The list of colleges for 1 lakh rank in JEE Main 2025 includes NIT Agartala, NIT Goa, NIT Sikkim, Sardar Vallabhbhai National Institute of Technology, and many others.The candidates who have qualified with a required cut off marks will be eligible to participate in the counselling process.
Table of Contents
Colleges for 1 lakh rank in JEE Main 2025 include the Indian Institute of Handloom Technology Salem; BITS, Deoghar; Institute of Engineering and Technology Dr. H. S. Gour University, etc. Students can expect admission to ECE, CSE and Energy Engineering branches with a 1 lakh rank in JEE Main 2025.
NTA administers the JEE Main exam twice a year. It is conducted for UG admissions in engineering, architecture, and planning programs. Students must meet the JEE Main cutoff criteria and have valid marks in the entrance exam. Even at a low rank in JEE Main 2025, many NITs, IIITs, GFTIs, and IIEST offer admissions to various UG courses. Thus, candidates must know about the list of colleges offering admission to 1 lakh rank and above.
List of Colleges for 1 Lakh Rank in JEE Main 2025
The allocation of colleges is done based on the JEE Main 2025 cutoff which will be released by NTA after the declaration of JEE Main 2025 result. Students must meet the cutoff criteria set by the college to seek admission to the colleges during the JEE Main counselling 2025.
Candidates will 1 lakh rank in JEE Main can expect admission to numerous NITs, IIITs and GFTIs across India. However, they might not get their desired branch. Check out the list of colleges for 1 lakh rank in JEE Main 2025:
|
Institute |
Course |
Closing Rank |
|
Biotechnology and Biochemical Engineering |
163769 |
|
|
Civil Engineering |
114874 |
|
|
Computer Science and Engineering |
162246 |
|
|
Civil Engineering |
126258 |
|
|
Chemical Engineering |
107655 |
|
|
Carpet and Textile Technology |
100912 |
|
|
Agricultural Engineering |
187883 |
|
|
National Institute of Food Technology Entrepreneurship and Management, Kundli |
Food Technology and Management |
105781 |
|
National Institute of Food Technology Entrepreneurship and Management, Thanjavur |
Food Technology |
100380 |
|
Institute of Engineering and Technology, Dr. H. S. Gour University. Sagar (A Central University) |
Dairy Engineering |
101252 |
|
Electrical and Electronics Engineering |
116152 |
|
|
Indian Institute of Handloom Technology, Salem |
Handloom and Textile Technology |
101502 |
|
Ghani Khan Institute of Engineering and Technology, Malda, West Bengal |
Computer Science and Engineering |
12264 |
Also Read: JEE Main Participating Institutes 2025
JEE Main 2025 Percentile vs Rank
The allocation of colleges is dependent on various factors including, the JEE marks, JEE Main rank, cutoff scores, availability of seats and more. The initial cutoff is released in the form of percentile. Candidates must know the JEE Main 2025 percentile vs rank comparison to anticipate their college allocation during the counselling. The following table provides a clear picture of JEE Main 2025 percentile vs rank:
|
JEE Main 2025 Percentile |
JEE Main 2025 Rank |
|
100 - 99.99989145 |
1 - 20 |
|
99.994681 - 99.997394 |
80 - 24 |
|
99.990990 - 99.994029 |
83 - 55 |
|
99.977205 - 99.988819 |
210 - 85 |
|
99.960163 - 99.975034 |
367 - 215 |
|
99.934980 - 99.956364 |
599 - 375 |
|
99.901113 - 99.928901 |
911 - 610 |
|
99.851616 - 99.893732 |
1367 - 920 |
|
99.795063 - 99.845212 |
1888 - 1375 |
|
99.710831 - 99.782472 |
2664 - 1900 |
|
99.597399 - 99.688579 |
3710 - 2700 |
|
99.456939 - 99.573193 |
5003- 3800 |
|
99.272084 - 99.431214 |
6706 - 5100 |
|
99.028614 - 99.239737 |
8949 - 6800 |
|
98.732389 - 98.990296 |
11678 - 9000 |
|
98.317414 - 98.666935 |
15501 - 11800 |
|
97.811260 - 98.254132 |
20164 - 15700 |
|
97.142937 - 97.685672 |
26321 - 20500 |
|
96.204550 - 96.978272 |
34966 - 26500 |
|
94.998594 - 96.064850 |
46076 - 35000 |
|
93.471231 - 94.749479 |
60147 - 46500 |
|
91.072128 - 93.152971 |
82249 - 61000 |
|
87.512225 - 90.702200 |
115045 - 83000 |
|
82.016062 - 86.907944 |
165679 - 117000 |
|
73.287808 - 80.982153 |
246089 - 166000 |
How to Calculate JEE Main 2025 Percentile?
To determine the percentile, divide the total number of candidates who appeared in the "Session" by the number of candidates whose raw score was equal to or lower than the candidate. Then, multiply the quotient by 100.
The following formula can be used to determine a candidate's percentile score:
JEE Main Percentile = 100 X (Total number of candidates who appeared in the "Session" divided by the number of candidates whose raw score was either less or equal to the candidate)
The following formula is used to determine each of the four JEE Main percentile scores:
- Mathematics Percentile
- Physics Percentile
- Chemistry Percentile
- Total Percentile
Factors Determining JEE Main Cutoff 2025
Individual institutes determine the admission cutoff each year, and it varies. Not all students can crack the cutoff. Listed below are the factors affecting the JEE Main cutoff 2025.
- Total number of candidates appearing for the JEE Main exam
- Number of seats available
- The difficulty level of the JEE Main exam
- Previous year JEE Main cutoff trends
Also Read: JEE Main 2025 Passing Marks
![Indian Institute of Technology, [IIT] Kanpur](https://media.getmyuni.com/azure/college-image/small/indian-institute-of-technology-iit-kanpur.jpg)
![Indian Institute of Technology, [IIT] Roorkee](https://media.getmyuni.com/azure/college-image/small/indian-institute-of-technology-iit-roorkee.jpg)
![Jawaharlal Nehru University, [JNU] New Delhi](https://media.getmyuni.com/azure/college-image/small/jawaharlal-nehru-university-jnu-new-delhi.jpg)
![University of Calcutta, [UC] Kolkata](https://media.getmyuni.com/azure/college-image/small/university-of-calcutta-uc-kolkata.jpg)
POST YOUR COMMENT