NCERT SOLUTIONS CLASS 9 SCIENCE

NCERT Solutions for Class 9 Science is provided here. Students can download the complete NCERT Solutions for Class 9 Science in the PDF format. This page provides the complete solution for all Chapters of Class 9 Science. Download the NCERT Solutions for Class 9 Science PDF for free on this page. Get the solutions for all chapters and exercises of Class 9 Science.

Last modified:2019-08-18

NCERT solutions for Class 9 Science

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NCERT solutions for Class 9 Science and download NCERT class 9 science pdf dd MM YYYY

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Are you looking for NCERT solutions for Class 9 Science? You at last reached a perfect destiny. Students who study under the CBSE board can download their NCERT solutions for Class 9 Science. Here the main chapters of Class 9 Science are given below:Chapter 1 - Matter in Our Surroundings

Chapter 2 - Is Matter Around Us Pure

Chapter 3 - Atoms and Molecules

Chapter 4 - Structure of The Atom

Chapter 5 - The Fundamental Unit of Life

Chapter 6 - Tissues

Chapter 7 - Diversity in Living Organisms

Chapter 8 - Motion

Chapter 9 - Force and Laws of Motion

Chapter 10 - Gravitation

Chapter 11 - Work and Energy

Chapter 12 - Sound

Chapter 13 - Why Do We Fall ill

Chapter 14 - Natural Resources

Chapter 15 - Improvement in Food Resources

 

Students can visit our freshers live site regularly and can download the solutions with detailed questions with answers. Students can download chapter wise content for free. Here we have provided questions with answers students can utilize for better preparation.

Chapter 1 -  Matter in Our Surroundings

NCERT Solutions for Class 9 Science and Download NCERT Class 9 Science Notes for free

 

Fresherslive provides a great resource of NCERT solutions for class 9th science. Students should download the syllabus and they can use for it for free. Students can get detailed answers in a jiffy way. Check the latest NCERT latest updates here.Prepare for your NCERT exam by referring the NCERT Solutions for Class 9 Science to score top marks in the examination.

 

NCERT Solutions for Class 9 Science Notes Free Download

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Questions and answers. We have provided questions and answers of all exercises in a detailed manner.

 

Question 1: Which of the following are matter? Chair, air, love, smell, hate, almonds, thought, cold, cold drink, smell of perfume.

 

Solution: Anything that occupies space and has mass is called matter. Matter can exist in three physical

states—solid, liquid, and gaseous.

Chair and almond are forms of matter in the solid state.

Cold drink is a liquid state of matter.

Air and smell of perfume are gaseous states of matter.

Note: The sense of smell is not matter. However, the smell or odour of a substance is

classified as matter. The smell of any substance (say, perfume) is the gaseous form of that

substance which our olfactory system can detect (even at very low concentrations). Hence,

smell of perfume is matter.

 

Question 2:

Give reasons for the following observation:

The smell of hot sizzling food reaches you several metres away, but to get the smell from

cold food you have to go close.

 

Solution 2:

Solids diffuse at a very slow rate. But, if the temperature of the solid is increased, then the

rate of diffusion of the solid particles into air increases. This is due to an increase in the

kinetic energy of solid particles. Hence, the smell of hot sizzling food reaches us even at a

distance, but to get the smell from cold food we have to go close

 

Question 3:

A diver is able to cut through water in a swimming pool. Which property of matter does this

observation show?

Solution 3

The ability of a diver to cut through water in a swimming pool shows that matter is made up of particles.

Question 4:

What are the characteristics of particles of matter?

Solution 4:

The characteristics of particles of matter are:

(i) Particles of matter have spaces between them.

(ii) Particles of matter are in continuous motion.

(iii) Particles of  matter attract each other.

Question 5:

The mass per unit volume of a substance is called density (density = mass/volume).

Arrange the following in order of increasing density − air, exhaust from chimney, honey,

water, chalk, cotton, and iron.

Solution 5
The given substances in the increasing order of their densities can be represented as:

Air < Exhaust from chimney < Cotton < Water < Honey < Chalk < Iron

Question 6:

(a) Tabulate the differences in the characteristics of states of matter.

(b) Comment upon the following: rigidity, compressibility, fluidity, filling a gas container,

shape, kinetic energy, and density.

 

Solution:

S.NO

Solid State

Liquid State

Gaseous State

1

Definite shape and

volume.

 

 

No definite shape. Liquids

attain the shape of the vessel in

which they are kept

 

Gases have neither a

definite shape nor a

definite volume

 

2

Incompressible

Compressible to a small

extent.

Highly compressible

 

 

3

The intermolecular

distance is least

 

The intermolecular distance is

greater

 

The intermolecular

distance is

maximum

 

4

These particles attract

each other very strongly.

 

The force of attraction

between liquid particles is less

than solid particles.

 

The force of

attraction is least

between gaseous

particles

 

5

Particles of solid cannot

move freely.

 

These particles move freely.

Gaseous particles

are in a continuous,

random motion.

 

 

(b) Rigidity can be expressed as the tendency of matter to resist a change in shape.

Compressibility is the ability to be reduced to a lower volume when force is applied.

Fluidity is the ability to flow.

By filling a gas container we mean the attainment of shape of the container by gas.

Shape defines a definite boundary.

Kinetic energy is the energy possessed by a particle due to its motion.

1 2

. . mv

2

K E

m= mass of the particle, v= velocity of the particle

Density is mass per unit volume.

Question 7:

Give reasons:

(a) A gas fills completely the vessel in which it is kept.

(b) A gas exerts pressure on the walls of the container.

(c) A wooden table should be called a solid.

(d) We can easily move our hand in air, but to do the same through a solid block of wood, we

need a karate expert.

Solution

(a) There is little attraction between particles of gas. Thus, gas particles move freely in all

directions. Therefore, gas completely fills the vessel in which it is kept.

(b) The gas particles are in random motion due to weak intermolecular force of attraction.

These gaseous molecules continuously collide among themselves and they hit the walls of

the container with a greater force.

Therefore, gas exerts pressure on the walls of the container.

(c) A wooden table has a definite shape and volume. It is very rigid and cannot be

compressed i.e., it has the characteristics of a solid. Hence, a wooden table should be called a

solid.

(d) Particles of air have large spaces between them. On the other hand, wood has little space

between its particles. Also, it is rigid. For this reason, we can easily move our hands in air,

but to do the same through a solid block of wood, we need a karate expert.

Question 8:

Liquids generally have lower density as compared to solids. But you must have observed that

ice floats on water. Find out why?

Solution:

The mass per unit volume of a substance is called density (density = mass/volume).

As the volume of a substance increases, its density decreases.

Though ice is a solid, it has large number of empty spaces between its particles.

These spaces are larger as compared to the spaces present between the particles of water.

Thus, the volume of ice is greater than that of water. Hence, the density of ice is less than that

of water. A substance with lower density than water can float on water. Therefore, ice floats

on water.

Question: 9

Convert the following temperature to Celsius scale:

(a) 300 K

(b) 573 K

Solution:

(a) 300 K = (300 − 273)°C

= 27°C

(b)573 K = (573 − 273)°C

= 300°C

Question 10:

What is the physical state of water at:

(a) 250°C

(b) 100°C

Solution:

(a) Water at 250°C exists in gaseous state.

