JEE Main Formulas 2026 - JEE Mains is the competitive entrance exams for engineering programs. To do well on this exam, you must understand the fundamental concepts and formulas of Mathematics, Physics, and Chemistry. Aspirants preparing for the JEE Mains can check the JEE Main formulas 2026 available on this page. JEE Mains is Conducted by the National Testing Agency (NTA), this exam evaluates a candidate's proficiency in Physics, Chemistry, and Mathematics. Since there are many formulas in these three subjects, candidates need a way to recall them for revision purposes. JEE Main important formulas in Math, Physics, and Chemistry are integral to calculating answers for numerical questions according to the JEE Main 2026 syllabus.

LiveJEE Main 2026 Registration LIVE: NTA demo link shows server error ahead of official application formOct 9, 2025 | 5:26 PM IST

No, the exam dates for the JEE Main 2026 examination has not been announced yet. However, as per the official notification, the National Testing Agency (NTA) will conduct the JEE Mains 2026 exam in two sessions, one in January 2026 and second in April 2026.

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JEE Main Formulas 2026 for Physics, Chemistry, Maths
JEE Main 2026 Exam Pattern
JEE Main Formulas for Physics 2026
JEE Main formulas for Chemistry 2026
JEE Main formulas for Maths 2026

JEE Main Important Formulas 2026 for Physics, Chemistry, Maths

JEE Main All Formulas 2026 for Physics, Chemistry, Maths

JEE Main Formulas 2026 for Physics, Chemistry, Maths

Candidates must make a handy note of all important formulas to revise frequently. Candidates must have a good command of each topic and the formulas to crack the JEE Main 2026 exam to ace the test. Through this article, students can find the provided JEE Main formulas for all three subjects. Knowing important formulas in depth can help you solve problems fast and accurately, which is important for scoring well in JEE Main. All the formulas given here are according to the JEE Main 2026 syllabus strictly.

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JEE Main 2026 Exam Pattern

The exam is divided into two main sections:

Section A: The quizzes contained in this section are 20 MCQs for each course. What the MCQ requires is four options and only one of them is the right answer.

Section B: This section has 5 numerical value questions for each of the subject areas, and the candidate only has to answer all five of these. These numerical value questions have to be answered accurately, sometimes to the second decimal place.

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JEE Main Formulas for Physics 2026

Aspirants preparing for JEE Mains must remember that along with concepts one needs to revise and remember the formulas, which are very important while solving any problems. As JEE Main Physics formulas are given below, these formulas need to be memorized daily as direct questions and formulas are asked in exams.

Below are a few important formulas for JEE Main Physics.

Newton’s first law of motion

If $F_{net} = O \Rightarrow a_{net} = O \Rightarrow$ forces in all directions are zero,i.e, $\sum {\vec{F}}_{x} = 0, \sum {\vec{F}}_{y} = 0, \sum {\vec{F}}_{z} = 0$

Work, Energy, and Power for Rotating Body

Work-
For translation motion $W = \int F d s$
So for rotational motion $W = \int τ d θ$
2. Rotational kinetic energy-

The energy of a body has by virtue of Its rotational motion is called its rotational kinetic energy.

3. Power =Rate of change of kinetic energy

For translation motion $P = \vec{F}, \vec{V}$
So for rotational motion

$P = \frac{d (K_{R})}{d t} = \frac{d (\frac{1}{2} J_{ι} 2)}{d t} = I ω \frac{d ω}{d t} = I α ω = τ \cdot ω$

Or $P = \vec{τ} \cdot \vec{ω}$

Newton's Law of Gravitation

$F \propto \frac{m_{1} m_{2}}{r^{2}}$

Or, $F = \frac{G m_{1} m_{2}}{r^{2}}$

Where
$F \to$ Force
$G \to$ Gravitational constant
$m_{1}, m_{2} \to$ Masses
$r \to$ Distance between masses

Gravitational Potential energy at a point

Then gravitational force on test mass m at a distance r from M is given by $F = \frac{G M m}{r^{2}}$
And the amount of work done in bringing a body from $\infty$ to $r$

