Class 12 Chemistry Chapter 10 Biomolecules ncert solutions

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Biomolecules Class 12 ncert solutions: Chemistry Class 12 Chapter 10 exercise solutions

TextbookNCERT
ClassClass 12
SubjectChemistry
ChapterChapter 10
Chapter NameBiomolecules ncert solutions
CategoryNcert Solutions
MediumEnglish

Are you looking for Class 12 Chemistry Chapter 10 Biomolecules ncert solutions? Now you can download Ncert Chemistry Class 12 Chapter 10 exercise solutions pdf from here.

Question 10.1: What are monosaccharides?

Solution 10.1: Monosaccharides are the simplest form of carbohydrates, consisting of a single sugar unit. They are the building blocks of more complex carbohydrates like disaccharides and polysaccharides. Monosaccharides have the general chemical formula ( CnH2nOn, where ( n ) typically ranges from 3 to 7.

Key characteristics of monosaccharides include:

  • Single sugar unit: They cannot be broken down into simpler sugars through hydrolysis.
  • Solubility: They are usually water-soluble due to their hydroxyl (-OH) groups.
  • Taste: Most monosaccharides are sweet-tasting.

Examples of monosaccharides:

  • Glucose: A primary energy source for cells.
  • Fructose: Found in fruits and honey, sweeter than glucose.
  • Galactose: Often found in milk, combined with glucose to form lactose.

Monosaccharides are classified based on the number of carbon atoms:

  • Hexoses (6 carbons): e.g., glucose, fructose, and galactose.
  • Trioses (3 carbons): e.g., glyceraldehyde.
  • Pentoses (5 carbons): e.g., ribose, which is part of RNA.

Question 10.2: What are reducing sugars?

Solution 10.2: Reducing sugars are sugars that can donate electrons to another chemical, thereby reducing it. This is possible because they have a free aldehyde group (-CHO) or a free ketone group (C=O) that can be oxidized. Reducing sugars can react with reagents like Benedict’s solution or Fehling’s solution, causing a color change.

Examples of reducing sugars include:

  • Monosaccharides: Glucose, fructose, and galactose.
  • Disaccharides: Lactose and maltose (but not sucrose).

Reducing sugars are important in biochemical tests and food chemistry.

Question 10.3: Write two main functions of carbohydrates in plants

Solution 10.3: Energy Storage: Carbohydrates, primarily in the form of starch, serve as an energy reserve in plants. Starch is stored in seeds, roots, and tubers, and can be broken down into glucose to provide energy during growth or periods of low photosynthesis.

Structural Support: Carbohydrates like cellulose form the structural component of plant cell walls. Cellulose gives rigidity and strength to the plant, helping it maintain its shape and withstand external pressures.

Question 10.4: Classify the following into monosaccharides and disaccharides.
Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.

Solution 10.4:

  • Monosaccharides: Ribose, 2-deoxyribose, Galactose, Fructose
  • Disaccharides: Lactose, Maltose

Question 10.5: What do you understand by the term glycosidic linkage?

Solution 10.5: A glycosidic linkage is a type of covalent bond that connects two sugar (carbohydrate) molecules through an oxygen atom. It forms when the hydroxyl group (-OH) of one sugar reacts with the anomeric carbon (the carbon attached to the oxygen in the ring form) of another sugar, releasing a water molecule (condensation reaction). This bond is key in forming disaccharides, polysaccharides, and other carbohydrate structures.

For example, the bond between glucose and fructose in sucrose is a glycosidic linkage.

Question 10.6: What is glycogen? How is it different from starch?

Solution 10.6: Glycogen is a polysaccharide that serves as the primary form of energy storage in animals and fungi. It is a highly branched polymer of glucose, stored mainly in the liver and muscles, and can be quickly broken down into glucose when energy is needed.

Differences between Glycogen and Starch:

1. Source:

  • Glycogen is found in animals and fungi.
  • Starch is the main energy storage polysaccharide in plants.

2. Structure:

  • Glycogen is more highly branched, with branches occurring every 8–12 glucose units.
  • Starch consists of two components: amylose (unbranched) and amylopectin (branched, but less so than glycogen).