(b) At 100°C, water can exist in both liquid and gaseous form. At this temperature, after

getting the heat equal to the latent heat of vaporization, water starts changing from liquid

state to gaseous state.

 

Question 11:

For any substance, why does the temperature remain constant during the change of state?

 

Solution:

During a change of state, the temperature remains constant. This is because all the heat

supplied to increase the temperature is utilised in changing the state by overcoming the forces

of attraction between the particles. Therefore, this heat does not contribute in increasing the

temperature of the substance.

 

Question 12:

Why does a desert cooler cool better on a hot dry day?

 

Solution:

When a liquid evaporates, the particles of the liquid absorb energy from the surroundings to

compensate the loss of energy during evaporation. This makes the surroundings cool.

In a desert cooler, the water inside it is made to evaporate. This leads to absorption of energy

from the surroundings, thereby cooling the surroundings. Again, we know that evaporation

depends on the amount of water vapour present in air (humidity). If the amount of water

vapour present in air is less, then evaporation is more. On a hot dry day, the amount of water.

 

Question 13:

 

How does water kept in an earthen pot (matka) become cool during summers?

 

Solution:

There are some pores in an earthen pot through which the liquid inside the pot evaporates.

This evaporation makes the water inside the pot cool. In this way, water kept in an earthen

pot becomes cool during summers.

Question 14:

Why does our palm feel cold when we put some acetone or petrol or perfume on it?

Solution:

Organic compounds are covalently bonded and are volatile in nature. When we put some

acetone or petrol or perfume on our palm, it evaporates. During evaporation, particles of the

liquid absorb energy from the surrounding or the surface of the palm to compensate for the

loss of energy, making the surroundings cool. Hence, our palm feels cold when we put some

acetone or petrol or perfume on it.

Question 15:

 

What type of clothes should we wear in summers?

 

Solution:

We should wear cotton clothes in summers. During summers, we sweat more. On the other

hand, cotton is a good absorber of water. Thus, it absorbs sweat from our body and exposes

the liquid to the atmosphere, making evaporation faster. During this evaporation, particles on

the surface of the liquid gain energy from our body surface, making the body cool.

 

Question 16:

Convert the following temperatures to Celsius scale.

(a) 300 K

(b) 573 K

 

Solution:

Kelvin is an SI unit of temperature, where 0°C = 273.16 K (approximately 273 K)

(a) 300 K = (300 − 273) °C

= 27 °C

(b) 573 K = (573 − 273) °C

= 300 °C

 

Question 17

 

Convert the following temperatures to Kelvin scale.

(a) 25°C

(b) 373°C

 

Solution:

Kelvin is an SI unit of temperature, where 0°C = 273.16 K (approximately 273 K)

(a) 25 °C = (25 + 273) K

= 298 K

(b) 373 °C = (373 + 273) K

= 646 K

 

Question 18:

Give reason for the following observations.

(a) Naphthalene balls disappear with time without leaving any solid.

(b) We can get the smell of perfume sitting several metres away

 

Solution:

(a) Naphthalene undergoes sublimation easily i.e., the change of state of naphthalene from

solid to gas takes place easily. Thus, naphthalene balls disappear with time without leaving

any solid.

(b) Gaseous particles possess high speed and large spaces between them. Particles of perfume

diffuse into these gaseous particles at a very fast rate and reach our nostrils. This enables us

to smell the perfume from a distance.

Question 19:

Arrange the following substances in increasing order of forces of attraction between particles

− water, sugar, oxygen.

Solution

Sugar is a solid; the forces of attraction between the particles of sugar are strong.

Water is a liquid; the forces of attraction here are weaker than sugar. Oxygen is a gas; the

forces of attraction are the weakest in gases.

Thus, the increasing order of forces of attraction between the particles of water, sugar and

oxygen is Oxygen < Water < Sugar

Question 20:

What is the physical state of water at

(a) 25°C

(b) 0°C

(c) 100°C

 

Solution:

(a) Water at 25°C is present in the liquid state.

(b) At 0 °C, water can exist as both solid and liquid. At this temperature, after getting the heat

equal to the latent heat of fusion, the solid form of water i.e., ice starts changing into its liquid

form i.e., water.

(c) At 100 °C, water can exist as both liquid and gas. At this temperature, after getting the

heat equal to the latent heat of vaporization, water starts changing from its liquid state to its

gaseous state, i.e., water vapours.

Question 21:

Give two reasons to justify−

(a) water at room temperature is a liquid.

(b) an iron almirah is a solid at room temperature.

Solution:

a) At room temperature (25 °C), water is a liquid because it has the following characteristic

of liquid:

(i) At room temperature, water has no shape but has a fixed volume that is, it occupies the

shape of the container in which it is kept.

(ii) At room temperature, water flows.

(b) An iron almirah is a solid at room temperature (25 °C) because:

(i) it has a definite shape and volume like a solid at room temperature.

(ii) it is rigid as solid at room temperature.

 

Question 22

Why is ice at 273 K more effective in cooling than water at the same temperature?

 

Solution:

Ice at 273 K has less energy than water (although both are at the same temperature). Water

possesses the additional latent heat of fusion. Hence, at 273 K, ice is more effective in

cooling than water.

 

Question 23

What produces more severe burns, boiling water or steam?

 

Solution:

Steam has more energy than boiling water. It possesses the additional latent heat of

vaporization. Therefore, burns produced by steam are more severe than those produced by boiling water.

 

Chapter 2 - Is Matter Around Us Pure NCERT Solutions - Class 9 Science

 

Question 1:

What is meant by a pure substance?

Solution

A pure substance is the one that consists of a single type of particles, i.e., all constituent

particles of the substance have the same chemical nature. Pure substances can be classified as elements or compounds.

Question 2:

List the points of differences between homogeneous and heterogeneous mixtures

Solution:

A homogeneous mixture is a mixture having a uniform composition throughout the mixture.

For example: salt in water, sugar in water, copper sulphate in water A heterogeneous mixture

is a mixture having a non-uniform composition throughout the mixture. For example: sodium

chloride and iron fillings, salt and sulphur, oil and water

Question 3

Differentiate between homogeneous and heterogeneous mixtures with examples.

Solution:

A homogeneous mixture is a mixture having a uniform composition throughout the mixture.

For example, mixtures of salt in water, sugar in water, copper sulphate in water, iodine in

alcohol, alloy, and air have uniform compositions throughout the mixtures.

On the other hand, a heterogeneous mixture is a mixture having a non-uniform composition

throughout the mixture. For example, composition of mixtures of sodium chloride and iron

fillings, salt and sulphur, oil and water, chalk powder in water, wheat flour in water, milk and

water are not uniform throughout the mixtures

Question 4:

How are sol, solution and suspension different from each other?

Solution:

Sol is a heterogeneous mixture. In this mixture, the solute particles are so small that they

cannot be seen with the naked eye. Also, they seem to be spread uniformly throughout the

mixture. The Tyndall effect is observed in this mixture. For example: milk of magnesia, mud

Solution is a homogeneous mixture. In this mixture, the solute particles dissolve and spread

uniformly throughout the mixture. The Tyndall effect is not observed in this mixture. For

example: salt in water, sugar in water, iodine in alcohol, alloy Suspensions are heterogeneous.mixtures. In this mixture, the solute particles are visible to the naked eye, and remain suspended throughout the bulk of the medium.The Tyndall effect is observed in this mixture. For example: chalk powder and water, wheat flour and water.

