# Posts Tagged: "Class 9 Science Solutions"

## NCERT Solutions for Class 9th Science Chapter 1 : Matter In Our Surroundings

- - CBSE

CBSE NCERT Solutions for Class 9th Science Chapter 1 : Matter in Our Surroundings . NCERT Class 9 Science Solutions for Chapter 1. Class IX Science Solutions NCERT

### NCERT Solutions for Class 9th Science Chapter 1 : Matter In Our Surroundings

In Text Questions

Page No: 3

1. Which of the following are matter?

Chair, air, love, smell, hate, almonds, thought, cold, cold drink, smell of perfume.

Chair, air, almonds and cold drink

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.

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.

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

This observation shows that the particles of matter have intermolecular spaces. The intermolecular spaces in liquids is fair enough to let the diver pass through it.

4. What are the characteristics of particles of matter?

The characteristics of particles of matter are:
→ Particles of matter have spaces between them.
→ Particles of matter are continuously moving.
→ Particles of matter attract each other.

Page No: 6

1. 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.

air, Exhaust from chimneys, cotton, water, honey, chalk, and iron.
2. (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.

(a)

 Property Solid state Liquid state Gaseous state 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 Slightly Compressible Highly compressible 3. Cannot flow Can flow Can flow 4. Particles don’t move freely Particles move freely but are confined within boundary. Particles move freely. 5. Force of attraction between particles are maximum. Force of attraction between particles is less than solid but more than that in gas Force of attraction between particles is least.

(b)
→ Rigidity: It is the property of matter to resist the change of its shape.→ Compressibility: It is the property of matter in which its volume is decreased by applying force.
→ Fludity: It is the ability of matter to flow.
→ Filling a gas container: On filling a gas takes the shape of the container.
→ Shape: Having definite boundaries.
→ Kinetic Energy: It is the energy possessed by the particles of matter due to its motion.
→ Density: It is the ratio of mass with per unit volume.

3. Give reasons:

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

► The force of attraction between particles of gas is negligible. Because of this, particles of gas move in all directions. Thus, a gas fills the vessel completely in which it is kept.

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

► Particles of gas move randomly in all directions at high speed. As a result, the particles hit each other and also hit the walls of the container with a force. Therefore, gas exerts pressure on the walls of the container.

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

► A wooden table has fixed shape and fixed volume, which are the main characteristics of solid. Thus 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.

► 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.

4. Liquids generally have lower density as compared to solids. But you must have observed that ice floats on water. Find out why.

Ice which is a solid has vacant spaces between water molecules thus making ice lighter than water. Thus ice floats on water.

Page No: 9

1. Convert the following temperature to Celsius scale:

(a) 300 K

► 300 K = (300 – 273)°C
= 27°C
(b) 573 K

► 573 K = (573 – 273)°C
= 300°C

2. What is the physical state of water at:

(a) 250°C

► Gaseous State (As Boiling temperature of water is 100° C).

(b) 100°C

► Since water boils at this temperature thus it 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.

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

During the change of state of any substance, the heat supplied or released is utilised in phase change. Such heat is called latent heat. So, the temperature of any substance remains constant during the change of state.

4. Suggest a method to liquefy atmospheric gases.

The gases can be converted into liquids by bringing its particles closer so atmospheric gases can be liquefied either by decreasing temperature or by increasing pressure.

Page No: 10

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

A desert cooler increases the humidity of the surrounding air. The water particles in the air take the heat from the surrounding objects and evaporates. In hot and dry days the moisture level is very low in atmosphere which increases the rate of evaporation. Because of faster evaporation, cooler works well. That’s why desert cooler cool better on a hot dry day.

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

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.

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

Acetone, petrol, and perfume evaporate at low temperatures. When some acetone, petrol, or perfume is dropped on the palm, it takes heat from the palm and evaporates, thereby making the palm cooler.

4. Why are we able to sip hot tea or milk faster from a saucer than a cup?

A liquid has a larger surface area in a saucer than in a cup. Thus, it evaporates faster and cools faster in a saucer than in a cup. Thus, we are able to sip hot tea or milk faster from a saucer than a cup.

5. What type of clothes should we wear in summers?

We should wear cotton clothes in summers as cotton is a good sweat absorber. Sweat is absorbed by the cotton and is exposed 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.

Page No: 12

Excercises

(For Conversion Process we must know,
Kelvin is an SI unit of temperature, where 0°C = 273 K approx.)

1. Convert the following temperatures to Celsius scale.

(a) 300 K
► 300 K = (300 – 273) °C
= 27 °C

(b) 573 K
► 573 K = (573 – 273) °C
= 300 °C

2. Convert the following temperatures to Kelvin scale.

(a) 25°C
►25 °C = (25 + 273) K
= 298 K

(b) 373°C
► 373 °C = (373 + 273) K
= 646 K

3. 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.

(a) Naphthalene balls disappear with time without leaving any solid because they undegoes sublimation easily i.e., the change of state of naphthalene from solid to gas takes place easily.

(b) Perfumes has high degree of vaporization and its vapour diffuse into air easily. Therefore, we can get the smell of perfume sitting several metres away.

4. Arrange the following substances in increasing order of forces of attraction between particles– water, sugar, oxygen.

Oxygen, Water, Sugar.

5. What is the physical state of water at-

(a) 25°C
► Liquid State

(b) 0°C
► Solid State, can also be in liquid state(conditions required).

(c) 100°C
► Gaseous State, can also be in liquid state(conditions required).

6. Give two reasons to justify-

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

(a) Water at room temperature is a liquid because it has fluidity also it has no shape but has a fixed volume that is, it occupies the shape of the container in which it is kept.

(b) An iron almirah is a solid at room temperature it has rigid and fixed shape.

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

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.

8. What produces more severe burns, boiling water or steam?

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.

9. Name A, B, C, D, E and F in the following diagram showing change in its state.

## NCERT Solutions for Class 9th Science Chapter 2 : Is Matter Around Us Pure

- - CBSE

CBSE NCERT Solutions for Class 9th Science Chapter 2 : Is Matter Around Us Pure . NCERT Class 9 Science Solutions for Chapter 2. Class IX Science Solutions NCERT

### NCERT Solutions for Class 9th Science Chapter 2 : Is Matter Around Us Pure

In Text Questions

Page No: 15

1. What is meant by a pure substance?

A material that is composed of only one type of particles is called pure substance. All the constituent particles of a pure substance have same chemical nature.

2. List the points of differences between homogeneous and heterogeneous mixtures.

 Homogeneous mixtures Heterogeneous mixtures Homogeneous mixtures have uniform composition. Heterogeneous mixtures have non uniform composition. It has no visible boundaries of separation between its constituents. It has visible boundaries of separation between its constituents.

Page No: 18

1. Differentiate between homogeneous and heterogeneous mixtures with examples.

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.

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

 Sol Solution Suspension They are heterogeneous in nature. They are homogeneous in nature. They are heterogeneous in nature. They scatter a beam of light and hence show Tyndall effect. They do not scatter a beam of light and hence do not show Tyndall effect They scatter a beam of light and hence show Tyndall effect. They are quite stable. Examples of solution are: salt in water, sugar in water. Examples of suspension are: sand in water, dusty air

3. 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.

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

Page No: 24

1. How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25°C), which are miscible with each other?

Kerosene and petrol are miscible liquids also the difference between their boiling point is more than 25 ºC so they can be separated by the method of distillation.

In this method, the mixture of kerosene and petrol is taken in a distillation flask with a thermometer fitted in it. We also need a beaker, a water condenser, and a Bunsen burner. The apparatus is arranged as shown in the above figure. Then, the mixture is heated slowly. The thermometer should be watched simultaneously. Kerosene will vaporize and condense in the water condenser. The condensed kerosene is collected from the condenser outlet, whereas petrol is left behind in the distillation flask.

2. Name the technique to separate

(i) butter from curd

► By Centrifugation

(ii) salt from sea-water

► By Evaporation
(iii) camphor from salt

► By Sublimation

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

The crystallisation method is used to purify solids.

1. Classify the following as chemical or physical changes:

• Cutting of trees
► Physical change

• Melting of butter in a pan
► Physical change

• Rusting of almirah
► Chemical change

• 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

Page No: 28

Excercises

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

(a) 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.
► Filtration or Centrifugation 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

2. Write the steps you would use for making tea. Use the words: solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.

First, water is taken as a solvent in a saucer pan. This water (solvent) is allowed to boil. During heating, milk and tea leaves are added to the solvent as solutes. They form a solution. Then, the solution is poured through a strainer. The insoluble part of the solution remains on the strainer as residue. Sugar added to thefiltrate, which dissolves in the filtrate. The resulting solution is the required tea.

3. Pragya tested the solubility of three different substances at different temperatures and collected the data as given below( results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).

(a)  What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?
(b)  Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.

(c)  Find the solubility of each salt at 293 K. What salt has the highest solubility at this temperature?

(d)  What is the effect of change of temperature on the solubility of a salt?

(a) Since 62 g of potassium nitrate is dissolved in 100g of water to prepare a saturated solution at 313 K, 31 g of potassium nitrate should be dissolved in 50 g of water to prepare a saturated solution at 313 K.

(b) The amount of potassium chloride that should be dissolved in water to make a saturated solution increases with temperature. Thus, as the solution cools some of the potassium chloride will precipitate out of the solution.

(c) The solubility of the salts at 293 K are:
Potassium nitrate – 32 g
Sodium chloride – 36 g
Potassium chloride – 35 g
Ammonium chloride – 37 g

Ammonium chloride has the highest solubility at 293 K.

(d) The solubility of a salt increases with temperature.
4. Explain the following giving examples:

(a) Saturated solution
(b) Pure substance
(c) Colloid
(d) Suspension

(a) Solution in which no more solute can be dissolved at a particular temperature is known as saturated solution. For example in aqueous solution of sugar no more sugar can be dissolved at room temperature.

(b) 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 water, sugar, salt etc.

(c) A colloid is a heterogeneous mixture whose particles are not as small as solution but they are so small that cannot be seen by naked eye. When a beam of light is passed through a colloid then the path of the light becomes visible. For example milk, smoke etc.

