Category: CBSE Class 12

NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter

NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Physics
Chapter	Chapter 5
Chapter Name	Magnetism and Matter
Number of Questions Solved	25
Category	NCERT Solutions

Question 1.
Answer the following questions regarding earth’s magnetism :
(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.
(b) The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain ?       (C.B.S.E. 1995)
(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground ?
(d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole ? (C.B.S.E. 1995)
(e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of
magnetic moment 8 x 10²² JT^_1 located at its center. Check the order of magnitude of this number in some way.
(f) Geologists claim that besides the main magnetic N­S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all ?
Answer:

(a) Magnetic elements

• Declination
• Dip and
• Horizontal intensity

(b) Greater in Britain (it is about 70°), because Britain is closer to the magnetic north pole.

(c) Field lines of B due to the earth’s magnetism would seem to come out of the ground.

(d) Compass needle can move only in the horizontal plane. Since the field is entirely vertical no direction is shown by the needle.
(e) Using the formula for magnetic field on the equatorial line of a magnetic dipole i.e.

This value tells the order of magnitude of magnetic field of earth.
(f) Geologists are correct to think so because it is an approximation to consider the magnetic field of earth to be a single dipole field. The magnetised mineral deposits can be treated as local dipoles on earth.

Question 2.
Answer the following questions :
(a) The earth’s magnetic field varies from point to point in space. Does it also change with time ? If so, on what time scale does it change appreciably ?
(b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why ?
(c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e. the source of energy) to sustain these currents ?
(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past ?
(e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion ?(f) Interstellar space has an extremely weak magnetic field of the order of 10¹² T. Can such a weak field be of any significant consequence ? Explain.
Answer:

(a) Yes, it changes with time. After a few hundred years, the earth’s magnetic field undergoes an appreciable change.

(b) The temperature inside the earth is so high that it is impossible for the iron to remain as a magnet and act as a source of the magnetic field. The magnetic field due to the earth is considered to be due to the circulating electric currents induced in the iron in the molten state and other conducting materials inside the earth.

(c) A possible explanation can be the phenomenon of radioactivity.

(d) Analysis of the rock magnetism /earth’s magnetic field gets recorded in certain rocks during solidification, (through weekly) provides clues to the geomagnetic history.

(e) At large distances, the earth’s magnetic field gets modified by the fields produced by the motion of ions in the earth’s ionosphere.

(f) At very-very large distances like interstellar distances the small fields can significantly affect the charged particles like that of cosmic rays. For small distances, the deflections are not noticeable for small fields but at very large distances the deflections are significant.

clearly small value of B gives a very large value of radius R

Question 3.
A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 X 10^-2 J. What is the magnitude of magnetic moment of the magnet ?
Answer:
Using τ- MB sin θ, we get

Question 4.
A short bar magnet of magnetic moment m = 0.32 JT^-1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium ? What is the potential energy of the magnet in each case ?
Answer:

Question 5.
A closely wound solenoid of 800 turns and area of cross section 2.5 x 10^-4 m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment ?
Answer:
When current is passed through the solenoid, the magnetic field is produced along with its axis. The magnetic field lines emanate from one end and enter the other just as in the case of a bar magnet. The two ends of the solenoid act as the two poles of a bar magnet.
Here, the number of turns in the solenoid = 800
I = 3A
A = 2.5 x 10^-4m²
The magnetic moment of the solenoid,
M = (IA) x number of turns
= 3 x 2.5 x 10^-4 x 800
= 0.6 Am²

Question 6.
If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of the torque on the solenoid when its axis makes an angle of 30° with the direction of the applied field ?
Answer:
Using x = MB sin θ, we get
x = 0.6 x 0.25 x sin 30
= 0.6 x 0.25 x \(\frac { 1 }{ 2 } \)
= 0.3 x 0.25 = 0.075 Nm
= 7.5 x 10-² Nm.

Question 7.
A bar magnet of magnetic moment 1.5 JT^-1 lies aligned with the direction of a uniform magnetic field of 0.22T.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment, (i) normal to the field direction, (ii) opposite to the field direction ?
(b) What is the torque on the magnet in cases (i) and (ii)?
Answer:

Question 8.
A closely wound solenoid of 2000 turns and area of cross-section 1.6 X 10^-4 m², carrying a current of 4.0 A, is suspended through its center allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid ?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 x 10^-2 T is set up at an angle of 30° with the axis of the solenoid?
Answer:
N =2000, A= 1.6 x 10^-4m², I = 4.0 A
(a) m = ANI = 1.6 x 10^-4 x 2000 x 4.0
= 1.28 Am², along the axis

(b) B = 7.5 x 10^-2T, θ = 30°
Net force = 0
τ = mB sin θ = 1.28 x 7.5 x 10^-2 x sin 30
= 0.64 x 7.5 x 10^-2
= 4.800 x 10^-2 Nm
By the action of this, the solenoid can come to the direction of the external field.

Question 9.
A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 x 10^-2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s^_1. What is the moment of inertia of the coil about its axis of rotation ?
Answer:

Question 10.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
Answer:

Question 11.
At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.
Answer:
Using B_H = B cos δ, we get

Direction of B is 12° west of geographic meridian making upward angle of 60° with horizontal.

Question 12.
A short bar magnet has a magnetic moment of 0.48 JT^-1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the center of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Answer:
On axial line

Question 13.
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the center of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null-points (i.e., 14 cm) from the center of the magnet ? (At null points, Held due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
Answer:
Magnetic field at the equatorial line of the magnet is given

Question 14.
If the bar magnet in Exercise 5.13 is turned around by 180°, where will the new null-points be located ?
Answer:
When magnet is turned around 180°, its south pole lies in the geographical south direction. Hence null point will lie on the equatorial line at a distance x from die center of the magnet.

Question 15.
A short bar magnet of magnetic moment 5.25 x 10^-2 JT^-1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the center of the magnet, the resultant field is inclined at 45° with the earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Answer:
Normal bisector
(a) Let resultant magnetic field of a magnet at point P makes an angle θ= 45° with the earth’s field. Therefore,



Question 16.
Answer the following questions:
(a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled? (C.B.S.E. 1991)
(b) Why is diamagnetism, in contrast, almost independent of temperature?
(c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?
(d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?
(e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point). Why?
(f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet ?
Answer:
(a) When cooled, the tendency of the thermal agitation to disrupt the alignment of magnetic dipoles decreases in the case of paramagnetic materials. Hence they display greater magnetisation.

(b) The atoms of a diamagnetic do not have an intrinsic magnetic dipole moment. On placing a diamagnetic sample in a magnetic field, the magnetic moment of the sample is always opposite to the direction of the field. It is not affected by the thermal motion of the dipoles.

(c) Since bismuth is diamagnetic, the field in the core coil be sightly less than that when a core is empty.

(d) Permeability of a ferromagnetic material depends on applied magnetic field. Permeability is more for lower magnetic field.

(e) One of the reasons for this fact is that when a material has µ_r > > 1, the field lines meet the material nearly normally.

(f) Yes, a paramagnetic sample with saturated magnetisation will have the same .order of magnetisation as the magnetisation of a ferromagnetic substance. However, the saturated magnetisation will require magnetising field too high to achive. Further, there may be a minor difference in the strengths of the atomic dipoles of paramagnetic and ferromagnetic materials.

Question 17.
Answer the following questions:
(a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.
(b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetization, which piece will dissipate greater heat energy ?
(c) ‘A system displaying a hysteresis loop such as a ferromagnet is a device for storing memory.’ Explain the meaning of this statement.
(d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for budding ‘memory stores’ in a modern computer ?
(e) A certain region of space is to be shielded from magnetic fields. Suggest a method.
Answer:
A piece of carbon steel will dissipate a greater amount of heat energy as its hysteresis loop has a greater area.
The magnetisation produced in a ferromagnet does not have a unique value corresponding to the applied magnetizing field.

In addition, the magnetisation produced depends on the history of the magnetisation i.e. the number of cycles of magnetisation, it has been taken through. In other words, the value of magnetisation of a ferromagnet is a record or memory of its magnetisation. If information bits can be made corresponding to the cycles of magnetization, the system displaying the hysteresis loop of the ferromagnet can act as a device for storing the information.

• ceramics are used for coating magnetic tapes in a cassette player or for building memory stores in a modem computer. Ceramics are specially treated barium iron oxides and are also called ferrates.
• The shielding of the region can be done by surrounding it with soft iron rings. The magnetic field lines will be drawn into the rings and the enclosed region will become free of the magnetic field.

Question 18.
A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (Ignore the thickness of the cable). (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)
Answer:
Let neutral point lies at a distance x from the cable. Now, at neutral point, magnetic field due to cable is equal in magnitude and opposite in direction of the earth’s magnetic field.

Question 19.
A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
Answer:

Question 20.
A compass needle free to turn in a horizontal plane is placed at the center of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
(a) Determine the horizontal component of the earth’s magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the place to be zero.
Answer:

Question 21.
A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 x 10^-2 T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field ?
Answer:
Here B, = 1.2 X 10-² T,θ₁= 15°, θ₂ = 45°.

The dipole will be in equilibrium, if torque acting on dipole due to B₁ is equal and opposite to the torque acting on dipole due to B₂.
That is, MB₁ sin = MB₂ sin θ₂

Question 22.
A monoenergetic (18 keV) electron beam intially in the horizontal direction is subjected to a horizontal magnetic field of 0.40 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (m_e = 9.11 x 10^-19 Q.
[Note. Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]
Answer:

^{Question 23.}
A sample of paramagnetic salt contains 2.0 x 10²⁴ atomic dipoles each of dipole moment 1.5 x 10^-23 JT^-1. The sample is placed under a homogeneous magnetic field of 0.64 T and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K ? (Assume Curie’s law).
Answer:
Magnetic dipole moment of sample,
M = 15% of M (1.5 x 10-²³) (2 x 10²⁴) = 30 JT-<sup>1

Question 24.</sup>
A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetizing current of 1.2 A?
Answer:


^{Question 25.}
The magnetic moment vectors μ_s and μ_l; associated with the intrinsic spin angular momentum S and orbital angular momentum Z, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by μ_s= -(e/m)S, μ_l= -(e/2m)l. Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
Answer:

We hope the We hope the NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 History Chapter 14 Understanding Partition Politics, Memories, Experiences are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 14 Understanding Partition Politics, Memories, Experiences.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	History
Chapter	Chapter 14
Chapter Name	Understanding Partition Politics, Memories, Experiences
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 12 History Chapter 14 Understanding Partition Politics, Memories, Experiences

Question 1.
What did the Muslim League demand through its resolution of 1940?
Solution :
The resolution of 23 March 1940, demanded a measure of autonomy for the Muslim- majority areas of the subcontinent. It never mentioned partition or Pakistan. Sikandar Hayat Khan, who had drafted the resolution was the Punjab Premier and leader of the Unionist Party. He declared in the Punjab Assembly on 1 March, 1941 that he was opposed to a Pakistan that would mean “Muslim Raj here and Hindu Raj elsewhere… “If Pakistan means unalloyed Muslim Raj in the Punjab then I will have nothing to do with it”. He reiterated his plea for a loose (united) confederation with considerable autonomy for confederating units.

Question 2.
Why did some people think of Partition as a very sudden development?
Solution :
Some people think that partition of India in 1947 was a sudden development. Many Muslim leaders were not serious in their demand for Pakistan as a separate nation. On many occasions, Jinnah used the idea of Pakistan to seek favours from the British and to block concessions into the Congress. Even the Muslims were confused about the idea of Pakistan. They could not think of their future in an independent country called Pakistan. Many people had migrated to the new country with the hope that they would soon come back to India as soon as the situation improved.
In fact, the partition was so sudden that nobody could imagine it.

Question 3.
How did ordinary people view Partition?
Solution :
Ordinary people did not know what the Partition was and it would affect their lives in the future. They even did not know about the different areas of the subcontinent. Migrants thought they would return to their original place as soon as peace prevailed again.

Question 4.
What were Mahatma Gandhi’s arguments against Partition?
Solution :

1. Mahatma Gandhi opposed the Partition by arguing that both Hindus and Muslims were bom of same soil and they had the same blood, ate the same food, drank the same water and spoke the same language. So, they were similar to each other.
2. He stated that the demand for Pakistan put forward by the Muslim League was un- Islamic and sinful because Islam stands for the unity and brotherhood of mankind and not for disrupting the oneness of the human family. So, those who wanted Partition were enemies alike of Islam and India.

Question 5.
Why is Partition viewed as an extremely significant marker in South Asian history?
Solution :
The following reasons can be put forward for the given view:

• The partition of India had a unique nature. This partition was based on religions. The partition took place in the name of the communities. History has never witnessed such type of partition.
• The partition marked a severe violence. Innumerable people were killed. People began to kill each other irrespective of their earlier relation. Earlier they lived with each other in harmony and peace but now started to kill each other. Government machinery failed to check this.
• People faced a lot of problems. Their life became miserable. Their near and dear ones were killed. Many people were abducted.
• People moved across the border. Most of the Muslims of India crossed over to Pakistan and almost all Hindus and Sikhs came to India from Pakistan. They were forced to start their life afresh.
• People lost all their movable and immovable property all of a sudden. They became homeless and forced to live in refugee camps.

Question 6.
Why was British India partitioned?
Solution :
Partition of India was not a sudden event because even in its resolution of March 1940, the Muslim League had only demanded a measure of autonomy for the Muslim majority areas of the subcontinent. It was a culmination of events such as communal politics that started developing in the opening decades of the twentieth century as mentioned below :

1. Government of India Act 1909 and 1919 – The British Government granted separate electorate for Muslims in 1909. These were expanded in 1919. Separate electorates implied that Muslims could elect their own representatives in designated constituencies. Thus, religious identities were encouraged. Community identities no longer indicated simple difference in faith and belief: but they led to active opposition and hostilities between communities.
2. Events during 1920s and 1930s – During the 1920s and 1930s, Muslims were agitated by the activities of the Hindus such as “music-before-mosque”, cow protection movement, and shuddhi movement of Arya Samaj. Similarly, Hindus were angered by the rapid spread of tabligh (propaganda) and tanzim (organisation). These activities led to riots at different places and deepened differences between two communities.
3. The provincial elections of 1937 and the Congress ministries – In the elections of 1937, Congress did well but Muslim League failed poorly in the constituencies reserved for Muslims. The Muslim League wanted to form a joint government with the Congress in United Provinces where Congress had won an absolute majority. The Congress had, therefore, rejected the offer. This led to drifting away of the Muslim League but thereafter Muslim League doubled its efforts at expanding its social support.
4. Policies of the Congress ministries – The Congress ministry in UP wanted to abolish landlordism which was supported by the Muslim League. The Congress also could gain much in its mass contact programme in UP. But its policies alarmed the conservative Muslims.
5. Rise of Hindu Mahasabha and Rashtriya Swayamsevak Sangh – The rise of Hindu Mahasabha and Rashtriya Swayamsevak Sangh which had over 100,000 trained and highly disciplined cadres pledged to an ideology of Hindu nationalism, convinced Muslims that India was a land of the Hindus.

The above factors created differences between two communities but inspite of this fact remains that the Cabinet Mission (1946) plan that recommended a loose three-tier confederation was accepted by all the major parties. It was due to later developments such as ‘Direct Action da/ (16 August, 1946), riots and violence, fear of Sikh leaders and Congressmen in the Punjab and a section of bhadralok. Bengali Hindus in Bengal which compelled the Congress to accept the partition of the country.

Question 7.
How did women experience Partition?
Solution :
For women, partition was horrible. Women were raped, abducted and many times forced to live with strangers and start a new life. They were deeply traumatised and began to develop new family bonds in the changed circumstances.

Women became victims on both the sides of the border. They were forced to live in a strange circumstances. But the government officials of both the countries did not take any serious step to consult those women. Women were left on their fate.

They were even murdered by their own family members. When the men realized that the women of their family would fall into the hands of the enemy, they killed their women with their own hands. To escape from the hands of enemy, in a Sikh village, ninety women were said to have voluntarily jumped into a well.

