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NCERT Solutions for Class 6 Science Chapter 2 Components of Food

NCERT Solutions for Class 6 Science Chapter 2 Components of Food

Edited By Irshad Anwar | Updated on Apr 09, 2025 09:53 AM IST

NCERT Solutions for Class 6 Science Chapter 2 Diversity in the Living World- On Earth, various living forms exist, ranging from microorganisms to large animals and plants. All living forms have their characteristics and play an essential role in maintaining the ecosystem and balance of nature. Chapter 2 of Class 6 Science, Diversity in the Living World, provides information about different plants and animals around us in a very interesting and fun manner.

This Story also Contains
  1. NCERT Solutions for Class 6 Science Chapter 2: Diversity in the Living World (Solved Exercise Questions)
  2. Important Topics in NCERT Solutions for Class 6: Diversity in the Living World
  3. NCERT Solutions for Class 6 Science- Chapter-wise
NCERT Solutions for Class 6 Science Chapter 2 Components of Food
NCERT Solutions for Class 6 Science Chapter 2 Components of Food

By studying Class 6 Diversity in the Living World NCERT Solutions, students can easily understand these living forms, as step-by-step explanations along with required diagrams are provided by our subject experts. The grouping of animals and plants based on features like root, stem, leaves, movement, etc., is well highlighted here. Different adaptations that plants and animals develop to survive in difficult environmental conditions are also well explained here. Regular practice of NCERT Solutions Class 6 Science allows students to solve their doubts easily and score better marks in the upcoming examination.

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NCERT Solutions for Class 6 Science Chapter 2: Diversity in the Living World (Solved Exercise Questions)

The solved exercise given below will help students learn the key topics of the chapter and increase their knowledge of biodiversity and classification.

Question 1. Here are two types of seeds. What differences do you find among the roots and leaf venation of their plants?

wheat

Answer:

The first seed given is of wheat, which is one of the most important stable crops worldwide.

Roots: Wheat plants have a fibrous root system. The roots arise from the base of the stem and are thin and spread out in all directions, forming a dense network of small roots.

Leaf Venation: Wheat leaves have parallel venation, in which the veins run parallel to each other in the leaf.

The second seed is kidney beans, which are rich in protein, fiber, and essential minerals.

Roots: Kidney beans have a tap root system, where the primary root grows thicker and deeper, and the small branches arise from this.

Leaf Venation: Kidney bean have reticulate type of leaf venation, where the veins form a network or a web-like structure across the leaf.

Question 2. The names of some animals are given below. Group them based on their habitats. Write the names of aquatic animals in the area marked ‘A’ and terrestrial animals in the area marked ‘B. ’ Enter the names of animals living in both habitats in part C. Horse, Dolphin, Frog, Sheep, Crocodile, Squirrel, Whale, Earthworm, Pigeon, Tortoise

venn diagram

Answer:

A (Aquatic Animals):

  • Dolphin

  • Crocodile

  • Whale

B (Terrestrial Animals):

  • Horse

  • Sheep

  • Squirrel

  • Pigeon

  • Earthworm

  • Tortoise

C (Animals living in both habitats):

  • Frog

  • Crocodile (can live in both water and on land)

Question 3. Manu’s mother maintains a kitchen garden. One day, she was digging out radishes from the soil. She told Manu that radish is a kind of root. Examine a radish and write what type of root it is. What type of venation would you observe in the leaves of the radish plant?

Answer:

The radish is an example of a taproot. In this, the plant has a thick main root called the primary root and bears many side branches called secondary roots. In the radish plant, it helps in the storage of various sugars and starch, making it an important food source for us.

The leaves of the radish plant show reticulate venation. In which veins are irregularly distributed in the lamina, forming a network. This reticulate venation is mostly found in dicot plants.

Question 4. Look at the image of a mountain goat and a goat found in the plains. Point out the similarities and differences between them. What are the reasons for these differences?

goats

Answer:

Similarities: Given below are some of the similarities found in the mountain goat and the goat found in the plains:

1- Both belong to the same kind of family; that is, both are goats.

2- Mountain goats and goats found in plains are herbivores, meaning they eat grass, leaves, and plants.

3- Both have horns on their head, despite their size and structure.

4- Both the two goats are mammals, meaning they give birth to their young ones and give milk.

Differences: There are some differences between these two types of goats, which are provided here:

1- Mountain goats have thick fur, which protects them from the cold and snow. Goats in the plains have thin fur as it is warm in the plains.

2- Mountain goats live alone or in small groups, whereas the goats in the plains live in larger groups.

3- The size of mountain goats is generally smaller than that of the goats in the plains.

4- Mountain goats can easily climb steep hills, but goats in the plains are more used to grazing on flat ground.

Question 5. Group the following animals into two groups based on any feature other than those discussed in the chapter— cow, cockroach, pigeon, bat, tortoise, whale, fish, grasshopper, lizard.

Answer:

Apart from the feature which is already been discussed in this chapter, we can classify these animals based on the presence or absence of a backbone.

