Class 9

Haryana State Board HBSE 9th Class Science Important Questions Chapter 1 Matter in Our Surroundings Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 1 Matter in Our Surroundings

Very Short-Answer Type Questions

Question 1.
What is meant by matter?
Answer:
Matter are substances that have mass and which occupy space (volume).

Question 2.
What is every object in the world made up of?
Answer:
Every object in the world is made up of matter.

Question 3.
How many basic elements have the ancient philosophers in India classified matter into?
Answer:
The ancient philosophers in India classified matter into five basic elements.

Question 4.
What is meant by ‘Panch Tatvas’?
Answer:
Five basic elements-air, earth, fire, water and sky are called as ‘Panch Tatvas’.

Question 5.
How many basic elements did the Greek philosophers agree that matter possesses?
Answer:
The greek philosophers agreed that matter is composed of four basic elements i.e., earth, fire, air and water.

Question 6.
What is the basis of the classification of matter in modern science?
Answer:
In modem science, the basis of classification of the matter is physical properties and chemical nature.

Question 7.
Write down four examples of solid materials.
Answer:
Stone, wood, iron and ice.

Question 8.
Write down four examples of liquid materials.
Answer:
Water, kerosene, spirit and petrol.

Question 9.
Write down four examples of gaseous materials.
Answer:
Oxygen, nitrogen, carbon dioxide and water vapour.

Question 10.
On the basis of physical state how many types of matter are there?
Answer:
On the basis of physical state, matter is available in three states
(i) solid
(ii) liquid
(iii) gas.

Question 11.
How many types of matter are there on the basis of chemical formation?
Answer:
On the basis of chemical formation matter is of three types –
(i) element
(ii) compound
(iii) mixture.

Question 12.
Write the name of the solid state of water.
Answer:
The solid-state of water is ice.

Question 13.
Which are the three different states of water?
Answer:
The three different states of water are –
(i) ice (solid)
(ii) liquid (water)
(iii) Eteam (gaseous).

Question 14.
Write down the physical properties of solids.
Answer:
Solids are completely incompressible, they have definite size, shape and volume.

Question 15.
What are the physical properties of liquids?
Answer:
Liquids are comparatively incompressible fluids. They have a definite volume, but the shape and size is indefinite.

Question 16.
What are the physical properties of gas?
Answer:
Gas is an excessively compressible fluid. The given quantity of gas will fill up in containers of any size and shape. .

Question 17.
What property of gas enables it to inflate maximum air in the type?
Answer:
Due to the property of compressibility of gas.

Question 18.
What is meant by CNG?
Answer:
CNG means Compressed Natural Gas.

Question 19.
What do you mean by intermolecular force?
Answer:
The forces of cohesion applicable in between the particles of substances (molecules or atoms) is called intermolecular force.

Question 20.
Which has the maximum intermolecular force: solid, liquid or gas?
Answer:
Solid has the maximum intermolecular force.

Question 21.
Which has the minimum intermolecular force: solid, liquid or gas?
Answer:
Gas has the minimum intermolecular force.

Question 22.
Why do solids have definite shapes?
Answer:
Due to random intermolecular force solids have definite shapes.

Question 23.
Do the molecules of matter have hollow spaces in between them?
Answer:
Yes, the molecules of matter have fair hollow space in between them.

Question 24.
What is meant by diffusion?
Answer:
Self-intermixing of molecules of two different substances into one another is called diffusion.

Question 25.
Why does the aroma of a burning incense stick spread at a far distance?
Answer:
Due to diffusion, the aroma of a burning incense stick spreads at a far distance.

Question 26.
What characteristic of gases enables us to find the leakage of LPG?
Answer:
Due to the diffusion characteristics of gases leakage of LPG is found out.

Question 27.
How does the flavour of ether and cooking reach us?
Answer:
Gases diffuse in air very quickly. Due to this very property, the flavour of ether and cooking reaches us.

Question 28.
How is the purity of honey tested?
Answer:
If on pouring down a drop of honey in a glass of water the drop of honey creeps down in the form of a coloured streak, then the honey is supposed to be pure otherwise the honey is supposed to be impure.

Question 29.
What effect does the temperature leave on the speed of molecules of matter?
Answer:
With the increase in temperature, the speed of the molecules accelerates.

Question 30.
Give an example of such a solid which changes its shape on stretching it.
Answer:
Rubber band.

Question 31.
Give an example of a compressible solid.
Answer:
The sponge is a compressible solid.

Question 32.
Which one is related to flowing, out of solid or liquid?
Answer:
The liquid is related to flow.

Question 33.
Write down names of two miscible gases.
Answer:
Oxygen and carbon dioxide are miscible gases.

Question 34.
Where do the aquatic animals receive oxygen from to breathe?
Answer:
The aquatic animals make use of oxygen dissolved in water to breathe.

Question 35.
Which gas does the balloon-seller fill into the balloon ?
Answer:
The balloon-seller fills hydrogen gas in the balloon.

Question 36.
Which gas is used in houses to cook?
Answer:
Liquefied Petroleum Gas (LPG).

Question 37.
Which gas cylinders are used in hospitals for artificial respiration?
Answer:
Oxygen gas cylinders are used in hospitals for artificial respiration.

Question 38.
In solids, liquids and gases which one has the maximum compressibility property?
Answer:
Gases have the maximum compressibility.

Question 39.
What is called the irregular speed of dust particles in air?
Answer:
The irregular speed of dust particles in air is called Brownian speed.

Question 40.
What happens when ice is heated up?
Answer:
When ice is heated up, it converts into (liquid) water.

Question 41.
What is formed when liquid (water) is heated up?
Answer:
On heating up liquid (water) vapours (steam) is formed.

Question 42.
Out of solid, liquid or gas in which case molecules are totally free to move?
Answer:
In gas the molecules are totally free to move.

Question 43.
What is the use of stirring a liquid while heating it up?
Answer:
While heating up if the liquid is stirred constantly it will get heated equally overall.

Question 44.
Write down name of the device to measure temperature.
Answer:
Thermometer.

Question 45.
What is meant by melting point?
Answer:
That fixed temperature at which a solid becomes liquid on melting is called melting point.

Question 46.
What is SI unit of temperature?
Answer:
SI unit of temperature is ‘Kelvin’ (K).

Question 47.
What is done to convert Kelvin temperature into Celsius temperature?
Answer:
In order to convert Kelvin temperature into Celsius temperature 273 is to be subtracted from the given temperature.

Question 48.
Convert 373 K into Celsius temperature.
Answer:
373 K = (373 – 273)° C = 100° C

Question 49.
How is celsius temperature converted into Kelvin temperature?
Answer:
To convert celsius temperature into Kelvin temperature 273 is added to the given temperature.

Question 50.
Convert 0° C into Kelvin temperature.
Answer:
0° C = (0 + 273) K = 273 K

Question 51.
What is the melting point of ice?
Answer:
The melting point of ice is 273.16 K.

Question 52.
What is meant by melting?
Answer:
The process of melting, i.e., conversion of a solid into a liquid state is called melting.

Question 53.
What is meant by the dormant heat energy of melting?
Answer:
At atmospheric pressure, the heat energy that is required to convert 1 kg of solid at its melting point into a liquid state, is called dormant heat energy of melting.

Question 54.
Which of the two, water or ice’s molecules will possess more energy at 0° C?
Answer:
At 0° C, the molecules of water will possess more energy than ice.

Question 55.
What is meant by boiling point ?
Answer:
At atmospheric pressure, the temperature at which the liquid starts boiling is called boiling point.

Question 56.
What is the boiling point of water?
Answer:
The boiling point of water is 373 K.

Question 57.
What is meant by in exposed evaporation or dormant heat energy of evaporation?
Answer:
At atmospheric pressure, the heat energy that is required to convert 1 kg of liquid at its boiling point into water vapours, is called inexposed evaporation or dormant heat.

Question 58.
Which of the two, steam or water will render more burning sensation at 373 K temperature?
Answer:
At 373 K temperature, steam will render more burning sensation than water, for its molecules possess yet more in exposed heat of evaporation.

Question 59.
What is meant by sublimation?
Answer:
Without changing into liquid state, the process of changing of solid, directly into gaseous and again in solid state is called sublimation.

Question 60.
Write names of two sublime substances.
Answer:
(i) Camphor
(ii) ammonium chloride.

Question 61.
What happens when pressure is increased on any gas?
Answer:
On increasing pressure on any gas it changes into liquid.

Question 62.
What is meant by dry ice?
Answer:
Dry carbon dioxide is called dry ice.

Question 63.
What is the SI unit of pressure?
Answer:
The SI unit of pressure is Pascal (Pa).

Question 64.
What is the atmospheric pressure on the sea surface?
Answer:
On the sea surface the atmospheric pressure used to be 1 atmosphere.

Question 65.
What is meant by ‘evaporation’?
Answer:
The process of changing of liquid into vapours below the temperature of boiling point is called evaporation.

Question 66.
Write down one advantage of evaporation in daily life.
Answer:
In daily life, wet clothes dry up because of evaporation.

Question 67.
What is the effect of the level region on evaporation?
Answer:
An increase in the level region, the evaporation rate increases.

Question 68.
What is the relation of evaporation with temperature?
Answer:
With the increase in temperature, the evaporation rate increases.

Question 69.
What is meant by humidity?
Answer:
The quantity of water vapours present in the air is called humidity.

Question 70.
What is the effect of humidity on evaporation?
Answer:
With the increase in humidity, the evaporation rate decreases.

Question 71.
Why do people sprinkle water on their roofs and in the open places after the scorching bright sunshine day comes to an end?
Answer:
Because the dormant heat of evaporation cools down the hot surface.

Question 72.
Which of the following substances will you expect to possess the strongest and the weakest intermolecular force: water, alcohol, sugar, sodium chloride, carbon dioxide?
Answer:
Sodium chloride will have the strongest and carbon dioxide will have the weakest intermolecular force.

Question 73.
Compression in gases is possible, whereas that of liquids is not. Why is it so?
Answer:
Because of too much intermolecular gap distance in gases, they can be compressed. But in liquids, due to less intermolecular gas the distance they cannot be compressed.

Short-Answer Type Questions

Question 1.
How has the classification of matter been done in ancient period in context with the modern science?
Answer:
Classification of matter:
(1) In ancient times, the ancient philosophers had the concept that matter is composed of five basic elements i.e.,
(i) air
(ii) earth
(iii) fire
(iv) water
(v) sky (all were said to be Panch Tatvas)
all the living or non-living things according to them were bom out of these five elements, whereas the Greek Philosophers are of the view that there have been only four elements in all i.e.,
(i) air
(ii) earth
(iii) fire and
(iv) water.
(2) The modem scientists on the other hand, on the bases of physical properties and chemical nature have classified matter into two kinds.

Question 2.
How is the classification of matter done according to the physical properties?
Answer:
According to physical properties matter are divided into three categories:
1. Solids: Solids are completely incompressible, of definite shape and volume; like stone, wood, iron, ice, salt, sugar, etc.
2. Liquids: Liquids are comparatively less incompressible. They have a definite volume but have an indefinite shape like water, kerosene, spirit, petrol, milk, etc.
3. Gases: Gas is an excessively compressible fluid. Its volume and shape is indefinite which means the gas of a given quantity can be filled in any container of any shape like oxygen, nitrogen, carbon dioxide, water vapours, methane, etc.

Question 3.
Write four special properties of solid matter.
Answer:
Special properties of solid matter are as follows:
(1) They can be collected in the form of piles.
(2) They have a definite volume.
(3) They can be scratched.
(4) Solid substances are hard to touch.
(5) They have definite shapes.

Question 4.
What are the general properties of liquid matter?
Answer:
The general properties of liquids are as follows:
(1) Liquids cannot be placed in piles.
(2) All liquids can flow.
(3) Liquids cannot be scratched.
(4) All liquids have definite volumes.
(5) All liquids form the shape of that very container in which they are kept that is they do not have a definite shape.

Question 5.
Which of the properties are common in all in solids, liquids and gases?
Answer:
The following properties are common in all in solids, liquids and gases:
(1) All are composed of molecules and atoms collectively.
(2) All occupy space.
(3) All have mass.
(4) All can be felt through our senses.

Question 6.
Give the general properties of gases.
Answer:
The general properties of gases are:
(1) Gases do not have definite shape and volume.
(2) Gases dissolve into air.
(3) Gases cannot be scratched.
(4) Gases can be collected only in enclosed containers.
(5) The gas spreads in the container in which it is kept, it spreads at that very place where it is released.
(6) Gases can flow.

Question 7.
Clarify your answer with suitable examples that the properties of all the liquid matter do not resemble one another.
Answer:
The following examples make it clear that the properties of all the liquid matter do not resemble with one another:
(1) Some liquids have low ignition point like petrol, but there are yet certain liquids which do not catch fire like water.
(2) Some liquids are comparatively heavier in weight like mercury. Some liquids are not too heavier like kerosene oil.
(3) Some liquids flow’ quickly like water. Some liquids flow slowly, for example, molten sugar and honey.
(4) Some liquids have the property to saturate salt, sugar, etc. in them. Some liquids cannot saturate these in them like mustard oil.
(5) Some liquids are colourless like kerosene oil, whereas some liquids are coloured like mustard oil.
(6) Some liquids evaporate earlier like spirit. Some liquids evaporate later like turpentine oil.

Question 8.
Compare the intermolecular force in molecules or atoms in solids, liquids and gases.
Answer:
In solids, molecules or atoms are densely and compactly arranged and they are tied up with one another by means of force of attraction whereas in liquids, atoms or molecules are thinly and loosely arranged. In solids, molecules or atoms oscillate about their respective positions, while in liquids, molecules or atoms can move about here and there within their fixed path. In gases, molecules or atoms are quite far away from one another and they move randomly. These molecules or atoms are free to move about within maximum space.

Question 9.
Clarify the reason of generating of pressure in gases.
Answer:
Since in a gaseous state, the motion of molecules is free at all. they can move about in a container, they can collide with one another and also, with the walls of the container. Due to the regular collision of them against the walls, a definite force gets produced. Consequently, pressure is generated in gases.

Question 10.
When we turn ice into water by heating process and further convert water into steam by heating it up, then what differences do we observe in the compactness of the molecules of water ?
Answer:
Ice is a solid substance and its molecules are associated with one another due to the force of cohesion. On heating ice, its molecules get active and start moving on receiving energy with the result, their force of cohesion gets loosened and molecules start flowing that means the solid ice forms the shape of liquid. When water is heated up the random speed of molecules speeds up very much and they generally leave apart from one another to a great extent and become free to move about that is, their intermolecular force weakens too much, such a state of water is said to be steam (gas).

Question 11.
Fill up the blanks choosing the suitable word or words from the list given below: too much, definite, another, very closely, freely, move, container, force, attraction, shape
(i) In gaseous state the molecules move.
(ii) In liquid state molecules can in it.
(iii) Molecules of a solid are packed and can move at a speed.
(iv) The containers in which the liquids are kept, they form the of them.
(v) The gap distance between the molecules of a gas is
(vi) In solids the molecules are joined together with strong forces of.
Answer:
(i) freely
(ii) move
(iii) very closely, definite
(iv) shape
(v) too much
(vi) attraction.

Question 12.
Match the statements given in column A by selecting appropriate words or phrases from column B:

Column A	Column B
(i) Gases can be compressed	(a) by increasing speed of its molecules
(ii) Liquids can be compressed	(b) easily
(iii) Solids cannot be	(c) do not remain tied up with one another
(iv) Solid converts into liquid when the force of cohesion between its molecules is made in effective	(d) less strongly
(v) In solids, molecules are interwoven with one another	(e) to some extent
(vi) In liquids, molecules are tied up with one another	(f) strongly
(vii) In gases, molecules	(g) compressed

Answer:

Column A	Column B
(i) Gases can be compressed	(b) easily
(ii) Liquids can be compressed	(e) to some extent
(iii) Solids cannot be	(g) compressed
(iv) Solid converts into liquid when the force of cohesion between its molecules is made in effective	(a) by increasing speed of its molecules easily
(v) In solids, molecules are interwoven with one another	(f) strongly
(vi) In liquids, molecules are tied up with one another	(d) less strongly
(vii) In gases, molecules	(c) do not remain tied up with one another

Question 13.
The shape of a rubber band can be changed by stretching it, but even then it is not considered as a solid, why ?
Answer:
The shape of the rubber band can be changed by stretching it, but even then it is not considered as a solid, it is because of the reason that on applying external force the rubber band changes its shape and on the removal of the external force it again retains its original shape, whereas it breaks up if the applied force is beyond a limit.

Question 14.
Sugar and salt form the shape exactly of the vessels of different shapes in which they are kept, but still they are considered as solids, why ?
Answer:
Sugar and salt form the shape of the vessels of different shapes in which they are kept, but still they are considered as solids, it is because the shape of their crystals do not change.

Question 15.
A sponge, even being a solid gets compressed, why?
Answer:
Even being a solid, a sponge gets compressed, it is because the sponge consists of small pores in it, which keep trapped air in them. When we press or compress it, the air escapes out from these pores and the sponge gets compressed.

Question 16.
In fluids, diffusion of three states (solid, liquid and gas) is possible, why?
Answer:
In fluids, diffusion of solids, liquids and gases is possible. In liquids the diffusion rate is comparatively higher, in a liquid state, the molecules of a substance move freely and in comparison to solids, the molecules of liquids occupy more vacant space.

Question 17.
The temperature remains equal when a solid undergoes the process of melting, where does the heat energy go during the process?
Answer:
During the melting process any solid (like ice), on reaching the melting point, till the solid melts as a whole, the temperature does not change. Despite providing heat to the beaker, it happens so. By bringing under control the intermolecular force of the molecules, while in changing the state of the matter this heat energy is utilised, because without indicating any increase in the temperature, this heat energy is absorbed by the solid (ice), this thing is assumed that it remains hidden in the material of the beaker, which is called latent heat. At atmospheric pressure, the amount of heat energy that is required to convert 1 kg of solid at its melting point into liquid, is called dormant heat energy of melting, at 0° centigrade (273 K) the energy of the molecules of water is more than that of the molecules of ice at a similar temperature.