(d) A suspension is a heterogeneous mixture in which solids are dispersed in liquids. The solute particles in suspension do not dissolve but remain suspended throughout the medium. For example Paints, Muddy water chalk water mixtures etc.

5. Classify each of the following as a homogeneous or heterogeneous mixture.
Soda water, wood, air, soil, vinegar, filtered tea

Homogeneous mixtures: Soda water, air, vinegar, filtered tea
Heterogeneous mixtures: Wood, soil
Note: Pure air is homogeneous mixture but Polluted air is heterogeneous mixture.

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

Take a sample of colourless liquid and put on stove if it starts boiling exactly at 100 ºC then it is pure water. Any other colourless liquid such as vinegar always have different boiling point. Also observe carefully that after some time whole liquid will convert into vapour without leaving any residue.

7. 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

The following materials fall in the category of a “pure substance”:
(a) Ice
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury

8. Identify the solutions among the following mixtures:

(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water

The following mixtures are solutions:
(b) Sea water
(c) Air
(e) Soda water

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

(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution

Tyndall effect is shown by colloidal solution. Here milk and starch solution are colloids therefore milk and starch solution will show Tyndall effect.

10. Classify the following into elements, compounds and mixtures:

(a) Sodium
(b) Soil
(c) Sugar solution
(d) Silver
(e) Calcium carbonate
(f) Tin
(g) Silicon
(h) Coal
(i) Air
(j) Soap
(k) Methane
(l) Carbon dioxide
(m) Blood

Elements: Sodium, Silver, Tin and Silicon.
Compounds: Calcium carbonate, Methane and carbon dioxide.
Mixtures: Soil, Sugar, Coal, Air, Soap and Blood.
11. 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

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

## NCERT Solutions for Class 9th Science Chapter 3 : Atoms and Molecules

- - CBSE

CBSE NCERT Solutions for Class 9th Science Chapter 3 : Atoms and Molecules . NCERT Class 9 Science Solutions for Chapter 3. Class IX Science Solutions NCERT

### NCERT Solutions for Class 9th Science Chapter 3 : Atoms and Molecules

In Text Questions

Page No: 32

1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g 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

In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon dioxide, and water.
Sodium Carbonate + Ethanoic acid → sodium ethanoate + carbon dioxide + water

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.

Page No: 33

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?

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.

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

The postulate of Dalton :”Atoms are indivisible particles, which can not be created or destroyed in a chemical reaction” is the result of the law of conservation of mass.

4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

The postulate of Dalton, “The relative number and kinds of atoms are constant in a given compound”, can explain the law of definite proportions.

Page No: 35

1. Define atomic mass unit.

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

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

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.

Page No: 39

1. Write down the formulae of

(i) sodium oxide
► Na2O
(ii) aluminium chloride
► AlCl3
(iii) sodium sulphide

► Na2S

(iv) magnesium hydroxide

► Mg(OH)2

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

(i) Al2(SO4)3

► Aluminium sulphate

(ii) CaCl2
► Calcium chloride

(iii) K2SO4

► Potassium sulphate

(iv) KNO3

► Potassium nitrate

(v) CaCO3

► Calcium carbonate

3. What is meant by the term chemical formula?

The chemical formula of a compound is a symbolic representation of its composition.

4. How many atoms are present in a
(i) H2S molecule and
(ii) PO43-ion?

(i) In an H2S molecule, three atoms are present; two of hydrogen and one of sulphur.

(ii) In a PO43-ion, five atoms are present; one of phosphorus and four of oxygen.

Page No: 40

1. Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.

► Molecular mass of H2= 2 × Atomic mass of H
= 2 × 1
= 2 u

► Molecular mass of O2= 2 × Atomic mass of O
= 2 × 16
= 32 u

► Molecular mass of Cl2= 2 × Atomic mass of Cl
= 2 × 35.5
= 71 u

► Molecular mass of CO2= Atomic mass of C + 2 × Atomic mass of O
= 12 + 2 × 16
= 44 u

► Molecular mass of CH4= Atomic mass of C + 4 × Atomic mass of H
= 12 + 4 × 1
= 16 u

► Molecular mass of C2H6= 2× Atomic mass of C + 6× Atomic mass of H
= 2 × 12 + 6 × 1
= 30 u

► Molecular mass of C2H4= 2 x Atomic mass of C + 4 × Atomic mass of H
= 2 × 12 + 4 × 1
= 28 u

► Molecular mass of NH3 = Atomic mass of N + 3 × Atomic mass of H
= 14 + 3×1
= 17 u

► Molecular mass of CH3OH= Atomic mass of C + 3 × Atomic mass of H + Atomic mass of O + Atomic mass of H
= 12 + 3×1 + 8 + 1
= 24 u

2. Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.

► Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O
= 65 + 16
= 81 u

► Formula unit mass of Na2O = 2 × Atomic mass of Na + Atomic mass of O
= 2 × 23 + 16
= 62 u

► Formula unit mass of K2CO3= 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O
= 2 × 39 + 12 + 2 × 16

= 78 + 12 + 32

= 122 u
Page No: 42

1. If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?

One mole of carbon atoms weighs 12 g (Given)
i.e., mass of 1 mole of carbon atoms = 12 g
Then, mass of 6.022 × 1023 number of carbon atoms = 12 g
Therefore, mass of 1 atom of carbon = 12 ÷ (6.022 × 1023)
= 1.9926 x 10-23 g

2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?

Atomic mass of Na = 23 u (Given)
Then, gram atomic mass of Na = 23 g
Now, 23 g of Na contains = 6.022 × 1023 g number of atoms
Thus, 100 g of Na contains = 6.022 × 1023 / 23×100 number of atoms
= 2.6182 × 1024 number of atoms

Again, atomic mass of Fe = 56 u (Given)
Then, gram atomic mass of Fe = 56 g

Now, 56 g of Fe contains = 6.022 × 1023 g number of atoms

Thus, 100 g of Fe contains = 6.022 × 1023 / 56 × 100 number of atoms
= 1.0753 × 1024 number of atoms

Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.

Page No: 43
Excercises

1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Total mass of Compound = 0.24 g (Given)
Mass of boron = 0.096 g (Given)
Mass of oxygen = 0.144 g (Given)

Thus, percentage of boron by weight in the compound = 0.096 / 0.24 × 100%
= 40%

And, percentage of oxygen by weight in the compound = 0.144 / 0.24 × 100% = 60%

2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combinations will govern your answer?

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.

If 3 g of carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left un-reactive.
In this case also, only 11 g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.
Page No: 44

3. What are polyatomic ions? Give examples?

A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, Nitrate (NO3-) , hydroxide ion (OH – ).

4. Write the chemical formulae of the following:

(a) Magnesium chloride

► MgCl2

(b) Calcium oxide

► CaO

(c) Copper nitrate

► Cu (NO3)2

(d) Aluminium chloride

► AlCl3

(e) Calcium carbonate

► CaCO3

5. Give the names of the elements present in the following compounds:

(a) Quick lime
► Calcium and oxygen

(b) Hydrogen bromide
► Hydrogen and bromine

(c) Baking powder
► Sodium, hydrogen, carbon, and oxygen

(d) Potassium sulphate
► Potassium, sulphur, and oxygen

6. Calculate the molar mass of the following substances:

(a) Ethyne, C2H2
► Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 26 g

(b) Sulphur molecule, S8
►Molar mass of sulphur molecule, S8 = 8 × 32 = 256 g

(c) Phosphorus molecule, P4 (atomic mass of phosphorus = 31)
► Molar mass of phosphorus molecule, P4 = 4 × 31 = 124 g

(d) Hydrochloric acid, HCl
► Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g

(e) Nitric acid, HNO3
► Molar mass of nitric acid, HNO3 = 1 + 14 + 3 × 16 = 63 g

7. What is the mass of-

(a) 1 mole of nitrogen atoms?

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na2SO3)?

(a) The mass of 1 mole of nitrogen atoms is 14 g.

(b) The mass of 4 moles of aluminium atoms is (4 × 27) g = 108 g

(c) The mass of 10 moles of sodium sulphite (Na2SO3) is
10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g

8. Convert into mole.

(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide

(a) 32 g of oxygen gas = 1 mole
Then, 12 g of oxygen gas = 12 / 32 mole = 0.375 mole

(b) 18 g of water = 1 mole
Then, 20 g of water = 20 / 18 mole = 1.111 mole

(c) 44 g of carbon dioxide = 1 mole
Then, 22 g of carbon dioxide = 22 / 44 mole = 0.5 mole

9. What is the mass of:

(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?

(a) Mass of one mole of oxygen atoms = 16 g
Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2 g

(b) Mass of one mole of water molecule = 18 g

Then, mass of 0.5 mole of water molecules = 0.5 × 18 g = 9 g
10. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

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 = 6.022 × 1023 / 256  = 16 molecules
= 3.76375 × 1022 molecules

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)

mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16
= 102 g
i.e., 102 g of Al2O3= 6.022 × 1023molecules of Al2O3
Then, 0.051 g of Al2O3contains = 6.022 × 1023 / 102 × 0.051 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 × 1020molecules (0.051 g ) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020
= 6.022 × 1020

## NCERT Solutions for Class 9th Science Chapter 4 : Structure Of The Atom

- - CBSE

CBSE NCERT Solutions for Class 9th Science Chapter 4 : Structure Of The Atom . NCERT Class 9 Science Solutions for Chapter 4. Class IX Science Solutions NCERT

### NCERT Solutions for Class 9th Science Chapter 4 : Structure Of The Atom

In Text Questions

Page No: 47

1. What are canal rays?

Canal rays are positively charged radiations that can pass through perforated cathode plate. These rays consist of positively charged particles known as protons.

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

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.

Page No: 49

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

As per Thomson’s model of the atom, an atom consists both negative and positive charges which are equal in number and magnitude. So, they balance each other as a result of which atom as a whole is electrically neutral.

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

On the basis of Rutherford’s model of an atom, protons are present in the nucleus of an atom.

3. Draw a sketch of Bohr’s model of an atom with three shells.

 Bohr’s Model of an atom with three shells

4. 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?

If α-particle scattering experiment is carried out using a foil of any metal as thin as gold foil used by Rutherford, there would be no change in observations. But since other metals are not so malleable so, such a thin foil is difficult to obtain. If we use a thick foil, then more α-particles would bounce back and no idea about the location of positive mass in the atom would be available with such a certainty.

