Category: Class 9

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

Improvement in Food Resources Intext Questions

Improvement in Food Resources Intext Questions Page No. 204

Question 1.
What do we get from cereals, pulses, fruits and vegetables?
Answer:
We get carbohydrates from cereals and proteins from pulses. Fruits and vegetables give us lots of vitamins and minerals.

Improvement in Food Resources Intext Questions Page No. 205

Question 1.
How do biotic and abiotic factors affect crop production?
Answer:
Biotic factors like pests, insects etc. reduce the crop production. Pests harm crops by feeding over them. Weeds also reduce crop productivity by competing with the main crop for nutrients and light. Abiotic factors like wind, rain, temperature etc. impact the overall crop production. For example, droughts and floods almost completely destroy the whole crop of a particular area.

Question 2.
What are the desirable agronomic characteristics for crop improvements?
Answer:
The desirable agronomic characteristics for crop improvements are:

1. Tallness and profuse branching in any fodder crop.
2. Dwarf characteristics in cereals.

Improvement in Food Resources Intext Questions Page No. 206

Question 1.
What are macro – nutrients and why are they called macro – nutrients?
Answer:
Macro – nutrients are those nutrients which are required in large quantities for growth and development of plants. Since they are required in large quantities, they are known as macro – nutrients. The six macro – nutrients required by plants are nitrogen, phosphorus, potassium, calcium, magnesium and sulphur.

Question 2.
How do plants get nutrients?
Answer:
Plants consume nutrients from air, water and soil. Soil is one of the important sources of nutrients. Plants get thirteen nutrients from soil. The remaining three nutrients (carbon, oxygen, and hydrogen) are obtained from air and water.

Improvement in Food Resources Intext Questions Page No. 207

Question 1.
Compare the use of manure and fertilizers in maintaining soil fertility.
Answer:
Manures are biodegradable substances which increase soil fertility by enriching the soil with organic matter and nutrients as it is prepared by the decomposition of animal excreta and plant wastes.

On the other hand, fertilizers are mostly artificially developed inorganic compounds whose excessive use is harmful to ecosystems components and soil fertility. Hence, fertilizers are considered good only for short term use.

Improvement in Food Resources Intext Questions Page No. 208

Question 1.
Which of the following conditions will give the most benefits? Why?
(а) Farmers use high – quality seeds, do not adopt irrigation or use fertilizers.
(b) Farmers use ordinary seeds, adopt irrigation and use fertilizer.
(c) Farmers use quality seeds, adopt irrigation, use fertilizer and use crop protection measures.
Answer:
(c) Farmers using good quality seeds, adopting irrigation, using fertilizers and using crop protection measures will get most benefits.
Reasons are as follows:

1. Increases the total crop production as good quality seeds, then a majority of the seeds will germinate properly, and will grow into a healthy plants.
2. Proper irrigation will improve the water availability to crops.
3. Fertilizers promote high growth and development in plants by providing the essential nutrients such as nitrogen, phosphorus, potassium etc.
4. Crop protection measures include various methods to control weeds, pests and infectious agents.

Improvement in Food Resources Intext Questions Page No. 209

Question 1.
Why should preventive measures and biological control methods be preferred for protecting crops?
Answer:
Preventive measures and biological control methods should be preferred for protecting crops because excessive use of chemicals leads to environmental problems. Biological methods harm neither crops nor environment.

Question 2.
What factors may be responsible for losses of grains during storage?
Answer:
Factors causing loss of grains during storage:

1. Biotic Factors: Insects, rodents, fungi and bacteria etc.
2. Abiotic Factors: moisture content and temperature etc.

Improvement in Food Resources Intext Questions Page No. 210

Question 1.
Which method is commonly used for improving cattle breeds and why?
Answer:
Cross breeding is commonly used for improving cattle breeds. Cross breeding between two good varieties of cattle will produce a new improved variety. For example, the cross between foreign breeds such as Jersey Brown, Swiss and Indian breeds such as Red Sindhi, Sahiwal produces a new variety having qualities of both breeds.

Improvement in Food Resources Intext Questions Page No. 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:
Poultry in India is the most efficient converter of low fibre food stuff in nutritious animal protein food. In poultry farming, domestic fowls are raised to produce eggs and chicken. For this, the fowls are given animal feeds in the form of roughage, which mainly consists of fibres. Thus, by feeding animals a fibre rich diet, the poultry gives highly nutritious food in the form of eggs and chicken.

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

1. Proper shelter facilities and their regular cleaning.
2. Some basic hygienic conditions such as clean water, nutritious food etc.
3. Animals are kept in spacious, airy and ventilated place.
4. Prevention and cure of diseases at the right time is ensured.

Question 3.
What are the differences between broilers and layers and in their management?
Answer:
Layers are meant for egg production, whereas broilers are meant for poultry meat. Nutritional, environmental and housing conditions required by broilers are different from those required by egg layers. A broiler chicken, for their proper growth, requires vitamin rich supplements especially vitamin A and K. Also, their diet includes protein rich food and enough fat. They also require extra care and maintenance to increase their survival rate in comparison to egg layers.

Improvement in Food Resources Intext Questions Page No. 213

Question 1.
How are fish obtained?
Answer:
Fish can be obtained by two ways:

1. Capture fishing: It is the process of obtaining fish from natural resources.
2. Culture fishery: It is the practice of farming fishes. Farming can be done in both freshwater ecosystem (which includes river water, pond water) and marine ecosystem.

Question 2.
What are the advantages of composite fish culture?
Answer:
The advantages of composite fish culture are:

1. Fish can be grown in crop fields especially paddy.
2. Intensive fish farming is possible because plenty of water is available during crop seasons.
3. In this system, both local and imported fish species can be cultivated.

Question 3.
What are the desirable characters of bee varieties suitable for honey production?
Answer:
Bee varieties having the following desirable characters are suitable for honey production:

1. They should yield high quantity of honey.
2. They should not sting much.
3. They should stay in the beehive for long durations.
4. They should breed very well.

Question 4.
What is pasturage and how is it related to honey production?
Answer:
Pasturage is the availability of flowers from which bees collect nectar and pollen. It is related to the production of honey as it determines the taste and quantity of honey.

Improvement in Food Resources NCERT Textbook Exercises

Question 1.
Explain any one method of crop production which ensures high yield.
Answer:
Inter – cropping method ensures high yield of crop production. It is a practice of growing two or more crops simultaneously in the same field in rows. In inter – cropping, definite row patterns are followed such as one row of main crop is followed by two rows of inter – crop. In inter – cropping, there is a greater utilisation of the inter – spaced area, light, nutrients, water and air. As a result, productivity per unit area is increased.

Question 2.
Why are manures and fertilizers used in fields?
Answer:
Manures and fertilizers are used in fields to enrich the soil with the required nutrients. Manure helps in enriching the soil with organic matter and nutrients. This improves the fertility and structure of the soil. On the other hand, fertilizers ensure a healthy growth and development in plants. They are a good source of nitrogen, phosphorus and potassium. To get an optimum yield, it is instructed to use a balanced . combination of manures and fertilizers in the soil.

Question 3.
What are the advantages of inter-cropping and crop rotation?
Answer:
Inter – cropping and crop rotation, both are used to get maximum benefit on limited land. Inter – cropping helps in preventing pests and diseases to spread throughout the field. It also increases soil fertility, whereas crop rotation prevents soil depletion, increases soil fertility and reduces soil erosion. Both of these methods reduce the need for fertilizers. They also helps in controlling weeds and the growth of pathogens and pests in crops.

Question 4.
What is genetic manipulation? How is it useful in agricultural practices?
Answer:
Genetic manipulation is a process where the gene for a particular character is introduced inside the chromosome of a cell. When the gene for a particular character is introduced in a plant cell, a transgenic plant is produced. These transgenic plants exhibit characters governed by the newly introduced gene.

Genetic manipulation is useful in developing varieties with higher yield, good quality, biotic and abiotic resistance, short maturity period, wider adaptability and desirable agronomic characteristics.

Question 5.
How do storage grain losses occur?
Answer:
There are many biotic and abiotic factors that harm stored grains and result in degradation, poor germinability, discolouration etc. which leads to storage grain losses. Biotic factors like insects or pests cause direct damage by feeding on seeds.

They also deteriorate and contaminate the grain, making it unfit for further consumption.  Abiotic factors such as temperature, light, moisture etc., also affect storage food. They decrease the germinating ability of the seeds and make them unfit for future use by farmers. Unpredictable occurrence of droughts and floods also causes destruction of crops.

Question 6.
How do good animal husbandry practices benefit farmers?
Answer:
Cattle farming is one of the methods of animal husbandry that is most beneficial for farmers. Better breeds of draught animals can be produced. Such draught animals are engaged in agricultural fields for labour such as carting, irrigation, tilling, etc.

Question 7.
What are the benefits of cattle farming?
Answer:
Benefits of cattle farming:

1. Good quality and quantity of dairy products can be produced.
2. Draught labour animals can be raised for agricultural work.
3. New breed of animals that are resistant to diseases can be raised by crossing two breeds with the desired traits.

Question 8.
For increasing production, what is common in poultry, fisheries and bee – keeping?
Answer:
Proper management techniques are the common factors for increasing production in poultry, fisheries and bee keeping. Regular cleaning of farms is of importance. Also, maintenance of temperature and prevention and cure of diseases in farming is also required to increase the growth of animals.

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

1. Capture fishing is the method to obtain fishes from natural resources like rivers, ponds, waterfalls etc.
2. Mariculture is the culture of marine fishes for commercial use.
3. Aquaculture involves the production of aquatic animals that are of high economic value such as prawns, lobsters, fishes, crabs etc.

Improvement in Food Resources Additional Questions

Improvement in Food Resources Multiple Choice Questions

Question 1.
Which one is an oil yielding plant among the following?
(a) Lentil
(b) Sunflower
(c) Cauliflower
(d) Hibiscus.
Answer:
(b) Sunflower

Question 2.
Which one is not a source of carbohydrate?
(a) Rice
(b) Millets
(c) Sorghum
(d) Gram.
Answer:
(d) Gram.

Question 3.
Find out the wrong statement from the following:
(a) White revolution is meant for increase in milk production.
(b) Blue revolution is meant for increase in fish production.
(c) Increasing food production without compromising with environmental quality is called as sustainable agriculture.
(d) None of the above.
Answer:
(d) None of the above.

Question 4.
To solve the food problem of the country, which among the following is necessary?
(a) Increased production and storage of food grains.
(b) Easy access of people to the food grain.
(c) People should have money to purchase the grains.
(d) All of the above.
Answer:
(d) All of the above.

Question 5.
Find out the correct sentence.
(i) Hybridisation means crossing between genetically dissimilar plants.
(ii) Cross between two varieties is called as inter specific hybridisation.
(iii) Introducing genes of desired character into a plant gives genetically modified crop.
(iv) Cross between plants of two species is called as inter varietal hybridisation.
(a) (i) and (ii)
(b) (ii) and (iv)
(c) (ii) and (iii)
(d) (iii) and (iv).
Answer:
(a) (i) and (ii)

Question 6.
Weeds affect the crop plants by
(a) killing of plants in field before they grow.
(b) dominating the plants to grow.
(c) competing for various resources of crops (plants) causing low availability of nutrients.
(d) all of the above.
Answer:
(c) competing for various resources of crops (plants) causing low availability of nutrients.

Question 7.
Which one of the following species of honey bee is an Italian species?
(a) Apis dorsata
(b) Apis florae
(c) Apis cerana indica
(d) Apis mellifera.
Answer:
(d) Apis mellifera.

Question 8.
Find out the correct sentence about manure
(ii) Manure contains large quantities of organic matter and small quantities of nutrients.
(ii) It increases the water holding capacity of sandy soil.
(iii) It helps in draining out of excess of water from clayey soil.
(iv) Its excessive use pollutes environment because it is made of animal excretory waste.
(a) (i) and (iii)
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (iii) and (iv).
Answer:
(b) (i) and (ii)

Question 9.
Cattle husbandry is done for the following purposes:
(i) Milk production
(ii) Agricultural work
(iii) Meat production
(iv) Egg production.
(a) (i), (ii) and (iii)
(b) (ii), (iii) and (iv)
(c) (iii) and (iv)
(d) (i) and (iv).
Answer:
(a) (i), (ii) and (iii)

Question 10.
Which of the following are Indian cattle?
(i) Bos indicus
(ii) Bos domestica
(iii) Bos bubalis
(iv) Bos vulgaris.
(a) (i) and (iii)
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (iii) and (iv).
Answer:
(a) (i) and (iii)

Question 11.
Which of the following are exotic breeds?
(i) Brawn
(ii) Jersey
(iii) Brown Swiss
(iv) Jersey Swiss.
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (ii) and (iv).
Answer:
(b) (ii) and (iii)

Question 12.
Poultry farming is undertaken to raise the following:
(a) Egg production
(b) Feather production
(c) Chicken meat
(d) Milk production.
Answer:
(a) Egg production

Question 13.
Poultry fowl are susceptible to the following pathogens:
(a) Viruses
(b) Bacteria
(c) Fungi
(d) All of the above.
Answer:
(d) All of the above.

Question 14.
Which one of the following fishes is a surface feeder?
(a) Rohus
(b) Mrigals
(c) Common carps
(d) Catlas.
Answer:
(d) Catlas.

Question 15.
Animal husbandry is the scientific management of ___________ .
(i) animal breeding
(ii) culture of animals
(iii) animal livestock
(iv) rearing of animals
(a) (i), (ii) and (iii)
(b) (ii), (iii) and (iv)
(c) (i), (ii) and (iv)
(d) (i), (iii) and (iv).
Answer:
(d) (i), (iii) and (iv).

Improvement in Food Resources Very Short Answer Type Questions

Question 1.
Match the column A with the column B

Answer:
(a) (ii)
(b) (iii)
(c) (i)
(d) (iv).

Question 2.
Fill in the blanks.

1. Pigeon pea is a good source of ___________ .
2. Berseem is an important ___________ crop.
3. The crops which are grown in rainy season are called ___________ crops.
4. ___________ are rich in vitamins.
5. ___________ crop grows in winter season.

Answer:

1. protein
2. fodder
3. Kharif
4. vegetables
5. Rabi.

Question 3.
State any one importance of photoperiod in agriculture.
Answer:
Photoperiod is important for growth of plants and flowering.

Question 4.
What are desirable traits for fodder crop?
Answer:

1. Profused branching.
2. Tall Plants.

Question 5.
White revolution in India has made us self – sufficient in a product. Name it.
Answer:
Milk.

Improvement in Food Resources Short Answer Type Questions

Question 1.
What is a GM crop? Name any one such crop which is grown in India.
Answer:
Crop developed artificially with the help of modern biotechnology to obtain the desired character is called as genetically modified(GM) crop. But, cotton and egg plant are examples of GM crops which are resistant to many harming biological factors.

Question 2.
List out some useful traits in improved crop.
Answer:
Useful traits in improved crops are as follows:

1. better yield.
2. improved quality.
3. resistance to biotic and abiotic factors.
4. better adaptability.
5. better economically.

