BITSAT-2020 Exam Dates Announced | Exam Pattern | Admit Card – Check Here

BITSAT-2020 Exam Dates Announced – Birla Institute of Technology and Science, Pilani has announced revised dates for conducting BITSAT 2020. The exam will be conducted on August 6 to 10, 2020 in online mode and BITSAT admit card will be released soon.

The authorities provided the facility to make corrections in BITSAT application form between May 15 and May 20 in online mode. The authorities had earlier extended the last date to submit the application form of BITSAT 2020 till May 11, 2020. Online registration and correction for the exam is over and now BITS Pilani will start test center allotment, slot booking, and issuance of admit card.

BITSAT 2020 will be conducted in online mode for admissions into undergraduate engineering courses at different campuses of BITS.

The result of the entrance test will be declared in two forms – the initial result of BITSAT 2020 will be announced after the completion of the entrance exam and the second in the form of admit list in wait list.

BITSAT is a student’s gateway to get admission to any of the 3 campuses of BITS namely BITS Pilani, BITS Goa and BITS Hyderabad.

The BITSAT 2020 Admit Card is issued to only those candidates who do the slot booking before the last date. Candidates can download the admit card using login credential like application number and password. BITSAT 2020 Hall ticket/admit card will carry the candidate’s name category, password and important details regarding the examination such as the slot booked and the important exam related instructions. The hard copy of the admit card is to be carried at the exam centers as the soft copy is not accepted at the centers.

### BITSAT 2020 Exam Pattern

For better preparation candidates must know the exam pattern and marking scheme of the exam. Check the pattern and Marking scheme from below:

• Mode of the exam: It is a Computer Based Test.
• Type of Questions: The exam is a multiple choice based test
• The Language of Exam: English
• The Duration of the Exam: 180 minutes
• Total Number of Questions: 150
• Marking Scheme: For every correct answer, candidates get 03 marks.
• Negative Marking: For every incorrect answer, 01 marks are deducted.
• Distribution table of questions:

### BITSAT 2020 Result

Candidates who have appeared for the BITSAT 2020 gets to know their score on the very next day. As the BITSAT 2020 is held in various slots, so side by side the conducting body keeps uploading the scores of the respective candidates who are taking the exam. Candidates can check their result by using their login credentials. The final BITSAT result is declared later in the form of the merit list after the candidates submit their 10+2 marks. Candidates whose names appear in the merit list, qualify the exam. BITS releases two merit lists.

Merit in BITSAT (BITS Online Admission Test): The BITSAT top scorers are offered scholarships as per the table below.

Start your JEE & NEET Exam Preparation with Kota’s top IITian & Doctor Faculties. Download eSaral App Now!

eSaral is offering revision videos with mind maps covering all important formulae and key points. The revision lectures are taken by Kota’s Top IITian Faculties. This revision series would prove to be immense helpful in your upcoming Exams like Boards, JEE or NEET.

We bring you the revision notes on Maths, Physics and Chemistry which include some practical guidelines as well some useful facts of solving some typical problems. We have also provided mimd-maps for every topic that works as a flashcard, so it will help you more in quick revision.

### Revision Video + Mind-Maps for –

#### Subject

PHYSICS Revise
CHEMISTRY Revise
MATHS Revise

Along with the formulae and facts, we have also compiled some interactive examples and illustrations which have been picked up from the previous year papers of IIT JEE and other engineering exams. We’ll keep on adding more topics on daily basis and you are also free to share your suggestions.

eSaral Revision Series Benefits for JEE & NEET Students:

• Prepared by Kota’s Top IITian Faculties
• Based on the latest syllabus of IIT JEE and other engineering exams.
• Cover almost all important facts and formulae.
• They have been presented in the most crisp and precise form.
• Easy to memorize through Mind-Maps.
• Supported with illustrations from past year papers.
• The solutions are elaborate and easy to understand.
• These notes can aid in your last minute preparation.

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• #### MATHEMATICS REVISION by N.K.Gupta Sir (IIT-Kanpur) & Vishnu Sir (IIT-BHU)

ICSE Class 10 & ISC Class 12 Exam Date-sheet Released | Exams Between July 1 And July 14

The CISCE announced the revised schedule & timetable for the remaining subjects of ICSE Class 10 & ISC Class 12 Exam for Year 2020 exams. According to the revised schedule, CISCE will be conducting the remaining exams for class 10 from July 2 to 12 and class 12 exams are scheduled to be held from July 1 to July 14, at various centres spread across the country.

The Council for the Indian School Certification Examinations (CISCE) on Friday announced the revised schedules for the remaining subjects/exams of the ICSE and ISC Year 2020 examinations. The ICSE class X exams will be held from July 2 to July 12 whereas the ICE exams for Class XII will be held from July 1 to July 14. CISCE has also released important guidelines for students to follow while appearing for their respective examinations.

For ISC or 12th students, exam for 8 papers remain while for ICSE or 10th, exams for 6 subjects had to be postponed in view of the coronavirus crisis. After the Council conducts the remaining exams, it will announce the results within a period of 6-8 weeks.

Here are the datesheets for Class 10th & 12th Exam released by CISCE-

The Council has asked candidates to reach well ahead of time to ensure students are allowed to enter the exam centre one-by-one smoothly (staggered movement) to avoid overcrowding. Candidates must maintain the social distancing norms and should cover their faces with masks and carry own hand sanitisers. The use of gloves is optional. The question paper will be distributed at 10:45 am and the exam will begin at 11 am. Students will be given 15 minutes to read the question paper.

