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You can Download An Empire Across Three Continents Questions and Answers, Notes, Plus One History Chapter Wise Questions and Answers Kerala Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Plus One History Chapter Wise Questions and Answers Chapter 3 An Empire Across Three Continents

Class 11 History Chapter 3 Extra Questions And Answers Question 1.
It is usual to divide the history of the Roman Empire into two phases. What are they? Explain.
Answer:
The history of the Roman Empire is usually divided into two phases the Early Empire and the Late
Empire. The Early Empire is the history up to the end of the 3^rd century. The later period from the 4^th to the 7^th century is the. Late Empire. It is also called the Late Antiquity.

An Empire Across Three Continents Question Answers Question 2.
Raju said that there were big differences between the Roman Empire and the Iran Empire. Do you agree with this opinion? Why?
Answer:
Yes, I do. There were many differences between the Roman Empire and the Iran Empire. In the Roman Empire, there was more cultural diversity than in the Iran Empire. The majority of the Iran people belonged to the ‘Iranian’ race. But in the Roman Empire, there were many regions and different cultures. In the Roman Empire, many different kinds of people stayed together under one common government.

In the Roman Empire, there was also much diversity in languages. The Iranians used Aramaic language. But in the Roman Empire, there were different languages. Latin and Greek were the administrative languages. The upper classes in the Eastern part of the Empire used Greek whereas those in the Western part used Latin in their writing. Different from Iran, all the people who lived in the Roman Empire were the subjects of a single Emperor.

Class 11 History Chapter 3 Extra Questions And Answers In English Question 3.
Augusts Caesar was the first Emperor of Rome. Explain.
Answer:
The Roman Republic lasted 500 years (BC 509 – 27). Towards the end period of the Republic, military officials like Julius Caesar had established their dominance in the Empire. In BC 27, Octavian, the adopted son, and heir of Julius Caesar overturned the Republic and got into power. He ruled by the name Augustus Caesar and he became the first Emperor of Rome.

The administration established by Augustus was known as the Principate. He ruled by taking the title ‘Princeps’which means the first citizen. The only ruler and centre of authority in the Roman Empire was Emperor Augustus. He took the title Princeps just because of his admiration for the Senate. It was a strategy to please the Senate and make it dance to his tunes.

An Empire Across Three Continents Extra Questions Question 4.
In the history of Rome, the Senate had much significance. Examine the validity of this statement.
Answer:
Senate was a body representing rich landowners and other wealthy people in Rome. In administration, the Senate had the second place after the Emperor. Most of Roman history is written by historians who were closely associated with the Senate. Even Emperors were evaluated on the basis of their relations with the senate.

All the emperors who were angry with the Senate or refused to cooperate with it were depicted as bad emperors. Some Senators were interested in going back to the old Republican times, but they knew it was an impossibility.

An Empire Across Three Continents Important Questions Question 5.
In the history of Rome, Army had much significance. Examine the validity of this statement.
Answer:
The army of Rome was quite different from that of the Persians, who were the adversaries of Rome. Recruitment to the Persian Army was done by force. For sections of the society, military service was compulsory. But the Roman army was a professional one.

The soldiers were appointed with the expectation that they would serve the army for at least 25 years and they were paid wages for their service. This army was a special feature for the Roman army. It was one of the most powerful and organized institutions in the Roman Empire. It had the power even to decide the fate of the emperors.

Class 11 History Chapter 3 Important Questions Question 6.
It was urbanization in the Roman Empire that enabled the Emperor to rule and control it. Prepare a seminar paper on this topic.
Answer:
It was urbanization that enabled the Emperor to rule and control the vast Roman Empire which had 60 million people and extensive and diverse regions. The real strength of the imperial administration was . the great urban centres in the Mediterranean shores. Carthage, Alexandria, and Antioch were big urban centres of those times. It was through the cities that taxes from the rural areas of the provinces were collected by the Government. A good measure of the income of the empire came from rural areas.

The Upper classes in the Provinces cooperated with the Roman Administration in administering their areas and to collect taxes. The Provinces and the Upper classes there were an integral part of the Roman administration. In the 2^ndand 3^rd centuries it was the Upper Classes in the Provinces that supplied the Cadres and army commanders for the Provincial administration.

Gradually, they became a new elite class of administrators and army commanders. Since they had the backing of the Emperors, they were able to become stronger than the senators. Emperors like Gallienus helped the new elite class to establish. themselves. Gallienus avoided senators from any position in the army. Thus he prevented the imperial administration falling into their hands. This strengthened the hands of the elite class.

In short, during the Early Roman Empire period, the importance of the Provinces increased tremendously. The majority of the Cadres in the administration as well as in the army were from the Provinces. As the power of the Provinces increased, the importance of Italy was lost.

Even her dominance in the Senate was lost. Until the 3^rd century Italy had dominated the Senator As the members from the Provinces increased in the Senate, Italy’s position became weak. Thus Italy degenerated politically and economically and an urbanized and new elite class began to come up.

