Class 9 Science

Structure Of The Atom

We have learned that atoms and molecules are the fundamental building blocks of matter. The existence of different kinds of matter around us is due to different types of atoms and molecules present in them.

• Dalton assumed that an atom is indivisible, i.e. it has no constituent particles. But, a series of experimental evidence revealed that an atom is not the smallest particle. Some other particles smaller than the atom are also present which are called sub-atomic particles, i.e. electrons, protons, and neutrons.
• The atoms of different elements differ in the number of electrons, protons, and neutrons.
• In this chapter, we will describe how electrons, protons, and neutrons were discovered and the various models that have been proposed to explain how these particles are arranged within the atom.

Charged Particles In Matter

The particles that carry an electric charge are called charged particles. Generally, on rubbing two objects together, they become electrically charged. It means that some charged particles are present within the atom or the atom is made up of some charged particles. Two such particles are electrons and protons.

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Discovery Of Electrons

• It was known by 1900, that the atom was not a simple, indivisible particle but contained at least one sub-atomic particle—the electron, which was identified by J. J. Thomson when he performed a cathode ray experiment using a discharge tube.
• In the experiment, a gas at low pressure was taken in a discharge tube made up of glass At the ends of the discharge tube two electrodes (metal plates) were placed, connected to a battery for high voltage supply.
• The electrode connected to the negative end was known as the cathode and that to the positive is the anode. During this experiment, he found a beam of negatively charged particles, called cathode rays, as they originated from the cathode. These negatively charged particles were called electrons.

Electrons are negatively charged particles and are denoted by ‘e^–‘ The charge present on an electron is equal to -1.6 × 10^-19 Coulomb.

Since this charge is considered to be the smallest, therefore, charge on e^– is taken as -1. The mass of an electron is equal to 9.1 × 10^-31 kg.

Discovery Of Protons

An atom is electrically neutral but the formation of cathode rays has shown that all the atoms contain negatively charged electrons. So, atoms must also contain some positively charged particles to balance the negative charge of electrons. This was the basis of the discovery of protons.

• Before the identification of electrons, E. Goldstein in 1886, discovered the presence of new radiations known as canal rays or anode rays. These rays were positively charged radiations which are seen moving from the anode towards.
• Cathode in a specially designed discharge tube (with a porous cathode), when a high voltage is applied across the electrodes. A porous cathode is used to provide the path for passing anode rays. It led to the discovery of another sub-atomic particle, the proton.

Protons are positively charged particles and are denoted by ‘p^+’. The charge present on a proton is equal to +1.6 × 10^-19 Coulomb and it is considered as +1.

The mass of a proton is equal to 1.6 × 10^-27 kg. The mass of a proton is approximately 2000 times that of an electron.

Conclusion

The mass of a proton is taken as one unit and its charge is (+1), whereas the mass of an electron is considered to be negligible and its charge is (-1). It seems that an atom is composed of protons and electrons, mutually balancing their charges.

The Structure Of An Atom

According to Dalton’s atomic theory, the atom was indivisible and indestructible. Now, the discovery of two fundamental particles (electrons and protons) inside the atom, led to the failure of this aspect of Dalton’s theory. To know the arrangement of electrons and protons within an atom, many scientists proposed various atomic models.

Thomson’s Model Of An Atom

• J.J. Thomson was the first scientist to propose a model for the structure of an atom. Thomson’s model of an atom was similar to Christmas pudding. The electrons in a sphere of positive charge were like currants (dry fruits) in a spherical Christmas pudding.
• It can also be compared to a watermelon, in which, the positive charge in an atom is spread all over like the red edible part, while the electrons are studded in the positively charged sphere, like the seeds in the watermelon.

The following are the postulates of this model:

1. Electrons are embedded in the sphere of positive charge.
2. The negative and positive charges are equal in magnitude. Therefore, the atom as a whole is electrically neutral.
3. The mass of an atom is assumed to be uniformly distributed throughout the atom.

Limitations of Thomson’s Model of an Atom

Limitations of J.J. Thomson’s model of an atom are:

1. J.J. Thomson’s model could not explain the experimental results of other scientists such as Rutherford, as there is no nucleus in the atomic model proposed by Thomson.
2. It could not explain the stability of an atom, i.e. how positive and negative charges could remain, so close together.

Rutherford’s Model Of An Atom

Ernest Rutherford designed an experiment to know how the electrons are arranged within an atom. He bombarded fast-moving a-particles (these are doubly charged helium ions having a mass of 4 u) on a thin sheet of gold foil. He selected a gold foil because he wanted a layer as thin as possible. This gold foil was about 1000 atoms thick.

The following observations were made by Rutherford:

1. Most of the fast-moving a-particles passed straight through the gold foil.
2. Some of the a-particles were deflected by the foil by small angles.
3. Very few a-particles (one out of 12000) appeared to rebound.

Based on his experiment, Rutherford concluded that:

1. Most of the space inside the atom is empty because most of the a-particles pass through the gold foil without getting deflected.
2. Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space.
3. A very small fraction of a-particles were deflected by 180° (i.e. they rebound), indicating that all the positive charge and mass of the atom were concentrated in a very small volume within the atom.

Based on his experiment, Rutherford put forward the nuclear model of an atom, having the following features:

1. There is a positively charged, highly dense center in an atom, called a nucleus. Nearly, the whole mass of the atom resides in the nucleus.
2. The electrons revolve around the nucleus in a circular path.
3. The size of the nucleus (10^-15m) is very small as compared to the size of the atom (10-10m).

Note: Rutherford suggested that his model of the atom was similar to that of the solar system. In the solar system, the different planets revolve around the Sun. Similarly, in an atom, the electrons are revolving around the nucleus. So, these electrons are also called planetary electrons.

Limitations Of Rutherford’s Model Of An Atom

Limitations of Rutherford’s model of an atom are:

1. Any charged particle when accelerated is expected to radiate energy. To remain in an ill circular orbit, the electron. Electrons would need to undergo acceleration. Therefore, it would radiate energy.
   1. Thus, the revolving electron would lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable. Therefore, matter would not exist, but we know matter exists. It means that atoms are quite stable.
   2. Thus, it could not explain the stability of an atom when charged electrons are moving under the attractive force of positively charged nucleus.
2. Rutherford’s model could not explain the distribution of electrons in the extra nuclear portion of the atom.

Bohr’s Model Of An Atom

To overcome the objections raised against Rutherford’s model of the atom, Neils Bohr put forward the following postulates about the model of an atom:

1. Atom consists of a positively charged nucleus around which electrons revolve in discrete orbits, i.e. electrons revolve in certain permissible orbits and not just in any orbit.
2. Each of these orbits are associated with certain value of energy. Hence, these orbits are called energy shells or energy levels. As the energy of an orbit is fixed (stationary), orbit is also called stationary state.
3. Starting from the nucleus, energy levels (orbits) are represented by numbers (1, 2, 3, 4, etc.) or by alphabets (K, Z, M, Netc.).
4. The electrons present in the first energy level (Z^) have the lowest energy. Energies increases on moving towards outer energy levels.
5. The energy of an electron remains the same as long as it remains in a discrete orbit and it does not radiate energy while revolving.
6. When energy is supplied to an electron, it can go to higher energy levels. Wfiile an electron falls to lower energy level, when it radiate energy.

Neutrons (n)

• Neutron is another sub-atomic particle, discovered by J. Chadwick in 1932. It is represented by n. Neutrons are electrically neutral particles and are as heavy as protons (i.e. their mass is 1.67493 × 10^-27kg) which is equal to that of protons.
• Neutrons are present in the nucleus of all atoms except hydrogen. The mass of an atom is given by the sum of the masses of protons and neutrons present in the nucleus.

Distribution Of Electrons In Different Orbits (Shells)

The distribution of electrons into different orbits of an atom was suggested by Bohr and Bury. For writing the number of electrons in different energy levels or shells, some rules are followed. These are:

1. The maximum number of electrons present in a shell is given by the formula 2n2, where, n is the orbit number or energy level, 1, 2, 3,….

Therefore, the maximum number of electrons in different shells is as follows:

First orbit or K-shell = 2 × (1)² =2

Second orbit or Z-shell = 2 × (2)² =8

Third orbit or Mshell = 2 × (3)² = 18

Fourth orbit or Af-shell = 2 × (4)² =32 and so on.

2. The maximum number of electrons that can be accommodated in the outermost orbit is 8.

3. Electrons are not accommodated in a given shell unless the inner shells are filled (i.e. the shells are filled in a stepwise manner).

Valency

The electrons present in the outermost shell of an atom are known as the valence electrons. They govern the chemical properties of atoms. The atoms of elements having completely filled outermost shell means which has eight electrons show little chemical activity, i.e. they are highly stable. Such elements are called inert elements.

• It means, their valency is zero. Of these inert elements, the helium atom has two electrons in its outermost shell and all other elements have atoms with eight electrons in the outermost shell.
• The tendency to react with atoms of the same or different elements to form molecules is an attempt to attain fully-filled outermost shell. It means, atoms react with other atoms in order to attain fully-filled outermost shell.
• An outermost-shell, which had eight electrons is called an octet. Atoms would thus react, so as to achieve an octet in the outermost shell. This was done by sharing, gaining or the loss of electrons. The number of electrons lost or gained or shared by an atom to become stable or to achieve an octet in the outermost shell is known as valency of that element.
• In other words, it is the combining capacity of the atom of an element with the atom(s) of other element(s) in order to complete its octet.

The valencies of elements of some groups are described below:

1. Hydrogen (H), lithium (Li), sodium (Na) and potassium (K) atoms contain one electron each in their outermost shell, therefore, each one of them can lose one electron to become stable. Hence, their valency is 1.
2. The valency of each of Mg, Ca and Be is 2 because all of these have 2 valence electrons and they can lose these 2 electrons to make the octet of electrons in the outermost shell or to become stable.
3. The valency of boron and aluminium is 3 because each has 3 valence electrons.
4. The valency of carbon and silicon is 4 because each has 4 valence electrons. Nitrogen and phosphorus each has 5 valence electrons, so their valency is 3 because they can gain 3 electrons (instead of losing five electrons) to become stable. Hence, their valency is determined by subtracting five electrons from the octet, i.e. 8-5 = 3. However, P can also share 5 electrons, hence it shows a valency of 5 along with 3.
5. Oxygen and sulfur each have 6 valence electrons, therefore, their valency is 2 because they can gain 2 electrons or share 2 electrons to complete their octet.
6. Similarly, fluorine and chlorine each has 7 valence electrons, their valency is 1 because they can gain 1 electron or share 1 electron to complete their octet.
7. All the inert elements, i.e. He, Ne, Ar, etc., have filled outermost shells. Therefore, their valency is zero.

Note: For metals, valency = Number of valence electrons and for non-metals, valency = 8 – number of valence electrons.

Composition of Atoms of the First Twenty Elements with Electron Distribution in Various Shells:

Atomic Number And Mass Number

Atomic Number

It is defined as the number of protons present in the nucleus of an atom. All the atoms of the same element have the same number of protons in their nuclei and hence, they have the same atomic number. It is denoted by Z and written as a subscript to the left of the symbol, for example., \({ }_2^4 \mathrm{He},{ }_3^7 \mathrm{Li} \text {, }\) Z =2 and 3 for He and Li respectively.

Note: In a neutral atom, atomic number = number of protons = number of electrons

Mass Number

It is defined as the sum of several protons and neutrons present in the nucleus of an atom. Protons and neutrons together are called nucleons. The mass number is denoted by A. Mass number = Number of protons + a number of neutrons, e.g. 2 He, 3 Li, A -4, and 7 for He and Li respectively.

Number of neutrons = Mass number – atomic number

(∵ Atomic number = Number of protons)

The mass number is written as a superscript to the left of the symbol.

In the notation for an atom, the atomic number, mass number, and symbol of the element are to be written as:

Question 1. An atom of an element A may be written as \({ }_{12}^{24} \mathrm{~A} \text {. }\)

1. What does the superscript 24 indicate?
2. What does the subscript 12 indicate?
3. What are the number of protons, neutrons, and electrons in atom A?
4. Write the symbol of an ion formed by an atom of element A.

Answer:

24 is the mass number of atom A.

12 is the atomic number of atom A.

Number of protons =12 Number of neutrons =12 Number of electrons = 12

Electronic configuration of given atom =\(\begin{aligned}
& K L M \\
= & 2, 8,2.
\end{aligned}\)

It can lose two electrons (and attain a stable configuration), therefore the symbol of its ion is A²⁺.

Different Atomic Species

Isotopes

These are defined as the atoms of the same element, having the same atomic number but different mass numbers, e.g. there are 3 isotopes of the hydrogen atoms, namely protium (¹H₁), deuterium (²H₁), and tritium (³H₁).

• In other words, it can be said that isotopes have same number of protons but differ in the number of neutrons. Each isotope of an element is a pure substance.
• Since, chemical properties of elements largely depend on their electronic configuration or outermost electrons and as the isotopes of an element have similar electronic configuration, therefore, isotopes of an element have same chemical properties.
• We know that, masses of isotopes of elements are different. Since, physical properties such as density, light scattering etc., depend on mass therefore, these are different for isotopes of an element.

Average Atomic Mass

If an element has no isotopes, the mass of its atom would be the same as the sum of masses of protons and neutrons in it. But if an element occurs in isotopic forms, then from the percentage of each isotopic form, the average mass is calculated as:

The average atomic mass of an element [(Atomic mass of isotope 1 × percentage of isotope 1) + (Atomic mass of isotope 2 × percentage of isotope 2) +… ] e.g. the two isotopic forms of chlorine atom with masses 35u and 37u occur in the ratio of 3: 1.

Therefore, the average atomic mass of a chlorine atom can be calculated as:

The average atomic mass of a chlorine atom

⇒ \(\left(35 \times \frac{75}{100}+37 \times \frac{25}{100}\right)\)

⇒ \(\left(\frac{105}{4}+\frac{37}{4}\right)=\frac{142}{4}=35.5 \mathrm{u}\)

Here, 35.5 u is not the atomic mass of any one atom of chlorine but it shows that its given amount contains both the isotopes and their average atomic mass is 35.5 u.

Note: The fractional atomic masses of elements are due to the existence of their isotopes having different masses.

Applications Of Isotopes

1. An isotope of uranium (U-235) is used as a fuel for the production of electricity in nuclear reactors.
2. U-238 is used to determine the age of very old rocks and even the age of the earth.
3. An isotope of cobalt (Co-60) is used in the treatment of cancer.
4. An isotope of carbon (C-14) is used to determine the age of old specimens of wood or old bones of living organisms.
5. An isotope of iodine (1-131) is used in the treatment of goiter.

Isobars

Atoms of different elements with different atomic numbers but the same mass number are known as isobars. In other words, isobars are the atoms of different elements that have the same number of nucleons (protons + neutrons) but differ in the number of protons, for example., \({ }_{18}^{40} \mathrm{Ar} \text { and }{ }_{20}^{40} \mathrm{Ca}\) are isobars.

Since isobars have different atomic numbers as well as different electronic configurations. Thus, they also have different chemical properties.

Question 2. Consider the following pairs,

⇒ \((1) ${ }_{26}^{58} A,{ }_{28}^{58} B$
(2) ${ }_{35}^{79} \mathrm{X},{ }_{35}^{80} \mathrm{Y}$\)

1. Which of the above pairs are isotopes and isobars?
2. What factors are responsible for the change in superscripts, 79, 80 (in case II), though the element is the same?
3. Give the nuclear composition of H A.

Answer:

Isobars:

1. \(\text { Isobars: }{ }_{26}^{58} A \text { and }{ }_{28}^{58} B \text { Isotopes: }{ }_{35}^{79} X \text { and }{ }_{35}^{80} Y\)
2. X and Y are pairs of isotopes. Isotopes have the same number of protons but differ in the number of neutrons (hence, their mass number differs, from each other because mass number is the sum of several protons and neutrons).
3. Number of protons = 26, Number of electrons = 26 Number of neutrons = 58 – 26 = 32

Activity 1

Objective

To show the presence of charged particles in matter.

Procedure

1. Comb dry hair.
   1. Bring the comb near the small pieces of paper.
2. Rub a glass rod with a silk cloth and bring the rod near an inflated balloon.

Observation

The comb will attract the pieces of paper. Rod will also attract the inflated balloon. In both cases, it is concluded that on rubbing two objects together, they become electrically charged.

Conclusion

This shows that an atom is divisible and consists of charged particles.

Question 1. What happens when we rub two objects?
Answer: On rubbing two objects together, they become electrically
charged.

Question 2. What conclusion can you draw from the above activity?
Answer: The activity shows that an atom is divisible and consists of
charged particles.

Question 3. Name the charged particles present in an atom.
Answer: Electrons and protons are the charged particles, present in an
atom.

Question 4. Name the neutral particle present in an element.
Answer: A neutron is a neutral particle present in an element.

Question 5. What happens if two oppositely charged substances are placed near each other?
Answer: They attract each other.

Activity 2

Objective

To understand the composition of atoms of the first twenty elements.

Procedure

Make a static atomic model displaying the electronic configuration of the first twenty elements.

Electronic Configuration Of Some Elements:

Question 1. What is the number of valence electrons in an atom of element having atomic number =16?
Answer: The electronic configuration of elements is 2, 8, 6. So, the number of
valence electrons is 6.

Question 2. Name the atom that shows two types of valencies.
Answer: P shows a valency of 3 and 5.

Question 3. The electronic configuration of an element Vis \(\begin{aligned}
& K L M N \\
& 2,8,8,1
\end{aligned}\). Name this element Y.
Answer: The element Y is potassium.

Question 4. There are four atoms A, B, C, and D with atomic numbers 9, 11,13, and 15 respectively. Which atom contains less than 8 electrons in an L-shell?
Answer: The electronic configuration of A (9) is 2, 7. So, atom A contains less than 8 electrons in the L-shell

Question 5. Calcium has twenty electrons which occupy K, L, M, and N-shells. Which of these is/are incomplete?
Answer: The electronic configuration of Ca is\(\begin{aligned}
& K L M N \\
& 2,8,8,2
\end{aligned}\). As N-shells can accommodate a maximum of 8 electrons thus, N-shells are incomplete.

Structure Of The Atom Summary

Discovery Of Electrons

• J.J. Thomson in 1990 discovered cathode rays (or electrons) originating or emitting from the cathode in a gas discharge tube. Electrons are the fundamental particles of all atoms.
• Cathode rays travel in a straight line. In the presence of an electric field, these get deflected towards the positive electrode. They produce fluorescence when strike on the walls of the discharge tube.
• The charge and mass of electrons are 1.6 ×10^-19 C and 9.11 × 10^-31 kg respectively.

Discovery Of Protons

• E. Goldstein in 1886, discovered the presence of new radiations known as canal rays or anode rays passing through holes or ‘canals’ of the cathode and moving towards the cathode in a discharge tube.
• Anode rays consist of positively charged particles, known as protons.
• Protons have a charge, equal in magnitude but opposite in sign to that of electrons. Its mass is about 1840 times that of the electron.

Thomson’s Model of an Atom: Postulates are:

• The mass of an atom is assumed to be uniformly distributed throughout the atom.
• An atom is considered to be a sphere of uniformly distributed positive charge in which electrons are embedded.
• The negative and positive charges balance each other, therefore, the atom as a whole is neutral.

Rutherford’s Model of an Atom After performing a-particle experiment, he suggested that:

• There is a positively charged, highly dense center in an atom, called the nucleus. Nearly the whole mass of the atom resides in it.
• The electrons revolve around the nucleus in circular paths. The size of the nucleus is very small compared to the size of the atom.

Bohr’s Model of an Atom: Postulates are:

• Only certain special orbits called discrete orbits or energy levels of electrons are allowed inside the atom.
• While revolving in discrete orbits, the electrons do not radiate energy.
• The orbits are represented by the letters K, L, M, and N or the numbers 1,2,3, and 4.

Neutrons

In 1932, J. Chadwick discovered another sub-atomic particle called neutrons. They are electrically neutral and are as heavy as protons. They are present in the nucleus of all atoms, except hydrogen.

Bohr and Bury Scheme for Distribution of Electrons in Different Energy Levels The maximum number of electrons in an energy level is equal to 2n² where ‘n’ is the energy level of orbits or shells.

Valency It is the combining capacity of an element with the atom(s) of another element (s) to complete its octet.

Atomic Number It is defined as the number of protons present in the nucleus of an atom. It is also equal to the number of electrons in the case of a neutral atom. It is denoted by Z and written as a subscript, for example., ₆C.

Mass Number It is defined as the sum of the numbers of protons and neutrons in the nucleus. It is denoted by A.

Isotopes They have the same atomic number but different mass numbers or the same number of protons but different numbers of neutrons, for example., ₁H¹,₁H², ₁H³. Their chemical properties are the same due to the same atomic number.

Isobars They have different atomic numbers but the same mass number. Their physical and chemical properties are different, for example., \({ }_{18}^{40} \mathrm{Ar},{ }_{20}^{4 a} \mathrm{Ca} .\)

Class 9 Science Chapter 5 The Fundamental Unit Of Life Long Answer Type Questions

Question 1. Describe an activity to demonstrate endosmosis and exosmosis. Draw the diagram also.
Answer:

Activity to show endosmosis and exosmosis Put dried raisins in plain water and leave them for some time. Then place them into a concentrated solution of salt.

Observation

1. When dried raisins are placed in plain water, raisins swell up due to the osmotic entry of water into raisins. Plain water is a hypotonic medium for raisins. Thus, endosmosis occurs.
2. When swollen raisins are transferred to a concentrated solution, raisins shrink. Concentrated solution is a hypertonic medium for swollen raisins.
   Thus, exosmosis occurs.

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Question 2. Explain the main functional regions of a cell with the help of a diagram.
Answer:

The plasma membrane, cytoplasm and nucleus are the three main functional regions of a cell.

1. Plasma membrane It is a thin, selectively permeable membrane, covering the cell and is made up of lipids and proteins.
2. Cytoplasm It is an aqueous material containing a variety of cell organelles along with non-living inclusions.
3. Nucleus It is the control centre of a cell. It contains the cell’s hereditary information (DNA).

Question 3. Given below statements have underlined words which may be incorrect. Rewrite these words and state one function for each of them other than those (if) given.

1. The fundamental organisational unit of life is an organ.
2. The cell wall is an active part of the cell and is selectively permeable.
3. The presence of plasma membrane enables the cells of plants and fungi to exist in hypotonic media without bursting.
4. The Golgi body functions both as a passageway for intracellular transport and as a manufacturing surface.
5. Leucoplasts contain carotenoids and their primary function is storage.

Answer:

1. Cells perform many important physiological functions in living organisms essential for life.
2. Plasma membrane It acts as a mechanical barrier, preventing the leakage of cellular contents to the outside.
3. Cell wall It helps in interaction among plant cells through cytoplasmic channels called plasmodesmata.
4. Endoplasmic reticulum It functions as a cytoplasmic framework providing a surface for some biochemical activities of the cell.
5. Chromoplasts impart colour to the parts of plants like flowers and fruits.

Question 4.

1. Draw a neat labelled diagram of an animal cell.
2. Name the structure, which helps in
   1. Energy production
   2. Exchange of materials between cytoplasm and nucleoplasm.
   3. Lipid synthesis

Answer:

1. For the diagram of an animal cell.
2. Mitochondria
3. Nuclear pore
4. Smooth endoplasmic reticulum

Question 5. Why are mitochondria called the powerhouse of the cell? Give three similarities and one difference between mitochondria and plastids.
Answer:

Mitochondria are often associated with cellular respiration and energy generation of the cell. The energy required for various chemical activities is released by the mitochondria in the form of ATP molecules. For this reason, mitochondria are known as the powerhouse of the cell.

Three similarities between mitochondria and plastids are as follows

1. Both have their own DNA and ribosomes.
2. External structures of mitochondria and plastids are similar.
3. Both have more than one membrane layer.

One major difference between mitochondria and plastids is that mitochondria are present in both plant and animal cells, whereas plastids are present only in plant cells.

Question 6. Write the main functions of at least ten cell components.
Answer:

1. Plasma membrane It acts as a semipermeable membrane and allows only selective substances to pass through it.
2. Chromosomes carry the hereditary characteristics of an organism from one generation to another.
3. Lysosomes Breakdown of unwanted macromolecules is the main function of these organelles.
4. Ribosomes These help in protein synthesis.
5. Nucleus Control centre of the cell. Contains cellular DNA (genetic information) in the form of genes.
6. Mitochondria The main function of mitochondria in aerobic cells is the production of energy by the synthesis of ATP.
7. Nucleolus Biosynthesis of ribosomal RNA (rRNA) and acts as a platform for protein synthesis.
8. Cell wall It provides protection and rigidity to the plant cell.
9. Chloroplasts These are the sites of photosynthesis within plant cells.
10. The endoplasmic reticulum Serves as a channel for the transport of materials.

Question 7. Grass looks green, papaya appears yellow. Which Cell organelle is responsible for this?
Answer:

• Plastids These are found in plant cells only. Plastids are the major cell organelles in plants. On the basis of pigments present in plastids, they are divided into two types; the colourless leucoplasts and the pigmented chromoplasts.
• The colourless leucoplasts store starch, oil and protein granules whereas the pigmented chroloplasts have different colours and can be of several types.
• The most important ones are those containing the pigment chlorophyll, known as chloroplasts, which is responsible for the preparation of food by photosynthesis.
• Other chromoplasts contain non-green pigments, which are responsible for the characteristic colours of fruits and flowers.

Question 8. How are the following related to each other?

1. Chromatin network and chromosomes
2. Chloroplast and chromosomes
3. Genes and DNA

Answer:

1. The cell contains nuclear material which can be seen as an entangled mass of thread-like structure when it is not dividing. The chromatin material gets organised into rod-like structures called chromosomes when the cell is about to divide.
2. Chloroplasts are green-coloured plastids which contain a green-coloured pigment called chlorophyll.
3. Genes are the functional segments of DNA (present on DNA) which control a specific trait by making specific proteins.

Class 9 Science Chapter 5 The Fundamental Unit Of Life Short Answer Type Questions

Question 1. A cell is a building block of an organism. Explain why.
Answer:

The body of an organism is made up of various organ systems and these organ systems are made up of various tissues and tissues are the groups of cells performing the same function. Hence, a cell is a building block of an organism.

Question 2. Every multicellular organism has arisen from a single cell. Justify this statement.
Answer:

Cells divide to produce cells similar to themselves. Thus, all cells are derived from pre-existing cells and every multicellular organism has arisen from a single cell, for example., the development of a human starts from a single diploid cell called a zygote formed by the fusion of an ovum and sperm.

Question 3. Illustrate the various functions performed by a cell.
Answer:

Cell performs various functions in our body such as

1. Synthesis of substances like proteins and lipids.
2. Secretion of enzymes.
3. Removal of dead materials.
4. Oxidation to generate energy.

