NEET Mock Test | PreMedicalBiology

Q1. A particle is projected from ground with velocity 20 m/s at 30°. At a certain instant, its velocity makes 45° with horizontal. Neglect air resistance. Find the time after projection when this occurs (g = 10 m/s²).

A. 0.5 s B. 2.73 s C. 1.5 s D. 2 s

Explanation:
Option B is correct. Horizontal velocity: vx = 20 cos 30° = 10√3 (constant) , Vertical velocity: vy = 20 sin 30° − gt = 10 − 10t , For 45° below horizontal: tan 45° = 1 = |vy| / vx , Since velocity is downward, vy is negative: (10t − 10) / (10√3) = 1 , 10t − 10 = 10√3 , t − 1 = √3 , t = 1 + √3 ≈ 2.73 s

Q2. The position of a particle moving along the x-axis is given by: x = t³ − 6t² + 9t , Determine the time(s) at which the velocity becomes zero and identify the nature of motion at those instants.

A. t = 1 s, 3 s B. t = 2 s only C. t = 1 s only D. t = 3 s only

Explanation:
Option A is correct. Velocity is the first derivative of position: v = dx/dt = 3t² − 12t + 9 , For velocity to be zero: 3t² − 12t + 9 = 0 , t² − 4t + 3 = 0 , (t − 1)(t − 3) = 0 , t = 1 s, 3 s , Now, acceleration: a = dv/dt = 6t − 12 , At t = 1 s: a = 6(1) − 12 = −6 → velocity changes from positive to negative ⇒ maximum position , At t = 3 s: a = 6(3) − 12 = +6 → velocity changes from negative to positive ⇒ minimum position .

Q3. A block of mass 5 kg slides down a rough inclined plane inclined at an angle of 37°. The coefficient of friction is 0.25. Calculate its acceleration along the plane. (g = 10 m/s², sin37° = 0.6, cos37° = 0.8)

A. 2 m/s² B. 3 m/s² C. 4 m/s² D. 5 m/s²

Explanation:
Option C is correct. For a block sliding down a rough inclined plane. Forces along the plane: Component of weight down the plane = mg sinθ , Friction (opposing motion) = μ mg cosθ , Net force along plane: F = mg sinθ − μ mg cosθ , Using Newton’s second law (F = ma): ma = mg (sinθ − μ cosθ) , a = g (sinθ − μ cosθ) , Now substitute values: a = 10 (0.6 − 0.25 × 0.8) , a = 10 (0.6 − 0.2) , a = 10 × 0.4 , a = 4 m/s²

Q4. A satellite revolves very close to the surface of the Earth. If the mass of the Earth remains the same but its radius becomes four times, what will be the new orbital velocity compared to the original value?

A. Same B. Half C. One-fourth D. One-third

Explanation:
Option B is correct. Orbital velocity of a satellite near Earth’s surface is given by: v = √(GM / R) , Thus v ∝ 1 / √R (since M is constant) , If new radius R' = 4R , v' / v = √(R / R') , v' / v = √(R / 4R) , v' / v = 1 / 2 , v' = v / 2 ,

Q5. A force vector F = (3i + 4j) N acts on a particle moving with velocity v = (2i − j) m/s. Calculate the instantaneous power delivered by the force.

A. 2 W B. 5 W C. 10 W D. 14 W

Explanation:
Option A is correct. Instantaneous power is given by the dot product: P = F · v , P = (3i + 4j) · (2i − j) , P = (3 × 2) + (4 × −1) , P = 6 − 4 , P = 2 W ,

Q6. A uniform circular ring of mass M and radius R rotates about an axis passing through its diameter. Determine its moment of inertia about this axis.

A. MR² B. 2MR² C. MR²/4 D. MR²/2

Explanation:
Option D is correct. Moment of inertia of a ring about an axis perpendicular to its plane through the centre: I_z = MR² , Using perpendicular axis theorem: I_z = I_x + I_y , For symmetry of a ring: I_x = I_y , So: MR² = 2 I_x , I_x = MR² / 2 ,

Q7. A particle is moving in uniform circular motion with constant speed 10 m/s. If the radius of the circle is 5 m, calculate the magnitude of centripetal acceleration acting on it.

A. 20 m/s² B. 10 m/s² C. 25 m/s² D. 50 m/s²

Explanation:
Option A is correct. Centripetal acceleration is given by: a = v² / r , Substitute values: a = (10)² / 5 , a = 100 / 5 , a = 20 m/s²

Q8. A body of mass m is raised vertically through a height h slowly and then allowed to fall freely back to the starting point. Compare the work done by gravity during upward and downward motion.

