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KSEEB Solutions For Class 9 Science Chapter 15 Improvement In Food Resources Important Concepts

Crop yields, crop variety, crop production manure and fertilizers, Irrigation, cropping patterns. Animal husbandry – cattle farming, poultry farming, Fish production, Beekeeping.

Crop production: Crops are plants cultivated by human beings for food, fodder, and other materials. The important types of crops are. Cereal crops – Wheat, rice, maize, Pulses – Pea, Greengram, Oil seed – Groundnut, soybean
Crop seasons:
1)Kharif season -These crops are grown in the rainy season.
Example: Paddy, soybean.
2)Rabi season – These crops are grown in the winter season.
Example: Wheat, peas.

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Macronutrients: The essential elements utilized by plants relatively in large quantities.
Examples: Nitrogen, phosphorous, potassium, and calcium.
Micronutrients: The essential elements utilized by plants in small quantities.
Examples: Manganese, Boron, Zinc, and Copper.
Manures: Manures are organic substances obtained through the decomposition of plant waste and animal excreta. Manure can be classified into
1)compost
2)Vermicompost
3)Green manure.
Fertilizers: Fertilisers are commercially produced plant nutrients.
Types of fertilizers:
1)Nitrogenous fertilizers
2)Phosphatic fertilizers
3)potassic fertilizers
Organic Farming: It is a farming system with minimal or no use of chemicals as fertilizers, herbicides, pesticides, etc., and with a maximum input of organic manures, recycled farm wastes, etc.
Irrigation: It is the process of supplying water to crop plants in the fields by means of canals, reservoirs, wells, tube wells, etc.,
Mixed cropping: It is the practice of growing two or more crops simultaneously on the same piece of land. Eg. Wheat + gram, ground nut T sunflower.
Intercropping: It involves growing two or more crops simultaneously on the same field in a definite pattern. A few rows of one crop alternate with a few rows of a second crop.
Example : soybean + maize + cowpea
Crop rotation: The growing of different crops on a piece of land in a preplanned succession is known as crop rotation.
Pest: Any destructive organism that causes great economic damage to crop plants.
Example:  Insects, mites, etc.,
Pesticide: It refers to a chemical that is used to kill a pest organism.

1. Insecticides – killing the insects
2. Weedicides – killing the weeds
3. Fungicides – killing the fungi
4. Rodenticides – killing rodents

Weeds: They are small-sized unwanted plants that grow along with a cultivated crop in a field.
Livestock: It refers to the domestic animals kept in use for milk, and flesh and includes cattle, buffaloes, sheep, and goats.
Animal husbandry: Animal husbandry is the scientific management of animal livestock. It includes various aspects such as feeding, breeding, and disease control.
Cattle farming: Cattle husbandry is done for two purposes milk and drought labor. Milk-producing females are called milch animals. Draught animals are used for farm labor.
Breeding: It means ‘to reproduce’ and is done to obtain animals with desired characteristics.
Poultry: It is the branch of animal husbandry concerned with rearing birds for eggs and meat. Poultry practices required good care for food, shelter, and disease control.
Fish production: There are two ways of obtaining fish. One is from natural resources, which is called capture fishing. The other way is by fish farming, which is called culture fishery.
Bee-keeping: Bee-keeping or Apiculture is the rearing, care, and management of honey bees for obtaining honey, wax, and other substances.

Improvement In Food Resources Exercises

Question 1. Explain any one method of crop production which ensures high yield.
Answer: Intercropping is one of the methods of crop production which ensures high yield because it ensures maximum utilization of the nutrients supplied and also prevents pests and diseases from spreading to all the plants belonging to one crop infield. In this way, both crops can give better returns.

Question 2. Why are manures and fertilizers used in fields?
Answer: Manures add a great amount of organic matter in the form of humus in the soil. Fertilizers are very rich in plant nutrients such as nitrogen, phosphorous, and potassium.

Improvement In Food Resources KSEEB Class 9 Question Answers 

Question 3. What are the advantages of intercropping and crop rotation?
Answer:
Advantages of using intercropping:
1)It helps to maintain soil fertility.
2)Crops can easily be harvested and trashed separately.
Advantages of crop rotation:
1)It avoids the depletion of a particular nutrient from the soil.
2)It helps in weed control.

Question 4. What is genetic manipulation? How is it useful in agricultural practices?
Answer: Genetic manipulation is a process of transferring genes or characteristics that are desirable from one plant to another plant for the production of varieties with desirable characteristics. Genetic manipulation is useful in increasing yield, better quality, and desirable characteristics.

Question 5. How do storage grain losses occur?
Answer:
There are mainly two types of factors
1)Biotic factors — Insects, rodents, birds & microorganisms.
2)Abiotic factors Moisture content, temperature, and humidity.

Question 6. How do good animal husbandry practices benefit farmers?
Answer:
Advantages of animal husbandry practices are.
1)Increasing the yield of foodstuffs such as milk, eggs, and meat.
2)It is beneficial for the farmers as increased yield brings more income to the farmer and raises his living standard.

Question 7. What are the benefits of cattle farming?
Answer:
Cattle fanning is beneficial in the following ways
1)Good quality of meat, fiber can be obtained.
2) Milk-yielding animals and a good breed of draught animals can be obtained.

Question 8. For increasing production, what is common in poultry, fisheries, and bee-keeping?
Answer: Through cross-breeding, the production of poultry, fisheries, and beekeeping can be increased.

Question 9. How do you differentiate between capture fishing, mariculture, and aquaculture?
Answer:
Capture fishing:
It is the process of obtaining fish from natural resources such as rivers, ponds, canals, etc.
Mariculture:
It is a practice of a culture of marine fish varieties in the open sea. 
Aquaculture:
It is the production of fish from freshwater resources.

Improvement In Food Resource Textual Questions

Question 1. What do we get from cereals, pulses, fruits, and vegetables?
Answer: Cereals give carbohydrates which provide energy. Pulses give proteins that build our bodies. Vegetables and fruits provide all essential minerals.

Question 2. How do biotic and abiotic factors affect crop production?
Answer:
Biotic and abiotic factors affect poor crop production by the following factors.
1)Weight loss
2)Poor germination ability
3)Infestation of insects.
4)Degradation in quality.

Class 9 Science Chapter 15 KSEEB Textbook Solutions 

Question 3. What are the desirable agronomic characteristics for crop improvements?
Answer:
Desirable agronomic characteristics are.
1)Tallness and profuse branching are desirable characteristics for fodder crops.
2)Dwarfness is desired in cereals so that less nutrients are consumed by these crops.

Question 4. What are macronutrients and why are they called macronutrients?
Answer: Macronutrients are essential elements utilized by plants relatively in large quantities. They require for the plants in greater amounts for proper growth and development.

Question 5. How do plants get nutrients?
Answer: The nutrients which are found in the soil get dissolved in the water and is absorbed by the roots of a plant

Question 6. Compare the use of manure and fertilizers in maintaining soil fertility.
Answer:
Manure:
1)Manure is a natural substance
2)It adds a great amount of organic matter
3)It is not nutrient specific
Fertilizers:
1)A fertilizer is a man-made substance.
2)It does not add any humus to the soil.
3)It is nutrient specific

Question 7. Which of the following conditions will give the most benefits? Why?
1)Farmers who use high-quality seeds do not adopt irrigation or use fertilizers.
2)Farmers use ordinary seeds, adopt irrigation and use fertilizers.
3)Farmers use quality seeds adopt irrigation use fertilizers and use crop protection measures.
Answer: Option (c) is the right answer.
The use of only quality seeds and irrigation and fertilizers are not sufficient for better yield, but also crop should be protected by biotic and abiotic factors.

Question 8. Why should preventive measures and biological control methods be preferred for protecting crops?
Answer: Diseases are spread by pathogens so to get rid of pathogens some preventive measures, as well as biological methods, are preferred.
Example: To kill the pathogen biological method will be used without affecting the plant or soil quality.

Question 9. What factors may be responsible for losses of grains during storage?
Answer:
1) Biotic factors – Insects, birds, and rodents
2) Abiotic factors — moisture content, temperature, and humidity.

Question 10. Which method is commonly used for improving cattle breeds and why?
Answer:
Cross-breeding of cattle is commonly used for improving cattle breeds because.
1)It helps in getting the desired traits.
2)To get disease-resistant and high-yielding cattle breeds.

Question 11. Discuss the implications of the following statement. “It is interesting to note that poultry is India’s most efficient converter of low-fiber food stuff (which is unfit for human consumption) into highly nutritious animal protein food.
Answer: Poultry farming is to raise domestic fowl for egg production and chicken meat. The poultry birds are also efficient converters of agricultural byproducts.

Question 12. What management practices are common in dairy and poultry farming?
Answer:

• Providing proper shelter.
• Feeding
• Caring for animal health.

Question 13. What are the differences between broilers and layers and in their management?
Answer: The poultry bird groomed for obtaining meat is called a broiler. The egg-laying poultry bird is called a layer. The proper nutrition and environmental requirements of broilers are somewhat different from those of egg layers.

Question 14. How are fish obtained?
Answer: The fish can be obtained by
1)Capturing
2)Culturing

KSEEB Solutions For Improvement In Food Resources Short Notes 

Question 15. What are the advantages of composite fish culture?
Answer: The composite fish culture is a combination of five or six fish species is used in a single pond.
Advantages
1. They do not compete for food among themselves.
2. The (fishes) have different types of food habits
3. Fishes feed in different zones.

Question 16. What are the desirable characteristics of the varieties suitable for honey production?
Answer:
1)The variety of bees should be able to collect a large amount of honey.
2)They (Bee) should stay in a given beehive for a long period.

Question 17. What is pasturage and how is it related to honey production?
Answer: Pasturage means the flowers available to the bees for nectar and pollen collection. In addition to an adequate quantity of pasturage, the kind of flowers available will determine the taste of the honey.

Improvement In Food Resources Additional Questions

Question 1. Name the crop whose production has increased by
1)blue revolution
2) yellow revolution
Answer:
1) Blue revolution—Fish production
2)Yellow revolution—Oil production

Question 2. Is organic farming beneficial and why?
Answer: Yes organic fanning is very beneficial because it is a farming system with no use of chemicals with a maximum input of organic manures, recycled from waste.

Question 3. Why removal of weeds is necessary?
Answer: Weeds should be removed because it takes up nutrients from soil and reduce the growth of crops.

Question 4. Which is the most advantageous fish culture system?
Answer: Composite fish culture because fishes do not compete for food among them.

Question 5. How genetically modified crops can be obtained?
Answer: Genetically modified crops can be obtained by introducing a gene that would provide the desired characteristics.

Question 6. How does catla differ from mrigal?
Answer: Catla is a surface feeder, while mrigal is a bottom-feeding fish.

Improvement In Food Resources High Order Thinking Questions

Question 1. On what factors does the growth of plants and flowering are dependent?
Answer: Temperature and photoperiod (duration of sunlight)

Question 2. Neem and turmeric powders are often used in grain storage.
1)What are they called?
2)What is the purpose of using need and turmeric?
Answer:
1) Neem and turmeric powders are biopesticides,
2) Neem and turmeric keep away insects rodents, fungi, bacteria, etc from the stored foods.

Question 3. A farmer had a plot just beside the bank of a river. Each time be planted Kharif crops, crops got damaged due to floods. He consulted the agricultural scientist who gave him a special variety of seeds and also advised him to practice fish farming.
1)What was the specialty of seed grains?
2)What name can be given to this type of fish farming?
Answer:
1) The special variety of seed grains are genetically modified to protect them from the damaging effect of floods,
2)Composite fish culture.

KSEEB Class 9 Science Chapter 15 Important Questions 

Question 4. What determines the quality of honey?
Answer:
1) The pasturage, i.e, the kind of flowers available.
2) Apiary location.

Question 5. Name one indigenous and one exotic breed of poultry.
Answer:
1) Indigenous—Aseel
2) Exotic—Leghorn.

Question 6. What are the desired agronomic characteristics for fodder and cereal crops?
Answer: Tallness and profuse branching are desirable characteristics for fodder crops. Dwarfhess is desired in cereals so that less nutrients are consumed by these crops.

Question 7. Excessive use of chemicals, such as insecticides and pesticides causes a threat to ecology.” Explain with reason.
Answer:
1) Excessive use of fertilizers and pesticides when washed away courses water pollution,
2) The pesticides accumulate in the soil for a long period.
3)They may lead to eutrophication,
4)They may result in biological magnification.

Improvement In Food Resources Unit Test

Question 1. To solve the food problem of the country, which among the following is necessary
1)Increased production and storage of food grains.
2)Easy access of people to the food grains
3)Proper storage of food grains.
4)All of these
Answer: (1) Increased production and storage of food grains.

Question 2. The place where bees are reared for commercial honey production is called
1)Beehive
2)Aviary
3)Apiary
4)collection unit
Answer: (3) Apiary

Question 3. The process of crossing genetically dissimilar plants of a species is called
1)Intervarietal cross
2) Interspecific cross
3)Intergeneric cross
4) Ail of these
Answer: (1) Intervarietal cross

Improvement In Food Resources Give One Word For The Following

1. Planting soybean and maize in alternative rows in the same field is called as Intercropping
2. Parthenium is commonly known as weed
3. Growing different crops on a piece of land in pre-planned succession is known as Crop rotation

Improvement In Food Resources Answer The Following Questions

one mark

Question 1. Why do organisms need food?
Answer: Food is required for body development, growth, and good health.

Question 2. Which technique is incorporated to obtain crop varieties?
Answer: Hybridisation.

Question 3. Which species of honey bee is used for the commercial production of honey?
Answer: Apis mellifera

Question 4. What is Aquaculture?
Answer: The production and management of fish is called aquaculture.

Question 5. Define the term hybridization.
Answer: Hybridisation refers to crossing between genetically dissimilar organisms.

Question 6. Give two examples of milch animals.
Answer: The cow and buffalo are milk animals.

Question 7. What is the advantage of green manure?
Answer: Green manure helps in enriching the soil with nitrogen and phosphorus.

Two And Four Marks

Question 1. Write any two methods of weed control.
Answer:
1) Weeds can be controlled by using pesticides.
2)Weeds can be removed by using a harrow, hoe or by cutting them.
3)The biological method involves growing crops like mustard along with the main crops which do not allow weeds to grow.

Question 2.What are th e ad vantages of hybridisation ?
Answer:
1) It makes crops disease resistant and a good response to fertilizers.
2) It gives a higher yield and improves product quality.

Question 3. Write any two characteristics of a storage structure for grains.
Answer:
1)Storage structures should be cleaned dried and airtight.
2)Storage structures should have inbuilt arrangements for aeration, temperature control, protection from pests, etc.

KSEEB Solutions Chapter 15 Agricultural Techniques Class 9 

Question 4. State three ways by which pests attack the plants and name the chemical used to control pests.
Answer:
1) The pests cut the root, stem, and leaf,
2)They suck the cell sap from plants.
3)They bore into the stem, root, and fruit. Pesticides should be used to control pests.

Question 5. How can poultry farming be integrated with crop production?
Answer:
1) Poultry birds can be fed with farm wastes like degraded grains and certain parts of a plant as food.
2) Bird wastes or excreta can be used as manures to the plants.

Question 6. How is fumigant different from pesticides?
Answer: Fumigant is used before grains are stored for future use. Pesticide is used after the germination of seed and it is sprayed over the crop plant.

Question 7. List three factors on which cultivation practices and crop yield depends.
Answer:
The cultivation practices and crop yield depends upon the following factors.
1)Availability of high-quality seeds.
2)Availability of waters
3)Availability of fertile soil
4)Access to new information and technologies.

KSEEB Class 9 Science Notes for Chapter 8 Motion Important Concepts

Rest
A body is said to be at rest if its position does not change with respect its surroundings

Motion
A body is said to be in motion if its position changes with respect to its surroundings.

Reference Point
It is a fixed point with respect to which a body is at rest or in motion.

Distance
It is the length of the actual path travelled by a body in a given interval of time.

Displacement
It is the shortest distance between the initial and final positions of a body in a specified direction.

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Scalar Quantities
These are the physical quantities which have only magnitude and no direction. Ex: speed, mass, temperature, energy, work, etc

Vector quantities
These are physical quantities with both magnitude and a specified direction. Ex: velocity, force, acceleration, momentum, etc

Uniform motion
The motion of a body is said to be uniform if it covers equal distances in equal intervals of time however small the time intervals.

Non-uniform motion
The motion of a body is said to be non-uniform if it covers unequal distances in equal intervals of time however small the time intervals

Speed
It is the distance travelled by the body divided by the time taken to cover the distance OR it is the distance travelled by a body in unit time (per second)

S.I unit is meter per second (m/s or\( \mathrm{ms}^{-1}\))

Class 9 KSEEB Science Chapter 8 Motion Notes 

Uniform speed
The speed of a body is said to be uniform if the body covers equal distances in equal intervals of time.

Non-uniform speed
The speed of a body is said be non-uniform if the body covers unequal distances in equal intervals of time.

Average speed
It is the total distance travelled by a body divided by total time taken by the body to cover the distance.

Velocity

• It is a displacement of a body divided by the time taken to cover the displacement Or
• It is the displacement of a body per second Or
• It is the distance travelled by the body in one second in a given direction

S .1 unit is metre per second (m/s or \(\mathrm{ms}^{-1}\))

Uniform Velocity
The velocity of a body is said to be uniform if the body covers equal distances in equal intervals of time however small the time intervals in a given direction.

Variable Velocity
The velocity of a body is said to be variable if magnitude or direction or both of magnitude and direction of velocity change.

Average velocity
It is not displacement covered by the body divided by the total time taken.

Acceleration
It is the rate of change of velocity of a body with respect to time Acceleration

S.I unit is  \(\mathrm{m} / \mathrm{s}^2 \text { or } \mathrm{ms}^{-2}\)

Positive Acceleration
Acceleration of a body is said to be positive if it takes place in the direction of the velocity of the body. In this case, velocity increases with time.

Negative Acceleration
Acceleration of the body is said to be negative if it takes place opposite to the direction of the velocity of the body. In this case, velocity decreases with time. It is also known as retardation or deceleration.

Uniform Acceleration
If the change in the velocity of a body in equal intervals of time is always the same, then an acceleration of the body is said to be the uniform acceleration

Variable Acceleration
If the change in the velocity of a body in equal intervals of time is not the same, then the acceleration of the body is said to be the variable acceleration

Graphs
It is a convenient method of presenting basic information of the motion of an object

Distance – Time Graph

• It is the graph which represents the change in the position of the object with respect to time. It may be linear or non-linear
• It is obtained by taking time (t) along the x-axis and distance (s) along the y- axis
• The slope of the graph gives the speed of the body in a given time interval

Displacement – Time Graph

• The graph was obtained by taking time along the x-axis and displacement along the y-axis. It is always a straight line (linear)
• The slope of the graph gives the velocity of the body

Velocity – Time Graph

• It represents the change in velocity with respect to time.
• It is obtained by taking time(t) along the x-axis and velocity (v) along the y-axis
• The graph may be linear or non-linear
• The slope of the graph gives the acceleration of the body and the area under the curve gives the distance travelled by the body in the given interval of time.

Equations Of Motion
The motion of a body with uniform acceleration can be described by the following three equations
(1) u = v + at
(2) s=ut +1/2 at²
(3) 2as = v2 – u2

where u – Initial velocity, v = final velocity, a ~ acceleration, s = distance and t = time

Uniform circular motion

• When a body moves in a circular path with constant speed, then the motion of the body is said to be uniform circular motion.
• In uniform mo ion, velocity is variable due to changes in direction continuously. The magnitude of the velocity is given by
  the circumference of the circular path.

where r=radius of the circular path and T – time taken to cover the path once.

KSEEB Science Chapter 8 Motion Explanations For Class 9 

KSEEB Class 9 Science Notes for Chapter 8 Motion

Exercises

Question 1. An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example.
Answer: Yes. Suppose a student throws a ball to a height and catches it.
Distance travelled by the ball = h + h = 2h m
Displacement = 0.

Question 2. A farmer moves along the boundary of a square field of side 10m in the 40s. What will be the magnitude of displacement of the
farmer at the end of 2 minutes 20 seconds from his position?

Answer: Consider the square field ABCD as shown in the figure. Given that fanner covers the boundary ABCD in 40 s. If he walks for 2 minutes and 20 seconds, he will be at point C.

Thus magnitude of the displacement of the farmer
= AC
=√(AB)² + (BC)²
=√10² +10²
= √100+100
= √200
=14.14 m

Question 3. Which of the following is true for displacement?
(1)It cannot be zero
(2)Its magnitude is greater than the distance travelled by the object
Answer: Both statements are false

Question 4. Distinguish between speed and velocity
Answer:
Speed 
1. It is the distance travelled by the body in one second irrespective of direction.
2. It is a scalar quantity
3. It is always positive

Velocity
1. It is the distance travelled by a body in one second in the given direction.
2. It is a vector quantity

Question 5.Under what condition(s) is the magnitude of the average velocity of an object equal to its average speed?
Answer: When a body moves along a straight path in the same direction, its total path length is equal to the magnitude of the displacement. Under this condition, average velocity is equal to average speed.

