CLASS 9 Science CHAPTER 15 Improvement in Food Resources (NCERT Solution)

CLASS 9 Science                                                                                                 

CHAPTER 15             

Improvement in Food Resources (NCERT Solution) 

Intext Questions

On Page 204

Question 1: What do we get from cereals, pulses, fruits and vegetables?

Answer: The things we get from cereals, pulses, fruits and vegetables are as follows:

Cereals(wheat, rice, maize, etc.): are the sources of carbohydrates which provide energy.

Pulses(pea, gram and soybean, etc.): are source of proteins.

Vegetables and fruits: provide us vitamins, minerals, carbohydrates, proteins and fats.

On Page 205

Question 1: How do biotic and abiotic factors affect crop production?

Answer: Factors which affect crop production are:

Biotic factors which cause loss of grains are rodents, pests, insects, etc. and abiotic factors, such as temperature, humidity, moisture, etc.

Affects of both biotic and abiotic factors on crop production are in the following ways:

Weight loss

Infestation of insects

Poor germination ability

Discoloration

Question 2: What are the desirable agronomic characteristics for crop improvement?

Answer: Desirable agronomic characteristics are:

Desirable characters for fodder crops are tallness and profuse branching

Dwarfness is desired in cereals, so that fewer nutrients are consumed by these crops.

On Page 206

Question 1: What are the macronutrients and why are they called macronutrients?

Answer: Macronutrients are essential elements which are required by plants in major quantities.

Some of the main macronutrients are:

(i) N, P, S which are found in proteins.

(ii) Ca is found in cell wall.

(iii) Mg is a part of chlorophyll.

Question 2: How do plants get nutrients?

Answer: Soil is the main source of nutrients for plants. Plants absorb the dissolved nutrients by the roots from the soil. This water absorbed by the roots is transported by the xylem tissue throughout the plant body.

On Page 207

Question 1: Compare the use of manure and fertilisers in maintaining soil fertility.

Answer: Use of manure in maintaining soil quality:

(i) Manures are the very rich source of organic matter (humus) for the soil. Humus helps to restore water retention capacity of sandy soil and drainage in clayey soil.

(ii) Manures are the sources of soil organisms like soil friendly bacteria.

Use of fertilisers on soil quality

(i) Excess use of fertilisers cause to dryness of soil and hence the rate of soil erosion increases.

(ii) Continuous use of fertilisers decreases the organic matter which reduce the porosity of the soil and the plant roots do not get sufficient oxygen.

On Page 208

Question 1: Which of the following conditions will give the most benefits? Why?

(i) Farmers use high-quality seeds, do not adopt irrigation or use fertilisers.

(ii) Farmers use ordinary seeds, adopt irrigation, use fertilisers and use crop protection measures.

(iii) Farmers use quality seeds, adopt irrigation use fertilizer and use crop protection measures.

Answer: The (iii) third option is the best option which provides the best conditions to get most benefits. For this, farmers use quality seeds, adopt irrigation, use fertilizers and use crop protection measures. This is because the use of only quality seeds is not sufficient until they are properly irrigated, enriched with fertilizers and protected from biotic factors.

On Page 209

Question 1: Why should preventing measures and biological control methods be preferred for protecting crops?

Answer: In plants, diseases are caused by Pathogens. To remove pathogens, some preventive measure and biological control methods are used which are :

Minimise pollution without affecting the soil quality.

Question 2: What factors may be responsible for the losses of grains during storage?

Answer: Following are the factors responsible for loss of grains during storage:

(i) Abiotic factors like humidity and temperature.

(ii) Biotic factors like insects, rodents, birds, mites and bacteria.

On Page 210

Question 1: Which method is commonly used for improving cattle breeds and why? How is cross breeding useful in animals?

Answer: To improve the cattle breed breeds, we generally use the cross breeding method. It is a process in which a cross is made between indigenous varieties of cattle by exotic breeds to get a cross breed which is high-yielding. During cross breeding, the desired characters taken into considerations are the offsprings should be high Yielding, should have early maturity and should be resistant to diseases and climatic conditions.

On Page 211

Question 1: Discuss the implications of the following statement It is interesting to note that poultry is India's most efficient converter of low fibre food stuff (which is unfit for human consumption) into highly nutritious animal protein food.

