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JEE Main 2026 Marking Scheme- The National Testing Agency (NTA) has released marking scheme for JEE Main 2026 on the official website, jeemain.nta.ac.in. Candidates preparing for the JEE Main 2026 exam must clearly understand the marking scheme and exam pattern to maximize their scores. The JEE Mains marking scheme 2026 explains the number of questions, marks distribution, and negative marking rules for each subject. The authority will conduct the JEE Main exam in two sessions. Session 1 exam will be held from January 21 to 29, 2026 and Session 2 is scheduled for April 2 to 9, 2026.
For JEE Main 2026 session 2, candidates who passed the Class 12 or equivalent examination in 2023 or earlier are not eligible to appear for JEE Main 2026 session 2. The passing year is counted as the year in which a candidate is declared ‘pass’ for the first time in Class 12. Subsequent attempts, subject additions, or improvement exams will not be considered as a fresh passing year for eligibility purposes.
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There will be a total of 75 questions of 300 marks in JEE Main 2026. Each question will have +4 marks and -1 will be deducted for incorrect response. The complete details of the updated JEE Mains marking scheme is updated on this page. Along with the JEE marking scheme, candidates should also be aware of all topics listed in JEE Main syllabus 2026.
Below are the key details of the marking scheme for JEE Main 2026:
| Particulars | Details |
|---|---|
Total Number of Questions | 75 (25 Questions in each section) |
Questions to be Attempted | 75 (25 per subject) |
Marks for Correct Answer | +4 |
Marks for Wrong Attempt | -1 (only for MCQs) |
Marks for Unattempted Question | 0 |
Question Types |
|
Total Marks | 300 |
The JEE Main 2026 paper will consist of two sections in each subject:
Section A (MCQs) : All Compulsory, with negative marking.
Section B (Numerical Value Questions) : All Compulsory, with no negative marking.
Subject | Physics | Chemistry | Mathematics |
|---|---|---|---|
Section A (MCQs) : All Compulsory | 20 | 20 | 20 |
Section (Numerical) | 05 | 05 | 05 |
Total Number of Questions | 25 | 25 | 25 |
Marks Distribution | 25 x 4 = 100 | 25 x 4 = 100 | 25 x 4 = 100 |
Total Marks | 300 | ||
Candidates should also understand the overall JEE Main 2026 exam pattern before preparation:
| Particulars | Details |
|---|---|
Exam Mode | Computer Based Test (CBT) |
Duration | 3 Hours (180 Minutes) |
Language | English, Hindi, and Regional Languages (as per choice) |
Type of Questions | MCQs & Numerical Value Questions |
Total Marks | 300 (Paper 1) |
Type of Questions | MCQs & Numerical Value Questions |
Each correct answer will award +4 marks.
Each incorrect MCQ will deduct -1 mark.
No negative marking for Numerical Value Questions.
In Section B, candidates need to attempt only 5 out of 10 numerical questions in each subject.
The total marks for JEE Main Paper 1 (B.E/B.Tech) is 300 marks.
Frequently Asked Questions (FAQs)
In JEE Main 2026, each correct answer gives +4 marks, while every wrong MCQ deducts 1 mark. Numerical questions have no negative marking.
Candidates have to attempt 75 questions in total – 25 each from Physics, Chemistry, and Mathematics.
The JEE Main 2026 Paper 1 will be for 300 marks with a duration of three hours.
On Question asked by student community
JEE Main 28 Jan shift 2 exam will end at 6 PM. The complete analysis and memory-based questions will solution will be updated in the below article. Keep checking the page-
Math and Chemistry was difficult and Physics was moderate. The complete analysis is available here- https://engineering.careers360.com/articles/jee-main-2026-january-28-shift-1-question-paper-with-solutions-pdf
You can also check the memory-based questions and detailed solutions for JEE Main Jan 28 shift 1 paper.
Hi Lucky,
Please refer to this link and you can download the free pdf.
HI Manisha Maharana
You can download the JEE Mains 10 Free Mock Test with Detailed solutions. Its a feely downloadable pdf.
https://engineering.careers360.com/download/sample-papers/jee-main-10-full-mock-test-and-explanations-pdf
Also, you can check ad attemp the online mock test on our platform.
https://learn.careers360.com/test-series-jee-main-free-mock-test/
A general equation of a circle is
$
x^2+y^2+2 g x+2 f y+c=0
$
Since it passes through $(0,0)$,
$
c=0
$
So the equation becomes
$
x^2+y^2+2 g x+2 f y=0
$
It cuts the x -axis at ( $a, 0$ ).
Substituting:
$
a^2+2 g a=0
$
$
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