Question 4. Describe the structural features of the cell membrane and cell wall. Why is a cell membrane called a selectively permeable membrane?
Answer:

• The cell membrane is a flexible, semipermeable and living portion of a cell, which is made up of lipids and proteins. On the other hand, a cell wall is a tough, rigid and non-living portion of a plant cell, which is made up of cellulose.
• The plasma or cell membrane permits the entry and exit of selected materials in and out of the cell. It also prevents the movement of cell content outside the cell. Hence, it is called a selectively permeable membrane.

Question 5. Explain in detail what you know about the structure of the nucleus.
Answer:

Robert Brown discovered the nucleus in the cell in 1831. The nucleus is the control centre of a cell.

The structure of the nucleus is composed of

1. The nuclear membrane encloses the nucleus in eukaryotes. The nuclear membrane is penetrated by large nuclear pore complexes, which selectively transport molecules into or out of the nucleus.
2. Nucleoplasm It is a kind of protoplasm found in the nucleus containing genetic material (DNA), chromosomes and nucleolus.
3. Chromatin The chromatin material inside the nucleus is an organisation of DNA and protein. As a cell prepares itself to divide, the chromatin condenses and becomes thick enough to form specialised structures called chromosomes.
4. Nucleolus It acts as the most important site of RNA synthesis. It was first recognised by Fontana in 1874.

Question 6.

1. Where are chromosomes located? What is chromatin material and how does it change just before the cell divides?
2. The functional segments of DNA are genes/ Give reason.

Answer:

1. Chromosomes are located in the nucleus of the cells. Chromatin is a mass of thread-like structures. It condenses to form chromosomes just before the cell divides.
2. Genes present in DNA segments carry the hereditary information in them, which is transferred from one generation to next. They determine the structural and functional aspects of the next generation.

Question 7. State some differences between cytoplasm and nucleoplasm.
Answer:

Differences between cytoplasm and nucleoplasm are as follows:

Question 8. Differentiate between diffusion and osmosis.
Answer:

Differences between diffusion and osmosis are as follows

Question 9. What do you mean by plasmolysis?
Answer:

When a living plant cell loses water through osmosis, shrinkage or contraction of the cell contents away from the cell wall occurs. This phenomenon is known as plasmolysis.

Question 10. What would happen, if a cell wall is not present in plant cells?
Answer:

1. The shape of a cell will not remain definite because the cell wall provides structural strength to plant cells.
2. The cell will not be able to withstand pressure and burst if placed in very dilute media.

Question 11. Differentiate between Rough Endoplasmic Reticulum (RER) and Smooth Endoplasmic Reticulum (SER). How endoplasmic reticulum is important for membrane biosynthesis?
Answer:

Differences between RER and SER are as follows:

Question 12. Differentiate between nucleus and nucleoid.
Answer:

Differences between nucleus and nucleoid are as follows:

Question 13. What does a chromosome contain?
Answer: Chromosomes contain genetic information to be transferred from one generation to the next generation in the form of DNA (Deoxyribonucleic Acid).

Question 14. Differentiate between Rough Endoplasmic Reticulum (RER) and Smooth Endoplasmic Reticulum (SER). How endoplasmic reticulum is important for membrane biosynthesis?
Answer:

Differences between RER and SER are as follows

Question 15. Describe the phenomenon of membrane biogenesis. Give one function of ER.
Answer:

• The smooth endoplasmic reticulum helps in the manufacture of lipid or fat molecules, important for cell function. Some of these lipids and proteins manufactured in RER help in building the cell membrane.
• This process is known as membrane biogenesis. ER functions as a cytoplasmic framework providing a surface for some of the biochemical activities of the cells.

Question 16. Name the organelle of the cell, which is involved in the formation of lysosomes. Write its functions in the cell.
Answer:

Golgi apparatus is the organelle involved in the formation of lysosomes.

The functions of the Golgi apparatus are

1. Storage, modification and packaging of products in vesicles.
2. Helps to make complex sugars from simple sugars.
3. Material synthesised near the ER is packaged and dispatched to various targets inside and outside the cell through the Golgi apparatus.

Question 17. Name two nucleic acids found in the cell. Write their functions.
Answer:

• DNA and RNA are two nucleic acids present in the nucleus of a cell.
• Functions DNA is responsible for the storage and transmission of hereditary information, while RNA helps in the synthesis of protein. RNA is the genetic material of some viruses.

Question 18. Categorise the cells on the basis of the presence or absence of a nuclear membrane.
Answer:

Cells, which have nuclear membranes are called eukaryotic cells. On the contrary, the cells without nuclear membrane are called prokaryotic cells.

Question 19. What are the different types of endoplasmic reticulum found in an eukaryotic organism? Also, write one major function of each.
Answer:

Question 20. Write two functions indicating that lysosomes are the suicidal bags of the cell.
Answer:

The following two functions show that lysosomes are the suicidal bags of the cell

1. Lysosomes are the waste disposal system of the cells. They help keep the cells clean by digesting cellular wastes generated in cell.
2. Lysosomes also aid in phagocytosis. They digest foreign material entering the cell and kill them.

Question 21. Give scientific reasons for the following

1. The inner membrane of mitochondria is deeply folded.
2. Mitochondria are able to make some of their proteins.

Answer:

1. The inner membrane of mitochondria is deeply folded as these folds create a large surface area for ATP-generating chemical reactions.
2. Mitochondria contain their own DNA and ribosomes and hence, can make their own proteins.

Question 22. Bacteria do not have chloroplast, but some bacteria are photoautotrophic in nature and perform photosynthesis. Which part of the bacterial cell performs this?
Answer:

Bacterial cell do not have chloroplast yet some photoautotrophic bacteria perform photosynthesis due to the presence of light-absorbing pigments and reaction centres embedded in the cell membrane. The reaction centres exist in forms of sacs, tubes or sheets depending on the amount of surface area needed.

Question 23. Name the two cell organelles, which are bound by a double-layered membrane. Give one function of each.
Answer: Mitochondria- site of ATP-generating reactions. Chloroplasts- site of photosynthesis.

Question 24. Give a one-word answer to the following:

1. Organelle containing chlorophyll.
2. An organelle with cristae.
3. An organelle with ribosomes attached to its surface.
4. Living matter of the cell.

Answer:

1. Chloroplast
2. Mitochondria
3. Rough endoplasmic reticulum
4. Protoplasm

Question 25. Why do some regions appear darker than other regions on staining the cell with methylene blue? Name the other solutions that can be used for staining.
Answer:

Different regions of cells get coloured differentially due to their chemical composition.
Other solutions that can be used for staining are iodine solution or safranin solution.

Question 26. How does a fungal cell differ from a bacterial cell?
Answer:

Differences between fungal cells and bacterial cells are as follows:

Question 27. Explain how do cell walls permit the cells of fungi to withstand very dilute external media without bursting.
Answer:

• Fungi withstand very dilute/hypotonic external media without bursting because of their cell walls.
• In such media, the cells swell up by taking up water through osmosis and hence, building up pressure against the cell wall. The wall exerts an equal pressure against the swollen cell, thus preventing it from bursting.

Question 28. Why does the skin of your fingers shrink when you wash clothes for a long time?
Answer:

• The solution of soaps and detergents is hypertonic as compared to the osmotic concentration of our skin.
• Therefore, washing clothes results in exosmosis in skin cells that come in contact with the soap solution. Due to this reason, the skin of the fingers shrinks while washing clothes for a long time.

Question 29. A person takes a concentrated solution of salt. After some time, he starts vomiting. What is the phenomenon responsible for such a situation? Explain.
Answer:

The solution of salt is hypertonic, so causes irritation and excessive dehydration due to exosmosis in the intestine. This makes the person uncomfortable causing reverse movements such as vomiting.

Question 30. If you are provided with some vegetables to cook, during the cooking process you generally add salt to vegetables. After adding salt, vegetables release water. Which mechanism is responsible for this?
Answer:

On adding salt to vegetables, they release water due to exosmosis. When the external medium is hypertonic as compared to the osmotic concentration inside living cells, exosmosis occurs.

Question 31. What happens to an animal cell when it is placed in a very dilute external medium? Explain. Can diffusion lead to the same consequences?
Answer:

• In such conditions, the animal cell will gain water and would swell up or may even burst.
  Due to endosmosis, water moves from the dilute external medium through the semipermeable cell membrane into the cell with low water concentration.
• No, diffusion can not lead to the same consequences. It causes equal distribution. So, there will be no swelling or shrinkage of the cell.

Question 32. Why is endocytosis found in animals only?
Answer:

• Endocytosis is the process of swallowing food and other substances from an external medium by a plasma membrane. This is possible only when the plasma membrane is in direct contact with the external medium.
• It occurs only in animal cells because they lack cell walls. Plant cells contain cell walls, due to which their plasma membrane fails to perform endocytosis.

Question 33. The Golgi apparatus is also called the secretory organelle of the cell’. Why?
Answer:

The Golgi apparatus is also called a secretory organelle because secretion is the main function of the Golgi apparatus. The secretory proteins and lipids are packed and released on the surface by Golgi via exocytosis.

Question 35.

1. Why lysosomes are known as ‘scavengers of the cell’?
2. Lysosomes are self-destructive. True/ false. Give reason.

Answer:

1. Lysosomes are called scavengers of the cell because they remove dead and worn-out cells by digesting them and act as a kind of waste disposal system of a cell.
2. Lysosomes are self-destructive. This is true as during the breakdown of cell structure, lysosomes may burst and the enzymes contained in it may eat up their own cells.

Question 36. How many membranes are present in mitochondria? Give the characteristic features of these membranes. What is the advantage of such features?
Answer:

Mitochondria have two membranes. The outer membrane is very porous, while the inner membrane is deeply folded. A porous membrane helps in getting oxygen and food, while the folds create a large surface area for ATP-generating chemical reactions.

Question 37. Name a cell organelle found only in a plant cell and mention its various types along with their functions and location.
Answer:

Plastids are found only in plant cells.

Types of plastids are as follows

Question 38. Which type of plastid stores starch, oil and proteins?
Answer:

Leucoplasts are the plastids that function to store starch, oil and proteins hence, it is three types

1. Amyloplasts – Store starch
2. Elaioplasts – Store oil
3. Aleuroplasts – Store proteins

Question 39. Why do plant cells possess large-sized vacuoles?
Answer:

Plant cell possesses large-sized vacuoles because

1. It stores salt, sugar, amino acid, organic acid and some proteins.
2. The vacuole contains cell sap and helps maintain the turgidity of a cell.
3. They store some metabolic byproducts or end products of plant metabolism.
4. lysosomal enzymes occur in the vacuole of plant cells.

Question 40. State the reason for the following

1. Mitochondria is known as the powerhouse of the cell.
2. Plant cell shrinks when kept in a hypertonic solution.

Answer:

1. Oxidation of food takes place in mitochondria, which results in the release of energy in the form of ATP. This energy helps in various chemical activities needed for life. Hence, mitochondria are known as the powerhouse of the cell.
2. A hypertonic solution has a lower concentration of water than the cell. When a plant cell is kept in it, water present in the cell leaves the cell due to exosmosis. Therefore, the cell shrinks.

Question 41. Name the organelles, which show analogy written as under.

1. Transporting channels of the cell.
2. Powerhouse of the cell.
3. Packaging and dispatching unit of the cell.
4. Digestive bag of cells.
5. Storage sac of the cell.
6. Control room of the cell.

Answer:

1. Endoplasmic reticulum
2. Mitochondria
3. Golgi apparatus
4. Lysosome
5. Vacuole
6. Nucleus

Question 42. Carry out the following osmosis experiment:

Take four peeled potato halves and scoop each one out to make potato cups. One of these potato cups should be made from a boiled potato. Put each potato cup in a trough containing water. Now,

Keep cup A empty

1. Put one teaspoon of sugar in cup B.
2. Put one teaspoon salt in cup C.
3. Put one teaspoon of sugar in the boiled potato cup D.
4. Keep them for two hours. Then observe the four potato cups and answer the following:
5. Explain, why water gathers in the hollowed portion of B and C.
6. Why is potato A necessary for the experiment?
7. Explain, why water does not gather in the hollowed-out portion of A and D.

Answer:

1. The water gathers in the hollowed portion of B and C due to the process of osmosis. The concentration of solute (sugar in cup B and salt in cup C) is higher inside the cup than in water.
2. Hence, water flows from a region of higher concentration to a region of lower concentration.
3. Potato A acts as a reference of control for the experiment, which helps in comparing results.
4. Water does not gather in the hollow portion of A and D for the following reasons:
   1. The hollow portion of potato A is empty and there is no concentration difference so, no osmosis occurs.
   2. The hollowed portion of potato D contains sugar, but the potato cup is boiled. Osmosis cannot occur as the semipermeable membrane is destroyed by boiling.

Question 43. Make a comparison and write down ways, in which plant cells are different from animal cells.
Answer:

Comparison of Plant Cells and Animal Cells:

Question 44. What would happen if the plasma membrane ruptures or breaks down?
Answer:

In case the plasma membrane ruptures or breaks down,

1. All the useful substances will move out of the cell because the membrane is selectively permeable.
2. The transportation of materials will be disturbed.
3. The cell will lose its normal shape.
4. This may lead ultimately to the death of the cell.

Question 45. What would happen to the life of a cell if there was no Golgi apparatus?
Answer:

Effects of the absence of Golgi apparatus on the life of a cell are as follows:

1. The packaging and dispatching of different types of proteins to various targets inside and outside the cell will be influenced.
2. The products of cells cannot be stored and modified later.
3. This will affect the lysosome’s formation. This will cause an accumulation of worn-out and dead cell organelles within the cell, which may cause cell death.

Question 46. Which organelle is known as the powerhouse of the cell? Why?
Answer:

• Mitochondria are called the powerhouse of the cell. It contains oxidative enzymes, which oxidise the food and convert it into the energy currency of the cell in the form of ATP (Adenosine Triphosphate).
• This energy is used by body for making new chemical compounds and for doing other work. This is the reason, mitochondria are called the powerhouse of the cell.

Question 47. Why is the cell called the structural and functional unit of life?
Answer:

All living organisms are made up of cells, which perform various functions essential for the survival of the organisms, for example., respiration, digestion, excretion etc. Thus, a cell is the functional unit of life.

In unicellular organisms, a single cell carries out all the functions, while in multicellular organisms, a group of cells carry out different functions. Thus, a cell is the structural and functional unit of all living organisms.

Question 48. How do substances like CO₂ and water move in and out of the cell? Discuss.
Answer:

• CO₂ and other gases move in and out of the cell by the process of diffusion. When the concentration of carbon dioxide is higher inside the cell than outside, CO₂ diffuses out of the cell.
• If the CO₂ concentration inside the cell is less, CO₂ moves inside the cell from outside. The water moves in and out of the cell by the process of osmosis.
• Osmosis is the diffusion of water from a region of its high concentration to a low concentration through a semipermeable membrane.

Class 9 Science Chapter 5 The Fundamental Unit Of Life Very Short Answer Type Questions

Question 1. Who coined the term ‘protoplasm’ for the fluid substance of the cell?
Answer: Purkinje in 1839 coined the term protoplasm’ for the fluid substance of the cell.

Question 2. Name two unicellular organisms.
Answer: Paramecium and Chlamydomonas are the two unicellular organisms.

Question 3. Name the smallest known cell.
Answer: Mycoplasma is the smallest known cell.

Question 4. Plant cells have large vacuoles each surrounded by a membrane. What is the name of this membrane?
Answer: The membrane that surrounds the vacuole is called tonoplast.

Question 5. Write an example of diffusion in a plant.
Answer: Excess of CO₂ is present in the air as compared to leaf so, due to the process of diffusion, plants take CO₂ from the atmosphere to carry out photosynthesis.

Question 6. Name the process in which diffusion takes place through a semipermeable membrane.
Answer: Osmosis takes place through a semipermeable membrane.

Question 7. During cooking, on adding salt to the vegetables, water is released. Name the mechanism involved.
Answer: Osmosis.

Question 8. What will happen if the already swollen raisin is kept in a salt solution?
Answer: The water flows out from the raisin and goes into the solution medium. Consequently, the raisin shrinks in size.

Question 9. Name the process by which unicellular freshwater organisms and most plant cells tend to gain water.
Answer: Endosmosis, i.e. inward movement of water into the cell from the surrounding medium.

Question 10. What is the function of cellulose in plant cells?
Answer: The plant cell wall is mainly composed of cellulose. Cellulose is a complex substance, which provides structural strength to plants.

Question 11. What does the nucleus contain?
Answer: The nucleus contains nucleolus and chromatin material.

Question 12. Why is the nucleus called the controller or brain of the cell?
Answer:

The nucleus coordinates and directs all the metabolic functions of the cell, which is why it is called the controller or brain of the cell.

Question 13. State two important functions of the nucleus of the cell.
Answer:

1. The nucleus is the control centre of a cell.
2. It consists of cells’ DNA (genetic information) in the form of genes to carry hereditary characteristics from one generation to another.

Question 14. What are chromosomes made up of?
Answer: Chromosomes are made up of DNA and proteins.

Question 15. What is DNA? Where is it present?
Answer: DNA is Deoxyribonucleic Acid. It is the genetic map of an organism, present in the nucleus.

Question 16. How DNA is present in a cell which is

1. Dividing?
2. Not dividing?

Answer:

1. Chromosomes
2. Part of chromatin material.

Question 17. Which organelle is called the factory of ribosomes?
Answer: Nucleolus is called the factory of ribosomes.

Question 18. Name two structures, which are found in plant cell, but not in animal cell.
Answer: Chloroplast and cell walls are found in plant cell, but not in animal cells.

Question 19. Give the name of a structure and its function, which is found only in animal cells, but not in plant cells.
Answer: Centrioles are present only in animal cells and help in cell division.

Question 20. Which of the following are present in animal cells? Chloroplast, nucleus, vacuoles, cell wall and mitochondria Nucleus, vacuoles and mitochondria are present in animal cells. What are dictyosomes?
Answer: In plants, Golgi bodies are called dictyosomes.

Question 21. Is there any animal cell that lacks lysosomes?
Answer: Mammalian RBCs (Red Blood Corpuscles) lack lysosomes.

Question 22. In which cell organelle, the complete breakdown of glucose in the presence of oxygen takes place?
Answer:

The complete breakdown of glucose in the presence of oxygen in a cell is called aerobic respiration. It takes place in mitochondria.

Question 23. Name the energy currency of the cell.
Answer: Adenosine Triphosphate (ATP) is considered by biologists to be the energy currency of cells.

Question 24. Where- does the ATP synthesis occur in mitochondria?
Answer: ATP synthesis occurs in folds of the inner membrane of mitochondria called cristae.

Question 25. Which is the most widely occurring plastid and where is it present?
Answer: The most widely occurring plastid is chloroplast. It is present in all green plants.

Question 26. Name the type of plastid that helps in the process of photosynthesis.
Answer: Chloroplast helps in the process of photosynthesis.

Question 27. Where stroma is present in a cell?
Answer: Stroma is present in the chloroplast of the cell.

Question 28. Which organelles are present only in plant cells and possess their own genome and ribosomes?
Answer: Plastids are found in plant cells only and contain DNA, RNA and ribosomes.

Question 29. Why plastids are called the ‘kitchens of the cell’?
Answer:

Because they contain pigments which can trap sun energy and convert it into chemical energy, i.e. food called glucose.

Question 30. Who discovered cells and how?
Answer:

Cells were discovered by Robert Hooke in 1665. He observed cells in a cork slice with the help of a primitive microscope. The cork slice resembled the structure of a honeycomb consisting of many small compartments or box-like structures. Hooke called these boxes as cells.

Question 31. Why is the plasma membrane called a selectively permeable membrane?
Answer:

The plasma membrane is called a selectively permeable membrane because it allows the entry and exit of some selected molecules only through the cells. It also prevents the movement of some other materials.

Question 32. How is a prokaryotic cell different from a eukaryotic cell? Or Fill in the gaps in the following table illustrating differences between prokaryotic and eukaryotic cells.
Answer:

Question 33. Can you name two organelles we have studied that contain their genetic material?
Answer: Mitochondria and plastids are the two cell organelles that contain their genetic material.

Question 34. If the organisation of a cell is destroyed due to some physical or chemical influence, what will happen?
Answer:

Living cells are capable of performing certain basic functions due to the presence of cell organelles present in it. If these are destroyed then cells will not be able to work properly and will die after some time.

Question 35. Why are lysosomes known as suicidal bags?
Answer:

Lysosomes contain powerful digestive enzymes. During the disturbance in cellular metabolism, lysosomes may burst and digest their cell. Therefore, they are called suicidal bags of the cell.

Question 36. Where are proteins synthesised inside the cell?
Answer: Ribosomes are the site of protein synthesis inside the cell.

Question 37. Where do the lipids and proteins constituting the cell membrane get synthesised?
Answer:

The synthesis of lipids occurs in the Smooth Endoplasmic Reticulum (SER). The proteins are synthesised in the ribosomes, which are attached to the Rough Endoplasmic Reticulum (RER).

Question 38. How does Amoeba obtain its food?
Answer:

Amoeba obtains its food through endocytosis. It is the process of ingestion of food through the plasma membrane. This occurs due to the flexibility of the plasma membrane, which enables the Amoeba to engulf food and other materials from its surroundings.

Question 39. What is osmosis?
Answer:

Osmosis is a process of diffusion of water from a region of its higher concentration to a region of lower concentration through a semipermeable membrane.

The Fundamental Unit Of Life Multiple Choice Questions

Question 1. The word ‘cell’ is derived from

1. The Greek word that means ‘small box-like structure’
2. The Latin word that means a little room’
3. The Greek word that means a little room’
4. The Latin word that means ‘small box-like structure’

Answer: 2. The Latin word that means a little room’

Robert Hooke in 1665, while examining a thin slice of cork, discovered dead cells. He used the word’ cell, which is derived from Latin, and means a little room’ to describe these units of cork slice.

Question 2. Choose the incorrect statement from the following options.

1. All cells arise from pre-existing cells only
2. Rudolf Virchow proposed the cell theory
3. The nucleus was discovered by Robert Brown in 1831
4. The nucleus and cytoplasm of a living cell, all together form the protoplasm

Answer: 3. Nucleus was discovered by Robert Brown in 1831

• The nucleus was discovered by Robert Brown in 1831. The cell theory was proposed by
  Schleiden and Schwann in 1839, postulated that all living beings are composed of cells and their products.
• In 1855, Rudolf Virchow modified the cell theory with his postulate, ‘Omnis cellula-e-cellula, which means, a new cell is derived from a pre-existing cell only.

Question 3. Every living animal cell is enclosed by an outermost thin and delicate covering that separates the contents of the cell from the external environment, is primarily composed of

1. Glycolipids
2. Glycoproteins
3. Lipids
4. Phospholipids

Answer: 4. Phospholipids

• The outermost thin and delicate covering of animal cells that separates the contents of a cell from the external environment is called the plasma membrane or cell membrane.
• The chemical analysis of the plasma membrane revealed that it is primarily composed of phospholipids. While the remaining minor fraction is constituted by proteins embedded in the phospholipid layer, polysaccharides glycolipids, etc.

Question 4. While preparing a temporary mount of onion peel, the peel taken from the onion bulb is placed immediately into the Petri dish containing water, because

1. It is an experimental procedure
2. It brings out plasmolysis due to osmosis of water into the cell
3. It prevents the peel from getting folded and dehydrated
4. It removes the impurities from the peel

Answer: 3. It prevents the peel from getting folded and dehydrated

While preparing a temporary mount of plant cells or animal cells, the specimen collected is immediately placed into the Petri dish/watch glass containing water, to prevent the cells from dehydrating and the peel from getting folded.

Question 5. Which one of the following terms describes a nucleus without a nuclear membrane?

1. Nucleolus
2. Primitive nucleus
3. Nucleoid
4. All of these

Answer: 3. Nucleoid

The primitive type of undeveloped nucleus which lacks a nuclear membrane is called nucleoid. A nucleoid is found in prokaryotic cells, for example., bacteria.

Question 6. Which one of the following cellular functions is performed by the endoplasmic reticulum?

1. Production of hydrolytic enzymes
2. Supply of energy to the cell
3. Formation of lysosomes
4. Production of vacuoles

Answer: 1. Production of hydrolytic enzymes

The digestive or hydrolytic enzymes (proteins) are produced by the rough endoplasmic reticulum for the synthesis of lysosomes.

Question 7. The function of biosynthesis of the plasma membrane and metabolism of toxic or poisonous substances is carried out by

1. Rough endoplasmic reticulum
2. Golgi apparatus
3. Smooth endoplasmic reticulum
4. Both (1) and (3)

Answer: 3. Smooth endoplasmic reticulum

Smooth endoplasmic reticulum is involved in the detoxification of drugs, insecticides, etc., inside a cell. It also carries out the biogenesis of plasma membranes and various steroid hormones.

Question 8. Which one of the following is not a functional role of the plasma membrane?

1. It forms the microvilli in the intestines of human beings
2. It maintains the cellular homeostasis
3. It protects the cell from plasmolysis
4. It projects small finger-like projects called pseudopodia for endocytosis

Answer: 3. It protects the cell from plasmolysis

Plasmolysis is the phenomenon of contraction or shrinkage of protoplasm in a plant cell, away from the cell wall, due to exosmosis of water, when the cell is placed in a hypertonic solution.

The plasma membrane cannot restrict the osmosis of water, thus it cannot protect cells from plasmolysis.

Question 9. Choose the incorrect statement in context with the cell.

1. A cell can replicate independently to produce daughter cells
2. They can perform all the metabolic activities of an individual
3. Every organism starts its life as a single-cell
4. Mycoplasma is the smallest cell but is larger than bacterial cell

Answer: 4. Mycoplasma is the smallest cell, but is larger than bacterial cell

Mycoplasma (PPLO) is the smallest cell, which is smaller than bacteria, but larger than viruses.

Question 10. Select the correct functional role of the vacuole in a cell.

1. Vacuoles do not help in maintaining the rigidity of the cell
2. Vacuoles form thread-like tails in sperms of many mammals
3. Vacuoles expel excess water and waste products from the cell
4. Vacuoles store only excretory products of a cell

Answer: 3. Vacuoles expel excess water and waste products from the cell

In unicellular organisms, vacuoles carry out the role of expelling excess water and wastes from the cell. They also store food and water in cells and help maintain the rigidity of cells by controlling osmosis. Thus, they are also known as dustbins or storehouses of cells.

Question 11. Choose the incorrect statement

1. Lysosomes from the garbage disposal system of animal cells
2. Ribosome carries out the synthesis of proteins
3. Liquid content of the vacuole in a plant cell is called cell sap
4. The colorless plastid that stores proteins, lipids, and starch is named xanthophyll

Answer: 4. The colorless plastid that stores proteins, lipids, and starch is named xanthophyll

Leucoplast is a colorless plastid that stores proteins, lipids, and starch, while xanthophylls are yellow, red, or orange pigments of green plants.

Question 12. Who coined the term ‘protoplasm’?

1. Purkinje
2. Robert Brown
3. Virchow
4. Palate

Answer: 1. Purkinje

Question 13. The longest cell of the human body is

1. RBC
2. Nerve cell
3. Muscle cell
4. Sperm

Answer: 2. Nerve cell

Question 14. The outermost thin, living membrane that separates the contents of the cell from its external environment is called as

1. Slime layer
2. Cell wall
3. Cell membrane
4. Capsule

Answer: 3. Cell membrane

Question 15. Select the incorrect pair.