Answer: The poultry birds are efficient converters of agricultural by products, particularly cheaper fibrous wastes into high quality meat and in providing egg, feathers and nutrient rich manure. So, the given statement is correctly said for the poultry birds.

Question 2: What management practices are common in dairy and poultry farming?

Answer: Well designed and hygenic shelter.

Proper food is required to get good yield of egg and meat.

Complete protection from diseases causing agents like virus, bacteria or fungi.

Question 3: What are the differences between broilers and layers and in their management?

Answer: The poultry bird giving meat are called broilers and the egg laying birds are called layers.

The ration required for broilers should be rich in protein with sufficient fat. The food should also have high vitamin-A and K. Layers require enough space and lighting.

On Page 213

Question 1: How are fish obtained?

Answer: Fish are obtained by two ways:

(i) Capture fisheries from natural resources

(ii) Fish farming by culture for commercial purposes.

Question 2: What is the advantage of composite fish culture?

Answer: Combination of five or six fish species in a single fish pond is known as composite fish culture. Basis of selection of species is food habits so that they do not compete for food among themselves. As a result, the food available in all parts of the pond is utilized without competing with each other. This increases the fish yield from the pond.

Question 3: What are the desirable characters of the varieties suitable for honey production?

Answer: Desirable characters in varieties for honey production are:

(a) Capacity to collect a large amount of honey.

(b)Should stay in beehive for a longer time.

(c) Should have good breeding capacity.

Question 4: What is pasturage and how is it related to honey producton?

Answer: Flowers available for nectar and pollen collection is known as pasturage . The quality and taste of honey depends on adequate quantity of pasturage and flowers available.

Exercises

Question 1: Explain anyone method of crop production which ensures high yield.

Answer: Inter cropping is a method used for a high yielding crop production. In this method, two or more crops are grown simultaneously on the same field in definite pattern. A few rows of one crop alternate with a few rows of second crop.

Example: soybean, maize or finger millet (bajara) and caw pea (labia).

The selected crops should have different nutrient requirements. This ensures maximum utilisatian of the nutrients supplied. It also prevents pests and diseases from spreading to all the plants belonging to one

crop in a field. This method gives a better crop yield.

Question 2: Why are manure and fertilizers used in fields?

Answer: The essential nutrients of the soil are full filled by manures and fertilizers supply. Therefore, they help in good vegetative growth, giving rise to healthy plants that rise in high crap production.

Question 3: What are the advantages of inter-cropping and crop rotation?

Answer: Advantages of inter-cropping

Maintain soil fertility.

Save time and labour.

Increases productivity per unit area.

Both crops can be easily harvested and threshed separately.

Advantages of crop rotation

Improves soil fertilitiy.

Reduces pest infestation and diseases. Helps in weed control.

Avoids depletion of a particular nutrient from soil.

Question 4: What is genetic manipulation? How is it useful in agricultural practices?

Answer: Transferring desirable genes from one plant to another plant far the production of varieties with desirable characters is known as genetic manipulation.

Example: Profuse branching in fodder craps, high yielding varieties in maize wheat, etc.

Uses in agricultural practices:

Better adaptability to adverse environmental conditions.

Contain desirable features.

Helps in increasing yield and quality.

Maturation period is shorter.

Question 5: How do storage grain losses occur?

Answer: The main reason for the losses of storage grain is abiotic and biotic factors.

The abiotic factors: Moisture and temperature

The biotic factors: Insects, rodents, birds, mites and bacteria.

Question 6: How do good animal husbandry practices benefit farmers?

Answer: Benefits of good farming practices:

(i) It provides improved breeds of domestic animals.

(ii)It increase the production of products like milk, egg and meat.

(iii) Proper shelter, feeding, care and protection against disesase help

the farmers to improve their economic conditions.

Question 7: What are the benefits of cattle farming?

Answer: Benefits of cattle farming

(i) Milk production is increased.

(ii) Good quality meat, fibre and skin is obtained.

(iii) Goad breed of draught animals can be obtained.

Question 8: For increasing production, what is common in poultry, fisheries and bee keeping?

