JEE Main Important 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 Maths, Physics, and Chemistry. Knowing important formulas in depth can help you solve problems fast and accurately, which is important for scoring well in JEE Main.

New 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.

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 $\mathrm{F}_{\text {net }}=\mathrm{O} \Rightarrow \mathrm{a}_{\mathrm{net}}=\mathrm{O} \Rightarrow$ 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 $W=\int \tau d \theta$
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

$$
P=\frac{d\left(K_R\right)}{d t}=\frac{d\left(\frac{1}{2} J_\iota{ }^2\right)}{d t}=I \omega \frac{d \omega}{d t}=I \alpha \omega=\tau \cdot \omega
$$

Or $P=\vec{\tau} \cdot \vec{\omega}$

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
$F \rightarrow$ Force
$G \rightarrow$ Gravitational constant
$m_1, m_2 \rightarrow$ Masses
$r \rightarrow$ 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 $=-\frac{G M m}{r}$
$U \rightarrow$ gravitational potential energy
$M \rightarrow$ Mass of source-body
$m \rightarrow$ mass of test body
$r \rightarrow$ 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

$$
U=-G\left[\frac{m_1 m_2}{r_{12}}+\frac{m_2 m_3}{r_{23}}+\cdots\right]
$$

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

Change of potential energy

If a body of mass $m$ is moved $\qquad$些， $r_1$ to $r_2$

Then Change of potential energy is given as

$$
\Delta U=G M m\left[\frac{1}{r_1}-\frac{1}{r_2}\right]
$$

$\Delta U \rightarrow$ change of energy
$r_1, r_2 \rightarrow$ 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

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

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

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

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

Using $G M=g R^2$

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

$U_h \rightarrow$ The potential energy at the height $h$
$R \rightarrow$ 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

$$
P_c=\frac{\left(\frac{E_c}{\sqrt{2}}\right)^2}{R}=\frac{E_c^2}{2 R}
$$

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

$$
P_{s t}=\frac{\left(\frac{m_0 E_c}{2 \sqrt{2}}\right)^2}{R}+\frac{\left(\frac{m_n E_c}{2 \sqrt{2}}\right)^2}{R}=\frac{m_a^2 E_c^2}{4 R}
$$

Where

$$
m_a=\text { modulation index }
$$

$E_c=$ the amplitude of carrier waves

$$
\mathrm{R}=\text { resistance }
$$

And this gives Total power of AM wave as

$$
\begin{aligned}
& P_{\text {tatal }}=P_c+P_{s b} \\
& =\frac{E_c^2}{2 R}\left(1+\frac{m_a^2}{2}\right)
\end{aligned}
$$

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 $P_{\mathrm{t}}=1.5 P=3 P_{\mathrm{s}}$

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 $=\hat{\delta}=\int-f_c$

$$
\begin{aligned}
& d_{\text {Wax }}=f_{\text {max }}-f_c \\
& = \pm K E_{\mathrm{m}}
\end{aligned}
$$

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

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

$$
\begin{aligned}
& m_f=\frac{\delta_{\mathrm{muxx}}}{f_m} \\
& = \pm \frac{K E_m}{f_m}
\end{aligned}
$$

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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

Molar Conductance

Equivalent Conductance

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

Method of Preparation of Carboxylic Acid

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:

$$
\mathrm{E}_{\mathrm{K}}=\frac{1}{2} m\left(\mathrm{~m}_1 \mathrm{v}_{\mathrm{S}}^2+\mathrm{n}_2 \mathrm{v}_2^2+\mathrm{n}_{\mathrm{s}} \mathrm{v}_{\mathrm{j}}^2+\ldots \ldots \ldots\right)
$$

where $m$ is the mass of the molecule. The corresponding average kinetic energy $\overline{E_k}$ of the gas will be:

$$
\overline{E_K}=\frac{1}{2} \frac{m_1\left(n_1 v_1^2+n_2 v_2^2+n_3 v_2^2+\ldots \ldots . .\right)}{\left(n_1+n_2+n_3+\ldots \ldots .\right)}
$$

If the $\operatorname{verm} \frac{\left\{\mathrm{n}_1 v_i^2+\mathrm{n}_2 \mathrm{v}_2^3+\mathrm{n}_{\mathrm{j}} \mathrm{v}_j^2+\ldots \ldots .\right)}{\left(\mathrm{n}_1+\mathrm{n}_2+\mathrm{n}_3+\ldots \ldots .\right)}=\mathrm{v}^2$
then the average kinetic energy is given by :

$$
\overline{\mathrm{E}_{\mathrm{K}}}=\frac{1}{2} \mathrm{~m} \overline{\mathrm{v}^2}
$$

where v is ghen by

$$
\mathrm{v}=\sqrt{\frac{\left(\mathrm{n}_1 \mathrm{v}_1^2+\mathrm{n}_2 \mathrm{v}_2^2+\mathrm{n}_1 \mathrm{v}_3^2+\ldots \ldots \ldots\right)}{\left(\mathrm{n}_1+\mathrm{n}_2+\mathrm{n}_5+\ldots \ldots \ldots\right)}}
$$

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

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

Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ 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. $\quad(x-h)^2+(y-k)^2=r^2$

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 ( $\mathrm{h}, \mathrm{k}$ ) 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 $\mathrm{S}: \mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$
Equation of line L: $y=m x+c$
To find their point(s) of intersection, we can solve these equations simultaneously $\mathrm{x}^2+(\mathrm{mx}+\mathrm{c})^2=\mathrm{a}^2$
$\left(1+m^2\right) x^2+2 m c x+c^2-a^2=0$

Exponential Limits

(i) $\lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{a}^{\mathrm{x}}-1}{\mathrm{x}}=\log _{\mathrm{e}} \mathrm{a}$

Proof:

$$
\lim _{x \rightarrow 0} \frac{a^x-1}{x}=\lim _{x \rightarrow 0} \frac{\left(1+\frac{x(\log a)}{1!}+\frac{x^2(\log a)^2}{2!}+\cdots\right)-1}{x}
$$

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

$$
\begin{aligned}
& =\lim _{x \rightarrow 0}\left(\frac{\log a}{1!}+\frac{x(\log a)^2}{2!}+\cdots\right) \\
& =\log _e a
\end{aligned}
$$

(ii) $\lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{e}^{\mathrm{x}}-1}{\mathrm{x}}=1$

In General, if $\lim _{x \rightarrow a} f(x)=0$, then we have
(a) $\lim _{x \rightarrow a} \frac{a^{f(x)}-1}{f(x)}=\log _e a$
(b) $\lim _{x \rightarrow a} \frac{e^{f(x)}-1}{f(x)}=\log _e e=1$

Logarithmic Limits

To evaluate the Logarithmic limit we use the following results:

$$
\lim _{x \rightarrow 0} \frac{\log _e(1+x)}{x}=1
$$

Proof:

$$
\lim _{x \rightarrow 0} \frac{\log _e(1+x)}{x}=\lim _{x \rightarrow 0} \frac{x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots}{x}
$$

[using Taylor series expansion of $\log _e(1+x)$ ]

$$
\begin{aligned}
& =\lim _{x \rightarrow 0}\left(1-\frac{x}{2}+\frac{x^2}{3}-\cdots\right) \\
& =1
\end{aligned}
$$

In General, if $\lim _{x \rightarrow a} f(x)=0$, then we have $\lim _{x \rightarrow a} \frac{\log _e(1+f(x))}{f(x)}=1$