Animals having backbone

  • Cow

  • Pigeon

  • Bat

  • Tortoise

  • Whale

  • Fish

  • Lizard

Animals without backbone

  • Cockroach

  • Grasshopper

Question 6. As the population grows and people want more comfortable lives, forests are being cut down to meet various needs. How can this affect our surroundings? How do you think we can address this challenge?

Answer:

As the population is growing, forests are being cut on a large scale, which is known as deforestation, to meet various needs. Here are some effects of this:

  • It leads to soil erosion, which results in the loss of the fertile layer of soil.

  • Deforestation also disturbs the habitat of many wild animals, birds, and insects.

  • Deforestation may lead to frequent floods as there are no trees and their roots to absorb rainwater.

There are some ways by which we can address this challenge:

  • We need to plant trees around us.

  • Natural resources like wood can be used carefully.

  • By using eco-friendly products that cannot harm the environment.

  • Avoid using plastic bags as they do not decompose easily and cause pollution.


Question 7. Analyze the flowchart. What can be examples of ‘A’ and ‘B’?

flow chart


Answer:

According to the above flow chart, plants having leaves and also the reticulate venation can be the dicot plants as this type of venation is unique in dicot plants.

Plants having leaves but not the reticulate venation must be the monocot plants because this venation cannot be seen in monocot plants.

A- Dicot Plant

B- Monocot Plant

Question 8. Raj argues with his friend Sanjay that the “Gudhal (hibiscus) plant is a shrub.” What questions can Sanjay ask for clarification?

Answer:

Sanjay can ask multiple questions from Raj to clarify his argument. Provided below are some of the questions:

1- What is the definition of shrub?

2- How is a hibiscus plant different from a tree?

3- Can a hibiscus plant grow tall or stay small?

4- Is the hibiscus plant woody or soft?

Question 9. Based on the information in the table, find examples of these plants for each group.

Group

Type of Seed

Type of Root

Examples

A

Dicot

Taproot


B

Monocot

Fibrous roots



(a) What other similarities do plants of group A have?

(b) What other similarities do plants of group B have?

Answer:

Group

Type of Seed

Type of Root

Examples

A

Dicot

Taproot

Kidney Bean

B

Monocot

Fibrous roots

Wheat


(a) Dicot plants having taproots have some similarities that include the seeds having two cotyledons and reticulate venation found in leaves.

(b) Group B, having Monocot seed and fibrous roots have one cotyledon in their seeds and leaves are having parallel venation.


Question 10. Observe the labeled part of a duck in the picture given below. What differences do you observe in the feet of the duck compared to the other birds? Which activity would the duck be able to perform using this part?

duck and pigeon

Answer:

The labeled part of the duck is its webbed feet, which are different from the feet of other birds. The toes of ducks have skin in between that form a ‘web, ’ which is not present in other birds. Other ones are having individual toes without any skin present.

This helps the duck to swim and move easily in water. When the duck moves with its webbed feet, a lot of water is pushed backward, which allows the duck to move quickly forward; that is why ducks are very good swimmers.

Also, check NCERT Books and NCERT Syllabus here

NCERT Books Class 6

NCERT Syllabus Class 6

Important Topics in NCERT Solutions for Class 6: Diversity in the Living World

1- It includes the diversity of plants and animals that are present around us.

2- Classification and grouping of organisms based on similarities and differences are highlighted in Chapter 2.

3- Grouping of plants based on their roots, stems, leaves, and flowers is well-explained, along with diagrams.

4- Adaptations in animals such as camels and plants such as Rhododendrons are also provided in this chapter.

NCERT Solutions for Class 6 Science- Chapter-wise

The list of all the chapters included in class 6 Science is given below:

NCERT Solutions for Class 6 Science Chapter 1 The Wonderful World of Science

NCERT Solutions for Class 6 Science Chapter 2 Diversity in the Living World

NCERT Solutions for Class 6 Science Chapter 3 Mindful Eating: A Path To A Healthy Body

NCERT Solutions for Class 6 Science Chapter 4 Exploring Magnets

NCERT Solutions for Class 6 Science Chapter 5 Measurement of Length and Motion

NCERT Solutions for Class 6 Science Chapter 6 Materials Around Us

NCERT Solutions for Class 6 Science Chapter 7 Temperature and its Measurement

NCERT Solutions for Class 6 Science Chapter 8: A Journey Through States of Water

NCERT Solutions for Class 6 Science Chapter 9 Methods of Separation in Everyday Life

NCERT Solutions for Class 6 Science Chapter 10 Living Creatures: Exploring their Characteristics

NCERT Solutions for Class 6 Science Chapter 11 Nature's Treasures
NCERT Solutions for Class 6 Science Chapter 12 Beyond Earth

We wish all Class 6 students the best of luck in their exams, and we hope the NCERT Solutions guide them toward scoring better marks and performing well in their exams!

Frequently Asked Questions (FAQs)

1. Where can I find the complete solutions of NCERT Class 6 Science ?

For complete solutions of NCERT Class 6 Science: https://school.careers360.com/ncert/ncert-solutions-Class-6-Science 

2. Does CBSE provides the solutions of NCERT Class 6 ?

No, CBSE doesn’t provide NCERT solutions for any class or subject

3. What is the official website of NCERT?

http://www.ncert.nic.in/ is the link to official website of NCERT

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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