB/1) with Solutions

Question 10.
At the time of short circuit, the electric current in the circuit:
(a) varies continuously
(b) does not change
(c) reduces substantially
(d) increases heavily
OR
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:
(a) 0.4 A
(b) 0.6 A
(c) 0.8 A
(d) 1 A
Answer:
(d) increases heavily
OR
(d) 1 A

Question 11.
Which one of the following is responsible for the sustenance of underground water?
(a) Loss of vegetation cover
(b) Division for high water demanding crops
(c) Pollution from urban wastes
(d) Afforestation
Answer:
(d) Afforestation

Question 12.
Incomplete combustion of coal and petroleum:
(A) increases air pollution
(B) increases efficiency of machines
(C) reduces global warming
(D) produces poisonous gases
The correct option is:
(a) (A) and (B)
(b) (A) and (D)
(c) (B) and (C)
(d) (C) and (D)
Answer:
(b) (A) and (D)

For questions number 13 and 14, 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:
(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.

Question 13.
Assertion (A): Esterification is a process in which a sweet smelling substance is produced. Reason (R): When esters react with sodium hydroxide an alcohol and sodium salt of carboxylic acid are obtained.
Answer:
(b); In esterification, an alcohol reacts with a carboxylic acid in the presence of a mineral acid.

Question 14.
Assertion (A): Energy flow in food chains is always unidirectional.
Reasofn (R): Only 10% energy is transferred to the next trophic level.
Answer:
(b); The flow of energy in food chains is said to be unidirectional because some energy is lost in the form of heat when moving from one trophic level to next.

Section – B

Question 15.
1 g of copper powder was taken in a China dish and heated. What change takes place on heating? When hydrogen gas 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.
Answer:
When copper powder is heated in a China dish, the copper powder surface becomes coated with black colour substance due to the formation of copper oxide by surface oxidation. This is because copper reacts with oxygen in the air upon heating and forms copper oxide.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 2
When hydrogen gas is passed over this heated copper oxide, then the black copper oxide is reduced and red-brown copper metal is obtained because copper oxide is reduced to copper metal whereas hydrogen is oxidised to water.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 3

Question 16.
List the important products of the Chlor-alkali process. Write one important use of each.
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 which can be removed by it.
Answer:
The products of chlor-alkali process:
(i) Sodium hydroxide (caustic soda) NaOH
Use: It is used for making soaps and detergents.

(ii) Chlorine gas (Cl2)
Use: Chlorine is used to sterilize drinking water supply and the water of swimming pools.

(iii) Hydrogen gas (H2)
Use: It is used in the hydrogenation of oils to obtain solid fats (vegetable ghee).
OR
Anhydrous sodium carbonate (soda ash) is dissolved in water and recrystallised to get washing soda crystals containing 10 molecules of water of crystallisation.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 4
The solution of washing soda in water is alkaline which turns red litmus to blue. Washing soda is used for removing permanent hardness of water.

Question 17.
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 first drop by drop to this solution, then in excess.
(i) How is 5% solution of KMnO4 prepared?
(ii) State the role of alkaline potassium permanganate in this reaction. What happens on adding it in excess?
(iii) Write chemical equation of this reaction.
Answer:
(i) For preparing 5% solution of KMnO4, 5 g of KMnO4 is dissolved into 95 mL of hot water.

(ii) Alkaline KMnO4 is a strong oxidising agent because it provides nascent oxygen, i.e.,
freshly generated atomic oxygen which is very-very reactive. On warming ethanol and alkaline KMnO4 solution gently, ethanol is oxidised to ethanoic acid and pink colour of the solution disappears. If excess of KMn04 is added, the purple colour will persist because there is no more alcohol left and hence there is no reaction.

(iii)
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 5

Question 18.
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?
Or
Why is chemical communication better than electrical impulses as a means of communication between cells in a multi-cellular organism?
Answer:
Adrenalin hormone is secreted in large amount in the body of the squirrel during scary situation. It speeds up the heart beat and breathing, raises blood pressure and allows more glucose (carbohydrate) to go into the blood to give it a lot of energy quickly to fight or to run away.
OR
Nerve impulses can reach only those cells which are connected by the nervous tissue. Cells cannot continuously generate and transmit nerve impulses. Thus most multi-cellular organisms use chemical coordination as means of communication between cells.

Question 19.
Define the term pollination. Differentiate between self pollination and cross pollination. What is the significance of pollination?
Answer:
The transfer of pollen grains from anthers (male sexual parts) of a flower to the stigma part of the pistil (female sexual part) is known as pollination. Pollination is done by insects, birds, wind and water.

Pollination can occur in two ways:
(i) Self pollination. When the pollen grains from the anther of a flower are transferred to the stigma of the same flower or another flower on the same plant, it is called self pollination.

(ii) Cross pollination. When the pollen grains from the anther of a flower on one plant are transferred to the stigma of a flower on another similar plant, it is called cross pollination. Significance of pollination. Pollination is important because it leads to the production of fruits and seeds.

Question 20.
State the importance of chromosomal difference between sperms and eggs of human beings.
Answer:
A male has one X chromosome and one Y chromosome. Thus half the sperms will have X chromosomes and the other half will have Y chromosomes.

A female has two X chromosomes. So all the female gametes will have only X chromosomes. If a sperm carrying X chromosome fertilizes an ovum then the child born will be a girl. If a sperm carrying Y chromosome fertilizes an ovum then the child born will be a boy. Thus the chromosomal difference between sperms and eggs of humans determines the sex of the child.

Question 21.
What is Tyndall effect shown by colloidal particles? State four instances of observing the Tyndall effect.
OR
Differentiate between a glass slab and a glass prism. What happens when a narrow beam of (i) a monochromatic light, and (ii) white light passes through (a) glass slab and (b) glass prism?
Answer:
The scattering of light by particles in its path is called Tyndall effect. In a colloidal solution (or colloid), the particles are big enough to scatter light. Four instances of observing the Tyndall effect:

  • The path of light beam becomes visible because the colloidal particles are big enough to scatter light falling on them in all the directions. This scattered light enters our eyes and we are able to see the path of light beam.
  • The way of beam of sunlight becomes visible due to Tyndall effect as it passes through dust particles in the air of a room.
  • Tyndall effect can also be observed when sunlight passes through the canopy of a dense forest.
  • Milk is a colloid that exhibits the Tyndall effect.

OR
Difference between glass slab and glass prism:

Glass slab Glass prism
1. Glass slab is a transparent object having six rectangular faces. Glass prism is a transparent object made of glass having two triangular ends and three rectangular sides.
2. The light emerges from a parallel-sided glass slab in a direction parallel with that in which it enters the glass slab. In refraction through a glass prism, the emergent ray is not parallel to the incident ray.
3. The angle of emergence is equal to the angle of incidence. The incident ray, when produced, cuts the emergent ray at an angle called angle of deviation.

(a) (i) When a narrow beam of monochromatic light passes through a glass slab, the emergent ray coming out of the glass slab is parallel to the incident ray and laterally displaced.
(ii) When a narrow beam of white light travelling in a glass slab comes out into air obliquely then the part of light wave on the left side of beam of light emerges out into the air first and right side of beam of light emerges out into air little later.

(b) (i) When a ray of monochromatic light passes through a prism, it bends towards the base of the prism (thicker part).
(ii) If a narrow beam of white light is passed through a prism, the white light splits to form a band of seven colours.

CBSE Class 10 Science Question Paper 2020 (Series: JBB/1) with Solutions

Question 22.
What eye defect is hypermetropia? Describe with a ray diagram how this defect of vision can be corrected by using an appropriate lens.
Answer:
Long sightedness is hypermetropia. Due to this defect, a person is not able to see the nearby objects clearly but can see the distant objects clearly. Causes of long-sightedness. It is caused due to the following reasons:

  • Normal increase in the focal length of the eye lens. The lens becomes less convergent.
  • Shortening of the eyeball size.