Answer: Cross breeding is a common practice between poultry, fisheries

and bee keeping for increasing production.

Question 9: How do you differentiate between capture fishing, mariculture and aquaculture?

Answer: Differences between capture fishing, mariculture and aquaculture

Fishing

1.Fish are obtained from natural resources, like ponds, canals, rivers,

2.I Locating fish is easy and can be captured by using fishing nets.

Mariculture

1.A method of marine fish culture in the open sea.

2.Fish can be located with the help of satellites and echosounders.

These can be caught by many kinds of fishing nets using fishing boats.

Aquaculture

1.Production of fish from freshwater and brackish water resources.

2.Can be located easily and caught using fishing nets.

( From www.ncrtsolutions.in )

CLASS 9 Science CHAPTER 2 Is Matter around Us Pure (NCERT Solution)

CLASS 9 Science                                                                                                 

CHAPTER 2                                                                                                                           

Is Matter around Us Pure (NCERT Solution)

Intext Questions

On Page 15

Question 1: What is meant by a pure substance?

Answer: Substance having single type of particles is known as pure substance.

For example: Hydrogen, Water etc., are pure.

Note: All elements and compounds are considered to be pure.

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

Answer: Homogeneous mixture

Its constituent’s particles cannot be seen easily.

There are no visible boundaries of separation in a homogeneous mixture.

Its constituents cannot be easily separated.

Examples: Alloys, solution of salt in water etc.

Heterogeneous mixture

Its constituent particles can be seen easily.

Have visible boundaries of separation between the constituents.

Its constituents can be separated by simple methods.

Examples: Mixture of sand and common salt, mixture of sand and water etc.

On Page 18

Question 1: Differentiate between homogeneous and heterogeneous mixture with example.

Answer: Homogeneous mixture

Its constituent’s particles cannot be seen easily.

There are no visible boundaries of separation in a homogeneous mixture.

Its constituents cannot be easily separated.

Examples: Alloys, solution of salt in water etc.

Heterogeneous mixture

Its constituent particles can be seen easily.

Have visible boundaries of separation between the constituents.

Its constituents can be separated by simple methods.

Examples: Mixture of sand and common salt, mixture of sand and water etc.

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

Answer: Solution or true solution is homogeneous.

NO Tyndall effect.

Solute particles cannot be filtered by using a filter paper

True solution is transparent.

Examples: Sea water, alloys, solution of lemon juice in water etc.

Sol (colloidal solution) :

Sol or colloidal solution is heterogeneous.

Tyndall effect.

Cannot be separated by ordinary filter paper.

It may be transparent or translucent.

Examples: Milk of magnesia, cough syrup, mist, fog, clouds, smoke, mud etc.

Suspension :

Suspension is also heterogeneous.

Tyndall effect.

It may be transparent or translucent.

Separated easily by filter paper.(because of large particles)

Examples: Mixture of sand in water, mixture of chalk in water.

Question 3: To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.

Answer: Mass of sodium chloride (solute) = 36 g

Mass of water (solvent) = 100 g

We know that, mass of solution = mass of solute + mass of solvent

= 36 g+ 100 g= 136 g

Concentration (mass percentage) of the solution 

      Mass of Solute

= ---------------------- x 100

     Mass of Solution

      36g

= --------- x 100 = 26.47%

    136 g

On Page 24

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

Answer: Simple distillation is the method which can separate the mixture of kerosene and petrol (b.p. differ by more than 25°C).

 Method: In a distillation flask, a mixture of kerosene and petrol is taken. The mixture is heated slowly and the temperature is recorded with the help of thermometer. Petrol (b.p. = 70° C to 1200 ° C) vaporizes first and the temperature becomes constant for some time (till all petrol evaporates from the mixture). Vapours of petrol are condensed and collected in another container while the kerosene remains in the distillation flask. As soon as the temperature starts’ rising again, the heating is stopped and both the components are collected separately.

Question 2: Name the technique to separate

(i) Butter from curd

(ii) Salt from sea water

(iii) Camphor from salt

Answer: (i) By using centrifugation method, butter can be separated from curd.

(ii) By using evaporation method, salt from sea water can be separated. Water vaporises on evaporation leaving behind the salt.