Question 5:

To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K.Find its concentration at this temperature.

Solution:

Mass of solute (sodium chloride) = 36 g (Given)

Mass of solvent (water) = 100 g (Given)

Then, mass of solution = Mass of solute + Mass of solvent

= (36 + 100) g

= 136 g

Therefore, concentration (mass by mass percentage) of the solution

Mass of solute 100%

Mass of solvent

=> 36 00%

=> 136

=> 26.47%


Question 6:

Name the technique to separate

(i) butter from curd

(ii) salt from sea-water

(iii) camphor from salt

Solution:

Butter can be separated from curd by centrifugation.

(ii) Salt can be separated from sea-water by evaporation.

(iii) Camphor can be separated from salt by sublimation.

Question 7:

What type of mixtures is separated by the technique of crystallization?

Solution:

By the technique of crystallization, pure solids are separated from impurities. For example,

salt obtained from sea is separated from impurities; crystals of alum (Phitkari) are separated

from impure samples.

Question 8:

Classify the following as chemical or physical changes:

• Cutting of trees

• Melting of butter in a pan

• Rusting of almirah

• Boiling of water to form steam

• Passing of electric current through water, and water breaking down into hydrogen and

oxygen gas

• Dissolving common salt in water

• Making a fruit salad with raw fruits

• Burning of paper and wood

 

Solution:

Cutting of trees → Physical change

• Melting of butter in a pan → Physical change

• Rusting of almirah → Chemical change

2. Is Matter around us Pure www.vedantu.com 3

Class IX - NCERT –Physics Is Matter around us Pure

• Boiling of water to form steam → Physical change

• Passing of electric current through water, and water breaking down into hydrogen and

oxygen gas → Chemical change

• Dissolving common salt in water → Physical change

• Making a fruit salad with raw fruits → Physical change

• Burning of paper and wood → Chemical change

 

Question 9:

Try segregating the things around you as pure substances or mixtures.

Solution:

Pure substance: Water, salt, sugar

Mixture: Salt water, soil, wood, air, cold drink, rubber, sponge, fog, milk, butter, clothes,

Food

 

Question 10:

Which separation techniques will you apply for the separation of the following?

(a) Sodium chloride from its solution in water.

(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.

(c) Small pieces of metal in the engine oil of a car.

(d) Different pigments from an extract of flower petals.

(e) Butter from curd.

(f) Oil from water.

(g) Tea leaves from tea.

(h) Iron pins from sand.

(i) Wheat grains from husk.

(j) Fine mud particles suspended in water.

Solution :

 Sodium chloride from its solution in water → Evaporation

(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride

→ Sublimation

(c) Small pieces of metal in the engine oil of a car → Centrifugation or filtration or

decantation

(d) Different pigments from an extract of flower petals → Chromatography

(e) Butter from curd → Centrifugation

(f) Oil from water → Using separating funnel

(g) Tea leaves from tea → Filtration

(h) Iron pins from sand → Magnetic separation

(i) Wheat grains from husk → Winnowing

(j) Fine mud particles suspended in water → Centrifugation

 

Question 11:

Explain the following giving examples:

(a) Saturated solution

(b) Pure substance

(c) Colloid

(d) Suspension

 

(a) Saturated solution

A saturated solution is a solution in which the maximum amount of solute has been dissolved

at a given temperature. The solution cannot dissolve beyond that amount of solute at that

temperature. Any more solute added will settle down at the bottom of the container as a

precipitate.

Suppose 500 g of a solvent can dissolve a maximum of 150 g of a particular solute at 40°C.

Then, the solution obtained by dissolving 150 g of that solute in 500 g of that solvent at 300

K is said to be a saturated solution at 300 K.

(b) Pure substance

A pure substance is a substance consisting of a single type of particles i.e., all constituent

particles of the substance have the same chemical properties. For example, salt, sugar, water

are pure substances.

(c) Colloid

A colloid is a heterogeneous mixture. The size of the solutes in this mixture is so small that

they cannot be seen individually with naked eyes, and seems to be distributed uniformly

throughout the mixture. The solute particles do not settle down when the mixture is left

undisturbed. This means that colloids are quite stable.

Colloids cannot be separated by the process of filtration. They can be separated by

centrifugation. Colloids show the Tyndall effect. For example, milk, butter, foam, fog,

smoke, clouds.

(d) Suspension

Suspensions are heterogeneous mixtures. The solute particles in this mixture remain

suspended throughout the bulk of the medium. The particles can be seen with naked eyes.

 

Question 12:

Classify each of the following as a homogeneous or heterogeneous mixture.

Soda water, wood, air, soil, vinegar, filtered tea

 

Solution:

Homogeneous mixtures: Soda water, air, vinegar

Heterogeneous mixtures: Wood, soil, filtered tea

 

Question 13:

How would you confirm that a colourless liquid given to you is pure water?

 

Solution

Every liquid has a characteristic boiling point. Pure water has a boiling point of 100°C (373

K) at 1 atmospheric pressure. If the given colourless liquid boils at even slightly above or

below 100°C, then the given liquid is not pure water. It must boil at sharp 100°C. Thus, by

observing the boiling point, we can confirm whether a given colourless liquid is pure water or

not.

Question 14:

Which of the following materials fall in the category of a “pure substance”?

(a) Ice

(b) Milk

(c) Iron

(d) Hydrochloric Acid

(e) Calcium oxide

(f) Mercury

(g) Brick

(h) Wood

(i) Air

 

Solution

The following materials fall in the category of a “pure substance”:

(a) Ice

(c) Iron

(d) Hydrochloric acid

(e) Calcium oxide

(f) Mercury

Question 15:

Identify the solutions among the following mixtures:

(a) Soil

(b) Sea water

(c) Air

(d) Coal

(e) Soda water

 

Solution

The following mixtures are solutions:

(b) Sea water

(c) Air

(e) Soda water

Question 16:

Which of the following will show the “Tyndall effect”?

(a) Salt solution

(b) Milk

(c) Copper sulphate solution

(d) Starch solution

 

Solution:

Milk and starch solution will show the “Tyndall effect”.

 

Question 17:

Which of the following are chemical changes?

(a) Growth of a plant

(b) Rusting of iron

(c) Mixing of iron fillings and sand

(d) Cooking of food

(e) Digestion of food

(f) Freezing of water

(g) Burning of candle

 

Solution:

The following changes are chemical changes:

(a) Growth of a plant

(b) Rusting of iron

(d) Cooking of food

(e) Digestion of food

(g) Burning of candle

 

Chapter 3 - Atoms and Molecules

Question 1:

In a reaction, 5.3g of sodium carbonate reacted with 6g of ethanoic acid. The products were

2.2g of carbon dioxide, 0.9g water and 8.2g of sodium ethanoate. Show that these

observations are in agreement with the law of conservation of mass.

Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water

Solution:

In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium

ethanoate, carbon dioxide, and water.

Sodium Ethanoic Sodium Carbon Water

carbonate acid ethanoate dioxide

2 3 3 3 2 2 Na CO CH COOH CH COONa CO H O  

Mass of sodium carbonate = 5.3 g (Given)

Mass of ethanoic acid = 6 g (Given)

Mass of sodium ethanoate = 8.2 g (Given)

Mass of carbon dioxide = 2.2 g (Given)

Mass of water = 0.9 g (Given)

Now, total mass before the reaction = (5.3 + 6) g

= 11.3 g

And, total mass after the reaction = (8.2 + 2.2 + 0.9) g

= 11.3 g

∴Total mass before the reaction = Total mass after the reaction

Hence, the given observations are in agreement with the law of conservation of mass.

 

Question 2:

Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of

oxygen gas would be required to react completely with 3 g of hydrogen gas?

 

Solution:

It is given that the ratio of hydrogen and oxygen by mass to form water is 1:8.

Then, the mass of oxygen gas required to react completely with 1 g of hydrogen gas is 8 g.

Therefore, the mass of oxygen gas required to react completely with 3 g of hydrogen gas is

8 × 3 g = 24 g

 

Question 3:

Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

 

Solution :

Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Atoms are indivisible particles, which can neither be created nor destroyed in a chemical

reaction.

The postulate of Dalton’s atomic theory which can explain the law of definite proportion is:

The relative number and kind of atoms in a given compound remains constant.

 

Question 4:

Define atomic mass unit.

 

Solution:

 

Mass unit equal to exactly one-twelfth th 1/12 the mass of one atom of carbon-12 is called one atomic mass unit. It is written as ‘u’.

 

Question 5:

Why is it not possible to see an atom with naked eyes?

 

Solution:

The size of an atom is so small that it is not possible to see it with naked eyes. Also, the atom

of an element does not exist independently.

 

Question 6:

Write down the formulae of

(i) sodium oxide

(ii) aluminium chloride

(iii) sodium suphide

(iv) magnesium hydroxide

Solution:

Sodium oxide →Na2O

(ii) Aluminium chloride → AlCl3

(iii) Sodium suphide → Na2S

(iv) Magnesium hydroxide → Mg(OH)2

 

Question 7:

Write down the names of compounds represented by the following formulae:

(i) Al2(SO4)3

(ii) CaCl2

(iii) K2SO4

(iv) KNO3

(v) CaCO3

 

Solution:

(i) Al2(SO4)3 → Aluminium sulphate

(ii) CaCl2 → Calcium chloride

(iii) K2SO4 → Potassium sulphate

(iv) KNO3 → Potassium nitrate

(v) CaCO3 → Calcium carbonate

Question 8:

What is meant by the term chemical formula?

Solution:

The chemical formula of a compound means the symbolic representation of the composition of a compound. From the chemical formula of a compound, we can know the number and kinds of atoms of different elements that constitute the compound. For example, from the chemical formula CO2 of carbon dioxide, we come to know that one carbon atom and two oxygen atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.

 

Question 9:

How many atoms are present in a (i) H2S molecule and (ii) 3 PO4  ion?

Solution:

(i) In an H2S molecule, three atoms are present; two of hydrogen and one of sulphur. (ii) In a 3 PO4  ion, five atoms are present; one of phosphorus and four of oxygen.

Question 10:

Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Solution:

1 mole of solid sulphur (S8) = 8 × 32 g = 256 g

i.e., 256 g of solid sulphur contains = 6.022 × 1023 molecules

Then, 16 g of solid sulphur contains

23 6.022 10 16

256

molecules

= 3.76 × 1022 molecules (approx)

 

Question 11:

Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.

(Hint: The mass of an ion is the same as that of an atom of the same element.

Atomic mass of Al = 27 u)

 

Solution:

1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102 g

i.e., 102 g of Al2O3 = 6.022 × 1023 molecules of Al2O3

Then, 0.051 g of Al2O3 contains =

23 6.022 10 0.051

102

molecules

= 3.011 × 1020 molecules of Al2O3

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al3+) present in 3.011 × 1020 molecules (0.051 g )

of aluminium oxide (Al2O3) = 2 × 3.011 × 1020

= 6.022 × 1020

 

Chapter 4 - Structure of Atoms

Question 1:

What are canal rays?

Solution:

Canal rays are positively charged radiations. These rays consist of positively charged

particles known as protons. They were discovered by Goldstein in 1886.

 

Question 2:

If an atom contains one electron and one proton, will it carry any charge or not?

Solution:

An electron is a negatively charged particle, whereas a proton is a positively charged particle.The magnitude of their charges is equal. Therefore, an atom containing one electron and one proton will not carry any charge. Thus, it will be a neutral atom.

 

Question 3:

On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.

Solution:

According to Thomson’s model of the atom, an atom consists of both negatively and

positively charged particles. The negatively charged particles are embedded in the positively

charged sphere. These negative and positive charges are equal in magnitude. Thus, by

counterbalancing each other’s effect, they make an atom neutral. Thomson’s model is also

known as Plum pudding model and Water melon model.

 

Question 4:

On the basis of Rutherford’s model of an atom, which subatomic particle is present in the

nucleus of an atom?

Solution:

On the basis of Rutherford's model of an atom, protons (positively-charged particles) are

present in the nucleus of an atom.

 

Question 5:

What do you think would be the observation if the α-particle scattering experiment is carried out using a foil of a metal other than gold?

Solution:

If the α-scattering experiment is carried out using a foil of a metal rather than gold, there would be no change in the observation. In the α-scattering experiment, a gold foil was taken because gold is malleable and a thin foil of gold can be easily made. It is difficult to make such foils from other metals

 

Question 6:

Name the three sub-atomic particles of an atom.

Solution:

The three sub-atomic particles of an atom are: (i) Protons (ii) Electrons (iii) Neutrons

 

Question 7:

 

Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?

Solution:

 

atomic number of the atom and (ii) what is the charge on the atom?

Solution:

(i) The atomic number is equal to the number of protons. Therefore, the atomic number of the atom is 8. (ii) Since the number of both electrons and protons is equal, therefore, the charge on the atom is 0.

 

Question 11:

With the help of Table 4.1, find out the mass number of oxygen and sulphur atom.

 

Solution:

Mass number of oxygen = Number of protons + Number of neutrons

= 8 + 8

= 16

Mass number of sulphur = Number of protons + Number of neutrons

= 16 +16

= 32

 

Question 12:

For the symbol H, D and T tabulate three sub-atomic particles found in each of them.

 

Solution:

Symbol

Proton

Neutron

Electron

H

1

0

1

D

1

1

1

T

1

2

1

 

Question 13:

Write the electronic configuration of any one pair of isotopes and isobars.

 

Solution:

Two isotopes of carbon are 12 14 6 6 C C and . The electronic configuration of 12 6 C is 2, 4. The electronic configuration of 14 6 C is 2, 4. [Isotopes have the same electronic configuration] 40 20Ca and 40 18 Ar are a pair of isobars The electronic configuration of 40 20Ca is 2, 8, 8, 2. The electronic configuration of 40 18 Ar is 2, 8, 8.

 

Question 14:

What are the limitations of J.J. Thomson’s model of the atom?

 

Solution:

According to J.J. Thomson’s model of an atom, an atom consists of a positively charged sphere with electrons embedded in it. However, it was later found that the positively charged particles reside at the centre of the atom called the nucleus, and the electrons revolve around the nucleus.

 

Question 15:

What are the limitations of Rutherford’s model of the atom?

 

Solution:

According to Rutherford’s model of an atom, electrons revolve around the nucleus in fixed

orbits. But, an electron revolving in circular orbits will not be stable because during

 

revolution, it will experience acceleration. Due to acceleration, the electrons will lose energy

in the form of radiation and fall into the nucleus. In such a case, the atom would be highly

unstable and collapse.