1. Powerhouse of cell—Mitochondria
2. Kitchen of cell—Chloroplast
3. Brain of cell—Ribosomes
4. Suicidal bag—Lysosomes

Answer: 3. Brain of cell—Ribosomes

Question 16. The example(s) of prokaryotic cells is/are

1. Blue-green algae
2. Fungi
3. Bacteria
4. Both (1) and (3)

Answer: 4. Both (1) and (3)

Question 17. Name the scientist who discovered the Golgi apparatus.

1. Camillo Golgi
2. Robert Brown
3. Leeuwenhoek
4. None of these

Answer: 1. Camillo Golgi

Question 18. Which of the following is/are semi-autonomous organelle(s)?

1. Nucleus
2. Mitochondria
3. Chloroplast
4. Both (2) and (3)

Answer: 4. Both (2) and (3)

Question 19. Which one of the following is known as the energy currency of the cell?

1. ATP
2. GTP
3. FTP
4. None of these

Answer: 1. ATP

Question 20. Oil or fat-storing plastids are called as

1. Elaioplast
2. Amyloplast
3. Aleuroplast
4. Tonoplast

Answer: 1. Elaioplast

Question 21. The function(s) of vacuoles is/are

1. To provide turgidity or rigidity to plant cell
2. Storage of food material
3. To expel excess of water
4. All of the above

Answer: 4. All of the above

The Fundamental Unit Of Life

All organisms including plants and animals are composed of cells. Every cell arises from a pre-existing cell. These cells become specialised to perform different specialised functions after division.

Cell is the basic fundamental structural and functional unit of living organisms. In this chapter, we will study the complex structure of a cell, its various organelles and their functioning inside the cell.

Discovery Of Cell

Robert Hooke (1665), examined a thin slice of cork under a primitive microscope. He observed that cork consists of small box-like structures resembling honeycomb. He called these boxes cells. The substance called cork comes from the bark of a tree.

Cell is a Latin word for a little room. The basic characteristics of cells are as follows:

1. They can replicate independently.
2. They contain hereditary information.
3. They can perform all the life-sustaining activities on their own.
4. They show similar chemical composition and metabolic activities.

Read and Learn  More Class 9 Science Notes

Major Landmarks Related To Cell Discovery:

Cellular Composition In Different Organisms

Based on the number of cells present in different organisms, they are classified into two types:

1. Unicellular organisms (having single cell)
2. Multicellular organisms (having many cells).

Differences Between Unicellular And Multicellular Organisms

Note:

• Every multicellular organism starts its life as a single cell (i.e. zygote), which divides and forms many cells. All cells thus, come from pre-existing cells.
• The invention of magnifying lenses made the discovery of single-celled microscopic organisms possible.

Microscopes

These are high-resolution instruments. They are used for observing the fine details of very minute objects, for example., cells. With the help of a microscope, the size of a small cell can be magnified up to 300-1500 times.

A simple microscope, which is often used in schools is a compound microscope. It uses sunlight for the illumination of objects to be seen, so it is called a light microscope. An electron microscope is used to observe complex structures of the cell.

Shape Of Cells

Some cells have fixed shapes (for example., most plant and animal cells), while some cells like WBCs and Amoeba keep changing their shapes.

Fixed-shaped cells may be of various types like elliptical (for example., fat cell), spherical (for example., ovum), spindle-shaped (for example., smooth muscle cell), knobbed thread (for example., sperm), discoidal (for example., RBC), elongated (for example., nerve cell), etc.

The following figures depict some cells from the human body:

Size Of Cells

The size of the cell varies significantly from the smallest cell of Mycoplasma (0.1-0.5 pm) to very large egg cells of the Ostrich (18 cm). The longest cells of the human body are the nerve cells, which may reach upto a length of 90 cm.

Functions Of Cells

Each living cell can perform some basic functions that characterise living organisms.

1. The shape and size of cells are related to the specific function they perform.
2. Multicellular organisms like human beings perform these functions by division of labour. Different parts of the human body are specialised to perform different functions such as the heart is made to pump blood, the stomach to digest food, the kidney to filter urine etc.
3. Division of labour is also seen within a single cell. Every cell possesses certain specific components known as cell organelles. These enable it to survive and perform special functions.
4. Cell organelles together with protoplasm constitute the basic unit of life called the cell. Each kind of cell organelle performs a specific function. For example, obtaining nutrition, respiration, clearing waste material or forming new progeny. Mitochondria is the organelle responsible for providing energy to the cell.

Note: The same cellular organelles are found in all the cells regardless of their function and the type of organism they are found in.

Structural Organisation Of A Cell

Microscopic studies revealed that every cell possesses three basic features in common, i.e. plasma membrane, nucleus and cytoplasm. Due to the presence of these features, all activities inside the cell and interaction of the cell with its environment are possible.

Plasma Membrane Or Cell Membrane

This is the outermost living, thin and delicate covering of cells. It separates the contents of the cell from its external environment.

• The presence of lipids and proteins (as phospholipids) provides flexibility to the plasma membrane. It enables cells to engulf food and other materials from the external environment.
• This process is called endocytosis, for example., Amoeba acquires food through this process, with the help of finger-like projections called pseudopodia.

Functions Of Plasma Membrane

1. It allows the entry and exit of some selective materials in and out of the cell. The cell membrane, therefore, acts as a semipermeable, selectively permeable, partially permeable and differentially permeable membrane.
2. It helps to maintain the shape of the cell.
3. It acts as a mechanical barrier and protects the internal contents of the cell from leaking out.
4. It protects against microbes and foreign substances.
5. It gets modified to perform different functions, for example., microvilli in the intestine of human beings for absorption.
6. Its semipermeability enables the cell to maintain cellular homeostasis. Amongst all the functions listed above, the transport of substances is the most important function.

It is, therefore, discussed in detail below:

Transport Across The Membrane

Transportation of substances across the membrane may take place with the expenditure of energy (active transport) and without the expenditure of energy (passive transport).

Transport Across The Membrane By Diffusion

The spontaneous movement of a substance (solid, liquid or gas) from a region of its higher concentration to a region of its lower concentration is called diffusion.

• For example, CO₂ (cellular waste, which needs to be excreted out) accumulates in higher concentrations inside the cell. In the cell’s external environment, the concentration of CO₂ is lowered compared to the inside of the cell.
• Due to this difference in the concentration, CO₂ moves out of the cell by the process of diffusion. Similarly, O₂ enters the cell by the process of diffusion, when the level or concentration of O₂ inside the cell decreases.
• Diffusion is faster in gases than in liquids and solids. It plays an important role in gaseous exchange between the cells and also between the cell and its -external environment.

In addition to gaseous exchange, diffusion also helps an organism in obtaining nutrition from the environment.

Transport Across The Membrane By Osmosis

The movement of water molecules through a selectively permeable membrane along the concentration gradient is called osmosis.

• The movement of water across the plasma membrane is also affected by the amount of substance dissolved in water.
• Osmosis is thus, also defined as the movement of water molecules from a region of its higher concentration to a region of lower concentration through a semipermeable membrane.
• Unicellular freshwater organisms and most plant cells tend to gain water through osmosis.

Absorption of water by plant roots is also an example of osmosis. The process of osmosis can be seen in a cell placed in a solution of different concentrations (such as hypotonic, isotonic and hypertonic).

1. Hypotonic Solution The medium or solution surrounding the cell has a high water concentration as compared to the inside of the cell (or the outside solution is very diluted).
   1. The cell gains water and swells up via endosmosis. This happens because the water molecules are free to pass through the cell membrane in both directions. More water however enters the cell than that leaving it.
2. Isotonic Solution The medium surrounding a cell has the same concentration of water as that present inside the cell.
   1. Water crosses the cell membrane in both directions, but the amount moving in remains the same as the amount moving out. So there is no overall movement of water. As a result, no overall change is observed and the cell size remains the same.
3. Hypertonic Solution The medium surrounding a cell has a lower concentration of water than the cell (i.e. outside solution is very concentrated).
   1. Water crosses the cell membrane in both directions, but this time more water leaves the cell than enters it. As a result, the cell protoplasm gets shrunk (exosmosis).

Cell Wall

It is a tough, non-living covering outside the plasma membrane. It is found in plant and fungal cells. It is freely permeable. It is mainly made up of cellulose, a complex substance that provides structural strength to plants.

Functions of Cell Wall

1. The cell wall permits the cells of plants, fungi and bacteria to withstand hypotonic conditions without bursting.
   1. In hypotonic media, the cells tend to take up water by osmosis. The cell swells up, building up pressure against the cell wall. The wall exerts an equal pressure against the swollen cell.
   2. Cell walls help plant cells to tolerate greater changes in surrounding medium than animal cells.
2. It has narrow pores, called pits. Through them, fine strands of cytoplasm (or cytoplasmic bridges) called plasmodesmata can cross the cell walls. Plant cells interact with each other through these cytoplasmic channels.

Plasmolysis

It is the phenomenon, in which a living plant cell loses water through osmosis when kept in a hypertonic solution. It results in either the shrinkage or contraction of protoplasm away from the cell wall.

Nucleus

It is popularly called the brain of a cell. It is composed of a double-layered covering called a nuclear membrane. It has numerous pores called nuclear pores. They transfer the materials from inside the nucleus to the cytoplasm.

The nucleus contains chromosomes. They are visible as rod-shaped structures only when the cell is about to divide. It encloses a liquid ground substance called nucleoplasm. It contains nucleolus and chromatin material.

The nucleolus is a more or less round structure found inside the nucleus. It does not have a covering of membranes. It is known as the factory of ribosomes.

Chromatin is an entangled network of long, thread-like structures. It condenses to form chromosomes during cell division.

Chromosomes contain information for the inheritance of features from parents to the next generation in the form of DNA (Deoxyribonucleic Acid). Chromosomes are composed of two components, i.e. DNA and protein.

The DNA molecules contain information necessary for constructing and organising cells. The functional segments of DNA are called genes. The nucleus also contains RNA that directs protein synthesis.

Functions Of Nucleus

1. The nucleus plays an important role in cellular reproduction. In this process, a cell divides to form two new cells.
2. It determines the cell’s development and maturity by directing the chemical activities of the cell.
3. It helps in the transmission of hereditary traits from parents to offspring.
4. It controls all metabolic activities of cells. If it is removed, the protoplasm dries up.

In some organisms like bacteria, the nuclear region of the cell is poorly defined because of the absence of a nuclear membrane. The nuclear region in these organisms contains only nucleic acid. Such an undefined nuclear region is called a nucleoid.

Organisms whose cells lack a nuclear membrane are called prokaryotes (pro = primitive, eukaryote ≈ karyon = nucleus).

• Prokaryotes also lack cytoplasmic organelles. Most functions are thus performed by poorly developed parts of the cytoplasm.
• For example, the chlorophyll in photosynthetic prokaryotic bacteria is associated with membrane vesicles. Plastids are not observed in it as in photosynthetic eukaryotes.
• The organisms with cells having a well-defined nucleus enclosed in a nuclear membrane are called eukaryotes. Eukaryotic cells are further categorised into plant and animal cells. These are also different from each other in many ways.

Differences Between Prokaryotic And Eukaryotic:

Cytoplasm

The large region of each cell enclosed by the cell membrane is called cytoplasm. It is the fluid content present inside the plasma membrane. It contains many specialised cell organelles, each of which performs a specific function for the cell.

Fundions Of Cytoplasm

1. It helps in the exchange of material between cell organelles.
2. It acts as a storehouse of vital molecules such as amino acids, glucose, vitamins, iron etc.
3. It acts as the site for certain metabolic pathways such as glycolysis etc.

Cell Organelles

Large and complex cells need a lot of chemical activities to support their complicated structure and function.

• To keep these activities separated from each other, these cells use membrane-bound structures.
• These structures perform specialised functions within themselves and are called cell organelles. This is the main characteristic feature that differentiates eukaryotic cells from prokaryotic cells.

Endoplasmic Reticulum (ER)

It is a large network of membrane-bound tubes and sheets. It extends from the outer nuclear membrane into the cytoplasm. It looks like long tubules of round and oblong bags (vesicles). The ER membrane is similar in structure to the plasma membrane. lt occurs in three forms, i.e. cisternae, vesicles and tubules. Depending upon the nature of its membrane, ER is of two types:

1. Rough Endoplasmic Reticulum (RER) It contains ribosomal particles, due to which its surface is rough. The ribosomes are the site of protein synthesis. RER is mainly formed of cisternae.
2. Smooth Endoplasmic Reticulum (SER) It helps in the manufacture of fat molecules or lipids. It is formed of vesicles and tubules. Its surface is smooth due to the absence of ribosomes. ER appears in varying forms in different cells. It always forms a network system of vesicles and tubules.

Functions Of Endoplasmic Reticulum

1. Ribosomes present in all active cells act as sites for protein synthesis. Proteins manufactured here are transported throughout the cell by the endoplasmic reticulum.
2. Fat and lipid molecules manufactured by SER help in building cell membranes. This process is called membrane biogenesis.
3. Some other proteins and lipids synthesised by ER function as enzymes and hormones.
4. SER plays a crucial role in the detoxification of poisons and drugs in the liver cells of vertebrates (a group of animals).
5. It forms a network system, providing channels for the transport of materials, especially proteins. It transports between various regions of the cytoplasm or between the cytoplasm and the nucleus.
6. It functions as a cytoplasmic framework. It provides a surface for some of the biochemical activities of the cell.
7. It gives mechanical support to the cells.

Golgi Apparatus

It consists of a system of membrane-bound, fluid-filled vesicles, large spherical vacuoles and smooth, flattened cisternae. These are stacked parallel to each other. Each of these stacks is called a cistern.

The Golgi apparatus (or dictyosomes) arises from the membrane of smooth ER. Therefore, it constitutes another portion of a complex cellular membrane system.

The material that is synthesised near the Endoplasmic Reticulum (ER) is packaged and dispatched to various parts of the cell through the Golgi apparatus.

Functions Of Golgi Apparatus

1. Golgi apparatus stores modifies and packs products in vesicles.
2. It is involved in the formation of lysosomes.
3. It forms complex sugars from simple sugars in some cases.
4. It is involved in the synthesis of cell wall and plasma membrane.

Note: The scientist, who described the Golgi apparatus for the first time was Camillo Golgi. Most of his investigations were concerned with the nervous system. His greatest work was a revolutionary method of staining individual nerve and cell structures.

This method is called ‘black reaction’. It uses silver nitrate solution to trace the most delicate ramification of cells. He shared the Nobel Prize in 1906 with Santiago Ramon y Cajal for their work on the structure of the nervous system.

Lysosomes

• These are a kind of waste disposal system of the cell. Lysosomes are membrane-bound sacs that are filled with digestive enzymes. These enzymes are made by the rough endoplasmic reticulum. Lysosomes are also called the suicidal bags of a cell.
• During the disturbance in cellular metabolism or when the cell gets damaged, lysosomes may burst and the enzymes can digest their cell. They are absent in RBCs.

Functions Of Lysosomes

1. They help to keep the cell clean by digesting any foreign material that enters the cell as well as worn-out cellular organelles. Hence, called scavengers and cellular housekeepers.
2. They remove foreign material by breaking it into small pieces through their powerful digestive enzymes. These enzymes can break down all organic materials.
3. During starvation, the lysosomes digest stored food contents by autophagy and supply energy to the cell.

Mitochondria

Mitochondria were first observed by Kolliker in 1880. It is a double membrane-bounded cell organelle. The outer membrane is very porous. The inner membrane is deeply folded into finger-like projections called cristae. It creates a large surface area for ATP-generating chemical reactions.

The space between the outer and inner membrane is called intermembranous space. The mitochondrion is a self-replicating (semiautonomous) organelle. It is the largest organelle in animal cells.

Functions Of Mitochondria

1. It generates energy for various activities of the cell. It is known as the powerhouse of the cell. Mitochondria are sites of cellular respiration. They release energy required by the cell in the form of ATP (Adenosine Triphosphate). This ATP is known as the energy currency of the cell.
2. Whenever the cell requires energy, the ATP molecule breaks down. It generates energy to be used for metabolic activities of the body.
3. Mitochondria are strange organelles in the sense that they have their DNA and ribosomes. Hence, they can make some of their proteins.
4. They provide intermediates for the synthesis of various chemicals like fatty acids, steroids, amino acids etc.

Plastids

• These are found only in plant cells. The internal organisation of plastids contains numerous membrane layers embedded in a material called the stroma.
• Plastids are similar to mitochondria in external structure. They are double-layered. They contain their DNA and ribosomes.

Types Of Plastids

Plastids are of three types:

1. Chloroplasts These are the plastids containing chlorophyll (a green pigment). They give a green colour to the plant. Chloroplasts also contain various yellow or orange pigments in addition to chlorophyll. It is a semiautonomous organelle. Chloroplasts are also known as the kitchen of cells.

Function These are important for photosynthesis in plants.

Note: Photosynthetic bacteria do not contain chloroplasts. They contain light-absorbing pigments and reaction centres, which make them capable of converting light energy into chemical energy.

2. Leucoplasts These are the white or colourless plastids. They can change into other types of plastids.

Function Leucoplasts store materials such as starch (amyloplasts), oils (elaioplasts) and protein granules (aleuroplasts).

3. Chromoplasts These are coloured plastids (except green).

Function Chromoplasts impart colour to flowers and fruits. They are rich in carotenoid pigments and lipids.

Vacuoles

These are the storage sacs for solid or liquid contents. In animal cells, vacuoles are small-sized, but in plants, the vacuoles are large-sized. Some may occupy 50-90% of the total cell volume. The vacuole is bounded by a membrane called the tonoplast.

Functions Of Vacuoles

1. Vacuoles are full of cell sap and provide turgidity and rigidity to cells in plants.
2. Many substances like amino acids, sugars, organic acids and proteins are stored in vacuoles.
3. In Amoeba, consumed food items are stored in food vacuoles.

Activity 1

Objective

Observation of plant cells.

Materials Required

Onion, knife, forceps, watch glass, water, dilute glycerine, safranine (safranin) solution, thin camel hair paint brush, dropper, mounting needle, glass slide, coverslip, blotting paper and microscope.

Procedure

Cut a small piece from an onion bulb. Separate a fleshy scale with the help of forceps and remove a transparent peel (called epidermis) from the inner concave side of the scale.

• Place it immediately in the watch glass containing a small quantity of water. Pour 2-3 drops of safranin solution into the watch glass. After 5-10 minutes, remove the stain, pour water into the watch glass and wash it again.
• Place a drop of water or dilute glycerine in the middle of a glass slide. Transfer a small piece of stained onion and peel onto it using a fine camel hair paintbrush.
• Place a cover slip over the peel with the help of a mounting needle, avoiding the appearance of air bubbles. Wipe out excess liquid by blotting paper. Now observe it under the compound microscope.

Observation

The peel is found to have several similar elongated rectangular cells joined with each other. Each cell consists of an outer cell wall, a filler material or cytoplasm with a covering of plasma membrane containing a small oval nucleus.

Question 1. Why glycerine is used in this experiment?
Answer:

Glycerine is used in this experiment because it acts as a preservative and has a high refractive index.

Question 2. Why safranin is used in this experiment?
Answer: Safranin is an azo dye commonly used for plant microscopy, especially to stain lignified tissues such as the xylem of plants, etc

Question 3. Why was the sample washed again with water after beeping in safranin solution for 5-10 minutes?
Answer: The sample was washed again with water to remove the extra stain of safranin.

Question 4. What will happen, if air bubbles appear while placing a cover slip?
Answer: If air bubbles appear, then the structure of the sample will not be observed properly.

Activity 2

Objective

Temporary mounts of cells from different sources. Prepare temporary mounts of onion peel, leaf peel and onion root tip. Study the shapes, sizes and internal structures of the cells.

Materials Required

Onion, knife, forceps, watch glass, water, dilute glycerine, safranine (safranin) solution, thin camel hair paint brush, dropper, mounting needle, glass slide, coverslip, blotting paper and microscope.

Procedure

Repeat the procedure described in Activity 1 for all three types of tissues onion peel, leaf peel and onion root tip.

Observation

The cells from different sources are of different shapes and sizes besides having certain specific structures.

Question 1. What conclusion can be drawn from this activity?
Answer: The cells from different sources are of different shapes and sizes, but have the same structure.

Question 2. Describe briefly the basic structure of cells as observed in this activity.
Answer: Cells consist of an outer cell wall having cytoplasm, plasma membrane and a nucleus.

Question 3. Why blotting paper is used in this activity?
Answer: Blotting paper wipes out excess liquid.

Question 4. What is the shape of cells in onion peel?
Answer: The cells of the onion peel are elongated and rectangular.

Activity 3

Objective

To observe the phenomenon of osmosis in a typical cell

Skill Developed

Observation skills, problem-solving and critical thinking.

Time Required

1 hour 30 minutes.

Materials Required

Containers dilute hydrochloric acid, eggs, concentrated sugar/salt solution and water.

Procedure

1. Place an egg in dilute hydrochloric acid.
2. The outer shell starts to dissolve in the acid and the egg membrane appears clearly. A thin outer skin now encloses the egg.
3. Place the egg into a container containing distilled water.
4. Record the observation.
5. Place another similar egg into a container containing a concentrated salt/sugar solution.
6. Record the observation.

Observation

1. The egg in distilled water swells up because water moves into the egg by osmosis (in this case solution inside the egg is more concentrated).
2. The egg in the concentrated solution shrinks because water moves out of the egg, into the concentrated sugar/salt solution (in this case solution outside the egg is more concentrated)

Question 1. Name the process that made the egg swell.
Answer:

The egg swelled up due to the process of endosmosis. The inward flow of a fluid through a semipermeable membrane towards a fluid of greater concentration is called endosmosis.

Question 2. Differentiate between exosmosis and endosmosis.
Answer:

Exosmosis The passage of water molecules out of the cell into a concentrated solution surrounding it. Endosmosis The passage of water molecules into the cytoplasm of a cell from a less concentrated solution surrounding it.

Question 3. What happens to an egg, when it is kept in hydrochloric acid?
Answer: Hydrochloric acid dissolves the outer shell of the egg and the thin semi-permeable membrane of an egg becomes visible.

Question 4. Name the process of diffusion through a semi-permeable membrane that occurs during certain conditions.
Answer: Osmosis is a special case in which diffusion takes place through a selectively permeable membrane.

Activity 4

Objective

To observe osmosis in raisins or apricots.

Time Required

1 hour 30 minutes.

Materials Required

Water, raisins or apricots, sugar or salt solution and beakers.

Procedure

1. Take two beakers and label one as ‘water’ and the other as ‘concentrated’.
2. Take some water in the beaker labelled ‘water’ (beaker A)
3. Put some raisins or apricots in this beaker A containing water for some time.
4. Add salt or sugar-concentrated solution in a beaker marked as ‘concentrated’ (beaker B).
5. Now add raisins or apricots to a concentrated beaker.
6. Observe the raisins or apricots in both beakers.

Observation

1. The raisins in beaker A swell up as water moves inside from outside due to the low concentration of water inside the cell. This process is called endosmosis.
2. The raisins in beaker B shrink because water moves out of the cell. Since the solution outside the cell is concentrated and the water concentration is lower than that present inside the cell, the water moves outside the cell and this process is called exosmosis.

Question 1. What does a concentrated solution mean?
Answer:

A concentrated solution is a solution with a high amount of solutes. When a large amount of solute (like sugar) is mixed in a solvent (like pure water) the solution becomes a concentrated solution (for example., sugar solution).

Question 2. Why did the raisins in Beaker A swell?
Answer:

• In beaker A, raisins were kept in water. The concentration of water outside the cell was more than the concentration of water inside the cell.
• Therefore, water moved from the region of high concentration to the region of low concentration (inside raisin) and caused swelling of the raisins.

Question 3. Why did the raisins in Beaker B shrink?
Answer: In beaker B, raisins were kept in a concentrated solution, where water concentration was lower than that of the cell. Hence, water moved from the cell to outside leading to the shrinkage of the cell (exomosis).

Question 4. What is the primary requirement of osmosis?
Answer: For osmosis, the presence of a semi-permeable membrane is the primary requirement.

Activity 5

Objective

To study plasmolysis using the cells of Rheo leaf.

Time Required

1 hour 30 minutes (approximately).

Materials Required

Microscope, glass slide, cover-slip, Rheo leaf, salt/sugar solution, water and burner.

Procedure

1. Mount an epidermal peel of a freshly plucked Rheo leaf in a drop of water on a glass slide.
2. Cover it with a cover slip and observe it under a microscope.
3. Remove the cover slip and add a drop of concentrated sugar/salt solution. Place the coverslip again onto the slide.
4. Observe the slide again under a microscope after 1 or 2 minutes.
5. Record your observation.
6. Remove the cover-slip and wipe out the salt/sugar solution, add one or two drops of water and observe the slide under a microscope after 1-2 min.
7. Record your observation.
8. Repeat the same experiment with a leaf that was kept in boiling water for a few minutes.
9. Record your observation.

Observation

1. When the peel is mounted first in water, pink colour is uniformly filling up the cells.
2. When sugar/salt solution is added, the pink colour (protoplasm) of the cell appears in small areas of the cell, cell, i.e. it has undergone plasmolysis (shrinking of cell protoplasm due to outward movement of water).
3. When water is added to replace the sugar/salt solution, the pink colour/protoplasm fills the cell again.
4. In the second leaf, which was kept in boiling water for a few minutes, no effect of sugar/salt solution was seen.

Question 1. Name the phenomenon which occurs due to shrinkage or contraction of the content away from the cell wall.
Answer: During osmosis, when a living plant cell shrinks or contracts with its contents away from the cell wall, the phenomenon is called plasmolysis.

Question 2. The boiled leaf showed no change with the sugar/salt solution. Why?
Answer: The cells of the leaf died when the leaf was kept in boiling water for a few minutes.

Question 3. Can osmosis occur in dead cells?
Answer: No, osmosis can occur only in living cells.

Question 4. What happens to the cell when kept in an isotonic solution?
Answer: In an isotonic solution, the concentration of the medium was the same as that of the cell, there is no net movement of water across the cell membrane. Thus, the cell will stay the same size.

Activity 6

Objective

Study of human cheek cells.

Materials Required

Sterilised toothpick or ice-cream spoon, glass slide, needle, cover-slip, water or dilute glycerine, dropper, methylene blue, blotting paper and microscope.

Procedure

1. Scrape a small piece of membrane from inside of your cheek lightly with the help of a clean sterilised toothpick or an ice-cream spoon.
2. Mount the membranous scraping in a drop of water or dilute glycerine over a clean glass slide. Spread the scraping with the help of a needle.
3. Put a drop of methylene blue over it. Wait for two minutes.
4. Now put a cover-slip gently over the slide avoiding entry of air bubbles.
5. Replace the stain by pouring water drop on one side and soaking the stain from the sides using blotting paper.
6. Observe it under the microscope first with normal or low power (1 Ox) and then with high power (45x).

