CBSE Sample Papers for Class 10 Science Set 10 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 10 Science with Solutions Set 10 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Science Set 10 with Solutions

Time : 3 Hr.
Max. Marks : 80

General Instructions:

• This question paper consists of 39 questions in 5 sections.
• Alt 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 02 marks each. Answers to these questions should in the range of 30 to 50 words.
• Section C consists of 7 Short Answer type questions carrying 03 marks each. Answers to these questions should in the range of 50 to 80 words.
• Section D consists of 3 Long Answer type questions carrying 05 marks each. Answer to these questions should be in the range of 80 to 120 words.
• Section E consists of 3 source-based/case-based units of assessment of 04 marks each with sub-parts.

SECTION – A (20 Marks)
(Select and write one most appropriate option out of the four options given for each of the questions 1-20)

Question 1.
In naturopathy, it is suggested that in the morning, we should drink water kept in a copper vessel overnight. However, copper vessel gets tarnished after a few days. Why do they get tarnished due to the formation of:
(a) green layer
(b) brown layer
(c) blue layer
(d) red layer (1)
Answer:
(a) green layer

Explanation: Copper vessels got tarnished due to the formation of green coloured layer of basic copper carbonate
(Cu(OH)₂.CuCO₃) when copper reacts with the water vapour and carbon dioxide present in air.

Caution
Students usually get confused as they related corrosion with brown layer only. When copper metal is attacked by substances around it like moisture, air, acids etc. a green coating is formed on its surface due to the process of corrosion.

Question 2.
A teacher demonstrated the usefulness of an indicator for testing acidity or basicity of a solution. The teacher performed an experiment and obtained the following graph between volume of alkali added and pH:

The teacher further remarked that to be useful, an indicator must change over the “vertical” section of the curve where there is a large change in pH for the addition of a very small amount of alkali.
In the above activity, the only suitable indicator is:
(a) methyl orange
(b) litmus
(c) phenolphthalein
(d) none of these (1)
Answer:
(c) phenolphthalien

Explanation: Both phenolphthalein and methyl orange are synthetic indicators whereas litmus and vanilla essence are natural indicators.

Question 3.
Identify chemical equation of the reaction involved when a rod of zinc metal is dipped into a solution of copper sulphate. (1)
(a) Zn + CuSO₄ -> 4Cu + ZnSO₄
(b) 2Zn + 2CUSO₄ -> 2Cu + ZnSO₄
(c) Zn + CuSO₄ -> Cu + ZnSO₄
(d) Zn + 2CuSO₄ -> 2ZnSO₄ + Cu
Answer:
(c) Zn + CuSO₄ -> Cu + ZnSO₄
Explanation: When a rod of zinc metal is dipped into a solution of copper sulphate then the zinc reacts with it and displaces the copper from copper sulphate solution as it is more reactive than copper. The blue colour of copper sulphate solution will change to colourless.
The chemical equation for the reaction taking place is:
Zn + CuSO₄ -> Cu + ZnSO₄

Question 4.
If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 6Q% of the same population, which trait is likely to have arisen earlier? (1)
(a) Trait A
(b) Trait B
(c) Both A and B
(d) Can not say
Answer:
(b) Trait B

Explanation: Trait B is likely to have arisen earlier because in asexual reproduction traits are carried from parents to offspring with least variations and as trait B has higher percentage (60%) as compared to trait A So, it is likely to have arisen earlier.

Question 5.
Which of the following is an example of a triple covalent bond?

Answer:

Explanation: A triple covalent bond is formed when two atoms share three pairs of electrons among them to complete their octet. For example, in order to attain an octet, each nitrogen atom (Atomic number = 7) in a molecule of nitrogen contributes three electrons giving rise to three shared pairs of electrons.

Question 6.
When we touch the leaves of a chhui-mui or touch-me-not plant of mimosa family, they begin to fold up and droop as shown in the figure. (1)

Plants show two types of movements. Which of the following movement is shown by touch-me-not in the figures?
(I) Movement dependent on growth
(II) Movement independent of growth
(III) Nastic movement
(IV) Tropic movement Select the correct option:
(a) (I) and (III)
(b) (I) and (IV)
(c) (II) and (III)
(d) (II) and (IV)
Answer:
(a) (I) and (III)

Explanation: The leaves of the sensitive plant move very quickly in response to tauch. There is no growth involed in this movement, the movement which are non- directional and occur due to turgor changes and reveal immediate response to stimuli but do not involve growth.