Question 8.
How did the Congress come to change its views on Partition?
Solution :
Initially, the proposals of the Cabinet Mission were accepted by all the major political parties but due to differences over interpretation of the plan, neither the Congress, nor the League agreed to the Cabinet Mission’s proposal. Thereafter, following developments took place:

• The Muslim League announced 16 August 1946 as “Direct Action Day” for winning its Pakistan demand.
• “Direct Action Day” led to riots at Calcutta and other places.
• At that time, many Sikh leaders and Congressmen in the Punjab were convinced that Partition was a necessary evil, otherwise they would be swamped by Muslim majority and Muslim leaders would dictate their terms to them.
• Similarly, a section of bhadralok, Bengali Hindus, who wanted political power to remain with them, began to fear the “permanent tutelage of Muslims”. They were in a numerical minority so only a division of the province could ensure their political dominance.

Thus, under these circumstances, the Congress had no option except to agree to the Partition.

Question 9.
Examine the strengths and limitations of oral-history. How have oral-history techniques furthered our understanding of Partition ?
Solution :
The strengths and limitations of oral-history are as mentioned below :
(a) Strengths :

• It helps us grasp experiences and memories in detail. It enables historians to write richly textured, vivid accounts of what happened to people during Partition.
• Oral-history enables historians to broaden the scope of their discipline by writing experiences of the poor and the powerless who have been generally ignored in mainstream history.

(b) Limitations :

• The oral-history lacks concreteness. Its chronology is not precise.
• The uniqueness of personal experience makes generalisation difficult because a large picture cannot be built from micro-evidence and one witness is no witness.
• Oral accounts deal with tangible issues. Small individual experiences are not relevant to unfold larger processes of history.

But inspite of above shortcomings the oral-history is important because it can be corroborated by other sources. The experiences of the people during Partition are significant and should be used to check other sources and vice-versa.

We hope the NCERT Solutions for Class 12 History Chapter 14 Understanding Partition Politics, Memories, Experiences help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 14 Understanding Partition Politics, Memories, Experiences, drop a comment below and we will get back to you at the earliest.

Here we are providing NCERT Solutions for Class 12 English Flamingo Chapter 8 Going Places. Students can get Class 12 English Going Places NCERT Solutions, Questions and Answers designed by subject expert teachers.

Going Places NCERT Solutions for Class 12 English Flamingo Chapter 8

Going Places NCERT Text Book Questions and Answers

Going Places Think as you read 

Question 1.
Where was it most likely that the two girls would find work after school?
Answer:
Jansie, being realistic, realized that both of them were likely to end up working for the biscuit factory. However, Sophie, who lived in a dream world, dreamt of starting a boutique, being an actress, or starting her career as a manager somewhere.

Question 2.
What were the options that Sophie was dreaming of? Why does Jansie discourage her from having such dreams?
Answer:
Sophie dreamt of starting a boutique with the money she would earn as a manager. She imagined herself as a fashion designer with the best shop in the city or of becoming an actress. She aspired after a career that would be considered “sophisticated”. Jansie was more practical and warned Sophie to come to terms with reality.

Question 3.
Why did Sophie w riggle when Geoff told her father that she had met Danny Casey?
Answer:
Sophie squirmed when Geoff told her father that she had met Danny Casey because she had lied about it. She felt uncomfortable about having lied to her brother, Geoff. She did not want her lies to be discovered.

Question 4.
Does Geoff believe what Sophie says about her meeting with Danny Casey?
Answer:
Geoff refused to believe Sophie but was later convinced on hearing the vivid description of their meeting. He told his dad and then Frank about the meeting.

Question 5.
Does her father believe her story?
Answer:
No, Sophie’s father did not believe her story. When Geoff told him about the meeting, he expressed disbelief. Her father warned her that she would talk herself “into a load of trouble”.

Question 6.
How does Sophie include her brother Geoff in her fantasy of her future?
Answer:
Sophie imagined her brother travelling to exotic and mysterious places and she craved to be taken along. She imagined a vast world that awaited her arrival. She saw herself riding there, behind Geoff. She imagined him wearing new, shining black leathers and she saw herself in a yellow dress with a kind of cape that flew out behind. She imagined being greeted to applause by the world.

Question 5.
Which country did Danny Casey play for?
Answer:
Danny Casey was a football player; he played for Ireland.

Question 6.
Why didn’t Sophie want Jansie to know about her story with Danny?
Answer:
Sophie expressed displeasure at Geoff telling Frank about how she had met the football star, Dafiny. Jansie, her friend, found out about the same from Frank. She felt that Geoff had betrayed her trust letting out their secret. She was also wary about the whole neighbourhood finding out about it from Jansie.

Question 7.
Did Sophie really meet Danny Casey?
Answer:
No, Sophie had not really met Danny Casey. Her father, who knew her well, realized that her story was a figment of her wild imagination. Sophie was in the habit of living in a world of fantasy that had no bearing with reality.

Question 8.
Which was the only occasion when she got to see Danny Casey in person?
Answer:
The only time she got to see Danny Casey was on Saturdays when she, along with her family, went to watch United at a match. They saw Casey play football.

Going Places Understanding the text

Question 1.
Sophie and Jansie were classmates and friends. What were the differences between them that show up in the story?
Answer:
Sophie and Jansie were classmates and friends. Yet, they were inherently different in their personalities. Sophie was a romantic, who seemed to have nothing much to do with the realistic world. Her ambitions and her ways to achieve them were unrealistic. Sophie dreamt of starting a boutique with the money she would earn as a manager. She dreamt about opening the best shop in the city or of becoming an actress.

Jansie was more practical and did not want Sophie to talk of such unrealistic ventures. She knew that both of them were destined to work at the biscuit factory. Sophie was childish and lived in a dream world where Danny, a football star, had met her and was likely to meet her yet again. Jansie disapproved of her telling such a story.

Question 2.
How would you describe the character and temperament of Sophie’s father?
Answer:
Sophie’s father was a hardworking man. He is described as a plump and a heavy-breathing man in a vest. His face was grubby and sweaty after the day’s hard work. When Geoff told him about Sophie’s meeting with Danny, he said nothing but looked at her disparagingly, knowing it to be untrue. When Sophie claimed that Danny had informed her that he was going to buy a shop, her father frowned knowingly. He dismissed Sophie’s claim as another of her “wild stories” and was afraid that she would talk herself “into a load of trouble”. This made her apprehensive of talking about such fantasies to him.

Her father, too, was a sports lover. This family of modest means depended on their only source of entertainment— watching football. They went to watch United play, each Saturday, as a “weekly pilgrimage”. As a sports enthusiast, he shouted encouragement to Danny and went to the pub to celebrate the sport team’s victory.

Question 3.
Why did Sophie like her brother Geoff more than any other person? From her perspective, what did he symbolise?
Answer:
Sophie liked her brother Geoff more than anyone else in the family. She confided in him and also idolised him. He was older, an apprentice mechanic, who travelled for his work to the other end of the city. She supposed that there were many more mysterious elements about him, than she knew of. He spoke little and she envied his silence. To her it seemed that when he was quiet, he travelled in his imagination to those places where she wanted to go. She suspected him of knowing interesting people and she longed to know them too.

She wanted to be closer to her brother and hoped that someday he might take her with him.Geoff symbolised freedom to Sophie’s limited experience. In her childish dreams, she imagined an exotic world beyond her knowledge, which awaited her arrival.

Question 4.
What socio-economic background did Sophie belong to? What are the indicators of her family’s financial status?
Answer:
Sophie belonged to a middle class socio-economic background. The reader receives the first hint from Jansie’s and Sophie’s conversation and how they were both earmarked for the biscuit factory. Jansie also prioritised buying a “decent house” to live in, if she was to come upon money. Sophie’s father’s appearance with dirt and sweat all over him after a day’s work, indicated that he was a hard-working man. She observed her mother stooping over the sink, her back bent from all the hard work.

The small room in which Sophie and her family lived was steamy from the stove and cluttered with the dirty washing piled up in the comer. Her brother Geoff was an apprentice mechanic, having left school. The humble status of Sophie’s family is brought out through the details worked into the story.

Going Places Talking about the text

Discuss in pairs.

Question 1.
Sophie’s dreams and disappointments are all in her mind. Discuss.
Answer:
Sophie’s views, her ambitions, her ideas about her brother Geoff s life, or her meeting with Danny, were all figments of her imagination. She was confident of becoming an actress and having a boutique on the side, though she neither had a decent house, nor any material comfort. The realistic view is reflected through Jansie who knew that they would both end up working in the biscuit factory.

Sophie imagined her brother, Geoff visiting places that she had never seen. She longed to be introduced to the vast world that she believed awaited her. She imagined herself riding there behind Geoff and being welcomed with thunderous applause.Sophie had never met Danny Casey, a sports star, but she told her family a concocted story of meeting. Her father recognised that it was one of her wild stories and was afraid that she would get herself into lot of trouble.She fantasised to such a level that her lies became a living reality and she went to wait for Danny to turn up. Even when she realized that Danny would not turn up to meet with her, she indulged in a similar fantasy.

Question 2.
It is natural for teenagers to have unrealistic dreams. What would you say are the benefits and disadvantages of such fantasising?
Answer:
Fantasy is the creative imagination or unrestrained fancy. Commonly known as daydreaming, this is a fairly common phenomenon—and though it is not harmful, it should not be indulged in excessively. In a world of fantasy, one can have whatever one likes, even things that he or she cannot have in reality. In other cases, the person sometimes becomes so obsessed with his own thoughts, that he is absolutely unconcerned about the happenings in the real environment around him. So, even though a person may be physically present at a certain place, he may be mentally absent and lose sight of everything around him. An example could be such students who find it difficult to grasp and retain concepts because they are daydreamers. On the other hand, Mark Twain too was a daydreamer.

Perhaps, the source of daydreaming may be the need to escape from unpleasant or stressful situations. A daydreamer would have sensitivity, depth and intelligence and hence should be given space and nurtured. But, it could also shut one out from reality. This is because the daydreamer, lost in thoughts, loses his bearings entirely and is oblivious to things around him. If a daydreamer happens to be on the road, this state of mind could prove dangerous, even fatal. Daydreaming can be very distracting. It affects one’s day-to-day work, and the person gets more and more absent- minded. After a point, he may start losing physical and social contact with his surroundings and become increasingly withdrawn.

Daydreaming has its advantages. Daydreaming fosters a child’s imagination, and enhances creativity. But, if one is a habitual daydreamer and tends to wander off a bit too often for comfort, it might be detrimental to how he comes to terms with the world around him.

Going Places Working with words

Notice the following expressions. The highlighted words are not used in a literal sense. Explain what they mean.

• Words had to be prized out of him like stones out of a ground — it was difficult to get him to talk or to get information out of him
• Sophie felt a tightening in her throat — tension or anxiety
• If he keeps his head on his shoulders — thinks intelligently
• On Saturday they made their weekly pilgrimage to the United — they went to the United dutifully, without fail and with a lot of devotion as one would go to a place of worship
• She saw. him ghost past the lumbering defenders — she viewed a faint image of him as he flashed past the awkward or clumsy defenders

Going Places Thinking about language

Notice these words from the story:

• chuffed’, meaning delighted or very pleased
• ‘nosey’, meaning inquisitive
• ‘gawky’, meaning awkward, ungainly

These are words that are used in an informal way in colloquial speech.

Make a list of ten other words of this kind.

• row – a noisy quarrel or dispute
• bad mouth – to insult
• bell – to telephone
• to belt – to hit
• blast – enjoyable experience
• blow one’s top – be very angry
• daft – silly, foolish
• dim – not intelligent
• savvy – well informed
• scaredy-cat – a person who is frightened

Going Places Extra Questions and Answers

Going Places Short Answer Questions

Question 1.
What was Sophie’s ambition? How did she plan to achieve it?
Answer:
Sophie was a romantic who was far removed from reality. She aspired to start a boutique after leaving school. She told her friend, Jansie, that to save money for the boutique she would work as a manager. She wanted a boutique like Mary Quant, a famous fashion designer. She also toyed with the idea of working as an actress and having a boutique on the side. She desired to have a career that was considered “sophisticated”.

Question 2.
Were Sophie’s ambitions were divorced from reality?
Answer:
Sophie was a schoolgirl—a teenager from a middle-class family. But she had exalted ambitions about starting a boutique or becoming an actress. Her dreams were divorced from reality. Her friend, Jansie, realized the irony of their situation and was realistic enough to know that they would end up working in a biscuit factory.

Question 3.
Sophie idolized her brother Geoff. Justify.
Answer:
Sophie’s brother, Geoff, had been three years out of school, an apprentice mechanic, who travelled to his work each day to the far side of the city. Sophie thought that Geoff lived the lives of her hopes and dreams that she wanted for her own self. She romanticized his life. She imagined that he visited faraway, exotic places and met interesting people. She wished to visit the places of her imagination and ride away with Geoff.

Question 4.
How did Sophie dream of herself in Geoff’s world?
Answer:
Sophie wished that her brother would take her to his world with him. She was conscious of a vast world that awaited her arrival. She saw herself riding there behind Geoff. She imagined him in new, shining black leather and herself in a yellow dress with a kind of cape that flowed out behind her. She imagined the world greeting them with wild applause.

Question 5.
What did Sophie tell Geoff about Danny Casey?
Answer:
Sophie told Geoff that she had met Danny Casey in the arcade. She told him how she was looking at the clothes in Royce’s window when Danny Casey came and stood beside her. She said that he had gentle, green eyes but was not very tall. She asked him for an autograph for little Derek, but neither of them had any paper or pen. She claimed that Danny had invited her to meet him next week.

Question 6.
What was her father’s reaction to her claim to have met Danny Casey?
Answer:
Geoff told his father that Sophie had met Danny Casey. Their father disbelieved Sophie’s claim; he looked at her with disdain. When she told her father that Danny said he was going to buy a shop, her father dismissed it as one of her wild stories. He felt that she would land herself into trouble because of her wild imagination.

Question 7.
What did Geoff warn Sophie about Danny?
Answer:
Geoff warned Sophie about how she was still in school, and Danny was likely to have lots of girls. Sophie tried to refute it, but Geoff insisted on how he knew better. When she told him that Danny was to meet her the following week, Geoff told her that Danny would never turn up.

Question 8.
Where did the family see Danny? What was Sophie’s reaction?
Answer:
Sophie, Derek, Geoff and their father went to watch United play on Saturday. It was a “weekly pilgrimage” for them. Sophie, her father and little Derek went down and sat near the goal while Geoff went higher up with his friends. They saw Danny score a goal, leading his team to victory. Sophie blushed with satisfaction as her hero won the match for them.

Question 9.
What was Sophie’s reaction when Jansie questioned her about Danny Casey?
Answer:
Sophie was upset when Jansie revealed that she had come to know about Sophie’s episode with Casey. She felt betrayed as Geoff had let out her secret. Jansie accused her of lying, but Sophie convinced her otherwise. However, she felt apprehensive that Jansie would spread the rumour around her neighbourhood.

Question 10.
Where did Sophie go to wait for Danny?
Answer:
After dark, Sophie walked by the canal, along a sheltered path lighted only by the glare of the lamps from the wharf across the water. It was a secluded place where she had often played as a child. She sat down on a wooden bench under a solitary elm, to wait for Danny Casey.

Question 11.
How did Sophie come to terms with the fact that Danny would not come?
Answer:
As Sophie waited for Danny to turn up, she even imagined him coming. But when some time had elapsed, she felt the pangs of doubt stirring inside her. She recalled that Geoff had said how he would never come. She waited and finally resigned herself to the truth that Danny would not come.

Question 12.
“I will have to live with this burden.” Why did Sophie feel this way?
Answer:
Sophie was a romantic. Like many other teenagers, she lived in her world of fantasy. She dreamt of Danny coming to meet her. Once Sophie realized that she had been deluding herself, she became sad and felt that it was a hard burden to carry. She was ashamed to face her brother, her family and her neighbours who would make fun of her whims.

Question 13.
“And she saw it all again…” What did Sophie see?
Answer:
Sophie saw a vision of meeting Danny Casey as she had imagined a week before. She visualized herself talking to him and asking for an autograph. She replayed the entire episode of a week before in her mind. To her, it was a lived reality. However, it was nothing but a figment of her imagination. The reader gets an insight into the dynamics of the situation. The cyclical narrative suggests that Sophie’s world was replete with her imagined reality.

Going Places Long Answer Questions

Question 1.
How did Sophie aspire to achieve her lofty ambitions? Why did they seem unrealistic to Jansie? How did her family react to her fantasies and ambitions?
Answer:
Sophie desired to set up a boutique after leaving school. She told her friend, Jansie, that to save that much money she would work as a manager. She wanted a boutique like Mary Quant. She wanted it to be one of its kind. She also contemplated working as an actress and having the boutique on the side.

Jansie, the grounded of the two, tried to reason with her saying that the boutique would require a lot of capital and nobody would employ her as a manager unless she had experience. Sophie’s father was disdainful of her behaviour. He felt that Sophie needed to keep her head on her shoulders. He dismissed her talk, knowing she lived in her own world of fantasy. He felt that she would land into trouble because of her immature talk.

Question 2.
Sophie’s fascination for Danny Casey stemmed from the fact that he had all that Sophie wanted for herself. Elaborate.
Answer:
Sophie desired an affluent and sophisticated lifestyle. She wished to save some money and start a boutique like Mary Quant had. Sophie’s fascination with Danny Casey stemmed from the fact that he had the lifestyle she aspired for. She dreamt of a world that awaited her arrival. Her fixation with fame and luxury is revealed as she fantasized about becoming an actress.

She weighed each of her unrealistic options as an avenue to fame, stardom and money. She hailed from a middle-class background, but dreamt of achieving fame and luxury, much beyond the humble reach of her station. Danny Casey represented an entry to such a lifestyle. Her infatuation had much more to do with her own ambition than any genuine adoration of Casey’s skills as a footballer.

Question 3.
Sophie is a middle-class girl who longs, like any other teenager, to reach out to the horizons. Justify.
Answer:
Sophie was a middle-class girl with lofty aspirations. To achieve a glamorous and sophisticated life, she wished to have a boutique after she finished school. Sophie was confident of saving the required amount of money by working as a manager. She also considered the idea of becoming an actress as there was real money in that, and having the boutique on the side. She longed to go with Geoff to places that she had never seen. These places held a lure for her because they were mysterious and distant.

She imagined herself riding there, behind Geoff. She pictured the world admiring and applauding her entry. Her infatuation with Danny Casey, a sports star, also stemmed from the fact that he was the epitome of glamour and sophistication. Like a regular teenager, she dreamt, uninhibited of her station and skills. Sophie was yet to experience the hardships of adult life, she was content to live in her world of fantasies where all was within her reach.

Going Places Value Based Question

Question 1.
Unrealistic dreams often lead to a great deal of unhappiness. Justify this on the basis of your reading of the story.
Answer:
Sophie who lived in the world of her dreams, found her reality quite suffocating. Sophie dreamt of owning a boutique one day or of being an actress or a fashion designer, but her friend Jansie believed that both of them were earmarked for the biscuit factory. Jansie, who was more realistic, tried to make Sophie accept the imminent reality, but Sophie continued with her make-believe ways. She imagined befriending Danny Casey, the sports star, only to be disillusioned.

She imagined that Danny had asked her to meet; she went there and waited for hours, believing that he would turn up. Sophie got sucked into the story of her own creation and began to believe that it was true. When Sophie realized that she had believed in a lie, her disappointment was painful and almost life-changing. She is seen moving from one dream to another in her mind. When the harsh reality stared her in the face, her disappointment was evident.

 

NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	History
Chapter	Chapter 7
Chapter Name	An Imperial Capital: Vijayanagara
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara

Question 1.
What have been the methods used to study the ruins of Hampi over the last two centuries ? In what way do you think they would have complemented the information provided by the priests of the Virupaksha temple ?
Solution :
(a) The methods used to study the ruins of Hampi over the last two centuries were as given below:

• Colonel Colin Mackenzie, an engineer, antiquarian and an employee of the English East India Company discovered the ruins at Hampi in 1800. He prepared the first survey map of site.
• From 1856, photographers began to record the monuments which enabled scholars to study them.
• Since 1836, epigraphists began collecting several dozen inscriptions found at the temples at Hampi.
• Thereafter the historians collated the information from above sources with accounts of foreign travellers and other literature written in Telugu, Kannada, Tamil and Sanskrit.

(b) The above methods would have complemented the information provided by the priests of the Virupaksha temple because that was based on the memories of the priests. The earlier information was corroborated by the inscriptions, photographs, maps and accounts of foreign travellers and other material.

Question 2.
How were the water requirements of Vijayanagara met ?
Solution :
The water requirements of Vijayanagara were met in the following ways :

1. Its location is the natural basin formed by the river Tungabhadra which flows in a north-easterly direction. The stunning granite hills form a girdle around the city. A number of streams flow down to the river from these rocky outcrops.
2. The embankments were built along these streams to create reservoirs of varying
   sizes.
3. Tanks were built to store rainwater and conduct it to the city. The most important tank built is now called Kamalapuram tank. Water from this tank irrigated fields nearby as well as was also conducted through a channel to the “royal centre”.
4. The Hiriya canal drew water from a dam across the Tungabhadra and irrigated the cultivated valley that separated the “sacred centre” from the “urban core”.

Question 3.
What do you think were the advantages and disadvantages of enclosing agricultural land within the fortified area of the city?
Solution :
Advantages of enclosing agriculture land within fortified area:
(i) It had an elaborate canal system which drew water from the Tungabhadra to provide irrigation facilities.
(ii) It enclosed agricultural tracts, cultivated fields, gardens, and forests.
(iii) This enclosure saved crops from being eaten by wild animals.
(iv) In the medieval period, sieges were laid to starve the defending armies into submission. These sieges lasted for many months or many years. So the rulers of Vijayanagara adopted and elaborated a strategy to protect the agricultural belt and built large granaries.

Disadvantages
(i) This system was very expensive.
(ii) During adverse, circumstances this system proved inconvenient to the farmers.
(iii) The farmers had to seek the permission of gate-keeper to reach their field.
(iv) If enemy encircled the field the farmer could not look after their field.

Question 4.
Figure given below is an illustration of another pillar from the Virupaksha temple. Do you notice any floral motifs? What are the animals shown? Why do you think they are depicted? Describe the human figures shown.

Solution :
(a) There are many floral motifs.
(b) Horses and elephants have been shown in the pillar.
(c) The Vijayanagara kings competed with contemporary rulers including the Sultans of the Deccan and the Gajapati rulers of Orissa – for control of the fertile river valleys and resources generated by lucrative overseas trade. The kingdom remained in constant state of military preparedness. So, the kings paid attention to improve harbours and encouraged its commerce so that horses, elephants, precious gems, etc. are freely imported. Thus, due to the importance of horses and elephants in the warfare, these animals had been depicted on the pillars.
(d) The images of gods have been shown in the pillar. One devotee has also been shown before a Shiva linga.

Question 5.
What do you think was the significance of the rituals associated with the mahanavami dibba?
Solution :
The mahanavami dibba was one of the most impressive platforms in the “king’s palace”. It was located on one of the highest points in the city. Rituals associated with the structure probably coincided with mahanavami (literally the great ninth day) of the ten day Hindu festival during the autumn months of September and October, known variously as Dusehra (northern India), Durga Puja (in Bengal) and Navaratri or Mahanavami (in peninsular India). The Vijayanagara kings displayed their prestige, power and suzerainty on this occasion.

The ceremonies such as worship of the image, worship of the state horse, the sacrifice of buffaloes and other animals were performed on this occasion. Dances, wrestling matches, processions of caparisoned horses, elephants, chariots, soldiers and ritual presentations were held before the kings, guests, the chief nayakas were held. On the last day, the king inspected his army and the armies of the nayakas who brought rich gifts for the king as well as the stipulated tribute. Thus, there was great significance of the rituals associated with the mahanavami dibba.

Question 6.
Discuss whether the term “royal centre” is an appropriate description for the part of the city for which it is used.
Solution :
The term “royal centre” is an appropriate description for the part of the city for which it is used because the Royal center had more than 60 temples. Most of these temples were constructed by the ruler of Vijayanagara Empire to express their supremacy. The royal centre had 30 palaces. These were made of perishable material. A brief description of the building of Royal centre are as given below:
(i) One of the most beautiful buildings in the royal centre is the Lotus Mahal. It was named by British travellers in the nineteenth century. While the name is certainly romantic, historians are not quite sure what the building was used for. One suggestion, found in a map drawn by Mackenzie, is that it may have been a council chamber, a place where the king met his advisers.
(ii) Most temples were located in the sacred centre. One of the most spectacular of these is the Hazara Rama Temple. This was probably meant to be used only by the king and his family.

Question 7.
What does the architecture of buildings like the Lotus Mahal and elephant stables tell us about the rulers who commissioned them ?
Solution :
The Lotus Mahal had nine towers – a high central one, and eight along the sides. Although it is not clear for what the building was used for but according to Mackenzie, it may have been a council chamber, place where the king met his advisers. Elephant stables were located close to the Lotus Mahal.

The architecture of Lotus Mahal tells us that the rulers used to consult their advisers on various issues and problems and meetings were held in the council chamber i.e., Lotus Mahal. The construction of “elephant stables” shows that the rulers took interest in the trade of elephants as well as in keeping them properly because elephants were very important factor in the warfare. It is perhaps one of reasons that elephants and horses have been depicted on the panels of the Hazara Rama temple.

Question 8.
What are the architectural traditions that inspired the architects of Vijayanagara ? How did they transform these traditions ?
Solution :
(a) The architectural traditions that inspired the architects of Vijayanagara were as given below :

1. Prior to Vijayanagara, Cholas in Tamil Nadu and the Hoysalas in Karnataka had extended patronage to elaborate temples such as the Brihadishvara temple, Thanjavur and the
   Chennakeshava temple at Belur. The rulers of Vijayanagara built on these traditions and carried them literally to new heights.
2. Like Indo-Islamic architecture, there was an arch on the gateway leading into fortified settlement as well as dome over the gate.
3. The architecture of tombs and mosques located in the urban core resembles that of the mandapas found in the temples of Hampi.
4. The Pallavas, Chalukyas, Hoysalas and Cholas encouraged temple building as a means of associating themselves with the divine – the deity was identified with the king. The choice of the site Vijayanagara was perhaps inspired by the existence of the shrines of Virupaksha and Pampadevi.
5. The arches in the Lotus Mahal were inspired by Indo-Islamic technique.

(b) They transformed these traditions in the followings ways :

1. In the fortifications, according to Abdur Razzaq, no mortar or cementing agent was employed. The stone blocks were wedge shaped, which held them in place, and the inner portion of the walls was of earth packed with rubble.
2. Royal portrait sculpture was innovated and developed. It was displayed in temples and the king’s visits to temples were treated as important state occasions.
3. In temple architecture, new features were structures of immense scale best exemplified by the Raya gopurams or royal gateways. They often dwarfed the towers on the central shrines and signalled the presence of the temple from a great distance. Other features include mandapas or pavilions and long, pillared corridors.

Question 9.
What impressions of the lives of the ordinary people of Vij ay an agar a can you cull from the various descriptions in the chapter ?
Solution :
The various descriptions in this chapter give the following impression of the lives of the ordinary people of Vijayanagara :

1. Horses were imported from Arabia and Central Asia. This trade was done by the traders and local communities of merchants i.e., kudirai chettis or horse merchants.
2. There were markets dealing in spices, textiles and precious stones.
3. Vijayanagara boasted of a wealthy population that demanded high-value exotic goods.
4. Portuguese traveller Barbosa described the houses of ordinary people, which have not survived as “the other houses of the people are thatched, but nonetheless well-built and arranged according to occupations, in long streets with many open places”. Besides this there is little archaeological evidence of the houses of ordinary people.
5. There were numerous shrines and small temples which implies that there were variety of cults, supported by different communities.
6. There were wells, rainwater tanks, temple tanks which may have served as sources of water to the ordinary town dwellers.
7. Paes gives us a vivid description of a bazaar. He states that the provisions, such as rice, wheat, barley, etc. were available cheaply and abundantly. This means that the life of the ordinary people was good and they did not suffer for want of essential things.

We hope the NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara, drop a comment below and we will get back to you at the earliest.

Here we are providing NCERT Solutions for Class 12 English Vistas Chapter 3 Journey to the end of the Earth. Students can get Class 12 English Journey to the end of the Earth NCERT Solutions, Questions and Answers designed by subject expert teachers.

Journey to the end of the Earth NCERT Solutions for Class 12 English Vistas Chapter 3

Journey to the end of the Earth NCERT Text Book Questions and Answers

Journey to the end of the Earth Reading with insight

Question 1.
How do geological phenomena help us to know about the history of humankind?
Answer:
Geological phenomena such as the drifting of land masses and their separating into countries help us to know about the history of humankind. A visit to Antarctica around which Gondwana once existed, is like going back to past as it gives us an understanding of evolution and extinction, ozone and carbon, where humankind came from, and where it is headed.

Question 2.
What are the indications for the future of humankind?
Answer:
All thoughtless activities of humankind such as increasing cities and megacities, cutting forests and turning those to concrete jungles, careless burning of fossil fuel, depleting ozone and increasing carbon dioxide, and global warming, melting ice caps and shields, our battle with other species for limited resources and other similar reckless activities point to a grim future for humankind. If concrete steps are not taken immediately, these drastic changes may lead to the end of the world.

Journey To The End Of The Earth Reading with Insight

Question 1.
‘The world’s geological history is trapped in Antarctica’. How is the study of this region useful to us?
Answer:
Antarctica holds half a million-year-old carbon track records in its layers of ice. It gives us an understanding of evolution and extinction, ozone and carbon. A visit to Antarctica, around which Gondwana once existed, is like going back to the past. Witnessing the geological phenomena, such as the drifting of land masses and their spreading into countries, help us to know about the history of humankind. These are visible signs of where humankind came from and it gives us a clear understanding of where human life is headed if we do not take care of the environment. Actually seeing with our own eyes all these changes, make us understand that global warming is a real threat.

Question 2.
What are Geoff Green’s reasons for including high school students in the Students on Ice Expedition?
Answer:
Geoff Green feels that students are the future generation of policy-makers. They should be provided an opportunity to have this life-changing experience at a young age in order to foster a new understanding and respect for our planet. It would help them to absorb, learn and act for the benefit of the planet. The youngsters still have the idealism to save the world and they need to understand that it belongs to them. So, to sensitize them, it is important to provide them the visible life changing experience.

Question 3.
‘Take care of the small things and the big things will take care of themselves.’ What is the relevance of the statement in the context of the Antarctica environment?
Ans. This statement means that if small things are taken care of, big things will take their own care. There are tall grasses, called phytoplankton, in the southern oceans that use the sun’s energy to assimilate • carbon and synthesize organic compounds by photosynthesis. Marine life and birds in the region sustain themselves on these tall grasses. Any disturbance in the environment in Antarctica might affect the activities of the phytoplankton, which, in turn, might affect the existence of the other life forms that depend on them. Small things like the phytoplankton are important in the food chain.

Question 4.
Why is Antarctica the place to go to understand the Earth’s present, past and future?
Answer:
The author states that to understand the earth’s present, past and future, Antarctica is the right place to go. Antarctica is relatively untouched in this respect as it has never had human population. It is relatively pristine. It holds in its ice cores half a million-year-old carbon records, trapped in the layers of ice. It embodies all that is pre-historic: cordilleran folds, pre-Cambrian granite shields ozone and carbon: evolution and extinction. The simple eco system and lack of biodiversity indicate how little changes in the environment can have big repercussions.

A visit to Antarctica and witnessing the geological phenomena, such as the drifting of land masses, glaciers receding and ice shelves collapsing makes us understand that global warming is a real threat. Hence, to study the earth’s past, present and future, these factors make Antarctica the best place to go.

Journey To The End Of The Earth Extra Questions and Answers

Journey To The End Of The Earth Short Answer Questions

Question 1.
When did the author start her journey to Antarctica and what had she to pass through?
Answer:
The author started her journey 13.09 degrees north of the Equator in Madras—she was on board a Russian research vessel—the Akademik Shokalskiy. She had to pass through nine time zones, six checkpoints, three bodies of water and at least as many ecospheres. After travelling over hundred hours in combination of a car, an aeroplane and a ship, she reached Antarctica.