$= W = \int_{\infty}^{r} \frac{G M m}{x^{2}} d x = - \frac{G M m}{r}$

And this is equal to gravitational potential energy
SoU $= - \frac{G M m}{r}$
$U \to$ gravitational potential energy
$M \to$ Mass of source-body
$m \to$ mass of test body
$r \to$ distance between two
Note- U is always negative in the gravitational field because Force is attractive in nature.
Means As the distance r increases U becomes less negative
l.e U will increase as r increases

And for $r = \infty, U = 0$ which is maximum

Gravitational Potential energy of discrete distribution of masses

$U = - G [\frac{m_{1} m_{2}}{r_{12}} + \frac{m_{2} m_{3}}{r_{23}} + \dots]$

$U \to$ Net Gravitational Potential Energy
$r_{12}, r_{23} \to$ The distance of masses from each other

Change of potential energy

If a body of mass $m$ is moved 些， $r_{1}$ to $r_{2}$

Then Change of potential energy is given as

$Δ U = G M m [\frac{1}{r_{1}} - \frac{1}{r_{2}}]$

$Δ U \to$ change of energy
$r_{1}, r_{2} \to$ distances
If $r_{1} > r_{2}$ then the change in potential energy of the body will be negative.
Le To decrease potential energy of a body we have to bring that body closer to the earth.

The relation between Potential and Potential energy

$\begin{aligned} As U = \frac{- G M m}{r} = m [\frac{- G M}{r}] \\ But V = - \frac{G M}{r} \\ So U = m V \end{aligned}$

Where $V \to$ Potential
$U \to$ Potential energy
$r \to$ distance

The gravitational potential energy at height 'h' from the earth's surface

$U_{h} = - \frac{G M m}{R + h}$

Using $G M = g R^{2}$

$\begin{aligned} U_{h} = - \frac{g R^{2} m}{R + h} \\ U_{h} = - \frac{m g R}{1 + \frac{h}{R}} \end{aligned}$

$U_{h} \to$ The potential energy at the height $h$
$R \to$ Radius of earth

Power in AM waves

If $V_{r m s}$ is root mean square value
and $R =$ Resistance
then Power dissipated in any circuit.

$P = \frac{V_{r m s}^{2}}{R}$

So Carrier Power will be given as

$P_{c} = \frac{{(\frac{E_{c}}{\sqrt{2}})}^{2}}{R} = \frac{E_{c}^{2}}{2 R}$

$E_{c} =$ The amplitude of the carrier wave
$R =$ Resistance
Similarly, Total Power of sidebands will be given as

$P_{s t} = \frac{{(\frac{m_{0} E_{c}}{2 \sqrt{2}})}^{2}}{R} + \frac{{(\frac{m_{n} E_{c}}{2 \sqrt{2}})}^{2}}{R} = \frac{m_{a}^{2} E_{c}^{2}}{4 R}$

Where

$m_{a} = modulation index$

$E_{c} =$ the amplitude of carrier waves

$R = resistance$

And this gives Total power of AM wave as

$\begin{aligned} P_{tatal} = P_{c} + P_{s b} \\ = \frac{E_{c}^{2}}{2 R} (1 + \frac{m_{a}^{2}}{2}) \end{aligned}$

where
$m_{a} =$ modulation index
$E_{c} =$ the amplitude of carrier waves
$R =$ Resistance
Note-maximum power in the AM wave without distortion Occurs when $m_{a} = 1$
Le $P_{t} = 1.5 P = 3 P_{s}$

Frequency modulation

Frequency modulation
- Frequency modulation deviation-The The amount by which carrier frequency is varied from its unmodulated value.

The deviation is made proportional to the instantaneous value of the modulating voitage.
- Value of frequency deviation $= \hat{δ} = \int - f_{c}$

$\begin{aligned} d_{Wax} = f_{max} - f_{c} \\ = \pm K E_{m} \end{aligned}$

$E_{x 0} =$ modulating amplitude
- The modulation index of frequency modulation-

It is defined as the ratio of maximum frequency deviation to the modulating frequency.

