Important Physics Formulas For NEET 2025 Exam- Topic-wise Formulas

Important Physics Formulas for NEET 2025: Are you preparing for the NEET 2025 exam? Have you strategized for physics preparation? Physics plays a game-changing role in the NEET UG exam. So, mastering the important formulas from physics is critical to scoring high in the NEET exam. Here, we are providing a physics NEET formula sheet 2025 that contains a detailed topic-wise breakdown of the most important physics formulas that every NEET aspirant should know. From mechanics and electrodynamics to thermodynamics and modern physics, we cover all the key formulas in physics formula sheet for NEET 2025 to enhance your preparation. Boost your confidence and improve your NEET exam preparation with the physics formula sheet NEET. Read on to find out the important formulas for NEET Physics 2025.

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This Story also Contains

1. Physics Formula Sheet for NEET PDF - Topic-wise and Chapterwise
2. 1. Series LCR Circuit
3. 2. Equations Of Motions Of SHM
4. 3. Electric Potential Due To Continuous Charge Distribution
5. 4. Resistance and Resistivity
6. 5. Parallel Grouping of Resistance
7. 6. Kirchoff's second law
8. 7. Nature of Electromagnetic Waves
9. 8. Total Internal Reflection:
10. 9. Young's Double Slit Experiment
11. 10. De-Broglie Wavelength Of An Electron
12. 11. Logic Gates

The National Eligibility cum Entrance Test (NEET) is expected to be conducted in the first week of May. Aspirants must understand that reviewing the most important NEET physics formulas during last-minute revision plays a significant role in the exam. Every mark matters in the fiercely competitive NEET exam. NEET Physics Formula-based questions are frequently simple and offer students an opportunity to secure marks easily if they have a firm understanding of the physics chapter-wise formulas for NEET.

Lastly, becoming proficient with the NEET physics formula improves accuracy, efficiency, confidence, and problem-solving abilities. These are essential traits for NEET exam success. We have compiled a NEET physics formula sheet 2025 from the top 11 most-scoring concepts in physics on the NEET syllabus. The Physics NEET formula sheet will be the guiding light for last-minute NEET exam preparation.

Also, check:

• NEET Application Form 2025
• NEET Eligibility Criteria 2025
• NEET Previous Year Question Paper

Physics Formula Sheet for NEET PDF - Topic-wise and Chapterwise

Here are some physics formulas to help you with the revision:

1. Series LCR Circuit

Let 'i' be the amount of current in the circuit at any time and VL, VC and VR be the potential drop across L, C, and R, respectively. Then,
$\mathrm{v}_{\mathrm{R}}=\mathrm{iR} \rightarrow$ Voltage is in phase with i
$\mathrm{v}_{\mathrm{L}=\mathrm{i} \omega \mathrm{L}} \rightarrow$ voltage is leading i by $90^{\circ}$
$\mathrm{v}_{\mathrm{c}}=\mathrm{i} / \omega \mathrm{c} \rightarrow$ voltage is lagging behind i by $90^{\circ}$

using all these, we can draw a phasor diagram as shown below -

So, from the above phasor diagram, V will represent the resultant of vectors VR and (VL -VC). So the equation becomes -

$
\begin{aligned}
&\begin{aligned}
V & =\sqrt{V_R^2+\left(V_L-V_C\right)^2} \\
& =i \sqrt{R^2+\left(X_L-X_C\right)^2} \\
& =i \sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2} \\
& =i Z
\end{aligned}\\
&\text { where, }\\
&Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}
\end{aligned}
$

Here, Z is called the impedance of this circuit.

Now come to the phase angle. The phase angle for this case is given as -

$\tan \varphi=\frac{V_L-V_C}{V_R}=\frac{X_L-X_C}{R}=\frac{\omega L-\frac{1}{\omega C}}{R}$

2. Equations Of Motions Of SHM

As we know, $a=-\omega^2 x$

• General equation of SHM

1. For Displacement:-

$x=A \operatorname{Sin}(w t+\phi) ;$ where is initial phase and $(\omega t+\phi)$ is called as phase.

Various displacement equations:-

(1) $x=$ ASinwt $\Rightarrow$ when particle starts from mean position towards right.
(2) $x=-$ ASinwt $\Rightarrow$ when particle starts from mean position towards left.
(3) $x=A \operatorname{Cos} \omega t \Rightarrow$ when particle starts from extr eme position towards
(4) $x=-$ ACoswt $\Rightarrow$ when particle starts from left extreme position towards Right.

