Hello Aspirant,

Congrats for getting good percentile. This year total 869010 students  attempted jee. You can calculate your approximate rank using :(100-percentile)*869010/100. That is 29,833.The chances for getting seat in NITC are below average. Last year cut offs for obc ncl of NITC are :

Civil:11631

CsE:3542

EEE:7482

ECE:4809

ME:7658

Aswin, NIT CALICUT offers admission to B.Tech as well as B. Arch courses through JEE MAIN PAPER1 AND PAPER2 respectively. The National Testing Agency , NTA conducts two papers viz. Jee Mains Paper 1 and Paper 2 for entrance to B.Tech and B.Arch respectively. In your case, you have not

Deepak, you can calculate your expected JEE Mains paper 2 rank for the January session using the following formula:

(100-your total percentile) x 112679/100

So with 94 percentile, your rank would be around 6760.

Also, if the number of students increases to 1.5 lakh in the April session, the formula

Hello,

The cut-off / minimum marks required to get admission in NIT for the year 2020 is still left to be declared and will come out after the JoSAA counselling. Till then, you can check out the previous year cut and also use the Career360 portal to check the score

Hi dear,

IIEST Shibpur is not national institute of technology right now. IIEST Shibpur takes admission on the the basis of jee main rank from last few YEAR but not a national institute of technology.You can check details here

https://www.google.co.in/url?q=https://www.careers360.com/university/indian-institute-of-engineering-science-and-technology-shibpur&sa=U&ved=2ahUKEwj69Pr70qbnAhUcyDgGHZvCA2EQFjAbegQIAhAB&usg=AOvVaw0cZ81qeproMLsEapRFp-Gb

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.

You can calculate your rank in JEE Main Paper 2 using the formula:Rank = (100-percentile)*(No. of students appeared for the examination)/100.

As the no. of students appeared for Paper 2 (Jan) are around 112679, your rank will be around 12600. However as the no. of students in April changes, your