1. Name the three sub-atomic particles of an atom.

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

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

Number of neutrons = Atomic mass – Number of protons
Therefore, the number of neutrons in the atom = 4 – 2 = 2

Page No: 50

1. Write the distribution of electrons in carbon and sodium atoms

► The total number of electrons in a carbon atom is 6. The distribution of electrons in carbon atom is given by:

First orbit or K-shell = 2 electrons
Second orbit or L-shell = 4 electrons

Or, we can write the distribution of electrons in a carbon atom as 2, 4.

► The total number of electrons in a sodium atom is 11. The distribution of electrons in sodium atom is given by:

First orbit or K-shell = 2 electrons
Second orbit or L-shell = 8 electrons
Third orbit or M-shell = 1 electron

Or, we can write distribution of electrons in a sodium atom as 2, 8, 1.

2. If K and L shells of an atom are full, then what would be the total number of electrons in the atom?

The maximum capacity of K shell is 2 electrons and L shell can accommodate maximum 8 electrons in it. Therefore, there will be ten electrons in the atom.

Page No: 52

1. How will you find the valency of chlorine, sulphur and magnesium?

If the number of electrons in the outermost shell of the atom of an element is less than or equal to 4, then the valency of the element is equal to the number of electrons in the outermost shell. On the other hand, if the number of electrons in the outermost shell of the atom of an element is greater than
4, then the valency of that element is determined by subtracting the number of electrons in the outermost shell from 8.
The distribution of electrons in chlorine, sulphur, and magnesium atoms are 2, 8, 7; 2, 8, 6 and 2, 8, 2 respectively.

Therefore, the number of electrons in the outer most shell of chlorine, sulphur, and magnesium atoms are 7, 6, and 2 respectively.

► Thus, the valency of chlorine = 8 -7 = 1

► The valency of sulphur = 8 – 6 = 2

► The valency of magnesium = 2
1. If number of electrons in an atom is 8 and number of protons is also 8, then (i) what is the atomic number of the atom and (ii) what is the charge on the atom?

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

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

Page No: 53

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

 Symbol Proton Neutron Electron H 1 0 1 D 1 1 1 T 1 2 1

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

12C6 and 14C6 are isotopes, have the same electronic configuration as (2, 4)22Ne10and 22Ne11 are isobars. They have different electronic configuration as given below:
22Ne10 – 2, 8
22Ne11 – 2, 8, 1

Page No: 54

Excercise

1. Compare the properties of electrons, protons and neutrons.

 Particle Nature of Charge Mass Location Electron Electrons are negatively charged. 9 x 10–31 kg Extra nuclear part distributed in different shell or orbits. Proton Protons are positively charged. 1.672 x 10–27kg (1 µ) (approx. 2000 times that of the electron) Nucleus Neutron Neutrons are neutral. Equal to mass of proton Nucleus

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

The limitations of J.J. Thomson’s model of the atom are:
→ It could not explain the result of scattering experiment performed by rutherford.
→ It did not have any experiment support.

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

The limitations of Rutherford’s model of the atom are
→ It failed to explain the stability of an atom.
→ It doesn’t explain the spectrum of hydrogen and other atoms.

4. Describe Bohr’s model of the atom.

→ The atom consists of a small positively charged nucleus at its center.
→ The whole mass of the atom is concentrated at the nucleus and the volume of the nucleus is much smaller than the volume of the atom.
→ All the protons and neutrons of the atom are contained in the nucleus.
→ Only certain orbits known as discrete orbits of electrons are allowed inside the atom.
→ While revolving in these discrete orbits electrons do not radiate energy. These orbits or cells are represented by the letters K, L, M, N etc. or the numbers, n = 1, 2, 3, 4, . . as shown in below figure.

5. Compare all the proposed models of an atom given in this chapter.

 Thomson’s model Rutherford’s model Bohr’s model → An atom consists of a positively charged sphere and the electrons are embedded in it. → The negative and positive charges are equal in magnitude. As a result the atom is electrically neutral. → An atom consists of a positively charged center in the atom called the nucleus. The mass of the atom is contributed mainly by the nucleus. →  The size of the nucleus is very small as compared to the size of the atom. → The electrons revolve around the nucleus in well-defined orbits. → Bohr agreed with almost all points as said by Rutherford except regarding the revolution of electrons for which he added that there are only certain orbits known as discrete orbits inside the atom in which electrons revolve around the nucleus. → While revolving in its discrete orbits the electrons do not radiate energy.

6. Summarize the rules for writing of distribution of electrons in various shells for the first eighteen elements.

The rules for writing of the distribution of electrons in various shells for the first eighteen elements are given below.
→ If n gives the number of orbit or energy level, then 2n2 gives the maximum number of electrons possible in a given orbit or energy level. Thus,
First orbit or K-shell will have 2 electrons,
Second orbit or L-shell will have 8 electrons,
Third orbit or M-shell will have 18 electrons.
→ If it is the outermost orbit, then it should have not more than 8 electrons.
→ There should be step-wise filling of electrons in different orbits, i.e., electrons are not accompanied in a given orbit if the earlier orbits or shells are incompletely filled.

7. Define valency by taking examples of silicon and oxygen.

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.→ Valency of Silicon: It has electronic configuration: 2,8,4
Thus, the valency of silicon is 4 as these electrons can be shared with others to complete octet.
→ Valency of Oxygen: It has electronic configuration: 2,6
Thus, the valency of oxygen is 2 as it will gain 2 electrons to complete its octet.

Page No: 55

8. Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars. Give any two uses of isotopes.

(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: These are atoms of the same element having the same atomic number, but different mass numbers. For example, chlorine has two isotopes with atomic number 17 but mass numbers 35 and 37 represented by

(iv) Isobars: These 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, Ne has atomic number 10 and sodium has atomic number 11 but both of them have mass numbers as 22 represented by –

Two uses of isotopes:
→ One isotope of uranium is used as a fuel in nuclear reactors.
→ One isotope of cobalt is used in the treatment of cancer.

9. Nahas completely filled K and L shells. Explain.

The atomic number of sodium is 11. So, neutral sodium atom has 11 electrons and its electronic configuration is 2, 8, 1. But Na+ has 10 electrons. Out of 10, K-shell contains 2 and L-shell 8 electrons respectively. Thus, Na+ has completely filled K and L shells.

10. 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.

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:

11. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 16 / 8 X and 18 / 8 X in the sample?

It is given that the average atomic mass of the sample of element X is 16.2 u.
Let the percentage of isotope 18 / 8 X be y%. Thus, the percentage of isotope 16 / 8 X will be (100 – y) %.

Therefore,

18y + 1600 – 16y = 1620
2y + 1600 = 1620
2y = 1620 – 1600
y= 10
Therefore, the percentage of isotope 18 / 8 X is 10%.
And, the percentage of isotope 16 / 8 X is (100 – 10) % = 90%.

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

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.

13.Composition of the nuclei of two atomic species X and Y are given as under

X              Y
Protons =   6              6
Neutrons = 6             8
Give the mass numbers of X and Y. What is the relation between the two species?

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

= 6 + 6
= 12

Mass number of Y = Number of protons + Number of neutrons
= 6 + 8
= 14

These two atomic species X and Y have the same atomic number, but different mass numbers. Hence, they are isotopes.

14. For the following statements, write T for ‘True’ and F for ‘False’.

(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
► False

(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
► False

(c) The mass of an electron is about 1 / 2000times that of proton.

► True

(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.

► False

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

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

17. Number of valence electrons in Cl ion are:
(a) 16
(b) 8
(c) 17
(d) 18
► (b) 8

Page No: 56

18. 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
► (d) 2, 8, 1

19. Complete the following table.

 Atomic number Mass number Number of Neutrons Number of protons Number of electrons Name of the Atomic species 9 − 10 − − − 16 32 − − − Sulphur − 24 − 12 − − − 2 − 1 − − − 1 0 1 0 −

 Atomic number Mass number Number of Neutrons Number of protons Number of electrons Name of the Atomic species 9 19 10 9 9 Fluorine 16 32 16 16 16 Sulphur 12 24 12 12 12 Magnesium 1 2 1 1 1 Deuterium 1 1 0 1 0 Hydrogen ion

## NCERT Solutions for Class 9th Science Chapter 5 : The Fundamental Unit of Life

- - CBSE

CBSE NCERT Solutions for Class 9th Science Chapter 5 : The Fundamental Unit of Life . NCERT Class 9 Science Solutions for Chapter 5. Class IX Science Solutions NCERT

### NCERT Solutions for Class 9th Science Chapter 5 : The Fundamental Unit of Life

In Text Questions

Page No: 59

1. Who discovered cells and how?

An English Botanist, Robert Hooke discovered cells. In 1665, he used self-designed microscope to observe cells in a cork slice.

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

Cells are called the structural and functional unit of life because all the living organisms are made up of cells and also all the functions taking place inside the body of organisms are performed by cells.

Page No: 61

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

The substances like CO2and water move in and out of a cell by diffusion from the region of high concentration to low concentration.

When the concentration of CO2and water is higher in external environment than that inside the cell, CO2and water moves inside the cell. When the concentration outside the cell becomes low and it is high inside the cell, they moves out.

2. Why is the plasma membrane called a selectively permeable membrane?

Plasma membrane called 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.

Page No: 63

1. Fill in the gaps in the following table illustrating differences between prokaryotic and eukaryotic cells.

 Prokaryotic cell Eukaryotic cell 1. Size: generally small ( 1-10 µm) 1 µm== 10-6 m 1. Size: generally large (5-100 µm) 2. Nuclear region: _____________________________ and is known as ________. 2. Nuclear region: well-defined and surrounded by a nuclear membrane 3. Chromosome: single 3. More than one chromosome 4. Membrane-bound cell organelles are absent 4. ___________________________________________

 Prokaryotic cell Eukaryotic cell 1. Size: generally small ( 1-10 µm) 1 µm== 10-6 m 1. Size: generally large (5-100 µm) 2. Nuclear region: poorly defined because of the absence of a nuclear membrane, and is known as nucleoid 2. Nuclear region: well-defined and surrounded by a nuclear membrane 3. Chromosome: single 3. More than one chromosome 4. Membrane-bound cell organelles are absent 4. Membrane-bound cell organelles such as mitochondria, plastids, etc., are present

Page No: 65
1. Can you name the two organelles we have studied that contain their own genetic material?