Observation

Several polygonal flat (squamous) cells are present in the scraped membrane. Each cell has a distinct boundary of plasma membrane, a central rounded or oval nucleus, many small dotted mitochondria, small vacuoles and cytoplasm.

Question 1. What is the purpose of putting methylene blue on the cell?
Answer: Methylene blue is poured on the slide to stain the cells.

Question 2. What is the shape of the observed cells?
Answer: The observed cells are polygonal, flat or squamous shaped.

Question 3. Why it is necessary to avoid air bubbles?
Answer: Air bubbles will not give the correct image to the observer.

Question 4. Why after staining, water is poured over the cell?
Answer: Water is poured after staining to remove the excess stain. Blotting paper is used for soaking this stain and water mixture.

The Fundamental Unit Of Life Summary

A cell is the basic structural and functional unit of all living organisms. It was discovered by Robert Hooke in the year 1665.

• Unicellular organisms are those organisms which are made up of a single cell only, for example., Amoeba, Chlamydomonas, bacteria, etc.
• Multicellular organisms are organisms made up of many cells. These cells group and assume different functions in the body to form various body parts, for example., plants and animals.
• Prokaryotic cells are cells lacking a well-defined nucleus enclosed by a nuclear membrane, for example., bacteria and cyanobacteria.
• Eukaryotic cells are those having a well-defined nucleus enclosed in a nuclear membrane, for example., plant cells and animal cells. Plant cells possess a cell wall and a vacuole that occupies most of the space. It lacks centrosome and centrioles.
• Animal cells do not have cell walls, they possess highly complex Golgi bodies, centrioles, etc.
• Structurally, a cell mainly consists of a plasma membrane, cytoplasm and nucleus. Cell organelles such as Golgi bodies, mitochondria, etc, are also present in cytoplasm.
• The plasma membrane is the outermost covering of the cell that is composed of proteins and lipids.
• It permits the entry and exit of some materials. It maintains the shape of the cell, acts as a mechanical barrier and protects the internal contents of a cell. Transport of substances across plasma membrane may take place by diffusion, i.e. process of movement of solutes or osmosis, i.e. process of movement of water.
• The nucleus is properly called as brain of the cells. It controls all functions of a cell. It also determines the development of cells by directing the chemical activities of cells.
• Cytoplasm is the fluid content present inside the plasma membrane that contains many specialised cell organelles and acts as a site for metabolic pathways such as glycolysis.
• The endoplasmic reticulum is a large network of membrane-bound tubules and sheets. It plays an important role in protein and lipid synthesis.
• Mitochondria is known as the powerhouse of cells that releases energy required by the cell in the form of ATP.
• Golgi apparatus consists of a system of membrane-bound vesicles called cisternae. It helps in the formation of lysosomes and in storing and packaging various molecules in a cell.
• Lysosomes are the waste disposal systems of a cell also called suicidal bags of cells.
• Plastids are found in plant cells as chloroplasts, chromoplasts and leucoplasts.
• Vacuoles are storage sacs of solids and liquids.

Class 9 Science Chapter 6 Tissues Long Answer Type Questions

Question 1. Write a note on the protective tissue in plants. (Give appropriate diagram also)
Answer:

The protective tissue or the outermost covering of cells in plants is known as the epidermis, which performs a protective function (protecting plants from adverse conditions). It is usually made up of a single layer of cells. In dry habitats, the epidermis gets thicker to protect the plant from undue loss of water.

• On aerial parts of plants, epidermal cells often secrete a waxy, water-resistant layer on their outer surface. This waxy covering aids in protecting the plant against loss of water, mechanical injury, and invasion by parasitic fungi. The cells of epidermal tissue are present in a continuous layer without intercellular spaces.
• Small pores are present on the epidermis of the leaf. These pores are called stomata. They are enclosed by two kidney-shaped cells called guard cells. They help in gaseous exchange and transpiration.

Read and Learn More Class 9 Science Solutions

As the plant grows older, a strip of secondary meristem replaces the epidermis of the stem. This forms a several-layer thick cork or bark of the tree in which cells are dead and compactly arranged without intercellular spaces.

Question 2. List the characteristics of cork. How is it formed? Mention its role. NCERT Exemplar
Answer:

The common characteristics of a cork are as follows

1. It is the outer protective tissue of older stems and roots.
2. It is formed by a secondary lateral meristem called cork cambium.
3. The mature cork becomes dead and filled with tannin, resin, and air.
4. The cells are arranged compactly without intercellular spaces and several layers become thick, which are impermeable due to the deposition of suberin in their wall.

Formation of cork

As the plant grows older, the outer protective tissue undergoes certain changes. A strip of secondary meristem replaces the epidermis of the stem. Cells on the outside are cut off from this epidermal layer. This forms several layers of thick cork or bark with no intercellular spaces.

Role of cork

1. It prevents the loss of water by evaporation.
2. It protects plants from the invasion of parasites and other harmful microorganisms.

Question 3. The transportation system of plants is composed of complex permanent tissue. They have their transportation system within themselves. Justify in detail with appropriate diagrams.
Answer:

• The transportation system of plants is composed of complex permanent tissue. These tissues are made up of more than one type of cells, and all these cells coordinate to perform a common specific function.
• These cells may appear structurally different, but they perform the same function. The permanent tissues are of two types

Xylem It helps in the transportation of water and minerals from roots to other parts of the plant.

Elements Of Xylem

1. Tracheids and Vessels Tubular structure, transport water and minerals vertically.
2. Parenchyma Stores food and helps in the sideways conduction of water.
3. Fibers are Supportive in function.

The phloem transports food from leaves to other parts of the plant. Food is prepared in leaves by the process of photosynthesis.

Elements of phloem

1. Sieve tubes Tubular cells with perforated walls. These consist of living cells.
2. Companion cells Small elongated cells with dense cytoplasm.
3. Phloem parenchyma Thin-walled cells. Mainly function in storage and transportation of food.
4. Phloem fibers Thick-walled cells. These are dead cells. Provide mechanical strength to tissue.

Both the xylem and phloem maintain a transportation system within the plants. There is continuous transportation of food, water, and minerals within the plant.

This transportation is necessary for the proper growth and maintenance of the plant.

Question 4. How is adipose tissue different from blood tissue?
Answer:

Differences between adipose and blood tissue are as follows:

Question 5. Differentiate between bone and cartilage concerning structure, function, and location.
Answer:

Differences Between Bone And Cartilage Are As Follows

Question 6. Explain the significance of the following

1. Hair-like structures on epidermal cells.
2. The epidermis has a thick waxy coating of cutin in desert plants.
3. Small pores in the epidermis of the leaf.
4. Numerous layers of epidermis in cactus.
5. Presence of a chemical suberin in cork cells.

Answer:

1. They increase the total absorptive surface area and help in absorption.
2. Cutin has a waterproof quality and helps in preventing water loss due to transpiration. It also protects plants from entry of pathogens, etc.
3. They help in the gaseous exchange and transpiration process.
4. To prevent water loss.
5. Suberin makes cork cells impervious to gases and water.

Question 7. ‘We can control some of the actions of our body, but some are not in our control’. Comment on this statement,
Answer:

Some of the actions like moving our limbs, fingers, neck, etc., can be controlled by our will. We can move these parts of our body whenever we want to, but some actions of our body like contraction and relaxation of the heart, blinking of an eye, etc., are not under our will, i.e. we cannot stop the functioning of the heart if we want to do so. The actions, which can be manipulated by our wishes are known as voluntary actions.

The muscles, which can perform voluntary actions are voluntary.

• These muscles are also called skeletal muscles or striated muscles. Spindle-shaped muscle cells (al Striated muscle Cb) Smooth muscle These muscles are mostly attached to bones and help in body movement.
• Their cells are long, cylindrical, unbranched, and multinucleate (having many nuclei).
• The actions, which are not under our control are known as involuntary actions. These actions are performed by smooth muscles or involuntary muscles. Their cells are long with pointed ends (spindle-shaped) and uninucleate (single nucleus).

Question 8.

1. What will happen if cells are not properly organized in tissue?
2. Under certain circumstances, the squamous epithelium is known as stratified squamous epithelium. Justify.

Answer:

1. Different organisms whether unicellular or multicellular need to perform many functions in the body such as respiration, digestion, and locomotion.
   1. In multicellular organisms, cells present in a group and specialized in one particular function form a tissue.
   2. Some tissues help in growth, while others in locomotion and some in body movement.
   3. So, if cells are not organized in these tissues, then highly organized and specialized processes will become disorganized.
   4. There will be no coordination in the functioning of the cells and body.
2. When the simple squamous epithelium is arranged in a pattern of multilayers to prevent wear and tear, the epithelium is called the stratified squamous epithelium, for example., skin.

Class 9 Science Chapter 6 Tissues Short Answer Type Questions

Question 1. Explain the basic criteria for the classification of permanent tissue in plants.
Answer:

The following points form the basis of criteria for classification of permanent tissue in plants

1. Whether the tissue is made up of one type of cell (simple) or more than one type of cell (complex).
2. Function Supportive (parenchyma), protective (epidermis) or conducting (xylem and phloem).
3. Whether the cell wall is thick or thin.
4. Whether the cells are living or dead.

Question 2. List any six characteristics of parenchyma.
Answer:

The major characteristics of parenchyma are as follows

1. The cells of the parenchyma are living and possess the power of division.
2. Each parenchyma cell is isodiametric in shape with a thin cell wall and encloses dense cytoplasm and a small nucleus.
3. The cells are loosely packed with large intercellular spaces between them.
4. It is found in soft parts of plants such as the cortex of roots, ground tissue in the stem, and mesophyll cells of leaves.
5. It serves as a packing tissue to fill the spaces between other tissues and maintains the shape of a plant.
6. It stores waste products of plants such as tannin, gum, crystals, etc.

Question 3. Name the tissue responsible for the flexibility in plants. How would you differentiate it from other permanent tissues?
Answer:

Collenchyma tissue is responsible for providing flexibility in plants. Differences between collenchyma tissue and other permanent tissues (parenchyma and sclerenchyma) are as follows

Question 4.

1. Identify the given figures.
2. Give any two major differences between the structures identified.
3. Describe the role performed by these two in the plant body.

Answer:

1. The given figures are of tracheids
   1. And Vessels
   2. Of xylem tissue.
2. The differences between tracheids
   1. And Vessels
   2. Are as follows

Tracheids and vessels are lignified tissues that provide mechanical strength to the plant body. They also function in carrying water and mineral salts from roots to different parts of the plant body vertically.

Question 5. Describe three functions of protective tissue in plants.
Answer:

Two types of protective tissues present in plants are epidermis and cork. The three common functions of these protective tissues are as follows

1. Cork protects plants from invasion of parasitic microorganisms and excessive heat and cold.
2. The cuticle of the epidermis checks the excessive evaporation of water.
3. The epidermis allows transpiration and gaseous exchange through stomata.

Question 6. Give the name of the following

1. Tissue is concerned with the conduction of food materials.
2. Tissue capable of cell division.
3. Multiple pores are present in the epidermis of the leaf.

Answer:

1. Complex tissue (phloem)
2. Meristematic tissue
3. Stomata

Question 7. List the constituents of phloem. What will happen if the phloem at the base of a branch is removed?
Answer:

• Major constituents of phloem include sieve tubes, companion cells, phloem fibers, and phloem parenchyma.
• If the phloem at the base of the branch is removed, then a lower area of the branch will not receive food from the leaves. But the plant will not die, as it will continue to receive food from other branches as food can move through phloem in both directions.

Question 8. Which is the simplest protective tissue present in the animal body? State its two functions.
Answer:

The most simple protective tissue present in the animal body is epithelial tissue. Its major functions are as follows

1. It protects underlying cells from drying, injury, and infections.
2. It helps in the elimination of waste products from the body.

Question 9. Name the type of epithelium present in the respiratory tract. What is its specialization?
Answer:

The ciliated columnar epithelium is present in the respiratory tract of humans. It has specialized hair-like projections called cilia which help in the movement of substances and mucus forward. Thus, the name ciliated columnar epithelium.

Question 10. Give three differences between epithelial tissue and connective tissue.
Answer:

Differences Between Epithelial And Connective Tissues Are As Follows:

Question 11. Mention three different types of blood cells with their functions. Draw diagrams also.
Answer:

Three different types of blood cells are as follows

1. Red Blood Cells (RBCs) Contains hemoglobin, help in the transportation of gases, digested food, hormones, etc.
2. White Blood Cells (WBCs) Integral part of the immune system, that help in fighting diseases by producing antibodies and engulfing germs and pathogens.
3. Platelets Help in the clotting of blood.

Question 12. Differentiate between bone and cartilage.
Answer:

Differences Between Bone And Cartilage Are As Follows:

Question 13. Name the tissue that smoothens bone’s surfaces at joints. Describe its structure with the help of a diagram.
Answer:

Cartilage is the tissue that smoothens bone surfaces at joints. Structure It is a specialized connective tissue, which is compact and less vascular.

Its extensive matrix is composed of proteins and slightly hardened by calcium salts and also has a delicate network of collagen fibers, living cells, and chondrocytes. The chondrocytes are present in lacunae. Cartilage provides support and flexibility to body parts. It is present in the nose, trachea, ear, and larynx.

Question 14. Give the differences between tendon and ligament.
Answer:

The Differences Between Tendon And Ligament Are As Follows:

Question 15. State the functions of skeletal connective tissue.
Answer:

1. The functions of skeletal connective tissue are as follows
2. It gives a definite shape to the body.
3. It protects the vital organs of the body, for example., the brain.
4. It provides a surface for the attachment of muscles to increase their efficiency.

Question 16. Write a note about the structure and significance of striated muscles with a diagram.
Answer:

Striated muscles are voluntary muscles, i.e. we can move these muscles according to our will.

• These are mostly attached to bones and help in body movement. They show alternate light and dark bands or striations (when stained appropriately).
• Cells of striated muscles are long, cylindrical, unbranched, and multinucleate, for example., muscles of limbs.

Question 17. Name the kinds of muscles found in your limbs and lungs. How do they differ from each other structurally and functionally?
Answer:

1. Limbs Striated muscle.
   1. Structure Cells are long and cylindrical in shape, presence of dark and light bands, multinucleate and unbranched.
   2. Function Voluntary movement.
2. Lungs Smooth/non-striated muscle.
   1. Structure Cells are spindle-shaped, absent of striation, uninucleate, and unbranched. Function Involuntary movement.

Question 18. Which type of muscle, smooth or striated is found in the iris of eye? Why are smooth muscles called involuntary muscles? In what way are they different from striated muscles concerning several nuclei?
Answer:

Smooth muscles are found in the iris of the eye. Smooth muscles are the muscles that cannot be moved or stopped according to our will. So, they are known as involuntary muscles. Smooth muscles are uninucleate, whereas striated muscles are multinucleate.

Question 19. Draw a labeled diagram of unstriated muscle tissue and mention its occurrence, features, and functions.
Answer:

The diagram of unstriated muscle tissue is given

Occurrence These tissues are found in the walls of the alimentary canal, urinary bladder, esophagus, iris, bronchi, etc.

Features

1. They are unbranched and non-striated.
2. Cells are long, thin, and spindle-shaped.
3. Each cell has a single central nucleus.

Functions

1. These muscles do not work as per our will.
2. These carry out the movement in the urinary bladder and gall bladder.

Question 20. What are neurons? Where are they found in the body? What function do they perform in the body of an organism?
Answer:

Neurons are the structural and functional units of the nervous system. These are found in the brain, spinal cord, and nerves.

Functions

1. These coordinate various body parts during any body function.
2. These control all the activities of the body.
3. These transmit messages in the form of nerve impulses to the brain and spinal cord.

Question 21. Differentiate between axon and dendrite.
Answer:

Differences Between axons and dendrites are As Follows:

Question 22. Give one function of each of the following.

1. Stomata
2. Cardiac muscle fibres

Answer:

1. Stomata They help in the gaseous exchange between the plant and the atmosphere. Transpiration takes place through the stomata.
2. Cardiac muscle fibers They help in rhythmic contraction and relaxation of heart throughout the life.

Question 23. Write functions of the following

1. Areolar connective tissues
2. Neurons
3. Adipose connective tissues
4. Cardiac muscles

Answer:

1. Areolar connective tissues These fill the space inside the organs and support internal organs of the body. These also help in the repair of tissues.
2. Neurons These are the fundamental cells of nervous tissue. These are highly specialized for being stimulated and then transmitting the stimulus very rapidly from one place to another within the body.
3. Adipose connective tissues These are fat-storing tissue found below the skin between internal organs. The cells of this tissue are filled with fat globules. Due to the storage of fats, these act as an insulator.

Question 24. What happens, when

1. Formation of cork in older stems does not occur.
2. Blood platelets are removed from the blood.

Answer:

• If the formation of cork does not occur in older stems then the outer tissue will rupture due to the increase in girth and the plant will get infected by the parasites.
• Blood clotting will not occur at the site of injury. Bleeding will continue which may lead to death.

Question 25. How are simple tissues different from complex tissues in plants?
Answer:

The main points of difference between simple tissues and complex tissues are:

Question 26. Differentiate between parenchyma, collenchyma, and sclerenchyma based on their cell wall.
Answer:

Differences Between Parenchyma, Collenchyma, And Sclerenchyma Are As Follows:

Question 27. Give reasons for the following

1. Meristematic cells have prominent nuclei and dense cytoplasm, but they lack vacuole.
2. Intercellular spaces are absent in sclerenchymatous tissue.
3. We get a crunchy and granular feeling when we chew pear fruit.
4. Branches of trees move and bend freely in high wind velocity.
5. It is difficult to pull out a husk of coconut.

Answer:

1. Meristematic cells undergo division and do not store food, thus lacking vacuole.
2. Because their walls are lignified and form bundles for mechanical functions.
3. Due to the presence of sclerenchymatous tissue (stone cell) or sclereids, we get a crunchy feeling when we chew pear fruit.
4. The presence of collenchyma tissue provides flexibility to the branches of a tree.
5. The husk of coconut is made up of sclerenchymatous fibers, which are closely packed.

Question 28. What are the functions of the stomata?
Answer:

The main functions of stomata are:

1. They help in the exchange of gases with the atmosphere.
2. They help in the transpiration of water.

Question 29. Diagrammatically show the difference between the three types of muscle fibers.
Answer:

Question 30. Differentiate between striated, unstriated, and cardiac muscles based on their structure and site/location in the body.
Answer:

Question 31. Draw a labeled diagram of a neuron

Question 32. Name the following.

1. Tissue that forms the inner lining of the mouth.
2. Tissue that connects muscle to bone in humans.
3. Tissue that transports food in plants.
4. Tissue that stores fat in our body.
5. Connective tissue with a fluid matrix.
6. Tissue present in the brain.

Answer:

1. Squamous epithelium
2. Tendons
3. Phloem
4. Adipose tissue
5. Blood
6. Nervous tissue.

Question 33. Identify the type of tissue in the following: Skin, the bark of a tree, bone, the lining of the kidney tubule, and vascular bundle.
Answer:

• Skin Squamous epithelium
• The bark of tree Epidermal tissue
• Bone Connective tissue
• The lining of kidney tubule Cuboidal epithelium
• Vascular bundle Conductive tissue (xylem and phloem).

Question 34. What is the role of the epidermis in plants?
Answer:

Role of epidermis in plants:

1. It helps in the protection of the internal organs of plants.
2. It becomes thick and prevents water loss in plants living in very dry habitats.
3. Its cell secretes a waxy, water-resistant layer on the outer surface, which protects against loss of water, mechanical injury, and infections.
4. Leaf epidermis have stomata to help in gas exchange and transpiration.
5. In old plants, the epidermal layer becomes thick and forms cork. Cork cells contain a chemical called suberin in their walls, which makes them impervious to gases and water.

Question 35. How does the cork act as a protective tissue?
Answer:

Cork is made up of several layers of epidermal cells. These cells of cork are dead and compactly arranged without intercellular spaces. They also have a chemical called suberin in their walls, which makes cork impervious to gases and water.

Class 9 Science Chapter 6 Tissues Very Short Answer Type Questions

Question 1. Name the regions in which parenchyma tissue is present.
Answer:

Parenchyma tissue is mainly found in soft parts of the plant such as roots, stems, and leaves and is also present in ground tissue of petioles.

Question 2.

1. Which type of tissue is found at the shoot apex? Name another part of the plant body, where this type of tissue is found,
2. If a plant is to show longitudinal growth, which kind of meristematic tissue will be promoted in the plants?

Answer:

1. Apical meristem is found at the shoot apex. It is also present in the growing tips of roots.
2. Intercalary meristem helps in the longitudinal growth of plants.

Question 3. What is the specific function of cardiac muscle?
Answer: Cardiac muscles contract and relax rhythmically throughout life.

Question 4.

1. Identify the region of the stem marked A in the given diagram and the type of simple permanent tissue found in this region.
2. Mention any two characteristic features of this tissue.

Answer:

1. The region marked as A is the pith and the tissue is parenchyma.
   1. The cells of parenchyma are living, thin-walled and loosely packed.
   2. This tissue provides support to plants and also stores food.

Question 5.

1. Plant tissue is observed under a microscope, as shown in the given figure. Identify the tissue.
2. State the characteristic features of these cells.
3. Name any two parts of the plant, where such cells are present.

Answer:

1. It is parenchyma.
2. It consists of thin-walled unspecialized cells, which are loosely packed, i.e. having intercellular spaces. Each cell has a prominent nucleus.
   1. Pith
   2. Cortex

Question 6.

1. Identify the tissue given in the following figure.
2. Mention the characteristic features of the cells.
3. Specify the function of this tissue.
4. Name any one part of the plant, where these cells are present.

Answer:

1. The tissue given in the figure is collenchyma.
2. The cells of collenchyma are living, elongated, thickened at the corners, and have very little intercellular space.
3. It provides mechanical support and flexibility to the plant.
4. It is present in leafstalks, below the epidermis.

Question 7. What are the functions of collenchyma in plants?
Answer:

The functions of collenchyma are as follows

1. It allows easy bending in various parts of a plant (leaf and stem) without breaking them.
2. It also provides mechanical support to plants.

Question 8. Write the difference between aerenchyma and chlorenchyma.
Answer:

Differences between aerenchyma and chlorenchyma are as follows

Question 9.

1. Identify the tissue shown in the figure.
2. Specify any part of the plant, where such cells are present.

Answer:

1. Sclerenchyma
2. Sclerenchymatous cells are found
   1. Around vascular bundles
   2. In veins of leaves

Question 10. Write the functions of the following chemical substances found in plant tissues.

1. Lignin
2. Cutin

Answer:

1. The functions of lignin are as follows
   1. It is a chemical substance, which acts as cement and provides strength to the plant.
   2. The presence of lignin in plants makes the walls of sclerenchyma thick.
2. The functions of cutin are as follows
   1. It is a chemical substance having a waterproof quality, present in the epidermis of desert plants.
   2. It prevents excessive loss of water through transpiration.

Question 11. What is the difference between sclerenchyma and collenchyma?
Answer:

Differences Between Sclerenchyma And Collenchyma Are As Follows:

Question 12. Answer the following

1. How is the epidermis of the plants living in very dry habitats adapted?
2. Write functions of guard cells of stomata in the leaf.

Answer:

1. The epidermis of the plants living in very dry habitats has a thick waxy coating of waterproof cutin over it. This prevents the loss of water.
2. Guard cells help in exchange of gases with the atmosphere, due to the opening and closing of the stomatal pore.

Question 13.

1. Epidermal cells help in the absorption of water and nutrients from soil. How?
2. Write a note on a vascular bundle.

Answer:

1. Epidermal cells in roots bear long hair-like outgrowth, i.e. root hairs that greatly increase the total absorptive surface area and help in increased absorption of water and nutrients from soil.
2. The vascular bundle is formed of two main components known as xylem and phloem. Both of them are the conducting tissues that help in the conduction of water, minerals (xylem), and food (phloem) throughout the plant body.

Question 14. The diagram shows the part of a xylem vessel.

What is the function of the spiral structure X? What is it made up of?

Answer:

The spiral structure X is made up of lignified cellulose, which provides mechanical support to the xylem vessel.

Question 15. Give differences between sclerenchyma fibers and sclereids.
Answer:

The Differences Between Fibres And Sclereids Are As Follows:

Question 16. What is epithelial tissue? State the type of epithelial tissue present in the lining of blood vessels.
Answer:

The outermost covering or protective tissue in the animal body is called epithelial tissue. Simple squamous epithelium covers the lining of blood vessels.

Question 17. What is the function of areolar tissues?
Answer:

1. They act as supporting and packing tissue as they are found between visceral organs.
2. They help in the repair of tissue.

Question 18. Name the tissue present between internal organs. What are its functions?
Answer:

The tissue present between internal organs is areolar tissue.

Its functions are as follows

1. To provide support to internal organs.
2. To repair body tissue.

Question 19.

1. What will happen if the ligament gets overstretched?
2. Why are skeletal muscles known as striated muscles?

Answer:

1. Overstretching of ligament leads to sprain.
2. Skeletal muscles are known as striated muscles because they show dark and light bands or stripes or striation.

Question 20. Determine the location of the following tissues:

1. Unstriated muscle fibers
2. Cuboidal epithelium
3. Adipose tissue
4. Striated muscle fibers

Answer:

1. Alimentary canal, iris of the eye, ureter, bronchi, etc.
2. Kidney tubules and ducts of salivary glands.
3. Below the skin and between internal organs.
4. Limbs, tongue, etc.

Question 21. Cutting of rose plants is done timely in gardens, but still, it regains its length. Give reason.
Answer:

The rose plant regains its length because of the presence of intercalary meristem at the base of leaves or internodes on twigs. The cells of these tissues divide and increase the length of the plant.

Question 22. Do the roots of a plant continue growing after their tips are removed? Explain giving reasons.
Answer:

• No, roots do not grow after the removal of their tips. Normally, the apical meristem is present at the tip of the root, which divides repeatedly to increase its length.
• When tips are removed, the apical meristem is also removed and no further growth in the length of roots occurs.

Question 23. Water hyacinth floats on the water’s surface. Explain.
Answer:

Aerenchyma (a type of parenchyma) is present in water hyacinth, which encloses a lot of air and makes the plant lighter than water so that it can float on the surface of the water.

Question 24. A mango tree has several branches. Which tissue helps in the sideways conduction of water in the branches?
Answer:

Xylem vessels are very long tube-like structures formed by a row of cells placed end to end. The transverse walls between these cells are partially or completely dissolved to form continuous water channels. Thus, xylem vessels help in the sideways conduction of water.

Question 25. It is known that the function of the heart is to pump blood throughout the body. Can you explain the cause of this peculiar ability of the heart?
Answer:

The heart is made up of cardiac muscles. These cardiac muscles are involuntary muscles (not under the control of our will). These muscles show continuous rhythmic contraction and relaxation throughout life thereby, helping the heart to pump blood in our body.

Question 26. What is a tissue?
Answer: The group of cells combined together to perform a common function is called a tissue, for example., blood, and muscle.