Long sightedness can be corrected by using a convex lens of suitable focal length in the spectacles of such a person.
When a convex lens of suitable power is placed in front of the hypermetropic eye then the diverging rays of light coming from the nearby object are first converged by this convex lens. Due to this, the convex lens forms a virtual image of the nearby object at a point near to the hypermetropic eye. Then the hypermetropic eye can easily focus the image formed by convex lens on the retina.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 6

Question 23.
A V-I graph for a nichrome wire is given below. What do you infer from this graph? Draw a labelled circuit diagram to obtain such a graph.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 7
Answer:
The V-I graph for a nichrome wire is found to be a straight line passing through the origin. This shows that Current (A) is directly proportional to the Potential Difference (V). In other words, as the potential difference V increases, the current I also increases and therefore ratio \(\left(\frac{\mathrm{V}}{\mathrm{I}}\right)\) remains constant.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 8

Question 24.
(a) Write the mathematical expression for Joule’s law of heating.
(b) Compute the heat generated while transferring 96,000 coulomb of charge in two hours through a potential difference of 40 V.
Answer:
(a) According to Joule’s law of heating given by the formula, H = I2 × R × t. It is clear that heat produced in a wire is directly proportional to:
(i) square of current (I2)
(ii) resistance of wire (R)
(iii) time (t), for which current is passed.
(b) Given. Charge, Q = 96,000 C; Time, t = 2 hrs.; Potential difference, V = 40 V
Heat generated, H = ?
As we know, Current, I = \(\frac { Q }{ t }\)
I = \(\frac{96,000}{2 \times 60 \times 60}\) ….. [∵ 1 hour = 60 × 60 seconds
I = 13.33 A
We will now calculate the Resistance by using Ohm’s law
R = \(\frac { Q }{ t }\) = \(\frac{40}{13.33}\) = 3
⇒ R = 3Ω
∴ Heat Produced, H = I2 × R × t
= (13.33)2 × 3 × 2 × 60 × 60 = 3838080.24 J = 3838.08 kJ
Thus, the heat produced is 3838.08 kilo joules.

Section – C

Question 25.
Carbon cannot reduce the oxides of sodium, magnesium and aluminium to their respective metals. Why? Where are these metals placed in the reactivity series? How are these metals obtained from their ores? Take an example to explain the process of extraction along with chemical equations.
Answer:
The oxides of highly reactive metals (like potassium, sodium, calcium, magnesium and aluminium) are very stable and cannot be reduced by the most common reducing agent ‘carbon’ to obtain free metals. This is because these metals have more affinity for oxygen than carbon. So, carbon is unable to remove oxygen from these metal oxides and hence cannot convert them into free metals.

Thus, the highly reactive metals cannot be extracted by reducing their oxides with carbon. The highly reactive metals such as potassium, sodium, calcium, magnesium and aluminium are placed high up in the reactivity series in its upper part. The highly reactive metals are extracted by the electrolytic reduction of their molten chlorides and oxides.

Extraction of sodium metal: Sodium metal is extracted by the electrolytic reduction (or electrolysis) of molten sodium chloride. When electric current is passed through molten sodium chloride, it decomposes to form sodium metal and chloride gas:
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 9
The formation of sodium and chlorine by the electrolysis of molten sodium chloride can be explained as follows:
Molten sodium chloride (NaCl) contains free sodium ions (Na+) and free chloride ions (Cl). During the electrolysis of molten sodium chloride, the following reactions take place at the two electrodes:

(i) The positive sodium ions (Na+) are attracted to the cathode (negative electrode). The sodium ions take electrons from the cathode and get reduced to form sodium atoms (or sodium metal):
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 10
Thus, sodium metal is produced at cathode (negative electrode).

(ii) The negative chloride ions (Cl) are attracted to the anode (positive electrode). The chloride ions give electrons to the anode and get oxidised to form chlorine gas:
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 11
Thus, chlorine gas is formed at the anode (positive electrode).

Question 26.
(a) In the formation of compound between two atoms A and B, A loses two electrons and B gains one electron.
(i) What is the nature of bond between A and B?
(ii) Suggest the formula of the compound formed between A and B.
(b) On similar lines explain the formation of MgCl2 molecule.
(c) Common salt conducts electricity only in the molten state. Why?
(d) Why is melting point of NaCl high?
OR
The metals extracted from their ores are not very pure. They contain impurities, which can be removed by the process of refining. Name the most widely used process of refining impure metals. Draw a diagram of the apparatus used for refining of copper metal and state:
(i) The name of the rods which are used as cathode and anode.
(ii) The electrolyte used during the process.
(iii) What happens to the pure metal when current passes through the electrolyte?
(iv) What happens to the soluble and insoluble impurities present in the impure copper?
Answer:
(a)
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 12
(i) Ionic bond is formed between A and B.
(ii)
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 13
(b) Mg has 2 electrons in its outermost shell so it loses its 2 electrons to achieve the inert gas configuration of eight valence electrons and forms positively charged ion or divalent cation.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 14
(ii) Cl has 7 electrons in its outermost shell so it gains one electron to achieve the stable inert gas configuration and forms negatively charged ion or monovalent anion.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 15
(c) Common salt (NaCl) is an ionic compound which conducts electricity only in molten state because in molten state the
electrostatic forces of attraction between oppositely charged ions (Na+ and Cl) are overcome due to heat. Thus the ions move freely and conduct electricity.