#### Here are some guidelines announced by CISCE for Class 10 & 12 Students appearing in Exams –

Once the exam starts, it is advised to begin with the easiest questions and then move forward  towards difficult ones! With this approcah you can save extra time to attempt long questions.

Don’t take stress stay calm and positive. Even if you don’t know the answers of the questions, don’t stressed out! Attempt easy questions first, that will help you in building confidence.

Class X CBSE Board Exam Datesheet
Class XII CBSE Board Exam Datesheet


All the Best Students!!!

HRD Minister Launches AI-powered Mock Test App for JEE & NEET Students

Union HRD Minister Mr Ramesh Pokhriyal Nishank launched an AI-powered mock test App for JEE and NEET 2020 Students. Its an android app ‘National Test Abhyas’ for students who are going to appear in JEE Main 2020 and NEET 2020 examinations.

NTA plans to release one new mock test on the app everyday, which students can then download and attempt offline. Once the test is completed, students can go online again to submit the test and view their test report.

Every student is unique and requires specific guidance to detect and overcome gaps in knowledge and test-taking strategy.

eSaral offers personalized learning platform where a student can study and attempt test at its own pace. eSaral also offers one to one Mentoring to help each and every student in academic and non-academic related issues.

#### Prepare for JEE & NEET with the Kota’s Top IITian & Doctor Faculties !

Quickly Revise All the Topics – Absolute FREE REVISION WITH MIND-MAPS!

This time is very crucial for all those who are preparing for the upcoming JEE and NEET Exams. To ensure that learnind doesn’t affect and no student is left behind in getting exposure to practice testing. Click Here to download eSaral App!

#### Why eSaral ?

eSaral offers a broad categories of mock tests for students preparing for JEE 2020 and NEET 2020 Examinations. With the Part and full syllabus mock tests, eSaral App also provides topic wise tests that are very helpful in analyzing weak and strong areas and also helpful in revising important concepts from each and every chapter individually.

### Importance of Practicing Mock Tests

1. Practicing Mock tests helps to re-examine the concepts that you have learned before.
2. It also helps in Analyzing the weak and strong areas. Always try to strengthen strong points and work optimistically on weaker sections.
3. Helps in improving Speed and presence of mind. As the students gets acclimatized to the types of questions asked in mock tests.
4. Time-Management: Time allotted for the exam is limited. Since the Exam will held in Online Mode, so practicing online mock tests will help the students to get acquainted with Exam duration and attempting exam in online mode rather than pen – paper. More practice will help the students in managing time between easy and difficult questions.
5. To Shed Exam Fear – To practice mock test works very efficiently in improving confidence level and eliminating exam fear. Practice 3 Hours at a stretch and solve questions for a sustained period of time without breaks.
6. Helps in developing right strategy to attempt exam.

Practicing mock tests helps the students in knowing where they stand so they can prepare better next time.

Now we hope that you have understood the importance of JEE Main & NEET Mock Tests, make sure you sign up for mock test.

Start Early to take a lead. Time waits for none.

CRACK JEE 2020 | How to Prepare for JEE Main in last 2 Months

Less than 2 months remaining for JEE Main 2020 (July Attempt), but at the same time it is still possible to secure good percentile and an All India Rank to get qualified for JEE advanced and also to get admissions in some Top NIT’s. Read the complete article till the end to know How to Prepare for JEE Main in 2 Months.

After a long wait the dates for JEE Main 2020 has finally been announced. The exam will be held between 18th and 23rd July 2020. Although students have had an extended preparation time ever since we got into preventive lockdown due to COVID 19 pandemic.

Students were finding it bit difficult to concentrate in the absence of a specific target exam date. Now the students do have a specific date in front of them to look forward to. Whenever there is less time remaining, you have to put more effort into your studies to practice enough questions for good JEE preparation.

### JEE Main 2020 Exam Preparation Tips for Last two remaining months:

#### 1. Practice mock tests and solve previous year question papers

Solving previous year question papers of JEE Main and taking mock tests is extremely important to get a good rank in the exam. Attempt the various free mock tests available on eSaral App and get the feel of the actual exam and improve your time management skills.

Solve at least ONE full syllabus online test every week. Solve questions by applying your reasoning and analytical skills. Develop test taking strategy, it will evolve with more tests that you attempt. Also solve previous years JEE Main question paper bank. While taking the tests students should look at building the right exam temperament, adjusting the body clock and getting comfortable with test environment.

#### 2. Practice more and more conceptual problems and Clear your Doubts.

No matter what how big or small your doubt is, get it cleared. Download eSaral App & we will assist you in clarifying your doubts about a particular topic related to JEE Main. For clearing doubts on important topics and concepts, you can join a crash course.

Crash Course will also help you to complete class 11 & 12 JEE Topics covering both Mains & Advanced syllabus with complete package of test series. To know more about JEE 2020 Crash Course, CLICK HERE

#### 3. Effective Revision | Technique

Ensure you make a checklist of JEE Main syllabus with the list of topics/units against each subject – Physics, Chemistry and Mathematics. Tick off the important topics as and when you are done with revising the particular topic.

For Quick Revision with Mind- Maps Visit the given Links Below:

JEE Main is not difficult, rather it is predictable and tricky. Hence, clarity of concept and familiarity with the question type is essential to do well, so revision of concepts is the key. Prepare a list of concepts, topic-wise. Keep marking the concepts which are your weak areas. Keep referring to the notes&important formulae.