Question 3.
Why is organic matter important for crop production?
Answer:
Organic matter is important for crops because a fertile and healthy soil is the basis for healthy plant’s growth and soil organic matter works as a foundation for healthy and productive soils.
It helps soil by:

1. improving soil structure.
2. increasing water holding capacity of soil.
3. drainage and in avoiding water logging.
4. giving shelter to number of animals which also improve soil’s nutrients value.

Question 4.
Why is excessive use of fertilizers detrimental for environment?
Answer:
Excessive use of fertilizers is harmful because it is artificial in origin and accumulate in biological cycle and disturb its balance which causes environmental pollution as their residual and unused amounts will become pollutants for air, water and soil.

Question 5.
Give one word for the following:

1. Farming without the use of chemicals as fertilizers, herbicides and pesticides is known as ___________ .
2. Growing of wheat and groundnut on the same field is called as ___________ .
3. Planting soyabean and maize in alternate rows in the same field is called as ___________ .
4. Growing different crops on a piece of land in preplanned succession is known as ___________ .
5. Xanthium and Parthenium are commonly known as ___________ .
6. Causal organism of any disease is called as

Answer:

1. organic farming
2. mixed farming
3. inter – cropping
4. crop rotation
5. weeds
6. pathogens

Question 6.
Match the following A and B:

Answer:
(a) (iii)
(b) (v)
(c) (iv)
(d) (i)
(e) (ii).

Question 7.
Define the term hybridization and photoperiod.
Answer:

1. Hybridization: Hybridization is the process in which two genetically different individuals are crossed to result generally into an individual with a desired trait.
2. Photoperiod: Photoperiod in plants is defined as the developmental responses towards the comparative lengths of light and dark periods in its region.

Question 8.
Fill in the blanks:

1. Photoperiod affects the ___________ .
2. Kharif crops are cultivated from ___________ to ___________ .
3. Rabi crops are cultivated from ___________ to ___________ .
4. Paddy, maize, green gram and black gram are ___________ crops.
5. Wheat, gram, pea, mustard are ___________ crops.

Answer:

1. flowering
2. June to October
3. November to March,
4. Kharif
5. Rabi

Question 9.
Cultivation practices and crop yield are related to environmental condition. Yes/No? Give reasons.
Answer:
Environmental conditions are the components which influence plant growth and genetic factors. Environmental factors include everything in our surrounding i.e. air, water, temperature, atmospheric pressure, surrounding gases etc. When we cultivate any plant or consider crop growth, we check all the above environmental factors before sowing. Hence yes, cultivation practices and crop yield are related to environmental condition.

Question 10.
Fill in the blanks:

1. A total of ___________ nutrients are essential to plants.
2. ___________ and ___________ are supplied by air to plants.
3. ___________ is supplied by water to plants.
4. Soil supplies ___________ nutrients to plants.
5. nutrients are required in large quantity and called as ___________ .
6. nutrients are needed in small quantity for plants and are called ___________ .

Answer:

1. sixteen
2. Oxygen and carbon
3. Hydrogen
4. thirteen
5. Six, macro – nutrients
6. Seven, micronutrients

Improvement in Food Resources Long Answer Type Questions

Question 1.
Define:

1. Vermicompost
2. Green manure
3. Bio fertilizer.

Answer:

1. Vermicompost: Vermicomposting is the process in which worms and micro – organisms are used to turn organic waste into a black, earthy – smelling, nutrient – rich humus.
2. Green manure: When crop or plant that is grown and then intentionally plowed under to improve the underlying soil, it is called green manure.
3. Bio fertilizer: Fertilizer created from biological components as small plants.

Question 2.
Discuss various methods for weed control.
Answer:
Unwanted plants which grow along with crops are called weeds. They compete with the crop plants for light, water, space and minerals, and harm their growth and yield. Some of them produce poisonous substances which can harm the growth of the crop or poison the produce.

Removing weeds from a field is called weeding. It can be done Manually or by using an implement like a trowel, hoe or rake. A third way of getting rid of weeds is by spraying chemicals called weedicides. These chemicals destroy the weeds but not the crop plants.

Improvement in Food Resources Higher Order Thinking Skills (HOTS)

Question 1.
Why should our food contain cereals, pulses, fruits and vegetables?
Answer:

1. Cereals provide carbohydrates for energy requirements.
2. Pulses give proteins.
3. Fruits and vegetables provide various vitamins and minerals.

Question 2.
Why moisture level of food grains should be checked before their storage?
Answer:
Before storage, moisture level of food grains should not be more than 9% otherwise growth of micro – organisms (bacteria and fungi) will take place causing either discolouration or complete spoilage of food grains.

Improvement in Food Resources Value Based Questions

Question 1.
A group of Eco Club students made a compost pit in the school. They collected all bio – degradable waste from the school canteen and used it to prepare the compost

1. Name two waste that can be used for the compost and two wastes obtained from canteen which cannot be used for the compost making.
2. What is the other important compound required for making the compost?
3. State the values of Eco Club students.

Answer:

1. Two waste used for compost are vegetable peels, fruit peels and pulps. Two waste materials that cannot be used as compost are pickels and curd.
2. Bacteria present in soil are used as component for making compost.
3. Eco club students reflect the value of group workers and responsible citizens.

Question 2.
In a poultry farm, large number of birds died. Farm owner was unable to identify the reason and thought someone has done it intentionally. Aryan living nearby his farm noticed that the conditions in the farm were very unhygienic and poor ventilation was there.

1. What do you think can be the reason for these deaths?
2. What measures can be followed to prevent this in future?
3. What values are shown by Aryan?

Answer:

1. The conditions of the farm were unhygienic and ventilation was also poor. These may be the possible reasons.
2. These conditions should be improved. Proper food, hygiene practices and good ventilation should be there to prevent this.
3. The values of awareness and carefullness are shown by Aryan.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.1

Question 1.
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

Solution:
(i) Base DC, Parallels DC and AB
(iii) Base QR, Parallels QR and PS
(v) Base AD, Parallels AD and BQ.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 8 Quadrilaterals Ex 8.2

Question 1.
ABCD is quadrilateral in which P, Q, R and S are mid-point^jf the sides AB, BC, CD and DA (see Fig.). AC is a diagonal. Show that:

1. SR ∥ AC and SR = \(\frac{1}{2}\) AC
2. PQ = SR
3. PQRS is a parallelogram.

Solution:
Given
ABCD is in which P, Q, R and S are the mid-points of sides AB,BC, CD and DA.
To prove.

1. SR ∥ AC and SR= \(\frac{1}{2}\) AC
2. PQ = SR
3. PQRS is a parallelogram.

Proof:
In ∆ABC. P is the mid point of AB and Q is the midpoint of BC.
∴ PQ ∥ AC and PQ = \(\frac{1}{2}\) AC (by MPT)…(1)
In ∆ADC, S is the mid-point of AD and R is the mid – point of DC.
SR ∥ AC and SR = \(\frac{1}{2}\) AC (by MPT)…(2)

1. SR ∥ AC and SR = \(\frac{1}{2}\) AC (proved)
2. PQ ∥ AC and SR = \(\frac{1}{2}\) AC
∴ PQ = SR
3. From (1) and (2), we get
PQ ∥ SR and PQ = SR
∴ PQRS is a parallelogram

Question 2.
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral P&RS is a rectangle.
Solution:
Given
ABCD is rhombus and P, Q and R and S are the mid-points of sides AB, BC, CD and DA.
To prove
PQRS is a rectangle.
Construction:
Join AC and BD).

Proof:
In ∆ABC, P is the midpoint of AB and Q is the midpoint of BC.
∴ PQ ∥AC (By MPT) …(1)
In ∆ADC, S is the midpoint of AD and R is the midpoint of DC.
SR ∥ AC (By MPT) …(2)
From (1) and (2), we get
PQ ∥ SR …(3)
In ∆ABD, P is the midpoint of AB and S is the midpoint of AD.
PS ∥ BD …(4)
In ABCD, Q is the midpoint of BC and R is the midpoint of CD
∴ QR ∥ BD (ByMPT) …(5)
From (4) and (5), we get
PS ∥ QR …(6)
In quadrilateral PQRS, PQ ∥ SR and PS ∥ QR
∴ PQRS is a parallelogram.
In quadrilateral OESF
SF ∥ EO (∴ SR ∥ AO …(7)
SE ∥ FO (∴ SP ∥ BD) …(8)
SEOF is a parallelogram.
We know that, that in a parallelogram opposite angles are equal.
∴ ∠ESF = ∠EOF= 90°
(∴ In a rhombus, diagonals intersect each other at right angles)
Hence, PQRS is a rectangle.

Question 3.
ABCD is a rectangle and P,Q,R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Given
ABCD is a rectangle P,Q,R and S are the mid-points of AB, BC, CD and DA.
To prove.
PQRS is a rhombus.
Proof:
In ∆ABC, P is the midpoint of AB and Q is the midpoint of BC.

PQ ∥ AC and PQ = \(\frac{1}{2}\) AC (By MPT) ….(1)
In ∆ADC, S is the midpoint of AD and R is the midpoint of DC.
∴ SR ∥ AC and SR = \(\frac{1}{2}\) AC (By MPT)…(2)
From (1) and (2),we get
PQ ∥ SB and PQ = SR
PQRS is a parallelogram
AD = BC (∴ ABCD is a rectangle)
⇒ \(\frac{1}{2}\) AD = \(\frac{1}{2}\) BC
AS = BQ
In ∆PAS and ∆PBQ, AS = BQ (proved)
AP = BP P is the mid-point of AB
∠A = ∠B (each 90°)
and so PS = PQ (ByCPCT)
In a ∥^gm, if adjacent sides are equal, then it a rhombus.
∴ PQRS is a rhombus.

Question 4.
ABCD is a trapezium in which AB ∥ DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC and F (see Fig. below). Show that F is the mid-point of BC.

Solution:
Given
AB ∥ DC, DE = AE and EF ∥ AS.
To prove
F is the mid-point of BC
Proof:
AB ∥ DC (given) …(1)
AB ∥ EF (given) …(2)

From (1) and (2) we get,
DC ∥ EF
In ∆ABD, E is the mid – point of AD
EP ∥ AB (∴ EF∥AB)
∴ P is the mid – point of BD (By CMPT)
In ∆BCD, P is the midpoint of BD.
PF ∥ DC (∴ EF ∥ DC)
F is the midpoint of BC (By CMPT).

Question 5.
In a parallelogram ABCD, E and F are the mid-point of sides AB and CD respectively (See Fig.). Show that the line segments AD and EC trisect the diagonal BD.

Solution:
Given
ABCD is a ∥^gm, E and F are the mid-point of AB and CD.
To prove:
BQ = PQ = DP
Proof:
AB ∥ DC and AB = DC (∴ ABCD is a ∥gm)
⇒ \(\frac{1}{2}\) ∥ AB \(\frac{1}{2}\) DC and \(\frac{1}{2}\) AB = \(\frac{1}{2}\)DC
⇒ AE ∥ FC and AE = FC (∴ \(\frac{1}{2}\) AB = AE and \(\frac{1}{2}\) DC = FC)
AECF is a parallelogram (∴ AECF is a ∥^gm)
In ∆ABP, E is the mid – point of AB,
EQ ∥AP (∴ AECF is a ∥<sup<gm)

∴ Q is the midpoint of BP (∴ by CMPT) …(1)
BQ = PQ
In ∆DQC, F is the midpoint of DC
EF ∥ CQ AECF is a ∥<sup<gm)
P is the midpoint of DQ
i.e., DP = PQ
From (1) and (2), we get BQ = PQ = DP.

Question 6.
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Solution:
Given
ABCD is a in which P, Q.R and S are the mid-points of AB, BC, CD and DA respectively.
To prove
PR and SQ bisect each other.
Construction:
Join AC. Join PQ, QR, RS and SP.

Proof:
In ∆ABC, P is the midpoint of AB and Q is the midpointof BC.
PQ ∥ AC and PQ = \(\frac{1}{2}\) AC (By MPT) …..(1)
In ∆ADC, S is the midpoint of AD and R is the midpoint of DC.
SR ∥ AC and SR = \(\frac{1}{2}\) AC (by MPT) …(2)
Form (1) and (2), we get
PQ ∥ SR and PQ = SR
∴ PQRS is a parallelogram in which PR and SQ are diagonals^ and so PR and SQ bisect each other (In a ∥^gm, diagnaols bisect each other)

Question 7.
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

1. D is the mid – point of AC
2. MD ⊥ AC
3. CM = MA = \(\frac{1}{2}\) AB.

Solution:
Given
ABC is a right ∆. ∠C = 90°
AM = BM and MD ∥ BC.
To prove:

1. D is the mid-point of AC, i.e., AD – CD
2. MD ⊥ AC
3. CM = MA = \(\frac{1}{2}\) AB

Proof:
1. In ∆ACB, M is the mid-point of AB and MD ∥ BC
∴ D is the mid – point of AC (By CMPT)
i.e., AD = CD

2. MD ∥ BC and AC is the transversal
∴ ∠ADM = ∠ACB (CA’s)
⇒ ∠ADM = 90° (∠ACB = 90°)
∴ MD ⊥ AC

3. In ∆ADM and ∆CDM,
AD = CD (proved)
∠ADM = ∠CDM (each 90°)
[∴ ∠ADM + ∠CDM = 180° (LPA’s); 90° + ∠CDM= 180°; ∠CDM = 90°]
MD = MD (common)
∴ ∆ADM = ∆CDM (By SAS)
and so MA = MC (By cpct) …(1)
MA = \(\frac{1}{2}\) AB (Given) …(2)
Form (1) and (2), we get MA = MC = \(\frac{1}{2}\) AB.

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 8 Motion

Motion Intext Questions

Motion Intext Questions 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, if an object start from a point and finally, reaches the same point. Then, it has distance but displacement is 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?
Answer:

Suppose farmer starts from A of the square field.
Now, he covered 10 m in 40 sec.
Now, 2 minutes 20 seconds =140 sec.
So, he cover in 140 seconds = \(\frac { 10 }{ 40 }\) × 140 m
= \(\frac { 140 }{ 4 }\)m
= 35 m.
Now, in square field = 35m
AB + BC + CD + DE (\(\frac { 1 }{ 2 }\)AD)
So, he will reach at E.
Then, his displacement is AE = 5 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:

1. False
2. False.

Motion Intext Questions Page No. 102

Question 1.
Distinguish between speed and velocity.
Answer:

Question 2.
Under what condition (s) is magnitude of average velocity of an object equal to its average speed?
Answer:
When a body is moving in a straight line in a particular direction.

Question 3.
What does the odometer of an automobile measure?
Answer:
It measures the distance travelled by the vehicle.

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

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.
Answer:
Time taken by signal to reach ground station = 5 minutes
= 5 × 60 sec.
= 300 seconds.
Speed of the signal = 3 × 10⁸ m/s.
∴ Distance between spaceship and ground station = speed × time
= 3 × 10⁸ × 300
= 9 × 10¹⁰ m.

Motion Intext Questions Page No. 103

Question 1.
When will you say a body is in

1. Uniform acceleration?
2. Non – uniform acceleration?

Answer:

1. A body is in uniform acceleration if its velocity changes uniformly with equal intervals of time. Example: Freely falling object.
2. A body is in non – uniform acceleration if its velocity changes non – uniformly with equal intervals of time. Example: A bus running in a city.