Question 18.
Convert the following Celsius measures into Kelvin measuring sequence:
-273°C; -100°C; -40°C; 30°C; 2000°C
Answer:
We know that K = °C + 273
Therefore,
(i) -273°C = -273 + 273 = 0 K
(ii) -100°C = -100 + 273 = 173K
(iii) – 40°C = – 40 + 273 = 233 K
(iv) 30°C = 30 + 273 = 303 K
(v) 2000°C = 2000 + 273 = 2273

Question 19.
Air, at 82 K temperature changes into liquid and at 61K temperature, converts into solid Convert these temperatures in Celsius measuring sequence:
Answer:
We know that
(i) While keeping K = 82
K = °C + 273
82 = °C + 273
⇒ °C = 82 – 273 -191
So, the temperature of the liquefied air = 82 K = -191°C

(ii) Assuming K = 61
61 = °C + 273
⇒ °C = 61 – 273 = -212
So, the temperature of solidified air = 61 K = – 212°C

Question 20.
Why is ice, with a temperature 0°C more effective for cooling than water at 0°C ?
Answer:
We know, the dormant heat energy of melting of ice is 335 J/g. Therefore, ice with 0°C does have comparatively more heat than water at 0°C. Hence, ice with 0PC is more effective than the water with 0°C for cooling.

Question .21.
Why does our body feel cooler in the Himalayan region during winters?
Answer:
In the Himalayan region during winters, the temperature of environment drops below 0°C, whereas the normal temperature of our body used to be 37°C or 98.6° F. Just because of this drastic difference in temperature we feel more chilled in those regions during winters. Covering the body with woollen clothes the inner heat of the body gets maintained and the low temperature outside does not affect our body.

Question 22.
What is meant by evaporation and boiling ?
Answer:
Evaporation: During this process the liquid converts into water vapours before it gets boiled. Evaporation generally occurs on the outer surface of a liquid at all temperatures.

Boiling: When the pressure of any liquids surface equalises the atmospheric pressure, then the liquid starts converting into vapours. This process is called the boiling of a liquid.

Question 23.
Why do the wet clothes dry up sooner if they are spread out ?
Answer:
The speed of evaporation depends upon the surface of the wet article. When the wet clothes are spread out to dry up, the size of the wet surface area gets increased with that evaporation takes place fastly. For that reason the wet clothes dry up earlier when they are spread out and put to dry.

Question 24.
When during winters the water in the tank thaws, then how the fishes living in the tank survive ?
Answer:
During chilly winter days the water in the tank thaws at 0°C into ice, but in this process, only the outer layer of the water surface freezes and the water below to that layer does not thaw. Due to this non-uniform expansion when water is cooled down below 4°C temperature, it instead of contracting expands up to 0°C. For that reason, water thaws into lighter ice and ice start floating on the water. The fish found in such water and the other aquatic animals move to the liquid water below to ice and they survive there.

Question 25.
Where do water vapours come from in the atmosphere?
Answer:
When we put wet clothes in the sun, they dry up sooner. The water present in the fibres of the clothes on evaporating becomes a part of the atmosphere. Similarly, water evaporates in a huge quantity from lakes, rivers and oceans Air cannot sustain water vapours at a given temperature more than a fixed amount. When there is the maximum amount of water vapours in air, the air is said to be quenched.

Question 26.
Why do people sprinkle water on floor and roofs during summers ?
Answer:
People sprinkle water on floor and roofs during summer because water evaporates for which it gets heat from the floor and the roof and thereby keeps them cool. Therefore to keep floor and roofs cool, people sprinkle water on them in summer.

Essay Type Questions

Question 1.
Write down the difference in the following :
(i) heat and temperature
(ii) boiling and evaporation
Answer:
(i) Following are the differences between heat and temperature
Heat:
1. Heat is that form of energy in which we feel the sensation of hot and cool.
2. This is a form of energy.
3. It is measured in calorie or kilocalorie. SI unit is joule (J).
4. It depends upon the form, temperature and nature of a substance.
5. It is a reason.
6. It is the amount of energy in a substance.
7. It is measured by calorimeter.

Temperature:
1. This is that property of a substance which fixes the flow of heat.
2. This is a state by which we come to know the direction of flow of heat.
3. It is measured in parts1 like centigrade and kelvin.
4. It does not depend upon these things.
5. It is the effect of heat.
6. It is a physical property of a substance.
7. It is measured with thermometer.

(ii) Following are the differences between boiling and evaporation
Boiling:
1. It is a random process.
2. It exists only at a fixed temperature.
3. It exists in the entire liquid.
4. This process produces sound.
5. Coolness does not produce in it.
6. Steam is produced in it which is visible.
7. The properties of liquids, open base and the direction of wind do not affect this process.
8. In this process, bubbles are seen moving up wards.

Evaporation:
1. It is a slow process.
2. It exists at all temperatures.
3. It exists only at the bottom level of the liquid.
4. This is a mute process.
5. Coolness produces in it.
6. Water vapours are formed which get mixed into air. Water vapours are invisible.
7. This process depends upon all these things.
8. In this process, bubbles are not produced.

Question 2.
On what factors does the process of evaporation depend ?
Answer:
The process of evaporation is affected by the following factors:
(1) On being the air become dr}’, the evaporation process goes rapidly.
(2) Also, due to fast speed of wind the process of evaporation speeds up.
(3) Even when the base of the liquid is open the process of evaporation will take place randomly.
(4) With the increase in the temperature of the liquid evaporation process will be faster.
(5) If the temperature of air increases the speed of the evaporation process will be rapid.
(6) In more volatile liquids like spirit, petrol, etc. evaporation takes place quickly.
(7) Due to low pressure evaporation process accelerates.

Question 3.
Why does coolness occur with the evaporation of water ? Give some of its uses in our daily life.
Answer:
Coolness occurs due to evaporation, because when water evaporates, it needs latent heat energy of evaporation. This heat can be obtained from that substance which is in the contact of water, thus that substance becomes somewhat cool.

Uses of Evaporation in Daily Life:
(1) Water remains cool in an earthen pot, but not in a bucket, because there are pores in the pot. Water seeps out of them and it evaporates which produces coolness. Thus, water remains cool.
(2) During summer the trees bear more leaves and from the stomata of these leaves evaporation takes place, as a result coolness produces and the trees remain cool.
(3) During perspiration, fanning brings the sensation of coolness, because by fanning the sweat evaporates which produces coolness. For evaporation of sweat, heat is obtained from the body itself.
(4) On sprinkling water on the ground there is coolness for when evaporation of water takes place, then it gains heat from the earth, so the earth becomes cool.
(5) After bathing we feel cool, it is because the body is wet. The evaporation of water takes place and for evaporation, heat is obtained from the body, hence we feel cold.

Practical Work

Experiment 1:
Prove through an activity that matter are composed of molecules and molecules have vacant space or gap in between them.

Procedure:
Take a graduated measuring cylinder. Fill % part of it with water and note down the surface level. Now add some sugar to it and note down its surface level. You will notice that the surface level of water has increased somewhat.

Now, stir well sugar in the water. Sugar will dissolve in water and the surface level of water will again reduce. This proves, that there is gap between molecules of water. While stirring the molecules of sugar-filled up that vacant space, consequently, the water level inside the cylinder was reduced. Since matter is composed of molecules, hence sugar got dissolved in the whole of the water.

Experiment 2:
Prove experimentally, the molecules of matter are so small in size, that we cannot even imagine.

Procedure:
Take a beaker, add 100 ml of water in it. Add two or three crystals of potassium permanganate and dissolve. Draw out approximately 10 ml of solution from the beaker, add 90 ml of pure water to it. Again draw out 10 ml of solution and add 90 ml of pure water to it. Repeat this process 5 to 8 times. You will notice, the water will still remain coloured. This experiment shows that with very small amount of crystals of potassium permanganate, the huge amount of water (1000 litres) too becomes coloured. Thus, we come to the conclusion that just one crystal of potassium permanganate will have a number of minute molecules. They are so minute that we cannot even imagine.

Experiment 3:
Prove experimentally that diffusion takes place more rapidly in gases than in liquids. Procedure: Make a preparation of concentrated film by adding a spoonful of potassium permanganate in 100 ml of water. Keep this solution in the beaker. Now, keep the beaker in a little bit tilted position and add 100 ml of water to it. While adding water to the beaker, be careful that the water should flow down along the wall of the beaker in such a way, so that the concentrated solution of potassium permanganate remains undisturbed. After some time you will observe two layers are formed separately which are quite clear.

Nevertheless, you will see the water gradually flowing into coloured film and the coloured film gradually flowing into water. After some time both the layers will not be seen isolated quite clearly. Note down the tune taken by both the layers to mix up with each other. Now, light a candle in one of the comers in your classroom. You stand far away at the other comer. Now, ask one of your friends to light a few incense sticks. You will very soon experience the fragrance of the incense sticks. Note down the time taken right from the time of lighting of the incense sticks, till the time taken by fragrance to reach to you. We will notice that in the latter stage total time taken was very less than that in the former stage. It proves that in gases diffusion was more rapid than in the liquids.

Experiment 4:
Illustrate experimentally, what effect does pressure have on solids, liquids and gases.

Procedure:
Take a syringe of 100 ml volume. Insert its needle (nozzle) into a rubber cork and close it up, as shown in the figure. Take away the piston and let air fill completely inside the syringe. Now, insert the piston back into the syringe carefully and make sure there is no leakage around the ends of the syringe. It would be better to apply a little bit of vaseline on the piston. Now, try to compress the air. You will notice that air will be compressed.

Now, fill the syringe with water and again repeat the same process. In the experiment done with water you will notice in water, compression is comparatively less than in the first one. Now, repeat the same experiment with piece of chalk in place of water, there will be no compressibility. On the bases of this experiment, we can say gases have maximum effect of pressure, liquids have lesser than that and on solids, it is almost negligible; that is, gases are the most compressible.

Experiment 5:
Experimentally show that with the change in temperature, the state of matter changes.

Procedure:
Take a piece of ice of 150 gm in a beaker and according to the figure hang the thermometer used in the laboratory into it such that the bulb of the thermometer should touch the ice. Start heating the beaker on mild flame. When the ice starts melting, then note down the temperature.
(1) When the whole of the ice change into water, then again note down the temperature.
(2) Note down the downfall in temperature in the change of solid into a liquid state.
(3) Now, place a glass rod into the beaker and heat it while stirring, till the water boils.
(4) Keep a constant eye on the degree of temperature in the thermometer, till most of the water converts into water vapours.
(5) Note down the drop in temperature right from the conversion of water in a liquid state into the gaseous state

Experiment 6:
Experimentally show that ammonium chloride is a sublime substance.

Procedure:
Take a small amount of powdered ammonium chloride in a dish. Place a funnel in an inverted position on this dish. Insert a cotton plug into the mouth of the stem of the funnel as shown in the figure. Now gently heat the dish carefully. We will notice, that ammonium chloride, without getting converted into a liquid state gets converted directly from a solid state to a gaseous state and gets deposited on the walls of the funnel. This proves that ammonium chloride is a sublime substance.

Quick Review of the Chapter

Question 1.
According to Creek philosophers, which fundamental element is not included in the matter?
(A) sky
(B) earth
(C) fire
(D) air and water
Answer:
(A) sky

Question 2.
The solution in the following is:
(A) ocean water
(B) soda water
(C) air
(D) pond water
Answer:
(B) soda water

Question 3.
On the basis of physical state, the types of matter are:
(A) elements
(B) compounds
(C) mixtures
(D) none of the above
Answer:
(D) none of the above

Question 4.
On the basis of chemical composition, the types of matter are:
(A) solids
(B) liquids
(C) gases
(D) noneoftheabove
Answer:
(D) none of the above

Question 5.
S.l. unit of weight is:
(A) Kelvin
(B) Newton
(C) Meter
(D) Pascal
Answer:
(B) Newton

Question 6.
A balloon-seller fills the balloons with:
(A) oxygen gas
(B) hydrogen gas
(C) nitrogen gas
(D) carbon gas
Answer:
(B) hydrogen gas

Question 7.
The property of compressibjIjt is highly present:
(A) in a wooden block
(B) in sponge
(C) in water
(D) in hydrogen gas
Answer:
(D) in hydrogen gas

Question 8.
Dry ice means:
(A) compressed oxygen gas
(B) compressed nitrogen gas
(C) solid carbon dioxide
(D) Acetone
Answer:
(C) solid carbon dioxide

Question 9.
To convert a kelvin temperature into a celsius temperature, what should we abstract from kelvin temperature?
(A) 80
(B) 212
(C) 273
(D) 373
Answer:
(C) 273

Question 10.
The melting point of Ice In kelvin is:
(A) 0K
(B) 80K
(C) 212K
(D) 273.16K
Answer:
(D) 273.16 K

Question 11.
Whose compression is possible?
(A) solid
(B) liquid
(C) gas
(D) none of the above
Answer:
(C) gas

Question 12.
Which cannot pile as a heap?
(A) ice
(B) salt
(C) water
(D) sugar
Answer:
(C) water

Question 13.
Which substance gets spread ¡n the complete/whole pot?
(A) ‘CC
(B) petrol
(C) kerosene
(D) methane
Answer:
(D) methane

Question 14.
Which liquid flows quickly?
(A) molasses
(B) mustard oil
(C) water
(D) honey
Answer:
(C) water

Question 15.
Which prepares liquid quickly?
(A) turpentine oil
(B) mustard oil
(C) spirit
(D) diesel
Answer:
(C) spirit

Question 16.
The normal temperature of our body Is:
(A) 370 F
(B) 37.6° F
(C) 80°F
(D) 98.6° F
Answer:
(D) 98.6° F

Question 17.
98.6°Fisequaito:
(A) 20°C
(B) 25°C
(C) 37°C
(D) 73°C
Answer:
(C) 37° C

Question 18.
Below which temperature at the cooling of water spreads instead of shrinking?
(A) 0°C
(B) 4°C
(C) 37° C
(D) none of the above
Answer:
(B) 4° C

Question 19.
Heat is measured:
(A) in centigrade
(B) in kelvin
(C) in calorie
(D) in Fahrenheit
Answer:
(C) in calorie

Question 20.
Diffusion is possible in:
(A) liquid
(B) solid
(C) gas
(D) all of the above
Answer:
(D) all of the above

Question 21.
Which substance will not cause a feeling of coolness when pouring on the palm?
(A) acetone
(B) petrol
(C) perfume
(D) honey
Answer:
(D) honey

Question 22.
The substance found in nature ¡n free state ¡n all the three states (solid, liquid, gas) is:
(A) petroleum
(B) sulphur
(C) water
(D) oxygen
Answer:
(C) water

Question 23.
Scientifically the materials by which all the things of the universe is formed, are called:
(A) wealthy
(B) resources
(C) substances
(d)) materials
Answer:
(C) substances

Question 24.
Intermixing of particles of two different substances by themselves is called:
(A) spreading
(B) diffusion
(C) compression
(D) smelling
Answer:
(B) diffusion

Question 25.
The gas which ¡s formed by compression of butane at high pressure for use in homes for making food, Is called:
(A) Liquefied Petroleum Gas
(B) Hydrogen Gas
(C) Compressed Natural Gas
(D) Natural Gas
Answer:
(A) Liquefied Petroleum Gas

Question 26.
The energy that is required to convert 1 kg of s solid st atmospheric pressure at its melting point into
liquid is called:
(A) latent heat of melting (B) latent heat of vaporisation
(C) latent heat of liquefication (D) none of these
Answer: (A) latent heat of meltmg

Question 27.
The energy that is required to convert 1 kg of liquid at atmospheric pressure at its boiling point is called:
(A) latent heat of fusion
(B) latent heat of vaporisation
(C) latent heat of Jiquefication
(D) none of the above
Answer:
(B) latent heat of vaporisation

Question 28.
Which energy Is present in particles of matter?
(A) magnetic energy
(B) static energy
(C) kinetic energy
(D) electrical energy
Answer:
(C) kinetic energy

Question 29.
What is called, that definite temperature at which a liquid starts changing into a solid?
(A) freezing point
(B) boiling point
(C) melting point
(D) ignition point
Answer:
(A) freezing point

Question 30.
The process ¡n a liquid changes into a vapour state below the temperature of its boiling point is called:
(A) meltinisation
(B) vaporisation
(C) fusionism
(D) sublimation
Answer:
(B) vaporisation

Question 31.
In which state the volume and size are not definite?
(A) solid
(B) liquid
(C) gas
(D) all of the above
Answer:
(C) gas

Question 32.
The property of hardness finds:
(A) in solids
(B) in liquids
(C) in gases
(D) in all of – liquid and gases
Answer:
(A) in solids

Question 33.
Whose compression is possible in the following?
(A) stone
(B) aluminium
(C) brick
(D) sponge
Answer:
(D) sponge

Question 34.
Lowest compression occurs:
(A) in solids
(B) in liquids
(C) in gases
(D) in gases and liquids
Answer:
(A) in solids

Question 35.
Highest compression occurs:
(A) in solids
(B) in liquids
(C) in gases
(D) in solids and liquids
Answer:
(C) in gases

Question 36.
After changing 25°C into kelvin we get:
(A) 248 K
(B) 298 K
(C) – 248 K
(D) -298 K
Answer:
(B) 298 K

Question 37.
Among water, sugar and oxygen which has the maximum intermolecular force?
(A) oxygen
(B) water
(C) sugar
(D) oxygen and water
Answer:
(C) sugar

Question 38.
What is the physical state of water at 100° C?
(A) Solid
(B) Liquid
(C) Gas
(D) Both (B) and (C)
Answer:
(C) Gas

Question 39.
Which of the following is sublimation matter?
(A) sodium chloride
(B) ammonium chloride
(C) calcium chloride
(D) all of the above
Answer:
(B) ammonium chloride

Haryana State Board HBSE 9th Class Science Solutions Chapter 15 Improvement in Food Resources Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 15 Improvement in Food Resources

HBSE 9th Class Science Improvement in Food Resources Intext Questions and Answers

Question from Sub-section 15.1

Question 1.
What do we get from cereals, pulses, fruits and vegetables?
Answer:
From these, we get proteins, carbohydrates, fats, vitamins and minerals, which are the main and required constituents of our food. These all constituents are helpful for our body development, growth and health.

Questions from Sub-section 15.1.1

Question 1.
How do biotic and abiotic factors affect crop production ?
Answer:
Biotic factors: Diseases, insects and nematodes.
Abiotic factors: Drought, salinity, water logging, heat, cold, frost, etc., decrease the production of crops.