It is extremely important to be in question solving mode. So, after revising every concept/topic, solve an unsolved question. One must NOT refer to solution without an attempt to solve the question. So, sufficient time must be devoted. Identify mistakes and/or weak area & fix them instantly.You have to be quick and accurate too.

#### 4. Effective Revision with the help of Mind Maps & Flashcards

Have a large stack of flashcards handy. It should contain all the important JEE Main formulae, concepts and diagrams. These could also be used in the last minutes before JEE Main.Here are the list of Complete Mind Maps for Organic Chemistry & Physics

#### 5. Analyse & Strengthen your weak areas

Every time you take a mock test, Analyze your mistakes and weak areas. Work on the subjects, topics and units which are you are weak at. Resolve and strengthen them well before the actual exam takes place.

#### 6. Performance Analysis during practice test

Analyze your performance after every test. Subject wise weak areas needs to be identified and revisit those concepts all over again to eradicate them. Attempt Mock Tests available on eSaral App to get an instant test analysis with performance report.

Don’t panic about less months remaining for JEE preparation. You just need to focus on more study hours and practice to crack the JEE exam in remaining time.

Your preparation in these final months before the exam will have a major impact on your performance and JEE Main percentile, ranks and scores. Furthermore, your time management skills clubbed with a positive attitude will make your chances of getting into the top tech institutes of the country better. Remember, it’s important to utilize each day efficiently to secure success in JEE Main 2020.

Stay away from social networking sites for a few month. Also maintain a good health, as fitness goal is paramount during this time. Light exercise, healthy diet, good sleep are the key aspects to remain fit. Take all necessary precaution as prescribed by our government & health department.

Whatever is the score in JEE Main 2020 January exam, there will be an option to increase score & percentile in the upcoming attempt of JEE Main 2020.

Always have a positive mindset and Believe it that whatever you have prepared is more than enough for exams. A positive mindset helps you to retain more and you will be able to score well in exams.

Kindly share your reviews and let us know the queries regarding JEE Main 2020 Exam. Drop your messages in the comment box below & let us help you with the correct guidance.

All the best 🙂

NTA JEE Main 2020 Result

Candidates are able to check the NTA JEE Main 2020 Result only through online mode. The result has been declared in the form of percentile (NTA Scores). Candidates Rank will be declared after the declaration of result for April session.

After the declaration of both sessions result, the rank will be announced. Candidates rank will be calculated by comparing both the attempts. NTA will not provide the result through any offline method.

The NTA Score of JEE Main 2020 January exam has been released on 17th January 2020. This score comprise of the actual marks obtained in Paper-I of all subjects. The result has been provided in the form of percentile. The result for second session will be announced in the month of August 2020.

The score & rank obtained by the candidate will be used for counseling & eligibility for JEE Advanced exam.

Rank will comprise of All India Rank and All India Category Rank. The authority will not dispatch the rank card to any candidates. Candidates will have to download it through the website.

## What After JEE Main 2020 Result?

NTA will announce the cut-off score for JEE Advanced. In 2019, the cut-off score for general category students was 89.75 while that of OBC – NCL students was 74. Check out JEE Main cut offs for other categories.

• Counselling/Selection Procedure

After JEE Main and JEE Advanced results will be declared, JoSAA conducts joint seat allocation process for admissions to 100 institutes including 23 IITs, 31 NITs, 25 IIITs and 28 Other GFTIs. Admission to all these JEE Main and JEE Advanced participating institutes will be done through a single counselling platform.

JEE Advanced 2020 Admit Card (Postponed) | Exam on 23rd August ’20

• Visit the official website of JEE Advanced 2020
• Enter your JEE Advanced 2020 registration number, Date of Birth, Mobile number and email id to login
• JEE Advanced 2020 hall ticket will be displayed on the screen
• Check all the details printed on admit card thoroughly

#### Dates

• Server issues: Once JEE Advanced hall tickets will be released, successful applicants can download their respective admit card. The server may be down due to a sudden traffic surge on the website.

#### Errors in JEE Advanced admit card and ways to resolve them

• Discrepancy/ies in JEE Advanced admit card 2020: For discrepancy in the particulars (photograph/signature) mentioned on the JEE Advanced hall ticket or on the confirmation page, candidates should immediately get in touch with the JEE Advanced authorities.

### Documents to Carry to JEE Advanced Test Centre

Along with JEE Advanced 2020 admit card, candidates need to carry one of the following original photo identity proofs:

• School/ College ID
• Pan Card
• Passport

#### JEE Advanced Previous Year Question Papers are available here.

NTA JEE Main 2020, NEET 2020 Exam dates Announced | Check here for details

The exam dates for NTA JEE Main 2020, NEET 2020 announced by the Union Minister of Human Resource Development Ramesh Pokhriyal Nishank on tuesday (5th May’20).

The JEE (Main) 2020 and NEET 2020 exams were earlier re-scheduled by the ministry, due to the nationwide lockdown in the wake of coronavirus outbreak. The new date for the JEE (Main) 2020 exams was since then awaited.

The JEE Main exam will be conducted from July 18 to July 23, 2020, Ramesh Pokhriyal Nishank said.

The NEET 2020 exam will be conducted on July 26, he said, while the JEE Advanced on 23rd August ’20.

### NEET 2020: Public Notice by NTA

According to the announcement, Union Minister of Human Resource Development Ramesh Pokhriyal Nishank said the examination process for the JEE (Main) 2020 will begin from July this year, while the session will begin by August.

The JEE Main exam will be conducted from July 18 to July 23, 2020, Ramesh Pokhriyal Nishank said.

