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

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

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

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 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.

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

1. Series LCR Circuit-


Let 'i' be the amount of current in the circuit at any time and VL, VC and VR be the potential drop across L, C, and R, respectively. Then,
\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 -


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


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.

  1. 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})

  1. 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:-


\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.


  • 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 )


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


Electric Potential due to Uniformly Charged Non conducting Sphere-

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

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


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

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


4. Resistance and Resistivity


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=\rhoResistivity 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



If two resistances are in Parallel:


  • Current through any resistance:


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

The required current of the first branch 1714399583527

The required current of the second branch 1714399583610

6. Kirchoff's second law

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

In closed-loop



7. Nature of Electromagnetic Waves

From Maxwell’s equations, we can observe that electric and magnetic fields in an electromagnetic wave are perpendicular to each other and the direction of propagation. Also, from our discussion of the displacement current, in that capacitor, the electric field inside the plates is directed perpendicular to the plates. The figure given below shows a typical example of a plane electromagnetic wave propagating along the z direction (the fields are shown as a function of the z coordinate at a given time t). The electric field Ex is along the x-axis and varies sinusoidally with z at a given time. The magnetic field By is along the y-axis and again varies sinusoidally with z. The electric and magnetic fields Ex and By are perpendicular to each other and the direction z of propagation.


Now from the Lorentz equation -

\begin{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.


Conditions for TIR :

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

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

9. Young's Double Slit Experiment

Young's double-slit experiment

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


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

So, the \ path \ differnece \ is = \frac{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.


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 A^0=Angstrom)

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.


A is input

Y is output


The truth table for a NOT gate is as follows:


AND Gate-

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


A and B are inputs

Y is output

Y= A\cdot B

The truth table for the AND gate is as follows


OR Gate-

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


A and B are inputs

Y is output

Relation between input and output

Y= A+B

The truth table for an OR gate is as follows:


NAND Gate-

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


Y= \overline{A\cdot B}

A and B are inputs

Y is output

NOT + AND gate

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


NOR Gate-

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


Y= \overline{A+ B}

A and B are inputs

Y is output

NOT + OR Gate

The truth table for a NOR gate is as follows:


D'morgan's Theorem -

A and B are input.

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

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

3) \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

A+ 0=A


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Questions related to NEET

Have a question related to NEET ?

No, you cannot get direct admission to a program in Anesthesiology without qualifying in the NEET exam in India.

To pursue an MD in Anesthesiology, which is the postgraduate qualification required for practicing as an Anesthesiologist, you must qualify in the NEET PG (National Eligibility cum Entrance Test for Postgraduate) exam. NEET PG is the mandatory entrance exam for admission to MD/MS programs in medical colleges across India.

There is no provision for direct admission to MD Anesthesiology programs without appearing for NEET PG. All admissions happen through a centralized counseling process based on NEET PG scores.

I hope it helps!

Hello aspirant

You should not worry as your father's name is correct. It's OK if In your class xii certificate it is written Late before your father's name .

In neet ug form all the names ( your name, father's name, mother's name) should be correct with right spellings . The names should be same as in aadhar card. Date of birth should be correct. Your photo must match with your photo in aadhar card so that you can be recognized easily,  for that always use recent photo copy of yourself.

You will be called for counseling and there will not be any issue .

All the best.

Hello aspirant

With 480 marks you have hogh chances of getting a medical seat through state quota but through All India quota, we can't predict regarding getting which seat and which course being obc category person.

This is because your admission depends upon neet cut off 2024. Cut off changes every year. Cut iff depends upon various factors such as students appeared, seat intake, previous year's cut off and difficulty level of exam etc.

Neet qualifying marks for obc category are 40 percentile .

Now looking at previous year's trends for obc category , with 480 marks you may get a prestigious college , it may be for mbbs or bds .

For rank prediction you may visit the link given below

Fir college prediction you may go through the following link

If you have an SC certificate issued by the competent authority in Delhi  where you have resided since birth, then your SC quota will be valid for Delhi. This is because Delhi considers residence a crucial factor for claiming reservation benefits.

There might be some ambiguity in this case. Ideally, for claiming a reservation benefit in a state, your SC certificate should be issued by that specific state's authority.

I hope it helps!

Hello Sana

For pursuing BPT , you need to appear for Neet.

Neet is National eligibility cum entrance test. It is thf b9ggest entrsncd test of India.

Neet is mandatory for various medical courses including mbbs, bpt, bds, bhms, bams, Ayush and even veterinary science.

Eligibility for appearing Neet

The candidate must be an Indian

The candidate must have completed 17 years of age. There is no upper age limit.

The student must havd passed class xii with 50% marks in physics, chemistry and biology .

To get more information. You must visit the link given below

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Column I ( Salivary gland)


Column II ( Their location)



Below tongue

Sub-maxillary / sub-mandibular


Lower jaw




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                            



    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) Organisation of cellular contents and further cell growth.  


b) Leads to formation of two daughter cells.


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


d)  synthesis and replication of DNA.

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




Option: 1


Option: 2


Option: 3



Option: 4


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


Option: 4


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


Option: 2


Option: 3


Option: 4


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


Option: 2

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

Option: 3

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

Option: 4


(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


Option: 2


Option: 3


Option: 4


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2 Jobs Available
Drug Inspector

A career as a Drug Inspector is regarded as one of the most diverse in the field of healthcare and pharmacy. Candidates must undergo a screening process administered by the UPSC and or SPSCs in order to become drug inspectors. Those who manage it through the selection process will have a rewarding career with a high salary.

2 Jobs Available

A Biotechnologist is a professional who possesses strong knowledge and techniques that are utilised in creating and developing innovative products that improve the quality of human life standards. A biochemist uses biological organisms to create and improve goods and procedures for agriculture, medicine, and sustainability. He or she researches the genetic, chemical, and physical characteristics of cells, tissues, and organisms to determine how they can be used industrially.

2 Jobs Available
R&D Personnel

A career as R&D Personnel requires researching, planning, and implementing new programs and protocols into their organization and overseeing new products’ development. He or she uses his or her creative abilities to improve the existing products as per the requirements of the target market.

2 Jobs Available
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