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:
Initial velocity of the bus, u = 80 km/h
= \(\frac { 80\times 100 }{ 60\times 60 } \) m/s
u = \(\frac { 800 }{ 36} \) m/s
Final velocity of the bus, v = 60 km/h
= \(\frac { 60\times 100 }{ 60\times 60 } \) m/s
v = \(\frac { 600 }{ 36 } \) m/s
Time taken by bus to change its speeds, t = 5 seconds
∴ Acceleration, a = \(\frac { v-u }{ t } \)
= \(\frac { 600 }{ 36 } \) – \(\frac { 800 }{ 36 } \) ÷ 5
= \(\frac { 600-800 }{ 36 } \) ÷ 5
= \(\frac { 200 }{ 36 } \) × \(\frac { 1 }{ 5 } \) m/s²
= \(\frac { -10 }{ 9 } \) m/s²
a = -1.11 m/s².

Question 3.
A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h^-1 in 10 minutes. Find its acceleration.
Answer:
Initial velocity, u = 0
Final velocity, v = 40 km/h
= \(\frac { 40\times 1000 }{ 60\times 60 } \) m/s
= \(\frac { 400 }{ 36 } \) m/s
= \(\frac { 200 }{ 18 } \) m/s.
Time taken, t = 10 minutes
= 10 × 10 seconds = 100 seconds
a = \(\frac { v-u }{ t } \)
= \(\frac { 200 }{ 18 } \) – 0 ÷ 100
= \(\frac { 200 }{ 18 } \) ÷ 100
= \(\frac { 200 }{ 18 } \) × \(\frac { 1 }{ 100 } \) m/s²
= \(\frac { 1 }{ 9 } \) m/s²
a = 0.018 m/s².

Motion Intext Questions Page No. 107

Question 1.
What is the nature of the distance – time graphs for uniform and non – uniform motion of an object?
Answer:
For uniform motion, it is a straight line.

For non – uniform motion, it is a curved line.

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:
It shows the body 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:
It means body has constant speed or uniform motion with zero acceleration.

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

Motion Intext Questions Page No. 109 – 110

Question 1.
A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes. Find
(a) the speed acquired,
(b) the distance travelled.
Answer:
Initial velocity, u = 0
Acceleration, a = 0.1 m/s²
Time, t = 2 min = 2 × 60 seconds
= 120 seconds
(a) Velocity, v = ?
We know, v = u + at
v = 0 + 0.1 × 120 m/s
= 12 m/s.

(b) Distance, s = ?
We Know, s = ut + \(\frac { 1 }{ 2 }\) at²
= 0 × 120 + \(\frac { 1 }{ 2 }\) × 0.1 × 120 × 120
= \(\frac { 1 }{ 2 }\) × \(\frac { 0.1 }{ 10 }\) × 120 × 120 m
= 720 m.

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/s². Find how far the train wilt go before it is brought to rest.
Answer:
Initial velocity, u = 90 km/h
= \(\frac { 90\times 5 }{ 18 } \) m/s
= 25 m/s
Final velocity, v = 0
Acceleration, a = -0.5 m/s².
Distance, s = ?
We know,
v² = u² + 2as
0² = (25)² + 2 × (-0.5) × s
0 = 625 – 1 × s
= 625 – 1 × s
= 625 – s
∴ s = 625 m.
∴ Distance travelled by the train = 625 m.

Question 3.
A trolley while going down an inclined plane, has an acceleration of 2 cm/s². What will be its velocity in 3 s after the start?
Answer:
Acceleration, a = 2 cm/s²
Time, t = 3 sec.
Initial velocity, u = 0
Final velocity, v = ?
Now, v= u + at
v = 0 + 2 × 3
= 6 cm/s.

Question 4.
A racing car has a uniform acceleration of 4 m/s². What distance will it cover in 10 s after start?
Answer:
Acceleration, a = 4 m/s²
Time, t = 10 sec,
Initial velocity, u = 0
Distance, s = ?
Now, s = ut + \(\frac { 1 }{ 2 }\) at²
s = 0 × 10 + \(\frac { 1 }{ 2 }\) × 4 × (10)²
s = \(\frac { 1 }{ 2 }\) × 4 × 100 m
s = 200 m.
So, the distance covered is 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:
Initial velocity, u = 5 m/s
Final velocity, v = 0
Acceleration, a = -10 m/s² (∴ It is from opposite direction)
Height attained, s = ?
Time taken, t = ?
Now, V² = u² + 2 as
(0)² = (5)² + 2 × (-10) × s
0 = 25 – 20s
⇒ 20s = 25
∴ s = \(\frac { 25 }{ 20 }\) = \(\frac { 5 }{ 4 }\) = 1025m
Now, v = u + at
0 = 5 + (-10) × t
0 = 5 – 10 × t
-5 = -10 × t
∴ t = \(\frac { -5 }{ -10 }\) = \(\frac { 1 }{ 2 }\) = 0.5 sec
So, height attained by the stone is 1.25 m and time taken to reach there is 0.5 sec.

Motion NCERT Textbook Exercises

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 the track = 200 m
So, radius = \(\frac { 200 }{ 2 }\) m
= 100m
Circumference = 2πr
= 2 × \(\frac { 22 }{ 7 }\) × 100 m
= \(\frac { 4400 }{ 7 }\) m.
Now, athlete covers 200 m in 40 s.
So, rounds covered in 40 seconds = 1
Rounds covered in 1 sec = \(\frac { 1 }{ 40 }\)
Rounds covered in 140 sec = \(\frac { 1 }{ 40 }\) × 140
= 3.5 rounds
So, distance covered = 3.5 × circumference
= \(\frac { 3.5 }{ 10 }\) × \(\frac { 4400 }{ 7 }\) m
= 2200 m.
And, displacement after 3.5 rounds
= Diameter of 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:

(a) A to B,
Time taken, t = 2 min, 30 sec
= 150 sec.

= \(\frac { 300 }{ 150 }\) m/s = 2 m/s

= \(\frac { 300 }{ 150 }\) m/s = 2 m/s.

(b) A to C,
Time taken = Time from A to B + Time from B to C
= 2 min 30 sec + 1 min
= 150 sec + 60 sec
= 210 sec.
Total distance covered (AB + BC) = (300 + 100) m
= 400 m
Displacement = AC = (300 – 100) m
= 200 m

= \(\frac { 400 }{ 210 }\) m/s = 1.9 m/s

= \(\frac { 200 }{ 210 }\) m/s = 0.95 m/s.

Question 3.
Abdul, while driving to school, computes the average speed for his trip to be 20 kmh^-1. On his return trip along the same route, there is less traffic and the average speed is 30 kmh^-1. What is the average speed for Abdul’s trip?
Answer:
Let the distance between Abdul’s home and school = x km.
Lets time taken from home to school = t₁ hr.
and time taken from school to home = t₂ hr.
Now,

∴ t₁ = \(\frac { x }{ 20 }\) hr.
t₂ = \(\frac { x }{ 30 }\) hr.
Now,

∴ Average Speed = \(\frac { 120 }{ 5 }\) 24 km/h.

Question 4.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms^-2 for 8.0 s. How far does the boat travel during this time?
Answer:
Initial velocity, u = 0
Acceleration, a = 3.0 m/s²
= 3 m/s²
Time, t = 8 sec
Now,
s = ut + \(\frac { 1 }{ 2 }\) at²
s = 0 × 8 + \(\frac { 1 }{ 2 }\) × 3 × (8)²
= 0 + \(\frac { 1 }{ 2 }\) × 3 × 64
s = 96 m.
∴ Distance Travelled by Boat is 96 m.

Question 5.
A driver of a car travelling at 52 kmh^-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 34 kmh^-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:
We have,
Speed of first driver = 52 km/h
= \(\frac { 52\times 1000 }{ 60\times 60 } \) m/s = 14.4 m/s
Speed of second driver = 3 km/h
= \(\frac { 3\times 1000 }{ 60\times 60 } \) m/s = 0.83 m/s
Now, distance covered by first driver = Area of ΔAOB
= \(\frac { 1 }{ 2 }\) × OA × OB
= \(\frac { 1 }{ 2 }\) × 14.4 × 5 = 36.1 m
Distance covered by second driver = Area of ΔMON
= \(\frac { 1 }{ 2 }\) × OM × ON
= \(\frac { 1 }{ 2 }\) × O.83 × 10 = 4.16 m
So, first driver covers more distance.

Question 6.
Figure below shows the distance-time graph of three objects A, B and C. Study the graph 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 far has C travelled when B passes A?
4. How far has B travelled by the time it passes C?

Answer:

1. B is travelling the fastest.
2. No.
3. C was at around 9 km from origin when B passes A.
4. B travelled around 5.5 km when it passes C.

Question 7.
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms^-2, with what velocity will it strike the ground? After what time will it strike the ground?
Answer:
Distance, s = 20 m
Initial velocity, u = 0
Acceleration, a = 10 m/s²
Now, s = ut +\(\frac { 1 }{ 2 }\) at²
20 = 0 × t + \(\frac { 1 }{ 2 }\) × 10 × t²
20 = 5 × t²
∴ t² = \(\frac { 20 }{ 5 }\) = 4
∴ t = 2 sec.
Also, v = u + at
v = 0 + 10 × 2
y = 20 m/s.
So, it will strike the ground after 2 seconds with the velocity of 20 m/s.

Question 8.
The speed – time graph for a car is shown in figure below:

(a) Find how far does the car travel 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 distance travelled by car in first 4 seconds
= Area of triangular like region OAB
= \(\frac { 1 }{ 2 }\) × AB × OB = \(\frac { 1 }{ 2 }\) × 6.9 × 4
= 12 m (approximately).

(b) Part of the graph after 5.5 seconds shows the uniform motion.

Question 9.
State which of the following situations are possible and give an example for each of these:

1. an object with a constant acceleration but with zero velocity.
2. an object moving in a certain direction with an acceleration in the perpendicular direction. Both are possible.

Answer:

1. Freely falling.
2. Motion of an object in a circular path.

Question 10.
An artificial, satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer:
Radius of the orbit = 42250 km = 42250 × 1000 m.
Time taken to complete one revolution = 24 hours.
= 24 × 60 × 60 sec
Distance covered by satellite to complete 1 revolution = 2πr
= 2 × \(\frac { 22 }{ 7 }\) × 42250 × 1000 m

Motion Additional Questions

Motion Multiple Choice Questions

Question 1.
Deceleration of a body is expressed in _______ .
(a) m
(b) ms^-1
(c) ms^-2
(d) -ms^-2
Answer:
(c) ms^-2.

Question 2. A car goes from town A to another town B with a speed of 40 km/h and returns back to the town A with a speed of 60 km/h. The average speed of the car during the complete journey is _______ .
(a) 48 km/h
(b) 50 km/h
(c) Zero
(d) None of these.
Answer:
(a) 48 km/h

Question 3.
A ball is thrown vertically upwards. It rises to a height of 50 m and comes back to the thrower, _______ .
(a) The total distance covered by the ball is zero.
(b) The net displacement of the ball is zero.
(c) The displacement is 100 m.
(d) None of the above.
Answer:
(b) The net displacement of the ball is zero.

Question 4.
The SI unit of velocity is _______ .
(a) ms^-1
(b) ms^-2
(c) ms^-3
(d) Nm^-1.
Answer:
(a) ms^-1

Question 5.
In 12 minutes a car whose speed is 35 km/h travels a distance of _______ .
(a) 7 km
(b) 3.5 km
(c) 14 km
(d) 28 km.
Answer:
(a) 7 km

Question 6.
A 50 m long train passes over a 250 m long bridge at a velocity of 60 km/h. How long will it take to pass completely over the bridge?
(a) 18 s
(b) 20 s
(c) 24 s
(d) none of these.
Answer:
(a) 18 s

Question 7.
The initial velocity of a train which is stopped in 20 s by applying brakes (retardation due to brakes being 1.5 ms^-2) is _______ .
(a) 30 ms^-1
(b) 30 cm^-1
(c) 20 cm^-1
(d) 24 ms^-1.
Answer:
(a) 30 ms^-1

Question 8.
A wooden slab starting from rest, slides down a 10 m long inclined plane with an acceleration of 5 ms². What would be its speed at the bottom of the inclined plane?
(a) 10 ms^-1
(b) 12 ms^-1
(c) 10 cm s^-1
(d) 12 cm s^-1.
Answer:
(a) 10 ms^-1

Question 9.
m/s² is the SI unit of _______ .
(a) Distance
(b) Displacement
(c) Velocity
(d) Acceleration.
Answer:
(d) Acceleration.

Question 10.
A car increases its speed from 20 km/h to 50 km/h in 10 seconds. Its acceleration is _______ .
(a) 30 m/s²
(b) 3 m/s²
(c) 18 m/s²
(d) 0.83 m/s².
Answer:
(d) 0.83 m/s².

Motion Very Short Answer type Questions

Question 1.
Define uniform motion.
Answer:
When an object covers equal distances in equal intervals of time: it is said to be in uniform motion.

Question 2.
Define speed.
Answer:
It is defined as the distance travelled by an object in unit time. Its unit is m/s.

Question 3.
Define average speed.
Answer:
The total distance travelled by an object divided by the total time taken.

Question 4.
Define velocity.
Answer:
Velocity is the speed of an object moving in definite direction.

Question 5.
Define acceleration.
Answer:
Change in the velocity of an object per unit time.
a = \(\frac { v-u }{ t } \), S.I. unit is m/s².

Motion Short Answer type Questions

Question 1.
While plotting a distance – time graph, why do we plot time on x – axis.
Answer:
While plotting a distance – time graph, we should plot time on x – axis as, all independent quantities are plotted on x – axis.

Question 2.
What is odometer?
Answer:
Odometer is a device fitted in the automobiles to show the distance travelled.

Question 3.
Draw a distance – time graph that represents uniform speed.
Answer:

Question 4.
With the help of distance – time graph show that the object is stationary.
Answer:

Question 5.
What does the area under velocity – time graph represent?
Answer:
The area under the velocity – time graph represents the displacement.

Question 6.
A body travels a distance from A to B, its physical quantity is measured to be -15 m/s. Is it speed or velocity? Give reason for the answer.
Answer:
The physical quantity is -15 m/s. It represents velocity because sign of speed cannot be a negative. Negative sign indicates the opposite direction.

Question 7.
What type of velocity – time graph will you get for a particle moving with a constant acceleration?
Answer:
For a particle moving with a constant acceleration velocity – time graph will be a straight line inclined to x – axis.

Question 8.
Define uniform circular motion.
Answer:
When a body moves in a circular path with uniform speed, its motion is called uniform circular motion.

Question 9.
What is deceleration?
Answer:
When the speed of an object decreases, the object is said to be decelerating. In this case, the direction of acceleration is opposite to the velocity of the object.

Question 10.
What conclusion do you draw about acceleration of the particle in motion from given velocity – time graphs?

Answer:
(a) No acceleration.
(b) Uniform acceleration.
(c) Non – uniform acceleration.

Question 11.
With the help of graph, show the uniform acceleration and uniform retardation of a body.
Answer:

• OA – Shows uniform acceleration.
• CB – Shows uniform retardation.

Question 12.
Give difference between acceleration and velocity.
Answer:
table

Motion Long Answer Type Questions

Question 1.
How can we get speed from distance – time graph?
Answer:
Let us assume that an object moves with a uniform speed. The distance – time graph will be a straight line.
Let d₁ be the distance covered in time t₁.
Let d₂ be the distance covered in time t₂.
∴ d₂ – d₁ will be the distance covered in t₂ – t₂.

But, here \(\frac { BC }{ AC }\) = slope of the graph.
Conclusion:
To find the speed from distance – time graph, we can get its slope which is equal to the speed of the object.

Question 2.
Represent a graph that shows the following:
(a) Uniform speed
(b) Non – uniform speed
(c) Stationary object.
Answer:
Time is taken on x – axis as it is independent quantity and all dependent quantities are taken along y – axis.

Question 3.
(a) What is motion?
(b) State types of motion.
(c) Derive the unit for acceleration.
Answer:
(a) Motion: An object is said to be in motion when its position changes with time. We describe the location of an object by specifying a reference point. Motion is relative. The total path covered by an object is said to be the distance travelled by it.

(b) There are two types of motion: Uniform motion and non – uniform motion

• Uniform motion: When an object covers equal distances in equal intervals of time, it is said to be in uniform motion.
• Non – uniform motion: Motion where object cover unequal distances in equal intervals of time.

(c) Acceleration, ‘a’ is change in velocity per unit time.

Question 4.
(a) What is uniform circular motion?
(b) An athlete runs on a circular track, whose radius is 50 m with a constant speed. It takes 50 seconds to reach point B from starting point A. Find:

1. the distance covered.
2. the displacement.
3. the speed.

Answer:
(a) When a body moves in a circular path with uniform speed, its motion is called uniform circular motion.

(b) Radius = 50 m
Time = 50 s
The distance covered by an athlete A to B i.e., semicircle of the track
∴ Circumference = 2πr
circumference half of = πr.

Question 5.
Derive the equation for velocity – time relation (v = u + at) by graphical method.
Answer:

Let, the initial velocity of a body be, u = OA = CD.
The final velocity of a body be, v = OE = CB
Time, t = OC = AD
AD is parallel to OC
BC = BD + DC
= BD + OA
v = BD + u
∴ BD =v – u …….. (1)
In velocity – time graph, slope gives acceleration.
∴ a = \(\frac { BD }{ AD }\) = \(\frac { BD }{ OC }\) = OC = t
we get, a = \(\frac { BD }{ t }\)
∴ BD = at …….. (2).

Question 6.
A car accelerates uniformly from 20 km/h to 35 km/h in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
Answer:
u =20 km/h = 5.5 m/s
v =35 km/h = 9.72 m/s
t = 5 s
a = \(\frac { v-u }{ t } \)
= \(\frac { 9.7-5.5 }{ 5 } \)
a = \(\frac { 4.2 }{ 5 } \)
= 0.84 m/s².

(ii) The distance covered by the car s = ut + \(\frac { 1 }{ 2 }\) at²
= 5.5 × 5 + \(\frac { 1 }{ 2 }\)(0.84)(5)²
= 27.5 + \(\frac { 1 }{ 2 }\) × 0.84 × 25
= 27.5 + 10.5
s = 38 m.

Question 7.
Draw graph to show the following:
(a) Uniform acceleration.
(b) Non – uniform acceleration.
(c) Uniform motion.
Answer:

Uniform acceleration – A body travels equal distances in equal intervals of time.

Non – uniform acceleration – A body travels unequal distances in equal intervals of time.

Uniform motion or zero acceleration A body moves with constant speed in same line.

Question 8.
How can you get the distance travelled by an object from its speed – time graph?
Answer:

Suppose an object is moving with a same speed v, the distance covered by this object when it was at (point A) t₁ to t₂ is:
AB = t₂ – t₁
AD = BC = v

= AD . AB
∴ s = Area of rectangle ABCD
∴ To find the distance travelled by a body from its speed – time graph, we need to find the area enclosed by the graph.

Motion Higher Order Thinking Skills (HOTS)

Question 1.
A blacksmith strikes a nail with a hammer of mass 500 g moving with a velocity of 20 ms^-1. The hammer comes to rest in 0.02 s after striking the nail. Calculate the force exerted by the nail on the hammer.
Answer:
m = 500 g = \(\frac { 1 }{ 2 }\) kg
u = 20 ms^-1 , v = 0
t = 0.02 s, F = ?

Therefore, force exerted by nail on the hammer = 500 N.

Motion Value Based Question

Question 1.
A jeep was moving at a speed of 120 km/h on an highway. All of a sudden tyre of the jeep got punctured and burst The driver of the jeep did not apply sudden brakes but he slowed down its speed and finally stopped it at the road side. All the passengers in the jeep were shocked but happy with the driver’s decison.

1. What would have happened if the driver applied sudden brakes?
2. On applying brakes in which direction would the passengers fall?
3. What value of driver is reflected in this act?

Answer:

1. If the driver would have applied sudden brakes, then all the passengers would get hurt and even other vehicles moving on the highway would have collided with this jeep from behind.
2. On applying brake, the passengers would fall in the forward direction.
3. The driver showed the presence of mind, concern and responsible behaviour and quick decision taking capability.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 8 Quadrilaterals Ex 8.1

Question 1.
The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all Ijbe angles of the quadrilateral.

Solution:
∠A = 3x, ∠B = 5x, ∠C = 9x, ∠D = 13x
In quadrilateral ABCD
∠A + ∠B + ∠C + ∠D = 360°
(∴ Sum of all the angles of ♢ is equal to 360°)

3x + 5x + 9x + 13x = 360°
30x = 360°
x = \(\frac{360^{\circ}}{30^{\circ}}\)
Let angle in ratio be x then angles are x= 12°
∠A = 3 x 12° = 36°
∠B = 5 x 12° = 60°
∠C = 9 x 12° = 108°
∠D = 13 x 12° = 156°

Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:
Given
ABCD is a parallelogram in which
AC =DB
To prove:
ABCD is a rectangle.
Proof
In ∆DAB and ∆CBA
DB = CA (given)
AB = BA (common)
AD = BC (∴ opposites sides of ∥^gm are equal)
∆DAB = ∆CBA (by SSS)
and so ∠DAB = ∠CBA
AD∥BC and AB is the transversal (by CPCT)
∴ ∠A + ∠B = 180° (CIA’s)
⇒ ∠A + ∠A = 180° (∴ ∠A = ∠B)
∴ ∠A = 90°
∠A = ∠C = 90°
and ∠B = ∠D = 90°
In ∥^gm ABCD, all the angles are right angles.
ABCD is a rectangle.

Question 3.
Show that if the diagonals ofa quadrilateral bisect each other at right angles, then it is a rhombus.
Solution:
Given
OA = OC, OB = OD and ∠AOD = 90°
To prove:
ABCD is a rhombus.
Proof:
In ∆AOD and ∆COB

OA = OC (given)
OD = OB (given)
∠1 = ∠2 (V.O.A.’s)
∴ ∆AOD = ∆COB (by SAS)
and so AD = CB (by CPC 7)
∠3 = ∠4 (by CPCT)
∠3 and ∠4 are A.I.A.’s and are equal
∴ AD ∥ PC (proved)
AD = BC
∴ ABCD is a parallelogram
In ∆AOD and ∆COD,
OA = OC (given)
OD = OD (common)
∠1 = ∠5 = 90°
(∴ ∠1 + ∠5 = 180° ⇒ 90° + ∠5 = 180° ∴ ∠5 = 90°)
∴ ∠AOD = ∠COD (by SAS)
and so AD = CD (by CPCT)
In ∥^gm APCD, all the sides are equal.
ABCD is rhombus.

Question 4.
Show that the diagonals of a square are equal and bisect each other at right angles.
Solution:
Given
ABCD is a square.
To prove:

1. AC = BD
2. OA = OC and OB = OD
3. ∠AOD = 90°

Proof:
In ∆DAB and ACBA,
DA = CB (given)
AB = BA (common)
∠A = ∠B (each 90°)
∴ ∆DAB = ∆CBA (by SAS)
and so BD =AC (by CPCT)

2. In ∆AOD and ∆COB, AD = CB(given)

∠4 = ∠5 (A.I.A.’s)
∠6 = ∠7 (A.I.A. ’s)
∆AOD = ∆COB (by ASA)
and so OA=OC , (byCPCT)
OD = OB (byCPCT)

3. In ∆AOD and ∆COD,
AO = CO (proved)
OD = OD (common)
AD = CD (given)
∆AOD = ∆COD (by SSS)
and so ∠1 = ∠3 (byCPCT)
∠1 + ∠3 = 180° (LPA’S)
⇒ ∠1 + ∠1 = 180° (∠1 = ∠3)
⇒ 2∠1 = 180°
∴ ∠1 = \(\frac{180^{\circ}}{2}\)

Question 5.
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution:

Give
ABCD is in which
AC = BD
OA = OC
and OB = OD
∠AOD = 90°
To prove
ABCD is a square
Proof:
In ∆AOD and ∆COB,
OA = OC (given)
OD = OB (given)
∠7 = ∠8 (V.CXA.’s)
∴ ∆AOD = ∠COB
and so AD = BC
and ∠3 = ∠1 (byCPCT)
∠3 and ∠1 are A.I.A.’s and are equal
∴ AD ∥ BC
Similarly, AB ∥ CD
∴ ABCD is a parallelogram.
In ∆AOD and ∆COD,
OA = OC (given)
OD = OD (common)
∠7 = ∠9 (each 90°)
∆AOD = ∠COD (by SAS)
AD = CD (byCPCT)
In ∥^gmABCD, adjacent sidesAD = CD
∴ ABCD is a rhombus
In ∆DAB and ∆CBA,
DA = CB (proved)
AB = BA (common)
DB = CA (given)
∆DAB = ∆CBA (bySSS)
∠A = ∠B (by CPCT)
∠A + ∠B = 180° (CIA’s)
2∠A = 180°
∠A =90°
ABCD is a square.

Question 6.
Diagonals AC of a parallelogram ABCD bisects ∠A (see Fig). Show that

1. it bisects ∠C also,
2. ABCD is a rhombus.

Solution:
Given
ABCD is a parallelogram in which ∠1 = ∠2
To prove:

1. ∠3 = ∠4
2. ABCD is a rhombus.

Proof
1. ∠1 = ∠4 (A.I.A.’s) ….(i)
(∴ AD ∥ SC and AC is the transversal)
∠2 = ∠3 (A.I.A.’s) …(ii)
(∴ AB ∥ DC and AC is the transversal)
∠1 = ∠2 (given) …(iii)

From (i), (ii) and (iii), we get
∠4 = ∠3
(ii) From (ii) and (iii), we get,
∠1 = ∠3
In ∆ADC,
∠1 = ∠3
∴ AD = DC (In a A, sides opposites to equal angles are equal) and so ABCD is a rhombus
(∴ In a ∥^gm, if adjacent sides are equal then it is a rhombus)

Question 7.
ABCD is a rhombus, show that diagonal AC biusects ∠A as well as ∠C and diagonal BD biusects ∠B as well as ∠D.
Solution:
Given
ABCD is a rhombus.
To prove
∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8.
Proof:
In ∆ADC and ∆ABC.

AD = AB (Adj. sides of a rhombus)
DC = BC (Adj. sides of a rhombus)
AC = AC (common)
∴ ∆ADC = ∆ABC (by SSS)
so ∠1 = ∠2 (by CPCT)
and ∠3 = ∠4 (by CPCT)
∴ AC bisects ∠A and ∠C
Similarly, ∠5 = ∠6 and ∠7 = ∠8
BD bisects ∠B and ∠D.

Question 8.
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that

1. ABCD is a square
2. diagonal BD bisects ∠B as well as ∠D.

Solution:
Given
ABCD is a rectangle.
∠1 = ∠2 and ∠3 = ∠4
To prove:

1. ABCD is a square
2. ∠5 = ∠6 and ∠7 = ∠8

Proof:
1. ∠A = ∠C
(∵ Rectangle is ∥^gm and in a ∥^gm opp. angles are equal.)

\(\frac{1}{2}\)∠A = \(\frac{1}{2}\)∠C
∠A = ∠C
∠2 = ∠4
In ∆ABC, ∠2 = ∠4
AB = BC
(In a A, sides opp. to equal angles are always equal)
ABCD is a rectangle in which adjacent sides are equal.
∴ ABCD is a square.

2. In ∆ABD,
AB = AD (∴ ABCD is a square)
∴ ∠5 = ∠7 (∴ In a A, angles opp. to equal sides are equal) ….(1)
AB ∥ DC and BD is the transversal
∴ ∠6 = ∠7 …(2)
AD ∥ BC and BD is the transversal.
∴ ∠5 = ∠8 …(3)
From (1) and (3), we get
∠7 = ∠8
From (1) and (2), we get
∠5 = ∠6
Diagonal BD bisects ∠B as well as ∠D.

Question 9.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig.). Show that

1. ∆APD = ∆CQB
2. AP = CQ
3. ∆AQB = ∆CPD
4. AQ = CP A
5. APCQ is a parallelogram

Given
ABCD is a parallelogram.
∴ AD = BC, AB= DC and DP = BQ
To prove:

1. ∆APD = ∆CQB
2. AP = CQ
3. ∆AQB = ∆CPD
4. AQ = CP
5. APCQ is a parallelogram

Proof:
In ∆APD and ∆CQB
PD = QB (given)
AD = CB (given)
∠2 = ∠1 (AIA’s)
∆APD = ∆CQB (by SAS)
and so AP = CQ (by CPCT)

In ∆AQB and ∆CPD,
AB = CD (given)
∠3 = ∠4 (AIA’s)
BQ = DP (given)
∆AQB = ∆CPD (by SAS)
and so AQ = CP (by CPCT)
In quadrilaterals AQCP,
AP = CQ
AQ = CP
AQCP is a parallelogram.

Question 10.
ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD (see Fig.). Show that

1. ∆APB = ∆CQD
2. AP = CQ

Solution:
Given
ABCD is a ∥^gm in which Ap ⊥ BD and CQ ⊥ BD.
To prove:

1. ∆APB = ∆CQD
2. AP = CQ

Proof:
In ∆APB and ∆CQD,
∠P = ∠Q (each 90°)
∠1 = ∠2 (AIA’s)
AB = CD (given)
∆APB ≅ ∆CQD (byAAS)
and so AP = CQ (by CPCT)

Question 11.
In ∆ABC and ∆DEF, AB = DE, AB ∥ DE, BC = EF and BC ∥ EE. Vertices A, B and C are joined to vertices, D,E and F respectively (see Fig.). Show that

1. quadrilateral ABED is a parallelogram
2. quadrilateral BEFC is a parallelogram
3. AD ∥ CF and AD = CF
4. quadrilateral ACFD is a parallelogram
5. AC = DF
6. ∆ABC = ∆DEF.