Urban centres which contained villages and which had their own magistrates and corporations were called “Cities” by the Romans. Villages were under, the jurisdiction of the Citie. Some Villages were able to raise themselves to the status of cities. The reverse also happened. Cities became villages.lt all depended on the sweet will of the government.

Living in cities had an advantage. Curing famine the distribution of food in the cities was far better than in villages. Public baths were another important feature of the. Roman urban life. The people in the Roman cities used to enjoy entertainment of high standards. One Roman calendar shows that there were shows of various kinds on 176 days a year.

Chapter 3 History Class 11 Questions And Answers Question 7.
The 3^rd century was a time of crisis for the Roman Empire. Do you agree with this view? Clarify.
Answer:
As far as the Roman Empire was concerned, the 1^st and 2^nd centuries were those of peace, development and economic growth. But in the 3^rd century the Empire began to show signs of trouble. It was foreign attacks that-caused the problem. In 225 AD the Sassanian Dynasty came to power in Iran and this was a great threat to the Roman Empire. When the Irani an army marched forward with Euphrates in sight, it became a big crisis for the Roman Empire.

In one of his famous stone inscriptions, it is written that Shapurl, who was the ruler of Iran, destroyed a  Roman army numbering 60,000 and captured Antioch, the Eastern capital of the Roman Empire. The Roman Empire faced attacks by Barbarians. The Romans scornfully called the Tribal people who lived in the northern border of the Roman Empire as Barbarians to mean that they were uncivilized. These Tribal Groups belonged to the Germanic race and inducted Alamanni’s, Franks and Goths. They started infiltrating into the Rhine-Danube boundaries.

Between 233 and 280 they attacked the Roman Provinces that lay between Black Sea and Alps in Southern Germany. The Romans were forced to quit from areas on the other side of River Danube. During this period Emperors had to spend a lot of time in the battlefronts. In 47 years, 25 emperors ascended the throne and this shows the extent of the Crisis the Roman Empire faced.

An Empire Across Three Continents Class 11 Important Questions Question 8.
“In the Roman Empire, women had a high place.” Anwar.
“In the Roman Empire, the condition of women was bad.” Hassan.
This was part of a classroom debate. Which one do you support? Why?
Answer:
The Roman society was a patriarchal one. Women were under the power and control of their husbands. Husbands would even physically abuse their women. Fathers had legal control over their children. They had even the right to Rill the children whom they did not like by leaving them in the terrible cold.

Women, however, had the right to own and handle property. Women did not give their family property which they got during marriage to their husbands. They would give whatever money they got as dowry to their husbands, but they retained their other properties as they had the right to keep them and use them the way they wanted.

An Empire Across Three Continents Class 11 Extra Questions Question 9.
In the Roman Empire, there were a lot of cultural diversities. Explain.
Answer:
In the Roman Empire, there were a lot of cultural diversities. We could see these cultural diversities in their different religious beliefs, their deities, their languages, their dresses, their food, their organizational forms whether Tribal or non-Tribal and their dwellings. In short the cultural diversities in the – Roman Empire were reflected in many things and in many ways.

Ch 3 History Class 11 Important Questions Question 10.
Present a seminar paper on the economic development of the Roman Empire.
Areas to be considered: Production, trade, agriculture, technology, and other factors.
Answer:
In the Roman Empire, there were many ports, mines, quarries, brick-making kilns, factories producing olive oil, etc. Things like wheat, wine, olive oil, etc. were manufactured in a large scale. But they also got more from outside. They imported them from Spain, the Gallic provinces, North-Africa, Egypt and Italy. Wines and olive oil were brought in huge amphorae (tall jar or jug). Plenty of broken pieces of these jars and jugs have been found by archaeologists.

In the Roman Empire trade in Spanish olive oil. had achieved great progress. In the 140-160 period, the trade had reached its peak. During this period the Spanish olive oil was taken in jars called Dressel 20. From the Mediterranean sites, plenty of such jars have been found.

It shows that Spanish olive oil was much traded. Evidence shows that the Spanish producers of olive oil were able to capture the market from Italian competitors. They succeeded as they gave high-quality olive oil at cheaper prices.

The success the Spanish producers of olive gained in the olive oil market was repeated by North African producers of olive oil. In the 3^rd and 4^th centuries, the olive oil market was under the control of the olive estates of this region. But in the 5^th and 6^th centuries the monopoly of the North African producers was lost. Oriental countries like Egypt, Southern Asia Minor (Turkey), Syria arid Palestine captured the olive oil and wine markets.

