Class 9

RBSE Solutions for Class 9 Social Science Chapter 12 Physical Features of India are part of RBSE Solutions for Class 10 Social Science. Here we have given Rajasthan Board RBSE Class 9 Social Science Chapter 12 Physical Features of India.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 9
Subject	Social Science
Chapter	Chapter 12
Chapter Name	Physical Features of India
Number of Questions Solved	77
Category	RBSE Solutions

Rajasthan Board RBSE Class 9 Social Science Chapter 12 Physical Features of India

We Provide No Men are Foreign MCQ Questions for Class 9 English Poems Chapter 6 with Answers To Increase Knowledge concept Very Well.

TEXTBOOK QUESTIONS SOLVED

Multiple Type Questions (MCQs)

Question 1.
The line that passes through mid India is
(a) Tropic of Cancer
(b) Central line
(c) Equator
(d) Tropic of Capricorn
Answer:
(a).

Question 2.
The North mountain region is a part of
(a) Aravalli Range
(b) Pamir Plateau
(c) Armenia Plateau
(d) Kailash Mountain
Answer:
(b).

Question 3.
The most dense physical part of India is
(a) Great Mid Plain
(b) Thar Desert
(c) Southern Plateau
(d) None of these
Answer:
(a).

Question 4.
India Southernmost point, Indira Point is situated
(a) in Andaman
(b) in Nicobar
(c) in Lakshadweep
(d) in Minicoy
Answer:
(b).

Question 5.
In view of the area, Rajasthan occupies the place in India
(a) first
(b) second
(c) third
(d) fourth
Answer:
(a).

Question 6.
The smallest district of the state is
(a) Karauli
(b) Dungarpur
(c) Dhaulpur
(d) Sikar
Answer:
(c).

Question 7.
The highest peak between Nilgiri and Himalayas is
(a) Gurushikhar
(b) Sers
(c) Mahabaleshwar
(d) Achalgarh
Answer:
(a).

Very Short Answer Type Questions

Question 1.
What is Dhand?
Answer:
In desert region of Rajasthan, low land between sand dunes allows water to collect and form lakes. Such lakes in local language in Rajasthan is called Dhand and Sur (Talai).

Question 2.
Which physical division of India is the most densely populated?
Answer:
The Great Northern Plains

Question 3.
Where are Bhabar found?
Answer:
Bhabar are found in Shivalik range of Himalayas in foothill region of Satluj to Teesta river in between 8 to 16 km belt.

Question 4.
What are marg found?
Answer:
Margs are found in lower range of Himalayas known as Himachal. They are slopes covered with conical trees at upper part while grass in lower part. Such slopes are called margs in Kashmir.

Question 5.
What is the total area of Rajasthan?
Answer:
Total area of Rajasthan is 3,42,239 sq. km.

Short Answer Type Questions

Question 1.
Name the main divisions of Himalayas.
Answer:
The three main divisions of Himalayas are:

1. Greater Himalayas,
2. Lesser Himalayas,
3. Sub Himalayas.

Question 2.
Give the importance of the Southern Plateau.
Answer:
Importance of Southern Plateau are:

1. It is rich in mineral deposits.
2. Black soil found here is good for growing cotton.
3. Monsoon forests covered with expensive shesham and sandalwood trees are found here.
4. Natural waterfalls are located which are the basis of hydro-electricity production.
5. Pachmadhi, Mahabaleshwar, Ooty are famous hill stations are located here.

Question 3.
Distinguish between Eastern Ghats and Western Ghats of India.
Answer:
Western Ghats:

1. The western edge of the southern plateau is known as Western Ghats.
2. It is known as Sahayadri in North and Nilgiri in south.
3. They have steep slope towards Arabian Sea while in East gentle slope.
4. They have regular extension from Sahayadri ranges which are 1000 meter in Average height from Tapti valley to Kumari Antarip.
5. Bhor Ghat, Thai Ghat and Pal Ghat are important passes of this range.
6. Annai Muddi is the highest peak of this range.

Eastern Ghats:

1. It is less higher than Western Ghat, irregular and situated far away from the East coast.
2. They are situated parallel to eastern coast in 800 km length.
3. They are spread from North of Mahanadi valley to south in Nilgiris.
4. They have average height of 600 m.
5. The highest peaks of this region is of 1051 m.
6. It is dissected due to all east flowing rivers.

Question 4.
Name those islands of India which are formed due to drainage.
Answer:
Andaman Nicobar, Lakshadweep, Panbon, Hari, Parikud and Shriharikota are islands of India formed due to drainage.

Question 5.
Give the characterstics of desert region of western Rajasthan.
Answer:

1. It is largest region in size.
2. It can be divided into three parallel belts as:
   1. It extends from west to east, it is rich in sand dunes.
   2. Bangar plains made of sandy soil.
   3. Rahi floodplains formed due to small rivers.
3. This region has approximately 300 m average height.
4. Eastern boundary forms the rain divider of 25 cms.
5. Here sand hills are found which are called ‘Dhore’ in local language.
6. Physical environment has changed due to Indira Gandhi canal, earlier known as Rajasthan canal.