Question 2.
What emotions did the author experience when she reached Antarctica at last?
Answer:
The author finally set foot on the Antarctica continent after travelling over 100 hours in combination of car, aeroplane and ship. Her first emotion on seeing the vast expansive white landscape and the blue horizon was of relief. She experienced the emotion of wonder at its immensity and isolation and its strange relationship with India.

Question 3.
How would you describe Gondwana?
Answer:
Gondwana was a giant amalgamated southern supercontinent, centering around present-day Antarctica. Humans had not arrived on the global scene. The climate was much warmer. There was a huge variety of flora and fauna. Gondwana thrived for 500 million years. When the age of the mammals got underway, the landmass was forced to separate into countries. Antarctica separated from the whole landmass shaping the globe as we know it today.

Question 4.
What is that thing that can happen in a million years and would be mind-boggling?
Answer:
The author says that in a million years India may push northwards, jamming against Asia. It will buckle its crust and form the Himalayas – South America may drift off to join North America. The Drake Passage may open up to create a cold circumpolar current. Antarctica may remain frigid, desolate and at the bottom of the world.

Question 5.
In what respect, Tishani Doshni’s encounter with Antarctica is a chilling prospect?
Answer:
The author remained there for two weeks. For a sun worshipper South Indian, being face to face with ninety per cent of earth’s total ice volume was a mind-boggling and chilling prospect. It was also a chilling experience for circulatory and metabolic functions and for imagination. It is like walking into a giant ping-pong ball with no human markers such as trees, billboards, and buildings.

Question 6.
What is the visual experience in Antarctica?
Answer:
In Antarctica the visual scale ranges from the microscopic to the mighty midgets and mites to blue whales and icebergs as big as countries. The writer refers to it as walking into a giant ping-pong ball devoid of any human markers, without trees, billboards, buildings. Days go on in 24 hours austral summer light. A ubiquitous silence, interrupted only by an occasional avalanche or calving ice sheet consecrates the place.

Question 7.
How, according to the author, has mankind etched its dominance over nature?
Answer:
According to the author, though civilizations have been around for barely a few seconds on the geological clock, yet they have created a ruckus by their various activities like exploiting the limited resources and careless burning of fossil fuels. In the short span of existence on the earth, they have already created a blanket of carbon dioxide and increased the average global temperature.

Question 8.
How has Antarctica sustained itself and managed to remain pristine?
Answer:
Antarctica, on account of being the coldest, windiest and driest continent in the world, has never sustained a human population and has thus managed to remain pristine. This has prevented man from being able to create ruckus in this part of the world by his thoughtless exploitation of the natural resources.

Question 9.
How is global temperature increasing? What are the immediate fears due to it?
Answer:
Global temperature is increasing due to the increasing burning of fossil fuels. It has now created a blanket of carbon dioxide around the world. This has given birth to questions like: Will the West Antarctica ice sheet melt entirely? Will the Gulf Stream Ocean current be disrupted? Will it be the end of the world as we know of? It may be. It may not be.

Question 10.
How is Antarctica a crucial element in the debate of climate change?
Answer:
Antarctica is a crucial element not because it has no human population but because it holds in its ice cores half a million year old carbon records. They are trapped in its layers of. ice. It will open up areas of knowledge about the past, present and future of the earth.

Question 11.
What are the reasons for the success of the Students on Ice programme?
Answer:
Sitting distant in the comfort zone of our houses, any talk about global warming looks so unreal and one can be unconcerned. But the visible experience of seeing glaciers retreating, ice caps melting and ice shelves collapsing makes one understand and realize what global warming is all about. The indications for the future of humankind become clear when one actually witnesses the geological phenomena.

Question 12.
The author says that her Antarctica experience was full of such epiphanies. What was that best epiphany that occurred there?
Answer:
The Akademik Shokalskiy got wedged into a thick white sheet of ice. The captain decided to turn around and asked the passengers to walk on the ocean. Underneath their feet they saw 180 metres of living, breathing salt water. Crab eater seals were stretching and sunning themselves on ice floes much like stray dogs under a banyan tree. It was a great epiphany, a revelation.

Question 13.
What is that beauty of balance that a trip to Antarctica unfolded to the author?
Answer:
The author was wonderstruck by the beauty of balance in play on our planet. Travelling across nine time zones, three bodies of water and as many ecospheres was an experience that unfolded a wide range of climate, geographical features, and flora and fauna. It was also a visible experience of the varied geographical phenomena.

Question 14.
Why does the author conclude the chapter by saying that a lot can happen in a million years, but what a difference a day makes?
Answer:
The author concludes the chapter by saying that much more can really happen in a million years as it happened in the case of Antarctica. But in this long period, changes even in a day make a great difference because global climate is changing. It is posing a threat to the beauty of balance on the earth.

Question 15.
What are phytoplanktons? What is their importance?
Answer:
Phytoplanktons, the grasses of the sea, are single-celled organisms living in the southern ocean. They nourish and sustain the entire ocean’s food chin, being first link in the food chain of ocean. Using sun’s energy, they assimilate carbon and synthesize organic compounds.
The diminishing number of these organisms due to the depletion of ozone layers affects other organisms of the ocean, finally leading to the extinction of life on earth.

Question 16.
Why does the author feel that the prognosis for the human beings is not healthy?
Answer:
The world is battling an ever increasing population, leading to burning of fossil fuels. This has created a blanket of carbon dioxide around the world thereby increasing global temperatures. All this is hazardous and life threatening for all flora and fauna. Hence the future of mankind in fact, all life on earth, is bleak. So, the author is correct in saying that the prognosis for man is not encouraging and healthy. . , j

Question 17.
Why is it necessary to remain fully equipped while walking on ice?
Answer:
While walking on ice, the troupe was fully kitted out in Gore-Tex (type of spiked boots that help in walking on ice) and glares. The spiked boots protect them from falling down on ice which might result in injury and the glares protect the eyes because the sunglasses can injure their eyes, particularly the ratina.

Question 18.
Do you think that programmes like the Students on Ice do more harm than good? Support your answer.
Answer:
I personally feel that such trips do more harm than good. We have ruined the earth as much as we could and as wide as we could go, because Antarctica was far away and extremely cold. But now we have so many reasons to go to this pristine continent. Let’s not encourage such trips. After all, what else do we have to learn about the earth than the fact that we have been running a business, not a service. Please spare Antarctica.

Student on Ice is an educational journey to Antarctica. It took high school students to Antarctica where they understood the seriousness of the threat that the end of the earth is quite near. By visiting Antarctica they would act their bit to save the planet from further deterioration. The educational youth of today is the hope for the earth and if they are more informed and more aware of the weakening strength of the earth, they will be able to steer the government machinery of their countries as they grow up.

Question 19.
Does the study of the lesson give you a feeling that man is his own great enemy?
Answer:
In his 12000-year-long stint on the earth so far man has caused untold harm to the planet, its environment and biodiversity. His activities in the name of development have spelt doom for the flora and fauna and his own existence is in danger. Man is to blame for all the havoc and ruckus created on earth. Thus it is quite right that man is his own great enemy.

Journey To The End Of The Earth Long Questions and Answers

Question 1.
What is the significance of the title ‘Journey to the End of the Earth’?
Answer:
The title ‘Journey to the End of the Earth’, has more than one meaning. It describes an educational journey to Antarctica undertaken by a group of high school students. To learn more about the real impact of global warming and future of the earth 52 students went to the coldest, driest, windiest continent in the world called Antarctica in Russian research vessel, the Akademik Shokalskiy.

The author calls it a journey to the end of the earth because it began 13:09 degrees North of Equator in Madras, involved crossing nine time zones, six checkpoints, three oceans and as many ecospheres. She travelled over 100 hours in combination of a car, an aeroplane and a ship. The journey being to the extreme south of the the earth, was really towards the end of it. Another meaning of this title is more significant as the warnings that Antarctica gives are shocking and much concerning the humanity and the millions of other species on the earth. The changes taking place in Antarctica are pointing a warning finger at the existence of of the earth; the earth is journeying to its end.

Question 2.
The author says, ‘It was nothing short of a revelation: everything does connect.’What does it mean?
Answer:
Antarctica is a perfect place to study how little changes in the environment can have big repercussions as far as Antarctica is concerned. Various human activities like exploiting the limited resources and careless burning of fossil fuel have already created a blanket of carbon dioxide, increased the average global temperatures and caused the retreating of glaciers, melting of ice caps and collapse of ice shelves as far as Antarctica. Global warming does not only change the geographical features, but also cause depletion in the ozone layer which will affect the activities of the phytoplanktons, the tall grasses which support the lives of marine animals and birds of the region. Hence, the author says everything does connect and all human activities are interlinked with the geological phenomena, whatever be the geological distance.

Question 3.
By whom and with what objective was Students on Ice programme started? How far has it achieved its goals?
Answer:
The Students on Ice programme was started by Canadian Geoff Green. He felt students are the future generation of policy-makers. They should be provided an opportunity to have this life¬changing experience at a young age in order to foster a new understanding and respect for our planet. It would help them to absorb, learn and, more importantly, act for the benefit of the planet.

Geoff Green was tired of taking celebrities and retired rich curiosity seekers who could only give back in a limited way. It means Geoff wanted something in return from his passengers to solve the problems relating to climate changes due to environmental pollution. It is difficult to imagine or be affected by the polar ice caps melting while sitting in our living rooms and so this visible life changing expence is important. Hence, this programme made the children learn that to save big things, small . things must be cared for.

Question 4.
What makes Antarctica an ideal subject of study?
Answer:
Antarctica is the only place in the world which has never sustained a human population. It thus remains relatively pristine in this respect. But, more importantly, it holds in its ice core, half a million- year-old carbon records trapped in its layers of life. Antarctica has a simple ecosystem and lack of biodiversity. It is, therefore, a perfect place to study how little changes in the environment can have big repercussions. Visiting Antarctica means knowing where we have come from and where we could possibly be heading. This place holds the key to know the geological evolution and it shall reveal the earth’s past, present and future.

Question 5.
The author states that her Antarctic experience was full of epiphanies, but the best occurred just short of the Antarctic Circle of 65-55 degrees south? Explain.
Answer:
Epiphanies is a Christian festival that celebrates the revelation or enlightenment. Here epiphanies are used metaphorically to suggest moments when the author suddenly becomes conscious of something that is very important to her.

The author experienced the rare of the rarest experiences there in Antarctica both in relation to beauty, wonder, and geological phenomena. Such masterly geological epiphany was experienced by her when the Akademik Shokalskiy got wedged into a thick white stretch of ice between the peninsula and Tadpole Island. The captain decided to turn around and asked the passengers to walk on the ocean. They kitted out in Gore-Tex and glares, walking on a white sheet of ice. Underneath their feet was a metre-thick ice pack. And underneath that, 180 metres of living breathing, saltwater lay before them. In the periphery, crabeater seals were stretching and sunning themselves on ice floes. They were doing so like stray clogs will do under the shade of a banyan tree. It was nothing short of revelation. The author saw in it that everything does indeed connect. This really proved to be the most wonderful experience of all experiences of Antarctica.

NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Diseases

These Solutions are part of NCERT Solutions for Class 12 Biology. Here we have given NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Diseases

Question 1.
What are the various public health measures, which you would suggest as a safeguard against infectious diseases?
Solution:
Prevention and control of infectious diseases

I. For water-borne diseases like typhoid, amoebiasis, etc.
Practice personal and public hygienic measures.

a. Personal hygienic measures

• Keeping the body clean
• Consumption of clean drinking water
• Eating fresh food

b. Public hygienic measures

• Proper disposal of waste and excreta
• Periodic cleaning and disinfection of water reservoirs, pool, tank etc.

II. For air-borne diseases like common cold, pneumonia

• Avoid close contact with infected persons.
• Avoid the use of belongings of the infected persons.

III. For vector-borne diseases like malaria

• Control and eliminate the vectors and their breeding places
• Introducing larvivorous fishes like Gambusia in ponds that feed on the larvae of the mosquito
• Avoid stagnation of water around the residential area.
• Spraying of insecticides in ditches, drainage areas, etc.
• Protection from a mosquito bite. Use mosquito nets in the doors and windows to prevent the entry of mosquitoes. It is very important in the light of recently widespread diseases like dengue fever, chikungunya etc.

The use of vaccines and immunization programmes has enabled us to eradicate smallpox. Diseases like polio, diphtheria, tetanus etc. have been controlled to an extent by the use of vaccines. Nowadays biotechnology is focussing on the preparation of newer and safer vaccines. A large number of antibiotics are available to treat many infectious diseases.

Question 2.
In which way has the study of biology helped us to control infectious diseases?
Solution:
Study of biology has helped us to know about causes of diseases, carriers of diseases (vectors), effects of diseases on different body functions and above all, means to control diseases. Our immune system plays a major role in preventing diseases.

Question 3.
How does the transmission of each of the following diseases take place ?

1. Amoebiasis
2. Malaria
3. Ascariasis
4. Pneumonia

Solution:

1. Through contaminated food and water.
2. Through Anopheles mosquito.
3. Through contaminated food and water.
4. By inhaling the droplets or aerosols released by infected persons.

Question 4.
What measure would you take to prevent water-borne diseases?
Solution:
Water-borne diseases can be prevented by drinking clean water. Water should be free from contamination, suspended and dissolved substances. If water is contaminated it should be boiled and filtered before drinking. Periodic cleaning and disinfection of water reservoirs, pools, and tanks should be done.

Question 5.
Discuss with your teacher what does ‘a suitable gene’ means, in the context of DNA vaccines.
Solution:
‘A suitable gene’ means the gene which is able to produce antigenic polypeptides of the pathogen in bacteria and yeast. Using recombinant DNA technology, it is possible to produce vaccines in large scale for immunisation. Hepatitis B vaccine is produced using this technology.

Question 6.
Name the primary and secondary lymphoid organs.
Solution:
Primary lymphoid organs are bone marrow and thymus. Secondary lymphoid organs are the spleen, lymph nodes, tonsils, Peyer’s patches of the small intestine, and mucosa-associated lymphoid tissues (MALT).

Question 7.
The following are some well-known abbreviations, which have been used in this chapter. Expand each one to its full form.

1. MALT
2. CMI
3. AIDS
4. NACO
5. HIV

Solution:

1. MALT – Mucosal-associated lymphoid tissue.
2. CMI – Cell-Mediated Immunity
3. AIDS – Acquired Immuno Deficiency Syndrome
4. NACO – National AIDS Control Organisation
5. HIV – Human immunodeficiency virus.

Question 8.
Differentiate the following and give examples of each.

1. Innate and acquired immunity,
2. Active and passive immunity

Solution:

1. : Differences between innate and acquired immunity are as follows:
   
2. Differences between active and passive immunity are as follows:
   

Question 9.
Draw a well-labeled diagram of an antibody molecule.
Solution:

Question 10.
What are the various routes by which transmission of the human immunodeficiency virus takes place?
Solution:
Various routes of entry of ADDS virus are:

• Sexual contact with the infected person.
• Through placenta (from infected mother to foetus).
• Transfusion of infected blood or blood products.
• Sharing infected needles by drug abusers.

Question 11.
What is the mechanism by which the AIDS virus causes a deficiency of the immune system of the infected person?
Solution:
After getting into the body, the virus enters into the macrophages and converts its RNA genome into DNA with the help of a reverse transcriptase enzyme. The viral DNA takes and directs the infected cells to produce more virus particles i.e., the infected macrophages act like an HIV factory. Simultaneously, the HIV attack the T- lymphocytes and replicate and produce more viruses. Then they are released into the blood and attack other T-lymphocytes.

This will lead to a decrease in the number of T-lymphocytes and the patient begins to show the symptoms such as fever, diarrhea, weight loss etc. Subsequently, his immune system weakens and becomes more prone to infections of bacteria (like Mycobacterium), viruses, fungi and even parasites like Toxoplasma. Finally, he is unable to protect himself.

Question 12.
How is a cancerous cell different from a normal cell?
Solution:
Cancerous cell and normal cell are different in the following aspects:

Question 13.
Explain what is meant by metastasis.
Solution:
The rapid growth of cancerous tumour causes overcrowding and disruption of normal cells. It extends to neighbouring tissues. In the last stage, bits of tumour tissue break off and are carried by the circulating blood or lymphs to other parts of the body, where they invade new tissues and start new tumors called secondary tumors. This property is called metastasis. It is fated due to increasing interference with the body’s life processes.

Question 14.
List the harmful effects caused by alcohol/drug abuse.
Solution:
Harmful effects caused by alcohol/drug abuse are as follows:

• Among youth there is drop in academic performance, lack of interest in personal hygiene, isolation, depression, fatigue, aggressive and rebellious behavior, deteriorating relationships with family and friends, loss of interest in hobbies, change in sleeping and eating habits, fluctuations in weight, appetite, etc.
• Excessive dose of drugs leads to coma and death due to respiratory failure, heart failure or cerebral haemorrhage.
• Abusers become mental and cause financial distress to their entire family and friends.
• They may acquire serious infections like AIDS and hepatitis by taking drugs intravenously.
• Intake of alcohol/drugs damages nervous system, liver (cirrhosis) and kidney.
• Drug abuse adversely affects foetus in case of pregnancy, leading to Foetal Alcohol Syndrome (FAS).
• Continuous use of narcotics and stimulants cause impotency and chromosomal aberrations.
• Heavy drinking can cause an acute alcoholic myopathy characterised by painful and swollen muscles and high levels of serum creatine phosphokinase (CK). Chronic alcoholic men may show testicular atrophy with shrinkage of the seminiferous tubules and loss of sperm cells.
• Heavy drinking causes acute and chronic pancreatitis.
• Alcohol increases RBC size causing a mild anemia.
• Legal problems occur, such as arrest by police for obtaining and keeping drugs unlawfully.

Question 15.
Do you think that friends can influence one to take alcohol/drugs? If yes, how can one protect himself/herself from such an influence?
Solution:
Yes. This can be avoided by

• Choosing a good peer group.
• Discussing ways and means to counteract the presence if any with family elders and teacher/counselors
• Telling the program of an outing to family.
• Keeping contact with family while outside the home.

Question 16.
Why is it that once a person starts taking alcohol or drugs, it is difficult to get rid of this habit? Discuss it with your teacher.
Solution:
Once a person starts taking alcohol or drugs, it is difficult to get rid of this habit because he becomes addicted to it. Addiction is a psychological attachment to certain effects such as euphoria and a temporary feeling of well-being. These drive people to consume drugs/alcohol even when these are not needed, or even when their use becomes self-destructive. With repeated use, the tolerance level of the receptors present in the body increases, which consequently leads to a higher dose of drugs/alcohol and addiction.

Thus, the addictive potential of drugs and alcohol pull the user into a vicious circle leading to their regular use from which he/she may not be able to get out.

Question 17.
In your view what motivates youngsters to take alcohol or drugs and how can this be avoided?
Solution:
There are many factors that motivate youngsters to take alcohol or drug. These include:

• Choosing a good peer group.
• Discussing ways and means to counteract the presence if any with family elders and teacher/counselors
• Telling the programme of an outing to family.
• Keeping contact with family while outside the home.

This can be avoided by the following measures:

1. Education and counseling: Educating and counseling people to face problems and stresses, and to accept disappointments and failures as a part of life.
2. Seeking help from parents and peers: Help from parents and peers should be sought immediately so that they can guide appropriately. Help may even be sought from close and trusted friends.
3. Looking for danger signs: Alert parents and teachers to look for and identify the danger signs. Even friends, if they find someone using drugs or alcohol, should not hesitate to bring this to the notice of parents or teachers in the best interests of the person concerned.
4. Seeking professional and medical help: Lots of help is available in the form of highly qualified psychologists, psychiatrists, and de-addiction and rehabilitation programmes to help individuals who have unfortunately got in the quagmire of drug/alcohol abuse.
5. Cross-checking before prescribing and selling drugs: The physicians should prescribe the habituating drugs only to genuine persons and only for the essential duration. Pharmacists should not sell these drugs without the physician’s prescription.
6. Discipline: Good nurturance with consistent discipline but without suffocating strictness reduces the risk of addictions.
7. Communication: The child must be able to communicate with the parents seeking clarification of all doubts and discussing problems that arise in studies or develop in the class, with friends, siblings and others.
8. Appreciation: For even the smallest achievement, good behavior and other activities, the child should be appreciated.
9. Independent working: Giving responsibility to the child for small tasks and allowing him/her to perform independently. However, guidance should be provided where required.
10. Avoid undue pressure: Every child has a specific personality with certain preferences and choices. They should be taken care of and respected. No child should be asked to perform beyond threshold limits whether in studies, sports or extracurricular activities.

We hope the NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Diseases help you. If you have any query regarding NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Diseases, drop a comment below and we will get back to you at the earliest.

Class 12 Chemistry NCERT Solutions Chapter 14 provides a detailed insight into the concepts related to the chapter “Biomolecules”. Chemistry is an important subject and the students need to be thorough with its concepts if they are preparing for boards or NEET and JEE.

NCERT Solutions also contains solutions to the questions provided in the textbook. The students can use these solutions to answer in the exam and score well.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 14
Chapter Name	Biomolecules
Number of Questions Solved	33
Category	NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules

Biomolecules is a very important chapter and requires detailed understanding of the concepts. This topic is often asked in the examination and the students need to be thorough with the concepts.

The students can learn the structure, properties and classification of various biomolecules such as carbohydrates, nucleic acids, etc. These concepts are also taught in further studies. Therefore, the students need to be thorough with the basics.

NCERT INTEXT QUESTIONS

Question 1.
Glucose and sucrose are soluble in water but cyclohexane and benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer:
Both glucose (C₆H₁₂O₆) and sucrose (C₁₂H₂₂O₁₁) are organic compounds and are expected to be insoluble in water. But quite surprisingly, they readily dissolve in water. This is due to the presence of a number of OH groups (five in the case of glucose and eight in sucrose) which are of polar nature. These are involved in the intermolecular hydrogen bonding with the molecules of H₂O (water). As a result, both of them readily dissolve in water.
Benzene (C₆H₆) and cyclohexane (C₆H₁₂) are hydrocarbons which don’t have any polar group. They, therefore, don’t dissolve in water since there is hardly any scope of hydrogen bonding in their molecules with those of H₂O (water).

Question 2.
What are the expected products of hydrolysis of lactose?
Answer:
The hydrolysis of lactose (disaccharide) can be done either with dilute HC1 or with enzyme emulsin. D-glucose and D-galactose are the products of hydrolysis. Both of them are monosaccharides with the molecular formula C6Hi₂0₆.

Question 3.
How do you explain the absence of an aldehydic group in the pentaacetate of D-glucose?
Answer:
Glucose, as we know is an aldohexose and it is expected to give the characteristic reactions of the aldehydic group e.g., action with NH₂OH, HCN, Tollen’s reagent, Fehling reagent etc. However, the pentadactyl glucose formed by the acylation of glucose with acetic anhydride does not give these reactions.

This means that the aldehydic group is either absent or is not available in the penta acetyl glucose for chemical reactions. In fact, the aldehydic group is a part of the hemiacetal structure which the penta acetyl derivative has. It is, therefore, not free or available to take part in these reactions.

Question 4.
The melting points and solubility of amino acids in water are generally higher than those of corresponding haloacids. Explain.
Answer:
The amino acids exist as zwitter ions, H₃N⁺ — CHR-COO-. Due to this dipolar salt like character, they have strong dipole-dipole attractions. Therefore, their melting points are higher than corresponding haloacids which do not have salt-like character.
Due to their salt-like character, amino acids interact strongly with water. As a result, their solubility in water is higher than corresponding haloacids which do not have a salt-like character.

Question 5.
Where does the water in the egg go after boiling the egg?
Answer:
Upon boiling the egg, denaturation of globular protein present in it occurs. Water present probably gets either absorbed or adsorbed during denaturation and disappears.

Question 6.
Explain why vitamin C can not be stored in the body.
Answer:
Vitamin C is mainly ascorbic acid which is water-soluble. It is readily excreted through urine and cannot be stored in the body.

Question 7.
What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer:
The products obtained are 2-deoxy-D-ribose, phosphoric acid, and thymine.

Question 8.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained? What does this fact suggest about the structure of RNA?
Answer:
As we know, a molecule of DNA has a double-strand structure, and the four complementary bases pair each other. Cytosine (C) always pairs up with guanine (G) while thymine (T) is paired up with adenine (A). Because of the presence of the double-strand structure, when a molecule of DNA is hydrolysed, in each pair the molar ratio of the bases remains the same. However, this is not seen when RNA is subjected to hydrolysis. This suggests that RNA has not a double-strand structure like DNA. It exists as a single strand.

NCERT Exercises

Question 1.
What are monosaccharides?
Answer:
Monosaccharides are carbohydrates which cannot be hydrolysed to smaller molecules. Their general formula is (CH₂O)_n where n = 3 → 7. These are of two types. Those which contain an aldehyde (-CHO) group are called aldose and those which contain a keto (C = 0) group are called ketose. They are further classified as triose, tetrose, pentose etc. according to the no. of carbon atoms present (3, 4, 5 respectively).

Question 2.
What are reducing sugars?
Answer:
Carbohydrates which reduce Fehling’s solution to red precipitate of Cu₂0 or Tollen’s reagent to metallic Ag are called reducing sugars. All monosaccharides (both aldoses and ketoses) and disaccharides except sucrose are reducing sugars. Thus, D – (+) – glucose, D-(-)-fructose, D – (+) – maltose and D – (+) – lactose are reducing sugars.

Question 3.
Write two major functions of carbohydrates in plants. (C.B.S.E. Delhi 2008)
Answer:

1. Structural material for plant cell walls: The polysaccharides cellulose acts as the chief structural material of the plant’s cell walls.
2. Reserve food material: The polysaccharide starch is the major reserve food material in the plants. It is stored in seeds and acts as the reserve food material for the tiny plant till its capable of making food on its own by photosynthesis.

Question 4.
Classify the following into monosaccharides and disaccharides:
Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Answer:
Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose.
Disaccharides: Maltose, lactose.

Question 5.
What do you understand by the term glycosidic linkage?
Answer:
Glycosidic linkage is used to link different monosaccharides in disaccharides and polysaccharides through an oxygen atoms. For example, the glycosidic linkage is present in sucrose, lactose, maltose, etc. These are all disaccharides.

Question 6.
What is glycogen? How is it different from starch?
Answer:

1. The carbohydrates are stored in the animal body as glycogen. It is present in the liver, muscles, and brain. Enzymes break the glycogen down to glucose when the body needs glucose.
2. Glycogen is more highly branched than amylopectin (starch) glycogen chain consists of 10-14 glucose units, whereas amylopectin (starch) glycogen chain consists of 20-25 glucose units.

Question 7.
What are the hydrolysis products of starch and lactose?
Answer:
Starch upon hydrolysis gives α – D(+) glucose which is the constituent of both amylose and amylopectin. Lactose upon hydrolysis gives galactose and glucose.
Upon hydrolysis, cellulose gives only D(+) glucose. This means that only D(+) glucose units are present in cellulose but unlike starch these are -D(+) glucose molecules and not a – D(+) glucose molecules. The X – ray analysis has shown that there are large linear chains of 3 – D( +) glucose molecules lying side by side in the form of bundles held together by hydrogen bonding in the neighbouring hydroxyl groups. 

Question 8.
What is the basic structural difference between starch and cellulose?
Answer:
Starch is not a single component. It consists of amylose and amylopectin. In contrast, cellulose is a single compound. Amylose is a linear polymer of α – D glucose while cellulose is a linear polymer of β -D glucose. In amylose, C1 – C4 α- glycosidic linkage is present, whereas in cellulose C1 – C4 β- glycosidic linkage is present. Amylopectin has a highly branched structure.

Question 9.
What happens when D-glucose is treated with
(i) HI
(ii) Bromine water
(iii) HNO₃?
Answer:
(i) Reaction with HI

(ii) Reaction with bromine water

(iii) Reaction with HNO₃

Question 10.
Enumerate the reactions of D-glucose which cannot be explained by its open-chain structure.
Answer:
The following reactions of D- glucose cannot be explained by its open-chain structure.
1. D – the glucose does not undergo certain characteristic reactions of aldehydes. For example, glucose does not form NaHSO₃ addition product, aldehyde-ammonia adduct, 2, 4, DNP derivative and does not respond to Schiff’s reagent test.

2. Glucose reacts with NH₂OH to form an oxime but glucose pentaacetate does not. This implies that the aldehyde group is absent in glucose pentaacetate.

3. D (+) – Glucose exists in two stereoisomeric forms ie. α – glucose and β- glucose, α – D (+) – glucose is obtained when a concentrated aqueous or alcoholic solution is crystallised at 303K. It has a melting point of 419K and has a specific rotation of +111° in a freshly prepared aqueous solution. However when glucose is crystallised from the water above 371 K β – D (+) glucose is obtained.

4. Both α – D glucose and β – D glucose undergoes mutarotation in an aqueous solution.

Question 11.
What are essential and non-essential amino acids?
Answer:
α – Amino acids which are needed for the health and growth of human beings but are not synthesised by the human body are called essential amino acids. For example, valine, leucine phenylalanine etc. On the other hand α – amino acids which are needed for the health and growth of human beings and are synthesised by the human body are called non-essential amino acids. For example glycine, alanine, aspartic acid etc.

Question 12.
Define the following as related to proteins
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.
Answer:
(i) Peptide Linkage: Proteins are the polymers of a-amino acids which are connected to each other by peptide bond or peptide linkage. Chemically, peptide linkage is an amide formed between -COOH group and -NH2 group. The reaction between two molecules of similar or different amino acids, proceeds through the combination of the amino group of one molecule with the carboxyl group of the other This results in the elimination of a water molecule and formation of a peptide bond -CO-NH-. The product of the reaction is called a dipeptide because it is made up of two amino acids. For example, when carboxyl group of glycine combines with the amino group of alanine we get a dipeptide, glycylalanine.

(ii) Primary Structure: Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino acids that is said to be the primary structure of that protein. Any change in this primary structure i.e.; the sequence of amine acids creates a different protein.

(iii) Denaturation: Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein. When a protein in its native form, is subjected to physical change like change in temperature or chemical change in pH, the hydrogen bonds are disturbed. Due to this; globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein. During denaturation 2° and 3° structures are destroyed but 1° structure remains intact. The coagulation of egg white on boiling is a common example of denaturation. Another example is the curdling of milk which is caused due to the formation of lactic acid by the bacteria present in milk.

Question 13.
What are the common types of secondary structures of proteins?
Answer:
Secondary structure of a protein refers to the shape in which a long polypeptide chain can exist. These are found to exist in two types :

• α-helix structure
• β-pleated sheet structure.

Secondary Structure of Proteins:
The long, flexible peptide chains of proteins are folded into the relatively rigid regular conformations called the
secondary structure. It refers to the conformation which the polypeptide chains assume as a result of hydrogen bonding
between the > C= O and > N-H groups of different peptide bonds.
The type of secondary structure a protein will acquire, in general, depends upon the size of the R-group. If the size of the
R-groups are quite large, the protein will acquire a ct-helix structure. If on the other hand, the size of the R-groups are relatively
smaller, the protein will acquire a β – flat sheet structure.

(a) α-Helix structure: If the size of the R-groups is quite large, the hydrogen bonding occurs between > C = O group
of one amino acid unit and the > N-H group of the fourth amino acid unit within the same chain. As such the polypeptide
chain coils up into a spiral structure called right-handed ct- helix structure. This type of structure is adopted by most of the
fibrous structural proteins such as those present in wool, hair, and muscles. These proteins are elastic i.e., they can be
stretched. During this process, the weak hydrogen bonds causing the α- helix are broken. This tends to increase the length of
the helix like a spring. On releasing the tension, the hydrogen bonds are reformed, giving back the original helical shape.