 

Question 16:

Summarize the rules for writing of distribution of electrons in various shells for the first

eighteen elements.

Solution:

The rules for writing of the distribution of electrons in various shells for the first eighteen

elements are given below.

(i) The maximum number of electrons that a shell can accommodate is given by the formula

‘2n2’, where ‘n’ is the orbit number or energy level index (n = 1, 2, 3…).

The maximum number of electrons present in an orbit of n = 1 is given by 2n2 = 2×12 = 2

Similarly, for second orbit, it is 2n2 = 2×22 = 8

For third orbit, it is 2n2 = 2×32 = 18

And so on……

(ii) The outermost orbit can be accommodated by a maximum number of 8 electrons.

(iii) Shells are filled with electrons in a stepwise manner i.e., the outer shell is not occupied

with electrons unless the inner shells are completely filled with electrons.

 

Question 17:

Define valency by taking examples of silicon and oxygen.

Solution:

The valency of an element is the combining capacity of that element. The valency of an

element is determined by the number of valence electrons present in the atom of that element.

If the number of valence electrons of the atom of an element is less than or equal to four, then

the valency of that element is equal to the number of valence electrons. For example, the

atom of silicon has four valence electrons. Thus, the valency of silicon is four.

On the other hand, if the number of valence electrons of the atom of an element is greater

than four, then the valency of that element is obtained by subtracting the number of valence

electrons from eight. For example, the atom of oxygen has six valence electrons. Thus, the

valency of oxygen is (8 − 6) i.e., two.

 

Question 18:

Explain with examples

(i) Atomic number,

(ii) Mass number,

(iii) Isotopes and

(iv) Isobars. Give any two uses of isotopes.

 

Solution:

(i) Atomic number The atomic number of an element is the total number of protons present in the atom of that element. For example, nitrogen has 7 protons in its atom. Thus, the atomic number of nitrogen is 7.

(ii) Mass number The mass number of an element is the sum of the number of protons and neutrons present in the atom of that element. For example, the atom of boron has 5 protons and 6 neutrons. So, the mass number of boron is 5 + 6 = 11.

(iii) Isotopes Isotopes are atoms of the same element having the same atomic number, but different mass numbers. For example, hydrogen has three isotopes. They are protium   1 1H , deuterium   2 1 H , and tritium   3 1H

(iv) Isobars.Isobars are atoms having the same mass number, but different atomic numbers i.e., isobars

are atoms of different elements having the same mass number. For example,

40 40

20 18 Ca Ar and

are isobars.

Two uses of isotopes are:

(i) One isotope of uranium is used as a fuel in nuclear reactors.

(ii) One isotope of cobalt is used in the treatment of cancer.

 

Question 19:

Na+ has completely filled K and L shells. Explain.

Solution:

An atom of Na has a total of 11 electrons. Its electronic configuration is 2, 8, 1. But, Na+ ion

has one electron less than Na atom i.e., it has 10 electrons. Therefore, 2 electrons go to Kshell

and 8 electrons go to L-shell, thereby completely filling K and L shells.

 

Question 20:

If bromine atom is available in the form of, say, two isotopes 79 35Br (49.7%) and 81 35Br (50.3%), calculate the average atomic mass of bromine atom.

Solution:

It is given that two isotopes of bromine are 79 35Br (49.7%) and 81 35Br (50.3%). Then, the average atomic mass of bromine atom is given by: 49.7 50.3 79 81 100 100 3926.3 4074.3 100 100 8000.6 100 80.006u = 80 u (approx.)

 

Question 21:

If Z = 3, what would be the valency of the element? Also, name the element.

Solution:

By Z = 3, we mean that the atomic number of the element is 3. Its electronic configuration is 2, 1. Hence, the valency of the element is 1 (since the outermost shell has only one electron). Therefore, the element with Z = 3 is lithium.

 

Question 22:

Rutherford’s alpha-particle scattering experiment was responsible for the discovery of (a) Atomic nucleus (b) Electron (c) Proton (d) Neutron

Solution:

Rutherford’s alpha-particle scattering experiment was responsible for the discovery of (a) Atomic nucleus.

 

Question 23:

Isotopes of an element have (a) the same physical properties (b) different chemical properties (c) different number of neutrons (d) different atomic numbers

Solution:

Isotopes of an element have

(c) different number of neutrons

 

Question 24:

Number of valence electrons in Clion are:

(a) 16

(b) 8

(c) 17

d) 18

 

Solution:

Number of valence electrons in Clion

are:

(b) 8

 

Question 25:

Which one of the following is a correct electronic configuration of sodium? (a) 2, 8 (b) 8, 2, 1 (c) 2, 1, 8 (d) 2, 8, 1

 

Solution:

(d) The correct electronic configuration of sodium is 2, 8, 1.

 

Chapter 5 - The Fundamental Unit of Life

Question 1:

Who discovered cells and how?

Solution:

Cells were discovered in 1665 by an English Botanist, Robert Hooke. He used a

primitive microscope to observe cells in a cork slice.

 

Question 2:

Why is the cell called the structural and functional unit of life?

Solution:

Cells constitute various components of plants and animals. A cell is the smallest unit

of life and is capable of all living functions. Cells are the building blocks of life. This is the

reason why cells are referred to as the basic structural and functional units of life. All cells

vary in their shape, size, and activity they perform. In fact, the shape and size of the cell is

related to the specific functions they perform.

 

Question 3:

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

Solution:

The cell membrane is selectively permeable and regulates the movement of

substances in and out of the cell.

Question 4:

Why is the plasma membrane called a selectively permeable membrane?

Solution:

The cell membrane or the plasma membrane is known as a selectively permeable

membrane because it regulates the movement of substances in and out of the cell. This means that the plasma membrane allows the entry of only some substances and prevents the

movement of some other materials.

Question 5:

Can you name the two organelles we have studied that contain their own genetic material?

Solution:

Mitochondria and plastids are the two organelles that contain their own genetic material. Both these organelles have their own DNA and ribosomes.

Question 6:

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

Solution:

 

Cell is the smallest unit of life, which is capable of all living functions. If the

organisation of a cell is destroyed due to some physical or chemical influence, then the ability

of the cell to perform all living functions such as respiration, nutrition, excretion, etc. would

be affected.

Question 7:

Why are lysosomes known as suicide bags?

Solution:

Lysosomes are membrane-bound vesicular structures that contain powerful digestive

enzymes. These enzymes are capable of breaking down any foreign food particle or microbes

entering the cell. Sometimes, lysosomes can cause self-destruction of a cell by releasing these digestive enzymes within the cells. Hence, they are also known as ‘suicidal bags’.

Question 8:

Where are proteins synthesized inside the cell?

Solution

Ribosomes are the site for protein synthesis. Ribosomes are very small structures found either in a free state, suspended in the cytoplasm, or attached to the surface of the endoplasmic reticulum. They are composed of ribonucleic acids and proteins.

Question 9:

What would happen if the plasma membrane ruptures or breaks down?

Solution:

If the plasma membrane of a cell is ruptured, then the cell will die. The plasma membrane regulates the movement of substances in and out of the cell by diffusion or osmosis. Thus, if the plasma membrane is ruptured, then the cell might leak out its contents.

Question 10:

What would happen to the life of a cell if there was no Golgi apparatus?

Solution:

If there was no Golgi apparatus in the cell, then most activities performed by the Golgi apparatus will not take place. (i) Membranes of the Golgi apparatus are often connected to ER membranes. It collects simpler molecules and combines them to make more complex molecules. These are then packaged in small vesicles and are either stored in the cell or sent out as per the requirement. Thus, if the Golgi apparatus is absent in the cell, then the above process of storage, modification, and packaging of products will not be possible. (ii) The formation of complex sugars from simple sugars will not be possible as this takes place with the help of enzymes present in Golgi bodies. (iii) The Golgi apparatus is involved in the formation of lysosomes or peroxisomes. Thus, if the Golgi body is absent in a cell, the synthesis of lysosomes or peroxisomes will not be possible in the cell.

Question 11:

Which organelle is known as the powerhouse of the cell? Why?

Solution:

Mitochondria are known as the powerhouse of cells. Mitochondria create energy for the cell, and this process of creating energy for the cell is known as cellular respiration. Most chemical reactions involved in cellular respiration occur in the mitochondria. The energy required for various chemical activities needed for life is released by the mitochondria in the form of ATP (Adenosine triphosphate) molecules. For this reason, mitochondria are known as the powerhouse of cells.

Question 12:

Where do the lipids and proteins constituting the cell membrane get synthesized?

Solution:

Lipids and proteins constituting the cell membrane are synthesized in the endoplasmic reticulum. SER (Smooth endoplasmic reticulum) helps in the manufacturing of lipids. RER (Rough endoplasmic reticulum) has particles attached to its surface, called ribosomes. These ribosomes are the site for protein synthesis.

 

Question 13:

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,

(a) Keep cup A empty

(b) Put one teaspoon sugar in cup B

(c) Put one teaspoon salt in cup C

(d) Put one teaspoon sugar in the boiled potato cup D.

Keep these for two hours. Then observe the four potato cups and Solution the following:

(i) Explain why water gathers in the hollowed portion of B and C.

(ii) Why is potato A necessary for this experiment?

(iii) Explain why water does not gather in the hollowed out portions of A and D

Solution:

Experimental set up (i) Water gathers in the hollowed portions of set-up B and C because water enters the potato as a result of osmosis. Since the medium surrounding the cell has a higher water concentration than the cell, the water moves inside by osmosis. Hence, water gathers in the hollowed portions of the potato cup. (ii) Potato A in the experiment acts as a control set-up. No water gathers in the hollowed portions of potato A. (iii) Water does not gather in the hollowed portions of potato A because potato cup A is empty. It is a control set-up in the experiment. Water is not able to enter potato D because the potato used here is boiled. Boiling denatures the proteins present in the cell membrane and thus, disrupts the cell membrane. For osmosis, a semi-permeable membrane is required, which is disrupted in this case. Therefore, osmosis will not occur. Hence, water does not enter the boiled potato cup.

Chapter 6 -Tissues

 

Question 1:

 

What is a tissue?

 

Solution:

 

Tissue is a group of cells that are similar in structure and are organised together to perform a specific task.

 

Question 2:

What is the utility of tissues in multi-cellular organisms?

 

Solution:

In unicellular organisms, a single cell performs all the basic functions such as respiration, movement, excretion, digestion, etc. But in multicellular organisms, cells are grouped to form tissues. These tissues are specialised to carry out a particular function at a definite place in the body. For example, the muscle cells form muscular tissues which helps in movement, nerve cells form the nervous tissue which helps in transmission of messages. This is known as division of labour in multicellular organisms. It is because of this division of labour that multicellular organisms are able to perform all functions efficiently.

 

Question 3:

Name types of simple tissues.

 

Solution:

Simple permanent tissues are of three types:

1.Parenchyma,

2.Collenchyma,

3.Sclerenchyma.

Parenchyma tissue is of further two types

1. Aerenchyma

2. chlorenchyma.

 

Question 4:

Where is apical meristem found?

Solution:

Apical meristem is present at the growing tips of stems and roots. Their main function is to initiate growth in new cells of seedlings, at the tip of roots, and shoots.

Question 5:

What are the functions of the stomata?

Solution:

Functions of the stomata: (i) They allow the exchange of gases (CO2 and O2) with the atmosphere. (ii) Evaporation of water from the leaf surface occurs through the stomata. Thus, the stomata help in the process of transpiration.

 

Question 6:

What is the specific function of the cardiac muscle?

Solution:

The specific function of the cardiac muscle is to control the contraction and relaxation of the

heart.

Question 7:

Identify the type of tissue in the following: skin, bark of tree, bone, lining of kidney tubule, vascular bundle

Solution:

Skin: Stratified squamous epithelial tissue Bark of tree: Simple permanent tissue Bone: Connective tissue Lining of kidney tubule: Cuboidal epithelial tissue Vascular bundle: Complex permanent tissue

Question 8 :

Name the regions in which parenchyma tissue is present.

Solution :

Leaves, fruits, and flowers are the regions where the parenchyma tissue is present.

Question 9:

What is the role of epidermis in plants?

Solution:

 

Epidermis is present on the outer surface of the entire plant body. The cells of the epidermal tissue form a continuous layer without any intercellular space. It performs the following important functions: (i) It is a protective tissue of the plant body (ii) It protects the plant against mechanical injury (iii) It allows exchange of gases through the stomata.

 

Question 10:

How does the cork act as a protective tissue?
Solution:

The outer protective layer or bark of a tree is known as the cork. It is made up of dead cells. Therefore, it protects the plant against mechanical injury, temperature extremes, etc. It also prevents the loss of water by evaporation.

 

Chapter 7 -Diversity in Living organisms

Question 1:

Why do we classify organisms?

Solution :

There are a wide range of life forms (about 10 million −13 million species) around us. These life forms have existed and evolved on the Earth over millions of years ago. The huge range of these life forms makes it very difficult to study them one by one. Therefore, we look for similarities among them and classify them into different classes to study these different classes as a whole. Thus, classification makes our study easier.

 

Question 2:

Give three examples of the range of variations that you see in life-forms around you.

Solution:

Examples of range of variations observed in daily life are:

(i) Variety of living organisms in terms of size ranges from microscopic bacteria to tall trees of 100 metres.

(ii) The colour, shape, and size of snakes are completely different from those of lizards.

(iii) The life span of different organisms is also quite varied.

For example, a crow lives for only 15 years, whereas a parrot lives for about 140 years.

 

Question 3:

Which do you think is a more basic characteristic for classifying organisms? (a) The place where they live. (b) The kind of cells they are made of. Why?

Solution:

The kind of cells that living organisms are made up of is a more basic characteristic for classifying organisms, than on the basis of their habitat. This is because on the basis of the kind of cells, we can classify all living organisms into eukaryotes and prokaryotes. On the other hand, a habitat or the place where an organism lives is a very broad characteristic to be used as the basis for classifying organisms. For example, animals that live on land include earthworms, mosquitoes, butterfly, rats, elephants, tigers, etc. These animals do not resemble each other except for the fact that they share a common habitat. Therefore, the nature or kind of a cell is considered to be a fundamental characteristic for the classification of living organisms.

Question 4:

What is the primary characteristic on which the first division of organisms is made?

