Updated on Apr 3, 2024 23:23 IST
JEE Main Cut off 2024 is the minimum mark required by the candidate in the exam to be eligible for admission in an institute. Check all the details related to the expected JEE Main 2024 qualifying marks for admission to NITs. Admission to NITs will be based on the All India Rank (AIR) in JEE Mains.

JEE Main cut-off 2024 for NITs will be released by JoSAA after the announcement of JEE Main 2021 results. Read the article below for all the BTech admissions-related updates offered by the National Institute of Technology (NIT).

JEE Main cut off for NITs depends on the marks obtained by the candidate in the JEE Main entrance exam and All India Ranks (AIR). The cutoff list varies as per the category and courses offered under the NITs and the seat allotment will be purely based on JEE Main cut off.

Admission to BTech courses in the National Institute of Technology (NIT) is completely based on qualifying JEE Main entrance exam. According to the NTA eligibility criteria, a candidate needs to score a minimum 75 percent aggregate score in the class 12 board examination or he/she must be among the top 20 percentile in 10+2 or any equivalent examination for admissions at the NITs and also for the other institutes of national importance such as IITs/IIITs and CFTIs.

For SC/ST candidates minimum of 65 percent required in +2 class by a recognised board or equivalent examination as per the relaxation norms for reserved category. Aspirants who will qualify the JEE Main or JEE Advanced have to register for JoSAA 2021 for further admission process for various renowned engineering institutes across India. The Joint Seat Allocation Authority (JoSAA) is responsible for admissions in 23 IITs, 31 NITs, 25 IIITs, and 28 Other- Government Funded Technical Institutes (Other-GFTIs). To know more read the article below.

JEE Main Question Paper PDF 2024-2013 JEE Mains 2024 Syllabus PDF 
How to Prepare for JEE Main 2024 in 1 Month JEE Main Sample Papers 2024
Table of contents
  • NIT 2024 Cut Offs for BTech Admission
  • Expected JEE Main 2024 Qualifying Marks for NITs
  • List top NITs in India

NIT 2024 Cut Offs for BTech Admission

  • Candidates need to qualify specific entrance exams to get admission to various UG Engineering courses under NITs.
  • For UG or BTech courses, candidates are admitted through JEE Main exam.
  • JoSAA will release the category-wise and round-wise opening and closing cut off/ranks for admission to BTech, Mech and dual degree courses in NITs. 
  • Candidates who qualify the JEE Main exam will have to register and participate in the JoSAA counselling cum seat allotment process. 
  • JoSAA (Joint Seat Allocation Authority) conducts a joint counselling session for candidates qualifying JEE Main and JEE Advanced exam.
  • JEE Main Cut off depends on several factors such as:
    • Total number of candidates appearing in the JEE Main entrance exam.
    • Total candidates applied for the admission in UG or PG degree courses in NITs.
    • Total seat intake and difficulty level of the examination.
Q:   Is 75 percent marks in class 12 required for JEE Mains?
A:

Candidates need not have 75 percent marks in class 12 for appearing in JEE Main. As per the official brochure released by the NTA, there is no minimum percentage required for appearing in the JEE Main. The requirement of 75 percent marks need not be associated with appearing in the JEE Main exam.

It is to be understood that 75 percent marks is required for admission to institutes which include NITs, IIITs, GFTIs and IITs. Many institutes offer admission at 60% and even less, along with JEE Main. Candidates must clearly understand that the requirement of 75 percent marks in class 12 is only for taking admission in the institutes.

Q:   What is the cutoff for BTech in Computer Science and Engineering at NIT Hamirpur for other than General categories?
A:

The last year 2022 cutoff ranks for BTech in Computer Science and Engineering at NIT Hamirpur - National Institute of Technology other than General category are - 17445 for OBC, 1663 for SC, 578 for ST, 16770 for EWS, 187395 for PWD. While the admissions are based on the scores/rank obtained in JEE exam, the last cut offs also change a lot depending on All India quota or Home State. It also varies if the applicant is a female.

Q:   What is NIT Rourkela cutoff for BTech?
A:

NIT Rourkela cutoff 2024 has been released across various specialisations in opening and closing ranks. NIT Rourkela BTech cutoff 2024 ranged between 2680 and 35593 for the General AI category candidates, wherein BTech in CSE is the most competitive course. The easiest BTech specialisation to secure admission in is B.Tech. in Ceramic Engineering + M.Tech. in Industrial Ceramic Engineering. For a detailed NIT Rourkela branch-wise cutoff 2024, candidates can refer to the table below for the Round 1 cutoff. The cutoff mentioned below is for the General category under the AI Quota.

Course2024
B.Tech. in Biomedical Engineering28053
B.Tech. in Ceramic Engineering34492
B.Tech. in Chemical Engineering16708
B.Tech. in Civil Engineering23425
B.Tech. in Computer Science and Engineering2680
B.Tech. in Electrical Engineering8770
B.Tech. in Electronics and Communication Engineering5053
B.Tech. in Mechanical Engineering13682
B.Tech. in Metallurgical and Materials Engineering27499
B.Tech. in Mining Engineering31497
B.Tech. in Biotechnology27533
B.Tech. in Electronics and Instrumentation Engineering8646
B.Tech. in Industrial Design27601
B.Tech. + M.Tech. in Chemical Engineering18843
B.Tech. in Ceramic Engineering + M.Tech. in Industrial Ceramic Engineering35593
B.Tech. + M.Tech. in Metallurgical and Materials Engineering33676
B.Tech. + M.Tech. in Mining Engineering32461
Bachelor of Architecture (B.Arch.)439
B.Tech. in Food Process Engineering34726
B.Tech. in Artificial Intelligence3880

Expected JEE Main 2024 Qualifying Marks for NITs

Below is the list of expected JEE Main Qualifying Marks for admission to NITs based on the previous year trends:

Category

Minimum Percentile 

General

78-80

OBC – NCL

73- 76

SC

54- 57

ST

43- 46

PwD

0- 2

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