Important Physics Formulas For NEET 2024 Exam- Topic-wise Formulas

Edited By Irshad Anwar | Updated on May 05, 2024 12:26 PM IST | #NEET

Syllabus

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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Important Physics Formulas For NEET 2024 Exam- Topic-wise Formulas

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 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):-

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

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

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

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 ?

MY NEET SCORE PREDICTION 320 MY AIR PERCENTILE IS 85.41. WHAT IS THE CHANCE OF GETTING MBBS SEAT IN PRIVATE MEDICAL INSTITUTIONS IN TAMIL NADU

Hello aspirant

You haven't mentioned your category to which you belong .

Considering you as general candidate it's not possible to get a mbbs seat with 320 score , though you may try for other medical courses , also it depends upon neet cut off 2024 . Cut off changes every year. It is dependent on factors including students appeared , seat intake , difficulty level of exam, normalisation process and previous year's cut off.

Qualifying percentile For general category is 50 percentile ie score between 720 - 137

As you seek admission in Tamil Nadu

You may try in following colleges

ACS medical college and Hospital, Chennai

Alagappan university

For rank prediction you may visit the link given below

https://medicine.careers360.com/neet-rank-predictor

For college prediction you must follow the link given below

https://medicine.careers360.com/neet-college-predictor

Read Complete Answer

Manisha 13 May,2024

can i get mbbs govt. college seat at 580 marks in neet 2024( UP home state) including state quota

Hello aspirant

Considering you as a general category candidate you have chances of getting mbbs seat In UP but whether you will get through state quota government college , seems a bit difficult.

Looking at 2 - 3 previous year's cut off , after analysis it is observed that for state quota seat in UP you must score 611+ marks , this year you need 620+ marks for state counseling in UP .

You may get medical seat in following colleges

GS medical college

Saraswati medical college

heritage Institute , Bhadawar

For rank prediction you may visit the link given below

https://medicine.careers360.com/neet-rank-predictor

For college prediction you may visit the following link

https://medicine.careers360.com/neet-college-predictor

Read Complete Answer

Manisha 13 May,2024

I have scored 591 in NEET 2024. Please list the names of the colleges (Govt/Private) where I may get a chance for MBBS under the General caste category under West Bengal State Quota.

Hello aspirant

591 is quite a good score . In private colleges in west Bengal you have chances of getting a mbbs seat though it depends upon cut off neet 2024.

Hence Which college you will get can't bd predicted now.

Now looking at previous year's cut off trends and also your score , You may try in following colleges in west Bengal

Medical college , kolkata

ESI PGIMSR, Esic medical college

All India Institute of Medical Sciences, Bankura

Aiims kalyani

Fir rank prediction please refer the link given below

https://medicine.careers360.com/neet-rank-predictor

For college prediction you must follow the link given below

https://medicine.careers360.com/neet-college-predictor

Read Complete Answer

Manisha 13 May,2024

meri beti ko 686 mark aane ki sambhawna hai, all india medical college koun milengi?

Hello aspirant

Apki beti ke yadi neet 2024 me 686 marks aate hai to bahut achhi baat hai .

686 score bahyt badiya score hai, in marks par apki beti ko achhe college me mbbs seat mil sakti hai lekin ALL India Quota se seat milegi ya nahj ye neet cut off 2024 par depend karta hai .

Neet qualifying marks general category ke liye hai 50 percentile, score 720 - 137

SC /ST /OBC ke liye hai 40 percentile 137 - 107

Aur pwd candidate ke liye hai 45 percentile

Pichhle varshon ki cut off ke adhar par nimnlikhit college me admission milne ki sambhavna hai

Grant medical college mumbai

Government medical college, Andhra pradesh

Maharaja Agrasen medical college

Madras medical college , Madras

Aiims colleges

Rank aur college prediction ke liye neeche di gayi link par contact kare

https://medicine.careers360.com/neet-rank-predictor

College predictor

https://medicine.careers360.com/neet-college-predictor

Read Complete Answer

Manisha 10 May,2024

I got 320 marks in neet 2024 iam an oc female from telangana can get seat in bds

Hello Akhila

320 marks are a moderate score , you may try but it won't be possible to get bds with this score .

Even if you get , you may get medical seat for other courses such as bams or bhms or physiotherapy as it depends upon cut off . Cut off neet 2024 will decide which course and college you will get .

Cut off changes every year , also it depends upon your category to which you belong, your state , number of students appeared in exam, Toughness level of exam, previous year's cut off etc .

Now looking at previous years cut off trends it seems a bit difficult to get bds seat but you must try .

Following links will help you out in rank and college prediction .

https://medicine.careers360.com/neet-rank-predictor

https://medicine.careers360.com/neet-college-predictor

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Manisha 10 May,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

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Option: 2

Option: 3

128

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256

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31.75

Option: 2

15.8

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47.4

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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

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0xone is

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The above statements/information are correct for:

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Morphine

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Heroin

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Cocaine

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