In the Roman Empire, there were extraordinarily fertile places. Strabo and Pliny point out that Compariia (Italy), Cicily, Faiyum (Egypt), Galilee, Byzantia (Tunisia), Southern Gaul and Baetica (Southern Spain) were very rich and they had dense populations. , The best wine came from Compania, Cicily, and Byzantia exported wheat to Rome in large quantities. People cultivated every inch of land in Galilee. Spanish olive oil came mainly from the estates in southern Spain.

Romans were much advanced in technology. They developed the technology to use water power to work mills. They also developed the water-energy technique to mine gold and silver in the mines of Spain. Roman Empire also had an organized commercial banking chain. Cash was extensively used, All these are proofs of the strength Of Roman economy. There were also problems like the exploitation of workers and the use of slaves.

Question 11.
Slavery was a very deep-rooted evil system that was prevalent in the ancient times. Based on this statement, write about the slavery in the Roman Empire.
Answer:
Slavery was a very deep-rooted evil system that was prevalent in the ancient times. In the Mediterranean region and in the Near East, slavery had deep roots. Even Christianity did not challenge slavery. But it is wrong to assume that all the work in the Roman Economic System was carried out by slaves.

During the Republican times, in most of the areas of Italy, slaves were made to do all the work. Under Augustus, there were 3 million slaves. In those days the Italian population was only 7.5 million. But slaves were not used in all areas of the Empire. In many places, work was got done by giving wages to people. Slaves were considered an investment. The upper classes of the Roman society did not show any mercy to the slaves.

question 12.
For managing labour, agricultural writers and owners gave much attention. Explain.
Answer:
For managing labour, agricultural writers and owners gave much attention. They gave the greatest importance to supervision. The owners of land believed that unless the workers are supervised. nothing would work out properly.

Writers like Cato, Columella, Varro, and Palladius wrote handbooks on farming practice. To make the supervision on the paid workers and slaves, they were divided into smaller teams called gangs. Columella recommended that workers should be divided into teams of 10. If you divided them into smaller teams, it would be easy to find out who are really working and who are not. It shows a lot of importance was given to the management of labour.

Question 13.
In the Roman society, there were different social. groups. On the basis of this statement, evaluate the social hierarchy in Rome.
Answer:
In the Roman society there were different social groups, Historian Tacitus divides the. main social groups into 5:

1. The Senators (Paters)
2. The top class cavalry men (Equites)
3. The Respectable Middle Class
4. The Lower Class people who were interested in circus and colorful shows (Plebs sordida or humiliores).
5. Slaves

Question 14.
There was no stable currency system in Rome. Discuss.
Answer:
In the first 3 centuries, the currencies used were based on silver. But this system failed completely in the later period of the Empire. The reason was the lack of silver in the Spanish mines. Because of the shortage of silver, the government could not maintain a stable silver currency. Emperor Constantine started a new currency system based on gold. During the Late Roman Empire, a lot of gold coins were in circulation throughout the Empire.

Question 15.
Corruption and dictatorship were the. trademarks of the Roman government. Evaluate this statement.
Answer:
Corruption was rampant in the empire. This was especially so in the judiciary and in the army administration. The greed of higher officials in the army and the governors of the provinces was notorious. The government had to frequently interfere to stop such corruption. Legislation against corruption and the criticisms against corruption made by historians and intellectuals help us in knowing more about the corruption prevalent in the empire. Criticism is an important aspect of the classical world. Roman Administration was a despotic one. Government never tolerated any criticism or opposition against it. Such criticisms or oppositions were brutally suppressed by the government.

Question 16.
Diocletian and Constantine were two rulers that brought revolutionary changes in the Roman Empire. This is what Honey wrote in her seminar paper. Justify this statement.
Answer:
During the time of Constantine, there were revolutionary changes in the religious life of the people in the Empire. He made Christianity the official religion of the empire. In the 7^th century, Islam came into being. There were great changes in the structure of the nation. It was Diocletian (244-305) who brought changes here.

The large areas created administrative inconveniences and therefore Diocletian took steps to solve the problem. He reduced the size of his Empire by removing the strategically and economically unimportant regions. He protected the boundaries by building fortresses. He reorganized the provincial boundaries. He exempted citizens from military service. The Duces (army commanders) were given autonomy.

Constantine (306-334) Was the successor of Diocletian. He brought great changes in the administrative setup. The most important among them were the new currency system, new capital, and economic reforms. He brought out new gold coins called Solidus which weighed 4 1/2 grams of gold. A lot of these coins were minted. Millions of such coins circulated in the empire.

Even after the fall of the Roman Empire, these coins remained valuable. Constantine made Constantinople (old Byzantium) his second capital. It was in the modem Istanbul in Turkey and it was covered on all the three sides by oceans. He also formed a new Senate for the new capital. The emperor invested heavily in the oil mills and crystal factories in villages. Screw-making machines and watermills were introduced. He also re-established the trade relations with the East.

Question 17.
The Romans were polytheists (worshippers of many gods). Based on this statement, write a note on the religious practices of the Romans.
Answer:
The Romans were polytheists. They worshipped many gods and goddesses like Jupiter, Juno, Minerva, and Mars. They built temples and other places of worship for their deities. Their faith did not have any special name or label.