Question 2.
What are the desirable agronomic characteristics for crop improvements ?
Answer:
Tallness and profuse branching are desirable characters for fodder crops. Dwarfness is required in cereals and less nutrients are consumed by these crops. Thus, developing varieties of desired agronomic characters help to give higher productivity.

Questions from Sub-section 15.1.2

Question 1.
What are macro-nutrients and why are they called macro-nutrients?
Answer:
Macro-nutrients :
Such nutrients which are required by the plants in large quantities are called macro-nutrients. Due to their requirement in large quantities, these are called macro-nutrients, e.g., sulfur, nitrogen, phosphorus potassium, calcium, magnesium, etc;

Question 2.
How do plants get nutrients?
Answer:
Plants get the nutrients in the form of liquid from earth. These nutrients are also available in manures and fertilizers, which plants get easily and in large quantities.

Questions from Sub-section 15.1.2

Question 1.
Compare the use of manures and fertilizers in maintaining soil fertility.
Answer:
The comparison of manures and fertilizers to maintain the fertility of soil is as following ;
Manures
1. Manures are required in large quantities.
2. Continuous use of manures is not harmful to the soil.
3. By the use of manures, production always increases.
4. Using process of manure and crop production is delayed.
5. Use of manures is not costly.
6. Transportation and storage is unconvenient.

Fertilizers:
1. Fertilizers are required in small quantities.
2. Continuous use of fertilizers is harmful to the soil.
3. Unnecessary use of fertilizers can decreased the production.
4. By using fertilizers, crop production can be get early.
5. Use of fertilizers is costly.
6. Transportation and storage is convenient.

Questions from Sub-section 15.1.3

Question 1.
Which of the following conditions will give the most benefits? Why?
(a) Farmers use high-quality seeds, do not adopt irrigation or use fertilizers.
(b) Farmers use ordinary seeds, adopt irrigation and use fertilizers.
(c) Farmers use good quality seeds, adopt irrigation, use fertilizers and use crop protection measures.
Answer:
In the above three conditions, the third condition (c) is best because in this condition, seeds are of
high quality, i.e., high-level production. Irrigation is helpful in the nutrition of plants, fertilizers supply the nutrients to plants that are required for the development, growth and production of the plants. Crop protection prevents the crops from various harms and in fact, production increases. Hence, condition (c) is best for farmers.

Questions from Sub-section 15.1.3

Question 1.
Why should preventive measures and biological control methods be preferred for protecting crops?
Answer:
The preventive measures for the protection of crops are: timely sowing of crops, proper seedbed preparation, intercropping and crop rotation also help in weed control. Any type of poisonous chemical such as pesticides, weedicides, etc. is not used. Some other preventive measures against pests are the use of resistant varieties and summer plowing, in which fields are plowed deep in summers to destroy weeds and pests. In this way, soil pollution is not taking place and no extra amount is to be consumed on these measures. Hence, these are also useful economically.

Question 2.
What factors may be responsible for the losses of grains during storage?
Answer:
During storage, two factors are responsible for the loss of grains:
(1) Biotic factors: Insects, rodents, fungi, mites, bacteria, etc.
(2) Abiotic factors: Lack of appropriate moisture and temperature.

Questions from Sub-section 15.2.1

Question 1.
Which method is commonly used for improving cattle breeds and why?
Answer:
The cross-breed method is used for improving cattle breeds so that the lactation and resisting power against diseases increase in such cattle.

Questions from Sub-section 15.2.2

Question 1.
Discuss the implications of the following statement:
“It is interesting to note that poultry is India’s most efficient converter of low fiber foodstuff (which is unfit for human consumption) into highly nutritious animal protein food.”
Answer:
Poultry has this quality that low fiber foodstuffs which are unfit for human consumption that means on the basis of quality these are totally unfit for human consumption, to those it has the ability to convert into highly nutritious animal protein food. Poultry eggs and meat are enriched with proteins and this diet is considered to be a protein-rich diet for humans

Questions from Sub-section 15.2.2

Question 1.
What management practices are common in dairy and poultry farming?
Answer:
There is a big similarity between animal husbandry and poultry farming :
(1) Both are required clean, well-ventilated, and moisture-less homes.
(2) Both are supplied well balanced and enough diet.
(3) Both suffer from a number of diseases that are prevented by appropriate vaccination and both are also treated by the physicians
(4) Both have economic importance, i.e., both are domesticated for economic profit.
(5) Both the high-quality breeds are domesticated to earn a big profit.
Hence, both are similar under the management system.

Question 2.
What are the differences between broilers and layers in their management?
Answer:
Broiler farming is done for meat purposes, whereas layer farming is undertaken for egg production. This is the main difference between them. Broilers and chickens are fed with protein, fats, and vitamin-rich supplementary food for good growth rate and better feed efficiency. Care is taken to avoid mortality and to maintain feathering and carcass quality, whereas the layer requires this type of feeding in less quantity. The housing, nutritional and environmental requirements of broilers are somewhat different from those of egg-producing layers.

Questions from Sub-section 15.2.3

Question 1.
How are fish obtained?
Answer:
Fish are obtained in two ways:
(1) From natural resources,
(2) From fish farming or culture fishery.
Fish are available both in freshwater (non-saline water) and seawater (saline water). Freshwater fish varieties are: catla, rohu, mullets, millions, etc. whereas marine fish varieties include: pomfret, mackerel, tuna, sardines, Bombay duck, etc. Mullets, bhetki, pearl spots, prawns, mussels, and oysters are cultured in seawater. Fish farming is done in freshwater or non-saline water.

Question 2.
What are the advantages of composite fish culture?
Answer:
More intensive fish farming can be done in composite fish culture systems. In such a system, a combination of five or six fish species is used in a single fishpond. These species are selected so that they do not compete for food among them having different types of food habits:
(1) The food available in all the parts of the pond is used.
(2) Weeds can be controlled by bio-practices in composite fish culture.
(3) Without competing with each other, fish yield increases in the pond.

Questions from Sub-section 15.2.4

Question 1.
What are the desirable characteristics of bee varieties suitable for honey production?
Answer:
It is the natural desirable character in variety that it is suitable for increasing honey production. Honey collection and the defense of the beehive are both qualities in Italian bees. They stay in a given beehive for long periods and are freed very well, it increases honey production.

Question 2.
What is pasturage and how is it related to honey production?
Answer:
Pasturage is the natural grassy field where grass for animals and other flowering plants are available. By these flowers, the bees make honey from nectar and pollen collection. The taste of honey depends upon the pasturage, or the flowers available to the bees.
For Examples; honey of bruce flowers, honey of neem flowers, and honey of eucalyptus flowers. The quality of honey is determined by the taste of honey.

HBSE 9th Class Science Improvement in Food Resources Textbook Questions and Answers

Question 1.
Explain any one method of crop production which ensures high yield.
Answer:
One way of incorporating desirable characters into crop varieties is by hybridization. Hybridization is the mating or crossing of genetically different plants. The progeny resulting from hybridization is known as a hybrid. Hybridization refers to crossing between genetically dissimilar plants. This crossing may be intervarietal (between different varieties), interspecific (between two different species of the same genus), or intergeneric (between different genera). Another way of improving the crop is by introducing a gene that would provide the desired characteristics. This results in genetically modified crops.

Question 2.
Why are manures and fertilizers used in fields?
Answer:
Manures and fertilizers have macro-nutrients. There is a lack of nutrients in the soil as crops absorb nutrients from the soil continuously. Manures and fertilizers are mixed in the fields to fulfill the loss of nutrients. The land proves to be barren and the fertility of the land decreases if we do not fulfill the loss. The loss of these nutrients can also be completed by crop rotation.

Question 3.
What are the advantages of intercropping and crop rotation?
Answer:
The following are the advantages of intercropping and crop rotation:
(1) It maintains the fertility of the land. It does not become barren.
(2) Two or more crops can be grown in a year with good harvests.
(3.) Weeds and insects are destroyed.
(4) With the increase of production, farmers’ income also increases.
(5) Threat of unsuccessful crop production also decreases.

Question 4.
What is genetic manipulation? How is it useful in agricultural practices?
Answer:
The high yield of the crops is dependent on the characteristics of their varieties. If we want to improve the crop varieties, then genes with desirable characteristics are used because only genes establish what characteristics of the plants will be, i.e., crop production will increase or decrease. Hence, a change in genes is called genetic manipulation. Today, by using this technique we have got high-quality seeds and crop production has increased the number of times. Especially farmers and gardeners have got an advantage by this technique.

Question 5.
How do storage grain losses occur?
Answer:
(A) Biotic Factors: Following biotic factors are responsible for storage losses of grain :
1. Rodents, birds, and other insects: Rats, squirrels, etc. make a loss to the cereals kept in fields, houses, and in warehouses. Loss is caused to foodgrains due to their hair, feathers, and excreta.
2. Micro-organisms: Various insects and molds change the structure of the storage even after the chemical changes and in this way they cause loss of foodgrains.
3. Insects: Various types of insects damage the raw food storage, e.g., cereals and pulses.
4. Enzymes: These are biocatalysts that are found in nerve cells. These destroy the fruits, and vegetables which are stored a long time.

(B) Abiotic Factors: Following abiotic factors are responsible in storage losses of grain:
1. Moisture: There should not be high moisture at the time of storage. Moisture in the foodgrains should not exceed 14% in relation to its weight.

At high moisture in the foodgrains:
(1) Size of the grains develop.
(2) Microorganisms and enzymes cause more activity.
(3) Possibility of insect-infection increases.
(4) Possibility of growing molds in the moist air increases.

2. Temperature: At low temperatures, enzymes, insects and other micro-organisms do not set more active. For this reason, profligate food materials are kept in cold storage.

3. Pots for storage: Pots used in storage should not be made of lead, metal or nickel because these are toxic and the possibility of toxicity in foodgrains increases.

Question 6.
How do good animal husbandry practices benefit farmers?
Answer:
Animal husbandry practices benefit farmers as follows:
(1) Farmers get milk from dairy animals (milch animals).
(2) In animal husbandry, draught animals are useful for agricultural work and carting.
(3) In animal husbandry, due to economical advantages, farmers get occupational work.
(4) Agriculture by animals provides an extra and round year income.
(5) Biogas and organic matter are prepared from the excreta of animals domesticated by the farmers.
(6) Animal husbandry provides wool, meat, skins, and bones.

Question 7.
What are the benefits of cattle farming?
Answer:
Two main benefits of cattle farming are as under:
1. To get foodstuffs:
The main objective of cattle farming is to produce food products in which the main production is of milk and meat. Furthermore, we get wool, skins and bones from cattle farming. In this way, the objective of cattle farming is related to economic reasons.

2. To get assistance in agricultural activities:
To get assistance in agricultural activities is the main second objective of cattle farming. Oxen, mule, camel, etc. are used in agricultural works such as tilling and carting. Even today low-level farmers are also dependent on animals for agricultural work.

Question 8.
For increasing production, what is common in poultry, fisheries, and bee-keeping?
Answer:
This is only the main common factor to increase the production that the varieties of organisms should be of having disease resistance efficiency, be favorable to the environment, and have high yielding capacity. The major objective for the domestication of these three organisms is to earn the money for the highest yielding. All three organisms are used in agricultural activities as well as in supplementary occupations where the farmer earns income and there is no need to devote their full time.

Question 9.
How do you differentiate between capture fishing, mariculture and aquaculture?
Answer:
Some marine fish of high economic value are farmed in seawater. It is called capture fishing mariculture. The provision of natural environment in the ponds for fish farming is called ‘aquaculture’. In ponds, fish farming is done in non-saline water or freshwater whereas in marine fisheries the fishing is done in saline water.

Haryana State Board HBSE 9th Class Science Solutions Chapter 14 Natural Resources Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 14 Natural Resources

HBSE 9th Class Science Natural Resources Intext Questions and Answers

Questions from Sub-section 14.1

Question 1.
How is our atmosphere different from the atmospheres on Venus and Mars?
Answer:
The components of life air (breathe of life – oxygen) are present in our atmosphere while this lacks on Venus and Mars.

Question 2.
How does the atmosphere act as a blanket?
Answer:
Air present in the atmosphere is a bad conductor of heat which keeps the average temperature of the atmosphere fairly steady during day-night and even during the course of the whole year, such that a blanket prevents the heat to go out, just like that atmosphere prevents the heat of earth surface to go out and also does not let the temperature to be increased, therefore, the atmosphere acts as a blanket.

Question 3.
What causes winds?
Answer:
With uneven movements, the warming of the atmosphere of the earth creates somewhere low-pressure area and somewhere high-pressure area. The difference of this pressure creates airflow which is called breeze or wind.

Question 4.
How are clouds formed?
Answer:
Water vapours in the atmosphere when a pass from the cold areas then water vapours are condensed in the form of droplets. These droplets grow bigger and form the shape of clouds.

Question 5.
List any three human activities that you think would lead to air pollution.
Answer:
Following are the human activities which cause air pollution :
(1) Burning of fossil fuels produces pollutants, which pollute the air after mixing with it.
(2) Smoke from vehicles also pollutes the air.
(3) Gases from the industries/factories established by man also pollute the air after mixing with it.

Questions from Sub-section 14.2

Question 1.
Why do organisms need water?
Answer:
Water is the fundamental need of all living organisms. Every organism’s body contains 75% amount of water. In organisms, all biogeochemical reactions take place with water. Water medium is compulsory for all cellular reactions. Transportation of the substances in the bodies of the organisms takes place in a water medium and excretion activities also take place by water. Terrestrial organisms require freshwater to survive. Water is a living place at a lot of organisms.

Question 2.
What is the major source of freshwater in the city/town/village where your live ?
Answer:
In villages, towns and cities sources of freshwater are underground water, ponds, lakes and rivers. After melting the snow, water is available in the rivers coming from the mountains. In ponds, lakes, etc., mostly we find rainwater.

Question 3.
Do you know of any activity which may be polluting this water source?
Answer:
Sewage waste is mostly dumped into water resources (rivers, canals, lakes) which causes water pollution. Sewage excreta of metropolitan cities and towns situated on the banks of rivers is being dumped into rivers directly. Likewise, half burnt or unburnt dead bodies are thrown in the water of the Ganga by the habitats of the banks of Ganga due to which the water of Ganga is being polluted.

Questions from Sub-section 14.3

Question 1.
How is soil formed?
Answer:
Soil is formed by the breaking and grinding-rubbing of rocks. Soil is formed by the mixing of small pieces of rocks, fresh and decomposition of leaves, dead decayed remains of organisms and sand particles etc. The formation of topsoil of the earth takes a minimum 300 to 800 years. Soil formation is a continuous process.

Question 2.
What is soil erosion?
Answer:
Cutting or flowing away of the soil by any of the reasons is called soil erosion. Soil erosion may take place by various reasons, for example:
1. Flood: Flood is the main reason of soil erosion by which soil flows away from one place to another.
2. Fast Winds: The second main reason of soil erosion is the fast-blowing winds.
3. Deforestation: Deforestation is on high scale due to growing population and industrialisation by which soil erosion increases.
4. Uncontrolled and Overgrazing: Uncontrolled and overgrazing increases soil erosion.
5. Unscientific Techniques of Agriculture: Mostly farmers use to grow one variety of crops continuously in the fields by which fertility of soil decreases and soil erosion increases.
6. Deeper Cultivation of the Land: Soil erosion increases due to deeper cultivation of the land for farming.

Question 3.
What are the methods of preventing or reducing soil erosion?
Answer:
Following are the methods to prevent soil erosion:
1. By Growing Grass: Leaves and roots of grass retard the flow of water by which soil particles may not be flown by water.
2. By Levelling the Fields: By distributing the tapered land into small levelled pieces for agriculture, the flow of water is checked by which cutting of land is also prevented.
3. By Growing Trees: Roots of the trees bind the soil by which water cannot flow away the soil.
4. Better Techniques of Farming: In spite of ploughing in the sloping direction, if ploughing is being done in left-right sides of the field, then flow of water decreases, and erosion of soil is checked.
5. By Making Small Dams: Small dams constructed on slopes decrease the flow of fast-moving water
and soil erosion is stopped.
6. By Making Ladder-type Field: In hilly areas, farming should be done by making steps opposite to
slope.
7. By Using Manure: Use of manure in field increases the water retention efficiency of the soil by which soil erosion is checked.
8. By Fixing Air-protectors: Where the land is sandy and winds flow fast, there plants should be grown profusely around the field, i.e., fast winds may not flow away the soil particles.
9. By Making High Fences: Fences of the fields should be high because the water of field will not flow out and cutting of land is not possible.
10. By Growing Particular Crops : Soil erosion can be decreased by growing particular crops such as groundnut.

Questions from Sub-section 14.4,14.5

Question 1.
What are the different states in which water is found during the water cycle ?
Answer:
In water cycle, three states of water are found :
(i) Water in liquid state,
(ii) Water in vapour state and
(iii) Ice as solid state.

Vapours form clouds and these clouds bring rain on the earth and sometimes water of clouds freezes in ice, i.e., converts into hailstone. Ice on the peaks of the mountains is also water in solid state, which after melting comes in rivers and there by converting into vapours, mix with air.

Question 2.
Name two biologically important compounds that contain both oxygen and nitrogen.
Answer:
Nitrates and nitrites.

Question 3.
List any three human activities which would lead to an increase in the carbon dioxide content in air.
Answer:
Following are the reasons of human activities which cause an increase in the carbon dioxide content in air.
(i) By combustion of fuel.
(ii) By using petroleum products in vehicles.
(iii) CO₂is also released by respiration.

Question 4.
What is the greenhouse effect?
Answer:
The earth receives different types of radiation from the sun. The ozone layer in the atmosphere absorbs ultraviolet radiation and let go to other radiations, but radiations which come on the earth’s surface, some part of it in the form of infrared radiation reflected back. Molecules of CO₂ have the efficiency to absorb infrared radiations, hence carbon dioxide present in the atmosphere can absorb radiation, that reason atmosphere becomes warmer. The warming of the atmosphere in this way is called the greenhouse effect or glasshouse effect.

Question 5.
What are the two forms of oxygen found in the atmosphere?
Answer:
CO₂ (Oxygen) and O₃ (Ozone).