Solution:
Given
AB = DE and AB ∥ DE
BC = EF and BC ∥ EF
To prove

1. ABED is a ∥^gm
2. BEFC is a ∥^gm
3. AD ∥ CF and AD – CF
4. ACFD is a ∥^gm
5. AC =DF
6. ∆ABC ≅ ∆DEF

Proof:
1. AB = DE and AB ∥ DE (given)
ABED is ∥^gm
and so AD ∥ BE and AD = BE …(1)

2. BC = EF and BC ∥ FC (given)
BEFC is a ∥^gm
and so BE ∥ CF and BE = CF …..(2)

3. From (1) and (2), we get
AD ∥ CF and AD = CF

4. AD ∥ CF and AD = CF (proved)
ACFD is a ∥^gm

5. and so AC = DF
(In a parallelogram, opp. sides are equal)

6. In ∆ABC and ∆DEF,
AB = DE (given)
BC = EF (given)
AC = DF (proved)
∆ABC = ∆DEF (by SSS)

Question 12.
ABCD is a trapezium in which AB ∥ CD and AD = BC (see Fig.). Show that:

1. ∠A = ∠B
2. ∠C = ∠D
3. ∆ABC ≅ ∆BAD
4. Diagonal AC = diagonal BF)

[Hint: Extend AB and draw line through C parallel to DA intersecting AB produced at E.]
Solution:
Given
AB ∥ CD, AD = BC
To prove:

1. ∠A = ∠B
2. ∠C = ∠D
3. ∆ABC ≅ ∆BAD
4. diagonal AC = diagonal BD

Construction:
Draw a line CE ∥ DA which intersect AB produced at E.
Proof:
1. In quadrilateral ADCE,
AD ∥ EC (by const)
and AE ∥ DC ( AB ∥ DC)
∴ ADCE is a parallelogram
and so AD = EC (opp. sides of a ∥ are equal)…(i)
AD = BC (given) ….(ii)
From (i) and (ii), we get
BC = EC
In ∆BCE BC = EC (proved)
∠4 = ∠3
(∴ In a ∆, angles opp. to equal sides are equal)
∠2 + ∠3 = 180° (LPA’s) …(iii)
∠1 + ∠4 = 180° (CIA’s) …(iv)
From (iii) and (iv), we get
∠2 + ∠3 = ∠1 + ∠4
∠2 = ∠1 (∠3 = ∠4)
i.e., ∠A = ∠B

2. ∠3 = ∠5 (AIA’s) …(v)
∠6 = ∠4
(∴ ADCE is a ∥^gm and in a ∥^gm opp. angles are equal) …(vi)
∠4 = ∠3 (proved) …(vii)
From (v), (vi) and (vii), we get
∠5 = ∠6
i.e., ∠C = ∠D

3. In ∆ABC and ∆BAD,
AB = BA (common)
BC = AD (given)
∠2 = ∠1 (proved)
∆ABC ≅ ∆BAD (by SAS)
and so AC =BD (by CPCT)

4. diagonal AC = diagonal BD (proved)

Mid Point Theorem:
The line segment joining the mid-points of the sides of a triangle is parallel to the third side and equal to half of it.
Given.
ABC is a A in which D and E are the mid-points of sides AB and AC respectively.
To prove.
DE ∥ BC and DE = \(\frac{1}{2}\) BC
Construction:
Extend DE uptoFsuch that DE = EF. Join CF.
Proof:
In ∆AED and ∆CEE
AE = CE (E is the mid – point of AC)
∠AED = ∠CEF (VOA’s)
DE = FE (By constriction)
∆AED = ∆CEF , (By SAS)
and so ∠DAE = ∠FCE (By CPCT)
AD = CF (By CPCT)
∠DAE and ∠FCE are alternate interior angles and are equal.

AD ∥ FC
⇒ DB ∥ FC
Now, AD = DB and AD = FC
DB = FC
In BCFD, DB ∥ FC and DF = BC
BCFD is a ∥^gm
and so DF ∥ BC and DF = BC
⇒ DF ∥ BC and 2DF = BC
DE ∥ BC and DE = \(\frac{1}{2}\) BC

Converse of mid point theorem:
The line drawn through the mid – point of one side of a triangle and parallel to another side, bisects the third side.
Given
ABC is a A in which D is the mid-point of AB and DE ∥ BC.
To Prove:
E is the mid – point of AC.
Construction:
Mark a point F on AC and join DF.
Proof:
Let E be not the mid – point of AC. Let us assume that F be the mid – point of AC.
Then by mid-point theorem

DF ∥ BC
DE ∥ BC (Given)
From (i) and (ii), we get
DE ∥ DF
But lines DE and DF are intersecting lines, intersecting at D. This is a contradiction. So our supposition is wrong. Hence E is the mid – point of AC.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.4

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:
Given
ABC is a right angle A.
To prove:
AC > AB and AC > BC

Proof:
In ∆ABC
∠B = 90°
∴ ∠A + ∠C = 90° (by ASP)
and so ∠B > ∠A and ∠B > ∠C
AC > BC and AC > AB
(In a A, sides opposite to large angle are always longer).

Question 2.
In Fig. given below, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Solution:
Given
∠PRC < ∠QCB To prove AC > AB

Proof:
In ∆ ABC
Exterior angle is equal to sum of two opposite angles
∠PBC = ∠1 + ∠3 and
∠QCB = ∠1 + ∠2
∠QCB > ∠PBC (given)
⇒ ∠1 + ∠2 > ∠1 + ∠3
⇒ ∠2 > ∠3
∴ AC > AB
(∴ In a A, side opposite to larger angle is always longer).

Question 3.
In Fig. below, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:
Given
∠B < ∠A i.e., ∠A > ∠B
∠C < ∠D i.e., ∠D> ∠C
To prove: AD > BC
i.e., BC > AD
Proof:
In ∆OCD
∠D > ∠C
OC > OD
(∴ In a ∆, sides opposite to larger angle are always longer) …(1)
In ∆OBA
∠A > AB
OB > OA
(∴ In a A, sides oppositedo larger angle are always longer) …(2)
Adding (1) and (2), we get
OC + OB > OD + OA
BC > AD

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. below). Show that ∠A > ∠C and ∠B > ∠D.
Solution:
Given
AB is the smallest siBe and CD is the longest side
To prove:

1. ∠A > ∠C and
2. ∠B > ∠D

Construction:
Join AC

1. Proof:
In, ∆ABC
BC > AB (∴ AB is the smallest side) ∠1 > ∠3
(∴ In a ∆ angle opposite to longer side is always larger) …..(1)
In ∆ACD
CD > AD (∴ CD is the largest side) ∠2 > ∠4
(∴ In a ∆ angles opposite to longer side are always larger) …(2)
Adding (1) and (2), we get
∠1 + ∠2 > ∠3 + ∠4
∠A > ∠C

2. To prove: AB > AD
Construction:
Join BD
Proof:
In ∆ABD
AD > AB (AB is the smallest side)
∠5 > ∠7 …(3)
In ∆BCD
CD > BC (CD is the longest side)

∠6 > ∠8 …(4)
Adding (3) and (4), we get
∠5 + ∠6 > ∠7 + ∠8
∠B > ∠D

Question 5.
In Fig. below, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Sol.
Given:
PR > PQ
∠1 = ∠2
To prove:
∠PSR > ∠PSQ
Proof:
In ∆PSQ
∠PSR = ∠1 + ∠Q (EAP)
In ∆PSR
∠PSQ = ∠2 + ∠R (EAP)
= ∠1 + ∠R (∠1 = ∠2)
In ∆PQR
PR > PQ (given)
∠Q > ∠R
(∴ In a ∆, angle opposite to longer side is always larger)
Adding ∠1 on both sides
∠Q + ∠l > ∠R + ∠1
∴ ∠PSR > ∠PSQ
(∴ ∠PSR = ∠1 + ∠Q and ∠PSQ = ∠1 + ∠R)

Question 6.
Show that of all the line segments drawn from a give point not on it, the perpendicular line segment is the shortest.
Solution:
Given
Let us consider the ∆PMN such that ∠M = 90°
Since, ∠M + ∠N + ∠P = 180° [Sum of angles of a triangle]

∠M = 90° [PM ⊥ l]
∠N < ∠M
PM < PN …..(1)
Similarly PM < PN₁ …..(2)
PM < PN₂ …..(3)
From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l.
Thus, the perpendicular segment is the shortest line segment drawn on a line from a point not on it.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.3

Question 1.
∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. below). If AD is extended to intersect BC at P, show that

1. ∆ABD = ∆ACD
2. ∆ABP = ∆ACP
3. AP bisects ∠A as well as ∠D.
4. AP is the perpendicular bisector of BC.

Solution:
Given
AB = AC
DB = DC
To prove:

1. ∆ABD = ∆ACD
2. ∆ABP = ∆ACP
3. ∠1 = ∠2 and ∠5 = ∠6
4. ∠3 = ∠4 = 90° and BP = PC

Proof
1. In ∆ABD and ∆ACD
AB = AC (given)
BD = CD (given)
AD =AD (common)
∆ABD = ∆ACD (by SSS)
and so ∠1 = ∠2 (by CPCT)

2. In ∆ABP and ∆ACP
AB = AC (given)
∠1 = ∠2 (proved)
AP = AP (common)
∆ABP ≅ ∆ACP (by SAS)
and so ∠3 = ∠4
and BP = CP (by CPCT)

∠1 = ∠2 AP bisects ∠A

3. In ∆DBP and ∆DCP
BP = CP (proved)
∠3 = ∠4 (proved)
DP = DP (common)

∆DBP ≅ ∆DCP (by SAS)
and so ∠5 = ∠6 (by CPCT)
∴ AP bisects ∠D.

4. BP = CP (proved)
∠3 = ∠4 (proved)
∠3 + ∠4 = 180° (LPA’s)
∠3 + ∠3 = 180° (∠3 = ∠4)
2∠3 = 180°
∠3 = 90°
∠3 = ∠4 (each 90°)
and therefore, AP is the perpendicular bisector of BC

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

1. AD bisects BC
2. AD bisects ∠A.

Solution:
Given
AB = AC and AD ⊥ BC
To prove
∠1 = ∠2 and
BD = CD
Proof:
In ∆ABD and ∆ACD
AB = AC
AD = AD
∠3 = ∠4
∆ABD = ∆ADC
and so BD = CD and ∠1 = ∠2

Question 3.
Two sides AS and BC and median AM on one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR (see-Fig. below). Show that:

1. ∆ABM ≅ ∆PQN
2. ∆ABC ≅ ∆PQR

Solution:
Given
AB = PQ
BC = QR
AM = PN
To prove:

1. ∆ABM ≅ ∆PQN
2. ∆ABC ≅ ∆PQR

Proof:
BC = QR (given)
\(\frac{1}{2}\) BC = \(\frac{1}{2}\)QR
BM = QN

1. In ∆ABM and ∆PQN
AB = PQ (given)
BM = QN (proved)
AM = PN (given)
∆ABM ≅ ∆PQN (by SSS)
and so ∠ABC = ∠PQR (by CPCT)

2. In ∆ABC and ∆PQR
AB = PQ (given)
BC = QR (given)
∠B = ∠Q (proved)
∆ABC ≅ ∆PQR by (SAS)

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that triangle ABC is isosceles.

Solution:
Given
∠E = ∠F
BE = CF
To prove
AB = AC
Proof:
In ∆FBC and ∆ECB
BE = CF (given)
∠F = ∠E (each 90°)
BC = CB (common)
∴ ∆FBC = ∆ECB (by RHS)
and so ∠B = ∠C (by CPCT)
In ∆ABC, ∠B = ∠C
AB = AC
(sides opposite to equal angles of a A are equal)
and so ABC is isosceles.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:
Given
AB = AC
∠APB = ∠APC = 90°
To prove:
∠B = ∠C

Proof:
In ∆ABP and ∆ACP
AP = AP (common)
∠APB = ∠APC (each 90°)
AB = AC (given)
∴ ∆ABP = ∆ACP (by RHS)
and ∠B = ∠C (by CPCT)

Theorem 7.6.
If two angles of a triangle are equal, then the sides opposite to them are also equal.
Given
In ∆ABC, ∠C = ∠B

To prove: AB = AC
Construction:
Draw the bisector of ∠A and let it meet BC an D
Proof:
In ∆s ABD and ACD, we have ∠B = ∠C (Given)
∠BAD = ∠CAD (Construction)
AD = AD (Common)
∴ ∆ABD ≅ ∆ACD (AAS Cong. Criterion)
Hence, AB = AC (CPCT)

Theorem 7.7.
If two sides of a triangle are unequal, the longest side has greater angle opposite to it.
Given:
In ∆ABC; AC > AB
To prove: ∠ABC > ∠ACB.

Construction:
Mark point D on AC such that AB = AD. Join BD.
Proof:
In ∆ABD,
AB = AD
∴ ∠1 = ∠2 (Const. As opp. equal sides) ….(1)
But ∠2 is an exterior angle of ABCD.
∠2 > ∠ACB (Exterior Angle Theorem) …(2)
From (1) and (2), we have
∠1 > ∠ACB (Const.)
But ∠ABC > ∠1
∴ ∠ABC > ∠ACB

Theorem 7.8. (Converse of Theorem 7.7)
In a triangle the greater angle has the longer side opposite to it.
Given:
In ∆ABC, ∠ABC > ∠ACB
To prove: AC > AB

Proof:
For ∆ABC, there are only three possibilities of which exactly one must be true.

1. AC = AB
2. AC < AB (iii) AC > AB.

Case 1:
If AC = AB, then ∠ABC = ∠ACB, which is contrary to what is given.
AB ≠ AC

Case 2:
If AC < AB, the longer side has the greater angle opposite to it.
∴ ∠ACB > ∠ABC.
This is also contrary to what is given.

Case 3:
We are left with the only possibility, namely AC > AB which is true.
AC > AB.

MP Board Class 9th Maths Solutions

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

Force and Laws of Motion Intext Questions

Force and Laws of Motion Intext Questions Page No. 118

Question 1.
Which of the following has more inertia:

1. a rubber ball and a stone of the same size?
2. a bicycle and a train?
3. a five – rupees coin and a one – rupee coin?

Answer:
As we know, inertia is the calculated value for the mass of the body. It is proportional to mass of the body:

1. Inertia of the stone is greater than that of a rubber ball as mass of a stone is more than the mass of a rubber ball for the same size.
2. Inertia of the train is greater than that of the bicycle. As mass of a train is more than the mass of a bicycle.
3. Mass of a five rupee coin is more than that of a one – rupee coin. Hence, inertia of the five – rupee coin is greater than that of the one – rupee coin.

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

• First, when a football player kicks to another player. Agent supplying the force: First case – First player. Second when that player kicks the football to the goalkeeper. Agent supplying the force. Second case – Second player.
• Third when the goalkeeper stops the football. Agent supplying the force: Third case – Goalkeeper.
• Fourth when the goalkeeper kicks the football towards a player of his own team. Agent supplying the force: Fourth case – Goalkeeper.

Question 3.
Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer:
When we shake any tree’s branches vigorously some leaves of that tree get detached because branches comes in motion while the leaves tend to remain at rest due to inertia of rest.

Question 4.
Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Answer:
Due to inertia of motion, we fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest.

1. Case I: Since the driver applies brakes and bus comes to rest. But, the passenger tries to maintain its inertia of motion. As a result, a forward force is exerted on him.
2. Case II: The passenger tends to fall backwards when the bus accelerates from rest because when the bus accelerates, the inertia of rest of the passenger tends to oppose the forward motion of the bus.