Question 7.
How will the image formed by a convex lens change when an object moves closer to the lens? (1)
(a) Image becomes highly magnified
(b) Image diminishes
(c) Size of image increases slightly
(d) Size remains unchanged
Answer:
(a) Image becomes highly magnified

Explanation: As the object moves closer to the lens, the image distance increases and size of the image increases from a point size to highly magnified image. When the object is placed with in the focal length of the lens, a virtual and magnified image is formed on the same side of the lens as the object.

8. What is the aim of the given experiment?

(a) To show that chlorophyll is necessary for photosynthesis
(b) To show that sunlight is necessary for photosynthesis
(c) To show that oxygen is released as a result of photosynthesis
(d) To show that leaves can be double coloured. (1)
Answer:
(a) To show that chlorophyll is necessary for photosynthesis

Explanation: The given figure is that of an experiment to demonstrate that chlorophyll is essential for photosynthesis.

Question 9.
A student carries out an experiment and plots the V-l graph of three samples of nichrome wire with resistances R₁, R₂ and R₃ respectively as shown in figure.

Identify the correct order:
(a) R₁ > R₂ > R₃
(b) R₁ > R₃ > R₂
(c) R₂ > R₃ > R₁
(d) R₃ > R₂ > R₁ (1)
Answer:
(d) R₃ > R₂ > R₁
Explanation: According to Ohm’s law
V = IR
R = \(\frac { V }{ I }\)
R ∝ \(\frac { 1 }{ l }\)
As can be seen from the graph, for a given value of V, the current for R3 is less than R2 which is less than R1.

This means that R₁ < R₂ < R₃.

Question 10.
The potential difference across the wire having fixed resistance is tripled. By how much does the electric power increase? (1)
(a) 9 times
(b) 3 times
(c) \(\frac { 1 }{ 3 }\)times
(d) \(\frac { 1 }{ 9 }\) times
Answer:
(a) 9 times

Explanation: The relation between electric power, potential difference and current is given by P = VI.
Further, according to Ohm’s law,
V = IR.
So, if V is tripled, I will also be tripled for a fixed resistance R.
This means power will be P = (3V)(3I)
= 9 times

Question 11.
Hydrotropism is the directional movement of plant part in response to water stimulus. Study the following figures and select which appears more accurate.

Answer:

Explanation: The root of seedling gets water that dozes out of clay-pot burred in the soil.
That is why, the root of the seedling grows by bending towards the direction of math (positive hydrotropism).

Question 12.
A child is curious to know the bending of plant towards light during growth as shown in the given figure.

Her friend gave her different reasons to explain, but one of the reasons is incorrect. Select that one.
(a) When a growing plant detects light, gibberellin hormone is synthesised at the shoot tip which helps the cells to grow longer.
(b) The hormone synthesised diffuses towards the shady side of the plant.
(c) High concentration of the hormone in a the shaded region stimulate the cells to grow longer as compared to the region exposed to light.
(d) The plant appears to bend to wards light. (1)
Answer:
(a) When a growing plant detects light, gibberellin hormone is synthesized at the shoot tip which helps the cells to grow longer.

Explanation: Auxin is a phytohormone which is synthesized at shoot this which helps the cells to grow longer. Gibberellin helps in the growth of the stem.

Question 13.
An astronaut in space finds sky to be dark. What is the reason for this observation?
(a) Due to absence of atmosphere
(b) Due to absence of scattering of light
(c) Both (a) and (b)
(d) Due to refraction of light. (1)
Answer:
(c) Both (a) and (b)

Explanation: An astronaut in space find sky to be dark because there is no atmosphere in space and hence no scattering of light takes place. The phenomenon of change in the direction of light on striking an obstacle like an atom, a molecule, dust particle, water droplet etc is known as scattering of light. It involves bouncing off of electromagnetic radiation by atoms/molecules of the medium through which they are travelling.

Related Theory
Very fine particles scatter mainly blue light while particles of larger size scatter light of longer wavelength. If the size of the scattering particle is large enough, then the scattered light may even appear white.

Question 14.
Directional movements in response to stimulus which are dependent on growth are tropic movements plants show tropic movements in response to various stimuli. Study the following flow chart and answer the marked I, II, III and IV boxes.