2. For Velocity (v):-

$\begin{aligned} x & =A \operatorname{Sin}(\omega t+\phi) \\ \Rightarrow v & =\frac{d x}{d t}=A \omega \operatorname{Cos}(\omega t+\phi)=A \omega \operatorname{Sin}\left(\omega t+\phi+\frac{\pi}{2}\right)\end{aligned}$

3. For Acceleration:-

$
\begin{aligned}
x & =A \operatorname{Sin}(\omega t+\phi) \\
\Rightarrow v & =\frac{d x}{d t}=A \omega \operatorname{Cos}(\omega t+\phi)=A \omega \operatorname{Sin}\left(\omega t+\phi+\frac{\pi}{2}\right) \\
\Rightarrow a & =\frac{d v}{d t}=-A \omega^2 \operatorname{Sin}(\omega t+\phi)=A \omega^2 \operatorname{Sin}(\omega t+\phi+\pi)=-\omega^2 x
\end{aligned}
$

So, here we can see that the phase difference between x and v is $\frac{\pi}{2}$
Similarly, the phase difference between v and a is $\frac{\pi}{2}$
Similarly, the phase difference between a and x is $\pi$

• Differential equation of SHM:-

$\begin{aligned} & \frac{d v}{d t}=-\omega^2 x \\ & \Rightarrow \frac{d}{d t}\left(\frac{d x}{d t}\right)=-\omega^2 x \\ & \Rightarrow \frac{d^2 x}{d t^2}+\omega^2 x=0\end{aligned}$

If the motion of any particle satisfies this equation, then that particle will do SHM.

Also read:

• How to prepare for NEET in 30 days?
• How to complete NEET syllabus on time?
• Preparation tips to crack NEET

3. Electric Potential Due To Continuous Charge Distribution

Electric Potential due to Hollow conducting, Hollow non-conducting, and Solid conducting Sphere-

In the case of Hollow conducting, Hollow non-conducting, and solid conducting spheres, charges always reside on the surface of the sphere.

If the charge on a conducting sphere of radius R is Q. We want to find V at point P at a distance r from the centre of the sphere.

• Outside the sphere (P lies outside the sphere. I.e r>R)

$
\begin{aligned}
& E_{\text {out }}=\frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}=\frac{\sigma R^2}{\epsilon_0 r^2} \\
& V(r)=-\int_{r=\infty}^{r=r} \vec{E} . d \vec{r}=\frac{1}{4 \pi \varepsilon_0} \frac{\vec{Q}}{r}
\end{aligned}
$

- Inside the sphere (P lies inside the sphere. I.e $\mathrm{r}<\mathrm{R}$ )

$
E_{\text {in }}=0
$

$V_{\text {in }}=$ constant and it is given as

$
\begin{aligned}
\boldsymbol{V}(\boldsymbol{r}) & =-\int_{r=\infty}^{r=r} \overrightarrow{\boldsymbol{E}} \cdot d \vec{r}=-\int_{\infty}^R \boldsymbol{E}_r(d \boldsymbol{r})-\int_R^r \boldsymbol{E}_r(\boldsymbol{d} \boldsymbol{r})=\frac{1}{4 \pi \boldsymbol{\varepsilon}_0} * \frac{\boldsymbol{q}}{\boldsymbol{R}}+0 \\
\Rightarrow V(r) & =\frac{1}{4 \pi \varepsilon_0} * \frac{\boldsymbol{q}}{\boldsymbol{R}}
\end{aligned}
$

• At the surface of the Sphere (I.e at r=R)

$\begin{aligned} & E_s=\frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2}=\frac{\sigma}{\epsilon_0} \\ & V_s=\frac{1}{4 \pi \epsilon_0} \frac{Q}{R}=\frac{\sigma R}{\epsilon_0}\end{aligned}$

• The graph between (E vs r) and (V vs r) is given below

Electric Potential due to Uniformly Charged Non conducting Sphere-

Suppose charge Q is uniformly distributed in the volume of a non-conducting sphere of Radius R.

And we want to find V at point P at a distance r from the centre of the sphere.