(d) NaCl is an ionic compound so there is a strong force of attraction between the positively charged sodium ion and negatively charged chloride ion. Therefore a considerable amount of energy is required to break the strong interionic attraction. Thus NaCl has high melting point.
OR
The most common method for refining of impure metals is electrolytic refining.
Electrolytic Refining of Copper:
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 16
(i) A thin strip of pure metal is made the cathode whereas impure metal is made the anode.
(ii) Aqueous solution of salt of the metal to be refined.
(iii) On passing the current through the electrolyte, pure metal from the anode dissolves into the electrolyte. An equivalent amount of pure copper metal from the electrolyte gets deposited on the cathode. Pure copper metal is collected at the cathode.
(iv) The soluble impurities go into the solution while the insoluble impurities settle down at the bottom of the anode and are known as anode mud.

Question 27.
(a) Why is there a difference in the rate of breathing between aquatic organisms and terrestrial Organisms? Explain.
(b) Draw a diagram of human respiratory system and label – pharynx, trachea, lungs, diaphragm and alveolar sac on it.
OR
(a) Name the organs that form the excretory system in human beings.
(b) Describe in brief how urine is produced in human body.
Answer:
(a) Animals have evolved different organs for the uptake of oxygen from the environment and for getting rid of the carbon dioxide produced. Terrestrial animals can breathe the oxygen in the atmosphere, but animals that live in water need to use the oxygen dissolved in water. Since the amount of dissolved oxygen is fairly low compared to the amount of oxygen in the air, the rate of breathing in aquatic organisms is much faster than that seen in terrestrial organisms.

Fishes take in water through their mouths and force it past the gills where the dissolved oxygen is taken up by blood. Terrestrial organisms use the oxygen in the atmosphere for respiration.
(b)
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 17
OR
(a) The excretory system of human beings consists of the following main organs:
Two kidneys, a pair of ureters, bladder and urethra.

(b) The dirty blood containing waste like urea (brought by renal artery) enters the glomerulus. The glomerulus filters the blood. During Alteration, the substances like glucose, amino acids, salts, water and urea etc. present in the blood pass into Bowman’s capsule. Waste substances pass through the tubule, then the useful substances like all glucose, all amino acids, most salts and most water etc. are reabsorbed into the blood through blood capillaries surrounding the tubule.

Only the waste substances urea, some unwanted salts and some water remain behind in the tubule. The liquid left behind in the tubule of nephron is urine. The nephron carries this urine into the collecting duct of the kidney from where it is carried to ureter. From the ureter, urine passes into urinary bladder. Urine is stored in the bladder for some time and is ultimately passed out of the body through urethra.

Question 28.
(a) What is the law of dominance of traits? Explain with an example.
(b) Write and explain law of segregation (separation) of traits with the help of a flow chart.
Answer:
(a) According to this law, the characteristics (or traits) of an organism are determined by internal ‘factors’ which occur in pairs. Only one of a pair of such factors can be presented in a single gamete. This law explains expression of only one of the parental character in F1 generation and expression of both in F2 generation.
Example:
(i) Mendel first crossed pure-bred tall pea plants with pure-bred dwarf pea plants and found that only tall pea plants were produced in the first generation or . F1 generation. No dwarf pea plants were obtained in the first generation of progeny. From this Mendel concluded that the F1 generation showed the traits of only one of the parent plants: tallness being the dominant trait. The trait of other parent plant, dwarfness, being recessive did not show up in the progeny of first generation.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 18
(ii) In the F2 generation, both the tall and dwarf traits are present in the ratio of 3 : 1. This showed that the traits for tallness and dwarfness are present in the F1 generation, Dwarfness being the recessive trait does not express itself in the presence of tallness, the dominant trait.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 19
We see trait (gene) for tallness and dwarfness were together in F1 generation but these got segregated or separated during gametes formation to produce F2 generation. This is known as law of segregation or separation of traits.

