NIT Patna All Questions

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%

There is no such test being conducted, the students can apply for scholarship on the basis of 10+2 marks being above 75%

These scholarships are provided by MHRD or other such NGOs

Scholarship can also be availed on the basis of minority

Hello,

There are eligibility criteria for placements that depends on company to company. One of the criteria is that the student should not have any backlog it is just an example. There are such different criteria  So the remaining students failed the eligibility criteria thats why there are only 80 students eligible for campus placement

Hello,

Try to score 80% or above marks to get an admission int NIT. The cutoff for NIT patna  in OBC category is very low so apply through general category.

for top branch like CSE the general category rank is :- female home state 24847 and for other state :- 19565. for obc category home state :-8392     other state:-8146

for more details and cutoff for different NITs campus and branch please visit the link given below:- https://engineering.careers360.com/articles/nits-cutoff-for-top-btech-branches

Hope it helps!

Hi there,

It is very difficult to predict the collage or your ranking. You can still calculate your rank by 100- total score multiplied by 869010/100 and if the number of candidates are increasing then multiply by 1200000/100.

Your predicted rank is 373,674.

Some of the colleges you can get admitted are

• G.H Raisoni Institute of Engineering
• D.R D.Y Patil Institute of Technology
• Jawaharlal Nehru Institute of Technology, Aurangabad
• KDK college of Engineering
• Bajaj Institute of Technology
• SPM Narhe Technical campus

With 57% in Jee mains it is very difficult to get admitted to any NIT even in ST category. You can definetly improve your score in Jee mains of April which starts its registration in February 7th.

All the best for your bright future, Thank you