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

Edited By Irshad Anwar | Updated on Nov 11, 2024 05:47 PM IST | #NEET

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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

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

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

Ready to take the National Eligibility cum Entrance Test (NEET) to get into one of the best medical colleges in India? If so, you 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.

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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

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,
$\begin{array}{l}{\mathrm{v}_{\mathrm{R}}=\mathrm{i} \mathrm{R} \rightarrow \text { Voltage is in phase with i }} \\ \\ {\mathrm{v}_{\mathrm{L}=\mathrm{i} \omega \mathrm{L}} \rightarrow \text { voltage is leading i by } 90^{\circ}} \\ \\ {\mathrm{v}_{\mathrm{c}}=\mathrm{i} / \mathrm{\omega} \mathrm{c} \rightarrow \text { voltage is lagging behind i by } 90^{\circ}}\end{array}\varepsilon$

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 -

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

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

1714399579263

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

General equation of SHM

Most Scoring concepts for NEET

This ebook serves as a valuable study guide for NEET exams, specifically designed to assist students in light of recent changes and the removal of certain topics from the NEET exam.

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For Displacement:-

$x=ASin(wt+\phi);\ where\ is\ initial\ phase\ and\ (\omega t+\phi)\ is\ called\ as\ phase.$

Various displacement equations:-

$(1)\ x=ASin\omega t \Rightarrow \ when\ particle\ starts\ from\ mean\ position\ towards\ right.$

$(2)\ x=-ASin\omega t \Rightarrow \ when\ particle\ starts\ from\ mean\ position\ towards\ left.$

$(3)\ x=ACos\omega t \Rightarrow \ when\ particle\ starts\ from \ extreme\ position\ towards$

$(4)\ x=-ACos\omega t \Rightarrow \ when\ particle\ starts\ from\ left\ extreme\ position\ towards\ Right.$

For Velocity (v):-

$x=A Sin(\omega t+\phi )\\ \Rightarrow v=\frac{dx}{dt}=A\omega\ Cos(\omega t+\phi )=A\omega\ Sin(\omega t+\phi +\frac{\pi}{2})$

For Acceleration:-

$x=A Sin(\omega t+\phi )\\ \Rightarrow v=\frac{dx}{dt}=A\omega\ Cos(\omega t+\phi )=A\omega\ Sin(\omega t+\phi +\frac{\pi}{2})$

$\Rightarrow a=\frac{dv}{dt}=-A\omega^2\ Sin(\omega t+\phi )=A\omega^2\ Sin(\omega t+\phi +\pi)=-\omega^2x$

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:-

$\frac{dv}{dt}=-\omega^2x$

$\Rightarrow \frac{d}{dt} \left( \frac{dx}{dt}\right )=-\omega^2x$

$\Rightarrow \frac{d^2x}{dt^2}+\omega^2x=0$

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

Also read:

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.

1714399588951

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

$\dpi{100} E_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r^{2}}=\frac{\sigma R^{2}}{\epsilon _{0}r^{2}}$

$V(r)=-\int_{r=\infty}^{r=r} \vec{E}.d \vec{r} =\frac{1}{4 \pi \varepsilon_{0}} \frac{\vec{Q}}{r}$

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

$E_{in}=0$

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

$\boldsymbol{V}(\boldsymbol{r})=-\int_{r=\infty}^{r=r} \overrightarrow{\boldsymbol{E}} . d \vec{r} =-\int_{\infty}^{R} \boldsymbol{E}_{r}(d \boldsymbol{r})-\int_{R}^{r} \boldsymbol{E}_{r}(\boldsymbol{d} \boldsymbol{r})=\frac{1}{4 \pi \boldsymbol{\varepsilon}_{0}} *\frac{\boldsymbol{q}}{ {\boldsymbol{R}}}+0 \\ \\ \Rightarrow V(r)=\frac{1}{4 \pi \boldsymbol{\varepsilon}_{0}} *\frac{\boldsymbol{q}}{ {\boldsymbol{R}}}$

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

$E_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R^{2}}=\frac{\sigma }{\epsilon _{0}}$

$V_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}=\frac{\sigma R}{\epsilon _{0}}$

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)

$E_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r^{2}}$ $V_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r}$

$E_{out}=\frac{\rho R^{3}}{3 \epsilon _{0}r^{2}}$ $V_{out}=\frac{\rho R^{3}}{3 \epsilon _{0}r}$

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

$E_{in}=\frac{1}{4\pi \epsilon _{0}}\frac{Qr}{R^{3}}$ $\dpi{100} V_{in}=\frac{Q}{4\pi \epsilon _{0}}*\frac{3R^{2}-r^{2}}{2R^{3}}$

$\dpi{100} E_{in}=\frac{\rho r}{3 \epsilon _{0}}$ $V_{in}=\frac{\rho \left ( 3R^{2}-r^{2} \right )}{6 \epsilon _{0}}$

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

$E_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R^{2}}$ $V_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}$

$\dpi{100} E_{s}=\frac{\rho R}{3 \epsilon _{0}}$ $V_{s}=\frac{\rho R^{2}}{3 \epsilon _{0}}$

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

$V_{centre}=\frac{3}{2}\times \frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}=\frac{3}{2}V_{s}$

i.e $\dpi{100} V_{c}> V_{s}$

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

1714399589301

4. Resistance and Resistivity

Resistance

For a conductor of resistivity $\rho$ having a length of a conductor= l

and Area of a crosssection of conductor= 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 $\rho=\frac{m}{ne^2\tau}$

for different conductors, n is different

And $\rho$ depends on the n

So R is also different.

Resistivity or Specific Resistance $(\rho)$

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

If l = 1 m and A= 1 m2

Then 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

1714399590617

If two resistances are in Parallel:

Current through any resistance:

1714399590271

$i'=i(\frac{Resistance\, of \, opposite \, Branch}{total\, Resistance})$

The required current of the first branch

The required current of the second branch

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

In closed-loop

1714399583779

Also read:

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.

1714399589057

Now from the Lorentz equation -

$\begin{array}{l}{\text { }} \\ {\qquad \begin{aligned} \vec{F}=& q(\vec{E}+\vec{v} \times \vec{B}) \\ \\ E_{z}=E z_{0} \sin (\omega t-k y) & \\ \\ B_{x}=B x_{0} \sin (\omega t-k y), & \text { where } \frac{\omega}{k}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} \end{aligned}}\end{array}$

since, $\omega=2 \pi f$ , where f is the frequency and $k= \frac{2 \pi }{\lambda}$ , where $\lambda$ is the wavelength.

$\text { Therefore, } \frac{\omega}{k}=\frac{2 \pi f}{2 \pi / \lambda}=f \lambda$

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 ε and magnetic permeability µ, 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 $90^{\circ}$ 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 :

$\mu _2 \sin C = \mu _1\sin r$

$\implies \mu _2 \sin C = \mu _1$ since, $\sin r = 1$ .

$\implies \sin C = \frac{\mu _1}{\mu _2}=\frac{\text{R.I of rarer medium }}{\text{R.I of denser medium }}$

or $\boxed{\mu=\frac{1}{\sin C}}$ when $\mu_1$ = 1 for air and $\mu_2=\mu$ .

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 .

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

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,

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

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

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

For Dark fringes -

By the principle of interference, the condition for destructive interference is the path difference = $\frac{(2n-1) \lambda}{2}$

Here, n = 1,2,3 … indicates the order of the dark fringes.

So,

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

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 -

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

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

Angular fringe width -

$\begin{array}{l} \\ \\ {\theta=\frac{\beta}{D}=\frac{\lambda D / d}{D}=\frac{\lambda}{d}}\end{array}$

10. De-Broglie Wavelength Of An Electron

De - Broglie wavelength of Electron-

De Broglie’s Equation is given as $\lambda = \frac{h}{p}= \frac{h}{mv}= \frac{h}{\sqrt{2mK}}$

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. $K=W_E\Rightarrow \frac{1}{2}m_ev^2=eV$

So, the De-Broglie wavelength of Electron is given as $\lambda_e = \frac{h}{m_ev}= \frac{h}{\sqrt{2m_eK}}=\frac{h}{\sqrt{2m_e(eV)}}$

using $h=6.626\times 10^{-34} \ Js$ and $m_e=9.1\times 10^{-31} \ kg$ and $e=1.6\times 10^{-19} \ C$

we get $\lambda _{e }= \frac{12.27}{\sqrt{V}}A^{\circ}$ ( i.e answer will be in )

Similarly, we can find De - Broglie wavelength associated with charged particle

De - Broglie wavelength with charged particle-

$\lambda = \frac{h}{\sqrt{2mK}}= \frac{h}{\sqrt{2mqV}}$

Where $K\rightarrow kinetic\: energy\: o\! f particle$

$q\rightarrow charged \: particle$

$V\rightarrow potential \: diffenence$

De - Broglie wavelength of the proton

using $m_p=1.67\times 10^{-27} \ kg$ and $q_p=e=1.6\times 10^{-19} \ C$

we get $\lambda _{proton}= \frac{0.286}{\sqrt{V}}A^{\circ}$

De - Broglie wavelength of Deuteron

using $m_D=2\times 1.67\times 10^{-27} \ kg$ and $q_D=e=1.6\times 10^{-19} \ C$

we get $\lambda _{deutron}= \frac{0.202}{\sqrt{V}}A^{\circ}$

De - Broglie wavelength of an Alpha particle (He2+)

using $m_ {\alpha ^{2+}}=4\times 1.67\times 10^{-27} \ kg$ and $q_ {\alpha ^{2+}}=2e=2\times 1.6\times 10^{-19} \ C$

we get $\lambda _{\alpha -partical}= \frac{0.101}{\sqrt{V}}A^{\circ}$

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.

1714399587890

A is input

Y is output

$Y=\bar{A}$

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

$Y= A\cdot B$

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 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

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

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

A and B are inputs

Y is output

NOT + AND gate

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

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

$Y= \overline{A+ B}$

A and B are inputs

Y is output

NOT + OR Gate

The truth table for a NOR gate is as follows:

1714399588691

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) $\overline{\bar A+ \bar B }= {A}\cdot {B}$

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

Some Important relation -

$A\cdot A=A$

$A\cdot 1=1$

$A\cdot 0=0$

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Questions related to NEET

Have a question related to NEET ?

sir my rank neet pg is 69320 i can get any clinical post in neet pg

Yes, you can get a seat depending on the cut-off mark but in non clinical post at this rank.if you are satisfied with this go ahead or you can do is try next year and achieve good rank for clinical.

Keep trying till the last round and you'll definitely make it. Make sure you enter the right colleges during option entry with the priority according to their cut-off rank and availability of seats.

Keep trying! Don't give up.

Best of luck :)

Read Complete Answer

sadafreen356 12 November,2024

It is possible to crack neet in 5 month with zero level preparation if yes please suggest me schedule or time table and strategy to get least 600 marks in 2025.. reply who really achieve this score..

While scoring 600+ in NEET isn't an easy achievement, especially from a backdrop of zero, its obtainability will depend on how hard one can work and plan through a strategy.

Here's how you do it:

1. Setting Realistic Goals:

Short-term Goals: Have a specific number in mind and deconstruct the target goal into smaller chunks, for example, complete one chapter or topic each day.

Long-term Goals: Target to cover entire syllabus with a specific date.

2. Study Plan Making:

Time Distribution: Give specific time slots to each subject with your strength and weakness.

Prioritization of Topics: Identify high-weightage topics and focus on those.

Regular Revision: Plan regular revision sessions to enforce your understanding.

3. Effective Techniques for Studying: Active involvement of the subject matter with the help of some techniques like summarizing, mind mapping, or even teaching it to others.

Practice Regularly: Practice several MCQs from previous year's papers and mock tests.

Join a Test Series: Join a test series so that you can track your progress and may know where exactly to improve.

4. Get Help:

Seek the help of perfect teachers or mentors who will give valuable insights and strategies.

Join online forums and communities: Interact with other NEET aspirants and share knowledge and experiences.