The NEET 2020 exam will be conducted on July 26, he said, while the JEE Advanced on 23rd August ’20.

CRASH COURSE to CRACK JEE MAIN 2020 | Register NOW

The date for the release of admit cards for JEE Main, NEET 2020 is yet not available.

This was the second interaction between Ramesh Pokhriyal Nishank and students. This session was earlier scheduled to be held on May 2, 2020. However, the same was postponed for May 5, for some reason.

Earlier he interacted on 27th April 2020 and answered some critical questions of students and their parents.

According to the National Testing Agency (NTA), for JEE Main over nine lakh candidates have registered, on the other hand over 15.93 lakh students are likely to take up the NEET 2020 exams.

#### JEE Advanced Previous Year Question Papers are available here.

NTA NEET 2020 Postponed due to Coronavirus Lockdown

Medical entrance exam, NTA NEET 2020 postponed due to COVID-19 outbreak. NEET 2020 Exam was scheduled to be held on May 3. The entrance exam is held annually for admission to undergraduate medical or MBBS, BDS and AYUSH programmes. Through this exam, admissions to veterinary courses are also done.

As per the orders of the HRD Minister, the NEET and JEE Main exams are postponed till last week of May. The admit cards will be issued after April 15. Here are the latest Update  by NTA through public notice issued on official website :

.
With this NEET is the second major exam to have been postponed due to the nationwide lockdown due to COVID-19 outbreak.

Earlier this week, following the orders of the HRD Minister, NTA had postponed the engineering entrance exam, JEE Main. The exam was scheduled to begin on April 5.

As of now, the examination is proposed to be held in the last week of May 2020. The exact date will be announced later on after assessing the situation.

Accordingly, the Admit Cards for the Examination which were to be issued on 27th March 2020 will now be issued later on after assessing the situation after 15th April 2020 only.

We understand that the academic calendar and schedule is important but equally important is well being of every citizen including students.

Students and parents are advised to not get worried about the Examination. Moreover, parents are requested to help them in utilizing this time very efficiently for preparing for the Examination and focus on critical concepts in order to close learning gaps if any, NTA would keep students updates about the latest developments and would inform about changes with ample time.

All the best!

Stay home, Stay safe and study from home.

Click Here to Know How to Study Well in 21 Days Lockdown Period!!!

NEET UG 2020 Admit Card | Exam to be held on 26th July

In view of this pandemic situation due to Coronavirus (COVID-19), the admit cards for National Eligibility cum Entrance Test (NEET-UG) will not be released today (on March 27, 2020), as mentioned in the brochure issued by National Testing Agency (NTA). NEET UG 2020 admit card new date will be announced after the lockdown.

NEET 2020 Exam Postponed

NEET 2020 is to be held on July 26th, however, the ongoing situation due to Coronavirus outbreak might affect the exam date as well.  The new dates for downloading the admit cards will be announced soon.

Students are advised to take advantage of the postponement and prepare well for the NEET 2020 Exam.

The NEET exam is for admission to the undergraduate medical courses in all medical institutions.

NTA NEET 2020 application process was conducted from December 2nd, 2019 to January 1st, 2020. The exam is conducted in 11 languages and will test candidates on Physics, Chemistry, and Biology subjects with 180 MCQ questions for a total of 720 marks and for 3 hours’ duration. For more details, Click Here.

Bihar Board 12th Result 2020 Out || Check Here

Much awaited Bihar Board 12th result 2020 has been declared. BSEB has announced the results for Class 12th Board Exams at BSEB Official Website. BSEB releases results through online mode. The Students who have appeared in Bihar 12th Board Exam can check their result and download it.

The Examination of Bihar Board Class 12 has completed on 13th February 2020.

Students can also check the Bihar Board Inter Topper List, Merit List, Percentage of Result status on official site.

Steps to Check Bihar 12th Board Result 2020:

Step 1:- Visit the Official Website or Click on direct link below.

Step 2:- Enter Roll Code and Roll Number.

Step 3:- Enter Code shown below on Result page.

Step 4:- Click on View Result Button.

Step 5:- BSEB 12th Result 2020 will display on screen.

Step 6:- Now you can check out Bihar 12th Result 2020

The students, after checking the Bihar Board Class 12 result 2020, will also be able to either download a PDF copy or take a printout of the scorecard. The students are advised to download / take a printout of the Bihar Board 12th Result 2020 mark sheet to use as a provisional result until official scorecards are issued by the BSEB. However, the provisional BSEB Result 2020 copy is only valid until the original mark sheets are issued by the board. Therefore, we advise all the students to collect their BSEB Board 12th Result 2020 mark sheets from their respective schools later.

## Re-evaluation and Rechecking

The performance of the students in the BSEB 12th Result 2020 will lay the foundation for their academic future. Taking into account the importance of the Bihar Board 12th Result 2020, if the students are unhappy with the outcome of their hard work, they can apply for re-evaluation / rechecking of the answer sheets. The Bihar Board provides this facility to all the concerned students at the cost of a nominal fee. More information in this regard can be found at the respective schools.

## Compartmental Exam

The BSEB also provides the students unable to pass in the Bihar Board Class 12 exam 2020 with an opportunity to prove themselves again. Students who failed in one or more exams in the board exam can opt for Bihar Board compartmental exam 2020. This exam will prevent an entire academic year from getting wasted. Students can appear in the compartmental exam of Bihar Board 2020 by filling in the online application form and paying the requisite application fee. For details, Visit BSEB Official Website.

Bihar 12th Board Exam Syllabus.