Judaism was another religion in Rome. It was also not monolithic as the different ancient Jewish communities followed different ways. By the 4^th and 5^th century Christianity began to spread in Rome. Constantine was the first Emperor to become Christian. Later Christianity was made the state religion.

Question 18.
The reason for the fall of the Western European Empire was internal differences. Do you agree with this statement? Examine.
Answer:
In the 4^th century AD, the Roman Empire was divided into two-the Eastern and the Western Empire. They were under two Emperors. In the Eastern Roman Empire, there was general prosperity. It not only survived the great plague of the 540s which made the Mediterranean area a vast graveyard, but the population went on increasing. But, at the same time, the Western Roman Empire faced political crises. The attacks of the germanic tribes were the reason for that.

Students can Download Chapter 3 Motion in a Straight Line Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 3 Motion in a Straight Line

Plus One Physics Motion in a Straight Line One Mark Questions and Answers

Plus One Physics Motion In A Straight Line Questions And Answers Chapter 3 Question 1.
Which of the followig curves does not represent motion in one dimension?

Answer:
(b) In one dimensional motion, the body can have at a time one value of velocity but not two values of velocities.

Plus One Physics Chapter 3 Questions And Answers  Question 2.
Free fall of an object (in vacuum) is a case of motion with
(a) Uniform velocity
(b) Uniform acceleration
(c) Variable acceleration
(d) Uniform speed
Answer:
(b) Uniform acceleration:
Free fall of an object (in vacuum) is a case of motion with uniform acceleration.

Motion In A Straight Line Questions And Answers Pdf Chapter 3 Question 3.
The area under velocity-time graph fora particle in a given interval of time represents
(a) velocity
(b) acceleration
(c) work done
(d) displacement
Answer:
(d) displacement:
Area under velocity-time graph represents displacement of a particle in a given interval of time.

Plus One Physics Chapter 3 Previous Year Questions And Answers Question 4.
The velocity-time graph of a body moving in straight line is shown in the figure. The displacement and distance travelled by the body in 6s are respectively

(a) 8, 16m
(b) 16m, 8m
(c) 16m, 16m,
(d) 8m, 8m
Answer:
(a) 8, 16m

Displacement is equal to area under the velocity time graph with proper sign.
∴ Displacement = 4 × 2 – 2 × 2 + 2 × 2 = 8m
Distance is equal to total area under the speed time graph.
∴ Distance = 4 × 2 + 2 × 2 + 2 × 2 = 16m.

Motion Straight Line Question Bank Pdf Download Chapter 3 Question 5.
A cartravels half the distance with constant velocity of 40 kmph and the remaining half with a constant velocity 60 kmph. The average velocity of the car in kmph is
(a) 40
(b) 45
(c) 48
(d) 50
Answer:
(c) 48
Average velocity = ν_av = \(\frac{s}{\frac{s}{80}+\frac{s}{120}}\) = 48kmph.

Plus One Physics Chapter Wise Questions And Answers Chapter 3 Question 6.
Two objects A and B travel from P to Q through two different paths as shown in figure. If both A and B takes the same time interval to travel from P to Q, then which of the following statements are true?

(a) A and B have same speed.
(b) A and B have same velocity
(c) A and B have same average velocity.
(d) The speed of A is greater than that of B
(e) The speed of B isgreaterthanthatofA
Answer:
(d) The speed of A is greater than that of B.

Plus One Physics Motion In A Straight Line Chapter 3 Question 7.
The acceleration of a moving object is equal to the
(a) gradient of a displacement-time graph
(b) gradient of a velocity-time graph
(c) area below a speed-time graph
(d) area below a displacement – time graph
(e) area below a velocity-time graph
Answer:
(b) Gradient of a velocity-time graph.

Plus One Physics Important Questions And Answers Pdf Chapter 3 Question 8.
A ball is thrown vertically upwards and comes back. Which of the following graph represents the velocity-time graph of the ball during its flight?

Answer:

Hsslive Plus One Physics Chapter Wise Questions And Answers Chapter 3 Question 9.
The magnitude of average velocity is equal to average speed. In which case this condition is satisfied?
Answer:
When a particle is moving with constant velocity, the magnitude of its average velocity is equal to average speed.

Motion In A Straight Line Previous Year Questions Chapter 3 Question 10.
Can a body be said to be at rest as well as in motion at the.same time?
Answer:
Yes, rest and motion are relative terms. A body at rest with respect to one body may be in motion with respect to another body.

Motion In A Straight Line Problems With Solutions Pdf Chapter 3 Question 11.
What conclusion can you draw if the average velocity is equal to instantaneous velocity?
Answer:
The particle is moving with constant velocity.

