About this NEET Physics practice test

Compared to JEE, NEET Physics is more direct: fewer multi-concept combinations, more straight formula-and-plug-in numericals — which means marks here come down to speed and accuracy, not cleverness. This set spans the full NTA syllabus from Mechanics and Thermodynamics through Electrostatics, Current Electricity, Magnetism, Optics and Modern Physics, with the same numerical style the real paper uses. Every answer shows the exact formula and the substitution, so a wrong answer tells you exactly which step to revisit. Run through a full set under a timer before exam day — NEET Physics rewards the student who never has to pause and think twice.

NEET Physics Practice Test sample questions

These starter questions help you launch a physics mock test quickly. Swap them with your own worksheet, notebook, or textbook questions any time.

1. 1. In an experiment, the mass of an object is measured with 2% error and its velocity with 3% error. What is the percentage error in the calculated kinetic energy?

2. 2. Planck's constant has the same dimensional formula as which other physical quantity?

3. 3. A projectile is fired with a speed of 20 m/s at 45° to the horizontal. Calculate its range on level ground. (g = 10 m/s²)

4. 4. A car moving at 20 m/s decelerates uniformly at 2 m/s². How far does it travel before coming to rest?

5. 5. A 5 kg block is pulled along a rough horizontal surface by a force of 20 N. If the coefficient of kinetic friction is 0.2, find the block's acceleration. (g = 10 m/s²)

6. 6. Two masses of 3 kg and 2 kg are connected by a light string over a frictionless pulley (Atwood machine). Calculate the tension in the string. (g = 10 m/s²)

7. 7. A 2 kg body is lifted vertically through 5 m and simultaneously given a final speed of 4 m/s. Calculate the total work done on the body. (g = 10 m/s²)

8. 8. A pump lifts 300 kg of water through a height of 10 m in 5 s. Calculate the power delivered by the pump. (g = 10 m/s²)

9. 9. A uniform disc of mass 2 kg and radius 0.5 m rotates about an axis through its centre, perpendicular to its plane. Calculate its moment of inertia.

10. 10. A torque of 10 N·m acts on a wheel of moment of inertia 2 kg·m², initially at rest. Find its angular velocity after 4 s.

11. 11. Calculate the escape velocity from the Earth's surface, given g = 9.8 m/s² and Earth's radius R = 6.4 × 10⁶ m.

12. 12. At what height above the Earth's surface does the acceleration due to gravity fall to one-fourth of its surface value?

13. 13. A wire of length 2 m and cross-sectional area 1 × 10⁻⁶ m² stretches by 0.5 mm under a load of 100 N. Calculate the Young's modulus of the wire.

14. 14. Water flows through a horizontal pipe that narrows from a cross-sectional area of 4 cm² to 2 cm². If the water's speed in the wider section is 3 m/s, find its speed in the narrower section.

15. 15. One mole of an ideal gas is heated at constant volume so its temperature rises from 300 K to 400 K. Calculate the heat absorbed by the gas. (Cv = 3/2 R, R = 8.31 J/mol·K)

16. 16. Calculate the rms speed of oxygen molecules (molar mass 32 g/mol) at a temperature of 300 K. (R = 8.31 J/mol·K)

17. 17. A simple pendulum has a length of 1 m. Calculate its time period of oscillation. (g = 9.8 m/s²)

18. 18. A transverse wave has a frequency of 500 Hz and a wavelength of 0.66 m. Calculate its speed.

19. 19. Two point charges of +2 μC and +3 μC are placed 30 cm apart in air. Calculate the electrostatic force between them. (k = 9 × 10⁹ N·m²/C²)

20. 20. A parallel plate capacitor has a plate area of 0.02 m² and a plate separation of 1 mm. Calculate its capacitance. (ε₀ = 8.85 × 10⁻¹² F/m)

21. 21. A wire of resistance 10 Ω is uniformly stretched until its length doubles, with its volume remaining constant. Calculate its new resistance.

22. 22. A battery of EMF 12 V and internal resistance 1 Ω is connected to an external resistor of 5 Ω. Calculate the current flowing in the circuit.

