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

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

Edited By Irshad Anwar | Updated on May 02, 2025 10:20 AM IST | #NEET
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NEET  Result Date : 14 Jun' 2025 - 14 Jun' 2025

Important Physics Formulas for NEET 2025: Preparation for the NEET 2025 exam demands thorough knowledge of Physics, which is generally regarded as a difficult but scoring subject. A clear knowledge of Physics formulas is important for solving questions fast. To help NEET candidates, a topic-wise physics NEET formula sheet 2025 has been prepared. It covers important formulas of most asked topics including mechanics, thermodynamics, tlectrodynamics, waves, and modern physics. Such formulas prove very helpful for instant revision and last-minute study.

This Story also Contains
  1. Physics Formula Sheet for NEET PDF - Topic-wise and Chapterwise
  2. Series LCR Circuit
  3. Equations Of Motions Of SHM
  4. Electric Potential Due To Continuous Charge Distribution
  5. Resistance and Resistivity
  6. Parallel Grouping of Resistance
  7. Kirchoff's second law
  8. Nature of Electromagnetic Waves
  9. Total Internal Reflection
  10. Young's Double Slit Experiment
  11. De-Broglie Wavelength Of An Electron
  12. Logic Gates
Important Physics Formulas For NEET 2025 Exam- Topic-wise Formulas
Important Physics Formulas For NEET 2025 Exam- Topic-wise Formulas

The NEET UG 2025 exam date is 4th May 2025. Revision gets easier if there is a good understanding of the physics formulas. Formula-based questions tend to be straightforward and provide an easy chance to get marks if the concept is understood properly. The Important NEET formula sheet pdf for physics includes formulas from the most relevant and regularly asked concepts from the syllabus. It helps in better accuracy, speed, and problem-solving skills which are important components of success in the NEET exam.

Also, check:

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

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

Series LCR Circuit

1714399578466

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,
vR=iR→ Voltage is in phase with i
vL=iωL→ voltage is leading i by 90∘
vc=i/ωc→ voltage is lagging behind i by 90∘

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

1714399578835

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

V=VR2+(VL−VC)2=iR2+(XL−XC)2=iR2+(ωL−1ωC)2=iZ where, Z=R2+(ωL−1ωC)2

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⁡φ=VL−VCVR=XL−XCR=ωL−1ωCR

Equations Of Motions Of SHM

1714399579263

As we know, a=−ω2x

  • General equation of SHM

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  1. For Displacement:-

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x=ASin(wt+ϕ); where is initial phase and (ωt+ϕ) is called as phase.

Various displacement equations:-

(1) x= ASinwt ⇒ when particle starts from mean position towards right.
(2) x=− ASinwt ⇒ when particle starts from mean position towards left.
(3) x=ACosωt⇒ when particle starts from extreme position towards
(4) x=− ACoswt ⇒ when particle starts from left extreme position towards Right.

  1. For Velocity (v):-

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x=ASin(ωt+ϕ)⇒v=dxdt=AωCos(ωt+ϕ)=AωSin(ωt+ϕ+π2)

  1. For Acceleration:-

x=ASin(ωt+ϕ)⇒v=dxdt=AωCos(ωt+ϕ)=AωSin(ωt+ϕ+π2)⇒a=dvdt=−Aω2Sin(ωt+ϕ)=Aω2Sin(ωt+ϕ+π)=−ω2x


So, here we can see that the phase difference between x and v is π2
Similarly, the phase difference between v and a is π2
Similarly, the phase difference between a and x is π

  • Differential equation of SHM:-

dvdt=−ω2x⇒ddt(dxdt)=−ω2x⇒d2xdt2+ω2x=0

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

Also read:

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.

1714399588951

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

Eout =14πϵ0Qr2=σR2ϵ0r2V(r)=−∫r=∞r=rE→.dr→=14πε0Q→r

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

Ein =0

Vin = constant and it is given as

V(r)=−∫r=∞r=rE→⋅dr→=−∫∞REr(dr)−∫RrEr(dr)=14πε0∗qR+0⇒V(r)=14πε0∗qR

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

Es=14πϵ0QR2=σϵ0Vs=14πϵ0QR=σRϵ0

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

1714399588805

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.

1714399589174

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

Eout =14πϵ0Qr2Vout =14πϵ0QrEout =ρR33ϵ0r2Vout =ρR33ϵ0r

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

Ein=14πϵ0QrR3Vin=Q4πϵ0∗3R2−r22R3Ein =ρr3ϵ0Vin=ρ(3R2−r2)6ϵ0

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

Es=14πϵ0QR2Vs=14πϵ0QREs=ρR3ϵ0Vs=ρR23ϵ0

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

Vcentre =32×14πϵ0QR=32Vs i.e Vc>Vs

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

1714399589301

Resistance and Resistivity

Resistance

For a conductor of resistivity ρ having a length of a conductor=1
and Area of a crosssection of conductor =A
Then, the resistance of a conductor is given as

