NEET 2026 Question Paper Step-by-Step Solutions

At any intermediate position x along the rough track, the normal reaction on the slab is N = M*g. The instantaneous kinetic friction opposing motion is given by:

• • Because the slab moves slowly at constant speed, external pulling force matches friction at every instant.
• • Infinitesimal work required to advance by dx is dW = f_k(x) dx.

Set up the definite integral from the entry point x = 0 to the exit point x = L:

• • Pull out constant scalar factors (mu_0, M, g, L) outside the integral sign.
• • Evaluate integral of x dx: [x^2 / 2] evaluated from 0 to L.

Substitute upper limit L and lower limit 0 into the algebraic expression:

• • The factor L in the denominator cancels with one factor of L in L^2.
• • Yields the final exact work value: (1/2) * mu_0 * M * g * L.

For an ideal gas with spherical molecules of diameter d and number density n, the mean free path lambda is given by:

• • The sqrt(2) factor accounts for the relative motion of colliding molecules.
• • pi * d^2 represents the effective spherical collision cross-section.

Rearranging the formula to isolate the number density n:

• • Number density is inversely proportional to the mean free path.
• • Number density is inversely proportional to the square of molecular diameter.

Calculate the ratio of number densities for Gas A relative to Gas B:

• • lambda_B / lambda_A = 2 (since lambda_A is half of lambda_B).
• • (d_B / d_A)^2 = (1/2)^2 = 1/4 (since d_A is twice d_B).
• • Multiplying gives n_A / n_B = 1/2 ==> n_A = (1/2) * n_B.

Carefully observe the diagram for P: two identical inductors are in series, but terminals a and b are located at the top and bottom of the FIRST inductor only.

• • Because the inductors are in series, the same current I(t) flows through both.
• • The induced EMF E_P measured strictly across a and b is the EMF of that single inductor: E_P = L * (dI/dt).

In configuration Q, the two identical inductors are connected in parallel between terminal nodes a and b.

• • Total current I(t) divides equally between two identical branches: I_1 = I_2 = I(t) / 2.
• • Induced EMF across terminals a and b: E_Q = L * (d(I/2)/dt) = (L/2) * (dI/dt).

Divide the magnitude of E_P by E_Q:

• • The factor L * (dI/dt) cancels cleanly from numerator and denominator.
• • Leaves the exact dimensionless ratio: 2.

In an open pipe of length L, both ends must be displacement antinodes (A). Standing wave modes require L = p * (lambda / 2), where p is the harmonic number.

• • Fundamental (1st harmonic, p=1): A - N - A ==> 1 node.
• • 2nd harmonic (p=2): A - N - A - N - A ==> 2 nodes.
• • 5th harmonic (p=5): exactly 5 displacement nodes.

In a closed pipe of length L, the closed end is a displacement node (N) and the open end is an antinode (A). Only odd harmonics exist: f_k = (2k - 1) * (v / 4L).

• • 1st harmonic (k=1): N - A ==> 1 node.
• • 3rd harmonic (k=2): N - A - N - A ==> 2 nodes.
• • 5th harmonic (k=3): 3 nodes.
• • 7th harmonic (k=4): 4 nodes.
• • 9th harmonic (k=5): exactly 5 displacement nodes (m = 5).

Substitute n = 5 and m = 5 into the ratio:

• • Both configurations have exactly 5 displacement nodes.
• • The ratio is identically 1, corresponding to Option (3).

For a long solenoid with n turns per unit length carrying current I, the uniform interior magnetic field is:

• • End effects are neglected for a long solenoid (l >> r).
• • Field lines are axial and uniform throughout the interior cross-section.

Cross-sectional area of one turn is A = pi * r^2. Total number of turns in length l is N = n * l. Total magnetic flux linkage Phi is:

• • Each turn links flux Phi_1 = B * A = mu_0 * n * I * pi * r^2.
• • Multiplying by total turns N = n * l gives total flux linkage.

By definition of self-inductance L = Phi / I:

• • Current I cancels completely, leaving self-inductance as a purely geometric property.
• • Matches Option (1) exactly.

Differentiate position function s(t) with respect to time to find instantaneous velocity:

• • Set v(t) = 0 to determine if the particle changes direction during the interval [0, 6] s:
• • 2t - 6 = 0 ==> t = 3 s.
• • Since t = 3 s lies strictly within [0, 6] s, the particle reverses direction!

Evaluate position at t = 0, t = 3, and t = 6 s:

• • Segment 1 (t = 0 to 3 s): Distance d_1 = |s(3) - s(0)| = |-4 - 5| = 9 m.
• • Segment 2 (t = 3 to 6 s): Distance d_2 = |s(6) - s(3)| = |5 - (-4)| = 9 m.
• • Total distance D = d_1 + d_2 = 9 + 9 = 18 m.

