HRD Minister Launches AI-powered Mock Test App for JEE & NEET Students


Union HRD Minister Mr Ramesh Pokhriyal Nishank launched an AI-powered mock test App for JEE and NEET 2020 Students. Its an android app ‘National Test Abhyas’ for students who are going to appear in JEE Main 2020 and NEET 2020 examinations.

NTA plans to release one new mock test on the app everyday, which students can then download and attempt offline. Once the test is completed, students can go online again to submit the test and view their test report.

Every student is unique and requires specific guidance to detect and overcome gaps in knowledge and test-taking strategy.

eSaral offers personalized learning platform where a student can study and attempt test at its own pace. eSaral also offers one to one Mentoring to help each and every student in academic and non-academic related issues.


Prepare for JEE & NEET with the Kota’s Top IITian & Doctor Faculties !

Quickly Revise All the Topics – Absolute FREE REVISION WITH MIND-MAPS!


This time is very crucial for all those who are preparing for the upcoming JEE and NEET Exams. To ensure that learnind doesn’t affect and no student is left behind in getting exposure to practice testing. Click Here to download eSaral App!

Why eSaral ?

eSaral offers a broad categories of mock tests for students preparing for JEE 2020 and NEET 2020 Examinations. With the Part and full syllabus mock tests, eSaral App also provides topic wise tests that are very helpful in analyzing weak and strong areas and also helpful in revising important concepts from each and every chapter individually.

For JEE 2020

For NEET 2020

Importance of Practicing Mock Tests

  1. Practicing Mock tests helps to re-examine the concepts that you have learned before.
  2. It also helps in Analyzing the weak and strong areas. Always try to strengthen strong points and work optimistically on weaker sections.
  3. Helps in improving Speed and presence of mind. As the students gets acclimatized to the types of questions asked in mock tests.
  4. Time-Management: Time allotted for the exam is limited. Since the Exam will held in Online Mode, so practicing online mock tests will help the students to get acquainted with Exam duration and attempting exam in online mode rather than pen – paper. More practice will help the students in managing time between easy and difficult questions.
  5. To Shed Exam Fear – To practice mock test works very efficiently in improving confidence level and eliminating exam fear. Practice 3 Hours at a stretch and solve questions for a sustained period of time without breaks.
  6. Helps in developing right strategy to attempt exam.

Practicing mock tests helps the students in knowing where they stand so they can prepare better next time.

Now we hope that you have understood the importance of JEE Main & NEET Mock Tests, make sure you sign up for mock test.

Start Early to take a lead. Time waits for none.


 

CRACK JEE 2020 | How to Prepare for JEE Main in last 2 Months


Less than 2 months remaining for JEE Main 2020 (July Attempt), but at the same time it is still possible to secure good percentile and an All India Rank to get qualified for JEE advanced and also to get admissions in some Top NIT’s. Read the complete article till the end to know How to Prepare for JEE Main in 2 Months.

After a long wait the dates for JEE Main 2020 has finally been announced. The exam will be held between 18th and 23rd July 2020. Although students have had an extended preparation time ever since we got into preventive lockdown due to COVID 19 pandemic.

Students were finding it bit difficult to concentrate in the absence of a specific target exam date. Now the students do have a specific date in front of them to look forward to. Whenever there is less time remaining, you have to put more effort into your studies to practice enough questions for good JEE preparation.

JEE Main 2020 Exam Preparation Tips for Last two remaining months: 

1. Practice mock tests and solve previous year question papers

Solving previous year question papers of JEE Main and taking mock tests is extremely important to get a good rank in the exam. Attempt the various free mock tests available on eSaral App and get the feel of the actual exam and improve your time management skills.

Solve at least ONE full syllabus online test every week. Solve questions by applying your reasoning and analytical skills. Develop test taking strategy, it will evolve with more tests that you attempt. Also solve previous years JEE Main question paper bank. While taking the tests students should look at building the right exam temperament, adjusting the body clock and getting comfortable with test environment.


CLICK HERE for JEE MAIN Previous Year Topicwise Questions. 


2. Practice more and more conceptual problems and Clear your Doubts.

No matter what how big or small your doubt is, get it cleared. Download eSaral App & we will assist you in clarifying your doubts about a particular topic related to JEE Main. For clearing doubts on important topics and concepts, you can join a crash course.

Crash Course will also help you to complete class 11 & 12 JEE Topics covering both Mains & Advanced syllabus with complete package of test series. To know more about JEE 2020 Crash Course, CLICK HERE

3. Effective Revision | Technique

Ensure you make a checklist of JEE Main syllabus with the list of topics/units against each subject – Physics, Chemistry and Mathematics. Tick off the important topics as and when you are done with revising the particular topic.

For Quick Revision with Mind- Maps Visit the given Links Below:

JEE Main is not difficult, rather it is predictable and tricky. Hence, clarity of concept and familiarity with the question type is essential to do well, so revision of concepts is the key. Prepare a list of concepts, topic-wise. Keep marking the concepts which are your weak areas. Keep referring to the notes&important formulae. 

It is extremely important to be in question solving mode. So, after revising every concept/topic, solve an unsolved question. One must NOT refer to solution without an attempt to solve the question. So, sufficient time must be devoted. Identify mistakes and/or weak area & fix them instantly.You have to be quick and accurate too. 

4. Effective Revision with the help of Mind Maps & Flashcards

Have a large stack of flashcards handy. It should contain all the important JEE Main formulae, concepts and diagrams. These could also be used in the last minutes before JEE Main.Here are the list of Complete Mind Maps for Organic Chemistry & Physics

5. Analyse & Strengthen your weak areas

Every time you take a mock test, Analyze your mistakes and weak areas. Work on the subjects, topics and units which are you are weak at. Resolve and strengthen them well before the actual exam takes place. 

6. Performance Analysis during practice test

Analyze your performance after every test. Subject wise weak areas needs to be identified and revisit those concepts all over again to eradicate them. Attempt Mock Tests available on eSaral App to get an instant test analysis with performance report.

Don’t panic about less months remaining for JEE preparation. You just need to focus on more study hours and practice to crack the JEE exam in remaining time.

Your preparation in these final months before the exam will have a major impact on your performance and JEE Main percentile, ranks and scores. Furthermore, your time management skills clubbed with a positive attitude will make your chances of getting into the top tech institutes of the country better. Remember, it’s important to utilize each day efficiently to secure success in JEE Main 2020.

Stay away from social networking sites for a few month. Also maintain a good health, as fitness goal is paramount during this time. Light exercise, healthy diet, good sleep are the key aspects to remain fit. Take all necessary precaution as prescribed by our government & health department. 

Whatever is the score in JEE Main 2020 January exam, there will be an option to increase score & percentile in the upcoming attempt of JEE Main 2020.

Always have a positive mindset and Believe it that whatever you have prepared is more than enough for exams. A positive mindset helps you to retain more and you will be able to score well in exams.

Kindly share your reviews and let us know the queries regarding JEE Main 2020 Exam. Drop your messages in the comment box below & let us help you with the correct guidance.

All the best 🙂

 

JEE Advanced 2020 Admit Card (Postponed) | Exam on 23rd August ’20


IIT Delhi has postponed the JEE Advanced 2020 Admit Card release along with other events. Candidates who will complete the registration process before the last date, will be able to access the JEE Advanced admit card 2020 in online mode by providing their registration number, date of birth, mobile number and email ID. Information regarding the examination such as the venue address, timings, date and more will be made known through JEE Advanced admit card 2020. Without the admit card of JEE Advanced 2020, the candidates will not be allowed to attempt the examination. Since it is an important document, the candidates are advised to keep the JEE Advanced 2020 admit card safely till the end of the admission procedure.