Question 29.
Draw a ray diagram in each of the following cases to show the formation of image, when the object is placed:
(i) between optical centre and principal focus of a convex lens.
(ii) anywhere in front of a concave lens.
(iii) at 2F of a convex lens.
State the signs and values of magnifications in the above mentioned cases (i) and (ii).
OR
An object 4.0 cm in size, is placed 25.0 cm in front of a concave mirror of focal length 15.0 cm.
(i) At what distance from the mirror should a screen be placed in order to obtain a sharp image?
(ii) Find the size of the image.
(iii) Draw a ray diagram to show the formation of image in this case.
Answer:
(i) Formation of image, when the object A’ is placed between optical centre and principal focus of a convex lens:
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 20
The image formed is virtual and erect, so the sign of magnification will be positive.

(ii) Formation of image, when the object is placed anywhere in front of a concave lens:
Now, m = \(+\frac{h_1}{h_2}\)
∴ m < + 1
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 21
The image formed is virtual and erect. Since image formed is diminished, therefore the value of magnification will be less than one.

(iii) Formation of image, when the object is placed at 2F of a convex lens:
Now, m = \(+\frac{h_1}{h_2}\)
∴ m = -1
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 22
OR
Given. Height of object, h1 = 4 cm, u = -25 cm; f = -15 cm; v = ?; h2 = ?
(i) By applying mirror formula,
\(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
⇒ \(\frac { 1 }{ v }\) + \(\frac { 1 }{ -25 }\) = \(\frac { 1 }{ -15 }\)
⇒ \(\frac { 1 }{ v }\) – \(\frac { 1 }{ 25 }\) = \(\frac { -1 }{ 15 }\)
⇒ \(\frac { 1 }{ v }\) = \(\frac { -1 }{ 15 }\) + \(\frac { 1 }{ 25 }\) = \(\frac{-5+3}{75}\) = \(\frac{-2}{75}\)
⇒ v = \(\frac { -75 }{ 2 }\) cm = -37.5 cm
A real sharp image is formed on the screen when it is placed at 37.5 cm in front of the mirror.

(ii) Now, \(\frac{h_2}{h_1}\) = \(\frac{-v}{u}\)
⇒ \(\frac{h_2}{4}\) = \(\frac{-\left(\frac{-75}{2}\right)}{-25}\)
∴ h2 = \(\frac{-75}{2 \times 25}\) × 4 = -6 cm
Therefore, a magnified inverted image of 6 cm is obtained.

(iii)
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 23
Object is placed between F and C in front of the concave mirror.

CBSE Class 10 Science Question Paper 2020 (Series: JBB/1) with Solutions

Question 30.
(a) What is an electromagnet? List any two uses.
(b) Draw a labelled diagram to show how an electromagnet is made.
(c) State the purpose of soft iron core used in making an electromagnet.
(d) List two ways of increasing the strength of an electromagnet if the material of the electromagnet is fixed.
Answer:
(a) An electromagnet is a magnet consisting of a long coil of insulated copper wire Wrapped around a soft iron core that is magnetised only when electric current is passed through the coil. Uses. Electric bells, magnetic levitation, etc.

(b) Diagram to show how an electromagnet is made:
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 24

(c) The core of an electromagnet must be of soft iron because soft iron loses all of its magnetism when current in the coil is switched off.
(d) The strength of an electromagnet depends on –

  • the number of turns in the coil. If we increase the number of turns in the coil, the strength of electromagnet increases.
  • the current flowing in the coil. If the current in the coil is increased, the strength of electromagnet increases.

SET – II Code No. 31/1/2
Except for the following questions, all the remaining questions have been asked in Set I.