Force and Laws of Motion Intext Questions Page No. 126

Question 1.
If action is always equal to the reaction, explain how a horse can pull a cart.
Answer:
According to Newton’s third law of motion, a force is exerted by the Earth on the horse in the forward direction while horse pushes the ground in the backward direction. As a result, the cart moves forward.

Question 2.
Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Answer:
According to Newton’s third law of motion, a reaction force is exerted over fireman by the ejecting water in the backward direction when a fireman holds a hose, which is ejecting large amounts of water at a high velocity. As a result of the backward force, the stability of the fireman get affected. Hence, it is difficult for him to remain stable while holding the hose.

Question 3.
From a rifle of mass 4kg, a bullet of mass 50g is fired with an initial velocity of 35 ms^-1. Calculate the initial recoil velocity of the rifle.
Answer:
Given,
Mass of the rifle, m₁ = 4kg
Mass of the bullet, m₂ = 50g = 0.05kg
Recoil velocity of the rifle = v₁
Bullet is fired with an initial velocity, v₂= 35 m/s
Condition:
Initially, the rifle is at rest.
Thus, its initial velocity, v = 0
Total initial momentum of the rifle and bullet system = (m₁+ m₂)v = 0
Total momentum of the rifle and bullet system after firing = m₁v₁ + m₂v₂
= 0.05 × 35 = 4v₁ + 1.75
As per law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing
⇒ 4v₁ + 1.75 = 0
v₁ = –\(\frac { 1.75 }{ 4 }\) = -0.4375 m/s
So, the rifle recoils backwards with a velocity of 0.4375 m/s because value is negative.

Question 4.
Two objects of masses 100g and 200g are moving along the same line and direction with velocities of 2 ms^-1 and 1  ms^-1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms^-1. Determine the velocity of the second object.
Answer:
Given,
m₁ = 100g = 0.1kg
m₂ = 200g = 0.2kg
Velocity of m₁ before collision, v₁ = 2 m/s
Velocity of m₂ before collision, v₂ = 1 m/s
Velocity of m₁ after collision, v₃ = 1.67 m/s
Velocity of m₂ after collision = v₄
As per the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
Hence,
m₁v₁ + m₂v₂ = m₁v₃+ m₂v₄
Putting values
2(0.1) + 1(0.2) = 1.67(0.1) + v₄(0.2)
0.4 = 0.167 + 0.2v₄
v₄ = 1.165 m/s
Velocity of the second object = 1.165 m/s.

Force and Laws of Motion NCERT Textbook Exercises

Question 1.
An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non –  zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Answer:
Yes, an object may travel with a non – zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the upthrust and the viscosity of air. The net force on the drop is zero.

Question 2.
When a carpet is beaten with a stick, dust comes out of it Explain.
Answer:
When we beat the carpet with a stick, it comes into motion. But the dust particles continue to be at rest due to inertia and get detached from the carpet.

Question 3.
Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:
Due to sudden jerks or due to the bus taking sharp turn on the road, the luggage may fall down from the roof because of its tendency to continue moving in the original direction. To avoid this, the luggage is tied with a rope on the roof.

Question 4.
A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would come to rest.
Answer:
(c) there is a force on the ball opposing the motion.

Question 5.
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg.)
Answer:
Here, u = 0, s = 400m, t = 20 s, a = ?, F = ?.
m = 7 tonnes
= 7 × 1000kg
= 7000kg
⇒ s = ut + \(\frac { 1 }{ 2 }\)at²
400 = (0 + 20) + \(\frac { 1 }{ 2 }\)a(20)²
a = \(\frac { 400\times 2 }{ { 20 }^{ 2 } } \)
∴ a = 2 m/s²
Force,
F = ma
= 7000 × 2 = 14,000 N.

Question 6.
A stone of 1kg is thrown with a velocity of 20 ms^-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50m. What is the force of friction between the stone and the ice?
Answer:
m = 1kg, u = 20 m/s, s = 50m, v = 0, F = ? a = ?.
⇒ v² – u² = 2as
(0)² – (20)² = 2a (50)
∴ – 400 = 100a
= \(\frac { 400 }{ 100 }\) – 4 m/s²
Force of friction, F = m × a
= 1kg × -4 m/s² = -4 N

Question 7.
A 8000kg engine pulls a train of 5 wagons, each of 2000kg, along a horizontal track. If the engine exerts a force of 40,000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2.
Answer:
m = 8,000 + 5 × 2,000 = 18,000kg
(a) The net accelerating force,
F = Engine force – friction force
= 40,000 – 5,000 = 35,000 N.

(b) The acceleration of the train,
a = \(\frac { F }{ m }\) = \(\frac { 35,000 }{ 18,000 }\) = \(\frac { 35 }{ 18 }\) = 1.94 ms^-2

(c) The force of wagon 1 on wagon 2
= The net accelerating force – mass of wagon × acceleration
= 35,000 – 2,000 × \(\frac { 35 }{ 18 }\)
= 35,000 – 3888.8 = 31,111.2 N.

Question 8.
An automobile vehicle has a mass of 1500kg. What must be the force between the vehicle and road if the vehicle is stopped with a negative acceleration of 1.7 ms^-2?
Answer:
Here, mass = 1500kg
a = -1.7 ms²
F = m × a
= 1500 × (-1.7)
= -2550 N
The force between the vehicle and the road is 2,550 is, m a direction opposite to the direction of the vehicle.

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

Question 10.
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer:
The cabinet will move with constant velocity only when the net force on it is zero.
Force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet.

Question 11.
Two objects, each of mass 1.5kg, ate moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms^-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Answer:
Here, m₁ = m₂ = 1.5kg,
u₁ = 2.5 ms^-1, u₂ = – 2.5 ms^-1
Let v be the velocity of the combined object after the collision.
By conservation of momentum,
Total momenta after collision = Total momenta before collision
= (m₁ + m₂) v = m₁u₁ + m₂u₂
= (1.5 + 1.5) v = 1.5 × 2.5 + 1.5 × (-2.5)
= 3.0 v = 0
or
⇒ v = 0 ms^-1

Question 12.
According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer:
Action and reaction always act on different bodies, so they cannot cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so, the truck does not move.

Question 13.
A hockey ball of mass 200g travelling at 10 ms^-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms^-1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer:
Here, m = 200g = 0.2kg,
u = 10 ms^-1,
v = -5 ms^-1
change in momentum = m (v – u)
= 0.2 (- 5 – 10) = -3kg ms^-1.

Question 14.
A bullet of mass 10g travelling horizontally with a velocity of 150 ms^-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer:
Here m = 10g = 0.01kg,
u = 150 ms^-1,
v = 0, t = 0.03 s
a = \(\frac { v-u\quad }{ t } \) = \(\frac { 0-150\quad }{ 0.03 } \) = -5,000 ms^-1
The distance of penetration of the bullet into the block,
s = ut + \(\frac { 1 }{ 2 }\)at²
150 × 0.03 + \(\frac { 1 }{ 2 }\) × (-5,000) × (0.03)
= 4.5 – 2.25 = 2.25
The magnitude of the force exerted by the wooden block on the bullet
= ma = 0.01 × 5,000 = 50 N.

Question 15.
An object of mass 1kg travelling in a straight line with a velocity of 10 ms^-1 collides with, and sticks to, a stationary wooden block of mass 5kg. Then, they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Answer:
Here, m₁ = 1kg, u₁ = 10 ms^-1, m₂ = 5kg, u₂ = 0
Let v be the velocity of the combined object after the collision.
Total momentum just before the impact
= m₁u₁ + m₂u₂  = 1 × 10 + 5 × 0 = 10kg ms^-1
Total momentum just after the impact = (m₁ + m₂)v = (1 + 5)v
= 6v kg ms^-1 by conservation of momentum,
6v = 10
= v = \(\frac { 10 }{ 6 }\) = \(\frac { 5 }{ 3 }\) ms^-1
∴ Total momentum just after the impact
= 6 × \(\frac { 5 }{ 3 }\) = 10 ms^-1

Question 16.
An object of mass 100kg is accelerated uniformly from a velocity of 5 ms^-1 to 8 ms^-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer:
Here, m = 100kg,
u = 5 ms^-1,
v = 8 ms^-1,
t = 6 s
Initial momentum, P₁ = mu = 100 × 5 = 500kg ms^-1
Final momentum, P₂ = mu = 100 × 8 = 800kg ms^-1
The magnitude of the force exerted on the object.
F = \(\frac { { P }_{ 2 }-{ P }_{ 1 } }{ t } \) = \(\frac { 800-500 }{ 6 }\) = \(\frac { 300 }{ 6 }\) = 50 N.

Question 17.
Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Answer:
Both, the motorcar and insect experience the equal force and hence, a same change in their momentum. So, we agree with Rahul. But due to smaller mass or inertia, the insect dies.

Question 18.
How much momentum will a dumb – bell of mass 10kg transfer to the floor if it falls from a height of 80cm? Take its downward acceleration to be 10 ms^-2.
Answer:
Here, m = 10kg, u = 0,
s = 80cm = 0.80m,
a = 10 m/s^-2
Let v be the velocity gained by the dumb-bell as it reaches the floor.
As v² – u² = 2as
v² – 0² = 2 × 10 × 0.80 = 16
or
v = 4 ms^-1
Momentum transferred by the dumb-bell to the floor
p = mv = 10 × 4 = 40kg ms^-1

Force and Laws of Motion Additional Questions

Force and Laws of Motion Multiple Choice Questions

Question 1.
Which of the following statements is not correct for an object moving along a straight path in an accelerated motion?
(a) Its speed keeps changing.
(b) Its velocity always changes.
(c) It always goes away from the earth.
(d) A force is always acting on it.
Answer:
(c) It always goes away from the earth.

Question 2.
According to the third law of motion, action and reaction ________ .
(a) Always act on the same body.
(b) Always act on different bodies in opposite directions.
(c) Have same magnitude and directions.
(d) Act on either body at normal to each other.
Answer:
(b) Always act on different bodies in opposite directions.

Question 3.
A goalkeeper in a game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goal keeper to ________ .
(a) Exert larger force on the ball.
(b) Reduce the force exerted by the ball on hands.
(c) Increase the rate of change of momentum.
(d) Decrease the rate of change of momentum.
Answer:
(b) Reduce the force exerted by the ball on hands.

Question 4.
The inertia of an object tends to cause the object ________ .
(a) To increase its speed.
(b) To decrease its speed.
(c) To resist any change in its state of motion.
(d) To decelerate due to friction.
Answer:
(c) To resist any change in its state of motion.

Question 5.
A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is ________ .
(a) Accelerated
(b) Uniform
(c) Retarded
(d) Along circular tracks.
Answer:
(a) Accelerated

Question 6.
An object of mass 2kg is sliding with a constant velocity of 4 ms^-1 on a frictionless horizontal table. The force required to keep the object moving with the same velocity is ________ .
(a) 32 N
(b) 0 N
(c) 2 N
(d) 8 N.
Answer:
(b) 0 N

Question 7.
Rocket works on the principle of conservation of ________ .
(a) Mass
(b) Energy
(c) Momentum
(d) Velocity.
Answer:
(c) Momentum

Question 8.
A water tanker filled up to \(\frac { 2 }{ 3 }\) of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would ________ .
(a) Move backward
(b) Move forward
(c) Be unaffected
(d) Rise upwards.
Answer:
(b) Move forward

Question 9.
Which of the following represents example(s) of potential energy?
(a) A moving car
(b) A moving fan
(c) A book resting on the table
(d) Both (a) and (c).
Answer:
(d) Both (a) and (c).

Question 10.
Unit of force is ________ .
(a) Ampere
(b) Volt
(c) Joule
(d) Hertz.
Answer:
(c) Joule

Question 11.
Product of mass and acceleration of a body is called ________ .
(a) Acceleration
(b) Work
(c) Power
(d) Energy.
Answer:
(b) Work

Question 12.
Which of the following is correct about energy?
(a) Energy is not required to do work.
(b) Work can be expressed as Force × Displacement.
(c) Unit of power is joule.
(d) Power is the amount of work done per unit time.
Answer:
(c) Unit of power is joule.

Question 13.
An object of mass 3kg is falling from the height of 1m. The kinetic energy of the body will be when it touches the ground ________ .
(a) 29.4 N
(b) 29.4 J
(c) 30 N
(d) 15 J
Answer:
(b) 29.4 J

Question 14.
Two objects with masses 1kg and 9kg, and equal momentum. Calculate the ratios of their kinetic energies ________ .
(a) 3 : 1
(b) 9 : 1
(c) 1 : 1
(d) 1 : 2
Answer:
(b) 9 : 1

Question 15.
Considering air resistance negligible, the sum of potential and kinetic energies of the free falling body would be ________ .
(a) zero
(b) infinite
(c) would decrease
(d) remains fixed.
Answer:
(d) remains fixed.

Force and Laws of Motion Very Short Answer Type Questions

Question 1.
What do we call to the product of mass and velocity of an object?
Answer:
Momentum.

Question 2.
Define inertia.
Answer:
The property by which an object tends to remain in the state of rest or of uniform motion unless acted upon by some force is called inertia.

Question 3.
Which property has S.I. unit kilogram metres per second i.e., 1kg m/s?
Answer:
Momentum.

Question 4.
Give an example of scalar quantity.
Answer:
Mass.

Question 5.
Give an example of vector quantity.
Answer:
Momentum.

Question 6.
Calculate the total momentum of the bullet and the gun before firing.
Answer:
For both, it would be zero because both of them are at rest.

Question 7.
Which force slows down a moving bicycle when we try to stop?
Answer:
The force of friction.

Question 8.
Which kind of force of gravity work when an object is under free fall?
Answer:
Unbalanced force.

Question 9.
Which property of an object resist a change in their state of rest or motion?
Answer:
Inertia.

Question 10.
Which law of Newton is also known as Galileo’s law of inertia?
Answer:
First law.

Question 11.
Is force a vector quantity?
Answer:
Yes.

Question 12.
Which force of motion opposes motion of an object?
Answer:
Force of friction.

Question 3.
When action and reaction forces act on two different bodies, what kind of magnitude they have?
Answer:
Action and reaction forces act on two different bodies but they are equal in magnitude.

Question 14.
When gun moves in the backward direction, which kind of velocity is generated?
Answer:
Recoil.

Question 15.
Which factor of body is dependent on its mass?
Answer:
Inertia of a body depends on its mass.

Force and Laws of Motion Short Answer Type Questions

Question 1.
There are three solids made up of aluminium, steel and wood, of the same shape and same volume. Which of them would have highest inertia?
Answer:
Steel has highest inertia because it has greatest density and greatest mass, therefore, it has highest inertia.

Question 2.
Two balls of the same size but of different materials, rubber and iron are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If so, in which direction? Will they move with the same speed? Give reasons for your answer.
Answer:
If the breaks are applied suddenly then, the balls will start rolling in the direction in which the train was moving. Due to the application of the brakes, the train comes to rest but due to inertia the balls try to remain in motion, therefore, they begin to roll. Direction and speed of all balls will not be same because the masses of the balls are not the same, therefore, the inertial forces are not same on both the balls. Thus, the balls will move with different speeds.