How to Download JEE Advanced 2020 Admit Card?

To download JEE Advanced hall ticket 2020, candidates need to follow the steps given below:

  • Visit the official website of JEE Advanced 2020
  • Navigate to the ‘Candidate Login’ link
  • Enter your JEE Advanced 2020 registration number, Date of Birth, Mobile number and email id to login
  • JEE Advanced 2020 hall ticket will be displayed on the screen
  • Check all the details printed on admit card thoroughly
  • Download and take a printout for future reference

In case of any discrepancy in downloading the admit card of JEE Advanced 2020, aspirants have to immediately contact the officials of their respective zonal IIT.  

JEE Advanced Admit Card Important Dates:

Events

Dates

Release of JEE Advanced 2020 admit card/hall ticket To be announced
Last date to download JEE Advanced 2020 admit card To be announced
JEE Advanced 2020 exam  23-Aug-2020

 

Issues you might face while downloading JEE Advanced admit card 2020

  • Server issues: Once JEE Advanced hall tickets will be released, successful applicants can download their respective admit card. The server may be down due to a sudden traffic surge on the website.
  • Incorrect credentials: Candidates who forgot their login credentials will not be able to download JEE Advanced admit card 2020. However, they can retrieve their password by clicking the ‘Forgot Password’ tab on the official website.
  • Slow internet: This is one of the most common issues faced by JEE Advanced aspirants while downloading the admit card.

Errors in JEE Advanced admit card and ways to resolve them

  • Discrepancy/ies in JEE Advanced admit card 2020: For discrepancy in the particulars (photograph/signature) mentioned on the JEE Advanced hall ticket or on the confirmation page, candidates should immediately get in touch with the JEE Advanced authorities.
  • Admit card not received: Please note that JEE Advanced authorities will not send the admit cards by post to candidates. Candidates can download the admit card only from the official website of JEE Advanced 2020.

Documents to Carry to JEE Advanced Test Centre

Along with JEE Advanced 2020 admit card, candidates need to carry one of the following original photo identity proofs:

  • Aadhaar Card
  • Driving License
  • School/ College ID
  • Pan Card
  • Passport

JEE Advanced Previous Year Question Papers are available here.


JEE Advanced 2020 Syllabus

Click Here for JEE Advanced Complete Syllabus


 

NTA JEE Main 2020, NEET 2020 Exam dates Announced | Check here for details


The exam dates for NTA JEE Main 2020, NEET 2020 announced by the Union Minister of Human Resource Development Ramesh Pokhriyal Nishank on tuesday (5th May’20).

The JEE (Main) 2020 and NEET 2020 exams were earlier re-scheduled by the ministry, due to the nationwide lockdown in the wake of coronavirus outbreak. The new date for the JEE (Main) 2020 exams was since then awaited.

  1. Public Notice by NTA for JEE 2020 Students
  2. Public Notice by NTA for NEET 2020 Students

The JEE Main exam will be conducted from July 18 to July 23, 2020, Ramesh Pokhriyal Nishank said.

The NEET 2020 exam will be conducted on July 26, he said, while the JEE Advanced on 23rd August ’20.

JEE Main 2020: Public Notice by NTA

NEET 2020: Public Notice by NTA

According to the announcement, Union Minister of Human Resource Development Ramesh Pokhriyal Nishank said the examination process for the JEE (Main) 2020 will begin from July this year, while the session will begin by August.

The JEE Main exam will be conducted from July 18 to July 23, 2020, Ramesh Pokhriyal Nishank said.

The NEET 2020 exam will be conducted on July 26, he said, while the JEE Advanced on 23rd August ’20.

CRASH COURSE to CRACK JEE MAIN 2020 | Register NOW 

The date for the release of admit cards for JEE Main, NEET 2020 is yet not available.

This was the second interaction between Ramesh Pokhriyal Nishank and students. This session was earlier scheduled to be held on May 2, 2020. However, the same was postponed for May 5, for some reason.

Earlier he interacted on 27th April 2020 and answered some critical questions of students and their parents.

According to the National Testing Agency (NTA), for JEE Main over nine lakh candidates have registered, on the other hand over 15.93 lakh students are likely to take up the NEET 2020 exams.


JEE Advanced Previous Year Question Papers are available here.


JEE Advanced 2020 Syllabus

Click Here for JEE Advanced Complete Syllabus

JEE Main 2020 April Exam Postponed due to Corona-virus


JEE Main 2020 April Exam has been postponed by NTA due to Corona-virus Outbreak. The exam was scheduled to be held on April 5, 7, 9 and 11. In recent notice at official website, National Testing Agency on Wednesday (18th march) took the decision to postpone the JEE Main April 2020 Examination in pursuance of MHRC letter. New dates of JEE Main likely to be decided on March 31, 2020.

The new date will be decided in accordance with the board exams schedule and other competitive exams to ensure there is no clash.

The Ministry of Human Resource Development (MHRD) on Wednesday directed the Central Board of Secondary Education (CBSE) and all educational institutions in the country to postpone all exams including JEE Main till March 31 in view of coronavirus outbreak.

MHRD also stated that “While maintenance of academic calendars and exam schedule is important, equally important is the safety and security of the students who are appearing in various exams as also that of their teachers and parents”.

The Public Notice by NTA:

 

To ensure, your Learning never gets affected, eSaral has taken a lead in this regard and has made eSaral App FREE till 15th April’20.  

This step is taken by eSaral keeping in mind Safety and Future of students amidst ongoing health crisis. Now students can study from home safely and continue their learning.

eSaral App comprises complete courses for Class 9 to 12 including Video Tutorials, Complete Study Material designed by Top IITians & Doctor Faculties.

DOWNLOAD eSaral App from here !!!

Share this Information among your dear and near ones and continue your learning with eSaral by being safe and keeping safe.

CRASH COURSE for JEE 2020

To know more about JEE Main 2020 CLICK HERE

Difference Between Potentiometer and Voltmeter – Current Electricity || Class 12, JEE & NEET


The Potentiometer and the Voltmeter both are the voltage measuring devices. The main Difference Between Potentiometer and Voltmeteris that the potentiometer measures the emf of the circuit whereas voltmeter measures the end terminal voltage of the circuit. The other differences between the Potentiometer and the voltmeter are explained below in the comparison chart.

  1. Standardization of Potentiometer
  2. Key Differences

Comparison Chart

Difference between Potentiometer and Ammeter

Browse More Topics Related to Potentiometer:

Standardization of Potentiometer

The process of determination of potential gradient on wire of potentiometer is known as standardisation of potentiometer. A standard cell is one whose emf remains constant. Cadmium cell with emf 1.0186 V at $20^{\circ}$C is used as a standard cell. In laboratory a Daniel cell with emf 1.08 V is usually used as a standard cell.

standardization of Patentiometer
If $\ell_{0}$ is the balancing length for standard emf $E_{0}$ then potential gradient $x=\frac{E_{0}}{l_{0}}$

Standardization of Potentiometer

Key Differences:

The following are the key differences between Potentiometer and voltmeter.

  1. The Potentiometer is an instrument used for measuring the emf, whereas the voltmeter is a type of meter which measures the terminal voltage of the circuit.
  2. The Potentiometer accurately measures the potential difference because of zero internal resistance. Whereas, the voltmeter has a high internal resistance which causes the error in measurement. Thus the voltmeter approximately measures the voltage.
  3. The sensitivity of the Potentiometer is very high, i.e. it can measure small potential differences between the two points. The voltmeter has low sensitivity.
  4. The Potentiometer uses the null deflection type instrument whereas the voltmeter uses the deflection type instrument.
  5. The Potentiometer has infinite internal resistance, whereas the Potentiometer has high measurable resistance.

Watch out the Video: Applications of Potentiometer & its Construction by Saransh Sir.

Conclusion

The Potentiometer and voltmeter both measures the emf in volts. The Potentiometer is used in a circuit where the accurate value of voltage is required. For approximate calculation, the voltmeter is used.

 

Physics Revision Series by Saransh Sir (AIR 41 in IIT-JEE)

Also Read:

 

 

 

JEE Main 2020: NTA reopens Application Process | Apply before 24th May


  JEE Main 2020 Application Form – NTA has again released JEE Main application form 2020. The forms are available online on JEE Main Official Website. Fresh registered are also open for eligible 10+2 appeared / passed candidates, until May 24, 2020. The process to submit application form of JEE Main 2020 is the same as before.

Candidate need to click on the apply link below and register by entering name, e-mail id, etc. information. Apart from the details, candidates also have to upload images and pay application fee. The amount of JEE Main 2020 application fee varies as per the category and number of papers.

JEE Main application form correction reopens between May 25 to 31. Know more details on JEE Main 2020 application form below.

DIRECT LINK TO REGISTER IN JEE MAIN 2020

The application process of JEE Main 2020 includes registration, filling the application form, uploading scanned images, payment of application fee and taking the printout of the filled-in form. For JEE Main 2020 registration, candidates must check the eligibility criteria as prescribed by the NTA. Candidates who have missed the JEE Main 2020 January exam can appear for the JEE Main 2020 2nd session. 

Public Notice by NTA regarding JEE MAIN 2020 Application Process

 

About JEE Main 2020 Exam:

Joint Entrance Examination is organized by National Testing Agency to offer admission into courses like B.Tech / B.E., B.Arch, B.Plan. JEE Main is a computer based test that takes place at national level. The duration of the exam is 3 hours. It consists of multiple-choice questions and numerical value type questions. There is negative marking in the case of multiple choice questions. After qualifying this exam, the candidates can take admission in IIITs / CFTIs / NITs or can sit for JEE Advance exam.

IMPORTANT DATES AND EVENTS TO REMEMBER

  • The JEE Main Exam 2020 will be held across the country from July 18 to July 23 this year.
  • The application form process for JEE-Mains 2020 will be available on or before May 24, 2020.
  • The forms can be completed and submitted only up to 5 pm on May 24, whereas the submission of the fees can be done until 11.50 pm on May 24, 2020.
  • Candidates have been asked to pay the application fee online via debit card, net banking, UPI, or Paytm.
  • Students qualifying in the JEE-Mains exams will appear for the JEE Advanced exam which will be held later

 

List of Documents Required for Filling Application Form

Candidate needs some documents to fill and submit the application form. Check the list of required documents here:

  • scanned image of photograph
  • scanned image of signature
  • a valid e-mail id
  • valid and active mobile number
  • past academic mark sheets and certificates

The candidates have to upload Photo and Signatue in the application form of JEE Main 2020 and as per the given specifications:

Scanned image Format Size Remarks
Photo JPG/JPEG 10 kb–200 kb Colored or black/white
(but clear contrast)
Signature JPG/JPEG 4 kb–30 kb Must be in Running hand

JEE Main 2020 Exam Pattern

As per the eligibility criteria for, a few changes in the pattern of the question paper(s) and number of question(s) for B.E./B.Tech has been approved by the JEE Apex Board (JAB) for the conduct of JEE (Main)-2020 Examination.

 

Paper Subject with Number of Questions Type of Questions TIMING OF THE EXAMINATION(IST)
B.tech Mathematics – 25(20+5)

Physics – 25(20+5)

Chemistry – 25(20+5)

20 Objective Type – Multiple Choice Questions (MCQs) & 5 Questions with answer as numerical value, with equal weightage to Mathematics, Physics & Chemistry 1st Shift: 09:30 a.m. to 12:30 p.m.

2nd Shift: 02:30 p.m. to 05:30 p.m.

IMPORTANT DATES FOR JEE Main 2020 (2nd Session)

JEE Main 2020 Important Dates – 2nd Session (July’20)

EVENTS

JEE Main 2020 Dates

Application form availability February 7, 2020
Last date to apply May 24, 2020 (reopens)
Last date to upload images and pay application fee May 24, 2020
Correction in Particulars (last date)
May 31, 2020
Exam date 18 July – 23 July ’20
Release date of answer key and recorded responses To be notified later 
Release of answer key To be notified later 
Declaration of result To be notified later 
Announcement of NTA score To be notified later 

 

Eligibility Criteria

There is no age limit for the candidates to appear in JEE Main 2020 Examination. The candidates who have passed their 12th Examination in 2018, 2019 and appearing in 2020 are eligible to JEE Main Examination 2020.

Those candidates who cleared the class 12 exam in 2017 or before 2017 are not eligible to appear for JEE Main 2020 exam.

Educational Qualification

Candidates must have at least 5 subjects in class 12 exam or equivalent exam. Where Math, Physics and Chemistry are the essential subjects.

JEE Main 2020 Detailed Syllabus

 

FREE Revision Series For JEE 2020 | Quick Revision Videos 

👉Physics Revision Series

👉Chemistry Revision Series 

👉Mathematics Revision Series
JEE Main 2020 New Dates Announced | Check Here for Details


The exam dates for NTA JEE Main 2020 announced by the Union Minister of Human Resource Development Ramesh Pokhriyal Nishank on tuesday (5th May’20). The JEE (Main) 2020 exam was earlier re-scheduled by the ministry, due to the nationwide lockdown in the wake of coronavirus outbreak. The new date for the JEE (Main) 2020 exams was since then awaited.

According to the announcement, Union Minister of Human Resource Development Ramesh Pokhriyal Nishank said the examination process for the JEE (Main) 2020 will begin from July this year, while the session will begin by August.

The JEE Main exam will be conducted from July 18 to July 23, 2020, while the JEE Advanced in August, Ramesh Pokhriyal Nishank said.

Through JEE Main entrance exam, students will be able to get admission to BE/B.tech, B.Plan, and B.Arch degree courses at the various IITs (Indian Institute of Information Technology), NITs (National Institute of Technology) and other Centrally Funded Technical Institutions (CFTIs) across India.

Click Here for January JEE Main 2020 Exam Paper Analysis

 

JEE Main 2020 Exam Pattern

As per the eligibility criteria for, a few changes in the pattern of the question paper(s) and number of question(s) for B.E./B.Tech has been approved by the JEE Apex Board (JAB) for the conduct of JEE (Main)-2020 Examination.

 

Paper Subject with Number of Questions Type of Questions TIMING OF THE EXAMINATION(IST)
B.tech Mathematics – 25(20+5)

Physics – 25(20+5)

Chemistry – 25(20+5)

20 Objective Type – Multiple Choice Questions (MCQs) & 5 Questions with answer as numerical value, with equal weightage to Mathematics, Physics & Chemistry 1st Shift: 09:30 a.m. to 12:30 p.m.

2nd Shift: 02:30 p.m. to 05:30 p.m.

 

There is no age limit for the candidates to appear in JEE Main 2020 Examination. The candidates who have passed their 12th Examination in 2018, 2019 and appearing in 2020 are eligible to JEE Main Examination 2020.

Those candidates who cleared the class 12 exam in 2017 or before 2017 are not eligible to appear for JEE Main 2020 exam.

 

JEE Main 2020 Syllabus

 

Quick Revision Videos

👉Physics Revision Series by Saransh Gupta Sir (AIR-41) 

 


 

Question papers JEE main 2020
JEE MAIN 2020 Question Paper PDF Download | All Shifts (7th, 8th & 9th January)


NTA has successfully conducted JEE Main Jan 2020 exam in all the test centers and we understand that students are waiting for JEE MAIN 2020 Question Paper PDF format. Based on the reviews by the students, it is concluded that the exam is comparatively easy than the previous year. Furthermore, the questions asked in the exam were more conceptual based. So Question Papers PDF of all shifts are available here to download.

