NIT Calicut All Questions

Hi there,

From my research, IIEST shibpur is recognized as an NIT by MHRD and becomes the 32nd NIT of India. College was established in the year 1856 and the second most oldest engineering college in India.collge offers 45 courses in 5 streams and accepts Jee mains, GATE, WBJEE, XAT, CAT, MAT etc.

Hope this will help you.

Thank you

Hi dear,

National institute of technology CALICUT offer engineering and architecture courses. You can check fee structure here for both courses

https://www.google.co.in/url?q=https://engineering.careers360.com/articles/nit-calicut-fee-structure&sa=U&ved=2ahUKEwj-9ZLQzKbnAhUb7nMBHVdjB8UQFnoECAoQAw&usg=AOvVaw3DSrTLGttUA9eLCN6oBRdf

Hope it helps

Hi dear,

National institute of technology Tiruchirapalli offer engineering and architecture courses. You can see fee details here

https://www.google.co.in/url?q=https://engineering.careers360.com/articles/nit-tiruchirappalli-fee-structure&sa=U&ved=2ahUKEwjsifz8y6bnAhXt63MBHZihANYQFjAQegQICRAB&usg=AOvVaw02ILjrp5HQRBU5GNBeZo10

Hope it helps.

Hi dear,

You have scored 88.82 percentile in JEE main paper 2 that means your rank should be around 12597. This is not your final rank.Your final rank will depend on number of student appear in April month exam. You can also calculate your rank as

Rank=(100-your percentile)×112679÷100

With this rank you have low chance to get NIT CALICUT. You can check cutoff of all college here

https://www.google.co.in/url?q=https://engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b-planning&sa=U&ved=2ahUKEwjI3Nqsk6TnAhXUYisKHdLTD88QFjAAegQICRAB&usg=AOvVaw3NUYS-M5KQ_IMws3rLkS0e

Hope it helps

Hello ashish ,
As you have secured 99.5 percentile in the jee mains  paper it is very good .
To calculate the rank for B.Arch:
100 - your total percentile score is the percentage of people ahead of you. 112679 are the number of candidates appeared for January exam.
So rank formula is (100 - your total score) X 112679 /100 .
Your estimated rank will be 563 crl rank .
Your satisfying the cut off of the NIT calicut in b.arch
You can easily grab a seat in b.arch of NIT calicut
All the best

Hello,
Your expected rank will be around 1205.You can calculate your expected rank using the formula given below
(100-your total percentile)*112679/100
112679 is the total number of candidates appeared for JEE main paper 2 in January session.
By Calculation it is (100-98.93)*112679/100=1205

Considering 2019 cut off a candidate with rank 1668 belonging to General category and holding a domicile of Kerala secured a seat in NIT, Calicut and a candidate with rank 2468 belonging to general category and holding a domicile of Kerala seacured a seat in NIT, Tiruchirapalli.
A candidate with rank 1362 belonging to general category from other state secured a seat in NIT, Calicut and a candidate with rank 1075 belonging to General category from other state secured a seat in NIT, Tiruchirapalli.

So, you may have chances of getting a seat in NIT, Calicut and NIT, Trichy if you hold domicile of Kerala.However, cut off is subjected to change it may increase or decrease slightly.

You can also predict the colleges which you may get using our college predictor
https://engineering.careers360.com/jee-main-college-predictor

Hello!!

To calculate your JEE main rank,you can use this formula ((100-your total percentile)*869010/100).This formula will give your general category rank(Common list rank).And your rank is approximately 39105

Since you are a not domicile of Kerala,you cannot avail the home state quota of NIT Calicut.You will come under the other state quota.The previous year's cutoff of NIT Calicut for different b.tech branches and categories is given in the below given website.

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

To know about all the possibilities of your admission into various colleges,visit the below given website.You will have to enter the details asked such as your percentile,category,home state etc..and the possibilities will be predicted.

https://engineering.careers360.com/jee-main-college-predictor

Hey there hope you're doing well.
Dear aspirant, congratulation on scoring a good percentile in the JEE MAIN examination paper 2 B ARCH. As you say your percentile is 97 you can calculate your rank by the formula

(100 - your percentile score) 112679/100
Also I m providing you with our college predictor link .Go through this to get a better idea.
Link- https://engineering.careers360.com/jee-main-architecture-college-predictor.
Thank you hope this helps you!

Hello!

Congratulations! With such a high percentile, I believe that you can get into NIT, Calicut!

To predict if you can get into your desired college, first calculate your rank!

To calculate the rank for B.Arch:

100 - your total percentile score is the percentage  of people ahead of you. 112679 are the number of candidates appeared for January exam.

So rank formula is (100 - your total score) X 112679 /100.

To check about the colleges:

Refer to the previous year cutoff after calculating the rank from the formula mentioned above.

https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b- planning/amp

Use our college predictor tool too:

https://engineering.careers360.com/jee-main-architecture-college-predictor

Hello Sourabh

Link: https://engineering.careers360.com/articles/jee-main-cutoff-for-manit-bhopal

Maulana Azad Institute of Technology, Bhopal offers 7 B.Tech courses for the students namely Chemical Engineering, Civil Engineering, Computer Science and Engineering, Electrical Engineering, Electronics and Communication Engineering, Mechanical Engineering and Materials Science and Metallurgical Engineering. The institute has a total of 817 seats for these engineering programmes.