HBSE 9th Class Science Natural Resources Textbook Questions and Answers

Question 1.
Why is the atmosphere essential for life?
Answer:
Life is dependent only on the atmosphere because
(i) The atmosphere keeps the average temperature of the earth fairly steady.
(ii) Oxygen is available in the atmosphere in the form of life air.
(iii) Ozone layer of the atmosphere protects from the harmful ultraviolet radiation coming from the sun.
(iv) Atmosphere protects the earth from meteors, by which life is also protected.
(v) CO₂ gas which is used by green plants in photosynthesis is also found in the atmosphere.
(vi) Water in the atmosphere is found in the form of water vapours which is the foundation of life.
Hence, the atmosphere is essential for life.

Question 2.
Why is water essential for life?
Answer:
Water is the fundamental need of all living organisms. Every organism’s body contains 75% amount of water. In organisms, all biogeochemical reactions take place with water. Water medium is compulsory for all cellular reactions. Transportation of the substances in the bodies of the organisms takes place in a water medium and excretion activities also take place by the water. Terrestrial organisms require freshwater to survive. Water is a living place for a lot of organisms.

Question 3.
How are living organisms dependent on the soil? Are organisms that live in water totally independent of soil as a resource?
Answer:
Soil is the base of life. All living organisms are dependent on soil directly or indirectly. Plants get nutrients from the soil which main are water and minerals and all animals are dependent on plants. Animals get food and oxygen to respire from plants. Life of living organisms is not possible without plants and plants cannot survive without soil. Aquatic animals are also dependent on soil because they fulfill their requirements from plants.

Question 4.
You have seen weather reports on television and in newspapers. How do you think we are able to predict the weather?
Answer:
Yes, we are able to predict the weather. By collecting the data through artificial satellites established in earth orbit, it is very easy to predict the forthcoming weather. After each half, hour reports and photographs are collected by INSAT-IB. Through these photographs, prediction is relayed which proves to be very useful.

Question 5.
We know that many human activities lead to increasing levels of pollution of the air, water bodies, and soil. Do you think that isolating these activities to specific and limited areas would help in reducing pollution?
Answer:
Yes, isolating these activities to specific and limited areas would help in reducing pollution, e.g.,
(i) Air pollution will decrease by running the vehicles based on solar system batteries rather than fossil fuel/ petroleum.
(ii) These will be a check on pollution if cooking of food, heating of water, etc. are done by the use of solar energy in place of coal, L.P.G., etc. fuels in houses.
(iii) These will be a check on soil pollution by using compost manure in the agricultural field rather than chemical fertilizers.
(iv) The water pollution will be decreased by using useful substances in the treatment plants inspite of throwing the industrial waste directly in the water bodies.

Question 6.
Write a note on how forests influence the quality of our air, soil and water resources.
Answer:
Forests are the main components in environmental balance. Forests equalize the formation (78% nitrogen, 21% oxygen) of air. This balance is disturbed by deforestation. Temperature increases with the increase of C02 and the content of condensed particles in air while the air becomes, dust free, pure and cool by the deep forestation. Forest decreases pollution. Forests help in maintaining the quality of soil by increasing its fertility and also preventing soil erosion. By deforestation land becomes barren. The temperature of the environment increases. Rains are excess by excess forestation. Water level increases in water bodies and makes the availability of useful fresh water in large quantities. Therefore, forests are helpful in improving the quality of air, soil and water resources.

Haryana State Board HBSE 9th Class Science Solutions Chapter 13 Why Do We Fall Ill Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 13 Why Do We Fall Ill

HBSE 9th Class Science Why Do We Fall Ill Intext Questions and Answers

Questions from Sub-section 13.1

Question 1.
State any two conditions essential for good health.
Answer:
Following are the two conditions essential for good health :
(i) Community cleanliness
(ii) Balanced diet.

Question 2.
State any two conditions essential for being free of diseases.
Answer:
Following are the two conditions essential for being free of diseases :
(1) Clean environment
(ii) Good habits and regular exercise.

Question 3.
Are the answers to the above questions necessarily the same or different ? Why ?
Answer:
The answers to the above questions are necessarily same because to keep good health or diseases free are almost same situation.

Questions from Sub-section 13.2

Question 1.
List any three reasons why you would think that you are sick and ought to see a doctor. If only one of these symptoms were present, would you still go to the doctor ? Why or why not ?
Answer:
There can be many
(i) cough
(ii) fever
(iii) weakness.

If the given symptoms appear for a small period of time, there is no need to go to a doctor, because the immune system of our body stops these quickly and makes the body free of the effects of any disease. But if persistent cough prevails, lungs may be infected. If fever persists, it can be due to a serious ailmant T.B. If there is nutritious deficiency, it can lead to general weakness. In conditions like these, it is proper to go to a physician, so that the real course of the symptoms may be as certained and it can be cured.

Question 2.
In which of the following case do you think the long term effects on your health are likely to be most unpleasant ?

• If you get jaundice.
• If you get lice.
• If you get acne. Why?

Answer:
If we’re suffering from jaundice, it is most likely to have ill effects on our health. Jaundice is a disease that effect the liver, which is an important organ of the body and which has an important role to play in our body. So the body of the person suffering from jaundice is predominantly affected. The effect of getting lice or acne are not so predominant.

Questions from Sub-section 13.3

Question 1.
Why are we normally advised to take bland and nourishing food when we are sick?
Answer:
Bland and nourishing food is recommended to sick people so that it is easily digested and there is reimbursement of the nutrients lost due to sickness.

Question 2.
What are the different means by which infectious diseases are spread?
Answer:
Infectious diseases spread through the following methods:
1. Through ai: Many disease particles spread in the air due to sneezing, coughing, talking, spitting or defecating. When a healthy person breathes in the air containing these disease particles, the person catches the disease.

2. Through water: Pathogens of cholera, tuberculosis etc., get mixed in drinking water and enter the body. The disease spreads by washing utensils or vegetables and fruits in polluted water. It also spreads when the clothes of an infected person are washed near a source of water or he defecates near it.

3. Through food:
Pathogens present in foodstuffs also enter the body and make the person infected. It is harmful to take stale or cold food.

4. Individual contact:
Coming in contact with the patient’s clothes, bedding, utensils, towel or directly with the patient also transmits the pathogens.

5. Through animals: Some animals like houseflies, mosquitoes, mice etc., also play the role of vectors. By polluting food the pathogens are left there which ultimately enter the human body. Malaria is spread by mosquito bite while cholera spreads through houseflies and plague through mice.

Question 3.
What precautions can you take in your school to reduce the incidence of infectious diseases?
Answer:
Following precautions should be taken:
(i) Chlorination of drinking water should be done.
(ii) All the eatables in the school canteen should be kept covered.
(iii) The sale of stale fruits and other eatables should be prohibited.
(iv) If a student catches an infectious disease he should not be allowed to mix up with his classmates in the classroom as well as in the playground till he fully recovers.
(v) Children should be properly educated about personal as well as community hygiene.
(vi) To prevent communicable diseases all children should be given proper vaccination.

Question 4.
What is immunisation?
Answer:
The strength produced in the body to resist diseases is called immunity for example; disease producer is got entered into the body. This disease producer combats with the blood cells. In reaction to that some specific chemical substances are produced. These various substances destroy the pathogens in the body and thus, produce the ability of resistivity to combat diseases. It is called as immunity of the body and this process is called as immunisation.

Question 5.
What are the immunisation programmes available at the nearest health centre in your locality? Which of these diseases are the major health problems in your area ?
Answer:
In our town health centre following vaccination programmes are available:

• National tuberculosis eradication programme.
• National, cholera eradication programme.
• National leprosy eradication programme.
• National polio eradication campaign.
• National child communicable disease eradication (vaccination of children).

Generally, in all cities and towns problems in relation to hygienic environment community hygienic and pure drinking water one quite common. In proper disposal of garbage in cities and towns is responsible for polluting the environment. Pure water is not being available. Air and water pollution have been adding to health problems.

HBSE 9th Class Science Why Do We Fall Ill Textbook Questions and Answers

Question 1.
How many times did you fall ill in the last one year? What were the illnesses?
(i) Think of one change you could make in your habits in order to avoid any of/most of the above illnesses.
(ii) Think of one change you would wish for in your surroundings in order to avoid any of/most of the above illnesses.
Answer:
Every person can suffer from a minor or major diseases at any time. Hardly a few persons can be there who do not develop any disease, but even then due to some reasons health can be affected. I also fell ill in the last year. I had suffered from malaria.
(i) Different safety measures can be adopted to get rid of malaria. It can be avoided by filling the pits and landfills around our neighbourhood with soil and sand, drain blockage should be removed, water should be drained out from the flower pots, desert cooler-tanks and empty vessels in our houses, at night net and mosquito repellent cream should be applied to the skin. The doors and windows in the houses should be equipped with wire-gauze.

(ii) The growth of mosquitoes can be checked by keeping our surroundings clean and tidy pits and drains must not be let overflow or choked with stagnant water. Unwanted bushes or weeds should be destroyed.

Question 2.
A doctor/nurse/health worker is exposed to more sick people than others in the community. Find out how she/he avoids getting sick herself/himself.
Answer:
Doctors, nurses and health workers immensely remain in contact with patients than other people. To protect themselves from harmful bacteria or microbes following safety measures are adopted by them :
(1) They are habitual wearing washed and clean clothes every day.
(2) While examining a patient, doctor use to cover their mouth and nose with a mask so that the least microbes carry ing infection should not enter their body through breathing.
(3) Nurses and other medical workers wear rubber gloves while injecting or dressing a patient, so that they may not come in direct contact with the patient.
(4) After examining a patient, a doctor or another person at help wash up their hands with an antiseptic solution.
(5) Whole of the staff in the hospital is much concerned about common hygiene. The floors of the wards and rooms are swabbed with disinfectant phenyl regularly.
(6) The garbage and waste should be destroyed with scientific methods.

Question 3.
Conduct a survey in your neighbourhood to find out what the three most common diseases are. Suggest three steps that could be taken by your local authorities to bring down the incidence of these diseases.
Answer:
On conducting a survey in the neighbourhood it was found that the three most common diseases in my locality are:
(i) Diarrhoea and expectoration (vomiting)
(ii) Malaria
(iii) Cough and cold.

Following are the three suggestions to immunise ourselves against these diseases :
(1) There is a great need to take care of the eatables and drinking water they must be protected against any kind of pollution.
(2) Ensure that the water is covered so that mosquitoes can’t breed and we can have safe drinking water.
(3) Maintaining cleanliness and proper disposal system is needed.

Question 4.
A baby is not able to tell her/his caretakers that she/he is sick. What would help us to find out
(a) that the baby is sick?
(b) what is the sickness?
Answer:
(a) If a baby is not able to tell his/her caretaker that he/she is sick then body symptoms should be observed by the caretaker. The caretaker should take the baby to the doctor if she or he finds a clue that the baby is sick. Some symptoms that can be observed are :
1. If the baby cries continuously.
2. If the baby doesn’t want to eat anything.
3. If the baby feels irritated all the time.

(b) Kind of sickness can be known by certain symptoms e.g. yellowness of skin and eyes, redness in eye may indicate conjunctivitis.

Question 5.
Under which of the following conditions is a person most likely to fall sick?
(a) when she is recovering from malaria.
(b) when she has recovered from malaria and is taking care of someone suffering from chicken pox.
(c) when she is on a four-day fast after recovering from malaria and is taking care of someone suffering from chicken pox. Why?
Answer:
A person is most likely to fall sick when she is on a four-day fast after recovering from malaria and is taking care of someone suffering from chicken pox. This is because since she has suffered from malaria and is in the recovery phase, it means that she is still not completely disease-free. Her chances of getting chicken pox are also high as her body has lowered.

Question 6.
Under which of the following conditions are you most likely to fall sick?
(a) when you are taking examinations.
(b) when you have travelled by bus and train for two days.
(c) when your friend is suffering from measles. Why?
Answer:
(a) I am most likely to fall sick when my friend is suffering from measles. Measles is a viral infection. It is a contagious disease. It spreads through contact with infected mucus and saliva. The virus can sustain for several hours. Even coming in contact or sharing items with an infected person results in the spread of disease.

Haryana State Board HBSE 9th Class Science Solutions Chapter 12 Sound Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 12 Sound

HBSE 9th Class Science Sound Intext Questions and Answers

Questions from Sub-section 12.2

Question 1.
How does the sound produce by a vibrating object in a medium reach your ear?
Answer:
The sound produced by a vibrating object reaches our ear through the medium.

Questions from Sub-section 12.2.1

Question 1.
Explain how sound is produced by your school bell.
Answer:
The school bell produces sound through vibration because when the hammer strikes the bell, it produces vibration in the bell.

Question 2.
Why are sound waves called mechanical waves?
Answer:
The sound waves are produced by the motion of the particles of the medium, and mechanical work is possible by them, so sound waves are called mechanical waves.

Question 3.
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
No, because sound needs a medium to propagate There is no medium on moon and hence it is not possible to hear any sound on Che moon.

Questions from Sub-section 12.2.3

Question 1.
Which wave property determines
(a) loudness
(b) patch?
Answer:
Pitch.

Question 2.
Guess which sound has a higher pitch : guitar or car horn?
Answer:
Car horn.

Questions from Sub-section 12.2.3

Question 1.
What arc wavelength, frequency, time period and amplitude of a sound wave?
Answer:
(i) Wavelength: The distance between two consecutive compresssions or two consecutive
rarefactions is called wavelength. It is denoted by a Greek alphabet, (lambda)
(ii) Frequency: The number of complete oscillations per unit second is called frequency.
(iii) TIme Period: The time taken by the wave for one complete oscillation is called time period.
(iv) AmplItude: The magnitude of maximum disturbance in the medium on either side of the mean value is called amplitude.

Question 2.
How are the wavelength and frequency of a sound wave related to its speed?
Answer:
υ = λv
or
Velocity wavelength × frequency.

Question 3.
Calculate the wavelength ora sound wave whose frequency is 220 Hz and speed Is 440 m/s in a given medium.
Solution:
Frequency of wave (υ) = 220 Hz
Velocity of wave (v) = 440 m/s
We know that, = λ × 220
λ = \(\frac {220}{440}\) = 2m

Question 4.
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Solution:
Frequency of wave (υ) = 500 Hz
Time (T) =?
We know that,
v = \(\frac {1}{T}\)
Or
T = \(\frac {1}{v}\)
T = \(\frac {1}{500}\) = 0.002 sec

Questions from Sith-section 12.2.3

Question 1.
Distinguish between loudness and intensity of sound.
Answer:
The loudness of sound: Loudness is a measure of the response of the ear to the sound. It depends upon low and high vibrations.
The intensity of sound: The amount of sound energy passing each second through unit area is called the intensity of sound.

Questions from Sub-section 12.2.4

Question 1.
In which of three media, air, water or iron, does sound travel the fastest at a particular temperature?
Answer:
The sound travels the fastest in iron. the speed of sound at 25°C in air is 346 m/s, in water 1531 m/s and in iron 5950 m/s.

Questions from Sub-section, 12.3.

Question 1.
An echo returned in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms^-1?
Solution:
Speed nf sound(v) = 342 ms^-1
lime taken by echo (t) = 3s
Distance covered by sound = velocity × time = 342 × 3 m = 1026 m
Thus. the distance of the reflecting surface from the source
\(\frac {1026}{2}\) m = 513 m

Questions from Sub-section 12.3.3

Question 1.
are the ceilings nl concert halls curved?
Answer:
The ceilings of concert halls arc curved so that sound after reflection reaches all corners of the hail.

Questions from Sub-section 12.4

Question 1.
What is the audible range of the average human ear?
Answer:
20 Hz to 20.000 Hz.

Question 2.
What is the range of frequencies associated with
(a) Infrasound?
(h) Ultrasound?
Answer:
(a) The range of infrasound is less than 20 Hz.
(b) The range of ultrasound is more than 20 kHz

Questions from Sub-section 12.5

Question 1.
A submarine emits a sonar pulse, which returns from an underwater cliff I.02s. If the speed of sound in salt water is 1531 mIs, how far awa is the cliff?
Solution:
Here, Time between transmission and detection (t) 1.02 s
Speed of sound in seawater (v) = 1531 m/s
Distance traveled by the ultrasound = 2 x distance of cliff from submarine = 2d
Where. d = distance of cliff from submarine
Total distance (2d) = speed x time = 153 1 m/s x 1.02 s = 1561.62rn
d = \(\frac {1561.62}{2}\) = 780.81m.

HBSE 9th Class Science Sound Textbook Questions and Answers

Question 1.
What is sound and how is it produced?
Answer:
Sound:
It is a form of energy that produces a sensation of hearing. Sound is produced by any vibrating material body. For example. the sound of the human voice is produced due to vibrations in the vocal cords. The strings of trumpets and shehnais vibrate when they are played. The vibration is also produced when the bell is rung. The vibration in bell can be felt by touching it. In this way sound is produced.

Question 2.
Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of the sound.
Answer:
When a vibrating object moves forward, it pushes and compresses the air in front of it creating a region of high pressure which is called compression (C). This compression starts to move away from the vibrating object. When the vibrating object moves backward, it creates a region of low pressure called rarefaction (R). Thus, due to the vibration of an object, a series of compressions and rarefactions is created in the air. These make the sound wave that propagates through the medium.

Question 3.
Cite an experiment to show that sound needs a material medium for its propagation.
Answer:
Sound is a mechanical wave. It requires a material medium for propagation. ¡t can
be proved by the following experiment:
Experiment:
Take an electric bell and an airtight glass bell jar. The electric bell is suspended inside the airtight bell jar. Connect the bell jar to a vacuum pump. Switch on the bell. The sound of bell is heard. Now pump out the air of bell jar by vacuum pump gradually. The sound becomes fainter. When there is full vacuum in bell jar, the sound of bell will not be heard. Thus, the experiment proves that sound cannot travel in vacuum. Sound need a material medium to travel. The air was the medium in bell jar so the sound of the electric bell was being heard.

Question 4.
Why is a sound wave called a longitudinal wave?
Answer:
In sound waves, the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance. The particles do not move from one place to another but they oscillate back and forth about their position of rest. Hence, sound waves are longitudinal waves.

Question 5.
Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer:
Pitch of sound. How the brain interprets the frequency of an emitted sound is called pitch. Pitch helps us to identify the sound.

