Indian Institute of Technology (IIT) Roorkee, has released a notification on its official website regarding the upcoming JEE Advanced 2019 examination. As per the official notice, the top 2,24,000 candidates who appeared for the JEE Main 2019 examination are eligible for JEE Advanced.
However, the total number of candidates may be slightly higher than 2,24,000 in the presence of ‘tied’ ranks/scores in each category.
In addition, the notice also contains details of eligibility, important dates, registration exam fee, registration process and other important instructions.
Eligibility Criteria for JEE Advanced 2019
According to the notice available on the website, candidates should meet the following eligibility criteria to appear for the entrance examination:
— Should be among the top 2,24,000 candidates in Paper-1 of JEE Main 2019.
— Should be born on or after 1 October, 1994. Five years relaxation is given to SC/ST and PwD category candidates.
— Should not have appeared in JEE Advanced in 2017 or earlier.
— Should have appeared in class XII examination for the first time in 2018 or 2019.
— Should not have accepted admission at any of the IITs earlier.
According to Careers360, IIT Roorkee will be conducting the JEE Advanced 2019 on 19 May in two shifts: Paper 1 from 9 am to 12 pm, and Paper 2 from 2 to 5 pm. The examination will be a computer-based test (CBT).
However, authorities have not yet disclosed other dates related to the examination.
Candidates can access the syllabus and other details on the official website jeeadv.ac.in
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Updated Date: Feb 21, 2019 18:12:51 IST