(iii) Camphor from salt can be separated by sublimation method. On subliming camphor will be converted into vapour leaving behind the salt.

Question 3: What types of mixtures are separated by the technique of crystallisation?

Answer: Crystallisation method can be used for the purification of those mixtures whichccontain insoluble and/or soluble impurities.

Have crystalline nature cannot be separated by filtration as some impurities are soluble.

Question 4: Classify the following as chemical or physical changes

(a) Cutting of trees,

(b) Melting of butter in a pan,

(c) Rusting of almirah,

(d) Boiling of water to form steam,

(e) Passing of electric current, through water and the water breaking down into hydrogen and oxygen gases,

(f) Dissolving common salt in water,

(g) Making a fruit salad with raw fruits, and

(h) Burning of paper and wood

Answer: Physical Change :

Cutting of trees

Melting of butter in a pan

Boiling of water to form steam

Dissolving common salt in water

Making a fruit salad with raw fruits

Chemical Change :

Rusting of almirah.

Passing of electric current, through water and the water breaking down

into hydrogen and oxygen gases.

Burning of paper and wood.

Question 5: Try segregating the things around you as pure substances or mixtures.

(a) Wood

(b) Coal

(c) Milk

(d) Sugar

(e) Common salt

(f) Soap

(g) Soil

(h) Rubber

Answer: (a) Mixture

(b) Mixture

(c) Mixture

(d) Pure substance

(e) Pure substance

(f) Compound/mixture

(g) Mixture

(h) Pure substance

Exercises

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

(a) Sodium chloride from its solution in water.

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

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

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

(e) Butter from curd.

(f) Oil from water.

(g) Tea leaves from tea.

(h) Iron pins from sand.

(i) Wheat grains from husk.

(j) Fine mud particles suspended in water.

Answer: (a) Evaporation

(b) Sublimation

(c) Filtration

(d) Chromatography.

(e) centrifugal machine or churning the curd by hand.

(f) Decantation

(g) Filtration.

(h) Magnetic Separation.

(i) Winnowing.

(j) Coagulation and decantation:

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

Solution: Method of preparation of tea

(i) Take some water (solvent) in a pan and heat it.

(ii) Add some sugar (solute) and boil to dissolve the sugar completely the

obtained homogeneous mixture is called solution.

(iii) Add tea leaves (or tea) in the solution and boil the mixture.

(iv) Now add milk and boil again.

(v) Filter the mixture through the tea stainer and collect the filtrate or soluble substances, i.e., tea in a cup. The insoluble tea leaves left behind as residue in the 8 trainer.

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

(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?

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

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

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

Answer: (a) Mass of potassium nitrate needed to produce its saturated solution

in 100 g of water at 313 K = 62 g

Mass of potassium nitrate needed to produce its saturated solution in 50 g of water at 313K

       62 

= ------- x 50g = 31g

     100

(b) Crystals of potassium chloride are formed. This happens as solubility of solid decreases with decreasing the temperature.

(c) Solubility of each salt at 293 K

Potassium nitrate 32 g per          100 g water

Sodium chloride 36 g per           100 g water

Potassium chloride 35 g per       100 g water

Ammonium chloride 37 g per    100 g water

Note: Solubility of a solid is that amount in gram which can be dissolved in 100 g of water (solvent) to make saturated solution at a particular temperature.

Ammonium chloride has the maximum solubility (37 g per 100 g of water) at 293 K.

(d) Solubility of a (solid) salt decreases with decrease in temperature while it increases with rise in temperature.

Question 4: Explain the following giving examples.

(a) Saturated solution

(b) Pure substance

(c) Colloid

(d) Suspension

Answer: (a) Saturated solution: A solution in which no more amount of solute can be dissolved at a particular temperature is called saturated solution. Example: when sugar is dissolved repeatedly in a given amount of water, a condition is reached at which further dissolution of sugar is not possible in that amount of water at room temperature.

(b) Pure substance: A substance made up of single type of particles (atoms and/or molecules) is called pure substance. All elements and compounds are said to be pure. Example: water, sugar etc.