Question 6.
Flash and thunder are produced simultaneously. But thunder ¡s heard a few seconds after the flash is seen, why?
Answer:
At room temperature. the speed of sound is 344 metres per second and the velocity of light is 300000 km per second. Flash and thunder are produced simultaneously but due to the difference in the speed of both, flash is seen first and thunder is heard after it.

Question 7.
A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 ms^-1.
Solution:
The velocity of sound in air (y) = 344 ms^-1
First frequency of person (v₁) = 20 Hz
Second frequency of person (v₂) = 20 kHz = 20,000 Hz
First wavelength (λ₁) = \(\frac{v}{v_1}=\frac{344}{20}\) = 17.2 m
Second wavelength (λ₂) = \(\frac{v}{v_2}=\frac{344}{20,000}\) = 0.0172 m

Question 8.
Two children are at opposite ends of an aluminum rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminum to reach the second child.
Solution:
Suppose the length of rod = 1m
Speed of sound in air (v₁) = 346 m/s
Time taken in Air = = \(\frac {1m}{346m}\)s 1/346s
The speed of sound in aluminium (v₂) = 6420 m/s
Time taken in Aluminium = = \(\frac {1m}{6420m}\) = I /6420s
Ratio of both = \(\frac{1}{346}: \frac{1}{6420}\) = 18.55:1

Question 9.
The frequency of a source of sound is loo Hz. How many times does it vibrate in a minute?
Solution:
Frequency of sound (v) = 100 Hz
Number of vibrations in 1 second = 100
Number of vibrations in 60 seconds = 100 x 60 = 6000

Question 10.
Does sound follow the same laws of reflection as light does? Explain.
Answer:
Yes, sound follows the same laws of reflection as light does. Sound is reflected from the surface of
Laws:
(1) The angle of incidence is equal to the angle of reflection.
(2) Incident wave, reflected wave and the normal lie in the same plane. An obstacle of larger size which may be polished or rough is needed for the reflection of sound waves.

Question 11.
When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear an echo sound on a hotter day?
Answer:
Yes, because the speed of sound is more on a hotter day than on a colder day.

Question 12.
Give two practical applications of reflection of sound waves.
Answer:
(1) During night, the bats feel their way by the reflection of sound.
(2) Ultrasound is used to break small ‘stones’ formed in the kidneys into fine grains.

Question 13.
A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m/s^-2 and speed of sound = 340 ms^-1.
Solution:
Here, u 0, s 500 m, g 10 rns2
Buts = ut + gt² = 0 + \(\frac {1}{2}\) x 1o × t² ⇒ 5t² = 500 or = 10s
Distance of the top of the tower from surface faster = 500 m
Speed of sound = 340 ms^-1

\(\frac {500}{340}\) = 1.47s
The time taken by the splash to reach at the top = 10 + 1 .47 = 11.47 s

Question 14.
A sound wave travels at a speed of 339 ms^-1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Solution:
Here, the Speed of the sound wave = 339 ms^-1
Wavelength = 1.5 cm = \(\frac {15}{10}\) = \(\frac {15}{1000}\) m
Frequency Wavelength = img = \(\frac {339}{15}\) x 1000 = 22600 Hz
This sound is not audible because a person cannot hear a sound more than 20,000 Hz.

Question 15.
What is reverberation? How can it be reduced?
Answer:
Reverberation:
The repeated reflection that results in the persistence of sound is called reverberation. Methods of reducing:
To reduce reverberation, the roof and walls of the auditorium are generally covered with sound absorbent materials like compressed fibreboard, rough plaster or draperies. The seat materials are also selected on the basis of their sound-absorbing properties.

Question 16.
What is the loudness of sound? What factors does it depend on ?
Answer:
The measure of the response of the ear to the sound is called the loudness of sound. It is the sensation of hearing. This level of sensation depends upon the loud or faint vibrations of the source. When vibrations are loud, the loudness of sound is more and when vibrations are faint, the loudness of sound is faint. The loudness or softness of the sound is found by the amplitude. The sound of high energy has more loudness.

Question 17.
Explain how bats use ultrasound to catch a prey.
Answer:
The bat searches out prey in dark night by emitting and detecting reflections of ultrasonic waves. The high pitch ultrasonic squeaks of the hat are reflected from the prey and returned to the bat’s ear as shown in fig. The nature of reflections tells that bat where the prey is and it catches it. In this way the bats use ultrasound to catch prey.

Question 18.
How is ultrasound used for cleaning?
Answer:
Objects to be cleaned are placed in a cleansing solution and ultrasonic waves are sent into the solution. Due to the high frequency. the particles of the dust, grease and dirt get detected and drop out. This technique can clean those objects which are difficult to clean due to their large size.

Question 19.
Explain the working and application of a sonar.
Answer:
Procedure:
SONAR-Sound Navigation And Ranging is such a device in which ultrasonic waves are used to measure the distance, direction and speed of underwater objects. Sonar consists of a transmitter and a detector and is installed in a boat or a ship as shown in fig. The transmitter produces and transmits ultrasonic waves and travel through water and after striking the object on the seaboat, get reflected back and are sensed by the detector. The distance of the object that reflected the sound wave can he calculated by knowing the speed of sound in water and the time interval between transmission and reception of the ultrasound.

Applications: Following are its applications:

(1) To determine the depth of the sea and oil wells.
(2) To find the distance of groups of fish.
(3) To locate the enemy submarine or torpedo.
(4) To cate underwater hill valleys. icebergs and sunken slp.

Question 20.
A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water, if the distance of the object from the submarine is 3625 m.
Solution:
Distance of the object from the submarine (d) = 3625 m
The echo covers the double distance.
2d = 2 x 3625 = 7250 m
Time (t) = 5 s
Speed of sound in water = \(\frac {2d}{t}\) = imh = \(\frac {7250}{5}\) = 1450 m/s
Thus, speed of sound in water 1450 m/s

Question 21.
Explain how defects in a metal block can be detected using ultrasound.
Answer:
Ultrasound can be used to detect cracks and flaws in metal blocks. The cracks or holes inside the metal blocks, which are invisible from the outside reduce the strength of the structure of the building or bridge.

Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected back indicating the presence of the flaw or defect. Ordinary sound of longer wavelengths cannot be used for such purpose as it will bend around the corners of the defective location and enter the detector and instead of detecting the flaw, they create a state of confusion.

Question 22.
E%plaìn how the human ear works.
Answer:
The outer ear of human being collects the sound from surroundings and passed through the auditory canal, and the pressure on the outside of the membrane increases. The pressure from inside is less. Due to the difference in pressure, the eardrum starts vibrating. The vibrations are amplified several times by three bones (the hammer, anvil and stirrup) in middle ear. In the inner ear, the pressure variations are turned into electric signals by the cochlea. These electrical signals are sent to the brain via the auditory nerve, and the brain interprets them as sound.

Haryana State Board HBSE 9th Class Science Solutions Chapter 11 Work and Energy Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 11 Work and Energy

HBSE 9th Class Science Work and Energy Intext Questions and Answers

Questions from Sub-Section 11.1

Question 1.
A force of 7N acts on an object. The displacement is, say 8 m in the direction of the force (Fig. 11.1). Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Solution:
Here,
Force (F) = 7N
Displacement (s) = 8m
Work done (W) =?
We know that, Work done (W) = Force (F) × Displacement (s) = 7 × 8N.m = S6 Joule

Questions from Sub-Section 11.1

Question 1.
When do we say that work is done?
Answer:
When an object is displaced by applying force, it is said to be work done.
Thus, work done (W) = Force (F) × Displacement (s)

Question 2.
Write an expression for the work done when a force is acting on an object in the direction of its displacement?
Answer:
When a force is acting on an object in the direction of its displacement. The equation of work will be as follows:
Work done (W) = Force (F) X Displacement (s)
Thus, work done ¡s positive when the force is ¡n the direction of displacement.

Question 3.
Define 1 J of work.
Answer:
Work is said to be done 1 joule when a force of 1 N acting on an object and it is displaced through 1 m in the direction of force.
Thus, 1 J = 1 N x lm

Question 4.
A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in plowing the length of the field?
Solution:
Here,
Force (F) = 140 N
Displacement (s) = 15 m
Work done (W) = F × S = 140 × 15 = 21001

Questions from Sub-Section 11.2

Question 1.
What is the kinetic energy of an object?
Answer:
The energy possessed by an object due to its motion is called kinetic energy. A body of mass m moving with velocity v has kinetic energy (E) equal to \(\frac {1}{2}\)mv².

Question 2.
Write an expression for the kinetic energy of an object.
Answer:
The kinetic energy of a body of mass m moving with velocity v is; K.E = \(\frac {1}{2}\) v²

Question 3.
The kinetic energy of an object of mass m moving with a velocity of 5 ms^-1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times ?
Answer:
The kinetic energy of an object of mass, m moving with a velocity of 5 ms^-1 is 25 J. If its velocity is doubled (10 ms^-1), its kinetic energy will increase four (2^-1) times i.e. 100 J. But if its velocity is tripled (15 ms^-1), its kinetic energy will increase nine (3²) times i.e. 225 J.

Questions from Sub-Section 11.3

Question 1.
What is power ?
Answer:
Power is defined as the rate of doing work or the rate of transfer of energy. If an agent does a work W in time t, then power is given by :

Unit of power is watt (W).

Question 2.
Define 1 watt of power.
Answer:
1 watt is the power of an agent, which does work at the rate of 1 joule per second.

Question 3.
A lamp consumes 1000 J of electric energy in 10 s. What is its power ?
Solution:
Here,
Work done (W) = 1000J Time(t) = 10 s Power (P) = ?
We know that,
Work done(W) = 1000 J
Time(t) = 10 s
Power(P) = ?
We know that,
P = \(\frac {W}{t}\) = \(\frac {1000}{10}\) Js^-1= 100 W(Watt)

Question 4.
Define average power.
Answer:
The ratio of total energy used to total time given is called average power. Thus,

HBSE 9th Class Science Work and Energy Textbook Questions and Answers

Question 1.
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
(i) Suma is swimming in a pond.
(ii) A donkey is carrying a load on its back.
(iii) A wind-mill is lifting water from a well.
(iv) A green plant is carrying out photosynthesis.
(v) A engine is pulling a train.
(vi) Food grains are getting dried in the sun.
(vii) A sailboat is moving due to wind energy.
Answer:
(i) There is work when suma is swimming in a pond, because she is displacing in the direction of force.
(it) No work is done because its displacement is zero.
(iii) Work is done in lifting the water because it displaces in the direction of force.
(iv) No work is done because displacement is zero.
(v) Work is done in pulling the train because the train is displacing in the direction of engine.
(vi) No work is done because the displacement is zero.
(vii) Work is done by the wind energy in producing motion in sailboat because sail boat is displacing in the direction of force.

Question 2.
An object was thrown at a certain angle to the ground moves in a curved path and falls back to. the ground. The initial and the final point of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer:
The work done by the force of gravity on the object is zero because the displacement of the object is zero.

Question 3.
A battery lights a bulb. Describe the energy changes involved in the process.
Answer:
When a battery lights a bulb, chemical energy is converted in electric energy.

Question 4.
The certain force acting on a 20 kg mass changes its velocity from 5 ms-1 to 2 ms^-1. Calculate the work done by the force.
Solution:
Here,
Mass of the body (m) = 20 kg
Initial velocity of the body (u) = 5 ms^-1
Final velocity of the body (v) = 2 ms^-1
Initial kinetic energy of the body (E₁) = \(\frac {1}{2}\) mu² = \(\frac {1}{2}\) × 20 × (5)² J = 250J
Final kinetic energy of the body (E₂) = \(\frac {1}{2}\)m(v)² = \(\frac {1}{2}\) × 20 × (2)² J = 40 J
Thus, Work done = Change in the Kinetic Enegry = E₁ – E₂
= 250 J – 40J = 210 J

Question 5
A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal. What is the work done on the object by the gravitational force ? Explain your answer.
Solution:
Here,
Mass of the object (m) = 10 kg
Force of gravity (g) = 9.8 ms^-2
Height (h) = 0
Work done by force of gravity (w) = mgh = 10 × 9.8 × 0 = 0
Thus, the work done by force of gravity is zero because displacement is horizontal.

Question 6.
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy ? Why ?
Answer:
The potential energy of a freely falling object decreases progressively but this does not violate the law of conservation of energy because this energy is transferred in kinetic energy. The total potential energy nearest to earth changes in kinetic energy and on reaching the earth, it converts in potential energy.

Question 7.
What are the various energy transformations that occur when you are riding a bicycle?
Answer:
When we ride a bicycle, our muscular energy changes in kinetic energy and we get muscular energy from the food after converting in chemical energy.

Question 8.
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it ? Where is the energy you spend going?
Answer:
When we push a huge rock with all our might and fail to move it, the transformation of energy in this stage wastes against friction and the work done is considered as zero.

Question 9.
A certain household has consumed 250 units of energy during a month. How much energy is , in joules?
Solution:
Here,
Energy consumed = 250 units = 250 kWh = 250 × 1000 × 3600s = 900000000 watt second = 9 × 108J

Question 10.
An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy ? If the object is allowed to fall, find its kinetic energy when it is halfway down, (g = 10 ms^-2 )
Solution:
Here,
The mass of the object (m) = 40 kg
Height of the object from the earth (h) = 5 m
Acceleration due to gravity (g) = 10 ms^-2
Potential energy of the object (w) = mgh
40 × 5 × 10 J = 200J
If the object is allowed to fall freely, the half of the potential energy of the object in the halfway will convert in kinetic energy. Therefore the kinetic energy of the object when it is half-way down = \(\frac {2000}{2}\) = 1000J

Question 11.
What is the work done by the force of gravity on a satellite moving around the earth? Justify your answer.
Answer:
The work done by the force of gravity on a satellite moving around the earth is zero because the displacement of both is zero.

Question 12.
Can there be displacement of an object in the absence of any force acting on it ? Think. Discuss this question with your friends and teacher.
Answer:
Ho, there cannot be displacement of an object in the absense of any force acting on it because the displacement is always due to unbalanced force.

Question 13.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not ? Justify your answer.
Answer:
The work done by the person is zero when he hold a bundle of hay over his head for 30 minutes because the displacement is zero.

Question 14.
An electric heater is rated 1500 W. How much energy does it use in 10 hours ?
Solution:
Here,
The power of the electric heater (P) = 1500W
Time (t) = 10 hours
Energy (w) = P × t = 15000 Wh = \(\frac {15000}{1000}\) kWh = 15 k Wh

Question 15.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest ? What happens to its energy eventually? Is it a violation of the law of conservation of energy.gy ?
Solution:
When we take the bob or pendulum from its middle position to one side (let left side) upto some height, the work done by it is conserved in the form of potential energy. When the bob is allow to swing, then the total potential energy of the bob changes into kinetic energy on reaching left-to-right and middle position. This kinetic energy takes the bob upto that height at which the kinetic energy is changed into potential energy. Due to this potential energy, the bob oscillates again from right to middle. This continuous and bob or pendulum moves from right-left.

After sometime bob comes in rest because frictional force of air works on it and the energy is wasted to work against it. This is not the violation of the law of conservation of energy.

Question 16.
An object of mass m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest ?
Answer:
An object of mass is moving with a constant velocity v, its kinetic energy is \(\frac {1}{2}\) mv². The work is equal to kinetic energy should be done on the object in order to bring the object to rest.

Question 17.
Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h ?
Solution:
Here,
Mass of the car (m) = 1500 kg
Velocity of the car (v) = 60 km/h
= \(\frac{60 \times 1000}{3600}\) ms^-1 = \(\frac{50}{3}\) ms^-1
Thus, kinetic energy of the car = \(\frac {1}{2}\) mv² = \(\frac {1}{2}\) = 1500 × \(\frac {1}{2}\) × \(\frac {1}{2}\)J = 208333.3 J
∴ The Work Required to be done to stop the car = 208333.3 J

Question 18.
In each of the following, a force F is acting on an object of mass m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

Answer:
The work done in the first stage is zero, in the second stage is positive and in the third stage is negative.

Question 19.
Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer:
Yes, we agree with Soni because when several forces are acting on an object, the total effect of the force becomes zero, Thus, F = 0, from this, ma = 0, but m cannot be zero, Therefore, the acceleration can be zero.

Question 20.
Find the energy in kW h consumed in 10 hours by four devices of Dower 500 W each.
Solution:
Here,
The total power of four devices (P) = 500 W × 4 = 2000W
Time (t) = 10 hours
The energy consumed = Power × Time
= 2000 W × 10 h
\(\frac {20000}{1000}\) k Wh = 20kWh

Question 21.
A freely falling object eventually stops on reaching the ground. What happenes to its kinetic energy ?
Answer:
Its kinetic energy is transferred in potential energy.

Haryana State Board HBSE 9th Class Science Solutions Chapter 10 Gravitation Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 10 Gravitation

HBSE 9th Class Science Gravitation Intext Questions and Answers

Questions from Sub-section 10.1

Question 1.
State the universal law of gravitation.
Answer:
Universal Law of Gravitation:
Every object in the universe attracts every other object with a force that is proportional to the product of their masses and inversely proportional to the square of the

distance between them. The force is along the line joining the centres of two objects. Let two objects A and B of masses M and ‘m lies at a distance ‘d’ from each other as shown in figure. Let the force of attraction between two objects be F. According to the universal law of gravitation, the force between two objects is directly proportional to the product of their masses, that is,
F ∝ M × m ……….(i)
And the force between two objects is inversely proportional to the square of the distance between them,
that is,
F ∝ \(\frac{1}{d^2}\) ………..(ii)
Combining eqs. (i) and (ii), we get
F ∝ \(\frac{M \times m}{d^2}\) or F = G \(\frac{M \times m}{d^2}\)
Where G is the constant of proportionality and is called universal gravitational constant.

Question 2.
Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Answer:
Suppose the mass of earth = M_e
Mass of the object on the surface of earth = m
The radius of earth = R
∴ The magnitude of the gravitational force between the earth and an object on the surface of the earth (F) = G \(\frac{M_e \times m}{R^2}\) (where G is gravitational constant)

Questions from Sub-section 10.2

Question 1.
What do you mean by free fall?
Answer:
When the objects fall towards earth only because of gravitation, then it is said that the objects are in free faIl.