Long Answer Type Questions

Question 1.
Divide India into physical divisions. Describe any one of them.
Answer:
Physical divisions of India:

1. The Northern Mountains range
2. The Northern Vast Plain
3. The Thar Desert
4. The Peninsular Plateau
5. The Coastal Plain and Island groups

Description of the Northern Plain:
The Northern plain is stretched in bow-shape between the Himalayan mountain and the Southern plateau. It is about 2400 km long and its area is about 7 lakh sq km. This plain also known as the Sutluj-Ganga-Brahmaputra plain covers the parts of Punjab, Haryana, Uttarakhand, North Rajasthan, Bihar, Odisha, West Bengal and Jharkhand.

Map Skill

Question 1.
Show the major physical divisions on an outline map of India.
Answer:

Question 2.
Show the major high peaks of India on an outline map of India.
Answer:

Question 3.
Locate the physical division of Rajasthan on an outline map of India.
Answer:

ADDITIONAL QUESTIONS SOLVED

Multiple Choice Questions (MCQs)

Question 1.
The line of 82Vs° E longitude passes through the city of
(a) Allahabad
(b) Bareilly
(c) Lucknow
(d) Meerut
Answer:
(d).

Question 2.
Which of the following lies to the West of India?
(a) The Bay of Bengal
(b) The Arabian Sea
(c) The Pacific Ocean
(d) The Himalaya ranges
Answer:
(b).

Question 3.
India is
(a) a continent
(b) a sub-continent
(c) an island
(d) a mountain range
Answer:
(c).

Question 4.
Which of the following is not the sub¬division of the Himalaya region?
(a) The Greater Himalayas
(b) The Lesser Himalayas
(c) The Coastal Plain
(d) The trans-Himalayas
Answer:
(c).

Question 5.
The imaginary line which divides India into two parts is
(a) Equator
(b) longitudes
(c) Tropic of Capricorn
(d) Tropic of Cancer
Answer:
(d).

Question 6.
What is the shape of India?
(a) Quadrangular
(b) Triangular
(c) Circular
(d) Arc
Answer:
(a).

Question 7.
The old alluvium area in the Thar desert is called
(a) Khadar
(b) Bangar
(c) Bhabhar
(d) Tarai
Answer:
(b).

Question 8.
Which of the following is a physical division of India?
(a) The Northern Vast Plain
(b) The Northern Mountain region
(c) The Northern Valleys region
(d) The Coastal Plain
Answer:
(c).

Question 9.
Which of the following is determined in India on the basis of 82V2° East longitude?
(a) Greenwich Meantime
(b) Local Time
(c) Indian Standard Time
(d) International Time
Answer:
(c).

Question 10.
The Coastal line of India is:
(a) 6100 km
(b) 56100 km
(c) 1600 km
(d) 65100 km
Answer:
(a).

Question 11.
The irrigation facility is made available in the Thar desert of India by
(a) Rajiv Gandhi canal
(b) Luni river
(c) Indira Gandhi Canal
(d) Saraswati river
Answer:
(c).

Question 12.
Which of the following is not the peak of the Greater Himalaya?
(a) Mt. Everest
(b) Nanda Devi
(c) Godwin Austin
(d) Pirpanjal peak
Answer:
(d).

Question 13.
The height of Gurushikhar peak is
(a) 1722 mt
(b) 1772 mt
(c) 2717mt
(d) 1373 mt
Answer:
(a).

Question 14.
Aadwala mountain is another name for the:
(a) Aravalli
(b) Himalayas
(c) Mt. Abu
(d) Nilgiri
Answer:
(a).

Question 15.
The total area of Rajasthan is
(a) 3,24,329 sq. km
(b) 2,93,423 sq. km
(c) 3,42,239 sq. km
(d) 4,23,932 sq. km
Answer:
(c).

Question 16.
The south-eastern plateau region of Rajasthan is also known as the
(a) Mewar Plateau
(b) Haraute Plateau
(c) Marwar Plateau
(d) Bangar Plateau
Answer:
(b).

Question 17.
Rajasthan shares its southern boundary with
(a) Gujarat
(b) Maharashtra
(c) Barmer
(d) Uttar Pradesh
Answer:
(a).

Very Short Answer Type Questions

Question 1.
Which line denotes standard time of Bharat and from where does it pass?
Answer:
The line of 82 V20 East longitude denotes the standard time of India. It passes through Allahabad.

Question 2.
Write the names of two peaks of the Greater Himalayas.
Answer:
Two peaks of Greater Himalayas:

1. Mount Everest, and
2. Nanda Devi.

Question 3.
Write the names of two famous tourist places situated in sub-Himalayas.
Answer:
Two tourist places situated in sub- Himalayas:

1. Dheradun and
2. Haridwar.

Question 4.
What do you understand by Khadar and Bangar?
Answer:

1. Khadar is the area replenished with new alluvium every year by floods.
2. Bangar is the old alluvium area.

Question 5.
Name any two major peaks of the trans-Himalayas.
Answer:
Two major peaks of trans-Himalayas are:

1. Karakoram, and
2. Kailash.

Question 6.
On what basis has Sidney Burrad divided the Himalaya?
Answer:
Sidney Burrad has divided the Himalaya on the regional basis into Punjab, Kumaon, Nepal and Assam Himalaya.