Section – A

Question 1.
Name the functional group present in propanone.
Answer:
Propanone functional group = Ketone (> C = 0)

Question 5.
The compound obtained on reaction of iron with steam is/are:
(a) Fe2O3
(b) Fe3O4
(c) FeO
(d) Fe2O3 and Fe3O4
OR
An element ‘X’ reacts with O2 to give a compound with a high melting point. This compound is also soluble in water. The element ‘X’ is likely to be:
(a) Iron
(b) Calcium
(c) Carbon
(d) Silicon
Answer:
(b) Fe3O4
OR
(b) Calcium

Question 11.
In an ecosystem, 10% of energy available for transfer from one trophic level to the next is in the form of:
(a) Heat energy
(b) Chemical energy
(c) Mechanical energy
(d) Light energy
Answer:
(b) Chemical energy

Question 12.
Soil fertility is determined by its ability to:
(a) Decay organic matter
(b) Hold organic matter
(c) Hold water
(d) Support life
Answer:
(d) Support life

For question number 13, 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:
(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.

Question 13.
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.
Answer:
(c); The difference between the molecular masses of the two consecutive numbers of a homologous series is CH2.

Section – B

Question 15.
What is ‘rusting’? Describe with a labelled diagram an activity to investigate the conditions under which iron rusts.
Answer:
When an iron object is left in damp air (or water) for a considerable time, it gets covered with a red-brown flaky substance called rust. Thus the corrosion of iron is called rusting. Chemically rust is hydrated iron (III) oxide, Fe2O3 . xH2O.

Activity to investigate the conditions under which iron rusts. Take three test tubes. Mark them as A, B and C respectively. Place few clean iron nails in each of these test tubes. Pour some ordinary water in test tube ‘A’. Pour some boiled and distilled water in test tube ‘B’ and put 1 ml of oil into it. Then put some anhydrous calcium chloride in test tube ‘C’. Cork all the three test tubes tightly. Anhydrous calcium chloride will absorb all the moisture from test tube ‘C’.

Leave these test tubes for some days and then observe. You will observe that nails of test tube ‘A’ get rusted and the nails of test tubes ‘B’ and ‘C’ do not get rusted. Nails of test tube ‘A’ got both moisture and oxygen of air. Nails of test tube ‘B’ got only pure water while the nails of test tube ‘C’ were only exposed to dry air. This activity shows that moisture and air both are needed for the rusting of iron metal articles.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 25

Question 19.
(a) List in tabulor form two differences between binary fission and multiple fission.
(b) What happens when a mature spirogyra filament attains considerable length?
Answer:
Difference between binary fission and multiple fission

Binary fission Multiple fission
In binary fission, the parent organism splits to form two organisms. In multiple fission, the parent organism splits to form many new organisms at the same time.
In binary fission, the division of nucleus occurs followed by the division of cytoplasm. In multiple fission, a cyst or protective wall is formed (during unfavourable condition) in which the nucleus splits into many smaller daughter nuclei having little bit cytoplasm around each daughter nuclei.
Example. This type of fission occurs in amoeba, paramecium, leishmania etc. Example. This type of fission occurs in plasmodium etc.

(b) When a mature spirogyra filament attains considerable length, it breaks into two or more fragments and each fragment then grows into a new spirogyra.

Question 23.
(a) State the relation correlating the electric current flowing in a conductor and the ‘ voltage applied across it. Also draw a graph to show this relationship.
(b) Find the resistance of a conductor if the electric current flowing through it is 0.35 A when the potential difference across it is 1.4 V.
Answer:
(a) At a constant temperature, the current flowing through a conductor is directly proportional to the potential difference (voltage) across its ends.
I ∝ V
⇒ V ∝ I
V = R × I
⇒ \(\frac { V }{ I }\) = R (resistance) which is a constant
The relationship between current and voltage is given by Ohm’s law.

The graph between Potential Difference (V) and corresponding current values (I) is a straight line passing through the origin.
Since the current potential difference graph is a straight line, it shows that current is directly proportional to the potential difference.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 26
(b) Given. R = ?; I = 0.35 A; V = 1.4 V
V = IR (Ohm’s law)
∴ Resistance, R = \(\frac { V }{ I }\) =\(\frac{14 \times 100}{10 \times 35}\) = \(\frac{14 \times 10}{35}\) = 4 Ω

Section – C

Question 25.
(a) What is thermit process? Where is this process used? Write balanced chemical equation for the reaction involved.
(b) Where does the metal aluminium, used in the process, occur in the reactivity series of metals?
(c) Name the substances that are getting oxidised and reduced in the process.
Answer:
(a) The reduction of a metal oxide to form metal by using aluminium powder as a reducing agent is called a thermite reaction (or thermit process).
A mixture of iron (III) oxide and aluminium powder is ignited with a burning magnesium ribbon. Aluminium reduces iron oxide to produce iron metal with the evaluation of lot of heat. Due to this heat, iron metal is produced in the molten state.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 27
The molten iron is then poured between the broken iron pieces to weld them (to join them).

