NEET 2025 Question Paper Step-by-Step Solutions

Because the tank is very wide in the z-direction, the liquid surface is cylindrical (curvature in z-direction is zero). The Laplace excess pressure across a surface with radius of curvature R is governed by surface tension S.

• • For a surface profile y(x) with small slope (dy/dx ≪ 1), the curvature simplifies to 1/R ≈ d²y/dx².

At any horizontal position x, the atmospheric pressure acts above the surface. At the flat baseline level (y = 0), the pressure is atmospheric. The hydrostatic pressure difference between the liquid at height y(x) and the flat surface is ρgy.

• • Equilibrium requires excess Laplace pressure to balance the gravitational head.

Equating the Laplace pressure to the hydrostatic pressure head yields the second-order linear differential equation governing the meniscus profile.

• • This is the classical capillary rise meniscus differential equation with characteristic capillary length λ_c = √(S/ρg).

For a compound microscope with tube length L much larger than the focal lengths of the lenses, the total magnifying power m for an image formed at infinity (normal adjustment, minimum eye strain) is the product of the magnifications of objective and eyepiece.

• • L is the distance between second focal point of objective and first focal point of eyepiece (tube length).

Substitute the given optical parameters into the formula:

• • Matches Option (2) with 100% precision.

For the electron to travel without any deflection, the net Lorentz force acting on it must be zero:

• • Because v and B are mutually perpendicular, the vector (v × B) is perpendicular to both v and B.
• • Therefore, E must be strictly perpendicular to B.

Equating the magnitudes of electric and magnetic forces gives the electric field magnitude required:

• • Thus, E ⟂ B with magnitude 27 × 10² V/m (2700 V/m).

On a frictionless incline of angle θ, the acceleration is simply the component of gravity along the incline. On a rough incline with coefficient of kinetic friction μ_k, kinetic friction opposes motion.

• • Normal reaction N = mg cos θ; limiting kinetic friction f_k = μ_k mg cos θ.

Starting from rest over length L, the distance formula gives L = (1/2) a t², meaning t ∝ 1/√a. Taking the ratio of times:

• • Given n = 2, so n² = 4.

Substitute the acceleration expressions into the ratio and solve for μ_k:

• • Matches Option (4) exactly.

When a car of initial kinetic energy KE is brought to rest by a constant braking force F over distance s, the net work done by friction/brakes equals the loss in kinetic energy.

• • Forces are assumed constant over the stopping distances.

Set up the ratio of the braking forces for cars A and B:

• • 225 / 1500 simplifies by dividing numerator and denominator by 75: 3 / 20 = 0.15.
• • Ratio is exactly 2/3, corresponding to Option (2).

In the circuit diagram, line AF is a plain conducting wire, which implies V_A = V_F. Similarly, line CD is a plain conducting wire, meaning V_C = V_D. Redrawing the internal mesh between nodes B and E:

• • Resistor R_FE (3 Ω) is effectively connected between node A and node E.
• • Resistor R_BC (2.5 Ω) is effectively connected between node B and node D.
• • Resistor of 6 Ω is connected diagonally across node A and node D.

Examine the ratios of the arms connected between terminals B and E:

• • Because the ratios are equal, potential at node A equals potential at node D (V_A = V_D).
• • No current flows through the 6 Ω bridge resistor; it can be removed from the circuit.

With the 6 Ω resistor removed, the upper series arm (5 + 3 = 8 Ω) is in parallel with the lower series arm (2.5 + 1.5 = 4 Ω):

• • The external circuit contains series resistors: (1.5 + 5.5) = 7 Ω and (1/3) Ω.
• • Total resistance: R_total = R_BE + 7 + 1/3 = 8/3 + 1/3 + 7 = 3 + 7 = 10 Ω.
• • Total current through battery: I = V / R_total = 5 V / 10 Ω = 0.5 A.

Point P is in the upper quadrant, where the string makes an angle θ above the horizontal. The component of gravity acting radially inward towards the suspension point O is mg sin θ.

• • When the string becomes slack, tension vanishes: T = 0.
• • Therefore, mg sin θ = m v² / l ⟹ v² = g l sin θ.

Taking the bottom-most point as reference potential energy (PE = 0), the height of point P is h = l + l sin θ. Conservation of mechanical energy between the bottom and point P gives:

• • Express gl in terms of v²: gl = v² / sin θ.

Substitute gl into the energy equation:

• • Inverting to find the ratio v / v₀:
• • v / v₀ = ( sin θ / (2 + 3 sin θ) )^(1/2), matching Option (4).

The logic circuit consists of an upper NOR gate, a lower NAND gate, both receiving identical binary inputs A and B, feeding into an AND gate.

• • Y = (\overline{A + B}) \cdot (\overline{A \cdot B})

Apply De Morgan’s first and second laws to express the terms in product-of-sums / sum-of-products format:

• • Using the involution rule, A' · B' is universally equivalent to (A + B)', which is the definition of a NOR gate.

Verify all four input combinations (A, B):

• • The truth table [1, 0, 0, 0] is uniquely identical to a NOR gate.

In a vacuum / free space, the ratio of electric field amplitude to magnetic field amplitude equals the speed of light c = ω / k:

• • B₀ = E₀ / c = 60 / (3 × 10⁸) = 2 × 10⁻⁷ T.

The direction of electromagnetic wave propagation is given by the unit vector along the Poynting vector Ŝ = Ê × B̂. Here, the wave travels in the negative x-direction (-î) and E is polarized along the z-axis (k̂):

• • Because k̂ × ĵ = -î, the magnetic field must be along the positive y-axis: B_y.

The magnetic field oscillates in phase with the electric field in a plane EM wave:

• • Corresponds uniquely to Option (1).

Using kinematic equation v² = u² + 2gh for free fall under gravity:

• • After collision, the ball rises to height h₂ = 10 m:
• • v₂ = √(2 g h₂) = √(2 × 9.8 × 10) = √196 = 14 m/s (upward: +14ĵ).

Impulse imparted by the ground equals the vector change in linear momentum of the ball:

• • Direction of impulse is vertically upward, opposing downward momentum.
• • Magnitude is exactly 21 N·s.

Writing the potential step equation starting from point A and proceeding to point B:

• • Because di/dt > 0, the inductor develops an opposing back EMF with positive terminal facing A: drop = -L di/dt.
• • The battery is traversed from its positive (long) plate to negative (short) plate: drop = -E = -5 V.
• • The resistor is traversed in the direction of current flow: drop = -iR.

Substitute the given values into the potential difference expression:

• • Potential difference V_A - V_B equals 10 volts, matching Option (4).

The magnetic dipole moment M of a planar loop of area A carrying current I is defined as:

• • M is directly proportional to the enclosed area.

For two circular coils carrying the same current I = 2 A with radii r₁ and r₂:

• • Thus the ratio of their respective magnetic moments is 1:4.

For n thin lenses placed axially in contact, the equivalent focal length F satisfies 1/F = 1/f₁ + 1/f₂ + ... + 1/f_n. Since power P = 1/f:

• • Power is strictly additive for lenses in contact.

The image formed by the first lens acts as the object for the second lens, and so forth. Consequently, the total lateral magnification produced by the lens system is the product of the magnifications produced by each individual lens:

• • Thus, the combination has power 4p and magnification m⁴, corresponding to Option (3).

Pressure gauges read pressure relative to ambient atmospheric pressure. Therefore, absolute pressure is the sum of gauge pressure and atmospheric pressure:

• • Failure to add 1 atm leads to 0.156 kg (the exact trap in Option 4).

Using the ideal gas equation P₂ V = n₂ R T with V = 0.03 m³ and T = 300 K:

• • Remaining moles in cylinder: n₂ ≈ 14.544 moles.

The number of moles withdrawn from the cylinder is the difference between initial and final moles:

• • Mass withdrawn: Δm = Δn × Molar Mass = 3.656 × 32 g = 116.99 g ≈ 0.117 kg ≈ 0.116 kg.
• • Matches Option (3) with precision.

Given the implicit trajectory relation t = x² + x, differentiate both sides with respect to time t using the chain rule:

• • v can be written in power notation: v = (2x + 1)⁻¹.

Differentiate velocity with respect to position x:

• • Now substitute into the kinematic identity for acceleration: a = v (dv/dx).

Multiply velocity by its spatial derivative:

• • The negative sign indicates retardation.
• • Matches Option (2) perfectly.

In a series LCR circuit, inductive reactance leads the current while capacitive reactance lags by 90°. The net reactance is the algebraic difference:

• • Total impedance Z = √(R² + (X_L - X_C)²) = √(20² + 20²) = 20√2 Ω ≈ 28.28 Ω.

The root-mean-square current supplied by the 220 V AC source is given by Ohm’s law for AC circuits:

• • Current is inductive (voltage leads current by phase angle ϕ).

The phase angle ϕ between voltage and current is determined from the impedance triangle:

• • Thus the current is 7.8 A and the phase angle is 45°, corresponding to Option (2).

Since the Sun expands solely due to internal forces with zero external torque acting on it, its total angular momentum is strictly conserved.

• • Angular speed is related to the period of rotation by ω = 2π / T.

For a solid spherical body of uniform density and constant mass M, the moment of inertia is proportional to the square of its radius:

• • Given R₂ / R₁ = 2.