23. 23. A straight current-carrying conductor carries a current of 5 A and lies perpendicular to a magnetic field of 0.2 T. Calculate the force per unit length on the conductor.

24. 24. A coil of 100 turns and area 0.01 m² lies with its plane perpendicular to a magnetic field that changes uniformly from 0.2 T to 0.8 T in 0.3 s. Calculate the induced EMF.

25. 25. A convex lens of focal length 20 cm forms a real image of an object placed 30 cm from the lens. Calculate the image distance.

26. 26. Two thin lenses of focal lengths +10 cm and −20 cm are placed in contact. Calculate the focal length of the combination.

27. 27. Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of 100 V. (h = 6.63 × 10⁻³⁴ J·s, m = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)

28. 28. Calculate the energy of a photon of wavelength 600 nm. (h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s)

29. 29. A p-n junction diode is used as a half-wave rectifier with an AC input of frequency 50 Hz. What is the frequency of the rectified output?

30. 30. A common-emitter transistor amplifier has a current gain β = 100. If the base current is 20 μA, calculate the collector current.

Answer key & quick explanations

Short answers for the sample questions above. Use this to self-check before generating a fresh AI-built mock test.

1. 1. In an experiment, the mass of an object is measured with 2% error and its velocity with 3% error. What is the percentage error in the calculated kinetic energy?

   8%

   Kinetic energy is given by KE = ½mv², so the fractional errors combine as ΔKE/KE = Δm/m + 2(Δv/v). Substituting the given values gives 2% + 2 × 3% = 8%. Notice the velocity term gets doubled because it appears squared in the formula.

2. 2. Planck's constant has the same dimensional formula as which other physical quantity?

   Angular momentum

   Planck's constant h appears in E = hν, so its units work out to J·s, which reduces to kg·m²·s⁻¹, i.e., [ML²T⁻¹]. Angular momentum L = Iω carries exactly this same dimensional formula. This equivalence is a frequently tested fact in NEET dimensional-analysis questions.

3. 3. A projectile is fired with a speed of 20 m/s at 45° to the horizontal. Calculate its range on level ground. (g = 10 m/s²)

   40 m

   The range formula for a projectile launched and landing at the same height is R = u²sin(2θ)/g. With θ = 45°, sin(2θ) = sin90° = 1, so R = (20)²(1)/10 = 40 m. Forty-five degrees always gives the maximum possible range for a given launch speed.

4. 4. A car moving at 20 m/s decelerates uniformly at 2 m/s². How far does it travel before coming to rest?

   100 m

   Using v² = u² − 2as with final velocity v = 0, we get 0 = (20)² − 2(2)s. Solving gives s = 400/4 = 100 m. The negative sign in the deceleration is already accounted for by subtracting it in the equation.

5. 5. A 5 kg block is pulled along a rough horizontal surface by a force of 20 N. If the coefficient of kinetic friction is 0.2, find the block's acceleration. (g = 10 m/s²)

   2 m/s²

   The friction force opposing motion is f = μmg = 0.2 × 5 × 10 = 10 N. This leaves a net force of 20 − 10 = 10 N acting on the block. Dividing by the mass gives a = 10/5 = 2 m/s².

6. 6. Two masses of 3 kg and 2 kg are connected by a light string over a frictionless pulley (Atwood machine). Calculate the tension in the string. (g = 10 m/s²)

   24 N

   For an Atwood machine, the common acceleration is a = (m₁ − m₂)g/(m₁ + m₂) = (3 − 2)(10)/5 = 2 m/s². The tension can then be found from the lighter mass's equation, T = m₂(g + a) = 2(10 + 2) = 24 N. Checking with the heavier mass, T = m₁(g − a) = 3(10 − 2) = 24 N, confirming consistency.

7. 7. A 2 kg body is lifted vertically through 5 m and simultaneously given a final speed of 4 m/s. Calculate the total work done on the body. (g = 10 m/s²)

   116 J

   By the work-energy theorem, total work done equals the gain in gravitational potential energy plus the gain in kinetic energy. The potential energy term is mgh = 2 × 10 × 5 = 100 J, while the kinetic energy term is ½mv² = ½ × 2 × 16 = 16 J. Adding these gives 100 + 16 = 116 J.

