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Top Colleges Accepting JEE Main Rank from 50000 to 2 Lakhs

Top Colleges Accepting JEE Main Rank from 50000 to 2 Lakhs

Edited By K Guna Sahiti | Updated on Jun 04, 2024 02:27 PM IST | #JEE Main

Top Colleges Accepting JEE Main Rank from 50000 to 2 Lakhs- There are 24 best colleges in India offering admission to students who scored valid rank between 50,000 and 2,00,000 in the JEE Main exam. As per NIRF ranking, BIT Mesra (rank 53) tops the list in the engineering domain. It is followed by, Shri Mata Vaishno Devi University, Katra (rank 101-150) and six more universities within the 151-200 range. Their fee typically varies between ₹15 K and ₹14.5 Lakhs.

Top Colleges Accepting JEE Main Rank from 50000 to 2 Lakhs
Top Colleges Accepting JEE Main Rank from 50000 to 2 Lakhs

Candidates must go through the JEE Main 2024 cutoff to get a detailed picture of the college admission criteria. Students whose JEE Main marks vs rank are not within the specified range can use the JEE Main college predictor to get the list of colleges best suited for them. Students who scored under 75% in 12th, can check the list of top engineering colleges without 75% criteria. We have compiled the JEE Main college data based on the past year's scores in this article.

Top Colleges Accepting JEE Main Rank Between 50000 to 2 Lakhs

Each JEE Main participating institute releases its admission criteria on its official website. Students can check the detailed list of colleges offering admission to students below 2 lakh rank in JEE Mains percentile here. Candidates with queries like which college can I get with 1 lakh rank in JEE Main can refer to the table below. The information provided on this page is the opening and closing ranks of Male applicants in the general category.

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Best Colleges Accepting JEE Main Rank from 50000 to 2 Lakhs

College Name

Opening Rank

Closing Rank

NIRF Ranking

Fees

Assam University, Silchar

53772

72777

_

_

BIT Deoghar

52728

69141

_

_

Birla Institute of Technology, Mesra (BIT Mesra)

50804

57103

53

₹ 14.42 Lakhs

BIT Patna

50920

71062

_

₹12.22 Lakhs

CIT Kokrajhar

51049

80426

_

₹1.81 Lakhs

Central University of Haryana, Mahendergarh

51085

77643

_

₹2.41 Lakhs

Central University Jammu

52434

72001

_

₹3.85 Lakhs

CURAJ

57868

68007

_

₹7.27 Lakhs

DHSGSU Sagar

50751

87737

_

_

GGU Bilaspur

50455

81582

_

₹1.10 Lakhs

GKV Haridwar

52288

78683

_

₹5.06 Lakhs

IICT Bhadohi

69747

84390

_

₹2.64 Lakhs

IIHT Salem

53499

86103

_

_

Indian Institute of Handloom Technology, Varanasi

62458

83642

_

₹ 70.50 K

Institute of Infrastructure Technology Research and Management, Ahmedabad

50132

60091

_

₹4.79 Lakhs

Mizoram University

52637

74625

_

₹1.70 Lakhs

National Institute of Advanced Manufacturing Technology

50649

74266

151-200

₹3.15 Lakhs

National Institute of Food Technology Entrepreneurship and Management

51423

90661

151-200

₹6.67 Lakhs

North Eastern Hill University

55240

77967

_

₹13.94 K

North Eastern Regional Institute of Science and Technology

53489

57317

151-200

₹46.18 K

Puducherry Technological University

50207

70293

151-200

_

Sant Longowal Institute of Engineering and Technology

50662

85401

151-200

₹ 2.83 Lakhs

Shri Mata Vaishno Devi University, Katra

53000

77552

101-150

₹6.46 Lakhs

Tezpur University, Tezpur

51742

85085

151-200

₹2.18 Lakhs

Jain University B.Tech 2024

100% Placement Record | Highest CTC 42 LPA | NAAC A++ Accredited | Ranked #68 in India by NIRF Ranking 2023 | Last Date to Apply - 14th June

Vijaybhoomi University B.E./B.Tech Admissions 2024

27+ Years of Legacy | India’s First Liberal University. Highest CTC 43 LPA | Average CTC 13.3 LPA 100% Placement Record Merit-Based Scholarships Available


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Questions related to JEE Main

Have a question related to JEE Main ?

Newton's Laws of Motion are fundamental to understanding the motion of objects. Here are the key formulas and concepts for JEE Mains:

Newton's First Law of Motion (Law of Inertia):

  • An object at rest will remain at rest, and an object in motion will continue to move in a straight line with constant speed, unless acted upon by an unbalanced force.

Newton's Second Law of Motion:

  • The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

  • Formula: F = ma
    • F: Net force (N)
    • m: Mass of the object (kg)
    • a: Acceleration of the object (m/s²)


    Newton's Third Law of Motion (Law of Action and Reaction):

    • For every action, there is an equal and opposite reaction.


    Important Formulas and Concepts:

    • Weight: W = mg
      • W: Weight of the object (N)
      • m: Mass of the object (kg)
      • g: Acceleration due to gravity (9.8 m/s²)
    • Friction: The force that opposes the relative motion between two surfaces in contact.
      • Static friction: Fs ≤ μsN
      • Kinetic friction: Fk = μkN
        • μs: Coefficient of static friction
        • μk: Coefficient of kinetic friction
        • N: Normal force
    • Tension: The force transmitted through a string, rope, or cable when it is pulled taut.
    • Normal force: The force exerted by a surface on an object in contact with it, perpendicular to the surface.
    • Centripetal force: The force that causes an object to move in a circular path.
      • Fc = mv²/r
        • Fc: Centripetal force (N)
        • m: Mass of the object (kg)
        • v: Velocity of the object (m/s)
        • r: Radius of the circular path (m)


    These formulas and concepts are essential for understanding Newton's Laws of Motion and solving problems related to mechanics in JEE Mains. Practice solving a variety of problems to solidify your understanding and improve your problem-solving skills.


    I hope it helps !!

Hello aspirant,

Here's a breakdown of your options and factors to consider:

Option 1: Continue with Sachin Rana's videos

Advantages:

  • Familiarity: You're already comfortable with his teaching style and approach.
  • Progress: You've made progress in GOC and isomerism, building a foundation.
  • Time Efficiency: You might be able to complete the remaining chapters faster since you've already covered some basics.

Disadvantages:

  • Uncertain Completion Time: Completing all of NS sir's lectures might take longer than you anticipate, potentially affecting your preparation for the JEE Advanced.
  • Lack of Focus: You might find it difficult to switch between two different teaching styles and approaches.

Option 2: Switch to NS sir's lectures

Advantages:

  • Comprehensive Coverage: NS sir's lectures are known for their depth and clarity, ensuring a thorough understanding of all organic concepts.
  • Advanced Preparation: Focusing on NS sir's lectures could help you prepare better for the JEE Advanced level.

Disadvantages:

  • Restarting from GOC: You'll need to revisit some concepts you've already covered with Sachin Rana.
  • Time Investment: Completing all of NS sir's lectures might take longer than you currently have.

Decision-Making Factors:

  1. Your Current Progress: If you feel confident with your understanding of GOC and isomerism from Sachin Rana, you can consider continuing with him. However, if you want a more comprehensive and in-depth understanding, switching to NS sir might be beneficial.
  2. Time Constraints: Assess how much time you have before the JEE Advanced. If time is a major factor, sticking with Sachin Rana's shorter videos might be more efficient.
  3. Learning Style: Consider which teacher's teaching style resonates better with you. If you prefer a more structured and detailed approach, NS sir might be a good fit.

Recommendation:

Ultimately, the decision should be based on your individual preferences, learning style, and time constraints. It might be helpful to watch a few lectures from both teachers to see which one aligns better with your learning needs.

Additional Tips:

  • Supplement with Practice: Regardless of which teacher you choose, practice regularly to solidify your understanding and problem-solving skills.
  • Mock Tests: Take mock tests frequently to assess your progress and identify areas where you need to improve.
  • Seek Clarification: Don't hesitate to ask questions or seek clarification from your teachers or classmates if you have any doubts.

I hope this guidance helps you make an informed decision!

The five-year combined B.Tech-M.Tech degree in Computer Science & Engineering (Cyber Security) is offered by the National Forensic Sciences University (NFSU) in Delhi. With topics like secure programming, network security, malware analysis, IoT security, and forensics, the curriculum offers an in-depth concentration on computer science, cybersecurity, and digital forensics.

In terms of cut-offs, NFSU bases its admissions decisions on the B.Tech program's JEE Mains rank. Because of the specialised nature of the program, competitive ranks for B.Tech-M.Tech in Cyber Security often fall within upper JEE Main rank bands. However, the actual cut-off for the Delhi campus changes based on criteria including category and seat availability.

There are 40 seats available for the Delhi campus, with reservations for members of the SC, ST, OBC, and EWS groups. For the first three years of the program, the tuition costs are approximately Rs65,000 each semester; after that, they rise to Rs 90,000 per semester.

Atleast you decide to prepare for boards finally no matter how late Just keep the same enthusiasm and zeal for the next three months and you will definitely achieve your goal.

Evaluate the options you have.


Select the best one.


Make a plan for each subject


Go with it till the end no matter how tough the situation is.

Completing organic chemistry (OC) from the Samarpan batch of MKA Sir on YouTube can be a good option for you, especially if you have already covered the basics of GOC and Isomerism. However, there are a few things to consider:

  • Scope of the batch: While the Samarpan batch covers most of the important topics in organic chemistry, it may not go into as much depth as a traditional classroom course. You may need to supplement your learning with additional resources, such as textbooks or online tutorials.
  • Your current level of understanding: If you have a strong foundation in GOC and Isomerism, you may be able to complete the batch relatively quickly. However, if you are struggling with these topics, it may take you longer.
  • Your learning style: If you are a self-directed learner who can work independently, you may be able to benefit from the flexibility and convenience of online learning. However, if you prefer a more structured learning environment, you may find it challenging to stay on track without a teacher or classmates.

Overall, completing organic chemistry from the Samarpan batch of MKA Sir on YouTube is a feasible option for class 12th JEE aspirants. However, it is important to be realistic about your expectations and to supplement your learning with additional resources as needed.


I hope this information helps you.






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