The JEE Main Cut Off will be soon released by the National Testing Agency on the official website. The JEE Mains exam is conducted annually to test the students of their skills. The students who clear the JEE Mains and get cut-off according to the different colleges get admission in the B. Tech course. In this article, I have discussed the JEE Main Cut Off, the process to arrive at the normalized score, and opening and closing JEE Main Cut Off for different colleges.
JEE Main Cut Off
The NTA will soon announce the JEE Main Cut Off for different colleges and different courses. The students will get a formula to calculate the normalized score for themselves. The NTA has decided to give fair chance to all the students hence a fair competition can be ensured in the whole exam. The students can also check the opening and closing JEE Main Cut Off for top colleges of India. A lot of factors also affect the JEE Main Cut Off for the students which I have mentioned at the end of the post.
Summary of JEE Main Cut Off
| Summary | Details |
| Recruiting Organisation | National Testing Agency |
| Notified Exam | JEE Mains |
| Notification Date | 07 February 2020- 12 March 2020 |
| Application correction window | 04 July 2020- 12 July 2020 |
| Application Process | Online |
| Latest Exam Dates | Earlier- July 2020 Computer Based Test- 01 September 2020 to 20 September 2020 |
| Total colleges participating | Top 31 National Institute of Technology participate in the exam |
| Qualification | The candidate needs to have completed his Class XII with first division from a recognised educational board. The candidate needs to complete his class XII with Physics, Chemistry and Maths. |
| Selection process | Computer Based Test Personal Interview and Counselling |
| Official notification | Read here |
| Official Website | Click Here |
Previous year JEE Main Cut Off
The JEE Main cut-off will be declared for the student’s category wise once the exam is conducted by the National Testing Agency. The JEE Main Cut-off will be released for all the NITs in the month of October if the examination is conducted in September.
JEE Main Cut-0ff 2019
| Category | Expected Cut-Off |
| Common Rank List | 89.75 |
| Gen EWS | 78.82 |
| OBC – NCL | 74.31 |
| ST | 54.01 |
| SC | 44.04 |
Expected Cut-off of JEE Main
| Category | Expected Cut-Off |
| Common Rank List | 86 – 91 |
| Gen EWS | 77-82 |
| OBC – NCL | 71-76 |
| ST | 39-44 |
| SC | 51-55 |
Factors responsible for JEE Main Cut Off
There are various factors which affect the JEE Main Cut Off 2020 every year. There are a lot of coaching classes which have deciphered the factors that affect the cut-off count:
- The number of candidates appearing in the JEE Main entrance exam
- The total applications received for the different courses by the NTA
- The difficulty level of the JEE Mains entrance exam
- The total number of seats available with each college and each course
How to calculate JEE Main Percentile Cut-off?
This year JEE Main will be conducted on multiple days and shifts. Each day will have around 4 shifts, the students often gets confused regarding the marks. The reason for this is that some students will have easy paper while some will have tough [a[er. Hence, to ensure equality in the examination of JEE Main and providing a fair competition to all the students. Hence, to ensure this fairness, the NTA has devised a normalization process for the students. This process will help the students in calculating the percentile score of a candidate appearing in the exam:
- JEE Main Cut-off 2020 = (100- NTA percentile score ) X 869010 /100
- If NTA percentile score is 90.70, JEE Main rank will be (100-90.70 ) X 869010/100 = 80818
- Consider the NTA score of 80.60 then JEE Main 2020 rank = (100-80.60) X 869010 /100. The rank will be 168588.
Normalization Process of NTA
The NTA has released a normalization process to ensure a fair chance to every student who appear for the JEE Main exam in 2020. The students can calculate their normalized marks by applying the above formula. However, the students often get confused regarding the tie-up of the marks of JEE Mains. The National Testing Agency has decided to issue percentile scores up to 7 decimal places to further avoid ties. The students who get higher marks will be available for the selection in any of the colleges according to their ranks. However, to resolve any end of tie-up amongst the students the NTA has given a series of formulas which is as follows:
- Candidates with higher percentile in Mathematics
- Candidates with higher percentile in Physics
- Candidates with higher percentile in Chemistry
- The age of the candidate, the one who will have an age lower will be rejected.
Opening and Closing rank of top Technological College
| Institute | Opening Rank | Closing Rank |
| IIIT Allahabad | 109 | 14,757 |
| NIT Calicut | 470 | 17,731 |
| NIT Delhi | 707 | 13,423 |
| NIT Kurukshetra | 874 | 8,916 |
| Faculty of Engg. Haridwar | 1,170 | 36,709 |
| IIIT Design & Manuf., Tamil Nadu | 1,542 | 17,503 |
| NIT Goa | 2,574 | 18,383 |
| AB Vajpayee IIITM, Gwalior | 3,419 | 20,927 |
| NIT Jamshedpur | 3,318 | 13,041 |
| NIT Puducherry | 5,377 | 22,276 |
| BIT Deoghar | 8,290 | 39,292 |
| BIT PATNA | 16,775 | 28,684 |
| BIT Ranchi | 4,818 | 13,285 |
| Dr Bheem Rao Ambedkar NIT, Jalandhar | 4,303 | 22,540 |
| Indian Institute of Eng. Science & Tech., Shibpur | 7,982 | 30,886 |
| IIIT Nagpur | 15,794 | 8,135 |
| IIIT Pune | 6,060 | 6,558 |
| Indian Institute of Information Tech. Design & Manuf. Kurnool, AP (Andhra Pradesh) | 10,736 | 12,639 |
| IIIT Srirangam, Tiruchirappalli | 18,473 | 31,157 |
| NIT Durgapur | 7,376 | 21,206 |
| NIT Goa | 2,574 | 18,383 |
| NIT Hamirpur | 5,200 | 12,743 |
| NIT Manipur | 10,577 | 38,742 |
| NIT Meghalaya | 12,185 | 34,830 |
| NIT Mizoram | 8,929 | 34,658 |
| NIT Nagaland | 12,466 | 40,511 |
| NIT Patna | 6,861 | 18,903 |
| NIT Raipur | 7,280 | 17,664 |
JEE Main Counselling
Counselling Process
The JEE Main will hold the counselling process once it has released the merit list for the students. They can appear in the counselling centre on the date allotted. The students will be called for the counselling process based on their ranks. Post-registration the students will be allotted branches and campus on the basis of availability and ranks. The aforesaid authority starts the Document Verification the panel checks all the certificate and documents presented bt the candidate. After this, the students can fill the application form of the desired college and branch.
Documents Required for Counselling
There is a prescribed list of documents that the students have to carry in the JEE Main Counselling process after the final SRMJEEE Result list of the students are declared. The students who fail to carry any of these documents can lose their seats.
- Class X mark sheet and Passing certificate
- Class XII mark sheet and Passing certificate
- ID proof such as Adhaar card or pan card
- Graduation degree certificates
- Work experience or internship certificate attended previously
- Reservation category certificates (if applicable)
- JEE Main Application form
- Interview Letter
Useful Links For JEE Main Exam