Question 7.
Which physical division is known for the black soil and how is it beneficial?
Answer:
The Peninsular plateau is known for the black soil which is beneficial for the production of cotton.

Question 8.
Name the sub-divisions of the coastal plains.
Answer:
Sub-divisions of the coastal plains * are:

1. The Western coastal plains are divided into Konkan coast and Malabar coast.
2. The Eastern Coastal plains are divided into Northern Circar and Coromandel coast.

Question 9.
Which are the major sub-divisions of the Peninsular Plateau?
Answer:
The major sub-division of the Peninsular Plateau are Malwa Plateau,
Chota Nagpur plateau, Karnatak, plateau, Telangana Plateau and Deccan Plateau.

Question 10.
What is the proof to the presence of the large rivers in the area presently in the Thar Desert?
Answer:
Discovery of remains of Saraswati river is a proof to the presence of large rivers.

Question 11.
What is the position of India in the world regarding its land area?
Answer:
India, as regards its area, holds seventh place in the world.

Question 12.
Give other two names for the Greater Himalayas.
Answer:
The other two names for the Greater Himalayas are:

1. Himadri and
2. Main Himalayas.

Question 13.
What is another name for the Western Ghats?
Answer:
Western Ghats are also called Sahayadri hills.

Question 14.
In which Zone does the major part of Rajasthan lie?
Answer:
Major part of Rajasthan lies in the Temperate sone.

Question 15.
What is the number of districts in the Rajasthan state?
Answer:
There are about 33 districts in Rajasthan.

Question 16.
Which new district has been recently created in Rajasthan?
Answer:
Pratapgarh is the newly created district of Rajasthan.

Question 17.
Name the seven divisions of Rajasthan.
Answer:
Seven divisions of Rajasthan are Ajmer, Jaipur, Bikaner, Jodhpur, Udaipur, Kota and Bharatpur.

Question 18.
Which is the largest and the smallest district of Rajasthan?
Answer:
The largest district of Rajasthan is Jaisalmer and the smallest district is Dhaulpur.

Question 19.
What is the average density of population of Rajasthan?
Answer:
The average density of population of Rajasthan is 200 persons per sq. m.

Question 20.
Name the three parallel belts of the western desert region ,of Rajasthan.
Answer:
Three parallel belts of the western desert region of Rajasthan are Rahi, Bangar and Marusthali.

Question 21.
What is the number of districts, divisions and tehsils in Rajasthan?
Answer:
There are: 33 districts, 7 divisions and 244 tehsils in Rajasthan.

Short Answer Type Questions

Question 1.
What is the difference between Khadar and Bangar in the context of the Northern plain?
Answer:
Khadar is the area replenished with new alluvium every year by flood water whereas Bangar is the area with the old alluvium.

Question 2.
Which are the bordering countries of India?
Answer:
The bordering countries of India: Pakistan and Afghanishtan in the North-West, China, Nepal and Bhutan in the North and Myanmar and Bangladesh in the East and Sri Lanka in the South.

Question 3.
What is special about Lakshadweep?
Answer:
This island situated in the Arabian Sea on the west coast of India. Its literal meaning is ‘one lakh dweeps’. The coconut trees are found here in abundance. In fact, these are the coral islands.

Question 4.
How can it be proved that the slope of the Peninsular Plateau is towards east?
Answer:
Eastward slope of the Peninsular Plateau: Most of the rivers of the plateau namely Mahanadi, the Godavari, Krishna, Kaveri, etc. flow eastwards and drain into the Bay of Bengal. This is a clear evidence to the eastward slope of the Penninsular Plateau.

Question 5.
Which are the salt water lakes of the Thar Desert? Give reason for their strategic importance.
Answer:
Major salt water lakes of the Thar Desert are Lunkarnsar, Sambhar, Deedwana, Pachpadra, etc. Salt is prepared in these lakes. They have their strategic importance because of their location between India and Pakistan.

Question 6.
Which are the major islands of India and where are they located?
Answer:
Major islands of India are Andaman- Nicobar, Lakshadweep, Maldives, Pamban, Heir, Parikud and Sriharikota. They are located on the eastern and the western coasts of India and in the Arabian Sea and Bay of Bengal.

Question 7.
Which is the highest peak of the world situated in the Greater Himalayas?
Answer:
The highest peak of the world situated in the Greater Himalayas is Mount Everest also known as Gauri Shankar peak. Its height is 8848 mt. Tenzing was the first Indian to climb its peak.

Question 8.
Give the classification of the physical divisions of India.
Answer:
Physical divisions of India:

1. The western desert region
2. The semi-arid region
3. The Aravalli region
4. The eastern region
5. The south-eastern plateau region.