(a) (I) – Phototropism, (II) – Geotropism, (III) – Chemotropism, (IV) – Hydrotropism
(b) (I) – Hydrotropism, (II) – Chemokropism, (III) – Geotropism, (IV) – Phototropism
(c) (I) – Chemotropism, (II) – Hydrotropism, (III) – Geotropism, (IV) – Phototropism
(d) (I) – Hydrotropism, (II) – Chemtropism, (III) – Phototropism, (IV) – Geotropism (1)
Answer:
(b) I – Hydrotropism, II – Chemotropism, III – Geotropism, IV – Phototropism

Explanation: Phototropism- Movement towards light. e.g., growth of shoot towards light.

Geotropism- Movement towards gravity e.g., growth of roots towards the soil.

Chemotropism- Movement towards or amay from chemicals. e.g., growth of pollen tube towards ovule.

Hydrotropism: Movement towards water e.g., growth of roots towards water.

It we keep the potted plant horizontally for 2-3 days, shoots grow upwards and away from the soil and show positive phototropic and negative geotropic movement. While growth roots is directed in downward direction and away from light. So roots growth shows positive geotropism and negative phototropism.

Question 15.
Why does the sun appear white at noon?
(a) Due to high scattering of white light
(b) Due to least scattering of white light
(c) Due to high dispersion of white light
(d) Due to least dispersion of white light (1)
Answer:
(b) Due to least scattering of white light

Explanation: The sun appears white at noon because when the sun is nearly overhead, the sunlight has to pass through much smaller portion of earth’s atmosphere due to which scattering is much less and the sun looks white.

Question 16.
Seed germination is affected by several environmental factors such as temperature, light, moisture, pH of soil, presence of ions etc. An experiment was conducted to study the effect of soil pH on the germination of seeds.

After analyzing the above graph, a student noted the following observations:

Select the correct observation
(a) pH of the soil has no effect on the growth of seedling.
(b) Best growth was observed in seeds which were watered with pH 3
(c) Best growth was observed in seeds which were watered with pH 6.5
(d) Seeds which were watered with pH 4 and 6.5 showed similar results. (1)
Answer:
(c) Best growth was observed in seeds which were watered with pH 6.5.

Explanation: Seed germination is affected by the pH of the soil as seeds require an optimum pH range between 6.5 – 7.5. Too much acidic soil or too much basic soil will adversely affect seed germination.

Q. no 17 to 20 are Assertion – Reasoning based questions.
These consist of two statements – Assertion (A) and Reason (R). Answer these questions 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 and 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): Covalent compounds are formed by sharing of valence electrons between the ions.
Reason (R): Hydrogen molecule is formed by mutual sharing of electrons between two hydrogen atoms. (1)
Answer:
(d) A is false but R is true

Explanation: Covalent compounds are those that are created when two or more atoms share their valence electrons, for example, two hydrogen atoms sharing an electron to create a hydrogen molecule.

Caution
Students should know that in contrast to ionic compounds, which are created by the complete transfer of electrons from one atom to another, covalent compounds are not ionic compounds.

Question 18.
Assertion (A): The opening and closing of the stomatal pore is a function of the guard cells.
Reason (R): Stomatal pores are the site for exchange of gases by diffusion. (1)
Answer:
(b) Both A and R are true and R is not the correct explanation of A.

Explanation: Both statements are correct but the given reason does not explain how guard cells open and close stomatal pores. When water flows into the guard cells, they swell, become curved and cause the pore to open. Similarly, when the guard cells lose water, they shrink and the stomatal pore closes.

Question 19.
Assertion (A): Ozone shields the surface of the earth from ultraviolet radiation from the sun.
Reason (R): Ozone at the higher levels of the atmosphere is a product of UV radiations acting on oxygen molecule. (1)
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation: Ozone is formed when the high energy UV radiations act on oxygen molecules (O₂) and split it into free oxygen atoms (O) which combine with the molecular oxygen to form ozone (O₃). Ozone thus shields the earth’s surface from the harmful effects of UV radiations.

Question 20.
Assertion (A): Condom protects a person from the sexually transmitted diseases.
Reason (R): Condom prevents the sperms from meeting the ovum by acting as a barrier. (1)
Answer:
(b) Both A and R are true and R is the correct explanation of A.

Explanation: Condoms are used by males. These are rubber or plastic sheaths which are put on penis before copulation. The condom prevents the sperms from meeting the ovum (or egg) by acting as a barrier between them. The benefit of condom is that it protects a person from syphilis and AIDS.

SECTION – B (12 Marks)
(Q. no. 21 to 26 are very short answer questions.)

