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JEE Main Physics Formulas 2025 - Topic wise Important Formulas
Edited By Team Careers360 | Updated on Dec 15, 2024 10:33 PM IST | #JEE Main
JEE Main Physics Formulas 2025 - JEE Main is one of the top Exams for engineering. Having a consolidated formula sheet makes your preparation for the JEE Main exam more manageable and efficient. Candidates should manage all three subjects properly, when candidates have a compiled list of important physics formulas candidates can revise the concepts faster and solve numerous problems easily. Knowing when and which formula to apply is crucial for aspirants. Here are some important JEE Main Physics formulas 2025 based on the JEE syllabus. Candidates can follow the Physics formulas to prepare for the IIT JEE exam:
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Few Important JEE Main Physics Formulas 2025
Chapter - Kinematics
Speed and Velocity
I. Speed Formula-
Speed $=\frac{\text { Change in distance }}{\text { change in time }}$
$v=\frac{\text { distance }}{\text { time }}$
- Tips of Speed-
1. S.I. unit $\longrightarrow \mathrm{ms}^{-1}$ or meters per second.
2. Dimensions $=L T^{-1}$
3. Speed is a scalar quantity.
- E.g: A body covers a distance of 18 m in 1 sec
Using the formula of speed. $v=\frac{\text { distance }}{\text { time }}=\frac{18 \mathrm{~m}}{1 \mathrm{~s}} \Rightarrow 18 \mathrm{~m} / \mathrm{s}$
II. Average Speed and Instantaneous Speed
Average Speed Formula-
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Use NowAverage speed $=\frac{\text { total distance covered }}{\text { total time taken }} v_{a v}=\frac{s}{t}$
E.g. A body covers a total distance of 50 m with variable speed in 5 sec.
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$$
\begin{aligned}
& v_{a v}=\frac{s}{t} \\
& \Rightarrow \frac{50 \mathrm{~m}}{5 \mathrm{~s}}=10 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Average Speed $=10 \mathrm{~m} / \mathrm{s}$
- Tips for average speed-
If an object or body covers s1 distance in $\mathrm{t1}$ time and s2 distance in 12 time then average speed is calculated by the Formula- $V_{\alpha v}=\frac{s_1+s_2}{t_1+t_2}$
2. Instantaneous Speed-
- It is the speed at that particular instant or small interval of time.
- Formula-
$$
v_{\text {inst }}=\lim _{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}
$$
III. Velocity
Rate of change of displacement with time.
Formula-
$$
V=\frac{\text { displacement }}{\text { time }}
$$
- Tips For Velocity-
1. S.I. unit $\rightarrow \mathrm{m} / \mathrm{s}$
2. Dimensions- $L T^{-1}$
3. Velocity is a vector quantity.
4. Velocity is also called speed in a definite (particular) direction.
IV. Average Velocity and Instantaneous Velocity
1. Average Velocity-Formula-
Average Velocity $=\frac{\text { Total Displacement }}{\text { Total time taken }} \vec{V}_{\text {avg }}=\frac{\vec{S}_{\text {net }}}{t}$
2. Instantaneous Velocity Formula-
$$
\vec{V}_{\text {inst }}=\lim _{\Delta t \rightarrow 0} \frac{\overrightarrow{\Delta x}}{\Delta t}
$$
Chapter - Work Energy and Power
- Work done by a constant force-
1. The scalar product of the force vector $(\vec{F})$ and the displacement vector $(\vec{S})$
$$
W=\vec{F} \cdot \vec{S}
$$
2. The product of the magnitude of force $(F)$ magnitude of displacement $(S)$ and cosine of the angle between them $(\Theta)$
$$
W=F S \cos \theta
$$
3. If the number of forces $\vec{F}_1, \vec{F}_2, \vec{F}_3 \ldots \ldots, \vec{F}_{\mathrm{n} \text {, are acting on a body and it shifts from position vector }} \vec{r}_1$ to position vector $\vec{r}_2$
Then
$$
W=\left(\vec{F}_1+\vec{F}_2+\vec{F}_3 \ldots \ldots \ldots+\vec{F}_n\right) \cdot\left(\vec{r}_2-\vec{r}_1\right)=\vec{F}_{n e t} \cdot \vec{r}_{n e t}
$$- Work done by a variable force
Force is a vector quantity. So it has a magnitude as well as direction. A variable force means when its magnitude its direction or both varies with position.