Question 6. What does the Odometer of an automobile measure?
Answer: The odometer measures the distance travelled by automobile

Question 7. What does the path of an object look like when it is in uniform motion?
Answer: The path looks like a straight line

Question 8. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light ie 3 x \0sm/s.
Answer: Given data, t = 5 minutes 300 second and
v = 3 x 10 m/s
Distance of the spaceship (s) = v
t = 3 x 108 x 300 = 9 x 10 10 m

Question 9. When will you say a body is in (1) uniform acceleration? (2) non-uniform acceleration?
Answer:
(1) If the change in velocity in equal intervals of time is always the same, then the body is said to be in uniform acceleration
(2) If the change in velocity in equal intervals of time is not the same, then the body is said to be moving with non-uniform acceleration

Question 10. A bus decreases its speed from 80kmh1 in 5 s. Find the acceleration of the bus.
Answer:
Acceleration = (Final velocity – Initial velocity)/(time interval)
Initial velocity = 80km/h

Final velocity=60km/h

Acceleration = \(\frac{16.67-22.22}{5}=-6.47 / 5=-1.11 \mathrm{~ms}^{-2}\)

Question 11. A train starting from a railway station and moving with uniform acceleration attains a speed of 40km 1 in 10 minutes. Find its acceleration
Answer:
Final Speed= 40 km/h= \(\frac{40 \times 1000}{3600}=\frac{100}{9}=11.11 \mathrm{~m} / \mathrm{s}\)

Initial Speed = 0 km/h
Acceleration = (Final velocity – Initial velocity)/(time velocity)

\(=\frac{11.11-0}{10 \times 60}=\frac{11.11}{600}=0.019 \mathrm{~m} / \mathrm{s}^2\)

Question 12. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

Answer: For uniform motion, the distance-time graph is a straight line. For non-uniform motion, the distance-time graph is a curved line

Question 13. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer: The object is at rest

KSEEB Science Notes Chapter 8 Motion With Solved Examples 

Question 14. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Answer: The body is moving with uniform motion ie uniform speed.

Question 15. What is the quantity which is measured by the area occupied below the velocity-time curve?
Answer: Displacement covered by the body in the given interval of time.

Question 16. A bus starting from rest moves with a uniform acceleration of 0.1 m/s2 for 2 minutes. Find
(1) the speed acquired
(2) the distance travelled
Answer: Given data: u=0, a=0.1 m/s2 t=2 minutes – 120 second
(1) The speed acquired (v) = u+at
\(=0+0.1 \times 120=12 \mathrm{~ms}^{-1}\)
(2) The distance travelled =\( u t+1 / 2 a t^2\)
\(=0 \times 120+1 / 2 \times 0.1 \times(120)^2\)
=720m

Question 17. A train is travelling at a speed of 90km/h. Brakes are applied so as to produce a uniform acceleration of 0.5 m/s2. Find how far the train will go before it is brought to rest.
Answer: Given data:
\(\begin{aligned}
&\mathrm{u}=90 \mathrm{~km} / \mathrm{h}=\frac{90 \times 1000}{3600}=25 \mathrm{~m} / \mathrm{s} \\
&\mathrm{a}=-0.5 \mathrm{~m} / \mathrm{s}^2, \mathrm{v}=0
\end{aligned}\)
We know that \(v^2-u^2=2 a s\)
\(\mathrm{s}=\frac{v^2-u^2}{2 a}=\frac{0^2-25^2}{2(-0.5)}=625 m\)

Question 18. A trolley while going down an inclined plane, has an acceleration of 2cm s’2. What will be its velocity 3s after the start?
Answer:
Given data: u = 0, a=3cm=0.03m, t=3sec.
Velocity after 3 sec = v=u+at \( =0+0.03 \times 3\)
\( =0.09 \mathrm{~m} / \mathrm{s}\)

Question 19. A racing car has a uniform acceleration of 4 m/s2. What distance will it cover in 10s after the start?
Answer:
Given data : \(\mathrm{u}=0, \mathrm{a} 4 \mathrm{~m} / \mathrm{s}^2, \mathrm{t}=10 \mathrm{~s}\)
Distance travelled after \( \begin{aligned}
&10 \mathrm{~s}=\mathrm{s}=u \mathrm{t}+1 / 2 \mathrm{at}^2 \\
&=0 \times 10+1 / 2(4)(10)^2 \\
&=200 \mathrm{~m}
\end{aligned}\)

Question 20. A stone is thrown in a vertically upward direction with a velocity of 5m/s. If the acceleration of the stone during its motion is 10ms’2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer:
Given data: \(\mathrm{u}=5 \mathrm{~m} / \mathrm{s}, \mathrm{a}=-10 \mathrm{~m} / \mathrm{s}^2\)
At the height point v=0
Thus using the formula
\(\mathrm{s}=\frac{v^2-u^2}{2 a}=\frac{0^2-5^2}{2(-10)}=\frac{25}{25}=1.25 \mathrm{~m}\)
Thus the height attained by the stone = 1.25m
Now, v=u+at \( \begin{aligned}
\Rightarrow 0=5-10 \mathrm{t} & \Rightarrow \mathrm{t}=5 / 10 \\
&=0.5 \mathrm{~s}
\end{aligned}\)

Question 21. An athlete completes one round of a circular track of diameter 200m in the 40s. What will be the distance covered and displacement at the end of Z minutes 20s?
Answer: Time is taken to complete 1 round =40s
Total time taken = 2minutes+20s=140s
Total distance covered = (circumference of circularpath*total time taken)/(time taken to complete one round)
\(=\frac{\pi d \times 140}{40} \mathrm{~m}\)
\(=\frac{22}{7} \times 200 \times \frac{7}{2}=2200 \mathrm{~m}\)
After the 140s, the athlete will be in the exactly diametrically opposite position. (Why?)
Hence displacement = 200m

Question 22. Joseph jogs from one A to the other end B of a straight 300m road in 2 minutes 30 seconds and then turns around and jogs 100m back to point Cin another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C
Answer:
Case (1): In jogging from A to B,
Distance covered = 300m and displacement = 300m(Why?)
Time taken = 2 minutes + 30s = 150s
Average Speed = (distance)/(time) =
\(\frac{300}{150}=2 \mathrm{~m} / \mathrm{s}\)
Average velocity = (displacement)/(time) =
\( \frac{300}{150}=2 \mathrm{~m} / \mathrm{s}\)

Case (2): In jogging from A to C
Distance travelled = 300 + 100 = 400m
Displacement = AB – BC = 300-100 = 200m
Time taken = 2 min 30s + 1 minute = 210s
Average Speed =(distance)/(time) =
\(=\frac{400}{210}=1.9 \mathrm{~m} / \mathrm{s}\)
Average velocity = (displacement)/(time)=
\(=\frac{200}{210}=0.95 \mathrm{~m} / \mathrm{s}\)

Question 23. Abdul while driving to school, computes the average speed for his trip to be 20kmh~1. On his return trip along the same route, there is less traffic and the average speed is 30km What is the average speed for Abdul’s trip?
Answer: Let the distance between Abdul’s house and school = X km
Time is taken in driving to school at a speed of
\(20 \mathrm{kmph}=\frac{\text { dis } \tan c e}{\text { time }}=\frac{x}{20} h\)

Time taken in returning at a speed 30kmph =\( \frac{x}{30} h\)

Total time is taken for the whole trip =

Average speed = (total distance)/(total time taken)

= \(\frac{2 x}{5 x / 60}=\frac{120}{5}=24 \mathrm{kmh}^{-1}\)

Free notes for KSEEB Class 9 Science Chapter 8 Motion 

Question 24. A driver of a car travelling at 52k mb1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another car driver going at 34 knlr1 in another car applies his brakes slowly and stops in 10s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after brakes were applied?
Answer:

In the figure, AB and CD represent the speed versus time graphs for the two cars.

Distance travelled by the 1st car
\(\begin{aligned}
& =1 / 2 \times \mathrm{AB} \times \mathrm{AO} \\
& =1 / 2 \times \frac{52 \times 1000}{3600} \times 5 \\
& =36.1 \mathrm{~m}
\end{aligned}\)

Similarly, the distance travelled by the 2nd car =
\(\begin{aligned}
& =1 / 2 \times \mathrm{CO} \mathrm{CD} \\
& =1 / 2 \times \frac{34 \times 1000}{3600} \times 10
\end{aligned}\)
= 47.2 m

It is clear that the 2nd car travelled farther than the 1st car after breaks were applied.

Question 25. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graphs
and answer the following questions :

(1) Which of the three is travelling the fastest?
(2) Are all three ever at the same point on the road?
(3) How has C travelled when B passes A?
(4) How far has B travelled by the time it passes C?
Answer:
We know the slope of the distance-time graph represents the speed of the object.
(1) Steeper the graph, the greatest and B travels with the fastest speed
(2) Since there is no common point of intersection of the three lines, they are not at the same point ever on the road.
(3) From the graph, C has travelled a distance of about 9km
(4) Again from the graph, B has travelled about 6km

Question 26. A ball is gently dropped from a height of 20m. If its velocity increases uniformly at the rate of 10m/s2, with what velocity will it strike the ground? After what time will it strike the ground?
Answer:
Given data, u=0, h=20m, a = \(10 \mathrm{~m} / \mathrm{s}^2\), v = ?, t = ?
We
know that \( v^2-u^2=2 a h\)
\( v^2-0^2=2 \times 10 \times 20\)
\(v^2=400 \Rightarrow v=20 \mathrm{~m} / \mathrm{s}\)
Now we know that \( \mathrm{v}=\mathrm{u}+\mathrm{at} \Rightarrow\)
\(t=\frac{v-u}{a}=\frac{20-0}{10}=2 s\)

Question 27. The speed-time graph for a car is shown in figure 8.12.

(1) Find how far the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by car during the period.
(2) Which part of the graph represents the uniform motion of the car?
Answer:
(1)Along time – axis, 5 squares = 2s
Along speed – axis, 3 squares = 2m/s
Area of 15 squares = Distance
\(=2 \mathrm{~s} \times 2 \mathrm{~m} / \mathrm{s}=4 \mathrm{~m}\)
Area of 1 square = distance pf 4/15m
The number of squares under the graph between 4s and 6m/s = \(\begin{gathered}
57+1 / 2(6) \\
=60
\end{gathered}\)
Hence distance travelled by the car = 60 x 4/15 = 16m

(2) The graph after 6s represents uniform motion since the graph remains flat

KSEEB Science Notes for 9th Standard Chapter 8 Motion 

Question 28.State which of the following situations are possible and give an example for each of these
(1) an object with constant acceleration but with zero velocity
(2) an object moving in a certain direction with an acceleration in the perpendicular direction.
Answer:
(1) When an object is thrown vertically upwards, at the highest point, velocity becomes zero but still it has acceleration which is equivalent to acceleration due to gravity.
(2) When an aeroplane moves horizontally, acceleration due to gravity acts on it perpendicular to the direction of the motion.

