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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.
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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.
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.
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.
$$
\begin{aligned}
& m_f=\frac{\delta_{\mathrm{muxx}}}{f_m} \\
& = \pm \frac{K E_m}{f_m}
\end{aligned}
$$
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
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 $\mathrm{P}(\mathrm{x}, \mathrm{y})$ 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 $\lim _{x \rightarrow a} \frac{\log _e(1+f(x))}{f(x)}=1$
Frequently Asked Questions (FAQs)
Revision is the best way to remember all the formulas. Practice more questions based on formulas and revise the formulas on a daily basis.
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.
General formula for alkanes is CnH2n+2 , alkenes is CnH2n and for alkynes is C
nH2n-2 respectively.
The formula of molecular mass in terms of vapor density is
Molecular mass = 2 * vapor density
On Question asked by student community
In Malda district, West Bengal, the JEE Main 2025 exam centre code is WB20. Candidates from the district can choose Malda as their preferred exam centre during registration. The final allotment of centres depends on availability and preferences submitted by students. The exact number of exam halls may vary each year based on the number of applicants. Students should confirm their allotted centre through the official JEE Main admit card.
Yes, you should apply EWS in JEE Main form if you are eligible. For 2026 attempt, it’s better to make your EWS certificate in advance and keep it ready, because at the time of counseling they ask for a valid certificate.
The approximate annual cost for 11th/12th PCM and JEE coaching is around 1.5 to 3.5 lakhs for institutes, excluding hostel and mess fees. The total fees including hostel as well as mess fees can rise upto 4.5 to 6.5 lakhs and above, depending on location and institute quality.
Hi dear candidate,
You can anytime visit our official website to find the previous 10 years JEE Mains question papers with solutions. Kindly refer to the link attached below to download them in PDF format:
JEE Main Last 10 Years Question Papers with Solutions (2025 to 2015)
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Hello,
For JEE Main and JEE Advanced , the cut-offs are lower for ST category students. Here is a simple idea based on recent trends:
JEE Main qualification for ST : Around 50–60 marks is usually enough to qualify for JEE Advanced.
JEE Advanced qualification for ST : You just need to clear the JEE Main cut-off, then appear for Advanced.
To get good NITs or IITs , you will need higher marks.
For NITs (Hyderabad or good branches), try for 120+ marks in JEE Main .
For IITs , even with ST quota, you should aim for at least 80–100+ marks in JEE Advanced for decent branches.
Since you are from ST category and Hyderabad , you don’t need 300 marks in JEE Main. Try to score as high as possible to get better branches, but even moderate marks can qualify you.
Hope it helps !
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