8. 8. A pump lifts 300 kg of water through a height of 10 m in 5 s. Calculate the power delivered by the pump. (g = 10 m/s²)

   6000 W (6 kW)

   The work done in raising the water is W = mgh = 300 × 10 × 10 = 30000 J. Power is the rate of doing this work, so P = W/t = 30000/5 = 6000 W. This equals 6 kW.

9. 9. A uniform disc of mass 2 kg and radius 0.5 m rotates about an axis through its centre, perpendicular to its plane. Calculate its moment of inertia.

   0.25 kg·m²

   For a uniform disc about a central axis perpendicular to its plane, I = ½MR². Substituting M = 2 kg and R = 0.5 m gives I = 0.5 × 2 × 0.25 = 0.25 kg·m². This is one of the standard moment-of-inertia formulas worth memorizing directly.

10. 10. A torque of 10 N·m acts on a wheel of moment of inertia 2 kg·m², initially at rest. Find its angular velocity after 4 s.

    20 rad/s

    The angular acceleration follows from τ = Iα, giving α = 10/2 = 5 rad/s². Since the wheel starts from rest, ω = ω₀ + αt = 0 + 5 × 4 = 20 rad/s. This mirrors the linear kinematics equation v = u + at, just rewritten in rotational variables.

11. 11. Calculate the escape velocity from the Earth's surface, given g = 9.8 m/s² and Earth's radius R = 6.4 × 10⁶ m.

    ≈ 11.2 km/s

    Escape velocity is given by v_e = √(2gR). Plugging in the values, v_e = √(2 × 9.8 × 6.4 × 10⁶) = √(1.2544 × 10⁸) = 11200 m/s. This works out to the well-known figure of about 11.2 km/s.

12. 12. At what height above the Earth's surface does the acceleration due to gravity fall to one-fourth of its surface value?

    h = R (one Earth radius)

    The variation of g with height is g' = g/(1 + h/R)². Setting g' = g/4 requires (1 + h/R)² = 4, so 1 + h/R = 2, giving h = R. So gravity drops to a quarter of its surface value at an altitude equal to Earth's own radius.

13. 13. A wire of length 2 m and cross-sectional area 1 × 10⁻⁶ m² stretches by 0.5 mm under a load of 100 N. Calculate the Young's modulus of the wire.

    4 × 10¹¹ Pa

    Young's modulus is Y = (F/A)/(ΔL/L) = FL/(AΔL). Substituting F = 100 N, L = 2 m, A = 1 × 10⁻⁶ m², and ΔL = 0.5 × 10⁻³ m gives Y = (100 × 2)/(1 × 10⁻⁶ × 0.5 × 10⁻³) = 200/(5 × 10⁻¹⁰) = 4 × 10¹¹ Pa. This value is close to that of a typical metallic wire.

14. 14. Water flows through a horizontal pipe that narrows from a cross-sectional area of 4 cm² to 2 cm². If the water's speed in the wider section is 3 m/s, find its speed in the narrower section.

    6 m/s

    For an incompressible fluid, the equation of continuity requires A₁v₁ = A₂v₂. Here 4 × 3 = 2 × v₂, so v₂ = 12/2 = 6 m/s. The speed doubles because the cross-section is halved, keeping the volume flow rate constant.

15. 15. One mole of an ideal gas is heated at constant volume so its temperature rises from 300 K to 400 K. Calculate the heat absorbed by the gas. (Cv = 3/2 R, R = 8.31 J/mol·K)

    ≈ 1246.5 J

    At constant volume, the heat absorbed is Q = nCvΔT. With n = 1 mol, Cv = 1.5 × 8.31 = 12.465 J/mol·K, and ΔT = 100 K, we get Q = 1 × 12.465 × 100 = 1246.5 J. No work is done in this process since the volume doesn't change, so this heat entirely raises the internal energy.