Bihar 12th Board 2020 Topper’s of Science Stream:

 RANK Reg. No. Candidates Name Total Marks 1 R-430700022-18 Neha Kumari 476 2 R-410440239-18 Vikky Kumar 474 2 R-360020136-18 Jahangeer Alam 474 3 R-230020391-18 Shivam Kumar Verma 473 3 R-710430106-18 Manish Kumar Jaiswal 473

JEE Main 2020 April Exam Postponed due to Corona-virus

JEE Main 2020 April Exam has been postponed by NTA due to Corona-virus Outbreak. The exam was scheduled to be held on April 5, 7, 9 and 11. In recent notice at official website, National Testing Agency on Wednesday (18th march) took the decision to postpone the JEE Main April 2020 Examination in pursuance of MHRC letter. New dates of JEE Main likely to be decided on March 31, 2020.

The new date will be decided in accordance with the board exams schedule and other competitive exams to ensure there is no clash.

The Ministry of Human Resource Development (MHRD) on Wednesday directed the Central Board of Secondary Education (CBSE) and all educational institutions in the country to postpone all exams including JEE Main till March 31 in view of coronavirus outbreak.

MHRD also stated that “While maintenance of academic calendars and exam schedule is important, equally important is the safety and security of the students who are appearing in various exams as also that of their teachers and parents”.

The Public Notice by NTA:

To ensure, your Learning never gets affected, eSaral has taken a lead in this regard and has made eSaral App FREE till 15th April’20.

This step is taken by eSaral keeping in mind Safety and Future of students amidst ongoing health crisis. Now students can study from home safely and continue their learning.

eSaral App comprises complete courses for Class 9 to 12 including Video Tutorials, Complete Study Material designed by Top IITians & Doctor Faculties.

JEE Advanced 2020 Exam Date Announced, Check Syllabus – Important Information

Joint Entrance Examination, JEE Advanced or IIT JEE Exam dates 2020 have been announced. Union HRD Minister Ramesh Pokhriyal today shared that the JEE Advanced 2020 would be conducted on August 23, 2020. The dates for JEE Main 2020 were announced on May 5 by the HRD Minister during a webinar with the students.

JEE Main 2020 would be conducted from July 18 to July 23. Given the date of the JEE Advanced 2020, it can be suggested that the results of JEE Main 2020 are expected to be announced by August 10. The online application process for JEE Advanced 2020 would begin only after the final results of JEE Main 2020 are announced along with the JEE Main 2020 Ranks

#### JEE Main 2020 (Revised dates by MHRD): July 18th to July 23rd ’20

JEE Advanced is organised by one of seven Zonal Coordinating IIT’s guided by the Joint Admission Board (JAB). It is a National Level Engineering Entrance Exam for admission in the Bachelors, Integrated Masters and Dual Degree Programs. The exam will be conducted in online mode. Candidate needs to clear the JEE Main Exam to be eligible for the JEE Advanced Examination.

Results for JEE Advanced 2020 will tentatively be released by the IIT Delhi in the second week of August. The result will comprise the exam scores, All India Rank (AIR) of candidates, subject-wise marks, etc.

Students may please note that the timeline provided above is only based on the usual time frame followed by the IITs in regards the examination. The actual schedule would be announced by IIT Delhi in a few days time. IIT JEE or JEE Advanced 2020 is conducted by the IITs while JEE Main, the qualifier is conducted by National Testing Agency. Both JEE Main and JEE Advanced 2020 would be online or computer based tests.

### Eligibility Criteria

Candidates must fulfill one of the following two criteria in order to be eligible for admission at IIT:

1. Candidates must have secured at least 75% aggregate marks in the Class XII (or an equivalent) Board examination. The aggregate marks for SC, ST and PwD candidates should be at least 65%. Physics, Chemistry, and Mathematics are required as compulsory subjects in Class XII (or equivalent) Board examination in 2019 or 2020.
2. Candidates must be within the category-wise top 20 percentile of successful candidates in their respective Class XII (or equivalent) board examination in 2019 or 2020 with Physics, Chemistry, and Mathematics as compulsory subjects.

The details regarding the JEE Advanced Exam pattern is given below:

• Online Mode: From 2020, the exam is conducted via online mode only.
• Number of Papers: JEE Advanced consists of two compulsory papers (Paper 1 and Paper 2).
• Duration of Exam: 3 hours/each paper. Some extra time will be given to the PwD candidates.
• Type of Questions: Objective type (MCQs).
• Language of Question Paper: English and Hindi.
• Subjects: Physics, Chemistry, and Mathematics subjects will be asked in the exam.
• Negative Marking: Yes, there is a provision of negative marking and different for paper 1 & 2.

#### JEE Advanced Previous Year Question Papers are available here.

All The Best 🙂

True Dip and Apparent Dip – Magnetism and Matter Class 12

Here we will study about True Dip and Apparent Dip.

### Apparent Dip

The dip at a place is determined by a dip circle. It consists of magnetized needle capable of rotation in vertical plane about a horizontal axis. The needle moves over a vertical scale graduated in degrees.

If the plane of the scale of the dip circle is not in the magnetic meridian then the needle will not indicate correct direction of earth’s magnetic field. The angle made by the needle with the horizontal is called Apparent Dip.

### True Dip

When the plane of scale of dip circle is in the magnetic meridian the needle comes to rest in direction of earth’s magnetic field. The angle made by the needle with the horizontal is called True Dip.

Suppose dip circle is set at angle $\alpha$ to magnetic meridian.