Question 27. What is the utility of tissue in multicellular organisms?
Answer: The human body works with the principle of division of labor. The cells specializing in one function are grouped to perform a common function.

Question 28. Name the types of simple tissue.
Answer: The three main types of simple tissues are:

1. Parenchyma
2. Collenchyma
3. Sclerenchyma.

Question 29. Where is apical meristem found?
Answer: Apical meristem is present in the growing tips of stems and roots of plants.

Question 30. Which tissue makes up the husk of coconut?
Answer: Sclerenchymatous tissue.

Question 31. What are the constituents of phloem?
Answer: The main constituents of phloem are:

1. Sieve tubes
2. Companion cells
3. Phloem parenchyma
4. Phloem fibers.

Question 32. Name the tissue responsible for movement in our body.
Answer: Muscular tissue.

Question 33. What does a neuron look like?
Answer:

A neuron possesses a cell body and various processes emerging from it. These processes include a long axon and short dendrites.

Question 34. Give three features of cardiac muscles.
Answer: Features of cardiac muscles are listed below:

1. These are involuntary muscles.
2. Cardiac muscle cells are cylindrical, branched, and uninucleated.
3. These muscles show rhythmic contraction and relaxation throughout life.

Question 35. What are the functions of areolar tissue?
Answer: Functions of areolar tissue are as follows:

1. Areolar tissue fills the space inside the organs and supports them.
2. It helps in the repair of the tissue.

Question 36. How many types of elements together make up the xylem tissue? Name them.
Answer:

Four elements make up xylem tissue which include:

1. Tracheids
2. Vessels
3. Xylem parenchyma
4. Xylem fibers.

Tissues Multiple Choice Questions

Question 1. At maturity, the sieve plates become impregnated with

1. Callose
2. Lignin
3. Suberin
4. Pectin

Answer: 1. Callose

Sieve plates are formed by two adjoining end walls of neighboring sieve elements of the sieve tube of phloem. At maturity, they become impregnated with a callose pad, which may be seasonal callose (for only a limited unfavorable period) or definitive (formed permanently in functionless old sieve tubes).

Question 2. Which of the following is not a part of this? epidermal tissue system?

1. Companion cells
2. Guard cells
3. Root hairs
4. Subsidiary cells

Answer: 1. Companion cells

Companion cells are characteristic elements of phloem tissue associated with the sieve tubes in the angiosperms. Thus, they are the part of vascular tissue system. They are absent in pteridophytes and gymnosperms. Guard cells and subsidiary cells are associated with stomata, found on the surface of the epidermal tissue system.

Question 3. A keratinized dead layer of skin is made of

1. Stratified columnar epithelium
2. Simple cuboidal epithelium
3. Stratified squamous epithelium
4. Simple columnar epithelium

Answer: 3. Stratified squamous epithelium

Stratified squamous epithelium is seen in the adult human body. It may be keratinized or non-keratinized. In keratinized stratified squamous epithelium, the outer few layers contain a hard waterproof protein in their cytoplasm, called keratin. It is found on the epidermis of skin, hair, and nails.

Question 4. Connective tissue is

1. Ectodermal in origin with intercellular spaces
2. Ectodermal in origin without intercellular spaces
3. Mesodermal in origin without intercellular spaces
4. Mesodermal in origin with intercellular spaces

Answer: 4. Mesodermal in origin with intercellular spaces

Connective tissue is mesodermal in origin with intercellular spaces, connective tissue cells, and fibers. The major functions of the connective tissue are binding, support, protection, transport, insulation, fat storage, and body defense.

Question 5. The striated appearance of a myofibril is due to the distribution pattern of

1. Meromyosin
2. Actin and myosin
3. Sarcoplasmic reticulum
4. Troponin and fascicles

Answer: 2. Actin and myosin

The striated appearance of a myofibril is due to the distribution pattern of both actin and myosin proteins, which are arranged as a rod-like structure, parallel to each other and also to a longitudinal axis of myofibrils. A myofibril has an alternative dark and light band. The dark band contains myosin, while the light band contains actin.

Question 6. Simple mechanical tissue devoid of lignin is

1. Chlorenchyma
2. Sclerenchyma
3. Collenchyma
4. Parenchyma

Answer: 3. Collenchyma

Collenchyma is a specialized supporting or simple mechanical permanent tissue of living cells possessing characteristically unevenly distributed thickenings of cellulose, pectin, and hemicellulose. Lignin is found in the cell walls of sclerenchyma.

Question 7. The type of tissue lining of the nasal passage, bronchioles, and Fallopian tubes is

1. Cuboidal epithelium
2. Columnar ciliated epithelium
3. Stratified squamous epithelium
4. Simple squamous epithelium

Answer: 2. Columnar ciliated epithelium

• Ciliated columnar epithelium comprises columnar cells, which have cilia on the free surface.
• This epithelium lines most of the respiratory tract and Fallopian tubes (oviducts). It also lines the ventricles of the brain and the central canal of the spinal cord.

Question 8. Healing of wounds is done in plants by the activity of

1. Lateral meristem
2. Primordial meristem
3. Intercalary meristem
4. Apical meristem

Answer: 1. Lateral meristem

Lateral meristems are present along the side of the organs, for example., vascular cambium in the plants. It is responsible for healing wounds of plants by its meristemic activity.

Question 9. The tissue whose cells are thin-walled and living, more or less isodiametric, and have intercellular spaces, is known as

1. Collenchyma
2. Sclerenchyma
3. Parenchyma
4. Chlorenchyma

Answer: 3. Parenchyma

Parenchymatous cells are isodiametric, spherical, oval, or polygonal with intercellular space or may be closely packed. They are living cells with thin cellulosic cell walls. Their main function is to store food and water. They are found in most of the plant body.

Question 10. Choose the correctly matched pair.

1. The inner lining of salivary ducts- Ciliated epithelium
2. Moist surface of buccal cavity- Glandular epithelium
3. Tubular parts of nephrons- Cuboidal epithelium
4. The inner surface of bronchioles- Squamous epithelium

Answer: 3. Tubular parts of nephrons- Cuboidal epithelium

• Cuboidal epithelium is present in the tubular parts of the nephron. It consists of short, cube-shaped cells with round nuclei located in the centre of cell. These cells often form microvilli to increase the absorptive surface area of a cell.
• Others are incorrectly matched pairs as, the moist surface of the buccal cavity and inner lining of the salivary duct is lined by compound epithelium, while the inner surface of bronchioles is lined by ciliated epithelium.

Question 11. Aerenchyma provides

1. Mechanical strength of plants
2. Extra space for photosynthesis and storage of food
3. Flexibility to plants
4. Buoyancy to hydrophytic plants

Answer: 4. Buoyancy to hydrophytic plants

Aerenchyma is a special tissue found in hydrophytes (for example., Hydrilla, Potamogeton, etc.) and some land plants (for example., petiole of banana, Canna), formed by parenchyma cells. They make a network by leaving wide air spaces (air cavities) for the gaseous exchange and make the aquatic plants light and buoyant so that they can easily afloat.

Question 12. Axons and dendrites are special features of

1. Cardiac muscle
2. Cartilage
3. Specialized epithelium
4. Neuron

Answer: 4. Neuron

Axons and dendrites are parts of neurons or nerve cells. In a neuron, dendrites may be one to several, but an axon is always one. They help in the conduction of impulses.

Question 13. A group of cells similar in structure, function, and origin is known as

1. Organ
2. Organ system
3. Tissue
4. Organism

Answer: 3. Tissue

Question 14. The tissue responsible for an increase in the girth of the stem is

1. Lateral cambium
2. Apical meristem
3. Intercalary meristem
4. None of these

Answer: 1. Lateral cambium

Question 15. Which of the following are living cells?

1. Sclereids
2. Sclerenchyma
3. Vessels
4. Collenchyma

Answer: 4. Collenchyma

Question 16. Aerenchyma tissue is found in

1. Halophytes
2. Hydrophytes
3. Xerophytes
4. Mesophytes

Answer: 2. Hydrophytes

Question 17. Which of the following is an example of permanent tissue?

1. Apical meristem
2. Parenchyma
3. Xylem
4. Both (2) and (3)

Answer: 4. Both (2) and (3)

Question 18. Simple squamous epithelium is found in

1. Liver
2. Fallopian tube
3. Lungs
4. Stomach

Answer: 3. Lungs

Question 19. Blood devoid of blood cells is called as

1. Serum
2. Plasma
3. Lymph
4. None of these

Answer: 2. Plasma

Question 20. The tissues, which join muscles to the bones are called as

1. Ligament
2. Tendons
3. Cartilage
4. Adipose tissue

Answer: 2. Tendons

Question 22. Name the type of tissue found in the iris of the eye, walls of the alimentary canal, etc.

1. Cardiac muscle
2. Smooth muscle
3. Involuntary muscle
4. Both (2) and (3)

Answer: 4. Both (2) and (3)

Question 22. The gap between two nerve cells is known as

1. Node of Ranvier
2. Synapse
3. Matrix
4. None of these

Answer: 2. Synapse

Tissues

All living organisms are composed of cells. In unicellular organism, a single cell performs all the basic functions, but in multicellular organisms, different functions are performed by different cells. Cells specializing in one function group together and form tissues.

• A group of cells similar in structure that work together to achieve a particular function forms a tissue. These cells are arranged and designed, so as to give the highest possible efficiency of the function they perform. All cells of a tissue have a common origin.
• A tissue may be a simple or complex type. Blood, phloem, and muscles are all examples of tissues. The structural and functional organization of cells in plants and animals is different. Plants remain stationary while animals move as per their needs.
• Each pursues different feeding methods and is differently adapted. In this chapter, we will study various types of tissues found in plants and animals along with their respective functions.

Plant Tissues

Based on dividing capacity, plant tissues can be classified into two fundamental types as follows:

1. Meristematic Tissue

The tissues in which cells always keep dividing giving rise to new cells are called meristematic tissues. These tissues are responsible for the growth of plants. Plants grow only in those regions where meristematic tissues are present, for example., root and shoot tips.

Read and Learn  More Class 9 Science Notes

• It is also called growth tissue. Cells forming this tissue are very active and have dense cytoplasm, thin cellulose walls, and prominent nuclei. They lack vacuoles. The new cells produced by meristem are initially like those of meristem.
• Their characteristics change once they grow and become differentiated as components of other tissues. Meristematic tissue is classified based on the regions, where they are present.

1. Apical Meristem
   1. These are present at the growing tips of stems and roots. This helps increase the length of the stems and the roots. It acts as a pro-meristem having actively dividing cells, giving rise to other meristems.
2. Intercalary Meristem
   1. These are present at the base of the leaves or internodes (on either side of the node) of twigs. It helps in the longitudinal growth (elongation) of plants.
3. Lateral Meristem (Cambium)
   1. These are present on the lateral sides of stems and roots. It helps in increasing the girth of the stem and root.

2. Permanent Tissue

• This tissue is formed from the cells of meristematic tissue when they lose their ability to divide and have attained a permanent shape, size, and function by the process called differentiation.
• As a result of differentiation, the meristematic tissues tend to form different types of permanent tissues as follows:

Simple Permanent Tissue

It is made up of only one type of cells, i.e. the cells forming these tissues are similar in structure and function.

Simple permanent tissue is further classified as:

Parenchyma: A few layers of cells form the basic packing tissue. They are present in the cortex and pith of stems and roots in the mesophyll of leaves.

Characteristics: These are simple living cells with little specialization and thin cell walls.

Simple Permanent Tissue Functions

• It serves as a food storage tissue.
• This tissue provides support to plants.
• When the parenchyma cell contains chlorophyll in some situations, it performs photosynthesis. Such type of parenchyma tissue is called chlorenchyma.
• In aquatic plants, large air cavities are present in parenchyma cells to give buoyancy to plants, which helps them to float. Such type of parenchyma tissue is called aerenchyma.
• Parenchyma of stems and roots also stores nutrients and water.

Collenchyma: These tissues are generally found in leaf stalks below the epidermis and leaf midribs.

Collenchyma Characteristics

• Cells are living, elongated, and irregularly thickened at the corners due to the deposition of pectin.
• They have very little intercellular spaces.

Collenchyma Functions

• It provides mechanical support and elasticity (flexibility) to plants.
• It also allows easy bending in various parts of a plant (leaf and stem) without breaking.

Sclerenchyma

This type of tissue is present in stems, around vascular bundles, in the veins of leaves, and in the hard covering of seeds and nuts.

Sclerenchyma Characteristics

• The cells of sclerenchymatous tissue are dead. The cells are long and narrow in appearance.
• Cell walls are thickened due to lignin (a chemical substance) deposition, which acts as cement and hardens them.
• Due to the presence of thick walls, there is no internal space between the cells.

Sclerenchyma Functions

• It is known to be the chief mechanical tissue, that makes plants hard and stiff, for example., the husk of the coconut is made up of sclerenchymatous tissue.
• It forms a protective covering around seeds and nuts. It gives rigidity, flexibility, and elasticity to the plant body.

Protective Tissues

• The protective tissue i; meant to protect the plants from undue loss of water.
• Thus, they retain adequate water in them. The two types of protective tissues present in plants are:

Protective Tissues Epidermis

• The outermost layer, i.e. epidermis in plants is made up of a single layer of cells. It protects all parts of the plant.
• On the aerial parts of the plant, epidermal cells often secrete a waxy, water-resistant layer on their outer surface. It protects against loss of water, mechanical injury, and invasion by microbes.
• Cells of epidermal tissue form a continuous layer. They have no intercellular spaces due to their protective role. Most epidermal cells are relatively flat. The outer wall and side walls are thicker than the inner wall.
• Epidermal cells of the leaf bear small pores known as stomata. These are enclosed by two kidney-shaped cells called guard cells.

Stomata are responsible for the exchange of gases with the atmosphere and for the process of transpiration (loss of water in the form of water vapors).

• Epidermal cells of the roots bear long hair-like outgrowths called root hairs. They greatly increase the total absorptive surface area and help the roots to absorb water and nutrients from the soil.
• In the case of desert plants, the epidermis of the aerial parts has a thick waxy coating of cutin (a chemical substance with waterproof quality) on its outer surface. It prevents water loss.

Protective Tissues Cork

It is the strip of secondary meristem, which replaces the epidermis of older stems. Cells of cork are dead, compactly arranged, and have no intercellular spaces. It forms the bark of the tree (several layers thick). A chemical called suberin is present in their walls. It makes them impervious to gases and water.

Complex Permanent Tissue: It is made up of more than one type of cells having a common origin. Regardless of different appearances, all the cells coordinate to perform a common function.

Types of complex permanent tissue are:

1. Xylem
2. Phloem

Both of them are conducting tissues and constitute vascular bundles. This is a distinctive feature of complex plants. It provides them with the possibility of surviving in the terrestrial environment.

Complex Permanent Tissue Xylem

It is responsible for the transport of water and minerals from roots to other parts of the plant. The cells of the xylem have thick walls and many of them are dead. Xylem consists of various types of elements, which are as follows:

1. Tracheids

• These are dead, long, tubular structures with tapering ends.
• They transport water and minerals vertically.

2. Vessels

• Long, tube-like structures, formed by a row of cells, placed end to end.
• These are also dead cells with lignified walls.
• They also help in the conduction of water.

3. Xylem parenchyma

• These are only living cells of the xylem with thin cell walls.
• It stores food and helps in the sideways conduction of water.

4. Xylem fibres

• They are elongated dead cells with tapering ends and thick cell walls.
• These are fibers associated with the xylem and supportive of the functioning of the xylem.

Phloem

It transports food from leaves to other parts of the plant. Materials can move in both directions in it. All phloem cells are living except phloem fibers.

Phloem is made up of the following four types of elements:

1. Sieve tubes
   1. These are tubular cells with perforated walls.
   2. They have a thin layer of cytoplasm.
2. Companion cells
   1. These are small elongated cells having thin walls that are not perforated and have active cytoplasm. They help sieve tubes in the translocation of food.
3. Phloem fibers
   1. They are thick-walled sclerenchyma cells that provide mechanical strength to the tissue.
4. Phloem parenchyma
   1. They are thin-walled cells that help in the storage and slow lateral conduction of food.

Differences Between Meristematic And Permanent Tissues:

Animal Tissues

Based on the functions they perform, animal tissues are classified into four basic types namely epithelial, connective, muscular, and nervous tissue.

1. Epithelial Tissue

The covering or protective tissues in the animal body are epithelial tissues. It is the simplest protective tissue of the animal body. It covers most organs and cavities of the body.

• It forms a barrier to keep different body systems separated from each other. In these tissues cells are tightly packed and form a continuous sheet.
• There is almost no intercellular space between them. They have a very small amount of cementing material between them.
• The epithelium is separated from underlying tissue by an extracellular fibrous basement membrane containing collagen.
• Based on the shape of the cells and their arrangement, epithelial tissues are further classified as follows:

Squamous Epithelium

Squamous epithelial tissue constitutes the skin that protects the body. It is further categorized as:

Simple Squamous Epithelium

• It is single-layered and closely fitted. The cells are very thin and flat and appear as tiles over a floor.
• It forms a delicate lining of blood vessels and lung alveoli, where substance transport occurs through a selectively permeable membrane.
• It also covers the esophagus and the lining of the mouth.

Stratified Squamous Epithelium

• It is found on the outer side of the skin as it is highly resistant to mechanical injury and is water-proof.
• Cells are arranged in many layers to prevent their wear and tear

Cuboidal Epithelium

• It is made up of cube-shaped cells, which have round nuclei.
• It forms the lining of kidney tubules and ducts of salivary glands, where it provides mechanical support. It also forms the germinal epithelium of gonads.
• It also helps in absorption, excretion, and secretion.

Columnar Epithelium

• It is made up of tall, pillar-like cells, with elongated nuclei.
• It is usually found in the inner lining of the intestine, where absorption and secretion occur.
• It facilitates movement across the epithelial barrier.

Ciliated Columnar Epithelium

• When columnar epithelial cells possess cilia (hair-like projections), it is called ciliated columnar epithelium.
• The cilia can move. Their movement pushes substances like mucus forward.
• It is found in the respiratory tract and also lines oviducts, sperm ducts, kidney tubules, etc.

Glandular Epithelium

• Gland cells secrete substances at the epithelial surface.
• Sometimes, a portion of epithelial tissue folds inward. This results in the formation of a multicellular gland. Its tissue is called glandular epithelium.

Functions of Epithelial Tissue

1. It protects the underlying cells from drying, injury, infections, and also from harmful effects of chemicals.
2. It plays a vital role in regulating the exchange of materials between the body and the external environment and between different body parts.
3. It helps in the absorption of water and nutrients and in the diffusion of gases.
4. It helps in the elimination of waste products from the body.

2. Connective Tissue

• This tissue is specialized to connect various body organs. For example., it connects two or more bones muscles to bones, binds different tissues together, and also gives support to various parts of the body.
• The cells of connective tissue are loosely packed, living, and embedded in an intercellular matrix that may either be jelly-like, fluid, dense, or rigid. The nature of the matrix differs in concordance with the function of the particular connective tissue.

Various types of connective tissues are:

Connective Tissue Blood

• It is a fluid connective tissue that links different parts of the body. It helps to maintain the continuity of the body. It contains a fluid matrix called plasma and blood cells such as RBCs (Red Blood Corpuscles or Cells), WBCs (White Blood Corpuscles), and platelets suspended in it.
• Plasma also contains proteins, salts, and hormones. Blood transports nutrients, gases, hormones, and vitamins to various tissues of the body. It carries excretory products from tissues to excretory organs. It also conducts heat and regulates body temperature.

Properties shown by different blood cells in the body are as follows:

• RBCs Help in the transport of respiratory gases, oxygen, and carbon dioxide with the help of hemoglobin to and from the various parts of our body. The average lifespan of RBCs is 120 days.
• WBCs Also called leucocytes, fight diseases by producing antibodies.
• Blood platelets Also called thrombocytes, help in the clotting of blood.

Connective Tissue Bone

• It is a very strong and non-flexible tissue. It is porous, highly vascular, mineralized, hard, and rigid. Its matrix is made up of proteins and is rich in salts of calcium and phosphorus.
• It forms the framework that supports the body. It also anchors the muscles and supports the main organs.

Ligaments: They connect one bone to another bone. A ligament is very elastic and has considerable strength. It contains very little matrix. Ligaments strengthen joints and permit normal movement. Their overstretching leads to sprain.

Tendons: They are strong and inelastic structures, which join skeletal muscles to bones. These are composed of white fibrous tissues with limited flexibility, but great strength.

Cartilage: It is a specialized connective tissue having widely spaced cells. It has a solid matrix called chondrin which is composed of proteins and sugars.

Cartilage provides smoothness to the bone surfaces at the joints. It is present in the nose, ear, trachea, and larynx. We can fold the cartilage of the ears, but we cannot bend the bones in our arms.

Areolar Tissue: It is a supporting and packing tissue found between the organs lying in the body cavity. It is located between skin and muscles, around blood vessels and nerves, and in the bone marrow.

It is a loose and cellular tissue. It fills the space inside the organs and supports internal organs. It helps in the repair of tissues.

Adipose Tissue: It serves as a fat reservoir, and keeps visceral organs in position. It acts as an insulator due to the storage of fats. It is located below the skin in between the internal organs.

3. Muscular Tissue

It consists of elongated cells, called muscle fibers. This tissue is responsible for the movement in our body. It contains a special type of proteins called contractile proteins which causes the movement of muscles by contraction and relaxation. Different types of muscular tissues are given below:

Striated Muscles

• The muscles present in our limbs which move or stop as per our will, are called striated muscles. These are also called as voluntary muscles as we can move them by conscious will. Mostly these are attached to bones and help in body movement, e.g, muscles of limbs.
• Hence, they are also called as skeletal muscles. The cells constituting their muscles are long, cylindrical, unbranched, and multinucleate (having many nuclei). Under microscope, striated muscles show alternate light and dark bands or striations.
  Thus, they are also known as striated muscles.
  Nuclei

Unstriated Muscles Or Smooth Muscles

• The muscles which we cannot move as per our will are called unstriated or smooth muscles.
• They are also called involuntary muscles. For example, movement of food in the alimentary canal, contraction, and relaxation of blood vessels, iris of the eye, and muscles present in ureters and in bronchi of the lungs.
• The cells constituting these muscles are long, with pointed ends (spindle-shaped) and uninucleate (single nucleus).
• These muscles do not show any dark or light bands. Hence, they are also called unstriated muscles.

Cardiac Muscles: These are involuntary muscles present only in our heart. They perform rhythmic contraction and relaxation throughout life.

The cells constituting cardiac muscles are cylindrical, uninucleate, and branched. Cardiac muscles have stripes of light and dark bands.

4. Nervous Tissue

• The tissue that receives a stimulus and transmits it from one part of the tissue to another, are nervous tissue.
• The cells that constitute nervous tissue are called nerve cells or neurons. These are highly specialized for receiving a stimulus and then transmitting it very rapidly from one place to another within the body itself.
• The brain, spinal cord, and nerves are composed of nervous tissues.

An individual nerve cell or a neuron may be upto a metre long and is composed of three major parts:

1. Cell body It consists of cytoplasm, nucleus, and cell membrane.
2. An axon is a single long conducting fiber extending from the neuron. It transmits impulses away from the cell body.
3. Dendrites These are short-branched fibers of neurons, which receive nerve impulses. Nucleus

Note: Synapse is a region of the union of axons of one neuron with the dendrite of the next. This allows the transfer of nerve impulses generated to and fro in the body.

• Many nerve fibers bound together by connective tissue make up a nerve. Nerve impulse allows us to move our muscles according to our will.
• Combination of nerve and muscle tissue in animals is of fundamental importance as causes rapid movement in response to stimuli.

Differences Between Plant Tissues And Animal Tissues

Activity 1

Objective

Apical meristem causes growth in the length of the plant.

Time Required

Five days or a week.

Materials Required

Two jars of the same size, two onion bulbs, water, scissors, and a measuring scale.

Procedure

1. Take two glass jars and fill them with water.
2. Now, take two onion bulbs and place one in each jar.
3. Observe the growth of roots in both the bulbs for a few days.
4. Measure the length of roots on days 1, 2, and 3.
5. On day 4, cut the root tips of the onion bulb in jar 2 by about 1 cm. After this, observe the growth of roots in both the jars and measure their lengths each day for five more days, and record the observations in a table.

Observation

Result/Conclusion

When root tips are removed from the onion of jar 2, it is observed that the growth of roots is more in the onion of jar 1, because apical meristem which is responsible for the increase in the length of roots is present only at the growing tips of roots. Thus, when we remove root tips, the apical meristem is also lost and the roots stop growing.

Question 1. What is meristematic tissue?
Answer: Meristematic tissues are those tissues that contain cells that are capable of dividing and forming new cells throughout their life.

Question 2. What will happen if the apical meristem is damaged or cut?
Answer: When apical meristem is damaged or cut, the growth or length of the growing tissue will be retarded.

Question 3. Why did the roots of the bulb of jar 2 stop growing after the fourth day?
Answer: Due to the removal of the apical portion of the roots in jar 2, the growth stops.

Question 4. What is the function of meristematic tissue?
Answer: It is responsible for increasing the length of the plant.

Question 5. Which of the two onions has longer roots? Why?
Answer: The onion in jar 1 has longer roots, as the growth of roots continues in it. This is because apical meristem which is responsible for the increase in the length of roots is present at the growing tips of roots.

Activity 2

Objective

To understand the structure, location, and arrangement of various types of cells of permanent tissue in plants.

Time Required

30-45 minutes.

Materials Required

A plant stem, blade, safranin, glycerine, coverslip, and a microscope.

Procedure

1. Take a plant stem and with the help of your teacher, cut it into very thin slices or sections.
2. Now, stain the slices with safranin. Place one needy cut section on a slide and put a drop of glycerine.
3. Cover the cut section of the plant stem with a cover slip and observe the arrangement of cells under a microscope.
4. Draw the figure you observed and label the parts.

Observation

The section of the stem appears as drawn below:

Result/Conclusion

The section of the stem depicts that various types of cells are arranged in specific ways. All cells are different in structure. The epidermis forms the outermost layer followed by the cortex. Vascular bundles are present encircled by the cortex.

Question 1. Which chemical is used to stain the plant sections in this activity?
Answer: Safranin is used to stain the plant sections in this activity.

Question 2. What is the role of glycerine in the above activity?
Answer:

Glycerine forms a protective covering on the plant section. This forms a barrier between the air and the plant section. Thus, the mount does not get dry and remains fresh for some time.

Question 3. Name the tissues that comprise vascular bundles.
Answer: The xylem and phloem comprise vascular bundles.

Question 4. What is the function of a vascular bundle?
Answer: The vascular bundle acts as a conducting tissue, i.e. they conduct water and food to all parts of the plant.

Activity 3

Objective

To understand the role of epidermis in plants. Or To understand the location and function of stomata.

Time Required

30-45 minutes.

Materials Required

Leaf of Rhoeo, petri dish, water, safranin, slides, coverslip, and microscope.