Question 29. An artificial satellite is moving in a circular orbit of radius 42250km. Calculate its speed it it takes 24 hours to revolve around the earth.
Answer:
Given data, Radius (R) = 42, 250km = 42250000m
Time period (T) = 24 h = \(24 \times 60 \times 60 \text { seconds }\)
Speed =

KSEEB Class 9 Science Notes for Chapter 8 Motion

Additional Questions

One Mark

Question 1. What is motion?
Answer: A body is said to be in motion if its position changes with respect to a reference point

Question 2. What is the simplest motion?
Answer: Motion along a straight line is the simplest motion

Question 3. Rest and motion are relative terms. Why?
Answer: An object at rest for one person is in motion for other people

Question 4. Give an example of directly perceivable
motion
Answer: Motion of a car on the road

Question 5. Give an example of an indirectly perceivable motion
Answer: Motion of the air observed due to the movement of dust or leaves of branches of a tree.

Question 6. Identify a scalar and vector from the
following
(1) distance
(2) displacement
Answer: distance – a scalar and Displacement – a vector

Question 7. Given an example for a uniform motion question
Answer:
1) Motion of the earth around the sun
2) Motion of a stone thrown vertically upwards
3) Motion of a train approaching a railway station

Question 8. Give two examples of non-uniform motion
Answer: It is the ratio of the total distance travelled to the time interval to cover the distance

Class 9 Motion KSEEB Notes With Numerical Solutions 

Question 9. Define average speed
Answer: It is the ratio of net displacement to the time interval

Question 10. What do you mean by the term reference point?
Answer: An object or place specified to locate the position of an object is called a reference point.

Question 11. Define displacement of a body.
Answer: The shortest distance measured from the initial to the final position of an object is known as the displacement of a body.

Question 12. What is an odometer?
Answer: It is a device used in automobiles to measure the distance travelled.

Question 13. What is a point object?
Answer: An object whose size is very small compared to the distance travelled by the object is called a point object.

Question 14. What is the S.I. unit of displacement?
Answer: metre (m)

Question 15. Can displacement be greater than the distance of an object?
Answer: No. Displacement may be equal to or less than distance.

Question 16. How is the speed of an object measured?
Answer:
\(\text { speed }=\frac{\text { distance travelled }}{\text { total time taken }}\)

Question 17. A body moves from rest with uniform speed. What type of distance-tome graph do we set?
Answer: A straight line passing through the origin.

Question 18. What does the area under velocity – time graph represent?
Answer: The area enclosed by the velocity-time graph and the time axis represent the magnitude of the displacement.

Question 19. Under what condition average velocity of an object equal to its average speed?
Answer: If the object moves along a straight path

Question 20. What can a speed meter measure?
Answer: Instantaneous speed ie speed at the moment of observation.

Question 21. Define average velocity
Answer: It is the ratio of net displacement to total time taken.

Two Marks

Question 22. Define the uniform speed of a body
Answer: The speed of a body is said to be uniform, it covers equal distances in equal intervals of times

Question 23. Define uniform velocity
Answer: The velocity of the body is said to be uniform if it covers equal displacements in equal intervals of time in the given direction

Question 24. Define variable velocity.
Answer: When the magnitude or direction or both of velocity change then the velocity is called variable velocity.

Question 25.Mention S.I units of
(1) velocity
(2)acceleration
Answer: S.I unit of velocity is  \( \mathrm{ms}^{-1}\)and the S.I unit of acceleration is \(\mathrm{ms}^{-2}\)

Question 26.Define uniform acceleration
Answer: Acceleration of a body is said to be uniform if changes in the velocity are equal in equal intervals of time

Question 27. Give an example of u uniformly acceleration motion
Answer: Motion of the freely falling body

Question 28. Define acceleration. Mention its S.I units
Answer: Acceleration is defined as the rate of change of velocity. S. I units are \(\mathrm{ms}^{-2}\)

Question 29. What do you understand by retardation of a body? Give an example
Answer: The rate of decrease of velocity is known as retardation or deceleration.
Example: When a stone is thrown upwards, its velocity decreases.

Question 30. Mention any two natural phenomena that occur due to motion.
Answer: Sunrise, Sunset, seasons etc.,

Question 31. Give an example for directly and indirectly perceivable motion.
Answer: The motion of a vehicle is directly perceivable motion. The motion of air is an indirectly perceivable motion

Question 32. When an object is thrown vertically upwards, it rises to a height of 10m – and returns to its initial position. What is the total distance travelled by the object? What is the displacement of the object?
Answer: Total distance travelled -10+10 = 20m
Net displacement = 0

Question 33. Name the type of the following motion
1)A man jogging in C part
2)Motion of around the earth.
Answer:
1) Non-uniform motion
2)uniform motion

KSEEB Chapter 8 Motion Revision Notes For Class 9 

Question 34. When is the acceleration is taken as (a) +ve (b)-ve
Answer: When acceleration occurs in the direction of motion, then it is said to be positive. When acceleration occurs against the direction of motion, then it is said to be negative.

Question 35. The time interval between lightning and thunder is 1.5 seconds. Find the distance of lightning from the earth (speed of sound in air 343 m/s)
Answer: Distance of lightning = speed of lightning x time informed = 343 x 1.5
= 515.5m

Question 36.Draw a typical distance-time graph of an object moving with a uniform speed
Answer:

Question 37.Draw a typic distance-time graph of an object moving hm – a uniform speed
Answer:

Question 38. What does the slope of the following graphs indicate?
1)displacement-time graph
2)velocity-time graph
Answer: Instantaneous speed is the speed at the moment of observation.

Three Marks Questions

Question 39.Mention any one use of
1)displacement time graph
2)velocity-time graph
Answer:
1) The slope gives a velocity of the body at any given instruction.
2) Area Enclosed by the graph gives the magnitude of the displacement of the curve.

Question 40. When the acceleration of a body is said to be
1) Positive
2) negative?
Answer: When the velocity of a body increases with time, then its acceleration is said to be positive. When the velocity of a body decreases with time, then its acceleration is said to be negative.

Question 41. Can an object have a constant acceleration with zero velocity? Give an example
Answer: Yes. An object can have a constant acceleration with zero velocity.
For example, when a stone is thrown up vertically, at the highest point velocity becomes zero but acceleration will be accelerated due to gravity.

Question 42. Uniform circular motion is an accelerated motion. Justify this statement
Answer: In a uniform circular motion, though the magnitude of velocity remains constant, its direction changes continuously. Hence, uniform circular motion is an accelerated motion.

Question 43. Write equations of motion
Answer:
(1) v = u + at
(2) s = \(u t+1 / 2 a t^2\)
(3)\( v^2=u^2+2 a s\)
where u – initial velocity, v = final velocity, a — acceleration, s = distance, t = time

Question 44. A moving car is brought to rest in 30 seconds after applying brakes. If the retardation rate is 2 m/s2, find the initial velocity of the car and the distance travelled by car.
Answer: Given t = 30s, v = 0, a = – \(2 \mathrm{~m} / \mathrm{s}^2\)
Using the equation v = u + at, we get 0 = u – 2
x 30 => u = 60 m/s

Class 9 KSEEB Science Motion Notes For Quick Learning 

Question 45. Give an example for
1)accelerated motion in the direction of motion
2)accelerated motion opposite to the direction of motion
3)zero acceleration
Answer:
1) accelerated motion in the direction of motion
2)accelerated motion opposite to the direction of motion
3)zero acceleration

Question 46. The velocity-time graph for the motion of an object is given

What type of motion the part
1)OA
2)AB
3)BC represent?
Answer:
1) OA represents uniformly accelerated motion
2)AB represents zero acceleration motion
3)non-uniformly accelerated motion.

Four Marks

Question 47.
1) Distinguish between distance and displacement.
2) Usha swims in a 90 m-long pool. She covers 180m in 1 minute by swimming from one end to another and back along the same straight path. Find the average speed and average velocity of Usha.
Answer:
Distance

• Distance is the actual path travelled by the body
• A scalar quantity
• Distance is always greater or equal to dis-placement
• The distance can be positive or zero

Displacement

• The shortest distance between an initial and final position of the body.
• A vector quantity
• Displacement is always equal to less than the distance
• Displacement can be +ve, -ve or zero.

b)\( \text { Average speed }=\frac{\text { Total distance travelled }}{\text { Total time taken }}\)
\( =\frac{180}{60} \mathrm{~ms}^{-1}\)
\( =3 \mathrm{~ms}^{-1}\)

Question 48. Prove that the area enclosed by the veloc¬ity-time graph and the time axis is equal to the magnitude of the displacement.
Answer:

Consider car moving with uniform velocity V. The velocity-time graph PQ is a straight line parallel to the time axis as shown in fig.

Let A and B be the position of the car at the time t. and t2 respectively. W.K.T.

distance travelled the car – velocity x time
\(=v\left(t_2-t_1\right)\)
= AC x CD
S = Area of the rectangle ABDC then the distance travelled by car is equal the area enclosed by the velocity-time graph and time axis.

Question 49. From the following velocity-time graph of a car, find the distance travelled by the car from t = 10 to t = 20

Answer: From the above velocity-time graph, the distance travelled by the car from t = 10S to 20S is given by
S = Area of rectangle ABCD + Area of triangle ADE
\(=(A B \times B C)+1 / 2 \times A D \times D E\)
AB = 20 \(\mathrm{~ms}^{-1}\)
BC = 10s
AD = AB = 10s
ED = CE – CD
= CE – AB
= 40 – 20
= \(20 \mathrm{~ms}^{-1}\)

\(S=(20 \times 10)+1 / 2 \times 10 \times 20\)
= 200 + 100
s = 300m

Question 50.Derive the equation of motion v = u + at using the v -1 graph

Answer: consider an object moving with uniform acceleration. Let ‘u’ be the initial velocity and 1 v’ be the final velocity of the object respectively.
The velocity-time graph is a straight line represented by AB. In the graph
OA = Initial velocity=u
OE – Final velocity = v
OC = Time interval = t
Draw AD || to OC
BC = BD + DC = BD + OA
substituting BC = V and OA=U
we get = BD + U
=> BD = V – U
the slope of the v -1 graph represents acceleration ie
\(a=\frac{\text { change in velocity }}{\text { time interval }}\)
\(a=\frac{v-u}{t}\)
at = v – u
\(\Rightarrow \quad v=u+\text { at }\)

KSEEB Science Chapter 8 Motion Explanations For Class 9 

Question 51. Derive the equation of motion s = ut + at2 using a velocity-time graph.