5. Take Care of Yourself:

Healthy Lifestyle: Get ample sleep, nutrition, and exercise to keep yourself fit both physically and mentally.

Stress Management: Try to relax by practicing meditation or yoga to avoid stress and anxiety.

Sample Daily Routine:

Presession (6-8 AM): Revision of what was learnt the previous day

Session (9 AM - 1 PM): Grasp a new concept or practice questions.

Afternoon Session Time (1-2 PM): Take rest, hit the gym, or a hobby.

Afternoon Session (2-6 PM): Revision or practice tests.

Evening Session (7-10 PM): Important concepts revision and mock tests

Actually, you should be devoted to your dreams. Focus hard, work hard, and believe in yourself.

More Tips:

Positive Thinking: This indeed is a major factor for performance.

Time Management: This would be very crucial to study the whole syllabus.

Healthy Lifestyle: Sleep, Nutrition, and Exercise.

Just don't waste time and search for some teachers, mentors or tutors without delay in case you are facing some issues.

Therefore, by adhering to these guidelines and sticking consistently to your studies, you would surely get a good score in NEET 2025.

Read Complete Answer

Samprikta Mondal 12 November,2024

When is the registration starting for NEET 2025

Hello aspirant,

NEET registration for the 2025 exam is open from December to January of that year. The official bulletin issued by NTA will announce the NEET 2025 registration date.

Additionally, if needed, candidates will have the chance to edit or change the information on their NEET application forms. NEET is the largest and only medical entrance exam in India, with an estimated 20 lakh students taking it, underscoring its importance in the field of medical education.

For complete information, you can visit our site through following link:

https://medicine.careers360.com/articles/neet-2025

Thank you

Hope it helps you

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Sajal Trivedi 11 November,2024

ive lost my neet scorecard 2019 and i need it for fmge ive mailed nta but theres no response what do i do ?

Here are the steps to retrieve your lost NEET 2019 scorecard:

Check your Email: Look through your email inbox. Don't forget the spam and promotions folder. There must be an email from NTA mentioning regarding NEET 2019 result or scorecard.

Visit the NTA Website:

Go the official website of NTA.

Look for the section called "NEET".

Look if link is there for download scorecards of previous years. You may be required to provide your application number, birth date and security pin.

Contact NTA Helpdesk:

Send your complaint to the NTA helpdesk through their website or helpline number. Brief your problem clearly and ask for their aid to obtain your scorecard. Don't forget to remember to be able to offer all your NEET application number, your roll number and other details to them.

If you still cannot retrieve your scorecard by above options, then, you may make use of these alternatives:

Apply through RTI Application: You can apply for an RTI, i.e., a Right to Information application to NTA to get a copy of your scorecard. This, in any case, would be a time-consuming process, but somehow, could be useful.

Contact your coaching institute or college: They would probably have a copy of your scorecard or could help you get it from the NTA.

Don't lose hope and keep trying an attempt to get back your scorecard. Keep trying different ways and don't shy away from taking legal help from the concerned authorities.

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Samprikta Mondal 11 November,2024

can i get admission to agmc after cracking neet with 450marks and if i get admission what will be the fees for it

Hello,

A score of 450 in the NEET might enable you to get admission to AGMC according to your category, state quota, and annual cutoffs. Generally, 450 marks could probably be competitive under state quotas; however, it might not necessarily guarantee a seat under the All India Quota.

Using the Careers360 NEET College Predictor is one way of ascertaining the chances. The application will consider NEET scores, category, home state, and previous years' trends to give an estimate of which colleges a student might qualify for. It is simply a case of filling in details on the Careers360 website, and it churns out a list of potential colleges - in this case, AGMC.

Regarding the fees, it is pretty low and being a government medical college, the tuition fee is about 30,000-50,000 for a year. Add to that all other expenses towards hostels, exams etc may hike it to around 60,000-70,000 per year. For more latest updates you may refer to the website of AGMC or call up their office for admission.

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Deepti Singh 10 November,2024

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Popular NEET Questions

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

Option: 2

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

Option: 2

Option: 3

Option: 4

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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