NCERT Solutions for Class 10 English Literature Chapter 6 Virtually True

NCERT Solutions for Class 10 English Literature Chapter 6 Virtually True are part of NCERT Solutions for Class 10 English. Here we have given NCERT Solutions for Class 10 English Literature Chapter 6 Virtually True.

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Board CBSE
Textbook NCERT
Class Class 10
Subject English Literature
Chapter Chapter 6
Chapter Name Virtually True
Category NCERT Solutions

NCERT Solutions for Class 10 English Literature Chapter 6 Virtually True

TEXTUAL EXERCISES
(Pages 71, 78)

Question 1.
Before reading the story, attempt the following working in groups of four or five.

(a) Do you play computer games ? How many hours do you spend playing games on the computer as compared to outdoor games ?

(b) Make a list of your favourite games. Have a class discussion on the advantages and disadvantages of computer games.

(c) Look in your dictionaries/computer to find synonyms for the word ‘virtual’.

(d) Look at the K.W.L. chart given on next page. Based on the information you have gathered till now, complete the K and W columns. You may work with your partner. After reading the story complete the third column.

K-What
I know
W-What
I want to know
L-What
I learnt
Virtual Reality (a) (b) (c)
Virtual Environment (d) (e) (f)
3-D/three-dimensional (g) (h) (i)
Simulation games (j) (k) (l)
Computer simulations (m) (n) (o)
Interactive psycho-drive games (p) (q) (r)
Teleporting (s) (t) (u)

Answer:

(a) Yes, I do play computer games. On an average, I spend almost 2-3 hours playing computer games everyday. I play outdoor games for two hours daily, r

(b) My favourite games are : Pokemon Go, Road Rash, Need For Speed, Resident Evil,Grano Tursimo, Forza, Skyrim … class discussion to take place.

Advantages : pleasure-giving, entertaining, mental exercise, growth in analytical power, knowledge, general awareness, competitive spirit etc.

Disadvantages : promotion of violence, aggression, pain in backbone, in shoulders, time wastage, detachment from and neglect of family, social relationship hampered, sedentary life style, change in thinking, hardening of sensitivity, growth of abnormal behaviour.

(c) Few synonyms of VIRTUAL are : computer generated, simulated, cybernetic and realistic.

(d) (a) I know that it is not reality per se, but reality created with the help of computer software and some database. The user feels that he is inside it and is participating in it.

(b) I need to know the names of those particular softwares which help in creating such a situation. Also, if it could really be that easy then why isn’t every game prepared in such a format ?

(c) I learnt that everything is possible. Since cost is involved, so not everyone prefers to use these softwares etc.

(d) By Virtual Environment I know that it is nearer-to-truth environment. However, it is generated only by computers and softwares for the user.

(e) I want to know how much time does it take to create such an environment. Also what are the pros and cons for it ?

(f) There are no negatives of such an environment. However, the biggest plus point of this would be that the day such a procedure is initiated, the current era wouldn’t even exist. It’s all about advanced technology. The more you use it… the better it is for mankind.

(g) 3-D is about creating some sort of special effects to an image. It is to make the user feel as if the image or the subject is actually/physically present in front of , him.

(h) I want to know if there is any difference between a 3-D and a Virtual Environment.

(i) There is perhaps a difference between both of them. On the one hand 3-D is all about creating special effects to an image or a moving subject on screen. On the other hand, ‘Virtual Environment’ is all about creating a completely different environment. In it it’s not about throwing special effects over to a particular image. Instead, it’s about giving a continuous effect to the entire programme.

(j) Simulation Games. These games are made to give the player a realistic experience. For example : There is a normal racing game which is available with just a fixed image of a car. And the car keeps moving with the help of a remote controller. On the other hand, there’s another car racing game. In it the player gets to select his choice of car with all the settings of the car and tracks. There is sound as well along with a complete “Virtual Environment’. Here, in simulation games there is more of excitement than the normal games.

(k) I want to know nothing.

(l) I learnt nothing new.

(m) By Computer Simulations I know that it is all about using those extra additional gadgets to a computer. These are like, steering wheel, a shooting gun, some extra tools and weapons to play a particular game.

(n) I want to know if it is possible to get the same experience without using all such gadgets.

(o) I learnt that it’s not possible. To get a realistic feeling of the game and to actually develop more interest into it, one must have all those extra additions to his computer. Else, it would be like any other normal game.

(p) By Interactive Psycho-drive Games I learn that these are those car racing games wherein the player gets to handle a lot of crazy stuff. For example: If he wants to win the game or a particular level, he must perform some special stunts using some cheat codes. It is to gain some additional points than won by any other player.

(q) No questions.

(r) I learnt nothing.

(s) By Teleporting I know that it means switching over from one point to another within a split second of time. For example : While playing a game if the player gets a special bonus score or if he finishes up his target then he gets a special benefit. It is of skipping a level ahead or to get transformed into something else within a split second.

(t) I want to know if it has anything to do with any kind of voice service or interaction in between the game.

(u) No, it doesn’t have anything to do with any such service or interaction.

Exercises (Page 78)

  1. According to the newspaper, what had happened to Sebastian Shultz ?
  2. ‘Dad’s nutty about computers.’ What evidence is there to support this statement
  3. In what way did the second game seem very real ? (V. Imp.)
  4. The last game has tanks, jeeps, helicopters, guns and, headings. Would you put this and the other games under ?
  5. What was Michael’s theory about how Sebastian had entered the games ? (V. Imp.)

Answer:

  1. According to newspaper, Sebastian Shultz was badly injured in a motorway accident almost six weeks back. Now after six weeks he woke up from a coma that doctors had feared might have lasted forever.
  2. He’s nutty about computers as he has a Pentium 150 Mhz processor with 256 of RAM, a 1.2 GB hard disk drive and a 16 speed CD ROM with speakers, printer and also a scanner.
  3. It looked real as it was a continuation game after the first game ‘Wildwest.’ Also the name of the player was as well the same “Sebastian”.
  4. This game with machine guns, bombs, helicopter, sniper fire, tanks, can be put under the heading ‘Wargames’.
  5. Michael believed that Sebastian was also another intelligent personality like him in this world. He was so much crazy about playing games on computers. Both of them were playing a real game but virtually all together.

B. Reference to context

Read these lines from the story, then answer the questions.

‘That was my idea’ said Sebastian excitedly. ‘If only it would go a bit faster.’

  1. Where was Sebastian when he spoke these words ?
  2. What was his idea, and what was he referring to ?
  3. Was the idea a good one, and did it eventually succeed ? How ?

Answer:

  1. He was on the roof with the narrator and was waiting for the helicopter.
  2. His idea was to make it to the staircase and move upwards to get the helicopter after its landing on the roof and escape.
  3. The idea proved out to be negative as they somehow made it to the roof. But they couldn’t escape using the helicopter. They had to face almost 12 guards there along with their dogs. Sebastian got scared and he slipped down from the roof.

Question 2.
Answer the following questions briefly.
(a) Why did the news of the ‘miracle recovery’ shock Michael ?
Or
Why did the news about Sebastian Shultz shock Michael ? (CBSE 2015)

(b) Michael’s meeting with Sebastian Shultz had been a chance meeting. Where had it taken place and how ? (V. Imp.)
Or
How and where did Michael meet Sebastian Shultz ? (CBSE 2016)

(c) What kind of computers fascinated Michael and his dad ? Why ?