Answer: consider an object moving with uniform acceleration ‘a’.
AB represents the velocity-time graph of the object in the time interval +’. In the graph
O A = Initial velocity=u
OE = Final velocity = v
OC = Time interval = t
If S is the distance travelled by the object isn’t a second, then
S = Area of trapezium OABC
=Area of the rectangle OADC + Area of the triangle ADB
\(=(\mathrm{OA} \times \mathrm{OC})+1 / 2(\mathrm{AD} \times \mathrm{BD})\)……….(1)
From the fig.
OA = u
OC = t
BC = BD + DC = BD + OA
v = BD + u \( \Rightarrow \mathrm{BD}=\mathrm{v}-\mathrm{u}\)
\(\text { acceleration }(a)=\frac{\text { change in velocity }}{\text { time }}\)
[/latex] a=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{\mathrm{BD}}{\mathrm{t}}[/latex]
[/latex] \Rightarrow \mathrm{BD}=a \mathrm{t}[/latex]
substituting these values in equation(1)
Then [/latex]\mathrm{S}=\mathrm{u} \times \mathrm{t}+1 / 2 a t^2[/latex]
\(\mathrm{S}=u t+1 / 2 a t^2\)

Question 52. Derive the equation of motion 2as = v2 – u2 using the velocity-time graph.

Answer: In the figure, AB represent the velocity-time graph for an object moving with uniform acceleration (a)
Let S be the distance travelled by the object in ‘t’ second.
S = Area enclosed by the v -t graph
ie S = Area of trapezium OABC
\(=1 / 2(\mathrm{OA}+\mathrm{BC}) \times \mathrm{OC}\)
Area of trapezium = \(1 / 2(a+b) h\)
\(S=1 / 2[u+v] \times t\) ………..(1)
From the velocity-time velocity ie v = u + at
\(\Rightarrow \mathrm{t}=\frac{v-u}{a}\)
substituting this value in Eqn (1)
\(\mathrm{S}=\frac{1}{2}(u+v) \times \frac{(v-u)}{a}\)
\(\Rightarrow 2 \mathrm{as}=\mathrm{u}^2-\mathrm{v}^2\)……….(2)

KSEEB Class 9 Science Notes for Chapter 8 Motion

Application Questions

Question 1. Can displacement of a body be zero?
Answer: Yes, If a body returns to its initial position, then
its displacement is said to be zero.

Question 2. The walls of your classroom are at rest. Justify.
Answer: When time passes, the walls do not change with respect to the floor or ceiling. Hence the walls at rest.

Question 3. A body moves along a circular path of radius R. What will be its distance and displacement when it sweeps 180°?
Answer: When the body sweeps 180°, it will be diametri¬cally opposite point. Hence.
\(.\text { distance }=1 / 2 \times \text { circumference }\).
\(=1 / 2 \times 2 \pi r=\pi r\)
displacement = diameter = 2r

Question 4. Can displacement be negative?
Answer: Yes, In retardation motion, displacement is said to be negative.

Free Notes For KSEEB Class 9 Science Chapter 8 Motion 

Question 5. During the rainy day, we see lightning first and then hear thunder. Why?
Answer: The speed of light is greater than the speed of sound.

Question 6. What happens to an acceleration of an object when the force acting on it is doubled?
Answer: Since acceleration and the applied force acceleration doubles when the force is doubled.

Question 7.Velocity – time graph is parallel to the x-axis. What does this mean?
Answer: This means that the object is moving with constant velocity is zero acceleration.

Question 8. Mention the type of motion of a body from the following velocity-time graph

Answer: 1) uniformly accelerated motion
2)non-uniform retardation
3)non-uniformly accelerated motion

Question 9. An object moves such that the magnitude of velocity does not change but the direction changes continuously. What type of motion the object has?
Answer: Uniform circular motion.

Question 10. A girl travels from A to B towards the north from B she moves to C towards the east. Draw the resultant displacement in the diagram.
Answer:

Question 11. Name the force that keeps an object in a uniform circular motion. What happens if the force vanishes suddenly?
Answer: Centripetal force keeps the object in a uniform circular motion. If the force vanishes, then the object moves in. a direction tangent to the circular path.

Why Do We Fall Ill KSEEB Class 9 Chapter 13 Important Concepts

Health and diseases, infectious and non-infectious disease, the disease caused by microbes and their prevention principles of treatment and principles of prevention

Health: A state of complete physical, mental and social well-being, and not merely an absence of disease, or infirmity-WHO
Disease: Any physical or functional change from the normal state that causes discomfort or impairs the health of the living organism.
Acute diseases: The diseases which last for only very short periods of time.
Chronic diseases: The diseases last for long time, even as much as a lifetime.
Example: AIDS
Causes of diseases: Infection by microorganisms, malfunctioning of body organs, deficiency of one or more nutrients, genetic factors, polluted environment.
Infectious disease: The disease which are caused immediately because of microbes are called infectious diseases.
Example: Dengue fever.
Non-infectious disease: The disease that are not caused by infectious agents like microbes and which do not spread in the community are called non-infectious diseases.
Example: Cancers.

Read and Learn More KSEEB Solutions for Class 9 Science 

Infectious agents: Organisms that can cause disease are found in a wide range of categories of classification some of them are viruses, Bacteria, Fungi, protozoans, helminths etc.
Spread of infectious diseases: Infectious diseases spread through Air – common cold, Tuberculosis Water – Cholera Sexual contact – AIDS, syphilis
Vectors: Vectors are the intermediaries which carry the infectious agents from a sick person to another potential host.
Example: Insects (Mosquitoes)
Communicable diseases: These diseases are caused by pathogens or infectious agents such as bacteria viruses, fungi, protons, worms etc. These diseases can spread from a diseased or affected person to a healthy person by means of air, water, food, insects, physical contact etc.,
Antibiotics: They are chemical substances produced by living organisms such a bacteria and fungi, which can kill or stop the growth of some pathogenic microorganisms.
Vaccine: It is a suspension of disease-producing microorganisms which is modified by killing so that the suspension will not cause disease but on entering the body initiates the immune system to produce antibodies against a particular disease.
Vaccination: It is the inoculation of a vaccine in the body of a healthy person in order to develop immunity against a particular disease.
Immunisation: It is the production of immunity in an individual by artificial means.
Common Vaccines: DPT, BCG, Polio vaccine, Typhoid Vaccine, Measles Vaccine, TT Vaccine, Hepatitis etc.,
Diseases caused by microbes: Protozoa – Malaria Viruses – Hepatitis, AIDS, Rabies, Polio, Herpes, Polio Bacteria – Tuberculosis, Typhoid, Diarrhoea

Why Dowe Fall Ill Exercises

Question 1. How many times did you fall ill in the last one year? What was the illness?
1)Think of one change you could make in your habits in order to avoid any of / most of the above illnesses?
2)Think of one change you would wish for in your surroundings in order to avoid any of / most of the above illnesses.
Answer:
I fell ill twice in the last year. The first time suffered from diarrhoea and at the second time suffered from viral fever.
1. Washing hands before meals and remaining away from people suffering from viral fever.
2. Drinking pure water, planting more trees and disposal of garbage.

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:
A doctor/nurse/health worker should take the following precautions.
1. Washing hands with soap thoroughly after serious examination of patients.
2. Keeping a place of work sterilized by using phenyl etc.
3. takes balanced food to develop a powerful immune system.
4. proper disposal of blood samples urine, and sputum.

Question 3. Conduct a survey in your neighbourhood to find out what the three most common diseases are and suggest three steps that could be taken by your local authorities to bring down the incidence of these diseases.
Answer: The three most common diseases in my neighbourhood are.
1. Hepatitis
2. Cold and cough
3. Diarrhoea
The three steps that should be taken by local authorities are.
1. Arranging immunisation programmes
2. Providing clean drinking water
3. Providing better sanitation.

Class 9 Science Chapter 13 KSEEB Textbook Solutions 

Question 4. A baby is not able to tell her/his caretakers that she/he is sick. What would help us to find out
1)that the baby is sick?
2)What is the sickness?
Answer:
The following symptoms will help us to find out that the baby is sick are,
1. Continuous crying.
2. High body temperature
3. Improper intake of food,
4. Loose motions.

Question 5. Under which of the following conditions is a person most likely to fall sick?
1)When she is recovering from malaria.
2)When she has recovered from malaria and is taking care of someone suffering from chickenpox.
3)When she is on a four-day fast after recovering from malaria and is taking care of someone suffering from chickenpox why?
Answer: A person is most likely to sick when she is on a four-day fast after recovering from malaria and is taking care of someone suffering from chicken pox because when she is on fasting, her body may weak due to insufficient food. So, she is more prone to infections. If she is taking core of someone suffering from chicken pox, then it is more likely that she may also get the disease.

Question 6. Under which of the following conditions are you most likely to fall sick?
1)When you are taking examinations.
2)When you have travelled by bus and train for two days.
3)When your friend is suffering from measles. Why?
Answer: We are most likely to fall sick when we are near to our friend who is suffering from measles. It is so because measles is an infectious disease.

Why Dowe Fall Ill Textual Questions

Question 1. State any two conditions essential for good health.
Answer: The conditions essential for good health are
1)Clean surroundings.
2)Adequate and nutritious food

Question 2. State any two conditions essential for being free of disease.
Answer:
1)Living in a hygienic environment
2)Getting vaccinated against common infectious diseases.

Question 3. Are the answers to the above questions necessarily the same or different? Why?
Answer: Answers to the above questions are interconnected but different. It is so because being disease-free does not mean one is healthy.

Question 4. 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:
The three reasons of feeling sick and ought to see a doctor are,
1)Having a fever (body temperature)
2)Having a cold and cough
3)Having diarrhoea.
These symptoms indicate that there may be a disease, so I would have gone to consult the doctor.

Question 5. 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 jaundiced,
• If you get lice,
• If you get acne. Why?

Answer: Lice and acne can be removed easily with short treatment, both these do not produce long-term effects on the body. Jaundice will have a drastic long-term effect.

Question 6. Why are we normally advised to take nourishing food when we are sick?
Answer: When a person is sick, then his normal body functions get disturbed. In such a situation, food is required which is easy to digest and contains adequate nutrients for recovery, thus nourishing food is required during sickness.