3. Energy Release:

  • Starch releases glucose more slowly because of its lower degree of branching.
  • Glycogen is more rapidly broken down due to its higher branching, providing quick energy.

Question 10.7: What are the hydrolysis products of
(i) sucrose and (ii) lactose?

Solution 10.7: (i) On hydrolysis, sucrose gives one molecule of ∝-D glucose and one molecule of β- D-fructose.

NCERT Solutions for Class 12  Chemistry Chapter 14 - Biomolecules

(ii) The hydrolysis of lactose gives β-D-galactose and β-D-glucose.

NCERT Solutions for Class 12  Chemistry Chapter 14 - Biomolecules

Question 10.8: What is the basic structural difference between starch and cellulose?

Solution 10.8: Starch consists of two components − amylose and amylopectin. Amylose is a long linear chain of ∝−D−(+)−glucose units joined by C1−C4 glycosidic linkage (∝-link).

NCERT Solutions for Class 12  Chemistry Chapter 14 - Biomolecules

Amylopectin is a branched-chain polymer of ∝-D-glucose units, in which the chain is formed by C1−C4 glycosidic linkage and the branching occurs by C1−C6 glycosidic linkage.

NCERT Solutions for Class 12  Chemistry Chapter 14 - Biomolecules

On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C1−C4 glycosidic linkage (β-link).

NCERT Solutions for Class 12  Chemistry Chapter 14 - Biomolecules

Question 10.9: What happens when D-glucose is treated with the following reagents?
(i) HI (ii) Bromine water (iii) HNO3

Solution 10.9: (i) When D-glucose is heated with HI for a long time, n-hexane is formed.

NCERT Solutions for Class 12  Chemistry Chapter 14 - Biomolecules

(ii) When D-glucose is treated with Brwater, D- gluconic acid is produced.

NCERT Solutions for Class 12  Chemistry Chapter 14 - Biomolecules

(iii) On being treated with HNO3, D-glucose get oxidised to give saccharic acid.

NCERT Solutions for Class 12  Chemistry Chapter 14 - Biomolecules

Question 10.10: Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.

Solution 10.10: (1) Aldehydes give 2, 4-DNP test, Schiff’s test, and react with NaHSO4 to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.

(2) The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free −CHO group is absent from glucose.

(3) Glucose exists in two crystalline forms − ∝ andβ. The ∝-form (m.p. = 419 K) crystallises from a concentrated solution of glucose at 303 K and the β-form (m.p = 423 K) crystallises from a hot and saturated aqueous solution at 371 K. This behaviour cannot be explained by the open chain structure of glucose.

Question 10.11: What are essential and non-essential amino acids? Give two examples of each type.

Solution 10.11: Amino acids are the building blocks of proteins and play crucial roles in various biological processes. They can be classified into two categories: essential and non-essential amino acids.

Essential Amino Acids: Essential amino acids are those that cannot be synthesized by the human body and must be obtained from the diet. There are nine essential amino acids:

  1. Leucine – Important for protein synthesis and muscle repair.
  2. Lysine – Plays a crucial role in calcium absorption and hormone production.

Non-Essential Amino Acids

Non-essential amino acids are those that can be synthesized by the human body, and thus do not need to be obtained from the diet. There are eleven non-essential amino acids:

  1. Alanine – Involved in energy production and metabolism of glucose.
  2. Aspartic Acid – Plays a role in the urea cycle and as a neurotransmitter.

Summary Table

TypeExamples
Essential Amino AcidsLeucine, Lysine
Non-Essential Amino AcidsAlanine, Aspartic Acid

Key Points

  • Essential Amino Acids: Must be obtained through diet (e.g., meat, dairy, and legumes).
  • Non-Essential Amino Acids: Can be produced by the body from other amino acids or nitrogen sources.

Question 10.12: Define the following as related to proteins
(i) Peptide linkage (ii) Primary structure (iii) Denaturation

Solution 10.12: Here’s a definition and explanation of each of the terms related to proteins:

(i) Peptide Linkage

Definition: A peptide linkage, also known as a peptide bond, is a covalent bond formed between the carboxyl group of one amino acid and the amino group of another amino acid, resulting in the release of a molecule of water (condensation reaction).

