JEE Mains PYQ Chapterwise PDF Available: Download Previous Year Questions with Solutions

JEE Mains PYQ Chapterwise PDF Download – Why It Matters

The JEE Mains 2026 syllabus is huge. Going through study material and trying to solve the previous years question papers all at once can be hectic. So, we have divided the material into all the chapters and added previous years question after each chapter. This technique will be very helpful as you can finish one chapter completely before moving into the next one. There are several reasons why chapterwise PYQs are important. Let's see the benefits of solving the JEE Mains chapterwise previous years questions below:

1. Previous year questions are very important for practicing. Practising for each chapter one at a time can help boost confidence and minimize confusion.

2. They will help you understand the exam pattern in a practical manner. Then you can also understand the type of questions asked from each chapter.

3. You will get a very good idea about the weightage and important topics as well.

The following link provides the PYQs that you must solve to practice for JEE Mains 2026.

JEE Main Physics PYQ chapterwise

It is very important to know which chapters to prioritize. This plan should be prepared according to the weightage of each chapter in the past years chpaterwise weightage. Our experts have prepared the JEE Mains chapterwise weightage of all subjects based on the total number of questions asked from them in the last 10 years. You can download JEE Main Physics PYQ chapterwise PDF from the link below:

In this section, we will be seeing some of the top weighted chapters and some questions asked from them.

JEE Main Physics Chapterwise Previous Years Questions

This table has the most important chapters of JEE Main Physics PYQ chapterwise PDF according to the last 10 years question papers.

JEE Main Physics PYQ

Some questions have been listed below from the top chapters:

1. Optics

Q: A convex lens of focal length 20 cm is placed at a distance of 30 cm from an object. Find the position and nature of the image formed.

Solution: Using the lens formula,

As we know,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
Put the values in the lens formula,

$\begin{aligned}
& \frac{1}{20}=\frac{1}{v}-\frac{1}{(-30)} \\
& \frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60} \\
& \mathrm{v}=60 \mathrm{~cm}
\end{aligned}$

The image is real, inverted, and formed 60 cm on the other side of the lens.

2. Electrostatics

Q: Two equal point charges +q are placed at a distance d apart. Find the electric field at the midpoint of the line joining them.

Solution: At the midpoint, the fields due to both charges are equal in magnitude and opposite in direction. Hence,

E net = 0

3. Properties of Solids & Liquids

Q: A sample of a liquid is kept at 1 atm . It is compressed to 5 atm which leads to change of volume of $0.8 \mathrm{~cm}^3$. If the bulk modulus of the liquid is 2 GPa , the initial volume of the liquid was _____ litre. $\left(\right.$ Take $\left.1 \mathrm{~atm}=10^5 \mathrm{~Pa}\right)$

Solution:

Given,

Initial pressure of liquid $\left(\mathrm{P}_{\mathrm{i}}\right)=1 \mathrm{~atm}$
Final pressure of liquid $\left(\mathrm{P}_{\mathrm{f}}\right)=5 \mathrm{~atm}$
Change in pressure $(d P)=P_f-P_i=4 \mathrm{~atm}$

$
=4 \times 10^5 \mathrm{~Pa}
$
Change in volume $(\mathrm{dV})=-0.8 \mathrm{~cm}^3$
Bulk modulus $(\mathrm{B})=2 \times 10^9 \mathrm{~Pa}$
Now,

$B=\frac{-d P}{(d V / V)} \Rightarrow V=-B\left(\frac{d V}{d P}\right)$

$
\begin{aligned}
\Rightarrow \mathrm{V} & =-2 \times 10^9 \times \frac{\left(-0.8 \times 10^{-6}\right)}{4 \times 10^5} =4 \times 10^{-3} \mathrm{~m}^3=4 \text { litre }
\end{aligned}
$

Hence, the answer is 4.