Question 2.
What do you mean by acceleration due to gravity?
Answer:
When a body falls towards earth, acceleration works. This acceleration is due to the force of gravity of earth. Therefore, this acceleration is called the acceleration due to the force of gravity of earth or acceleration due to gravity. it is denoted by ‘g’ and its unit is ms².

Questions from Sub-sections 10.3, 10.4

Question 1.
What are the differences between the mass of an object and its weight?
Answer:
The differences between the mass and weight of an object are as follows:

Mass:
1. Mass of a body is the quantity of matter contained in it.
2. The mass of an object cannot be zero.
3. Mass of an object is constant.
4. The mass of an object is found by physical balance.
5. Mass is measured ¡n kilogram (kg).
6. Mass is a scalar quantity.

Weight:
The force by which the earth attracts an object
towards itself is called the weight of that object on earth.
2. The weight of an object is zero on the centre of earth.
3. Weight is not constant. The weight of an object is more on poles as compared to equator.
4. The weight of an object is found by spring balance.
5. Weight is measured in newton.
6. Weight is a vector quantity.

Question 2.
Why is the weight of an object on the moon th its weight on the earth?
Answer:
We know that the value of acceleration due to gravity (g) on earth is 9.8 ms^-2. The value of g on moon is \(\frac {1}{6}\) th time of that on earth. Therefore, the weight of objects on moon is less than that on earth. That is g \(\frac {1}{6}\)th times.

Questions from Sub-section 10.5

Question 1.
Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer:
It is difficult to hold a school bag having a strap made of a thin and strong string because its thrust affects a smaller area (in the center of the string).

Question 2.
What do you mean by buoyancy?
Answer:
When an object is immersed in a fluid, the fluid exerts a force on the body in an upward direction which is called buoyant force. The result of buoyant force depends upon the density of the fluid. This is known as buoyancy.

Question 3.
Why does an object float or sink when placed on the surface of the water?
Answer:
Objects of density less than that of a liquid float on the liquid. The objects of density greater than that of a liquid sink in the liquid. For example, when a nail and a cork of the same mass are put in water, the nail will sink because its density is more than water, whereas the cork will float because its density is less than that of water.

Questions from Sub-section 10.6

Question 1.
You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:
if the mass is 42 kg on a weighing machine, then our mass will be 42 kg because mass remains constant.

Question 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 ¡s heavier and why?
Answer:
Iron bar will be heavier than a bag of cotton because the density of iron is more than cotton.

HBSE 9th Class Science Gravitation Textbook Questions and Answers

Question 1.
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
If the distance between two objects is reduced to half, then the gravitational force between them will increase 4 times, because gravitational force is inversely proportional to the square of the distance between them.

Question 2.
Gravitational force acts on all objects is proportional to their masses. Why then, a heavy object does not fall faster than a light object?
Answer:
Gravitational force acts on all objects is proportional to their masses, yet a heavy object will not fall faster than a light object because the mass of objects is negligible in comparison to that of earth.

Question 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 x 10²⁴ kg and radius of the earth is 6.4 x 10⁶m.)
Solution:
Here, Mass of earth (Me) = 6 × 10²⁴kg
Mass of object (m) = 1 kg
Radius of earth (R) = 6.4 × 10⁶m
Gravitational constant (G) = 6.673 × 101 Nm² kg²
Force of gravity (g) = ?
We know that,
Force of gravity (g) = \(\frac{G M_e \times m}{R^2}\) = \(\frac{6.673 \times 10^{-11} \times 6 \times 10^{24} \times 1}{\left(6.4 \times 10^6\right)^2}\) = 9.8N

Question 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?
Answer:
The earth attracts the moon with greater force because the mass of earth is more than that of moon.

Question 5.
If the moon attracts the earth, why does the earth not move towards the moon?
Answer:
The moon attracts the earth, but the earth does not move towards the moon because the mass of the moon is much less than that of earth.

Question 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?
Answer:
(i) If the mass of one object is doubled, then the force between two objects will be doubled because gravitational force is proportional to mass.
(ii) If the distance between two objects is doubled, the gravitational force will become times and if the distance is tripled, the gravitational force will become times, because gravitational force is inversely proportional to the square of the distance.
(iii) As the force between two bodies is proportional to the product of masses of the bodies, so if the masses of both the bodies are doubled, then force between them will become four times.

Question 7.
What i.s the importance of universal law of gravitation?
Answer:
The importance of universal law of gravitation is as follows:
(i) This force binds us to the earth.
(ii) The motion of the moon around the earth is due to this force.
(iii) The motion of planets around the sun is due to this force.
(iv) The tides due to the moon and the sun are due to this force.

Question 8.
What is the acceleration of free fall?
Answer:
When an object falls towards the earth, then acceleration works. Acceleration of free fall is, acceleration due to the gravitational force of earth and is equal to g. Its value is 9.8 ms².

Question 9.
What do we cafl the gravitational force between the earth and an object?
Answer:
The gravitational force between the earth and an object is weight.

Question 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 g is greater at the poles than at the equator.)
Answer:
No, the friend will not agree with the weight of gold bought, because the weight of gold is less on the equator than on the poles.

Question 11.
Why will a sheet of paper fall slower than one of that ¡s crumpled into a ball?
Answer:
The sheet of paper will fall slower than one of that is crumpled into a ball because the area of sheet is more than the area of ball, due to this, the pressure decreases.

Question 12.
The gravitational force on the surface of the moon is only 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?
Solution:
Here,
Mass of the object on earth (m) = 10kg
Acceleration due to gravity on earth (g) = 9.8 m/s²
Weight of the object on earth (w) m x g = 10 x 9.8 N = 98 N
Mass of object on moon (m) = 10 kg
Acceleration due to gravity on moon (g₁) = \(\frac {9.8}{6}\)m/s²
∴ Weight of the object on moon (w) = mg₁ = 10 x \(\frac {9.8}{6}\) N = 16.3 N

Question 13.
A ball ¡s thrown vertically upwards with a velocity 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.
Solution:
Here,
Initial velocity of ball (u) = 49 m/s
Final velocity of ball (v) = 0 m/s
Acceleration due to gravity (g) = -9.8 m/s²
(i) We know that,
v² – u² = 2gs
s = \(\frac{v^2-u^2}{2 g}=\frac{(0)^2-(49)^2}{2(-9.8)}\) = \(\frac{-49 \times 49}{-19.6}\) = 122.5m
∴ The maximum height to which ball rises is 122.5 m

(ii) Time taken (i) = \(\frac{v-u}{g}\) = \(\frac{0-49}{-9.8}\) = 5s
∴ The total time it takes to return to the surface of earth 5s + 5s = 10

Question 14.
A stone is released from the top of a tower of a height of 19.6 m. Calculate Its final velocity just before touching the ground.
Solution:
Here,
Initial velocity of stone (u) = 0 m/s
The final velocity of stone (v) =?
Height of the top of the tower (s) = 19.6 m
Acceleration due to gravity (g) = 9.8 ms^-2
We know that,
v² – u²= 2gs
v² – (0)² = 2 x 9.8 x 19.6
or v² = 19.6 x 19.6
or v = \(\sqrt{19.6 \times 19.6}\)= 19.6 ms^-1
∴ Thus velocity of stone will be 19.6 ms^-1.

Question 15.
A stone is thrown vertically upward with an initial velocity of 40 ms. 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?
Solution:
Here,
Initial velocity of stone (u) = 40 m/s
Final velocity of stone (v) = 0 m/s
Acceleration due to gravity (s) = -10 m/s²

We know that,
v² – u² = 2gs
s = \(\frac{v^2-u^2}{2 g}=\frac{(0)^2-(40)^2}{2(-10)}\) = \(\frac{-40 \times 40}{-2 \times 10}\) = 80m
∴ The maximum height reached by the stone = 80 m
Total distance covered by the stone = 80 m + 80 m = 160m
TotaÌ displacement of the stone = Zero

Question 16.
Calculate the foíce of gravitation between the earth and the Sun, given that the mass of the earth = 6 x 10²⁴ kg and of the Sun 2 x 10³⁰ kg. The average distance between the two Is 1.5 x 10¹¹ m.
Solution:
Here, The mass of the earth (M_e) = 6 x 10²⁴ kg
Mass of the sun (M_e) = 2 x 10³⁰ kg
The average distance between the two (R_s) = 1.5 x 10¹¹ m
Gravitational constant (G) = 6.673 x 10¹¹ Nm²/kg²
∴ Force of gravitation between the earth and sun (F_s) = \(\frac{G M_e M_s}{\left(R_s\right)^2}\)
= \(\frac{6.673 \times 10^{-11} \times 6 \times 10^{24} \times 2 \times 10^{30}}{\left(1.5 \times 10^{11}\right)^2}\) N
= \(\frac{6.673 \times 12 \times 10^{43}}{2.25 \times 10^{22}}\) N = 35.59 x 10²¹ N

Question 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.
Solution:
Suppose the two stones will meet at a distance of m.
For the stone tell from the top;
initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Distance covered (s) = x m
We know that
S = ut + \(\frac {1}{2}\) gt²
x = (0) (t) + \(\frac {1}{2}\) (9.8)t²
or x = 4.9 t² ………….(i)

For the stone projected upwards;
Initial velocity (u) = 25 m/s
Acceleration due to gravity (g) = -9.8 ms²
Distance covered (s) = (100 – x)m
We know that,
(s) = ut+ \(\frac {1}{2}\) gt²
100 – x 251 + \(\frac {1}{2}\) (-9.8) t²
or 100 – 4.9t² = 25 t – 4.9 t² [From (i)]
or 100 = 25 t
or t = \(\frac {100}{25}\) = 4s

Both stones will meet after 4 seconds Ans
Height from top (x) = 4.9 (4)²
4.9 x 16 = 78.4 m
∴ Distance from down = 100 – 78.4 = 21.6 m

Question 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 4s.
Solution:
Here,
Initial velocity (u) = ?
Final velocity (v) = 0 m/s
Time taken by ball to cover maximum height (t) = \(\frac {6s}{2}\) = 3s
Upward acceleration due to gravity (g) = – 9.8 m/s²
(a) We know that
v = u + gt
u = v – gt = 0 – (-.9.8)(3) = 29.4 m/s
Therefore, initial velocity of ball (u) = 29.4 m/s

(b) We know that
s = ut + \(\frac {1}{2}\) gt²
= (29.4) (3) + \(\frac {1}{2}\) (-9.8) (3)²
88.2 – 44.1 = 44.1 m
∴ Maximum height it reaches = 44.1 m

(c) For this state, distance will be found from up to down
u = 0
t = 4s – 3s = 1s
g = 9.8 m/s²
s = ut + \(\frac {1}{2}\) gt²
0(1) + \(\frac {1}{2}\) (9.8)(1)²
= 4.9m
∴ The ball will be at a height of 4.9 m from up.
Height of ball from down = 44.1 – 4.9 = 39.2 m

Question 19.
In what direction does the buoyant force on an object Immersed in a liquid act?
Answer:
The buoyant force on an object immersed in a liquid acts in the vertically upward direction.

Question 20.
Why does a block of plastic released under water come up to the surface of the water?
Answer:
The block of plastic when left under water comes to the surface because the thrust due to water is greater than the weight of the block itself.

Question 21.
The volume of 50 g of. the substance is 20. If the density of water is 1 gcm⁴, will the substance float or sink?
Solution:
Here,
Mass of the substance = 50g
Volume of the substance = 20 cm³
∴Density of substance = = \(\frac {50}{20}\) g cm^-3
= 2.5g cm^-3
Since the density of the substance is greater than that of water, hence, the substance will sink in water.

Question 22.
The volume of. 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water Is 1g cm³? What will be the mass of the water displaced by this packet?
Solution:
Here,
Mass of the packet = 500g
Volume of the packet = 350 cm³
Density of packet =
= \(\frac {500}{350}\) g cm^-3 = 1.43 gcm^-3
Since the density of packet is greater than that of water, hence, it will sink in water.
Mass of water displaced = volume x density = 350 x 1 g = 350 g

Haryana State Board HBSE 9th Class Science Solutions Chapter 9 Force and Laws of Motion Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 9 Force and Laws of Motion

HBSE 9th Class Science Force and Laws of Motion Intext Questions and Answers

Questions from Sub-section 9.1, 9.2, 9.3

Question 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-rupee coin and a one-rupee coin?
Answer:
(a) The inertia of stone will be more, in a rubber ball and a stone of the same size.
(b) In a bicycle and a train, the inertia of the train will be more.
(c) In a five-rupee coin and a one-rupee coin, the inertia of a five-rupee coin will be more.

Question 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.
Answer:
The velocity of the ball changes four times as explained below:
(i) When the player kicks the ball.
(ii) When the second player kicks the ball towards the goal.
(iii) When the goalkeeper collects the ball.
(iv) When the goalkeeper kicks the ball towards the player of his own team.
The agent of force in each case is the muscular effort of the respective player.

Question 3.
Explain why some of the leaves may get detached from a tree if we vigorously shake its branch ?
Answer:
When branches of a tree are shaken vigorously, the leaves get detached from the tree. This is because the branches of the tree come into motion whereas the leaves tends to remain at rest due to inertia of rest. Thus, the leaves get detached from the branches and fall down under the action of gravity.

Question 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 ?
Answer:
When a moving bus stops suddenly, the lower part of our body in contact with the bus comes to rest immediately, whereas the upper part of the body tends to move in forward direction due to inertia of motion and we fall in the forward direction.
But, when the bus accelerates from rest, the lower part of our body comes into motion alongwith the bus but the upper part of the body has tendency to remain in the state of rest due to inertia of rest and we fall in backward direction.

Questions from Sub-section 9.4, 9.5, 9.6

Question 1.
If action is always equal to the reaction, explain how a horse can pull a cart ?
Answer:
According to third law of motion, the action is always equal to reaction. When the horse pulls the cart, it acts on earth by a force F, the earth offers it reaction R, which has two components R cos A and R sin A. The horizontal part R cos A pulls the cart in forward direction, whereas the vertical part R sin A reduces the frictional force between the road and cart. In this way, the horse can pull the cart.

Question 2.
Explain, why it is difficult for a Fireman to hold a hose, which ejects large amounts of water at a high velocity ?
Answer:
According to third law of motion, the action is always equal to reaction. When the excess water comes out of the hose with high velocity, then it recoils the hose with same amount of velocity in backward direction. Therefore, it becomes difficult for a fireman to hold a hose.

Question 3.
From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 ms^-1. Calculate the initial recoil velocity of the rifle.
Solution:
Here,
Mass of the bullet (m₁) = 50g = 0.05 kg
Mass of the rifle (m₂) = 4 kg
The initial velocity of bullet (u₁) and the initial velocity of the rifle (u₂) is zero.
or u₁ = u₂ = 0

Velocity of bullet (v,) = + 35 ms^-1
The direction of bullet, from left to right is taken + ve according to figure.
Let the recoil velocity of rifle = v ms^-1
The total momentum of bullet and rifle before firing = (4 + 0.05) kg x 0 ms^-1 = 0
Total momentum of bullet and rifle after firing = 0.05 kg × 35 ms^-1 + 4 kg x v ms^-1 = (1.75 + 4 v) kg ms^-1
According to the law of conservation of momentum,
Total momentum after firing = Total momentum before firing
1.75 + 4 v = 0
v = –\(\frac{-1.75}{4}\) = -0.4375 = -0.44 ms^-1
Negative sign shows that rifle will recoil in the opposite direction or right to left.

Question 4.
Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms-1 and 1 ms”1, respectively. They collide and after collision, the first object moves at a velocity of 1.67 ms”1. Determine the velocity of the second object.
Solution:
Here,
Mass of the first object (m₁) = 100g = 0.1 kg
Mass of second object (m₂) = 200 g = 0.2 kg
Initial velocity of first object (u₁) = 2 ms^-1
Initial velocity of second object (u₂) = 1 ms^-1
Total momentum of two objects before collision = m₁u₁ + m₂u₂
= (0.1 x 2 + 0.2 x 1) kg ms^-1 = 0.4 kg ms^-1
The velocity of first object after collision (v₁) = 1.67 ms^-1
Let, the velocity of second object after collision (v₂) = v ms^-1
Total momentum of two objects after collision = m₁v₁ + m₂v₂
= [0.1 x 1.67 + 0.2 x v₂] kg ms^-1
= [0.167 + 0.2 v₂] kg ms^-1
According to the law of conservation of momentum
Total momentum after collision = Total momentum before collision
0.167 + 0.2 v₂ = 0.4
or 0.2v₂ = 0.4 – 0.167
0.2v₂ = 0.233
or v₂ = \(\frac {0.233}{0.2}\) = 1.165 ms^-1
The velocity of second object after collision = 1.165 ms^-1

HBSE 9th Class Science Force and Laws of Motion Textbook Questions and Answers

Question 1.
An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling 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.
Answer:
Here, the external unbalanced force on an object is zero. It means,
F = 0
⇒ m a = 0
But, m ≠ 0 ∴ a = 0
It is possible to move in this state, but the motion will be in the same magnitude and in one direction or straight path.

Question 2.
When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer:
When a carpet is beaten with a stick, it moves forward, whereas dust particles have a tendency to remain at rest due to the inertia of rest. Thus, dust particles come out and the carpet gets clean.

Question 3.
Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:
It is advised to tie any luggage kept on the roof of a bus with a rope because when the moving buses or cars suddenly stop or change direction, the luggage becomes safe from falling due to inertia of motion.

Question 4.
A batsman hits a cricket ball which then rolls on level ground. After covering a short distance, the ball comes to rest The ball slows to stop because:
(а) 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
(Choose the correct option)
Answer:
(c) There is a force on the ball opposing the motion.

Question 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 tonnes. [Hint: 1 tonne = 1000 kg]
Solution:
Here, Initial velocity in first stage (u) = 0
Distance covered (s) = 400 m
Acceleration (a) = ?
Time (t) = 20s
According to formula, s = ut + \(\frac {1}{2}\) at²
ut + \(\frac {1}{2}\) at² = s
0 x (20) + \(\frac {1}{2}\) a x (20)² = 400
200 a = 400
a = \(\frac {400}{200}\) = 2 ms^-2
Mass in second stage (m) = 7 tones = 7 x 1000 = 7000kg
force (F) = m x a = 7000 kg x 2 ms^-2 = 14000 N

Question 6.
A stone of 1 kg is thrown with a velocity of 20 ms^-1 across the frozen surface of a lake and it comes to rest after traveling a distance of 50 ni. What i the force of friction between the stone and the ice?
Solution:
Here,
Mass of stone (m) = 1 kg
Initial velocity of stone (u) = 20 ms^-1
Final velocity of stone (v) = 0
Distance covered by stone (s) = 50 m
Acceleration (a) = ?

