

B.E. in Computer Engineering
- A NAAC accredited
- Private Institute
- Estd. 2008
B.E. in Computer Engineering at SITRC Overview
Sandip Institute of Technology and Research Centre offers a 4 years B.E. in Computer Engineering course at the UG level. To get admitted to this course, applicants must meet the entry requirements and must have a valid MHT CET score. The total tuition fee for this course is INR 4,19,212 for the entire duration. In addition to the tuition fee, Sandip Institute of Technology and Research Centre charges students for other applicable fee components. The university has a seat intake of 180 for this course. Check out other specialisation of Sandip Institute of Technology and Research Centre B.E. / B.Tech.
Total Tuition Fees | ₹4.19 Lakh Get Fees details |
Duration | 4 years |
Course Level | UG Degree |
Mode of Course | Full Time |
Average package | ₹ 5.00 Lakh |
Official Website | Go to Website |
Seat breakup | 180 |
Type of University | Private |
B.E. in Computer Engineering at SITRC Fees
Sandip Institute of Technology and Research Centre B.E. in Computer Engineering fee is cumulative of several elements, such as Tuition Fees, One-Time Fees, Registration Fees, Exam Fees, etc. The B.E. in Computer Engineering at Sandip Institute of Technology and Research Centre tuition fee is INR 4,19,212. Candidates might have to pay specific extra charges to the institution. More information around Sandip Institute of Technology and Research Centre B.E. in Computer Engineering fee breakup is as follows:
| Fee components | Amount (4 years) |
|---|---|
Tuition fee is calculated on the basis of 1st year/semester. Actual amount may vary.
Mentioned fee is as per Fee Regulating Authority, Maharashtra State. | ₹ 4.19 Lakh |
₹ 4.19 Lakh |
B.E. in Computer Engineering at SITRC Frequently Asked Questions
Download exam sample paper
B.E. in Computer Engineering at SITRC Placements
| Particulars | Statistics (2021) |
|---|---|
| Average Salary | INR 5.00 Lakh |
| Highest Salary | INR 12.00 Lakh |
| Median Salary | INR 1.92 Lakh |
B.E. in Computer Engineering at SITRC Entry Requirements
B.E. in Computer Engineering at SITRC Admission Process
- CounsellingAdmission is through counselling based on the rank obtained in Maharashtra Health and Technical Common Entrance Test (MHTCET). Vacant seats will be filled by the institute based on the rank obtained in JEE-Mains.
Important Dates
B.E. in Computer Engineering at SITRC Cutoffs
JEE Main round-wise cutoff Rank: (General-All India)
| Round | 2023 | 2024 | 2025 |
|---|---|---|---|
| 1 | 160527 | 204541 | 215742 |
| 2 | 175248 | 222122 | 219171 |
| 3 | 169509 | 186337 | 219580 |
| 4 | - | - | 180370 |
JEE Main round-wise cutoff Rank: (OBC-All India)
| Round | 2021 | 2022 | 2023 |
|---|---|---|---|
| 1 | 274112 | 255212 | 160527 |
| 2 | 272159 | 166923 | 175248 |
| 3 | - | - | 169509 |
MHT CET round-wise cutoff Rank: (General-All India)
| Round | 2019 | 2020 | 2021 |
|---|---|---|---|
| 1 | 85.54 | 72.12 | 80.23 |
| 2 | 77.34 | 69.76 | 77.73 |
| 3 | 80.33 | - | - |
MHT CET round-wise cutoff Rank: (General-Home State)
| Round | 2021 | 2022 | 2023 |
|---|---|---|---|
| 1 | 77.36 | 81.97 | 83.44 |
| 2 | 74.52 | 79.59 | 81.24 |
| 3 | - | 83.15 | 83.01 |
MHT CET round-wise cutoff Rank: (OBC-All India)
| Round | 2019 | 2020 | 2021 |
|---|---|---|---|
| 1 | 80.39 | 70.97 | 76.49 |
| 2 | 72.58 | 64.68 | 75.44 |
| 3 | 66.54 | - | - |
MHT CET round-wise cutoff Rank: (OBC-Home State)
| Round | 2021 | 2022 | 2023 |
|---|---|---|---|
| 1 | 72.28 | 80.37 | 82.21 |
| 2 | 72.12 | 79.53 | 80.08 |
| 3 | - | 81.51 | 81.86 |
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| Courses | Eligibility |
|---|---|
| B.E. / B.Tech | Candidate must have completed Diploma Course in Engineering and Technology in relevant stream with minimum 45% marks (40% marks in case of candidates of Backward class categories and Persons with Disability belonging to Maharashtra State only) from an All India Council for Technical Education or Central or State Government approved Institution or its equivalent. OR Candidate must have completed B.Sc. from UGC/ Association of Indian Universities recognised University with at least 45% marks (40% in case of candidates of Backward class categories and Persons with Disability belonging to Maharashtra State only) and passed 10+2 with Mathematics as a subject. |
| M.E./M.Tech | Sponsored candidate must have minimum of two years of full time work experience in a registered firm/ company/ industry/ educational and/ or research institute/ any Government Department or Government Autonomous Organization in the relevant field in which admission is being sought. |
| MBA/PGDM | Candidate must have completed minimum three year Bachelor's Degree in any discipline or its equivalent with above mentioned marks (45% in case of candidates of backward class categories and Persons with Disability belonging to Maharashtra State only) from any of the Universities recognised by UGC or AIU. |
Student Forum
Answered 2 years ago
It is better to personally contact Sandip Foundation or consult their official website for their complete admission requirements and cutoff marks to find out if you may get admitted there with a JEE score of 62.45 and a 12th grade score of 83.17%.
The criteria for admission to colleges can change dep
D
Contributor-Level 8
Answered 5 years ago
P
Beginner-Level 1
Answered 9 years ago
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B.E. in Computer Engineering at SITRC News & Updates




B.E. in Computer Engineering at SITRC Contact Information
Sandip Institute of Technology & Research Center, Trimbak Road
Nashik ( Maharashtra)
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Can I get admission in Sandip Foundation with a JEE score of 62.45 and 12th score of 83.17%?