Question 3.
Two identical bullets are fired, one by a light rifle and another by a heavy rifle with the same force. Which rifle will hurt the shoulder more and why?
Answer:
According to law of conservation of momentum or explanation by Newton’s laws of motion, light rifle will hurt the shoulder more.

Question 4.
A horse continues to apply a force in order to move a cart with a constant speed. Explain why?
Answer:
The force applied by the horse balances the force of friction

Question 5.
Suppose a ball of mass m is thrown vertically upward with an initial speed v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attains the speed v again before striking the ground. It implies that the magnitude of initial and final momentums of the ball are same. Yet, it is not an example of conservation of momentum. Explain.
Answer:
Law of conservation of momentum is applicable to isolated system (no external force is applied). In this case, the change in velocity is due to the gravitational force of earth.

Question 6.
In which of the following conditions work done will be equal to zero?
Answer:
In the absence of any one of the two conditions given below, work done will be equal to zero, that is work is not considered to be executed:

1. Force should act on the object.
2. Object must be displaced.

Question 7.
Define energy and explain its forms.
Answer:

1. Energy: Energy is the capacity of doing work. More the power, more will be energy and vice – versa. For example, a motorcycle has more energy than a bicycle.
2. Forms of energy: There are many forms of energy, such as kinetic energy, potential energy, mechanical energy, chemical energy, electrical energy etc.

Force and Laws of Motion Long Answer Type Questions

Question 1.
Give the formulation of work. In which conditions work can occur?
Answer:
Work = Force × Displacement
or W = F × s
where, W is work ‘F’ is force and ‘s’ is displacement.
If force, F = 0
Therefore, work done, W = 0, s = 0
If displacement, s = 0
Therefore, Work done, W = F × 0 = 0
It proves that, there are two conditions for work to occur or be done:

1. Force should act on the object.
2. Object must be displaced.

Question 2.
Give the conditions when work done become positive and negative.
Answer:
When force is applied in the direction of displacement, the work done is considered as positive.
i.e., W = F × s
When force is applied in opposite direction of displacement, the work done is considered as negative.
i.e., W = -F × s = -Fs.

Question 3.
Explain positive and negative work.
Answer:
1. Positive work:
If a force displaces the object in its direction, then the work done is positive.
Here,
W = Fd
Example:
Motion of ball falling towards ground where displacement of ball is in the direction of force of gravity.

2. Negative work. If the force and the displacement are in opposite directions, then the work is said to be negative.
Here,
W = -Fd.
Example:
If a ball is thrown in upward direction but the force due to earth’s gravity is in the downward direction.

Question 4.
A cyclist moving along a circular path of radius 63m completes three rounds in 3minutes.
1. The total distance covered by him during this time.
2. Net displacement of cyclist.
3. The speed of the cyclist
Answer:
1. Total distance covered
s = 2πr × t
s = 2πr × 3
= 2 × \(\frac { 22 }{ 7 }\) × 63 × 3 = 1188m

2. Displacement = Zero

3. Speed = \(\frac { Distance }{ Time }\)
= \(\frac { 1188 }{ 180 }\)
= 6.6 m/s.

Force and Laws of Motion Higher Order Thinking Skills (HOTS)

Question 1.
As per Newton’s third law, every force is accompanied by equal and opposite force. How then can anything move?
Answer:
According to the Newton’s third law, action and reaction are two equal and opposite forces but they act on different bodies. This make the motion of a body possible.

Question 2.
The passengers travelling in a bus fall ahead when a speeding bus stops suddenly. Why?
Answer:
When the speeding bus stops suddenly lower part of the body, a long with the bus comes to rest while the upper part tends to remain in motion due to inertia of motion. That is why passengers fall ahead.

Question 3.
A player always runs some distance before taking a jump. Why?
Answer:
A player always runs for some distance before taking a jump because inertia of motion helps him to take a longer jump.

Force and Laws of Motion Value Based Question

Question 1.
Sushil saw his karate expert breaking a slate. He tried to break the slate but Sushil’s friend stopped him from doing so and told him that it would hurt, one needs lot of practice in doing such activity.

1. How can a karate expert break the slate without any injury to his hand?
2. What is Newton’s third law of motion?
3. What value of Sushil’s friend is seen in the above case?

Answer:

1. A karate expert Sushil applies the blow with large velocity in a very short interval of time on the slate, therefore large force is exerted on the slate and it breaks.
2. To every action, there is an equal and opposite reaction, both act on different bodies.
3. Sushil’s friend showed the value of being responsible and caring friend.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:

1. OB = OC
2. AO bisect ∠A.

Solution:
Given
AB = AC
∠1 = ∠2, ∠3 = ∠4
To prove:

1. OB = OC
2. ∠5 = ∠6

Proof:
In ∆ABC,
AB = AC
∠B = ∠C
\(\frac{1}{2}\) ∠B = \(\frac{1}{2}\) ∠C
∠1 = ∠3 or ∠2 = ∠4
In ∆OBC
∠2 = ∠4
and so OB = OC
(In a A, sides opposite to equal angles are equal)
In ∆ABO and ∆ACO
BO = CO (proved)
∠1 = ∠3 (proved)
AB = AC (given)
∆ABO = ∆ACO (by SAS)
and so ∠5 = ∠6 (by CPCT)

Question 2.
In ∆ABC, AD is the perpendicular bisector of BC (see Fig. below). Show that AABC is an isosceles triangle in which AB =AC.

Solution:
Given
∠ABC = ∠ADC
To prove:
AB = AC
Proof:
In A ABD and A ACD
BD = CD (given)
∠ADB = ∠ADC (given each 90°)
AD = AD (common)
∴ ∆ABD = ∆ACD (BySAS)
and so AB = AC (by CPCT)

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. below). Show that these altitudes are equal.

Solution:
Given
AB = AC
∠E = ∠F (each 90°)
To prove: BE = CF
Proof:
In ∆ABE and ∆ACE
∠A = ∠A (common)
∠E = ∠F (each 90°)
AB = AC (given)
∆ABE = ∆ACE (byAAS)
and so BE = CF (by CPCT)

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. below). Show that:

1. ∆ABE ≅ ∆ACF
2. AB = AC,

i. e., ABC is an isosceles triangle.

Solution:
Given
BE = CF
∠E = ∠F (each 90°)
To prove:

1. ∆ABE = ∆ACE
2. AB = AC

Proof:
In ∆ABE and ∆ACF
∠A = ∠A (common)
BE = CF (given)
∠E = ∠F (each 90°)
∆ABE = ∆ACF (by AAS)
and so AB = AC (by CPCT)

Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see Fig. below). Show that ∠ABD = ∠ACD.

Solution:
Given
AB = AC
BD = CD
To prove
∠ABD = ∠ACD
Construction: Join AD
Proof:
In ∆ABD and ∆ACD
AD = AD (common)
AB = AC (given)
BD = CD (given)
∆ABD = ∆ACD (by SSS)
and so ∠ABD = ∠ACD (by CPCT)

Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. below). Show that ∠BCD is a right angle.
Solution:
Given: AB = AC
AD = AB
To show: ∠BCD = 90°
i.e., ∠2 + ∠3 = 90°

Proof:
AB = AC …..(1)
AB = AD …..(2)
From (1) and (2), we get
AC = AD
In ∆ABC
AB = AC
∠1 = ∠2
(In a A, angles opposite to equal sides are always equal) …p) …(3)
In ∆ACD
AC = AD
∠3 = ∠4
(In a A, angles opposite to equal sides are always equal) …(4)
In ∆BCD
∠1 + ∠2 + ∠3 + ∠4 = 180° (ASP)
∠2 + ∠2 + ∠3 + ∠3 = 180°
(∴ ∠1 = ∠2, ∠3 = ∠4)
2 (∠2 + ∠3) = 180°
(∠2 + ∠3) = 90°
∠BCD = 90°

Question 7.
ABC is a right angled triangle in which ∠A – 90° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆BAC
AB =AC
∠B = ∠C = x
∠A + ∠B + ∠C= 180°
∠B + ∠C = 180° – 90°
∠B + ∠C = 90°
2x = 90°
x = \(\frac{90^{\circ}}{2}\)

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
Given
ABC is an equilateral ∆
i. e., AB = BC = AC
To prove
∠A = ∠B = ∠C = 60°
Proof:
In ∆BAC
AB = AC
∠B = ∠C
(In a A, angles opposite to equal sides are always equal) ……(1)
AC = BC
∠A = ∠B
(In a A, angles opposite to equal sides are always equal) …..(2)
From (1) and (2), we get,

∠A = ∠B = ∠C = x (say)
∠A + ∠B + ∠C = 180° (ASP)
⇒ x + x + x = 180°
⇒ 3x = 180°
⇒ x = \(\frac{180^{\circ}}{3}\) = 60°
∴ ∠A = ∠B = ∠C = 60

Theorem 7.4.
SSS (Side-Side-Side) Congruence Theorem:
Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.
Given
In ∆s ABC and DEF we have,

AB =DE
BC = EE
and AC = DF
To prove:
∆ABC = ∆DEF
Construction:
Suppose BC is the longest side.
Draw EF such that EE = AB and FEG = ∠CBA.
Join GF and DG.

Proof:
In ∆s ABC and GEE, we have
AB = GE (Const.)
∠ABC = ∠GEF (Const.)
and BC = EF (Given)
∴ ∆ABC = ∆GEF (SAS Cong. Axiom)
∠A = ∠G (CPCT) …..(1)
and AC = GF (CPCT) …..(2)
Now AB = EG (Const.)
AB = DE (Given)
∴ DE = EG ……(3)
Similarly, DF = GF ……(4)
In ∆EDG
DE = EG (Proved above)
∴ ∠A = ∠2 (∠s opp. equal side) …..(5)
In ∆DFG
FD = FG (Proved above)
∴ ∠3 = ∠4 (∠s ppp. equal side) …..(6)
∴ ∠1 + ∠3 – ∠2 + ∠4 [From (5) and (6)]
i. e. ∠D = ∠G …..(7)
But ∠G = ∠A [From (1)]
∴ ∠A = ∠D …..(8)
In ∆s ABC and DEF,
AB – DE (Given)
AC = DF (Given)
∠A = ∠D [From (8)]
∆ABC ≅ ∆DEF (SAS Cong. Axiom)

Theorem 7.5.
RHS (Right Angle Hypotenuse Side) Congruence Theorem:
Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle.
Given
In ∆s ABC and DEF,
∠B = ∠E = 90°
AC = DF
BC = EF.

∆ABC ≅ ∆DEF
Construction:
Produce DE to M so that
EM = AB, Join ME.
Proof:
In ∆s ABC and MEF
AB = ME (Const.)
BC = EE (Given)
∠B = ∠MEF (each 90°)
∴ ∆ABC = ∆MEF (SAS Cong. Axiom)
Hence ∠A = ∠M (CPCT) …(1)
AC = MF (CPCT) …(2)
Also AC =DF (Given)
∴ DF = MF
∴ ∠D = ∠M (∠s opp. equal side of ADFM) …(3)
From (1) and (3), we have
∠A = ∠D …..(4)
Now, in ∆s ABC and DEF, we have
∠A = ∠D [From (4)]
∠B = ∠E (Given)
∴ ∠C = ∠F …..(5)
Again, in ∆s ABC and DEF, we have
BC = EF (Given)
AC = DF (Given)
∠C = ∠F [From (5)]
∴ ∆ABC = ∆DEF (SAS Cong. Axiom)

MP Board Class 9th Maths Solutions

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

Diversity in Living Organisms Intext Questions

Diversity in Living Organisms Intext Questions Page No. 80

Question 1.
Why do we classify organisms?
Answer:
Classification of organism make it easy to study the millions of organisms on this earth. Similarities among them is the basis to classify them into different classes. Classification makes study easier.

Question 2.
Give three examples of the range of variations that you see in life – forms around you.
Answer:
Variations observed in life are:

1. Size: Organisms vary greatly in size – from microscopic bacteria to elephants, whales and large trees.
2. Appearance: The colour of various animals is quite different. Number of pigments are found in plants. Their body – built also varies.
3. Life time: The life span of different organisms is varied.

Diversity in Living Organisms Intext Questions Page No. 82

Question 1.
Which do you think is a more basic characteristic for classifying organisms?
(a) the place where they live.
(b) the kind of cells they are made of. Why?
Answer:
(a) Different organisms may share same habitat but may have entirely different form and structure. So, the place where they live cannot be a basis of classification.

(b) The kind of cells they are. made of. Because placement of organism to other destination can create a easy confusion.

Question 2.
What is the primary characteristic on which the first division of organisms is made?
Answer:
The primary characteristic on which the first division of organisms is made is the nature of the cell – prokaryotic or eukaryotic cell.

Question 3.
On what basis are plants and animals put into different categories?
Answer:
Plants and animals are very different from each other but main basis to differentiate is “Mode of nutrition’’. Plants are autotrophs. They can make their food own while animals are heterotrophs which are dependent on others for food. Locomotion, absence of chloroplasts etc. also make them different.

Diversity in Living Organisms Intext Questions Page No. 83

Question 1.
Which organisms are called primitive and how are they different from the so-called advanced organisms?
Answer:
A primitive organism is the one which has a simple body structure and ancient body design or features that have not changed much over a period of time. As per the body design, the primitive organisms which have simple structures are different from those so – called advanced organisms which have complex body structure and organization.

Question 2.
Will advanced organisms be the same as complex organisms? Why?
Answer:
Yes, they are developed from same ancestor once. They have relatively acquired their complexity recently. There is a possibility that these advanced or ‘younger’ organisms acquire more complex structures during evolutionary time to compete and survive in the changing environment.

Diversity in Living Organisms Intext Questions Page No. 85

Question 1.
What is the criterion for classification of organisms as belonging to kingdom Monera or Protista?
Answer:
The organisms belonging to kingdom Monera are unicellular and prokaryotic whereas the organisms belonging to Kingdom Protista are unicellular and eukaryotic. This is the main criterion of their classification.

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

Question 3.
In the hierarchy of classification, which grouping will have the smallest number of organisms with a maximum of characteristics in common and which will have the largest number of organisms?
Answer:
In the hierarchy of classification, “species” will have the smallest number of organisms with a maximum of characteristics in common whereas “the kingdom” will have the largest number of organisms a Arthropoda.

Diversity in Living Organisms Intext Questions Page No. 88

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

Question 2.
How are pteridophytes different from the phanerogams?
Answer:
1. Pteridophyta: They have inconspicuous or less differentiated reproductive organs. They produce naked embryos called spores.
Examples:

• Ferns
• marsilea
• equisetum, etc.

2. Phanerogams: They have well developed reproductive organs. They produce seeds.
Example:

• Pinus
• cycas
• fir etc.

Question 3.
How do gymnosperms and angiosperms differ from each other?
Answer:
Gymnosperm:

1. They are non – flowering plants.
2. Naked seeds not enclosed inside fruits are produced.
3. Examples:
   • Pinus
   • Cedar
   • Fir
   • Cycas etc.