Motion Straight Line Question Bank Pdf Chapter 3 Question 12.
Two cars are moving in such a way that their relative velocity is zero. Which of the following graph represent this situation?

Answer:
(a) b

Motion In A Straight Line Class 11 Solutions Chapter 3 Question 13.
The speed-time graph is shown in figure. Is it possible.

Answer:
No speed cannot be negative.

Motion In A Straight Line Objective Questions Pdf Chapter 3 Question 14.
Why the speed of the object can never be negative?
Answer:
Speed is distance covered per unit time. Since distance cannot be negative, speed cannot be negative.

Motion In A Straight Line Class 11 Pdf Chapter 3 Question 15.
Is it possible that the velocity of an object be in a direction other than the direction of acceleration? If yes, give an example.
Answer:
Yes. A body is moving with decreasing velocity.

Plus One Physics Chapter Wise Questions And Answers Pdf Hsslive Question 16.
Is it possible to have the rate of change of velocity constant while the velocity itself changes both in magnitude and direction? If yes, give an example.
Answer:
Yes. Projectile motion.

Plus One Physics Chapter Wise Questions And Answers Pdf Question 17.
If the acceleration of the particle is constant in magnitude but not in direction, what type of path does the body flow?
Answer:
Circular path.

Class 11 Physics Chapter 3 Extra Questions Question 18.
Two stones of different sizes are dropped simultaneously from the top of a building. Which stone would reach earlier? Why?
Answer:
Both reach at ground simultaneously. Acceleration is same for both stones.

Question 19.
A piece of paper and iron piece are dropped simultaneously from the same point in vacum. Which one will reach at ground earlier?
Answer:
Both reach at ground simultaneously.

Question 20.
Is it possible that your cycle has a southward velocity but northward acceleration? If yes, give an example.
Answer:
Yes, when brakes are applied to a moving cycle, the directions of velocity and acceleration becomes opposite.

Plus One Physics Motion in a Straight Line Two Mark Questions and Answers

Question 1.
An ant is moving through a graph paper along x-axis. A boy observes that the ant covers 1mm in every second.

1. What type of motion is this?
2. When the boy is in school bus he observe the speedometer of the bus. Which speed is observed by the speedometer?

Answer:

1. Uniform motion or uniform velocity
2. Instantaneous speed (Ratio of the displacement to small interval of time).

Question 2.
Match the following

Answer:
a. (c)
b. (d)
c. (b)
d. (a)

Question 3.
Some examples of motion are given below. State in each case if the motion is one, two or three dimension

1. A Kite flying on a windy day.
2. A speeding car on a long straight highway.
3. A carrom coin rebounding from the side of the board.
4. A planet revolving around its star.

Answer:

1. 3 Dimensional motion
2. 1 Dimensional motion
3. 2 Dimensional motion
4. 2 Dimensional motion

Plus One Physics Motion in a Straight Line Three Mark Questions and Answers

Question 1.
Two bodies start moving in the same straight line at the same instant of time from the same origin. The first body moves with a constant velocity of 40 m/s and the second starts from rest with a constant acceleration of 4 m/s².

1. What is uniform speed?
2. Find the time that elapses before the second catches the first body.

Answer:
1. A body is said to be uniform if it travels equal displacements in equal intervals of time.

2. Distance travelled by first body in a time t
S₁ = Vt
S₁ = 40 × t
Distance travelled by second body in a time t
S₂ = ut + \(\frac{1}{2}\) at²
S₂ = \(\frac{1}{2}\) × 4 × t²
When these two bodies meet,
S₁ = S₂
40 × t = \(\frac{1}{2}\) × ²
t = 20 s.

Question 2.
Velocity time graph of a moving object is shown below.

1. What is the acceleration of the object?
2. Draw displacement – time graph for the above motion shown in the graph.

Answer:
1. Acceleration = 0

2.

Question 3.
Two straight lines drawn on the same displacement time graph make angles 30° and 60° with time axis respectively in the figure.

1. Which line represents greater velocity?
2. What is the ratio of the velocity of line A to the velocity of line B?

Answer:
1. B

2.

Question 4.
A particle starts from rest and its acceleration a plotted against time t is shown below.

1. This body is at
   • constant acceleration
   • variable acceleration
   • constant velocity
   • rest
2. Plot the corresponding velocity (V) against time (t)
3. Plot the corresponding displacement (S) against time (t)

Answer:
1. Constant acceleration

2.

3.

Question 5.
Displacement is a vector quantity which distance is a scalar quantity.

1. Distinguish between scalar and vector quantities.
2. An athlete runs along a circular track of radius 50m. Find the distance travelled and the displacement of the athlete when he coveres % of the circle.
3. What is the distance travelled by a body in a time t having an initial velocity u and moving with uniform acceleration ‘a’?

Answer:
1. A physical quantity having both magnitudes and direction is called vecter. A physical quantity having only magnitude is called scalar.

2.

