i am from Telangana,i have scored 95.67 percentile in jee mains 2020. my CRL is 47k and 1332 sc category rank , in which IITs and NITs can i get with cse branch???

It is quite difficult to recommend colleges for you. There is a chance that you get NIT Durgapur with this rank if the cutoff goes below previous year. If you think other than CSE, you have plenty of chances in various NITs like NIT Bhopal, NIT Calicut, NIT Delhi etc.

Check out this link for previous year NIT cutoffs.

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

It is advisable to use JEE MAIN COLLEGE PREDICTOR tool to know about your admission chances.

https://engineering.careers360.com/jee-main-college-predictor?icn=QnA&ici=qna_answer

Hello Manisha

Through JEE Mains, you can only get admission in NITs, IIITs, CFTIs, and other colleges which takes admission through JEE Mains marks. To get admission in IITs, you have to give JEE Advanced and you have to score a good rank to get into one out of 23 IITs. Now, with your category rank there are good chances that you get admission in NITs and that too with CSE branch. To check which colleges you can get you can use Careers360 college predictor tool. You can check your expected colleges at:

https://engineering.careers360.com/jee-main-college-predictor?ici=rp_result_Paper-1&icn=cp_input_form_Paper-1

The tool uses previous years opening and closing ranks of colleges. Also, as the opening and closing ranks changes every year, you can only get predicted colleges and not exact because the opening and closing rank for this year is yet to be announced by the institutes.

Hope this helps and all the best.

Hello aspirant

95.67 is quite good score.with SC category rank 1332 you have decent chances of getting seat in top NIT or IIT colleges.

SC cut off Jee main 2020 is 50.176025 percentile.

For College prediction following link s may help you

https://engineering.careers360.com/jee-main-cutoff--for-top-iits

https://engineering.careers360.com/jee-main-college-predictor?icn=QnA&ici=qna-answer