Mitochondria and plastids

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

If the organisation of a cell is destroyed due to some physical or chemical influence then cell will not be able to perform the basic functions like respiration, nutrition, excretion etc. This may stop all the life activities and may result in its death.

3. Why are lysosomes known as suicide bags?

Lysosomes are called suicide bags because in case of disturbance of their cellular metabolism they digest their own cell by releasing own enzymes.

4. Where are proteins synthesized inside the cell?

The proteins are synthesized in the Ribosome inside the cell.

Page No: 66

Exercise

1. Make a comparison and write down ways in which plant cells are different from animal cells.

 Animal cell Plant cell The do not have cell wall. They have cell wall made up of cellulose. They do not have chloroplast. They contain chloroplast. They have centrosome. They do not have centrosome. Vacuoles are smaller in size. Vacuoles are larger in size. Lysosomes are larger in number. Lysosomes are absent or very few in number Prominent Golgi bodies are present. Subunits of Golgi bodies known as dictyosomes are present.

2. How is a prokaryotic cell different from a eukaryotic cell?

 Prokaryotic cell Eukaryotic cell Most prokaryotes are unicellular. Most eukaryotes are multicellular. Size of the cell is generally small (0.5- 5 µm). Size of the cell is generally large (50- 100 µm). Nuclear region is poorly defined due to the absence of a nuclear membrane or the cell lacks true nucleus. Nuclear region is well-defined and is surrounded by a nuclear membrane, or true nucleus bound by a nuclear membrane is present in the cell. It contains a single chromosome. It contains more than one chromosome. Nucleolus is absent. Nucleolus is present. Membrane-bound cell organelles such as plastids, mitochondria, endoplasmic reticulum, Golgi apparatus, etc. are absent. Cell organelles such as mitochondria, plastids, endoplasmic reticulum, Golgi apparatus, lysosomes, etc. are present. Cell division occurs through binary fission Cell division occurs by mitosis. Prokaryotic cells are found in bacteria and blue-green algae. Eukaryotic cells are found in fungi, plants, and animal cells.

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

If the plasma membrane ruptures or breakdown then the cell will not be able to exchange material from its surrounding by diffusion or osmosis. Thereafter the protoplasmic material will be disappeared and the cell will die.

Page No: 67

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

Golgi apparatus has the function of storage modification and packaging of the products. If there is no Golgi apparatus then the packaging and transporting of materials synthesized by cell will not happen.

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

Mitochondria are known as the powerhouse of cells because energy required for various chemical activities needed for life is released by mitochondria in the form of ATP (Adenosine triphosphate) molecules.

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

Lipids are synthesized in Smooth endoplasmic reticulum (SER) and the proteins are synthesized in rough endoplasmic reticulum (RER).

7. How does an Amoeba obtain its food?

Amoeba takes in food using temporary finger-like extensions of the cell surface which fuse over the food particle forming a food-vacuole as shown in figure. Inside the food vacuole, complex substances are broken down into simpler ones which then diffuse into the cytoplasm. The remaining undigested material is moved to the surface of the cell and thrown out.

8. What is osmosis?

Osmosis is the process in which water molecules moves from the region of high concentration to a region of low concentration through a semi permeable membrane.

9. 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 answer 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.

(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.

## NCERT Solutions for Class 9th Science Chapter 6 : Tissues

- - CBSE

CBSE NCERT Solutions for Class 9th Science Chapter 6 : Tissues. NCERT Class 9 Science Solutions for Chapter 6. Class IX Science Solutions NCERT

### NCERT Solutions for Class 9th Science Chapter 6 : Tissues

In Text Questions

Page No: 69

1. What is a tissue?

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

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

In multicellular organisms, the different types of tissues perform different functions. Since a particular group of cells carry out only a particular function, they do it very efficiently. So, multicellular organisms possess a definite division of labour.

Page No: 74

1. Name types of simple tissues.

Simple permanent tissues are of three types:→ Parenchyma
→ Collenchyma

→ Sclerenchyma
Parenchyma tissue is of further two types:

• Aerenchyma

• Chlorenchyma

2. Where is apical meristem found?

Apical meristem is present at the growing tips of stems and roots.

3. Which tissue makes up the husk of coconut?

Sclerenchyma tissue makes up the husk of coconut.

4. What are the constituents of phloem?

The constituents of phloem are:
→ Sieve tubes
→ Companion cells
→ Phloem parenchyma
→ Phloem fibres

Page No: 78

1. Name the tissue responsible for movement in our body.

► Muscular tissue
2. What does a neuron look like?

Neuron look like a star shaped cell with a tail.

3. Give three features of cardiac muscles.

Three features of cardiac muscles are:
→ Cardiac muscles are involuntary muscles that contract rapidly, but do not get fatigued.
→ The cells of cardiac muscles are cylindrical, branched, and uninucleate.
→ They control the contraction and relaxation of the heart.

4. What are the functions of areolar tissue?

Functions of areolar tissue:
→ It helps in supporting internal organs.
→ It helps in repairing the tissues of the skin and muscles.

Page No: 79

Excercise

1. Define the term “tissue”.

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

2. How many types of elements together make up the xylem tissue? Name them.

Xylem is composed of following elements:
→ Tracheids
→ Vessels
→ Xylem parenchyma
→ Xylem fibres

3. How are simple tissues different from complex tissues in plants?

 Simple tissue Complex tissue These tissues consist of only one type of cells. These tissues are made up of more than one type of cells. The cells are more or less similar in structure and perform similar functions. Different types of cells perform different functions. For example, in the xylem tissue, tracheids help in water transport, whereas parenchyma stores food. Three types of simple tissues in plants are parenchyma, collenchyma, and sclerenchyma. Two types of complex permanent tissues in plants are xylem and phloem.

4. Differentiate between parenchyma, collenchyma and sclerenchyma, on the basis of their cell wall.

 Parenchyma Collenchyma Sclerenchyma Cell walls are relatively thin, and the cells in parenchyma tissues are loosely packed. The cell wall is irregularly thickened at the corners, and there is very little space between the cells. The cell walls are uniformly thickened, and there are no intercellular spaces. The cell wall in this tissue is made up of cellulose. Pectin and hemicellulose are the major constituents of the cell wall. An additional layer of the cell wall composed mainly of lignin is found.

5. What are the functions of the stomata?

The functions of stomata are:
→ The exchange of gases (CO2 and O2) with the atmosphere.
→ The loss of excess water in the form of water vapour which is known as transpiration.
6. Diagrammatically show the difference between the three types of muscle fibres.

The three types of muscle fibres are: Striated muscles, smooth muscles (unstriated muscle fibre), and cardiac muscles.

7. What is the specific function of the cardiac muscle?

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

8. Differentiate between striated, unstriated and cardiac muscles on the basis of their structure and site/location in the body.

 Striated muscle Unstriated muscle Cardiac muscle On the basis of structure: Cells are cylindrical Cells are long Cells are cylindrical Cells are not branched Cells are not branched Cells are branched Cells are multinucleate Cells are uninucleate Cells are uninucleate Alternate light and dark bands are present There are no bands present Faint bands are present Its ends are blunt Its ends are tapering Its ends are flat and wavy On the basis of location: These muscles are present in body parts such as hands, legs, tongue, etc. These muscles control the movement of food in the alimentary canal, the contraction and relaxation of blood vessels, etc. These muscles control the contraction and relaxation of the heart

9. Draw a labelled diagram of a neuron.

10. Name the following:

(a) Tissue that forms the inner lining of our mouth.
► Epithelial tissue

(b) Tissue that connects muscle to bone in humans.
► Tendon

(c) Tissue that transports food in plants.
► Phloem

(d) Tissue that stores fat in our body.

(e) Connective tissue with a fluid matrix.
► Blood

(f) Tissue present in the brain.

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

→ 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

12. Name the regions in which parenchyma tissue is present.

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

13. What is the role of epidermis in plants?

Epidermisis present on the outer surface of the entire plant body which perform following role:
→ It is a protective tissue of the plant body.
→ It protects the plant against mechanical injury.
→ It allows exchange of gases through the stomata.

14. How does the cork act as a protective tissue?

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.

15. Complete the table:

## NCERT Solutions for Class 9th Science Chapter 7 : Diversity In Living Organisms

- - CBSE

CBSE NCERT Solutions for Class 9th Science Chapter 7 : Diversity In Living Organisms. NCERT Class 9 Science Solutions for Chapter 7. Class IX Science Solutions NCERT

### NCERT Solutions for Class 9th Science Chapter 7 : Diversity In Living Organisms

In Text Questions

Page No: 80

1. Why do we classify organisms?

There are millions of organisms on this earth. So, it is harder 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. Classification makes our study easier.

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

Examples of range of variations observed in daily life are:
→ Organisms vary greatly in size-from microscopic bacteria to elephants, whales and large trees.
→ The colour of various animals is quite different. Some worms are even colourless or transparent. Various types of pigments are found in plants.
→ 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.

Page No: 82

1. 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?

The more basic characteristic for classifying organisms is the kind of cells they are made of because different organisms may share same habitat but may have entirely different form and structure. So, the place where they live cannot be a basis of classification.

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

The primary characteristic on which the first division of organisms is made is the nature of the cell – prokaryotic or eukaryotic cell.

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

Plants and animals are put into different categories on the basis of Mode of nutrition.Plants are autotrophs. They can make their food own while animas are heterotrophs which are dependent on others for food. Also, locomotion, absence of chloroplasts etc. make them different.

Page No: 83

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

A primitive 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.As per the body design, the primitve organisms which hav simple structures are different from those so-called advanced organisms which have complex body structure and organization.

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

Yes, because the advanced organisms also were like the primitive ones once. They have acquired their complexity relatively recently. There is a possibility that these advanced or ‘younger’ organisms acquire more complex structures during evolutionary time to compete and survive in the changing environment.

Page No: 85

1. What is the criterion for classification of organisms as belonging to kingdom Monera or Protista?

The organisms belonging to Kingdom Monera are unicellular and prokaryotic whereas the organisms belonging to Kingdom Protista are unicellular and eukaryotic.

