CBSE Sample Papers for Class 10 Science Set 11 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 10 Science with Solutions Set 11 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Science Set 11 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.
The graph below shows the variation of average pressure exerted bg blood as blood flows in the body. (1)

A student analyzed the above graph and noted the following observations:
(I) The force exerted by blood on walls of arteries is greater than in veins.
(II) The force exerted by blood on walls of veins is greater than in arteries.
(III) Systolic pressure is the pressure of blood in the artery during contraction of ventricles.
(IV) Diastolic pressure is the pressure of blood in the veins during relaxation of ventricles
Select the correct observation(s):
(a) Only (I)
(b) Both (II) and (III)
(c) Both (I) and (III)
(d) Both (II) and (IV)
Answer:
(c) Both (I) and (III)

Explanation: The force exerted by blood on walls of arteries is greater than in veins. The pressure of blood in the artery during ventricular systole (contraction) is called systolic pressure and pressure in artery during ventricular diastole (relaxation) is called diastolic pressure. So, the correct observations are statements (I) and (IN).

Question 2.
Student performed the following activity: He took 3 ml of sodium sulphate solution in a conical flask (or test tube) and added 3 ml barium chloride solution. (1)

He noted down the following observations: Identify X, Y and Z ?

	Observation	Chemical equation of reaction taking place
(a)	No visible change observed as both the products formed are soluble in water	Na2SO4(aq) + BaCl2(aq) → BaSO4(aq) + 2NaCl(aq)
(b)	A white precipitate of sodium chloride is formed as barium sulphate is soluble in water	Na2SO4(aq) + BaCl2(aq) → BaSO4(aq) + 2NaCl(s)
(c)	A white precipitate of barium sulphate is formed as sodium chloride is soluble in water	Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)
(d)	White precipitates of barium sulphate and sodium chloride formed as both are insoluble in water	Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(s)

Answer:
(b) Observation: A white precipitate of barium sulphate is formed as sodium chloride is soluble in water; Chemical equation of reaction taking place:
Na₂SO_4(aq) + BaCl_2(aq) → BaSO_4(s) + 2NaCl_(aq)

Explanation : The reaction between barium chloride and sodium sulphate is a double displacement reaction in which exchange of ions takes place between the reactants. It is also a precipitation reaction in which a white precipitate of barium sulphate is formed, whereas sodium chloride formed is soluble in water.

Question 3.
Colonies of yeast fail to multiply in water, but multiply in sugar solution. Identify the reason for this.(1)
(a) Sugar solution provides nutrition
(b) Water provides nutrition
(c) Sugar solution provides moisture
(d) Both (b) and (c)
Answer:
(a) Sugar solution provides nutrition

Explanation: Energy is required for the growth and other activities of any organism. Colonies of yeast fail to multiply in water, but multiply in sugar solution as they get nutrition and hence energy from the sugar solution whereas there is no such food source to provide energy in water.

Question 4.
Human urine smell of ammonia if allowed to stand for sometime because: (1)
(a) it gets decomposed into ammonia by bacteria
(b) ammonia is absorbed
(c) it gets decomposed into area by the action of bacteria
(d) it reacts with ammonia
Answer:
(a) it gets decomposed into ammonia by bacteria

Explanation: Urea is present in human urine which gets decomposed into ammonia by the action of bacteria. So it smells strongly of ammonia if allowed to stand for sometime.

Question 5.
A stduent studied the following graph:

After studying the above graph, a student noted down the following observations:
(I) The serum uric acid level is same for both men and women for risk below 10.
(II) Serum uric acid level should be under 8.0 mg/dl for both men and women for relative risk below 10.
(III) The normal serum uric acid level is more for women and less for men for the same relative risk.
(IV)The normal serum uric acid level is more for men and less for women for the same relative risk.
Select the correct statement(s):
(a) Only (II)
(b) Both (I) and (IV)
(c) Both (II) and (III)
(d) Both (I) and (III) (1)
Answer:
(c) Both (II) and (III)

Explanation: As observed from the graph, the normal level of serum uric acid is higher in women as compared to men for the same relative risk.

Question 6.
Identify the organism exhibits the property of regeneration.
(a) Aschelminthes
(b) Tapeworms
(c) Flukes
(d) Planaria (1)
Answer:
(d) Planaria

Explanation: Regeneration is the ability of many fully differentiated organisms to give rise to new individual organisms from their body parts, if the individual is somehow cut or broken up into many pieces.
Planaria is an example of organism that has the ability to regenerate.

Related Theory
Regeneration is carried out by specialized cells. From this mass of cells, different cells undergo changes in an organised sequence referred to as development to become various cell types and tissues. Planarian flatworms are highly adapted with regeneration capabilities because of their asexual reproduction method. Star fishes also have the same ability to regenerate their arm, but unlike tailed amphibians and lizards, lost arms of star fishes could regenerate a complete new organism.

Question 7.
Prism ABC is positioned in various directions with BC serving as its base. As depicted in the figure, a small white light beam is incident on the prism. Which of the following situations, after dispersion, has the third colour from the top match the sky’s colour? (1)

Answer:

Explanation: The base BC of the prism is at the top in the figure (b), violet would be at the top after dispersion, and third colour would be blue. This contrasts with figure (a), in which the base BC of the prism is at the bottom and violet is at the bottom.

Question 8.
Figure shows a ray of light as it travels from medium A to medium B. (1)

Refractive index of medium B relative to medium A is:
(a) \(\frac{\sqrt{3}}{\sqrt{2}}\)
(b) \(\frac{\sqrt{2}}{\sqrt{3}}\)
(c) \(\frac{1}{\sqrt{2}}\)
(d) \(\sqrt{2}\)
Answer:
(a) \(\frac{\sqrt{3}}{\sqrt{2}}\)

Explanation : Here, angle of incidence (i) = 60° and angle of refraction (r) = 45°. According to Snell’s law, the refractive index of medium B with respect to medium A
n_BA = \(\frac { sini }{ sinr }\) = \(\frac { sin60° }{ sin45° }\)
= \(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}\) = \(\frac{\sqrt{3}}{\sqrt{2}}\)

Question 9.
A student arranged the following apparatus for finding the focal length and hence power of a convex lens. (1)

Object distance, u/cm	image dis­tance, v/cm	\(\frac { 1 }{ u }\)/cm-1	\(\frac { 1 }{ v }\)/cm-1	\(\frac { 1 }{ u }\)+\(\frac { 1 }{ v }\)/cm-1
40.0	24.0	0.025	0.042	0.067
35.0	26.3	0.029	0.038	0.067
30.0	30.0	0.033	0.033	0.066
25.0	37.5	0.040	0.027	0.067
20.0	60.0	0.050	0.017	0.067
18.0	90.0	0.056	0.011	0.067

He plotted a graph between 1/u and 1/v:

The approximate value of power of the lens is:
(a) + 15 D
(b) – 15 D
(c) + 6.7 D
(d) – 6.7 D
Answer:
(c) + 6.7 D

Explanation: The value of \(\frac { 1 }{ u }\) + \(\frac { 1 }{ v }\) comes out to 0.067 cm-1. According to the lens formula,
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ u }\) + \(\frac { 1 }{ v }\) = 0.067
As power is the reciprocal of focal length (in m), P = +6.7 D

Question 10.
The graph between power dissipated and current flowing is drawn below:

From the graph, the resistance is calculated and equal to
(a) 10 Ohm
(b) 5 Ohm
(c) 2 Ohm
(d) 1 Ohm (1)
Answer:
(c) 2 Ohm

Explanation: Power P = VI = I₂R, where I is the current flowing through a resistor having resistance R.
From the graph, corresponding to
I = 5 A,
P = 50 W.
P = I₂R
R = \(\frac{P}{1^2}\)
R = \(\frac{50}{5^2}\) Ω
R = \(\frac{50}{25}\) Ω
R = 2Ω

Question 11.
The normal diastolic blood pressure in a normal healthy adult human is:
(a) 80 mm Hg
(b) 60 mm Hg
(c) 90 mm Hg
(d) 120 mm Hg (1)
Answer:
(a) 80 mm Hg

Explanation: The normal diastolic blood pressure in a normal healthy adult human is 80 mm Hg whereas the normal systolic pressure is about 120 mm Hg.

Question 12.
Observe the reaction shown:

Will Y dissolve in an organic solvent or not?
(a) Yes
(b) No
(c) Insufficient information
(d) Soluble sometimes (1)
Answer:
(b) No

Explanation: When a metal combines with a non-metal, the compound formed will be an ionic compound. Ionic compounds are soluble in water but insoluble in organic compounds such as alcohol. Therefore, the compound Y will not dissolve in alcohol.

Related Theory
The general properties of ionic compounds are summarized below:

S.No.	Property name	Property shown by ionic compounds
(1)	Physical nature	Ionic compounds are generally  crystalline solids and hard due to the strong force of attraction between the positive and negative ions. They are generally brittle
(2)	Melting and boiling points	They have high melting and boiling points as a large amount of energy is required to break the strong inter-ionic attraction
(3)	Solubility	These are generally soluble in water but insoluble in organic solvents like           ether, kerosene, petrol etc.
(4)	Conduction of electricity	These conduct electricity in the molten state as the electrostatic forces of attraction between the oppositely charged ions are overcome due to the heat. Moreover, they also conduct electricity when dissolved in water as its solution in water contains ions. However, these do not conduct electricity in the solid state due to their rigid structure.

Question 13.
The oxide formed in the reaction shown below dissolves in dilute hydrochloric acid.

The oxide formed also turns a solution of red litmus blue. What is the element ‘X’?
(a) Metal
(b) Non-metal
(c) Metalloid
(d) None of these (1)
Answer:
(a) Metal

Explanation: As the oxide of the given element dissolves in dilute hydrochloric acid and also turns a solution of red litmus blue, it is basic in nature. It is a metal oxide as oxides of metals are basic in nature.

Question 14.
The thickness of blood capillaries is:
(a) 1 cell thick
(b) 10 cells thick
(c) 100 cells thick
(d) 100,000 cells thick (1)
Answer:
(a) 1 cell thick

Explanation: The blood capillaries are about one cell thick so that exchange of materials between the blood and surrounding cells takes place across this thin wall.

Question 15.
Select the incorrect statement (s):
(I) Arteries carry blood away from the heart to different body organs
(II) Arteries have thin, elastic walls.
(III) Veins collect blood from different body organs and bring it back to heart.
(IV) Veins have thick walls.
(a) Both (I) and (II)
(b) Both (II) and (IV)
(c) Both (I) and (III)
(d) Both (II) and (III) (1)
Answer:
(b) Both (II) and (IV)

Explanation: Arteries carry blood from the heart to different organs and hence have thick elastic walls. Veins carry blood from different organs to the heart and are thin walled blood vessels.

Question 16.
Which of the following is not an example of abiotic factors of an ecosystem?
(a) Temperature
(b) Rainfall
(c) Wind
(d) Bacteria (1)
Answer:
(d) Bacteria

Explanation: The examples of abiotic factors of an ecosystem are temperature, rainfall, wind, soil and minerals.

Related Theory
An ecosystem is a self-contained unit of living things and their non-living environment. All the interacting organisms in an area together with the non-living constituents of the environment form an ecosystem. It therefore consists of biotic components and abiotic components.

There are two types of ecosystem:
(1) Terrestrial ecosystem: These are land based ecosystems such as forest, grassland, desert, mountains etc.

(2) Aquatic ecosystem: These are water based ecosystems such as ponds, lakes, river, sea, aquarium etc.
Ecosystems can also be classified as natural ecosystem and artificial ecosystem. Examples of natural ecosystem are forests, ponds etc whereas examples of artificial ecosystem are gardens, aquarium and crop-fields.

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): Carbon forms strong bonds with other elements making these compounds exceptionally stable.
Reason (R): The formation of strong bonds by carbon is due to its small size. (1)
Answer:
(a) Both A and R are true and R is correct explanation of A.

Explanation: The atomic number of carbon is 6 and its electronic configuration is 2, 4. It forms covalent compound by sharing 4 valence electrons. Because of small size of carbon, nucleus holds the shared pairs of electrons strongly.

Question 18.
Assertion (A): Electrical impulses are used by neurons to transmit messages across the junction.
Reason (R): Synapse refers to the intersection or space between two points. Neurotransmitters are released as chemicals when an electrical signal reaches its terminal. (1)
Answer:
(d) A is false but R is true

Explanation: Through chemical signals, neurons transmit information across the junction. Electrical impulses are used to transport messages within neurons.

Question 19.
Assertion (A): Positive phototropism is defined as a movement toward light.
Reason (R): The auxin diffuses toward the sunny side of the shoot when sunlight hits a particular side of the plant. Cells are stimulated to grow longer by auxin concentration, and a stem that is bent toward the sun is produced. (1)
Answer:
(c) A is true but R is false

Explanation: Auxin diffuses toward the shaded side of a shoot when a plant receives sunlight on one side only.

Question 20.
Assertion (A): The flow of energy in an ecosystem is unidirectional.
Reason (R): Green plants in a terrestrial ecosystem capture about 10% of the energy of sunlight that falls on their leaves. (1)
Answer:
(c) A is true but R is false.

Explanation: The flow of energy in an ecosystem is unidirectional as the energy that is captured by the autotrophs does not go back to the solar input and the energy which passes to the herbivores does not come back to the autotrophs. The green plants in a terrestrial ecosystem capture about 1% of the energy of sunlight that falls on their leaves and convert it into food energy.

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

Question 21.
Every day, Ayush enjoys cold beverages, chocolates, and sweets. He has toothache. After each meal and after indulging in sweets, his science teacher instructed him to brush his teeth.

(A) How are teeth harmed by eating sweets and chocolate?
(B) Brushing your teeth helps to protect against tooth decay. Justify. (2)
Answer:
(A) When eating sweets or chocolate, the pH of the mouth falls below 5.5, which causes tooth enamel to corrode and tooth decay to begin.

(B) By using toothpaste, which is naturally basic, you can stop tooth decay by neutralising the extra acid your mouth produces.

Question 22.
Rahul took the following elements:
C₂H₆,C₂H₂,C₃H₈,C₄H₈.
Which two compounds could belong to the same homologous series? Give reason for your answer. (2)
Answer:
Out of C₂H₆, C₂H₂, C₃H₈, C₄H₈, the compounds C₂H₆ and C₃H₈ belong to the same homologous series as the difference between their molecular formula is – CH₂. Whereas, the difference in the molecular formulae of C₂H₂ and C₄H₈ is neither – CH₂ nor a multiple of – CH₂.

Related Theory
Characteristics of Homologous series:
(1) All the members of a homologous series can be represented by the same general formula.
(2) Any two adjacent homologues differ by 1 carbon atom and 2 hydrogen atoms in their molecular formula.
(3) The difference in the molecular masses of any two adjacent homologues is 14 u.
(4) All the compounds belonging to the same homologous series have similar chemical properties since these are determined solely by the functional group.
(5) The members of a homologous series show a gradual change in their physical properties with increase in molecular mass. This is because the melting and boiling points increase with increasing molecular mass.

Question 23.
(A) Define dispersion of light. How does a prism disperse white light?
(B) Which colour bends the most and the least? (2)
Answer:
(A) Dispersion of light is defined as the splitting of light into its component colours when light passes through a glass prism. A prism disperses white light as light rays of different colours travel with different speeds in a prism as refractive index of glass is different for different colours.

(B) Violet colour bends the most and red colour bends the least as the refractive index of glass is largest for violet colour and least for red colour due to which violet colour is dispersed the maximum and red colour the least.

Question 24.
Define parasitic nutrition. Name two organisms having parasitic mode of nutrition. (2)
Answer:
Parasitic mode of nutrition is defined as the mode of nutrition in which an organism derives its food from the body of another living organism without killing it. Organisms having parasitic mode of nutrition are Plasmodium and roundworms.

Question 25.
How does trait get expressed?
OR
Why do all the gametes formed in human females have an X chromosome? (2)
Answer:
A trait in an organism is the character which is transmitted from parents to offsprings i.e, from one generation to the next generation. Cellular DNA is the information source for making proteins in the cell. A section of DNA called gene provides information for one protein. A particular trait depends on the specific protein synthesized as it decides the efficiency of the process for synthesis of particular hormone.
OR
Women have a perfect pair ofsex chromosomes, both called X. Women are XX, while men are XY. All children will inherit an X chromosome from their mother regardless of whether they are boys or girls. Thus, the sex of the children will be determined by what they inherit from their father. A child who inherits an X chromosome from her father will be a girl. During meiosis, one X chromosome enters each gamete. Hence, all the gametes formed in human females have an X chromosome.

Question 26.
The chromosome number of zygote, embryonal cells and adult of a particular organism is always constant. How is the constancy maintained in these three stages?
OR
justify why the male reproductive system is called “urinogenital system”. (2)
Answer:
Sexual reproduction includes gamete formation through meiosis and fertilization. Meiosis reduces the number of chromosomes to half in male and female gametes. This reduced chromosome number is then restored to normal during fertilization of male and female gametes. This is how constant chromosome number is maintained in sexually reproducing

organisms. Growth and development of zygote into embryonic cell and then into adult one takes place by mitosis (equatorial cell division) which produce the daughter cells carrying same chromosome number as that of parent cell.

OR

The male reproductive system is called “urogenital system” as the urethra forms a common passage for both the sperms and urine. The sperms formed are delivered through the vas deferents which unites with a tube coming from the urinary bladder which stores the urine formed in each kidney.

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

Question 27.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:

1 (Amperes)	0.5	1.0	2.0	3.0	4.0
V (Volts)	1.6	3.4	6.7	10.2	13.2

Plot a graph between V and i and also calculate the resistance of that resistor. (3)
Answer:
The values of the current for different values of the voltage are shown in the given table.

I (Amperes)	0.5	10	20	3.0	4.0
V (volts)	1.6	3.4	6.7	10.2	13.2

The graph between voltage (plotted on X-axis) and current (plotted on Y-axis) is shown below:

Slope of the graph = Resistance
R = \(\frac { V }{ I }\)
= \(\frac { 10.2-6.7 }{ 3-2 }\)
= \(\frac { 3.5 }{ 1 }\)
= 3.5 Ohm

Question 28.
My mom baked a cake but the cake did not rise as she had forgotten to add an important ingredient.
(A) Which ingredient has she forgotten to add that would have made the cake soft and fluffy? Give reason and chemical equation, if any.
(B) Write two other uses of this compound.
OR
Dry pellets of a base ‘X’ when kept in open absorbs moisture and turns sticky. The compound is also formed by chlor- alkali process. Write the chemical name and formula of X. Describe chlor-alkali process with balanced chemical equation. Name the type of reaction that occurs when X is treated with dilute hydrochloric acid. Write the chemical equation. (3)
Answer:
(A) She has forgotten to add baking powder while making the dough for the cake. Sodium hydrogencarbonate present in baking powder releases carbon dioxide on baking. The bubbles of the gas evolved leave behind pores which make the cake soft and fluffy.

(B) Uses of baking powder:
(1) It is used as an ingredient in antacids. Being alkaline, it neutralises excess acid in the stomach and provides relief.
(2) It is also used in soda-acid fire extinguisher.
OR
The common base which absorbs moisture and becomes sticky and is also a by-product of chloralkali process is sodium hydroxide (NaOH).

When electricity is passed through an aqueous solution of sodium chloride (called brine), it decomposes to form sodium hydroxide. The process is called the chlor- alkali process because of the products formed: chlor for chlorine and alkali for sodium hydroxide.

Following is the chemical equation for the process:
2NaCl_(aq) + 2H₂O_(l) → 2NaOH_(aq) + Cl_2(g) +
H_2(g)
When NaOH is treated with HCl, NaCl is formed.
NaOH + HCl → NaCl + H₂O
It is a type of neutralization reaction.

Question 29.
Mohan had been complaining of frequent weight loss, leg pain, and frequent urination. He went to the doctor right away. The physician determined that Mohan had high blood sugar levels. To keep his blood sugar levels within the normal range, he should watch what he eats and exercise frequently.

(A) Identify the illness he has and the hormone whose shortage is to blame for it.
(B) Name the gland that secretes it and describe this hormone’s purpose.
(C) Describe how the human body controls the amount and timing of this hormone’s secretion. (3)
Answer:
(A) Disease-Diabetes, Hormone: Insulin (deficiency of insulin hormone causes diabetes)
(B) Gland-Pancreas: The blood glucose level is regulated by insulin hormone secreted by the pancreas.
(C) When the blood’s level of blood glucose rises, pancreatic cells secrete the hormone insulin. Blood glucose levels are regulated by insulin, and when levels drop, less insulin is secreted.

Question 30.
In the formation of the compound A₂B, atoms of A lose one electron each while atom of B gained two electrons. Show the formation of the compound A₂B and write the nature of bond present. Predict any two properties of A₂B. (3)
Answer:
Atoms of ‘A’ lose one electron each, so ‘A’ is a metal having valency 1 and forms a positively charged ion A⁺. ‘B’ gains two electrons, so ‘B’ is a non-metal having valency 2 and forms negatively charged ion B^2-.
As the bond between a metal and non-metal is always ionic, so nature of bond is ionic or electrovalent.

Two properties of the ionic compound A₂B are:
(1) It is soluble in water but insoluble in organic solvent such as alcohol.
(2) It does not conduct electricity in solid state but conducts electricity in solution or in molten state.

Question 31.
A person A has only B chromosomes in all its gametes. On the other hand, another person C has chromosome D in half of gametes and chromosome E in another half of gametes. When chromosomes B and D combine during fertilization, a female zygote results. On the other hand, combination of B and E chromosomes produces a male zygote. What are chromosomes B, D and E? Give reason for your answer. (3)
Answer:
(A) As a woman has only X chromosomes in all her gametes and it is given that person A has only B chromosomes in all its gametes, B is ‘X’ chromosome. I t is also given that another person C has chromosome D in half of gametes and chromosome E in another half of gametes. As we know that males have X chromosomes in half gametes and Y chromosomes in another half, person C is a male. When chromosome B and D combine during fertilization, a female zygote results. It means that both B and D are X chromosomes.

When B and E chromosomes combine, a male zygote results which means that while B is X chromosome, E is Y chromosome. Therefore, B is X chromosome, D is X chromosome and E is Y chromosome. Out of A and C, A is a female as it has only X chromosomes in all gametes and C is a male as it has X chromosomes in half of the gametes and Y in the other half.

Question 32.
Answer the following:
(A) Why does a ray passing through the centre of curvature of a concave mirror retraces its path?
(B) Draw the given diagram in your answer book and complete it for the path of a ray of light after passing through the lens. (3)

Answer:
(A) A ray of light passing through the centre of curvature of a concave mirror retraces its path because when the ray of light passes through centre of curvature of a concave mirror it strikes the mirror at 90° i.e., the incident ray coincides with the normal. Therefore angle of incidence = 0°. We know that according to Laws of reflection angle of incidence is equal to the angle of reflection; therefore angle of reflection will also be

(B) A ray of light from the object, parallel to the principal axis, after refraction from a convex lens, passes through the principal focus on the other side of the lens, as shown in the figure given below.
A ray of light parallel 0°. Thus ray of light retraces its path.

Question 33.
What do you understand by biological magnification? Which trophic level organisms would have highest concentration of chemicals? Give reasons for your answer. (3)
Answer:
The increase in concentration of harmful chemical substances like pesticides in the body of living organisms at each trophic level of a food chain is called biological magnification. The topmost level organisms of trophic level would have highest concentration of chemicals.

The organisms at each higher trophic level will have more concentration of chemicals as the toxic substances move up the food chain and become more concentrated at each trophic level. These chemicals are absorbed from the soil by the plants and from the water bodies by the aquatic plants and animals. As these are not biodegradable, they get accumulated progressively at each trophic level.

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

Question 34.
State what happens when:
(A) Gypsum is heated at 373 K
(B) Blue crystals of copper sulphate are heated
(C) Excess of carbon dioxide gas is passed through lime water
(D) Zinc granules are heated with sodium hydroxide solution
(E) Dilute HCl is added to sodium hydrogen carbonate.(5)
Answer:
(A) On heating gypsum at 373 K, it loses water molecules and becomes calcium sulphate hemihydrate (CaSO₂. \(\frac { 1 }{ 2 }\)H₂O).

(B) When we heat the blue crystals of copper sulphate, the molecules of water of crystallization is removed and the salt turns white.
CuSO₄. 5H₂O → CuSO₄ + 5H₂O

(C) When we pass excess carbon dioxide through lime water, then the white precipitate formed first (calcium carbonate) dissolves due to the formation of a soluble salt calcium hydrogencarbonate and the solution becomes clear again.
CaCO_3(s) + CO_2(g) + H₂O_(l) → Ca(HCO₃)_2(aq)

(D) When zinc granules are heated with sodium hydroxide solution, a salt sodium zincate is formed along with the evolution of hydrogen gas.
2NaOH_(aq) + Zn_(s) → Na₂ZnO_2(aq) + H_2(g)

(E) Hydrochloric acid reacts with sodium hydrogen carbonates to form sodium chloride, water and carbon dioxide gas.
NaHCO_3(s) + HCl_(aq) → NaCl_(aq) + H₂O_(l) + CO_2(g)

Question 35.
Sita studied in her class about vegetative propagation in Science class. But she had some doubts which she wanted to clear. What is vegetative propagation? List with brief explanation three advantages of practicing this process for growing some types of plants. Select two plants from the following which are grown by this process: Banana, Wheat, Mustard, Jasmine, Gram.
OR
Name the reproductive parts of an angiosperm. Where are these parts located? Explain the structure of its female reproductive parts with the help of a labelled diagram. (5)
Answer:
Vegetative propagation is a method of reproduction in some higher plants in which a new plant develops from the vegetative parts of a plant such as root (as in Dahlia, sweet potato), stem (as in ginger, potato, onion) or Leaf (as in Bryophyllum).

Three advantages of practicing this process for growing some types of plants:
(1) Plants raised by this method can bear flowers and fruits earlier than those produced from seeds.
(2) It is cheaper, easier and more rapid method of propagation in pLants as compared to growing pLants from their seeds.
(3) The traits or characters of the parent plant are preserved. That is, all plants produced are genetically similar enough to the parent plant to have all its characteristics.
(4) Better quality of the plants can be maintained. Two plants out of the given plants that can be grown by this process are banana and jasmine as mustard, wheat and gram can be grown from seeds.

OR

The reproductive parts of angiosperms are stamens and carpeLs and are Located in the flower. Stamen is the maLe reproductive part and produces pollen grains. Carpel is the female reproductive part and contains the ovary, styLe and stigma. Carpel is the femaLe reproductive part of a flower and consists of a swollen ovary at the base, an elongated middle style and a terminal stigma. The ovary contains ovules and each ovule has an egg.

Question 36.
A student fixes a sheet of white paper on a drawing board. He places a bar magnet in its centre. He sprinkles some iron filings uniformly around the bar magnet. Then he taps the board gently and observes that the iron filings arrange themselves in a particular pattern.
(A) Why do the iron filings arrange in a pattern?
(B) What do the lines along which the iron filings align represent?
(C) What does the crowding of iron filings at the end of the magnet indicate?
(D) List the properties of magnetic field lines.
OR
AB is a current carrying conductor in the plane of the paper as shown in figure.

(A) What are the directions of magnetic fields produced by it at points P and Q?
(B) State the rule to find direction of magnetic field around a current carrying conductor.
(C) Draw a diagram showing the pattern of magnetic field.
(D) Given r₁ > r₂, where will the strength of the magnetic field be larger? Give reasons for your answer. (5)
Answer:
(A) Iron filings arrange in a pattern because a magnetic field exists around a magnet and force is exerted by the magnet within its magnetic field.

(B) The lines represent magnetic field lines.

(C) The crowding of iron filings at the end of the magnet indicates that the strength of magnetic field is maximum near the poles of the magnet.

(D) Properties of magnetic field lines:
(1) Magnetic field lines are closed continuous curves.
(2) The tangent at any point on the magnetic field lines gives the direction of magnetic field at that point.
(3) No two magnetic field lines can intersect each other.
(4) They are crowded in a region of strong magnetic field and are far from each other in a region of weak magnetic field.

OR

(A) By applying Right-hand thumb rule, the direction of magnetic field at P will be into the plane of the paper. At point Q, the direction of magnetic field will be out of the plane of the paper.

(B) The rule to find direction of magnetic field around a current carrying conductor is right hand thumb rule which states that “Imagine that you are holding a current carrying wire in your right hand such that the thumb is stretched along the direction of the current, then, the direction in which the fingers wrap around the conductor wiLL give the direction of the field lines of the magnetic field.”

(C) Diagram showing the pattern of magnetic field:

(D) The magnitude of magnetic field B is inversely proportional to distance r. Now r₁ > r₂, or point Q is closer than point P, so the magnetic field is stronger at point Q and weaker at point P.

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.
During a chemical reaction both the form and composition of matter are changed. Old substances are converted to new substances, which have unique physical and chemical properties of their own. Chemical reactions can be double displacement reactions which have the general form:
AB + CD → AD + CB
Here AB and CD are usually aqueous ionic compounds (or acids) consisting of aqueous ions (A⁺ and B^–, C⁺ and D^–). When a double displacement reaction occurs, the cations and anions switch partners, resulting in the formation of two new ionic compounds AD and CB, one of which is in the solid state. This solid product is an insoluble ionic compound called a precipitate. (4)

(A) 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. What type of reaction is this? Explain.

(B) What happens when reaction occurs between potassium iodide and lead (II) nitrate?

OR

(B) Two metals X and Y form the salts XSO₄ and Y₂SO₄ respectively. The solution of salt XSO₄ is blue in colour whereas that of Y₂SO₄ is colourless. When barium chloride solution is added to XSO₄ solution, then a white precipitate Z is formed along with a salt which turns the solution green. And when barium chloride solution is added to Y₂SO₄ solution, then the same white precipitate Z is formed along with colourless common salt solution. Identify X, Y and Z. (4)
Answer:
The reaction between copper sulphate and hydrogen sulphide gas is a double displacement reaction as the positive and negative ions exchange their places.
CuSO_4(aq) + H₂S_(g) → CuS_(s) + H₂SO_4(aq)
As a black precipitate of copper sulphide is formed, the physical state of CuS is shown as(s).

Related Theory
Any reaction that produces a precipitate (insoluble solid) that separates from the solution can be called a precipitation reaction.

For example, when potassium iodide is added to lead nitrate solution, a yellow precipitate of lead iodide is formed along with potassium nitrate solution.
Pb(NO₃)_2(aq) + 2KI_(aq) → PbI_2(s) + 2KNO_3(aq)

(B) When potassium iodide is added to Lead nitrate solution, a yellow precipitate of Lead iodide is formed aLongwith potassium nitrate soLution.
Pb(NO₃)_2(aq) + 2KI_(aq) → PbI_2(s) + 2KNO_3(aq)
This is a double displacement reaction in which an exchange of ions is taking place between the reactants. It is aLso a precipitation reaction as a yellow precipitate of lead iodide is formed. Potassium nitrate remains in the solution.

OR

(B) It is given that the salt solution XSO₄ is blue in colour and as we know solution of copper sulphate is blue in colour, therefore X is copper. On adding barium chloride solution to CuSO₄ solution, a white precipitate of Barium sulphate is formed, which is also formed when barium chloride solution is added to a solution of sodium sulphate. Therefore, Y is sodium and Z is barium.

Question 38.
Total internal reflection is defined as the phenomenon which occurs when the light rays travel from a more optically denser medium to a less optically denser medium. Consider a ray of light passing from a medium of water to that of air. Light ray will be refracted at the junction separating the two media.

Since it passes from a medium of a higher refractive index to that having a lower refractive index, the refracted light ray bends away from the normal. At a specific angle of incidence, the incident ray of light is refracted in such a way that it passes along the surface of the water. This particular angle of incidence is called the critical angle as shown in the figure below

When the angle of incidence is greater than the critical angle, the incident ray is reflected back to the medium. We call this phenomenon total internal reflection. An application of this phenomenon is found in optical fibres.

An optical fibre consists of:
(1) Core – Thin glass center of the fiber where the light travels.
(2) Cladding – Outer optical material surrounding the core that reflects the light back into the core
(3) Buffer coating – Plastic coating that protects the fiber from damage and moisture

The core has higher refractive index than the cladding. When the incident ray falls on the cladding, it suffers total internal reflection as the angle formed by the ray is greater than the critical angle.

(A) How will the speed of light change when light is incident on cladding of an optical fibre?
(B) The speed of light in a certain material is 50% of its speed in a vacuum. What is the refractive index of this material?
(C) Whenever light travels from one medium to another, what is the angle of refraction?
OR
(C) What are the two conditions for total internal reflection of light? (4)
Answer:
(A) The refractive index of cladding is less than that of the core. As refractive index of a medium varies inversely as the speed of light in that medium, the speed of light will be less in core than in cladding.

(B) The refractive index of a medium

= \(\frac{c}{\frac{c}{2}}\) = 2
as speed of Light in the medium is 50% of the speed of Light in vacuum.

(C) When Light travels from one medium to another, both having the same refractive index, angle of refraction will be equal to the angle of incidence. Also, when Light is incident along the normal to the refracting surface, angle of refraction will be equal to the angle of incidence, both being equal to zero.

OR

(C) The two conditions for total Internal reflection of Light are that Light must travel from a denser medium to a rarer medium and the angle of incidence in the denser medium must exceed the critical angle for the given pair of media. The critical angle for a given pair of media is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium = 90°.

Question 39.
We alt are aware of the heating effects of electric current. The heat is produced in a conductor due to the collision of electrons in the conductor. Have you ever wondered about the amount of heat generated during the flow of current through a wire and the parameters and conditions it is based upon?

It was the English physicist James Prescott Joule who discovered in 1840 that the amount of heat per second that develops in a wire carrying a current is proportional to the electrical resistance of the wire and the square of the current. He determined that the heat evolved per second is equivalent to the electric power absorbed, or the power loss.

(A) How is the heat produced in a resistor related to T, ‘R’ and ‘t’ according to Joule’s law of heating?
(B) Find the heat energy produced in resistance of 5 Q when 3 A current flows through it for 2 minutes.
(C) Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in a circuit across the same potential difference. What is the ratio of heat produced in series and parallel combination?
OR
(C) If the heat produced in each second in a 8 Ohm resistance is 800 J, find the difference across the resistor. (4)
Answer:
(A) H = I² x R x t. According to Joule’s Law of heating, heat produced in a conductor of resistance R is given by H = I²RT, where I is the current flowing in the conductor and t is the time.
That is, H ∝ I², H ∝ R, H ∝ t.

(B) Here,
R = 5Ω,
I = 3A ,
t = 2 minutes = 120 s
H = I² x R x t
= 3 x 3 x 5 x 120
= 5400 J

(C) As the two wires are made of the same material, have the same Lengths and diameters, their resistances will also be equal Let the resistance in each wire be R So in series, R_s = R + R = 2R
\(\frac{1}{R_p}\) = \(\frac { 1 }{ R }\) + \(\frac { 1 }{ R }\) = \(\frac { 2 }{ R }\)
Or,
R_s = \(\frac { 2 }{ R }\)
We know that heat produced is given by:

OR
(C) According to JouIe’s law of heating, heat produced in a wire of resistance R is given by

Putting the values, we get \(\frac{v^2}{8}\)
V² = 6400 or V = 80V