We know that,
v² – u² = 2as .
a = \(\frac{v^2-u^2}{2 s}=\frac{(0)^2-(20)^2}{2(50)}\)
= \(\frac{-400}{100}\) = -4 ms^-1
A negative (-ve) sign shows that acceleration is working against the motion.
The frictional force between the stone and the ice I (-4) = – 4N

Question 7.
An 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 (be 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 I on wagon 2.
Solution:
Here,
The mass of train engine (m₁) = 8000 kg
Mass of 5 wagons(m₂) 2000kg x 5 = 10000 kg
Mass of train engine with wagons (m) = m₁ + m₂ = (8000 + 10000)kg = 18000 kg
The force applied by the engine on wagons (F₁) = 40000 N
The friction force applied by track on wagons (F₂) = 5000 N
(a) The net acceleration force (F) = F₁ – F₂
40000N – 5000N = 35000N

(b) The acceleration of the train (a) = \(\frac{F}{m_2}=\frac{35000}{10000}\) ms^-2 = 3.5 ms^-2

(c) The force of wagon I ou wagon 2 35000 N – \(\frac{35000}{5}\) = (35000 – 7000) N = 28000N Ans

Question 8.
An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is tobe stopped with a negative acceleration of 1.7 ms^-2?
Solution:
Here,
Mass of automobile vehicle (m) = 1500 kg
The acceleration on vehicle (a) = 17 ms^-2
Final velocity (v) = 0
The force between the vehicle and road (F) = m x a = 1500 × (-1.7) N = -2550 N
Negative (-ve) sign shows that the force is working against the motion of vehicle.

Question 9.
What is the momentum of an object of mass m, moving with a velocity v ?
(a) (mv)²
(b) mv²
(c) \(\frac {1}{2}\) mv²
(d) mv (choose correct option from above)
Answer:
(d) mv.

Question 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?
Answer:
If we push the wooden cabinet with constant velocity on the floor by applying a force of 200 N, then the frictional force of the wooden cabinet will also be 200 N.

Question 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?
Answer:
Here, the mass and velocity of both objects are the same, but both are moving in opposite directions, so they bend after the collision. Therefore, their combined velocity will be zero.

Question 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.
Answer:
The truck parked along the roadside will not move on applying force by us because the force applied by us is much less than the frictional force applied by the truck between the truck and the road. If our force increases to frictional force, the truck will move.

Question 13.
A hockey ball of mass 200 g traveling at 10 ms-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms^-1. Calculate the change of momentum that occurred in the motion of the hockey ball by the force applied by the hockey stick.
Solution:
Here,
Mass of hockey ball (m) = 200 g = 0.2 kg
Initial velocity of hockey ball (v₁) = 10 ms
The initial momentum of hockey ball travels in the opposite direction with velocity (v₂) = 5 ms^-1
Final Movemntum Of Hockey Boll = mv₂
Change in momentum = mv₁ – mv₂ = 0.2 [10 – (-5)] = 0.2(15) = 3 kg ms^-1
Thus, the change in momentum of the hockey ball is 3 kg ms^-1

Question 14.
A bullet mass Logis traveling horizontally with a velocity of 150 m1 strikes a stationary wooden block and comes to at rest in 0.03 s. Calculate the distance of penetration of the bullet into the Mock. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Solution:
Here,
Mass of bullet (m) = 10g
= \(\frac {10}{1000}\) = 0.01kg
Initial velocity of bullet (u) = 150 ms^-1
Final velocity of bullet (v) = 0 ms^-1
Time (t) = 0.03 s
Acceleration (a) = ?
Distance (s) = ?

We know that,
a = \(\frac{v-u}{t}=\frac{0-150}{0.03}\) ms^-2 = -5000 ms^-2
and s = ut + at²
= (150)(0.03) + \(\frac {π}{2}\) × (-5000) x (0.03)²
= 4,5 – 2.25 = 2.25m
Force(F) = ma
= 0.01 x (-5000)N = -50 N
The negative (-ve) sign shows that force is working in the opposite direction of motion.

Question 15.
An object mass 1 kg traveling In a straight line with a velocity of 10 ms 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.
Solution:
Here,
Mass of object (m) = 1 kg
Velocity of object (v₁) = 10 ms^-1
∴Momentum before collision = m x v₁
= 1 x 1o kg ms^-1 = 1o kg ms^-1
The momentum after collision = 10 kg ms^-1
Combined mass of wooden block and object (m₁) = 5 + 1 = 6kg
Suppose. the final velocity of object and wooden block = v₂ ms^-1
According to the law of conservation of momentum,
m₁ v₂ = 10
or 6 x v₂ = 10
0r v₂ = \(\frac{10}{6}\) ms^-1 = \(\frac{10}{6}\) ms^-1

Question 16.
An object of mass loo kg Is accelerated uniformly from a velocity of S mr1 to 8 mr’ in 6s. object. Also, find the magnitude of the force exerted on the object
Solution:
Here.
Mass of object (m) = 100 kg
Initial velocity (u) = 5 ms^-1
Final velocity (y) = 8 ms^-1
Time(t) = 6s
Initial momentum = mu = 100 x 5 = 500 kg ms^-1
Final momentum = mv = 100 x 8 = 800 kg ms^-1
We know that,
a = \(\frac{v-u}{t}=\frac{8-5}{6}=\frac{3}{6}\) = 0.5 ms^-2
The magnitude of the force (F) = m x a
= l00 x 0.5N = 1.50 N

Question 17.
Akhtar, ((Iran and Rbul were riding in a motorcar 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 motorca (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.
Answer:
Akhtar was right because both the velocity and mass of the motorcar were moveso the momentum of the motorcar was more than that of flying insect. Therefore, the motorcar imposed much force on flying insect due to the insect died.

Question 18.
How much momentum will a dumbbell of mass 10 kg transfer to the floor if It falls from a height of 80cm? Take its downward acceleration to be 10 ms² ?
Solution:
Here,
Mass of dumb-bell (m) = 10 kg
Initial velocity of dumb-bell (u) = 0 ms^-1
Final velocity of dumbbell (v) = ?
Height of dumb-bell (s) = 80 cm = 0.8 m
Acceleration (a) = 10^-2
We know that.
v² – u² = 2as
v² – (0)² = 2(10) (0.8)
v² = 16
v = \(\sqrt{16}\) = 4 ms
Momentum transferred to the floor by dumb-bell (p) = mv = 10 x 4 = 40 kg ms

Additional Exercise Questions

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

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

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, orzero?
(b) What do you infer about the force acting on the object 1
Answer:
(a) It is clear from the distance-time table that acceleration is continuously increasing.
(b) The force is also increasing due to the increase in acceleration on body.

Time (t)	Distance (s)	Velocity (v) = \(\frac {s}{t}\)	Acceleration (a) = \(\frac {v – u}{t}\)
t1 = 0s	s1 = 0 m	v<1 = 0 ms-1	a1 = 0 ms-2
t2 = 1 s	s2 = 1 m	v2 = 1 ms-1	a2 = 1 ms-2
t3 = 2 s	s3 = 8 m	v3 = 4 ms-1	a3 = 3 ms-2
t4 = 3 s	s4 = 27 m	v4 = 9 ms-1	a4 = 5 ms-2
t5 = 4S	s5 = 64 m	v5 = 16 ms-1	a5 = 7 ms-2
t6 = 5 s	s6 = 125 m	v6 = 25 ms-1	a6 = 9 ms-2
t7 = 6 S	s7 = 216 m	v7 = 36 ms-1	a7 = 11 ms-2
t8 = 7 s	s8 = 343 m	v8 = 49 ms-1	a8 = 13 ms-2

Question 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 ms^-2.With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
Solution:
Here,
Mass of motorcar (m) = 1200 kg
The acceleration produced on pushing by three persons
(a) = 0.2 ms^-2
∴ The force imposed on car by each person (F) = m a = 1200 x 0.2N. = 240N Ans

Question 3.
A hammer of mass 500 g, moving at 50 m1, to strike a nail. The nail stops the hammer In a very short time of 0.01s. What Is the force of the nail on the hammer?
Solution:
Here,
Mass of hammer (m) = 500 g = \(\frac {500}{1000}\) kg = 0.5kg
Initial velocity of hammer (u) = 50 ms^-1
Time (t) = 0.01 s
Final velocity of hammer (v) = 0
We know that,
Acceleration (a) = \(\frac{v-u}{t}=\frac{0-50}{0.01}\) = 0.01 ms^-2
= -5000 ms^-2
(-ve sign shows that acceleration is decreasing, i.e., retarded)
The force of the nail on the hammer (F) = m x a
= O.5 x (5000)N
= 2500N

Question 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 ¡n momentum. Also, calculate the magnitude of the force required.
Solution:
Here,
The mass of a motorcar (m) is 1200 kg
Initial velocity (u) = 90 km h^-1= \(\frac{90 \times 1000}{3600}\) = 25 ms^-1
Final velocity (v) = 18 km/h = \(\frac{18 \times 1000}{3600}\) = 5 ms^-1
Tíme(t) = 4s
Acceleration (a) = \(\frac{v-u}{t}=\frac{5-25}{4}=\frac{-20}{4}\)
= -5 ms^-2
Change ¡n momentum mv – mu
= m(v – u) = 1200(5 – 25) = 1200 x (-20) = -24000 kg ms^-1
The magnitude of the force required (F) = m a
1200 x – 5 = – 6000N

Question 5.
A large truck and a car, both moving with a velocity of magnitude w, have a head-on collision and both of them come to a halt after that. If the collision for is:
(a) Which vehicle experiences the greater force of impact?
(b) Which vehicle experiences the greater change ¡n momentum?
(C) Which vehicle experiences the greater acceleration?
(d) Why is the car likely to suffer more damage than the truck?
Answer:
(a) Here, the mass of the truck is more than that of car, and the force of the truck will be more which affects the car more.
(b) Truck will experience a greater change in momentum.
(c) The acceleration of the car will be greater.
(d) Car will suffer from more damage because the mass of truck is more.

Haryana State Board HBSE 9th Class Science Solutions Chapter 8 Motion Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 8 Motion

HBSE 9th Class Science Motion Intext Questions and Answers

Question 1.
An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example.
Answer:
Yes, displacement can be zero, if an object covers some distance, e.g., let an object moves from the origin point along a straight line to point ‘A’ by covering a distance of 55 km and covers the same distance of 55 km from ‘A’ to ‘O’ on return as shown in figure In this state-

Distance travelled by the object = 55 km + 55 km = 110 km
Displacement = Zero (0)

Question 2.
A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Solution:
Side of the square field ABCD = 10 metre
The perimeter of the square field ABCD = 4 × side
= 4 × 10 = 40 metres

Time taken for one round around the boundary = 40 s
Total time = 2 minutes 20 s (2 × 60 + 20) s = 140 s
Now, the distance covered by the farmer in 40 s = 40 m
The distance covered by the farmer in 1s m = \(\frac {40}{40}\) = 1 m
The distance covered by the farmer in 140 s 1 × 140 = 140 metres
If, the farmer starts moving from origin point A, then he will be at point C after covering a distance of 140 metres. In this way, the displacement of the farmer will be AC (the diagonal of the square.).
∴ AC = \(\sqrt{(\mathrm{AB})^2+(\mathrm{BC})^2}\)
= \(\sqrt{(10)^2+(10)^2}\)
= \(\sqrt{100+100}\)
= \(\sqrt{200}\)
= \(\sqrt{2 \times 100}\)
= 10\(\sqrt{2}\) m
= 10 × 1.414 = 14.14 m
= 10 × 1.414 = 14.14 m
= 10 × 1.414 = 14.14 m
= 10 × 1.414 = 14.14 m
= 10 × 1.414 = 14.14 m

Question 3.
Which of the following is true for displacement ?
(a) It cannot he zero.
(b) Its magnitude is greater than the distance travelled by the object.
Answer:
Both (a) and (b) are wrong for displacement.

Questions from Sub-section 8.2

Question 1.
Distinguish between speed and velocity.
Answer:
The differences between speed and velocity are as follows:

Speed:
1. It is the distance travelled by the body in unit time interval in any direction.
2. It is a scalar quantity, which has only magnitude.
3. It is always positive.

Velocity
1. It is the rate of distance travelled by the body in unit time interval in specified direction.
2. It is a vector quantity, which has both magnitude as well as direction.
3. It can be positive and negative.

Question 2.
Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer:
If the object is moving with uniform motion in a definite direction, then the magnitude of the average velocity of that object will be equal to its average speed.

Question 3.
What does the odometer of an automobile measure ?
Answer:
The odometer of the automobile measures the distance covered by the automobile.

Question 4.
What does the path of an object look like when it is in uniform motion ?
Answer:
When the object is in uniform motion, then its path seems as a straight line.

Question 5.
During an experiment, a signal from a spaceship tTactretf the ground-statianju 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 ms^-1.
Solution:
The time taken by the signal from spaceship to earth = 5 minutes = 5 x 60 = 300 s
Speed of signal = 3 × 10⁸ms^-1

∴ The distance of spaceship from ground station = Speed × Time = 3 × 10s × 300m = 9 × 1010m

Questions from Sub-section 8.3

Question 1.
When will you say a body is in (i) uniform acceleration (ii) non-uniform acceleration ?
Ans.
(i) Uniform acceleration : A body is said to be moving with uniform acceleration if there is equal change in the velocity of the body in equal interval of time.
(ii) Non-uniform acceleration : A body is said to be moving with non-uniform acceleration, if its velocity changes unequally.

Question 2.
A bus decreases its speed from 80 kmh^-1 to 60 kmh^-1 in 5 s. Find the acceleration of the bus.
Solution:
Initial speed of the bus (u) = 80 kmh^-1
= \(\frac{80 \times 1000}{3600}\)ms^-1 = \(\frac{200}{9}\)ms^-1

Final speed of the bus (v) = 60 kmh^-1
\(\frac{60 \times 1000}{3600}\)ms^-1
\(\frac{50}{3}\)ms^-1
Time (t) = Ss
Acceleration of the bus (a) = \(\frac{v-u}{t}\) = \(\frac{\left(\frac{50}{3}-\frac{200}{9}\right) \mathrm{ms}^{-1}}{5 \mathrm{~s}}\) = \(\frac{150-200}{9 \times 5}\)ms^-2
\(\frac{-50}{45}=\frac{-10}{9}\)ms^-2 = -1.1 ms^-2

Question 3.
A train starting from a railway station and moving with uniform acceleration attains a speed of 40 kmh1 in 10 minutes. Find its acceleration.
Solution:
The initial speed of the train (u) = 0
The final speed of the train (v) = 40 kmh^-1
= \(\frac{40 \times 1000}{3600}\) = \(\frac{40 \times 1000}{3600}\)
Time (t) = 10 minutes = 10 x 60s = 600s
∴ Acceleration (a) = \(\frac{v-u}{t}\) = \(\frac{\frac{100}{9}-0}{600}\)ms^-1
= \(\frac{100}{9 \times 600}=\frac{1}{54}\) ~ 0.02 ms^-2

Questions from Sub-section 8.4

Question 1.
What is the nature of the distance-time graps for uniform and non-uniform motion of an object?
Ans.
For a uniform motion, the graph of the distance covered with time is a straight line. In graph, part OB shows that the distance is increasing with uniform rate.

Distance-time graph for an object possessing non-uniform motion is a curved line as shown in the graph. .

Question 2.
What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer:
If the distance-time graph of an object, parallel to time axis is a straight line, then the object will move with uniform speed.

Question 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 ?
Answer:
If the speed-time graph of an object, parallel to time axis is a straight line, then the object will move with uniform speed.

Question 4.
What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:
The quantity measured by the area occupied below the velocity-time graph shows the distance covered by the body.

Questions front Sub-section 8.5

Question 1.
Abus starting from rest moves with a uniform acceleration of 0.1ms-2for 2 minutes. Find (a) the speed acquired (b) the distance travelled.
Solution:
Here,
Initial speed of the bus (u) = 0
Final speed of the bus (v) = ?
Acceleration (a) = 0.1 ms
Time (t) = 2 minutes = 2 x 60s = 120 s
(a) The final speed (v) of the bus = u + at = 0 + (120) + \(\frac {1}{2}\) = 12 ms^-1
(b) The distance travelled by the bus, (s) = ut + \(\frac {1}{2}\) at²
= 0(120) + \(\frac {1}{2}\)(0.1)(120)²

Question 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 ms^-2 Find how far the train will go before it is brought to rest?
Solution:
Here,
Initial speed of the train (u) = 90 km h^-1
\(\frac{90 \times 1000}{3600}\) ms^-1= 25 ms^-1

Final speed of the train (v) = 0
Acceleration (a) = – 0.5 ms²
Distance (s) = ?
We know that,  v² – u² = 2as
s = \(\frac{90 \times 1000}{3600}\)
= \(\frac{(0)^2-(25)^2}{2(-0.5)}=\frac{-625}{-1.0}\) = 625 metres
∴ The train will cover a distance of 625 metres, before it is brought to rest.

Question 3.
A trolley, while going down on inclined plane, has an acceleration of 2 ms^-2. What will be is velocity 3s after the start?
Solution:
Here,
The initial speed of the trolley (u) = 0
Final speed of the trolley (v) = ?
Acceleration (a) = 2 ms^-2
Time (t) = 3s
We know that, v = u + at
= 0 + 2(3) = 6 ms^-1
∴ The velocity of the trolley after 3 s will be 6 ms^-1.

Question 4.
A racing car has a uniform acceleration of 4ms2. What distance will it cover in 10 s after start?
Solution:
Here,
The initial speed of the car (u) = 0
Acceleration (a) = 4 ms
Time (t) = 10s
Distance (s) = ?

