Amity University-Noida B.Tech Admissions 2026
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Hello Aditya,
As I can check previous year cut off trends for NIT Hamirpur, Yes, you can get admission in this NIT.
For OBC cut off is about to 9k to 12k.
You can check here also
https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-nit-hamirpur/amp
Thank You.
As I can check previous year cut off trends for NIT Hamirpur, Yes, you can get admission in this NIT.
For OBC cut off is about to 9k to 12k.
You can check here also
https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-nit-hamirpur/amp
Thank You.
Your Predicted All India Rank is 27539
You will get NIT Hamirpur CSE branch at the second round of counselling.
The fee structure is 3.64lakh for 4 year duration of course.
To know more about your chances of getting other colleges check out the college predictor link below:
https://engineering.careers360.com/jee-main-college-predictor
Best of Luck for futureSooraj,
The closing rank for NIT HAMIRPUR is
Chemical Engineering --- 88553
Civil Engineering. ---81441
Computer Science and Engineering-----30447
Electrical Engineering------77485
Electronics and Communication Engineering-----47571
Materials Science and Engineering-----91979
Mechanical Engineering-----72435
If you are ready to take any branch in NIT HAMIRPUR then 112 marks are very good you will easily get Mechanical, Material, Civil and Chemical Engineering.
Hope it helps
The closing rank for NIT HAMIRPUR is
Chemical Engineering --- 88553
Civil Engineering. ---81441
Computer Science and Engineering-----30447
Electrical Engineering------77485
Electronics and Communication Engineering-----47571
Materials Science and Engineering-----91979
Mechanical Engineering-----72435
If you are ready to take any branch in NIT HAMIRPUR then 112 marks are very good you will easily get Mechanical, Material, Civil and Chemical Engineering.
Hope it helps
Hi dear,
You have scored 100 marks in jee main that means your rank should be around 55k in general and 1700 in sc category.you can check here
https://engineering.careers360.com/jee-main-rank-predictor
With this rank you will definitely get NIT HAMIRPUR.
Hope it helps and congratulations
Hii Ishaan,
your rank acording to your NTA percentile is 21485.
and last year cutoff of NIT Hamirpur ws 49313.
so you can easiliy get admisson in this college.
for more information please go through the link-
https://engineering.careers360.com/articles/jee-main-cutoff-for-nit-hamirpur
hope you understand
thank you
Hello!Priti
You are in 1st year in NIT Hamirpur.Lot of syllabus of 1st year tally with jee mains syllabus.Math,chem and physics syllabus matches with 1st and 2nd semester syllabus.You will have lot of idea about syllabus and questions type of jee since you are in NIT.Just focus in jee mains preparation.Don't worry about semester syllabus and pointer.Pay your total attention to in crash course and practice as much possible.And memorize all formula.You will pass in semester because questions of semester repeat each year and just go through organizer.You will get lot idea about previous year paper of semester.Just little effort require in passing in semester examination.So focus in jee mains preparation.
GOOD LUCK!
You are in 1st year in NIT Hamirpur.Lot of syllabus of 1st year tally with jee mains syllabus.Math,chem and physics syllabus matches with 1st and 2nd semester syllabus.You will have lot of idea about syllabus and questions type of jee since you are in NIT.Just focus in jee mains preparation.Don't worry about semester syllabus and pointer.Pay your total attention to in crash course and practice as much possible.And memorize all formula.You will pass in semester because questions of semester repeat each year and just go through organizer.You will get lot idea about previous year paper of semester.Just little effort require in passing in semester examination.So focus in jee mains preparation.
GOOD LUCK!
Dear aspirant,
Yeah ofcourse you will get a seat in NIT Hamirpur based on previous year cutoff trends.According to your score and previous year statistics,Your rank will be nearly (100−95.45)×1,45368/100=6614.And last year cutoff for NIT Hamirpur for general category is 11202.Thus you have chances with NIT Hamirpur.And you have chances with NIT Raipur also. For previous year NIT cutoff, go to
https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b-planning/amp
Good luck
Hello, dear.
Well you predictable rank is below 20k , so there is chances of getting CSE in NIT-H , if you are in home quouta then almost sure you will get ,closing rank of home quota of 2018 is 49313 . Well visit the below link to see catogery wise last year cut of of nit-H.
https://engineering.careers360.com/articles/jee-main-cutoff-for-nit-hamirpur
hope this will help you!!
Feel free to ask more queries
Hello Siddharth. Welcome to Careers360. First of all, cogratulations for such a good score. I would like to tell you that the final results are yet to be released by NTA. These are just approximate ones. After the second exam of JEE Mains, You will get to know your final percentile. Cutoffs will also be released after the second exam results are out. So you have to wait till April to know the final results and cutoffs. For your reference, I am sharing a link with you which consists of previous years NITs cutoffs
https://www.google.co.in/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-top-nits/amp
Hope this answer helps you. Good luck.
Hi there.
The final cut off list will be declared by NTA after the April session of the JEE mains examination.
As for now, your probable rank can be calculated.
Now, A total of 874469 candidates wrote the exam. Since the percentile is calculated from a scale ranging from 100 to 0 for each session of examinees, the formula applies the total number of students in the January session which is 874469 /100 multiplied by the percentage of people who are ahead of you to arrive at the rank.
Therefore, the formula to calculate the JEE Main 2019 rank for January 2019 is
JEE Main 2019 rank (probable) = (100-80.9) X 874469 /100 = 167023(approximately)
This rank may inflate or deflate depending on the total number of candidates who write the examination. in the April session.
You could take a probable calculation and see where you stand. This means that you take a hypothetical number of the candidates who will appear in April. If you go by past trends, the figures are somewhat between 11 lakhs to 12 lakhs. Using this as a basis, use the NTA score and see which JEE Main rank you will get.
Using the same formula,
If NTA percentile score is 80.9, JEE Main rank will be (100-80.9) X 1200000/100 = 229200
Now, using this estimated calculation, you can check previous year cut off trends and think of an idea of the probable colleges.
Check the following links for previous year trends in cut off :
https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff/amp
For college predictor of JEE mains, check the link below:
https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-college-predictor/amp
Hope this helps.
Good luck!
The final cut off list will be declared by NTA after the April session of the JEE mains examination.
As for now, your probable rank can be calculated.
Now, A total of 874469 candidates wrote the exam. Since the percentile is calculated from a scale ranging from 100 to 0 for each session of examinees, the formula applies the total number of students in the January session which is 874469 /100 multiplied by the percentage of people who are ahead of you to arrive at the rank.
Therefore, the formula to calculate the JEE Main 2019 rank for January 2019 is
JEE Main 2019 rank (probable) = (100-80.9) X 874469 /100 = 167023(approximately)
This rank may inflate or deflate depending on the total number of candidates who write the examination. in the April session.
You could take a probable calculation and see where you stand. This means that you take a hypothetical number of the candidates who will appear in April. If you go by past trends, the figures are somewhat between 11 lakhs to 12 lakhs. Using this as a basis, use the NTA score and see which JEE Main rank you will get.
Using the same formula,
If NTA percentile score is 80.9, JEE Main rank will be (100-80.9) X 1200000/100 = 229200
Now, using this estimated calculation, you can check previous year cut off trends and think of an idea of the probable colleges.
Check the following links for previous year trends in cut off :
https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff/amp
For college predictor of JEE mains, check the link below:
https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-college-predictor/amp
Hope this helps.
Good luck!
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College Infrastructure
IIT Mandi boasts a stunning Himalayan campus with state-of-the-art labs, smart classrooms, and well-equipped research centers. Hostels are modern, Wi-Fi enabled, and eco-friendly. The campus is clean, sustainable, and ideal for focused learning.
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Posted on 21 Oct'25 by Anonymous
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Batch of 2028-01-01 | B.Tech Computer Science and Engineering
College Infrastructure
NIT Hamirpur has one of the most beautiful campuses in India, surrounded by hills and greenery. Modern labs, smart classrooms, a central library, and good hostels with Wi-Fi ensure a great learning environment.
Our college is one of the best nit.
College Infrastructure
This college has good infrastructure. It has great scenery in college, it is in the mountains. Building of each department is too good with proper management and all. Hostels also have better space management and mess systems for hostelers, along with night canteen facilities. Separate hostels are available for both boys and girls.
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