Displacement AD

Distance, ABCD = AB + BC + CD

Distance = \(\frac{3}{2}\) πr

3. c = ut + \(\frac{1}{2}\) at²

Question 6.
“The aerial distance between two towers is 4km. But speedometer of car shows 5.6km when travel from one tower to another”

1. By reading this statement explain the concept of distance and displacement.
2. What is the numerical ratio of displacement of object to distance? Explain.
3. A particle is moving along a circular trace of radius ‘R’. What is the distance travelled and displacement of the particle in half revolution?

Answer:
1. Distance os the length of the path covered by the object. It is a scalar quantity. Displacement is the length between initial point and final point.

2. \(\frac{\text { displacement }}{\text { distance }} \leq 1\)
For the straight-line path, displacement is equal to the distance travelled. But for the curved path displacement is less than the distance travelled.

3. distance = πR
displacement = R + R = 2R.

Plus One Physics Motion in a Straight Line Four Mark Questions and Answers

Question 1.
A stone is thrown upwards from the ground with a velocity ‘u’.

1. What is the maximum height attained by the stone?
2. Check the correctness of the equation obtained in (a) using the method of dimensional analysis.
3. Draw the position-time graph of the stone during its return journey. (g = 10m/s²)

Answer:
1. V² = u² + 2as
0 = u² + 2gH
H = u²/2g

2. H = \(\frac{u^{2}}{2 g}\)
When we write the above equation in terms of dimension, we get

This means that, H= \(\frac{u^{2}}{2 g}\) is dimensionally correct.

3.

Question 2.
Gopal dropped an apple from the top of his flat at a height of 10m. He told his sister Seetha on the ground below that it will reach the ground in 2 seconds after he drops it.

1. Can she catch it after 2 seconds?
2. Derive suitable relation for time of fall.
3. Draw the velocity-time graph of the above body (assume the body rebounds from the floor)

Answer:
1. No.

2. S = ut + 1/2at²
h = 0 + 1/2gt²
If ball is dropped from a height ‘h’, we can write.
\(\sqrt{\frac{2 h}{g}}\) = t.

3.

Question 3.
A car of mass 1000 kg starts from rest at t = 0 and under goes acceleration as shown in figure.

1. Draw the corresponding velocity-time graph.
2. What is the retarding force acting on the car?
3. What is the total distance travelled by the car during t = 0 to t = 4 sec.

Answer:
1.

2. Retarding acceleration, a = -3m/s²
∴ Retarding force F = ma
= 1000 × 3 = 3000N.

3. Area of velocity time gives distance.
∴ Area = \(\frac{1}{2}\)bh = \(\frac{1}{2}\) × 4 × 6
Distance = 12m.

Question 4.
A tow rope used to pull the car of mass 700kg will break if the tension exceeds 1500N.

1. Calculate the maximum acceleration with which the car can be pulled through a level road
2. Calculate the minimum time required to bring the car to work station 500m away from the break point

Answer:
1. T = ma
1500= 700 × a
a = \(\frac{1500}{700}\) = 2.14 m/s².

2. S = ut + 1/2 at²
500= 0 + 1/2 × (2.14) × t²
t = \(\sqrt{\frac{2 \times 500}{2.14}}\) = 21.61sec.

Question 5.
1. Figure shows the position-time graph of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t ≤ 0 and on a parabolic path fort> 0? Justify your answer.

2. Can a body have an acceleration without velocity? Justify your answer with a physical situation.
3. The table given below shows the velocity of a car at different times.

a. Draw acceleration time graph
b. Find the distance travelled by the car in 6 sec.
Answer:
1. No. initially the body remains at rest and then the body moves with constant acceleration in a straight line.

2. Yes, considerthe oscillation of simple pendulum. At extreme position, velocity becomes zero and acceleration is non – zero value.

3. The velocity of a car at different times:
a. acceleration a = \(\frac{16-4}{1-0}\) = 5 m/s²
This value (acceleration is constant through-out the motion. Hence the acceleration time graph will be a straight line parallel to time axis.

b. S = ut + \(\frac{1}{2}\) + 2at²
= 11 × 6 + \(\frac{1}{2}\) × 5 × 6²
S = 156m.

Question 6.
The relative velocity of a body A with respect to a body B is the time rate at which body A changes its position with respect to body B.

1. If V_A and V_B are the velocities of A and B moving in opposite directions, what is the relative velocity of A with respect to B?
2. Two trains along the same straight rails are moving with constant velocity of 60 km/h and 30 km/b towards each other. If at time t = 0, the distance between them is 90km, find the time when they collide.
3. The velocity-time graph of two bodies A and B make angles of 30° and 60° with the time axis, what is the ratio of their acceleration?

Answer:
1. V_BA = V_A + V_B

2. Relative velocity = 60 + 30
= 90km/h
∴ Hence t = \(\frac{\text { displacement }}{\text { relative velocity }}\)
= \(\frac{90}{90}\) = 1h.