JEE Main 2020 Question Paper – Candidates can download JEE Main question paper for January 07, 08, 09, 2020, for Shift 1 and Shift 2 from this page.

 

JEE Main 2020 (January) – All Shifts Question Papers

S. No.  JEE Main 2020 (January) Shifts Question Papers
1.        January 7, 2020 Shift-1  Download PDF
2.        January 7, 2020 Shift-2 Download PDF
3.        January 8, 2020 Shift-1 Download PDF
4.        January 8, 2020 Shift-2 Download PDF
5.        January 9, 2020 Shift-1 Download PDF
6.        January 9, 2020 Shift-2 Download PDF

 

Click Here for January JEE Main 2020 Exam Paper Analysis

 

The candidates can challenge the answer keys online. The window to raise objections will be available for a week. In case, the NTA finds the objections raised are incorrect, they will publish the result of JEE Main. The tentative result declaration date is by January 31.

Candidates may check their JEE Main question papers from Here!!!

 

 

oday on 7th January 2020, it was the day, when JEE MAIN was held in the morning shift from 9:30-12:30 across India.JEE MAin 2020 Exam Analysis 
JEE Main 2020 Exam | January Attempt | Students’ Reactions and Reviews


JEE MAIN 2020 Students Reactions | 7th January (SHIFT-1)

JEE MAIN, one of the biggest examination in the country was scheduled today. The difficulty level and selection ratio are the well-known aspects of this exam but the number of candidates that appear every year, is also the factor that makes it so special. Every year more than ten lakhs of aspirants apply and compete for just only 30-35 thousand seats (approximately) of National Institute of Technologies and IIITs. Today on 7th January 2020, it was the day, when JEE Main 2020 Exam was held in the morning shift from 9:30-12:30 across India.

Many of you are also appeared or going to appear in near future and want to know the overall level of today’s exam. So, dedicated to students and in order to help them, eSaral is again here, providing you the details about today’s exam. In this video, our team members have gathered the information about the questions, paper pattern and the difficulty level of the exam. We went to some of the centers in KOTA city, yes!! The coaching city.

Get to know about the students reviews and reactions regarding JEE Main 2020 Exam:

 

So, watch the video to know how they feel about the exam. We asked them regarding the easy to tough subjects based on the questions and topics, also the marking scheme of the numerical type questions and the ratio of 11th and 12th standards topics in the exam.

So if you have these kind of doubts or want to know about the detail watch the video and based on the conversation with the aspirants we developed a PDF too. You can download the same to know in detail.

Here is the complete analysis of JEE Main 2020, 7th Jan SHIFT-1 Exam by Saransh Gupta Sir with Students who have appeared in Exam.

 

JEE MAIN 2020 Students Reactions | 7th January (SHIFT-2)

 

So, here are the students reviews on the Second shift of 7th January JEE Main Exam. As per the students reviews it can be concluded that both shifts of the day comprised of easy to moderate level questions. Watch out the video and comment your review of JEE Main Exam

 

JEE MAIN 2020 Students Reactions | 8th January (SHIFT-1)

 

 

JEE MAIN 2020 Students Reactions | 8th January (SHIFT-2)

 

JEE main 2020 paper analysis
JEE Main 2020 Paper Analysis | Discussion with Students


JEE MAIN 2020 Paper Analysis for 7th January SHIFT-1

Today on 7th January, the national level competition exam JEE MAIN was held across the nation and with the completion of the exam, there are many queries in the mind of aspirants. It is natural to feel a certain kind of anxiety after the exam like how others feel, the difficulty level of the exam and the expected cut-offs. Here, at eSaral we are providing you the first detailed analysis that is available to all the students. Watch the video to now about the questions’ level (according to eSaral students) and the topics wise questions from 11th and 12th standards.

The physics faculty at eSaral, Saransh Gupta sir and chemistry faculty Prateek Gupta sir are doing the detailed analysis of the JEE MAIN QUESTION PAPER as per the reviews by the students of eSaral. In the video, the questions from each subject and the topics from where they were asked are explained using prepometer, the tool designed by the eSaral. In the discussion session the students of eSaral told about the questions and difficulty level and the question that they faced in the examination.

JEE Main 2020 Paper Analysis for January 07th exam is updated here. Students are sharing their reviews of JEE Main Exam here. Watch out the complete video till the end to know in detail!

 

Know the Number of questions asked per chapter:

(Shift-1) – 7th Jan Download PDF

Watch the Reactions and Reviews of Students outside JEE Main Exam Center

 

In physics, it is found that many of the questions were direct and formula based related to the memory of the candidate. There were approximately 10 questions form the class 11th syllabus out of 25 questions.

Watch out the video if you want to know about the same kind of detailed analysis for chemistry and mathematics. You will also get some idea about specific questions and their solutions that were discussed in the analysis video.

So from our analysis we found the overall exam as easy to moderate level. Only a few questions were placed in the difficult to very difficult category and it was found that the question were not much confusing.

We have developed a PDF regarding the detailed analysis. Download the file and know about the exam.

If you think that this time you could not attempt your best attempt then don’t worry we are going to start our BOUNCE BACK CRASH COURSE for the April month JEE MAIN exam. Enroll and ace the exam.

The above PDF contains weightage of number of questions per chapter.

 

JEE MAIN 2020 Paper Analysis for 7th January SHIFT-2

 

 

P
 

 

Know the Number of questions asked per chapter:

(Shift-2) 7th Jan Download PDF 

JEE MAIN 2020 Paper Analysis for 8th January SHIFT-1

(Shift-1) 8th Jan Download PDF 

JEE MAIN 2020 Paper Analysis for 8th January SHIFT-2

(Shift-2) 8th Jan Download PDF

JEE MAIN 2020 Paper Analysis for 9th January SHIFT-1

(Shift-1) 9th Jan Download PDF 

JEE MAIN 2020 Paper Analysis for 9th January SHIFT-2

(Shift-2) 9th Jan Download PDF

 

JEE Main 2020 Question Paper JEE Main question paper for January 07, 08, 09, 2020, for Shift 1 and Shift 2 are given below. Download or View from here!!!

 

S. No.  JEE Main 2020 (January) Shifts Question Papers
1.        January 7, 2020 Shift-1 Download PDF
2.        January 7, 2020 Shift-2 Download PDF
3.        January 8, 2020 Shift-1 Download PDF
4.        January 8, 2020 Shift-2 Download PDF
5.        January 9, 2020 Shift-1 Download PDF
6.        January 9, 2020 Shift-2 Download PDF

 

👉 Click to Join Free Physics Revision Series by Saransh Gupta Sir (AIR-41, IIT-Bombay)

Liquid Solution – JEE Main Previous Year Question of with Solutions


JEE Main Previous Year Question of Chemistry with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Chemistry will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Chemistry Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

Simulator

 

Previous Years AIEEE/JEE Mains Questions

Q. A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the folloowing statements is correct regarding the behaviour of the solution ?

(1) The solution is non-ideal, showing –ve deviation from Raoult’s law

(2) n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s law

(3) The solution formed is an ideal solution.

(4) The solutionis non-ideal, showing +ve deviation from Raoult’s law

[AIEEE-2009]

Sol. (4)

(A) n–heptone : Non Polar

(B) Ethanol : Polar

$\mathrm{F}_{\mathrm{A}-\mathrm{B}}<\mathrm{F}_{\mathrm{A}-\mathrm{A}}, \mathrm{F}_{\mathrm{B}-\mathrm{B}} \Rightarrow+$ deviation


Q. Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively :-

(1) 400 and 600 (2) 500 and 600 (3) 200 and 300 (4) 300 and 400

[AIEEE-2009]

Sol. (1)

$550=\mathrm{P}_{\mathrm{A}}^{\circ} \times \frac{1}{4}+\mathrm{P}_{\mathrm{B}}^{\circ} \times \frac{3}{4}$

$560=\mathrm{P}_{\mathrm{A}}^{\circ} \times \frac{1}{5}+\mathrm{P}_{\mathrm{B}}^{\circ} \times \frac{4}{5}$

$\mathrm{P}_{\mathrm{A}}^{\circ}=400

\quad \mathrm{P}_{\mathrm{B}}^{\circ}=600$ torr


Q. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of

the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively.

Vapour pressure of the solution obtained by mixing 25.0 of heptane and 35 g of octane

will be (molar mass of heptane = 100 g $\mathrm{mol}^{-1}$ and of octane = 114 g $\mathrm{mol}^{-1}$) :-

(1) 144.5 kPa           (2) 72.0 kPa             (3) 36.1 kPa            (4) 96.2 kPa

[AIEEE-2010]

Sol. (2)

$\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{X}_{\mathrm{A}}=105 \times \frac{1 / 4}{1 / 4+0.307}$

$=105 \times 0.449=47.13 \mathrm{K} \mathrm{Pa}$

$\mathrm{P}_{\mathrm{B}}=\mathrm{P}_{\mathrm{B}}^{\circ} \mathrm{X}_{\mathrm{B}}=45 \times 0.551=24.795$

$\mathrm{P}_{\mathrm{T}}=\mathrm{P}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}=71.925 \mathrm{atm}$


Q. If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water $\left(\Delta \mathrm{T}_{\mathrm{f}}\right)$, when 0.01 mol of sodium sulphate isdissolved in 1 kg of water, is $\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right):$ :-

(1) 0.0186 K            (2) 0.0372 K              (3) 0.0558 K              (4) 0.0744 K

[AIEEE-2010]

Sol. (3)

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \cdot \mathrm{m}$

$=3 \times 1.86 \times 0.01 / 1$

$=0.0558 \mathrm{K}$


Q. The molality of a urea solution in which 0.0100g of urea, $\left.\left[\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\right]$ is added to 0.3000 $\mathrm{dm}^{3}$ of water at STP is :-

(1) 0.555 m

(2) $5.55 \times 10^{-4} \mathrm{m}$

(3) 33.3 m

(4) $3.33 \times 10^{-2} \mathrm{m}$

[AIEEE-2011]

Sol. (2)

$\mathrm{m}=\frac{\mathrm{n}}{\mathrm{W}(\mathrm{kg})}=\frac{0.01 / 60}{0.3 \mathrm{kg}}=5.55 \times 10^{-4} \mathrm{mol} / \mathrm{kg}$


Q. A 5% solution of cane sugar (molar mass 342) is isotonic with 1% of a solution of an unknown solute. The molar mass of unknown solute in g/mol is :-

(1) 136.2           (2) 171.2           (3) 68.4            (4) 34.2

[AIEEE-2011]

Sol. (3)

$\pi_{\mathrm{c.s}}=\pi_{\mathrm{Unk}}$

$\left(\frac{\mathrm{n}}{\mathrm{V}}\right)_{\mathrm{c.s.}} \mathrm{RT}=\left(\frac{\mathrm{n}}{\mathrm{V}}\right)_{\mathrm{unk} .} \mathrm{RT}$

$\frac{5 \times 10}{342}=\frac{1 \times 10}{\mathrm{M}}$

M = 68.4 gm/mol


Q. Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at – $6^{\circ} \mathrm{C}$ will be :

$\left(\mathrm{K}_{\mathrm{f}} \text { for water }=1.86 \mathrm{K} \mathrm{kgmol}^{-1}, \text { and molar mass of ethylene glycol }=62 \mathrm{gmol}^{-1}\right)$

(1) 400.00 g             (2) 304.60 g            (3) 804.32 g            (4) 204.30 g

[AIEEE-2011]

Sol. (3)

$6=1.86 \times \frac{\mathrm{w} / 62}{4} \Rightarrow \mathrm{w}=800 \mathrm{gm}$


Q. The degree of dissociation () of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the expression :-

$(1) \alpha=\frac{\mathrm{x}+\mathrm{y}-1}{\mathrm{i}-1}$

(2) $\alpha=\frac{\mathrm{x}+\mathrm{y}+1}{\mathrm{i}-1}$

(3) $\alpha=\frac{\mathrm{i}-1}{(\mathrm{x}+\mathrm{y}-1)}$

(4) $\alpha=\frac{\mathrm{i}-1}{\mathrm{x}+\mathrm{y}+1}$

[AIEEE-2011]

Sol. (3)


Q. $\mathrm{K}_{\mathrm{f}}$ for water is 1.86 K kg $\mathrm{mol}^{-1}$. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)$ must you add to get the freezing point of the solution lowered to –$-2.8^{\circ} \mathrm{C} ?$

(1) 27 g            (2) 72 g            (3) 93 g             (4) 39 g

[AIEEE-2012]

Sol. (3)

$2.8=1.86 \times \frac{\mathrm{w} / 62}{1} \Rightarrow \mathrm{w}=93.33 \mathrm{gm}$


Q. A solution containing 0.85 g of $\mathrm{ZnCl}_{2}$ in 125.0 g of water freezes at $-0.23^{\circ} \mathrm{C}$ . The apparent degree of dissociation of the salt is :

($\mathbf{k}_{f}$ for water = 1.86 K kg $\mathrm{mol}^{-1}$, atomic mass ; Zn = 65.3 and Cl = 35.5)

(1) 1.36%            (2) 2.47%              (3) 73.5%             (4) 7.35%

[Jee (Main)-2012 online]

Sol. (3)

$0.23=(1+2 \alpha) \times 1.86 \times \frac{0.85 / 134.5}{0.125}$

$\alpha=0.735=73.5 \%$


Q. Liquids A and B form an ideal solution. At $30^{\circ}$C, the total vapour pressure of a solution containing 1 mol of A and 2 moles of B is 250 mm Hg. The total vapour pressure becomes 300 mm Hg when 1 more mol of A is added to the first solution. The vapour pressures of pure A and B at the same temperature are

(1) 450, 150 mm Hg

(2) 250, 300 mm Hg

(3) 125, 150 mm Hg

(4) 150, 450 mm Hg

[Jee (Main)-2012 online]

Sol. (1)

$250=\mathrm{P}_{\mathrm{A}}^{0} \times \frac{1}{3}+\mathrm{P}_{\mathrm{B}}^{0} \times \frac{2}{3}$

$300=\mathrm{P}_{\mathrm{A}}^{0} \times \frac{1}{2}+\mathrm{P}_{\mathrm{B}}^{0} \times \frac{1}{2}$

$\mathrm{P}_{\mathrm{A}}^{0}=450 \mathrm{mm}$

$\mathrm{P}_{\mathrm{B}}^{0}=150 \mathrm{mm}$


Q. The freezing point of a 1.00 m aqueous solution of HF is found to be $-1.91^{\circ} \mathrm{C}$. The

freezing point constant of water, $\mathrm{K}_{\mathrm{f}}$, is 1.86 K kg $\mathrm{mol}^{-1}$. The percentage dissociation of HF at this concentration is

(1) 2.7%             (2) 30%            (3) 10%             (4) 5.2%

[Jee (Main)-2012 online]

Sol. (1)

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \mathrm{m}$

$1.91=(1+\alpha) \times 1.86 \times 1$

$\alpha=0.027$


Q. How many grams of methyl alcohol should be added to 10 litre tank of water to prevent its freezing at 268 K ?

$\left(\mathrm{K}_{f} \text { for water is } 1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)$

