NEET 2024 Question Paper Step-by-Step Solutions

For any rigid body undergoing pure planar rolling without slipping on a stationary surface with linear velocity v and angular velocity ω = v/R, the total velocity of any point on the rim is the vector sum of the translational velocity of the center of mass (v_cm) and the tangential rotational velocity (v_rot = ωR):

At the uppermost point P, the translational velocity is forward (+v î) and the rotational velocity is also in the forward direction (+ωR î = +v î). Adding both vectors yields:

At the bottom contact point Q, the translational velocity is directed forward (+v î), whereas the rotational tangential velocity is directed backward (-ωR î = -v î). Under the condition of pure rolling without slipping, these two components cancel each other out identically:

Comparing the speeds of the two points: v_P = 2v, while v_Q = 0. Therefore, point P moves significantly faster than point Q, matching Option (4).

As the bar magnet moves to the right with velocity v, the North pole moves away from Solenoid 1, causing a decrease in magnetic flux through Solenoid 1 directed to the left. According to Lenz’s law, the induced current must oppose this decrease by creating an attractive force (i.e. generating a South pole on its right face facing the magnet).

A South pole on the right end means that when viewed from the right, the current circulates clockwise. Tracing the turns wound over the core, current enters through terminal A, passes over the top turns towards the right, and emerges through terminal B. Therefore, current flows from A to B.

The South pole of the bar magnet is moving closer to Solenoid 2. To oppose this approach, the left face of Solenoid 2 facing the magnet must develop a South magnetic pole to exert a repulsive force on the approaching South pole.

A South pole on the left end means that when viewed from the left, current circulates clockwise. Tracing the winding on Solenoid 2, the current enters from terminal D, goes up the back, over the top forward, and exits at terminal C. Hence, the current flows in the direction DC.

A solar cell is a p-n junction diode operated without any external bias voltage. When sunlight illuminates the depletion region, electron-hole pairs are photogenerated. The internal junction electric field sweeps electrons to the n-side and holes to the p-side, setting up a photo-voltage. When connected across a load, it supplies current. Because it supplies power rather than drawing power, the current flows out of the positive terminal, resulting in a negative current convention for positive voltage. Hence, the I-V curve lies in the IV quadrant.

When a p-n junction is reverse biased (p-side connected to negative terminal, n-side to positive terminal), the potential barrier height increases, completely blocking majority carriers from diffusing across the junction. The only current that flows is the reverse saturation current (order of microamperes, μA), which is caused solely by the drift of thermally generated MINORITY carriers (electrons from p-side and holes from n-side) swept across the depletion layer by the junction field. Therefore, the reverse current is due to minority carriers, not majority carriers.

Statement A is completely correct. Statement B falsely claims the reverse current is due to majority carriers. Therefore, "A is correct but B is incorrect", matching Option (3).

Since resistance of a uniform wire is directly proportional to its length (R = ρl/A), dividing the wire into 10 equal parts divides its total resistance equally among the 10 segments:

The first 5 parts are connected end-to-end in series. For series combination, individual resistances simply add up:

The next 5 parts are connected in parallel across common terminals. For 5 identical resistors in parallel:

The problem states that these two combinations are now connected in series with each other. Therefore, the total equivalent resistance is:

The given circuit consists of five capacitors each having capacitance C = 2 μF arranged in a Wheatstone bridge configuration. The condition for balance in a capacitance bridge is:

Because the ratio of adjacent arms is equal, the electric potential at the upper junction and lower junction is identical. No charge flows into or through the central 2 μF capacitor, so it can be completely removed from the circuit without affecting the equivalent capacitance.

With the central capacitor removed, the circuit simplifies to two parallel branches, each containing two 2 μF capacitors in series:

The top and bottom branches are connected in parallel across terminals A and B. Their capacitances add directly:

The top gate is a two-input NAND gate whose inputs are shorted together to receive a common signal A. The Boolean operation of a NAND gate on identical inputs simplifies directly to an inverter (NOT gate):

The bottom gate is a two-input NOR gate whose inputs are shorted together to receive a common signal B. The Boolean operation on identical inputs also simplifies to an inverter (NOT gate):

The signals Y_A = Ā and Y_B = B̄ serve as inputs to the final NOR gate. The output Y is the NOR operation on Ā and B̄:

Applying De Morgan’s law to the complement of a sum yields the product of the complements:

According to Ohm’s law applied to the complete closed circuit consisting of the real battery and the external resistor, the current is:

During discharge through an external load, the terminal voltage V across the battery terminals is reduced from its open-circuit EMF by the internal voltage drop Ir:

The terminal voltage must also identically equal the potential difference across the external 4 Ω resistor:

The acceleration due to gravity on the surface of a spherical celestial body of mass M and radius R is given by Newton’s law of gravitation:

Writing the ratio of the planet’s surface gravity g_p to Earth’s surface gravity g_e:

Given that M_p = M_e / 10 and diameter is halved (meaning radius R_p = R_e / 2):

Multiplying by standard Earth surface gravity g_e = 9.8 m s⁻²:

The magnetic needle performs 20 complete oscillations in 5 seconds. The time period T for one complete cycle is:

In a uniform horizontal magnetic field B, a magnetic needle of dipole moment M and moment of inertia I executes angular simple harmonic motion with time period:

Isolating M from the time period equation:

Substitute I = 9.8 × 10⁻⁶ kg m², B = 0.049 T, and T = 1/4 s:

Simplifying the numerical coefficients:

Comparing with M = x × 10⁻⁵ A m², we get x = 1280π², which corresponds to Option (2).

According to Planck’s quantum hypothesis as formulated by Einstein, electromagnetic radiation is emitted and absorbed in discrete packets called quanta or photons. The energy of a photon of frequency ν is given by:

In free space (vacuum), all photons, irrespective of their frequency or wavelength, travel at the universal constant speed of light c = 3 × 10⁸ m/s:

From relativistic kinematics and the de Broglie relation, the momentum p of a photon is related to its energy E by p = E/c:

In particle-like collisions such as photon-electron collisions (e.g. Compton scattering), the interaction is perfectly elastic, meaning both total energy and total momentum are conserved:

Photons are electrically neutral. They are not deflected by electric or magnetic fields and possess zero rest mass and zero electrical charge:

Up to the elastic limit, deformation is completely reversible, and Hooke’s law holds rigorously. Young’s modulus Y is defined as the ratio of tensile stress σ to tensile strain ε:

To find the maximum elongation ΔL_max without exceeding the elastic limit, we use the maximum allowable stress σ_max:

Substituting σ_max = 8 × 10⁸ N m⁻², L = 1 m, and Y = 2 × 10¹¹ N m⁻²:

Since 1 m = 1000 mm:

Examine the given truth table row by row: - Row 1: A = 0, B = 0 -> Y = 1 - Row 2: A = 0, B = 1 -> Y = 0 - Row 3: A = 1, B = 0 -> Y = 1 - Row 4: A = 1, B = 1 -> Y = 0

Writing the canonical minterms for the rows where the output Y is 1 (Row 1 and Row 3):

Factor out the common term B̄ using the distributive law:

For a uniform thin straight rod of mass M and length L, the moment of inertia about an axis perpendicular to the rod and passing through its center of mass (midpoint) is given by integration as:

Substituting the given values I = 2400 g cm² and M = 400 g directly into the formula (in CGS units):

Simplifying the equation:

Taking the square root of 72:

The Vernier Constant (also known as the Least Count) of a Vernier caliper is the difference between the length of one main scale division (MSD) and one vernier scale division (VSD):

We are given that (N + 1) divisions of the vernier scale coincide with N divisions of the main scale:

Substituting 1 VSD into the definition of VC:

We are given 1 MSD = 0.1 mm. Expressing this in centimeters:

Substituting 1 MSD = 1/100 cm into the VC equation:

For transitions in the hydrogen atom ending at n1 = 2 (the Balmer series in the visible spectrum), the wavelength λ is given by the Rydberg formula:

The transition from n2 = 3 to n1 = 2 has the smallest energy difference:

The transition from n2 = 4 to n1 = 2 has intermediate energy:

The transition from n2 = 5 to n1 = 2:

The highest energy transition in the set from n2 = 6 to n1 = 2 produces the shortest wavelength:

In Young’s double slit experiment, the path difference between light waves emerging from slits S1 and S2 arriving at the central point O on the perpendicular bisector is identically zero:

Since Δx = 0 satisfies the condition for constructive interference (Δx = nλ with n = 0) for every wavelength in the visible spectrum simultaneously, all color components (VIBGYOR) constructively combine with maximum intensity. Their additive combination yields a bright white central fringe.