Procedure

1. Take a freshly plucked leaf of the Rhoeo plant.
2. Stretch and break it by applying pressure.
3. While breaking it, keep it stretched gently so that some peel or skin projects out from the cut.
4. Remove this peel and put it in a petri dish filled with water.
5. Add a few drops of safranin to it.
6. Wait for a couple of minutes and then transfer it onto a slide. Gently place a cover slip over it.
7. Observe it under a microscope.

Observation

The cells appear as shown below:

Result/Conclusion

1. The section shows the outermost layer called the epidermis. It is a single layer of cells with stomata embedded in it. It protects against water loss and mechanical injury.
2. Stomata are small pores present here and there in the epidermis of the leaf. These are enclosed by two kidney-shaped cells called guard cells.

Question 1. Name the outermost layer of cells in old plants.
Answer: The outermost layer of cells in old plants is called epidermis.

Question 2. Name the cells that enclose the stomata.
Answer: Stomata are enclosed by two kidney-shaped cells called guard cells.

Question 3. State the location of stomata in plants.
Answer: Stomata are small pores present in the epidermis of the leaf.

Tissues Summary

Tissues are group of cells that are similar in structure and work together to achieve a particular function, for example., blood, phloem, and muscles.

• Tissues are broadly classified into plant and animal tissues.
• On the basis of dividing capacity, plant tissues can be classified into two fundamental types, i.e. meristematic tissue and permanent tissue.
• Meristematic tissues divide actively throughout life. They are found in growing regions of plants like root and shoot tips. These tissues are mainly of three types, i.e. apical meristem, intercalary meristem, and lateral meristem.
• Apical meristem are present at the growing tips of stems and roots. These are helpful in increasing the length of the stems and the roots. Intercalary meristems are present at the base of the leaves or internodes of the twigs.
• Lateral meristems are present on the lateral sides of stems and roots. It helps in increasing the girth of the stem or root.
• Permanent tissue is formed from the cells of meristematic tissue when they lose their ability to divide and have attained a permanent shape, size, and function by the process called differentiation. These are mainly of two types, i.e. simple and complex permanent tissue. “
• Parenchyma tissue forms the basic packing tissue. These tissues are present in the cortex, the pith of stem, roots and also in the mesophyll of leaves.
• Collenchyma cells are living, elongated, and irregularly thickened at the corners, generally found in leaf stocks below the epidermis. These provide mechanical support and elasticity to plant tissues.
• Sclerenchyma tissue is present in stems around vascular bundles, in veins of leaves, and in hard covering of seeds and nuts. These provide strength and enable the plant to bear various stresses.
• Complex permanent tissues are made up of more than one type of cell.
• The xylem is a vascular and mechanical conducting tissue. It is responsible for transport of food from roots to other parts of a plant.
• Phloem is a vascular tissue, responsible for transport of food from leaves to other parts of the plant.
• Animal tissues are classified on the basis of the functions they perform, i.e. epithelial, connective, muscular, and nervous tissue. Epithelial tissue is a protective tissue. It is tightly packed. It is present in the skin and lining of the mouth.
• Squamous epithelium cells are flat, it form the delicate lining of the esophagus and mouth. It may be several layers thick as in skin, known as stratified squamous epithelium.
• Columnar epithelium cells are tall, pillar-like, and have elongated nuclei. It is usually found in the inner lining of the intestine, where absorption and secretion occur.
• Ciliated columnar epithelium cells have cilia, hair-like projections found on the outer surface of columnar epithelial cells found in the trachea, bronchi, etc.
• Glandular epithelium cells acquire additional specialization known as gland cells that can secrete substances at the epithelial surface.
• Connective tissue connects various body organs, i.e. blood, bone, tendon, areolar, adipose, cartilage, etc.
• Muscular tissue consists of elongated cells and is responsible for movement. Striated muscles are mostly attached to bones and help in body movement. Unstriated muscles cannot be moved according to will. Cardiac muscles present in the heart, show rhythmic contraction and relaxation throughout life.
• Nervous tissue enables the body to respond to stimuli. They transmit stimuli from one place to another within the body, through neurons.
• Neuron forms the functional unit of nervous tissue.

 Class 9 Science Chapter 7  Motion Very Short Answer Type Questions

Question 1. What is the importance of reference points?
Answer:

The reference point is important because it states the position of the object correctly, as motion is relative.

Question 2. Is it possible that the train in which you are sitting appears to move while it is at rest?
Answer:

Yes, if another train is moving in an adjacent line to the train we are sitting on, then it seems that our train is moving in the opposite direction.

Question 3. When is an object in motion considered to be a point object?
Answer:

If the object in motion has a dimension very small as compared to the distance travelled, then the object is considered to be a point object.

The odometer measures the displacement of the vehicle. Correct this statement.

No, the Odometer measures the distance covered by a vehicle.

Question 5. Suppose Ram throws a ball vertically upwards from a position P above the ground. It rises to the highest point Q and returns to the same point P. What is the net displacement and distance travelled by the ball?
Answer:

The net displacement is zero. The distance travelled by the ball is twice the distance between positions P and Q.

Read and Learn More Class 9 Science Solutions

Question 6. A particle is moving in a circular path of radius r. What will be the displacement after half a circle?

Answer:

So After half the circle, the particle will reach the diametrically opposite point, i.e. from point A to B. Displacement after half circle = AB = r + r = 2r

Question 7. A particle moves 3m North, 4m East and finally 6m South. Calculate the displacement.
Answer:

According to the question,

Now, by applying Pythagoras theorem in \(\triangle\) ODC,

|O C| =\(\sqrt{O D^2+D C^2}\)

=\(\sqrt{A B^2+(B C-B D)^2}\)

=\(\sqrt{4^2+(6-3)^2}\)

=\(\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5 \mathrm{~m}\)

Question 8. In which condition, will the magnitude of the displacement be equal to the distance travelled by an object?
Answer:

If an object moves in a straight line from one point to another, then the magnitude of displacement and distance will be equal.

Question 9. Give one condition for the motion of a body to be uniform.
Answer:

The body should travel along a straight line path in one direction.

Question 10. Does the speedometer of a car measure its average speed?
Answer:

No, the speedometer of a car does not measure its average speed. It measures only instantaneous speed.

Question 11. What do you mean by an acceleration of 5 m/s2?
Answer:

The velocity of the body increases by 5 m/s after every second.

Question 12. A physical quantity measured is -10 m/s. Is it a speed or a velocity?
Answer:

It is a velocity because velocity can be positive, zero or negative while speed is always positive.

Question 13. A person while travelling with his car found that the magnitude of average velocity was equal to its average speed. Identify the path formed by him.
Answer:

If a person moves in a straight line, then their average velocity will be equal to the average speed.

Question 14. What is the numerical ratio of average velocity to average speed of an object when it is moving along a straight path?
Answer:

The numerical ratio of the average velocity and average speed of an object when it is moving along a straight path is 1: 1.

Question 15. A person standing at A goes to B by following any of the paths 1, 2 and 3. Which path can we use to find the average velocity?

Can a particle be accelerated?

1. Is its speed constant?
2. Is its velocity constant?

Answer: Path-1

1. Yes, a particle moving with constant speed can be accelerated.
2. No

Question 16. When is the acceleration taken as negative?
Answer:

Acceleration is taken as negative if it is in the opposite direction of motion of the object.

Question 17. A body accelerates uniformly so that its velocity changes from u to v. What will be its average velocity?
Answer:

The average velocity is \(\frac{u+v}{2}\).

Question 18. What kind of motion of a body is represented by the graphs given below?

Answer:

1. The object is at rest
2. Non-uniform motion.

Question 19. The distance-time graph for the motion of Ram and Shyam is shown alongside. Which of them has greater acceleration? Justify your answer.

Answer:

Both Ram and Shyam have zero acceleration as they are moving with constant velocity. Since the distance-time graph is a straight line.

Question 20. Figure shows the displacement-time graph of four children A, B, C and D. Which child has the highest velocity?

Answer:

As we know, velocity = slope of the displacement-time graph. Here, the slope of child B is greater than all other children. Hence, child B has the highest velocity.

Question 21. From the given v-t graph (see figure), what can be inferred?

Solution:

From the given v-t graph, it is clear that the velocity of an object does not change with time, i.e. the object is in a uniform motion.

Question 22. How do we measure the magnitude of the displacement from a v-t curve?
Answer:

By measuring the area under the v-t curve along the time axis.

Question 23. Give one example of uniform circular motion.
Answer:

A satellite in a circular orbit around the Earth is an example of uniform circular motion.

Question 24. Is the motion of a body uniform or accelerated, if it goes round the sun with constant speed in a circular orbit?
Answer:

In this case, the motion of a body is accelerated as its velocity changes due to the change in direction.

Question 25. Why is the motion of a circulating fan non-uniform?
Answer:

The motion of a circular fan is non-uniform because the direction of motion changes at every point.

Question 26. Give an expression for the speed of an athlete if he takes time t to go around a circular track, of radius r.
Answer:

v=\(\frac{2 \pi r}{t}\)

Question 27. Find the angular velocity of a satellite which revolves in a circular orbit of radius 35000 km and completes one round in 12 h.
Answer:

Angular velocity =\(\frac{\text { Angle subtended at centre }}{\text { Time taken }}\)

= \(\frac{2 \pi}{12}\)

= \(\frac{\pi}{6} \mathrm{rad} / \mathrm{h}\)

 Class 9 Science Chapter 7  Motion Short Answer Type Questions

Question 1. (1) What is motion?

(2) State the types of motion.

(3) Write the unit of acceleration.

Answer:

(1) If a body changes its position concerning time and its surroundings, then it is said to be in motion.

(2) Generally, there are two types of motion:

(a) Uniform motion (b) Non-uniform motion

(3) The SI unit of acceleration is m/s².

Question 2. A cow and a bird both A 15 m B travelled from point A to point B. The cow travelled in a straight line but the bird travelled along the curved path as shown alongside:

1.  What is the distance travelled by the cow?
2. What is the distance travelled by the bird?
3.  Which one of them has more displacement?

Answer:

1. Distance travelled by the cow = 15 m
2. Distance travelled by the bird = 30 m
3. Displacement of the cow = 15 m

Displacement of the bird = 15 m

So, cows and birds have the same displacement.

Question 3. Give an example of a body which may appear to be moving for one person and stationary for the other.
Answer:

For the passengers in a moving bus, the trees, buildings as well as the people on the roadside appear to be moving backwards. Similarly, a person standing on the roadside observes that the bus (along with its passengers) is moving in a forward direction.

But at the same time, each passenger in a moving bus or train observes that his fellow passengers are at rest. Thus, we can tell that motion is relative.

Question 4. How can we describe the location of an object?
Answer:

To describe the location of an object we need to specify a reference point called origin, e.g. suppose that a library in a city is 2 km North of railway station. We have specified the position of the library with respect to the railway station, i.e. in this case, the railway station acts as the reference point.

Question 5. Differentiate between distance and displacement.
Answer:

Differences between distance and displacement are as given below:

Question 6. Express average velocity when the velocity of a body changes at a non-uniform rate and a uniform rate.
Answer:

When the velocity of a body changes at a non-uniform rate, then its average velocity is calculated by dividing the net displacement covered by the total time taken.

Average velocity =\(\frac{\text { Net displacement }}{\text { Total time taken }}\)

In case, the velocity of a body changes at a uniform rate, the average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time, i.e.

Average velocity= \(\frac{\text { Initial velocity }+ \text { Final velocity }}{2}\)

Question 7. Give one example each of type of motion when

1. acceleration is in the direction of motion.
2.  acceleration is against the direction of motion.

Answer:

1. A train starts moving from a station.
2. Brakes applied to a moving car.

Question 8. Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.
Answer:

When a stone is thrown vertically upwards, then it has some initial velocity (u). As the stone goes up its velocity decreases (since it is moving against gravity) and at the highest point, i.e. maximum height,) its velocity becomes zero.

Let the stone take time the t second to reach the highest point. After that stone begins to fall (with zero initial velocity) and its velocity goes on increasing (since it is moving with gravity) and it reaches its initial point of projection with the velocity (—u) at the same time (with which it was thrown). So,

Question 9. A biker rides 700 m north, 300 m east, 400 m north, 600 m west, 1200 m south, 300 m east and finally 100 m north. Draw the path of motion of the biker. What distance did he cover? What was his displacement?
Answer:

(2) Distance travelled by the biker

= OA + AB + BC + CD + DE + EF + FO = 700 + 300 + 400 +600 + 1200 + 300 + 100 = 3600 m

(3) Displacement of the biker = 0.

Question 10. An object P is moving with a constant velocity for 5 min. Another object O is moving with changing velocity for 5 min. Out of these two objects, which one has acceleration? Explain.
Answer:

Acceleration =\(\frac{\text { Change in velocity }}{\text { Time taken }}\)

Since the velocity of object P is not changing or the change in velocity of the object is gro, therefore object-t Phas no acceleration. On the other hand, there is i change in velocity of the object Q, so it has acceleration.

Question 11. A car covers the first 50 km at a uniform velocity of 45 km/h and the next 50 km at 90 km/h. Find the average velocity of the car.
Answer:

Total displacement = 50 + 50 = 100 km

Total time =\(\frac{50}{45}+\frac{50}{90} =50\left(\frac{1}{30}\right)=\frac{5}{3} \mathrm{~h}\)

∴ \({\left[\text { Time }=\frac{\text { Displacement }}{\text { Velocity }}\right] }\)

Average velocity =\(\frac{\text { Total displacement }}{\text { Time taken }}=\frac{100}{5 / 3}\)

= 60 km/h

The average velocity of the car is 60 km/h.

Question 12. A girl moves at a speed of 6 km/h for two hours and with a speed of 4 km/h for the next three hours. Find the average speed of the girl and the total distance moved.
Answer:

Here, \(v_1=6 \mathrm{~km} / \mathrm{h}, t_1=2 \mathrm{~h}, v_2=4 \mathrm{~km} / \mathrm{h}, t_2=3 \mathrm{~h}\)

Total distance =\(v_1 t_1+v_2 t_2=6 \times 2+4 \times 3=24 \mathrm{~km}\)

Average speed =\(\frac{\text { Total distance }}{\text { Total time }}\)

= \(\frac{24}{2+3}=4.8 \mathrm{~km} / \mathrm{h}\)

Question 13. A person travelling in a bus noted the timings and the corresponding distances as indicated on the km stones.

1. Name this type of table.
2. What conclusion do you draw from this data?

Answer:

1.  It is a distance-time table.
2. According to the given table, we conclude that the given motion is uniform motion as in every 15 the body covers 10 km.

Question 14. Represent the given data graphically.

 

Answer:

From the given distance-time table, the graph can be drawn as

Question 15. Two trains A and B start moving at the same time. The distances travelled by them in given internals of time are shown below. State which train has uniform motion and which train has non-uniform motion.
Answer:

Since, train A travels equal distances in equal intervals of time, i.e. every 15 min, the motion of train A is uniform motion.

On the other hand, train B travels unequal distances in equal intervals of time, i.e. every 15 min, so the motion of train B is non-uniform motion.

Question 16. Make a velocity-time graph from the following displacement-time graph:

Answer:

From the graph,

velocity after 2 \(\mathrm{~s}=\frac{10}{2}=5 \mathrm{~m} / \mathrm{s}\)

Velocity after 4 \(\mathrm{~s}=\frac{20}{4}=5 \mathrm{~m} / \mathrm{s}\) and velocity after 6 \(\mathrm{~s}=\frac{30}{6}=5 \mathrm{~m} / \mathrm{s}\)

For Example. Velocity is constant, acceleration = 0

So, the velocity-time graph will be a straight line parallel to the time axis as shown below:

Question 17. Find the displacement of a body whose velocity-time graph is shown as:

Answer: 

Displacement = Area under v-t curve (i.e. the area of trapezium ABCD )

= \(\frac{1}{2}(\text { sum of parallel sides }) \times \text { height }\)

= \(\frac{1}{2}(A D+B C) \times C F=\frac{1}{2}(10+6) \times 10=80 \mathrm{~m}\)

Question 18. The graph given below is the distance-time graph of an object.

1. Find the speed of the object during the first four seconds of its journey and how long was it stationary.
2. Does it represent a real situation? Justify your answer.

Answer:

(1) Speed during first 4s = Slope of OP

⇒ \(\tan \theta=\frac{75}{4}=18.75 \mathrm{~m} / \mathrm{s}\)

From 4 s to 14 s, the distance does not change with time, so it is stationary for 10 s.

(2) No, it is not a real situation because distance travelled cannot decrease with time. Moreover, time cannot flow backwards as shown by the QR part of the graph.

Question 19. The velocity-time graph (see figure) shows the motion of a cyclist. Find

1. its acceleration,
2. its velocity and the distance covered by the cyclist in 15 seconds.

Answer:

1. From the graph, it is clear that velocity does not change with time, i.e. acceleration is zero.
2. Again from the graph, we can see that there is no change in the velocity with time, so velocity after 15 seconds will remain the same as 20 m/s.

Distance covered in 15 s = Velocity x Time = 20×15 = 300 m

Question 20. Find the total displacement of the body from the following graph

Answer:

Total displacement = Sum of areas of (\(\triangle\) A B F+ Rectangle BCGF + \(\triangle\) CDE +{ Rectangle CEIG)

=\([\frac{1}{2}(A F \times F B)+(B C \times C G)\).

⇒ \(\left.\quad+\frac{1}{2}(C E \times D J)+(C E \times E I)\right]\)

= \(\left[\frac{1}{2}(2 \times 5)+(4 \times 5)+\frac{1}{2}(4 \times 5)+(5 \times 4)\right]\)

=55 m

Question 21. The speed-time graphs of the two cars are represented by P and Q as shown below:

1. Find the difference in the distance travelled by the two cars (in m) after 4 s.
2.  Do they ever move at the same speed? If, so when?

Answer:

(1) Total distance travelled by P (area of A POR)

= 1/2 x4x 6²

= 12 m

Total distance travelled by Q (area of RQSO)

= 4×3 = 12m

Difference in the distance = 12-12=0

(2) Yes, they move with the same velocity equal to 3 m/s at time 2 s. It can be determined, by viewing the intersection point of two v-t graphs.

Question 22. Study the v-t table and answer the following:

1. What is the value of an in 0 to 15 s?
2. Predict the nature of acceleration in different parts of the graph.

Answer:

(1) a =\(\frac{\text { Change in velocity }}{\text { Time interval }}\)

=\(\frac{20-0}{15-0}=\frac{20}{15}=\frac{4}{3}\)

=1.33 \(\mathrm{~m} / \mathrm{s}^2\)

(2) The body is in uniform motion from 0 to 15 s, then non-uniform related motion from 15 s to 30 s. Here, OA = uniformly accelerated AB, BC, CD = uniformly related.

Question 23. What is the difference between uniform motion in a straight line and circular motion?
Answer:

Question 24. Study the following distance-time graph and compare the velocity of the two objects

Answer:

Since A and B have parallel slopes for distance-time graphs, their velocities are equal. However, A starts from the 10 m mark and B starts from the origin, (i.e. A is ahead of B by 10 m)

Question 25. What do the graphs (1) and (2) shown here indicate?

Answer:

Graph (1) indicates that a body has a certain initial speed and then its speed decreases to zero.

Graph (2) shows that a body first retards and then accelerates non-uniformly.

Question 26. Can an object be accelerated, if it is moving with constant speed? Justify your answer with an example.

Or Can a body have a constant speed and still be accelerating? Give an example.

Or Explain, how is it possible for an object to move with a constant speed but with uniform acceleration.

Answer:

An object moving with a constant speed can be accelerated, if its direction of motion changes, For Example. An object moving with a constant speed in a circular path has acceleration because its direction of motion changes continuously. Its acceleration is always directed toward the centre of the circular path.

Question 27. A bridge is 500 m long. A 100 m long train crosses the bridge at a speed of 30 m/s. Find the time taken by train to cross it.
Answer:

Total length of path covered by train = 500 + 100 =600 m

Speed of train = 30 m/s

Time taken by train to cover the bridge

Time taken by train to cover the bridge

= \(\frac{\text { Distance }}{\text { Speed }}=\frac{600}{30}=20 \mathrm{~s}\)

Question 28. When two bodies move uniformly towards each other, then the distance between them decreases by 8 m/s. If both the bodies move in the same direction at the same speeds, then the distance between them increases by 4 m/s. What are the speeds of two bodies?
Answer:

Let u and v be the speeds of two bodies. According to the question, when two bodies move towards each other the distance between them decreases by u + v = 8 and when they are moving in the same direction,

then the distance between them increased by u – v = 4

Solving the above equations, we get 2u=12 ⇒ u — G m/s

and v =8 —u =8 -6 = 2 m/s The speed of two bodies are 6 m/s and 2 m/s.

Question 29. A fan rotates at 100 rpm. Find its angular velocity and linear velocity, if the tip of its blades is 0.20 m away from the axis of rotation.
Answer:

Given, angular velocity, \(\omega\)=100 rpm=100 \(\times \frac{2 \pi}{60}\)

= \(\frac{10 \pi}{3} \mathrm{rad} / \mathrm{s}\)

Radius, r=0.20 \(\mathrm{~m}\)

Linear velocity =\(\text { Radius } \times \text { Angular velocity }\)

v=r \(\omega\)

Linear velocity =0.20 \(\times \frac{10 \pi}{3}\)

= \(\frac{2 \pi}{3} \mathrm{~m} / \mathrm{s}\)

Question 30. An object is moving with a uniform speed in a circle of radius r. Calculate the distance and displacement,

1. when it completes half the circle,
2. when it completes full circle,
3. what type of motion does the object possess?

Answer:

(1) When an object completes half the circle, then the distance travelled by an object

= \(\frac{1}{2} \times \text { circumference of a circle }\)

= \(\frac{1}{2} \times 2 \pi r=\pi r\)

(2) When an object completes a full circle, then the distance travelled by an object = \(2 \pi\) r

Displacement travelled by an object = 0 [since initial and final positions are the same]

(3) The direction of motion of an object changes continuously in the circle, hence its velocity changes and its motion is accelerated motion.

Question 31. An athlete runs on a circular track of a radius of 50 m from the starting point A with a constant speed. It takes 50 seconds for him to reach point B which is at the other end of the diameter drawn from A. Calculate the distance covered, displacement and speed.
Answer:

Distance covered by the athlete =\(\pi r\)

= \(\frac{22}{7} \times 50=157.14 \mathrm{~m}\)

Displacement of the athlete

=2 r=2 \(\times 50=100 \mathrm{~m}\)

The speed of the athlete is given by

= \(\frac{\text { Distance }}{\text { Time }}\)

= \(\frac{\pi r}{50}\)

= \(\frac{22}{7} \times \frac{50}{50}=3.14 \mathrm{~m} / \mathrm{s}\)

 Class 9 Science Chapter 7  Motion Long Answer Type Questions

Question 1. Give one example of each of the following situations:

1. Uniformly accelerated motion.
2. Motion with uniform retardation.
3. Accelerated motion with a uniform magnitude of velocity.
4. Motion in a direction with acceleration in a perpendicular direction.
5. Motion in which the v-t graph is a horizontal line parallel to the X-axis.

Answer:

1. An object dropped down from a height towards the surface of the earth.
2. The object thrown up with a velocity retards uniformly.
3. Uniform circular motion, say planetary motion.
4. A bullet fired horizontally from a rifle, has acceleration in a downward direction (due to gravity).
5. A car moving with uniform velocity along a straight line

Question 2. The position-time graphs of two objects A and B in three different situations for a particular duration are shown below:

1. In which situation they are moving in opposite directions?
2. Is the velocity of object A positive or negative in the situation (2)?
3. Are they crossing each other in any situation (s)? If so, why?In which situation the distance between them will remain the same?

Answer:

1. In graph (3), the distance between them will be the same.
2. In the graph (1), they are moving in opposite directions.
3. The velocity of object A is positive in situation (1) because the slope of the s-t graph is positive.
4. Yes, in situations (1) and (2) they are crossing each other because in these two cases, the s-t graph of A and B intersect each other.

Question 3. On a 100 km track, a train travels the first 30 km at a uniform speed of 30 km/h. How fast must the train travel the next 70 km, so as to average 40 km/h for the entire trip?
Answer: 

Total time, t=\(t_1+t_2=\frac{30 \mathrm{~km}}{30 \mathrm{~km} / \mathrm{h}}+\frac{70 \mathrm{~km}}{x(\text { say })}\)

t=1+\(\frac{70}{x}=\left(\frac{x+70}{x}\right) \mathrm{h}\)

Now, average speed, v=\(\frac{s}{t}=40 \mathrm{~km} / \mathrm{h}\)

According to question, \(\frac{100}{\underline{x+70}}=40\)

10=4\(\left(\frac{x+70}{x}\right)^x\)

10 x=4 x+280

6 x=280

x=46.67 \(\mathrm{~km} / \mathrm{h}\)

The speed for the next 70 km to maintain an average of 40 km/h is 46.67 km/h.

Question 4. Study the speed-time graph of a body given here and answer the following questions:

1. What type of motion is represented by OA?
2. What type of motion is represented by AB?
3. What type of motion is represented by BC?
4. Find out the acceleration of the body.
5. Calculate the retardation of the body.
6. Find out the distance travelled by the body from A to B.

Answer:

1. OA is a straight line graph between speed and time and it is sloping upward from O to A. Here, OA represents uniform acceleration.

2. AB is a straight line graph between speed and time which is parallel to the time axis (X-axis). So, AB represents uniform speed. There is no acceleration from A to B.

3. BC is a straight line graph between speed and time which is sloping downwards from B to C. Therefore, BC represents uniform retardation or negative acceleration.

4. Acceleration of the body as we see from graph line OA represents it.

So, the slope of the velocity-time graph OA will give the acceleration of the body.

Thus, acceleration = slope of line OA=\(\frac{A D}{O D}\)

We have, AD=6 \(\mathrm{~m} / \mathrm{s} and OD=4 \mathrm{~s}\)

So, acceleration =\(\frac{6 \mathrm{~m} / \mathrm{s}}{4 \mathrm{~s}}=1.5 \mathrm{~m} / \mathrm{s}^2\)

5. The slope of line graph BC represents the retardation of the body.

So, retardation = slope of line BC = \(\frac{B E}{E C}\)

We have, BE= 6 \(\mathrm{~m} / \mathrm{s}\),

E C=16-10=6 \(\mathrm{~s}\)

Retardation =\(\frac{6 \mathrm{~m} / \mathrm{s}}{6 \mathrm{~s}}=1 \mathrm{~m} / \mathrm{s}^2\)

6. The distance travelled by the body is equal to the area enclosed between the speed-time graph and the time axis.

Distance travelled from A area under the line AB and the time axis = area of rectangle DABE — DA x DE.

Here, DA = 6 m/s and DE = 10 — 4 = 6 s Distance travelled from A to B=6×6 = 36 m

Question 5. The v-t graph of an object is shown below:

1. Identify the type of motion by lines OA and BC.
2. Velocity at t = 8s.
3. Calculate acceleration
   1.  between 3rd and 10th second.
   2.  last 2s.

