JEE Main Answer Key 2024 - Date, Steps to Download PDF

Physics Wallah Academic Expert
February 05, 2024

The answers to the JEE Mains exam in 2024 will be officially provided by the National Testing Agency (NTA) on its website. The NTA JEE Mains Answer Key for both the morning and afternoon sessions will include the correct solutions to the questions in the test. This helps candidates check their answers and estimate their scores. To get the official Answer Key for JEE Main 2024 session 1, candidates just need to enter their application number and date of birth. Unofficial answer keys for JEE Main 2024 shift 2 on January 29, and Shift 1 with memory-based questions, are also available in PDF format. 

JEE Main Answer Key 2024

The National Testing Agency is going to tell us the date for releasing the JEE Mains Answer Key 2024 soon. Based on last year's routine, we can expect to see the individual responses sheet and provisional answer key for JEE Mains 2024 Session 1 in the first week of February 2024.

If candidates find any mistakes in the JEE Answer Key 2024 for the January Session, they can speak up. Those not happy with the NTA Answer Key 2024 can share their concerns online through the JEE Main 2024 answer key challenge login page. To challenge the provisional answer key, candidates will need to pay INR 500 for each question. The NTA team will check the objections, and if needed, they might make changes to the JEE Main Final Answer Key 2024.

JEE Main Answer Key 2024 Overview

The provisional answers for JEE Main 2024 are available on NTA's official website, To check them, candidates just need to enter their application number and date of birth. By using the JEE Main 2024 answer key, candidates could estimate their probable scores. However, the final scores were officially revealed along with the JEE Main 2024 results. Check out the key points of the JEE Main Answer Key 2024 in the table below.

JEE Main Answer Key 2024 Overview



Examination Name

Joint Entrance Examination (JEE Main 2024)

Exam Conducting Body

National Testing Agency


Session 1

JEE Main Session 1 Exam Date

24 January 2024 to 01 February 2024

NTA Answer Key Release Date for 2024

First week of February 2024

Credentials Required to Check

Application number and date of birth

Duration of Raise Objection Against Answer Key

To be Notified Soon

Final JEE Mains Answer Key 2024

March 2024

Official Website

How to download JEE main 2024 answer key

The JEE Main 2024 shift 2 answer key, as well as for shift 1, can be accessed only on the NTA official website after all exam shifts in the JEE Main 2024 January Session have concluded. Candidates can follow these steps to check the JEE Mains answer key in PDF format:

Step 1: Visit the official JEE Main answer key website at 2024.

Step 2: On the homepage, click on the link that says “JEE Main answer key pdf download 2024”.

Step 3: Enter candidates' login credentials—JEE Main application number and date of birth.

Step 4: After clicking the submit button, the JEE Mains answer key will appear on the screen.

Step 5: Download the NTA JEE Main Answer key 2024 PDF for future reference.

How to challenge JEE Main 2024 Answer Key

After candidates download the provisional answer key for NTA JEE Main 2024, candidates can raise objections if they believe any answer is incorrect. To challenge an answer, candidates need to note the question ID and the answer provided in the JEE Main answer key, along with the correct answer. Here's how candidates can proceed:

  1. Visit the official NTA website at

  2. Click on the link for ‘challenges regarding answer key’.

  3. Log in using their application number and password.

  4. The JEE Main answer key will display the question ID and the correct option for both papers 1 and 2.

  5. If there is an error, candidates can challenge the question by selecting one or more IDs.

  6. Click on ‘Save your claim’ and then proceed to the next step.

  7. The challenged question IDs will appear on the screen.

  8. Candidates must provide necessary supporting documents, which should be uploaded in PDF format. Use the ‘Choose file’ button on the screen.

  9. The next step involves candidates paying a processing fee of Rs 200 for each question objected against. Payment options include credit card, debit card, or net banking.

How to calculate probable scores using JEE Main Answer Key

Candidates can figure out their expected score in JEE Main 2024 by using the answer key. To guess their possible scores in the JEE Main 2024 exam, candidates can use the formula below:

Probable Score = (Number of right answers × 4) - (Number of wrong answers)

Keep in mind that we say "probable" because authorities might cancel a question or change the correct response in the JEE Main 2024 answer key. So, the calculated score may change depending on how JEE Mains marks are distributed and the final JEE Main result.

Since the JEE Main exam has a system where candidates lose marks for wrong answers but gain them for correct ones, the table below shows how this scoring works:

Marking Scheme of JEE Main 2024

Type of Response

Marking Scheme

Correct response

Four marks awarded

Incorrect response

One mark deducted

No answer

No marks

Marked for review

No Marks

JEE Main Answer Key 2024 (January 29)

Q. A stationary hydrogen atom deexcites from the first excited state to the ground state. Find the recoil speed of the hydrogen atom up to the nearest integer value. (Take the mass of hydrogen atom = 1.8 x 10-27 *kg)

Answer: 3

Q. If an electric current passing through a conductor varies with time as / = I0 + βt, where l0 = 20 A and β = 3 A/s, then find the charge flow through the conductor in the first 10 seconds.

Answer: 350 C

Q. A square loop of resistance 16 Ω is connected with a battery of 9 V and an internal resistance of 1 2 Ω in a steady state. Find the energy stored in the capacitor of capacity C = 4 µF as per its position shown in the diagram. (The capacitor is connected diagonally in a separate loop attached to the circuit.)

Answer: 25.92𝜇𝐽

Q.If a biconvex lens of a material of refractive index 1.5 has a focal length of 20 cm in the air, then what will be its focal length when it is submerged in a medium of refractive index 1.6?

Answer: – 160 cm

Q. In a container, 1 g of hydrogen and 1 g of oxygen are taken. Find the ratio of hydrogen pressure to oxygen pressure inside the container.

Answer: 16

Q. The potential energy function corresponding to conservative force is given as U(x, y, z) = 3x²/2 + 5y + 6z, then the force at x = 6 is pN. The value of p. (Round off to the nearest integer.)

Answer: 20

Q. A solid sphere of radius 4a units is placed with its centre at the origin. Two charges -2q at (- 5a, 0) and 5q at (3a, 0) are placed. If the flux through the sphere is xq/ε0, then find x.

Answer: x = 5

Q. Consider the two statements (Assume the density of water to be constant):

Statement 1: A capillary tube is first dipped in hot water and then dipped in cold water. The rise is higher in hot water.

Statement 2: The capillary tube is first dipped in cold water and then in hot water. The rise is higher in cold water.

i. Statement 1 is true and Statement 2 is false.

ii. Statement 1 is false, and Statement 2 is true.

iii. Both statements are true.

iv. Both statements are false.

Answer: Statement 1 is false, and Statement 2 is true.

Q. A rod of length 2m moving with velocity 2mn/sec along the positive z-axis and B = 2T along the negative side x-axis. Find the emf induced in the rod.

Answer:  8mv

Q. A series of steps has 0.5 m tread and 0.5 m riser. If a ball is thrown from a point on the ground beside the first step, then find the minimum speed required by the ball to directly jump to the 5th step.

Answer: 5 (√√2 + 1) 𝑚/𝑠

Q. An electron is moving with the speed of 1 m/s at a distance of 1 m, from a large sheet of charge with density σ C/m2. Find the maximum value of σ such that the electron hits the sheet after 1 sec. Take the mass of electron = 9 × 10 -31 kg and the permittivity of free space = 9 x 10 -12 C²/Nm².

Answer: 4.5 × 10−22 𝐶/𝑚²

Q. In a convex mirror having a radius of curvature of 30 cm, the height of the image is half of the height of the object. What will be the distance of the object from the mirror in cm?

Answer: – 15 cm

Q. In a given voltage regulator circuit, the reverse breakdown voltage of the Zener diode is 3V. Find the current through the Zener diode. (A circuit diagram was given.)

Answer: 5.5mA

Q. In a given circuit, the galvanometer resistance is 10Ω and the current through the galvanometer is 3 mA. Find the resistance of the shunt. (Circuit diagram was given.)

Answer: 3.75 × 10 −3Ω

Q, A body of mass 100 kg travelled 10 m before coming to rest. If μ = 0.4, then find the work done against friction. Assume that the motion is happening on a horizontal surface and g = 10 m/s².

Answer: 4000 𝐽

JEE Main Chemistry Answer Key

Q1. Among the given ions which is the best-reducing agent (Lu3+, Nd3+, Ce4+, Gd2+)

Answer: Gd2+

Q2. What will be the oxidation state of Iron in the complex formed in the brown ring test?

Answer: +1/+3

Q3. What will be the reason behind the oxygen showing anomalous behaviour?

Answer: Small size, high electronegativity, absence of vacant d- orbital

Q4. Which of the following pairs will be formed by the decomposition of KMnO4?

i. MnO4 – , MnO2

ii. K2MnO4, MnO2

iii. KMnO4, MnO2

iv. MnO2, H2O

Answer:  K2MnO4, MnO2

Q5. Calculate the Molarity of a solution having a density of 1.5 g/ml, percentage of (w/w) of solute as 36%, and molecular weight of solute 36 g/mol.

Answer:  15

Q6. What is the energy difference between the actual structure and its most stable resonating structure having the least energy is called as?

1) Electromeric Effect 2) Resonance Energy 3) Inductive Effect 4) Hyperconjugation

Answer: Resonance Energy

Q7. If alkaline KMnO4 is oxidised iodide to give a particular product (A), then determine the oxidation state of iodine in the compound (A).


Q5: What is the effect that occurs between lone pair and pi bond?

Answer: Resonance Effect

Q6: Statement 1: Electronegativity of group 14 elements decreases from Si to Pb.

Statement 2: Group 14 has metals, metalloids, and non-metals

Both statement 1 and 2 are correct

Both statement 1 and 2 are incorrect

Statement 1 is correct and Statement 2 is incorrect

Statement 1 is incorrect and statement 2 is correct

Answer: Statement is incorrect and statement 2 is correct

Q.7: Hydrolysis of protein gives which type of amino acids

Answer: 𝛼 − 𝑎𝑚𝑖𝑛𝑜 𝑎𝑐𝑖d.

JEE Main Maths Answer Key

Q1. What is the rank of the word GTWENTY in the dictionary?

Answer: 553

Q2. If a die is rolled until 2 is obtained, then what is the probability that 2 is obtained on an even-numbered toss?

Answer: 5/ 11

Q3. A GP has 64 terms such that (Sn) total = 7(Sn) odd. Find the common ratio r.

Answer: R = 6

Best Books for JEE Main 2024

As mentioned before, there are many books in the market, making it hard for JEE candidates to decide which ones to pick. Buying separate books for each subject can be expensive. To address this, the Physics Wala team, keeping JEE in mind, has put together a combo of the Best Books for JEE 2024 in English. This combo is made to be budget-friendly for all JEE candidates. To buy directly from the PW Store, click the link below. 

Best Books for JEE Main 2024

JEE 5 Years

JEE Main & Advance Combo

JEE Main Answer Key 2024 FAQs

Q1. When will the final JEE Main Answer Key 2024 be released?

Ans. The final JEE Main Answer Key 2024 is expected to be released in March 2024, along with the declaration of JEE Main 2024 results.

Q2. How can candidates calculate their probable scores using the answer key?

Ans. Candidates can estimate their probable scores using the formula: Probable Score = (Number of correct answers × 4) - (Number of incorrect answers).

Q3. Why is the calculated score termed as "probable"?

Ans. The term "probable" is used because authorities may cancel or modify any question in the JEE Main 2024 answer key. The calculated score may vary based on the final JEE Main result.

Q4. Can candidates raise objections against the provisional answer key?

Ans. Yes, candidates have the opportunity to raise objections against the answers in the provisional JEE Main Answer Key 2024. To do so, they must note the question ID, provide the correct answer, and submit their challenges through the provided link. A processing fee of Rs 200 is applicable for each question challenged.

Q5. What is the duration for raising objections against the answer key?

Ans. The exact duration for raising objections against the JEE Main 2024 answer key will be notified soon by the authorities.

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