What is the eligibility criteria for B.E. / B.Tech at Sriram Engineering College?

<p>Hello! Greetings everyone</p><p>The eligibility criteria for the B.E. / B.Tech course at Sriram Engineering College are as follow these several steps:</p><p><strong>Eligibility:</strong> Candidates must have completed 10+2 (Higher Secondary Certificate) with a minimum of 45% aggregate marks in Physics, Chemistry, and Mathematics.</p><p><strong>Entrance Exams: </strong>Admission is based on scores from entrance exams such as JEE Main or TNEA (Tamil Nadu Engineering Admissions).</p><p>For more accurate and the latest information please visit the official website of Sriram Engineering College or by contacting the admission office. It is imperative to follow all the guidelines to get accurate information.</p><p>Thanking you</p>

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<p>Hi, &nbsp;To be eligible for the B.E. / B.Tech programs at Sriram Engineering College, candidates must have completed their 10+2 (or equivalent) education with at least 50% aggregate marks in Physics, Chemistry, and Mathematics. Additionally, candidates need to have appeared for the relevant entrance exam, such as the Tamil Nadu Engineering Admissions (TNEA) or other state-level entrance exams, and obtained a valid score. Admission is typically based on merit and counseling?</p><p>others . also&nbsp;</p><p>Sriram Engineering College offers admissions to the B.E./B.Tech courses under the lateral entry scheme for candidates with a Diploma or B.Sc. qualification.</p><ul><li><strong>Diploma Holders</strong>: Eligible candidates must have passed a Diploma with a minimum of 45% marks (40% for reserved categories) in any branch of Engineering or Technology.</li><li><strong>B.Sc. Graduates</strong>: Candidates must have passed a B.Sc. degree with at least 45% marks (40% for reserved categories) and should have studied Mathematics in 10+2.</li></ul><p>These candidates are eligible for admission into the second year of the B.E./B.Tech programs.</p><p>For more details, you can visit their official site&nbsp;</p><p>Thanks&nbsp;</p>

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