16. 16. Calculate the rms speed of oxygen molecules (molar mass 32 g/mol) at a temperature of 300 K. (R = 8.31 J/mol·K)

    ≈ 483 m/s

    The rms speed formula is v_rms = √(3RT/M), where M must be expressed in kg/mol. Substituting gives v_rms = √(3 × 8.31 × 300 / 0.032) = √(233718.75) ≈ 483 m/s. Heavier gas molecules at the same temperature move slower than lighter ones, since v_rms scales as 1/√M.

17. 17. A simple pendulum has a length of 1 m. Calculate its time period of oscillation. (g = 9.8 m/s²)

    ≈ 2.0 s

    The time period of a simple pendulum is T = 2π√(L/g). Substituting L = 1 m and g = 9.8 m/s² gives T = 2π√(0.1020) = 2π(0.319) ≈ 2.0 s. This is why a 1 m pendulum is often used as a rough benchmark for a two-second clock.

18. 18. A transverse wave has a frequency of 500 Hz and a wavelength of 0.66 m. Calculate its speed.

    330 m/s

    Wave speed relates frequency and wavelength through v = fλ. Multiplying the given values, v = 500 × 0.66 = 330 m/s. This happens to match the familiar speed of sound in air at room temperature.

19. 19. Two point charges of +2 μC and +3 μC are placed 30 cm apart in air. Calculate the electrostatic force between them. (k = 9 × 10⁹ N·m²/C²)

    0.6 N (repulsive)

    Coulomb's law gives F = kq₁q₂/r². Substituting q₁ = 2 × 10⁻⁶ C, q₂ = 3 × 10⁻⁶ C, and r = 0.3 m, the numerator becomes 9 × 10⁹ × 6 × 10⁻¹² = 0.054, and dividing by r² = 0.09 gives F = 0.6 N. Since both charges are positive, the force is repulsive.

20. 20. A parallel plate capacitor has a plate area of 0.02 m² and a plate separation of 1 mm. Calculate its capacitance. (ε₀ = 8.85 × 10⁻¹² F/m)

    ≈ 177 pF

    Capacitance for a parallel plate capacitor is C = ε₀A/d. Plugging in A = 0.02 m² and d = 1 × 10⁻³ m gives C = 8.85 × 10⁻¹² × 0.02 / 0.001 = 1.77 × 10⁻¹⁰ F. Converting to picofarads, this is about 177 pF.

21. 21. A wire of resistance 10 Ω is uniformly stretched until its length doubles, with its volume remaining constant. Calculate its new resistance.

    40 Ω

    Since R = ρL/A and volume V = AL is fixed, resistance can be rewritten as R = ρL²/V, so R is proportional to L². Doubling the length therefore quadruples the resistance: 4 × 10 Ω = 40 Ω. This quadratic dependence on stretching is a common trap in NEET resistance questions.

22. 22. A battery of EMF 12 V and internal resistance 1 Ω is connected to an external resistor of 5 Ω. Calculate the current flowing in the circuit.

    2 A

    Total resistance in the circuit is the sum of internal and external resistance, R + r = 5 + 1 = 6 Ω. Applying Ohm's law to the full loop, I = EMF/(R + r) = 12/6 = 2 A. Internal resistance always reduces the current below what an ideal 12 V source across 5 Ω alone would give.

23. 23. A straight current-carrying conductor carries a current of 5 A and lies perpendicular to a magnetic field of 0.2 T. Calculate the force per unit length on the conductor.

    1 N/m

    The force per unit length on a current-carrying conductor in a magnetic field is F/L = BIsinθ. Since the conductor is perpendicular to the field, θ = 90° and sinθ = 1, so F/L = 0.2 × 5 = 1 N/m. This is the maximum possible force per unit length for these given values of B and I.

24. 24. A coil of 100 turns and area 0.01 m² lies with its plane perpendicular to a magnetic field that changes uniformly from 0.2 T to 0.8 T in 0.3 s. Calculate the induced EMF.