(b) β—Flat sheet or β—Pleated sheet structure: If R-groups are relatively small, the peptide chains lie side by side in a zig
zag manner with alternate R-groups on the same side situated at fixed distances apart. The two such neighbouring chains are held together by intermolecular hydrogen bonds. A number of such chains can be inter-bonded and this results in the formation of a flat sheet structure These chains may contract or bend a little in order to accommodate moderate-sized R-groups. As a result, the sheet bends into parallel folds to form a pleated sheet structure known as β – pleated sheet structure. These sheets are then stacked one above the other like the pages of a book to form a three-dimensional structure. The protein fibrion in silk fibre has a β – pleated sheet structure. The characteristic mechanical properties of silk can easily be explained on the basis of its β – sheet structure. For example, silk is non-elastic since stretching leads to pulling the peptide covalent bonds. On the other hand, it can be bent easily like a stack of pages because, during this process, the sheets slide over each other.

Question 14.
What type of bonding helps in stabilising the α-helix structure of proteins?
Answer:
The α-helix structure of proteins is stabilized by intramolecular H-bonding between C = O of one amino acid residue and the N – H of the fourth amino acid residue in the chain. This causes the polypeptide chain to coil up into a spiral structure called the right-handed α- helix structure.

Question 15.
Differentiate between globular proteins and fibrous proteins. (Jharkhand Board 2014; C.B.S.E. Delhi 2015)
Answer:

Globular proteins	Fibrous proteins
1. Polypeptide chains are arranged as coils.	1.Polypeptide chains run parallel to each other.
2. They have a spherical shape.	2. They have a thread-like structure.
3. These are water-soluble.	3. These are insoluble in water.
4. These are sensitive to a small change in temperature and pH.	4. These are not affected by a small change in temperature and pH.
5. They possess biological activity.	5. They don’t have any biological activity but serve as the chief structural material of animal tissues.

Question 16.
How do you explain the amphoteric behaviour of amino acids?
Answer:
Amino acids have basic (NH₂) and acidic (COOH) groups. These are, therefore, amphoteric in nature. However, they exhibit these characters in the amino acid molecule itself i.e., NH₂ group (basic) accepts a proton from COOH group (acidic) in the same molecule. Therefore, a-amino acid exists as a dipolar ion.

Question 17.
What are enzymes?
Answer:
We have learned in the study of carbohydrates that these are body fuels i.e., they provide the necessary energy to the body and
keeps it working. Actually, the human body is just like a furnace in which chemical reactions take place and are responsible for the digestion of food, absorption of appropriate molecules, and production of energy. The entire process involves a series of reactions that are catalyzed by biocatalysts known as enzymes. Thus, enzymes may be defined as:
biological or biocatalysts which catalyse the reactions in living beings.
All enzymes are basically globular proteins. Enzymes are very specific for a particular reaction as well as for a particular
substrate. These are generally named after the compounds or the class of substances with which they are linked or work.

Question 18.
What is the effect of denaturation on the structure of proteins?
Answer:
As a result of denaturation, globules get unfolded and helixes get uncoiled. Secondary and tertiary structures of the protein are destroyed, but the primary structures remain unaltered. It can be said that during denaturation, secondary and tertiary – structured proteins get converted into primary – structured proteins. Also, as the secondary and tertiary structures of a protein are destroyed, the enzyme loses its activity.

Question 19.
How are vitamins classified? Name the vitamin responsible for the coagulation of blood? (C.B.S.E. Outside Delhi 2015)
Answer:
On the basis of their solubility in water or fat, vitamins are classified into two groups:
1. Fat-soluble vitamins:
Vitamins that are soluble in fat and oils, but not in the water, belong to their group.
For example Vitamins A, D, E, and K.

2. Water-soluble vitamins:
Vitamins that are soluble in water belong to their group.
For example, B group vitamins (B₁, B₂, B₆, B₁₂, etc) and vitamin C.
However, biotin or vitamin H is neither soluble in water nor in fat.
Vitamin K is responsible for the coagulation of blood.

Question 20.
Why are vitamin A and vitamin C essential to us? Mention their sources.
Answer:
vitamin A: Soluble in water but insoluble in oils and fas. Destroyed by cooking or prolonged exposure to air. it increases the resistance of the body towards diseases. Maintains healthy skin and helps in the healing of cuts and abrasions. It is available in

vitamin C:  Soluble in oils and fats but insoluble in water, stable to heat. Promotes growth and improves vision. ¡t also increases resistance to disease. It is available in Citrus fruits (oranges, lemon, grapefruit, lime, etc.), amia, cabbage. guava etc.

Question 21.
What are nucleic acids? Mention their two important functions.
Answer:
Nucleic acids are biologically important polymers which are present in all living cells.

• Nucleic acids play a vital role in the transmission of heredity characteristics.
• Nucleic acids help in the biosynthesis of proteins.

Two important biological functions of nucleic acids are (i) Replication and (ii) protein synthesis.
These are briefly discussed.
Replication: Replication may be defined as the process by which a single DNA molecule produces two identical copies of itself.

Replication is an enzyme-catalyzed process.  The process of replication starts with the partial unwinding of the two strands of the DNA double helix through the breaking of the hydrogen bonds between pairs of bases. Each strand then acts as the template (or pattern) for the synthesis of two new strands of DNA in the celllix environment. The specificity of base-pairing ensures that each new strand is complementary to its old template strand. As a result, two identical copies of DNA from the original DNA are produced. Each of these two copies is then passed on to the two new cells resulting from cell division. In this way, hereditary characters are transmitted from one cell to another.

Question 22.
What is the difference between a nucleoside and a nucleotide?
Answer:
A nucleoside contains a pentose sugar and base (purine or pyrimidine) while in the nucleotide, a phosphoric acid component is also present.
Arrangement of constituents in Nucleic Acids. These are, intact, three building blocks in nucleic acid. A combination ol
base and sugar are known as a nucleoside. Similarly, base, sugar, and phosphates from nucleotides while nucleic acids are
polynucleotides which means that these are the polymers of nucleotides.

Question 23.
The two strands in DNA are not identical but are complementary. Explain.
Answer:
In the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosine forms a hydrogen bond with guanine, while adenine forms a hydrogen bond with thymine. As a result, the two strands are complementary to each other.

Question 24.
Write the important structural and functional differences between DNA and RNA.
Answer:

Ribonucleic acid (RNA)	Deoxyribonucleic acid (DNA)
1. The pentose sugar present in RNA is D-ribose	1. The pentose sugar present in DNA is D-2-deoxyribose
2. RNA contains cytosine and uracil as pyrimidine bases and guanine and adenine as purine bases.	2. DNA contains cytosine and thymine as pyrimidine bases and guanine and adenine as purine bases.
3. It is a single chain of polynucleotides.	3. It is a double chain of polynucleotides.
4. It is formed by DNA and cannot replicate itself.	4. It can replicate itself.
5. Its molecule is relatively short with low molecular mass.	5. Its molecule is relatively long with a high molecular mass.
6. It regulates protein synthesis.	6. It controls structure, metabolism, differentiation and transfer the characters from one generation to the other.
7. It is an essential genetic material of plant viruses.	7. It is an essential genetic material of eukaryotic cells.

Question 25.
What are the different types of RNA found in the cell?
Answer:

• Messenger RNA (m – RNA)
• Ribosomal RNA (r – RNA)
• Transfer RNA (t – RNA)

We hope the NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules, drop a comment below and we will get back to you at the earliest.

 

Class 12 NCERT Solutions for Chemistry Chapter 13 contains solved answers for the questions provided in the textbook. These answers are provided by the subject experts and are accurate to the best of our knowledge. The students can use these answers along with the diagrammatic representation in the examination and score well.

NCERT Solutions are beneficial for the students appearing for UP board, MP Board, CBSE, Gujarat board, etc., and also for competitive exams such as NEET and JEE.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 13
Chapter Name	Amines
Number of Questions Solved	23
Category	NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 13 Amines

This chapter explains the importance, structure, physical and chemical properties, and the methods of preparation of amines. This chapter also explains the diazonium salts and the methods of their preparation. This will help you to understand the importance of these salts in the synthesis of aromatic compounds.

The students can use NCERT Solutions for better preparations of the concepts mentioned in this chapter. Basic understanding is very important for advanced concepts.

NCERT INTEXT QUESTIONS

Question 1.
Classify the following amines as primary, secondary, and tertiary amines

Answer:
(i) primary
(ii) tertiary
(iii) primary
(iv) secondary.

Question 3.
(i)Write the structures of different isomeric amines corresponding to the molecular formula, C₄H₁₁N.
(ii) Write 1UPAC names of all the isomers.
(iii) What type of isomerism is exhibited by different pairs of amines?
Answer:

Chain isomers: (a) and (c); (a) and (d); (b) and (c); (b) and (d)
Metamers: (e) and (g); (f) and (g)
Functional isomers: All 10 amines are functional isomers of 2° and 3° amines and vice-versa.

Question 3.
How will you convert:
(i) Benzene into aniline
(ii) Benzene into N, N-dimethylaniline
(iii) C1(CH₂)₆Cl into hexane-1, 6-diamine
Answer:
(i) Benzene into aniline:

(ii) Benzene into N, N-dimethylaniline:

(iii) Cl(CH₂)₆Cl into hexane-1, 6-diamine:

Question 4.
Arrange the following in increasing order of their basic strength:
(i) C₂H₅NH₂,C₆H₅NH₂,NH₃,C₆H₅CH₂NH₂ and(C₂H₅)₂ NH
(ii) C₂H₅NH₂,(C₂H₅)₂NH,(C₂H₅)₃N,C₆H₅NH₂
(iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂
Answer:
(i) C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ <  C₂H₅NH₂<(C₂H₅)₂NH
(ii) C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N <(C₂H₅)₂NH
(iii) C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃ N < CH₃NH₂<(CH₃)₂NH

Question 5.
Complete the following acid-base reactions and name the products
(i) CH₃CH₂CH₂NH₂ + HC1 →
(ii) (C₂H₅)₃N + HC1  →
Answer:

Question 6.
Write the reactions of the final alkylation product of aniline with excess methyl iodide in the presence of sodium carbonate solution.
Answer:
Aniline is a primary amine. It will react with excess methyl iodide to form quaternary ammonium salt as the final product. The reaction is known as Hoffmann’s ammonolysis.

Question 7.
Write the chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Answer:
Aniline will undergo benzoylation to form a benzoyl derivative. The reaction will take place in the presence of aqueous alkali.

Question 8.
Write the structures of the different isomers corresponding to the molecular formula C₃H₉N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.
Answer:
Four isomeric aliphatic amines are represented by the molecular formula C₃H₉N. These are:

Only the primary amines will evolve N₂ gas on reacting with nitrous acid (HONO) and form corresponding primary alcohols.

Question 9.
How will you convert:
(i) 3-Methylaniline into 3-nitrotoluene
(ii) Aniline into 1,3,5-tribromobenzene ?
Answer:
(i) 3-Methylaniline into 3-nitrotoluene

(ii) Aniline into 1,3,5-tribromobenzene

NCERT EXERCISE

Question 1.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH₃)₂CHNH₂
(ii) CH₃(CH2)₂NH₂
(iii) CH₃NHCH(CH₃)₂
(iv) (CH₃)₃CNH₂
(v) C₆H₅NHCH₃
(vi) (CH₃CH₂)₂NCN₃
(vii) m-BrC₆H₄NH₂
Answer:

Question 2.
Give one chemical test to distinguish between the following pairs of compounds :
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline. (C.B.S.E. Sample Paper 2015)
Answer:
(i) Methylamine on reaction with nitrous acid evolves N₂ gas with brisk effervescence while dimethylamine does not. Methylamine also gives carbylamine reaction upon warming with chloroform and alcoholic KOH while dimethylamine does not.
(ii) Secondary amines, both aliphatic and aromatic respond to Libermann’s nitroso reaction while tertiary amines do not.
(iii) Aniline responds to diazotisation and coupling reactions to form a dye while ethylamine does not.
(iv) Aniline gives diazotisation coupling reaction while benzylamine does not.
(v)  Aniline gives carbyl amine test with an extremely unpleasant smell while N-Methyl aniline does not.

Question 3.
Account for the following :
(i) pK_b of aniline is more than that of methylamine. (C.B.S.E. Delhi 2008, 2011)
(ii) Ethylamine is soluble in water whereas aniline is not. (C.B.S.E. Delhi 2008, 2011)
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. (C.B.S.E. Delhi 2008)
(iv) Although the amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. (C.B.S.E. Sample Paper 2010)
(v) Aniline does not undergo Friedel Crafts reaction. (C.B.S.E. Delhi 2008, Sample Paper 2010, C.B.S.E. Outside Delhi 2015)
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Answer:
(i) pK_b of aniline is more than that of methylamine because aniline is less basic. In aniline, the electron pair on the nitrogen atom is involved in conjugation with the ring and is less available for protonation than in methylamine. Therefore, aniline has more pK_b.

(ii) Ethylamine is water-soluble due to hydrogen bonding. However, in aniline, the phenyl (C₆H₅) group is bulky in size and has -I effect. As a result, its hydrogen bonding with water is negligible and is therefore not soluble or miscible with water.

(iii) Methylamine forms a soluble hydroxide on reacting with water. The OH^– ions released by the hydroxide combine with  Fe³⁺ ions of ferric chloride to give ferric hydroxide or hydrated ferric oxide which is brown in colour.

(iv) Amino group (-NH₂) is an electron releasing or activating group when present on the benzene ring. It activates the ortho and para positions in the ring towards electrophilic substitution due to its +M or +R effect. The nitration of aniline carried by nitrating mixture (cone. HN0₃ + cone. H₂SO₄) is electrophilic in nature. The expected product of nitration is a mixture of ortho and para nitroaniline. However, in this case, a substantial amount of metanitroaniline is also formed. In fact, aniline being a base gets protonated in the acidic medium to form anilinium cation which is no longer activating. Rather, it is deactivating in nature and deactivates the ring. The substitution takes place at the meta position.

Thus, the nitration of aniline as such gives a significant amount of m-nitroaniline (47%). In addition to this, p-nitroaniline
is the major constituent (51%) while ortho isomer is in negligible amount (2%) mainly due to the reason that the ortho position
is sterically hindered because of the —NH2 group.

In order to check the activation of the ring by an amino group, the nitration of aniline is carried out indirectly by first
acetylating with acetic anhydride (or acetyl chloride) to form acetanilide. The compound formed is nitrated by the nitrating
mixture and the isomeric nitro derivatives are then hydrolysed in the acidic medium as discussed under halogenation.

(v) Aniline does not undergo Friedel Crafts reaction. Actually, aniline being a Lewis base forms a complex with AICI3 which is a Lewis acid. The amino group is not in a position to activate the benzene ring towards electrophilic substitution i.e., alkylation or acylation which leads to Friedel Crafts reaction. Therefore, the reaction is not possible. The same problem arises in phenols as well.

(vi) The diazonium salts of aromatic amines are more stable than those of aliphatic amines because these are resonance stabilised while no such resonance stabilisation is possible in the corresponding diazonium salts of aliphatic amines.

(vii) Gabriel phthalimide synthesis is generally preferred over other methods for the synthesis of primary aliphatic amines. Potassium phthalimide formed by reacting phthalimide with alcoholic KOH reacts with an alkyl halide such as C₂H₅-I to form N-alkyl derivative which undergoes hydrolysis to form the primary amine. However, no reaction is possible with aryl halide such as C₆H₅-I. Therefore, primary aromatic amines are not formed in the reaction.

Question 4.
Arrange the following:
(a) In decreasing order of the pK_b values:
C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂

(b) In decreasing order of basic strength:
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂ (C.B.S.E. Delhi 2011, Haryana Board 2013, C.B.S.E. Outside Delhi 2015)

(c) Increasing order of basic strength
Aniline, p-nitroaniline and p-toluidine

(d) Decreasing order of basic strength in gas phase
C₂H₅NH₂, (C₂H_s)₂NH, (C₂H5)₃N and NH₃

(e) Increasing order of boiling point
C₂H₅OH, (CH₃)₂NH, C₂H_sNH₂

(f) Increasing order of solubility in water
C6H₅NH₂, (C₂H_s)₂NH, C₂H5NH₂.
Answer:
From K_b  and PK_b values of some Amines:

(a) The decreasing order of pK_b values or increasing order of basic strength is:
C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH

(b) The decreasing order of basic strength is:
(C₂H₅)₂NH > CH₃NH₂ > C₆H₅N(CH₃)₂ > QH5NH2

(c) The increasing order of basic strength is:
p-nitroaniline < Aniline < p-Toluidine

(d) The decreasing order of basic strength in gaseous phase.
(C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃

(e) Increasing order of boiling point is:
(C₂H₃)₂NH < C₂H₅NH₂ < C₂H₅OH

(f) Increasing order of solubility in water is :
C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂.

Question 5.
How will you convert
(i) Ethanoic acid to methanamfrie,
(ii)Hexanenitrile to pentan-l-amine,
(iii)Methanol to ethanoic acid,
(iv) thatfaminft to methanamine,
(v)Ethanoic acid to propanoic acid,
(vi)Methanamine to ethanamine,
(vii)Nitromethane into dimethylamine,
(viii)Propanoic acid into ethanoic acid ?
Answer:
(i) Ethanoic acid to methanamfrie

(ii) Hexanenitrile to pentan-l-amine

(iii) Methanol to ethanoic acid

(iv) thatfaminft to methanamine

(v) Ethanoic acid to propanoic acid

(vi) Methanamine to ethanamine

(vii) Nitromethane into dimethylamine

(viii) Propanoic acid into ethanoic acid

Question 6.
Describe a method for the identification of primary, secondary, and tertiary amines. Also, write chemical equations of the reactions involved.
Answer:
The three type of amines can be distinguished by Hinsberg test. In this test, the amine is shaken with benzene sulphonyl chloride (C₆H₅SO₂Cl) in the presence of excess of aqueous NaOH or KOH. A primary amine reacts to give a clear solution, which on acidification yields an insoluble compound.

Question 7.
Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) Hoffmann’s bromamide reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi)Acetylation
(vii)Gabriel phthalimide synthesis. (C.B.S.E. Delhi 2011, C.B.S.E. Outside Delhi 2015)
Answer:
(i) Carbyl amine reaction:
Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul-smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amines.

(ii) Diazotisation:
Primary aromatic amines such as aniline react with nitrous acid under ice-cold conditions (273 – 278 K) to form benzene diazonium salt. The reaction is known as diazotisation reaction.

In case, the temperature is allowed to rise above 278 K, benzene diazonium chloride is decomposed by water to form phenol.

Aliphatic primary amines also react with nitrous acid to form alkyl diazonium salts in a similar manner. But these are quite unstable and decompose to form a mixture of alcohols, alkenes, and alkyl halides along with the evolution of N₂ gas.

Aliphatic primary amines also react with nitrous acid to form alkyl diazonium salts in a similar manner. But these are quite unstable and decompose to form a mixture of alcohols, alkenes, and alkyl halides along with the evolution of N₂ gas.

(iii) Hoffmann’s bromamide reaction:
By Hoffmann degradation of Acid Amides. (Hoffmann Bromamide Reaction). When a primary acid amide is heated with an aqueous or ethanolic solution of sodium hydroxide and bromine, it gives a primary amine with one carbon
atom less.

The reaction is, therefore, regarded as a degradation reaction. For example.

(iv) Coupling reaction:
The reaction of diazonium salts with phenols and aromatic amines to form azo compounds having an extended conjugated system with both aromatic rings joined through the — N = N — bond, is called coupling reaction. In this reaction; the nitrogen atoms of the diazo group are retained in the product. The coupling with phenols takes place in a mildly alkaline medium while that with amines occurs under faintly acidic conditions. For example;

Coupling generally occurs at the p-position with respect to the hydroxyl or the amino group, if free, otherwise it takes place at the o-position.

(v) Ammonolysis:
The mechanism involves the nucleophilic attack of NH₃ molecule (through lone pair) on alkyl halide by an SN₂ mechanism. Amine salt is formed which reacts with ammonia to give primary amine and ammonium halide as follows:

The primary amine formed now acts as the nucleophile and reacts with another molecule of the alkyl halide to form secondary amine.

The reaction is repeated to form tertiary amine and quaternary ammonium salt as follows :

(vi) Acetylation:
Acylation of Amines Both aliphatic and aromatic amines form acyl derivatives (substituted acid amides) with reagents such as acid chlorides, esters, or acid anhydrides. The acylation is carried out in the presence of a base stronger than pyridine (e.g., NaOH) which can remove the acid formed in the reaction by neutralising it.
(a) Acylation of Aliphatic Amines: Both primary and secondary aliphatic amines from acyl derivatives as follows:

(b) Acylation of Aromatic Amines: Aromatic amines such as aniline can be acylated in the same manner with both acid
chloride and acid anhydride.

(vii) Gabriel’s phthalimide synthesis:
In this reaction phthalimide is converted into its potassium salt by treating it with alcoholic potassium hydroxide. Then potassium phthalimide is heated with an alkyl halide to yield an N-alkylpthalimide which is hydrolysed to phthalic acid and primary amine by alkaline hydrolysis

This synthesis is very useful for the preparation of pure aralkyl and aliphatic primary amines. However, aromatic primary amines cannot be prepared by this method.

Question 8.
Accomplish the following conversions:
(i) Nitrobenzene to benzoic acid
(ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline
(iv) Aniline to 2, 4, 6-tribromofluorobenzene
(v) Benzyl chloride to 2-phenylethanamine
(vi) Chlorobenzene to p-chloroaniline
(vii) Aniline to p-bromoaniline
(viii)Benzamide to toluene
(ix)Aniline to benzyl alcohol.
Answer:
(i) Nitrobenzene to benzoic acid: 

(ii) Benzene to m-bromophenol: 

 

(iii) Benzoic acid to aniline:

(iv) Aniline to 2, 4, 6-tribromofluorobenzene: 

(v) Benzyl chloride to 2-phenylethanamine: 

(vi) Chlorobenzene to p-chloroaniline:

(vii) Aniline to p-bromoaniline: 

(viii)Benzamide to toluene :

(ix)Aniline to benzyl alcohol:

 

Question 9.
Give the structures of A, B, and C in the following reactions:

Answer:

Question 10.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B, and C.
Answer:
Since the compound ‘C’ with molecular formula C₆H₇N is formed from compound ‘B’ on treatment with Br₂ KOH, therefore, compound ‘B’ must be an amide and ‘C’ must be an amine.
The only amine having the molecular formula C₆H₇N, i. e., C₆H₅NH₂ is aniline.
Since ‘C’ is aniline, therefore, the die amide from which it is formed must be benzamide (C₆H₅CONH₂). Thus, compound ‘B’ is benzamide. Since compound ‘B’ is formed from compound ‘A’ with aqueous ammonia and heating, therefore, compound ‘A’ must be benzoic acid.

Question 11.
Complete the following reactions:

Answer:

Question 12.
Why cannot aromatic amines be prepared by Gabriel’s phthalimide reaction? (C.B.S.E. Sample Question Paper 2012, H.P. Board2017)
Answer:
In Gabriel phthalimide reaction, the potassium salt of phthalimide is formed. It readily reacts with an alkyl halide to form the corresponding alkyl derivative.

But it is not in a position to react with the aryl halide in case primary aromatic amine is to be prepared. Actually, the cleavage of C – X bond in haloarene or aryl halide is quite difficult due to partial double bond character. Therefore, aromatic amines cannot be prepared by this method.

Question 13.
How do aromatic and aliphatic primary amines react with nitrous acid?
Answer:
Reaction with nitrous acid. All three types of amines, aliphatic as well as aromatic, react with nitrous acid under different conditions to form a variety of products. Since nitrous acid is highly unstable, it is prepared in situ by the action of dilute hydrochloric acid on sodium nitrite.

(a) Primary aliphatic amines react with nitrous acid at low temperature (cold conditions) to form primary alcohol and nitrogen gas accompanied by brisk effervescence. Nitrous acid is unstable in nature and is prepared in situ by reacting sodium nitrite with dilute hydrochloric acid. For example,

The reaction ¡s used as a rest for primary aliphatic amines as no other amine evolves nitrogen with nitrous acid.

(b) Primary aromatic amines such as aniline react with nitrous acid under ice cold conditions (273—278 K) to form benzene diazonium salt. The reaction is known as diazotisation reaction.

In case, the temperature is allowed to rise above 278 K, benzene diazonium chloride is decomposed by water to form phenol.

Aliphatic primary amines also react with nitrous acid to form alkyl diazonium salts in a similar manner. But these are quite unstable and decompose to form a mixture of alcohols, alkenes, and alkyl halides along with the evolution of N2 gas.

Question 14.
Give plausible explanation for each of the following :
(i) Why are amines less acidic than alcohols of comparable molecular masses ?
(ii) Why are primary amines higher boiling than tertiary amines ?
(iii) Why are aliphatic amines stronger bases than aromatic amines ? (H.P. Board 2008)
Answer:
(i) The acidic character in the both in cases is due to the release of H⁺ ion. Now, the anion in case of amine

has a negative charge on the nitrogen atom while the anion formed in case of alcohol has negative charge on the oxygen atom. Since oxygen is more electronegative than nitrogen atom, the negative charge can be accomodated easily on oxygen than on nitrogen in these anions. In other words, RO” ion is more stable than RNH“ ion. Consequently, alcohol is a stronger acid than amine. Please remember that even alcohols are very weakly acidic so much so that they donot turn blue litmus red.

(ii) Primary amines are higher boiling than tertiary amines due to the presence of intermolecular hydrogen bonding in their molecules. Since tertiary amines (R₃N) have no hydrogen atom present, these are not involved in any such hydrogen bonding. For example, the boiling point of n-butylamine (CH₃CH₂CH₂CH₂NH₂) is 322 K while that of trimethylamine (CH₃)₃ N is 276 K.

(iii) In the aromatic amines, (lie secondary and tertiary amines are more basic than aniline.
Actually, the basic strength or the electron releasing tendency of an amine depends upon the following factors.

1. The ability of the nitrogen atom to donate a pair of electrons.
2.  The stability of cation by accepting the pair of electrons.

Any factor which tends lo increase the electron releasing tendency of amine or increase the stability of the cation, will tend to increase the basic strength of amine.

We hope the NCERT Solutions for Class 12 Chemistry Chapter 13 Amines help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 13 Amines, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

These Solutions are part of NCERT Solutions for Class 12 Biology. Here we have given NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Question 1.
Name the parts of an angiosperm flower in which development of male and female gametophyte take place.
Solution:
Inside the anther, the cells of microsporangia develop as male gamete. Inside the ovary megasporangial cells develop as female gametes.

Question 2.
Differentiate between microsporogenesis and megasporogenesis. Which type of cell division occurs during these events? Name the structures formed at the end of these two events.
Solution:
Differences between microsporogenesis and megasporogenesis are as follows :

During microsporogenesis and Megas-megasporogenesis meiotic cell division occurs which results in haploid gametes – the microspores or pollen grains and megaspores.

Question 3.
Arrange the following terms in- the correct developmental sequence : Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male gametes.
Solution:
Sporogenous tissue → Pollen mother cell → microspore tetrad → pollen grain → male gamete.

Question 4.
With a neat, labelled diagram, describe the parts of a typical angiosperm ovule.

Solution:
An angiosperm ovule consists of the following parts:

• The ovule is attached to placenta by means of a stalk called funicle or funiculus.
• The point of attachment of funiculus to the body of ovule is called hilum.
• The main body of ovule is made of parenchymatous tissue called nucellus.
• Nucellus is covered on its outside by one or two coverings called integuments and hence ovule is rightly called as integument megasporangium.
• The integuments cover entire nucellus except for a small pore at upper end, which is called the micropyle. Micropyle is formed generally by inner integument or by both integuments.
• The place of junction of integuments and nucellus is called chalaza.
• In inverted ovules (most common type), the stalk or funiculus is attached to the main body of ovule for some distance to form a ridge like structure, called- raphe.
• In the nucellus of ovule, a large oval cell is present at micropylar end, which is known as embryo sac (female gametophyte), which develops from the megaspore.

Question 5.
What is meant by monosporic development of female gametophyte?
Solution:
The female gametophyte or the embryo sac develops, from a single functional megaspore. This is known as the monosporic development of the female gametophyte. In most flowering plants, a single megaspore mother cell present at the micropylar pole of the nucellus region of the ovule undergoes meiosis to produce four haploid megaspores. Later out of these 4 megaspores, only one functional megaspore develops into a female gametophyte, while the remaining 3 degenerates.

Question 6.
With a neat diagram explain the 7-celled, 8 nucleate nature of the female gametophyte
Solution:

The female gametophyte (embryo sac) develops from a single functional megaspore. Thus, the megaspore undergoes three successive mitotic divisions to form 8 nucleate embryo sac. The first mitotic division in the megaspore forms 2 nuclei. One nucleus moves towards the micropylar end while the other nucleus moves towards the chalazal end. Then these nuclei divide at their respective ends and redivide to form 8 nucleate stages.

As a result there are 4 nuclei each at both the ends i.e., at the micropylar and the chalazal end in the embryo sac. At the micropylar end, out of 4 nuclei only 3 differentiate into 2 synergids and one egg cell. Together they are known as egg apparatus. Similarly, at the chalazal end 3 out of 4 nuclei differentiates as antipodal cells. The remaining 2 cells (of the micropylar and chalazal end) move towards the centre and are known as the polar nuclei, which are situated in the centre of the embryo sac. Hence, at maturity, the female gametophyte appears as a 7 celled structure, though it has 8 nucleate.

Question 7.
What are chasmogamous flowers? Can cross-pollination occur in cleistogamous flowers? Give reasons for your answer.
Solution:
Chasmogamous flowers or open flowers in which anther and stigma are exposed for pollination. Cross-pollination cannot occur in cleistogamous flowers. These flowers remain closed thus causing only self-pollination. In cleistogamous flowers, anthers dehisce inside the closed flowers. So the pollen grains come in contact with stigma. Thus there is no chance of cross¬pollination, e.g., Oxalis, Viola.

Question 8.
Mention two strategies evolved to prevent self pollination in flowers.
Solution:
Two strategies evolved to prevent self-pollination are:

• Pollen release and stigma receptivity are not synchronized.
• Anthers and stigma are placed at such positions that pollen doesn’t reach stigma.

Question 9.
What is self-incompatibility? Why does self-pollination not lead to seed formation in self-incompatible species?
Solution:
When the pollen grains of an anther do not germinate on the stigma of the same flower, then such a flower is called self-sterile or incompatible and such condition is known as self¬incompatibility or self-sterility.
The transference of pollen grains shed from the anther to the stigma of the pistil is called pollination. This transference initiate the process of seed formation. Self-pollination is the transfer of pollen grain shed from the anther to stigma of pistil in the same flower. But in some flower self¬pollination does not lead to the formation of seed formation because of the presence of same sterile gene on pistil and pollen grain.

Question 10.
What is bagging technique? How is it useful in a plant breeding programme?
Solution:
It is the covering of female plants with butter paper or polythene to avoid their contamination from foreign pollens during the breeding programme.

Question 11.
What is triple fusion? Where and how does it take place? Name the nuclei involved in triple fusion.
Solution:
Inside the embryo sac, one male gamete fuses with egg cells to form a zygote (2n) and this is called syngamy or true act of fertilisation. This result of syngamy, i.e., zygote (2n) ultimately develops into an embryo.

The second male gamete fuses with 2 polar nuclei or secondary nucleus to form triploid primary endosperm nucleus and this is called triple fusion. The result of triple fusion, i.e., primary endosperm nucleus (3n) ultimately develops into a nutritive tissue for developing embryo called endosperm.

The nuclei involved in this triple fusion are the two polar nuclei or secondary nucleus and the second male gamete.

Question 12.
Why do you think the zygote is dormant for sometime in a fertilised ovule?
Solution:
The zygote is dormant in fertilized ovule for some time because, at this time, endosperm needs to develop. As endosperm is the source of nutrition for the developing embryo, nature ensures the formation of enough endosperm tissue before starting the process of embryogenesis.

Question 13.
Differentiate between:

1. Epicotyl and hypocotyl;
2. Coleoptile and coleorhiza;
3. Integument and testa;
4. Perisperm and pericarp

Solution:

1. Differences between epicotyl and hypocotyl are as follows :
   
2.  Differences between coleoptile and coleorhiza are as follows :
   
3. Differences between integument and testa are as follows :
   
4. Differences between perisperm and pericarp are as follows :
   

Question 14.
Why is apple called a false fruit? Which part (s) of the flower forms the fruit?
Solution:
Apple is called a false fruit because it develops from the thalamus instead of the ovary (the thalamus is the enlarged structure at the base of the flower).

Question 15.
What is meant by emasculation? When and why does a plant breeder employ this technique?
Solution:
Emasculation is the removal of stamens mainly the anthers from the flower buds before their dehiscence. This is mainly done to avoid self-pollination. Emasculation is one of the measures in the artificial hybridization. Plant breeders employed this technique to prevent the pollination within same flower or to pollinate stigmas with pollens of desired variety.

Question 16.
If one can induce parthenocarpy through the application of growth substances, which fruits would you select to induce parthenocarpy and why ?
Solution:
Oranges, lemons, litchis could be potential fruits for inducing the parthenocarpy because a seedless variety of these fruits would be much appreciated by the consumers.

Question 17.
Explain the role of tapetum in the formation of pollen-grain wall.
Solution:
Tapetum is the innermost layer of the microsporangium. The tapetal cells are multinucleated and polyploid. They nourish the developing pollen grains. These cells contain ubisch bodies that help in the ornamentation of the microspores or pollen grains walls. The outer layer of the pollen grain is called exine and is made up of the sporopollenin secreted by the ubisch bodies of the tapetal cells. This compound provides spiny appearance to the exine of the pollen grains.

Question 18.
What is apomixis and what is its importance ?
Solution:
Apomixis is the process of asexual production of seeds, without fertilization.
The plants that grow from these seeds are identical to the mother plant.

Uses:

• It is a cost-effective method for producing seeds.
• It has great use for plant breeding when specific traits of a plant have to be preserved.

We hope the NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants help you. If you have any query regarding NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Biology Chapter 7 Evolution

These Solutions are part of NCERT Solutions for Class 12 Biology. Here we have given NCERT Solutions for Class 12 Biology Chapter 7 Evolution

Question 1.
Explain antibiotic resistance observed in bacteria in light of Darwinian selection theory.
Solution:
Penicillin when discovered was used as an antibiotic against all bacteria. Soon many of these became resistant. This is because alleles of resistance which are already present in bacteria are of no importance in absence of antibiotics. Adjustment to change in environment due to genetic variation is adaptation.

Question 2.
Find out from newspapers and popular science articles any new fossil discoveries or controversies about evolution.
Solution:
Chimps are more evolved than humans (The Times of India):
Chimpanzees are more evolved than humans, a study suggests. There is no doubt that humans are the more advanced species. But a comparison of 14,000 human and chimpanzee genes shows the forces of natural selection have and the greatest impact on our ape cousins.

The researchers’ discovery challenges the common assumption that our large brains and high intelligence were the gifts of natural selection. Humans and chimps followed different evolutionary paths from a common ape ancestor about 5 million years ago. Both underwent changes as the fittest survived to pass their genes on to future generations. But the US study shows that humans possess a ‘substantially smaller’ number of positively-selected genes than chimps.

Question 3.
Attempt giving a clear definition of the term species.
Solution:
A species generally includes a similar organism. Members of this group can show interbreeding. A similar group of genes are found in the members of the same species and this group has the capacity to produce new species. Every species has some cause of isolation which interrupted the interbreeding with the nearest reactional species which is referred to as reproductively isolated.

Question 4.
Try to trace the various components of human evolution (hint: brain size and function, skeletal structure, dietary preference, etc.)
Solution:
Human evolution shows the following trends:
A. Brain size: It increased gradually along with evolution. The brain capacity of Australopithecus africanus – 500 cc, Homo habilis – 700 cc, Homo eredus – 800 – 1300 cc, Homo sapiens sapiens – 1450 cc.

B. Skeletal structure:

• Dryopithecus was ape-like, without brow ridges, had semierect posture, and prognathous face (having a projecting jaw).
• Ramapithecus had jaws and teeth like humans (small canines and large molars), prognathous face, and walked on legs
• Australopithecus africanus had erect posture, human-like teeth, was without chin, with brow ridges, and had a prognathous face.
• Homo habilis walked nearly erect, had human-like teeth, with brow ridges face was slightly prognathous.
• Homo erectus had an erect posture, prognathous face, with projecting brow ridges, small canines, and large molar teeth and had a small chin.
• Homo sapiens had four curves in the vertebral column, orthognathous face (without projecting jaw), forehead broad, chin well developed, walked on the sole.

C. Dietary preference: Dryopithecus and Ram-apithecus were herbivores, Australopithecus africanus and Homo habilis were carnivores, Homo erectus and Homo sapiens were omnivores.

Question 5.
Find out through the internet and popular science articles whether animals other than man have self-consciousness.
Solution:
There are many animals other than humans, which have self-consciousness. An example of an animal being self-conscious is dolphins. They are highly intelligent. They have a sense of self and, they also recognize others among themselves and others. They communicate with each other by whistles, tail-slapping, and other body movements, not dolphins, there are certain other animals such as Crow, Parrot, chimpanzees, Gorilla, Orangutan, etc., which exhibit self-consciousness.

Question 6.
List 5-6 modern-day animals and using the internet resources link it to a corresponding ancient fossil. Name both.
Solution:
The list of few modern-day animals and their corresponding ancient fossils is as follows:

Question 7.
Describe one example of adaptive radiation.
Solution:
Adaptive radiation – Formation of different species from a common ancestor with new species adapting to different geological niches.
Example: Darwin’s finches are Galapagos island have wolves from mainland finches. They underwent changes in the shape, size of beaks, food habits, feathers.

Question 8.
Can we call human evolution adaptive radiation?
Solution:
No, we can not be called human evolution as adaptive evolution.

Question 9.
Using various resources such as your school library or the Internet and discussions with your teacher, trace the evolutionary stages of any one animal say horse.
Solution:
The evolutionary stages of the modern horse are listed in the table given below:

We hope the NCERT Solutions for Class 12 Biology Chapter 7 Evolution help you. If you have any query regarding NCERT Solutions for Class 12 Biology Chapter 7 Evolution, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations

These Solutions are part of NCERT Solutions for Class 12 Biology. Here we have given NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations

Question 1.
How is diapause different from hibernation ?
Solution:
Diapause is different from hibernation. The table below shows the differences between them :

Question 2.
If a marine fish is placed in a freshwater aquarium/will the fish be able to survive? Why or why not?
Solution:
If a marine fish is placed in a freshwater aquarium, then its chances of survival will diminish. This is because their bodies are adapted to high salt concentrations in the marine environment. In freshwater conditions, they are unable to regulate the water entering the body (through osmosis). Water enters their body due to the hypotonic environment outside. This results in the swelling up of the body, eventually leading to the death of the marine fish.

Question 3.
Define phenotypic adaptation. Give one example.
Solution:
Phenotypic adaptation involves non-genetic changes in individuals such as physiological modifications like acclimatization or behavioural changes.

Question 4.
Most living organisms cannot survive at temperatures above 45°C. How are some microbes able to live in habitats with temperatures exceeding 100°C?
Solution:
organisms survive at a temperature range of 0° to 40°C or less. However, there are some notable exceptions. Certain microorganisms live in hot springs and deep-sea hydrothermal vents where temperature far exceeds 100°C. They survive at the high temperature due to the occurrence of branched-chain lipids in their cell membrane that reduces the fluidity of cell membranes and the occurrence of the minimum amount of free water in their cells that provides resistance to high temperature

Question 5.
List the attributes that populations but not individuals possess.
Solution:

1. Natality
2. Mortality
3. Growth forms
4. Population density
5. Population dispersion
6. Population age distribution

Question 6.
If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?
Solution:
The intrinsic rate of increase(r), can be calculated by the following exponential growth equation:

Question 7.
Name important defence mechanisms in plants against herbivory.
Solution:

1. Modification of leaves into thorns.
2. Development of spiny margins on leaves.
3. Development of sharp silicated edges on leaves.

Question 8.
An orchid plant is growing on the branch of the mango tree. How do you describe this interaction between the orchid and the mango tree?
Solution:
An orchid growing as an epiphyte on a branch of mango tree is an example of commensalism. Commensalism is the relationship between individuals of two species of which one is benefited and the other is almost unaffected, i.e., neither benefited nor harmed. A commensal may get shelter (protection), or ride, or support instead of or in addition to food. Epiphytes are space parasites, they use trees only for attachment and manufacture their own food by photosynthesis. In Vanda, an epiphytic orchid, a special kind of aerial roots (hanging roots) hang freely in the air and absorb moisture with the help of their special absorptive tissue called velamen.

Question 9.
What is the ecological principle behind the biological control method of managing pest insects?
Solution:
Predation is the means of biological control to manage pest insects where predators prey upon pests and regulate their numbers in the habitat.

Question 10.
Distinguish between the following:

1. Hibernation and Aestivation
2. Ectotherms and Endotherms

Solution:

1. Differences between hibernation and aestivation are as follows :
   
2. Differences between ectotherms and endotherms are as follows:
   

Question 11.
Write a short note on :
(a) Adaptations of desert plants and animals
(b) Adaptations of plants to water scarcity
(c) Behavioral adaptations in animals
(d) Importance of light to plants
(e) Effect of temperature or water scarcity and the adaptations of animals.
Solution:
a. Desert plants are called xerophytes. They have adaptations for increased water absorption, reduction in transpiration and water storage. Many desert plants have a thick cuticle on their leaf surfaces and have their stomata arranged in deep pits to minimise water loss through transpiration. They also have a special photosynthetic pathway that enables their stomata to remain closed during day time. In desert plants like Opuntia, leaves are reduced to spines. Animals of dry areas may use metabolic water and reduce water loss bypassing nearly solid faeces and urine.

b. Xerophytes have special adaptations to withstand prolonged periods of drought. These are of four types – ephemerals, annuals, succulents and non-succulent perennials.

• Ephemerals (drought escapers): Plants which live for a brief period and complete their life cycle during the rains.
• Annuals (drought evaders): Plants which continue to live for a few
  months even after rains in hot dry conditions. They have modifications to reduce transpiration.
• Succulents (drought resistants): Plants have fleshy organs to store large amounts of water. They have a very thick cuticle, sunken stomata which open during night only.
• Non-succulent perennials: These are true xerophytes. They have an extensive root system to absorb the maximum amount of water. They possess waxy coatings on leaves, sunken stomata, reduced leaf blades etc. to reduce transpiration.

c. The animals with variable temperatures called poikilotherms are affected by temperature variations. They are also called ectotherms. They show different adaptations like hibernation, aestivation, periodic activity, winter eggs, and migration.

d. Sun is the ultimate source of energy for most of the organisms on this earth. Light is the visible range of the electromagnetic spectrum. Light (400 nm-700nm) is effective in photosynthesis and is called photo-synthetically active radiation or PAR. The intensity of light, duration of light, etc. are also influencing the growth of plants.

e. Animals live in arid regions show two kinds of adaptations

1. Reducing loss of water from their bodies.
2. Ability to tolerate arid conditions.

Question 12.
List the various abiotic environmental factors.
Solution:
Abiotic factors are non-living factors and conditions of the environment which influence the survival, function and behaviour of organisms. Various abiotic factors are :

(i) Temperature – Temperature is one of the most important environmental factors. The average temperature varies seasonally. It ranges from subzero level in polar areas and high altitudes to more than 50°C in tropical deserts in summer and exceeds 100°C in thermal springs and deep-sea hydrothermal vents.

(ii) Water – Next to temperature, water is the most important factor which influences the life of organisms. The productivity and distribution of land plants are dependent upon the availability of water. Animals are adapted according to water availability. E.g., aquatic animals are ammonotelic while xerophytic animals excrete dry feces and concentrated urine.

(iii) Light – Plants produce food through photosynthesis for which sunlight is essential to the source of energy. Light intensity, light duration and light quality influences the number of life processes in organisms, such as – photosynthesis, growth, transpiration, germination, pigmentation, movement and photoperiodism.

(iv) Humidity – Humidity refers to the moisture (water vapour) content of the air. It determines the formation of clouds, dew and fog. It affects the land organisms by regulating the loss of water as vapour from their bodies through evaporation, perspiration and transpiration.

(v) Precipitation – Precipitation means rainfall, snow, sleet or dew. Total annual rainfall, seasonal distribution humidity of the air and amount of water retained in the soil are the main criteria that limit the distribution of plants and animals on land.

(vi) Soil – The soil is one of the most important ecological factor called the edaphic factor. It comprises of different layers called horizons. The upper weathered humus containing part of soil sustains terrestrial plant life.

Question 13.
Give an example for:

1. An endothermic animal
2. An ectothermic animal
3. An organism of the benthic zone.

Solution:

1. Hedgehog
2. Frog
3. Sponges

Question 14.
Define population and community.
Solution:
Population: A population is a group of individuals of the same species, which can reproduce among themselves and occupy a particular area in a given time.

Community: It is an assemblage of several populations in a particular area and time and exhibits interaction and interdependence through trophic relationship.

Question 15.
Define the following terms and give one example for each.
(a) Commensalism
(b) Parasitism
(c) Camouflage
(d) Mutualism
(e) Interspecific competition
Solution:

a. Commensalism is an interspecific interaction between individuals of two species where one species is benefitted and the other is not affected.
e. g. Orchid and mango tree.

b. Parasitism is an interspecific interaction between individuals of two species where generally small species is benefitted and the large species are affected, e.g. Malarial parasite and human beings.

c. Camouflage: It is the ability of the animals to blend with the surroundings or background. In this way, animals remain unnoticed for protection or aggression. An example is a stick insect.

d. Mutualism is an interspecific interaction between individuals of two species where both the interacting species are benefitted in an obligatory way. e.g. Pollination in plants by animals.

e. Interspecific competition: It is an interaction between individuals of two species where both the interacting species are affected, e.g. Monarch butterfly and Queen monarch.

Question 16.
With the help of a suitable diagram describe the logistic population growth curve.
Solution:
Logistic population growth curve or S-shaped or sigmoid growth curve is shown by the populations of most organisms. It has the following phases: lag phase, log phase, exponential phase and stationary phase. In lag phase there is little or no increase in population. In log phase increase in population starts and occurs at a slow rate in the beginning. During exponential phase, increase in population becomes rapid and soon attains its full potential rate. This is due to the constant environment, availability of food and other requirements of life in plenty, absence of predation and interspecific competition and no serious intraspecific competition so that the curve rises steeply upward. The growth rate finally slows down as environmental resistance increases.

Finally, the population becomes stable during the stationary phase because now the number of new cells produced almost equals to the number of cells that die. Every population tends to reach a number at which it becomes stabilized with the resources of its environment. A stable population is said to be in equilibrium, or at saturation level. This limit in population is a constant K and is imposed by the carrying capacity of the environment. The sigmoid growth form is represented by the following equation :

r = intrinsic rate of natural increase
N = population density at time t; K = carrying capacity.

Question 17.
Select the statement which best explains parasitism.
(a) One organism is benefited.
(b) Both the organisms are benefited
(c) One organism is benefited, other is not affected
(d) One organism is benefited, other is affected.
Solution:
(d) One organism is benefited, other is affected,

Question 18.
List any three important characteristics of a population and explain.
Solution:
The three important characteristics of a population are:

1. Birth and death rate
2. Age structure
3. Sex ratio

(i) The birth rate (natality) of a population refers to the average number of young ones produced per unit time (usually per year). In the case of humans, it is commonly expressed as the number of births per 1,000 individuals in the population per year. The death rate (mortality) of a population is the average number of individuals that die per unit time (usually per year). In humans, it is commonly expressed as the number of deaths per 1,000 persons in a population per year.

(ii) The age structure of a population is the percentage of individuals of different ages such as young, adult and old. Age structure is shown bv organisms in which individuals of more than one generation coexist. The ratio of various age groups in a population determines the current reproductive status of the population. It also indicates what may be expected in the future. The population is divided into three age groups; pre-reproductive, reproductive and post-reproductive.

(iii) The sex ratio of a population refers to the number of females per thousand male individuals. There were 933 females per 1,000 males in our country in the 2001 census. The number of females in a population is very important as it is often directly related to the number of births. The number of males may be less significant because in many species a single male can mate with several females.

We hope the NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations help you. If you have any query regarding NCERT Solutions for Class 12 Biology Chapter 13 Organisms and Populations, drop a comment below and we will get back to you at the earliest.