Solution:

The primary characteristic on which the first division of organisms is made is the nature of

the cell. It is considered to be the fundamental characteristic for classifying all living

organisms. Nature of the cell includes the presence or absence of membrane-bound

organelles. Therefore, on the basis of this fundamental characteristic, we can classify all

living organisms into two broad categories of eukaryotes and prokaryotes. Then, further

classification is made on the basis of cellularity or modes of nutrition.

Question 5:

On what basis are plants and animals put into different categories?

Solution:

Plants and animals differ in many features such as the absence of chloroplasts, presence of cell wall, etc. But, locomotion is considered as the characteristic feature that separates animals from plants. This is because the absence of locomotion in plants gave rise to many structural changes such as the presence of a cell wall (for protection), the presence of chloroplasts (for photosynthesis) etc. Hence, locomotion is considered to be the basic characteristic as further differences arose because of this characteristic feature.

Question 6:

Which organisms are called primitive and how are they different from the so-called advanced

organisms?

Solution:

A primitive organism or lower organism is the one which has a simple body structure and

ancient body design or features that have not changed much over a period of time. An

advanced organism or higher organism has a complex body structure and organization. For

example, an Amoeba is more primitive as compared to a starfish. Amoeba has a simple body

structure and primitive features as compared to a starfish. Hence, an Amoeba is considered

more primitive than a starfish.

 

Question 7:

Will advanced organisms be the same as complex organisms? Why?

 

Solution:

It is not always true that an advanced organism will have a complex body structure. But, there

is a possibility that over the evolutionary time, complexity in body design will increase.

Therefore, at times, advanced organisms can be the same as complex organisms.

 

Question 8:

What is the criterion for classification of organisms as belonging to kingdom Monera or

Protista?

 

Solution:

The criterion for the classification of organisms belonging to kingdom Monera or Protista is the presence or absence of a well-defined nucleus or membrane-bound organelles. Kingdom Monera includes organisms that do not have a well-defined nucleus or membrane-bound organelles and these are known as prokaryotes. Kingdom Protista, on the other hand, includes organisms with a well-defined nucleus and membrane-bound organelles and these organisms are called eukaryotes.

 

Question 9:

In which kingdom will you place an organism which is single-celled, eukaryotic and photosynthetic?

Solution:

Kingdom Protista includes single celled, eukaryotic, and photosynthetic organisms.

 

Question 10:

In the hierarchy of classification, which grouping will have the smallest number of organisms with a maximum of characteristics in common and which will have the largest number of organisms?

 

Solution:

In the hierarchy of classification, a species will have the smallest number of organisms with a

maximum of characteristics in common, whereas the kingdom will have the largest number

of organisms

 

Question 11:

Which division among plants has the simplest organisms?

Solution:

Thallophyta is the division of plants that has the simplest organisms. This group includes plants, which do not contain a well differentiated plant body. Their body is not differentiated into roots, stems, and leaves. They are commonly known as algae.

 

Question 12:

What are the advantages of classifying organisms?

Solution:

There are a wide range of life forms (about 10 million-13 million species) around us. These

life forms have existed and evolved on the Earth over millions of years ago. The huge range

of these life forms makes it very difficult to study them one by one. Therefore, we look for

similarities among them and classify them into different classes so that we can study these

different classes as a whole. This makes our study easier.

Therefore, classification serves the following advantages:

(i) It determines the methods of organising the diversity of life on Earth.

(ii) It helps in understanding millions of life forms in detail.

(iii) It also helps in predicting the line of evolution.

 

Question 13:

How would you choose between two characteristics to be used for developing a hierarchy in classification?

 

Solution:

For developing a hierarchy of classification, we choose the fundamental characteristic among several other characteristics. For example, plants differ from animals in the absence of locomotion, chloroplasts, cell wall, etc. But, only locomotion is considered as the basic or

fundamental feature that is used to distinguish between plants and animals. This is because

the absence of locomotion in plants gave rise to many structural changes such as the presence

of a cell wall for protection, and the presence of chloroplast for photosynthesis (as they

cannot move around in search of food like animals). Thus, all these features are a result of

locomotion. Therefore, locomotion is considered to be a fundamental characteristic. By

choosing the basic or fundamental characteristic, we can make broad divisions in living

organisms as the next level of characteristic is dependent on these. This goes on to form a

hierarchy of characteristics.

 

Question 14:

Explain the basis for grouping organisms into five kingdoms.

 

Solution:

R.H. Whittaker proposed a five kingdom classification of living organisms on the basis of Linnaeus’ system of classification. The five kingdoms proposed by Whittaker are Monera, Protista, Fungi, Plantae, and Animalia. The basis for grouping organisms into five kingdoms is as follows: (i) On the basis of the presence or absence of membrane-bound organelles, all living organisms are divided into two broad categories of eukaryotes and prokaryotes. This division lead to the formation of kingdom Monera, which includes all prokaryotes. (ii) Then, eukaryotes are divided as unicellular and multicellular, on the basis of cellularity. Unicellular eukaryotes form kingdom Protista, and multicellular eukaryotes form kingdom Fungi, Plantae, and Animalia. (iii) Animals are then separated on the basis of the absence of a cell wall. (iv) Since fungi and plants both contain a cell wall, they are separated into different kingdoms on the basis of their modes of nutrition. Fungi have saprophytic mode of nutrition, whereas plants have autotrophic mode of nutrition. This results in the formation of the five kingdoms.

 

Question 15:

What are the major divisions in the Plantae? What is the basis for these divisions?

Solution:

The kingdom Plantae is divided into five main divisions: Thallophyta, Bryophyta, Pteridophyta, Gymnosperms, and Angiosperms. The classification depends on the following criteria: • Differentiated/ Undifferentiated plant body • Presence /absence of vascular tissues • With/without seeds • Naked seeds/ seeds inside fruits (i) The first level of classification depends on whether a plant body is well differentiated or not. A group of plants that do not have a well differentiated plant body are known as Thallophyta. (ii) Plants that have well differentiated body parts are further divided on the basis of the presence or absence of vascular tissues. Plants without specialised vascular tissues are included in division Bryophyta, whereas plants with vascular tissues are known as Tracheophyta. (iii) Tracheophyta is again sub-divided into division Pteridophyta, on the basis of the absence of seed formation. (iv) The other group of plants having well developed reproductive organs that finally develop seeds are called Phanerogams. This group is further subdivided on the basis of whether the seeds are naked or enclosed in fruits. This classifies them into gymnosperms and angiosperms. Gymnosperms are seed bearing, non-flowering plants, whereas angiosperms are flowering plants in which the seeds are enclosed inside the fruit.

 

Chapter 8 -Motion

 

Question 1:

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 minutes 20 seconds?

 

Solution:

The farmer takes 40 s to cover 4 × 10 = 40 m. In 2 min and 20 s (140 s), he will cover a distance 40 140 140 40    m Therefore, the farmer completes 140 3.5 40  rounds (3 complete rounds and a half round) of the field in 2 min and 20 s.

 

That means, after 2 min 20 s, the farmer will be at the opposite end of the starting point. Now, there can be two extreme cases. Case I: Starting point is a corner point of the field. In this case, the farmer will be at the diagonally opposite corner of the field after 2 min 20 s. Therefore, the displacement will be equal to the diagonal of the field. Hence, the displacement will be 2 2 10 10.

 

 

Case I: Starting point is a corner point of the field.

In this case, the farmer will be at the diagonally opposite corner of the field after 2 min 20 s.