Question 21.
When is recessive trait capable of expressing itself? Write its expression with respect to height of plant (genotype).(2)
Answer:
Recessive trait is expressed only when the recessive alleles are present in a homozygous condition, i.e., the individual has both the alleles as recessive alleles. Short plant height (t) is the recessive trait in a plant. It is expressed when the individual plant has both the alleles as ‘t’. The genotype of a plant with short height will be ‘tt’. A recessive trait can express itself only when both the alleles carry the recessive genes. On the other hand, a plant will be tall even when it has a single ‘T’, which is a dominant allele.

Question 22.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us? (2)
Answer:
The two properties of carbon that give rise to a large number of carbon compounds or organic compounds are as follows:

Catenation: Carbon has the unique ability to form bonds with other atoms of carbon, giving rise to large molecules. This property is called catenation. These compounds may have long chains of carbon, branched chains of carbon or even carbon atoms arranged in rings.

Tetravalency: Since, carbon has a valency of four, it is capable of bonding with four other atoms of carbon or atoms of some other mono-valent element. Compounds of carbon are formed with oxygen, hydrogen, nitrogen, sulphur, chlorine and many other elements.

Question 23.
In two different beakers, a student made solutions of (i) an acid and (ii) a base.

He neglected to label the solutions, and the lab was lacking in litmus paper. How could he tell one solution from the other when they were both colourless? (2)
Answer:
In order to properly respond to this question, we must make some assumptions. Let’s assume the lab has all the necessary equipment, but there isn’t any litmus paper. To determine which of the beakers contains acid and which one contains a base, we can use phenolphthalein. In addition, there are other natural indicators we can use, such as the China rose or turmeric.

Question 24.
The refractive indices of four media P, Q, R and S are given in the following table:

Medium	P	Q	R	S
Refractive Index	1.33	1.50	1.52	2.40

(A) If light, travels from one medium to another, in which case the change in speed will be minimum.
(B) If light, travels from one medium to another, in which case the change in speed will be maximum. (2)
Answer:
(A) Minimum change is seen as light moves between 1.50 and 1.52, i.e. Q and R. This is because the refractive index of a medium is inversely proportional to the speed of light in that medium. As refractive index of both Q and S are nearly equal, minimum change in speed of light will be seen here.

(B) Maximum change is seen when light moves between 1.33 and 2.40, i.e., P and S due to the big difference in their refractive indices.

Question 25.
What are the requisites of any respiratory surface to become highly efficient?
OR
Major amount of water is selectively reabsorbed by the tubular part of nephron in humans. What are the factors on which the amount of water reabsorbed depends? (2)
Answer:
The requisites of any respiratory to become highly efficient are:
(1) Large surface area which is in contact with the oxygen-rich atmosphere.
(2) The surface should be fine and delicate so that exchange of oxygen and carbon dioxide gases can take place by diffusion.

OR

The factors on which the amount of water reabsorbed depends:
(1) how much excess water is there in the body.
(2) how much of dissolved waste is there to be excreted.

Question 26.
Akash took a solution of CuSO₄ and kept in an iron pot. After a few days, the iron pot was found to have a number of holes in it. Explain the reason in terms of reactivity. Write the equation of the reaction involved.
OR
Explain the formation of ionic compound, Al₂O₃ with electron-dot structure. (Given: atomic no. of Al and O are 13 and 8 respectively) (2)
Answer:
When copper sulphate solution is placed in an iron pot, iron reacts with copper sulphate and forms iron sulphate and copper. The blue colour of copper sulphate solution fades to light green due to the formation of iron sulphate and holes are produced at places where iron metal has reacted.

In reactivity series, iron is placed above copper. So, we can say that iron is more reactive as compared to copper and it can displace copper from its compounds.

The equation of the displacement reaction taking place is:
Fe + CuSO₄ → FeSO₂ + Cu

OR

The electronic configuration of oxygen is (2, 6) and that of aluminium is (2, 8, 3). Oxygen requires only 2 electrons in the valence shell to acquire the nearest noble gas (Neon) configuration and form O^2- ion. Al has to lose 3 electrons to attain the nearest inert gas configuration and form Al³⁺ ion. As one oxygen atom can gain only 2 electrons, therefore 2 atoms of Al and 3 atoms of oxygen are required to form Al₂O₃ by the transfer of electrons as shown below:

SECTION – C (21 Marks)
(Q.no. 27 to 33 are short answer questions.)

