Engineering Entrance Exam

Admission to SASTRA University for engineering programs is highly competitive and follows a unique admission process. Here's an analysis of your chances based on your JEE Mains percentile (89) and CBSE board marks (72%):SASTRA University Admission Process:SASTRA University admits students through two streams:Stream 1:70% of seats are filled based on a combined weightage of:JEE Main score (75% weightage)Class 12th marks (25% weightage).Stream 2:30% of seats are filled based on Class 12th board exam marks alone.This stream is typically for candidates who have not appeared for JEE or have lower scores in JEE.Your Case:Stream 1:Your 89 percentile in JEE Main is a strong point.However, your 72% in CBSE boards might slightly lower your overall weighted score for Stream 1.Admission to core branches like CSE, ECE, and Mechanical is highly competitive, and a higher combined score is typically required for these branches.You may have better chances in branches like Civil Engineering, Biotechnology, or Chemical Engineering.Stream 2:For admission through Stream 2 (based on board marks only), your 72% marks may make it challenging to secure admission in highly sought-after branches. However, you might have a chance in lesser competitive branches.Factors Influencing Admission:State of Residence:SASTRA reserves seats for candidates from Tamil Nadu and neighboring states. If you are from these states, your chances may improve.Category:SASTRA does not provide specific reservations for categories, as admissions are based purely on merit.Demand for Branches:Highly demanded branches like CSE, ECE, and IT usually require higher scores.Next Steps:Calculate Weighted Score for Stream 1:Use the formula: Weighted Score= (0.75*JEE Score)+ (0.25*Class 12 Percentage)\text {Weighted Score} = (0.75 \times \text {JEE Score}) + (0.25 \times \text {Class 12 Percentage})Weighted Score= (0.75*JEE Score)+ (0.25*Class 12 Percentage) Example: Assuming 89 percentile in JEE (converted to marks ~220/300): Weighted Score= (0.75*220)+ (0.25*72)=180+18=198\text {Weighted Score} = (0.75 \times 220) + (0.25 \times 72) = 180 + 18 = 198Weighted Score= (0.75*220)+ (0.25*72)=180+18=198Check Previous Year Cutoffs:Look for cutoffs specific to the branch and stream you're targeting. Cutoffs vary each year based on the competition and demand.Apply Early:Submit your application as early as possible, as SASTRA follows a transparent and rank-based allocation process.Likely Outcome:With 89 percentile in JEE Main, you have a reasonable chance of securing admission in non-core branches through Stream 1.Through Stream 2, admission is less likely due to relatively lower board marks unless applying for less competitive branches.For precise details, visit the official SASTRA University admissions page or contact their admissions office.

When selecting subjects for the GATE entrance examination, choose the subject that is most closely related to your engineering discipline and career aspirations, as you can only choose one or two subjects relevant to your field from the available options, depending on your specific goals and the GATE paper combination rules; always consult the official GATE syllabus and paper codes to confirm which subjects are available for your chosen branch. Key factors to consider when choosing GATE subjects:Your Engineering Degree:Choose a subject that is directly related to your undergraduate engineering degree.Career Goals:If you want to pursue a certain career route, such as M.Tech .

The University of Lucknow offers a four-year, full-time BTech course based on candidates' performance in the JEE Main entrance exam. For Mechanical Engineering, the JEE Main last round closing rank stood at 270103 for the General AI category students. The first-round cutoff for the same was 225423. So, considering the first and last-round cutoff ranks, the minimum rank required is 225423 to be eligible for admission to BTech in Mechanical Engineering. Tabulated below is the JEE Main round-wise cutoff ranks for the General AI category students. Round JEE Main Cutoff 2024 (Closing Rank General AI quota)1225423226380832348354270103

As an SC candidate with a 51 percentile in JEE Mains, you have some advantages for IPU B.Tech admission:Eligibility HighlightsSC candidates get a 5% relaxation in minimum marks requirementMinimum marks requirement reduced from 55% to 50%Eligible for B.Tech admission through JEE MainsAdmission DetailsAdmission based on JEE Mains scoreCounselling will determine final college allocationCandidates can get admission in various B.Tech specializations like:Computer ScienceArtificial Intelligence & Data ScienceAutomation & RoboticsIndustrial Internet of ThingsKey RecommendationsParticipate in IPU counselling processBe prepared for multiple rounds of seat allocationKeep documents ready for verificationConsider all available specializations. 

The cutoff ranks for the Tamil Nadu Engineering Admissions (TNEA) for the A.R.J College of Engineering and Technology (ARJCET) in the year 2024 are different for various courses and categories. For the Open Category, the cutoff ranks are given below:BE Electronics & Communication Engineering: 57,427 in Round 2, 192,791 in Round 3.BE Computer Science and Engineering: 69,970 in Round 2, 195,202 in Round 3.B.Tech Artificial Intelligence and Data Science: 101,248 in Round 2, 173,455 in Round 3.BE Agricultural Engineering: 120,667 in Round 2.BE Electrical and Electronics Engineering: 164,690 in Round 2.BE Mechanical Engineering: 178,143 in Round 2.BE Biomedical Engineering: 193,507 in Round 2.