And work done by the variable force is given by -$$
W=\int \vec{F} \cdot \overrightarrow{d s}
$$
Where $\vec{F}$ is a variable force and $\vec{d}$ is a small displacement
- Kinetic energy$$
K . E .=\frac{1}{2} m v^2
$$
Where $m \rightarrow$ mass$$
v \rightarrow \text { velocity }
$$- Work-energy theorem-
Net work done by all the forces acting on a particle is equal to the change in its kinetic energy.
$$
\begin{aligned}
W & =\frac{1}{2} m v^2-\frac{1}{2} m v_0^2 \\
W & =k_f-k_i
\end{aligned}
$$
Where $m=$ mass of the body
$v_0=$ initial velocity
$v=$ final velocity
- Relation of kinetic energy with linear momentum
$$
k=\frac{p^2}{2 m}
$$
Where $p \rightarrow$ momentum
$$
m \rightarrow \text { mass }
$$
- Change in potential energy -
The change in potential energy between any two points is defined as the work done by the associated conservative force in displacing the particle between these two points without any change in kinetic energy.
$$
U_i-U_f=\int_{r_i}^{r_f} \vec{f} \cdot \overrightarrow{d s}
$$
Where, $U_f$ - final potential energy
$U_i$ - initial potential energy
$f$-force
ds - small displacement
$r_i$ - initial position
$r_f-f i n a l$ position
We can define a unique value of potential energy only by assigning some arbitrary value to a fixed point called the reference point.
- Power
Average power-
$$
P_{a v}=\frac{\Delta w}{\Delta t}=\frac{\int_0^t p \cdot d t}{\int_0^t d t}
$$
Instantaneous power-
$$
P=\frac{d w}{d t}=P=\vec{F} \cdot \vec{v}
$$
Where, $\vec{F} \rightarrow$ force
$$
\vec{v} \rightarrow \text { velocity }
$$
i.e. power is equal to the scalar product of force with velocity
Experts of Careers 360 have compiled a list of important formulas for JEE Mains Physics to make the preparation easy for aspirants. Candidates preparing for the JEE Main exam should remember all the major formulas.
Frequently Asked Questions (FAQs)
1. Which chapter has more weightage in Physics in the JEE Main Exam?
Current electricity, optics, and Properties of solids have more weightage in Physics JEE Main.
2. How many questions of Physics are in the JEE Main Exam?
In the JEE Main Exam, there are 25 questions of physics out of which 20 questions in section A will be MCQ and 5 questions in section B will be questions with numerical value answers.
3. What is the formula in physics which relates speed, distance traveled and time taken by the body to travel that distance?
Speed= (distance) / (time)
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Have a question related to JEE Main ?
With a 72 percentile in JEE Main 2025 and an expected 80%+ in your 12th board exams , securing admission to B.Tech in Computer Science & Engineering (CSE) at Guru Nanak Dev University (GNDU), Amritsar may be challenging.
A 72 percentile typically translates to an AIR between 250,000–300,000 , which is significantly higher than the previous year's closing ranks for CSE and related branches at GNDU.
- Consider applying for branches with higher closing ranks, such as Computer Engineering or Cyber Security , though admission may still be competitive.
- Look into other universities or colleges
- For more detailed information, you can visit GNDU's official admission page: GNDU Admissions (https://www.gnduadmissions.org/)
Rohan 13 May,2025
With 48% in JEE Mains 2025 and an All India SC rank of 65,896, it is unlikely that you will qualify for JoSAA (Joint Seat Allocation Authority) counselling. Usually, candidates need to score above the cutoff marks and ranks to be eligible for JoSAA. However, you can still explore other options like state counselling, private engineering colleges, or universities that accept low JEE ranks. Don’t lose hope—many good colleges offer admission outside of JoSAA too.
Akansha Sharma 12 May,2025
To get into B.Tech in Cyber Security at IIT Gandhinagar (or similar top IITs) with OBC-NCL category, your JEE Main percentile should ideally be above 90. For top-tier colleges, the closing rank for OBC-NCL students usually falls between 10,000 and 20,000, depending on the competition and year. In terms of percentile, aiming for 90-95% would give you a competitive chance. You should also check JoSAA cutoffs after JEE Main results to know more precise details.
Akansha Sharma 12 May,2025
With a JEE Main 2025 rank of 1.8 lakh and no reservation (General category, Home State quota), your chances for admission to Odisha University of Technology and Research (OUTR), Bhubaneswar are limited for most core B.Tech branches.
-
Apply through the Odisha Joint Entrance Examination (OJEE) counseling process.
-
Be open to branches like Civil Engineering, Biotechnology, Textile Engineering, or Fashion & Apparel Technology.
-
Look into other engineering colleges in Odisha.
For more detailed information on cutoffs and counseling procedures, you can visit the OJEE official website (https://ojee.nic.in) and OUTR official website (https://outr.ac.in) .
Rohan 13 May,2025
- CSAB will use your OBC-NCL Category Rank (16,800) if you apply as OBC-NCL.
- JAC Delhi uses your OBC-NCL Category Rank (16,800) if you are applying under OBC-NCL category.
For more Details, Refer to : CSAB Counselling 2025 Process (Careers360) .
Rohan 13 May,2025
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