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JEE Main Rank Predictor 2024 helps students know their JEE Main Rank based on their marks obtained in the exam. The rank predictor of JEE Main 2024 is a tech-enabled tool to predict the JEE Mains rank with the help of scores in the JEE Main exam . JEE Main 2024 Rank Predictor is a tool designed to provide aspirants with a probable rank based on their performance in the JEE Main exam. Candidates can use the JEE Main official answer key to predict the probable score. To use the JEE Main rank predictor free 2024 candidates have to provide scores out of 300. By inputting your JEE Mains 2024 scores, you can receive an estimate of your rank, helping you strategize for college admissions and counselling sessions. Read the full article to learn more about NTA JEE Main rank predictor 2024, how to register, how to use it, college predictor, and more.
Also Read: JEE Main 2024 answer key | JEE question paper | JEE Mains marks vs rank vs percentile

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How Does JEE Main Rank Predictor 2024 Work?

Careers360's JEE Main Rank Predictor helps candidates predict their probable ranks before the JEE Main results . Using sophisticated algorithms and data analytics, the JEE Main 2024 Rank Predictor analyzes your JEE Main scores and compares them with historical data and trends. By evaluating different parameters such as the difficulty level of the exam, the number of test takers, and past JEE Main cutoffs , the predictor generates a projected rank that closely aligns with your performance. The tool is created to resolve queries like how to calculate the JEE Main 2024 ranks before the results.

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How to use JEE Main Rank Predictor 2024?

Candidates can use the JEE Mains rank predictor by registering and providing the details required to predict rank. Aspirants can predict rank using the JEE Main rank predictor from percentile or score. The steps for using the JEE Mains rank predictor 2024 are-

Frequently Asked Question (FAQs) - JEE Main 2024 Rank Predictor - Calculate JEE Mains Rank from Percentile, Score

Question: How is JEE Main rank predictor useful?

Answer:

Candidates can predict their rank of JEE Main 2024 using the rank predictor tool.

Question: How can I predict my probable colleges as per my JEE Main rank 2024?

Answer:

Candidates can predict the list of colleges they can be eligible for admission using the JEE Main college predictor tool.

Question: How can I register on the JEE Main 2024 rank predictor?

Answer:

Candidates will be able to register online for using rank predictor of JEE Main 2024 by entering details like name, email Id, mobile number, current city, etc.

Question: How can I use the JEE Main rank predictor 2024?

Answer:

Candidates can simply register on the JEE Main 2024 rank predictor tool, enter the required details like scores, category, etc and determine their probable ranks.

Question: What is a JEE Main rank predictor tool?

Answer:

JEE Main rank predictor 2024 is a tool designed to calculate the probable ranks of the candidates on the basis of their probable scores in the exam.

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2 Views

Admission for next year 2025 Class 9 for JEE scholarship and 80 marks of test

Kanishka kaushikii 19th Sep, 2024

The admission process for Class 9 in 2025 for JEE scholarship and 80 marks of test can vary depending on the specific school or institution you are interested in. However, here are some general steps you can expect:

1. Research and Shortlist:

  • Identify schools or institutions that offer JEE scholarship programs for Class 9 students.
  • Consider factors such as location, academic reputation, facilities, and the specific JEE scholarship requirements.

2. Gather Information:

  • Visit the school's website or contact their admissions office to obtain the necessary application forms, deadlines, and eligibility criteria.
  • Inquire about the specific requirements for the JEE scholarship, including the test format, syllabus, and marking scheme.

3. Prepare for the Test:

  • Start preparing for the 80-mark test well in advance.
  • Focus on the relevant subjects and topics covered in the syllabus.
  • Practice solving sample questions and previous year papers to improve your test-taking skills.

4. Submit Application:

  • Complete the application form accurately and provide all the required documents, such as academic transcripts, certificates, and test scores.
  • Ensure that the application is submitted before the deadline.

5. Appear for the Test:

  • On the designated test date, arrive at the test center on time and follow the instructions provided.
  • Give your best effort to perform well in the test.

6. Await Results:

  • After the test, wait for the results to be announced.
  • Check the school's website or contact the admissions office for updates on the admission process.

7. Interview (if applicable):

  • If shortlisted, you may be invited for an interview to assess your suitability for the program.
  • Prepare for the interview by researching the school and practicing your responses to potential questions.

8. Enrollment:

  • If you are offered admission and have secured the JEE scholarship, complete the enrollment formalities as per the school's guidelines.

Remember to check the specific requirements and deadlines for the schools you are considering. It's also advisable to start the application process early to give yourself ample time for preparation and avoid missing any deadlines.

I hope it helps !!






2 Views

Newtons laws of motion imp formulas for jee mains

Kanishka kaushikii 18th Sep, 2024

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 !!

12 Views

im in class 12th and i had whole organic chemistry backlog and i covered goc and isomerism through sachin ranas yt playlist of 62 videos (not from unacademy) but now im thinking to start 12th oc through NS sir on mohit tyagi channel so what should i do cauz NS sir content will take whole november to cover it and my aim is adv

sadafreen356 18th Sep, 2024

If you have already finished whole organic chemistry from a particular playlist you should stick to that only dont change your plan again and again.

Set aside dedicated hours for studying your backlog and do a regular study schedule. You can divide your subjects into weeks and clear your backlogs subject.avoid studying piles of books just one book is enough to clear the exam.

10 Views

Nfsu Delhi Jee main cut off for btech mtech cse cyber security

Juee Hote 15th Sep, 2024

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.

10 Views

Im currently in class 12th and as like most of the students I wasted 2 years of my crucial time now I only got like 3 months max but Im confused on what to do or what not to can someone please help me!?

sadafreen356 15th Sep, 2024

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.

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