(d) Describe the first place where Michael was virtually transported.

(e) What help did Sebastian Shultz ask Michael for ? How did he convey this message ? (V. Imp.)
Or
How did Sebastian Shultz request Michael to help him ? (CBSE 2015)

(f) Why did Michael fail in rescuing Sebastian Shultz the first time ?
Or
Why couldn’t Michael help Sebastian the first time ? (CBSE 2015)

(g) The second attempt to rescue Sebastian Shultz too was disastrous. Give reasons. (V. Imp.)

(h) Narrate the accident that injured Sebastian Shultz.

(i) How had Sebastian Shultz entered the games ?

(j) How was Sebastian Shultz’s memory stored on Michael’s disk ? How did Michael discover that ? (V. Imp.)

Answer:

(a) Michael was shocked after reading the article as the name of the person involved in the article was Sebastian Shultz. He was someone whom Michael had recently ‘met’. Secondly, it was also because it was Sebastian with whom he had been playing all those games till then.

(b) Both of them played virtual video games on the computer screen. They thus met each other while playing virtual games on their respective computers.

(c) They were fascinated with the computers which could do almost everything from painting, playing music, creating displays etc. Both of them were very fond of playing games and doing all usual stuff on computers.

(d) The first place that Michael was virtually transported to was a complete dusty track through the centre of a town. He had a sheriffs badge pinned to his shirt.

(e) Michael received a printout from Sebastian saying that he was stuck. He wanted him to retrieve him by playing DRAGONQUEST. And he made a request for it through the computer. Precisely, it came through the computer printout.

(f) Michael was unable to save him the first time as both of them were trying to escape on a horse. However, ‘Sebastian’ was shot dead by the other ‘enemies’ running after them on their horses.

(g) The second attempt was negative as well, as he went down the dungeons with his sword drawn. But all of a sudden there appeared the dragon at the end of the corridor. He tried using his sword to kill the dragon. But it didn’t do any good and he was ‘killed.’

(h) It happened almost six weeks ago that Sebastian was badly injured in a motorway accident. His condition on arrival at the General Hospital was declared as critical, though stable. He was unable to regain his consciousness and his parents were informed that he was in a coma.

(i) Like Michael, even Sebastian was an intelligent person. He was very fond of computers and also playing games on them. This is how Sebastian entered the gaming world through playing psycho-drive video games.

(j) Michael used to play games on computers which had the capability to save the entire game in its memory while playing it. Based on this fact Sebastian’s memory had also got saved in his computer. Michael’s dad always used to tell him that a computer memory can never forget anything. Also no kind of data can get lost from it.

Michael discovered this after recollecting his father’s remarks. He also recollected Shultz’s mother stating that they stocked some games. But someone stole the lot. Shultz didn’t know what had happened to them. But Michael found them in the Computer Fair.

Question 3.
Sebastian Shultz had a close brush with death. After he recovers, he returns to school and narrates his experience to his classmates. As Sebastian Shultz, narrate your experience.
Answer:
Hi everyone ! I know you all must have been wondering about my whereabouts since so long. For all of you it’s been six weeks but for me it feels like as if everything just happened yesterday. I couldn’t even imagine that I would have to ever encounter such a situation. You may be amazed to know what had actually happened to me. Well! almost six weeks back I was on my way home riding a motorbike. All of a sudden there were these two speeding trucks which somehow hit each other and they collided. It took a split second for the two trucks to crash then and there themselves. I was riding my motorbike a little faster. On top of that the trucks were almost fifteen steps in front of me itself. In order to save myself from the crash I applied the brakes. The bike slipped and I fell over a huge pile of rocks on the pavement. I got hurt on my head and fell unconscious. It’s been almost six weeks that I’ve been hospitalized as I was in a coma. The doctors feared that it could last forever. However, by the grace of God I woke up from the coma and am fine now after having been saved by Michael, my friend. I am standing in front of you all narrating the most scariest experience of my life.

Thank you very much for a patient hearing.

Question 4.
Continue the story.

Will Michael and Sebastian Shultz meet in real life ? Will they be friends ? Will they try to re-enter the virtual world together ? You may use the following hints :

  • How the accident occurred
  • Transfer of memory
  • Meeting with Michael
  • Appeals for help
  • Rescue and recovery

Answer:

No, it’s impossible that both of them would ever meet each other. After all, it’s just a game which has got no connection with the real world.

Perhaps, they can even be good friends to each other but only virtually and electronically. In reality there is no such real person by the name of Sebastian who is playing this game with Michael. They can never enter the virtual world together as Sebastian is simply the name of a player in the game and Michael is a real human being.

They would never be able to meet each other except in this virtual world created by these gizmos only.

If we go as per the points given above we may relate the story like this : The accident occurred which led Sebastian to slip in a coma. He had played the video game through 3-D and had got stored in the memory of his computer. It had been stolen and ended at the Computer Fair. From there Michael with his father had got the interactive psycho¬drive video game which had earlier been with Sebastian.

Appeals were sent for saving Sebastian from the coma. Michael got this information from the newspaper article. He played the video game himself and got hooked with Sebastian Shultz, the character in the video game. Thus he was rescued and saved.

Question 5.
Put the following sentences in a sequential order to complete the story.

(a) Sebastian Shultz was badly injured in a motorway accident and went into a coma.
(b) Sebastian’s memory was saved in the computer when he banged his head on it during the accident.
(c) When Michael played the game, he entered Sebastian’s memory.
(d) Michael bought the latest psycho-drive games from the Computer Fair.
(e) Sebastian Shultz was the second sheriff in the ‘Dragonquest’.
(f) Michael pulled Sebastian into the helicopter and the screen flashed a score of 40,000,000.
(g) Sebastian requested Michael to try ‘Jailbreak’.
(h) Sebastian failed to save the boy who fell through the air.
(i) Sebastian thought the helicopter was the right idea and they should go into the ‘Warzone’.
(j) The games were stolen from Shultz’s house.
(k) Sebastian thanks Michael for saving his life and asks him to keep the games.

Answer:

The sequential order to complete the story is : a, b, j, d, c, e, h, i, g, f, k

WRITING TASK
Question 6.
Do you think it is a true story ? Could it happen to you one day ? Here are some opinions about computer games in general.

NCERT Solutions for Class 10 English Literature Chapter 6 Virtually True 1

Answer:

Yes, I do believe that it’s a true story, as such a thing can happen with almost anyone in this world. A similar instance could happen with me as well if I would ever think of playing games on computers like Michael did. But it all depends if I play interactive psycho-drive video games and have all the accessories needed.

Question 7.
Do you think these opinions are biased ? Write an article entitled ‘Virtual games are a reality’.
Answer:
Yes, I think these opinions are biased.

Virtual Games are a Reality

It’s a fact that virtual games are a reality. They are computer simulated environments that can simulate physical presence in places in the real world as well as in the imaginary world. It’s just a matter of a few gadgets that are required to create such an environment while playing games on computers. Hi surround speakers, digital crystal clear sound output, a 3-D screen and few other hand-led devices are all it takes to view a picture image on a computer screen or an LCD/LED. Virtual games provide players a completely different experience compared to the kind of experience they get while playing normal games. It is because one gets almost physically involved in them through 3-D and becomes an integral part of what is seen on the screen. They are far much ahead of those normal animated games which are simple and have only a limited excitement. However, one needs both time and a lot of money to indulge in such activities because these modern gadgets are very costly.