(c) Colloid: A heterogeneous mixture in which the solute particle size is too small to be seen with the naked eye, but is big enough to scatter light is known as Colloid. There are two phases in colloidal solution Dispersed phase: solute particles are said to be dispersed phase Dispersion medium: the medium in which solute particles are spread is called the dispersion medium. Example: Milk, clouds etc., are the example of colloid.

(d) Suspension: A suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium. Particles of suspension are visible to the naked eye. Example: Mixture of sand, Water and Muddy water etc.

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

Answer: Homogeneous mixtures: Air, soda water, vinegar, filtered tea.

Heterogeneous mixtures: Wood, soil,

Question 6: How would you confirm that a colorless liquid given to you is

pure water?

Answer: If the given colorless liquid boils at 100°C sharp, it is pure water, otherwise not.

Question 7: Which of the following materials fall in the category of a "pure substance"?

(a) Ice (b) Milk (c) Iron

(d) Hydrochloric acid (e) Calcium oxide (f) Mercury

(g) Brick (h)Wood (i) Air

Answer: Ice, iron, calcium oxide, mercury are pure substance as they have definite composition.

Milk is a colloid, so it is a heterogeneous mixture.

Hydrochloric acid is also a mixture of hydrogen chloride gas and water.

Question 8: Identify the solutions among the following mixtures.

(a) Soil

(b) Sea water

(c) Air

(d) Coal

(e) Soda water

Answer: Sea water, air and soda water: Homogeneous mixture

Coal, Soil: Heterogeneous solution.

Question 9: Which of the following will show "Tyndall effect"?

(a) Salt solution

(b) Milk

(c) Copper sulphate solution

(d) Starch solution

Answer: Milk and starch solution will show "Tyndall effect" as both of these are colloids.

Question 10: Classify the following into elements, compounds and

mixtures.

(a) Sodium (b) Soil (c) Sugar solution

(d) Silver (e) Calcium carbonate (f) Tin

(g) Silicon (h) Coal (i) Air

(j) Soap (k) Methane (l) Carbon dioxide

(m) Blood

Answer: Elements : Sodium, silver, tin and silicon

Compounds : Calcium carbonate, methane, and carbon dioxide

Mixtures : Soil, sugar solution, coal, air, soap and blood.

Question 11: Which of the following are chemical changes?

(a) Growth of a plant

(b) Rusting of iron

(c) Mixing of iron filings and sand

(d) Cooking of food

(e) Digestion of food

(f) Freezing of water

(g) Burning of a candle

Answer: Growth of a plant, rusting of iron, cooking of food, digestion of food, burning of a candle are chemical changes, because here the chemical composition of substance changes.

( From www.ncrtsolutions.in )

CLASS 9 Science CHAPTER 11 Work And Energy (NCERT Solution)

CLASS 9 Science 
CHAPTER 11 
Work And Energy (NCERT Solution)

Question 1: A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (Fig. 11.3). Let us take it that the force acts on the object through the displacement. What is the work done in this case?                                                                                                                            

Answer: When a force F acts on an object to displace it through a distance S in its direction, then the work done W on the body by the force is given by:                                                                             Work done = Force × Displacement                                                                                                         W = F × S                                                                                                                                              Where,
F = 7 N

S = 8 m

Therefore, work done, W = 7 × 8

= 56 Nm

= 56 J

Question 1: When do we say that work is done?

Answer: Work is done whenever the given conditions are satisfied:

(i) A force acts on the body.

(ii) There is a displacement of the body caused by the applied force along the direction of the applied force.

Question 2: Write an expression for the work done when a force is acting on an object in the direction of its displacement.

Answer: When a force F displaces a body through a distance S in the direction of the applied force, then the work done W on the body is given by the expression:

Work done = Force × Displacement

W = F × s

Question 3: Define 1 J of work.

Answer: 1 J is the amount of work done by a force of 1 N on an object that displaces it through a distance of 1 m in the direction of the applied force.

Question 4: A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?

Answer: Work done by the bullocks is given by the expression:

Work done = Force × Displacement

W = F × d

Where,

Applied force, F = 140 N

Displacement, d = 15 m

W = 140 x 15 = 2100 J

CLASS 9 Science CHAPTER 10 Gravitation (NCERT Solution)

CLASS 9 Science  
CHAPTER 10 
Gravitation (NCERT Solution)

Question 1: State the Universal law gravitation?

