Most Repeated Questions in JEE Mains: JEE Mains is one of the toughest tests for students who want to get admission to engineering institutions in India. Analyzing different past year papers of JEE Main, there is always a tendency for some questions and topics to be repeated over and again. This article focuses on the most repeated questions in JEE Mains, including those with answers, to assist aspirants in effective preparation.
NTA will close the JEE Main 2026 application form correction window from December 1 to 2 on the official website, jeemain.nta.nic.in
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This article on the most repetitive questions in JEE Mains sorted out in the three parts; Physics, chemistry, and Math. By analyzing most repeated questions in JEE Mains Physics, Chemistry, and Maths, students can streamline their revision and practice. JEE Main 2026 registration is now open from 31 October to 27 November 2025, and Session 1 will be held from 21 to 30 January 2026. This year, new features include dark mode, adjustable font size, and screen zoom options.
In JEE Mains, Physics questions often come from a few important topics that are repeated every year. Knowing these topics helps you focus your preparation and score better. Here the most repeated Physics questions that you should definitely practise before the exam and prepared according to JEE Main Latest Syllabus 2026 are given:
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Some of the JEE Mains most repeated questions of Physics are given below, students can also refer JEE Main 2026 Sample Paper:
Question: A plane electromagnetic wave propagates along the $+x$ direction in free space. The components of the electric field, $\vec{E}$ and magnetic field, $\vec{B}$ vectors associated with the wave in Cartesian frame are :
(1) $\mathrm{E}_y, \mathrm{~B}_{\mathrm{x}}$
(2) $\mathrm{E}_{\mathrm{y}}, \mathrm{B}_{\mathrm{z}}$
(3) $E_x, B_y$
(4) $\mathrm{E}_z, \mathrm{~B}_y$
Solution:
Direction of propagation $=\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$
$E \rightarrow y, B \rightarrow z, c \rightarrow x$
$\hat{\mathrm{E}} \times \hat{\mathrm{B}}=\hat{\mathrm{j}} \times \hat{\mathrm{k}}=\hat{\mathrm{i}}=\hat{\mathrm{c}}$
Hence, the answer is the option (2).
Question: Given below are two statements :
Statement I: For transmitting a signal, the size of the antenna (I) should be comparable to the wavelength of the signal (at least $l=\frac{\lambda}{4}$ in dimension)
Statement II: In amplitude modulation, the amplitude of the carrier wave remains constant (unchanged).
In the light of the above statements, choose the most appropriate answer from the options given below.
(1) Statement I is correct but Statement II is incorrect
(2) Both Statement I and Statement II are correct
(3) Statement I is incorrect but Statement II is correct
(4) Both Statement I and Statement II are incorrect
Solution:
Statement 1 is correct.
In Modulation Amplitude of the carrier wave is increased.
Statement 2 is incorrect, in amplitude modulation amplitude of the wave is varied.
Hence, the answer is the option (1).
Chemistry includes conceptual as well as numerical questions. Among the most repeated questions in JEE Mains Chemistry, Organic and Physical Chemistry dominate.The most repeated questions in JEE Mains chemistry help students to score good marks. Alsoo they can refer JEE Main 2026 - A complete preparation strategy
Question:
Some of the most asked questions in JEE Mains Chemistry are given below:
Question: Concentrated nitric acid is labelled as $75 \%$ by mass. The volume in mL of the solution which contains 30 g of nitric acid is $\_\_\_\_$ . Given : Density of nitric acid solution is $1.25 \mathrm{~g} / \mathrm{mL}$
(1) 45
(2) 55
(3) 32
(4) 40
Solution :
$75 \%$ by mass means :
$\because \quad 75 \mathrm{gHNO}_3$ in 100 g solution.
$\therefore \quad 30 \mathrm{gHNO}_3$ in $\frac{100}{75} \times 30 \mathrm{~g}$ solution.
$\mathrm{m}_{\text {solution }}=\frac{100 \times 30}{75} \mathrm{~g}$
$$
\because \quad \mathrm{V}_{\text {sol }}=\frac{\mathrm{m}_{\text {sol }}}{\mathrm{d}_{\text {sol }}}=\frac{100 \times 30}{75 \times 1.25}=32 \mathrm{~mL}
$$
Hence, the correct answer is option (3).
Question:
Given below are two statements :
Statement I : $\mathrm{CrO}_3$ is a stronger oxidizing agent than $\mathrm{MoO}_3$
Statement II : $\mathrm{Cr}(\mathrm{VI})$ is more stable than $\mathrm{Mo}(\mathrm{VI})$
In the light of the above statements, choose the correct answer from the options given below
(1) Statement I is false but Statement II is true
(2) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are true
(4) Both Statement I and Statement II are false
Solution:
Statement-I is true but statement II is false.
$\mathrm{Cr}(\mathrm{VI})$ is less stable than $\mathrm{Mo}(\mathrm{VI})$
Stability : $\mathrm{Cr}^{6+}<\mathrm{Mo}^{6+}$
Hence, $\mathrm{CrO}_3$ easily reduce into $\mathrm{Cr}^{-3}$ as compared to $\mathrm{MoO}_3$ and show stronger oxidizing nature.
Hence, the correct answer is option (2).
Mathematics questions often involve direct applications of formulas and standard problem-solving techniques. Here are the JEE Mains most repeated questions of Maths. The most asked questions in JEE Mains of maths are predominantly involved in formulas and standard techniques, making them critical for scoring well.
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Some Important Questions from JEE Main Mathematics are given below:
Question: If in the expansion of $(1+x)^p(1-x)^q$, the coefficients of $x$ and $x^2$ are 1 and -2 , respectively, then $\mathrm{p}^2+\mathrm{q}^2$ is equal to:
(1) 8
(2) 18
(3) 13
(4) 1
Solution:
$$
(1+x)^p(1-x)^q=\left({ }^p C_0+{ }^p C_1 x+{ }^p C_2 x^2+\ldots\right)\left({ }^q C_0-{ }^q C_1 x+{ }^q C_2 x^2-\ldots\right)
$$
To find the coefficient of $x$, we consider terms whose degrees add up to 1 :
Coefficient of $x={ }^p C_0 \cdot\left(-{ }^q C_1\right)+{ }^p C_1 \cdot{ }^q C_0=-{ }^q C_1+{ }^p C_1=1$
Given that $p-q=1$, so:
$$
p=q+1
$$
Now for the coefficient of $x^2$, the relevant terms are:
Coefficient of $x^2={ }^p C_0 \cdot{ }^q C_2-{ }^p C_1 \cdot{ }^q C_1+{ }^p C_2 \cdot{ }^q C_0=-2$
Now express binomial coefficients using formulas:
$$
{ }^q C_2=\frac{q(q-1)}{2}, \quad{ }^p C_2=\frac{p(p-1)}{2}, \quad{ }^p C_1=p, \quad{ }^q C_1=q
$$
Substitute these into the expression:
$$
\frac{q(q-1)}{2}-p q+\frac{p(p-1)}{2}=-2
$$
Multiply the entire equation by 2 to eliminate denominators:
$$
q(q-1)-2 p q+p(p-1)=-4
$$
$$
q^2-q-2 p q+p^2-p=-4
$$
Now substitute $p=q+1$ into the equation:
$$
q^2-q-2 q(q+1)+(q+1)^2-(q+1)=-4
$$
Expand:
$$
q^2-q-2 q^2-2 q+q^2+2 q+1-q-1=-4
$$
Simplify:
$$
\left(q^2-2 q^2+q^2\right)+(-q-2 q+2 q-q)+(1-1)=-4 \Rightarrow 0-2 q=-4 \Rightarrow q=2
$$
Now:
$$
p=q+1=3
$$
Therefore:
$$
p^2+q^2=3^2+2^2=9+4=13
$$
Hence, the correct answer is option (3).
Question: $\int_0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x$ is equal to
(1) $1 / 12$
(2) $1 / 9$
(3) $1 / 6$
(4) $1 / 3$
Solution:
$\int_0^{\pi / 4} \frac{\tan ^2 x \sec ^2 x d x}{\left(1+\tan ^3 x\right)^2} d x$
$\begin{aligned} & \text { Let } 1+\tan ^3 \mathrm{x}=\mathrm{t} \\ & \tan ^2 x \sec ^2 x d x=\frac{d t}{3} \\ & \frac{1}{3} \int_1^2 \frac{d t}{t^2}=\frac{1}{6}\end{aligned}$
Hence, the correct answer is Option 3.
By focusing on the most repeated questions in JEE Mains physics, chemistry, and maths, aspirants can significantly improve their chances of scoring well in the exam.
High Probability of Appearance:
Questions from certain chapters or concepts tend to appear consistently every year. For example, in Physics, topics like Electrostatics, Modern Physics, and Mechanics often have recurring questions. Similarly, in Chemistry, Chemical Bonding, Coordination Compounds, and Organic Reactions are frequently tested. By focusing on these repeated topics, candidates increase their chances of attempting questions they are already familiar with, reducing uncertainty during the exam.
Efficient Time Management:
When you practice repeated questions, you solve questions in less time. This improves speed and accuracy, which are critical in a time-bound exam like JEE Main. Attempting familiar questions first during the exam can help you secure easy marks quickly, allowing more time for tougher problems.
Better Understanding of Question Patterns:
Repeated questions help candidates recognize patterns of question and trends in the exam. Like, the way a particular concept is framed in multiple-choice questions, or the typical twists applied to numerical problems, becomes predictable with practice. This pattern recognition reduces the chances of errors and builds confidence during the exam.
Smart Revision Strategy:
Focusing on repeated questions allows for revision of selected section. Instead of attempting all topics equally, candidates can prioritize high-yield areas that historically appear in the exam. This strategy ensures that the limited revision time is used efficiently and helps in retaining important formulas, concepts, and shortcuts.
Boosts Confidence and Reduces Exam Stress:
Solving repeated questions from previous years provides a sense of confidence in our mind. Candidates feel more prepared and confident, knowing that they have already practiced questions that are likely to appear. This reduces exam anxiety and helps maintain a calm and focused mindset on the day of the exam.
Also read
Frequently Asked Questions (FAQs)
Practicing these questions covers topics with high weightage, helps in understanding question patterns and difficulty levels, improves speed and accuracy, and boosts confidence, as similar questions often appear in subsequent exams.
Repeated questions are rarely identical. Usually, the underlying concept and question structure are repeated, but numbers, coefficients, or contexts may change.
Though the focus on the majority of the frequently asked questions gives an advantage, one needs to aim at utilizing all his/her potential, acquirements, and notions, as well as rehearse consistently for all the questions that might be included in the final check.
On Question asked by student community
Hey there,
Your OBC-NCL certificate will not be valid for the JEE Mains 2026 application as it is from June 2024. You can apply for a new one as soon as possible. For the JEE Mains, the OBC-NCL certificate should be issued on or after 1 April 2025.
And as for your Aadhar, the mistakes in your name and Date of Birth will be an issue during the document verification process. The details should match your 10th-class certificate. So you will also need to update it as soon as possible.
Hope it helps!!!
Hello Satyam,
Yes, your EWS certificate is valid.
In Bihar, the EWS certificate can be issued by the Circle Officer, BDO, SDO, or DM. So a certificate signed by the Circle Officer (Revenue Officer) is acceptable.
If the format of your certificate is the same as the Central Government EWS format, then it is valid for JEE Main registration and also for JoSAA counselling. The heading “Government of Bihar” does not create any problem.
Just remember one point:
For
JoSAA 2026
, you will need an EWS certificate issued on or after 1 April 2025. Even if your current certificate works for registration, you must update it before counselling.
Hope it helps !
Hello murali
No, your son is not eligible for OBC NCL for IIT JEE because you fall in the "creamy layer" occupational category, regardless of your current employment status or family income. Students whose family income is less than Rs. 8 lakhs annually and they are not belong to the "creamy layer".
Note -
Thank You
Hello,
You can fill the JEE Main form even if you are a private candidate
Write the
name of the school/board from where you are appearing as a private candidate
.
If your Class 12 admit card or registration slip shows a school/centre name, use that exactly.
If your board lists you as a “Private Candidate” under the board name , then write:
CBSE – Private Candidate
(or your board name – Private Candidate)
Use the pin code of the examination centre/school mentioned on your Class 12 private candidate admit card or registration details.
If your board does not give any school address and only shows the regional office address, then use the regional office address pin code given by your board.
Hope it helps !
Hello Aspirant
You should not leave the OBC-NCL certificate ID blank in the JEE Main form it can create problems later.
NTA wants the certificate details while filling the form, not just at counselling. If you can, apply for the OBC-NCL certificate immediately so you get the ID on time.
If you fail to submit the certificate during counselling, your category will shift to General. It’s safer to enter OBC-NCL only if you’re sure you’ll get the certificate before counselling.
Hope it will help you
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