3. Slope of velocity time graph gives acceleration. Hence

Plus One Physics Motion in a Straight Line Five Mark Questions and Answers

Question 1.
A balloon is ascending at the rate of 14 ms^-1 and at a height of 98 m above the ground. A stone is dropped from it.

1. State whether the motion of the balloon is accelerated or retarded.
2. After how much time does the stone reach the ground?
3. Determine the velocity with which the stone strikes the ground.

Answer:
1. The motion of balloon is uniform motion. It has neither acceleration nor retardation.

2. u = 14m/s, a = -9.8, S = -98 m.
S = ut + \(\frac{1}{2}\) at²
98 = 14 t – \(\frac{1}{2}\) × 9.8 t²
4.9 t² – 14 t – 98 = 0
Solvingthisweget t = 6.123 sec

3. ν² = u² + 2as
^ν2 = (14)² + 2 x ^–9.8 x ^–98
= 196 + 1920.8
^ν2 = 2116.8
^ν2 = \(\sqrt{2116.8}\) = 45.99 m/s.

Question 2.
A particle is moving along x-axis with a uniform positive acceleration.

1. Draw the position-time graph for its motion.
2. Obtain the expression for the displacement by drawing velocity-time graph.
3. A ball is thrown vertically upwards with a velocity of 20 ms^-1 from the top of the tower of height 25m from the ground. How long does it remain in air?(g = 10 ms^-2)

Answer:
1.

2.

Consider a body moving with an acceleration ‘a’. Let ‘u’ be the initial velocity at t = 0 and final velocity ‘v’ at t = t. The area of velocity-time graph gives displacement. This is a quadratic equation. Hence t can be found using this formula

Question 3.

1. State the difference between speed and velocity. Can a body move with uniform speed but with variable velocity? Explain with the help of an example.
2. Show that a body thrown vertically upwards returns with the same magnitude of velocity.

Answer:
1. Speed is scalar quantity but velocity is a vector quantity. If a body moving along the circumstance of a circle with uniform speed, its velocity changes continuously with time.

2.

Consider a body projected upward from a point A with a velocity ‘u’. If the body reaches at B, the displacement becomes zero. Hence time taken to reach at B can be found.
S = ut + 1/2 at²
0 = ut + 1/2 × ^–10 t²
ut = 5t²
t = \(\frac{u}{5}\) ____(1)
The velocity at B can be found using the formula
V_B = u + at _____(2)
substitute eq(1) in eq(2)
V_B = u + ^–10\(\frac{u}{5}\)
= u – 2u
V_B = -u
∴ V_A = -V_B

Plus One Physics Motion in a Straight Line NCERT Questions and Answers

Question 1.
In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
Answer:
(a) (b)

Question 2.
The position time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in the following figure Choose the correct entries in the brackets as follows:

1. (A/B) lives closer to the school than (B/A)
2. (A/B) starts from the school earlier than (B/A)
3. (A/B) walks faster than (B/A)
4. A and B reach home at the (same/different) time
5. (A/B) overtakes (B/A) on the road (once/twice).

Answer:
1. It is clear from the graph that OQ > OP. So, A lives closer to the school than B.

2. The position-time graph of A starts from the origin (t = 0) while the position-time graph of B starts from C which indicates that B started later than A after a time interval OC. So, A started earlier than B.

3. The speed is represented by the steepness (or slope) of the position-time graph. Since the position-time graph of B is steeper than the position-time of graph A, therefore, we conclude that B is faster than A.

4. Corresponding to both P and Q, the time interval is the same, i.e., OD. This indicates that both A and B reach their homes at the same time.

5. The position-time graphs intersect at the point K. This indicates that B crosses A. Since there is only one point of intersection, therefore, the two cyclists cross each other only once.

Question 3.
A car moving along a straight highway with speed of 126 kmh^-1 is brought to a stop within a distance of 200m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Answer:
Initial velocity,
u = 126kmh^-1 = 126 × \(\frac{5}{18}\) ms^-1 = 35m^-1
Final velocity, ν = 0;
Distance, S = 200m
Using ν² – u² = 2aS,
0² – 35 × 35 = 2a × 200
or
a = \(-\frac{35 \times 35}{400}\)ms^-2
= 3.06ms^-2
So, the retardation of the car is 3.06 ms^-2
Using ν = u + at,
0 = 35-3.06 × t
or
3.06t = 35
t = \(\frac{35}{3.06}\)s = 11.4s.

