Hey there, The JEE mains cutoff for 2018 is listed below :

General : 74

OBC-NCL : 45

SC : 29

ST : 24

If you want to pursue B.tech in CS branch from NIT Tiruchirapalli then you have to get a rank between 324 and 2120 as this was the opening and closing rank for this NIT.

Although you can apply for other NIT's which have comparatively higher opening ranks but try for the best one and prepare accordingly.

All the best!!

Hello Jayant!

Last year cuttoff for various reservation quota were:

1.General Quota =74

2.OBC-NCL=45

3.SC=29

4.ST=24

5.PWD=-4

These are previous year cutoff for various category .Cuttoff marks depends upon level of paper and number of participants.Go through link to Know more about cuttoff of 2017,2016.

Copy the link from answer board and paste it in your search engine.

https://www.google.com/url?sa=t&source=web&rct=j&url=https://engineering.careers360.com/articles/jee-main-cutoff/amp&ved=2ahUKEwjG1aCTi4PgAhXLSH0KHZ_2AXoQFjAAegQIBBAB&usg=AOvVaw0ay0p7xaI0Sc82xeEH0ED5&ampcf=1

You can ask if you have any other queries

GOOD LUCK!.

hello jayanta das,

i suggest you to go through this link to get the previous year cut-off's for nit's,

• https://engineering.careers360.com/articles/jee-main-cutoff-for-top-nits

all the best!

note:if any links available,copy and paste it in the url bar of your browser.

I Hope this information will be useful to you.

if you need additional information,let us know to help you.

Thank you!!!:)