Answer:

1. OA represents uniformly accelerated motion, and BC represents uniformly retarted motion.

2.  At t =8 s, the velocity of the object is 4.6 m/s.

3. (1) Between 3rd and 10th second, a – 0.

(2) During last 2s,

Acceleration, a=\(\frac{\text { change in velocity }}{\text { time interval }}\)

=\(\frac{0-4.6}{12-10}=\frac{-4.6}{2}=-\frac{4.6}{2}=-2.3 \mathrm{~m} / \mathrm{s}^2\)

Question 6. An insect moves along a circular path radius of 10 cm with a constant speed. It takes 1 minute to move from a point on the path to the diametrically opposite point, find

1. the distance covered,
2. the speed,
3.  the displacement and
4. the average velocity. Suppose the insect was at A initially and it moves along ACB to reach the diametrically opposite point B in 1 min.

Answer:

1. The distance moved in 1 min = nr = 3.14×10 =31.4 cm
2. Speed =\(\frac{\text { Distance }}{\text { Time }}=\frac{31.4}{1}=31.4 \mathrm{~cm} / \mathrm{min}\)
3.  Displacement, AB = 2r = 2 x 10 = 20cm
4.  Average velocity,

• • \(v_{\mathrm{av}} =\frac{\text { Displacement }}{\text { Time }}\)
  • =\(\frac{20 \mathrm{~cm}}{1 \mathrm{~min}}=20 \mathrm{~cm} / \mathrm{min}\)

Question 7. The driver of train A travelling at a speed of 54 km/h applies brakes and retards the train uniformly. The train stops in 5 s. Another train B is travelling on the parallel with a speed of 36 km/h.

Its driver applies the brakes and the train retards uniformly, train B stops in 10 s. Plot speed-time graphs for both trains on the same axis. Which of the trains travelled farther after the brakes were applied?

Answer:

For train A, the initial velocity,

u=54 \(\mathrm{~km} / \mathrm{h}=54 \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s}\)

Final velocity, v=0 and time, t=5 s

For train B, u=36 \(\mathrm{~km} / \mathrm{h}=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{s}\)

⇒ \(\nu\)=0, t=10 \(\mathrm{~s}\)

Speed-time graphs for trains A and B are shown

Distance travelled by train A= Area under straight line graph

RS = area of \(\triangle\) ORS

= \(\frac{1}{2} \times O R \times O S\)

= \(\frac{1}{2} \times 15 \times\) 5=37.5 m

Distance travelled by train B Area under PQ

= \(\frac{1}{2} \times O P \times OQ \)

= \(\frac{1}{2} \times 10 \times 10=50 \mathrm{~m}\)

Distance travelled by train B is more than train A.

Motion Multiple Choice Questions

Question 1. The minute hand of a wall clock is 14 m long. The distance and displacement covered by the tip of the minute hand from 9: 00 am to 9: 30 am, are

1. 88 cm, 28 cm
2. 28 cm, 44 cm
3. 28 cm, 88 cm
4. 44 cm, 28 cm

Answer: 4. 44 cm, 28 cm

Distance =\(\pi r=\frac{22}{7} \times 14=44 \mathrm{~cm}\)

Displacement = 2r = 2×14 = 28 cm

Question 2. If the velocity of an object decreases with time, then it is called

1. retardation
2. de-acceleration
3. negative acceleration
4. All of the above

Answer: 4. All of the above

When the velocity of an object decreases with time, then it is said to be retardation, de-acceleration or negative acceleration.

Question 3. A speedometer is a device which is used to measure

1. average speed
2. average acceleration
3. instantaneous speed
4. instantaneous acceleration

Answer: 3. instantaneous speed

A speedometer is used to measure instantaneous speed.

Question 4. An object travels 24 m with a speed of 4 m/s and then another 32 m with a speed of 8 m/s. The average speed of the object is

1. 6 m/s
2. 4 m/s
3. 5.6 m/s
4. 6.5 m/s

Answer: 3. 5.6 m/s

Time travel to cover 24 m with a speed of 4 m/s,

Time Travel, \(t_1=\frac{24}{4}=6 \mathrm{~s}\)

Time travel to cover 32 \(\mathrm{~m}\) with speed 8 \(\mathrm{~m} / \mathrm{s}\),

⇒ \(t_2=\frac{32}{8}=4 \mathrm{~s}\)

⇒ \(\text { Average } \text { speed }=\frac{\text { Total distance }}{\text { Total time }}=\frac{24+32}{t_1+t_2}\)

= \(\frac{56}{6+4}=5.6 \mathrm{~m} / \mathrm{s}\)

Question 5. A body moves on a circular path with a uniform speed of 6 m/s. If the body completes one revolution in 22 seconds, then the radius of the circular path is

1. 21 m
2. 7 m
3. 14 m
4. 28 m

Answer: 1. 21 m

Given, speed, v=6 m/s, time period, T=22 s

v=\(\frac{2 \pi r}{T}\)

r =\(\frac{v T}{2 \pi}=\frac{6 \times 22 \times 7}{2 \times 22}=21 \mathrm{~m}\)

Question 6. A bridge is 400 m long. A 150 m long train crosses the bridge at a speed of 50 m/s. Time taken by the train to cross it.

1. 5 s
2. 8 s
3. 6 s
4. 11 s

Answer: 4. 11 s

Total length of path covered by train = 400 + 150 = 550m

Time taken by train to cross the bridge

= \(\frac{\text { Total distance }}{\text { Speed }}=\frac{550}{50}\)=11 s

Question 7. If a body is moving on a circular path of radius 21 cm with a velocity of 2 m/s, then the time taken by the body to complete half revolution is

1. 11s
2. 22 s
3. 44 s
4. 33 s

Answer: 4. 33 s

Given, radius, r=21 cm and velocity, v=2 m/s

Time period (T)=\(\frac{2 \pi r}{v}\)

Half-time period (T)=\(\frac{\pi r}{v}=\frac{\frac{22}{7} \times 21}{2}=33 \mathrm{~s}\)

Question 8. A velocity-time graph for a moving object is found to be a curved line, then its acceleration is

1. constant
2. variable
3. zero
4. None of these

Answer: 2. variable

Since the velocity-time graph for a moving object is a curved line, this means that the object is moving with non-uniform accelerated motion, therefore the body is moving with variable acceleration.

Question 9. In the following figure of the velocity-time graph for the motion of the body, the total distance covered by the body from 3 s to 7 s is

1. 28 m
2. 56 m
3. 14 m
4. 35 m

Answer: 1. 28 m

Total distance moved by the body from 3 s to 7 s

= Area of shape ABCD =\(\frac{(A B+D C) \times A D}{2}\)

=\(\frac{(4+10)(7-3)}{2}=\frac{14 \times 4}{2}=28 \mathrm{~m}\)

Question 10. The speed-time graph in the following figure indicates the body starts from a certain initial velocity and accelerates

1. the body has a certain initial speed and then its speed decreases to zero
2. the body has constant acceleration
3. None of the above

Answer: 2. The body has a certain initial speed and then its speed decreases to zero.

Question 12. In a uniform circular motion, angular displacement is measured in

1. radian
2. steradian
3. meter
4. radian/second

Answer: 1. radian

Angular displacement is measured in radians.

Question 13. When two bodies move uniformly towards each other, then they cross each other at the speed of 10 m/s. If both the bodies move in the same direction, then they cross each other at the speed of 6 m/s. The speed of both bodies are

1. 8 m/s, 2 m/s
2. 8 m/s, 4 m/s
3. 6 m/s, 2 m/s
4. 6 m/s, 4 m/s

Answer: 1. 8 m/s, 2 m/s

Let \(v_1\) and \(u_2\) be the speeds of two bodies, then

⇒ \(u_1+u_2\)=10

(when they are moving in the opposite direction)  →  Equation 1

⇒ \(u_1-u_2\)=6

(when they are moving in the same direction)  →  Equation 2

Solving Eqs. (1) and (2), we get

∴  \(u_1=8 \mathrm{~m} / \mathrm{s} \text { and } u_2=2 \mathrm{~m} / \mathrm{s}\)

Question 14. A cyclist appears to be stationary with respect to another moving cyclist, then both cyclists are moving

1. In the opposite direction with the same velocity
2. In the same direction with the same velocity
3. In the perpendicular direction with any velocity
4. In the perpendicular direction with the same velocity

Answer: 2. In the same direction with the same velocity

When both cyclists are moving in the same direction with the same velocity, then they appear to be stationary.

Question 15. The ratio of displacement and distance is

1. <1
2. >1
3. >1
4. None of these

Answer: 2. >1

Question 16. A body moves in a circular path radius of 10 cm. If it completes two revolutions along the circular path, then the displacement of a body is

1. zero
2. 40 cm
3. 20 cm
4. None of these

Answer: 1. zero

Question 17. If a car moves along a straight line path, that covers an equal distance in equal intervals of time, it is said to be

1. In non-uniform motion
2. In uniform motion
3. Both (1) and (2)
4. None of the above

Answer: 2. In uniform motion

Question 18. The SI unit of acceleration is

1. m/s
2. m²/s
3. m/s²
4. s/m

Where m is the meter and s is the second.

Answer: 3. m/s²

Question 19. Which of the following is an example of a uniform acceleration?

1. The motion of a free-falling body
2. The motion of a ball rolling down on an inclined plane
3. Both (1) and (2)
4. None of the above

Answer: 3. Both (1) and (2)

Question 20. Which of the following statements is correct?

1. The average speed of an object can never be zero
2. The average velocity of a moving object can be zero
3. The motion of the moon and the earth is a uniform circular motion
4. All of the above

Answer: 4. All of the above

Question 21. Uniform circular motion

1. Is an accelerated motion
2. Is an unaccelerated motion
3. Have uniform velocity
4. None of the above

Answer: 1. Is an accelerated motion

Question 22. A wheel completes 2000 revolutions to cover the 9.5 km distance, then the diameter of the wheel is

1. 1.5 m
2. 1.5 cm
3. 7.5 cm
4. 7.5 m

Answer: 1. 1.5 m

Class 9 Science Notes For Chapter 7 Motion

• In everyday life, we observe several types of motions like vehicles moving on a road, flying birds, movement of needles of a watch, movement of blades of a fan, circulation of blood through veins and arteries, etc.
• Atom molecules, planets, stars and galaxies are all in motion. If a body changes its position concerning time and its surroundings, it is said to be in motion. Motion and rest are always relative but never absolute.

Various Terms Related To Motion

Position

The location of an object concerning a particular point is known as the position of the object. The particular point about which the position of the object is defined, is called reference point or origin.

Scalar and Vector Quantities

• Physical quantities with which we can associate only magnitude, i.e. numbers are called scalar quantities, For Example, mass, time, distance, speed, etc.
• Physical quantities with which we can associate magnitude, i.e. numbers as well as direction are called vector quantities, For Example, weight, displacement, velocity, etc.

Read and Learn  More Class 9 Science Notes

Distance

• The distance travelled by a body is the actual length of the path covered by it, irrespective of the direction in which the body travels. It is a scalar quantity. Its SI unit is metre, For Example, Consider the motion of an object moving along a straight path.
• Let the object start its motion from point O and move through points A, B, and C and reach upto point D.
• Then, the total distance covered by the object = actual length upto D of the path travelled

= OA + AB + BC + CD = 5m + 10m + 8m+20m = 43m

Displacement

• The displacement of an object is the change in the position of the object when it moves from a given position to another position. It is equal to the length of the shortest path measured in the direction from the initial position to the final position of the object. It is a vector quantity. Its SI unit is metre (m).
• Suppose an object starts to move from point O and reaches point B, passing through A.
• Here, the initial position of the object is O and the final position is B. So, displacement of the object = length of the shortest path between the initial position (O) and the final position (B).

= OB = \(\sqrt{O A^2+A B^2}\) [from Pythagoras theorem] }

= \(\sqrt{(3)^2+(4)^2}\)

= \(\sqrt{9+16}=\sqrt{25}\)

= 5 m

But, distance in this case = length of actual path = OA + AB = 3 m +4 m = 7 m

Distance versus Displacement

1. Displacement of a moving object can never be greater than the distance travelled by it.

Displacement \(\leq\) Distance

∴ \(\frac{\text { Displacement }}{\text { Distance }} \leq 1\)

i.e. Ratio of displacement and distance is always less than or equal to 1.

2. If a body moves along a straight line (only in one direction), then distance and displacement will be equal.

3. Displacement of the object can be positive, negative or zero but distance can never be negative or zero.

Example 1. A Jogger jogs along one length and breadth of a rectangular park. If the dimensions of park are 150 m x 120 m, then find the distance travelled and displacement of the Jogger.
Answer:

According to question,

Length, l = 150 m, Breadth, b =120 m Distance travelled, s =?

Displacement of the Jogger = ?

Suppose, the Jogger starts from point A and after covering one breadth and one length reaches at point C.

Distance, 5 = Length of the total path covered = AB + BC s = 150 + 120 = 270 m

Displacement = Minimum distance between initial and final position = AC

From Pythagoras theorem, A C=\(\sqrt{(A B)^2+(B C)^2}\)

i.e. Displacement =\(\sqrt{(150)^2+(120)^2}\)=30 \(\sqrt{41} \mathrm{~m}\)

Thus, the Jogger travels a distance of 270 m and his displacement is 30 \(\sqrt{41} \mathrm{~m}\).

Example 2, A body moves in a circular path of radius 20 cm. If it completes two and half revolution along the circular path, then find the distance and displacement of the body.
Answer:

According to question, body moves in a circular path of radius 20 cm. So, during one complete revolution the distance (d) moved by the body is equal to the circumference of the circular path, i.e. 2\(\pi r\). where, r = radius of circular path

So, for two and half revolution, body covers two complete revolutions, i.e. 2×2\(\pi r\) distance and half revolution, i.e. \(\frac{2 \pi r}{2}=\pi r\) distance.

Total distance (s) covered by the body = (2 x 2\(\pi r\)) + \(\pi\)r = 5\(\pi\)r

= 5 x 3.14 x 20

= 314 cm

Now, as we know displacement (d) is the straight line distance between the initial and final position of the body. So, after two and half revolutions, total displacement (d) of the body will be 2r, i.e. displacement (d) = 2×20 = 40 cm.

Uniform And Non-Uniform Motion

Uniform Motion

1. A body is said to have a uniform motion, if it travels equal distances in equal intervals of time, no matter how small these intervals may be. The distance travelled by an object in uniform motion increases linearly.
2. For Example. If a car moving along a straight line path, covers equal distances in equal intervals of time, it is said to be in uniform motion.

Non-uniform Motion

1. A body is said to have a non-uniform motion, if it travels unequal distances in equal intervals of time, no matter how small these intervals may be.
2. For Example, A car moving through a crowded market has non-uniform motion.

Rate of Motion

The ratio of distance travelled by an object to the time taken is called rate of motion. The various terms required to measure the rate of motion are as given below:

Speed

• Speed of an object is defined as the distance travelled by it per unit time.
• Speed or an object (v} = \(\frac{\text { Distance }(s)}{\text { Time }(t)}\)
• Speed is a scalar quantity. The SI unit of speed is metre per second (m/s). The distance travelled by an object is either positive or zero, so the speed may be positive or zero but never negative.

Speed can be classified as:

1. Uniform speed If a moving body covers equal distances in equal intervals of time, then the speed of the body Is said to be uniform, i.e. constant speed.
2. Non-uniform speed If a moving body covers unequal distances in equal intervals of time, then the speed of the body is said to be non-uniform, i.e. variable speed.
3. Average speed It is defined as the ratio of the total distance travelled by a body to the total time taken. It is expressed as the Total distance travelled

• • Average speed =\(\frac{\text { Total distance travelled }}{\text { Total time taken }}\)
  • =\(\frac{s_1+s_2+s_3+\ldots}{t_1+t_2+t_3+\ldots}\)

1. Instantaneous speed: The speed of an object at any particular instant of time or a particular point of its path is called the instantaneous speed of the object.

Example 3. The odometer of a bike reads 1600 km at the start of the trip and 2000 km at the end of the trip. If the bike took 16 h, calculate the average speed of the bike in km/h and m/s.
Answer:

Distance covered by bike, s = 2000 km -1600 km = 400 km

Time taken, t =16 h

Average speed \(v_{\mathrm{av}} =\frac{s}{t}\)

=\(\frac{400}{16}=25 \mathrm{~km} / \mathrm{h}\)

\(v_{\mathrm{av}} =25 \times \frac{5}{18}\)

= 6.9 m/s

Therefore, the average speed of the bike is 6.9 m/s.

Example 4.  An object travels 14 m and then another 16m in 2s. What is the average speed of an object?
Answer:

According to the question,

First distance, sx =14 m

According to the question,

First distance, \(s_1=14 \mathrm{~m}\)

Second distance, \(s_2=16 \mathrm{~m}\), Times, \(t_1=4 \mathrm{~s}, t_2=2 \mathrm{~s}\)

Average speed =\(\frac{\text { Total distance }}{\text { Total time }}=\frac{s_1+s_2}{t_1+t_2}\)

=\(\frac{14+16}{4+2}=\frac{30}{6}=5 \mathrm{~m} / \mathrm{s}\)

Therefore, the average speed of an object is 3 m/s.

Speed with Direction: Velocity

1. Velocity of an object is defined as the displacement of the body per unit time. i.e. velocity is the speed of an object moving in a definite direction.
2. It is expressed as \(\text { Velocity of an object }(v)=\frac{\text { Displacement }(d)}{{Time}(t)}\)
3. Velocity is a vector quantity. The SI unit of velocity is metres per second (m/s). The velocity of an object can be positive, zero or negative. The velocity of an object can be changed by changing the object’s speed, the direction of motion or both.

Velocity can be classified as:

1. Uniform velocity If an object covers equal displacements in equal intervals of time without changing direction, then its velocity is known as uniform velocity, i.e. constant velocity.

2. Non-uniform velocity If an object covers unequal displacements in equal intervals of time, then its velocity is known as non-uniform velocity, i.e. variable velocity.

3. Average velocity It is defined as the ratio of the total displacement of the object to the total time taken.

It is expressed as  \(v_{\text {av }}=\frac{\text { Total displacement }}{\text { Total time taken }}\)

If the velocity of an object changes at a uniform rate, then the average velocity,

⇒ \(\left(v_{\text {av }}\right)=\frac{{Initial} \text { velocity }(u)+\text { Final velocity }(v)}{2}\)

4. Instantaneous velocity The velocity of an object at a particular instant of time or at a particular point of its path is called its instantaneous velocity.

Note : (1) If a body is moving in a straight line, then the magnitude of its speed and velocity will be equal.

(2) Average speed of an object can never be zero but the average velocity of a moving object can be zero.

Example 5. Rajeev went from Delhi to Chandigarh and returned to Delhi on his motorbike. The odometer of that read 4200 km at the start of trip and 4460 km at the end of his trip. If Rajeev took 4 h 20 min to complete his trip, then find the average speed and average velocity in km/h as well as in m/s.
Answer:

As we know that, the total distance covered,

s = final reading of odometer – initial reading of odometer

= (4460 -4200) km = 260 km

Total time taken, t = 4 h 20 min = 4.33 h

Average speed =\(\frac{\text { Total distance covered }(s)}{\text { Total time taken }(t)}\)

=\(\frac{260 \mathrm{~km}}{4.33 \mathrm{~h}}=60 \mathrm{~km} / \mathrm{h} or \frac{60 \times 5}{18}=16.67 \mathrm{~m} / \mathrm{s}\)

Average velocity =\(\frac{\text { Total displacement }(d)}{\text { Total time taken }(t)}\)

= \(\frac{0}{4.33}=0 \mathrm{~m} / \mathrm{s} \text { or } 0 \mathrm{~km} / \mathrm{h}\)

Average speed of Rajeev is 16.67 m/s and average velocity is 0.

Rate of Change of Velocity: Acceleration

Acceleration is defined as the rate of change of velocity with respect to time.

Mathematically, it is expressed as \(\text { Acceleration }(a)=\frac{\text { Change in velocity }(\Delta v)}{\text { Change in time }(\Delta t)}\)

If in a given time interval t, the velocity of a body changes from u to v, then acceleration a is expressed as

a=\(\frac{{Final} \text { velocity }(v)-{Initial}{velocity}(u)}{\text { time interval }(t)}\)

a=\(\frac{v-u}{t}\)

This kind of motion is known as accelerated motion. The SI unit of acceleration is \(\mathrm{m} / \mathrm{s}^2\). It is a vector quantity.

The acceleration is taken to be positive, if it is in the direction of velocity, negative if it is opposite to the direction of velocity and zero when it is moving with a constant velocity.

Note: If velocity of an object decreases with time, then it is said to have negative acceleration. Negative acceleration is also called deceleration or retardation.

Acceleration can be classified as:

(1) Uniform acceleration If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the object is said to be in a uniform acceleration.

For Example:

1. The motion of a freely falling body.
2. The motion of a ball rolling down on an inclined plane.

(2) Non-uniform acceleration If the velocity of an object increases or decreases by unequal amounts in equal intervals of time, then the object is said to be in a non-uniform acceleration.

For Example:

1. The movement of a car on a crowded city road.
2. The motion of the train leaving or entering the platform.

Example 6. Starting from a stationary position, a car attains a velocity of 5 m/s in 20 s. Then, the driver of the car applies a brake such that the velocity of the car comes down to 3 m/s in the next 6 s. Calculate the acceleration of the car in both the cases.
Answer:

Case 1 Initial velocity of the car, u = 0  it starts from stationary position]

Final velocity, v = 5 m/s, Time taken, t = 20 s .

a=\(\frac{v-u}{t}=\frac{5-0}{20}=0.25 \mathrm{~m} / \mathrm{s}^2\)

Case 2 Initial velocity, u=5 m/s

Final velocity, v=3m/s, Time taken, t=6 s

Acceleration =\(\frac{\text { Change in velocity }}{\text { Time }}\)

= \(\frac{\text { Final velocity }- \text { Initial velocity }}{\text { Time }}=\frac{v-u}{t}\)

a =\(\frac{3-5}{6}=-0.33 \mathrm{~m} / \mathrm{s}^2\)

Thus, the acceleration in both cases are 0.25 \(\mathrm{~m} / \mathrm{s}^2\) and -0.33 \(\mathrm{~m} / \mathrm{s}^2\).

Graphical Representation Of Motion

To describe the motion of an object, we can use line graphs.

In this case, line graphs show the dependence of one physical quantity, such as distance or velocity, on another quantity, such as time.

Note: For a graphical description of a motion, it is convenient to take time along the A-axis, whereas distance, speed or velocity is taken along the Y-axis.

Types of Graph

There are two main types of graphs which we will be studying as given below:

1. Distance-Time Graph

The change in the position of an object with time can be represented on the distance-time graph adopting a convenient scale of choice.

To draw a distance-time graph, time is plotted along the X-axis distance of the body is plotted along the T-axis.

In this case, the slope of the distance-time graph is equal to the speed of the object.

Distance-time graphs under various conditions are explained below:

(1) Distance-Time Graph for Uniform Motion

If an object travels equal distances in equal intervals of time, then it moves with uniform speed.

For uniform speed, a graph of distance travelled against time is a straight line as shown:

Distance-time graph for uniform motion

Interpretation

From the graph, it is clear that in equal intervals of time, i.e. 2s, the object covers an equal distance of 4m, so the motion is uniform and the graph is a straight line.

Calculation of speed To calculate the speed of the object from a distance-time graph, choose any two points say A and B on a straight line. From points A and B, draw perpendiculars AE and BC respectively, on the time axis. Now, draw a perpendicular AD on BC. The distance travelled by the object from point A to B is given by

⇒ \(\Delta x=B C-C D=s_2-s_1\)

Time taken by the object to cover this distance

= \(\Delta t=t_2-t_1\) .

Speed, v=\(\Delta x / \Delta t=\left(s_2-s_1\right) /\left(t_2-t_1\right)\)

i.e. \(\Delta x / \Delta t\)= Slope of distance-time graph.

(2) Distance-Time Graph for Non-uniform Motion

• If a body travels unequal distances in equal intervals of time, then the motion of the body is known as non-uniform motion. Non-uniform motion is of two types such as:
• When the speed of the body increases with the passage of time, then the distance-time graph will be a curve with a positive slope as shown alongside:

Interpretation

1. From the graph, it is clear that in equal intervals of time of two seconds, the body is covering unequal distances and this distance goes on increasing. That means, with the passage of time, the body is covering more and more distance in equal time, i.e. the speed of the body is increasing.
2. When the speed of the body decreases with the passage of time, then the distance-time graph will be a curve with a negative slope

Interpretation

From the graph, it is clear that in equal intervals of time of one second, the body is covering unequal distances and this distance goes on decreasing. That means, with the passage of time, the body is covering lesser and lesser distance in equal time, i.e. the speed of the body is decreasing.

The graph below shows the positions of a body at different times. Calculate the speed of the body as it moves from (1) A to B, (2) B to C and (3) C to D.

(1) For the motion from points to B, time taken, t = 5 – 2 = 3 s Distance covered = 3 — 0 = 3 m

Speed during the motion from point to B

= \(\frac{\text { Distance }}{\text { Time }}\)

= \(\frac{3}{3}=1 \mathrm{~m} / \mathrm{s}\)

(2) For the motion from point B to C, time is taken, t-7 — 5 = 2 s

Distance covered = 3 – 3 = 0 m [Body does not change its position from point B to C]

Speed during the motion from point B to C _ Distance Time

=\(\frac{\text { Distance }}{\text { Time }}\)

=\(\frac{0}{2}\)=0

(3) For the motion from point C to D,

time taken, r=10-7=3s

Distance covered = 8 – 3 = 3 m

Speed during the motion from point C to D

=\(\frac{\text { Distance }}{\text { Time }}\)

=\(\frac{5}{3}\)

Example 7. The following table gives the data about motion of a car:

Plot the graph and

1. find the speed of the car between 12:00 h and 12:30 h,
2. what is the average speed of the car?
3. Is the car’s motion an example of uniform motion? Justify.

Answer:

(1) Speed of the car between 12: 00h and 12: 30 h is given by

v=\(\frac{65-30}{\frac{30}{60}}=\frac{35}{\left(\frac{1}{2}\right)}=70 \mathrm{~km} / \mathrm{h}\)

(2) Average speed, \(v_{\mathrm{av}}=\frac{100}{2}=50 \mathrm{~km} / \mathrm{h}\)

(3) No, because the car covered unequal distances in equal intervals of time.

2. Velocity-Time Graph

The velocity-time graph shows how the velocity of a body changes with the passage of time. To draw a velocity-time graph, the velocity of the body is plotted along the T-axis and the time taken by the body is plotted along X-axis.

The area under the velocity-time graph gives the displacement. Velocity-time graphs under various conditions are explained as below:

(1) Velocity-Time Graph for a Body Moving with Constant Velocity

When a body moves with constant velocity, i.e. its motion is uniform, then its velocity does not change with time.

The graph will be a straight line parallel to the time axis.

Interpretation

From the graph, it is clear that with time, there is no change in the velocity, i.e. the body is moving with constant velocity.