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

► Kingdom Protista

3. 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?

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.

Page No: 88

1. Which division among plants has the simplest organisms?

► Division Thallophyta

2. How are pteridophytes different from the phanerogams?

 Pteridophyta Phanerogams They have inconspicuous or less differentiated reproductive organs. They have well developed reproductive organs. They produce naked embryos called spores. They produce seeds. Ferns, Marsilea, Equisetum, etc. are examples of pteridophyta. Pinus, Cycas, fir, etc. are examples of phanerogams.

3. How do gymnosperms and angiosperms differ from each other?

 Gymnosperm Angiosperm They are non-flowering plants. They are flowering plants. Naked seeds not enclosed inside fruits are produced. Seeds are enclosed inside fruits. Pinus, Cedar, fir, Cycas, etc. are some examples of gymnosperms. Coconut, palm, mango, etc. are some examples of angiosperms.

Page No: 94

1. How do poriferan animals differ from coelenterate animals?

 Porifera Coelenterate They are mostly marine, non-motile, and found attached to rocks. They are exclusively marine animals that either live in colonies or have a solitary life-span. They show cellular level of organisation. They show tissue level of organisation. Spongilla, Euplectella,etc. are poriferans. Hydra, sea anemone, corals, etc. are coelenterates.

2. How do annelid animals differ from arthropods?

 Annelids Arthropods The circulatory system of annelids is closed. Arthropods have an open circulatory system. The body is divided into several identical segments. The body is divided into few specialized segments.

3. What are the differences between amphibians and reptiles?

 Amphibian Reptiles They have a dual mode of life. They are completely terrestrial. Scales are absent. Skin is covered with scales. They lay eggs in water. They lay eggs on land. It includes frogs, toads, and salamanders. It includes lizards, snakes, turtles, chameleons, etc.

4. What are the differences between animals belonging to the Aves group and those in the mammalia group?

 Aves Mammals Most birds have feathers and they possess a beak. They do not have feathers and the beak is also absent. They lay eggs. Hence, they are oviparous. Some of them lay eggs and some give birth to young ones. Hence, they are both oviparous and viviparous.

Page No: 97

Excercise

1. What are the advantages of classifying organisms?

Following are the advantages of classifying organisms:
→ It makes us aware of and gives us information regarding the diversity of plants and animals.
→ It makes the study of different kinds of organisms much easier.
→ It tells us about the inter-relationship among the various organisms.
→ It helps us understanding the evolution of organisms.
→ It helps in the development of other life sciences easy.

→ It helps environmentalists to develop new methods of conservation of plants and animals.

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

We choose that characteristics which depends on the first characteristics and determines the rest variety.

3. Explain the basis for grouping organisms into five kingdoms.

The basis for grouping organisms into five kingdoms are:
→ Complexity of cell structure – There are two broad categories of cell structure: Prokaryotic and Eukaryotic. Thus, two broad groups can be formed, one having prokaryotic cell structure and the other having eukaryotic cell structure. Presence or absence of cell wall is another important characteristic.
→ Unicellular and multicellular organisms – This characteristic makes a very basic distinction in the body designs of organisms and helps in their broad categorizations.

→ Cell Wall: Presence and absence of cell wall leads into grouping.

→ Mode of nutrition -Organisms basically have two types of nutritions – autotrophic who can manufacture their own food and heterotrophic who obtain their food from external environment, i.e., from other organisms). Thus, organisms can be broadly classified into different groups on the basis of their mode of nutrition.

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

The major divisions in Kingdom Plantae are:
→Thallophyta
→ Bryophyta
→ Pteridophyta
→ Gymnosperms
→ Angiosperms

The following points constitute the basis of these divisions:
→ Whether the plant body has well differentiated, distinct components.
→ whether the differentiated plant body has special tissues for the transport of water and other substances.
→ The ability to bear seeds.
→ Whether the seeds are enclosed within fruits.

5. How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals?

The characteristics used to classify plants is different from animals because the basic design are different, based on the need to make their own food (plants) or acquire food (animals).
Criteria for deciding divisions in plants are:
→ Differentiated/ Undifferentiated plant body
→ Presence/ absence of vascular tissues
→With/without seeds
→ Naked seeds/ seeds inside fruits

But the animals can’t be divided into groups on these criteria. It is because the basic designs of animals are very different from plants. They are divided on the basis of their body structure.

6. Explain how animals in Vertebrata are classified into further subgroups.

Animals in Vertebrata are classified into five classes:

(i) Class Pisces: This class includes fish such as Scoliodon, tuna, rohu, shark, etc. These animals mostly live in water. Hence, they have special adaptive features such as a streamlined body, presence of a tail for movement, gills, etc. to live in water.

(ii) Class Amphibia: It includes frogs, toads, and salamanders. These animals have a dual mode of life. In the larval stage, the respiratory organs are gills, but in the adult stage, respiration occurs through the lungs or skin. They lay eggs in water.

(iii) Class Reptilia: It includes reptiles such as lizards, snakes, turtles, etc. They usually creep or crawl on land. The body of a reptile is covered with dry and cornified skin to prevent water loss. They lay eggs on land.

(iv) Class Aves: It includes all birds such as sparrow, pigeon, crow, etc. Most of them have feathers. Their forelimbs are modified into wings for flight, while hind limbs are modified for walking and clasping. They lay eggs.

(v) Class Mammalia: It includes a variety of animals which have milk producing glands to nourish their young ones. Some lay eggs and some give birth to young ones. Their skin has hair as well as sweat glands to regulate their body temperature.

## NCERT Solutions for Class 9th Science Chapter 8 : Motion

- - CBSE

CBSE NCERT Solutions for Class 9th Science Chapter 8 : Motion. NCERT Class 9 Science Solutions for Chapter 8. Class IX Science Solutions NCERT

### NCERT Solutions for Class 9th Science Chapter 8 : Motion

In Text Question

Page No: 100

1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Yes,an object can have zero displacement even when it has moved through a distance.This happens when final position of the object coincides with its initial position. For example,if a person moves around park and stands on place from where he started then here displacement will be zero.

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 minutes 20 seconds from his initial position?

Given, Side of the square field= 10m
Therefore, perimeter = 10 m x 4 = 40 m
Farmer moves along the boundary in 40s.
Displacement after 2 m 20 s = 2 x 60 s + 20 s = 140 s =?
Since in 40 s farmer moves 40 m

Therefore, in 1s distance covered by farmer = 40 / 40 m = 1m

Therefore, in 140s distance covered by farmer = 1 × 140 m = 140 mNow, number of rotation to cover 140 along the boundary= Total Distance / Perimeter
= 140 m  / 40 m  = 3.5 round
Thus, after 3.5 round farmer will at point C of the field.

Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from intial position.

3. 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.

None of the statement is true for displacement First statement is false because displacement can be zero. Second statement is also false because displacement is less than or equal to the distance travelled by the object.

Page No: 102

1. Distinguish between speed and velocity.

 Speed Velocity Speed is the distance travelled by an object in a given interval of time. Velocity is the displacement of an object in a given interval of time. Speed = distance / time Velocity = displacement / time Speed is scalar quantity i.e. it has only magnitude. Velocity is vector quantity i.e. it has both magnitude as well as direction.

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

The magnitude of average velocity of an object is equal to its average speed, only when an object is moving in a straight line.

3. What does the odometer of an automobile measure?

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

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

An object having uniform motion has a straight line path.

5. 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.

Speed= 3 × 108 m s−1
Time= 5 min = 5 x 60 = 300 secs.

Distance= Speed x Time
Distance= 3 × 108 m s−1 x 300 secs. = 9 x 1010 m

Page No: 103

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

(i) A body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time.
(ii) A body is said to be in nonuniform acceleration if the rate of change of its velocity is not constant.

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

3. 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.

Page No: 107

1. What is the nature of the distance – ‘time graphs for uniform and non-uniform motion of an object?

When the motion is uniform,the distance time graph is a straight line with a slope.

When the motion is non uniform, the distance time graph is not a straight line.It can be any curve.

2. What can you say about the motion of an object whose distance – time graph is a straight line parallel to the time axis?

If distance time graph is a straight line parallel to the time axis, the body is at rest.

3. What can you say about the motion of an object if its speed – ‘time graph is a straight line parallel to the time axis?

If speed time graph is a straight line parallel to the time axis, the object is moving uniformly.

4. What is the quantity which is measured by the area occupied below the velocity -time graph?

The area below velocity-time graph gives the distance covered by the object.

Page No: 109

1. 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.

Initial speed of the bus, u= 0

Acceleration, a = 0.1 m/s2
Time taken, t = 2 minutes = 120 s

(a) v= u + at
v= 0 + 0.1 × 120
v= 12 ms–1

(b) According to the third equation of motion:
v2 – u2= 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.

Page No: 110

2. 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.

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:
v2= u2+ 2 as
(0)2= (25)2+ 2 ( – 0.5) s

Where, s is the distance covered by the train

The train will cover a distance of 625 m before it comes to rest.
3. 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?

Initial Velocity of trolley, u= 0 cms-1
Acceleration, a= 2 cm s-2
Time, t= 3 s
We know that final velocity, v= u + at = 0 + 2 x 3 cms-1
Therefore, The velocity of train after 3 seconds = 6 cms-1
4. A racing car has a uniform acceleration of 4 m s – ‘2. What distance will it cover in 10 s after start?

Initial Velocity of the car, u=0 ms-1
Acceleration, a= 4 m s-2
Time, t= 10 s
We know Distance, s= ut + (1/2)at2
Therefore, Distance covered by car in 10 second= 0 × 10 + (1/2) × 4 × 102
= 0 + (1/2) × 4× 10 × 10 m
= (1/2)× 400 m
= 200 m

5. 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?

Given Initial velocity of stone, u=5 m s-1
Downward of negative Acceleration, a= 10 m s-2
We know that 2 as= v2– u2

Page No: 112

Excercise

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 minutes 20 s?

Diameter of circular track (D) = 200 m
Radius of circular track (r) = 200 / 2=100 m
Time taken by the athlete for one round (t) = 40 s
Distance covered by athlete in one round (s) = 2π r
= 2 x ( 22 / 7 ) x 100
Speed of the athlete (v) = Distance / Time
= (2 x 2200) / (7 x 40)
= 4400 / 7 × 40
Therefore, Distance covered in 140 s = Speed (s) × Time(t)
= 4400 / (7 x 40) x (2 x 60 + 20)
= 4400 / ( 7 x 40) x 140
= 4400 x 140 /7 x 40
= 2200 m

Number of round in 40 s =1 round
Number of round in 140 s =140/40
=3 1/2
After taking start from position X,the athlete will be at postion Y after 3 1/2 rounds as shown in figure

Hence, Displacement of the athlete with respect to initial position at x= xy

= Diameter of circular track
= 200 m

2.  Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Total Distance covered from AB = 300 m
Total time taken = 2 x 60 + 30 s
=150 s

Therefore, Average Speed from AB = Total Distance / Total Time
=300 / 150 m s-1
=2 m s-1
Therefore, Velocity from AB =Displacement AB / Time = 300 / 150 m s-1
=2 m s-1
Total Distance covered from AC =AB + BC
=300 + 200 m
Total time taken from A to C = Time taken for AB + Time taken for BC
= (2 x 60+30)+60 s
= 210 s
Therefore, Average Speed from AC = Total Distance /Total Time
= 400 /210 m s-1
= 1.904 m s-1
Displacement (S) from A to C = AB – BC
= 300-100 m
= 200 m

Time (t) taken for displacement from AC = 210 s
Therefore, Velocity from AC = Displacement (s) / Time(t)
= 200 / 210 m s-1
= 0.952 m s-1
3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h−1. On his return trip along the same route, there is less traffic and the average speed is 40 km h−1. What is the average speed for Abdul’s trip?

The distance Abdul commutes while driving from Home to School = S
Let us assume time taken by Abdul to commutes this distance = t1
Distance Abdul commutes while driving from School to Home = S
Let us assume time taken by Abdul to commutes this distance = t2
Average speed from home to school v1av = 20 km h-1
Average speed from school to home v2av = 30 km h-1
Also we know Time taken form Home to School t1 =S / v1av
Similarly Time taken form School to Home t2 =S/v2av
Total distance from home to school and backward = 2 S
Total time taken from home to school and backward (T) = S/20+ S/30
Therefore, Average speed (Vav) for covering total distance (2S) = Total Distance/Total Time
= 2S / (S/20 +S/30)
= 2S / [(30S+20S)/600] = 1200S / 50S
= 24 kmh-1

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s−2 for 8.0 s. How far does the boat travel during this time?

Given Initial velocity of motorboat,  u = 0
Acceleration of motorboat,  a = 3.0 m s-2
Time under consideration,  t = 8.0 s
We know that Distance, s = ut + (1/2)at2
Therefore, The distance travel by motorboat = 0 x 8 + (1/2)3.0 x 8 2
= (1/2) x 3 x 8 x 8 m
= 96 m
5. A driver of a car travelling at 52 km h−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h−1 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?

As given in the figure below PR and SQ are the Speed-time graph for given two cars with initial speeds 52 kmh-1 and 3 kmh-1 respectively.

Distance Travelled by first car before coming to rest =Area of △ OPR
= (1/2) x OR x OP
= (1/2) x 5 s x 52 kmh-1
= (1/2) x 5 x (52 x 1000) / 3600) m
= (1/2) x 5x (130 / 9) m
= 325 / 9 m
= 36.11 m
Distance Travelled by second car before coming to rest =Area of △ OSQ
= (1/2) x OQ x OS
= (1/2) x 10 s x 3 kmh-1
= (1/2) x 10 x (3 x 1000) / 3600) m
= (1/2) x 10 x (5/6) m
= 5 x (5/6) m
= 25/6 m
= 4.16 m

6. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?

(c) How far has C travelled when B passes A?
(d)How far has B travelled by the time it passes C?

(a) Object B
(b) No
(c) 5.714 km
(d) 5.143 km

Therefore, Speed = slope of the graph
Since slope of object B is greater than objects A and C, it is travelling the fastest.

(b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road.

On the distance axis:
7 small boxes = 4 km
Therefore,1 small box = 4 / 7 Km
Initially, object C is 4 blocks away from the origin.
Therefore, Initial distance of object C from origin = 16 / 7 Km
Distance of object C from origin when B passes A = 8 km
Distance covered by C

Page No: 113

7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s−2, with what velocity will it strike the ground? After what time will it strike the ground?

Let us assume, the final velocity with which ball will strike the ground be ‘v’ and time it takes to strike the ground be ‘t’
Initial Velocity of ball,  u =0
Distance or height of fall,  s =20 m
Downward acceleration,  a =10 m s-2
As we know, 2as =v2-u2
v2 = 2as+ u2
= 2 x 10 x 20 + 0
= 400
∴ Final velocity of ball, v = 20 ms-1
t = (v-u)/a
∴Time taken by the ball to strike = (20-0)/10
= 20/10
= 2 seconds

8. The speed-time graph for a car is shown is Fig. 8.12.

(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

(a)

The shaded area which is equal to 1 / 2 x 4 x 6 = 12 m represents the distance travelled by the car in the first 4 s.

(b)

The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.

9. State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity.
(b) an object moving in a certain direction with an acceleration in the perpendicular direction.

(a) Possible
When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2.

(b) Possible
When a car is moving in a circular track, its acceleration is perpendicular to its direction.

10. 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.

Radius of the circular orbit, r= 42250 km
Time taken to revolve around the earth, t= 24 h
Speed of a circular moving object, v= (2π r)/t
=[2× (22/7)×42250 × 1000] / (24 × 60 × 60)
=(2×22×42250×1000) / (7 ×24 × 60 × 60) m s-1
=3073.74 m s -1

## NCERT Solutions for Class 9th Science Chapter 9 : Force and Laws of Motion

- - CBSE

CBSE NCERT Solutions for Class 9th Science Chapter 9 : Force and Laws of Motion . NCERT Class 9 Science Solutions for Chapter 9. Class IX Science Solutions NCERT

### NCERT Solutions for Class 9th Science Chapter 9 : Force and Laws of Motion

In Text Questions

Page No: 118

1. Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five-rupees coin and a one-rupee coin?

Inertia is the measure of the mass of the body. The greater is the mass of the body; the greater is its inertia and vice-versa.

(a) Mass of a stone is more than the mass of a rubber ball for the same size. Hence, inertia of the stone is greater than that of a rubber ball.

(b) Mass of a train is more than the mass of a bicycle. Hence, inertia of the train is greater than that of the bicycle.

(c) Mass of a five rupee coin is more than that of a one-rupee coin. Hence, inertia of the five rupee coin is greater than that of the one-rupee coin.

2. In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.
Also identify the agent supplying the force in each case.

The velocity of football changes four times.
First, when a football player kicks to another player, second when that player kicks the football to the goalkeeper. Third when the goalkeeper stops the football. Fourth when the goalkeeper kicks the football towards a player of his own team.

Agent supplying the force:
→ First case – First player

→ Second case – Second player
→ Third case – Goalkeeper
→ Fourth case – Goalkeeper

3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Some leaves of a tree get detached when we shake its branches vigorously because branches comes in motion while the leaves tend to remain at rest due to inertia of rest.

4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

In a moving bus, a passenger moves with the bus due to inertia of motion. As the driver applies brakes, the bus comes to rest. But, the passenger tries to maintain to inertia of motion. As a result, a forward force is exerted on him.

Similarly, the passenger tends to fall backwards when the bus accelerates from rest because when the bus accelerates, the inertia of rest of the passenger tends to oppose the forward motion of the bus. Hence, the passenger tends to fall backwards when the bus accelerates forward.

Page No: 126

1. If action is always equal to the reaction, explain how a horse can pull a cart.

A horse pushes the ground in the backward direction. According to Newton’s third law of motion, a reaction force is exerted by the Earth on the horse in the forward direction. As a result, the cart moves forward.

2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then a reaction force is exerted on him by the ejecting water in the backward direction. This is because of Newton’s third law of motion. As a result of the backward force, the stability of the fireman
decreases. Hence, it is difficult for him to remain stable while holding the hose.

3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s−1. Calculate the initial recoil velocity of the rifle.

Mass of the rifle, m1= 4 kg

Mass of the bullet, m2= 50g= 0.05 kg
Recoil velocity of the rifle= v1

Bullet is fired with an initial velocity, v2= 35m/s
Initially, the rifle is at rest.
Thus, its initial velocity, v= 0
Total initial momentum of the rifle and bullet system= (m1+m2)v= 0

Total momentum of the rifle and bullet system after firing:

= m1v1 + m2v2= 0.05 × 35= 4v1 + 1.75

According to the law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing  4v1 + 1.75= 0
v1= -1.75 / 4= -0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.

Page No: 127

4. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s−1 and 1 m s−1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m s−1. Determine the velocity of the second object.

Mass of one of the objects, m1 = 100 g = 0.1 kg
Mass of the other object, m2 = 200 g = 0.2 kg
Velocity of m1 before collision, v1= 2 m/s

Velocity of m2 before collision, v2= 1 m/s

Velocity of m1 after collision, v3= 1.67 m/s

Velocity of m2 after collision= v4

According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision

Therefore, m1v1 + m2v2 = m1v3 + m2v4
2(0.1) + 1(0.2) = 1.67(0.1) + v4(0.2)
0.4 = 0.167 + 0.2v4
v4= 1.165 m/s

Hence, the velocity of the second object becomes 1.165 m/s after the collision.

Page No: 128

Excercises

1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

Yes, an object may travel with a non-zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the up thrust and the velocity of air. The net force on the drop is zero.

2. When a carpet is beaten with a stick, dust comes out of it. Explain.

When the carpet is beaten, it is suddenly set into motion. The dust particles tend to remain at rest due to inertia of rest, therefore the dust comes out of it.

3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?

When a bust starts suddenly, the lower part of the luggage kept on the roof being in contact with the bus begins to move forward with the speed of bus, but the upper part tends to remain at rest due to inertia of rest. Therefore, the upper part is left behind and hence luggage falls backward. So, it is advised to tie any luggage kept on the roof of a bus with a rope.

4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.

(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.

The ball slows down and comes to rest due to opposing forces of air resistance and frictional force on the ball opposing its motion. Therefore the choice (c) there is a force on the ball opposing the motion is correct.

5.  A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg).

Initial velocity, u = 0

Distance travelled, s = 400 m
Time taken, t = 20 s

We know, s = ut + ½ at2

Or, 400 = 0 + ½ a (20)2
Or, a = 2 ms–2
Now, m = 7 MT = 7000 kg, a = 2 ms–2

Or, F = ma = 7000 × 2 = 14000 N Ans.

6. A stone of 1 kg is thrown with a velocity of 20 m s−1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

Initial velocity of the stone, u= 20 m/s
Final velocity of the stone, v= 0
Distance covered by the stone, s= 50 m

Since, v2u2

2 = 2as,
Or, 0 – 202 = 2a × 50,
Or, a = – 4 ms-2
Force of friction, F = ma = – 4N

7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force;
(b) the acceleration of the train; and
(c)the force of wagon 1 on wagon 2.

(a) Force exerted by the engine, F = 40000 N
Frictional force offered by the track, Ff = 5000 N
Net accelerating force, Fa = FFf = 40000 − 5000 = 35000 N
Hence, the net accelerating force is 35000 N.

(b) Acceleration of the train = a
The engine exerts a force of 40000 N on all the five wagons.
Net accelerating force on the wagons, Fa = 35000 N
Mass of the wagons, m = Mass of a wagon x Number of wagons
Mass of a wagon = 2000 kg
Number of wagons = 5
m = 2000 × 5 = 10000 kg
Total mass, M = m  = 10000 kg
From Newton’s second law of motion:
Fa= Ma
a=Fam   = 35000 10000    = 3.5 ms-2
Hence, the acceleration of the wagons and the train is 3.5 m/s2.

(c) Mass of all the wagons except wagon 1 is 4 × 2000 = 8000 kg
Acceleration of the wagons = 3.5 m/s2
Thus, force exerted on all the wagons except wagon 1
= 8000 × 3.5 = 28000 N
Therefore, the force exerted by wagon 1 on the remaining four wagons is 28000  N.
Hence, the force exerted by wagon 1 on wagon 2 is 28000  N.

8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s−2?

Mass of the automobile vehicle, m= 1500 kg
Final velocity, v= 0 (finally the automobile stops)
Acceleration of the automobile, a= −1.7 ms−2
From Newton’s second law of motion:
Force = Mass x Acceleration = 1500 x (−1.7) = −2550 N
Hence, the force between the automobile and the road is −2550 N, in the direction opposite to the motion of the automobile.

9. What is the momentum of an object of mass m, moving with a velocity v?

 (a) (mv)2 (b) mv2 (c) ½ mv2 (d) mv

(d) mv
Mass of the object = m
Velocity = v
Momentum = Mass x Velocity
Momentum = mv

10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

The cabinet will move with constant velocity only when the net force on it is zero.
Therefore, force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet.

11. Two objects, each of mass 1.5 kg are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms−1 before the collision during which they stick together. What will be the velocity of the combined object after collision?

Mass of one of the objects, m1 = 1.5 kg

Mass of the other object, m2 = 1.5 kg
Velocity of m1 before collision, u1 = 2.5 m/s
Velocity of m2, moving in opposite direction before collision, u2 = −2.5 m/s
Let v be the velocity of the combined object after collision. By the law of conservation of momentum,
Total momentum after collision = Total momentum before collision,
Or, (m1 + m2) v = m1u1 + m2u2
Or, (1.5 + 1.5) v = 1.5 × 2.5 +1.5 × (–2.5) [negative sign as moving in opposite direction] Or, v = 0 ms–1

Page No: 129

12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

The logic is that Action and Reaction always act on different bodies, so they can not cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so the truck does not move.

13. A hockey ball of mass 200 g travelling at 10 m s−1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s−1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Mass of the hockey ball, m = 200 g = 0.2 kg

Hockey ball travels with velocity, v1 = 10 m/s
Initial momentum = mv1
Hockey ball travels in the opposite direction with velocity, v2 = −5 m/s
Final momentum = mv2
Change in momentum = mv1mv2 = 0.2 [10 − (−5)] = 0.2 (15) = 3 kg m s−1
Hence, the change in momentum of the hockey ball is 3 kg m s−1.

14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s−1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Initial velocity, u= 150 m/s
Final velocity, v= 0 (since the bullet finally comes to rest)
Time taken to come to rest, t= 0.03 s
According to the first equation of motion, v= u + at
Acceleration of the bullet, a
0 = 150 + (a × 0.03 s)a = -150 / 0.03 = -5000 m/s2

According to the third equation of motion:
v2= u2+ 2as
0 = (150)2+ 2 (-5000)
= 22500 / 10000
= 2.25 m

Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Force, F = Mass × Acceleration
Mass of the bullet, m = 10 g = 0.01 kg
Acceleration of the bullet, a = 5000 m/s2
F = ma = 0.01 × 5000 = 50 N
Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.

15. An object of mass 1 kg travelling in a straight line with a velocity of 10 m s−1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Mass of the object, m1 = 1 kg
Velocity of the object before collision, v1 = 10 m/s
Mass of the stationary wooden block, m2 = 5 kg
Velocity of the wooden block before collision, v2 = 0 m/s
∴ Total momentum before collision = m1 v1 + m2 v2
= 1 (10) + 5 (0) = 10 kg m s−1

It is given that after collision, the object and the wooden block stick together.
Total mass of the combined system = m1 + m2
Velocity of the combined object = v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m1 v1 + m2 v2 = (m1 + m2) v
1 (10) + 5 (0) = (1 + 5) v
v = 10 / 6
= 5 / 3

The total momentum after collision is also 10 kg m/s.
Total momentum just before the impact = 10 kg m s−1
Total momentum just after the impact = (m1 + m2) v = 6 × 5 / 3 = 10 kg ms-1
Hence, velocity of the combined object after collision = 5 / 3 ms-1

16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s−1 to 8 m
s−1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Initial velocity of the object, u = 5 m/s
Final velocity of the object, v = 8 m/s
Mass of the object, m = 100 kg
Time take by the object to accelerate, t = 6 s
Initial momentum = mu = 100 × 5 = 500 kg m s−1
Final momentum = mv = 100 × 8 = 800 kg m s−1
Force exerted on the object, F = mv – mu / t
= m (v-u) / t
= 800 – 500
=  300 / 6
= 50 N

Initial momentum of the object is 500 kg m s−1.
Final momentum of the object is 800 kg m s−1.
Force exerted on the object is 50 N.

17. Akhtar, Kiran and Rahul were riding in a motorocar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.

The suggestion made by Kiran that the insect suffered a greater change in momentum as compared to the change in momentum of the motor car is wrong.

The suggestion made by Akhtar that the motor car exerted a larger force on the insect because of large velocity of motor car is also wrong. The explanation put forward by Rahul is correct. On collision of insect with motor car, both experience the same force as action and reaction are always equal and opposite. Further, changes in their momenta are also the same. Only the signs of changes in momenta are opposite, i.e., change in momenta of the two occur in opposite directions, though magnitude of change in momentum of each is the same.
18. How much momentum will a dumbbell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s−2.

Mass of the dumbbell, m = 10 kg
Distance covered by the dumbbell, s = 80 cm = 0.8 m
Acceleration in the downward direction, a = 10 m/s2
Initial velocity of the dumbbell, u = 0
Final velocity of the dumbbell (when it was about to hit the floor) = v
According to the third equation of motion:
v2 = u2 + 2as
v2 = 0 + 2 (10) 0.8
v = 4 m/s
Hence, the momentum with which the dumbbell hits the floor is
= mv = 10 × 4 = 40 kg m s−1

Page No: 130

1. The following is the distance-time table of an object in motion:

 Time in seconds Distance in metres 0 0 1 1 2 8 3 27 4 64 5 125 6 216 7 343

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
(b)What do you infer about the forces acting on the object?

(a) There is an unequal change of distance in an equal interval of time.
Thus, the given object is having a non – uniform motion. Since the velocity of the object increases with time, the acceleration is increasing.
(b) The object is in accelerated condition. According to Newton’s second law of motion, the force acting on an object is directly proportional to the acceleration produced in the object. So, we can say unbalanced force is acting on the object.

2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s−2. With what force does each person push the motorcar?
(Assume that all persons push the motorcar with the same muscular effort)

Mass of the motor car = 1200 kg
Only two persons manage to push the car. Hence, the acceleration acquired by the car is given by the third person alone.
Acceleration produced by the car, when it is pushed by the third person,
a = 0.2 m/s2
Let the force applied by the third person be F.
From Newton’s second law of motion:
Force = Mass × Acceleration
F = 1200 × 0.2 = 240 N
Thus, the third person applies a force of magnitude 240 N.
Hence, each person applies a force of 240 N to push the motor car.

3. A hammer of mass 500 g, moving at 50 m s−1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

Mass of the hammer, m = 500 g = 0.5 kg
Initial velocity of the hammer, u = 50 m/s
Time taken by the nail to the stop the hammer, t = 0.01 s
Velocity of the hammer, v = 0 (since the hammer finally comes to rest)
From Newton’s second law of motion:
Force, = m(v-u) / t
= 0.5(0-50) / 0.01
= -2500 N

The hammer strikes the nail with a force of −2500 N. Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e., +2500 N.

4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

Mass of the motor car, m = 1200 kg
Initial velocity of the motor car, u = 90 km/h = 25 m/s
Final velocity of the motor car, v = 18 km/h = 5 m/s
Time taken, t = 4 s
According to the first equation of motion:
v = u + at
5 = 25 + a (4)
a = − 5 m/s2
Negative sign indicates that its a retarding motion i.e. velocity is decreasing.
Change in momentum = mv mu = m (vu)
= 1200 (5 − 25) = − 24000 kg m s−1
Force = Mass × Acceleration
= 1200 × − 5 = − 6000 N
Acceleration of the motor car = − 5 m/s2
Change in momentum of the motor car = − 24000 kg m s−1
Hence, the force required to decrease the velocity is 6000 N.