Question 9.
Write about the location of Rajasthan.
Answer:
It is situated in the North-West of India and is in kite shape. It lies between 23° 3’ to 30° 12’ N latitude and from 69° 30’ to 78° 17’ East longitudes. The Tropic of Cancer passes through the southern tip.

Question 10.
Name the districts of Rajasthan.
Answer:
The districts of Rajasthan are as follows:
Ajmer, Alwar, Baran, Banswara, Barmer, Bharatpur, Bhilwara, Bikaner, Bundi, Chittore, Churu, Dungarpur, Dhaulpur, Dausa, Ganganagar, Hanumangarh, Jaipur, Nagaur, Pali, Rajsamand, Sawai Madhopur, Sikar, Sirohi, Tonk, Udaipur, Jaisalmer, Jallore, Jhalawar, Jhunjhunu, Jodhpur, Karauli, Kota and Pratapgarh.

Question 11.
What is the significance of the Northern Plain of India?
Answer:
Northern Plain of India is significant for many reasons as given below:

1. Due to the formation of alluvial soil, it is the most fertile plain.
2. Water for drinking and irrigation is available in abundance due to network of many rivers.
3. Rivers of this plain facilitate navigation.
4. Places of waterfalls are good for producing hydro-electricity.
5. It has a network of roads and rail routes.
6. This northern plain abounds with many large cities, industrial, commercial and religious cities e.g. Delhi, Kanpur, Haridwar, Mathura, Varanasi, Amritsar, Lucknow, Agra, Patna, Kolkata, etc.

Question 12.
Discuss the physical dimensions of India.
Answer:
India is located in the Northern Hemisphere. Its shape is quadrangular. It extends between 8°4’ to 37° 4’ North latitude and from 68° 7’ to 97° 25’ to East longitude. Its extension from North to South is 3214 km and from East to West is 2933 km. Its area is 32,87,267 sq. km. Its landline boundary is 15,200 km and the coastal boundary is 6100 kms. India is the seventh largest country of the world. It has 2.4% of the total area of the world.

Question 13.
What is the location of the Northern Plain? Why is it also called the Satluj-Ganga and Brahmaputra Plain?
Answer:
The Northern plain is in bow-shape about 2400 km in length, located * between the Himalayan mountain and the Southern Plateau. This is also known as the Ganga-Satluj and Brahmaputra plain because the rivers- Ganga, Satluj and Brahmaputra rivers flow here.

Question 14.
Express geologists’ views about the position of the area prior to the formation of the Thar Desert.
Answer:
Prior to the formation of the Thar Desert, the land in the West of the Aravalli mountain and upto the Indus plain in north-west, was fertile. A large number of rivers were flowing here. Discovery of Saraswati river is an evidence to it. According to the geologists, due to the rising up of this area, the drainage here merged with either the river Ganga or the Indus. Later due to scarcity of rainfall aridity increased and it turned into a desert.

Question 15.
Write about the area of Rajasthan and also name its bordering areas.
Answer:
The total area of Rajasthan is 3,42,239 sq. km which is 10.43% of the whole country. Its maximum length from East to West is 869 kms and maximum breadth from North to South is 826 km. 1070 km area of the state forms international border with Pakistan. In East are the districts of Uttar Pradesh and Madhya Pradesh, in North are Punjab and Haryana and in the South are Gujarat and Madhya Pradesh. Total population of the state according to census 2011 is 6,85,48,437.

Question 16.
Write the characteristics of the South¬Eastern plateau region of Rajasthan.
Answer:
Characteristics of the South-Eastern Plateau of Rajasthan:

1. This is also known as the ‘Harauti Plateau’.
2. It includes the districts of Baran, Bundi, Kota and Jhalawar.
3. It has black fertile soil in abundance.
4. It has the Mukandra and Bundi hills.
5. The known Chulia waterfall is located near Bhainsrodgarh on the Chambal.
6. The plateau part lying between Bhainsrodgarh and Bijolia is called ‘Uppermal’.
7. The alluvial basin is formed in Baran and Kota from Chambal and its Kalisindh and Parvati tributaries.

Long Answer Type Questions

Question 1.
Discuss briefly the coastal plains and group of islands as one of the physical divisions of India.
Answer:
The Coastal plains of our country are situated with width in variations on both sides of the peninsular plateau. Such plains are the work of either rivers or sea.
They are divided into two groups:

1. The Western coastal plain
2. The Eastern coastal plain.

(a) The Western Coastal Plain: This extends from the Gulf of Cambay to Cape Comerin. Its length is 1600 km and the average width is 64 km In this Coastal plain the faster short rivers flow. This plain is larger in North. The Narmada, the Tapti and the Mandvi rivers flow here. Its northern part is called the Konkan coast and the southern is the Malabar coast. The major ports of this river are the Kandla, Mumbai, Marmagao Cochin and Mangalore.
(b) The Eastern Coastal Plain: It extends from the Delta of the Ganga to Cape Comerin. Its length is 1500 km and width is 16 to 480 km. It is divided into Northern Circar and Coromandel coast. The major rivers flowing here are the Mahanadi, the Krishna, the Godavari and Kaveri and the major ports are Vishakhapatnam, Chennai, Paradeep and Tuticorin. Important lakes here are Chilka, Pulikat and Kolleru.
Group of Islands:
Several islands of our country are situated on the Western and the Eastern Coasts and the Arabian Sea and Bay of Bengal. Major islands are Andaman-Nicobar, Lakshadweep, Maldives, Pamban, Heir, Parikud and Sriharikota. The southern most edge of India, i.e. the Indira Point is in Nicobar Island which was affected by tsunami waves in Dec. 2004. The Andaman-Nicobar island is situated in the south 1248 km far from Kolkata in the Bay of Bengal. In this group of islands there are about 205 islands.
In the coastal edges are found the Sundari trees. The major large sized islands are Northern Andaman, Central Andaman, Southern Andaman, Baratung and Ruthland. From the Andaman the Nicobar Islands is 128 km away in the south. Its northern islands are called Kar Nicobar, central is Khmorata and Tankadri and the southern islands are called Nicobar.
The Lakshadweep is situated in the Arabian Sea on the west coast of India. It means one lakh islands. The coconut trees are found here in abundance. In fact, these are coral islands.

Question 2.
Discuss the physical divisions of Rajasthan.
Answer:
Physical Divisions of Rajasthan:

1. The Western Desert region
2. The Semi-Arid region
3. The Aravalli region
4. The Eastern region
5. The South-Eastern Plateau region

(a) The Western Desert region: It comprised of comprising of Bikaner, Barmer, Jaisalmer, Churu and Western Nagaur the largest region with average height of 300 cm, has three parallel belts of Marusthali (sand dunes), Bangar (large plains) and Rahi (flood plains). Hills of sands are known as Dhore. A big change was brought about with Indira Gandhi Canal. 25 cms. Isohyte forms Eastern boundary.
(b) The Semi-Arid region: From North¬East to South to West in the Aravalli mountain ranges, average height 450 mt, Isohyte of 25 cm forms Western boundary. Pali, Jallore, Sikar, Jhunjhunu, Churu and Hanumangarh fall in the region. Its sub-region has Luni Basin, Shekhawati, Nagaur high and Ghaggar plain.
(c) The Aravalli region (Aadwala mountain): Extending in 692 kms length, the oldest mountain ranges of tfye world, fall in height from South¬West to North-East, Guru-Shikhar
(1722) mts. Its highest peak is also called ‘Santonka Shikhar’ by Col. Tod. Its sub-regions- North-East hills, Central Aravalli, Mewar hill or Bhorat and Abu mountain. Dsuri and Hathi are two important passes.
(d) The Eastern Plain Region: It is spread in the eastern part includes lower parts of Chambal basin is called Chappan, Plain, Banas and Mahi basin. Two parts of Banas Basin- Malpurakarauli and Mewar plain. Mahi and Banas are the major rivers.
(e) The South-Eastern Plateau region or the Harauti region includes Baran, Bundi, Kota and Jhalawar districts. It is rich in black soil and has famous Chulia waterfall, the Mukandra and Bundi hills.

We hope the given RBSE Solutions for Class 9 Social Science Chapter 12 Physical Features of India will help you. If you have any query regarding Rajasthan Board RBSE Class 9 Social Science Chapter 12 Physical Features of India, drop a comment below and we will get back to you at the earliest.

Rajasthan Board RBSE Solutions for Class 9 Science Chapter 8 Motion Textbook Exercise Questions and Answers.

RBSE Class 9 Science Solutions Chapter 8 Motion

RBSE Class 9 Science Chapter 8 Motion InText Questions and Answers

Page No. 100

Question 1.
An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer:
Yes, an object can have zero displacement provided that final position of the object coincides with its initial position.
For example : If a person moves from home to store and stands on place from where he started then here displacement will be zero.

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

Boundary of a square field of side 10 m
Given,
Side of the square field = 10 m
So, perimeter = 10 m × 4 = 40 m
Farmer moves along the boundary in 40 s.
Displacement after 2m 20s = 2 × 60s + 20s = 140 s = ?
Since in 40 s farmer moves 40 m.
Therefore, in 1 s distance covered by farmer = \(\frac{40}{40}\)m = 1 m
Therefore, in 140 s distance covered by farmer = 1 × 140 m = 140 m
Now, number of rotation to cover 140 along the boundary =

= \(\frac{140}{40}\) = 3.5 round
Thus, after 3.5 round farmer will reach at point C of the field.
Therefore, Displacement AC = \(\sqrt{(10 \mathrm{~m})^{2}+(10 \mathrm{~m})^{2}}\)
= \(\sqrt{100 m^{2}+100 m^{2}}\)
= \(\sqrt{200 \mathrm{~m}^{2}}\)
= 10\(\sqrt{2}\)m
= 10 × 1.414 = 14.14 m
Thus, after 2 minutes 20 seconds the displacement of farmer will be equal to 14.14 m northeast from initial position.