Explanation:

  • When two amino acids react, the carboxyl group (-COOH) of one amino acid reacts with the amino group (-NH₂) of another, forming a peptide bond and releasing a water molecule (H₂O).
  • The resulting bond is -CO-NH-, which links the two amino acids together.
  • Peptide bonds can link together to form chains of amino acids, creating polypeptides and proteins.

(ii) Primary Structure

Definition: The primary structure of a protein refers to the specific linear sequence of amino acids that make up the protein.

Explanation:

  • This sequence is determined by the genetic code and is unique for each protein.
  • The primary structure is crucial because it dictates how the protein will fold into its secondary, tertiary, and quaternary structures, ultimately determining the protein’s function.
  • The primary structure is stabilized by peptide bonds and is typically represented as a string of amino acid abbreviations (e.g., Ala-Gly-Leu).

(iii) Denaturation

Definition: Denaturation is the process by which a protein loses its native structure and, consequently, its biological activity due to external factors such as heat, pH changes, or chemical agents.

Explanation:

  • During denaturation, the weak interactions (such as hydrogen bonds, ionic bonds, and hydrophobic interactions) that maintain the protein’s structure are disrupted.
  • Denaturation can lead to the unfolding of the protein, altering its secondary, tertiary, or quaternary structures.
  • While denaturation does not affect the primary structure (the amino acid sequence remains intact), it often renders the protein nonfunctional. Denaturation is often reversible, but in some cases, it can be irreversible.

Summary Table

TermDefinition
Peptide LinkageA covalent bond formed between the carboxyl group of one amino acid and the amino group of another.
Primary StructureThe specific linear sequence of amino acids in a protein, dictating its subsequent folding and function.
DenaturationThe process of losing a protein’s native structure due to external factors, leading to loss of function.

Question 10.13: What are the common types of secondary structure of proteins?

Solution 10.13: There are two common types of secondary structure of proteins: 

  • (i) ∝-helix structure
  • (ii) β-pleated sheet structure

∝- Helix structure:

In this structure, the −NH group of an amino acid residue forms H-bond with the group of the adjacent turn of the right-handed screw (∝-helix).

NCERT Solutions for Class 12  Chemistry Chapter 14 - Biomolecules

β-pleated sheet structure:

This structure is called so because it looks like the pleated folds of drapery. In this structure, all the peptide chains are stretched out to nearly the maximum extension and then laid side by side. These peptide chains are held together by intermolecular hydrogen bonds.

NCERT Solutions for Class 12  Chemistry Chapter 14 - Biomolecules

Question 10.14: What type of bonding helps in stabilising the ∝-helix structure of proteins?

Solution 10.14: The H-bonds formed between the −NH group of each amino acid residue and the group of the adjacent turns of the ∝-helix help stabilise the helix.

Question 10.15: Differentiate between globular and fibrous proteins.

Solution 10.15: Globular and fibrous proteins are two major classes of proteins that differ in structure, function, and properties. Here’s a detailed comparison of their characteristics:

1. Structure

  • Globular Proteins:
    • Shape: Spherical or globular shape, often compact and folded.
    • Secondary Structure: They typically contain a mix of α-helices and β-sheets.
    • Tertiary Structure: Highly folded and complex 3D structures due to interactions between amino acid side chains.
  • Fibrous Proteins:
    • Shape: Elongated, rod-like, or filamentous shape.
    • Secondary Structure: Primarily composed of one type of secondary structure, often α-helices or β-sheets, forming long strands.
    • Tertiary Structure: Generally have a more simple and repetitive structure, with limited folding.

2. Function

  • Globular Proteins:
    • Role: Involved in a wide range of biological functions including enzymes (catalysts), antibodies (immune response), hormones (signaling), and transport proteins (e.g., hemoglobin).
    • Solubility: Generally soluble in water due to their hydrophilic side chains on the surface, allowing them to function in aqueous environments.
  • Fibrous Proteins:
    • Role: Provide structural support and strength to tissues, such as collagen in connective tissues, keratin in hair and nails, and elastin in skin.
    • Solubility: Generally insoluble in water due to their hydrophobic nature and tightly packed structure, which minimizes exposure to the aqueous environment.