Question 27.
To demonstrate a chemical reaction, Rahul took 2 g of silver chloride in a china dish and placed in sunlight.

Answer the following questions based on the above activity:
(A) What will you observe after some time?
(B) What is this reaction known as?
(C) Write the balanced chemical equation for this reaction. Give reason for the observation. (3)
Answer:
(A) When silver chloride is kept in sunlight for some time, we will observe white crystals of silver chloride turning grey in sunlight.
(B) This reaction is known as photo decomposition reaction.
(C) Reaction taking place is:

Decomposition of silver chloride into silver and chlorine takes place due to sunlight.

Question 28.
A red haired woman marries a brown haired man, and all the children are brown haired. Explain this genetically.
Answer:
A red haired woman marries a brown haired man, and all the children are brown haired. The brown hair colour genes are dominant to the red hair colour genes. A brown haired man can have BB factors or Bb as only one dominant factor expresses itself in the next generation.

Situation I:

Situation II:

(1) In situation I brown haired man has a pure strain i.e., BB and all the children are brown haired.

(2) In situation II brown haired man carries recessive red coloured hair trait gametes which fertilized and results into 50% brown and 50% red haired children. So in the given situation, brown haired man has passed the pure factors i.e, BB because all the children are brown haired. That is why the children were having brown coloured hairs.

Question 29.
Is the circuit given below correct? Give an explanation for your response.

Answer:
The provided circuit is incorrect because a circuit’s ammeter and voltmeter are always connected in series and parallel to resistor R₁, respectively. The following figure depicts the proper circuit.

Question 30.
Answer the following:
(A) What will happen to a plant if its xylem is removed?
(B) Given below is the diagram of a stomatal apparatus.

Label the parts (i), (ii), (iii) and (iv). (3)
Answer:
(A) the xylem of the plant is removed, upward movement of water will stop leading to wilting of leaves and ultimately causes the death of a plant. In the absence of water, the plant will not be able to prepare food and also perform other essential activities.

(B)
(i) Epidermal cell
(ii) Subsidiary cell
(iii) Guard cell
(iv) Stomatal aperture

Question 31.
List the properties of magnetic field lines. (3)
Answer:
Properties of magnetic field Lines:
(1) They emerge from the north pole of a magnet and enter at the south pole of the magnet.

(2) Inside the magnet, the direction of field lines is from south pole to its north pole. Hence, magnetic field lines are closed curves. The relative strength of the magnetic field is shown by the degree of closeness of the field lines. Crowded field lines represent the strong magnetic field. They are dense close to the poles and sparse away from them. It means magnetic field is strongest around poles of the magnet.

(3) The magnetic field at any point is represented by the tangent at that point. No two field lines intersect each other. If they intersect, it would mean that at point of intersection there would be two field directions, which is not possible.

Question 32.
A man placed an object of 6 cm in height at 20 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed to obtain a sharp image of the object. Calculate the height of the image.
OR
The radius of curvature of a convex mirror used as a rear view mirror in a moving car is 2.0 m. The truck is coming from behind it at a distance of 3.5 m. Calculate position and size of the image relative to the size of truck. What will be the nature of the image? (3)
Answer:
Distance of object
u = -20 cm.
Focal length of concave mirror
f = -15 cm.
Height of object
h = + 6 cm.
Distance of image
v = ?
Height of image h’ = ?
According to mirror formula:

Thus, a screen placed infront of mirror at a distance of 60 cm from it.
Magnification m = \(\frac { h’ }{ h }\) = \(\frac { -v }{ h }\)
\(\frac { h’ }{ +6 }\) = \(\frac { (-60) }{ (20) }\)
h’ = \(\frac { (-60×6) }{ 20 }\) = 18cm

Thus, an inverted image of height 18 cm is formed.