For points away from the center (Δx ≠ 0), the fringe width β depends directly on wavelength:

Because λ_violet ≈ 400 nm is smaller than λ_red ≈ 700 nm, the violet bright fringes form closest to the central white fringe, with red forming farther out. After a few colored fringes, the overlapping of different bright and dark bands of different colors produces uniform white illumination.

For a non-relativistic free particle of mass m moving with momentum p and kinetic energy E, the momentum is expressed in terms of kinetic energy as p = √(2mE). The de Broglie wavelength is:

Squaring both sides of the relation:

Taking the reciprocal gives the function plotted on the vertical axis (1/λ²) in terms of the horizontal axis variable E:

Since 2m/h² is a positive constant, the relation has the mathematical form y = k · x with zero intercept (c = 0). The corresponding graph is a straight line passing through the origin with positive slope.

Magnetic susceptibility χ is a dimensionless proportionality constant that indicates the degree of magnetization M of a material in response to an applied magnetic field H:

Diamagnetic materials develop magnetization opposite to the applied field. Their susceptibility is negative and small, with superconductor limit χ = -1 (ideal diamagnetism):

Ferromagnetic materials possess spontaneous domain alignment and magnetize strongly in the direction of the applied field. Their susceptibility is positive and very large (typically 1000 to 100,000):

Paramagnetic materials possess permanent atomic dipole moments that weakly align with the applied field. Their susceptibility is positive, small, and temperature-dependent:

A completely non-magnetic substance (or vacuum) cannot be magnetized by an applied magnetic field, so its magnetization M = 0 and susceptibility is identically zero:

When an unpolarised light beam strikes a transparent refracting medium at Brewster’s angle i_B, the angle of incidence satisfies:

By Snell’s law, sin i_B / sin r = μ = sin i_B / cos i_B, which implies cos i_B = sin r = cos(90° - r). Therefore:

The oscillating electric dipoles in the medium produce the reflected wave. Because dipoles cannot radiate electromagnetic energy along their oscillation axes, the electric field oscillations parallel to the plane of incidence are completely absent in the reflected ray. Thus, the reflected light is completely (100%) linearly polarised with electric vectors perpendicular to the plane of incidence (dots).

The refracted ray transmits into the medium. Since only a fraction of the perpendicular component is reflected, the refracted beam contains the entire parallel component and the remaining fraction of the perpendicular component. It is therefore partially polarised.

Velocity is a vector quantity having both magnitude and direction. In uniform circular motion of radius R and angular velocity ω, the magnitude of velocity (speed) is constant, but its direction is along the instantaneous tangent to the circle:

Because the unit vector direction changes at every instant of time, the velocity vector is NOT constant:

Differentiating the velocity vector with respect to time yields the centripetal acceleration vector:

While the magnitude of centripetal acceleration is constant (|a| = v²/R = ω²R), its direction always points radially towards the instantaneous position of the center, which rotates in space. Because its direction changes continuously:

Solid angle Ω subtended by an area element dA at distance r is defined as the ratio of area to the square of distance:

Strain ε is defined as fractional change in dimension (e.g. change in length per unit original length):

Plane angle θ is defined as the ratio of arc length to radius:

Evaluating the other quantities present in distractors: - Arc: [L] - Stress: [Force / Area] = [M L⁻¹ T⁻²] - Angular speed: [Angle / Time] = [T⁻¹] Only the pair "strain and angle" are both dimensionless [M⁰L⁰T⁰].

For a mass m whirled in a horizontal plane of radius r on a frictionless surface, the tension T in the string provides the necessary centripetal force directed toward the center of the circle:

When the speed is increased to ω_new = 2ω while keeping the radius r unchanged, the new tension T_new becomes:

Expanding the squared term:

The general displacement equation of a particle executing simple harmonic motion (SHM) along the x-axis is given by:

Comparing the given equation x = 5 sin(πt + π/3) m with the standard form: - Amplitude A = 5 m - Angular frequency ω = π rad/s - Initial phase angle φ = π/3 rad

The time period T for one complete oscillation is related to the angular frequency ω by:

In any quasi-static thermodynamic process, the mechanical work done by a gas during transition from an initial state to a final state is represented by the area under the process curve on a P-V diagram:

From the given P-V diagram: - Volume at state b: V_b = 400 cm³ - Volume at state c: V_c = 400 cm³ The process bc is represented by a vertical straight line perpendicular to the volume axis. The volume remains strictly constant throughout the entire transformation.