A. Same B. Double C. Zero D. Depends

Explanation:
Option A is correct. Work done by gravity depends only on the change in vertical displacement and is independent of the path taken. Work done by gravity during upward motion: W₁ = −mgh , Work done by gravity during downward motion: W₂ = +mgh , Thus, magnitudes are equal: |W₁| = |W₂| = mgh , Gravity is a conservative force, so work done depends only on initial and final positions.

Q9. A figure skater pulls her arms inward while spinning. Which physical quantity remains conserved and what change occurs in angular velocity?

A. Kinetic energy conserved B. Angular momentum conserved, ω increases C. Torque conserved D. Linear momentum conserved

Explanation:
Option B is correct. No external torque acts on the system, so angular momentum is conserved. L = Iω = constant , When the skater pulls her arms inward, moment of inertia decreases (I ↓). , To keep L constant: I ↓ ⇒ ω ↑ , Hence, angular velocity increases.

Q10. A solid sphere rolls without slipping on a horizontal surface. Find the ratio of translational kinetic energy to total kinetic energy of the sphere.

A. 5/7 B. 2/7 C. 7/5 D. 1/2

Explanation:
Option A is correct. For a solid sphere rolling without slipping: Translational kinetic energy: K_trans = (1/2) m v² , Moment of inertia of solid sphere: I = (2/5) mR² , Rotational kinetic energy: K_rot = (1/2) Iω² , Using v = ωR: , K_rot = (1/2) × (2/5)mR² × (v²/R²) , K_rot = (1/5) m v² , Total kinetic energy: K_total = K_trans + K_rot , K_total = (1/2 + 1/5) m v² , K_total = (7/10) m v² , Now, ratio: K_trans / K_total = (1/2) / (7/10) , = (1/2) × (10/7) , = 5/7

Q11. One mole of an ideal gas expands isothermally at temperature T from volume V to 2V. Calculate the work done by the gas in terms of R and T.

A. RT ln 2 B. 2RT C. RT D. 1/2 RT

Explanation:
Option A is correct. Given: n = 1 mol , Vi = V , Vf = 2V , Process is Isothermal. Formula used: W = nRT ln(Vf/Vi) , Calculation: W = (1)RT ln(2V/V) , W = RT ln(2) ,

Q12. At absolute zero temperature, the kinetic energy of molecules becomes zero according to classical theory. What limitation of classical theory does this highlight?

A. Failure at high temperature B. Failure at low temperature C. Failure at high pressure D. None

Explanation:
Option B is correct. According to classical theory, the average kinetic energy of gas molecules is directly proportional to absolute temperature: E ∝ T , At T = 0 , KE = 0 , However, according to quantum theory, molecules possess zero-point energy even at absolute zero. Therefore, classical theory fails at low temperatures.

Q13. An ideal gas undergoes a cyclic process. The net heat supplied to the gas is equal to:

A. Zero B. Change in internal energy C. Entropy D. Work done

Explanation:
Option D is correct. In a cyclic process, the system returns to its initial state. Therefore, the net change in internal energy is zero: ΔU = 0 , From the first law of thermodynamics: Q = ΔU + W , Substituting ΔU = 0: Q = W , The Net heat supplied = Work done by the gas.

Q14. Two point charges +q and +4q are placed at a distance d apart. Find the position along the line joining them where the net electric field is zero.

A. d/3 B. d/2 C. d/4 D. No point

Explanation:
Option A is correct. Let the point where the electric field is zero be at a distance x from charge +q, between the two charges. Electric field due to charge +q: E₁ = kq / x² , Electric field due to charge +4q: E₂ = 4kq / (d − x)² , For net electric field to be zero: E₁ = E₂ , kq / x² = 4kq / (d − x)² , 1 / x² = 4 / (d − x)² , Taking square root: 1 / x = 2 / (d − x) , d − x = 2x , d = 3x , x = d/3 , hence the Electric field is zero at a distance d/3 from charge +q (between the two charges).

Q15. A wire of resistance R is stretched to double its length, keeping its volume constant. Find the new resistance in terms of R.

A. 2R B. 3R C. 4R D. R/2

Explanation:
Option C is correct. Given: L' = 2L, volume constant ⇒ L × A = constant ⇒ A' = A/2 , R = ρL/A , R' = ρ(2L)/(A/2) , R' = 4(ρL/A) , R' = 4R

Q16. In Young’s double slit experiment, the slit separation is doubled and the distance between the slits and the screen is halved. Determine the effect on fringe width.

A. Same B. Double C. Half D. One-fourth

Explanation:
Option D is correct. Fringe width in YDSE is given by: β = λD / d , Given: d' = 2d, D' = D/2, New fringe width:β' = λ(D/2) / (2d) , β' = (λD) / (4d) , β' = β / 4 , Fringe width becomes one-fourth of original.

Q17. In a photoelectric effect experiment, the stopping potential depends on the frequency of incident radiation. What happens if the intensity is increased while keeping the frequency constant?

A. Stopping potential increases B. Kinetic energy increases C. Photoelectric current increases D. None

Explanation:
Option C is correct. In the photoelectric effect, maximum kinetic energy and stopping potential depend only on frequency: Kmax ∝ f , Vs ∝ f , When frequency is constant, increasing intensity increases the number of incident photons. Therefore, the number of emitted photoelectrons increases, so photoelectric current increases.

Q18. A photon of wavelength 400 nm strikes a metal surface. Calculate the energy of the photon. (Given: h = 6.6 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s)

A. 5 × 10⁻¹⁹ J B. 4 × 10⁻¹⁹ J C. 3 × 10⁻¹⁹ J D. 2 × 10⁻¹⁹ J

Explanation:
Option A is correct. Given: λ = 400 nm = 400 × 10⁻⁹ m , Energy of photon is given by: E = hc / λ , Substituting values: E = (6.6 × 10⁻³⁴ × 3 × 10⁸) / (400 × 10⁻⁹) , E = 19.8 × 10⁻²⁶ / 4 × 10⁻⁷ , E = 4.95 × 10⁻¹⁹ J , E ≈ 5 × 10⁻¹⁹ J

Q19. The half-life of a radioactive substance is 10 days. Calculate the time required for 75% decay of the original nuclei.

A. 10 days B. 20 days C. 30 days D. 40 days

Explanation:
Option B is correct. Half-life, T₁/₂ = 10 days , 75% decay means 25% nuclei remain. So, N / N₀ = 1/4 , Using radioactive decay law: N / N₀ = (1/2)ⁿ , (1/2)ⁿ = 1/4 = (1/2)² , So, n = 2 half-lives , Time required: t = n × T₁/₂ , t = 2 × 10 = 20 days

Q20. A convex lens forms an image at twice the focal length. Identify the object distance.

A. 4f B. 3f C. 2f D. 1f

Explanation:
Option C is correct. Given: Image distance, v = 2f , Lens formula: 1/f = 1/v − 1/u , Substituting v = 2f: 1/f = 1/(2f) − 1/u , 1/u = 1/(2f) − 1/f , 1/u = (1 − 2) / (2f) , 1/u = −1/(2f) , u = −2f , Magnitude of object distance = 2f

Q21. A body of mass 2 kg is moving with a velocity of 3 m/s. The velocity of the body is suddenly doubled by applying a constant force in the direction of motion. What will be the increase in kinetic energy of the body?

A. 9 J B. 18 J C. 27 J D. 36 J

Explanation:
Option C is correct. Initial KE = (1/2)×2×3² = 9 J , Final velocity = 6 m/s , Final KE = (1/2)×2×6² = 36 J , Increase = 36 − 9 = 27 J ,

Q22. A 5 kg block is placed on a smooth horizontal surface. Two horizontal forces act on it simultaneously: 10 N towards east and 2 N towards west. What is the acceleration of the block considering resultant motion?

A. 1.2 m/s² B. 1.6 m/s² C. 2.0 m/s² D. 2.4 m/s²

Explanation:
Option B is correct. Net force = 10 − 2 = 8 N , Acceleration = F/m = 8/5 = 1.6 m/s²

Q23. A constant force of magnitude 10 N is applied on a body which moves 2 m in a direction making an angle of 60° with the force. What is the work done by the force on the body?

A. 5 J B. 20 J C. 15 J D. 10 J

Explanation:
Option D is correct. W = F s cosθ , W = 10 × 2 × cos60° , W = 20 × 1/2 W = 10 J

Q24. A ball of mass 1 kg moving with a speed of 4 m/s strikes a rigid wall and rebounds with the same speed in the opposite direction. What is the magnitude of change in momentum of the ball?

A. 4 kg·m/s B. 6 kg·m/s C. 8 kg·m/s D. 16 kg·m/s

Explanation:
Option C is correct. Initial momentum = +4 , Final momentum = −4 , Change = −4 − 4 = −8 → magnitude = 8 kg·m/s

Q25. Which of the following statements correctly explains why acceleration due to gravity does not depend on the mass of the falling object near Earth’s surface?