Therefore, the displacement will be equal to the diagonal of the field.

Hence, the displacement will be

2 2 10 10 14.1   m

Case II: Starting point is the middle point of any side of the field. In this case the farmer will be at the middle point of the opposite side of the field after 2 min 20 s. Therefore, the displacement will be equal to the side of the field, i.e., 10 m. For any other starting point, the displacement will be between 14.1 m and 10 m.

 

Question 2:

Which of the following is true for displacement?

(a) It cannot be zero.

(b) Its magnitude is greater than the distance travelled by the object.

 

Solution :

a) Not true Displacement can become zero when the initial and final position of the object is the same. (b) Not true Displacement is the shortest measurable distance between the initial and final positions of an object. It cannot be greater than the magnitude of the distance travelled by an object. However, sometimes, it may be equal to the distance travelled by the object.

 

Question 3:

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?


Solution:

 

Radius of the circular orbit, r = 42250 km Time taken to revolve around the earth, t = 24 h Speed of a circular moving object, 2 r v t   3.14 42250 4 1.105 10 / 3.069 / . 24 km h km s       Hence, the speed of the artificial satellite is 3.069 km/s.

 

Question 4:

Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Solution :

Total distance covered Average Speed = Total time taken Displacement Average velocity = Total time taken If the total distance covered by an object is the same as its displacement, then its average speed would be equal to its average velocity.

 

Question 5:

What does the odometer of an automobile measure?

Solution :

The odometer of an automobile measures the distance covered by an automobile.

 

Question 6:

What does the path of an object look like when it is in uniform motion?

 

Solution :

An object having uniform motion has a straight line path.

 

Question 7:

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, that is, 3 × 108 m s-1 .

 

Solution :

Time taken by the signal to reach the ground station from the spaceship = 5 min = 5 × 60 = 300 s Speed of the signal = 3 × 108 m/s

Distance travelled Speed= Time taken ∴Distance travelled = Speed × Time taken = 3 × 108 × 300 = 9 × 1010 m Hence, the distance of the spaceship from the ground station is 9 × 1010 m.

Question 8:

When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Solution:

(i) A body is said to have uniform acceleration if it travels in a straight path in such a way

that its velocity changes at a uniform rate, i.e., the velocity of a body increases or decreases

by equal amounts in an equal interval of time.

(ii) A body is said to have non-uniform acceleration if it travels in a straight path in such a

 

way that its velocity changes at a non-uniform rate, i.e., the velocity of a body increases or

decreases in unequal amounts in an equal interval of time.

 

Question 9:

A bus decreases its speed from 80 km h-1 to 60 km h-1 in 5 s. Find the acceleration of the bus

Solution:

Initial speed of the bus, u = 80 km/h 5 80 22.22 m/s 18   

Final speed of the bus, v = 60 km/h 5 60 16.66 m/s 18  

Time take to decrease the speed, t = 5 s

Acceleration, 16.66 22.22 2 1.112 m/s 5 v u a t      

Here, the negative sign of acceleration indicates that the velocity of the car is decreasing.

Question 10:

A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h-1 in 10 minutes. Find its acceleration.

Solution :

Initial velocity of the train, u = 0 (since the train is initially at rest)

Final velocity of the train, v = 40 km/h 5 40 11.11m/s 18   

Time taken, t = 10 min = 10 × 60 = 600 s

Acceleration, 11.11 0 2 0.0185 m/s 600 v u a t     

Hence, the acceleration of the train is 0.0185 m/s2 .

Question 11:

A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Solution :

(a) 12 m/s (b) 720 m (a) Initial speed of the bus, u = 0 (since the bus is initially at rest) Acceleration, a = 0.1 m/s2 Time taken, t = 2 minutes = 120 s Let v be the final speed acquired by the bus. 0 0.1 120 v u a t v      ∴ v = 12 m/s

 

(b) According to the third equation of motion: v 2 – u 2 = 2as Where, s is the distance covered by the bus (12)2 − (0)2 = 2(0.1) s s = 720 m Speed acquired by the bus is 12 m/s. Distance travelled by the bus is 720 m.

 

Question 12:

A train is travelling at a speed of 90 km h-1 . Brakes are applied so as to produce a uniform acceleration of −0.5 m s-2 . Find how far the train will go before it is brought to rest.

Solution :

Initial speed of the train, u = 90 km/h = 25 m/s Final speed of the train, v = 0 (finally the train comes to rest) Acceleration = −0.5 m s-2 According to third equation of motion: v 2 = u2 + 2 as (0)2 = (25)2 + 2 (−0.5) s Where, s is the distance covered by the train 2 (25) 625 2(0.5) s m   The train will cover a distance of 625 m before it comes to rest.

 

Question 13:

A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2 . What will be its velocity 3 s after the start?

 

Solution :

Initial velocity of the trolley, u = 0 (since the trolley was initially at rest) Acceleration, a = 2 cm s-2 = 0.02 m/s2 Time, t = 3 s According to the first equation of motion: v = u + at Where, v is the velocity of the trolley after 3 s from start v = 0 + 0.02 × 3 = 0.06 m/s Hence, the velocity of the trolley after 3 s from start is 0.06 m/s

Question 14:

A racing car has a uniform acceleration of 4 m s-2 . What distance will it cover in 10 s after start?

Solution :

 

Initial velocity of the racing car, u = 0 (since the racing car is initially at rest) Acceleration, a = 4 m/s2 Time taken, t = 10 s According to the second equation of motion: 1 2 2 s ut at   Where, s is the distance covered by the racing car

S = 0 + 1 / 2 * ( 4) X (10) ^2  = 400 / 2 =200 m

Hence, the distance covered by the racing car after 10 s from start is 200 m.

Question 15:

A stone is thrown in a vertically upward direction with a velocity of 5 m s-1 . If the acceleration of the stone during its motion is 10 m s-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

 

Solution :

Initially, velocity of the stone, u = 5 m/s Final velocity, v = 0 (since the stone comes to rest when it reaches its maximum height) Acceleration of the stone, a = acceleration due to gravity, g = 10 m/s2 (in downward direction) There will be a change in the sign of acceleration because the stone is being thrown upwards. Acceleration, a = −10 m/s2 Let s be the maximum height attained by the stone in time t.

According to the first equation of motion:

v = u + at 0 = 5 + (−10) t 5 0.5 10 t s     

According to the third equation of motion:

v 2 = u2 + 2 as (0)2 = (5)2 + 2(−10) s 2 5 1.25 20 s m  

Hence, the stone attains a height of 1.25 m in 0.5 s.

Question 16:

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 minutes 20 s?

Solution:

Diameter of a circular track, d = 200 m

Radius of the track, r = 100 2 d  m

Circumference = 2πr = 2π (100) = 200π m In 40 s, the given athlete covers a distance of 200π m. In 1 s, the given athlete covers a distance = 200 40 m

The athlete runs for 2 minutes 20 s = 140 s

∴Total distance covered in 140.

Question 16:

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 minutes 20 s?

 

Solution:

Diameter of a circular track, d = 200 m

Radius of the track, r = 100 2 d  m

Circumference = 2πr = 2π (100) = 200π m In 40 s, the given athlete covers a distance of 200π m. In 1 s, the given athlete covers a distance = 200 40 m

The athlete runs for 2 minutes 20 s = 140 s

∴Total distance covered in 140


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