    2 V

    By Faraday's law, the induced EMF is ε = N(ΔΦ/Δt) = NA(ΔB/Δt). Here ΔB = 0.8 − 0.2 = 0.6 T over 0.3 s, giving ΔB/Δt = 2 T/s. Multiplying, ε = 100 × 0.01 × 2 = 2 V.

25. 25. A convex lens of focal length 20 cm forms a real image of an object placed 30 cm from the lens. Calculate the image distance.

    60 cm (real image)

    Using the lens formula 1/v − 1/u = 1/f with the sign convention u = −30 cm and f = +20 cm, we get 1/v = 1/20 − 1/30 = (3 − 2)/60 = 1/60. This gives v = 60 cm. Since the object lies between f and 2f, the resulting real image forms beyond 2f, consistent with this result.

26. 26. Two thin lenses of focal lengths +10 cm and −20 cm are placed in contact. Calculate the focal length of the combination.

    +20 cm (converging)

    For thin lenses in contact, the powers add: 1/f = 1/f₁ + 1/f₂. Substituting f₁ = +10 cm and f₂ = −20 cm gives 1/f = 1/10 − 1/20 = 1/20, so f = 20 cm. The positive sign shows the combination still behaves as a converging lens overall.

27. 27. Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of 100 V. (h = 6.63 × 10⁻³⁴ J·s, m = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)

    ≈ 1.23 Å (0.123 nm)

    For an electron accelerated through a potential V, the de Broglie wavelength is λ = h/√(2meV). Substituting the given values, 2meV = 2 × 9.1 × 10⁻³¹ × 1.6 × 10⁻¹⁹ × 100 = 2.912 × 10⁻⁴⁷, whose square root is about 5.396 × 10⁻²⁴. Dividing h by this gives λ ≈ 1.229 × 10⁻¹⁰ m, matching the familiar shortcut formula λ = 12.27/√V Å.

28. 28. Calculate the energy of a photon of wavelength 600 nm. (h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s)

    ≈ 3.3 × 10⁻¹⁹ J (≈ 2.07 eV)

    Photon energy is found from E = hc/λ. Substituting the values, E = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(600 × 10⁻⁹) = 1.989 × 10⁻²⁵/6 × 10⁻⁷ = 3.315 × 10⁻¹⁹ J. Dividing by the electronic charge converts this to about 2.07 eV, a value typical of visible light near the orange-red end of the spectrum.

29. 29. A p-n junction diode is used as a half-wave rectifier with an AC input of frequency 50 Hz. What is the frequency of the rectified output?

    50 Hz

    A half-wave rectifier conducts during only one half-cycle of the AC input and blocks the other, so it produces one output pulse per input cycle. This means the output frequency stays the same as the input frequency, 50 Hz. This is different from a full-wave rectifier, which would double the ripple frequency to 100 Hz.

30. 30. A common-emitter transistor amplifier has a current gain β = 100. If the base current is 20 μA, calculate the collector current.

    2 mA

    In a common-emitter configuration, the current gain relates collector and base currents through Ic = βIb. Substituting the given values, Ic = 100 × 20 × 10⁻⁶ A = 2000 × 10⁻⁶ A. This equals 2 mA, showing how a small base current controls a much larger collector current.

Curriculum Mapping & Learning Guide

Use this breakdown to identify which skills each question tests and guide post-test review.

NEET Physics units covered

1. Chapter 1: Units and Measurements
2. Chapter 2: Kinematics
3. Chapter 3: Laws of Motion
4. Chapter 4: Work, Energy and Power
5. Chapter 5: Rotational Motion
6. Chapter 6: Gravitation
7. Chapter 7: Properties of Matter
8. Chapter 8: Thermodynamics and Kinetic Theory
9. Chapter 9: Oscillations and Waves
10. Chapter 10: Electrostatics
11. Chapter 11: Current Electricity
12. Chapter 12: Magnetism and Electromagnetic Induction
13. Chapter 13: Optics (Ray and Wave)
14. Chapter 14: Dual Nature of Matter, Atoms, Nuclei
15. Chapter 15: Electronic Devices (Semiconductors)

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