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JEE Main NTA Session 2 Online Form 2022
Post Update:- 06/06/2022 | 03:30 PM
The National Testing Agency (NTA) Invites Online Application Form For Joint Entrance Examination JEE (Main) Session-2 2022 For Taking Admission In B.E/ B.Tech, B. Arch And B. Planning Courses.
Interested Candidates Read the Full Notification Before Apply.
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News:- Session 2 (two) Registration for JEE(MAIN) 2022 is Started Now. News Date: 01-06-2022 |
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National Testing Agency Joint Entrance Examination (Main) Session-2 Application 2022 |
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Important Dates
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Application Fee
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Eligibility Criteria For JEE Main▶ Age Limit :- For appearing in the JEE (Main) - 2022, there is no age limit for the candidates. The candidates who have passed the class 12/equivalent examination in 2020, 2021, or appearing in 2022 irrespective of their age can appear in JEE (Main) - 2022 examination. However, the candidates may be required to fulfill the age criteria of the Institute(s) in which they are desirous of taking admission. ▶ List of Qualifying Examinations (QE) :- i. The final examination of the 10+2 system, conducted by any recognized Central/ State Board, such as Central Board of Secondary Education, New Delhi; Council for the Indian School Certificate Examinations, New Delhi; etc. ii. Intermediate or two-year Pre-University examination conducted by a recognized Board/ University. iii. Final examination of the two-year course of the Joint Services Wing of the National Defense Academy v. Senior Secondary School Examination conducted by the National Institute of Open Schooling with a minimum of five subjects v. Any Public School/ Board/ University examination in India or any foreign country recognized as equivalent to the 10+2 system by the Association of Indian Universities (AIU). vi. Higher Secondary Certificate Vocational Examination. vii. A Diploma recognized by AICTE or a State board of technical education of at least 3 years duration. viii. General Certificate Education (GCE) examination (London/Cambridge/Sri Lanka) at the Advanced (A) level. ix. High School Certificate Examination of the Cambridge University or International Baccalaureate Diploma of the International Baccalaureate Office, Geneva. ▶ Year of Appearance in Qualifying Examination :- Only those candidates who have passed the Class 12/equivalent examination in 2020, 2021, or those who are appearing in Class 12/equivalent examination in 2022, are eligible to appear in JEE (Main) - 2022. Candidates who passed the Class 12/equivalent examination in 2019 or before as well as those who will appear in such examination in 2023 or later are not eligible to appear in JEE (Main) - 2022. |
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Some Important Useful Links |
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The Joint Entrance Examination, JEE (Main) comprises two papers. Paper 1 is conducted for admission to Undergraduate Engineering Programs (B.E/B.Tech.) at NITs, IIITs, other Centrally Funded Technical Institutions (CFTIs), Institutions/Universities funded/recognized by participating State Governments. JEE (Main) is also an eligibility test for JEE (Advanced), which is conducted for admission to IITs. Paper 2 is conducted for admission to B. Arch and B. Planning courses in the Country. From the year 2021, Dr. APJ Abdul Kalam Technical University (AKTU), Lucknow and Madan Mohan Malviya University of Technology, Gorakhpur will be using the score sheet of JEE (Main) for admission to the B.Tech./M.Tech. (Integrated)/B.Arch. Courses in its affiliated colleges across the State.
The JEE (Main) - 2022 will be conducted in 02 (two) sessions for admissions in the next academic session.
JEE (Main) Selection Process
The process of Normalization is an established practice for comparing candidate scores across multi-session papers and is similar to those being adopted in other large educational selection tests conducted in India. For normalization across sections, NTA shall use the percentile equivalence.
Percentile Scores: Percentile scores are scores based on the relative performance of all those who appear for the examination. The marks obtained are transformed into a scale ranging from 100 to 0 for each session of examinees.
JEE (Main) Syllabus
Mathematics - 30 Questions With 100 Marks, Physics- 30 Questions With 100 Marks And Chemistry- 30 Questions With 100 Marks For Paper1: B.E./ B. Tech. in Computer Based Test (CBT) mode.
To cater to the decision of different Boards across the country regarding the reduction in the syllabus, the NTA has decided to provide choice in one Section of each Subject of Paper 1 and Part I of Paper 2A and 2B. However, the total number of questions to be attempted will remain the same (Physics -25, Chemistry – 25, and Mathematics – 25), wherever applicable.
NTA has decided that there will be a total of 90 Questions in the Question Paper and the candidates will be required to attempt 75 questions only
Joint Entrance Examination, JEE (Main) Admit Card
The e-Admit Card would be issued provisionally to the candidates through the NTA website: https://jeemain.nta.nic.in/, subject to the fulfillment of the eligibility conditions and receipt of the prescribed application fee by NTA
The candidate has to download the Admit Card from the NTA website. The candidate will appear in the examination at the given Centre on the date and shift/timing as indicated in his/her e-Admit Card.
How To Apply For JEE (Main) Exam
Candidates can apply for JEE (Main) - 2022 through the “Online” mode only. The Application Form in any other mode will not be accepted.
Submission of Application Form could be made by the candidate online through the NTA website: https://jeemain.nta.nic.in/
- Step 1: Registration Form: Register for the Online Application Form and note down the system-generated Application Number. The candidate should supply the required details while filling the Online Application Form and is also required to create PASSWORD and choose Security Question and enter his/her Answer. After successful submission of the personal details, an Application number will be generated and it will be used to complete the remaining Steps of the Application Form and will also be required for all future reference/correspondence. This application number will also be used/referred for both Sessions of JEE (Main) - 2022. For subsequent logins, the candidate will be able to login directly with the respective system-generated Application Number and created Password.
- Step 2: Application Form: The Candidates can log in with the system-generated Application Number and pre-created Password for completing the Application Form including filling up of personal details, applying for the Paper, choosing the Examination Cities, providing the details of Educational Qualifications, and uploading the images and documents (if any).
- Step 3: Fee Payment: After completing Step 1 and Step 2, the candidates have to pay the requisite examination fee. The fee can be submitted only online through Net Banking, Credit Card, Debit Card, UPI, or Paytm Services. Processing charges and GST as applicable are chargeable to the candidate (in addition to the examination fee) by the concerned Bank/Payment Gateway Integrator.
- Download, save and print a copy of the Confirmation Page of the Application Form (which would be downloadable only after successful remittance of fee) for future reference.
JEE (Main) Contact Details
- Phone No. : 01169227700, 011-40759000
- Email- N/A
- Address: NSIC-MDBP Building First Floor, Okhla Industrial Estate New Delhi-110020.
JEE (Main) FAQs
1. What Is The Full Form Of JEE (Main)?
-JEE (Main) Stands For Joint Entrance Examination JEE (Main).
2. What Is JEE (Main) 2022 Season 2 Application Last Date?
- Last Date For Filling Of Online Application Form Is 30 June 2022.
3. Who Can Apply For Joint Entrance Examination JEE (Main)?
- Only those candidates who have passed the Class 12/ equivalent examination.
4. How Much Is JEE (Main) Application Fee?
- For Single Paper SC/ ST/ PWD (Male & Female): Rs.325/- And General/ EWS/ OBC- NCL: Rs.650/-.
5. When Will JEE (Main) - 2022 Examination Be Conducted?
- Session- 2: 21, 22, 23, 24, 25, 26, 27, 28, 29, and 30 July 2022.
6. What Is JEE (Main) Helpline/ Contact Number?
- Phone No. - 01169227700, 011-40759000