can I get nit college with 88 percentile in jee main in obc category

It depends upon the JEE Main and JEE Advance cut offs, If You manage to qualify the exam by achieving the cut off, You can apply for JOSAA Counselling. It should be noted the cut off for EWS/SC/ST/FEMALES/OBC category is lower than the general category

If you manage to qualify JEE Main Cutoff, You would be eligible for JEE Advance which is the intake exam for IITs as mentioned above.


However It is usually advised to fill the choices as per the preference of colleges and departments, for example CSE in NIT Trichy is your priority followed by CSE in NIT Warangal, you can fill CSE at NIT Trichy at serial number 1 and CSE at NIT Warangal at serial number 2 and so on and so forth. However to get any NIT, you must have qualified JEE Mains and to get any IIT, you must have qualifies JEE Advanced

But the previous cut offs of colleges should be taken in consideration, it is better to research over the cut off of colleges then make a preference list of colleges and fill accordingly

For the college preferences and the courses provided by these colleges and to know which colleges participate in the JOSAA counselling, please go to the below mentioned link: https://engineering.careers360.com/articles/josaa-participating-institutes/amp

For college prediction as per your marks and rank, please go through the below mentioned link: https://www.careers360.com/college-predictor

Rest for information on any particular college, you can dig their official site for better information

Dear Aspirant

There are less chances to get nit at this rank because the Candidates who acheive percentile above 98 are only have chances to get a seat in nit .
A formula is given by National Testing agency to calculate rank the formula is no of candidates apperared in exam/100 *(100- Nta score ).
The no of Candidates apperared in January session was 869010.So your rank will be 104281.
But your rank can vary if the no of candidates in April session changes to 12 lakh then your rank will calculated accordingly.
Thanks
All the very best