## NCERT Solutions for Class 9th Science Chapter 10 : Gravitation

- - CBSE

CBSE NCERT Solutions for Class 9th Science Chapter 10 : Gravitation . NCERT Class 9 Science Solutions for Chapter 10. Class IX Science Solutions NCERT

### NCERT Solutions for Class 9th Science Chapter 10 : Gravitation

In Text Questions

Page No: 134

1. State the universal law of gravitation

The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
For two objects of masses m1and m2 and the distance between them r, the force (F) of attraction acting between them is given by the universal law of gravitation as:

2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

Let ME be the mass of the Earth and mbe the mass of an object on its surface. If R is the radius of the Earth, then according to the universal law of gravitation, the gravitational force (F) acting between the Earth and the object is given by the relation:

Page No: 136

1. What do you mean by free fall?

Gravity of earth attracts every object towards its center. When an object is dropped from a certain height , it begins to fall towards Earth’s surface under the influence of gravitational force. Such a motion of object is called free fall.

2. What do you mean by acceleration due to gravity?

When an object falls freely towards the surface of earth from a certain height, then its velocity changes. This change in velocity produces acceleration in the object which is known as acceleration due to gravity denoted bu letter g . The value of acceleration due to gravity is g= 9.8 m/s2.

Page No: 138

1. What are the differences between the mass of an object and its weight?

 Mass Weight Mass is the quantity of matter contained in the body. Weight is the force of gravity acting on the body. It is the measure of inertia of the body. It is the measure of gravity. Mass is a constant quantity. Weight is not a constant quantity. It is different at different places. It only has magnitude. It has magnitude as well as direction. Its SI unit is kilogram (kg). Its SI unit is the same as the SI unit of force, i.e., Newton (N).

2.  Why is the weight of an object on the moon 1/6th its weight on the earth?

The mass of moon is 1/100 times and its radius 1/4 times that of earth. As a result, the gravitional attraction on the moon is about one sixth when compared to earth. Hence, the the weight of an object on the moon 1/6th its weight on the earth.

Page No: 141

1. Why is it difficult to hold a school bag having a strap made of a thin and strong string?

It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large. This is because the pressure is inversely proportional to the surface area on which the force acts. The smaller is the surface area; the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulder is very large.

2. What do you mean by buoyancy?

The upward force exerted by a liquid on an object that is immersed in it is known as buoyancy.

3. Why does an object float or sink when placed on the surface of water?

→ An object sink in water if its density is greater than that of water.
→ An object floats in water if its density is less than that of water.

Page No: 142

1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

When we weigh our body, an upward force acts on it. This upward force is the buoyant force. As a result, the body gets pushed slightly upwards, causing the weighing machine to show a reading less than the actual value.

2. You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?

The cotton bag is heavier than the iron bar. The cotton bag experiences larger up thrust of air than the iron bar. So, the weighing machine indicates a smaller mass for cotton bag than its actual mass.

Page No: 143

Excercises

1. How does the force of gravitation between two objects change when the distance between them is reduced to half?

According to Universal Law of gravitation , the gravitational force of attrection between any two objects of mass
M and m is proportional to the product of their masses
and inversly proportional to the square of distance r between them
So, force F is given by

Hence, if the distance is reduced to half, then the gravitational force becomes four times larger than the previous value.

2. Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?

All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.

3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 106 m).

Given that,
Mass of the body, m= 1 kg
Mass of the Earth, M= 6 x 1024 kg
Radius of the earth, R= 6.4 x 106 m
Now magnitude of the gravitational force (F) between the earth and the body can be given as,

4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

According to the universal law of gravitation, two objects attract each other with equal force, but in opposite directions. The Earth attracts the moon with an equal force with which the moon attracts the earth.

5. If the moon attracts the earth, why does the earth not move towards the moon?

The Earth and the moon experience equal gravitational forces from each other. However, the mass of the Earth is much larger than the mass of the moon. Hence, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. For this reason, the Earth does not move towards the moon.

Page No: 144

6. What happens to the force between two objects, if
(i) the mass of one object is doubled?
(ii) the distance between the objects is doubled and tripled?
(iii) the masses of both objects are doubled?

From Universal law of , force exerted on an object of mass m by earth is given by

So as the mass of any one of the object is doubled the force is also doubled

(ii) The force F is inversely proportional to the distance between the objects. So if the distance between two objects is doubled then the gravitational force of attraction between them is reduced to one fourth of its original value. Similarly f the distance between two objects is tripled , then the gravitational force of attraction becomes one ninth of its original value.

(iii) Again fron Universal law of attraction from equation 1 force F is directly proportional to the product of both the masses. So if both the masses are doubled then the gravitational force of attraction becomes four times the original value.

7. What is the importance of universal law of gravitation?

Universal law of Gravitation is important because it it tells us about:
→ the force that is responsible for binding us to Earth.
→ the motion of moon around the earth
→ the motion of planets around the sun
→ the tides formed by rising and falling of water level in the ocean are due to the gravitational force exerted by both sun and moon on the earth.

8. What is the acceleration of free fall?

Acceleration of free fall is the acceleration produced when a body falls under the influence of the force of gravitation of the earth alone. It is denoted by g and its value on the surface of the earth is 9.8 ms-2.

9. What do we call the gravitational force between the Earth and an object?

Gravitational force between the earth and an object is known as the weight of the object.

10. Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of gis greater at the poles than at the equator].

Weight of a body on the Earth is given by:
Wmg
Where,
m= Mass of the body
g= Acceleration due to gravity
The value of gis greater at poles than at the equator. Therefore, gold at the equator weighs less than at the poles. Hence, Amit’s friend will not agree with the weight of the gold bought.

11. Why will a sheet of paper fall slower than one that is crumpled into a ball?

When a sheet of paper is crumbled into a ball, then its density increases. Hence, resistance to its motion through the air decreases and it falls faster than the sheet of paper.
12. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the Earth. What is the weight in newtons of a 10 kg object on the moon and on the Earth?

Weight of an object on the moon = 1/6 x Weight of an object on the Earth
Also,
Weight = Mass x Acceleration
Acceleration due to gravity, g = 9.8 m/s2
Therefore, weight of a 10 kg object on the Earth = 10 x 9.8 = 98 N
And, weight of the same object on the moon= 1.6 x 9.8 = 16.3 N.
13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

(i) the maximum height to which it rises.
(ii) the total time it takes to return to the surface of the earth.

According to the equation of motion under gravity:
v2 – u2= 2 gs
Where,
u= Initial velocity of the ball
v= Final velocity of the ball
s= Height achieved by the ball
g= Acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., v= 0
u= 49 m/s
During upward motion, g= – 9.8 m s-2
Let h be the maximum height attained by the ball.
Hence,

Let t be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:
vu + gt
We get,
0= 49 + t x (- 9.8)
9.8t= 49
t= 49 / 9.8= 5s
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 s

14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

According to the equation of motion under gravity:
v2 − u2 = 2 gs
Where,
u = Initial velocity of the stone = 0
v = Final velocity of the stone
s = Height of the stone = 19.6 m
g = Acceleration due to gravity = 9.8 m s−2
∴ v2 − 02 = 2 × 9.8 × 19.6
v2 = 2 × 9.8 × 19.6 = (19.6)2
v = 19.6 m s− 1
Hence, the velocity of the stone just before touching the ground is 19.6 m s−1.

15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

According to the equation of motion under gravity:
v2 − u2 = 2 gs
Where,
u = Initial velocity of the stone = 40 m/s
v = Final velocity of the stone = 0
s = Height of the stone
g = Acceleration due to gravity = −10 m s−2
Let h be the maximum height attained by the stone.
Therefore,
0 – (40)2 = 2 x h x (-10)
h= 40 x 40 / 20 = 80 m

Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m
Net displacement of the stone during its upward and downward journey
= 80 + (−80) = 0

16. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 1024 kg and of the Sun = 2 × 1030 kg. The average distance between the two is 1.5 × 1011m.

According to question,
MSun = Mass of the Sun = 2 × 1030 kg
MEarth = Mass of the Earth = 6 × 1024 kg
R = Average distance between the Earth and the Sun = 1.5 × 1011m
From Universal law of gravitation,

17. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

Let t be the point at which two stones meet and let h be their height from the ground. It is given in the question that height of the tower is H= 100m

Now first consider the stone which falls from the top of the tower. So distance covered by this stone at time t can be calculated using equation of motion

18. A ball thrown up vertically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.

(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey.
Hence, it has taken 3 s to attain the maximum height.
Final velocity of the ball at the maximum height, v = 0
Acceleration due to gravity, g = −9.8 m s−2
Equation of motion, v = u + gt will give,
0 = u + (−9.8 × 3)
u = 9.8 × 3 = 29.4 ms− 1
Hence, the ball was thrown upwards with a velocity of 29.4 m s−1.

(b) Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, u = 29.4 m s−1
Final velocity, = 0
Acceleration due to gravity, g = −9.8 m s−2
From the equation of motion, s= ut + 1/2 at2
h= 29.4 x 3 + 1/2 x -9.8 x (3)2 = 44.1 m

(c) Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.
In this case,
Initial velocity, u = 0
Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in 4 s − 3 s = 1 s.
Equation of motion, s= ut + 1/2 gt2 will give,
s= 0 x t + 1/2 x 9.8 x 12 = 4.9 m
Total height = 44.1 m
This means that the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4 seconds.

Page No: 145

19. In what direction does the buoyant force on an object immersed in a liquid act?

An object immersed in a liquid experiences buoyant force in the upward direction.

20. Why does a block of plastic released under water come up to the surface of water?

For an object immersed in water two force acts on it
→ gravitational force which tends to pull object in downward direction
→ buoyant force that pushes the object in upward direction
here in this case buoyant force is greater than the gravitational pull on the plastic block. This is the reason the plastic block comes up to the surface of the water as soon as it is released under water.

21. The volume of 50 g of a substance is 20 cm3. If the density of water is 1 g cm−3, will the substance float or sink?

If the density of an object is more than the density of a liquid, then it sinks in the liquid. On the other hand, if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid.
Here, density of the substance = Mass of the substance / Volume of the substance
= 50 / 20
= 2.5 g cm-3
The density of the substance is more than the density of water (1 g cm−3). Hence, the substance will sink in water.

22. The volume of a 500 g sealed packet is 350 cm3. Will the packet float or sink in water if the density of water is 1 g cm−3? What will be the mass of the water displaced by this packet?