Question 3.
Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.
Answer:
None of the statement is true for displacement. First statement is false because displacement can be zero. Second statement is false as displacement is less than or equal to the distance travelled by the object.

Page No. 102

Question 1.
Distinguish between speed and Velocity.
Answer:

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

Question 2.
Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer:
The magnitude of average velocity of an object is equal to its average speed when an object is moving in a straight line in a given direction only.

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

Question 4.
What does the path of an object look like when it is in uniform motion?
Answer:
When an object having uniform motion, it moves along a straight line path.

Question 5.
During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10⁸ ms^-1.
Answer:
Speed = 3 × 10⁸ ms^-1
Time = 5 min = 5 × 60 = 300 seconds
Distance = Speed × Time
Distance = 3 × 10⁸ ms^-1 × 300 seconds = 9 × 10¹⁰ m

Page No. 103

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

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

Question 2.
A bus decreases its speed from 80 km h^-1 to 60 km h^-1 in 5 s. Find the acceleration of the bus.
Answer:
Initial speed of the bus, u = 80 km/h = 80 × \(\frac{5}{18}\) = 22.2 m/s
Final speed of the bus, ν = 60 km/h
= 60 × \(\frac{5}{18}\) = 16.6 m/s
Time taken to decrease the speed, t = 5 s
Acceleration, a = \(\frac{v-u}{t}=\frac{16.66-22.22}{5}\) = -1.112 m/s²

Question 3.
A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h^-1 in 10 minutes. Find its acceleration.
Answer:
Initial velocity of the train, u = 0
Final velocity of the train, ν = 40 km/h = 40 × \(\frac{5}{18}\) = 11.11 m/s
Timetaken, t = 10 min = 10 × 60 = 600 s
Acceleration, a = \(\frac{v-u}{t}=\frac{11.11-0}{600}\) = 0.0185 m/s²
Hence, the acceleration of the train is 0.0185 m/s².

Page No. 107

Question 1.
What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Answer:
When the motion is uniform, the distance-time graph is a straight line having a definite slope. The slope tells us the value of constant velocity of the object.

When the motion is non-uniform, the distance-time graph is a curve whose slope varies from point to point.

Question 2.
What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer:
If distance-time graph is a straight line parallel to the time axis, the body is at rest and the distance of the object does not change with time.

Question 3.
What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Answer:
If speed-time graph is a straight line parallel to the time axis, the object is moving uniformly.

Question 4.
What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:
Area occupied below the velocity-time graph of an object gives the magnitude of displacement or the distance covered during the given time interval.

Page No. 109-110

Question 1.
A bus starting from rest moves with a uniform acceleration of 0.1 m s^-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Answer:
Initial speed of the bus, u = 0
Acceleration, a = 0.1 m/s²
Time taken, t = 2 minutes = 120 s
(a) ν = u + at
ν = 0 + 0.1 × 120
ν = 12 ms^-1
Speed acquired by the bus is 12 m/s.

(b) According to the third equation of motion :
ν² = u² = 2 as
Where, s is the distance covered by the bus
(12)² – (0)² = 2(0.1) s
s = 720 m
Distance travelled by the bus is 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 s^-2. Find how far the train will go before it is brought to rest.
Answer:
Initial speed of the train, u = 90 km/h = 25 m/s
Fined speed of the train, ν = 0 (finally the train comes to rest)
Acceleration = – 0.5 m s^-2
According to third equation of motion :
ν² = u² + 2 as
(0)² = (25)² + 2 (- 0.5) s
Where, s is the distance covered by the train
s = \(\frac{25^{2}}{2(0.5)}\) = 625 m
The train will cover a distance of 625 m before it comes to rest.

Question 3.
A trolley, while going down an inclined plane, has an acceleration of 2 cm s^-2. What will be its velocity 3 s after the start?
Answer:
Initial velocity of trolley, u = 0 cm s^-1
Acceleration, a = 2 cm s^-2
Time, t = 3 s
We know that final velocity, ν = u + at = 0 + 2 × 3 cm s^-1
Therefore, the velocity of train after 3 seconds = 6 cm s^-1

Question 4.
A racing car has a uniform acceleration of 4 ms^-2. What distance will it cover in 10 s after start?
Answer:
Initial velocity of the car, u = 0 ms^-1
Acceleration, a = 4 m s^-2
Time, t = 10 s
We know, distance, s = ut + (\(\frac{1}{2}\))at²
Therefore, distance covered by car in 10 seconds = 0 × 10 + (\(\frac{1}{2}\)) × 4 × 10²
= 0 + (\(\frac{1}{2}\)) × 4 × 10 × 10 m
= (\(\frac{1}{2}\)) × 400 m = 200 m