(b) Aluminum is used as a reducing agent in those cases where the oxide is of a comparatively lesser reactive metal than zinc which cannot be satisfactorily reduced by carbon, as more reactive metal (aluminium) can displace a comparatively less reactive metal from its metal oxide to give free metal. For example, manganese and chromium which are moderately reactive metals are extracted by the reduction of their oxides with aluminum powder.

(c) Aluminium powder reduces the metal oxide to metal and is itself oxidised to aluminium oxide.

CBSE Class 10 Science Question Paper 2020 (Series: JBB/1) with Solutions

Question 28.
(a) What is genetics?
(b) What are genes? Where are the genes located?
(c) Why is DNA copying necessary during reproduction?
Answer:
(a) Gentics is a branch of biology concerned with the study of genes, genetic variation and heredity in organisms. In other words, genetics is the scientific study of the way that the development of living things is controlled by qualities that have been passed from parents to children.

(b) Genes are units of heredity which transfer characteristics (or traits) from parents to their offsprings during reproduction. A gene is a unit of DNA on a chromosome which governs the synthesis of one protein that controls a specific characteristics of an organism.

(c) DNA copying is essential during reproduction for the inheritance of features from parents to the next generation.

SET – III Code No. 31/1/3
Except for the following questions, all the remaining questions have been asked in Set I & Set II.

Section – A

Question 2.
State an important advantage of using alternating current (a.c.) over direct current (d.c.).
Answer:
Alternating current can be transmitted over long distances without much loss of electrical energy.

Question 7.
When sodium hydrogen carbonate is added to ethanoic acid a gas evolves. Consider the following statements about the gas evolved?
(A) It turns lime water milky.
(B) It is evolved with a brisk effervescence.
(C) It has a smell of burning sulphur.
(D) It is also a by-product of respiration.
The correct statements are:
(a) (A) and (C) only
(b) (B) and (D) only
(c) (A), (C) and (D)
(d) (A), (B) and (D)
Answer:
(d) (A), (B) and (D)

Question 8.
When a small amount of acid is added to water, the phenomena which occur are:
(A) Dilution
(B) Neutralisation
(C) Formation of H3O+ ions
(D) Salt formation
The correct statements are:
(a) (A) and (C)
(b) (B) and (D)
(c) (A) and (B)
(d) (C) and (D)
Answer:
(a) (A) and (C)

Question 9.
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?
(a) (A) and (D)
(b) (B) and (D)
(c) (A) and (B)
(d) (B) and (C)
OR
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:
(a) (A) and (D)
(b) (A) and (B)
(c) (A), (B) and (C)
(d) (A), (B) and (D)
Answer:
(C) are always erect.
OR
(c) (A), (B) and (C)

For question number 14, 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:
(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.

Question 14.
Assertion (A): The power of convex lens is positive and that of concave lens is negative. Reason (R): The degree of convergence or divergence of light rays achieved by a lens is called its power.
Answer:
(b); The focal length of convex surface is positive whereas the focal length of concave surface is negative.

Section – B

Question 17.
(a) Draw the structure for (i) ethanol, (ii) ethanoic acid.
(b) Why is the conversion of ethanol to ethanoic acid considered an oxidation reaction? Write the oxidising agent used in the reaction involved.
Answer:
(a)
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 28
(b) Conversion of ethanol to ethanoic acid is considered as oxidation reaction because addition of oxygen takes place.
Alkaline KMnO4 or Acidified K2Cr2O7 is used as an oxidising agent in the conversion of ethanol to ethanoic acid.

Question 18.
Name the parts (a) to (e) in the following diagram.
What is the term given to the sequence of events occurring in the diagram?
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 29
(a) What is tropism?
(b) How do auxins promote the growth of a tendril around a support?
Answer:
(a) Stimulus
(b) Sensory neuron
(c) Spinal cord
(d) Relay neuron
(e) motor neuron
Reflex action is the term given to the sequence of events in the diagram.
OR
(a) A growth movement of a plant part in response to an external stimulus in which the direction of stimulus determines the direction of response is called tropism. Tropism is a directional movement of a plant caused by its growth.

(b) Tendrils are sensitive to touch (or contact) of other objects. When a tendril touches an object, then the distribution of auxin at the tip of the tendril gets disturbed. The auxin hormone collects the side of the tendril which is not touching the solid object. As a result the side of the tendril in contact with the object grows slowly than its other side. This causes the tendril to bend towards the object by growing towards it, wind around the object and cling to it.

Question 23.
The near point of the eye of a person is 50 cm. Find the nature and power of the corrective lens required by the person to enable him to see clearly the objects placed at 25 cm from the eye.
Answer:
Given. Near point of the eye of a person = 50 cm
v = -50 cm;
u = -25 cm;
f = ?
Using lens formula, \(\frac { 1 }{ v }\) – \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
⇒ \(\frac { 1 }{ -50 }\) – \(\frac { 1 }{ -25 }\) = \(\frac { 1 }{ f }\)
⇒ \(\frac { -1 }{ 50 }\) + ⇒ \(\frac { 1 }{ 25 }\) = \(\frac { 1 }{ f }\)
⇒ \(\frac { 1 }{ f }\) = \(\frac{-1+2}{50}\) = \(\frac { 1 }{ 50 }\)
∴ f = +50 cm
Now, power of lens, P = \(\frac{1}{f \text { (in metres) }}\)
∴ P = \(\frac{1 \times 100}{+50}\) = + 2D
Therefore, a convex lens of power +2 diopter is required by the person to see clearly.

Section – C

Question 27.
(a) A gas is released during photosynthesis. Name the gas and also state the way by which the gas is evolved.
(b) What are stomata? What governs the opening and closing of stomata?
Or
(a) Draw a diagram of human alimentary canal and label – gall bladder, pancreas, liver and small intestine on it.
(b) Give two reasons to explain why absorption of digested food occurs mainly in the small intestine.
Answer:
(a) Oxygen gas is released during photosynthesis. Oxygen gas is released due to the splitting of water into hydrogen and oxygen by light energy in the chlorophyll.

(b) Stomata is a pore present on the epidermis of leaves and mostly found on a leaf’s lower surface. Stomata are capable of opening and closing as per the surrounding environmental conditions. They help in exchange of gases during respiration and photosynthesis. Opening and closing of stomata: The opening and closing of stomatal pores is controlled by guard cells. When water flows into the guard cells, they swell, become curved and cause the pore to open.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 30
On the other hand, when the guard cells lose water, they shrink, become straight and close the stomatal pore. A large amount of water is also lost from the cells of the plant leaves through open stomatal pores. So, when the plant does not need carbon dioxide and wants to conserve water, the stomatal pores are closed.
OR
(a)
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 31
(b) Reasons to explain why absorption of digested food occurs mainly in the small intestine:
(i) The small intestine is the main region for absorption of digested food.
(ii) The inner surface of small intestine has millions of tiny, finger-like projections called villi. The presence of villi gives the inner walls of the small intestine a very large surface area and the large surface area of small intestine helps in the rapid absorption of digested food.

CBSE Class 10 Science Question Paper 2020 (Series: JBB/1) with Solutions

Question 30.
(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 ‘n’ times as large as that produced by a single turn (loop)?
Answer:
(a) The pattern of magnetic field due to current carrying circular loop are circular near the current carrying loop but the concentric circles representing magnetic field lines become bigger and bigger. At the centre of the circular loop, the magnetic field lines are straight. Therefore on moving away from the centre of circular loop, magnetic field lines keep on diverging.
CBSE Class 10 Science Question Paper 2020 Series JBB 1 with Solutions Img 32
(b) If there is a circular coil having n turns, the magnetic field produced by this current carrying circular coil will be n times as large as that produced by a circular loop of a single turn of wire, because the current in each circular turn of coil flows in the same direction and magnetic field produced by each turn of circular coil then just adds up.


Show More
यौगिक किसे कहते हैं? परिभाषा, प्रकार और विशेषताएं | Yogik Kise Kahate Hain Circuit Breaker Kya Hai Ohm ka Niyam Power Factor Kya hai Basic Electrical in Hindi Interview Questions In Hindi