Horizontal component $${B_H}’ = {B_H}\cos \alpha$$

Vertical component $${B_V}’ = {B_V}$$ (remains unchanged)

Apparent dip is $${\theta ^\prime }\tan {\theta ^\prime } = {{{B_V}’} \over {{B_H}’}} = {{{B_V}} \over {{B_H}\cos \alpha }} = {{\tan \theta } \over {\cos \alpha }}\left( {\tan \theta = {{{B_V}} \over {{B_H}}} = {\rm{ true dip }}} \right)$$

1. For a vertical plane other than magnetic meridian $\alpha>0$ or $\cos \alpha<1$ so $\theta^{\prime}>\theta$ In a vertical plane other than magnetic meridian angle of dip is more than in magnetic meridian.
2. For a plane perpendicular to magnetic meridian $\alpha=\frac{\pi}{2}$ $\therefore \tan \theta^{\prime}=\infty \quad$ so $\quad \theta^{\prime}=\frac{\pi}{2}$ So in a plane perpendicular to magnetic meridian dip needle will become vertical.

At magnetic equator :

1. Angle of dip is zero.
2. Vertical component of earths magnetic field becomes zero $B_{V}=B \sin \theta=B \sin 0=0$
3. A freely suspended magnet will become horizontal at magnetic equator.
4. At equator earth’s magnetic field is parallel to earth’s surface i.e., horizontal.

At magnetic poles :

1. Angle of dip is $90^{\circ}$
2. Horizontal component of earth’s magnetic field becomes zero. $B_{H}=B \cos \theta=B \cos 90=0$
3. A freely suspended magnet will become vertical at magnetic poles.
4. At poles earth’s magnetic field is perpendicular to the surface of earth i.e. vertical.

Ex. If $\theta_{1}$ and $\theta_{2}$ are angles of dip in two vertical planes at right angle to each other and $\theta$ is true dip then prove $\cot ^{2} \theta=\cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}$.

Sol. If the vertical plane in which dip is $\theta_{1}$ subtends an angle $\alpha$ with meridian than other vertical plane in which dip is $\theta_{2}$ and is perpendicular to first will make an angle of $90-\alpha$ with magnetic meridian. If $\theta_{1}$ and $\theta_{2}$ are apparent dips than

$\tan \theta_{1}=\frac{B_{V}}{B_{H} \cos \alpha}$

$\tan \theta_{2}=\frac{B_{V}}{B_{H} \cos (90-\alpha)}=\frac{B_{V}}{B_{H} \sin \alpha}$

$\cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}=\frac{1}{\left(\tan \theta_{1}\right)^{2}}+\frac{1}{\left(\tan \theta_{2}\right)^{2}}=\frac{B_{H}^{2} \cos ^{2} \alpha+B_{H}^{2} \sin ^{2} \alpha}{B_{V}^{2}}=\frac{B_{H}^{2}}{B_{V}^{2}}=\left(\frac{B \cos \theta}{B \sin \theta}\right)^{2}=\cot ^{2} \theta$

So $\quad \cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}=\cot ^{2} \theta$

Properties of Paramagnetic & Diamagnetic Materials

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Neutral Point of Magnet – Magnetism and Matter Class 12 Physics Notes

A neutral point of Magnet is a point at which the resultant magnetic field is zero. In general the neutral point is obtained when horizontal component of earth’s field is balanced by the produced by the magnet. When the N pole of the magnet points South and the magnet in the magnetic meridian.

When we plot magnetic field of a bar magnet the curves obtained represent the superposition of magnetic fields due to bar magnet and earth.

DEFINITION
A neutral point in the magnetic field of a bar magnet is that point where the field due to the magnet is completely neutralized by the horizontal component of earth’s magnetic field. At neutral point field due to bar magnet (B) is equal and opposite to horizontal component of earth’s magnetic field $\left( B _{ H }\right) or \quad B = B _{ H }$

• #### Neutral point when north pole of magnet is towards geographical north of earth.

The neutral points $N_{1}$ and $N_{2}$ lie on the equatorial line. The magnetic field due to magnet at neutral point is $B =\frac{\mu_{0}}{4 \pi} \frac{ M }{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}$ Where M is magnetic dipole moment of magnet, $2 l$is its length and r is distance of neutral point.

At neutral point $B = B _{ H } \cdot \quad$ so $\quad \frac{\mu_{0}}{4 \pi} \frac{ M }{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}= B _{ H }$

For a small bar magnet $\left(\ell^{2}<< r ^{2}\right)$ then $\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}}= B _{ H }$

• #### Neutral point when south pole of magnet is towards geographical north of earth.

The neutral points $N_{1}$ and $N_{2}$ lie on the axial line of magnet. The magnetic field due to magnet

at neutral point is $B =\frac{\mu_{0}}{4 \pi} \frac{2 Mr }{\left( r ^{2}-\ell^{2}\right)^{2}}$

At neutral point $B =\frac{\mu_{0}}{4 \pi} \frac{2 Mr }{\left( r ^{2}-\ell^{2}\right)^{2}}$ so $\frac{\mu_{0}}{4 \pi} \frac{2 Mr }{\left( r ^{2}-\ell^{2}\right)^{2}}= B _{ H }$

For a small bar magnet $\left(\ell^{2}<< r ^{2}\right)$ then $\frac{\mu_{0}}{4 \pi} \frac{2 M }{ r ^{3}}= B _{ H }$

#### Special Point

When a magnet is placed with its S pole towards north of earth neutral points lie on its axial line. If magnet is placed with its N pole towards north of earth neutral points lie on its equatorial line. So neutral points are displaced by $90^{\circ}$ on rotating magnet through $180^{\circ}$ In general if magnet is rotated by angle $\theta$ neutral point turn through an angle $\frac{\theta}{2}$

#### Neutral Point in Special Cases

(a) If two bar magnets are placed with their axis parallel to each other and their opposite poles face each other then there is only one neutral point (x) on the perpendicular bisector of the axis equidistant from the two magnets.