A. Earth’s mass changes B. Gravitational force cancels mass term in acceleration C. Air resistance is always zero D. Velocity of object is constant

Explanation:
Option B is correct. F = GMm / R² , So acceleration due to gravity, a = F / m , a = (GMm / R²) / m , a = GM / R² , Mass (m) cancels out, so acceleration due to gravity does not depend on the mass of the object.

Q26. A particle is moving in a circle with constant speed. The force responsible for keeping the particle in circular motion always acts in which direction relative to velocity?

A. Same direction as velocity B. Opposite to velocity C. Perpendicular and toward center D. Tangential outward

Explanation:
Option C is correct. Centripetal force is always perpendicular to velocity and directed toward center.

Q27. The time period of a simple pendulum is given by: T = 2π √(L / g) . The time period depends only on the length (L) of the pendulum and the acceleration due to gravity (g). Which of the following changes will NOT affect the time period of oscillation?

A. Change in mass of the bob B. Change in length of the pendulum C. Change in acceleration due to gravity D. Change in amplitude (small oscillations)

Explanation:
Option A is correct. From the formula T = 2π √(L / g), the time period does not depend on the mass of the bob. It depends only on the length of the pendulum and acceleration due to gravity (for small oscillations).

Q28. A gas is compressed slowly in a cylinder fitted with a frictionless piston. During compression, heat is supplied to the gas. Which fundamental principle governs this process?

A. Conservation of momentum B. Conservation of energy C. Conservation of mass D. Newton’s law

Explanation:
Option B is correct. The first law of thermodynamics states that energy cannot be created or destroyed, only transformed from one form to another. It is expressed as: ΔQ = ΔU + W , This represents the conservation of energy in thermodynamic processes.

Q29. A hot object is placed inside an evacuated chamber where no air or matter is present. In such a case, heat transfer from the object to the surroundings will occur only through which mode?

A. Conduction B. Convection C. Radiation D. All of these

Explanation:
Option C is correct. Conduction and convection require a material medium for heat transfer. In a vacuum, no medium is present, so heat transfer can occur only through radiation.

Q30. Two point charges are placed at a certain distance in air. If the distance between them is doubled, how does the electrostatic force between them change?

A. Becomes half B. Doubles C. Becomes one-fourth D. Remains same

Explanation:
Option C is correct. According to Coulomb’s law, electrostatic force is inversely proportional to the square of the distance between charges: F ∝ 1/r² . If the distance is doubled, the force becomes 1/(2²) = 1/4 of the original value.

Q31. An electric field at a point in space is defined as the force experienced per unit positive test charge placed at that point. Which of the following correctly represents its SI unit?

A. N B. N/C C. C/N D. J/C

Explanation:
Option B is correct. The electric field is defined as force per unit charge, i.e., E = F/q. Since force is measured in newtons (N) and charge in coulombs (C), the SI unit of electric field is N/C (newton per coulomb).

Q32. According to Gauss’s law, the net electric flux through a closed surface depends only on the total charge enclosed within the surface and not on its shape. Which of the following is correct?

A. Depends on area only B. Depends on shape only C. Depends on enclosed charge only D. Depends on medium only

Explanation:
Option C is correct. According to Gauss’s law, the total electric flux through a closed surface is given by Φ = Q_enclosed / ε₀. This shows that the flux depends only on the total charge enclosed within the surface and is independent of the shape, size, or orientation of the surface.

Q33. A parallel plate capacitor is connected to a battery. A dielectric slab is inserted completely between its plates. What happens to its capacitance?

A. Decreases B. Remains the same C. Increases D. Becomes zero

Explanation:
Option C is correct. When a dielectric material is inserted between the plates of a capacitor, the permittivity of the medium increases. Since capacitance is given by C=εA/d, an increase in permittivity leads to an increase in capacitance.

Q34. A student studies different materials and finds that only some show a linear relationship between current and voltage. Which type of material obeys Ohm’s law?

A. Semiconductors B. Insulators C. Electrolytes D. Ohmic conductors

Explanation:
Option D is correct. Ohm’s law states that the current through a conductor is directly proportional to the applied voltage, provided the temperature and other physical conditions remain constant. Materials that show a linear V–I relationship are called ohmic conductors, and they obey Ohm’s law.

Q35. The resistance of a wire depends on its length, cross-sectional area, and material. If the length is doubled and the area is halved, what happens to the resistance?