Substituting dV = 0 into the work integral:

An isolated neutral atom of atomic number Z contains exactly Z positively charged protons in its nucleus (charge +Ze) and Z negatively charged extranuclear electrons (charge -Ze). The algebraic sum of charges is zero:

In the opening section of NCERT Physics, the textbook introduces atomic models by noting that ordinary matter is stable and that atoms of each chemical element emit a unique, characteristic line emission spectrum when energized. Under this textbook definition, Statement II is treated as correct, leading to Option (3): "Both Statement I and Statement II are correct".

Following the examination, subject matter experts and candidates formally challenged Statement II on two solid physical grounds: 1. Not all atoms are stable: elements with Z > 82 (e.g., Uranium, Polonium, Radium) as well as radioactive isotopes of lighter elements (e.g., Carbon-14, Tritium) undergo spontaneous nuclear decay and are inherently unstable. 2. Atoms in their ground state are stable but do NOT emit radiation; emission occurs only when atoms transition downward from excited states. Under this scientific rigor, Statement II is incorrect, leading to Option (1): "Statement I is correct but Statement II is incorrect".

After expert committee review, the National Testing Agency (NTA) officially upheld both perspectives and awarded full credit for BOTH Option (1) and Option (3) in the NEET 2024 Final Answer Key.

In an ideal transformer with negligible winding resistance and zero magnetic flux leakage, the same alternating magnetic flux Φ passes through each turn of both the primary and secondary coils. By Faraday’s law of electromagnetic induction:

Dividing the secondary voltage equation by the primary voltage equation gives the fundamental voltage transformation ratio:

We are given that N_p / N_s = 1 / 2. Inverting gives N_s / N_p = 2 / 1:

Before the collision, body A of mass m moves with velocity v_1 while body B of identical mass m is at rest. The total initial linear momentum of the system is:

In a completely inelastic one-dimensional collision, coefficient of restitution e = 0. The two bodies coalesce into a single combined mass of 2m moving together with common velocity v_2:

In the absence of external impulsive forces, linear momentum is conserved:

Dividing both sides by m v_2 gives the ratio v_1 : v_2:

An alpha particle is a helium-4 nucleus (⁴₂He). Emission of an alpha particle reduces the mass number by 4 and atomic number by 2:

In positron emission, a proton inside the nucleus transforms into a neutron, emitting a positron (⁰₊₁e) and a neutrino. The mass number remains unchanged, while the atomic number decreases by 1:

In beta-minus emission, a neutron transforms into a proton, emitting an electron (⁰₋₁e) and an antineutrino. The mass number is unchanged, while the atomic number increases by 1:

The final emission is an electron (e⁻, identical to β⁻). The mass number remains unchanged, and the atomic number increases by another 1:

The resulting product nucleus Q has mass number A = 286 and atomic number Z = 81.

The electrostatic potential due to a short electric dipole of dipole moment P at a distance r along its axial line is given by:

Substitute 1/(4πε₀) = 9 × 10⁹ N m²/C², P = 4 × 10⁻⁶ C m, and r = 2 m:

The calculated potential matches ±9 × 10³ V exactly. Therefore, Assertion A is completely TRUE.

Reason R claims that the potential formula is V = ± (2P) / (4πε₀ r²). This is scientifically incorrect. The factor of 2 appears in the axial electric field formula E = 2P / (4πε₀ r³), but NOT in the potential formula V = P / (4πε₀ r²). Therefore, Reason R is FALSE.

Consider a Gaussian spherical surface of radius r < R concentric with the shell. Because all charge q resides on the outer surface of the thin shell, the charge enclosed within the Gaussian sphere is zero. By Gauss’s law:

The electric field is the negative gradient of the electrostatic potential:

Because potential is constant everywhere from the center up to the surface, the potential at the center C is identically equal to the potential at point P on the surface:

The potential difference between points C and P is therefore identically zero: