NCERT Solution Class 10 Science

Science is a major subject for Class 10 students as it lays the foundation for higher education in various fields such as engineering, medicine, and technology. The NCERT textbook for Class 10 Science is comprehensive, covering key topics in physics, chemistry, and biology. 

Do you want to ace your CBSE board exams with confidence and ease? If yes, then you have come to the right place. The solutions are curated while keeping the latest CBSE pattern and will help students clear their fundamentals.

‍Where to Find NCERT Solutions for Class 10

NCERT solutions are prepared by subject experts following the latest CBSE syllabus and exam patterns. Simply scroll down to find the download links for each chapter, ranging from Chemistry’s "Chemical Reactions" to Biology’s "Life Processes" and Physics’ "Light: Reflection and Refraction." 

Having access to these PDF solutions makes it easier for students to study at their own pace and clear any doubts.

NCERT Science Solutions for Class 10 - Chapter-wise PDF

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NCERT Solutions Class 10 Science Complete PDF

13 chapters are categorized into four units: Physics, Chemistry, Biology, and Environmental Science as per the latest CBSE 2025 Science Syllabus.

Chapter-Wise NCERT Solutions for Class 10 Science

Physics

Physics in Class 10 covers fundamental principles of light, electricity, magnetism, and energy. The chapters provide a blend of theory and practical application, essential for understanding the natural world.

Light – Reflection and Refraction

• Understanding the laws of reflection and refraction, ray diagrams for mirrors and lenses, mirror formula, and magnification.
• Detailed study of refraction through glass slabs and prisms, and application in optical instruments.

Examples:

Q1. In the following cases, a ray is incident on a concave mirror. In which case is the angle of incidence equal to zero? 

A. A ray parallel to the principal axis. 

B. A ray passing through the centre of curvature and incident obliquely. 

C. A ray passing through the principal focus and incident obliquely. 

D. A ray incident obliquely to the principal axis, at the pole of the mirror

Answer: Option B - A ray passing through the centre of curvature and incident obliquely.

Explanation: The angle of incidence is defined as the angle between the incoming ray and the surface normal at the point where the ray strikes the surface. In the case of a concave mirror, the normal at any point on the mirror’s surface is the line that passes through the centre of curvature. When a ray strikes the mirror along this normal (i.e., along the line passing through the centre of curvature), the angle of incidence is zero because the incoming ray is parallel to the normal.

Q2. Choose the correct option for the colour of rays for A and B.

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Answer: Option C - The color of Ray A is Red and of Ray B, it is Violet

Explanation: When white light passes through a glass prism, it splits into its different colours due to dispersion. Violet light, having the shortest wavelength, bends the most, while red light, with the longest wavelength, bends the least. 

Looking at the diagram:
Ray A, closer to the base of the prism, bends less, meaning it is a red light.

Ray B, farther from the base, bends more, meaning it is violet light. 

Step-by-Step Explanation:
1. Dispersion of Light: When white light enters a prism, it splits into its constituent colours because each colour bends differently based on its wavelength.
2. Bending of Colours: Violet bends the most (shortest wavelength), and red bends the least (longest wavelength).
3. Diagram Analysis: Ray A is closer to the base (less bending = red), and Ray B is farther from the base (more bending = violet). 

Q3. What is the fundamental difference between hypermetropia and myopia in terms of the optical experience of a person?

Answer: The main difference between hypermetropia and myopia is that light focuses on the retina, which affects what a person can see clearly. 

‍Hypermetropia: Commonly referred to as farsightedness, this condition occurs when light from distant objects is focused behind the retina, causing nearby objects to appear blurry. It can result from an eye that is too short or a cornea that is too flat.

‍Myopia: Also known as nearsightedness, myopia causes light from distant objects to focus in front of the retina, making faraway objects appear blurry. This typically happens when the eye is too long from front to back.

Q4. The diagram below shows a special case of an eye defect. 

(i) What is the defect that is shown in the figure? 

(ii) State one cause for such a defect. 

(iii) Explain with reason if a concave lens can be used to correct the defect.

Answer: (i) The ray diagram shows a condition where light rays from a nearby object focus behind the retina, indicating a vision defect called hypermetropia, or farsightedness. 

(ii) This issue happens when either the eye lens has a focal length that is too long or the eyeball is shorter than normal, making it difficult to focus on nearby objects.

(iii) To correct hypermetropia, a convex lens is used. This lens helps by converging the incoming light rays so they focus directly on the retina, allowing for clear vision.

Electricity

• Concept of electric current, potential difference, resistance, and Ohm’s Law.
• Calculating resistance in series and parallel, heating effects of electric current, and practical applications in daily life.

Examples:

Q5. (i) State Joules law of heating and write its mathematical expression. 

(ii) Two resistors of resistances 2𝛺 and 4𝛺 are connected in a) series b) parallel with a battery of given potential difference. Compute the ratio of the total quantity of heat produced in the combination of the two cases if the total voltage and time are kept the same for both

Answer  (i) Joule's Law of Heating states that the heat generated in a conductor is directly proportional to the square of the current flowing through it, the resistance of the conductor, and the duration for which the current flows. The mathematical expression for Joule's law of heating is 𝐻=𝐼²𝑅𝑡

Where:

H: Heat produced by the conductor

I: Current passing through the conductor

R: Resistance of the conductor

T: time for which the current flows

(ii) Let the equivalent resistance in series be denoted by RS and that for parallel be denoted by R_p. The total voltage of the circuit is given in both cases and the time is denoted by

• R_s = 2 + 4 = 6  Ω
• 1/R_p = ½ + ¼ = ¾  Ω
• Therefore, R_p = 4/3 Ω
• H = V2/R_s* t , H = V2/R_p * t 
• Hs/Hp = R_p/R_s = 2/9 Ω

Q6. The above circuit is a part of an electrical device. Use the information given in the question to calculate the following.

(i) Potential Difference across R2. (ii) Value of the resistance R2. (iii) Value of resistance R1

Answer (i) P.D. across 4 Ω resistor = P.D. Across R2 as both are in parallel.  

1.5(A) × 4 (Ω) = 6 V 

(ii) Total Current through 4 Ω and R2 = 2.0 A (given). 

Current through 4 Ω = 1.5 A (given)   

Hence current through R2 = 2-1.5= 0.5 A  

Using Ohm’s law for R2 we get  6 V= 0.5 A x R2  

Hence R2 = 6/0.5 = 12 Ω 

(iii) P.D. across R1 = Total P.D. - (P.D. across R2) - (P.D. across 2.0 Ω) 

P.D. across 2.0 Ω = 2x2 = 4 V 

P.D. across R2 = 6 V (calculated before) 

Hence P.D. across R1 = 12 − 6 − 4 = 2 V  

Current through R1 = 2A  

Using Ohm’s Law, we get  R1 = 2V/2A = 1Ω

Magnetic Effects of Electric Current

• Magnetic fields and field lines, effects of current on a conductor, solenoids, and electromagnetic induction.
• Applications of electromagnetic effects like electric motors, generators, and transformers.

Example:

Q7. Mona was doing an experiment with a magnetic compass and a straight current-carrying wire. She observed that as she moved the compass away from the current-carrying wire, the deflection of the compass needle reduced. 

A. Explain why the deflection of the compass needle reduced as Mona moved away the compass needle from the current carrying wire. 

B. Mention one thing that could have changed in the circuit of the wire that could increase the deflection of the needle. 

C. Explain with reason what will be the direction of the magnetic field associated with the wire for the case described by the above figure.

Answer: A. The strength of the magnetic field is inversely proportional to the distance from the current-carrying wire. Therefore, when Mona moved the compass farther from the wire, the magnetic effect weakened, resulting in a smaller deflection of the compass needle.

B. The magnetic field strength is directly proportional to the current passing through the wire. To observe a larger deflection in the compass needle, Mona could increase the current in the circuit.

C. The battery indicates that the current flows from the top to the bottom of the plane. Using the right-hand thumb rule, we can determine that the magnetic field around the wire will be in a clockwise direction.

Sources of Energy

• Classification of different energy sources: renewable (solar, wind) and non-renewable (coal, petroleum).
• Advantages and limitations of each energy source, their environmental impact, and future energy sustainability.

Examples:

Q8. Which of the following is not a role of decomposers in the ecosystem?

A. Natural replenishment of soil.

B. Enrichment of oxygen in the atmosphere.

C. Waste decomposition.

D. Break-down of dead remains.

Answer: B. Enrichment of oxygen in the atmosphere is not a role of decomposers in the ecosystem. 

Explanation: A significant portion of the oxygen in today’s atmosphere is produced by photosynthesis carried out by microorganisms, such as cyanobacteria. Among them is Prochlorococcus, a recently identified microorganism known for its remarkable efficiency in capturing light for photosynthesis.

Q9. Identify the incorrect statement ‘The energy available to the producers is maximum’ because:

A. It is the first trophic level which absorbs 1% of light energy directly from the source.

B. It utilizes most of the chemical energy for its own respiration, growth, reproduction, movement etc.

C. It utilizes 10% of light energy and transfers the rest to the next trophic level.

D. It transfers only 10% of light energy to the next trophic level.

Answer: (C) It utilizes 10% of light energy and transfers the rest to the next trophic level.

Explanation: The incorrect statement in the first part is C. In the second part, the statement that does not describe the role of decomposers is B.

First Part: 

The question asks to identify the incorrect statement regarding energy available to producers.

Analyse the Statements:
Producers are at the first trophic level and absorb about 1% of light energy directly from the source.
True, as this is a well-established fact in energy transfer.
Producers utilise most of the chemical energy they capture for respiration, growth, reproduction, and movement.
True, this aligns with the energy usage within producers.
Producers utilise 10% of light energy.
Incorrect, as they only absorb about 1% of light energy from the source and transfer about 10% of their energy to the next trophic level.

Conclusion:
The incorrect statement is C.

Second Part:
The question asks which statement does not represent the role of decomposers in the ecosystem.

Analyse the Roles:
Decomposers are primarily responsible for breaking down dead organic matter, recycling nutrients back into the ecosystem, and helping maintain soil fertility.
Any statement outside these primary roles would be incorrect.

Conclusion:
The statement that is not a role of decomposers is B.

Chemistry

The chemistry section focuses on reactions, elements, and the behavior of various chemical compounds. Students learn about the periodic table, the reactivity of metals, and the chemistry of carbon compounds.

Chemical Reactions and Equations

• Explanation of chemical reactions, writing and balancing equations.
• Different types of reactions such as combination, decomposition, displacement, and redox reactions.

Examples:

Q10. When 50g of lead powder is added to 300 ml of blue copper sulphate solution, after a few hours, the solution becomes colourless. This is an example of 

A. Combination reaction 

B. Decomposition reaction 

C. Displacement reaction 

D. Double displacement reaction 

Answer: Option C - Displacement reaction

Explanation: Lead is more reactive than copper, so it replaces copper in the copper sulfate solution. This displacement reaction results in the formation of lead(II) sulfate and the deposition of copper metal. As copper ions are displaced, the blue color of the copper sulfate solution fades.

Pb + CuSO₄ → PbSO₄ + Cu.

Q11. Identify the type of each of the following reactions stating the reason for your answers. 

A. Fe₂O₃ + 2Al → Al₂O₃ + 2Fe + heat 

B. Pb(NO₃)₂ + 2KI → PbI²⁺ 2KNO₃

Answer: A. Exothermic/Displacement/Redox reaction: In this type of reaction, heat is released, or a more reactive element displaces a less reactive one. For example, aluminium reacts with iron(II) oxide, reducing it to form iron.

B. Double Displacement/Precipitation reaction: This reaction involves the exchange of ions between the reactants and the products. A yellow precipitate of lead iodide (PbI₂) is formed as a result of this exchange.

Acids, Bases, and Salts

• Properties of acids, bases, and salts, their reactions, and everyday uses.
• pH scale, neutralization reactions, and the importance of pH in the environment and industries.

Examples:

Q12. You are given 3 unknown solutions A, B, and C with pH values of 6, 8 and 9.5 respectively. In which solution will the maximum number of hydronium ions be present? Arrange the given samples in the increasing order of H⁺ ion concentration.

Answer: The number of hydronium ions (H₃O⁺ or H⁺) in a solution is directly related to its pH value. The pH scale is inversely proportional to the concentration of hydronium ions. A lower pH indicates a higher concentration of hydronium ions, while a higher pH indicates a lower concentration of hydronium ions.

Given that:

• Solution A has a pH of 6 (more acidic, higher H⁺ ion concentration).
• Solution B has a pH of 8 (mildly basic).
• Solution C has a pH of 9.5 (more basic, lower H⁺ ion concentration).

Therefore, Solution A (with a pH of 6) will have the highest concentration of hydronium ions.

Arranging the solutions in increasing order of H⁺ ion concentration:

The order of increasing H⁺ ion concentration (or decreasing acidity) is:

Solution C (pH 9.5) < Solution B (pH 8) < Solution A (pH 6)

Q13. Comment on the following statements: 

(i) Bee sting is treated with baking soda paste whereas wasp sting is treated with dilute vinegar. 

(ii) Farmers treat soil with quicklime when tilling. 

(iii) Ancient sculptures and marble structures are conserved by treating them with certain chemicals.

Answer: (i) A bee sting being acidic is treated with baking soda which is basic to neutralize the reaction and similarly, a wasp sting being basic is treated with dilute vinegar which is acidic. 

(ii) Quicklime, or calcium oxide, is basic which is used to make the soils less acidic by increasing their pH. This process helps improve nutrient availability for plants, promoting better crop yields.

(iii) Chemicals like calcium carbonate are used to neutralize acidic rain damage on marble. Treatments often include applying protective coatings or using solutions that bind to and stabilize the marble, preserving its integrity.

Metals and Non-Metals

• Physical and chemical properties of metals and non-metals, reactivity series, and displacement reactions.
• Methods of extraction of metals and formation of alloys, corrosion, and prevention.

Examples:

Q14. Give reasons for the following 

i. Certain metals are used for making cooking utensils. 

ii. Hydrogen gas does not evolve when certain metals except Mg & Mn react with nitric acid. 

Answer: (i) Metals are excellent conductors of heat, allowing them to transfer heat quickly and uniformly. This is why metals like copper and aluminium are commonly used in cooking utensils, as they are very efficient at conducting heat.

(ii) Nitric acid is a powerful oxidizing agent, capable of oxidizing hydrogen to form water. As a result, hydrogen gas does not evolve when most metals react with nitric acid, except for magnesium (Mg) and manganese (Mn), which do not follow this behavior.

Q15. Anirudh took two metal oxides; aluminium oxide and magnesium oxide as shown in the pictures given below. But he forgot to label them. How will you guide/ help Anirudh to identify the oxides and label them?

Answer: MgO(aq) turns red litmus blue. Al₂O₃ does not change litmus colour as it is amphoteric and insoluble in water. 

Or we can say it as, 

MgO reacts with acids only, but Al₂O₃ reacts with both acids and bases as it is amphoteric.

Carbon and its Compounds

• Covalent bonding in carbon, hydrocarbons (alkanes, alkenes, alkynes), and functional groups.
• Concepts of isomerism, homologous series, and the basics of organic chemistry, focusing on soaps and detergents.

Examples:

Q16. (i) “Keerthi thinks that Substitution reaction occurs in saturated Hydrocarbons, on the contrary Krishi thinks, it occurs in unsaturated Hydrocarbons.” Justify with valid reasoning whose thinking is correct. 

(ii) “Methane and Propane and their Isomers are used as fuels” Comment. Draw the electron dot structure of the immediate lower homologue of Propane. Give any two characteristics of homologues of a given homologous series. 

(iii) A mixture of oxygen and ethyne is burnt for welding. Can you predict why a mixture of ethyne and air is not used?

Answer: (i) She is correct, as substitution reactions take place in saturated hydrocarbons, where hydrogen atoms are replaced by heteroatoms. In contrast, unsaturated hydrocarbons undergo addition reactions, where simple molecules are added across the double or triple bonds.

(ii) Methane and propane go through a combustion reaction in the presence of oxygen and produce a large amount of energy. 

The lower homologue of propane ethane has the following electron dot structure:

Two Characteristics  

- Same general formula/ functional group 

- Similar chemical properties 

(iii) When a mixture of ethyne (acetylene) and oxygen undergoes complete combustion, it burns with a clean, blue flame. However, when there is insufficient oxygen or when air is present, ethyne undergoes incomplete combustion, resulting in a sooty, yellow flame.

Q17. Give the IUPAC name of the first member of Alkene which is formed by addition of conc. sulphuric acid to it. Illustrate the change with the help of a chemical equation. 

Answer: The IUPAC name of the first alkene formed when concentrated sulfuric acid is added to it is ethyl hydrogensulfate: 

Chemical equation: CH₂=CH₂ + →CH₃CH₂HSO₄

_‍Periodic Classification of Elements

• The modern periodic table, trends like atomic size, valency, electronegativity, and ionization energy.
• How the arrangement of elements helps predict their chemical and physical properties.

Examples:

Q18. The electronic configuration of three elements X, Y and Z are X- 2, 8, 7; Y- 2, 8, 2; and Z - 2, 8 

A. Y and Z are metals 

B. Y and X are non-metals 

C. X is a non -metal and Y is a metal 

D. Y is a non-metal and Z is a metal

Answer: Option D is correct.

Explanation: X = 2,8 = 10 - It is the electronic configuration of NEON, which is a non-metal.

Y = 2,8,7 = 17 - It is the electronic configuration of CHLORINE, which is a non-metal.

Z = 2,8,2 = 12 - It is the electronic configuration of Magnesium, which is a metal.

So, Y is a non metal and Z is a metal.

‍Biology

‍Biology emphasizes the life processes essential for living organisms, covering reproduction, evolution, and ecosystems. Students will learn about both plant and human biology in detail.

‍Life Processes

• Essential processes for living organisms: nutrition, respiration, transportation, and excretion.
• Detailed study of the human digestive system, respiratory system, and circulatory system with key diagrams.

Examples:

Q19. Explain the role of the following enzymes in the process of digestion of food in humans: 

(i) Salivary amylase (ii) Pepsin (iii) Trypsin (iv) Lipase

Answer: (i) Salivary amylase - Secreted by the salivary glands, salivary amylase breaks down starch into smaller carbohydrate molecules like maltose and dextrin in the mouth.

‍(ii) Pepsin - Secreted by the stomach lining in an inactive form called pepsinogen, pepsin breaks down proteins into smaller peptides.

‍(iii) Trypsin - Secreted by the pancreas and released into the small intestine, trypsin breaks down peptides into smaller amino acids.

‍(iv) Lipase - Secreted by the pancreas and released into the small intestine, lipase breaks down fats into fatty acids and glycerol. 

Q20. During which of the following stages of the circulation of blood in a normal human being, the oxygenated blood is pumped to all parts of the body? 

A. contraction of the left atrium 

B. contraction of left ventricle 

C. relaxation of the right atrium 

D. relaxation of the right ventricle

Answer: Option B is correct. 

Explanation: The contraction of the left ventricle is the stage of the cardiac cycle when oxygenated blood is pumped to all parts of the body

‍Control and Coordination

• Human nervous system and hormonal control, with emphasis on reflex actions, synapses, and endocrine glands.
• Control mechanisms in plants, such as tropism and hormones regulating growth and responses to stimuli.

Examples:

Q21. Mohan and Rohit observed that shoots of a plant growing in shade bend towards the sunlight. Whereas, leaves of ‘Touch me not’ plant fold and droop soon after touching. They were curious to know how these movements occur in plants. 

In order to help them understand the movements in the plants, answer the following questions:

 Attempt either subpart A or B. 

A. What causes the bending of shoots in the plants as shown in figure A? 

OR 

B. What causes the folding of the leaves in ‘Touch me not’ plant as shown in figure B? 

C. Compare the movement of growth of the pollen tube towards ovule with the movements shown in part A of the above figure. 

D. Compare the movement shown in figure B with the movement of body parts in the animals.

Answer: Ans A.

• The bending of plant shoots is a response to external stimuli and is a directional growth movement.
• When plants detect sunlight, a hormone called auxin, produced at the tip of the shoot, helps the cells elongate.
• If light is coming from one side, auxin moves toward the shaded side of the shoot.
• The increased concentration of auxin on the darker side stimulates the cells to elongate more, causing the shoot to bend toward the light.

Ans B.

• The leaves of the "Touch-me-not" plant respond to stimuli through a growth-independent movement.
• These plants use electrical and chemical signals to transmit information between cells. 
• The movement occurs at a location different from where the plant is touched. 
• The shape of the plant cells changes by adjusting the water content, causing them to either swell or shrink and alter their shape.

Ans C. The growth of pollen tubes toward the ovule is an example of chemotropism, while the bending of shoots toward sunlight is an example of phototropism.

Ans D. i) While both plants and animals use electrical and chemical methods to transmit signals between cells, plants do not have specialized tissues like nerve cells in animals to conduct this information.  

ii) In animals, changes in cell shape are driven by specialized proteins in muscle cells, whereas in plants, cell shape changes occur by altering the water content within the cells. 

How Do Organisms Reproduce?

• Asexual and sexual reproduction in animals and plants, with illustrations of reproductive organs.
• Topics like human reproductive health, contraception, and methods of birth control.

Examples:

Q22. The image below shows a developing fetus in the mother's womb. The developing fetus is connected to the placenta by means of umbilical cord. The Umbilical vein and artery run inside the umbilical cord. 

(i) Name two substance that moves through the blood vessels. 

(ii) If the placenta has less villi how will it affect the baby’s growth? 

(iii) Name the region where the embryo develops inside the female body. Explain how this region is adapted for nourishing the baby. 

(iv) Some of the fetal cells fall off into the amniotic fluid and can be collected by careful procedure. The cells were screened and found to contain XY chromosome. 

a) What is the sex of the foetus?   b) How is this prenatal sex determination misused? 

Answer: (i) Oxygen / Glucose 

(ii) Less villi means less surface area for nutrients to pass from mother to embryo which leads to slower growth of the baby. 

(iii) Uterus. It supplies the blood through the thick lining which helps with nourishing the embryo.

(iv) a) XY chromosomes = Male

b) It is misused when the gender is found out to be female and people due to their discriminative mindset engage in aborting the child (female foeticide)

Heredity and Evolution

• Basics of genetics, Mendel’s experiments, dominant and recessive traits, and the mechanism of inheritance.
• Evolutionary theories, Darwin’s theory of natural selection, and evidences of evolution.

Examples:

A. In a family of four individuals, the father possessed long ears and the mother possessed short ears. If the parents had pure dominant and recessive traits respectively, then calculate the ratio of genetic makeup of F2 generation. Show a suitable cross. 

B. If father had short ears and the mother had long ears, explain what effect it will have on the ratio of genetic makeup in F2 generation.

Answer: (a) Cross of Pure Dominant and Recessive Traits for Ear Length

Let's assume long ears (L) is the dominant trait and short ears (l) is the recessive trait. The father's genotype will be LL (pure dominant), and the mother's genotype will be ll (pure recessive). When these two individuals are crossed, the F1 generation will inherit one allele from each parent.

Parental Cross:

• Father (LL) × Mother (ll)

F1 Generation Genotypes:

• All offspring will have the genotype Ll because they inherit one L allele from the father and one l allele from the mother.
• So, all F1 individuals will have long ears (since L is dominant).

F2 Generation Cross:

To get the F2 generation, two F1 individuals (both with genotype Ll) will be crossed:

• F1 Cross: Ll × Ll

This will give the following genotypic ratio in F2 generation:

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So, the genotypic ratio in the F2 generation will be 1:2:1 (LL : Ll : ll), and the phenotypic ratio will be 3:1 (long ears : short ears).

(b) Effect of the Father with Short Ears and the Mother with Long Ears

If the father has short ears (genotype ll) and the mother has long ears (genotype LL), the cross would be:

• Father (ll) × Mother (LL)

In this case, all F1 offspring will inherit one l allele from the father and one L allele from the mother, so the genotype of all F1 individuals will be Ll. Since L is dominant, all F1 individuals will have long ears.

When these F1 individuals (Ll) are crossed with each other to produce the F2 generation:

• F1 Cross: Ll × Ll

The genotypic ratio in F2 will again be 1:2:1 (LL : Ll : ll), and the phenotypic ratio will be 3:1 (long ears : short ears), as the L allele still dominates over l.

Conclusion:

• F2 Generation from a LL × ll cross: Genotypic Ratio = 1:2:1 (LL : Ll : ll), Phenotypic Ratio = 3:1 (long ears : short ears).
• F2 Generation from a ll × LL cross: Genotypic Ratio = 1:2:1 (LL : Ll : ll), Phenotypic Ratio = 3:1 (long ears : short ears).

Q24. In snails individuals can begin life as male and depending on environmental conditions they can become female as they grow. This is because 

A. male snails have dominant genetic makeup. 

B. female snails have dominant genetic makeup. 

C. expression of sex chromosomes can change in a snail’s life time. 

D. sex is not genetically determined in snails. 

Answer: Option D is correct. 

Explanation: In snails, individuals can begin life as male and become female as they grow because sex is not genetically determined in snails. This is known as sequential hermaphroditism, which is when an organism can change its sex during its lifetime.

Our Environment

• The concept of ecosystems, food chains and food webs, biological magnification, and the importance of waste management.
• Human impact on the environment, conservation methods, and the role of individuals in environmental protection.

Examples:

Q25. What are the trophic levels? Give an example of a food chain and state the different trophic levels in it.

Answer In the food chain, the transfer of food or energy takes place at various levels, and these levels are known as trophic levels.

Example:

Grass → Goat → Man

In the food chain,

• Grass represents the first trophic level.
• The goat represents the second trophic level.
• Man represents the third trophic level.

Q26. What is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?

Answer: Biological magnification can be defined as the progressive increase in the concentration of non-biodegradable wastes in the food chain. As there is an increase in the magnification at the successive trophic levels of the ecosystems, all the other levels do get affected, and the concentration may vary when compared to the first level.

Management of Natural Resources

• Conservation of water, forests, wildlife, and non-renewable resources like coal and petroleum.
• Sustainable management strategies and government policies on the preservation of biodiversity and ecosystems.

Examples:

Q27. What changes can you make in your habits to become more environment-friendly?

Answer In order to become more environment-friendly, the following practices  can be incorporated into our day-to-day lives:

• Turning off any electrical appliance (such as TVs, water heaters, lights, fans, and air conditioners) when they are not in use.
• Avoiding the wastage of water by fixing any leaking taps or pipes as soon as possible. Also, the amount of water consumed must be controlled. For example, the tap should not be left running while brushing teeth.
• Disposing of plastic and glass wastes in recycling bins (many plastics take a long time to decompose and can have adverse effects on the environment).
• Using recyclable and eco-friendly products instead of convenient plastic products. For example, using paper or cloth bags instead of polythene bags is an environment-friendly habit.

Q28. List five things you have done over the last week to:

(a) conserve our natural resources.

‍(b) increase the pressure on our natural resources.

Answer Practices for the conservation of natural resources include the following:

• Use of bicycles and public transport while travelling to reduce fuel consumption.
• Use of recycled paper and other recycled products.
• Segregation of garbage into biodegradable and non-biodegradable bins.
• Avoiding the wastage of water by using it judiciously and fixing any leaking taps/pipes.

Practices that deplete natural resources include the following:

• Wastage of electrical energy by leaving electrical appliances on after use.
• Wastage of water by ignoring any leaking pipes/taps and leaving the tap on while brushing.
• Excessive use of plastic products such as polythene bags.

NCERT Solutions provides detailed explanations for each question in the textbook, making it easier for students to understand tough concepts, practice frequently asked questions, and get an in-depth understanding of all topics.

Importance of NCERT Solutions for Class 10 Science

NCERT Solutions for Class 10 Science is an invaluable resource for students as they prepare for their board exams. Designed according to the latest CBSE syllabus, these solutions provide clear, step-by-step explanations for all the topics covered in the textbook. From complex physics equations to intricate biology processes and essential chemistry reactions, NCERT Solutions simplifies learning, offering students a strong foundation in scientific concepts. They not only help students understand difficult ideas but also enable effective revision, making them an essential tool for mastering the subject.

Clear Conceptual Understanding: Each solution is explained in a simple and concise manner, ensuring that students can understand complex topics with ease.

Step-by-Step Solutions: Detailed, step-by-step solutions are provided for every question in the NCERT textbook. This helps students learn the correct approach to solving problems.

Board Exam Preparation: The solutions are aligned with the CBSE exam pattern, ensuring students are well-prepared for their board exams.

Time Management: Practicing these solutions helps students improve their problem-solving speed and manage time effectively during exams.

Comprehensive Coverage: The solutions cover all the chapters in the Class 10 Science textbook, ensuring that no topic is left out.

NCERT Solutions for Class 10 Science serve as a reliable guide for students aiming to excel in their board exams. They provide well-structured answers that align with the exam pattern and enhance conceptual understanding. By integrating these solutions into their study plan, students can ensure thorough preparation, improve problem-solving skills, and boost their overall confidence in tackling science topics.

With consistent practice and the help of Class 10 Science NCERT Solutions, students can overcome challenges and significantly improve their performance in board exams. The solutions are an essential tool to boost confidence, enhance problem-solving skills, and score above 90%. Download the chapter-wise PDFs today and start your journey toward academic success!

Frequently Asked Questions:

Where can I download NCERT Class 10 Science Solutions?

Students can download the chapter-wise NCERT Science Solutions PDF from the given links just by clicking on them anytime. 

How do NCERT Science solutions help in board exam preparation?

NCERT Science solutions are essential for board exam preparation, as they closely follow the CBSE exam pattern. Educart’s solutions break down difficult concepts into simpler explanations, helping students understand topics such as electricity, chemical reactions, and life processes. 

Do NCERT solutions cover the latest syllabus?

Yes, Educart’s NCERT Solutions are regularly updated to ensure they reflect the latest CBSE Class 10 Science Syllabus.