

B.E. in Computer Engineering
- A NAAC accredited
- Private Institute
- Estd. 2008
3.9
4.6
3.8
4.1
3.6
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. A total of 180 students can enrol in this course. The total tuition fee for the entire duration of this course is INR 4,19,212. Find out more about other courses under 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
The total tuition fee for Sandip Institute of Technology and Research Centre B.E. in Computer Engineering is INR 4,19,212. Besides, the fee also includes other components, such as exam fee, caution deposit, enrollment fee, etc. Some of the fee components, such as caution deposit are one-time payable. Candidates after getting admission in B.E. in Computer Engineering need to pay the Sandip Institute of Technology and Research Centre fee to confirm their seat in the institute. Below is fee breakup for Sandip Institute of Technology and Research Centre B.E. in Computer Engineering:
| 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
B.E. in Computer Engineering at SITRC Students also asked
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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Answered 3 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 6 years ago
P
Beginner-Level 1
Answered 2016-05-31 16:05:07
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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)
(For general query)
(For admission query)
(For general query)
(For admission query)
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Can I get admission in Sandip Foundation with a JEE score of 62.45 and 12th score of 83.17%?