Answer: According to Universal Law of Gravitation every object in universe attracts every other object with a force known as Gravitational force. It states that F Force be the force of attraction between two objects P and Q of masses M and m is given by 

F = GMm / r2

Here G is the constant known as Universal Gravitational constant and its value is 6.67 X 10-11 Nm2kg-2.

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

Answer: Let F be the be the force of attraction between earth and an object on its surface. Again assume that Me be the mass of earth and m be the mass of object on the surface of earth. Formula for the magnitude of the gravitational force between the earth and an object on the surface of the earth is given by using Universal Law of Gravitation. So required formula is F = GMm / r2

Page 136

Question 1: What do you mean by free fall?

Answer: Gravity of earth attracts every object towards its center. When an object is dropped from a certain height , it begins to fall towards

Earth’s surface under the influence of gravitational force. Such a motion of object is called free fall.

Question 2: What do you mean by acceleration due to gravity?

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

Page 138

Question 1: what are the differences between mass of an object and its weight?

Answer: Mass:-

1. Mass is the quantity of matter contained in the body

2. It is the measure of inertia of the body

3. It is a constant quantity for any object.

4. It is a scalar quantity and only only has magnitude

5. SI unit of mass is Kg

Weight :-

1. Weigh is the gravitational force acting on the body

2. It is the measure of gravity

3. Weight of an object is not a constant quantity it is different at different laces.

4. It is a vector quantity as it has both magnitude and direction.

5. Its SI unit is N (newton) same as the SI unit of force.

Question 2: Why is the weight of an object on the moon (1/6)th its weight on the earth?

Answer: Gravity is directly related to mass. The more mass an object has, the more gravitational pull it has. Now, the moon is significantly smaller than the earth (in fact, it is about the size of the earth’s core). Gravity is dependent on the size of the object. Your weight on the moon is 1/6 of that on earth because the moon has 1/6 the mass of the earth.

Page 141

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

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

Question 2: What do you mean by buoyancy?

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

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

Answer: If the density of an object is more then the density of the liquid then the object would sink this happens because in this case buoyant force that is acting on the object due to liquid is less than the force of gravity acting on the object. If the density of object is less than the density of the liquid , then it floats on the surface of the liquid , this happens because the buoyant force acting on the object is greater than the force of gravity.

Page 142

Question 1: You find your mass to be 42 Kg on a weighting machine. Is your more or less than 42 K?

Answer: On a weighing machine mass is measured by comparing the weights. So when you weigh your body an upward force (buoyant force) acts on it which causes machine to show your weight less than the actual weight.

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

Answer: Actual weight of any object is equal to the sum of its measured weight and the buoyant force. The bag of cotton is heavier since it is larger than iron bar, so the buoyant force is larger in case of cotton hence actual weight of cotton bag is more and it is heavier.

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CLASS 9 Science CHAPTER 8 Motion (NCERT Solution)

CLASS 9 Science  
CHAPTER 8 
Motion (NCERT Solution)

Page No: 100

Question 1:  An object has moved through a distance. Can it have zero

displacement? If yes, support your answer with an example.

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

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

Answer:

 Given, Side of the square field= 10m

Therefore, perimeter = 10 m x 4 = 40 m

Farmer moves along the boundary in 40s.

Displacement after 2 m 20 s = 2 x 60 s + 20 s = 140 s =?

Since in 40 s farmer moves 40 m

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

Therefore, in 140s distance covered by farmer = 1 × 140 m = 140 m

number of rotation to cover 140 along the boundary= Total Distance / Perimeter

= 140 m / 40 m = 3.5 round

Thus, after 3.5 round farmer will at point C of the field.

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

Question 3: Which of the following is true for displacement?

(a) It cannot be zero.

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

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

Page No: 102

Question 1: Distinguish between speed and velocity.

Answer: Speed

Speed is the distance travelled by an object in a given interval of time.

Speed = distance / time

Speed is scalar quantity i.e. it has only magnitude.

Velocity

Velocity is the displacement of an object in a given interval of time.