3. Examples

  • Globular Proteins:
    • Examples:
      • Hemoglobin: Transports oxygen in the blood.
        • Enzymes: Such as lactase, which catalyzes the breakdown of lactose.
  • Fibrous Proteins:
    • Examples:
      • Collagen: Provides structural support in connective tissues.
      • Keratin: Found in hair, nails, and the outer layer of skin.

4. Stability

  • Globular Proteins:
    • More sensitive to changes in pH, temperature, and other environmental factors, which can lead to denaturation (loss of structure and function).
  • Fibrous Proteins:
    • Generally more stable and resilient to denaturation due to their structural properties and fewer interactions that can be disrupted.

Summary Table

FeatureGlobular ProteinsFibrous Proteins
ShapeSpherical or globularElongated or filamentous
Secondary StructureMixed (α-helices and β-sheets)Predominantly one type (α or β)
Tertiary StructureHighly folded, complexSimple, repetitive
FunctionEnzymatic, transport, signaling, immuneStructural support
SolubilityGenerally soluble in waterGenerally insoluble in water
ExamplesHemoglobin, enzymesCollagen, keratin
StabilitySensitive to denaturationMore stable and resilient

Question 10.16: How do you explain the amphoteric behaviour of amino acids?

Solution 10.16: In aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton to give a dipolar ion known as zwitter ion.

NCERT Solutions for Class 12  Chemistry Chapter 14 - Biomolecules

Therefore, in zwitter ionic form, the amino acid can act both as an acid and as a base.

NCERT Solutions for Class 12  Chemistry Chapter 14 - Biomolecules

Thus, amino acids show amphoteric behaviour.

Question 10.17: What are enzymes?

Solution 10.17: Enzymes are proteins that catalyse biological reactions. They are very specific in nature and catalyse only a particular reaction for a particular substrate. Enzymes are usually named after the particular substrate or class of substrate and some times after the particular reaction. 

For example, the enzyme used to catalyse the hydrolysis of maltose into glucose is named as maltase.

NCERT Solutions for Class 12  Chemistry Chapter 14 - Biomolecules

Again, the enzymes used to catalyse the oxidation of one substrate with the simultaneous reduction of another substrate are named as oxidoreductase enzymes.

The name of an enzyme ends with ‘− ase’.

Question 10.18: What is the effect of denaturation on the structure of proteins?

Solution 10.18: As a result of denaturation, globules get unfolded and helixes get uncoiled. Secondary and tertiary structures of protein are destroyed, but the primary structures remain unaltered. It can be said that during denaturation, secondary and tertiary-structured proteins get converted into primary-structured proteins. Also, as the secondary and tertiary structures of a protein are destroyed, the enzyme loses its activity.

Question 10.19: How are vitamins classified? Name the vitamin responsible for the coagulation of blood.

Solution 10.19: Vitamins are essential organic compounds that the body needs in small amounts for various physiological functions. They are classified based on their solubility and function. Here’s a detailed classification:

Classification of Vitamins

1. Water-Soluble Vitamins

These vitamins dissolve in water and are not stored in large amounts in the body. They need to be consumed regularly in the diet.

  • Examples:
    • Vitamin B Complex:
    • Vitamin B1 (Thiamine)
    • Vitamin B2 (Riboflavin)
    • Vitamin B3 (Niacin)
    • Vitamin B5 (Pantothenic acid)
    • Vitamin B6 (Pyridoxine)
    • Vitamin B7 (Biotin)
    • Vitamin B9 (Folate)
    • Vitamin B12 (Cobalamin)
    • Vitamin C (Ascorbic Acid)

2. Fat-Soluble Vitamins

These vitamins dissolve in fats and oils and can be stored in the body’s fatty tissues and liver. They are absorbed along with dietary fat.

  • Examples:
    • Vitamin A (Retinol)
    • Vitamin D (Cholecalciferol)
    • Vitamin E (Tocopherol)
    • Vitamin K (Phylloquinone and Menaquinone)

Vitamin Responsible for Coagulation of Blood

The vitamin primarily responsible for blood coagulation is Vitamin K. It plays a crucial role in the synthesis of certain proteins (known as clotting factors) that are essential for blood clotting.