• Outside the sphere (P lies outside the sphere. I.e r>R)

$
\begin{aligned}
& E_{\text {out }}=\frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2} V_{\text {out }}=\frac{1}{4 \pi \epsilon_0} \frac{Q}{r} \\
& E_{\text {out }}=\frac{\rho R^3}{3 \epsilon_0 r^2} V_{\text {out }}=\frac{\rho R^3}{3 \epsilon_0 r}
\end{aligned}
$

- Inside the sphere (P lies inside the sphere. I.e r<R )

$
\begin{aligned}
& E_{i n}=\frac{1}{4 \pi \epsilon_0} \frac{Q r}{R^3} V_{i n}=\frac{Q}{4 \pi \epsilon_0} * \frac{3 R^2-r^2}{2 R^3} \\
& E_{\text {in }}=\frac{\rho r}{3 \epsilon_0} V_{i n}=\frac{\rho\left(3 R^2-r^2\right)}{6 \epsilon_0}
\end{aligned}
$

• At the surface of Sphere (I.e at r=R)

$\begin{aligned} E_s & =\frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2} V_s=\frac{1}{4 \pi \epsilon_0} \frac{Q}{R} \\ E_s & =\frac{\rho R}{3 \epsilon_0} V_s=\frac{\rho R^2}{3 \epsilon_0}\end{aligned}$

Note - If P lies at the centre of the uniformly charged non-conducting sphere (I.e at r = 0),

$\begin{aligned} & V_{\text {centre }}=\frac{3}{2} \times \frac{1}{4 \pi \epsilon_0} \frac{Q}{R}=\frac{3}{2} V_s \\ & \text { i.e } V_c>V_s\end{aligned}$

• The graph between (E vs r) and (V vs r) is given below

4. Resistance and Resistivity

Resistance

For a conductor of resistivity $\rho$ having a length of a conductor=1
and Area of a crosssection of conductor $=\mathrm{A}$
Then the resistance of a conductor is given as

$$
R=\rho \frac{l}{A}
$$

Where $\rho \rightarrow$ Resistivity
And For a conductor, if $n=$ No. of free electrons per unit volume in the conductor, $\tau=$ relaxation time then the resistance of the conductor
Then $^\rho=\frac{m}{n e^2 \tau}$
for different conductors, n is different
And $\rho$ depends on the $n$
So R is also different.
Resistivity or Specific Resistance $(\rho)$

• As

$
R=\rho \frac{l}{A}
$

$|f|=1 \mathrm{~m}$ and $A=1 \mathrm{~m} 2$
Then $\mathrm{R}=\rho$

Resistivity is numerically equal to the resistance of a substance having a unit area of cross-section and unit length.

5. Parallel Grouping of Resistance

Potential is the Same across each resistor and current is different

$
\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n}
$

If two resistances are in Parallel:

$
R_{e q}=\frac{R_1 R_2}{R_1+R_2}
$

• Current through any resistance:

$
i^{\prime}=i\left(\frac{\text { Resistance of opposite Branch }}{\text { total Resistance }}\right)
$

The required current of the first branch

$
i_1=i\left(\frac{R_2}{R_2+R_2}\right)
$

The required current of the second branch

$
i_2=i\left(\frac{R_1}{R_1+R_2}\right)
$

6. Kirchoff's second law

The algebraic sum of all the potential across a closed loop is zero. This law is also known as Kirchhoff's Voltage law (KVL)

In closed-loop

$-i_1 R_1+i_2 R_2-E_1-i_3 R_3+E_2+E_3-i_4 R_4=0$

Also read:

• Important Chemistry Formulas for NEET 2025
• NEET Formula Sheet

7. Nature of Electromagnetic Waves

From Maxwell’s equations, we can observe that electric and magnetic fields in an electromagnetic wave are perpendicular to each other and the direction of propagation. Also, from our discussion of the displacement current, in that capacitor, the electric field inside the plates is directed perpendicular to the plates.

The figure given below shows a typical example of a plane electromagnetic wave propagating along the z direction (the fields are shown as a function of the z coordinate at a given time t). The electric field Ex is along the x-axis and varies sinusoidally with z at a given time. The magnetic field By is along the y-axis and again varies sinusoidally with z. The electric and magnetic fields Ex and By are perpendicular to each other and the direction z of propagation.

