Institute of Technology Eligibility

The minimum requirement to be eligible for B.E/ B.Tech course at World Institute of Technology (WIT) is a candidate must have passed the 10 + 2 examination from a recognised board and have obtained a minimum of 45% marks in Physics, Mathematics & Either Chemistry, Computer Science or Biology. In addition, candidates need to have passed their JEE Main examination plus have a minimum passing percentile of 50. College also provides lateral entry admissions in Diploma with least rating of 50% of distinction. Total intake capacity for B.E/B.Tech is 540 students total and 25% fees is reserved for students from Haryana.

If you meet the eligibility requirements for the B.E/B.Tech  of the institute then that is the selection criteria. The minimum requirement is 10th, and 12th should pass with a minimum 45% . This is the basic criteria. You must have decent marks in JEE exam.

The eligibility criteria for BE/B.Tech is students must have passed in the Higher Secondary Examination with Physics, Chemistry, and Mathematics ( compulsory subjects.) With minimum 45%, you can get admission through entrance exams like JEE, other than this students belonging to the State of Maharashtra are required to secure a valid score in the State Common Entrance Test.

The eligibility criteria for B.E. / B.Tech at NIT - Narula Institute of Technology are as follows.Educational Qualification: Candidate should have completed 10+2 with a minimum of 45% marks in Physics and Mathematics along with any one of the following subjects: Chemistry, Biotechnology, Biology, Computer Science, or Technical Vocational subjects. For the reserved categories, the percentage required is 40%.Entrance Exam: For seeking admission to B.E. / B.Tech courses offered at the college, an applicant has to be eligible in either of JEE Main or WBJEE.