Types of Vitamin K:

  • Vitamin K1 (Phylloquinone): Found in green leafy vegetables, it is the primary form of vitamin K in the diet.
  • Vitamin K2 (Menaquinone): Found in fermented foods and animal products, it is produced by bacteria in the gut and also plays a role in coagulation.

Question 10.20: Why are vitamin A and vitamin C essential to us? Give their important sources.

Solution 10.20: These two vitamins are essential to us because the deficiency of these two vitamins causes us harmful diseases like xerophthalmia (which hardens the cornea of the eye) and night blindness. While the deficiency of Vitamin C causes scurvy (bleeding gums).

  •  sources of tThe sources of these two vitamins are given below:
    • Vitamin A: Carrots, fish liver oil, milk and butter.
    • Vitamin C: Amla, citrus fruits and green leafy vegetables.

Question 10.21: What are nucleic acids? Mention their two important functions.

Solution 10.21: Nucleic acids are large biomolecules essential for all forms of life. They are polymers made up of monomer units called nucleotides, which consist of three components: a nitrogenous base, a five-carbon sugar (ribose or deoxyribose), and a phosphate group. The two primary types of nucleic acids are deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).

Key Characteristics of Nucleic Acids

  • Structure:
    • DNA: Typically double-stranded, forming a double helix structure. The two strands are complementary and held together by hydrogen bonds between paired nitrogenous bases (adenine with thymine, and guanine with cytosine).
    • RNA: Usually single-stranded and can fold into various shapes. It contains ribose as its sugar and uracil instead of thymine (adenine pairs with uracil in RNA).

Important Functions of Nucleic Acids

  1. Storage and Transmission of Genetic Information:
  • DNA serves as the genetic blueprint for all living organisms. It contains the instructions necessary for the development, functioning, growth, and reproduction of organisms. The sequence of nucleotides in DNA encodes genes, which determine specific traits and are passed from one generation to the next during reproduction.
  1. Protein Synthesis:

RNA plays a crucial role in translating the genetic information stored in DNA into proteins. This process occurs through two main stages:

  • Transcription: The process of synthesizing messenger RNA (mRNA) from a DNA template. This mRNA carries the genetic code from the nucleus to the cytoplasm.
  • Translation: The ribosome reads the sequence of the mRNA and, with the help of transfer RNA (tRNA) and ribosomal RNA (rRNA), assembles the corresponding amino acids to form a protein.

Question 10.22: What is the difference between a nucleoside and a nucleotide?

Solution 10.22: Nucleosides and nucleotides are both essential components of nucleic acids, but they have distinct structures and functions. Here’s a breakdown of the differences between them:

1. Definition

  • Nucleoside: A nucleoside consists of a nitrogenous base (either a purine or a pyrimidine) attached to a five-carbon sugar (ribose in RNA and deoxyribose in DNA). It does not contain a phosphate group.
  • Nucleotide: A nucleotide is composed of a nucleoside (a nitrogenous base and a sugar) plus one or more phosphate groups attached to the sugar. Nucleotides are the building blocks of nucleic acids.

2. Structure

  • Nucleoside Structure:
    • Composed of:
      • Nitrogenous Base: (e.g., adenine, guanine, cytosine, thymine, or uracil)
      • Sugar: (ribose for RNA, deoxyribose for DNA)
  • Chemical Representation: Nucleosides are generally represented as:
    Nucleoside = Nitrogenous Base + Sugar
  • Nucleotide Structure:
    • Composed of:
      • Nitrogenous Base: (same as in nucleosides)
      • Sugar: (same as in nucleosides)
      • Phosphate Group: One or more phosphate groups can be attached to the sugar.
  • Chemical Representation: Nucleotides are generally represented as:
    Nucleotide = Nucleoside + Phosphate(s)

3. Function

  • Nucleoside:
    • Nucleosides are involved in various biological processes and can serve as precursors for nucleotides. Some nucleosides also play roles in signaling (e.g., adenosine is involved in energy transfer and signaling).
  • Nucleotide:
    • Nucleotides are the fundamental units of nucleic acids (DNA and RNA) and are essential for:
      • Storing and transferring genetic information.
      • Serving as energy carriers (e.g., ATP – adenosine triphosphate).
      • Acting as cofactors in enzymatic reactions (e.g., NAD⁺, FAD).