Now from the Lorentz equation -

$
\begin{aligned}
& \vec{F}=q(\vec{E}+\vec{v} \times \vec{B}) \\
& E_z=E z_0 \sin (\omega t-k y)
\end{aligned}
$

$
B_x=B x_0 \sin (\omega t-k y), \text { where } \frac{\omega}{k}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}
$

since, $\omega=2 \pi f$, where f is the frequency and $k=\frac{2 \pi}{\lambda}$, where $\lambda$ is the wavelength.
Therefore, $\frac{\omega}{k}=\frac{2 \pi f}{2 \pi / \lambda}=f \lambda$
But $f \lambda$ gives the velocity of the wave. So, $f \lambda=c=\omega k$. So we can write -

$
c=\frac{\omega}{k}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}
$

It is also seen from Maxwell’s equations that the magnitude of the electric and magnetic fields in an electromagnetic wave are related as -

$
B_0=\frac{E_o}{c}
$

In a material medium of permittivity $\varepsilon$ and magnetic permeability $\mu$, the velocity of light becomes,

$
v=\frac{1}{\sqrt{\mu \varepsilon}}
$

8. Total Internal Reflection:

When a ray of light goes from a denser to a rarer medium, it bends away from the normal, and as the angle of incidence in a denser medium increases, the angle of refraction in a rarer medium also increases and at a certain angle, the angle of refraction becomes 90 degree this angle of incidence is called critical angle (C).

When the Angle of incidence exceeds the critical angle then light ray comes back into the same medium after reflection from the interface. This phenomenon is called Total internal reflection (TIR).

Using Snell's law :

$\begin{aligned} & \mu_2 \sin C=\mu_1 \sin r \\ & \Longrightarrow \mu_2 \sin C=\mu_1 \text { since, } \sin r=1 . \\ & \Longrightarrow \sin C=\frac{\mu_1}{\mu_2}=\frac{\text { R.I of rarer medium }}{\text { R.I of denser medium }} \\ & \text { or }^{\mu=\frac{1}{\sin C}}{ }_{\text {when }} \mu_{1=1} \text { for air and } \mu_2=\mu .\end{aligned}$

Conditions for TIR :

(i) The ray must travel from a denser medium to a rarer medium.

(ii) The angle of incidence 'i' must be greater than the critical angle 'C' i.e i>C.

9. Young's Double Slit Experiment

Young's double-slit experiment

Let d be the distance between two coherent sources A and B with wavelength λ. A screen XY is placed parallel to an opaque screen at a distance D. O is a point on the screen equidistant from A and B. P is a point at a distance x from O

From the above figure, we can see that the waves from A and B meet at P. It may be in phase or out of phase, depending upon the path difference between the two waves,

So, the path differnece is $=\frac{x d}{D}$

Assumptions in this experiment -

1. D> d: Since D > > d, the two light rays are assumed to be parallel, then the path difference,

2. d/λ >> 1: Often, d is a fraction of a millimetre and λ is a fraction of a micrometre for visible light.

For Bright Fringes -

By the principle of interference, the condition for constructive interference is the path difference = nλ

$
\frac{x d}{D}=n \lambda
$

Here, $n=0,1,2 \ldots \ldots$ indicate the order of bright fringes
So, $x=\left(\frac{n \lambda D}{d}\right)$
This equation gives the distance of the $n^{\text {th }}$ bright fringe from the point O .

For Dark fringes -

By the principle of interference, the condition for destructive interference is the path difference $=\frac{(2 n-1) \lambda}{2}$ Here, $n=1,2,3 \ldots$ indicates the order of the dark fringes.

So,

$
x=\frac{(2 n-1) \lambda D}{2 d}
$

The above equation gives the distance of the nth dark fringe from the point O.

So, we can say that the alternately dark and bright fringe will be obtained on either side of the central bright fringe.

Band Width (β) -

The distance between any two consecutive bright or dark bands is called bandwidth.

Take the consecutive dark or bright fringe -

$
\begin{aligned}
& x_{n+1}-x_n=\frac{(n+1) \lambda D}{d}-\frac{(n) \lambda D}{d} \\
& x_{n+1}-x_n=\frac{\lambda D}{d}
\end{aligned}
$

$
\beta=\frac{\lambda D}{d}
$

Angular fringe width -

$
\theta=\frac{\beta}{D}=\frac{\lambda D / d}{D}=\frac{\lambda}{d}
$

10. De-Broglie Wavelength Of An Electron

De - Broglie wavelength of Electron-

De Broglie’s Equation is given as $\lambda=\frac{h}{p}=\frac{h}{m v}=\frac{h}{\sqrt{2 m K}}$

So for an electron having velocity v attained by it when it is accelerated through a potential difference of V.