(1) 899.04 g              (2) 886.02 g            (3) 868.06 g                 (4) 880.07 g

[Jee (Main)-2013 online]

Sol. (2)

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{f}}^{0}-\mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}$

$273.15-268=1.86 \times \frac{\mathrm{w} / 32}{10}$

$\mathrm{w}=886.02 \mathrm{g}$


Q. Vapour pressure of pure benzene is 119 torr and that of toluene is 37.0 torr at the same temperature. Mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.50, will be :

(1) 0.137           (2) 0.205            (3) 0.237            (4) 0.435

[Jee (Main)-2013 online]

Sol. (3)

Benzen $\rightarrow 4$

Toluene $\rightarrow B$

y $_{B}=\frac{P_{B}^{0} \times X_{B}}{P_{B}^{0} X_{B}+P_{A}^{0} X_{A}}=\frac{37 \times 0.5}{37 \times 0.5+119 \times 0.5}=0.237$


Q. A molecule M associates in a given solvent according to the equation M  $(\mathrm{M})_{\mathrm{n}}$. For a certain concentration of M, the van’t Hoff factor was found to be 0.9 and the fraction of associated molecules was 0.2. The value of n is :

(1) 2              (2) 4               (3) 5               (4) 3

[Jee (Main)-2013 online]

Sol. (1)

$\mathrm{M}=\mathrm{M}_{\mathrm{n}}$

$1-0.2 \quad 0.2 / \mathrm{n}$

$0.9=\frac{1-0.2+0.2 / \mathrm{n}}{1}$

$0.9=0.8+\frac{0.2}{\mathrm{n}}$

$0.1=\frac{0.2}{\mathrm{n}}$

$\mathrm{n}=2$


Q. 12g of a nonvolatile solute dissolved in 108g of water produces the relative lowering of vapour pressure of 0.1. The molecular mass of the solute is :

(1) 60           (2) 80             (3) 40            (4) 20

[Jee (Main)-2013 online]

Sol. (4)

$\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}=0.1=\frac{12 / \mathrm{m}}{108 / 18} \Rightarrow \mathrm{m}=20$


Q. The molarity of a solution obtained by mixing 750 mL of 0.5(M)HCl with 250 mL of 2(M)HCl will be :-

(1) 0.875 M           (2) 1.00 M            (3) 1.75 M            (4) 0.975 M

[Jee (Main)-2013]

Sol. (1)

$\mathrm{M}_{\mathrm{f}}=\frac{\mathrm{M}_{1} \mathrm{V}_{1}+\mathrm{M}_{2} \mathrm{V}_{2}}{\mathrm{V}_{1}+\mathrm{V}_{2}}=0.875 \mathrm{M}$


Q. The observed osmotic pressure for a 0.10 M solution of Fe$\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2}$ at $25^{\circ} \mathrm{C}$ is 10.8 atm. The expected and experimental (observed) values of Van’t Hoff factor (i) will be respectively : $\left(\mathrm{R}=0.082 \mathrm{L} \mathrm{atm} \mathrm{k}^{-} \mathrm{mol}^{-1}\right)$

(1) 3 and 5.42          (2) 5 and 3.42           (3) 4 and 4.00           (4) 5 and 4.42

[Jee (Main)-2014 online]

Sol. (4)

$\pi_{\mathrm{ob}}=\mathrm{i} \frac{\mathrm{n}}{\mathrm{V}} \mathrm{RT}$

$10.8=\mathrm{i} \times 0.1 \times 0.082 \times 298$

$\mathrm{i}=4.42$


Q. For an ideal Solution of two components A and B, which of the following is true ?

(1) $\Delta \mathrm{H}_{\text {mixing }}<0$ (zero)

(2) $\mathrm{A}-\mathrm{A}, \mathrm{B}-\mathrm{B}$ and $\mathrm{A}-\mathrm{B}$ interactions are identical

(3) $\mathrm{A}-\mathrm{B}$ interaction is stronger than $\mathrm{A}-\mathrm{A}$ and $\mathrm{B}-\mathrm{B}$ interactions

(4) $\Delta \mathrm{H}_{\text {mixing }}>0$ (zero)

[Jee(Main)-2014 online]

Sol. (2)

$\Delta \mathrm{H}_{\operatorname{mix}}=0$


Q. Consider separate solution of $0.500 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq}), 0.100 \mathrm{MM} \mathrm{g}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{aq}), 0.250 \mathrm{M} \mathrm{KBr}(\mathrm{aq})$ and 0.125 M $\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq})$ at $25^{\circ} \mathrm{C}$. Which statement is true about these solutions, assuming all salts to be strong electrolytes ?

(1) 0.125 M $\mathrm{Na}_{3} \mathrm{PO}_{4}$ (aq) has the highest osmotic pressure.

(2) 0.500 M $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ (aq) has the highest osmotic pressure.

(3) They all have the same osmotic pressure.

(4) 0.100 M $\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ (aq) has the highest osmotic pressure.

[Jee (Main)-2014]

Sol. (3)


Q. Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by :

(1) association

(2) dissociation

(3) complex formation

(4) partial ionization

[Jee (Main)-2015 online]

Sol. (1)

Acetic acid in non polar solvent (benzene) associates.


Q. A solution at $20^{\circ} \mathrm{C}$ is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively :

(1) 38.0 torr and 0.589

(2) 30.5 torr and 0.389

(3) 35.8 torr and 0.280

(4) 35.0 torr and 0.480

[Jee (Main)-2015 online]

Sol. (1)

$\begin{aligned} \mathrm{P}_{\mathrm{T}} &=\mathrm{P}_{\mathrm{A}}^{0} \mathrm{X}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{0} \mathrm{X}_{\mathrm{B}} \\ &=747 \times \frac{1.5}{5}+22.3 \times \frac{3.5}{5} \\ &=38 \mathrm{torr} \end{aligned}$


Q. The vapour pressure of acetone at $20^{\circ}$C is 185 torr. When 1.2 g of non-volatile substance was dissolved in 100 g of acetone at $20^{\circ}$$20^{\circ}$C, its vapour pressure was 183 torr. The molar mass $\left(\mathrm{g} \mathrm{mol}^{-1}\right)$ of the substance is :

(1) 128 (2) 488 (3) 32 (4) 64

[Jee (Main)-2015]

Sol. (4)

$\begin{array}{rl}{\frac{185-183}{185}} & {=\frac{1.2 / \mathrm{m}}{100 / 58}} \\ {\mathrm{m}=64} & {\mathrm{gm} / \mathrm{mol}}\end{array}$


Q. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point ?

(1) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}$

(2) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} .2 \mathrm{H}_{2} \mathrm{O}$

(3) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O}$

(4) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$

[Jee (Main)-2018]

Sol. (3)


JEE MAIN 2020 Admit Card for January Attempt | Download Hall Ticket for JEE Main 2020 Exam


The new release date of NTA JEE Main 2020 Admit Card for April session to be held from July 18 to 23 to be declared soon by NTA.

Candidates can download the JEE Main 2020 admit card from the official website by entering their JEE Main login details like their application number and date of birth or password and security pin.

The admit card of JEE Main 2020 is an essential document that the candidates must carry to the examination centre.

Details including name, roll number, state of eligibility, JEE Main exam centre address, etc. are printed on the JEE Main admit card

When you visit on the official page to download admit card, a new login page will open just like given below:

NOTE: Candidates must preserve the JEE MAIN 2020 admit card safely as it will be required at different stages of the admission and counselling process. Applicants who do not carry admit card will not be allowed to write the exam at the test center.

 

Details Required to Download JEE MAIN 2020 Admit Card

The hall ticket is only uploaded on the official JEE Main login and to download it, following details are needed:

  • Application number.
  • Password or date of birth.

 

JEE Main 2020 Admit Card will look like the image given below:

JEE Main 2020 Admit Card

What to Carry in Examination Hall

On the exam day, candidate have to carry following documents to the exam hall:

1. JEE Main 2020 hall ticket – Print it on an A4 sheet and make sure all the information should be clear and correct.

2. One passport size photograph – Along with the JEE Main admit card, candidates also have to carry one passport size photograph. This should be the same photo as the one uploaded in the form.

3. Valid Original ID Proof – As per information brochure, candidates have to carry one id proof. However, it has to be carried in original and should contain photograph of the candidate.

It is better that candidates carry the same id proof, the details of which were entered in the application form. List of valid id proofs for JEE Main 2020 is as follows:

  • PAN card.
  • Driving license.
  • Voter id card.
  • Aadhaar card.
  • E-aadhaar card with photograph.
  • Ration card.
  • Passport.
  • 12th Class admit card with photograph.
  • Bank passbook with photograph.

4. PwD Certificate – Candidates who applied for scribe facility have to carry PwD Certificate with the admit card.

What Not to Carry in the Examination Hall

Following items are not allowed inside the exam hall:

  • Text material.
  • Calculator.
  • Docu pen.
  • Log tables.
  • Silde rules.
  • Electronic watches.
  • Mobile phone.
  • Paper,
  • Metallic objects etc.

Important Instructions – Candidates should note that if they are planning to carry metallic objects such as Kara and Kirpan, etc. should report to the center at least 1 hour 30 minutes before closing of the gate. NTA might ask the candidates to not take it inside the exam hall.

Exam Time Schedule(Tentative)

Particulars Shift 1 Shift 2
Entry in the exam hall 7.30 am – 9.00 am 1.00 pm – 2.00 pm
Instruction by invigilators 9.00 am – 9.20 am 2.00 pm – 2.20 pm
Login and read the instructions 9.20 am 2.20 pm
Exam starts at 9.30 am 2.30 pm
B.E.. / B.Tech Exam 9.30 am – 12.30 pm 2.30 pm – 5.30 pm
B.Arch Exam 9.30 am – 12.30 pm 2.30 pm – 5.30 pm
B.Planning Exam 2.30 pm – 5.30 pm
B.Arch & B.Planning Exam (Both) 2.30 pm – 6.00 pm

 

JEE Main 2020 Sample Questions released by NTA are available here:

CHEMISTRY Sample Questions based on Numerical value by NTA 

PHYSICS Sample Questions based on Numerical value by NTA

MATHEMATICS Sample Questions based on Numerical value by NTA

 

Stay tuned with eSaral for more Updates.

Monotonicity – JEE Main Previous Year Question with Solutions


JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Mathematics

Maxima-Minima – JEE Main Previous Year Question with Solutions


JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Mathematics

Tangent & Normal – JEE Main Previous Year Question with Solutions


JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Mathematics

Ray Optics – JEE Main Previous Year Questions with Solutions


JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Communication System – JEE Main Previous Year Questions with Solutions


JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

 

Simulator

Previous Years AIEEE/JEE Mains Questions

Q. This questions has Statement-1 and Statement-2. Of the four choice given after the statements, choose the one that best describes the two statements.

Statement-1 : Sky wave signals are used for long distance radio communication. These signals are in general, less stable than ground wave signals.

Statement-2 : The state of ionosphere varies from hour to hour, day to day and season to season.

(1) Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of statement-1

(2) Statement-1 is true, Statement-2 is true, Statrment-2 is not the correct explanation of Statement-1

(3) Statement-1 is false, Statement-2 is true

(4) Statement-1 is true, Statement-2 is false

[AIEEE-2011]

Sol. (2)


Q. A radar has a power of 1 kW and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. The maximum distance up to which it can detect object located on the surface of the earth (Radius of earth $\left.=6.4 \times 10^{6} \mathrm{m}\right)$ is :

(1) 40 km (2) 64 (3) 80 km (4) 16km

[AIEEE-2012]

Sol. (3)

$\mathrm{d}=\sqrt{2 \mathrm{Rh}}=\sqrt{2 \times 6400 \times 0.5}=80 \mathrm{km}$


Q. A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it.

(1) 10.62kHz (2) 5.31 MHz (3) 5.31 kHz (4) 10.62MHz

[JEE Main-2013]

Sol. (3)

$\mathrm{f}_{\mathrm{max}} \leq \frac{\sqrt{\frac{1}{\mathrm{m}^{2}}-1}}{2 \pi \mathrm{RC}}$

$\mathrm{f}_{\max } \leq \frac{8}{6 \times 2 \pi \times 100 \times 10^{3} \times 250 \times 10^{-12}}$

$\mathrm{f}_{\max } \leq \frac{8 \times 10^{6}}{12 \pi \times 25}$

$\mathrm{f}_{\max } \leq 8.4925 \mathrm{kHz}$


Q. A single of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are –

(1) 2005 kHz, 2000 kHz and 1995 kHz

(2) 2000 kHz and 1995 kHz

(3) 2 MHz only

(4) 2005 kHz and 1995 kHz

[JEE Main-2015]

Sol. (1)

Frequency present after modulation

$\mathrm{f}_{\mathrm{c}}, \mathrm{f}_{\mathrm{c}} \pm \mathrm{f}_{\mathrm{s}}$

$\Rightarrow 2000 \mathrm{KHz}, 2005 \mathrm{KHz}$ and $1995 \mathrm{KHz}$


Q. Choose the correct statement :

(1) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal.

(2) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

(3) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

(4) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

[JEE Main-2016]

Sol. (2)


Q. In amplitude modulation, sinusoidal carrier frequency used is denoted by $\omega_{\mathrm{c}}$ and the signal frequency is denoted by $\omega_{\mathrm{m}}$. The bandwidth $\left(\Delta \omega_{\mathrm{m}}\right)$ of the signal is such that $\Delta \omega_{\mathrm{m}}<<\omega_{\mathrm{c}}$. Which of the following frequencies is not contained in the modulated wave ?

[JEE Main-2017]

Sol. (3)

Refer NCERT Page No. 526 Three frequencies are contained $\omega_{\mathrm{m}}+\omega_{\mathrm{c}}, \omega_{\mathrm{c}}-\omega_{\mathrm{m}} \& \omega_{\mathrm{c}}$


Q. A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz ?

[JEE Main-2018]

Sol. (2)

Since the carrier frequency is distributed as band width frequency, so 10% of 10 GHz = n × 5 kHz

where n = no of channels $\frac{10}{100} \times 10 \times 10^{9}=n \times 5 \times 10^{3} \mathrm{n}=2 \times 10^{5}$ telephonic channels


Kinematics 2D – JEE Main Previous Year Questions with Solutions


JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Simulator

Previous Years AIEEE/JEE Mains Questions

Q. A particle is moving with velocity $\overrightarrow{\mathrm{v}}=\mathrm{K}(\mathrm{y} \hat{\mathrm{i}}+\mathrm{x} \hat{\mathrm{j}})$ where K is a constant. The general equation for its path is :

(1) $\mathrm{y}^{2}=\mathrm{x}^{2}+$ constant

(2) $\mathrm{y}=\mathrm{x}^{2}+$ constant

(3) $\mathrm{y}^{2}=\mathrm{x}+$ constant

(4) xy = constant

[AIEEE – 2010]

Sol. (1)

$\overrightarrow{\mathrm{v}}=\mathrm{k}(\mathrm{y} \hat{\mathrm{i}}+\mathrm{x} \hat{\mathrm{j}})$

$\overrightarrow{\mathrm{v}}=\mathrm{v}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{v}_{\mathrm{y}} \hat{\mathrm{j}}$

$\overrightarrow{\mathrm{v}}=\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}} \hat{\mathrm{i}}+\frac{\mathrm{dy}}{\mathrm{dt}} \hat{\mathrm{j}}$

$\frac{d x}{d t}=k y \& \frac{d y}{d t}=k x$

$\Rightarrow \frac{d x}{d y}=\frac{y}{x}$

$\Rightarrow \mathrm{y}^{2}=\mathrm{x}^{2}+$ constant


Q. A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is :-

(1) $\frac{\pi}{2} \frac{v^{4}}{g^{2}}$

(2) $\pi \frac{\mathrm{v}^{2}}{\mathrm{g}^{2}}$

( 3)$\pi \frac{\mathrm{v}^{2}}{\mathrm{g}}$

( 4)$\pi \frac{\mathrm{v}^{4}}{\mathrm{g}^{2}}$

[AIEEE – 2011]

Sol. (4)

$\mathrm{r}=\mathrm{R}_{\max }=\frac{\mathrm{v}^{2}}{\mathrm{g}}$

area $=\pi \mathrm{r}^{2}$

$=\pi\left(\frac{\mathrm{v}^{2}}{\mathrm{g}}\right)^{2}$

$=\pi \frac{\mathrm{v}^{4}}{\mathrm{g}^{2}}$


Q. A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force $\mathrm{F}(\mathrm{t})=\mathrm{F}_{0} \mathrm{e}^{-\mathrm{bt}}$ in the x direction. Its speed v(t) is depicted by which of the following curves ?

[AIEEE – 2012]

Sol. (3)

$\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=\frac{\mathrm{F}_{0}}{\mathrm{m}} \mathrm{e}^{-\mathrm{bt}}$

$\int_{0}^{\mathrm{v}} \mathrm{d} \mathrm{v}=\frac{\mathrm{F}_{0}}{\mathrm{m}} \int_{0}^{\mathrm{t}} \mathrm{e}^{-\mathrm{bt}} \mathrm{dt}$

$\mathrm{v}=\frac{-\mathrm{F}_{0}}{\mathrm{m} \mathrm{b}}\left[\mathrm{e}^{-\mathrm{bt}}-1\right]$

$=\frac{\mathrm{F}_{0}}{\mathrm{m} \mathrm{b}}\left[1-\mathrm{e}^{-\mathrm{bt}}\right]$


Q. A projectile is given an initial velocity of $(\hat{i}+2 \hat{j}) m / s$ where $\hat{\mathbf{i}}$ is along the ground and $\hat{j}$ is along the vertical. If g = 10 m/s2, the equation of its trajectory is :

(1) $\mathrm{y}=\mathrm{x}-5 \mathrm{x}^{2}$

(2) $\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$

(3) $4 y=2 x-5 x^{2}$

(4) $4 y=2 x-25 x^{2}$

[AIEEE – 2013]

Sol. (2)

$\mathrm{u}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}$

$\mathrm{u}_{\mathrm{x}}=1$

$\mathrm{u}_{\mathrm{y}}=2$

$\tan \theta=\frac{2}{1}$

$\mathrm{y}=\mathrm{x} \tan \theta-\frac{1}{2} \mathrm{g} \frac{\mathrm{x}^{2}}{\mathrm{u}^{2} \cos ^{2} \theta}$

$\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$


Q. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ?

(Assume stones do not rebound after hitting the ground and neglect air resistance, take

g = 10 $\mathrm{M} / \mathrm{S}^{2}$) (The figure are schematic and not drawn to scale)

[JEE Mains – 2015]

Sol. (1)

$\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$

For particle 2

$-240=40 \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2}$

$5 t^{2}-40 t-240=0$

$\mathrm{t}_{2}=12 \mathrm{sec}$

for $0<\mathrm{t}<8$ sec $\rightarrow \mathrm{a}_{\text {rel }}=0$

straight line x-t graph

for $8<\mathrm{t}<12$ sec $\rightarrow \mathrm{a}_{\mathrm{rel}}=-\mathrm{g}$

downward parabola

for $\mathrm{t}>12 \mathrm{sec} \rightarrow$ Both particles comes to rest


Kinematics 2D – JEE Main Previous Year Questions with Solutions


JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Simulator

 

Previous Years JEE Advanced Questions

Q. A train is moving along a straight line with a constant acceleration ‘a’. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of $60^{\circ}$ to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in $\mathrm{m} / \mathrm{s}^{2}$, is –

[IIT-JEE 2011]

Sol. 5


Q. A rocket is moving in a gravity free space with a constant acceleration of 2 $\mathrm{ms}^{-2}$along + x direction (see figure). The length of a chamber inside the rocket is 4m. A ball is thrown from the left end of the chamber in + x direction with a speed of 0.3 $\mathrm{ms}^{-1}$ relative to the rocket. At the same time, another ball is thrown in –x direction with a speed of 0.2 $\mathrm{ms}^{-1}$ from its right end relative to the rocket. The time in seconds when the two balls hit each other is –

[JEE Advanced 2014]

Sol. 8 or 2

Assuming open chamber

$\mathrm{V}_{\text {relative }}=0.5 \mathrm{m} / \mathrm{s}$

$S_{\text {relative }}=4 \mathrm{m}$

time $=\frac{4}{0.5}=8 \mathrm{s}$

Alternate

Assuming closed chamber

In the frame of chamber :

Maximum displacement of ball A from its left end is $\frac{\mathrm{u}_{\mathrm{A}}^{2}}{2 \mathrm{a}}=\frac{(0.3)^{2}}{2(2)}=0.0225 \mathrm{m}$

This is negligible with respect to the length of chamber i.e. 4m. So, the collision will be very close to the left end.

Hence, time taken by ball B to reach left end will be given by

$\mathrm{S}=\mathrm{u}_{\mathrm{B}} \mathrm{t}+\frac{1}{2} \mathrm{at}^{2}$

$4=(0.2)(t)+\frac{1}{2}(2)(t)^{2}$

Solving this, we get

$\mathrm{t} \approx 2 \mathrm{s}$


Q. Airplanes A and B are flying with constant velocity in the same vertical plane at angles $30^{\circ}$ and $60^{\circ}$ with respect to the horizontal respectively as shown in figure. The speed of A is $\mathrm{ms}^{-1}$. At time t = 0 s, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t = $t_{0}$, A just escapes being hit by B, $t_{0}$ in seconds is

[JEE Advanced 2014]

Sol. 5

As observed from A, B moves perpendicular to line of motion of A. It means velocity of B along A is equal to velocity of A

$\mathrm{V}_{\mathrm{B}} \cos 30=100 \sqrt{3}$

$\mathrm{V}_{\mathrm{B}}=200$

If A is observer A remains stationary therefore

$\mathrm{t}=\frac{500}{\mathrm{V}_{\mathrm{B}} \sin 30}=\frac{500}{100}=5$


Q. A ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of $30^{\circ}$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is

[JEE Advanced 2018]

Sol. 30

$\mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} 45}{2 \mathrm{g}}=120$

$\Rightarrow \frac{\mathrm{u}^{2}}{4 \mathrm{g}}=120$ ….(i)

when half of kinetic energy is lost $\mathrm{v}=\frac{\mathrm{u}}{\sqrt{2}}$

$\mathrm{H}_{2}=\frac{\left(\frac{\mathrm{u}}{\sqrt{2}}\right)^{2} \sin ^{2} 30}{2 \mathrm{g}}=\frac{\mathrm{u}^{2}}{16 \mathrm{g}}$

from (i) & (ii)

$\mathrm{H}_{2}=\frac{\mathrm{H}_{1}}{4}=30 \mathrm{m}$ on 30.00