(b) If two bar magnets are placed with their axis parallel to each other and their like poles face each other then there are two neutral points on a line equidistant from the axis of the magnets.

Properties of Paramagnetic & Diamagnetic Materials

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Bar Magnet as an Equivalent Solenoid – Magnetism || Class 12 Physics Notes

### Bar Magnet as an Equivalent Solenoid

In solenoid each turn behaves as a small magnetic dipole having dipole moment $\mathrm{I} \mathrm{A}$. A solenoid is treated as arrangement of small magnetic dipoles placed in line with each other. The number of dipoles is equal to number of turns in a solenoid. The south and north poles of each turn cancel each other except the ends. So solenoid can be replaced by single south and north pole separated by distance equal to length of solenoid. The magnetic field produced by a bar magnet is identical to that produced by a current carrying solenoid.

Derivation of Bar Magnet as an Equivalent Solenoid

To demonstrate the similarity of a current carrying solenoid to a bar magnet, let us calculate axial field of a finite solenoid carrying current.

Consider:
$a=$ radius of solenoid

$2 l=$ length of solenoid with centre O

$n=$ number of turns per unit length
$I=$ current passing through solenoid

$O P=r$

Consider a small element of thickness $d x$ of solenoid at distance $x$ from
O. and number of turns in element $=n d x$

We know magnetic field due to n turns coil at axis of solenoid is given
by

$d B=\frac{\mu_{0} n d x I a^{2}}{2\left[(r-x)^{2}+a^{2}\right]^{\frac{3}{2}}}$

The magnitude of the total field is obtained by summing over all the elements $-$ in other words by integrating from $x=-1$ to $x=+1 .$ Thus,

$B=\frac{\mu_{0} n I a^{2}}{2} \int_{-1}^{l} \frac{d x}{\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}$

This integration can be done by trigonometric substitutions. This exercise, however, is not
necessary for our purpose. Note that the range of $x$ is from $-1$ to $+1 .$ Consider the far axial field of the solenoid, i.e., $r>>$ a and $r>>1 .$ Then the
denominator is approximated by

\begin{aligned}\left[(r-x)^{2}+a^{2}\right]^{3 / 2} &=r^{3} \\ \text { and } B &=\frac{\mu_{0} n I a^{2}}{2 r^{3}} \int_{-1}^{1} d x \\ &=\frac{\mu_{0} n I}{2} \frac{2 l a^{2}}{r^{3}} \end{aligned}

Note that the magnitude of the magnetic moment of the solenoid is, (total number of turns $\times$ current $\times$ cross-sectional area). Thus,

$B=\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}$

It is clear from the above expression that magnetic moment of a bar magnet is equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field.

Biot Savart’s Law

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Terrestrial Magnetism – Earth’s Magnetism || Class 12 Physics Notes

Do you know that the Earth is also a magnet? Yes!!! How do you think then that the suspended bar magnet always points in the north-south direction? Adware about the concept of the Terrestrial Magnetism that we are going to discuss in this chapter. It is really interesting to study and analyze this concept of earth’s magnetism.

The branch of Physics which deals with the study of earth’s magnetic field is called terrestrial magnetism.

1. William Gilbert suggested that earth itself behaves like a huge magnet. This magnet is so oriented that its S pole is towards geographic north and N pole is towards the geographic south.
2. The earth behaves as a magnetic dipole inclined at small angle $11.5^{\circ}$ to the earth’s axis of rotation with its south pole pointing geographic north.
3. The idea of earth having magnetism is supported by following facts.
4. A freely suspended magnet always comes to rest in N-S direction.
5. A piece of soft iron buried in N-S direction inside the earth acquires magnetism.
6. Existence of neutral points. When we draw field lines of bar magnet we get neutral points where magnetic field due to magnet is neutralized by earth’s magnetic field.
7. The magnetic field at the surface of earth ranges from nearly 30 $\mu T$ near equator to about 60$\mu T$ near the poles. The magnetic field on the axis is nearly twice the magnetic field on the equatorial line.

Biot Savart’s Law

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Magnetic Elements – Magnetism Class 12 || Physics Notes

The physical quantities which determine the intensity of earth’s total magnetic field completely both in magnitude and direction are called magnetic elements.

### Angle of Declination $(\phi)$:

The angle between the magnetic meridian and geographic meridian at a place is called angle of declination.

(a) Isogonic Lines : Lines drawn on a map through places that have same declination are called isogonic lines.

(b) Agonic Line : The line drawn on a map through places that have zero declination is known as an agonic line.

### Angle of Dip or Inclination

The angle through which the N pole dips down with reference to horizontal is called the angle of dip. At magnetic north and south pole angle of dip is $90^{\circ}$. At magnetic equator the angle of dip is zero.

OR The angle which the direction of resultant field of earth makes with the horizontal line of magnetic meridian is called angle of dip.

(a) Isoclinic Lines : Lines drawn up on a map through the places that have same dip are called isoclinic lines.

(b) Aclinic Line : The line drawn through places that have zero dip is known as aclinic line. This is the magnetic equator.

