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IIIT Sonepat JEE Main Cutoff 2024 - Previous Year JEE Mains Cut off for IIIT Sonepat

IIIT Sonepat JEE Main Cutoff 2024 - Previous Year JEE Mains Cut off for IIIT Sonepat

Edited By Samridhi Mishra | Updated on Apr 25, 2024 06:14 PM IST | #JEE Main

IIIT Sonepat JEE Main Cutoff 2024- JoSAA will release the JEE Main 2024 cutoff for IIIT Sonepat at josaa.nic.in. Candidates who wish to get admission in the B.Tech courses offered at IIIT Sonepat have to meet the cutoff percentile. JEE Mains 2024 cutoff for IIIT Sonipat will include the opening and closing ranks required for admission to the institute. Admission to the IIIT Sonipat is based on Joint Entrance Examination Main.

IIIT Sonepat JEE Main Cutoff 2024 - Previous Year JEE Mains Cut off for IIIT Sonepat
IIIT Sonepat JEE Main Cutoff 2024 - Previous Year JEE Mains Cut off for IIIT Sonepat

The authorities will release separate JEE Main cutoff for each course offered by the institute and each category of candidates. JoSAA will consider several factors to determine the IIIT Sonepat cutoff for the home state and all of India. Candidates who qualify JEE Main exam with marks over and above the IIIT Sonepat 2024 admission cutoff will be eligible for admission.

Previous Year JEE Main Cutoff for IIIT Sonepat

Candidates can refer to the given tables to know about the IIIT Sonepat cutoff JEE Mains for the years 2023, 2022, 2021, 2020, 2019, 2018, 2017 and 2016. The previous year JEE Main cutoffs of IIIT Sonepat will help candidates understand the cutoff trends.

JEE Main Cutoff 2023 for IIIT Sonepat

Course NameGender
Opening rank
Closing rank
Computer Science and Engineering
Male13718
20515
Female13718
20515
Information Technology
Male20547
22295
Female20547
22295
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JEE Main Cutoff 2022 for IIIT Sonepat

Course NameCategoryMaleFemale
Computer Science and EngineeringOBC79967996
EWS31563156
General2007220072
SC52395239
ST25642564
Information TechnologyOBC89198919
EWS34893489
General2167021670
SC59395939
ST28612861
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JEE Main Cutoff 2021 for IIIT Sonepat

Course NameCategoryMaleFemale
Computer Science and EngineeringOBC69766976
EWS28192819
General1652516525
SC49644964
ST33823382
Information TechnologyOBC79027902
EWS31773177
General1776317763
SC54475447
ST37023702
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JEE Main Cutoff for IIIT Sonepat 2020

Course Name

Category

Male

Female

Computer Science and Engineering

OBC

80538053

EWS

32933293

General

1762817628

SC

56505650

ST

37323732

Information Technology

OBC

1006410064

EWS

39533953

General

2235822358

SC

71067106

ST

47954795

JEE Main Cutoff 2019 for IIIT Sonepat

Course Name

Category

Gender Neutral

Female Only

Computer Science and Engineering

OBC

9463

-

EWS

3359

-

General

20758

-

SC

6509

-

ST

4028

-

Information Technology

OBC

11932

-

EWS

3578

-

General

25334

-

SC

8056

-

ST

4585

-

JEE Main Cutoff 2018 for IIIT Sonepat

Course Name

Categories

Opening Rank

Closing Rank

Computer Science and Engineering

General

714

20650

OBC-NCL

647

10570

SC

2464

7367

ST

2649

4307

Electronics and Communication Engineering

General

1428

34086

OBC-NCL

7617

13436

SC

4030

8890

ST

2647

4309

Information Technology

General

1332

29729

OBC-NCL

7943

13884

SC

4544

7932

ST

2963

5982

JEE Main Cutoff 2017 for IIIT Sonepat

Name of the Programme

Categories

Opening Rank

Closing Rank

Computer Science and Engineering

General

10240

21979

OBC NCL

5159

9301

SC

3028

6912

ST

2734

5537

Electronics and Communication Engineering

General

13771

31372

OBC-NCL

7115

12512

SC

4373

7615

ST

2370

4197

Information Technology

General

7165

26817

OBC- NCL

6501

11954

SC

3394

7755

ST

2750

6046

JEE Main 2016 Cutoff for IIIT Sonepat

Name of the Programme

Categories

Opening Rank

Closing Rank

Computer Science and Engineering

General

17099

28667

OBC NCL

8454

11113

SC

4836

5811

ST

2267

3551

Electronics and Communication Engineering

General

14572

26431

OBC-NCL

7682

11535

SC

4621

6799

ST

2754

3281

Information Technology

General

9004

22651

OBC- NCL

8489

9207

SC

4738

6564

ST

2340

3828

IIIT Sonepat JEE Main Cutoff 2024

JoSAA will release the IIIT Sonepat cutoff 2024 JEE Mains after each round of JoSAA counselling. The JEE Main 2024 cutoff for IIIT Sonepat is the minimum rank required by the candidates to be eligible for admission at IIIT Sonepat. It must be noted that merely qualifying the cutoff does not guarantee admission, it only increases the chances of admission. The IIIT Sonepat JEE Mains 2024 cutoff will be available here as soon as it is released by the authorities.

Factors Determining IIIT Sonepat JEE Main Cutoff 2024

The authorities will consider the following factors while determining the IIIT Sonepat cutoff JEE Mains 2024-

  • Difficulty level of JEE Main 2024 exam

  • Previous year JEE Main cutoff trends

  • Number of candidates who seek admission at the institute

  • Availability of seats

Frequently Asked Question (FAQs)

1. How will JEE Main cutoff for IIIT Sonepat be prepared?

The authorities will take into consideration various exam related details like number of applicants, difficulty level of the exam, number of available seats, previous year cutoff trends and more while preparing the cutoff.

2. Who will release the cutoff for IIIT Sonepat?

 Joint Seat Allocation Authority (JoSAA) will release the cutoff for IIIT Sonepat.

3. Which programmes are offered through IIIT Sonepat?

IIIT Sonepat offers two B.Tech programs- Computer Science and Engineering and Information Technology.

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Questions related to JEE Main

Have a question related to JEE Main ?

Hello aspirant,

The test will be administered online and consist of two exams, Paper 1 and Paper 2, in accordance with the JEE Advanced 2024 paper format. Candidates will be allowed three hours for each of the two papers. The JEE Advanced 2024 question paper format includes matching-style questions, multiple-choice questions with multiple responses, and questions with numerical values.

To know complete information related to jee exam, you can visit our website by clicking on the link given below.

https://engineering.careers360.com/articles/jee-advanced-exam-pattern

Thank you

Hope this information helps you.

JEE Advanced mein ST category mein 1216 rank ke saath, aapko kuch acche NITs, IIITs aur GFTIs (Government Funded Technical Institutes) mein admission mil sakta hai. Halaanki, IITs mein top branches milna mushkil ho sakta hai, lekin kuch naye ya kam prestigious IITs mein admission ki sambhavna ho sakti hai.

Sambhavit college aur branches:

1. NITs (National Institutes of Technology):

* NIT Warangal

* NIT Trichy

* NIT Surathkal

* NIT Rourkela

* NIT Calicut

* Anya NITs mein bhi admission ki sambhavna hai


2. IIITs (Indian Institutes of Information Technology):

* IIIT Allahabad

* IIIT Gwalior

* IIIT Jabalpur

GFTIs (Government Funded Technical Institutes):

* BIT Mesra

* Assam University

* PEC Chandigarh

* Anya GFTIs



Yes, you can register for Round 2 of JEE Main counselling even if you didn't register for Round 1. The JoSAA counselling process allows registration for subsequent rounds even if you missed the previous ones.

Each round of JEE Main counselling functions independently. Missing Round 1 doesn't disqualify you from participating in Round 2.

In Round 2, you have a fresh opportunity to fill out your choices and preferences for colleges and programs based on the seat allotment results and cutoffs from Round 1 (which will help you make more informed decisions).

https://engineering.careers360.com/download/ebooks/jee-main-2022-preparation-tips

I hope it helps!


Here is a details regarding the JEE EXAM

Purpose of JEE (Joint Entrance Examination):


Conducted for admission to premier engineering colleges like IITs and NITs in India.

This exam is conducted by the NTA (National Test Agency).

there is Two Phases of the Exam:

JEE Main: First phase of the examination.

JEE Advanced: Second phase for those who qualify in JEE Main.


Exam Difficulty:


This exam is Known for its challenging nature, requiring thorough preparation.  and for preparation, there are

multiple institutes that offer both online and offline modes of preparation. but if you ask me, I will suggest for that online is better than offline because now we don't have that much time for offline coaching


Home Learning Option:


ARIVIHAN APP provides an opportunity to learn from home.

In the Arivihan app Online personalised lectures are  available  and also this is totally

Budget-friendly, making it accessible to a wide range of students.

You can participate in MHT CET Counselling for admission into engineering colleges in Maharashtra if you:

  • Appeared for and qualified in JEE Main 2024.
  • Are a resident of Maharashtra or any other state in India. (There are a specific number of seats reserved for Maharashtra state and MHT CET qualified candidates though).

    The MHT CET Counselling for JEE Main scores is typically conducted online by the State Common Entrance Test Cell, Maharashtra (SCET Cell).


    The online registration window usually opens in the last week of June (expected in the last week of June 2024 for the 2024 admissions). You'll need to register on the official website of SCET Cell and provide details like your JEE Main roll number, score, and other required information.

    After registration, there will be a document verification process where you'll need to upload scanned copies of your documents (e.g., JEE Main scorecard, admit card, class 10th and 12th mark sheets, domicile certificate, etc.) online.

    During this stage, you can choose the engineering colleges and programs you are interested in based on your JEE Main score, seat availability, and preferences. You need to lock your choices by a specific deadline.

    Based on your JEE Main score, preferences, and available seats, the SCET Cell will allot seats through multiple rounds of counselling.

    If you are allotted a seat in a desired college, you'll need to pay the admission fees, report to the college for document verification, and complete the admission formalities.

    https://law.careers360.com/exams/mhcet-law

    I hope it helps!
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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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