Candidates will be awarded four marks for each correct answer, and one mark will be deducted for an incorrect answer. The total score will be calculated using the given formula.

Total Marks = (Correct Answers X 4) - (Incorrect MCQs X 1).

The NTA is likely to announce the JEE Main session  2 BE/BTech paper results and All India Rank (AIR) list in the evening today, April 20, 2026. Though, no official confirmation has been provided by the agency. 

The NTA will soon release the final answer key along with scorecard for the session 2 BE/BTech paper, likely in the evening. 

The evaluation will be done using the final answer key, and the raw (actual) marks obtained by a candidate will be calculated. In the Session 2 result declaration, the calculated NTA scores for all the registered candidates will comprise the two NTA scores for each of the three subjects (Mathematics, Physics and Chemistry).

Candidates will be awarded four marks for each correct answer, and one mark will be deducted for an incorrect answer. The total score will be calculated using the given formula.

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The JEE Main exam is conducted in multiple shifts with varying difficulty levels. So to ensure fairness across all shifts, NTA use normalisation procedure to calculate the percentile.

Percentile formula :

Percentile = (Number of candidates with score <= yours / Total candidates in session) x 100

The difficulty level of shifts is balanced through normalisation

Yes, NTA is most likely to release the BE/BTech paper scorecards today, April 20, 2026, as per its earlier announcement.

Based on previous years' trends, the qualifying cut-off for the General category is expected to be around 93-95 percentile, while for GEN-EWS, it may range between 80-82 percentile.

Category-wise expected cut-offs based on previous trends are as follows:

The JEE Main scorecard of candidates will be generated as per their names in Class 10th certificate or marksheet.

NTA will release the JEE Main session 2 paper 1 scorecards shortly. Direct link to download will be notified here. 

The NTA will most likely release the BE/BTech paper 1 final answer key first, followed by scorecards. 

The NTA will activate two to three scorecard download links for the Session 2 B.E./B.Tech paper to streamline access, allowing students to download their scores smoothly while reducing heavy traffic on a single link. Nearly 11 lakh students are awaiting the results.

Students will be able to download their session 2 BE/BTech paper results via jeemain.nta.nic.in using their application number and password. 

The JEE Main scorecard of candidates will be generated as per their names in Class 10th certificate or marksheet. A copy of the final NTA score of JEE Main 2026 will be sent to the registered e-mail address of the candidates.

The JEE Main 2026 final marksheet for all candidates who appeared in Session 1 and Session 2 will be displayed on the official website. The scorecard will contain the following information:

Based on past-year trends, the NTA is expected to release the Session 2 results in the evening.

According to the official document, the total NTA scores of candidates in both sessions will be merged for the compilation of the JEE Main 2026 result and preparation of the final merit list or the JEE Main 2026 ranking. Candidates who have appeared in both sessions of the JEE Main must note that the best of the two NTA Scores in total (and not in individual subject) will be considered for the final rank list. 

The Paper 1 scorecard will display the NTA scores for each of the three subjects (Mathematics, Physics and Chemistry) and the total in Paper 1 (BE/BTech) of JEE Main 2026.

Visit the official website, jeemain.nta.nic.in.

Click on "JEE (Main) Session 2 BE/BTech Paper 1 Result 2026" under the "Public Notices" section.

Enter your application number and password.

Your scorecard will appear on the screen.

Download and save it for future reference.

The JEE Main exam is conducted in multiple shifts with varying difficulty levels. So to ensure fairness across all shifts, NTA use normalisation procedure to calculate the percentile.

Percentile formula :

Percentile = (Number of candidates with score <= yours / Total candidates in session) x 100

The difficulty level of shifts is balanced through normalisation

Candidates must note that there will be no re-evaluation or re-checking of their scores, and no correspondence regarding the same will be entertained, the NTA has said. Those who were found using Unfair Means Practices (UMP) during the examination will not have their results issued.

Based on previous years' trends, the qualifying cut-off for the General category is expected to be around 93-95 percentile, while for GEN-EWS, it may range between 80-82 percentile.

Category-wise expected cut-offs based on previous trends are as follows:

Yes, NTA is most likely to release the BE/BTech paper scorecards today, April 20, 2026, as per its earlier announcement. 

11.06 lakh students await their JEE Main April session paper 1 result, which is most likely to be announced today, April 20, 2026.

The National Testing Agency (NTA) is most likely to release the JEE Main session 2 scorecard today, April 20, 2026.