NIT Patna All Questions

Hello Aspirant, If you got 88.91 percentile in JEE Main then your rank after JEE Main April will be around 1.3 lakhs,Now last year closing Cutoff rank for NIT Patna for SC Home state candidate


Civil Engineering 131159

Computer Science and Engineering 89441

Electrical Engineering 147345

Electronics and Communication Engineering 145631

Mechanical Engineering 140797


So, from the 2019 cutoff, we can say that you will get NIT Patna as being a SC Home state candidate.

Hello,

Any percentile around 98  is decent and have probability of getting good branches in NIT Patna. If you score around 98 percentile then your rank will be around 18000-19000 and you will have chances of getting all top branches in NIT Patna. You can get Branches like Computer Science and Electronics and Communication through OBC Category, Home State in NIT Patna

click at this link to know more about cutoff - https://www.careers360.com/university/national-institute-of-technology-patna/cut-off?college_id=143&exam_sub_exam_id=2&counselling_id=548&category_of_admission_id=3&caste_id=3&gender_id=1

Hope this was helpful

Hi Ayush,

It is not possible to calculate the exact rank from given percentile so there is a formula for calculating near about rank which is given as below

(100 - your percentile) *869010/100

According to this formula your rank would be around 15,700. According to the last year's trend there are little bit chances that you may get CSE branch in NIT Patna.

I would suggest you to appear in second attempt of JEE-Main exam in month of April. You should score above 99 percentile to get confirm chances of CSE branch in NIT Patna.

You can use JEE-Main College Predictor to predict the college according to your percentile and category.

I hope this answer helps you.

All The Best!!

Hello Aspirant.

As you've mentioned that you've scored 74.02 in JEE Main examination 2020. Then your chances of getting into NIT Patna are quite difficult but still if you're able to get into NIT Patna then you might face a problem of not getting the demanded branch( If it's CS or IT) as your rank won't match upto that level.

Best of luck !

Hello,

Your expected rank will be around 22759.You can calculate your expected rank using the formula (100-your total percentile)*869010/100

where 869010 is the total number of candidates appeared in January session.In April session 1200000 candidates are expected to appear then you can calculate your expected rank using the formula (100-your total percentile)*1200000/100.

Now, your category rank will be in between 5514-7164.Considering 2019 cut off trends you may get seat in the following branches

In NIT, Agartala

• Biotechnology and Biochemical Engineering
• Chemical Engineering
• Civil Engineering
• Production Engineering
• CSE
• EEE
• ECE
• ME
• EIE

In NIT, Silchar

• Civil Engineering
• EEE
• ME
• EIE

In NIT, Patna

• Chemical engineering
• EEE
• ECE
• ME

These are the branches in which you may get seat.However, cut off is subjected to change it may increase or decrease slightly.

Hello Aspirant,

NIT patna has not released its cut off yet. However The following parameters will be taken into account by the official authorities for determining the NIT Patna cutoff for JEE Main.

• Difficulty level of JEE Main 2020
• Cutoff trends of the previous years
• Total number of candidates applying
• Total number of seats available

However, as per previous year cut off, it is little difficult for you to get into the college. You can check the past year cut off details for CS from the following link:

https://engineering.careers360.com/articles/jee-main-cutoff-for-nit-patna

All the best

Hello student you can easily calculate and get your rank by using this formula
(100-your total percentile) 869010 /100.
In your case you rank must be around. 88,812.

Being a obc category candidate its difficult to get admission in nit patna as we all know that seats are very limited and admission is directly based on merit rank.

If you want to appear in jee main April exam, application shall start from 07th of February and shall remain open till 07th of March 2020.

Feel free to comment if you have any doubt
Good luck

Hello,

yes you are eligible as your parents must be a resident of bihar and you are a minor so you will get the domicile certificate.

there are some rules by bihar government to be eligible for domicile certificate Those are:-

1. applicant must have lived for past 3 years
2. applicant must have owned any property
3. applicant must have name in voter list
4. if minor then apply using parents residence basis

so you go in minor category. if your parent satisfy any one of above 3 then you can apply for domicile certificate and use home state quota for admission.

here is the link for NIT campus cutoff:-

https://engineering.careers360.com/articles/nits-cutoff-for-top-btech-branches

Hope it helps!

Hello Aspirant.
Congratulations for your success.

Getting any College all depend on your rank and preference of College by other people.
Predicting exact rank is difficult but we have formula to calculate the expected rank with you percentile.
(100-your percentile)*869010/100
And is found to be 43450.
With this expected rank it seems to have less chance in getting nitp as per last year scenario.

In NITP you may get integrated Msc course.

Hope this helps you

NIT Patna evaluates you on the basis of your GATE scores and then there is CCMT councling so there are no such entrance examinations

make sure your 10+2 overall score is above 75%