then (Kinetic energy gain by the electron) = (work is done on an electron by the electric field)

i.e. $K=W_E \Rightarrow \frac{1}{2} m_e v^2=e V$

So, the De-Broglie wavelength of Electron is given as

$
\lambda_e=\frac{h}{m_e v}=\frac{h}{\sqrt{2 m_e K}}=\frac{h}{\sqrt{2 m_e(e V)}}
$

using $h=6.626 \times 10^{-34} \mathrm{~J}$ and $m_e=9.1 \times 10^{-31} \mathrm{~kg}$ and $e=1.6 \times 10^{-19} \mathrm{C}$
we get

$
\lambda_e=\frac{12.27}{\sqrt{V}} A_{\text {(i.e answer will be in }} A^0=\text { Angstrom }_{\text {) }}
$
Similarly, we can find De - Broglie wavelength associated with charged particle
De - Broglie wavelength with charged particle-

$
\lambda=\frac{h}{\sqrt{2 m K}}=\frac{h}{\sqrt{2 m q V}}
$
Where $K \rightarrow$ kinetic energy of particle
$q \rightarrow$ charged particle
$V \rightarrow$ potential diffenence

• De - Broglie wavelength of the proton

using $m_p=1.67 \times 10^{-27} \mathrm{~kg}$ and $q_p=e=1.6 \times 10^{-19} \mathrm{C}$
${ }_{\text {we get }} \lambda_{\text {proton }}=\frac{0.286}{\sqrt{V}} A^{\circ}$
- De - Broglie wavelength of Deuteron
using $m_D=2 \times 1.67 \times 10^{-27} \mathrm{~kg}$ and $q_D=e=1.6 \times 10^{-19} \mathrm{C}$
we get $\lambda_{\text {deutron }}=\frac{0.202}{\sqrt{V}} A^{\circ}$
- De - Broglie wavelength of an Alpha particle (He2+)
using $m_{\alpha^{2+}}=4 \times 1.67 \times 10^{-27} \mathrm{~kg}$ and $q_{\alpha^{2+}}=2 e=2 \times 1.6 \times 10^{-19} \mathrm{C}$
we get $\lambda_{\alpha-\text { partical }}=\frac{0.101}{\sqrt{V}} A^{\circ}$

11. Logic Gates

Logic Gates-

The five most commonly used logic gates are:

• NOT

• AND

• OR

• NAND

• NOR

NOT Gate -

A NOT gate is also known as an inverter because it simply inverts the input signal. It is a simple gate with one input and one output. So, the output is ‘0’ when the input is ‘1’ and vice-versa.

A is input

Y is output

$Y=\bar{A}$

The truth table for a NOT gate is as follows:

AND Gate-

An AND gate has two or more inputs and a single output. In this gate, the output is 1 (high) only when all the inputs are 1 (high). The most commonly used symbol for an AND gate is as follows:

A and B are inputs

Y is output

Y=A.B

The truth table for the AND gate is as follows

OR Gate-

Like AND Gate, OR gate also has two or more inputs and one output. For this Gate, the logic is that the output would be 1 when at least one of the inputs is 1. It means when the output is high, any of the inputs are high. The commonly used symbol for an OR gate is as follows:

A and B are inputs

Y is output

Relation between input and output

Y=A+B

The truth table for an OR gate is as follows:

NAND Gate-

A NAND gate is an arrangement of AND gate followed by a NOT gate. The output is 1 only when all inputs are NOT 1, or the output is high when at least one is low. These are also called Universal gates. The commonly used symbol for a NAND gate is as follows:

$Y=\overline{A \cdot B}$

A and B are inputs

Y is output

NOT + AND gate

And the truth table for a NAND gate is as follows:

NOR Gate-

Like the NAND Gate, the NOR Gate is also an arrangement of an OR Gate followed by a NOT Gate. In this, the output is 1 (High) only when all inputs are 0 (Low). These are also called Universal gates. The commonly used symbol for a NOR gate is as follows:

$Y=\overline{A+B}$

A and B are inputs

Y is output

NOT + OR Gate

The truth table for a NOR gate is as follows:

D'morgan's Theorem -

A and B are input.

1) $\overline{A+B}=\bar{A} \cdot \bar{B}$
2) $\overline{A \cdot B}=\bar{A}+\bar{B}$
3) $\bar{A}+\bar{B}=A \cdot B$
4) $\bar{A} \cdot \bar{B}=A+B$

Some Important relation -

$\begin{aligned} & A+A=A \\ & A \cdot A=A \\ & A+1=1 \\ & A \cdot 1=1 \\ & A \cdot 0=0 \\ & A+0=A\end{aligned}$

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