What rank is required for nsit CSE?

All the mentioned ranks are JEE (Main) 2019 All India CRL Ranks….NSUT Delhi (Formerly NSIT Delhi) Cutoff 2019.

Round	Branch	Max JEE(M) Rank
2	Computer Engineering	3818
2	Computer Science and Engineering with specialization in Artificial lntelligence	3038
2	Electrical Engineering	23903
2	Electronics and Communication Engineering	15632

What rank is required for CSE in DTU?

Here you will get DTU Delhi 2019 shorts, from first slice off to last cut off for every one of the 2 departments/universities and their courses. For CSE cutoff rank is 11564 in the 2019 where as in 2018 cutoff rank is 9194.

What is the cutoff for NSIT Delhi?

NSUT Delhi (Formerly NSIT Delhi) Cutoff 2020

Round	Branch	Max JEE(M) Rank
1	Computer Science and Engineering with specialization in Data Science	1211
1	Electrical Engineering	3842
1	Electronics and Communication Engineering	2985
1	Electronics and Communication Engineering with specialization in Internet of Things	3248

What is the minimum marks to get admission in NIT through JEE?

The minimum marks to get admission in NIT through JEE Main 2020 for candidates belonging to the General Category were 89.7 marks and for OBC-NCL category candidates were 75.3 marks.

How many candidates score 100 percentile in JEE Main 2021 Phase 3?

9 August, 2021 : 17 candidates score 100 percentile in JEE Main 2021 Phase 3. Read More To address all such questions, we have prepared this article on JEE Main Rank Vs Marks Vs Percentile based on previous years’ analysis.

What is clearing JEE Main cut off marks vs rank?

Clearing JEE Main cut off marks enables students to apply for admission in NITs, IIITs and GFTIs. As the analysis of JEE Main Marks vs Rank for 2020 is yet to be released, the students can check previous years JEE Main Marks Vs Rank for references.

How many students have secured more than 240 marks in JEE Mains?

If Student 1 with 240 marks is on Rank 1, that means there are no students above him who have secured more than 240 in JEE Main. Including Student 1 and other 199 students below him/her, the total number of candidates appeared for the test is 200. So, by putting in the numbers in formula: 200 (students who secured less than or equal to Student 1)