Calculation of distance or magnitude of displacement

Let us calculate the distance or magnitude of the displacement of a body between time tx and time t₂ • Draw perpendiculars A Cand BD from the points corresponding to time tx and time t₂ on the graph.

Now, AD – BC = velocity of the body

CD =(t₂ -t₁) = time interval

Thus, AD x CD = area of rectangle ABCD

Also, AD x CD = velocity x time

= distance or magnitude of displacement Thus, the magnitude of displacement

= area under velocity-time graph

(2) Velocity-Time Graph for Uniform Accelerated Motion

In uniform accelerated motion, the velocity changes be equal amounts in equal intervals of time. In this case, the velocity-time graph is a straight line passing through the origin.

Interpretation

The nature of the graph shows that velocity changes by equal amounts in equal intervals of time. In equal intervals of time, i. e. 10 s, the change in velocity is 18 m/s, which remains the same, which means the acceleration of the body is constant. Thus, for all uniformly accelerated motion, the velocity-time graph is a straight line.

Calculation of distance or magnitude of displacement To determine the distance moved by the car from its velocity-time graph. The area under the velocity-time graph gives the distance (magnitude of displacement) moved by the car in a given interval of time. Therefore, S=area of ABCDE

=Area of the rectangle ABCD area of triangle ADE

∴ \(A B \times B C+\frac{1}{2}(A D \times D E)\)

(3) Velocity-Time Graph for Non-uniform Accelerated Motion

A velocity-time graph for non-uniform accelerated motion is given below:

It shows that the velocity of a body (or object) varies non-uniformly with time.

Example 8. The velocity-time graph of an ascending passenger lift is shown in the figure below. What is the acceleration of the lift?

1. During the first two seconds?
2. Between 2nd and 10th second?
3. During the last two seconds?

Answer:

Case 1. \(\Delta v=4-0=4 \mathrm{~m} / \mathrm{s}, \Delta t=2-0=2 \mathrm{~s}, a_1\)=?

⇒ \(a_1=\frac{\Delta v}{\Delta t}=\frac{4}{2}=2 \mathrm{~m} / \mathrm{s}^2\)

Case 2. \(\Delta v=4.6-4=0.6 \mathrm{~m} / \mathrm{s}, \Delta t=10-2=8 \mathrm{~s}, a_2\)= ?

⇒ \(a_2=\frac{\Delta v}{\Delta t}=\frac{0.6}{8}=0.075 \mathrm{~m} / \mathrm{s}^2\)

Case 3.  \(\Delta v=0-4.6=-4.6 \mathrm{~m} / \mathrm{s}, \Delta t=12-10=2 \mathrm{~s}, a_3\)=?

⇒ \(a_3=\frac{\Delta v}{\Delta t}=\frac{-4.6}{2}=-2.3 \mathrm{~m} / \mathrm{s}^2\)

A negative sign shows retardation.

Example 9. A body moves with a velocity of 2 m/s for 5 s, then its velocity increases uniformly to 10 m/s in the next 5 s. Thereafter, its velocity begins to decrease at a uniform rate until it comes to rest after 5 seconds.

1. Plot a velocity-time graph for the motion of the body.
2. From the graph, find the total distance covered by the body after 2 s and 12 s.

Answer:

As we know, the distance moved by the body after 2 s = area OAB’C’ = 2 m/s x 2 s = 4 m.

Again distance covered by the body after 12 s = Area OAED + Area of ABEF

=2 \(\mathrm{~m} / \mathrm{s} \times 10 \mathrm{~s}+\frac{1}{2} \times 5 \mathrm{~s} \times 8 \mathrm{~m} / \mathrm{s}+6 \mathrm{~m} / \mathrm{s} \times 2 \mathrm{~s} +\frac{1}{2} \times 2 \mathrm{~s} \times 4 \mathrm{~m} / \mathrm{s}\)

= 20m + 20m+12m + 4m = 56m

Uniform Circular Motion

If an object moves in a circular path with uniform speed, then its motion is called uniform circular motion.

When an object moves along a circular path, its direction of motion keeps changing continuously. The velocity changes due to continuous change in direction and thus motion along a circular path is said to be accelerated.

When a body takes one round of a circular path, then it travels a distance equal to its circumference which is \(2 \pi\) r, where r is the die radius of the circular path.

Then, the speed of the body moving in a circular path, v =2 \(\pi r / t\), where t is the time taken for one round of circular path and K is constant having a value of 22/7.

Some of the examples of uniform circular motion are as follows:

A piece of stone tied to a thread and rotated in a circle with a uniform speed.

The motion of blades of an electric fan around the axle.

The motion of the moon and the earth.

A satellite in a circular orbit around the earth.

A car is moving on a circular path with constant speed.

Example 10. The minute hand of a wall clock is 10 cm long. Find its displacement and the distance covered from 10:00 am to 10:30 am.
Answer:

Given, the length of the minute hand, l = 10 cm

Displacement from 10:00 am to 10:30 am is given by diameter AOB =21 = 20 cm

Total distance covered by the minute hand

ACB = \(\pi l=\frac{22}{7} \times 10 \mathrm{~cm}=\frac{220}{7}=31.43 \mathrm{~cm}\)

The displacement of the minute hand is 20 cm and the distance is 31.43 cm.

Activity 1

Objective: To measure the distance covered and magnitude of displacement

Materials Required: A metre scale and a long rope.

Procedure

1. Take a metre scale and a long rope.

2. Walk from one corner of a basketball court to its opposite corner along its sides.

3. Measure the distance covered by you and the magnitude of the displacement.

4. Note the difference between the two in this case.

Discussion/Conclusion

The distance covered while walking from one comer of the basketball court to its opposite comer along its sides is more, i.e. A to B and B to C is 2x.

The magnitude of the displacement is from A to Q which is less than the distance travelled.

Question 1. What do you understand by the displacement of an object?
Answer:

The displacement of an object is the shortest distance between the initial and final positions of the moving object.

Question 2. Which is least in this activity distance or displacement?
Answer:

Displacement is the least in this activity.

Question 3. Can displacement be negative?
Answer:

Yes, displacement can be negative.

Activity 2

Objective: To find the magnitude of the displacement using a road map.

Materials Required: Road map of India.

Procedure

1. A car fitted with an odometer is driven from Bhubaneshwar to New Delhi. The difference between the final reading and the initial reading of the odometer is 1850 km.

2. Find the magnitude of the displacement between Bhubaneshwar and New Delhi by using the road map of India.

Discussion Or Conclusion

We find the aerial distance from the atlas. Then, we measure the straight distance between Bhubaneshwar to New Delhi and convert the. measurement as given in the map.

Question 1. Name the device that is used to measure distance in a vehicle.
Answer:

An odometer is a device used to measure distance in a vehicle.

Question 2. What is the difference between the final reading and the initial reading of the odometer?
Answer:

The difference between the final reading and the initial reading of the odometer is called displacement.

Question 3. How do we find the displacement using a road map?
Answer:

We find the displacement by calculating aerial distance using an atlas map.

Activity 3

Objective: State whether the motion of the objects is uniform or non-uniform.

Materials Required: Given data table.

Procedure

1. The data regarding the motion of two different objects A and B are given in the table.
2.  Examine them carefully and state whether the motion of the objects is uniform or non-uniform.
3. Distances travelled by objects A and B at different times are as given below:
4. Time Distance travelled by object A (in m) Distance travelled by object B (in m)

Discussion Or Conclusion

1. Object A travels equal distances in equal intervals of time. Hence, object A shows uniform motion.
2. Object B travels unequal distances in equal intervals of time. Hence, object B is showing non-uniform motion.

Question 1. How will you identify the non-uniform motion of an object?
Answer:

An object is said to be in non-uniform motion if it covers unequal distances in equal intervals of time.

Question 2. What is the path of uniform motion?
Answer:

The path of uniform motion is a straight line.

Question 3. Give two examples of non-uniform motion.
Answer:

1. A racing horse
2. Throwing a softball

Activity 4

Objective: To show that light travels faster than sound.

Materials Required: Digital wristwatch or a stopwatch.

Data Required: Speed of sound in air = 346 m/s Speed of light = 3 x 108 m/s

Procedure

1. At a time when it is cloudy, there may be frequent thunder and lightning. The sound of thunder takes some time to reach you after you see the lightning.
2. Can you answer why this happens?
3. Measure this time interval using a digital wristwatch or a stopwatch.
4.  Calculate the distance of the nearest point of lightning.

Discussion Or Conclusion

1. Lightening travels with the speed of light, i.e. 3 x 10⁸ m/s and thunder travels with the speed of sound in air, i.e. 346 m/s. That is why the sound of thunder reaches us some time later than we see the lightning. Using a digital wristwatch or a stopwatch, we measure this time interval.
2. Let us consider 2 s. Thus, distance = Speed x Time = 346 x 2 =692 m Since the speed of light is almost infinite and takes almost no time to reach us. The measured time interval is the time taken by the sound of thunder to reach us. Thus, light travels much faster than sound.

Question 1. Name the instrument used to measure time intervals.
Answer:

A digital wristwatch/stopwatch is used to measure time intervals.

Question 2. At what speed does the lightning travel?
Answer:

Lightening travels with the speed of light, i.e. 3 x10⁸ m/s.

Question 3. Why do we see lightning first during thunderstorms?
Answer:

We see lightning first because light travels faster than sound.

Activity 5

Objective: To identify a range of motions.

Procedure

In your everyday life, you come across a range of motions in which

1. acceleration is in the direction of motion.
2.  acceleration is against the direction of motion.
3. acceleration is uniform.
4. acceleration is non-uniform.

Can you identify one example of each of the above types of motion?

Discussion Or Conclusion

1. When the speed of a car on the road is increasing, then the acceleration of the car is in the direction of motion.
2.  When we apply brakes, the speed of the car is decreasing. The acceleration produced in the car is against the direction of motion.
3.  When the body is falling freely under the action of gravity, it has a uniform acceleration, i.e. 9.8 m/s (speed of the falling body is increasing at a constant rate.)
4. When a car is passing through city limits on a highway, its acceleration/retardation is non-uniform depending on traffic congestion.

Question 1. Name the instrument used to measure time intervals.
Answer:

A digital wristwatch/stopwatch is used to measure time intervals.

Question 2. At what speed does the lightning travel?
Answer:

Lightening travels with the speed of light, i.e. 3 x10⁸ m/s.

Question 3. Why do we see lightning first during thunderstorms?
Answer:

We see lightning first because light travels faster than sound.

Activity 5

Objective: To identify a range of motions.

Procedure:

In your everyday life, you come across a range of motions in which

1. Acceleration is in the direction of motion.
2.  Acceleration is against the direction of motion.
3. Acceleration is uniform.
4. Acceleration is non-uniform.

Discussion Or Conclusion

1.  When the speed of a car on the road is increasing, then the acceleration of the car is in the direction of motion.
2.  When we apply brakes, the speed of the car is decreasing. The acceleration produced in the car is against the direction of motion.
3. When the body is falling freely under the action of gravity, it has a uniform acceleration, i.e. 9.8 m/s (speed of the falling body is increasing at a constant rate.)
4.  When a car is passing through city limits on a highway, its acceleration/retardation is non-uniform depending on traffic congestion.

Activity 6

Objective: To plot and interpret the distance-time graph.

Materials Required: (1) Graph paper (2) Data given in the table

Procedure

1. Study the times of arrival and departure of a train at three stations A, B and C and the distance of stations B and C from station A as given in the table.
2. Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform.Station Distance from A (km) Time of arrival (h) Time of departure (h)

Discussion Or Conclusion

1. The distance-time graph for the train is shown below. From 8:00 am to 8:15 am, the train is standing on station A. From 8:15 am to 11:15 am, the train moves with uniform speed and reaches station B at a distance of 120 km from A.
2. From 11:15 am to 11:30 am, the train is at rest at station B. From 11:30 am to 13:00 pm, the train moves uniformly from station B to station C covering a distance of 60 km (180 -120) and finally halts at point C.
3. Thus, when motion is uniform, the distance-time graph is a straight line and the slope of this straight line represents the velocity of the body.

Question 1. While plotting a distance-time graph, why do we plot time on the x-axis?
Answer:

While plotting a distance-time graph, we should plot time on the x-axis as all the independent quantities are plotted on the x-axis.

Question 2. What does the slope of the distance-time graph represent?
Answer:

The slope of the distance-time graph represents the velocity of the body.

Question 3. What is the geometrical shape of the distance-time graph?
Answer:

The geometrical shape of the distance-time graph is a straight line.

Activity 7

Objective: To plot distance-time graph on the same scale and interpret.

Materials Required: (1) Graph paper (2) Data given in the table

Procedure

1. Feroz and his sister Sania go to school on their bicycles. Both of them start at the same time from their home but take different times to reach the school although they follow the same route. The table shows the distance travelled by them at different times.
2. Plot the distance-time graph for their motions on the same scale and interpret.

Discussion Or Conclusion

1. The distance-time graph of both Feroz and Sania shows we have taken time on the x-axis and distance along the y-axis choosing a suitable scale.
2. The distance-time graph of Feroz lies above the distance-time graph of Sania.
3. This shows that Feroz is moving faster than Sania. However, speed is non-uniform.

Question 1. Which one of them is moving faster?
Answer:

Feroz is moving faster than Sania.

Question 2. What is the nature of motion?
Answer:

The nature of motion is non-uniform.

Question 3. While plotting a distance-time graph, what will you choose on the X-axis?
Answer:

We will choose distance on the y-axis.

Activity 8

Objective: To describe a circular path with constant speed.

Materials Required: (1) Piece of thread (2) Stone

Procedure:

1. Take a piece of thread and tie a small piece of stone at one of its ends. Move the stone to describe a circular path with constant speed by holding the thread at the other end.
2.  Now, let the stone go by releasing the thread.
3. Can you tell the direction in which the stone moves after it is released?
4. By repeating the activity a few times and releasing the stone at different positions of the circular path, check whether the direction in which the stone moves remains the same or not.

Discussion Or Conclusion

1. As soon as the stone is released, it moves along the tangent to the circular path at that moment.
2. By releasing the stone at different positions of the circular path, we will find the direction in which the stone moves is always different, but it is along the tangent to the circular path at that instant.
3. The stone is moving along a circular path due to the centripetal force provided by the pull of our hand on the stone.
4. A stone describing a circular path with a . velocity of constant magnitude

Question 1. Which force is required to change the direction in a circular motion?
Answer:

Centripetal force is required to change the direction.

Question 2. What is the direction of the stone which is released?
Answer:

The direction is tangential to the circular path.

Question 3. What is uniform circular motion known as?
Answer:

Uniform circular motion is known as accelerated motion.

Motion Question And Answers

Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer:

Yes, the displacement can be zero even, if the object has moved through a distance, e.g. a boy starts from his home to market and comes back. He has covered a distance but his displacement is zero.

Question 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 min 20 s from his initial position?
Answer:

The farmer takes 40 s to move along the boundary of the square field, i.e. after 40 the farmer is again at his initial position, so his displacement is zero.

Time is given = 2 min 20 s = (2 x 60 + 20) s = 140 s Displacement of farmer after 2 min 20 s, i.e. after 140 s = Displacement after (3 x 40 + 20) s = 0 + displacement after 20 s [v after each 40 s displacement is zero]

The farmer completes one round in 40 s, so he will complete 1/2 round in 20 s, i.e. after the 20s the final position of a farmer is C.

Displacement Of Farmer,

AC =\(\sqrt{A B^2+B C^2}\)

= \(\sqrt{10^2+10^2}\)

=10 \(\sqrt{2}=10 \times\) 1.414

=14.14 m

Question 3. Which of the following is true for displacement?

1. It cannot be zero.
2. Its magnitude is greater than the distance travelled by the object.

Answer:

1. The displacement can be zero, so the first statement is false.
2. The magnitude of displacement can never be greater than the distance travelled by the object. So, the second statement is also false.

Question 4. Distinguish between speed and velocity.
Answer:

Differences between speed and velocity are as given below:

Question 5. Under what condition(s) is the magnitude of the average velocity of an object equal to its average speed?
Answer:

When the object travels in one direction along a straight line path then its average velocity will be equal to the average speed.

Question 6. What does the odometer of an automobile measure?
Answer:

It measures the distance travelled by an automobile.

Question 7. What does the path of an object look like when it is in a uniform motion?
Answer:

An object having uniform motion has a straight line path.

Question 8. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, i.e. 3 x 108 m/s.
Answer:

Given, speed of signal = 3 x 10⁸ m/s

Time taken by the signal in reaching the earth = 5 min = 5 x 60 = 300s

Distance of spaceship from ground station

= Distance travelled by the signal in 5 min

= Speed x Time = 3 x 10⁸ x 300

= 9 x 10¹⁰m

Question 9. When will you say a body is in (1) uniform acceleration? (2) non-uniform acceleration?
Answer:

• A body is in uniform acceleration if it travels in a straight path when its velocity increases or decreases by an equal amount in equal time intervals.
•  A body is in non-uniform acceleration, if it travels in a straight path when its velocity increases or decreases by unequal amount in equal time intervals.

Question 10. A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.
Answer:

Given, the initial speed of the bus,

u=80 \(\mathrm{~km} / \mathrm{h}=80 \times \frac{5}{18}=22.22 \mathrm{~m} / \mathrm{s}\)

The final speed of the bus,

v=60 \(\mathrm{~km} / \mathrm{h}=60 \times \frac{5}{18}=16.67 \mathrm{~m} / \mathrm{s}\)

Time taken to decrease the speed, t=5 s

Acceleration of the bus,

a =\(\frac{v-u}{t}\)

= \(\frac{16.67-22.22}{5}\)

=-1.11 \(\mathrm{~m} / \mathrm{s}^2\)

The negative sign of acceleration indicates that the velocity of the bus is decreasing, i.e. the bus retards.

Question 11. A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 min. Find its acceleration.
Answer:

The initial velocity of the train, u = 0

Final velocity of train, v=40 \(\mathrm{~km} / \mathrm{h}=40 \times \frac{5}{18}\)

=11.11 m/s

Time taken, t=10 \(\mathrm{~min}=10 \times 60=600 \mathrm{~s}\)

Acceleration, a=\(\frac{v-u}{t}\)

= \(\frac{11.11-0}{600}\)

=0.0185 \(\mathrm{~m} / \mathrm{s}^2\)

=1.85 \(\times 10^{-2} \mathrm{~m} / \mathrm{s}^2\)

Hence, acceleration of the train is =1.85 \(\times 10^{-2} \mathrm{~m} / \mathrm{s}^2\).

Question 12. What is the nature of distance-time graphs for uniform and non-uniform motion of an object?
Answer:

The distance-time graph for the uniform motion of an object is a straight line.

The distance-time graph for the non-uniform motion of an object is a curved line.

Question 13. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer:

When an object is at rest, then its distance-time graph is a straight line parallel to the 8-time axis. Thus, it indicates 3 that with a change in time, 5 there is no change in the position of the object, i.e. the Time — object is at rest.

Question 14. What can you say about the motion of an object, if its speed-time graph is a straight line parallel to the time axis?
Answer:

A straight line parallel to the time axis in a speed-time graph indicates that a change in time, there is no change in the speed of the object. This indicates the speed of the object is constant.

Question 15. What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:

The displacement is measured by the area occupied below the velocity-time graph.

Exercises

Question 1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 min 20 s?
Answer:

According to the question,

diameter of the circular track = 200 m

Hence, radius =100 m

In 40 s, an athlete completes 1 round.

In 140 s, an athlete completes \(\frac{140}{40}=3 \frac{1}{2}\) rounds.

Hence, the distance covered by the athlete

= 3 \(\times 2 \pi r+\frac{1}{2} \times 2 \pi r\)

= 3 \(\times 2 \times \frac{22}{7} \times 100+\frac{1}{2} \times 2 \times \frac{22}{7} \times 100\)

= 2200 m

Displacement = Shortest path between the initial position and the final position = 2 r = 2 x 100 = 200 m

Question 2. Joseph jogs from one end A to the other end B  of a straight 300 m road in 2 min 30 s and then turns around and jogs 100 m back to point C in another 1 min. What are Joseph’s average speeds and velocities in jogging (1) from A to B and (2) from A to C?
Answer:

1. From A to B, Joseph covers distance = 300 m

Time = 2 min 30 s = 2 x 60 + 30

Hence, average speed =\(\frac{\text { total distance }}{\text { total time }}\)

=\(\frac{300}{150}=2 \mathrm{~m} / \mathrm{s}\)

Average velocity =\(\frac{\text { Total displacement }}{\text { Total time }}=\frac{300}{150}\)

= 2 m/s

Au in both cases, the distance covered and direction are the same.

2. From A to C, Joseph covers distance = 400 m

Time = 150 + 60 – 210 s

Hence, average speed =\(\frac{400}{210}=1.9 \mathrm{~m} / \mathrm{s}\)

Displacement =200 m

Time =150+60=210 s

Hence, average velocity = \(\frac{200}{210}=0.952 \mathrm{~m} / \mathrm{s}\)

Question 3. Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul’s trip?
Answer:

Let the distance covered by Abdul while driving to school =x

While going \(v_1=20=\frac{x}{t_1}\) [where, \(t_1\)= time taken to cover distance x ]

While returning \(v_2=30=\frac{x}{t_2}\) [where, \(t_2\)= time taken to cover x while returning]

Hence, \(t_1=\frac{x}{20}, t_2=\frac{x}{30}\)

Average speed  =\(\frac{\text { Total distance }}{\text { Total time }}=\frac{2 x}{\frac{x}{20}+\frac{x}{30}}=\frac{2 x}{\frac{5 x}{60}}\)

=\(\frac{2 x \times 60}{5 x}=24 \mathrm{~km} / \mathrm{h}\)

Question 4. A driver of a car travelling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km/h in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Answer:

Initial speed of car A =52 \(\mathrm{~km} / \mathrm{h}=52 \times \frac{5}{18} \)

=14.44 m/s

The car stops in 5 s, i.e. final speed of the car, v=0, time, t=5 s. For a speed-time graph of car A,

Initial speed of car B=3 km/h=3 \(\times \frac{5}{18}=0.83 \mathrm{~m} / \mathrm{s}\)

The car stops in 10s, i.e. final speed of the car, v=0, time, t=10s.

For a speed-time graph of car B,

The speed-time graph of both cars A and B is shown below:

Distance travelled by car A= Area of \(\triangle\) OCD

=\(\frac{1}{2} O C \times O D=\frac{1}{2} \times 14.44 \times\) 5=36.1 m

Distance travelled by car B= Area of \(\triangle\) OEF

= \(\frac{1}{2} O E \times E F \)

= \(\frac{1}{2} \times 0.83 \times 10\)=4.15 m

Thus, car A travelled farther than car B after the brakes were applied.

Question 5. The figure shows the distance-time graph of three objects A B and C. Study the graph and answer the following questions:

1. Which of the three is travelling the fastest?
2. Are all three ever at the same point on the road?
3. How far has C travelled when B passes A?
4. How far has B travelled by the time it passes C?

Answer:

1. The object for which the slope of the distance-time graph is maximum will have maximum speed, i.e. will travel the fastest. Here, for object B, the slope is maximum, so it is travelling the fastest.
2. All three objects will be at the same point on the road if the speed-time graph intersects each other at any point. Here, all three graphs do not intersect each other, so these three will never be at the same point on the road.
3. When B passes A, then the distance travelled by C =9.6 – 2 = 7.6 km
4. Distance travelled by B when it passes C =6 km.

Question 6. The speed-time graph for a car is shown in the below figure:

1. Find how far the car travels in the first 4 s. Shade the area on the graph that represents the distance travelled by the car during the period.
2. Which part of the graph represents the uniform motion of the car?

Answer:

The area under the slope of the speed-time graph gives the distance travelled by an object.

(1) We will calculate the distance represented by 1 square of the graph.

This can be done as follows. If 5 square on X-axis = 2 s,

1 square on X-axis = 2/5 s,

3 square on Y-axis = 2 m/s,

1 square on Y-axis = 2/3 m/s

So, area of 1 square on graph =\(\frac{2}{5} \times \frac{2}{3}\)

= \(\frac{4}{15} \mathrm{~m}\)

distance =\(\frac{4}{15} \mathrm{~m}\)

1 square represents distance =\(\frac{4}{15} \mathrm{~m}\)

Since approximately 62 squares come under the area of slope for the time of 4 s.

So, distance travelled in 4 s=\(\frac{4}{15} \times 62=16.53 \mathrm{~m}\)

(2) For uniform motion, the speed-time graph is a straight line parallel to the time axis. So, the straight part of the curve parallel to the time axis represents the uniform motion of the car.

Question 7. State which of the following situations are possible and give an example for each.

1.  An object with a constant acceleration but with zero velocity.
2. An object is moving in a certain direction with an acceleration in the perpendicular direction.

Answer:

1. When an object is thrown vertically upward, then at the highest point its velocity is zero but it has a constant acceleration of 9.8 m/s² (acceleration due to gravity).
2. An aeroplane flies in a horizontal direction but the acceleration due to gravity acts on it in a vertically downward direction, i.e. along the direction perpendicular to the direction of motion.

Question 8. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed, if it takes 24 hours to revolve around the earth.
Answer:

Given, the radius of the orbit, r=42250 km

Distance covered in one revolution,

d =2 \(\pi r\) [circumference of the orbit]

=2 \(\times \frac{22}{7} \times 42250=265571.43 \mathrm{~km}\)

Time taken in one revolution, t=24 h

Speed of satellite =\(\frac{\text { Distance }}{\text { Time }}=\frac{265571.43}{24}\)

=11065.48 \(\mathrm{~km} / \mathrm{h}\)

Summary

1. An object is said to be in motion if its position changes with time.
2. The distance travelled by a body is the actual length of the path covered by it, irrespective of the direction in which the body travels.
3. It is a scalar quantity. Its SI unit is metre.
4. The shortest distance between the initial and final position of the moving object is called the displacement of the object. It is a vector quantity. Its SI unit is metre.
5. The ratio of distance travelled by an object to the time taken is called the rate of motion.
6. The speed of an object is defined as the distance travelled by it in unit time.
7. Speed of an object (v) = \(\frac{\text { Distance }(s)}{\text { Time }(t)}\)
8. It is a scalar quantity. Its SI unit is metres per second (m/s).
9. The velocity of an object is defined as the displacement of the body per unit time.
10. Velocity of an object (v)=\(\frac{\text { Displacement }(d)}{\text { Time }(t)}\)
11. It is a vector quantity. Its SI unit is metres per second (m/s).
12. Acceleration is defined as the rate of change of velocity with respect to time.
13. Acceleration (a)=\(\frac{\text { Change in velocity }(\Delta v)}{\text { Change in time }(\Delta t)}\)
14. It is a vector quantity. Its SI unit is m/s².

Types of acceleration

1. Uniform acceleration If the velocity of an object changes by an equal amount in equal intervals of time, then the acceleration of the object is known as uniform acceleration.
2. Non-uniform acceleration If the velocity of an object changes by an unequal amount in equal intervals of time, then the acceleration of the object is known as non-uniform acceleration.
   1. When an object moves in a circular path with a uniform
   2. speed, its motion is called uniform circular motion.