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

Edited By Irshad Anwar | Updated on Jul 22, 2024 04:01 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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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.

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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

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 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

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

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 ?

I have got 256 marks in neet can I get a private college at this score in Uttar Pradesh??

Hello student

With a NEET score of 256, you may be eligible for admission to private medical colleges in Uttar Pradesh . Here are some points to consider:

- *Private Medical Colleges*: All private medical colleges in Uttar Pradesh have one type of cutoff .

- *100% All India Management Quota Cutoff*: The Directorate of Medical Education and Training (DMET), Lucknow, releases the Uttar Pradesh MBBS Management Quota Cutoff college-wise and category-wise .

- *No NRI seats*: There are no NRI seats available in Uttar Pradesh .

- *Cutoff Ranks and Scores*: The cutoff ranks and scores vary for different categories and colleges .

- *Eligibility*: You may be eligible for admission to private medical colleges with a NEET score of 256, but the eligibility criteria may vary for each college .

Some examples of private medical colleges in Uttar Pradesh and their cutoff scores are :

- *Autonomous State Medical College, Ghazipur*: 612 (OPEN category)

- *MAHARANI LAXMI BAI MEDICAL COLL, JHANSI*: 610 (OBC category)

- *Autonomous State Medical College Society, Mirzapur*: 610 (EWS category)

- *Govt Medical College Basti*: 485 (SC category)

- *Autonomous State Medical College, Pratapgarh*: 449 (ST category)

Please note that these cutoff scores are subject to change and may vary for the current academic year. It's always best to check with the colleges directly for the most up-to-date information.

Read Complete Answer

Vaibhavi 26 July,2024

can I get a GOVERNMENT BDS seat , my rank in neet 2024 is 71659 with a score of 605 marks anywhere in india (GEN category with no reservation )

Hello student

Congratulations on your NEET 2024 score! With a rank of 71659 and a score of 605 marks in the GEN category with no reservation, your chances of getting a government BDS seat are possible, but it depends on various factors:

1. *Cutoff ranks*: Last year's cutoff ranks for government BDS seats varied across India, ranging from 30,000 to 1,50,000.

2. *Seat matrix*: The number of BDS seats available in government colleges and their distribution across states.

3. *State-wise counseling*: Counseling processes and seat allotment vary across states.

4. *Category-wise cutoffs*: GEN category cutoffs might be higher than reserved categories.

Based on previous years' trends, here are some possible scenarios:

- *Kerala*: You might get a government BDS seat in Kerala, as their cutoff ranks have been around 70,000-80,000 in previous years.

- *Tamil Nadu*: You have a chance in Tamil Nadu, where cutoff ranks have been around 60,000-70,000.

- *Other states*: In states like Maharashtra, Gujarat, or Rajasthan, the competition is higher, and cutoff ranks might be lower (around 40,000-50,000).

To increase your chances:

1. *Participate in state-wise counseling*: Register for counseling in multiple states to explore opportunities.

2. *Check seat matrices*: Look for states with a higher number of BDS seats in government colleges.

3. *Stay updated*: Monitor counseling schedules, cutoffs, and seat allotment lists regularly.

Remember, these are general insights based on previous years' trends. The actual cutoffs and seat allotment may vary. Best of luck!

Read Complete Answer

Vaibhavi 26 July,2024

with 425 can I get private seat in medical collage

Hello aspirant,
The competition for medical seats, especially in private colleges, is extremely intense. Cutoff marks have been rising steadily over the years. To get a seat in a private medical college, you would generally need a score significantly higher than 425.
But, you should not lose hope here, instead, you can consider other healthcare-related fields like BDS, BAMS, BHMS, or paramedical courses. If you're determined to pursue MBBS, focus on improving your score for the next NEET exam.
Remember, it's essential to stay positive and explore alternative paths.

I hope this information helps you.

Read Complete Answer

kanishkakaushikii2004 26 July,2024

I got 478 marks in neet 2024 can i get bams in csmss

Hello aspirant,
While the exact cutoff for BAMS in CSMSS can fluctuate yearly based on various factors, a score of 478 is generally considered good for getting admission to a government Ayurveda college like CSMSS.
Factors Affecting Admission:

Category: If you belong to a reserved category (SC, ST, OBC), your chances of getting a seat increase significantly.
State Quota: Competition is generally lower for state quota seats compared to All India Quota seats.
Counseling Process: Your final rank after the counseling process will determine your admission.

To increase your chances, you can participate in counselling or you can also explore other options too.

I hope this information helps.

Read Complete Answer

kanishkakaushikii2004 26 July,2024

Can ijoin Bsc biotechnology with Neet card

Hello aspirant
For pursuing bsc in biotechnology You don't need neet score but if you have appeared for neet then you can submit , it will make impact.
Eligibility for pursuing bsc in biotechnology You must have passed our class xii from science stream with minimum 50% marks aggregate and must have physics, chemistry and biology or maths as core subjects with 60% marks in each core subject
. For sc/ st it is 45 %
Both pcb and pcm candidates can apply for bsc biotechnology . Universities conduct their own followed by personal interview.
Duration of the Course is 3 years .
Top universities offering bsc biotechnology
Christ university
Chandigarh University
Jamia milia Islamia
St. Xavier's college, Ahmedabad
SRM university chennai

Average fees ranges from 11,000 to 85,000 depending upon college.

Read Complete Answer

Manisha 26 July,2024

View All

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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