JEE Main 2019 Question Paper With Answer Key For 10 January 2019 Exam is made available after the exam. The answer key is available in the memory based format. Since the JEE Main 2019 exam is in online mode, the question is not available after the exam. The candidates after the exam provide the answers and questions based on their memory. Later after the exam is over, NTA uploads the JEE Main 2019 Question Paper With Answer Key For 09 January 2019 under the candidates’ login. Check this page below for memory based answers and questions.
Latest: JEE Main 2019 Response Sheet and Question paper by NTA – Check Here.
JEE Main 2019 Question Paper With Answer Key For 10 January 2019 Exam
The memory-based answer keys are provided right after the exam is over based on the candidates’ feedback on the exam. Thus, candidates who wish to know the probable answers to the questions in the exam can check this page below.
|JEE Main 2019 Question Paper With Answer Key For 10 January 2019 Exam||Important Dates|
|Exam date||10 Jan 2019|
|Releasing of Unofficial Answer key||On the exam day|
|Releasing of official answer key||1 week after the exam is over for all 5 days|
JEE Main 2019 Question Paper With Answer Key For 10 January 2019 Exam By Coaching Centers
Click on the direct links below to get the JEE Main 2019 questions and solutions by the coaching centers:
|Coaching Center||Shift 1||Shift 2|
JEE Main 2019 Questions For 10 Jan Shift 2
Q. Another question was from the Structure of Atoms. The students had to find the radius of the atom 0.0529 n^2/z.
Q. There were 2 rods and 2 balls the mass of the rods were given m1 and m2. The radius of the balls were given as r. Find the moment of inertia.
Q. A plane and a line was given and 4 point were give. The point of interaction was to be found.
Q. There was a question to find the molarity in a solution when the molecular mass was given.
JEE Main 2019 Questions For 10 Jan Shift 1
Q. Which molecule has 2 sigma and 1/2 PI bond? Options are N2, N2+, O2, O2+.
Q. Al203 + co2 ===> Al +c? The students were given reaction and were asked to choose which option it belonged to?
Q. A mass (m) placed on a plank has acceleration (g/2) upward. Find the work done by normal force in time (t).
Q. A and B are two bullets fired with velocities 1km/h and 2km/h respectively. Find the ratio of maximum range covered by these bullets.
Q. There is a wire of resistance 12 ohms it is made into an equilateral triangle what will be resistance across any two vertices.
Q. A body was revolving around earth with a velocity v if it was to escape earths feild what will ots total kinetoc energy on earth be?
Q. A resistance wire of resistance 18 ohm was bent to into an equilateral triangle. Find the resistance between two of its vertices?
Q. A wavelength of L1 was incident on a ydse system it was that a bright fringe was found at an angle of 1/40 rad further another light of wavelength L2 was incident it was found that the bright fringe again occurs at the same angle if the distance between the slits was 0.01mm then find the wavelengths.
Q. Another question was that in the communication system the students were given the height of both transmitting and receiving antenna and were asked to find the area covered.
Q. There was one question from Chemical Bonding, Maleic acid, and on order of pka values
Q. Number of lone pair of electron and hybridisation of XEOF4.
Ans. Sp3d2 and 1 lp
Q. In which of following there is 2 pie bond and 0.5 sigma bond options are N2+/N2/O2 /O2+.
Q. The area bounded by the parabolas x=ky^2 and y=kx^2 is 1 square unit. What is the value of k?
Ans. k =1/√3
Q: A question on pie bond and sigma was asked.
Q: There was a question to find the partial pressure.
Q: There were a few memory based questions
Q: Question from d and f block were asked.
Q: Question was asked from Aldehydes
Q: Carbolic acid based question was asked.
Q: Potential meter was given, resistance was given find the reading of galvanometer.
Q: Potential meter was given, resistance was given. Find the reading of galvanometer
Q: What is X-Ray tube made of?
Q: Capacitor was given, there was 3 mm difference between the 2 and it was divided into 3 columns. K2 and K3 were constant and their value was given. Find equivalent capacitance.
How To Use The JEE Main 2019 Question Paper With Answer Key For 10 January 2019 Exam
To use the JEE main 2019 answer key and question papers it is important to know the marking scheme. The marking scheme of JEE Main 2019 exam is that +4 marks is added for each correct answer and -1 is deducted for each wrong answer. The JEE Main 2019 marks distribution for paper 1 is given below:
For Paper 1: For B.E./B.Tech
|Sections||Number of Questions||Maximum Marks|
Using the above marking scheme, now the formula one can use to calculate their probable score is:
OUR SCORE IN JEE MAIN = NUMBER OF CORRECT ANSWERS X 4 – NUMBER OF INCORRECT ANSWERS X 1
How to Challenge JEE Main 2019 Question Paper With Answer Key for 10 January 2019 Exam?
Candidates will be able to challenge the official answer key and question paper of JEE Main 2019 for exam held on 10 Jan 2019. The steps to challenge the JEE Main 2019 official answer key is given in the points below:
- First, click on the link that is available in the article above.
- Login with valid credentials ie registration number and the password/date of birth.
- There you can raise the objection by choosing the appropriate tab.
- Enter question number and correct answer of the same.
- Select the types of doubt for your objection.
- Submit the queries and wait for the final answer key to release.
To cite your challenge using the steps above, candidates have to also pay a requisite amount of Rs. 1000/-. The challenge window is open for a limited period of time. After the challenges are accepted NTA will release the JEE main 2019 Official Answer Key.
Importance Of JEE Main 2019 Question Paper With Answer Key for 10 January 2019 Exam
Now, why is it important to have the JEE Main 2019 question paper with answer key with you? Well, it is nature of the candidates they are very curious about their marks in the exam and their chances of getting good college.
Using the JEE Main 2019 Answer Key and Question paper they can get their probable scores much before the result is announced and know what are their chances of admission and plan things accordingly.
It is an added advantage also for the second attempt candidates, as they will have the resources from first attempt exam to practice for the second attempt and know what was the nature of questions.
JEE Main 2019 Result
Based on the official answer key of JEE Main 2019, the scoring in JEE Main 2019 is done. Thus, the official answer key is one concrete source to get the JEE Main 2019 most correct scores. In addition to that, the JEE Main 2019 AIR will be also based on the scores of the candidates. AIR is All India Rank of the candidates. Based on the AIR the candidates will be called for counselling and admission.
The JEE Main 2019 Paper 1 Result is available under the login of the candidates. Candidates will need to enter their application number and date of birth/password to access their result.
JEE Main 2019 Question Paper With Answer For Second Attempt
The JEE Main 2019 second attempt is scheduled in the month of April 2019 between April 06-20, 2019. As the first attempt in January month, the answer key and question paper for the second attempt will be available first in the memory based format after the exam day. Later after the exam is over for all the 5 days the answer key and question paper will be available officially by NTA over the official website.