Question 7. What are the different means by which infectious diseases are spread?
Answer:
Infectious diseases are generally spread through,
1)Air
2)Water
3)Sexual contact
4)Physical contact with the infected person
5)Vectors

Question 8. What precautions can you take in your school to reduce the incidence of infectious diseases?
Answer:
The following precautions can be taken in the school to reduce the incidence of infectious diseases;
1)Clean surroundings.
2)Educating students about the causes of infectious diseases.
3)Immunisation.
4)Isolating the affected students to not attend classes till they recover from infectious diseases.

Question 9. What is immunisation?
Answer:
Immunisation is a technique in which people are given a particular vaccine so that they develop temporary or permanent immunity against a particular infectious disease.

KSEEB Solutions For Why Do We Fall Ill Short Notes 

Question 10. 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:
The following are the immunisation programmes available at the nearest health centre in our locality.
1) Polio
2) DPT
3)Measles
4)MMR
5)Smallpox
6)TT
7)Hepatitis B.

Why Dowe Fall Ill Additional Questions

Question 1. What is an antibiotic? Give two examples.
Answer: Antibiotic is a chemical secreted by microorganisms (bacteria and fungi) tliat kills the growth of some kinds of germs.
Examples: penicillin and streptomycin.

Question 2. Name any three diseases transmitted through vectors.
Answer: Malaria, typhoid, and plague.

Question 3. List some preventive measures against communicable diseases.
Answer:
1)Health Education
2)Isolation
3)Proper sanitation
4)Immunisation

Question 4. Distinguish between acute and chronic diseases.
Answer: Acute diseases last for short period and cause major effects on general health ina very short time. Chronic diseases lasts for long period, it takes a long time to have major effects on general health.

Question 5. Define health.
Answer: Health is described as the state of complete physical mental and social well-being, and not merely an absence of disease.

Question 6. What are contagious diseases? Give one example.
Answer: The diseases which are spread by actual contact between an infected person and healthy persons are called contagious diseases. Eg. Chickenpox.

Why Dowe Fall Ill High Order Thinking Questions

Question 1. Why do parents immunize their newborn babies?
Answer:
The parents immunise the newborn baby because.
1) To provide immunisation to a specific diseases.
2) To protect children from deadly diseases.

Question 2. Reema is suffering from cholera. It is likely that the children sitting around her will be exposed to the infection.
1) Do all the students get infected and suffer from disease?
2) List any two major symptoms of cholera.
Answer:
1. No, all students will not get infected and suffer from the disease because a large inoculum is required to transmit the disease.
2. Symptoms of cholera are.

• Rapid heartbeat
• Low blood pressure.

Question 3. Why are antibiotics not effective for viral disease?
Answer: Antibiotics generally block the biosynthetic pathways and they block the pathways of bacteria or microbes. However, viruses have very few biochemical mechanisms of their own and hence are unaffected by antibiotics

Question 4. Why did female Anopheles mosquito feed on human blood?
Answer: FemaleAnopheles mosquito feeds on blood because it requires proteins of human blood to lay eggs.

Question 5. What is meant by ‘symptoms’ of a disease? What do they indicate?
Answer: These are manifestations or evidence of the presence of diseases. These indicate that there is some abnormality in the body.

Why Dowe Fall Ill Unit Test

Question 1. Repeated exposure to a pathogen leads to
the development of ______
1) disease
2) immunity
3) Cancer
4) weakness
Answer: (2) immunity

Question 2. A child who has had an infection of chicken pox does not suffer from the disease for the second time, because_____
1) The body has developed immense strength to fight the infection.
2) The immune system keeps a memory of the specific disease and responds by killing the infectious agent, in the second exposure.
3) The boy is advised to take vitamin tablets daily after the disease
4)The child is given clean and nutritious food after the disease.
Answer: (2) The immune system keeps a memory of the specific disease and responds by killing the infectious agent, in the second exposure.

KSEEB Class 9 Science Chapter 13 Important Questions 

Question 3. Which one of the following diseases is not transmitted by mosquitoes?
1)Brain fever
2)Malaria
3)Typhoid
4)Dengue
Answer: (3) Typhoid

Why Dowe Fall Ill Fill In The Blanks

1.The full form of BCG Bacillus calmette and Guerin.
2. Mosquito is the vector of Dengue and Malaria.

Why Dowe Fall Ill Answer the following Questions

One Mark

Question 1. Name two diseases that spread through water.
Answer: Cholera, amoebic dysentery.

Question 2. Name two vaccines given to children below 5 years of age.
Answer:
1. Polio Vaccine
2. Hepatitis

Question 3. After eating contaminated food, a number of people complained of nausea, vomiting, abdominal pain and loose stools.
1)Name the disease they are suffering from
2)Name the causative organism.
Answer:
1) Diarrhoea
2)Escherichia coli or salmonella

Question 4. Name any two diseases against which vaccinations are available.
Answer:
1)Tuberculosis
2)Typhoid

Question 5. What do the physicians do on the basis of symptoms?
Answer: Physicians look for the signs of a particular disease.

Question 6. What are pathogens?
Answer: Disease-causing microorganisms are called pathogens.

Two Marks

Question 1. Differentiate between congenital diseases and acquired diseases.
Answer:
Congenital diseases are anatomical or physiological abnormalities present since birth.
Example: Haemophilia, colourblindness
Acquired diseases are those that develop after birth.
Example: AIDS.

Question 2. What is the difference between a virus and a bacteria?
Answer:
Viruses
1. Non-cellular
2. Have no metabolism of their own
3. Command the host cell to produce a virus.
4. can be crystallised
Bacteria
1. Single-celled
2. Have a metabolism of their own
3. They can reproduce by their own.
4. cannot be crystallised

Question 3. Give two examples for each of the following.
1)Infectious diseases
2)non-infectious diseases
3)Acute diseases
4)Chronic diseases
Answer:
1) Infectious diseases— Smallpox, chicken pox
2)Non-infectious diseases— Diabetes, simple Goitre
3)Acute diseases — Viral fever, flu
4)Chronic diseases—Tuberculosis, Elephantiasis

Question 4. List the following diseases into communicable and non-communicable diseases.
Answer:

1. Cancer: Non-Communicable
2. High blood pressure: Non-Communicable
3. Common cold: Communicable
4. Diabetes: Non-Communicable
5. Tuberculosis: Communicable
6. Night blindness: Non-Communicable
7. SARS: Communicable
8. Typhoid: Communicable
9. Cholera: Communicable
10. Dengue: Communicable

KSEEB Solutions Chapter 13 Diseases And Prevention Class 9 

Question 5. What are the modes of HIV transmission?
Answer:
1)Unprotected sexual contact with an infected partner.
2)Use of contaminated needles and syringes.
3)Transfusion of infected blood.
4)Mother to baby due to rupturing of blood vessels at the time of birth.

Why Dowe Fall Ill  Activity

Question 1. Conduct a survey in your locality. Talk to 10 families who are well-off and ten who are very poor (in your estimation) Both sets of families should have children who are below five years of age. Measure the heights of these children. Draw a graph of the height of each child against its age for both sets of families.
1) Is there a difference between the groups? If yes, why?
2) If there is no difference, do you think that your findings mean that well-off or poor does not matter for health?

Conclusion:
After conducting the survey and preparing the graphs of the height of each child against his age, you would find that there is a difference in the heights of children of two groups Children of rich families have better growth than children of very poor families. It is so because children of well-off families take adequate balanced diets, live in clean environments and better conditions of living than children of very poor families

KSEEB Solutions For Class 9 Science Chapter 10 Gravitation Important Concepts

Gravitation: It is the force of attraction between any two bodies in the universe
Gravity: It is the force of attraction between a planet and an object on its surface
The universal law of gravitation: Every object in the universe attracts every other object with a force which proportional to the product of their masses and inversely proportional to the square of the distance between them.
Universal gravitational constant: It is the gravitational force between two objects of unit mass with unit separation. It’s value is 6.67 x 10^-11 Nm²/kg²
Acceleration due to gravity: It is the acceleration of an object tailing towards the earth due to the earth’s gravity. It’s value on the surface is 9.8m/s²
Free fall: The motion of an object under the influence of earthing’s gravity only is called tree fall
Mass: It is the quantity of matter contained in a body. It’s value does not change with places and its S. I unit is a kilogram (kg)
Weight: It is the force with which a body is attracted by the earth and it’s value varies from place to place. Its S.I. unit is newton (N). Weight is given by W = mg
Density: It is the ratio of mass by the volume of a substance is density^ mass/volume and the S.I unit is kg/m³

^{Read and Learn More KSEEB Solutions for Class 9 Science} 

Relation between g and G: g = GM/R²
where g – acceleration due to gravity, G = universal gravitational constant, M = mass of the earth and R = mean radius of the earth
Equations of objects under Earth’s gravity:
(1) v = u + gt
(2) s = ut + ½ gt²
(3) v² = u² + 2gs
Fluid: A substance that can flow is called a fluid. Example: gases and liquids.
Thrust and Pressure: The force acting on an object perpendicular to the surface is called thrust and the thrust per unit area is called pressure. S.I unit pressure is N/m² = pascal
Buoyancy: When an object is immersed in a liquid, then the upward force exerted by a liquid on the object is called upthrust or buoyancy. The magnitude of buoyancy depends on the density of the liquid
Archimedes’s Principle: When a body is immersed completely or partially in a fluid, it experiences an upward force that is equal the weight of the fluid displaced by the body.
Relative density: Relative density of a substance is defined as the ratio of the density of the substance to the density of water. It has no unit.

Gravitation Exercises

Question 1. State universal law of gravitation
Answer: Every object in the universe attracts every other object with a force which proportional to the product of their masses and inversely proportional to the square of the distance between them.

Question 2. Write the formula to find the magnitude of the gravitational force between the earth and an object.
Answer: F = G Mm/R²

Gravitation KSEEB Class 9 Question Answers 

Question 3. What do you mean by free fall?
Answer: The motion of a body under the influence of the earth’s gravity

Question 4. What do you mean by the acceleration due to gravity?
Answer: It is the acceleration of an object falling to ward’s earth’s center due to the earth’s gravity

Question 5. What are the differences between the mass of an object and its weight
Answer:
Mass:

1. It is the quantity of matter contained in a body
2. Its value does not change
3. It is a scalar quantity
4. It is expressed in kg
5. It can never be zero

Weight:

1. It is the force with which a body is attracted toward the center of the earth
2. Its value changes from place to place
3. It is a vector quantity
4. It is expressed in newton
5. It is zero at the center of the earth

Question 6. Why is the weight of an object on the moon is 1 /6th its weight on the earth?
Answer: This is because the gravitational attraction on the moon is 1/6 th of that on the earth.