Velocity = displacement / time

Velocity is vector quantity i.e. it has both magnitude as well as direction.

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

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

Question 3: What does the odometer of an automobile measure?

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

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

Answer: An object having uniform motion has a straight line path.

Question 5: During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 m s −1 .

Answer: Speed= 3 × 108 m s−1

Time= 5 min = 5 x 60 = 300 sec Distance= Speed x Time

Distance= 3 × 108 m s−1 x 300 sec = 9 x 1010 m

Page No: 103

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

Answer: (i) A body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time.

(ii) A body is said to be in non- uniform acceleration if the rate of change of its velocity is not constant.

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

Answer:  

 

Question 3: A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h−1 in 10 minutes. Find its acceleration.

Answer:

Page No: 107

Question 1: What is the nature of the distance - 'time graphs for uniform and non

uniform motion of an object?

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

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

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

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

Question 3: What can you say about the motion of an object if its speed - 'time graph is a straight line parallel to the time axis?

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

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

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

Page No: 109

Question 1: A bus starting from rest moves with a uniform acceleration of 0.1 m

s −2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Answer: Initial speed of the bus, u= 0

Acceleration, a = 0.1 m/s2

Time taken, t = 2 minutes = 120 s

(a) v= u + at

v= 0 + 0×1 × 120

v= 12 ms–1

(b) According to the third equation of motion:

v2 - u2= 2as

Where, s is the distance covered by the bus

(12)2 - (0)2= 2(0.1) s

s = 720 m

Speed acquired by the bus is 12 m/s.

Distance travelled by the bus is 720 m.

Page No: 110

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

Answer: Initial speed of the train, u= 90 km/h = 25 m/s

Final speed of the train, v = 0 (finally the train comes to rest)

Acceleration = - 0.5 m s-2

According to third equation of motion:

v2= u2+ 2 as

(0)2= (25)2+ 2 ( - 0.5) s

Where, s is the distance covered by the train

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

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

Answer: Initial Velocity of trolley, u= 0 cms-1

Acceleration, a= 2 cm s-2

Time, t= 3 s

We know that final velocity, v= u + at = 0 + 2 x 3 cms-1

Therefore, The velocity of train after 3 seconds = 6 cms-1

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

Answer: Initial Velocity of the car, u=0 ms-1

Acceleration, a= 4 m s-2

Time, t= 10 s

We know Distance, s= ut + (1/2)at2

Therefore, Distance covered by car in 10 second= 0 × 10 + (1/2) × 4 ×102

= 0 + (1/2) × 4× 10 × 10 m

= (1/2)× 400 m

= 200 m

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

Answer: Given Initial velocity of stone, u=5 m s-1

Downward of negative Acceleration, a= 10 m s-2

We know that 2 as= v2- u2

Page No: 112

Excercise

Question 1: An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Answer: Diameter of circular track (D) = 200 m

Radius of circular track (r) = 200 / 2=100 m

Time taken by the athlete for one round (t) = 40 s

Distance covered by athlete in one round (s) = 2π r

= 2 x ( 22 / 7 ) x 100

Speed of the athlete (v) = Distance / Time

= (2 x 2200) / (7 x 40)

= 4400 / 7 × 40

Therefore, Distance covered in 140 s = Speed (s) × Time(t)

= 4400 / (7 x 40) x (2 x 60 + 20)

= 4400 / ( 7 x 40) x 140

= 4400 x 140 /7 x 40

= 2200 m

Number of round in 40 s =1 round

Number of round in 140 s =140/40

= 3 1/2

After taking start from position X ,the athlete will be at position Y after 3 1/2 rounds as shown in figure

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

= Diameter of circular track

= 200 m

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

Answer: Total Distance covered from AB = 300 m

Total time taken = 2 x 60 + 30 s

=150 s

Therefore, Average Speed from AB = Total Distance / Total Time

=300 / 150 m s-1

=2 m s-1

Therefore, Velocity from AB =Displacement AB / Time = 300 / 150 m s-1

=2 m s-1

Total Distance covered from AC =AB + BC

=300 + 200 m

Total time taken from A to C = Time taken for AB + Time taken for BC

= (2 x 60+30)+60 s

= 210 s

Therefore, Average Speed from AC = Total Distance /Total Time

= 400 /210 m s-1

= 1.904 m s-1

Displacement (S) from A to C = AB - BC

= 300-100 m

= 200 m

Time (t) taken for displacement from AC = 210 s

Therefore, Velocity from AC = Displacement (s) / Time(t)

= 200 / 210 m s-1

= 0.952 m s-1

Question 3: Abdul, while driving to school, computes the average speed for his trip to be 20 km h−1. On his return trip along the same route, there is less traffic and the average speed is 40 km h−1. What is the average speed for Abdul’s trip?