Substitute the given expansion factor and initial period:

• • Matches Option (4) exactly.

The problem defines that the magnetic flux passing through the circular orbit of the electron is quantized as Φ = n(h/e). For the ground state (lowest energy state, n = 1):

• • A = πr² is the orbital area enclosed by the electron path.

Under the magnetic Lorentz force F = e v B = m v² / r, the angular frequency of the electron is the cyclotron frequency ω = eB / m. The equivalent steady electric current loop formed by the circulating electron is:

• • T is the cyclotron orbital period.

The magnetic dipole moment μ of a planar current loop is the product of current and enclosed area:

• • Magnetic field B cancels cleanly from numerator and denominator.
• • Notice this fundamental result is directly proportional to the Bohr magneton μ_B = eh / (4πm) with factor of 2.

Using the thermal resistance formula R_th = L / (k A), let R₀ = L / (K A) denote the thermal resistance of the middle rod:

• • Total thermal resistance in series: R_total = R₁ + R₂ + R₃ = R₀/2 + R₀ + R₀/2 = 2R₀.

In steady state, the same rate of heat flow H passes through all three rods in series:

• • Heat flow rate H is constant throughout the composite rod.

Apply heat conduction across each individual section:

• • Ratio T₁ / T₂ = (5/2 T) / (3/2 T) = 5/3, corresponding to Option (3).

For a parallel plate capacitor of area A and plate separation d containing dielectric slabs in series, the capacitance is:

• • t₁ = (3/8) d, t₂ = (1/2) d = (4/8) d.
• • Air gap thickness: t_air = d - (3/8 d + 4/8 d) = (1/8) d.

The problem states that C = 2 C₀, where C₀ = ε₀ A / d:

• • Subtract 1/8 from both sides: 3/(8 K₁) + 4/(8 K₂) = 1/2 - 1/8 = 3/8.
• • Multiplying throughout by 8 gives the simplified linear form: 3/K₁ + 4/K₂ = 3.

Given K₁ = 1.25 K₂ = (5/4) K₂, which implies K₂ = (4/5) K₁:

• • Matches Option (1) with absolute precision.

Buses depart in either direction every T minutes at constant speed v_b. The spatial separation between any two successive buses traveling in the same direction is constant:

• • T is in minutes; v_b is in km/h.

For motion in the same direction, the relative speed is (v_b - 60) km/h and the time taken to cover distance d is 30 minutes. For motion in the opposite direction, relative speed is (v_b + 60) km/h and time is 10 minutes:

• • 3(v_b - 60) = v_b + 60 ⟹ 3v_b - 180 = v_b + 60 ⟹ 2v_b = 240 ⟹ v_b = 120 km/h.

Substitute v_b = 120 km/h back into either distance expression:

• • Bus speed = 120 km/h, Bus interval T = 15 min.
• • Corresponds uniquely to Option (4).

Four forces act on the uniform rod in equilibrium: (1) Normal reaction at the smooth wall N_w acting horizontally to the right. (2) Static friction force f from the rough floor acting horizontally to the left. (3) Normal reaction from the floor N_f acting vertically upward. (4) Weight of the rod W = mg = 20 × 10 = 200 N acting downward at the midpoint.

• • The wall is smooth, so there is no vertical friction at the wall.

Let the rod have length L. The angle with the vertical wall is θ = 60°. The top of the rod is at height h = L cos 60°, and the center of mass is at horizontal distance x_cm = (L/2) sin 60° from the base. Taking torque about the base point on the floor:

• • N_w = (mg / 2) tan 60°.

Substitute the given values mg = 200 N and tan 60° = √3:

• • Thus the friction force exerted by the floor on the rod is 100√3 N, matching Option (2).

For a mass M attached to an ideal spring of stiffness k, the natural angular frequency of oscillation is:

• • As sand leaks slowly out of the box, M(t) decreases monotonically over time towards the empty box mass M_box.
• • Therefore, ω(t) increases monotonically, approaching a constant upper asymptote.

When sand leaks out vertically, the escaping grains have horizontal velocity equal to the instantaneous velocity v of the box at the moment of exit. The escaping sand exerts no horizontal force on the remaining box, but carries away its share of kinetic energy:

• • The total mechanical energy E(t) of the oscillator continuously decreases.
• • Since total energy is related to amplitude by E = (1/2) k A², we have A(t) = √(2 E(t) / k).
• • Consequently, as E(t) decreases, the average amplitude A(t) decreases with time.

The correct schematic shows ω(t) starting from initial value and curving upwards towards an asymptote, while A(t) starts from initial amplitude and curves downwards:

• • Matches Option (2) unambiguously.

The excess pressure inside a spherical elastic balloon of instantaneous radius r with effective surface tension S is given by Laplace’s law:

• • Applying Bernoulli’s theorem between the high-pressure gas inside the balloon and the atmospheric exhaust outside through the small opening of area A:
• • ΔP = (1/2) ρ v² ⟹ v(r) = √(2ΔP / ρ) ∝ √(S / (ρ r)) = (S / ρ)^(1/2) r^(-1/2).
• • Comparing with v(r) ∝ r^a directly gives: a = -1/2.

By conservation of mass / volume continuity, the rate of decrease of the balloon volume equals the volume flux exiting through the aperture:

• • Rearranging to separate variables r and t:
• • -r^2 \cdot r^{1/2} dr = -r^{5/2} dr = \left(\frac{A}{4\pi}\sqrt{\frac{8S}{\rho}}\right) dt.

Integrate from the initial radius r = R at t = 0 to r = 0 at t = T:

• • Solving for total deflation time T:
• • T ∝ S^(-1/2) A^(-1) ρ^(1/2) R^(7/2).
• • Comparing with T ∝ S^α A^β ρ^γ R^δ:
• • α = -1/2, β = -1, γ = 1/2, δ = 7/2.
• • Along with a = -1/2, this corresponds uniquely to Option (3).

The least count (LC) of a vernier calliper is the difference between one main scale division and one vernier scale division:

• • LC = 1 MSD - 1 VSD = 0.1 cm - 0.09 cm = 0.01 cm.

The total observed reading before zero error correction is given by the main scale reading plus the vernier scale contribution:

• • Raw measurement on the scales reads 5.08 cm.

When the jaws are closed, the zero mark of the vernier scale lies at +0.1 cm to the right of the main scale zero. This is a positive zero error (+0.1 cm). The true reading is obtained by subtracting the zero error:

• • Matches Option (3) with absolute precision.

For circular parallel plates of radius R, the electric field between the plates is E = σ / ε₀. Since dσ/dt is constant, the time derivative of the electric field is uniform:

• • Displacement current density J_d = ε₀ (dE/dt) is uniform throughout the cylindrical region between the plates.

Construct a circular Ampèrian loop of radius r ≤ R coaxial with the plates. Applying the Ampère-Maxwell law:

• • The magnetic field grows linearly from zero on the central axis (r = 0) up to the edge r = R.

For an Ampèrian loop of radius r ≥ R, the enclosed displacement current is the full total displacement current I_d = ε₀ π R² (dE/dt):

• • Because B increases for r ≤ R and decreases for r ≥ R, it attains its global maximum at r = R (the peripheral boundary).
• • Thus B is non-zero everywhere and reaches maximum at the imaginary cylindrical surface connecting the peripheries, matching Option (3).

Brewster’s law states that when light is incident at a polarizing angle i_B, the tangent of the polarizing angle is equal to the refractive index μ of the reflecting medium:

• • By the fundamental law of reflection, angle of reflection equals angle of incidence: r' = i_B = 60°.

At Brewster’s angle, the reflected and refracted rays are mutually perpendicular (i_B + r = 90°). Electrons oscillating parallel to the refracted ray cannot radiate along their axis of oscillation towards the reflected direction.

• • Therefore, the reflected light contains only vibrations perpendicular to the plane of incidence (s-polarization) and is completely plane polarized.
• • The transmitted light contains both parallel and residual perpendicular components, making it only partially polarized.

By Coulomb’s law, the initial repulsive force between the two identical spheres A and B carrying charge q separated by distance r is:

• • Spheres are treated as point charges because radii are negligible compared to distance r.

An identical uncharged conducting sphere C (initial charge = 0) is brought into physical contact with sphere A (charge q). Due to identical capacitance and symmetry, charge redistributes equally:

• • Sphere A now retains charge q/2, and sphere C carries charge q/2.

Sphere C (charge q/2) is now brought into contact with sphere B (charge q). The total charge divides equally between the two identical spheres:

• • Sphere C is now removed.
• • The new force between A (charge q/2) and B (charge 3q/4) is:
• • F' = k (q/2)(3q/4) / r² = (3/8) [k q² / r²] = (3/8) F.
• • Corresponds uniquely to Option (4).

The two chambers are enclosed in a container made of a thermal insulator, so no heat is exchanged with the surroundings (Q = 0). The walls of the container are rigid, so no external work is done (W = 0). According to the First Law of Thermodynamics:

• • The total internal energy of the system remains conserved upon removing the partition.

For an ideal gas, internal energy is proportional to the product of pressure and volume: U = (f/2) nRT = (f/2) PV, where f is the degrees of freedom: