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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.
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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
Hi,
In your JEE application, you have to write the application number of your c entral caste certificate because for central government admissions like JEE, you have to use details of your central caste certificate. NTA asks for the application number, which is written on the upper side of the central caste certificate.
Hope it helps!!!
Hello,
Yes, you should select OBC-NCL while filling the JEE MAIN application form, because NTA only recognizes the central government categories that is GEN, GEN-EWS, OBC-NCL, SC, ST, PwD. As VJA is a state specific sub- category it is not separately listed in JEE MAIN forms.
Ensure your OBC-NCL certificate is valid under Central Government list, carry the certificate during counselling/admission for verification.
To know more access below mentioned link:
https://engineering.careers360.com/articles/jee-main-reservation-criteria
Hope it helps.
HELLO,
JEE Mains study materials are currently available in only English and Hindi, so if you need the material in Hindi and English, then you can check for it in Careers360. All the materials are available for all subjects for JEE Mains.
Hope it helps!!!
The date of JEE Advanced 2026 is not announced yet.
But according to previous years records we can assume that the registration for JEE Advanced 2026 exam will likely start in April and the date of exam will lie in the month of May.
For your upcoming JEE Advanced 2026 will need to study more harder than you did in your JEE Mains Exam.
Also you need to check the syllabus which is being asked in the exam. Here is the detailed information about JEE Advanced 2026 ---- Check Here
Hello,
Yes, you need an OBC-NCL certificate if you want to claim the OBC-NCL reservation in JEE Main .
Here are the key points:
If you want the OBC-NCL quota, you must upload a valid OBC-NCL certificate.
The certificate should be issued by the government and must follow the latest format.
It should usually be issued on or after 1 April of the current admission year.
If you do not have a valid certificate, you will be treated as a General (Unreserved) candidate.
If you apply under General category, then no OBC certificate is needed.
Hope it helps !
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