R=ρlA


Where ρ→ Resistivity
And for a conductor, if n= No. of free electrons per unit volume in the conductor, τ= relaxation time and the resistance of the conductor
Then ρ=mne2τ
for different conductors, n is different
And ρ depends on the n
So R is also different.
Resistivity or Specific Resistance (ρ)

  • As

R=ρlA

|f|=1 m and A=1 m2
Then R=ρ

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

Parallel Grouping of Resistance

Potential is the Same across each resistor, and current is different

1714399590617

1Req=1R1+1R2+⋯+1Rn


If two resistances are in Parallel:

Req=R1R2R1+R2

  • Current through any resistance:

1714399590271

i′=i( Resistance of opposite Branch total Resistance )


The required current of the first branch

i1=i(R2R2+R2)


The required current of the second branch

i2=i(R1R1+R2)

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

1714399583779

−i1R1+i2R2−E1−i3R3+E2+E3−i4R4=0

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.

1714399589057

Now from the Lorentz equation -

F→=q(E→+v→×B→)Ez=Ez0sin⁡(ωt−ky)


Bx=Bx0sin⁡(ωt−ky), where ωk=1μ0ε0

since, ω=2πf, where f is the frequency and k=2πλ, where λ is the wavelength.
Therefore, ωk=2πf2π/λ=fλ
But fλ gives the velocity of the wave. So, fλ=c=ωk. So we can write -

c=ωk=1μ0ε0

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

B0=Eoc


In a material medium of permittivity ε and magnetic permeability μ, the velocity of light becomes,

v=1με

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 :

μ2sin⁡C=μ1sin⁡r⟹μ2sin⁡C=μ1 since, sin⁡r=1.⟹sin⁡C=μ1μ2= R.I of rarer medium R.I of denser medium or μ=1sin⁡Cwhen μ1=1 for air and μ2=μ.

1714399585384

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.

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


1714399585552


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 difference is =xdD

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

xdD=nλ


Here, n=0,1,2…… indicates the order of bright fringes
So, x=(nλDd)
This equation gives the distance of the nth bright fringe from the point O.

For Dark fringes -

By the principle of interference, the condition for destructive interference is the path difference =(2n−1)λ2.here, n=1,2,3… indicates the order of the dark fringes.

So,

x=(2n−1)λD2d

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 -

xn+1−xn=(n+1)λDd−(n)λDdxn+1−xn=λDd


β=λDd


Angular fringe width -

θ=βD=λD/dD=λd

De-Broglie Wavelength Of An Electron

De Broglie’s Equation is given as λ=hp=hmv=h2mK

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

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

i.e. K=WE⇒12mev2=eV

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

λe=hmev=h2meK=h2me(eV)

using h=6.626×10−34 J and me=9.1×10−31 kg and e=1.6×10−19C
we get

λe=12.27VA(i.e answer will be in A0= Angstrom )
Similarly, we can find De - Broglie wavelength associated with charged particle
De - Broglie wavelength with charged particle-

λ=h2mK=h2mqV
Where K→ kinetic energy of particle
q→ charged particle
V→ potential difference

  • De - Broglie wavelength of the proton

using mp=1.67×10−27 kg and qp=e=1.6×10−19C
we get λproton =0.286VA∘
- De - Broglie wavelength of Deuteron
using mD=2×1.67×10−27 kg and qD=e=1.6×10−19C
we get λdeutron =0.202VA∘
- De - Broglie wavelength of an Alpha particle (He2+)
using mα2+=4×1.67×10−27 kg and qα2+=2e=2×1.6×10−19C
we get λα− partical =0.101VA∘

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.

1714399587890

A is input

Y is output

Y=A¯

The truth table for a NOT gate is as follows:

1714399588065

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:

1714399588156

A and B are inputs

Y is output

Y=A.B

The truth table for the AND gate is as follows

1714399588373

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 that when the output is high, any of the inputs are high. The commonly used symbol for an OR gate is as follows:

1714399588458

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:

1714399578750

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:

1714399579010

Y=A⋅B―

A and B are inputs

Y is output

NOT + AND gate

The truth table for a NAND gate is as follows:

1714399579349

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:

1714399579545

Y=A+B―

A and B are inputs

Y is output

NOT + OR Gate

The truth table for a NOR gate is as follows:

1714399588691

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Questions related to NEET

Have a question related to NEET ?

Medical Courses After 12th Without NEET

Hello,

Students can choose medical-related courses such as B.Sc. Nursing, BPT, B.Sc. in Medical Lab Technology, Radiology, Operation Theatre Technology, and Dialysis Technology after completing their 12th grade without taking the NEET. The duration of these courses is three to four years. Private college fees can reach Rs2,00,000 annually, while government college fees range from Rs30,000 to Rs70,000. There may be additional expenses for things like exam fees and lodging. Jobs in clinics, labs, and hospitals are available through these programs. Those without a NEET score who are interested in healthcare can use them. Science courses in the 12th grade and strong academic standing are typically prerequisites for admission.