4. Magnetic Effects of Current & Magnetism

Q: The percentage increase in magnetic field (B) when space within a current carrying solenoid is filled with magnesium (magnetic susceptibility $\left.\chi_{\mathrm{mg}}=1.2 \times 10^{-5}\right)$ is :

1) $\frac{6}{5} \times 10^{-3} \%$

2) $\frac{5}{6} \times 10^{-5} \%$

3) $\frac{5}{6} \times 10^{-4} \%$

4) $\frac{5}{3} \times 10^{-5} \%$

Solution:

$\%$ change in $\mathrm{B}=\frac{\mathrm{B}_{\text {new }}-\mathrm{B}_{\text {old }}}{\mathrm{B}_{\text {old }}} \times 100 \%$

$
\begin{aligned}
& =\frac{\mu \mathrm{ni}-\mu_0 \mathrm{ni}}{\mu_0 \mathrm{ni}} \times 100 \%=\frac{\left(\mu-\mu_0\right)}{\mu_0} \times 100 \% \\
& =\frac{\left(\mu_0 \mu_{\mathrm{r}}-\mu_0\right)}{\mu_0} \times 100 \% \\
& =\left(\mu_{\mathrm{r}}-1\right) \times 100 \% \\
& =\chi_{\mathrm{n}} \times 100 \% \\
& =1.2 \times 10^{-3} \%
\end{aligned}
$

Hence, the answer is option (1).

5. Rotational Motion

Q: A solid sphere of mass 2 kg and radius 0.2 m rolls without slipping with a linear speed of 5 m/s. Find its total kinetic energy.

Solution: $\begin{gathered}K_{\text {total }}=K_{\text {trans }}+K_{\text {rot }} \\ K_{\text {trans }}=\frac{1}{2} m v^2 \\ K_{\text {rot }}=\frac{1}{2} I \omega^2\end{gathered}$

$I=\frac{2}{5} m r^2$

$K_{\text {trans }}=\frac{1}{2} \times 2 \times 5^2$

$K_{\text {trans }}=25 \mathrm{~J}$

$\begin{gathered}K_{\mathrm{rot}}=\frac{1}{2}\left(\frac{2}{5} m r^2\right)\left(\frac{v}{r}\right)^2 \\ K_{\mathrm{rot}}=\frac{1}{2}\left(\frac{2}{5} m r^2\right)\left(\frac{v^2}{r^2}\right) \\ K_{\mathrm{rot}}=\frac{1}{5} m v^2\end{gathered}$

$\begin{gathered}K_{\mathrm{rot}}=\frac{1}{5} \times 2 \times 5^2 \\ K_{\mathrm{rot}}=\frac{1}{5} \times 50 \\ K_{\mathrm{rot}}=10 \mathrm{~J}\end{gathered}$

$\begin{gathered}K_{\text {total }}=K_{\text {trans }}+K_{\text {rot }} \\ K_{\text {total }}=25 \mathrm{~J}+10 \mathrm{~J} \\ K_{\text {total }}=35 \mathrm{~J}\end{gathered}$

JEE Main Chemistry PYQ chapter wise

As you already know, Chemistry is the most scoring subject in JEE Mains exam. Studying JEE Main Chemistry PYQ chapter wise PDF download will avoid confusion and help you prepare better. Let's look at some of the most important chapters in Chemistry and understand the type of questions asked.

JEE Main Chemistry Previous Years Questions

Let's see the type of questions asked in chemistry in previous years JEE Mains exam.

1. Co-ordination Compounds

Question: Match the following:

- Co-Wilkinson catalyst

- Zn - Carbonic anhydrase

- Rh - Vitamin B12

- Mg - Chlorophyll

Solution:

- Co forms Wilkinson catalyst (Rh-based, but Co often confused in matching themes)

- Zn associates with enzyme carbonic anhydrase

- Rh is central to Wilkinson catalyst (common match)

- Mg is central in chlorophyll

2. Chemical Thermodynamics

Question: Calculate ΔG for a reaction at 298 K if ΔH=−100 kJ and ΔS=−200 J/K.