4. Examples

  • Nucleoside Examples:
    • Adenosine (adenine + ribose)
    • Cytidine (cytosine + ribose)
    • Deoxyadenosine (adenine + deoxyribose)
  • Nucleotide Examples:
    • Adenosine triphosphate (ATP: adenosine + 3 phosphate groups)
    • Cytidine monophosphate (CMP: cytidine + 1 phosphate group)
    • Deoxyadenosine triphosphate (dATP: deoxyadenosine + 3 phosphate groups)

Summary Table

FeatureNucleosideNucleotide
CompositionNitrogenous base + five-carbon sugarNitrogenous base + five-carbon sugar + one or more phosphate groups
Phosphate GroupNoneOne (monophosphate), two (diphosphate), or three (triphosphate)
ExamplesAdenosine, Guanosine, Cytidine, Thymidine, UridineAdenosine triphosphate (ATP), Guanosine triphosphate (GTP), Cytidine monophosphate (CMP)
FunctionActs as a building block for nucleotides and plays roles in various biological processes (e.g., signaling)Serves as the basic building block of nucleic acids (DNA and RNA) and plays critical roles in energy transfer (e.g., ATP) and metabolic pathways
Role in Nucleic AcidsPrecursor to nucleotidesFundamental unit of DNA and RNA, involved in the formation of nucleic acid strands

Question 10.23: The two strands in DNA are not identical but are complementary. Explain.

Solution 10.23: In the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosine forms hydrogen bond with guanine, while adenine forms hydrogen bond with thymine. As a result, the two strands are complementary to each other.

Question 10.24: Write the important structural and functional differences between DNA and RNA.

Solution 10.24: DNA (Deoxyribonucleic Acid) and RNA (Ribonucleic Acid) are both nucleic acids essential for life, but they differ in several structural and functional aspects. Here’s a detailed comparison:

Structural Differences

FeatureDNARNA
SugarDeoxyribose (lacks one oxygen atom compared to ribose)Ribose (contains an oxygen atom at the 2′ carbon)
StrandsTypically double-stranded, forming a double helixTypically single-stranded (can form complex shapes)
Nitrogenous BasesContains adenine (A), thymine (T), cytosine (C), and guanine (G)Contains adenine (A), uracil (U), cytosine (C), and guanine (G)
LengthGenerally longer; can be several million base pairs longGenerally shorter; varies widely (from a few hundred to several thousand nucleotides)
StabilityMore stable due to the double-stranded structure and deoxyribose sugarLess stable than DNA; more susceptible to hydrolysis
LocationPrimarily located in the nucleus of eukaryotic cells; also in mitochondriaFound in the nucleus and cytoplasm of both prokaryotic and eukaryotic cells

Functional Differences

FeatureDNARNA
FunctionStores and transmits genetic information; serves as the blueprint for all living organismsInvolved in protein synthesis, gene regulation, and catalyzing biochemical reactions
TypesOne main type (though exists in different forms, such as chromosomal DNA and mitochondrial DNA)Several types: mRNA (messenger RNA), tRNA (transfer RNA), rRNA (ribosomal RNA), and others (like snRNA and miRNA)
ReplicationReplicates through semi-conservative mechanism during cell divisionSynthesized from DNA through the process of transcription
Protein SynthesisDoes not directly participate in protein synthesisDirectly involved in protein synthesis (mRNA carries the code, tRNA brings amino acids, rRNA is part of the ribosome)
MutabilityLess prone to mutation due to error-checking mechanisms during replicationMore prone to mutations, as it is synthesized quickly and can be degraded rapidly

Question 10.25: What are the different types of RNA found in the cell?

Solution 10.25:

  • (i) Messenger RNA (m-RNA)
  • (ii) Ribosomal RNA (r-RNA)
  • (iii) Transfer RNA (t-RNA)

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