Question 5.
A stone is thrown in a vertically upward direction with a velocity of 5 m s^-1. If the acceleration of the stone during its motion is 10 m s^-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer:
Given initial velocity of stone, u = 5 m s^-1
Downward of negative acceleration, a = – 10 m s^-2
(acceleration is taken negative because it is in downward direction, i.e. the direction of acceleration is opposite to that of velocity.)
We know that 2as = ν² – u²
or, s = \(\frac{v^{2}-u^{2}}{2 a}\)
Therefore, height attained by the stone, s = \(\frac{0^{2}-5^{2}}{2 \times(-10)}\)m
= \(\frac{-25}{-20}\)m = 1.25 m
Also we know that final velocity, ν = u + at
Time, t = \(\frac{v-u}{a}\)
Therefore, time, t taken by stone to attain the height s = \(\frac{0-5}{-10}\) = 0.5 S

RBSE Class 9 Science Chapter 8 Motion Textbook Questions and Answers

Question 1.
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer:
Diameter of circular track (D) = 200 m
Radius of circular track (r) = \(\frac{200}{2}\) = 100 m
Time taken by the athlete for one round (t) = 40 s
Distance covered by athlete in one round (s) = 2πr = 2 × \(\frac{22}{7}\) × 100

After taking start from position x, the athlete will be at position y after 3\(\frac{1}{2}\) rounds as shown in figure

Hence, displacement of the athlete with respect to initial position at x = xy
= Diameter of circular track
= 200 m

Question 2.
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Answer:
(a) Total distance covered from AB = 300 m
Total time taken = 2 × 60 + 30 s = 150 s

(b) Total distance covered from AC = AB + BC
= 300 + 100 m = 400 m
Total time taken from A to C = Time taken for AB + Time taken for BC
= (2 × 60 + 30) + 60 s = 210 s

Displacement (s) from A to C = AB – BC
= 300 – 100 m = 200 m
Time (t) taken for displacement from AC = 210 s

Question 3.
Abdul, while driving to school, computes the average speed for his trip to be 20 km h^-1. On his return trip along the same route, there is less traffic and the average speed is 30 km h^-1. What is the average speed for Abdul’s trip?
Answer:
Let the distance of Abdul’s school from his residence be s km.
As during his trip from his residence to school, his average speed,
ν₁ = 20 km h^-1
Hence, in covering a distance s km, he takes the time
t₁ = \(\frac{s}{v_{1}}=\frac{s}{20} \mathrm{~h}\)
Similarly, dining the return trip, he takes the time
t₂= \(\frac{s}{v_{2}}=\frac{s}{30} \mathrm{~h}\)

Question 4.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s^-2 for 8.0 s. How far does the boat travel during this time?
Answer:
Given, initial velocity of motorboat, u = 0
Acceleration of motorboat, a = 3.0 m s^-2
Time under consideration, t = 8.0 s
We know that distance, s = ut + \(\frac{1}{2}\)at²
Therefore, the distance travel by motorboat = 0 × 8+ (\(\frac{1}{2}\))3.0 × 8²
= \(\frac{1}{2}\) × 3 × 8 × 8m = 96m

Question 5.
A driver of a car travelling at 52 km h^-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h^-1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Answer:
PR and SQ are the speed-time graph for given two cars with initial speeds 52 km h^-1 and 3 km h^-1 respectively (given in figure).

Speed-time graph for given two cars
Distance travelled by first car before coming to rest = Area of ΔOPR

Distance travelled by second car before coming to rest = Area of ΔOSQ

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

Distance-time graph of three objects A, B and C
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?
Answer:
(a)

Therefore, speed = slope of the graph
Since slope of object B is greater than objects A and C, it is travelling the fastest.
(b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road.
(c)

7 square box = 4 km
∴ 1 square box = \(\frac{4}{7}\) km
C is 4 blocks away from origin, therefore, initial distance of C from origin = \(\frac{16}{7}\) km
Distance of C from origin when B passes A = 8 km
Thus, distance travelled by C when B passes A = 8 – \(\frac{16}{7}\) = \(\frac{(56-16)}{7}\) = \(\frac{40}{7}\)
= 5.714 km

(d)

Distance travelled by B by the time it passes C = 9 square boxes
9 × \(\frac{4}{7}\) = \(\frac{36}{7}\) = 5.143 km

Question 7.
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s^-2, with what velocity will it strike the ground? After what time will it strike the ground?
Answer:
Let us assume, the final velocity with which ball will strike the ground be ‘ν’ and time it takes to strike the ground be ‘t’
Initial velocity of ball, u = 0
Distance or height of fall, s = 20 m
Downward acceleration, a = 10 m s^-2
As we know, 2as = ν² – u²
ν² = 2as + u²
= 2 × 10 × 20 + 0 = 400
∴ Final velocity of ball, ν = 20 ms^-1
t = \(\frac{(v-u)}{a}\)
∴ Time taken by the ball to strike = \(\frac{(20-0)}{10}\) = \(\frac{20}{10}\) = 2 seconds

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

(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Answer:
(a)

The shaded area which is equal to \(\frac{1}{2}\) × 4 × 6 = 12 m represents the distance travelled by the car in the first 4 s.
(b) The part of the graph between time 6 s to 10 s represents uniform motion of the car.

Question 9.
State which of the following situations are possible and give an example for each of these :
(a) an object with a constant acceleration but with zero velocity.
(b) an object moving in a certain direction with an acceleration in the perpendicular direction.
Answer:
(a) Possible, a ball has zero velocity when a ball is thrown up at maximum height.
It will have constant acceleration due to gravity, which is equal to 9.8 m/s².
(b) Possible, a car’s acceleration is perpendicular to its direction when a car is moving in a circular track.

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 circular orbit, r = 42250 km
Time taken to revolve around the Earth, t = 24 h
Speed of a circular moving object, ν = (2πr)/t

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 7 Triangles Exercise 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 figure). If AD is extended to intersect BC at P, show that:
(i) ∆ABD ≅ AACD
(ii) ∆ABP ≅ AACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.

Answer:
(i) In ∆s ABD and ACD, we have :
AB = AC (Given)
BD = CD (Given)
and AD = AD (Common)
∴ By SSS criterion of congruence, we have:
∆ABD ≅ ∆ACD.

(ii) In ∆s ABP and ACP, we have :
AB = AC (Given)
∠BAP = ∠CAP
[∵ ∆ABD ≅ ∆ACD ⇒ ∠BAD = ∠DAC ⇒ ∠BAP = ∠CAP (CPCT)]
and AP = AP (Common)
∴ By SAS criterion of congruence, we have :
∆ABP ≅ ∆ACP.

(iii) Since ∆ABD ≅ ∆ACD, therefore
∠BAD = ∠CAD …….. (1)
⇒ AD bisects ZA
⇒ AP bisects ZA.
Now, in ∆s BDP and CDP, we have:
BD = CD (Given)
BP = CP (∵ ∆ABP ≅ ∆ACP ⇒ BP = CP)
and DP = DP (Common)
∴ By SSS criterion of congruence, we have:
∆BDP ≅ ∆CDP
∴ ∠BDP = ∠CDP
⇒ DP bisects ∠D ⇒ AP bisects ∠D. ……… (2)
Combining (1) and (2), we get:
AP bisects ∠A as well as ∠D.

(iv) Since AP stands on BC
∴ ∠APB + ∠APC = 180° (Linear pair)
But ∠APB = ∠APC (∵ ∆APB ≅ ∆APC ⇒ ∠APB = ∠APC)
∴ ∠APB = ∠APC = \(\frac{180^{\circ}}{2}\) = 90°.
Also, BP = CP (Proved above)
So, AP is perpendicular bisector of BC.
Hence proved.

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that:
(i) AD bisects BC
(ii) AD bisects ZA.
Answer:

AD is the altitude drawn from vertex A of an isosceles ∆ABC to the opposite base BC so that AB = AC,
∠ADC = ∠ADB = 90°.
Now, in As ADB and ADC, we have :
Hyp. AB = Hyp. AC (Given)
AD = AD (Common)
and ∠ADB = ∠ADC
∴ By RHS criterion of congruence, we have:
∆ADB ≅ ∆ADC
∴ BD = DC and ∠BAC = ∠DAC
(∵ Corresponding parts of congruent triangles are equal.)
Hence, AD bisects BC, which proves (i), and AD bisects ∠A, which proves (ii).

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

(i) ∆ABM ≅ ∆PQN
(ii) ∆ABC ≅ ∆PQR
Answer:
Two ∆s ABC and PQR in which AB = PQ, BC = QR and AM = PN.
Since AM and PN are medians of As ABC and PQR respectively.
Now, BC = QR
BM = QN
Now, in ∆s ABM and PQN, we have :
AB = PQ (Given)
BM = QN [From (1)]
and, AM = PN (Given)
∴ By SSS criterion of congruence, we have:
∆ABM ≅ ∆PQN, which proves (i)
So, ∠B = ∠Q ………. (2)
(∵ Corresponding parts of congruent triangles (CPCT) are equal.)
Now, in ∆s ABC and PQR, we have :
AB = PQ (Given)
∠B = ∠Q [From (2)]
and BC = QR (Given)
∴ By SAS criterion of congruence, we have:
∆ABC ≅ ∆PQR, which proves (ii).
Hence proved.

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

In ∆s BCF and CBE, we have :
∠BFC = ∠CEB
Hyp. BC = Hyp. BC
CF = BE
∴ By RHS criterion of congruence, we have :
∆BCF ≅ ∆ CBE
So, ∠FBC = ∠ECB
(∵ Corresponding parts of congruent triangles are equal.)
Now, in ∆ABC, ∠ABC = ∠ACB (∵ ∠FBC = ∠ECB)
∴ AB = AC
(∵ Sides opposite to equal angles of a triangle are equal.)
∴ ∆ABC is an isosceles triangle.
Hence proved.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Answer:

In ∆s ABP and ACP, we have:
AB = AC (Given)
AP = AP (Common)
and ∠APB = ∠APC (∵ Each = 90°)
∴ By RHS criterion of congruence, we have:
∆ ABP ≅ ∆ ACP
So, ∠B = ∠C
(∵ Corresponding parts of congruent triangles are equal.)
Hence Proved.