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B.E. in Computer Science and Engineering
at Jairupaa College of Engineering, Tamil Nadu - Other

Private Institute
Estd. 2011

B.E. in Computer Science and Engineering

Table of contents

Overview
Fees
Placements
Entry Requirements
Admission Process
+ 1 more item

B.E. in Computer Science and Engineering
at
Jairupaa College of Engineering
Overview

Jairupaa College of Engineering offers 4 years B.E. in Computer Science and Engineering to the students. The course is offered at the UG level. The total cost of studying B.E. in Computer Science and Engineering is INR 2,00,000. Besides tuition fee, students also need to pay caution deposit, exam fee, etc. for admission at Jairupaa College of Engineering. The institute offers admission to 60 students in Jairupaa College of Engineering B.E. / B.Tech. Below are the details for the programme:

Total Tuition Fees	₹2.00 Lakh Get Fees details
Duration	4 years
Course Level	UG Degree
Mode of Course	Full Time
Official Website	Go to Website
Seat breakup	60
Type of University	Private

B.E. in Computer Science and Engineering
at
Jairupaa College of Engineering
Fees

Jairupaa College of Engineering B.E. in Computer Science and Engineering fee is to be paid over 4 years. The fees structure is inclusive of multiple components, such as One-Time Fees, Registration Fees, Exam Fees, etc. Candidates have to pay the course fee to finalise their admission into the university. Candidates can have a look at the table below to get an understanding of the Jairupaa College of Engineering fee structure:

Fee components

Amount (4 years)

Tuition Fees

The fees might include components other than tuition fees. Mentioned fee is as per Tamil Nadu Engineering Admission (TNEA- 2023). The Fee is for the candidate selected through Counselling. The mentioned fee is for Non-Accredited courses. Only Accredited Fee will be applicable if the program/courses is currently accredited by NBA. If not, then above mentioned fee will be applicable. Total Accredited fee- INR 220000

₹ 2.00 Lakh

Total fee

₹ 2.00 Lakh

These are based on information available on various sources. Latest values may differ.

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B.E. in Computer Science and Engineering
at
Jairupaa College of Engineering
Placements

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B.E. in Computer Science and Engineering
at
Jairupaa College of Engineering
Entry Requirements

General

Requirements for 12th grade Up Arrow Icon

45%
Candidate must have passed HSC (10+2) or its equivalent with above mentioned marks from any recognized Board/council. Candidate who have passed HSC i.e., 10+2 in any one of the Vocational Subject (Basic Mechanical Engineering/ Basic Electrical Engineering/ Basic Electronics Engineering/ Basic Civil Engineering/ Basic Automobile Engineering/ Textile Technology) with any one of the Engineering related subjects namely Mathematics, Physics or Chemistry with above mentioned minimum average percentage put together are also eligible to apply.
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Tamilnadu 12th

ISC

CBSE 12th

Other eligibility criteria Up Arrow Icon

Candidate must have passed VIII, IX, X, XI & XII Std. from schools situated in Tamil Nadu. Tamil Nadu native candidate who have passed any of the VIII, IX, X, XI & XII Std. or their equivalent exam from schools situated outside Tamil Nadu can also apply. They should produce Original Nativity Certificate. Children of Central Government Employees, irrespective of their nativity, provided their parents or guardians (only if both parents are not alive) have been serving in Tamil Nadu continuously for the past five years can be eligible to apply. Sons & daughters of All India service Tamil Nadu cadre officers are eligible to apply. A Self declaration from the parent in this regard is to be produced. Children of employees in the Public Sector or Government recognized institutions had been working in their organisations for the past five years are also eligible, irrespective of their domicile.