A. Becomes half B. Becomes double C. Becomes four times D. Remains the same

Explanation:
Option C is correct. Resistance is given by R = ρL/A. If the length is doubled (2L) and the area is halved (A/2), then R' = ρ(2L)/(A/2) = 4ρL/A = 4R. Therefore, the resistance becomes four times.

Q36. At any junction in an electrical circuit, the algebraic sum of currents entering and leaving the junction is zero. This law is based on the conservation of which physical quantity?

A. Energy B. Charge C. Momentum D. Power

Explanation:
Option B is correct. Kirchhoff’s junction law is based on the conservation of charge. It states that the total current entering a junction is equal to the total current leaving it, ensuring that no charge is lost or accumulated at the junction.

Q37. A charged particle enters a magnetic field with a velocity making different angles with the field. At which angle will the magnetic force be maximum?

A. 0° B. 180° C. 30° D. 90°

Explanation:
Option D is correct. The magnetic force on a moving charge is given by F=qvBsinθ, where θ is the angle between the velocity and the magnetic field. The force is maximum when sinθ=1, i.e., when θ = 90°.

Q38. Magnetic field lines represent the direction and strength of a magnetic field. Which of the following statements is correct regarding them?

A. They intersect each other B. They form closed continuous loops C. They start from the negative pole D. They are always straight lines

Explanation:
Option B is correct. Magnetic field lines form closed continuous loops. Outside a magnet, they emerge from the north pole and enter the south pole, and inside the magnet, they return from the south pole to the north pole. They never intersect each other.

Q39. In Young’s double slit experiment, constructive interference occurs at points where the path difference between two waves is an integer multiple of wavelength. Which condition is correct?

A. (2n+1)λ/2 B. nλ C. λ/3 D. λ/4

Explanation:
Option B is correct. Constructive interference occurs when the path difference between two waves is an integer multiple of the wavelength, i.e., Δx = nλ, where n = 0, 1, 2, 3, ...

Q40. In photoelectric emission, electrons are emitted from a metal surface when light of sufficient frequency falls on it. Which factor determines the maximum kinetic energy of emitted electrons?

A. Frequency of light B. Intensity of light C. Area of metal D. Distance from source

Explanation:
Option A is correct. In the photoelectric effect, the maximum kinetic energy of emitted electrons depends on the frequency of incident light. According to Einstein’s photoelectric equation, Kmax = hf−ϕ, where h is Planck’s constant, f is the frequency of light, and φ is the work function of the metal.

Q41. The Bohr model of hydrogen atom successfully explains which of the following phenomena?

A. Nuclear stability B. Radioactive decay C. Hydrogen spectral lines D. Molecular bonding

Explanation:
Option C is correct. The Bohr model explains the discrete spectral lines of the hydrogen atom by assuming that electrons revolve in fixed energy orbits and emit or absorb energy only when they jump between these orbits.

Q42. Least count of a measuring instrument determines its smallest measurable value. Which of the following is most closely related to least count?

A. Accuracy B. Precision C. Resolution D. Error magnitude

Explanation:
Option C is correct. Least count is the smallest measurement that an instrument can read. It is most closely related to the resolution of the instrument, which describes its ability to distinguish between two close values.

Q43. An emf is induced in a coil when the magnetic flux linked with it changes. Which factor directly determines the magnitude of the induced emf?

A. Absolute value of flux B. Rate of change of flux C. Resistance of coil D. Length of wire

Explanation:
Option B is correct. According to Faraday’s law of electromagnetic induction, the induced emf in a coil is directly proportional to the rate of change of magnetic flux linked with it, i.e., emf ∝ dΦ/dt. Therefore, a faster change in magnetic flux produces a larger induced emf.

Q44. In a purely inductive AC circuit, the alternating current does not remain in phase with voltage. What is the phase relationship between current and voltage?

A. Current lags voltage B. Current leads voltage C. Both are in phase D. No current flows

Explanation:
Option A is correct. In a purely inductive AC circuit, the current lags behind the applied voltage by 90°. This is because an inductor opposes the change in current due to self-induction, causing the current to reach its maximum after the voltage.

Q45. A NOT gate is a basic logic gate used in digital electronics. What is its primary function?

A. Adds two inputs B. Multiplies inputs C. Inverts input signal D. Stores data

Explanation:
Option C is correct. A NOT gate performs logical negation. It inverts the input signal, meaning it changes 0 to 1 and 1 to 0.

Test Summary