How much percentile is required in jee mains to get cs in SVNIT for home state (obc) candidate?

Hello Sitanshu

Based on previous year data For OBC opening rank is 1100 and closing rank is 3500. So you must have percentile 99.95 or more than it to get admission in CSE branch of SNVIT with home state quota.

For more details about SNVIT do visit

https://engineering.careers360.com/articles/jee-main-cutoff-for-svnit-surat

Hello aspirants.

The minimum marks candidates need to score for admission to various courses at the institute are covered in JEE Main 2019 cutoff for SVNIT Surat.

*Candidates who meet the JEE Main cutoff for SVNIT Surat will be allowed to participate in further admission process like counselling.

*Alternately, candidates scoring below the cutoff will no longer be considered for admission.

*It must be noted that just meeting the cutoff of JEE Main for SVNIT is no assurance for admission, as it is only eligibility to move to the next stage of the admission process.

*To determine JEE Main cutoff for SVNIT Surat, the admission committee will take into consideration a host of factors like difficulty level ofJEE Mainexam, number of candidates applying for admission, and also cutoff trends of previous years.

*While the SVNIT Surat JEE Main cutoff of the present year will be available only after JoSAA counselling is concluded, by going through the cutoffs of previous years the candidate will be able to get a fair idea what the cutoff may look like this year.

*Candidates must note that the JEE Main cutoff 2019 for SVNIT Surat will be available category-wise for all the courses offered by the institute. Read the full article for more information about JEE Main Cutoff for SVNIT Surat.

*The admission authority is likely to release the cutoff of JEE Main 2019 for SVNIT Surat only after the exam is over.

*The SVNIT Surat JEE Main cutoff when released will contain the opening and closing ranks for admission to various courses offered by the institute.

*Also, the cutoff for SVNIT Surat will be category-specific and will be determined based on marks obtained by the candidate in the exam.JEE Main 2019will be conducted by NTA twice a year, once in January and then in April in computer-based mode.

All the best.

Thank you.

Hello sir

Well, it's very difficult as of now,to predict the cutoff for colleges of JEE mains and for jee advance based on percentile.You can first calculate your rank by rank=(100-percentile)*874469/100.With this rank,you can compare previous year's cutoff for different colleges and can predict some of the colleges you may be getting and also the advance cut off.But remember this is not the final rank.It will change based on the april session of mains.To further predict colleges visit https://engineering.careers360.com/jee-main-college-predictor  .Hope this helps.

Good Luck