IIT Roorkee - Indian Institute of Technology

Students can join IIT Roorkee, firstly they have to pass the JEE MAINS exam and qualify for the JEE ADVANCED exam.

• They should aslo have minimum 75% in 12th boards with PCM.
• They have to get minimum 95percentile to qualify in any branch of engineering in IIT Roorkee.

And rest of basic information is provided in SHIKSHA itself you can get details from there. 

Candidates who have passed a master's degree are eligible to apply for PhD programme at DoMS IIT Roorkee. Besides, aspirants must also pass any of the accepted entrance exams for admission. The institute considers scores in GATE, CAT, UGC-JRF, UGC-NET or CSIR JRF for admission. Selected candidates have to further the pass the personal interview round for final admission.

The IIT Roorkee UCEED cutoff 2025 was released for different categories of students belonging to the All India quota. The institute accepts the UCEED entrance exam, followed by counselling rounds for B.Des admission. For the General category students belonging to the AI quota, the UCEED last round closing rank stood at 111. The category-wise cutoff details for the last round are tabulated below:

For IIT Roorkee B.Des admission, a UCEED rank of 75 is considered to be a good rank in the General AI category. In 2025, the IIT Roorkee UCEED cutoff ranged between 91 and 111. Therefore, candidates with a cutoff rank of 75 have high chances of getting a seat at this institute. Check out the table below to view the B.Des round-wise UCEED cutoff ranks for the General AI category students. 

IIT Roorkee UCEED cutoff 2025 concluded with the release of the Round 5 cutoff list. If you belong to the General AI quota, the last round closing rank for B.Des admission was 111. During Round 1 admission, the closing rank stood at 91 for the same category and quota. So, candidates seeking admission to the B.Des programme at IIT Roorkee must target to secure a cutoff rank of up to 111 or below. The cutoff ranks vary for different rounds and categories. 

To get into IIT Roorkee, students must target a maximum rank of up to 3569 for BTech admission for the OBC AI category. The lower the cutoff rank, the higher the probability of getting the top B.Tech. branch for admission. The required cutoff rank varies for different rounds, courses, specialisations and categories. The cutoff rank mentioned here is for the last round. Candidates can refer to the table below to view the IIT Roorkee branch-wise maximum cutoff required for admission at IIT Roorkee. 

For CSE admission at IIT Roorkee, the JEE Advanced Round 4 closing rank was 555 for the students belonging to the OBC category under the AI quota. For the same specialisation and round, the closing rank was 481 in 2024, whereas in 2023, the cutoff rank stood at 405. Candidates seeking admission to BTech in CSE must target to secure a JEE Advanced rank of up to 550 or below. Please note that the cutoff ranks vary for different rounds, specialisations, and categories. 

IIT Roorkee cutoff 2025 has been announced for admission across multiple specialisations for the Round 4 seat allotment results based on opening and closing ranks. The opening and closing rank depends on the category and quota of candidates. For general category under All India, opening rank range at 41 and closing rank is 229. Admission to MSc specialisations at IIT Roorkee - Indian Institute of Technology is based on the scores obtained in the entrance exam IIT Jam, which should be between 41 and 229, scoring a rank of 400 in IIT Jam in not sufficient to get admission to the institute.

DOMS IIT Roorkee is known for its top placement packages. Annually, leading companies take part in placing graduating MBA students. For more information on placement packages for the DOMS IIT Roorkee placements between 2022-2024, refer to the table below: