NEET exam includes Physics, Chemistry, and Biology sections. With the highest weightage in the Biology section, Physics and Chemistry contain equal weightage.

Although practising to crack the NEET exam with full concentration and dedication is necessary, some smart tools and techniques also play a miraculous role. Other than these, if you are aware of the important concepts and questions from each chapter, you can predict the difficulty level of the exam and how to handle their solutions.

This article will help you learn some important Physics questions from all the chapters for the NEET exam. To help you understand the NEET exam’s Physics exam pattern, you can always Find Aakash Coaching Institute Near You and get guidance from our experts.

**Chapter-wise Important Questions of Physics for NEET**

**Chapter 1: Physical World and Measurement**

1. What are the dimensions of Planck’s constant?

E = h, i.e., h = E

Hence, [h] = [E][] = ML²T⁻²T⁻¹ = ML²T⁻¹

Therefore, the dimensions of Planck’s constant are ML²T⁻¹.

2. Why is velocity not considered a fundamental quantity?

It is because length and time can also calculate velocity. And both these are not fundamental quantities.

**Chapter 2: Kinematics**

1. What happens if the displacement of the particle varies with time as x =t+7?

x = t+7

x = t²+14t+49

V = dxdt = 2t + 14

a = dVdt = 2 m/s²

Therefore, the particles move with a constant acceleration.

2. A ball thrown by a player reaches the other in 2 seconds. Calculate the maximum height attained by the ball above the point of projection.

T = 2u sing = 2 seconds

u sin = 10

Now, H = u² sin²2g = 10 x 102 x 10 = 5

Therefore, the height will be 5 m.

**Chapter 3: Laws of Motion**

1. If the force of a rocket moving with a velocity of 300 m/s is 270 N, then calculate its fuel combustion rate.

Force applied by rocket, F = v dmdt

Therefore, Rate of combustion is dmdt = Fv = 270300 = 0.9 kg/s

2. Suppose a lady of weight 60 kg is standing in an elevator. Suddenly the wires of the lift break, and it begins to fall freely towards the ground. Calculate the force exerted by the floor of the lift on the lady.

The apparent weight of the freely falling object is always considered zero. Therefore, the force exerted by the elevator floor on the lady will be zero.

**Chapter 4: Work, Energy, and Power**

1. A body of mass 10 kg is rotated in a horizontal circle of radius 5 m. It is moving with a velocity of 10 m/s. What will be the work done in 50 such revolutions?

Here the value of potential energy (P.E.) and kinetic energy (K.E.) is constant. Therefore, the total work done by the body will be zero.

2. What happens to the kinetic energy and momentum of the body if it is moving with some friction on a surface?

Friction is a non-conservation external force on the body. It decreases the kinetic energy and momentum of the body.

**Chapter 5: Motion of System of Particles and Rigid Body**

1. A disc of mass 1 kg and radius of 15 cm is rotating about its fixed axis with an angular velocity of 4 radian/s. What will be its linear momentum?

Its momentum will be zero. It is because the disc is rotating around its fixed axis. As a result, the velocity of the centre of mass is zero.

2. A mass is rotated in a plane about a fixed point. In which axis will be the direction of its angular momentum?

The body’s angular momentum will direct along the axis of rotation.

**Chapter 6: Gravitation**

1. The body’s weight increases when taken from the equator to the poles. Why?

On moving the body from the equator to the poles, its acceleration increases due to gravity.

2. What is the value of ‘g’ on poles?

g = GMₑRₑ²

On the poles, the value of Rₑ becomes minimum. Therefore, the value of ‘g’ gets maximum on poles.

**Also Read:**

NEET Physics Preparation: How to prepare for NEET Physics Exam?

How to Prepare for NEET Physics?

Where to get NCERT Physics notes for NEET preparation?

**Chapter 7: Properties of Bulk Matter**

1. Why does the viscosity of a liquid decrease with the increase in temperature?

As the temperature rise, the liquid starts to flow more freely. As a result, the viscosity of the liquid decreases.

2. A manometer connected to a closed tap reads 3.5 x 10⁵ N/m². On opening the tap, the manometer reading becomes 3.0 x 10⁵ N/m². Calculate the velocity of the flow of water.

According to Bernoulli’s Theorem,

P + 12 v² = constant

With the flow of liquid, P (pressure energy) decreases. Hence,

12 v² = P₀ – P₁

12 v² = 3.5 x 10⁵ -3.0 x 10⁵ 10³

v² = 2 x 0.5 x 10⁵10³

v² = 100

v = 10 m/s

Hence, the velocity of flow of water is 10 m/s.

**Chapter 8: Thermodynamics**

1. The enthalpy of atomisation of carbon and hydrogen q₁ and q₂, respectively. If q₃ is the bond enthalpy of the C-H bond, then calculate the H for the given reaction:

C(s) + 2H₂ (g) → CH₄ (g)

Here,

C(s) → C(g); H₁ = +q₁

H₂(g) → 2H(g); H₂ = +2q₂

C(g) + 4H(g) → CH4(g); H₃ = -4q₃

H = q₁ + 4q₂ – 4q₃

2. Which thermodynamic quantity results from the second law of thermodynamics?

The term entropy is the direct outcome of the second law of thermodynamics. It is the result obtained through the Carnot cycle approach.

**Chapter 9: Behaviour of Perfect Gas and Kinetic Theory**

1. Name the motion responsible for the internal energy of monatomic and diatomic gases.

The internal energy of monatomic gases is due to linear motion only, while the internal energy of diatomic gases is due to rolling motion.

2. What are the average velocities of particles in a gas?

It is zero because the molecules of a gas do not favour any particular direction.

**Chapter 10: Oscillations and Waves**

1. A particle moves such that its acceleration is given by a = – (x -2). Calculate its time of oscillation.

a = – (x -2)

Also, a = –² (x – x₀)

Therefore, ² =

T = 2 1

Hence, the time period of oscillation is 2 1 .

2. The equation y = A sin(kx – t) describes a traveling wave in a stretched string. Calculate its maximum particle velocity.

V = dydt

V = – A cos(kx – t)

Vmax = A

Therefore, the maximum particle velocity is A.

**Chapter 11: Electrostatics**

1. If a point charge q is placed at the cube’s centre, what is the flux linked with the cube?

From Gauss’s law,

Flux linked with the closed body is (1₀) times the charge enclosed.

The cube encloses the charge q, then the flux linked with the cube will be, =1₀q .

2. Two points, P and Q, are maintained at the potential of 10V and -5V, respectively. What will be the value of work done in moving 100 electrons from P to Q?

W = q x V = q x (Vf– Vi)

W = – 100 x 1.6 x 10⁻¹⁹ x (-5-10)

W = -1.6 x 10⁻¹⁷ x (-15)

W = 24 x 10⁻¹⁷ = 2.4 x 10⁻¹⁶ J

Therefore, work done in moving 100 electrons from P to Q is 2.4 x 10⁻¹⁶ J.

**Chapter 12: Current Electricity**

1. In a metallic conductor, what is the direction of electrons’ movement in the presence of an electric field?

Electrons will move randomly but slowly. They slowly drifted in the direction opposite to the electric field.

2. The charge flown through a resistance R in time t varies according to Q = at – bt². Calculate the total heat produced in R by the time current becomes zero.

Q = at – bt²

i = dQdt = a – 2bt

Now, i = 0

a – 2bt = 0

t = a2b

Total heat produced = 0a2bi²Rdt = 0a2b(a – 2bt)²Rdt = a²R6b .

**Chapter 13: Magnetic Effects of Current and Magnetism**

1. A charged particle enters a uniform magnetic field with a velocity vector making an angle of 30° with the magnetic field. A particle describes a helical trajectory of pitch x. Calculate the radius of the helix.

Pitch, x = 2mv cos30° qB = 2mv qB x 32

Radius, r = mv cos30° qB = mv qB x 12

Therefore, r = x23 is the radius of helix.

2. The magnetic susceptibility of a paramagnetic substance at 200℃ is 0.0060. Then calculate its value at 100℃.

m ∝ 1T

₁₂ = T₂T₁ = 100200

₁₂ = 12

₂ = 2₁ = 2 x 0.0060

₂ = 0.0120

Therefore, the value of the magnetic susceptibility of the paramagnetic substance at 100℃ is 0.0120.

**Chapter 14: Electromagnetic Induction and Alternating Currents**

1. If the particle’s kinetic energy increases by 16 times, what will be the percentage change in the de Broglie wavelength of the particle?

= h2mK

₁ = h2m x 16K = 4

Percentage change in de Broglie wavelength = – ₁ x 100

– ₁ x 100 = (1 – ₁ ) x 100

– ₁ x 100 = (1 – 14 ) x 100 = 75%

Therefore, the percentage change in the de Broglie wavelength of the particle is 75%.

2. A current I = 10sin(100 ) amp. is passed in the first coil, which induces a maximum emf of 5 volts in a second coil. Calculate the mutual inductance between the coils.

Let I = I₀ sin ,

Where, I₀ = 10 and = 100

Now, = M dIdt

= M ddt I₀ cos

Then, max = MI₀

5 = M x 10 x 100

M = 5 mH

Therefore, the mutual inductance between the coils is 5mH.

**Watch our visual story on NEET Physics Syllabus & Important Topics**

**Chapter 15: Electromagnetic Waves**

1. An electromagnetic wave propagates along the Y-axis. What will be the direction of its oscillating electric field and oscillating magnetic field?

An electromagnetic wave propagates along the Y-axis, then the oscillating electric field will be along the Z-axis, and an oscillating magnetic field will be along the X-axis.

2. What is the ratio between the energy density of electromagnetic waves and energy density due to the magnetic field?

The energy density of electromagnetic wave, EMW = ₒE²

The energy density due to magnetic field, B = B²2₀ = ₒE²2

The ratio between them = EMW B

EMW B = ₒE²ₒE²2 = 2:1

Therefore, the ratio between these two is 2:1.

**Chapter 16: Optics**

1. If the tube length of an astronomical telescope is 105 cm and magnifying power is 20 for a normal setting, calculate the focal length of the object.

L = fₒ + fₑ = 105

M = fₒfₑ = 20

From there, fₒ = 100 cm and fₑ = 5 cm.

Therefore, the focal length of the object is 100 cm.

2. Radio intensities between point A and central fringe are 0.853. What will be the path difference between the two waves at point A?

R² = a² + b² + 2ab cos

IRImax = 0.853

IR = 0.853 x Imax = 0.853 x 4I

IR = I + Iₒ + 2I cos = 2I (1 + cos) = 0.853 x 4I

= 4 = 8

Therefore, the path difference linking the two waves at point A is 8 .

**Chapter 17: Dual Nature of Matter and Radiation**

1. If 1.5 mW of 400 nm light is directed at a photoelectric cell. If o.1% of the incident photons produce photoelectrons, then calculate the current in the cell.

= Phc

ₑ = x %

ₑ = Phc x 100

ₑ = 1.5 x 10⁻³ x 400 x 10⁻⁹6.6 x 10⁻³⁴ x 3 x 10⁸ x 0.1100

Now, I = ₑ x e = 0.48 A

Therefore, the current in the cell is 0.48 A.

2. Give the formula to calculate the mass of a photon.

E = mc²

m = Ec² = hc²

Hence, the mass of a photon is calculated as m = hc² .

**Chapter 18: Atoms and Nuclei**

1. Consider aiming a beam of free electrons toward free protons. An electron and a proton can not combine to produce an H-atom when they disband. Why?

It is because of the energy conservation principle and without simultaneously releasing energy in the form of radiation.

2. When a nucleus in an atom undergoes radioactive decay, what changes occur in its electronic energy levels?

and electronic energy levels of an atom will decay while the energy level will remain the same.

**Chapter 19: Electronic Devices**

1. How many NAND gates are used to form an AND gate?

Two NAND gates are used to form an AND gate.

2. What happens when an NPN transistor is used as an amplifier?

The majority of charge carrier electrons of N-type emitters move from emitter to base and then base to collector. It happens when an NPN transistor is used as an amplifier.