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SASTRA University B. Tech Admission 2023 – Round 4 Counselling Result (Out), Registration, Admission Procedure

SASTRA University B. Tech Admission 2023 – Round 4 Counselling Result (Out), Registration, Admission Procedure

Edited By Team Careers360 | Updated on Jul 20, 2023 04:09 PM IST | #JEE Main

SASTRA University B. Tech Admission 2023: Sastra University (Deemed) has announced the B.Tech Admission round 4 counselling results online July 20. Candidates can check the round 4 counselling result using their login details on the official website at sastra.edu. The authorities are conducting the B.Tech admission counselling in online mode. SASTRA University (Deemed) offers admission on the basis of class 12th equivalent marks and JEE Main scores. 70% of the total B. Tech admissions will take place on the basis of JEE Main score and the rest 30% on the basis of class 12th marks for SASTRA B. Tech Admission 2023.

Candidates who are interested in SASTRA University B. Tech Admission 2023 could fill out the application form on the official website. Before applying for SASTRA University B. Tech Admission 2023, candidates are advised to check the B.Tech admission eligibility criteria first. Candidates must go through this article to get the complete details of the SASTRA University B. Tech Admission 2023.

SASTRA University B. Tech Admission 2023 Highlights

Exam Name

SASTRA University B. Tech Admission 2023

Conducting Body

SASTRA (Deemed to be University), Tamil Nadu

Level of Admission

State Level

Category of Admission

Undergraduate (UG)

Course

B. Tech

SASTRA University official website

www.sastra.edu

Exam Purpose

To offer admission in B. Tech programme offered by SASTRA University, Tamil Nadu

Admission Basis

Class 12th qualifying marks and JEE Main 2023 score

SASTRA University B. Tech Admission 2023 – Important Dates

With the help of these important dates, candidates can keep track of all the important events regarding the SASTRA B. Tech Admission 2023. Candidates are advised to go through the table mentioned below:

SASTRA University B. Tech Admission 2023 Events

SASTRA University B. Tech Admission 2023 Dates

Application form availability

April 23, 2023

Last date of application form

June 24, 2023

Release of rank list

June 24, 2023

Counselling procedure

June 24 to 28, 2023 (Till 9 am)

Round 1 counselling resultJune 28, 2023

B. Tech Programmes Offered by SASTRA Deemed University

Serial No.

B. Tech Programs

1

Aerospace Engineering

2

Bioengineering

3

Bioinformatics

4

Biotechnology

5

Chemical Engineering

6

Civil Engineering

7

Computer Science & Business System

8

Computer Science & Engineering

9

Computer Science & Engineering (with Spl in Artificial Intelligence & Data Science)

10

Computer Science & Engineering (with Spl in Cyber Security & Blockchain Technology)

11

Computer Science & Engineering (with Spl in IoT & Automation)

12

Electrical & Electronics Engineering

13

Electrical & Electronics Engineering (with Spl in Smart Grid & Electric Vehicles)

14

Electronics & Communication Engineering

15

Electronics & Communication Engineering (with Spl in Cyber-Physical Systems)

16

Electronics & Instrumentation Engineering

17

Information & Communication Technology

18

Information Technology

19

Mechanical Engineering

20

Mechanical Engineering (with Spl in Digital Manufacturing)

21

Mechatronics

SASTRA University B. Tech Eligibility Criteria 2023

Interested candidates must check the eligibility criteria before applying for the SASTRA University B. Tech Admission 2023. Candidates are advised to follow the eligibility criteria mentioned below:

  • Candidates should have a passing certificate of class 12th or any other equivalent board with Physics and Mathematics; both being compulsory subjects. Along with Physics and Mathematics, the candidate must have Chemistry/Bio-Technology/Computer Science/Biology as a subject.
  • It is important that a candidate should have scored a minimum of 60% marks in Physics, Mathematics, and Chemistry/Bio-Technology/Computer Science/Biology in class 12th or any other equivalent board in one sitting.
  • The candidate shouldn’t be more than 20 years of age as of August 1, 2023 (21 years for lateral entry).
  • It is compulsory for the candidates to appear in JEE Main 2023 in order to apply for stream 1 in SASTRA University B. Tech Admission 2023.
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SASTRA University B. Tech Admission Procedure 2023

Candidates will be given an opportunity to apply for SASTRA B. Tech Admission 2023 through two streams:

Stream 1

Candidates who are applying for stream 1 have to submit their IIT JEE Main 2023 score and class 12th or equivalent exam marks to the authority. Stream 1 consists of 70% seats out of the total seats.

75% weightage will be given to the marks that a candidate will score in class 12th or any other equivalent exam. Whereas, the remaining 25% weightage will be given to JEE Mains 2023 score of the candidate.

Both these scores will be combined together and a rank list will form out of these. Candidates will get invited for admission on the basis of the rank list.

Stream 2

The remaining 30% out of the total will get filled through stream 2 which will demand class 12th or equivalent exam marks.

Note: Candidates are eligible to apply through both streams. If a candidate didn’t appear for JEE Main 2023 but still applies for stream 1, the 25% weighted score in stream 1 will get converted to zero but the candidate will be given 75% weightage of his class 12th or equivalent exam marks. However, in the case of stream 2, the candidate will get 100% weightage for his/her qualifying exam scores.

SASTRA University B. Tech Application Form 2023

The concerned authority of SASTRA Deemed University released the application form of SASTRA University B. Tech Admission 2023. All the eligible candidates are required to fill out the application form only in online mode. Although, the form can be submitted in both online and offline mode.

For Offline Application

In order to submit an offline SASTRA B. Tech Application Form 2023, candidates need to download the application form first from the official website. After downloading the form, candidates will have to fill out the form and send it along with the application fee of Rs. 650 in the form of DD drawn in favour of Sastra payable at Thanjavur.

Steps to Fill Online SASTRA University B. Tech Application Form 2023

Here are the following steps that candidates must follow to fill out the application form for SASTRA B. Tech Admission 2023.

1. Registration – Visit the official website of SASTRA University Deemed– www.sastra.edu and click on the “SASTRA B. Tech Admission 2023” link. Candidates can register by submitting the following personal details like name of the candidate, D.O.B, age (as of 1.07.2021), email ID, contact number, campus, and course applying for.

Once the registration is done, candidates will receive their registration number and password on their registered email ID. With the help of the registration number and password, candidates will have to fill out the application form and use it for future reference.

2. Login – After registration, candidates have to log in to their account with the help of the given ID and password. Apart from personal details, candidates also have to enter their academic details like class 12th or equivalent exam marks, JEE Mains 2023 score, subjects, school details.

3. Application fee payment – Candidates will be required to pay the application fee of SASTRA University B. Tech Admission 2023 through any of the payment modes like net banking, debit, or credit card.

4. Uploading essential documents – A couple of relevant documents will be required like a recent passport-sized photograph along with a scanned signature in the specified size and format.

5. Submission of the application form – In the final step, candidates are required to click the “Submit” button in order to finally submit their application form for SASTRA University Admission 2023. However, before submitting candidates are advised to check all the entered details.

SASTRA University B. Tech Rank List 2023

The candidates can view the rank list of SASTRA B. Tech Admission 2023 through online mode. The authorities has released a separate rank list for streams 1 and 2. The rank list is going to be based on class 12th or equivalent exam marks and IIT JEE Main 2023 score. Whereas, stream 2 will be based on class 12th or equivalent exam marks only.

In order to check the rank list, candidates have to enter their application number or D.O.B. Based on the rank list of SASTRA University B. Tech Admission 2023, candidates will be called for SASTRA B. Tech counselling 2023.

Note: If a candidate experiences any discrepancies regarding marking structure or any other data, then the candidate can reach out to the concerned authority by raising an objection within 48 hours. Candidates have to send the details to registrar@saastra.edu with ‘Application Error’ as a subject in the mail.

SASTRA University B. Tech Counselling 2023

Candidates will get selected for the SASTRA B. Tech counselling 2023 based on their ranking in the rank list. Both streams, that is stream 1 and stream 2 will have separate counselling of SASTRA University B. Tech Admission.

Candidates who will be attending the counselling for stream 1 will be allotted seats on the basis of class 12th or equivalent exam marks and JEE Main 2023 score. Whereas, candidates appearing in the counselling for stream 2 will be allotted seats on the basis of class 12th or equivalent exam marks.

SASTRA University B. Tech Admission Fee Structure 2023

The fee structure of SASTRA University B. Tech Admission 2023 is given below in a tabular form. Candidates are advised to go through all the fee structure details to get an idea about the admission process.

Serial No.

Type

Amount (in INR)

1

Tuition fee

Rs. 75,000 (each semester)

2

Development fee

Rs. 10,000 (one-time payment)

3

Sports and unit test fee

Rs. 1,000 per annum

4

Registration, medical, special fee, etc.

Rs. 6,000 (one-time payment)

SASTRA University B. Tech Admission 2023 Contact Details

  • Office of Admissions
  • SASTRA Deemed University
  • Thanjavur – 613 401
  • Phone: 91 4362 264101‐108, 304000‐010
  • E-mail: admissions@sastra.edu
  • URL: www.sastra.edu
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Frequently Asked Question (FAQs)

1. Can a seat be reserved before admission under the merit category of SASTRA Deemed University?

No, that is not possible as SASTRA University B. Tech Admission 2023 is based only on merit after the release of the rank list by the university.

2. Is there a management quota in SASTRA Deemed to be University?

No, the management quota process doesn’t exist in SASTRA University as admissions are based on merit lists only.

3. Is SASTRA a good University?

Yes, SASTRA Deemed to be University offers excellent placements, great career opportunities, and exposure to grow academically.

4. Which is better SSN or SASTRA?

SSN University is best for CSE and ECE programs, whereas, SASTRA is good for Mechanical and Biotech programs.

5. When will the authorities release the SASTRA University BTech 2023 application form?

The application form of SASTRA University BTech 2023 was available online on the official website. 

6. When will the rank list be released?

The rank list has released on the official website. 

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Hello Student,
Your chances after considering your category as general, at this given score will be there in some Institutes for CS Branch.
Like: Puducherry Technological University, Puducherry , Indian Institute of Information Technology, Manipur

To know your chances in detail, kindly check our College Predictor Tool: https://engineering.careers360.com/jee-main-college-predictor

Hello aspirant,

If you fall into the general category, securing admission through JoSAA Counselling might pose a challenge. However, you can explore alternative options such as applying to state and university counsellings or retaking the exam to improve your score.

You can Target following colleges with this percentile:

1) KIIT University

2) BIT, Ranchi

3) University of Engineering and Management, Jaipur

4) Lovely Professional University, Jalandhar

5) College of Engineering, Pune

6) Jaipur Engineering College, Jaipur

7) Dream Institute of Technology, Kolkata

Thank you

Hope this information helps you.

The requirement for the JEE Main exam can vary depending on the specific AI course and the institution offering it. Some AI courses may require JEE Main scores as part of their admission criteria, especially if they have a strong focus on engineering or computer science fundamentals.

However, many AI courses, particularly at the postgraduate level, may have their own entrance exam or may not require JEE Main score at all. It's essential to check the admission requirements of the specific AI course and institutions you're interested in to determine whether JEE Main is mandatory.

Hello Student,


As per your given rank your chances will be tough to get admission into NITs.
Still you can try for State Counselling where the JEE Mains Score is accepted.

To know in detail kindly check our college predictor tool: https://engineering.careers360.com/jee-main-college-predictor

Hello aspirant,

With your JEE Main score considerably below the previous year's cutoff for general category students, eligibility to apply to NITs, IIITs, or GFTIs through JoSAA counseling is unlikely. Considering the exam's high difficulty level and stringent cutoffs, gaining admission to these top colleges would be challenging at this percentile.

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Hope this information helps you.

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 5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.820C.  If Na2SO4 is 81.5% ionised, the value of x (Kf for water=1.860C kg mol−1) is approximately : (molar mass of S=32 g mol−1 and that of Na=23 g mol−1)
Option: 1  15 g
Option: 2  25 g
Option: 3  45 g
Option: 4  65 g  
 

 50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl.  If pKb of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75
Option: 3 8.25
Option: 4 9.25
 

CH_3-CH=CH-CH_3+Br_2\overset{CCl_4}{\rightarrow}A

What is A?

Option: 1

CH_3-CH(Br)-CH_2-CH_3


Option: 2

CH_3-CH(Br)-CH(Br)-CH_3


Option: 3

CH_3-CH_2-CH_2-CH_2Br


Option: 4

None


\mathrm{NaNO_{3}} when heated gives a white solid A and two gases B and C. B and C are two important atmospheric gases. What is A, B and C ?

Option: 1

\mathrm{A}: \mathrm{NaNO}_2 \mathrm{~B}: \mathrm{O}_2 \mathrm{C}: \mathrm{N}_2


Option: 2

A: \mathrm{Na}_2 \mathrm{OB}: \mathrm{O}_2 \mathrm{C}: \mathrm{N}_2


Option: 3

A: \mathrm{NaNO}_2 \mathrm{~B}: \mathrm{O}_2 \mathrm{C}: \mathrm{Cl}_2


Option: 4

\mathrm{A}: \mathrm{Na}_2 \mathrm{OB}: \mathrm{O}_2 \mathrm{C}: \mathrm{Cl}_2


C_1+2 C_2+3 C_3+\ldots .n C_n=

Option: 1

2^n


Option: 2

\text { n. } 2^n


Option: 3

\text { n. } 2^{n-1}


Option: 4

n \cdot 2^{n+1}


 

A capacitor is made of two square plates each of side 'a' making a very small angle \alpha between them, as shown in the figure. The capacitance will be close to : 
Option: 1 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{\alpha a }{4 d } \right )

Option: 2 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 + \frac{\alpha a }{4 d } \right )

Option: 3 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{\alpha a }{2 d } \right )

Option: 4 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{3 \alpha a }{2 d } \right )
 

 Among the following compounds, the increasing order of their basic strength is
Option: 1  (I) < (II) < (IV) < (III)
Option: 2  (I) < (II) < (III) < (IV)
Option: 3  (II) < (I) < (IV) < (III)
Option: 4  (II) < (I) < (III) < (IV)
 

 An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant.  If during  this process the relation of pressure P and volume V is given by PVn=constant,  then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively)
Option: 1  n=\frac{C_{p}}{C_{v}}


Option: 2  n=\frac{C-C_{p}}{C-C_{v}}


Option: 3 n=\frac{C_{p}-C}{C-C_{v}}

Option: 4  n=\frac{C-C_{v}}{C-C_{p}}
 

As shown in the figure, a battery of emf \epsilon is connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = tc (tc is the time constant of the circuit ) is : 


Option: 1 \frac{\epsilon L }{R^{2}} \left ( 1 - \frac{1}{e} \right )
Option: 2 \frac{\epsilon L }{R^{2}}


Option: 3 \frac{\epsilon R }{eL^{2}}

Option: 4 \frac{\epsilon L }{eR^{2}}
 

As shown in the figure, a particle of mass 10 kg is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed  of the particle at B is x m/s. (Take g = 10 m/s2 ) The value of 'x' to the nearest is ___________.
Option: 1 10
Option: 2 20
Option: 3 40
Option: 4 15

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