### Horizontal component of earth’s magnetic field

The total intensity of the earth’s magnetic field makes an angle  with horizontal. It has

(i) Component in horizontal plane called horizontal component $B_{H}$.

(ii) Component in vertical plane called vertical component $B_{V}$

$B _{ V }= B \sin \theta \quad B _{ H }= B \cos \theta$

So $\frac{ B _{ V }}{ B _{ H }}=\tan \theta \quad$ and $\quad B =\sqrt{ B _{ H ^{2}}+ B _{ V ^{2}}}$

#### IMPORTANT POINTS

1. If $\theta$ and $\phi$ are known we can find direction of B.
2. If $\theta$ and $B_{H}$ are known we can find magnitude of B.
3. So if $\theta$, $\phi$ and $B_{H}$ are known we can find total field at a place. So these are called as Elements of earth’s magnetic field.
4. The declination gives the plane, dip gives the direction and horizontal component gives magnitude of earth’s magnetic field.
5. If declination is ignored, then the horizontal component of earth’s magnetic field is from geogrophic south to geographic north.
6. Angle of dip is measured by instrument called dip circle.

Biot Savart’s Law

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Bohr Magneton: Unit of Bohr Magneton – Magnetism || Class 12 Physics Notes

Bohr Magneton is defined as the magnetic dipole moment associated with an atom due to orbital motion of an electron in the first orbit of hydrogen atom. This is the smallest value of magnetic moment. Unit of Bohr Magneton is-

• In CGS units it is defined by the equation:

$\mu \mathrm{B}=(e \cdot h) / 2 m_{\mathrm{e}} \mathrm{c}$

• In SI units it is defined by the equation:

$\mu_{\mathrm{B}}=e \cdot h / 2 m_{e}$

1. The electron possesses magnetic moment due to its spin motion also $\overrightarrow{ M }_{ s }=\frac{ e }{ m _{ e }} \overrightarrow{ s }$ .where $\overrightarrow{ s }$ is spin angular momentum of electron and $S=\pm \frac{1}{2}\left(\frac{h}{2 \pi}\right)$
2. The total magnetic moment of electron is the vector sum of its magnetic moments due to orbital and spin motion.
3. The resultant angular momentum of the atom is given by vector sum of orbital and spin angular momentum due to all electrons. Total angular momentum $\vec{J}=\vec{L}+\vec{S}$
4. The resultant magnetic moment $\overrightarrow{ M _{j}}=- g \left(\frac{ e }{2 m }\right) \overrightarrow{ J }$

where g is Lande’s splitting factor which depends on state of an atom.

For pure orbital motion g = 1 and pure spin motion g = 2.

Biot Savart’s Law

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Current loop as a Magnetic Dipole – Magnetism | Class 12th Physics Notes

#### Current loop as a Magnetic Dipole

Ampere found that the distribution of magnetic lines of force around a finite current carrying solenoid is similar to that produced by a bar magnet. This is evident from the fact that a compass needle when moved around these two bodies show similar deflections. After noting the close resemblance between these two, Ampere demonstrated that a simple current loop behaves like a bar magnet and put forward that all the magnetic phenomena is due to circulating electric current. This is Ampere’s hypothesis.

We consider a circular coil carrying current I. When seen from above current flows in anti clockwise direction.

1. The magnetic field lines due to each elementary portion of the circular coil are circular near the element and almost straight near center of circular coil.
2. The magnetic lines of force seem to enter at lower face of coil and leave at upper face.
3. The lower face through which lines of force enter behaves as south pole and upper face through which field lines leave behaves as north pole.
4. A planar loop of any shape behaves as a magnetic dipole.
5. The dipole moment of current loop $(\mathrm{M})=$ ampere turns (nI) $\times$ area of coil (A) or $\mathrm{M}=\mathrm{nIA}$.
6. The unit of dipole moment is ampere meter $^{2}\left( A – m ^{2}\right)$
7. Magnetic dipole moment is a vector with direction from S pole to N pole or along direction of normal to planar area.

### Atoms as a Magnetic Dipole

In an atom electrons revolve around the nucleus. These moving electrons behave as small current loops. So atom possesses magnetic dipole moment and hence behaves as a magnetic dipole.

The angular momentum of electron due to orbital motion $L = m _{ e } vr$

The equivalent current due to orbital motion $I =-\frac{ e }{ T }=-\frac{ ev }{2 \pi r }$

–ve sign shows direction of current is opposite to direction of motion of electron.

Magnetic dipole moment $M=I A=-\frac{e v}{2 \pi r} \cdot \pi r^{2}=-\frac{e v r}{2}$

Using $L=m_{e}$ vr we have $\quad M=-\frac{e}{2 m_{e}} L$

In vector form $\overrightarrow{ M }=-\frac{ e }{2 m _{ e }} \overrightarrow{ L }$

The direction of magnetic dipole moment vector is opposite to angular momentum vector.

According to Bohr’s theory $L=\frac{n h}{2 \pi} \quad n=0, \quad 1, \quad 2 \ldots \ldots$

So $M=\left(\frac{e}{2 m_{e}}\right) \frac{n h}{2 \pi}=n\left(\frac{e h}{4 \pi m_{e}}\right)=n \mu_{B}$

Where $\mu_{ B }=\frac{ eh }{4 \pi m _{ e }}=\frac{\left(1.6 \times 10^{-19} C \right)\left(6.62 \times 10^{-34} Js \right)}{4 \times 3.14 \times\left(9.1 \times 10^{-31} kg \right)}=9.27 \times 10^{-24} Am ^{2}$ is called Bohr Magneton. This is natural unit of magnetic moment.