All the best.

Hello Khushi,

As per your mentioned query, yes you can start your NEET preparation. NEET syllabus comprises of both class 11th and 12th syllabus. So you can start with basics of your class 12th. Many students also take coaching for exam and some of them start their own.

I hope this helps you!

Revert for further query!

For the 2025 NEET AIIMS exam, the expected SC category cutoff is likely to be around 400 to 450 marks. This estimate is based on previous years’ trends and may vary slightly depending on exam difficulty and seat availability.

If you have one back paper in 12th, your eligibility to participate in NEET counseling rounds (including mop-up rounds) depends on the rules of the specific state or counseling authority.


Generally:

You must have cleared all 12th subjects (no backlogs) to participate in the main NEET counseling.

If you still have a back in one subject and haven’t cleared it yet, you are usually not eligible for counseling.

In some states, you can participate in the mop-up round only if you have cleared your back and have a passing result before the counseling starts.

So, you can take part in all NEET counseling rounds only after your back paper is cleared and you have passed 12th.

I can also check the specific rules for your state if you want!

With a NEET score of 490 in Gujarat state quota:

General category: Usually needs 600+ score for govt colleges, so 490 may be low for top govt seats.

Reserved categories (SC/ST/OBC): You may have a chance at some government colleges.

Private colleges: You can explore private medical colleges with this score.

Check the latest Gujarat NEET counseling cutoff and participate in counseling for exact options.

View All

Column I ( Salivary gland)

 

Column II ( Their location)

Parotids

I

Below tongue

Sub-maxillary / sub-mandibular

Ii

Lower jaw

Sub-linguals

Iii

Cheek

Option: 1

a(i), b(ii) , c(iii)

 


Option: 2

a(ii), b(i), c(iii)

 


Option: 3

a(i), b(iii), c(ii)


Option: 4

a(iii), b(ii), c(i)


Ethyl \; ester \xrightarrow[(excess)]{CH_{3}MgBr} P

the product 'P' will be ,

Option: 1


Option: 2


Option: 3

\left ( C_{2}H_{5} \right )_{3} - C- OH


Option: 4


 

    

           

 Valve name                            

             

Function

    I   Aortic valve     A

Prevents blood from going backward from the pulmonary artery to the right ventricle.

    II   Mitral valve     B

 Prevent blood from flowing backward from the right ventricle to the right atrium.

    III   Pulmonic valve     C

 Prevents backward flow from the aorta into the left ventricle.

    IV   Tricuspid valve     D

 Prevent backward flow from the left ventricle to the left atrium.

 

Option: 1

I – A , II – B, III – C, IV – D


Option: 2

 I – B , II – C , III – A , IV – D


Option: 3

 I – C , II – D , III – A , IV – B


Option: 4

 I – D , II – A , III – B , IV – C 

 

 


Column A Column B
A

a) Organisation of cellular contents and further cell growth.  

B

b) Leads to formation of two daughter cells.

C

c) Cell grows physically and increase volume proteins,organells.

D

d)  synthesis and replication of DNA.

Match the correct option as per the process shown in the diagram. 

 

 

 

Option: 1

1-b,2-a,3-d,4-c
 


Option: 2

1-c,2-b,3-a,4-d


Option: 3

1-a,2-d,3-c,4-b

 


Option: 4

1-c,2-d,3-a,4-b


0.014 Kg of N2 gas at 27 0C is kept in a closed vessel. How much heat is required to double the rms speed of the N2 molecules?

Option: 1

3000 cal


Option: 2

2250 cal


Option: 3

2500 cal


Option: 4

3500 cal


0.16 g of dibasic acid required 25 ml of decinormal NaOH solution for complete neutralisation. The modecular weight of the acid will be

Option: 1

32


Option: 2

64


Option: 3

128


Option: 4

256


0.5 F of electricity is passed through 500 mL of copper sulphate solution. The amount of copper (in g) which can be deposited will be:

Option: 1

31.75


Option: 2

15.8


Option: 3

47.4


Option: 4

63.5


0.5 g of an organic substance was kjeldahlised and the ammonia released was neutralised by 100 ml 0.1 M HCl. Percentage of nitrogen in the compound is

Option: 1

14


Option: 2

42


Option: 3

28


Option: 4

72


0xone is

Option: 1

\mathrm{KO}_{2}


Option: 2

\mathrm{Na}_{2} \mathrm{O}_{2}


Option: 3

\mathrm{Li}_{2} \mathrm{O}


Option: 4

\mathrm{CaO}


(1) A substance  known as "Smack"

(2) Diacetylmorphine

(3) Possessing a white color

(4) Devoid of any odor

(5) Crystal compound with a bitter taste

(6) Obtained by extracting from the latex of the poppy plant

The above statements/information are correct for:

Option: 1

Morphine


Option: 2

Heroin


Option: 3

Cocaine


Option: 4

Barbiturates


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