$\begin{aligned} m_{f} = \frac{δ_{muxx}}{f_{m}} \\ = \pm \frac{K E_{m}}{f_{m}} \end{aligned}$

JEE Main Syllabus: Subjects & Chapters

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JEE Main formulas for Chemistry 2026

Candidates while studying the chemistry, they need to revise and practice the chemical equations and symbols, to some chemistry is tough subject but when candidates practices chemical equations, revises the properties, formulas and symbols they will have command over the subject Candidates can check the JEE Main Chemistry formulas below

Physical Properties of Alkali Metals

1701851766948

Molar Conductance

Equivalent Conductance

$\begin{aligned} Equivalent conductance = \frac{Molar conductance}{x} \\ where x = \frac{Molecular mass}{Equivalent mass} = n - factor \end{aligned}$

Method of Preparation of Carboxylic Acid

1701851769429

Kinetic Energy

If at a given temperature, $n_{3}$ molecules have speed $u_{1}, n_{2}$ , malecules have speed $u_{2}, n_{3}$ molecules have speed $u_{3}$ and so an Then, the total kinetic energy ( EK ) of the gas at this temperature is given ly:

$E_{K} = \frac{1}{2} m (m_{1} v_{S}^{2} + n_{2} v_{2}^{2} + n_{s} v_{j}^{2} + \dots \dots \dots)$

where $m$ is the mass of the molecule. The corresponding average kinetic energy $\overset{―}{E_{k}}$ of the gas will be:

$\overset{―}{E_{K}} = \frac{1}{2} \frac{m_{1} (n_{1} v_{1}^{2} + n_{2} v_{2}^{2} + n_{3} v_{2}^{2} + \dots \dots . .)}{(n_{1} + n_{2} + n_{3} + \dots \dots .)}$

If the $verm \frac{{n_{1} v_{i}^{2} + n_{2} v_{2}^{3} + n_{j} v_{j}^{2} + \dots \dots .)}{(n_{1} + n_{2} + n_{3} + \dots \dots .)} = v^{2}$
then the average kinetic energy is given by :

$\overset{―}{E_{K}} = \frac{1}{2} m \overset{―}{v^{2}}$

where v is ghen by

$v = \sqrt{\frac{(n_{1} v_{1}^{2} + n_{2} v_{2}^{2} + n_{1} v_{3}^{2} + \dots \dots \dots)}{(n_{1} + n_{2} + n_{5} + \dots \dots \dots)}}$

This 'v' is known as root-mean-square speed $u_{mins}$

Molarity(M): No. of moles of solute/ volume of solution in liter

Molality (m): No. of moles of solute/weight of solvent in kg

JEE Main formulas for Maths 2026

Candidates must go through all the formulas and practice the mathematical problems. Without formulas you cannot solve any problem though you know how to solve it. Revising the formulas daily is very important. Here we have provided Mathematics formulas for JEE Mains.

Equation of Circle

Centre-Radius Form

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Let $P (x, y)$ be any point on the circle. Then, by definition, $| C P | = r$ .
Using the distance formula, we have

$\sqrt{(x - h)^{2} + (y - k)^{2}} = r$

i.e. $(x - h)^{2} + (y - k)^{2} = r^{2}$

If the centre of the circle is the origin or $(0, 0)$ then equation of the circle becomes $(x - 0)^{2} + (y - 0)^{2} = r^{2}$
i.e. $x^{2} + y^{2} = r^{2}$

- General equation of a circle

The equation of a circle with centre at ( $h, k$ ) and radius r is

$\begin{aligned} \Rightarrow (x - h)^{2} + (y - k)^{2} = r^{2} \\ \Rightarrow x^{2} + y^{2} - 2 h x - 2 k y + h^{2} + k^{2} - r^{2} = 0 \end{aligned}$

Which is of the form :

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

This is known as the general equation of the circle.

Line and Circle
S is a circle with center O and radius r , and L is a straight line in the plane of the circle.
Equation of circle $S : x^{2} + y^{2} = a^{2}$
Equation of line L: $y = m x + c$
To find their point(s) of intersection, we can solve these equations simultaneously $x^{2} + (mx + c)^{2} = a^{2}$
$(1 + m^{2}) x^{2} + 2 m c x + c^{2} - a^{2} = 0$

Exponential Limits

(i) $lim_{x \to 0} \frac{a^{x} - 1}{x} = \log_{e} a$

Proof:

$lim_{x \to 0} \frac{a^{x} - 1}{x} = lim_{x \to 0} \frac{(1 + \frac{x (\log a)}{1!} + \frac{x^{2} (\log a)^{2}}{2!} + \dots) - 1}{x}$

[using Taylor series expansion of $a^{x}$ ]

$\begin{aligned} = lim_{x \to 0} (\frac{\log a}{1!} + \frac{x (\log a)^{2}}{2!} + \dots) \\ = \log_{e} a \end{aligned}$

(ii) $lim_{x \to 0} \frac{e^{x} - 1}{x} = 1$

In General, if $lim_{x \to a} f (x) = 0$ , then we have
(a) $lim_{x \to a} \frac{a^{f (x)} - 1}{f (x)} = \log_{e} a$
(b) $lim_{x \to a} \frac{e^{f (x)} - 1}{f (x)} = \log_{e} e = 1$

Logarithmic Limits

To evaluate the Logarithmic limit we use the following results:

$lim_{x \to 0} \frac{\log_{e} (1 + x)}{x} = 1$

$lim_{x \to a} \frac{\log_{e} (1 + f (x))}{f (x)} = 1$

Frequently Asked Questions (FAQs)

Q: How do I remember all the formulas for JEE Main?

Revision is the best way to remember all the formulas. Practice more questions based on formulas and revise the formulas on a daily basis.

Q: Can I derive the formula during the exam?

Yes, you can derive the formula during the exam but it is very time-consuming so candidates must learn all the formulas to save time during the exam.

Q: What is the general formula for alkanes, alkenes, and alkynes respectively?

General formula for alkanes is CnH2n+2 , alkenes is CnH2n and for alkynes is C

nH2n-2 respectively.

Q: What is the formula of molecular mass in terms of vapor density?

The formula of molecular mass in terms of vapor density is

Molecular mass = 2 * vapor density

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How to study efficiently in last 3 months of jee

Hello

Here is the short guidance for you to score well and prepare for JEE

Stick to a smart routine – Prioritize high-weightage topics and revise daily with a focused timetable.
Solve PYQs & mock tests – Practice like it’s the real thing; analyze mistakes without panic.
Quality over quantity – Understand concepts deeply rather than rushing through chapters.
Stay healthy & rested – Your brain needs sleep, water, and breaks to perform at its best.
Trust your journey – Stay calm, believe in your preparation, and avoid comparing with others.

Read Complete Answer

SEJAL JAIN

8 Oct'25

i gave my 12th boards in 2025 feb, in which i got a compartment in result. i had an option to reappear in all subjects, I'll be reappearing in 12th boards 26, so how many jee adv attempts do i have and will there be any issues in registration of mains and advance as i will pass my12th in may 2026?

Hello

Since you’re reappearing for your Class 12 boards in 2026, that will be counted as your official passing year. So, you will be eligible to write JEE Advanced in 2026 and 2027. Although you submitted the boards in 2025, that attempt won’t be considered due to the compartment. JEE Advanced rules allow 2 attempts in 2 consecutive years after passing 12th. So you still have both chances left, which is great! Just make sure you meet the other eligibility conditions too. Keep your focus strong, you’ve got this!

Read Complete Answer

SEJAL JAIN

8 Oct'25

How to prepare for jee paper 2

Hello aspirant,

Like all competitive exams, JEE MAINS requires careful preparation and strategy to pass. The JEE Main exam is extremely competitive, so it's critical to carefully plan and strategy. Students can increase their speed, accuracy, and confidence by adhering to a systematic study schedule and practicing frequently.

To know complete preparation tips, you can visit our site through following link:

https://engineering.careers360.com/articles/jee-main-preparation-tips

Thank you

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Sajal Trivedi

8 Oct'25

Preparing books for learning JEE main

Hello aspirant,

Start with NCERT in every subject for JEE Main. For physics, use H.C. Verma and D.C. Pandey; for chemistry, use R.C. Mukherjee and O.P. Tandon; and for mathematics, use R.D. Sharma or Cengage. J.D. Lee assists with inorganic chemistry, while Morrison & Boyd assists with organic chemistry. Practice chapter-by-chapter Arihant or MTG past year's papers. For efficient preparation, concentrate on clear concepts, frequent revision, and passing practice exams.

Thank you

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Sajal Trivedi

8 Oct'25

Is income certificate required during registration of jee mains of current year??

Hello,

Generally an income certificate isn't required for the JEE Main registration, but if you want to claim the EWS quota, then you need this. You must provide the certificate, issued by a government authority, as proof of your family's income being below the specified limit for the reservation category you wish to apply under.

I hope it will clear your query!!

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