Question 7. Why is it difficult to hold a school bag with a thin and strong string?
Answer: Thinner the string, the smaller is the area of the cross-section, and the greater will be the pressure.

Question  8. Why does an object float or sink when placed on the surface of the water?
Answer: A cork placed on water floats while the nail sinks because of their difference in their densities. The density of the cork is less than the density of water. This means that the upthrust of water on he cork is greater than the weight of the cork. So it floats. The density of iron nail is more than that of water. This means that the upthrust by water on he iron nail is less than the weight of the nail. Hence the nail sinks.

Question 9. You find your mass to be 42kg on a weighing machine. Is your mass more or less than 42kg?
Answer: Slightly more than 42 kg. The upward thrust of air acts on our body which reduces our original mass.

Question 10. You have a bag of cotton and an iron bar, each indicating a mass of 100kg when measured on a weighing machine. In reality, one is heavier than the other. Can you say which one is heavier and why?
Answer: The bag of cotton is heavier because the cotton bag has a larger area and it experiences a larger upthrust of air than the iron bar. This means that the weighing machine shows a smaller mass for the cotton bag than the original mass which is more ‘ than 100kg.

Gravitation Textual Questions

Question 1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer: Since F∞1/r², when the distance is reduced to r/2, then the new force ∞ 4/r^2. The fore becomes 4 times the initial force

Question 2. Gravitational force acts on all objects in proportion to their masses. Why then a heavy object does not fall faster than a light object?
Answer: Although F cc m, g is independent of the mass of the body. Hence all the bodies reach the ground at the same time in absence of air resistance.

Question 3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface ( Mass of earth = 6 x 1024 kg and R = 6.4 x 106 m)
Answer:
F = G Mm/R²

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: Both the earth and moon attract each other with the same force. This is because neither the earth nor the moon is toward each other.

Question 5. If the moon attracts the earth, why does the earth not move toward the moon?
Answer: Based on  Newton’s 3rd law of motion, the earth also attracts the moon with equal and opposite force.

Question 6. What happens to the force between two objects, if
(1)The mass of one object is doubled?
(2)The distance between the objects is doubled and trebled?
(3)masses of both objects are doubled
 Answer:
(1) Since force is directly proportional to mass, force is doubled
(2) Since the force(F) is inversely proportional to the square of the distance between the objects, he force becomes F/4 when the distance is doubled and F/9 when the force is trebled
(3) Since the force is proportional to the product  of masses, the new force will be 4F

Class 9 Science Chapter 10 KSEEB Textbook Solutions 

Question 7. What is the importance of the universal law of gravitation?
 Answer: The universal law of gravitation explains
1) The force that binds us to earth
2) The motion of the moon around he earth
3) The motion of the planets around the sun
4) The tides in sea and ocean

Question 8. What is the acceleration of free fall?
Answer: It is the acceleration produced on a body due to
the earth’s gravity.

Question 9. What do we call the gravitational force between the earth and an object?
 Answer: Weight of the object

Question 10. Amit buys few grams of gold at 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?
Answer: No. Due to the change in the value of g, the weight of a body is always less than its weight at the equator. Hence the gold weight will be less then its weight at the equator.

Question 11. Why will a sheet of paper fall slower than one that is crumpled into a ball? 
Answer:  Due larger surface area, the paper sheet experiences more air resistance and hence it falls slowly.

Question 12. The gravitational force on he surface of the moon is only 1/6 th of the earth. What is the weight in new tons of a 10kg object on the moon and on the earth?
Answer:

• On the earth, the weight of the object=10 g= 10x 9.8 = 98N
• On the moon, weight of the object = 10 x g/6 = 10 x 9.8/6 = 16.3N

Question 13. A ball is thrown vertically upwards with a velocity of 49m/s. Calculate
(1)The maximum height to which it risees?
(2)The total tiem it takes to return to the surface of the earth?
Answer:
(1) Here u = 45m/s and g = – 9.8 m/s². At the maximum height v = 0

Using the equation v² – u² = 2gs,

we get 0² – (45)² = 2 (-9.8) s => s – 122.5m

Thus maximum height the ball rises is 122.5m

(2) Again using the equation v=u + at, we get 0 = 45 – 9.8 t => t = 5s

Since time to reach maximum height = time to reach the ground from a maximum height

Total time is taken by the ball to return to the earth’s surface = 5 + 5 = 10s

Question 14. A stone is released from he top of a tower of height 19.6m. Calculate its final velocity just before touching the ground
Answer: Given u = 0, g = 9.8m/s2,s= 19.6m, v = ?

Using the equation

v² – u² = 2gs => v²– 0 = 2(9.8) (19.6) = (19.6)² (why?)

Hence \(v=\sqrt{(19.6)^2}=19.6 \mathrm{~m} / \mathrm{s}\)

KSEEB Solutions For Gravitation Short Notes 

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

Answer: We have u = 40m/s, g = – 10m/s² and at the maximum height v = 0

Using the equation  v²-u² = 2gs , we get v²– (40)²

= 2(- 10) s => – 20s = – 1600 => s = 80m

The total distance covered by the stone = 80 + 80= 160m

Net dispalcement = 0

Question  16. Calculate the force of gravitation between the earth and the sun given that the mass of the earth is 6 x 1024 kg and of the sun is 2 x 1(Pkg. The average distance between the two is 1.5 x 10n m
Answer:
F =\(\mathrm{G} \frac{M_E M_S}{r^2}\)

On simplification, we get F = 3.56 x 10²² N

Question 17. A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25m/s. Calculate when and where the stones meet.
Answer:
Let the stones meet at a height x from the ground after time ‘t’ second

For the falling stone, u = 0, g = l0 m/s² g = 100 -x

Using the equation s = u t + 1/2 gt², we get

100- X = (0) t + 1/2 (10) t²  …………….. (1)

For the upward stone,

u = 25m/s, g = -10m/s2 s = x

Hence x = 25t – 1/2(10) t²   …………….(2)

Adding equations (1) and (2), we get 100 = 25t => t = 100/25 = 4 sec

The two stones meet each other after 4 second

When t = 4, x = 25 (4) – 1/2 (10) (4)² x =

100-80-20m

I.e, The stones meet at a height 20m from the ground.

Question 18. A ball thrown up vertically returns to the thrower after 6 sec. Find
1) Velocity with which it was thrown up?
2) The maximum height it reaches, and
3) The position after 4 second
Answer:
Time to reach maximum height = 6/2=3 second

1) Now, t = 3s, v = 0, g = – 9.8 m/s² and u =?

v = u + gt  0 = u — 9.8 (3) u  => 29.4m/s

2) S _max = ut + 1/2 gt² = 29.4 (3) – 1/2(9.8) (3)²

= 44.1 m

3) S _{(t = 4)}= 29.4 (4) – 1/2 (9.8) (3)² = 39.2m from the ground

Question 19. In what direction does the buoyant force on an object immersed in a liquid act?
Answer: In the vertical direction through the center of gravity.

Question 20. Why does a block of plastic released under water come up to the surface of the water?
Answer: It is due to the upthrust exerted by the water the block of plastic.

Question 21. The volume of a substance is 20 cm-1. If the density of water is 1 g/cm3, will the substance float or sink?
Answer: Density of substance = 50/20 = 2.5g.cm3 > the density of water. Hence the substance will sink into the water.

Question 22. The volume of a 500g 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?
Answer:
Density of packet = 500/350 = 1.43g/cm³ > density of water (1.2g/cm³).
Hence the packet will sink.

The volume of water displaced = volume of the sealed packet = 350cm³

Therefore, a mass of water displace = 350 x 1 = 350g

KSEEB Class 9 Science Chapter 10 Important Questions 

Gravitation Additional Questions

Question 1. What is the S.I. unit of universal gravitational constant
Answer: Nm² kg ^-2

Question 2. What is the accepted value of the universal gravitational constant?
Answer: 6.673 x 10^-4 Nm²kg^-2

Question 3. What is free fall?
Answer: When an object falls under the earth’s gravitational force above, then the object is said to be in free fall

Question 4. The mass of an object is 19kg. What is its weight on the earth?
Answer:
Weight of the object (w) = mg

= 10 x 9.8 = 98 N

Question 5. What is weightlessness?
Answer: A body is said to be in a state of weightlessness when the reactionary force of the supporting surface is zero.

Question 6. One kg – wt is equal to how many Newton’s?
Answer:   
1 kg wt = 1kg x 9.8ms^-2
              = 9.8 N

Question 7. State two factors on which the gravitational force between two objects depends
Answer:
1) Masses of the objects
2) Distance between the objects

Question 8. Who formulated the universal law of gravitational
Answer: Issac Newton

Question 9. What is the value of the universal gravitational constant
Answer: G =6.67 x 10^-11 Nm²kg^-2

Question 10. On what factors weight of a body depend?
Answer: Mass and acceleration due to gravity

Question 11. Why does a man feel lighter on the moon?
Answer: On the moon, the gravity is less than that of the earth.

Question 12. What is weightlessness?
Answer: When the surface offers zero reactionary force on a body kept on it is said to be weightless.

Question 13. The value of g is not same at all places on the earth. Why?
Answer: This is due to the shape of the earth and its rotation about its own axis

Question 14. Is weight a scalar or a vector?
Answer: Weight is a vector quantity

Question 15. How many newtons are equal to one kg – wt?
Answer: 9.8N

Question 16. What is the value of g at the center of the earth?
Answer: zero

Question 17. What is the mass of an object whose weight is 19.6N
Answer: m = W/g = 19.6/9.8 = 2kg

Question 18. An astronaut feels weightlessness in space graft?
Answer: It is because no gravitational force acts on the astronaut

Question 19. What happens to the mass and weight of a stone broght fron the moon to the earth?
Answer: Mass remains the same while the weight increases

Question 20. What is fluid?
Answer: A substance that can flow is called a fluid.

Two Marks Questions

Question 21. How does gravitation differ from gravity?
Answer: Gravitation refers the force of attraction between any two bodies in the universe whereas gravity refers the force of attraction between the earth and any object on its surface.

Question 22. What do you understand by buoyancy?
Answer: The phenomenon of an upward force acting on a body partially or completely immersed in a fluid is called buoyancy

Question 23. What is the value of
1) On the surface of the earth
2) At the center of the earth
Answer:
1) on the surface g = 9.8ms’2
2) At the center, g = 0.

KSEEB Solutions Chapter 10 Gravitational Force Class 9 

Question 24. How does the value of g vary with
1) altitude
2) latitude
Answer:
1) g decreases with an increase in altitude
2) g is m make at poles and minimum at the equator.

Question 25. Mean’s weight on the surface of the earth is SOON. What will be her weight at the height 2R (where R = Radius of the earth)
Answer:
Weight on the earth W = \(G \frac{M m}{R^2}\)

Weight on the height 2R =\(\mathrm{W}^i \)=\( G \frac{G M m}{(2 R)^2}\)

Question 26. You feel less weigher in water than outside the water. Why?
Answer: Because inside the water, the body experiences an upthrust by water.

Question 27. State Archimedes’ principle
Answer: When a body is immersed fully or partially in a fluid it experiences an upward thrust equal to the weight of the fluid displaced by it

Question 28. Name any devices that work on Archimedes’s principle
Answer:
1) Lactometer – used to measure the purity of milk
2) Hydrometer – used to measure the density of liquids

Question 29. Define the relative density of a substance
Answer: It is the ratio of the density of the substance to the density of water at 4°C

Question 30. On what factors the buoyant force depends on?
Answer:
1) Density of the liquid
2) The volume of the body immersed in the liquid

Question 31. Why does an army tank rest upon a continuous heavy chain?
Answer: Due to the larger area of the chain, the pressure exerted by the tank will be less and this prevents the sinking of the ground.

Question 32. A sharp knife cut better than a blunt one. Why?
Answer: Sharper the edge, the lesser will be the area of the edge. Hence effective force per unit area ie pressure is more.

Question 33. State the conditions for
(1) floatation of a body on a liquid
(2) sinking of a body in a liquid
Answer:
(1) A body will float if its density is less than the density of the liquid
(2) The body will sink if the density of the body is greater than the density of the liquid.

Question 34. When a fresh egg is placed in water it sinks. On dissolving a large amount of salt in the water, the g begins to rise and float. Why?
Answer: The density of the fresh egg is more than that of water. Hence it sinks. When salt is added to water, the density of water becomes more than that of egg. Hence the egg begins to float on water.

Gravitation KSEEB Class 9 Detailed Solutions 

Question 35. The weight of an object on the earth is 12N. What would be its weight on the moon?
Answer:
We Know That

Weight on the moon =1/6 x weight on the earth

= (1/6) x 12

= 2N

Three Marks Questions

Question 36. Give reasons for the following:
(1)Pins and nails have pointed ends
(2) Building has wide foundation
(3)A camel walks easily on the sand but a man cannot
Answer:
(1) The pointed ends have very small area and hence the force per unit area ie pressure will be greater on the surface which makes them to penetrate the surface
(2)Due to larger surface area of the larger foundation, the effective force/area of the building on the ground will be less. This prevents the sinking of the ground.
(3)Feet of the camel are larger than that of man. Hence camels experience lower pressure than the man. This makes camel to walk easily on the sand.

Question 37. The mass of the earth and the moon are 6 x 10²⁴ kg and 7.4 x 10²² kg respectively. The distance between the earth and the moon is 3.84 x 104 km. Calculate the force exerted by the earth on the moon. G= 6.7 x 10 ^-11 Nm²kg^-2.
Answer: Mass of the earth (M) = 6 x 10²⁴ kg, Mass of the moon (m) = 7. 4 x 10²² kg, and distance between the earth and the moon (d) = 3.84 x 10⁵km = 3.84 x 10⁸ m

According to Newton’s universal law of gravitation, the force between the earth and the moon is

F = G Mm/d² = \( \frac{6.7 \times 10^{-11} \times 6 \times 10^{28} \times 7.4 \times 10^{22}}{\left(3.84 \times 10^8\right)^2}\)

Question 38. Derive an expression for acceleration due to gravity acting on an object on the surface of the earth.
Answer: According to Newton’s second law of motion, Force acting on an object of mass (m) and tailing with acceleration (g) is given by F = mg …………………(1)

According to Newton’s universal gravitational law of gravitation, the force of attraction between the object and earth is given by F₈ = G Mm/R²…………………(2)

Comparing equations (1) and (2), F = F₈ mg = G Mm/R² => g = GM/R2

Question 39.
1) Define Thrust and pressure. Mention their S.I units
2) A block of wood is kept on a table top. The mass of the wooden block is 5 kg and its dimensions are 40cm x 20cm x 10cm. Find the pressure exerted by the wooden block on the table top if it is made to lie on  the table top with its sides of dimensions (a) 20cm x 10cm and b) 40cm x 20cm
Answer:
1) Thrust: It is the force acting on an object perpendicular to the surface.

Pressure: It is thrust per unit area of contact ie P = F/A

S.I unit thrust is newton (N) and S.I unit of pressure = Nm²

2) Case (1) Thrust on the table (F) = mg – 5 x 9.8 = 49N

Area of the face on the table = 20 x 10 cm²

= 0.02m²

Hence pressure = F/A = 49/0.02 = 2450N/m²

Case (2) Area of the face on the table=40 x 20 cm² = 0.08 m²

Hence pressure = 49/0.08 = 612.5 N/m²

Four Marks Questions

Question 40. Give reasons:
1) You apply more pressure on loose sand when you stand than when you Me down
2) An iron nail floats on mercury but sinks in water
3) It is easier to swim in sea water than in river water
Answer: Pressure is inversely proportional to the area of contact. The area of contact is less when we stand on the loose sand. Hence we apply more pressure.

Question 43. State and explain the universal law of gravitation. Hence obtain the S.I. unit of the universal gravitational constant
Answer:
Statement: Every object in the universe attracts every other object with a force that is proportional to their masses and inversely proportional to the square of the distance between them. The force acts along the lining the centers of the objects.

Explanation:

consider two objects A and JB of masses M and M lie at a distance ’d’ from each other . Let F be the force of attraction between the two objects. According to the universal law of gravitation

1) \( F \propto \frac{M m}{q^2}\)

2) \( F \alpha \frac{\Phi}{d^2}\)

combining equations (1) and (2), we get

\(F \alpha \frac{M m}{d^2}\)   or

where G is the constant of proportionality and is called the universal gravitation constant.

To find the S.T. unit of G

From Equation (3)

substituting units of F, d, M ie N, m, and kg, we get

S.L unit of G= Nm² kg ²

Question  41.
1) state universal law of gravitation
2) Mention any four phenomena explained by the universal law of gravitation.
Answer:
1) Every object in the universe attracts every other object with a force that is proportional to their masses and inversely proportional to the square of the distance between them.
2) Universal law of gravitation explained the following phenomenon.

• The force that binds us to the earth
• The motion of the moon around the earth
• The motion of the planets around the sun
• The tides due to the moon and the sun.

Gravitation KSEEB Class 9 MCQ Solutions 

Question  42. using the following data, prove that weight of the object on the moon = (1/6)th of its weight on the earth

celestial body     Mass (kg)                Radi (in)

Earth                5.98 x10²⁴                  6.37 x10⁶

Moon               7.36 x10²²                1.74 x106

Answer: According to the universal law of gravitation, the weight  of an object on the moon is given by

\(W_m=G \frac{M_m m}{R_m^2}\)   ………………………………(1)

where m = mass of the body the weight of the same object on the earth is given by

\(  W_E=G \frac{M_e \times m}{R_e^2}\)  …………………………..(2)

Dividing Eqn (1) by Eqn (2), we get

Thus, the weight of an object on the moon is (1/6)th of its weight on the earth

Question 43. Distinguish between Mass and Weight
Answer:
Mass

1. Mass is the quality of material present in a body
2. The value of mass remains the same at all places.
3. It is a scalar quantity
4. S.I. unit is kilogram
5. Measured by a pan balance
6. Mass can never be zero

Weight

1. It is the gravitational force exerted by the earth on the body.
2. The value of weight changes as the ‘g’ value changes from place to place
3.  It is a vector quantity
4. S.I. unit is kilogram weight or newton
5. Measured by a spring balance
6. Weight is zero at the center of the earth.

Question 44.
1) Define acceleration due to the gravity of earth
2) On what factors acceleration due to gravity depends
3) Where is cg’ greater – at poles or equator
Answer:
1) The acceleration produced in the motion of a body under the gravitational force of the earth is called acceleration due to gravity.
2) Acceleration due to gravity depends on

• The shape of the earth
• Altitude from the surface of the earth.
• Depth from the surface of the earth
• Latitude of the place.

3) g is greater at poles

Question 45.
(1) What is gravity?
(2) Mention any three differences between g and G
Answer:
1) Gravity is the force of attraction between the earth and any object on the surface.
2) acceleration due to gravity (g)

1. It is acceleration acquired by a body by virtue of the earth’s gravity
2. It is equal to the force of attraction between two unit masses separated by unit distance
3. Its value depends on altitude, latitude shape of the earth, etc.,
4. S.I.unit is ms^-2

Universal gravitation constant IG)

1. It is equal to the force of attraction between two unit masses separated by unit distance
2. It is a quantity
3. It is a universal Constance
4. Nm²kg²

Gravitation Application Questions

Question 1. The weight of an object on the earth is I2N. What would be its weight on the moon?
Answer:
We Know That
Weight on the moon = 1/6  x weight on the earth

= 1/6 x12

= 2N

Question 2. A stone is brought from the moon to the earth? What happens to the mass and weight of the stone on the earth?
Answer: Mass remains the same Weight increases.

Question 3. What is the value of the universal gravitational constant of the center of the earth?
Answer: G = \(6.673 \times 10^{11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}\)

Question 4. The moon attracts the earth. Why does the earth not move toward the moon?
Answer: acceleration \( \alpha \frac{1}{m a s s}\)

Since the mass of the earth is very large, the acceleration produced in the earth is negligible.

Gravitation KSEEB Class 9 Question Answers 

Question 5. If the earth shrinks and mass remains the same, what happens to the weight of a person?
Answer:
We Know That           

When the earth shrinks, R decreases. This leads to an increase in g and therefore weight = mg also increases.

Question 6. When a piece of paper and a stone are dropped from the top of a building, the paper takes longer time. Why?
Answer: Since the paper has more surface area it experiences more air friction than the stone.

Question 7. On the earth’s surface mass remain constant while weight changes from the equator to the pole Why?
Answer: Since ‘g’ varies from equator to pole weight = mg also varies.

Question 8. Mass of a planet is (1/3)rd of the earth its radius is half of the earth. What will be acceleration due to gravity
Answer:
M1 – Mass of the planet
R1 = Radius of the planet k-riME and Rl = VL RE

Question 9. Buying gold at the equator and selling the same at the poles is profitable. Why?
Answer: The weight of gold at the poles will be higher than its weight at the equator.