We know that,
s = ut + \(\frac {1}{2}\) at²
= 0 x (10) + \(\frac {1}{2}\) (4) (10)²
= \(\frac {1}{2}\) × 4 × 100 = 200m
So, the covered distance will be 200m

Question 5.
A stone is thrown in a vertically upward direction with a velocity of 5 ms. If the acceleration of the stone during its motion is 10 ms2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Solution:
Here,
The initial velocity of the stone (u) = 5 ms^-1
Final velocity of the stone (v) = 0
Acceleration (a) = – 10 ms^-2
Maximum height (s) = ?
We know that,
v²– u² = 2as
⇒ S = \(\frac{v^2-u^2}{2 a}=\frac{(0)^2-(5)^2}{2(-10)}\) = \(\frac {-25}{20}\) = 1.25m
Applying, v = u + at
⇒ t = \(\frac{v-u}{a}=\frac{0-5}{-10}\) = 0.5 s

HBSE 9th Class Science Motion Textbook Questions and Answers

Question 1.
An athlete completes one round of a circular track of diameter 200 metre in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Solution:
Here,
The diameter of the circular track = 200 m
Radius (r) of the circular track = \(\frac {200}{2}\) = 1oo m
The circumference of the circular track = 2πr
= 2 x \(\frac {22}{7}\) × 1oo m
= \(\frac {4400}{7}\)m
Total time = 2 minutes 20 se
= (2 x 60 + 20)s (120 + 20)s = 140s

∴ The distance covered by the athlete in 40 s = \(\frac {4400}{7}\) m
The distance covered by the athlete in 1 s = \(\frac{4400}{7 \times 40}\) m
The distance covered by the athlete in 140 s = \(\frac{4400}{7 \times 40}\) x 140m = 2200 metres
Here, the total time is 140 s, in which the athlete will complete three complete rounds and one half round. If he will start from point A of the circumference and reach at point B.
In this way, displacement (AB) = The diameter of the circular track = 200 metres

Question 2.
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 s and then returns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speed and velocitie in jogging (a) from point A to B (b) from point A to C?
Solution:
(a) The total distance covered from point A to B = 300 m
The total time taken from point A to B = 2 min 30 s = (2 x 60 + 30)s = 150 s
Total displacement from point A to B = 300 – 0 = 300 m

= \(\frac {4400}{7}\) = 2 ms
(b) Total distance covered from point A to C = AB + BC = (300 + 100)m = 400 m
Total time taken fr om point A to C = 2 minutes 30 s + 1 minute
= (2 x 60 + 30)s + 60s = (150 + 60)s = 210s
Total displacement from point A to C = 300 – 100 = 200 m

= \(\frac {400}{210}\) = 1.90 ms^-1
= \(\frac {200}{210}\) = 0.952 ms^-1
= 0.952 ms^-1

Question 3.
Abdul, while driving to school, computes the average speed for his trip to be 20km h^-1 on his return trip along the same route there is less traffic and the average speed is 30km h^-1 . What is the average speed for Abdul’s trip?
Solution:
Average speed during the school trip v₁ = 20 km h^-1
Average speed for return trip v₂ = 30 km h^-1
Let, the distance from home to school = x km
Time taken for going school Speed = =\(\frac {x}{20}\) h
Time taken to return from school = =\(\frac {x}{30}\) h
Total distance covered for both (going and returning) trips = x + x = 2x km
Total time taken = \(\left(\frac{x}{20}+\frac{x}{30}\right) h\) = \(\left(\frac{3 x+2 x}{60}\right) h=\frac{5 x}{60} h=\frac{x}{12} h\)
∴ Average speed of Abdul during whole trip = = \(\frac{\frac{2 x}{x}}{12}=\frac{2 x \times 12}{x}=24 \mathrm{kmh}^{-1}\)

Question 4.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3 mr² for 8 seconds. How far does the boat travel during this time?
Solution:
Here,
Initial speed of motorboat on lake (u) = 0
Acceleration (a) = 3.0 m^-2
Time (t) = 8.0 s
Distance (s) = ?
We know that, s = ut + at²
= 0(8) + (3) (8)²
= \(\frac {1}{2}\) × 3 × 64 = 96m
∴ Motorboat will cover a distance of 96m on lake. Ans.

Question 5.
A driver of a car travelling at 52km h applies the brakes and accelerates uniformly in the opposite direction. The car stops in Ss. Another driver going at 30km h in another car applies his brakes slowly and stops in lOs. On the same graph paper, plot the speed versus time graph for two cars. Which of the two cars travelled farther after the brakes were applied?
Solution:
Speed-time graph for both drivers is shown in the figure. Suppose.
first driver starts from point A and second driver starts from point B.
Distance covered by first car before rest = area of ∆ AOC
= \(\frac {1}{2}\) × base × height
\(\frac {1}{2}\) × 5s × 52km/h
\(\frac {1}{2}\) × 3600 × 52 km
=\(\frac {1}{2}\) × 300 × 52 x 1000 m
= 36.11 metres

In same way, the distance covered by second car before
rest = area of ∆BOD
= \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × 10 s × 30 km/h
= \(\frac {1}{2}\) × \(\frac {10}{3600}\) × 30km
= \(\frac {1}{2}\) × \(\frac {10}{3600}\) × 30 × 1000 m = 41.67m
It is clear from above solution that second car travels larger distance after applying brakes.

Question 6.
Figure 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?
(e) How far has C travelled whenB passes A?
(d) How far has B travelled by the time it passes C?
Solution:

(a) B is travelling fastest as the slope in the graph of B is maximum as compared to A and C.
(b) Three can never be at the same point on the road because the three graphs do not meet at any single point.
(c) C has travelled the distance of 9 km, when B passes A.
(d) B has travelled a distance more than 4 km during the time he passed C.

Question 7.
A ball is gently dropped from a height of 20m. If its velocity increases uniformly at the rate of 10ms2, with what velocity will it strike the ground ? After what time will it strike the ground?
Solution:
Here,
Initial velocity of ball (u) = 0
Height (s) = 20 m
Acceleration (a) = 10 ms^-2
Final velocity of ball (v) = ?
Time (t) = ?
We know that,
v² – u² = 2as
v² = u² + 2as
= (0)² + 2(10)(20) = 400
or v = \(\sqrt{400}\) = 20 ms^-1
Now, acceleration, a = \(\frac{v-u}{t}\)
or t = \(\frac{v-u}{a}\) = \(\frac{20-0}{10}\) = 2s
∴The velocity of the ball will be 20 ms’ before it strike the ground and it will strike the ground in 2s.

Question 8.
The speed-time graph for a car is shown in figure.

(a) Find how far does the car travel in 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?
Solution:
(a) The area of the distance covered by the car in first 4 s is OAB, which is almost a right-angled triangle.

∴The distance covered by the car in first 4s = \(\frac {1}{2}\) × OA × AB
The distance covered by the car in first 4s = \(\frac {1}{2}\) × 4 × 6 = 12m
The distance covered by the car OABin the figure.

(b) In the graph, the speed after 6 s, shows the uniform motion of the car.

Question 9.
State which of the following situations are possible and give an eamnle for eich 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.
Solution:
(a) Yes, this situation is possible. When a body is thrown up with some velocity, velocity is zero at the highest point but acceleration is non-zero and constant.
(b) Yes, this is possible. A body moving with uniform velocity on circular path is its example.

Question 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.
Solution:
Here,
Radius of circular orbit of artificial satellite (r) = 42250 km
‘Time taken by the satellite to revolve around the earth (t) = 24 hours
= 24 × 3600s = 86400s
Speed of artificial satellite (v) = \(\frac {2πr}{t}\) = \(\frac{2 \times 3.14 \times 42250}{86400}\) k ms^-1 = 3.07 k ms^-1

Haryana State Board HBSE 9th Class Science Solutions Chapter 7 Diversity in Living Organisms Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 7 Diversity in Living Organisms

HBSE 9th Class Science Diversity in Living Organisms Intext Questions and Answers

Questions from Sub-section 7

Question 1.
Why do we classify organisms?
Answer:
Various kind of animals and plants occur on earth/world. For example, on one hand in world, we have microscopic bacteria of a few micrometres in size, on the other hand we have long organisms like blue whale of approximate sizes of 30 metres. While redwood trees of size 100 metres are found in California. Their age and structures are different. Some organisms have simple structures while.some have complex. Therefore, to make easy the study of various forms of life, they are classified on the basis of similarities and dissimilarities.

Question 2.
Give three examples of the range of variations that you see in life forms around you.
Answer:
(1) On one hand in the world, we have microorganisms like bacteria which can be viewed under a microscope, on the other hand we have 30 metres long blue whale.
(2) 100 metres long red wood trees are found in California whereas lichen, are visible merely like small spots.
(3) Tortoise remains alive for three hundreds years. While insects like mosquitoes have a life-span of a few days.

Questions from Sub-section 7.1

Question 1.
Which do you think is a more basic characteristic for classifying organisms ?
(i) the place where they live. (A) the kind of cells they are made. Why ?
Answer:
(A) The kind of cells they are made. It is a basic characteristic of classification because organisms with, nucleated cells have their pathways organised in very different ways than the organisms without nucleated cells.

Question 2.
What is the primary characteristic on which the first division of organisms is made?
Answer:
Greek thinker Aristotle classified organisms according to whether they lived on land, water and air.

Question 3.
On what bases are plants and animals put into different categories?
Answer:
Plants and animals are classified on the basis of cell wall found in their cells. Plants contain cell wall while animals are not.

Questions from Sub-section 7.2

Question 1.
Which organisms are called primitive and how are they different from the so-called advanced organisms?
Answer:
The earlier developed organisms on the earth are called primitive organisms. Another kind of organism means developed organisms from lower organisms are called advanced organisms.

Question 2.
Will advanced organisms be the same as complex organisms? Why?
Answer:
Yes, advanced organisms are complex organisms because earlier organisms are developed by complexity. So simple organisms changed into advanced organisms.

Questions from Sub-section 7.3

Question 1.
What is the criterion for the classification of organisms as belonging to the kingdom Monera or Protista?
Answer:
Cell structure, mode of nutrition and body organisation are the basis of the classification of Monera or Protista.

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

Question 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?
Answer:
The group ‘species’ will have the smallest number of organisms with maximum common characteristics. The largest number of organisms is placed in the ‘kingdom.’

Questions from Sub-section 7.4

Question 1.
Which division among plants has the simplest organisms?
Answer:
Thallophyta.

Question 2.
How are pteridophytes different from the phanerogams?
Answer:
Pteridophytes:
These are lower categories plants. Their body divided into roots, stem and leaves. They have well-developed vascular tissues, the xylem and phloem. The mature leaves bear spores inside sporangia. After fertilization, the embryo developed inside them, e.g. fern.

Phanerogams:
These are higher categories plants. Roots, stem, leaves, flower, fruit and seeds are found in them. They are produced sexually by means of seeds, e.g. wheat, maize, etc.

Question 3.
How do gymnosperms and angiosperms differ from each other?
Answer:
There are the following differences between gymnosperms and angiosperms
Gymnosperms:
1. They have naked seeds.
2. Seeds are not enclosed in the fruit.
3. They are mostly found in mountain plants.
4. Pine and Cycas are its examples.

Angiosperms:
1. They have not naked seeds.
2. Seeds are enclosed in the fruit.
3. These are almost found in all groups of land plants.
4. Pea and wheat are its examples.

Questions from Sub-section 7.5

Question 1.
How do poriferan animals differ from Coelenterate animals ?
Answer:
There are following differences between Poriferan animals and Coelenterate animals :
Poriferan:
1. They are the simplest multicellular animals.
2. They bear pores all over the body.
3. They are flat, round sac or branched in shape like vase.
4. They are sessile.
5. They possess skeleton made up of spongin.
For example: Sycon, Spongilla.

Coelenterate:
1. They are complex multicellular animals.
2. They do not bear pores on the body.
3. They bear tentacles in the body.
4. They are sessile but some are free swimming.
5. They possess outer skeleton made up of lime.
For example: Hydra, Physalia.

Question 2.
How do Annelid animals differ from Arthropods ?
Answer:
There are following differences between Annelid and Arthropods animals :
Annelida:
1. It is the smallest phylum.
2. They are segmented externally.
3. They have closed circulatory system.
4. They have no compound eyes.
5. Their body is not divided into head, thorax and abdomen.
6. They are bisexual.
Examples: Earthworm, Leach and Nereis.

Arthropoda:
1. It is the largest phylum in the animal kingdom.
2. They are covered externally by jointed exoskeleton.
3. They have an open circulatory system.
4. They have compound eyes.
5. Their body is divided into three parts head, thorax and abdomen.
6. They are unisexual.
Examples: Butterfly, Spider, Housefly and Prawn etc.

Question 3.
What are the differences between Amphibians and Reptiles ?
Answer:
There are following differences between Amphibians and Reptiles:
Amphibians:
1. They live in or near freshwater and moist land.
2. Their body lacks scales.
3. Respiration occurs by gills, lungs or skin.
4. They can move their eyelids.
5. They are cold blooded vertebrates.
Examples: Frog, Bufo and Hyla etc.

Reptiles:
1. They mostly live on land.
2. Their body have scales.
3. Respiration occurs by lungs only.
4. They cannot do so.
5. They are cold blooded terrestrial or aquatic.
Examples: Lizard, Chameleon and Snake etc.

Question 4.
What are the differences between animals belonging to the Aves group and those in the mammalia group ?
Answer:
There are following differences between aves group and mammalia group:
Aves:
1. Their body is covered by feathers.
2. They lay eggs.
3. Teeth are absent in their beak.
4. They have wave like body.
5. Their bones have air cavities.
6. Air sacs are present in their lungs.
Examples: Parrot, Pigeon and Crow.

Mammalia:
1. Their body is covered by hair.
2. They give birth to their young ones.
3. Teeth are present in their mouth.
4. They have not.
5. They have solid bones.
6. Air sacs are not present in their lungs.
Examples: Human, Cow and Bat.

HBSE 9th Class Science Diversity in Living Organisms Textbook Questions and Answers

Question 1.
What are the advantages of classifying organisms ?
Answer:
There are following advantages of classifying organisms:
(1) Classification makes the study of organisms easy.
(2) It helps to understand the origin of organisms and their interrelationship.
(3) It gives an idea of similarities and dissimilarities between organisms.
(4) It works as a base for other branches of biology.
(5) It projects before us a picture of all life forms clearly.

Question 2.
How would you choose between two characteristics to be used for developing hierarchy in classification?
Answer:
There are two characteristics to be used for developing hierarchy in classification: cell structure and body organisation. They are developing a hierarchy in classification. Body organisation is the main character in which body parts do work.

Question 3.
Explain the basis for grouping organisms into five kingdoms.
Answer:
There are the following main criteria on the basis of Whittaker’s five kingdoms :
1. Monera: Cell wall present in these unicellular prokaryotic organisms. They can be both autotrophic or heterotrophic. Examples: Bacteria, Blue-green algae, Mycoplasma.

2. Protista: These are unicellular eukaryotic, autotrophic, heterotrophic or both. Cilia or flagella are found in them for moving. Examples: Paramecium, Unicellular algae, Protozoans, Diatoms, etc.

3. Fungi: These are heterotrophic eukaryotic organisms. They use decaying organic material as food. So they called saprophytes. They have cell walls made of a tough complex sugar called chitin. Examples: Yeast, Mushrooms, Agaricus, etc.

4. Plantae: They are multicellular eukaryotes and autotrophs organisms having cell wall. They use chlorophyll for photosynthesis. Example: All green plants.

5. Animalia: They all are multicellular eukaryotic heterotrophic organisms. Cell wall is not present in them.

Question 4.
What are the major divisions in the Plantae? What is the basis for these divisions?
Answer:
The major divisions in Plantae are :
(i) Thallophyta
(ii) Bryophyta
(iii) Pteridophyta
(iv) Gymnosperms
(v) Angiosperms.

The basis of these divisions are –
(i) Differentiation of plant body into distinct components
(ii) Special tissues for the transportation of water and other substances
(iii) Ability to bear seeds, naked or enclosed in fruits.

Question 5.
How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals ?
Answer:
Mostly the body of all the plants is developed according to their capacity to make food. Although, the growth of animals’ bodies depends on the food getting from other sources. It is the main basis of classification between animals and plants because plants are autotrophs and animals are heterotrophs.

Question 6.
Explain how animals in Vertebrata are classified into further subgroups.
Answer:
All chordates possess the following features:
(i) Notochord
(ii) dorsal nerve cord
(iii) triploblastic
(iv) paired gill pouches
(v) coelomate.

Vertebrates are grouped into five classes:
1. Pisces:
(i) They are water-living animals found in both saline and freshwater.
(ii) They lay eggs and poikilothermic animals.
(iii) Their skin is covered with scales.
(iv) Their body is streamlined.
(v) Respiration occurs by gills.
(vi) They have two-chambered hearts.
(vii) They have fins on their body.
Examples:
Torpedo, Rohu, Shark, Tuna, Scoliodon, Sea horse, Electric ray.

2. Amphibia:
(i) They live in freshwater and moist land.
(ii) Their body is without scales.
(iii) Most of them have two pairs of pentadactyl (five digits) limbs.
(iv) Respiration occurs by gills, lungs or skin.
(v) They have a three-chambered heart.
Examples:
Frog, Toad, Hyla.

3. Reptilia:
(i) They are mostly land vertebrates.
(ii) Their skin is covered by scales.
(iii) Respiration occurs by lungs only,
(iv) They have a three-chambered heart. Only crocodiles have a four-chambered heart.
(v) They are unisexual.
Examples:
Hemidactylus (Lizard), Chameleon, Draco (Flying Lizard).

4. Aves:
(i) They are found all over the world.
(ii) Forelimbs are modified into wings.
(iii) Body covered with feathers.
(iv) Endoskeleton is made up of light bones with air cavities.
(v) Teeth are not present in their beak.
(vi) Respiration is by lungs only. Lungs have air sacs.
(vii) Heart is completely four-chambered.
Examples: Pigeon, Peacock, Sparrow, Ostrich, etc.

5. Mammalia:
(i) They are often found inland areas.
(ii) Their body is covered by hair.
(iii) Respiration in them occur by lungs only.
(iv) They have a completely four-chambered heart.
(v) They have mammary glands to nourish their young ones.
(vi) They are unisexual.
Examples: Kangaroo, Rat, Whale, Squirrel, Elephant, Rabbit, Bat, Chimpanzee, Man, etc.