OR

Given:
Radius of curvature,
R = 2m
Focal length, f = \(\frac { R }{ 2 }\) = \(\frac { 2 }{ 2 }\)
= + 1m (convex mirror)
(A convex mirror has a virtual focus and hence focal length is + ve)
Object distance,u = -3.5 m
We know that
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ v }\) = \(\frac { 1 }{ f }\) – \(\frac { 1 }{ u }\)
= \(\frac { 1 }{ 1 }\) – \(\frac { 1 }{ (-3.5) }\)
= 1 + \(\frac { 1 }{ 3.5 }\)
= \(\frac { 4.5 }{ 3.5 }\)
v = \(\frac { 3.5 }{ 4.5 }\) = 0.77 m
Therefore image distance v = 0.77 m
Image will be formed at a distance of 0.77 m from the mirror.
Now, magnification (m)
m = –\(\frac { v }{ u }\)
= \(\frac { 0.77 }{ (-3.5) }\) = 0.22 m
Thus, we see that the magnification is less than 1.
Therefore the image will be diminished, virtual and erect.
Therefore image distance v = 0.77 m
Image will be formed at a distance of 0.77 m from the mirror.
Thus, we see that the magnification is less than 1.
Therefore the image will be diminished, virtual and erect.

Question 33.
What are trophic levels? Draw an example of a food chain.
OR
(A) Akash buried a waste paper under ground. After some days, he digged ground again to see the paper and the paper vanished. When he told this to his teacher, his teacher told him that the paper has been decomposed. What is the role of decomposers in the ecosystem?

(B) Explain why the number of trophic levels in a food chain is limited? (3)
Answer:
The various Levels or steps in a food chain are called trophic Levels. They represent the successive stages of nourishment Like primary producer, primary consumer, secondary consumer, tertiary consumer etc.
Below is an example of food chain:

(A) Decomposers are microorganisms which breakdown complex organic substances into simple inorganic substances and help in recycling of nutrients. They feed on the dead and decaying bodies of plants and animals. They return the nutrients back to the soil and thus help in making this ecosystem stable e.g. fungi, bacteria.

(B) Only 10% of the energy gets transferred from one trophic level to the next. So after 3 or 4 trophic levels, the energy available for passing on is too less to support another trophic level. Very little usable energy remains after 4 trophic levels. Hence, the number of trophic levels in a food chain is limited.

SECTION – D (15 Marks)
(Q.no. 34 to 36 are Long answer questions.)

Question 34.
Answer the following:
(A) Define oxidizing agent.

(B) Translate the following statements into chemical equations and balance them.
(i) Hydrogen gas combines with nitrogen to form ammonia.
(ii) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.
(iii) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
(iv) Potassium metal reacts with water to give potassium hydroxide and Hydrogen gas.

OR

(A) Write the balanced chemical equations for the following reactions:
(i) Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water
(ii) Zinc + Silver nitrate → Zinc nitrate + Silver
(iii) Aluminium + Copper chloride → Aluminium chloride + Copper

(B) Why is respiration considered an exothermic reaction? Explain.
(C) Why should magnesium ribbon be cleared before burning in air? (5)
Answer:
(A) Oxidizing agent: It is a substance which gives oxygen or gains hydrogen and in the process it itself gets reduced.
For example, in the given reaction:
CuO + H₂ → Cu + H₂O
CuO has given oxygen atom to H₂, hence it is an oxidizing agent.

(B) (i) Hydrogen gas combines with nitrogen to form ammonia.
H₂S + N_2(g) → NH₂
Balanced chemical equation:
3H_2(g) + N_2(g) → 2NH_3(g)

(ii) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.
H₂S + 3O₂ → H₂O + SO₂
Balanced chemical equation:
2H₂S_(g) + 3O_2(g) → 2H₂O_(l) + 2SO_2(g)

(iii) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
BaCl_2(aq) + Al₂ (SO₄)_3(aq) → 2AlCl_3(aq) +
3BaSO_4(s)
Balanced chemical equation:
3BaCl_2(aq) + Al₂ (SO₄)_3(aq) → 2AlCl_3(aq) + 3BaSO_4(s)

(iv) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.
K + H₂O → KOH + H₂
Balanced chemical equation:
2K_(s) + 2H₂O_(l) → 2KOH_(s) + H_2(g)
OR
(A) Balanced chemical equations for the reactions taking place are:
(i) Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water
Ca (OH)₂ + CO₂ → CaCO₃ + H₂O

(ii) Zinc + Silver nitrate ^ Zinc nitrate + Silver
Zn + 2AgNO₃ → Zn(NO₃)₂ + 2Ag

(iii) Aluminium + Copper chloride → Aluminium chloride + Copper
2Al + 3CuCl₂ → 2AlCl₃ + 3Cu

(B) Respiration is considered an exothermic reaction because a large amount of heat is produced in respiration by the oxidation of glucose. The chemical equation for the process of respiration is shown below:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + Energy

(C) Magnesium is a reactive metal. When kept exposed in air for a long time, a layer of MgO (magnesium oxide) is formed on the surface of metal. The oxide Layer does not burn when flame is brought in contact with metal.

Question 35.
The following table given below shows the information about two heaters A and B. Analyze the table and answer the following questions:

	Heater A	Heater V
Power	100 w	150 W
Voltage	220 V	220 V
Resistance	?	?
Current	?	?

(A) Which heater has high resistance?
(B) If lkWh is priced at ? 5 which heater will be costlier if they run for 1 hour each?
OR
(A) What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
(B) Two resistors of 20 W and 40 W are connected in parallel in an electric circuit. Compare the current passing through the two resistors. (5)
Answer:

	Heater A	Heater B
Power P	100 W	220 V
Voltage V	220 V	220 V
Current I = \(\frac { P }{ V }\)	.45 A	.68 A
Resistance R = \(\frac { V }{ I }\)	489 Ohms	324 Ohms

(A) Electric Power,
P = VI
l = \(\frac { P }{ V }\)
Also, according to Ohm’s Law,
V = IR
R = \(\frac { V }{ I }\)
Using these relations, the values of current and resistance of heaters A and B have been calculated:
For heater A:
P = 100 W
V = 220 V
i = P
I = \(\frac { 100 }{ 220 }\)
l = 0.45 A
R = \(\frac { 220 }{ 0.45 }\)Ohm
= 489 Ohm
For heater B:
P = 150W
V = 220 V
I = \(\frac { P }{ V }\)
I = \(\frac { 150 }{ 220 }\)
I =0.68A
R = \(\frac { 220 }{ 0.68 }\) Ohm
= 324 Ohm
Resistance of heater A is 489 Ohm and that of heater B is 324 Ohms. Therefore, heater A has higher resistance.

(B) Electrical Energy consumed by heater-A:
E = P x t
= 100 x 1
= 1oo Wh
= 0.1 KWh
Heater-B:
E = P x t = 150 x 1
= 150 Wh = 0.15 KWh
Cost of operating heater-A for 1 hour
= ₹ 5 x 0.1 = ₹ 0.50
Cost of operating heater-B for 1 hour
= ₹ 5 x 0.15 = ₹ 0.75
So heater B will be costly as it consumes more energy.

OR

(A) Advantages of connecting electrical devices in parallel are:
(1) Different appliances need different values of current for their proper operation. Whereas, in a series circuit, the current is constant throughout the circuit.

(2) When one component fails in a series circuit, the entire circuit is broken and none of the components work. On the other hand, a parallel circuit divides the current through the electrical gadgets.

(3) There is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage.

(B) For parallel combination of resistors, the potential difference will be same. Let the potential difference be V
As I=R
Then the current passing through the resistance 20 Ohm will be,
l₁ = \(\frac { V }{ R }\)
The current passing through the resistance 40 Ohm will be,
l₂ = \(\frac { V }{ 40 }\)
Clearly we see that,
l₁ = 2l₂
Hence current in 20W is double as compared to the current in 40 W resistor.

Question 36.
(A) Rekha studied human reproduction in her class. She wants to draw a figure to memorise. What she learnt? Draw a neat diagram of human male reproductive system and label the parts performing the following functions:
(i) Production of sperms
(ii) Gland which provides fluid
(iii) Provides low temperature for the formation of sperms

(B) Name the human male reproductive organ that produces sperm and secretes a hormone. What is the name of the hormone secreted and state its functions?

(C) Name a sexually transmitted disease and a method to avoid it. (5)
Answer:
(A) The parts of human male reproductive system performing the following functions:

(i) Production of sperms: Testis
(ii) Gland which provides fluid: Prostate gland
(iii) Provides Low temperature for the formation of sperms: Scrotum
(iv) Common passage for sperm and urine: Urethra

(B) Testis is the human male reproductive organ that produces sperm and secretes a hormone. The name of the hormone secreted is testosterone.
Functions of hormone:
(1) It regulates development of secondary sexual characters.
(2) It regulates the formation, development and maturation of sperm
(C) A disease which is sexually transmitted disease or STD includes warts or syphilis or HIV-AIDS. They can be prevented by using condoms during sexual intercourse.

SECTION – E (12 Marks)
(Q.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.
Indicators are used everyday both in our lives and in our laboratories. Most acid and base indicators are large organic molecules that contain alternating double and single carbon-to-carbon bonds. Many types of indicators work, throughout the pH scale. One common acid and base indicator is litmus paper created by treating filter paper with a dye obtained from Lichens.