Question 8.
In groups of four, design a new computer game.
Decide on the

  • Setting
  • Plot
  • Characters
  • Objectives

Answer:
For self-attempt at class level.

LISTENING TASK
Question 9.
Listen carefully to a text on ‘Tour of Body’ and answer the questions that are given below.

On the basis of your listening to the passage complete the following statements by choosing the answers from the given options :

1. The Cave-Automatic Virtual Environment is

(a) a modern surgical procedure
(b) a three-dimensional virtual reality room
(c) an accurate projection of the eye and the brain
(d) a technique for developing anatomical pictures

2. Projected image on the four walls of a room enables researchers to

(a) carry out micro surgery
(b) understand the functioning of the brain
(c) virtually get inside the molecular structure of cells and parts of human body
(d) reconstruct damaged parts of human body

3. The ‘CAVE’ is a boon to surgeons because

(a) they can treat diseases located in unreachable parts of the human body
(b) it has made X-ray and MRI unnecessary
(c) it helps them avoid surgical procedures in most cases
(d) it enables surgeons to use very small surgical instruments

4. For the CAVE to develop a virtual environment it is essential

(a) to apply mathematical formulae
(b) to project three-dimensional images on the walls
(c) to obtain two-dimensional MRI data first
(d) to understand the nature of the diseased cells and parts

5. Once inside a three-dimensional representation of an anatomical structure, surgeons can

(a) ‘move’ through and ‘peel away’ its layers
(b) shrink themselves and travel inside the body parts
(c) convert the data into a flat picture for detailed study
(d) locate the diseased parts of the body quickly

6. Dr. Szilard Kiss used CAVE to

(a) travel inside the eye of his patient
(b) identify the scar tissue growing over the retina
(c) go inside the layers of the retina
(d) isolate the ridge of the scar tissue

Answer:

  1. → (h) a three-dimensional virtual reality room
  2. → (c) virtually get inside the molecular structure of cells and parts of human body
  3. → (a) they can treat diseases located in unreachable parts of the human body
  4. → (a) to apply mathematical formulae
  5. → (a) ‘move’ through and ‘peel away’ its layers
  6. → (c) go inside the layers of the retina.

We hope the NCERT Solutions for Class 10 English Literature Chapter 6 Virtually True help you. If you have any query regarding NCERT Solutions for Class 10 English Literature Chapter 6 Virtually True, drop a comment below and we will get back to you at the earliest.

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NCERT Solutions for Class 12 Physics Chapter 6

NCERT Solutions for Class 12 Physics Chapter 6

NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction is important for the students of CBSE, UP, MP, Gujarat Boards. NCERT physics class 12 chapter 6 Solutions PDF is presented here to aid students grasp the Electromagnetic Induction concepts in an interesting way. All Chapter 6 Electromagnetic Induction Exercise Questions and answers help 12th standard candidates to revise the entire syllabus & secure high.

You can rely on these NCERT Solutions to clear all your complex questions doubts and understand them clearly. All these solutions are answered by subject experts after doing thorough research. So, you can Download the Class 12 Physics NCERT Solutions Ch 6 Electromagnetic Induction PDF through quick links prevailing here without any hassle.

Class 12 Physics NCERT Solutions Chapter 6 Electromagnetic Induction

Firstly, you have to know about Electromagnetic Induction that means the production of an electromotive force across an electrical conductor. In Electromagnetic Induction Chapter, you will come to learn about some important & useful topics like electromotive force motions, magnetic flux, eddy currents, quantitative study on consideration of energy, and inductance.

Electromagnetic Induction also takes place in a changing magnetic field. Moreover, Fleming’s right-hand rule and Lenz’s law are also covered in this 12th class Physics chapter Electromagnetic Induction. So, practice more Numerical problems of Electromagnetic Induction chapter from NCERT Books & NCERT Solutions & score well in the examinations.

Class 12
Subject Physics
Book Physics
Chapter Number 6
Chapter Name Electromagnetic Induction

NCERT Solved Questions & Exercises for Class 12 Physics Chapter 6

Class 12 Physics NCERT Solved Questions & Exercises of Ch 6 will increase your subject knowledge by offering the strong basics on Electromagnetic Induction. All the answers provided for each question will raise your confidence to solve any questions in the actual examination. 12th Class NCERT Physics Solutions of Chapter 6 is a valuable resource for students during their homework & assignments too.

Question 1.
Predict the direction of induced current in the situations described by the following Fig. 2(a) to (f).
NCERT Solutions for Class 12 physics Chapter 6
NCERT Solutions for Class 12 physics Chapter 6.1
Answer:
(a) According to Lenz’s law south pole forms at q and north pole a to p. Therefore, induced current is from p to q.
(b) South pole develops at end q and at end x of the two coils as per Lenz’s law. Therefore, induced current is from q to p in coil PQ and x to y in coil
(c) The induced current should flow clockwise in the right loop (when seen from left) to oppose the current in the left loop. Hence, induced current flows along yzx
(d) When rheostat is adjusted to increase the current, magnetic flux through the neighbouring coil increases. As per Lenz’s law, induced current in the neighbouring coil should produce magnetic flux in opposite direction to oppose the original magnetic flux. Hence, induced current flows along
(e) When circuit breaks, magnetic flux decreases. The induced current should flow along xry to increase the magnetic flux.
(f) The magnetic flux threading the coil in the perpendicular direction is zero. Any change in current will not change this magnetic flux. Hence, no induced current is set up in the coil.

Question 2.
Use Lenz’s law to determine the direction of induced current in the situations described by Figure :
(a)
 A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.
NCERT Solutions for Class 12 physics Chapter 6.2
NCERT Solutions for Class 12 physics Chapter 6.3
Answer:
(a) The flux linked with the loop increases in this case so flux produced by the induced current should decrease it i.e. the direction of induced current should produce magnetic field in a direction outwards the plane of the paper. The direction of current as per right hand thumb rule comes out to be along adcba. (i.e. anti­clockwise)

(b) In this case, the flux linked with the loop is decreasing so the direction of induced current is anticlockwise so as to oppose the decrease in flux. Thus, direction of induced current is along a’d’c’b’.

Question 3.
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1s, what is the induced e.m.f. in the loop while the current is
changing ?
Answer:
byjus class 12 physics Chapter 6.4

Question 4.

A rectangular wire loop of sides 8 cm and 2 cm with a small cut ¡s moving out of a region of uniform magnetic field of magnitude O.3 T directed normal to the mop. What ¡s the e.m.f. developed across the cut if the velocity of the loop is 1 cm s-2 in a direction normal to the (a) longer side,
(b) shorter side of the loop ? For how long does the induced voltage last in each case ?
Answer:
e = B
NCERT Solutions for Class 12 physics Chapter 6.5
Question 5.
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s_1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic Field of 0.5 T parallel to the axis exists everywhere. Calculate the e.m.f. developed between the center and the ring.
Answer:
NCERT Solutions for Class 12 physics Chapter 6.6
Question 6.
A circular coil of radius 8.0 cm and 20 turns rotates about its vertical diameter with an angular speed of 500 rad s_1 a uniform horizontal magentic field of magnitude 3.0 x 10-2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from ?
Answer:
byjus class 12 physics Chapter 6.7
(v) Source of power loss is the external rotor which provides the necessary torque to rotate the coil.

Question 7.
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 ms-1, at right angles to the horizontal component of the earth’s magnetic field,
0.30 x 10-4 Wb m-2.
(a) What is the instantaneous value of the e.m.f. induced in the wire ?
(b) What is the direction of the e.m.f. ?
(c) Which end of the wire is at the higher electrical potential ?
Answer:
NCERT Solutions for Class 12 physics Chapter 6.8
(b) Using Fleming’s Right hand rule, the direction of induced e.m.f. is from West to East.
(c) Since the rod will act as a source, the Western end will be at higher electrical potential.

Question 8.
Current in a circuit falls from 5.0 A to 0.0 As. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit.
Answer:
NCERT Solutions for Class 12 physics Chapter 6.9

byjus class 12 physics Chapter 6.10
Question 9.

A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil ?
Answer:
Using dΦ = Mdl, we get
dΦ= 1.5 X (20 – 0) = 30 Wb

Question 10.
A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s
magnetic field at the location has a magnitude of 5 x 10-4 T and the dip angle is 30°? Answer:
NCERT Solutions for Class 12 physics Chapter 6.12

Question 11.
Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet
that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s-1. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat ? What is the source of this power ?
Answer:
NCERT Solutions for Class 12 physics Chapter 6.13
Source of this power is the external agency which brings change in magnetic field.

Question 12.
A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s_1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10-3 T cm-1 along the negative x- direction (that is it increases by
10-3 T cm-1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10-3 T s_1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.
Answer:
byjus class 12 physics Chapter 6.14
NCERT Solutions for Class 12 physics Chapter 6.15
Direction of induced current is such that it increase the magnetic flux linking with the loop in positive 2-direction.

Question 13.
It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction. The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of magnet.
Answer:
NCERT Solutions for Class 12 physics Chapter 6.16
Question 14.
Figure shows a metal rod PQ resting on the rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.
NCERT Solutions for Class 12 physics Chapter 6.17
(a) Suppose K is open and the rod is moved with a speed of 12 cm s-1 in the direction shown. Give the polarity and magnitude of the induced emf.
(b) Is there an excess charge built up at the ends of the rods when K is open ? What if K is closed ?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed ?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed (= 12 cm s_1) when K is closed ? How much power is required when K is open ?
(f) How much power is dissipated as heat in the closed circuit ? What is the source of this power ?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular ?
Answer:
(a) Using e = Bυl, we get
e = 0.5 x 12 x 10.2 x 15 x 102
= 9 x 10-3 V
The electrons in the rod will experience force along PQ, so end P becomes positive and Q becomes negative.
(b) On closing the key K, the number of electrons become more at end Q. Therefore, excess charge is maintained by the continuous current.
(c) Magnetic force gets cancelled by electric force due to excess charges of opposite sign at the ends of the rod.
(d) Retarding force, F = BIl =B eR l
byjus class 12 physics Chapter 6.18

Question 15.
An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10-3 s. How much is the average back emf induced across the ends of the open switch in the circuit ? Ignore the variation in magnetic field near the ends of the solenoid.
Answer:
NCERT Solutions for Class 12 physics Chapter 6.28

Question 16.
(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Figure
NCERT Solutions for Class 12 physics Chapter 6.29
(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, υ = 10 m/s. Calculate the induced e.m.f. in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.
Answer:
(a) Consider a small portion of the coil of thickness dt at a distance t from the current carrying wire.Then the magnetic field strength experienced by this portion
NCERT Solutions for Class 12 physics Chapter 6.30

NCERT Solutions for Class 12 physics Chapter 6.31NCERT Solutions for Class 12 physics Chapter 6.32

Question 17.

A line charge X per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Figure.) A uniform magnetic field extends over a circular region within the rim. It is given by
byjus class 12 physics Chapter 6.24
What is the angular velocity of the wheel after the field is suddenly switched off ?
NCERT Solutions for Class 12 physics Chapter 6.25
Answer:
Change in magnetic field is given by,
NCERT Solutions for Class 12 physics Chapter 6.26
NCERT Solutions for Class 12 physics Chapter 6.27

So, Download 12th Class Physics NCERT Exercise Questions with Solutions of Chapter 6 Electromagnetic Induction PDF and kickstart your exam preparation. For your convenience, we have tried to list out the NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction in Hindi & English Mediums. Brainstorm Important Questions, Formulae, Notes, Exemplar Problems concerning the Class 12 Physics Electromagnetic Induction NCERT Solutions PDF.

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NCERT Solutions for Class 12 Physics Chapter 5 Magnetism And Matter

NCERT Solutions for Class 12 Physics Chapter 5

NCERT Solutions for Class 12 Physics Chapter 5 Magnetism And Matter provides all needed topics and subtopics questions and answers in a detailed way for a better understanding of the concept. Thus, aims students score high in their board exams and competitive exams. Candidates who are pursuing class 12 can practice these NCERT Physics Solutions of Chapter 4 Magnetism and matter from this page for free.

Have a thorough subject knowledge from the solutions provided here and solve all types of problems and theoretical questions in the final exams with ease. For more practice, students can make use of some extra preparation resources like NCERT intext, exercises at the end of chapter 5 questions. You can download this helpful & needed NCERT Class 12 Physics Solutions for Ch 5 Magnetism and Matter PDF from here & access it in online & offline modes.

Class 12 Physics NCERT Solutions Chapter 5 Magnetism And Matter

In Chapter 5, Students of Plus Two will deal with the concept of Physics called Magnetism And Matter. The previous concept magnetism expands to the fullest in this 5th chapter Magnetism And Matter. The topics covered in this NCERT Physics chapter Magnetism And Matter are Gauss’s law & magnetism, a bar magnet & earth’s Magnetism. A total of 7 topics are included that discuss the several materials’ magnetic properties & magnetic field lines properties.

Check out the CBSE NCERT Solutions of Class 12 Physics Ch 5 Magnetism and Matter Exercises in PDF format, and score high in the board exam and competitive examination like JEE, NEET, etc.

Class 12
Subject Physics
Book Physics
Chapter Number 5
Chapter Name Magnetism And Matter

NCERT Solutions for Class 12th Physics Ch 5 Magnetism and Matter – Solved Exercises

Chapter 5 Magnetism And Matter NCERT Class 12 Physics Questions and Solutions are explained by subject experts and it is arranged based on CBSE Syllabus and NCERT guidelines. However, these NCERT Exercise Questions and answers of Class 12 Physics Chapter 5 Magnetism And Matter are most helpful to the students for their Homeworks and Assignments also.

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)

  1. Angle of dip or magnetic inclination, δ Galvanometer resistance, G = 12 Q The galvanometer can be converted into the voltmeter of range 0 to V (here V = 18 V) by connecting a high series resistance R given by
  2. Magnetic declination, θ
  3. Horizontal component of magnetic field of earth BH.

(b) It is about 70° in Britain because it is closer to the magnetic north pole of earth.
(c) The magnetic field lines due to the magnetism of earth would seem to come out of the ground at Melbourne in Australia because this region is in the Southern hemisphere of earth where north pole of earth’s magnet lies.
(d) Compass can point in any direction because it is free to move in horizontal plane but at magnetic poles the magnetic field of earth is exactly vertical so there will be no horizontal component of the magnetic field to effect the compass.
(e) Using the formula for magnetic field on the equatorial line of a magnetic dipole i.e.
NCERT Solutions for Class 12 physics Chapter 5
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) Although appreciable change can take a few hundred years yet change with time is very much there.
(b) Molten iron has a temperature more than Curie temperature so it is not ferromagnetic in nature.
(c) Probably radioactivity (i.e. emission of a, P and y particles from the nuclei) present in the interior of earth is responsible for it.
(f) During solidification of some rocks, the magnetic field of earth gets weakly recorded in these rocks. Geomagnetic history of these rocks can be traced by making analysis of such rocks.
(e) Ionosphere of earth consists of charged ions. Motion of these ions causes a magnetic field which affects the magnetic field of earth at large distances from earth. The magnetic field due to ions depends upon extra terrestrial disturbances like solar wind.
(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.
NCERT Solutions for Class 12 physics Chapter 5.1
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
vedantu class 12 physics Chapter 5.2

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:
NCERT Solutions for Class 12 physics Chapter 5.3

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:
Using M = NIA, we get
M = 800 x 3 x 2.5 x 10-4 = 0.60 JT-1.
The sense is determined by the direction of the current.

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 12
= 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:
vedantu class 12 physics Chapter 5.4

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:
(a) Using M = NIA, we get
M = 2000 x 4 x 1.6 x 10-4
= 1.28 Am2
Direction of M is as per sense of current using the right handed rule.
(b) Torque, x = MB sin 0
= 1.28 x 7-5 x 10-2 x sin 30°
= 4.8 x 10-2 N m
Force is zero because of uniform 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:
NCERT Solutions for Class 12 physics Chapter 5.5
NCERT Solutions for Class 12 physics Chapter 5.6

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:
NCERT Solutions for Class 12 physics Chapter 5.7
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 BH = B cos δ, we get
NCERT Solutions for Class 12 physics Chapter 5.8

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
vedantu class 12 physics Chapter 5.9

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
NCERT Solutions for Class 12 physics Chapter 5.10

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.
NCERT Solutions for Class 12 physics Chapter 5.11

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,
NCERT Solutions for Class 12 physics Chapter 5.26
vedantu class 12 physics Chapter 5.13
NCERT Solutions for Class 12 physics Chapter 5.14
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) At lower temperatures, the tendency of dealignment of magnetic dipoles with magnetising field is reduced because of decrease in their random thermal motion.
(b) The induced dipole moment in a diamagnetic sample is opposite to the magnetising field irrespective of the internal thermal motion of the atoms. Thus, diamagnetism is independent of the changes in temperature.
(c) It will be slightly less because bismuth is a diamagnetic substance.
(d) Permeability of a ferromagnetic material depends on applied magnetic field. Permeabliltiy is more for lower magnetic field.
(e) When two media are considered and one of these has μ > > 1, then magnetic field lines meet normally to this medium. Proof of this fact is available from the boundary conditions of magnetic fields (strength of magnetic field B and magnetising intensity H) at the interface of two media.
(f) The maximum possible magnetisation of a paramagnetic sample would be of the same order of magnitude as the magnetisation of a ferromagnetic sample but the paramagnetic sample requires very high magnetising field which may not be practically possible.

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) It can be observed from the given curve that magnetisation persists even when the external field is removed. This reflects the irreversibility of a ferromagnet.
(b) Heat dissipated per second is directly proportional to the area of the hysteresis loop so carbon steel piece will dissipate greater heat energy.
(c) Magnetisation of a ferromagnet depends upon the applied field as number of cycles of magnetisation given to it. Thus the value of magnetisation can indicate the number of cycles of magnetisation a sample has undergone or in other words the memory with respect to the cycles stands stored in the sample.
(d) Ceramics are used for this purpose. Ferrites which are specially treated barium iron oxides are commonly used.
(e) The space to be shielded is enclosed by soft iron rings. The magnetic field lines remain confined in the ferromagnetic material of the rings and the enclosed space will be free from magnetic effect.
Note. Magnetic shielding is not as effective as electrostatic shielding is.

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.
NCERT Solutions for Class 12 physics Chapter 5.15

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:
NCERT Solutions for Class 12 physics Chapter 5.16
NCERT Solutions for Class 12 physics Chapter 5.17

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:
NCERT Solutions for Class 12 physics Chapter 5.33

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°.
vedantu class 12 physics Chapter 5.19
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, MBsin = MB2 sin θ2
NCERT Solutions for Class 12 physics Chapter 5.20

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 (me = 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:
NCERT Solutions for Class 12 physics Chapter 5.27
NCERT Solutions for Class 12 physics Chapter 5.28

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-1
NCERT Solutions for Class 12 physics Chapter 5.30
Question 24.
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 magnetising current of 1.2 A?
Answer:
vedantu class 12 physics Chapter 5.31

Question 25.
The magnetic moment vectors μs and μl; 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 :
μs= -(e/m)S,
μl= -(e/2m)l
Which of these relations is in accordance with the result expected classically ? Outline the derivation of the classical result.
Answer:
NCERT Solutions for Class 12 physics Chapter 5.32

The NCERT Solutions For Class 12 Physics Chapter 5 PDF is covered with all the answers to the complicated & important questions. It helps students to ace up their exam preparation. Moreover, these NCERT solutions are simple & easy to understand as they are designed with detailed diagrams, formulas, equations, and neat explanations. Hence, Download CBSE NCERT Solutions For Class 12 Physics Chapter 5 Magnetism And Matter PDF & study offline.

Summary

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NCERT Solutions for Class 12 Physics Chapter 15 Communication Systems

NCERT Solutions for Class 12 Physics Chapter 15

NCERT Solutions for Class 12 Physics Chapter 15 Communication Systems are available here in PDF Format. So, Candidates of state and central boards like UP, MP, Gujarat, Uttarakhand, CBSE should follow them to ace up your preparation. These solutions are specially designed in both Hindi and English mediums to make it easy while learnings.

You all can avail NCERT Class 12 Physics Ch 15 Solutions in PDF Download links for easy download & access offline. The smartest way to get familiar with the concept of Communication systems by preparing with the NCERT Solutions material. These prevailing NCERT Solutions covers the practice problems so that you can understand all the concepts easily.

Class 12 Physics NCERT Solutions Chapter 15 Communication Systems

In this chapter, the 12th class students will come to learn about the communication systems concept. The introduction of the communication system is nothing but sending and receiving the data and information from one place to another. This last chapter in NCERT Solutions Class 12 Physics textbook talks all about covered 10 topics of communication systems.

Besides, this chapter goes in detail about the elements called the channel, the transmitter, and the receiver. Also, they will study some other important concepts like the bandwidth of various transmission mediums, the bandwidth of signals, amplitude modulation, the necessity of modulation, propagating electromagnetic waves, etc.

Class 12
Subject Physics
Book Physics
Chapter Number 15
Chapter Name Communication Systems

NCERT Exercise Questions with Solutions of 12th Physics Ch 15 PDF

These NCERT Solutions for Class 12 Physics Chapter 15 aims to introduce the learners to the communication concept. Besides, an explanation of each and every question in NCERT Solutions PDF includes charts, diagrams, examples, and illustrations to understand the concept in an interesting and easy to learn.

Question 1.
At which of the following frequency/frequencies the communication will not be reliable for a receiver situated beyond horizon:
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Answer:
(b) is correct. Here (c) and (d) frequencies have high penetration power so the earth will absorb them. Radiation (a) of 10 kHz will suffer from the problem of size of antenna.

Question 2.
Frequencies in the UHF range normally propagate by means of
(a) ground waves
(b) sky waves
(c) surface waves
(d) space waves.
Answer:
(d) space waves.

Question 3.
Digital signals (i) do not provide a continuous set of values, (ii) represent values as discrete steps, (Hi) can utilize only binary system, and (iv) can utilize decimal as well as binary system. Which of the following options is true :
(a) Only (i) and (ii).
(b) Only (ii) and (iii).
(c) Only (i), (ii) and (iii), but not (iv).
(d) AH the above (i) to (iv).
Answer:
(c) is correct because decimal system is concerned with continuous values (i) to (iii).

Question 4.
Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line- of-sight communication ? A TV transmitting antenna is 81 m tall. How much service area can it cover if the receiving antenna is at the ground level?
Answer:
For line-of-sight communication it is necessary that the transmitting antenna and receiving antenna should be eye to eye but it is not necessary that they should be at the same height.
NCERT Solutions for Class 12 physics Chapter 15
Question 5.
A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75% ?
Answer:
NCERT Solutions for Class 12 physics Chapter 15 Communication System.1
Question 6.
A modulation signal is a square wave as shown in figure. The carrier wave is given by
C(t) = 2 sin(8πt) V
NCERT Solutions for Class 12 physics Chapter 15 communication System.2
(a) Sketch the amplitude modulated waveform.
(b) What is the modulation index ?
Answer:
tiwari academy class 12 physics Chapter 15 Communication System.3
NCERT Solutions for Class 12 physics Chapter 15 Communication System.4

Accordingly, the amplitude modulated waveform is shown ahead:

Question 7.
For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2 V. Determine the modulation index μ.What would be the value of μ if the minimum amplitude is zero V ?
Answer:
NCERT Solutions for Class 12 physics Chapter 15.5
Question 8.
Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.
Answer:
Let there be two signals represented by
Ac cos ωct and A0 cos(ωc + ωm)t where Ac is the
amplitude, ωc is the angular frequency of a carrier wave at the receiving end and A0 is the amplitude, (ωc+ ωm) is the angular velocity of the modulated wave.
Multiplying these signals, we get
tiwari academy class 12 physics Chapter 15.6

Our experts designed each topic in the NCERT Solutions for Class 12 Physics Chapter 15 Communication Systems in a more simplistic manner. Additionally, we have also combined some extra traits that add to your learning like valuable notes, question papers, live classes, proficient help, etc for better preparation. So, kickstart your learnings & score high marks.

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NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

NCERT Solutions for Class 12 Physics Chapter 14

Free PDF Downlaod links of NCERT Solutions for Class 12 Physics Chapter 14 – Semiconductor Electronics Materials Devices And Simple Circuits are available in this article. Refer to this page and start downloading 12th class physics NCERT Ch 14 Solutions for free of charge in PDF Format.

These solutions are designed by our experienced subject teachers as per the latest NCERT Textbook syllabus based om the CBSE Curriculum. Ace up your preparation with these NCERT Exercise questions and solutions PDF for class 12 physics and improve your subject knowledge on Chapter Semiconductor Electronics Materials Devices And Simple Circuits.

Class 12 Physics NCERT Solutions Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuits

Students of 12th class can learn all the topics and subtopics of Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuits from the NCERT Solutions. Semiconductor Electronics Materials Devices And Simple Circuits Concept speak about the types of semiconductors and insulators.

Moreover, it also includes various other important topics like conductors, classification of metals, and semiconductors. Discussion of the difference between the intrinsic semiconductor and extrinsic semiconductor, semiconductor diode, application of junction diode, etc will be seen in this chapter 14 NCERT Questions & Solutions.

Refer to the 12th Class Physics NCERT Solutions of Ch 14 Solved Exercise and Miscellaneous Exercise Questions at your preparation & revision time to clear all your doubts on Chapter Semiconductor Electronics Materials Devices And Simple Circuits.

Class 12
Subject Physics
Book Physics
Chapter Number 14
Chapter Name Semiconductor Electronics Materials Devices And Simple Circuits

NCERT 12th Physics Ch 14 Solved & Miscellaneous Exercises Questions with Solutions PDF

Students who are preparing for their board exams or any other competitive exams can refer to these Class 12th Physics NCERT Solutions of Ch 14 Semiconductor Electronics Materials Devices And Simple Circuits. These solutions are taken from the NCERT Textbooks and our subject experts prepared these NCERT Solutions in a comprehensive way.

Question 1.
In an n-type silicon, which of the following statement is true :
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the doplants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Answer:
(c) ‘Holes are minority carriers and pentavalent atoms are the dopants in n type semiconductor.’

Question 2.
Which of the statements given in Exercise 1 is true for p-type semiconductors ?
Answer:
(d) Holes are majority carriers and trivalent atoms are the dopants in p-type semiconductors.

Question 3.
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)c, (Eg)si and (Eg)Ge-
Which of the following statements is true ?
(a) (Eg)Si < (Eg)Ge < (Eg)c
(b)(E)c<(Eg)Ge>(Eg)si
(c) (Eg)c > (Eg)si > (Eg)Ge
(d) (Eg)c = (Eg)si = (Eg)Ge
Answer:
(C) (Eg)c > (Eg)Si > (Eg)Ge. Energy band gap is maximum in carbon and least in germanium among the given elements.

Question 4.
In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.
Answer:
(c) hole concentration in p-region is more as compared to n-region because hole diffusion takes place from higher concentration to lower concentration.

Question 5.
When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier
(b) reduces the majority carrier current to zero
(c) lowers the potential barrier
(d) none of the above.
Answer:
(c) lowers the potential barrier by cancelling the depletion layer.

Question 6.
For transistor action, which of the following statements are correct :
(a) Base, emitter and collector regions should have similar size and doping concentrations.
(b) The base region must be very thin and lightly doped.
(c) The emitter junction is forward biased and collector junction is reverse biased.
(d) Both the emitter junction as well as the collector junction are forward biased.
Answer:
(b) and (c) : The base region must be very thin, lightly doped and the emitter junction is forward biased whereas collector junction is reverse biased to avoid unnecessary diffusion of charge carrier in the base and also for proper amplification.

Question 7.
For a transistor amplifier, the voltage gain :
(a) remains constant for all frequencies.
(b) is high at high and low frequencies and constant in the middle frequency range.
(c) is low at high and low frequencies and constant at mid frequencies.
(d) None of the above.
Answer:
(c) is low at high and low frequencies and constant at mid frequencies as per frequency response of a transistor.

Question 8.
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency.
Answer:
A half wave rectifier rectifies only one half cycle of input A.C.
.’. frequency of the output A.C.
= frequency of input A.C. = 50 Hz A full wave rectifier rectifies both halve cycles
of the A.C. input
.’. frequency of output A.C. = 2 x frequency of input A.C. = 2 x 50 = 100 Hz

Question 9.
For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kQ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.
Answer:
NCERT Solutions for Class 12 physics Chapter 14 Electronic Devices.1

Question 10.

Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is
0.01 volt, calculate the output ac signal.
Answer:
NCERT Solutions for Class 12 physics Chapter 14 Electronic Devices.2

Question 11.

A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm ?
Answer:
NCERT Solutions for Class 12 physics Chapter 14 Electronic Devices.3
Since the energy of the light photon is less than the band gap energy of p-n diode, it can not be detected.

Question 12.
The number of silicon atoms per m3 is 5 x 1028. This is doped simultaneously with 5 x 1022 atoms per mof Arsenic and 5 x 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that ni = 1.5 x 1016 m-3. Is the material n-type or p- type ?
Answer:
byjus class 12 physics Chapter 14 Electronic Devices.4

Question 13.

In an intrinsic semiconductor the energy gap Eg is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and that at 300 K ? Assume that the temperature dependence of intrinsic carrier concentration ni is given by
NCERT Solutions for Class 12 physics Chapter 14 Electronic Devices.5
Answer:
NCERT Solutions for Class 12 physics Chapter 14 Electronic Devices.6

Thus, conductivity of a semiconductor increases with rise in temperature.

Question 14.
In a P-n junction diode, the current I can be expressed as
NCERT Solutions for Class 12 physics Chapter 14 Electronic Devices.7
where I0 is reverse saturation current. V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, Kis the Boltzmann constant (8.6 x 10-5 eV/K) and T is the absolute temperature. If for a given diode I0 = 5 x 10-12 A and T = 300 K, then
(a) What will be the forward current at forward voltage of 0.6 V ?
(b) What will be the increase in current if voltage across diode is increased to 0.7 V ?
(c) What is the dynamic resistance ?
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V ?
Answer:
Statement of the given question is incorrect. The relation should be

Question 15.
You are given the two circuits as shown in Figure. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.
byjus class 12 physics Chapter 14 Electronic Devices.11
Answer:
NCERT Solutions for Class 12 physics Chapter 14 Electronic Devices.12

Question 16.

Write the truth table for a NAND gate connected as given in Fig. Hence identify the exact logic operation carried out by these circuits.
NCERT Solutions for Class 12 physics Chapter 14 Electronic Devices.13
Answer:
The NAND gate shown in the truth table has only one input. Therefore, the truth table is

A A y = A.A¯
0 0 1
1 1 0

Since Y = A¯ in this case, the circuit is actually a NOT gate with the truth table

A Y
0 1
1 0

Question 17.
You are given two circuits as shown in Fig., which consist of NAND gates. Identify the logic operation carried out by the two circuits.
NCERT Solutions for Class 12 physics Chapter 14 Electronic Devices.14
Answer:
NCERT Solutions for Class 12 physics Chapter 14 Electronic Devices.15
Question 18.
Write the truth table for circuit given in the Fig. below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.
byjus class 12 physics Chapter 14 Electronic Devices.16
Answer:
Let y1 be the output which appears at the first operation of NOR gate.
NCERT Solutions for Class 12 physics Chapter 14 Electronic Devices.17

A B Y
0 0 0
1 0 1
0 1 1
1 1 1

Question 19.
Write the truth table for the circuits given in the Fig., consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits
byjus class 12 physics Chapter 14 Electronic Devices.18
Answer:
NCERT Solutions for Class 12 physics Chapter 14 Electronic Devices.19

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NCERT Solutions for Class 12 Physics Chapter 13 – Nuclei

NCERT Solutions for Class 12 Physics Chapter 13

NCERT Solutions for Class 12 Physics Chapter 13 Nuclei PDF is one of the smartest study resources for all students out there during their preparation. It not only helps CBSE students but also state boards students like UP, Bihar, HP, MP, Gujarat, and many others. It supports you all in understanding the concepts clearly as they are written in a simple language.

So, we as a team from Ncertbook.guru advice you to refer this article and download NCERT Questions and solutions of ch 13 Nuclei for free in PDF format. You can access them anywhere and everywhere you wish during the preparation and get a great command on the physics Nuclei concept.

Class 12 Physics NCERT Solutions Chapter 13 Nuclei

In this chapter, you will deal with the important concept which is the core of atoms ie., Nuclei. From the Nuclei introduction to various other topics are discussed efficiently in the NCERT Solutions for class 12 physics chapter 13 Nuclei. Students will gain complete knowledge about the topics and subtopics of Nuclei from the important questions listed over in the NCERT Solutions PDF.

Class 12
Subject Physics
Book Physics
Chapter Number 13
Chapter Name Nuclei

NCERT Exercise Questions & Answers of 12th Class Physics Chapter 13 – Nuclei

12th Class Physics NCERT Solutions of Chapter 13 Nuclei available over here and they are prepared by subject experts as per the CBSE Board Guidelines. You will find detailed answers to the class 12 physics textbook questions along with ch 13 nuclei exemplary problems, worksheets, and exercises here which aids you to grasp the complete knowledge on the topics of Nuclei.

Question 1.
(a) Two stable isotopes of lithium 63Li and 73Li have respective abundance of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u respectively. Find the atomic weight of lithium.
(b) Boron has two stable isotopes 105Li and 115Li .Their respective masses are 10.01294 u and 11.00931 u and the atomic weight of boron is 10.811 u. Find the abundances of 105Li and
115Li
Answer:
(a) Atomic weight of lithium
NCERT Solutions for Class 12 physics Chapter 13

Question 2.

The three stable isotopes of neon :1020Ne and 1022Ne  have respective abundance of 90.51%, 0.27% and 9.22%. The atomic masses of three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.
Answer:
NCERT Solutions for Class 12 physics Chapter 13.1

Question 3.
Obtain the binding energy of a nitrogen nucleus (147N) from the following data :
mH = 1.00783 u
mn = 1.00867 u
mn = 14.00307 u
Give your answer in MeV.
Answer:
vedantu class 12 physics Chapter 13.2

Question 4.
Obtain the binding energy of the nuclei 5626Fe and
in units of 20983Bi from the following data:
mH =1007825u
mn =1008665u
m (5626Fe)= 55.934939 u
m (20983Bi)
Which nucleus has greater binding energy per nucleon ?
Answer:
NCERT Solutions for Class 12 physics Chapter 13.3

Question 5.
A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 6329Cu atoms (of mass 62.92960 u). The masses of proton and neutron are 1.00783 u and 1.00867 u, respectively.
Answer:
NCERT Solutions for Class 12 physics Chapter 13.4
vedantu class 12 physics Chapter 13.5

Question 6.
Write nuclear equations for :
(a) the α-decay of 22686Ra
(b) the β-decay of 3215p
(c) the β+-decay of 116p
Answer:
NCERT Solutions for Class 12 physics Chapter 13.6

Question 7.
A radioactive isotope has a half-life of T years. After how much time is its activity reduced to 3.125% of its original activity (b) 1% of original value ?
Answer:
NCERT Solutions for Class 12 physics Chapter 13.7
vedantu class 12 physics Chapter 13.8

Question 8.
The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive 146C present with the stable carbon isotope
612C When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of 146C , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of 146C dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.
Answer:
NCERT Solutions for Class 12 physics Chapter 13.9

Question 9.
Obtain the amount of 6027Co necessary to provide a radioactive source of 8.0 mCi strength. The half­ life of 6027Co is 5.3 years.
Answer:
NCERT Solutions for Class 12 physics Chapter 13.10
NCERT Solutions for Class 12 physics Chapter 13.11

Question 10.
The half life of 9038Sr is 28 years. What is the disintegration rate of 15 mg of this isotope ?
Answer:
vedantu class 12 physics Chapter 13.12

Question 11.
Obtain approximately the ratio of the nuclear radii of the gold isotope 19779Au and silver isotope 10747Au.
Answer:
NCERT Solutions for Class 12 physics Chapter 13.13

Question 12.
NCERT Solutions for Class 12 physics Chapter 13.14

Answer:
NCERT Solutions for Class 12 physics Chapter 13.15
NCERT Solutions for Class 12 physics Chapter 13.16

Question 13.
NCERT Solutions for Class 12 physics Chapter 13.17
Answer:
vedantu class 12 physics Chapter 13.18
Here mN stands for the nuclear mass of the element or particle. In order to express the Q value in terms of the atomic masses, 6 me mass has to be subtracted
from atomic mass of 116Au and 5 me mass has to beNCERT Solutions for Class 12 physics Chapter 13.19

Question 14.
The nucleus 2310Ne decays by β~ emission. Write down the p-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
m(2310Sr) = 22.994466 u
m(2311Sr) = 22.989770 u.
Answer:
NCERT Solutions for Class 12 physics Chapter 13.20

Question 15.
The Q value of a nuclear reaction A + b -> C + d is defined by
[Q = mA + mb-mc– md] cwhere the masses refer to nuclear rest masses. Determine from the given data whether the following reactions are exothermic or endothermic.
NCERT Solutions for Class 12 physics Chapter 13.21
Answer:
NCERT Solutions for Class 12 physics Chapter 13.22
vedantu class 12 physics Chapter 13.23
NCERT Solutions for Class 12 physics Chapter 13.24

Question 16.
Suppose, we think of fission of a 5626Fe nucleus into two equal fragments, if 2813Al. Is the fission energetically possible ? Argue by working out Q of the process. Given, m (5626Fe) = 55.93494 u and m (2813Al)= 27.98191
Answer:
NCERT Solutions for Class 12 physics Chapter 13.25
NCERT Solutions for Class 12 physics Chapter 13.26
Question 17.
The fission properties of 23994Pu are very similar to those of 23592uu. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure
23994Pu undergo fission ?
Answer:
vedantu class 12 physics Chapter 13.27

Question 18.
A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much
23592u did it contain initially ? Assume that all the energy generated arises from the fission of 23592u and that this nuclide is consumed by the fission process.
Answer:
NCERT Solutions for Class 12 physics Chapter 13.28
NCERT Solutions for Class 12 physics Chapter 13.29

Question 19.
How long an electric lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium ? The fusion reaction can be taken as
NCERT Solutions for Class 12 physics Chapter 13.30
Answer:
NCERT Solutions for Class 12 physics Chapter 13.31

Question 20.
Calculate the height of Coulomb barrier for the head on collision of two deuterons. The effective radius of deuteron can be taken to be 2.0 fm.
Answer:
The initial mechanical energy E of the two deutrons before collision is given by
E = 2 K.E.
vedantu class 12 physics Chapter 13.32

Question 21.
From the relation R = RA1/3, where R is a constant and A is the mass number of a nucleus, show that nuclear matter density is nearly constant (i.e. independent of A)
Answer:
NCERT Solutions for Class 12 physics Chapter 13.33

Question 22.
For the β+ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K- shell, is captured by the nucleus and a neutrino is emitted).
NCERT Solutions for Class 12 physics Chapter 13.34
Show that if β+ emission is energetically allowed, electron capture is necessarily allowed but not vice-versa
Answer:
NCERT Solutions for Class 12 physics Chapter 13.35
vedantu class 12 physics Chapter 13.36

Question 23.
In a Periodic Table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on Earth. The three isotopes and their masses are 2412Mg (23.98504u), ? 2512Mg (24.98584) and 2612Mg (25.98259u). The natural abundance of 2412Mg is 78.99% by mass. Calculate the abundances of the other two isotopes.
Answer:
NCERT Solutions for Class 12 physics Chapter 13.37
NCERT Solutions for Class 12 physics Chapter 13.38

Question 24.
The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei 2412Ca and 2713Al from the following data :
NCERT Solutions for Class 12 physics Chapter 13.39
Answer:
NCERT Solutions for Class 12 physics Chapter 13.40

Question 25.
vedantu class 12 physics Chapter 13.63

Answer:
NCERT Solutions for Class 12 physics Chapter 13.41
NCERT Solutions for Class 12 physics Chapter 13.42

Question 26.
Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes :
NCERT Solutions for Class 12 physics Chapter 13.43
(a) Calculate the Q values for these decays and determine that both are energetically possible.
(b) The Coulomb barrier height for α-particle
NCERT Solutions for Class 12 physics Chapter 13.44
Answer:
vedantu class 12 physics Chapter 13.45
NCERT Solutions for Class 12 physics Chapter 13.46
NCERT Solutions for Class 12 physics Chapter 13.47
NCERT Solutions for Class 12 physics Chapter 13.48

Question 27.
Consider the fission of 23992u by fast neutrons. In one fission event, no neutrons are emitted and the final stable end products, after the beta-decay of the primary fragments, are 14058Ce and 9944Ru. Calculate Q for this fission process. The relevant atomic and particle masses NCERT Solutions for Class 12 physics Chapter 13.49
vedantu class 12 physics Chapter 13.50
Answer:
NCERT Solutions for Class 12 physics Chapter 13.51

Question 28.
Consider the D-T reaction (deuterium-tritium-fusion) given in eqn. :
NCERT Solutions for Class 12 physics Chapter 13.52
(b) Consider the radius of both deuterium and tritium to be approximately 1.5 fm. What is the kinetic energy needed to overcome the Coulomb repulsion ? To what temperature must the gases be heated to inititate the reaction ?
Answer:
From the equation given in the question,
NCERT Solutions for Class 12 physics Chapter 13.53
mN refer to nuclear mass of the element given in the brackets and mn = mass of neutron. If in represents the atomic mass, then
NCERT Solutions for Class 12 physics Chapter 13.54
vedantu class 12 physics Chapter 13.55
Question 29.
Obtain the maximum kinetic energy of p-particles, and the radiation frequencies to y decays in the following decay scheme. You are given that
m (198Au) = 197.968233 u
m (198Hg) = 197.966760 u
Answer:
The total energy released for the transformation of
19879Au to 19880u can be found by considering the energies of ϒ-rays. We first find the frequencies of the ϒ-rays emitted.
NCERT Solutions for Class 12 physics Chapter 13.56
NCERT Solutions for Class 12 physics Chapter 13.57
NCERT Solutions for Class 12 physics Chapter 13.58

Question 30.

Calculate and compare the energy released by (a) fusion of 1.0 kg of hydrogen deep within the sun and (b) the fission of 1.0 kg of 235U in a fission reactor.
Answer:
NCERT Solutions for Class 12 physics Chapter 13.59
NCERT Solutions for Class 12 physics Chapter 13.60

Question 31.
Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plant. Suppose we are given that, on an average, the efficiency of utilisation (i.e., conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium did our country need per year by 2000 ? Take the heat energy per fission of 235U to be about 200 MeV. Avogadro’s number = 6.023 x 1023 mol-1.
Answer:
NCERT Solutions for Class 12 physics Chapter 13.61
vedantu class 12 physics Chapter 13.62

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