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JEE Main Question Paper with Solutions and Answer Keys- Download PDF 2024-2014

JEE Main Question Paper with Solutions and Answer Keys- Download PDF 2024-2014

Edited By Team Careers360 | Updated on Jul 17, 2024 05:11 PM IST | #JEE Main
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The JEE Main question paper is important study material to check the pattern and level of exam and previous year papers are effective study material to prepare for the national level engineering entrance exam. JEE Main Previous Year Question Papers pdf with solutions are provided here. Candidates appearing for the JEE Main exam can download the previous year's JEE Main question paper to check the difficulty of the entrance test. Moreover, candidates can solve the JEE Main previous year question paper to enhance the level of preparation. Careers360 has provided here the JEE Main Question Paper With Solution PDFs for the last 10 years.

JEE Main Question Papers with Solutions

The JEE Main question paper of previous year have been provided here. Students can download and view the question paper. Solutions of previous year question papers have also been provided here which refer to check the answers of questions either correct or incorrect.

Download Last 5 Years IIT JEE Mains Subject Wise PYQ PDFs

JEE Main 2024 Questions with Answers

JEE Main 2024 paper was conducted in 2 sessions. Let's have a look at the details of all shift papers from both sessions.

JEE Main 2024 Questions with Answers (Officials)

The JEE Main 2024 question paper for the January attempt has been provided here. Students can view and download the question paper. Students will also find the solutions for the questions which they can refer to check the answer either correct or incorrect.

JEE Main 2024 Questions and Paper Analysis (April Session)

JEE Main 2024 Questions and Paper Analysis (January Session)

You can check questions from all these shifts with detailed solutions in the below-given e-books

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From all 20 shifts of JEE Main 2024 paper, you will have a total of 1800 previous year questions to practice which are based on the same syllabus and exam pattern as that of JEE Main 2025.

JEE Main Previous Year Question Paper

Aspirants can check the previous year's JEE Main question paper below to prepare for the entrance exam.

JEE Main 2023 Question Paper with Solution PDF (April & January Sessions)

JEE main 2023 paper was conducted in 2 sessions. Candidates can download the JEE Mains 2023 Question Paper with Solutions PDF for both the January and April Sessions. It is advised to solve as many JEE Main 2023 all-shift question paper pdf as you can. These JEE Mains 2023 question papers consist of JEE Main chapter-wise pyq.

JEE Main 2023 Question Paper With Solution (April Session)

JEE Main 2023 Question Paper With Solution (January Session)

You can check questions from all these shifts with detailed solutions in the below-given e-books

JEE Main Exam's High Scoring Chapters and Topics
This free eBook covers JEE Main important chapters & topics to study just 40% of the syllabus and score up to 100% marks in the examination.
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In JEE Main 2023, a total of 24 shifts were there from both the January and April sessions. Thus, students can practice from a total of 2160 previous year question papers of JEE Main for a better understanding of the difficulty level and exam pattern of JEE Main 2025 paper.

JEE Main 2022 Question Paper PDF With Solution (Session 1 and 2)

JEE main 2022 paper was conducted in 2 sessions. Let's have a look at the details of all shift papers from both sessions.

JEE Main 2022 Question Paper Session 2 (July)

JEE Main 2022 Question Paper Session 1 (June)

You can check questions from all these shifts with detailed solutions in the below-given e-books

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In JEE Main 2022, a total of 22 shifts were there from both the June and July sessions. Aspirant preparing for JEE Main 2025 must go through these 1980 previous year question papers of JEE Main 2022 for better understanding and scoring higher marks in JEE Main 2025 exam.

JEE Main 2021 Question Paper PDF With Solution (All Sessions)

JEE main 2021 paper was conducted in 4 sessions. You can check questions from all these shifts with detailed solutions in the below-given e-books

In JEE Main 2021, a total of 26 shifts were there from all 4 sessions. Thus, students preparing for the JEE Main 2025 exam can practice from a total of, 2340 previous year question papers of JEE Main 2021. This activity is very helpful for the students.

JEE Main 2021 Question Paper, July session

JEE Main 2021 Question Paper PDF With Solution February Session

JEE Main 2020 Question Paper PDF with Solution (All Sessions)

JEE main 2020 paper was conducted in 2 sessions. You can check questions from all these shifts with detailed solutions in the below-given e-books

JEE Main 2020 Question Paper PDF With Solution January Session

JEE Main 2019 Question Papers PDF With Solution (April & January Session)

JEE main 2019 paper was conducted in 2 sessions involving 1440 question from 16 different shifts held on different days. You can check questions from all these shifts with detailed solutions in the below-given e-books

JEE Main 2019 Question Paper PDF With Solution April Session

JEE Main 2019 Question Paper PDF With Solution January Session

JEE Main 2018 Question Paper

JEE Main 2017/ 2016 Question Paper

JEE Main 2015 Question Paper

JEE Main 2014 Question Paper

Download- JEE Main 2013 Solved paper

JEE Main Previous Year Important Questions With Solutions

1. (Subject - Physics, Chapter - Properties of Solids and Liquids, Shift - 27 July-Evening Shift 2022)

Question:- A steel wire of length 3.2 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{s}}=2.0 \times 10^{11} \mathrm{Nm}^{-2}\right)and a copper wire of length 4.4 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{c}}=1.1 \times 10^{11} \mathrm{Nm}^{-2}\right), both of radius, 1.4 \mathrm{~mm} are connected end to end. When stretched by a load, the net elongation is found to be 1.4 \mathrm{~mm}. The load applied, in Newton, will be: \quad\left(\right.$ Given $\left.\pi=\frac{22}{7}\right)

  1. 360

  2. 180

  3. 1080

  4. 154

Answer- 4) 154

Solution:-

\mathrm{\ell_{\mathrm{s}}=3.2 \mathrm{~m}}

\mathrm{Y_s=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2}

\mathrm{\ell_c=4.4 \mathrm{~m}}

\mathrm{Y_c=1.1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2}

\mathrm{r=1.4 \mathrm{~mm}=1.4 \times 10^{-3} \mathrm{~m}}

\mathrm{\Delta \ell=1.4 \mathrm{~mm}=1.4 \times 10^{-3} \mathrm{~m}}

\mathrm{F=\left(\frac{Y A}{\ell}\right) \Delta \ell}

1720517024596

\mathrm{K=\left ( \frac{YA}{\ell} \right )\Delta \ell}

\mathrm{\frac{1}{k_{e q}}=\frac{1}{k_1}+\frac{1}{k_2} }

\mathrm{\frac{1}{K_{e q}}=\frac{\ell_s}{Y_s A_s}+\frac{\ell_c}{Y_c A_C}}

\mathrm{\frac{1}{K_{\text {eq }}}=\frac{3.2}{A \times 2 \times 10^{11}}+\frac{4.4}{A \times 1.1 \times 10^{11}}}

\mathrm{taking\: A_{s}=A_{c}=A}

\mathrm{\frac{1}{K_{e q}} =\frac{1.6 \times 10^{-11}}{A}+\frac{4 \times 10^{-11}}{A} }

\mathrm{K_{e q} =\frac{A}{5.6 \times 10^{-11}} }

\mathrm{K_{e q} =\frac{A \times 10^{11}}{5.6} }

\mathrm{F =K_{e q}(\Delta \ell) }

\mathrm{=\frac{A \times 10^{11}}{5.6} \times 1.4 \times 10^{-3} }

\mathrm{=\frac{3.14 \times\left(1.4 \times 10^{-3}\right)^3 \times 10^{11}}{5.6} }

\mathrm{F =\frac{22\times(1.4)^2 \times 10^2}{7\times 4} }

\mathrm{F =154 \mathrm{~N}}

Hence, 4 is correct option


2. (Subject - Maths, Chapter - Trigonometry, Shift - 15 Apr - Morning Shift 2023)

Question:- If the domain of the function f(x)-\log _e\left(4 x^2+11 x+6\right)+\sin ^{-1}(4 x+3)+\cos ^{-1} \frac{10 x+6}{3} is (\alpha, \beta), then 36|\alpha+\beta| is equal to

  1. 72

  2. 63

  3. 45

  4. 54

Answer- 3) 45

Solution:-

\begin{aligned} &f(x)=\ln \left(4 x^2+11 x+6\right)+\sin ^{-1}(4 x+3)\\ & +\cos ^{-1}\left(\frac{10 x+6}{3}\right) \\ & \text { (i) } 4 x^2+11 x+6>0 \\ & 4 x^2+8 x+3 x+6>0 \\ & (4 x+3)(x+2)>0 \\ & x \in(-\infty,-2) \cup\left(-\frac{3}{4}, \infty\right) \\ & \text { (ii) } 4 x+3 \in[-1,1] \\ & x \in[-1,-1 / 2] \\ & \text { (iii) } \frac{10 x+6}{3} \in[-1,1] \\ & x \in\left[-\frac{9}{10},-\frac{3}{10}\right] \\ & x \in\left(-\frac{3}{4},-\frac{1}{2}\right] \\ & \alpha+\beta=-\frac{5}{4} \\ & 36|\alpha+\beta|=45 \end{aligned}

Hence, the answer is the option (3).

3. (Subject - Chemistry, Chapter - Hydrocarbon, Shift - 9 Apr - Evening Shift 2024)

Question:- Number of compounds from the following which cannot undergo Friedel-Crafts reactions is :____ toluene, nitrobenzene, xylene, cumene, aniline, chlorobenzene, m-nitroaniline, m-dinitrobenzene

Answer- 3) 4

Solution:-

Compounds which can not undergo Friedel Crafts reaction are

1720517024083

JEE Main Question Paper with Solutions PDF Download Links

IIT JEE Mains previous year paper pdf is available at careers360.com. Students can download these IIT JEE Mains previous year question papers to prepare for the exam. These IIT JEE Mains question papers guide students about the difficulty level of the exam, and the weightage of Class 11 and 12 subjects. JEE Mains previous year questions pdf download link is easily accessible here. It is suggested to solve JEE Mains previous paper after preparing all topics to check preparedness. NTA JEE Mains previous papers are a helpful resource for exam preparation.

How to download the JEE Main Question Papers With Solutions?

Careers360 has collected different sets of JEE Main question papers which their experts have solved to provide the right solution for preparation. Below are the steps to download the JEE Mains previous year paper with solutions.

  • Go to the website- engineering.careers360.com

  • Click on the “Sample Paper" section

  • Now, select the "View All" option

  • From the exam filter, Select "JEE Main"

  • This will list down all the JEE Main previous year question papers and preparation e-books

  • Download the JEE Mains question paper pdf and solve them to know your preparation level

Benefits of Solving JEE Main Previous Year Question Paper

Practicing the previous year JEE Main question paper is an important part of exam preparation. Students can avail of several benefits of solving these JEE Main previous year paper pdf that are listed below.

  • Understand paper pattern: By solving the JEE Main question paper of the previous year pdf, candidates can get an idea of the type of questions asked in the exam, the weightage of different topics, and the overall JEE Main exam pattern. This helps them prepare a better strategy for the exam.

  • Improves time management: Solving the previous year JEE Main question paper helps candidates get familiar with the exam format and manage their time effectively. It also helps them identify which sections they need to focus on to improve their speed and accuracy.

  • Boosts confidence: Regularly practicing the previous year JEE Main question paper pdf with solutions can help candidates build their confidence and reduce anxiety before the actual exam. They can identify their strengths and weaknesses and work on improving their weak areas.

  • Provides practice for the actual exam: JEE Mains previous year papers provide candidates with an opportunity to practice the type of questions they are likely to encounter in the actual exam. This helps them familiarize themselves with the exam format and better prepare for the actual exam.

  • Helps in revision: Practicing JEE Main previous question papers is an effective way to revise the concepts learned during the preparation. By solving the JEE Main question paper of the previous year pdf, candidates can reinforce their understanding of the topics and identify any gaps in their knowledge.

Frequently Asked Question (FAQs)

1. Can I get JEE Main 2024 question papers in Hindi language?

Yes, JEE Main 2024 question papers are available in both English and Hindi languages.

2. Is it necessary to solve JEE Main previous year question papers?

Yes, solving JEE Main previous year question papers can help in identifying the important topics, understanding the exam pattern, and improving time management skills.

3. Can I challenge the JEE Main 2025 answer key?

Yes, NTA provides the facility to challenge the JEE Main 2025 answer key within a specified time period. Candidates have to pay a certain fee for each question they challenge.

4. How can I use JEE Main 2024 question papers for practice?

Candidates can use JEE Main 2024 question papers for practice by solving them in a timed manner, analysing their performance, and identifying areas where they need improvement.

5. Are JEE Main 2024 question papers available offline?

No, JEE Main 2024 question papers are not available offline. Candidates can download them in PDF format from the official website of NTA.

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Bhai ye to hona hi tha song ft of Baroda Gujarat and your family a very
Mujhe kuch samajh nahi aaya tha tab Tak hai Jaan the number of the kncjdsh

Hello,


Yes, even after clearing JEE, you need to pay fees to study in a college. The fee structure varies depending on the college and course you choose. Here's a general idea:


1. Government Colleges: Fees are relatively low, around 5,000 to 50,000 per year.

2. Government-Aided Colleges: Fees are moderate, around 50,000 to 1,50,000 per year.

3. Private Colleges: Fees can be higher, ranging from 1,50,000 to 5,00,000 or more per year.

4. IITs (Indian Institutes of Technology): Fees are around 2,00,000 to 3,00,000 per year.

5. NITs (National Institutes of Technology): Fees are around 1,50,000 to 2,50,000 per year.


Additionally, you may need to pay for:


· Hostel fees (if you opt for on-campus accommodation)

· Mess fees (for food)

· Other expenses (books, materials, etc.)


Hope this helps,

Thank you

Hi there,

Congratulations on getting a percentile of 96.21 in the jee mains examination.

At this percentile you will be certainly in the top list of candidates participating in the exam and the counselling. You can also check about the previous year cutoffs and the opening and closing ranks by visiting the official website of the nta and then checking about the score required to get an admission in the nits.

Also, you can take the help of our college predictor which will definitely help you in finding the exact list of the colleges which are suitable for admission based upon your jee mains rank and score.

The link of the predictor is mentioned below. You can take  the help from there.

https://engineering.careers360.com/jee-main-college-predictor?utm_source=qna&utm_medium=jee_cp

Hope this resolves your query.

Yes you can participate in Centralized Admission Process (CAP) conducted by the State Common Entrance Test Cell, Maharashtra with your JEE mains score under 15% jee quota.some private universities such as Pict ,Spit, vit ,rocem take admission under this quota but Vjti,Coep state government dont accept JEE marks.for counselling you need to report at the time of counselling and get your documents verified then the merit list will be displayed and admission process is done by the authority.for more refer link

https://engineering.careers360.com/articles/mht-cet-counselling-through-jee-main

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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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