Phenolphthalein is commonly used as an indicator in acid- base titration experiments in the chemical laboratory. Methyl red may be used as an acid and base indicator in the laboratory and as an azo dye, the largest group of synthetic dyes, commonly used to treat textiles. A . universal indicator is a solution that contains a mixture of indicators, often phenolphthalein, methyl red and bromothymol blue.
(A) What colour is shown by methyl orange in acidic and basic medium? An aqueous solution turns blue litmus solution red. Name a solution which can be added in excess to this solution to reverse the change?

(B) Which of the following are synthetic and natural indicators? Litmus, Phenolphthalein, Methyl orange, Vanilla essence

OR

(B) Which indicator can and cannot be used by visually impaired persons? (4)
Answer:
(A) Methyl orange is a synthetic indicator which is red in acidic medium and yellow in basic medium. An aqueous solution turns blue Litmus solution red. So, the given solution is acidic in nature. In order to reverse the change, an excess solution of a base should be added which will neutralize the acid.

(B) Both phenoLphthaLein and methyL orange are synthetic indicators whereas Litmus and vanilla essence are natural indicators.

OR

(B) Onion extract, vanilla essence and clove oil are olfactory indicators which can be used by visually impaired persons. On the other hand, red cabbage Leaf extract is red in acidic medium but changes to green in basic medium which cannot be seen by a visually impaired person.

Question 38.
Pollination is the transfer of pollen from a male part of a plant to a female part of a plant., later enabling fertilization and the production of seeds. Pollination may be biotic or abiotic. Biotic pollination relies on living pollinators to move the pollen from one flower to another. Abiotic pollination relies on wind, water or even rain. About 80% of

Seed formation begins with the combination of a male and female gamete: a process known as fertilization. Fertilization, or syngamy, can occur when both male and female gametophytes are fully mature.

(A) Pollination management is a branch of agriculture that seeks to protect and enhance present pollinators and often involves the culture and addition of pollinators in monoculture situations, such as commercial fruit orchards. The following bar graph shows how fruit setting rate varies with the frequency of pollination in an apple tree.

How is the fruit setting rote related to pollination frequency ?
(B) Where is embryo sac located?

(C) Draw structure of a gram seed and label cotyledon, plumule and radicle.

OR

(C) Define self and cross pollination.(4)
Answer:
(A) As the pollination frequency increases, the fruit setting rate also increases as it increases the probability of fertilization.

(B) Embryo sac is the structure within a plant ovule that contains the egg cell which deveLops from the megaspore and contains the embryo plant and endosperm after fertilization.

(C) The cotyledon is the food store for the growing embryo, the pLumuLe is the future shoot and radicLe is the future root. The given diagram shows the structure of a gram seed:

OR

(C) Self pollination: The transfer of pollen grains from the anther of a flower to the stigma of the same flower or another flower of the same plant.

Cross pollination: The transfer of pollen grains from the anther of a flower to the stigma of another flower of a different plant of the same species.

Question 39.
When a beam of light passes through a colloid, the colloidal particles present in the solution do not allow the beam to completely pass through. The light collides with the colloidal particles and is scattered (it deviates from its normal trajectory, which is a straight line). This scattering makes the path of the light beam visible, as illustrated below.

(A) What is Tyndall effect?
(B) Which wavelength of light is scattered by bigger size particles present in atmosphere?
(C) A star appears twinkling in the sky. Why?
OR
(C) Which colour of light scatters maximum due to atmosphere? Explain. (4)
Answer:
(A) The Tyndall effect is scattering of light by particles in a colloid or in a very fine suspension. The fine particles present in a true soLution or a colloid scatter the beam of Light thereby making the path of Light visible.

(B) The scattering of light and the colour of the scattered light depends upon the size of the particles. Larger particles scatter longer waveLength Like, red and orange.

(C) When the star light enters the earth’s atmosphere, where its refractive index is increasing gradually, the star Light is bent towards the normal. Moreover, this Light entering our eyes fluctuates randomly with time as the physical conditions of our atmosphere is changing continuously.

OR

(C) The blue colour of sky is due to the scattering of Light by the Large number of molecules present in the earth’s atmosphere. As the size of the scatterer is much smaller than the wavelength of light, light of smaller wavelength is scattered the most. If the earth had no atmosphere, the sky would appear black in the day time as no colour of sunlight would be scattered then.