Answer:  The distance Abdul commutes while driving from Home to School = S

Let us assume time taken by Abdul to commutes this distance = t1

Distance Abdul commutes while driving from School to Home = S

Let us assume time taken by Abdul to commutes this distance = t2

Average speed from home to school v1av = 20 km h-1

Average speed from school to home v2av = 30 km h-1

Also we know Time taken form Home to School t1 =S / v1av

Similarly Time taken form School to Home t2 =S/v2av

Total distance from home to school and backward = 2 S

Total time taken from home to school and backward (T) = S/20+ S/30

Therefore, Average speed (Vav) for covering total distance (2S) = Total

Distance/Total Time

= 2S / (S/20 +S/30)

= 2S / [(30S+20S)/600]

= 1200S / 50S

= 24 kmh-1

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

Answer: Given Initial velocity of motorboat, u = 0

Acceleration of motorboat, a = 3.0 m s-2

Time under consideration, t = 8.0 s

We know that Distance, s = ut + (1/2)at2

Therefore, The distance travel by motorboat = 0 x 8 + (1/2)3.0 x 8 2

= (1/2) x 3 x 8 x 8 m

= 96 m

Question 5: A driver of a car travelling at 52 km h−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h−1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

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

Distance Travelled by first car before coming to rest =Area of △ OPR

= (1/2) x OR x OP

= (1/2) x 5 s x 52 kmh-1

= (1/2) x 5 x (52 x 1000) / 3600) m

= (1/2) x 5x (130 / 9) m

= 325 / 9 m

= 36.11 m

Distance Travelled by second car before coming to rest =Area of △ OSQ

= (1/2) x OQ x OS

= (1/2) x 10 s x 3 kmh-1

= (1/2) x 10 x (3 x 1000) / 3600) m

= (1/2) x 10 x (5/6) m

= 5 x (5/6) m

= 25/6 m

= 4.16 m

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

(a) Which of the three is travelling the fastest?

(b) Are all three ever at the same point on the road?

(c) How far has C travelled when B passes A? 

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

Answer: 

(a) Object B

(b) No

(c) 5.714 km

(d) 5.143 km

Therefore, Speed = slope of the graph

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

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

On the distance axis:

7 small boxes = 4 km

Therefore,1 small box = 4 / 7 Km

Initially, object C is 4 blocks away from the origin.

Therefore, Initial distance of object C from origin = 16 / 7 Km

Distance of object C from origin when B passes A = 8 km

Distance covered by C

Page No: 113

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

Answer: Let us assume, the final velocity with which ball will strike the ground be 'v' and time it takes to strike the ground be 't'

Initial Velocity of ball, u =0

Distance or height of fall, s =20 m

Downward acceleration, a =10 m s-2

As we know, 2as =v2-u2

v2 = 2as+ u2

= 2 x 10 x 20 + 0

= 400

∴ Final velocity of ball, v = 20 ms-1

t = (v-u)/a

∴Time taken by the ball to strike = (20-0)/10

= 20/10

= 2 seconds

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

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

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

Answer:  

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

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

Question 10: State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity.

(b) an object moving in a certain direction with an acceleration in the perpendicular direction.

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

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

Question 11: An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Answer: Radius of the circular orbit, r = 42250 km

Time taken to revolve around the earth, t = 24 h

Speed of a circular moving object, v = (2π r)/t

= [2× (22/7)×42250 × 1000] / (24 × 60 × 60)

= (2×22×42250×1000) / (7 ×24 × 60 × 60) m s-1

= 3073.74 m s -1

( From www.ncrtsolutions.in )