Question 4.
On a two-lane road, car A is travelling with a speed of 36kmh^-1 Two cars B and C approach car A in opposite directions with a speed of 54 kmh^-1 each. At a certain instant, when the distance AB is equal to AC, both being I km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Answer:
ν_A = 36kmh^-1
= 36 × \(\frac{5}{18}\)ms^-1 = 10ms^-1
ν_B = ν_C = 54ms^-1
= 54 × \(\frac{5}{18}\)ms^-1 = 15ms^-1

Relative velocity of B w.r.t. A, ν_BA = 5 ms^-1
Relative velocity of C w.r.t.A, ν_CA= 25ms^-1
Time taken by C to cover distance AC
= \(\frac{1000 m}{25 m s-1}\) = 40s
Now, forB, 1000 = 5 × 40 + \(\frac{1}{2}\)a × 40 × 40
On simplification, a = 1 ms^-2.

Question 5.
Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion.

1. with zero speed at an instant may have non-zero acceleration at that instant,
2. with zero speed may have non-zero velocity,
3. with constant speed must have zero accelera¬tion,
4. with positive value of acceleration must be speed-ing up.

Answer:

1. True
2. False
3. True
4. False

For (1), consider a ball thrown up. At the highest point, speed is zero but the acceleration is non-zero, For (2), if a particle has non-zero velocity, it must have speed, For (3), if the particle rebounds instantly with the same speed, it implies infinite acceleration which is physically impossible. For (4), true only when the chosen positive direction is along the direction of motion.

Question 6.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5kmh^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5kmh^-1. What is the magnitude of average velocity, and average speed of the man over the following intervals of time:

1. 0 to 30 min,
2. 0 to 50 min,
3. 0 to 40 min?

Answer:
Average speed overthe interval of time from 0 to 30min
= \(\frac{2.5 \mathrm{km}}{30 \mathrm{min}}=\frac{2.5 \mathrm{km}}{\frac{1}{2} \mathrm{h}}\)
= 5kmh^-1
Magnitude of average velocity overthe interval of time from 0 to 30 min is 5kmh^-1. This is because the “distance travelled” and the ‘magnitude of displacement’ over the interval of time from 0 to 30 min are equal. Distance covered from 30 to 50 minutes
= 7.5kmh^-1 × \(\frac{20}{60}\) h = 2.5km
Total distance covered from 0 to 50 minute
= 2.5 km+ 2.5 km = 5km
Total time = 50min = \(\frac{50}{60}\) h = \(\frac{5}{6}\) h
Average speed overthe interval of time from 0 to 50min
= \(\frac{5 \mathrm{km}}{5 / 6 \mathrm{h}}\) = 6kmh^-1
The displacement overthe interval of time from 0 to 50 min is zero. So the magnitude of average velocity is zero. Distance covered from 30 to 40 min
= 7.5kmh^-1 × \(\frac{1}{6}\) h = 1.25km
Total distance covered from 0 to 40 minute
= 2.5 km + 1.25 km = 3.75km
Average speed overthe interval of time from 0 to 40min
= \(\frac{3.75 \mathrm{km}}{\frac{40}{60} \mathrm{h}}\) = 5.625kmh^-1
The “magnitude of displacement” is (2.5 -1.25) km, i.e., 1.25 km.
Time interval = \(\frac{2}{3}\)h
The ‘magnitude of average velocity’ = \(\frac{1.25 \mathrm{km}}{2 / 3 \mathrm{h}}\),
= 1.875 kmh^-1.

Question 7.
Look at the graphs (a) to (d) in the following figure carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of the particle.

Answer:
None of the four graphs can represent one-dimensional motion of the particle. In fact, all the four graphs are impossible.

1. A particle cannot have two different positions at the same time.
2. A particle cannot have velocity in opposite directions at the same time.
3. Speed is always positive (non-negative).
4. Total path length of a particle can never decrease with time.

Note: The arrows on the graphs are meaningless.

Question 8.
The x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.

Answer:
No, wrong, x-t plot does not show the trajectory of a particle. Context: A body is dropped from a tower (x = 0) at t = 0.

Question 9.
The velocity-time graph of a particle in one-dimensional motion is shown below. Which of the following formulae are correct for describing the motion of the particle over the time interval from t₁ to t₂?

(a) x(t₂) = x(t₁) + ν(t₁)(t₂ – t₁) + \(\frac{1}{2}\) a(t₂ – t₁)²
(b) ν(t₂) = ν(t₁) + a(t₂ – t₁)
(c) ν_average = [x(t₂) – x(t₁)]/(t₂ – t₁)
(d) a_average = [ν(t₂) – ν(t₁)]/(t₂ – t₁)
(e) x(t₂) = x(t₁)+ ν_av (t₂ – ₁) +\(\frac{1}{2}\) a_av(t₂ – t₁)²
(f) x(t₂) – x(t₁) = Area underthe ν – 1 curve bounded by t-axis and the dotted lines.
Answer:
(c), (d), (f)
Explanation: It is clear from the shape of ν – t graph that acceleration of the particle is not uniform between time intervals t₁ and t₂. [Note that the given ν – t graph is not straight.] The equations (a), (b) and (e) represent uniform acceleration.

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