JEE Main 2024 Question Paper With Solutions PDF

JEE Main 2024 Question Paper With Solutions PDF

Vikash Kumar VishwakarmaUpdated on 01 Nov 2024, 09:28 AM IST

JEE Main 2024 question paper - Download the 2024 question paper with the solution in pdf format from this article. JEE Main ( Joint Entrance Examination) is one of the most common exams of engineering. JEE Main Exam is conducted in two sessions one in January and the other In April. Herein, in this article, we provide a link to download the JEE Main 2024 question paper with solutions for both sessions. NTA will publish the JEE Main Question Paper 2025 within a week of the JEE Main exam conclusion. The JEE Main answer key along with the question paper will be released at jeemain.nta.ac.in.

LiveJEE Main 2026 Registration LIVE: NTA JEE application form this month; latest update on documentsOct 6, 2025 | 10:11 PM IST

The JEE Main 2026 eligibility criteria are as follows.

Age limit

No age limit

Academic eligibility

Class 12 or equivalent level

Year of the qualifying exam

2024/ 2025 / 2026

Limit of attempts

Three consecutive years

Class 12th subjects

  • BE/BTech- Physics, Mathematics and Chemistry/ Biology/ Biotechnology/ Technical Vocational Subjects
  • B.Arch- Maths, Physics and Chemistry
  • B.Plan- Maths

Required 12th class percentage

  • BTech in NITs/ IIITs/ GFTIs- 75% (general) or 65% (SC/ST)
  • B.Arch/ B. Plan- 50% aggregate scores
Read More
JEE Main 2024 Question Paper With Solutions PDF
JEE Main 2024 Question Paper PDF With Solution

To download the official JEE Mains 2024 question paper pdf, candidates can go through this article and download the corresponding pdf by clicking on the corresponding links provided in the article.

JEE Main Question Paper 2024 Session 1

Below are links to download the 2024 session 1 question paper with the answer key. Using these links you can easily access the question paper and answer key. Below is the link for the JEE Main Question Paper 2024 pdf.

JEE Main 2024 Questions Paper Analysis with Answers

JEE Main 2024 paper was conducted in 2 sessions. Let's have a look at the detailed analysis of both session question papers by experts of careers360. JEE Main 2024 question paper based on the latest syllabus and latest pattern which was changed by NTA.

JEE Main 2024 Questions and Paper Analysis (April Session)

JEE Main 2024 Questions and Paper Analysis (January Session)

How to download the JEE Main 2024 Question Papers With Solutions?

Here are the steps to download the JEE Mains 2024 previous year paper with solutions.

  • Click on the links provided in the tables above.

  • Register using your email ID.

  • The JEE question paper pdf download link will be sent to the registered email ID.

  • Simply download it and view the file.

Also Check:

JEE Main Last 10 Years Question Papers with Solutions

JEE Main Question Paper with Solutions and Answer Keys

JEE Main Syllabus: Subjects & Chapters
Select your preferred subject to view the chapters

Benefits of Solving JEE Main 2024 Question Paper

The previous year's JEE Main question papers are an important resource for exam preparation. The JEE Main 2024 question paper based on the latest syllabus and latest pattern offers the following benefits-

  • Helps in revising and identifying topics that need more practice and conceptual clarity.

  • Students gain a better understanding of the JEE Main exam pattern, frequently appearing topics and more. It acts as an effective preparation strategy for the exam.

  • Candidates can improve their time management skills. They can also use it to determine which sections to concentrate on for better accuracy and speed.

  • Regular practice of JEE Main question paper pdf with solutions can help aspirants gain confidence and reduce anxiety before the exam.

Amity University Noida B.Tech Admissions 2025

Among Top 30 National Universities for Engineering (NIRF 2024) | 30+ Specializations | AI Powered Learning & State-of-the-Art Facilities

NIELIT University(Govt. of India Institution) Admissions

Campuses in Ropar, Agartala, Aizawl, Ajmer, Aurangabad, Calicut, Imphal, Itanagar, Kohima, Gorakhpur, Patna & Srinagar

JEE Main Previous Year Questions 2024

Chemistry

1. Cyclohexene 1721277874250is __________ type of an organic compound.

a)Benzenoid non-aromatic

b)Benzenoid aromatic

c)Alicyclic

d)Acyclic

Solution : 1721277874171 is alicyclic compound

2. Which of the following statements are correct?

A. Glycerol is purified by vacuum distillation because it decomposes at its normal boiling point.

B. Aniline can be purified by steam distillation as aniline is miscible in water.

C. Ethanol can be separated from the ethanol-water mixture by azeotropic distillation because it forms azeotrope.

D. An organic compound is pure, if mixed M.P. remains the same.

Choose the most appropriate answer from the options given below :

Solution: Option (B) is incorrect because aniline is immiscible in water.


Physics

1. By what percentage will the illumination of the lamp decrease if the current drops by 20%?

a)56% b) 26% c) 36% d)46%

Solution: Heat and Power developed in a resistor

Heat developed in a resistor: When a steady current flows through a resistance R for time t, the loss in electric potential energy appears as increased thermal energy(Heat H) of the resistor and H=i^2RtThe power developed = \frac{energy}{time}=i^2R=iR=\frac{V^2}{R} (from Ohm's law)

The unit of heat is the joule (J)

The unit of power is the watt (W)

2. A block of mass 5 kg is there on a smooth inclined plane which is making 30 degrees angle with the horizontal, And a hanging mass of 6 kg is attached with it by a string which passes through a pulley; the acceleration of the system is

  1. 4m/s-2 b)1.8m.s-2 c) 4.5m/s-2 d) 5m/s-2

JEE Main 2026: Preparation Tips & Study Plan
Download the JEE Main 2026 Preparation Tips PDF to boost your exam strategy. Get expert insights on managing study material, focusing on key topics and high-weightage chapters.
Download EBook

Solution: Let us see the fbd for the system


1721277875916

For acceleration, we can write the equation as


a=\frac{6g-5gsin30^{o}}{6+5}

\Rightarrow a=\frac{60-(50\times \frac{1}{2})}{11}=3.18ms^{-2}

Maths

1. If the mean and variance of five observations are \frac{24}{5} and \frac{194}{25} respectively and the mean of the first four observations is \frac{7}{2}, then the variance of the first four observations in equal to

a)\frac{4}{5} b)\frac{5}{4} c)\frac{3}{5} d)\frac{105}{4}

Solution: MEAN (Arithmetic Mean)

The mean is the sum of the value of each observation in a dataset divided by the number of observations.

Mean of the Ungrouped Data

If n observations in data are x1, x2, x3, ……, xn, then arithmetic mean \mathit{\bar x} is given by

\mathit{\bar x}=\frac{x_1+x_2+x_3+\ldots\dots +x_n }{n}=\frac{1}{n}\sum^{n}_{i=1}x_i

Variance and Standard Deviation

The mean of the squares of the deviations from the mean is called the variance and is denoted by σ2 (read as sigma square).

Variance is a quantity that leads to a proper measure of dispersion.

The variance of n observations x1 , x2 ,..., xn is given by

\sigma^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}

\bar{X}=\frac{24}{5} ; \sigma^2=\frac{194}{25}

Let the first four observations be \mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4

Here, \frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}

Also, \frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4}{4}=\frac{7}{2}\Rightarrow \mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4=14

Now from eqn -1

\mathrm{x}_5=10 Now, \sigma^2=\frac{194}{25}

\begin{aligned} & \frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}{5}-\frac{576}{25}=\frac{194}{25} \\ & \Rightarrow x_1^2+x_2^2+x_3^2+x_4^2=54 \end{aligned}

Now, the variance of the first 4 observations

\mathrm{Var}=\frac{\sum_{\mathrm{i}=1}^4 \mathrm{x}_{\mathrm{i}}^2}{4}-\left(\frac{\sum_{\mathrm{i}=1}^4 \mathrm{x}_{\mathrm{i}}}{4}\right)^2=\frac{54}{4}-\frac{49}{4}=\frac{5}{4}

2. If the general term of the first AP = 3n-5 and that of the second AP is 3n+5. Then the 50th term of AP where both the AP are added

a)155 b) 145 c) 300 d)150

Solution: As we learned,

Properties of an AP:

If a_{1},a_{2},a_{3}\: \: and\: b_{1},b_{2},b_{3} are two APs,

then a_{1}+ b_{1},a_{2}+ b_{2,}a_{3}+ b_{3} is also an AP.

In this Question,

T_n=3n+5+3n-5= 6n

T_{50}=6\times 50=300

Articles
|
Upcoming Engineering Exams
Certifications By Top Providers
Basic Programming using Python
Via Indian Institute of Technology Bombay
Introduction to Aerospace Engineering
Via Indian Institute of Technology Bombay
Fundamental Concepts of Electricity
Via Indian Institute of Technology Bombay
Artificial Intelligence
Via Columbia University, New York
Computer Fundamentals
Via Devi Ahilya Vishwavidyalaya, Indore
B.Sc in Programming and Data Science
Via Indian Institute of Technology Madras
Udemy
 1525 courses
Swayam
 817 courses
NPTEL
 773 courses
Coursera
 697 courses
Edx
 608 courses
Explore Top Universities Across Globe

Questions related to JEE Main

On Question asked by student community

Have a question related to JEE Main ?

Hello,

You cannot "renew" your September 2025 certificate. You must apply for a completely new EWS certificate after April 1, 2026.

EWS certificates are linked to a specific financial year.

Your certificate issued in September 2025 is valid for the financial year 2025-26. It is based on your family's income from the previous financial year (i.e., 2024-25).

The certificate required for JEE counselling in mid-2026 must be valid for the financial year 2026-27. This certificate must be based on your family's income from the financial year 2025-26 and must be issued on or after April 1, 2026.

Since they are based on two different income periods, they are considered completely different documents.


Hope it's helpful to you.

For JEE Main 2026, NTA has said that students should update their Aadhaar, category certificate, and UDID if applicable before applying. During registration, you will need to upload your 10th marksheet, photo, signature, and ID proof. The registration is expected to start in October or November 2025.

https://jeemain.nta.ac.in/


Hi! Don’t worry, you can correct your category this year while filling out the JEE Main 2025 application form. If you belong to BC/OBC category and have the valid caste certificate, you can apply under OBC NCL now. It will not be a problem if last year you had applied as General/EWS, since each year’s application is considered separately.
OBC NCL is for candidates from socially and educationally backward communities (non creamy layer) with income below 8 lakh, while EWS is for students from the general category (not SC/ST/OBC) with family income below 8 lakh. Since your family income is less than 2 lakh and you are BC, applying under OBC NCL is the right choice and will give you reservation benefits.

Hello,

To get below AIR 100 in JEE Advanced 2027, you need a strong and focused plan. Here’s a simple guide:

  1. Clear Concepts : Make sure all NCERT basics are strong. Focus on Physics, Chemistry, and Maths concepts deeply.

  2. Advanced Practice : Solve previous years’ JEE Advanced papers and high-level questions from books like HC Verma, I.E. Irodov, OP Tandon, and Cengage.

  3. Regular Mock Tests : Take full-length tests every week. Analyse mistakes and improve weak areas.

  4. Time Management : Learn to solve questions quickly and accurately. Work on speed and accuracy together.

  5. Short Notes : Make notes/formulas for quick revision.

  6. Consistency : Study daily and avoid long breaks. Consistent practice is key.

  7. Focus on Weak Areas : Identify topics you struggle with and spend extra time on them.

  8. Stress Management : Stay healthy, sleep well, and avoid last-minute panic.

If you follow a structured plan and stay consistent, AIR below 100 is achievable.

Hope it helps !