Solution:

Convert units: ΔS=−0.200 kJ/K

$\Delta G=\Delta H-T \Delta S=-100-(298 \times-0.200)=-100+59.6=-40.4 \mathrm{~kJ}$

3. Some Basic Principles of Organic Chemistry

Question: Which of the following statements about organic compounds is correct?

Options:

A) Organic compounds always contain metals.

B) Carbon can form four covalent bonds.

C) Organic compounds do not contain hydrogen.

D) Carbon cannot form rings.

Solution:

B) Carbon can form four covalent bonds

4. Hydrocarbons

Question: Which of the following is a saturated hydrocarbon?

Options:

A) Ethene

B) Ethyne

C) Ethane

D) Benzene

Answer:

C) Ethane

5. Organic Compounds Containing Oxygen

Question: Which functional group is present in alcohols?

Options:

A) –COOH

B) –OH

C) –CHO

D) –CO

Solution:

B) –OH

Also Read:

• JEE Main- Top 30 Most Repeated Questions & Topics
• Most Scoring Concepts For JEE Mains 2025 April Session
• Most Scoring Concepts For JEE Mains 2025 January Session

JEE Main Maths Chapterwise PYQ

Maths is a tricky subject. It requires a lot of practice. Your main aim in maths should be to increase speed and accuracy. Let's see the Maths most important chapters and some previous year questions. JEE Main Maths chapterwise PYQ PDF, along with chapterwise distribution, is given as follows:

JEE Mains Maths previous Year questions:

Let's see some previous years' questions from JEE Mains maths.

1. Co-ordinate Geometry

Q: Let $\mathrm{A}(4,-2), \mathrm{B}(1,1)$ and $\mathrm{C}(9,-3)$ be the vertices of a triangle ABC . Then the maximum area of the parallelogram AFDE, formed with vertices $\mathrm{D}, \mathrm{E}$ and F on the sides $\mathrm{BC}, \mathrm{CA}$ and AB of the triangle ABC respectively, is________

Solution:

Given:

$A(4,-2), B(1,1), C(9,-3)$

Area of $\triangle ABC = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

$= \frac{1}{2} |4(1 + 3) + 1(-3 + 2) + 9(-2 - 1)|$

$= \frac{1}{2} |4 \times 4 + 1 \times (-1) + 9 \times (-3)|$

$= \frac{1}{2} |16 - 1 - 27| = \frac{1}{2} |-12| = 6$

Use the property: Area of parallelogram formed on same base and height $= 2 \times \text{Area of triangle}$

So, the maximum area of parallelogram AFDE inscribed with vertices on sides BC, CA, AB

$= 2 \times $Area of $\triangle ABC = 2 \times 6 = 12$

Hence, the answer is $12$ sq. units.

2. Integral Calculus

Q: $\begin{aligned}
& \text { If } \int\left(\frac{1}{x}+\frac{1}{x^3}\right)\left(23 \sqrt{3 x^{-24}+x^{-26}}\right) d x \\
& =-\frac{\alpha}{3(\alpha+1)}\left(3 x^\beta+x^\gamma\right)^{\frac{\alpha+1}{\alpha}}+C, x>0,(\alpha, \beta, \gamma \in Z),
\end{aligned}$

where $C$ is the constant of integration, then $\alpha+\beta+$ $\gamma$ is equal to ______ .

Solution:

$\begin{aligned} & I=\int\left(\frac{1}{x}+\frac{1}{x^3}\right)\left(\frac{3}{x^{24}}+\frac{1}{x^{26}}\right)^{\frac{1}{23}} d x \\ & =\int\left(\frac{1}{x^2}+\frac{1}{x^4}\right)\left(\frac{3}{x}+\frac{1}{x^3}\right)^{\frac{1}{23}} d x \\ & \text { Put } \frac{3}{x}+\frac{1}{x^3}=t \Rightarrow\left(-\frac{3}{x^2}-\frac{3}{x^4}\right) d x=d t \\ & \Rightarrow \quad I=-\frac{1}{3} \int t^{\frac{1}{23}} d t=-\frac{1}{3} \frac{t^{\frac{1}{23}}+1}{\frac{1}{23}+1}+C\end{aligned}$
$\begin{aligned} & =-\frac{1}{3} \times \frac{23}{24}\left(3 x^{-1}+x^{-3}\right)^{\frac{24}{23}}+C \\ \Rightarrow & \alpha=23, \beta=-1, \beta=-3 \\ \Rightarrow & \alpha+\beta+\gamma=19\end{aligned}$

Hence, the answer is (19).

3. Limit, Continuity \& Differentiability

Q: If $\lim _{x \rightarrow 0} \frac{\cos (2 x)+a \cos (4 x)-b}{x^{4}}$ is finite, then $(a+b)$ is equal to :

1) $\frac{1}{2}$

2) 0

3) $\frac{3}{4}$

4) -1

Solution:

Given that the limit

$\lim_{x \to 0} \frac{\cos(2x) + a \cos(4x) - b}{x^4}$

is finite, we need to find $a + b$.

Using the Maclaurin series expansions near $x=0$:

$\cos(2x) = 1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{24} + \cdots = 1 - 2x^2 + \frac{4x^4}{3} + \cdots$

$\cos(4x) = 1 - \frac{(4x)^2}{2} + \frac{(4x)^4}{24} + \cdots = 1 - 8x^2 + \frac{32x^4}{3} + \cdots$

Substitute these into the numerator:

$\cos(2x) + a \cos(4x) - b = (1 - 2x^2 + \frac{4x^4}{3}) + a(1 - 8x^2 + \frac{32x^4}{3}) - b + \cdots$

Simplify the expression:

$(1 + a - b) + (-2 - 8a) x^2 + \left(\frac{4}{3} + \frac{32a}{3}\right) x^4 + \cdots$

For the limit to be finite, the coefficients of the $x^0$ and $x^2$ terms must be zero, otherwise the numerator will not vanish at order $x^4$.

Set the constant term to zero:

$1 + a - b = 0 \implies b = 1 + a$

Set the coefficient of $x^2$ to zero:

$-2 - 8a = 0 \implies a = -\frac{1}{4}$

Use $a = -\frac{1}{4}$ to find $b$:

$b = 1 - \frac{1}{4} = \frac{3}{4}$

Therefore,

$a + b = -\frac{1}{4} + \frac{3}{4} = \frac{1}{2}$

Hence, the correct answer is option (1).

4. Sets, Relations and Functions

Q:The number of real roots of the equation $\mathrm{x}|\mathrm{x}-2|+3|\mathrm{x}-3|+1=0$ is :

1) 4

2) 2

3) 1

4) 3

Solution:

$\begin{aligned} & x|x-2|+3|x-3|+1=0 \\ & \text { Case I: } x<2 \\ & -x(x-2)-3(x-3)+1=0 \\ & -x^2+2 x-3 x+9+1=0 \\ & x^2+x-10=0 \\ & x=\frac{-1-\sqrt{41}}{2} \text { or } \frac{-1+\sqrt{41}}{2} \text { (rejected) }\end{aligned}$
1 solution

Case II: $2 \leq x<3$

$\begin{aligned}
& x(x-2)-3(x-3)+1=0 \\
& x^2-5 x+8=0
\end{aligned}
$

No solution

Case III: $x \geq 3$

$\begin{aligned}
& x(x-2)+3(x-3)+1=0 \\
& x^2+x-8=0 \\
& x=\frac{-1 \pm \sqrt{33}}{2}\left(\text { no solution}\right) \\
& \therefore x=\frac{-1-\sqrt{41}}{2} \text { is the only solution }
\end{aligned}$
Hence, the correct answer is option (3).