Angiosperm:

1. They are flowering plants.
2. Seeds are enclosed inside fruits.
3. Examples:
   • Coconut
   • Palm
   • Mango etc.

Diversity in Living Organisms Intext Questions Page No. 94

Question 1.
How do poriferan animals differ from coelenterate animals?
Answer:

Question 2.
How do annelid animals differ from arthropods?
Answer:

Question 3.
What are the differences between amphibians and reptiles?
Answer:

Question 4.
What are the differences between animals belonging to the Aves group and those in the mammalia group?
Answer:
Most birds have feathers and they possess a beak.Mammals do not have feathers and the beak is also absent. Birds lay eggs. Hence, they are oviparous. Some mammals lay eggs and some give birth to young ones. Hence, they are both oviparous and viviparous.

Diversity in Living Organisms NCERT Textbook Exercises

Question 1.
What are the advantages of classifying organisms?
Answer:
Advantages of classification:

1. Better categorization of living beings based on common characters.
2. Easier study for scientific research.
3. Better understanding of human’s relation and dependency on other organisms.
4. Helps in cross breeding and genetic engineering for commercial purposes.

Question 2.
How would you choose between two characteristics to be used for developing a hierarchy in classification?
Answer:
Gross character will form the basis of start of the hierarchy and fine character will form the basis of further steps of single hierarchy.
Examples:

• Presence of vertebral column in human beings can be taken under vertebrata.
• Presence of four limbs makes them members of Tetrapoda.
• Presence of mammary glands keeps them under mammalia.

Question 3.
Explain the basis for grouping organisms into five kingdoms.
Answer:
Basis of classification:

1. Number of cells: unicellular or multicellular.
2. Complexity of cell structure: Prokaryote and Eukaryote.
3. Presence or absence of cell wall.
4. Mode of nutrition.
5. Level of organization.

Question 4.
What are the major divisions in the Plantae? What is the basis for these divisions?
Answer:
Major divisions of Kingdom Plantae:

Question 5.
How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals?
Answer:
In plants, basic structure of their body is a major criteria based on which thallophytes are different from bryophytes. Apart from this, absence or presence of seeds is another important criteria. Gymnosperms and angiosperms are further segregated based on if seeds are covered or not. It is clear that it is the morphological character which makes the basis for classification of plants.

In animal, classification is based on more minute structural variations. So in place of morphology, cytology forms the basis. Animals are classified based on layers of cells, presence or absence of coelom. Further, higher hierarchy animals are classified based on the presence or absence of smaller features, like presence or absence of four legs.

Question 6.
Explain how animals in Vertebrata are classified into further subgroups.
Answer:
Vertebrata is divided into two super classes, viz. Pisces and Tetrapoda. Animals of pisces have streamlined body with fins and tails to assist in swimming. Animals of tetrapoda have four limbs for locomotion.
Tetrapoda is further classified into following classes:

1. Amphibia: Amphibians are adapted to live in water and on land. They can breathe oxygen through kin when under water.
2. Reptilia: These are crawling animals. Skin is hard to withstand extreme temperatures.
3. Aves: Forelimbs are modified into wings to assist in flying. Beaks are present. Body is covered with feathers.
4. Mammalia: Mammary glands are present to nurture young ones. Skin is covered with hair. Most of the animals are viviparous.

Diversity in Living Organisms Additional Questions

Diversity in Living Organisms Tissues Multiple Choice Questions

Question 1.
Find out incorrect sentence.
(a) Protista includes unicellular eukaryotic organisms.
(b) Whittaker considered cell structure, mode and source of nutrition for classifying the organisms in five kingdoms.
(c) Both Monera and Protista may be autotrophic and heterotrophic.
(d) Monerans have well defined nucleus.
Answer:
(d) Monerans have well defined nucleus.

Question 2.
Which among the following has specialised tissue for conduction of water?
(i) Thallophyta
(ii) Bryophyta
(iii) Pteridophyta
(iv) Gymnosperms
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv).
Answer:
(c) (iii) and (iv)

Question 3.
Which among the following produce seeds?
(a) Thallophyta
(b) Bryophyta
(c) Pteridophyta
(d) Gymnosperms.
Answer:
(d) Gymnosperms.

Question 4.
Which one is a true fish?
(a) Jellyfish
(b) Starfish
(c) Dogfish
(d) Silverfish.
Answer:
(c) Dogfish

Question 5.
Which among the following is exclusively marine?
(a) Porifera
(b) Echinodermata
(c) Mollusca
(d) Pisces.
Answer:
(b) Echinodermata

Question 6.
Which among the following animals have pores all over their body?
(a) Porifera
(b) Aves
(c) Mollusca
(d) Pisces.
Answer:
(a) Porifera

Question 7.
Which among the following have chi tin as cell wall?
(a) Sycon
(b) Yeast
(c) Jelly fish
(d) Euplectella.
Answer:
(c) Jelly fish

Question 8.
Which among the following is not a Monocotyledonous plant?
(a) Wheat
(b) Rice
(c) Maize
(d) Gram.
Answer:
(d) Gram.

Question 9.
Which among the following is not a dicotyledonous plant?
(a) Wheat
(b) Sunflower
(c) Mango
(d) Gram.
Answer:
(a) Wheat

Question 10.
An organism with a single cell is called _______ .
(a) Thallophyta
(b) Bryophyta
(c) Unicellular
(d) Multicellular.
Answer:
(c) Unicellular

Question 11.
The amphibians of the plant is _______ .
(a) Thallophyta
(b) Bryophyta
(c) Unicellular
(d) Multicellular.
Answer:
(b) Bryophyta

Question 12.
Plant bearing naked seeds are _______ .
(a) Thallophyta
(b) Bryophyta
(c) Unicellular
(d) Gymnosperm.
Answer:
(d) Gymnosperm.

Diversity in Living Organisms Very Short Answer Type Questions

Question 1.
Name a saprophyte and also tell, why are they called so.
Answer:
Aspergillus: They are called so because they obtain their nutrition from dead and decaying matter.

Question 2.
Why are lichens called dual organisms?
Answer:
Lichens are permanent symbiotic association between algae and fungi. Therefore, they are called dual organisms.

Question 3.
State the phylum to which centipede and prawn belong.
Answer:
Arthropoda.

Question 4.
Name one reptile with four – chambered heart.
Answer:
Crocodile.

Question 5.
Identify kingdom in which organisms do not have well defined nucleus and do not show multicellular body designs.
Answer:
Monera.

Diversity in Living Organisms Short Answer Type Questions

Question 1.
Why do we differentiate organism, give two main basis?
Answer:
Due to variation in various characteristics, we differentiate organism. Two main basis are mode of nutrition and habitat.

Question 2.
Which kingdom generate food on earth and initiate food chain?
Answer:
Plantae.

Question 3.
Which kingdom do not have cell wall to their cell?
Answer:
Animalia.

Question 4.
What do you understand by biodiversity?
Answer:
Biodiversity: The variety of living beings found in a particular geographical area is called biodiversity of that area. Amazon rainforests is the largest biodiversity hotspot in the world.

Question 5.
Why classification is required?
Answer:
Classification is necessary for the study of living beings in easy way. Without proper classification, it would be impossible to study millions of organisms which exist on this earth.

Question 6.
What was the basis of classification of Ancient Greek philosopher Aristotle?
Answer:
Aristotle classified living beings on the basis of their habitat. He classified them into two groups, i.e. those living in water and those living on land.

Question 7.
How can we divide organism on the basis of mode of nutrition ?
Answer:
On this basis, organisms can be divided into two broad groups, i.e. autotrophs and heterotrophs.

Question 8.
Define Monocotyledonous plants. Give examples.
Answer:
Monocotyledonous: There is single seed leaf in a seed. A seed leaf is a baby plant.
Examples:

• wheat
• rice
• maize, etc.

Question 9.
Give example of Dicotyledonous plants.
Answer:
Dicotyledonous plants: Mustard, gram, mango etc.

Question 10.
Give one difference between prokaryotes and eukaryotes
Answer:

1. Prokaryotes: When nucleus is not organized, i.e., nuclear materials are not membrane bound; the organism is called prokaryote.
2. Eukaryotes: When nucleus is organized, i.e., nuclear materials are membrane bound; the organism is called eukaryote.

Question 11.
What is the difference between unicellular and multicellular organism?
Answer:

1. Unicellular organism: An organism with a single cell is called unicellular organism.
2. Multicellular organism: An organism with more than one cell is called multicellular organism.

Question 12.
Write short notes on the following:
(a) Thallophyta
(b) Bryophyta
Answer:
(a) Thallophyta: The plant body is thallus type. The plant body is not differentiated into root, stem and leaves. They are known as algae also.
Examples:

• Spirogyra
• chara
• volvox
• ulothtrix etc.

(b) Bryophyta: Plant body is differentiated into stem and leaf like structure. Vascular system is absent, which means there is no specialized tissue for transportation of water, minerals and food. Bryophytes are known as the amphibians of the plant kingdom, because they need water to complete a part of their life cycle.
Examples:

• Moss
• marchantia.

Question 13.
What are cryptogams and phanerogams?
Answer:
Plant body is differentiated into root, stem and leaf. Vascular system is present. They do not bear seeds and hence are called cryptogams. Plants of rest of the divisions bear seeds and hence are called phanerogams.
Examples:

• Marsilear
• ferns
• horse tails etc.

Question 14.
How gymnosperms are different from angiosperms?
Answer:

1. Gymnosperms: They bear seeds. Seeds are naked i.e., are not covered. The word ‘gymnos’ means naked and ‘sperma’ means seed.
2. Angiosperms: The seeds are covered. The word ‘angios’ means covered. There is great diversity in species of angiosperm.

Question 15.
What is porifera?
Answer:
Porifera: These animals have pores all over their body. The pores lead into the canal system. They are marine animals. Examples:

• Sycon
• Spongilla
• Euplectella, etc.

Question 16.
What is coelenterata?
Answer:
Coelenterata: The body is made up of a coelom (cavity) with a single opening. The body wall is made up of two layers of cells (diploblastic).
Examples:

• Hydra
• jelly fish
• sea anemone, etc.

Question 17.
What is Platyhelminthes?
Answer:
The body is flattened from top to bottom and hence the name platyhelminthes. These are commonly known as flatworms. The body wall is composed of three layers of cells (triploblastic).
Example:

• Planaria
• liver fluke
• tapeworm etc.

Question 18.
What is Nematohelminthes and Annelida?
Answer:
Nematohelminthes: Animals are cylindrical in shape and the body is bilaterally symmetric and there are three layers in the body wall.
Example:

• Roundworms
• pinworms
• filarial parasite (Wuchereria) etc.

Annelida: True body cavity is present in these animals. The body is divided into segments and hence the name annelida.
Example:

• Earthworm
• leech etc.

Question 19.
Explain the followings:
(a) Arthropoda
(b) Mollusca
(c) Echinodermata
(d) Protochordata
(e) Chordata.
Answer:
(a) Arthropoda: Animals have jointed appendages which gives the name arthropoda. Exoskeleton is present which is made of chitin. This is the largest group of animals; in terms of number of species.
Examples:

• cockroach
• housefly
• spider
• prawn
• scorpion etc.

(b) Mollusca: The animal has soft body; which is enclosed in a hard shell. The shell is made of calcium carbonate.
Examples:

• Snail
• mussels
• octopus etc.

(c) Echinodermata: The body is covered with spines, which gives the name echinodermata. Body is radially symmetrical. The animals have well developed water canal system, which is used for locomotion.
Examples:

• Starfish
• sea urchins etc.

(d) Protochordata: Animals are bilaterally symmetrical, triploblastic and ceolomate. Notochord is present at least at some stages of life.
Examples:

• Balanoglossus
• herdmania
• amphioxus etc.

(e) Chordata: Animals have notochord, pharyngeal gill slits and post anal tail; for at least some stages of life. Phylum chordata is divided into many sub – phyla; out of which we shall focus on vertebrata.

Diversity in Living Organisms Long Answer Type Questions

Question 1.
What is the different levels of organizations in case multicellular organism?
Answer:
Level of organization: Even in case of multicellular organisms, there can be different levels of organization:
(a) Cellular level of organization: When a cell is responsible for all the life processes, it is called cellular level of organization.

(b) Tissue level of organization: When some cells group together to perform specific function, it is called tissue level of organization.

(c) Organ level of organization: When tissues group together to form some organs, it is called organ level of organization. Similarly organ system level of organization is seen in complex organisms.

Question 2.
“Classification of living organism is based on evolution.” Explain.
Answer:
It is a well – established fact that all the life forms have evolved . from a common ancestor. Scientists have proved that the life begun on the earth in the form of simple life forms. During the course of time, complex organism evolved from them. So, classification is also based on evolution. A simple organism is considered to be primitive while a complex organism is considered to be advanced.

Question 3.
Explain five kingdom classification by Robert Whittaker (1959).
Answer:
Five Kingdom Classification by Robert Whittaker (1959):
This is the most accepted system of classification. The five kingdoms and their key characteristics are given below:

1. Monera: These are prokaryotes; which means nuclear materials are not membrane bound in them. They may or may not have cell wall. They can be autotrophic or heterotrophic. All organisms of this kingdom are unicellular. Examples: bacteria, blue green algae (cyanobacteria) and mycoplasma.

2. Protista: These are eukaryotes and unicellular. Some organisms use cilia or flagella for locomotion. They can be autotrophic or heterotrophic. Examples: unicellular algae, diatoms and protozoans.

3. Fungi: These are heterotrohic and have cell wall. The cell wall is made of chitin. Most of the fungi are unicellular. Many of them have the capacity to become multicellular at certain stage in saprophytic. Some fungi live in symbiotic relationship with other organisms, while some are parasites as well.
Examples:

• Yeast
• penicillum
• aspergillus
• mucor etc.

4. Plantae: These are multicellular and autotrophs. The presence of chlorophyll is a distinct characteristic of plants, because of which they are capable of doing photosynthesis. Cell wall is present.

5. Animalia: These are multicellular and heterotophs. Cell wall is absent. They feed on decaying organic materials.

Diversity in Living Organisms Higher Order Thinking Skills (HOTS)

Question 1.
What are the differences between Platyhelminthes and Nematohelminthes?
Answer:

Question 2.
Differentiate between animals belonging to the Mammalia group and those in the Aves group.
Answer:
Differences between mammals and aves.

Diversity in Living Organisms Value Based Questions

Question 1.
Ashish, a IX class student, was studying chapter, ‘Diversity in Living Organisms’. He thought that all the fungi are harmful as these spoil food and cause various diseases. However, his elder sister Dimple told him that not all fungi are harmful as these are also used in making bread, vitamins and medicines.

1. Name any fungus which is the source of some medicine.
2. Name any fungus which is used in bread making.
3. What value are displayed by Ashish’s sister?

Answer:

1. Pencillium.
2. Yeast.
3. Dimple acted as elder sister and enhanced his younger brother’s scientific knowledge about fungi and their functions.

Question 2.
Coral is getting diminished in all the oceans due to global warming. People in Goa island protects their coral by not allowing people / tourist to take it away.

1. What is the phylum of coral?
2. What is coral made up of?
3. What value of people in Goa island is reflected here?

Answer:

1. Coelenterates is the phylum of coral.
2. It is made up of calcium carbonate.
3. They reflect the value of being responsible citizen, respecting environment and nature.

MP Board Class 9th Science Solutions