5. Complex Numbers and Quadratic Equations

Q : If the set of all $\mathrm{a} \in \mathrm{R}-\{1\}$, for which the roots of the equation $(1-a) x^{2}+2(a-3) x+9=0$ are positive is $(-\infty,-\alpha] \cup[\beta, \gamma)$, then $2 \alpha+\beta+\gamma$ is equal to________

Solution:

Given the equation: $(1 - a)x^2 + 2(a - 3)x + 9 = 0$

Let $f(x) = (1 - a)x^2 + 2(a - 3)x + 9$

To ensure both roots are positive:

(i) Discriminant
$\begin{align*}
D &\geq 0 \\
D &= [2(a - 3)]^2 - 4(1 - a)(9) \\
&= 4(a - 3)^2 - 36(1 - a) \\
&= 4[a^2 - 6a + 9 + 9a - 9] \\
&= 4(a^2 + 3a) \geq 0 \\
&\Rightarrow a(a + 3) \geq 0 \Rightarrow a \in (-\infty, -3] \cup [0, \infty)
\end{align*}$

(ii) Leading coefficient $(1 - a) > 0 \Rightarrow a < 1$

(iii) $f(0) = 9 > 0$, always true

(iv) Vertex should lie to the left of y-axis:

$\begin{align*}
-\frac{b}{a} &= -\frac{2(a - 3)}{1 - a} > 0 \\
\Rightarrow \frac{3 - a}{1 - a} > 0 \Rightarrow a \in (-\infty, 1)
\end{align*}$

Final intersection of conditions:
$a \in (-\infty, -3] \cup [0, 1)$

This is of the form $(-\infty, -\alpha] \cup [\beta, \gamma)$

$\Rightarrow \alpha = 3, \beta = 0, \gamma = 1$

$\Rightarrow 2\alpha + \beta + \gamma = 2(3) + 0 + 1 = 7$

Hence, the answer is $7$.

JEE Mains Previous years Questions

Use the link below to download all the question papers of JEE Mains in the last 10 years:

How to Use JEE Main Chapterwise PYQs for Revision

Some tips for your revision are given below:

1. Plan your chapters out wisely.

2. After the completion of 3 chapters from every subject, start giving mock tests on a weekly basis.

3. After finishing the syllabus, plan our mock tests on a daily basis.

4. Keep separate time for revision and separate for mock tests.

5. Analyze your answers in mock tests and identify your weak and strong chapters.

6. Strengthen and practise your weaker chapters.

7. Try to increase speed while solving topics that you are strong in.

8. Try to cover the entire syllabus so you have 2 to 3 months for only revision and solving mock tests.

How to Use PYQs for Scoring 150+ in JEE Main

It is by far the best source to improve your JEE Main score using Previous Years' Questions (PYQs). Once you figure out what the exam demands, scoring 150+ becomes much easier and that is precisely what PYQs help you decode. It tells you the patterns of questions, concepts most repeated, and the perfect precision and quickness required for scoring well.

The smart strategy isn’t just solving PYQs but learning from them:

Learn Important Topics: These are usually the high-weightage chapters which the previous year questions suggest, Kinematics, Chemical Bonding, Current Electricity, and Coordinate Geometry.

Trends in Questions: Find out how questions change every year. Sometimes the same concept is asked in a twisted way.

Understanding the JEE Pattern: You will actually feel how difficult it is, how questions are framed, and then some tricky distractors will be there in MCQs.

Speed & Accuracy: Time-bound practice through original papers helps you in speeding up their solutions for a gain with no loss of marks.

Constructing Exam Temperament: Solving previous years' papers gives due practice, decreases anxiety, boosts confidence, and improves decision-making in the exam.

Going through every paper thoroughly marking the errors, going through and revising less strong concepts, and practicing like problems should in itself boost your score to move beyond 150 with PYQs-difficult as starting late or going over information again could well be-but without having to study.

JEE Main PYQ Best Books