Divide total distance by total time elapsed:

• • Matches Option (3) exactly.
• • Notice: Average velocity = Delta s / Delta t = (5 - 5) / 6 = 0 m/s.

Sum the masses of the thorium daughter nucleus and the alpha particle:

• • Carry all decimal places carefully: 234.043 + 4.003 = 238.046 u.

Subtract product mass from the parent uranium mass:

• • Positive mass defect confirms the alpha decay is energetically spontaneous (exothermic).

Multiply mass defect by energy equivalent per atomic mass unit:

• • Convert MeV to keV: 1 MeV = 1000 keV.
• • Q = 3.726 * 1000 keV = 3726 keV.
• • Matches Option (1) exactly.

The number of incident photons per second is given by n = P(t) / (h*nu). Since each absorbed photon can eject at most one photoelectron (with quantum efficiency eta):

• • Because power P(t) decreases linearly with time (P(t) = P_0 - k*t), photocurrent I must also decrease linearly with time.
• • The graph of I vs t is a straight line sloping downward to zero.

Apply Einstein's photoelectric equation for maximum kinetic energy of emitted electrons:

• • Stopping potential depends solely on the frequency nu of incident radiation and the work function Phi.
• • Stopping potential is completely INDEPENDENT of light intensity and source power.
• • Therefore, |V| vs t is a horizontal straight line with zero slope.

Inspect the four graph sets in the question crop:

• • Matches Set (1) perfectly, corresponding to Option (1).

Capacitors P and Q are connected in series across the battery voltage V. Since they have identical capacitance C, the battery voltage divides equally across them:

• • Capacitor Q also stores exactly U_Q = (1/8) * C * V^2.

Capacitor S is connected in parallel directly across the battery terminals, experiencing the full potential difference V:

• • Total energy stored in the entire system is the sum of energies in all three capacitors:
• • U_T = U_P + U_Q + U_S = \frac{1}{8}CV^2 + \frac{1}{8}CV^2 + \frac{4}{8}CV^2 = \frac{6}{8}CV^2 = \frac{3}{4}CV^2.

Divide the energy stored in P by the total system energy:

• • The factor C * V^2 cancels cleanly.
• • Leaves the exact fraction 1/6, matching Option (4).

For an incompressible fluid in steady streamline flow, volume flow rate Q is constant:

• • Because the cross-section at Q is halved, water flows twice as fast through the throat.

Since the pipe is horizontal, potential energy terms cancel (h_P = h_Q):

• • Substitute v_Q = 2*v_P: v_Q^2 - v_P^2 = (2*v_P)^2 - v_P^2 = 4*v_P^2 - v_P^2 = 3*v_P^2.
• • P_P - P_Q = \frac{1}{2} (1000) (3 v_P^2) = 1500 v_P^2.

Substitute the given pressure difference Delta P = 15 N/m^2:

• • Calculate rate of flow: Q = A_P * v_P = (40 cm^2) * (10 cm/s) = 400 cm^3/s.
• • Cross-check at throat Q: Q = A_Q * v_Q = (20 cm^2) * (20 cm/s) = 400 cm^3/s.
• • Matches Option (4) exactly.

Semicircular segments ABC and ADC are connected in parallel between nodes A and C:

• • Given R_{ABC} = \frac{1}{2} R_{ADC} \implies I_{ABC} = 2 I_{ADC}.
• • Kirchhoff's Current Law at node A: I_{ABC} + I_{ADC} = I_0 \implies 2 I_{ADC} + I_{ADC} = 3 I_{ADC} = I_0.
• • Therefore: I_{ADC} = \frac{1}{3} I_0, \quad I_{ABC} = \frac{2}{3} I_0.

Use the Biot-Savart law for a circular arc spanning angle pi:

• • Branch ABC (current flows A -> B -> C, clockwise): Magnetic field B_{ABC} points perpendicularly INTO the page (\otimes): B_{ABC} = \frac{\mu_0 (2/3 I_0)}{4r} = \frac{2\mu_0 I_0}{12r}.
• • Branch ADC (current flows A -> D -> C, counter-clockwise): Magnetic field B_{ADC} points perpendicularly OUT OF the page (\odot): B_{ADC} = \frac{\mu_0 (1/3 I_0)}{4r} = \frac{\mu_0 I_0}{12r}.

Subtract the smaller field from the larger field to obtain net magnitude:

• • The net magnetic field points into the page with magnitude mu_0 * I_0 / (12r).
• • Matches Option (1) exactly.

When a sphere reaches terminal velocity v_t in a viscous medium, net downward acceleration is zero:

• • Weight of sphere: W = sigma * V * g = sigma * (4/3 pi r^3) * g.
• • Upthrust (Buoyant force): F_B = rho * V * g = rho * (4/3 pi r^3) * g.
• • Stokes' viscous drag force: F_v = 6 pi eta r v_t.

Equate net gravitational force to viscous drag and solve for v_t:

• • Let x = sigma / rho and constant k = (2 r^2 g rho) / (9 eta).
• • Then v_t = k * (x - 1).
• • This is a straight line equation of the form y = m(x - x_0) with x-intercept x_0 = 1.

Evaluate physical behavior across different density regimes:

• • If sigma < rho, the ball floats upward (or remains at rest on the surface); falling terminal velocity is 0.
• • At sigma / rho = 1, downward terminal velocity is exactly 0.
• • For sigma / rho > 1, terminal velocity increases linearly with slope k.
• • Matches Graph (1) perfectly, corresponding to Option (1).

Escape velocity from the surface of a spherical celestial body of mass M and radius R is given by conservation of mechanical energy:

• • G is the universal gravitational constant.
• • M is the planet mass, and R is the planetary surface radius.

Since both planets have identical mass M_1 = M_2 = M:

• • Constants 2, G, and M cancel out completely.

Substitute R_2 = R_1 / 2 into the square root expression:

• • Confirms v_2 / v_1 = sqrt(2).
• • Matches Option (3) exactly.

For a planet of mass m orbiting a star of mass M at radius R, gravitational attraction provides the necessary centripetal force:

• • Mass of the planet m cancels from both sides.
• • G is the universal gravitational constant, and M is the solar mass.

Substitute omega = 2pi / T into the dynamic balance:

• • Cross-multiply to isolate T^2: T^2 = \left(\frac{4\pi^2}{G M}\right) R^3.
• • This is Kepler's Third Law: T^2 \propto R^3.

Take the square root of both sides to obtain the explicit power of R:

• • The orbital period scales as R^(3/2) (or R^1.5).
• • Matches Option (2) exactly.

According to Ampere's force law, two parallel conductors carrying currents in the same direction attract each other. The magnetic force per unit length exerted on wire A by fixed wire B is:

• • Because wire B is situated vertically above wire A, this attractive magnetic force acts vertically UPWARDS on wire A.

Two vertical forces act on wire A resting on the insulated floor:

• • N is the normal reaction from the floor per unit length.
• • Downwards gravitational force per unit length is lambda * g.

For wire A not to rise from the floor, normal reaction must be non-negative (N >= 0):

• • The minimum permissible height is h_{min} = \frac{\mu_0 I^2}{\pi \lambda g}.
• • Matches Option (2) exactly.

Compare the magnitude of the applied reverse bias voltage |V_i| with the Zener breakdown rating V_Z:

• • Because the input reverse voltage exceeds breakdown, avalanche/Zener breakdown occurs.
• • The ideal Zener diode clamps the potential difference between terminals C and B to exactly V_Z = 3 V.

The complete circuit loop consists of the DC voltage source, the Zener diode, and series resistor R:

• • V_R is the voltage difference across resistor R, which is connected between terminals B and A.
• • V_R = |V_B - V_A|.

Isolate the resistor voltage:

• • Matches Option (2) exactly.

In an adiabatic process, there is no heat exchange with the surroundings (Q = 0):

• • Internal energy change of an ideal gas depends solely on temperature: Delta U = n * C_v * (T_f - T_i).

For an ideal monatomic gas with 3 translational degrees of freedom:

• • Given gamma = 5/3, C_v = R / (gamma - 1) = R / (5/3 - 1) = (3/2) R.

Substitute known values into the work equation:

• • W = \frac{3}{2} \times 8.3 \times 10 = 15 \times 8.3 = 124.5\text{ J}.
• • Matches Option (3) exactly.

For photoelectric emission to occur, incident photon energy must equal or exceed the work function:

• • For Cell 1, we are given lambda_1 < lambda, which means lambda > lambda_1.
• • The incident wavelength exceeds threshold wavelength; photon energy is strictly below the work function.
• • No photoelectrons are ejected, so the stopping potential is identically zero: V_1 = 0.

For Cells 2 and 3, lambda < lambda_2 and lambda << lambda_3, so photoelectric emission takes place in both cells:

• • eV_2 = hc (1/lambda - 1/lambda_2).
• • eV_3 = hc (1/lambda - 1/lambda_3).
• • Given lambda_3 >> lambda_2 > lambda, we have 1/lambda_3 < 1/lambda_2.
• • Subtracting a smaller number yields a larger difference: (1/lambda - 1/lambda_3) > (1/lambda - 1/lambda_2).

Compare the stopping potential magnitudes:

• • Combining with V_1 = 0 yields: V_1 = 0, V_2 < V_3.
• • Matches Option (1) exactly.

For a massless photon of energy E traveling at speed c:

• • Given E = 20 eV = 20 * 1.6 * 10^-19 J = 3.2 * 10^-18 J.
• • p_{Ph} = (3.2 * 10^-18 J) / (3 * 10^8 m/s) = (3.2 / 3) * 10^-26 kg m/s.

For an electron with kinetic energy E = 20 eV, non-relativistic momentum is:

• • Inside square root: 2 * 9 * 3.2 * 10^-49 = 57.6 * 10^-49 = 5.76 * 10^-48.
• • Taking square root: sqrt(5.76 * 10^-48) = 2.4 * 10^-24 kg m/s.

Divide electron momentum by photon momentum:

• • Confirms the exact integer ratio 225.
• • Matches Option (3) exactly.

In an optical bench experiment, uprights supporting the lens and object/image pins slide along a graduated wooden/metallic bed:

• • Because the optical center of the thick lens and the sharp tips of the needles do not lie exactly above the reference fiducial marks on the upright bases, observed scale readings differ from true physical separation.
• • This systematic discrepancy is determined using a thin knitting needle of known length and is called the Index Correction.

Examine the systematic corrections required in the other three experiments:

• • Meter bridge: Resistance of end copper strips and solder joints requires "End Correction".
• • Resonance tube: Displacement antinode lies at a distance e = 0.6 * r outside the open tube end; this is called "End Correction".
• • Simple pendulum: Uses vernier callipers to measure bob radius; no index correction is involved.

Confirm the experimental pairing:

• • Matches Option (3) exactly.

By Gauss's Law, outside a uniformly charged dielectric sphere of radius R and charge density rho, the electric field and potential are identical to a point charge Q at the origin:

• • Because both point A (r = 2R) and point B (r = 3R) lie outside the sphere, this potential formula applies directly to both endpoints.

Substitute r_A = 2R and r_B = 3R into the exterior potential formula:

• • Common denominator between 6 and 9 is 18.

Because the electrostatic field is conservative, work done on unit positive charge is given by the potential difference:

• • Comparing with the given expression |W| = \frac{\rho R^2}{n \epsilon_0}:
• • n = 18.
• • Matches Option (4) exactly.

The slab of medium Q has a thickness of 5 cm. Since object O is located at its center:

• • d_1 is the real depth from the upper P-Q interface.
• • d_2 is the real depth from the lower Q-R interface.

Observer is in medium P (n_P = 1), looking at object O in medium Q (n_Q = 1.25):

• • The object appears shifted closer to the interface by 0.5 cm.

Observer is in medium R (n_R = 1.5), looking at object O in medium Q (n_Q = 1.25):

• • Because n_R > n_Q, the object appears further from the interface.

Subtract the two apparent depths:

• • Matches Option (2) exactly.

Both particles move along the unit circle with speed v = omega * R = omega (since R = 1). Let angle theta be measured from the positive x-axis:

• • Particle 1 moves counter-clockwise along the left semicircle.
• • Particle 2 moves clockwise along the right semicircle.

Sum the individual momenta (with unit mass m = 1):

• • Horizontal momentum components cancel out completely at every angle due to bilateral symmetry.
• • Magnitude of total momentum is P(\theta) = 2v |\cos\theta|.

Evaluate P(theta) as theta progresses from pi/2 down to -pi/2:

• • The graph is a smooth positive cosine arch starting at zero, peaking at 2v, and terminating at zero.
• • Matches Figure (3) perfectly.

For an isotropic metallic sphere of initial radius R:

• • gamma is the coefficient of cubical (volume) expansion, which equals 3 times the linear coefficient alpha.

The increase in volume for a small temperature rise Delta T is given by thermal expansion theory:

• • Valid for small temperature increments (Delta T << 1/alpha).

Substitute V = (4/3) * pi * R^3 into the expansion formula:

• • The factor of 3 in the numerator cancels the factor of 3 in the denominator.
• • Leaves the exact result: 4 * pi * R^3 * alpha * Delta T.
• • Matches Option (3) exactly.

Consider a cylinder of mass m and cross-sectional area A floating in equilibrium in a liquid of density rho. When depressed downward by distance y from equilibrium, additional buoyant force provides a restoring force:

• • Because restoring force is directly proportional to displacement y and directed toward equilibrium, the motion is Simple Harmonic Motion (SHM).
• • Effective restoring force constant: k = A * rho * g.

Apply the standard harmonic oscillator time period formula:

• • Mass m and area A are constant since the same cork is transferred between liquids.
• • Squaring gives: T^2 \propto \frac{1}{\rho} \implies \rho \propto \frac{1}{T^2}.

Substitute T_1 = T and T_2 = 2T into the inverse square relationship:

• • Confirms rho_2 / rho_1 = 1/4.
• • Matches Option (4) exactly.

For a standard Vernier calliper where 10 Vernier scale divisions equal 9 main scale divisions:

• • 1 MSD = 1 mm.
• • LC = 0.1 mm = 0.01 cm.

When the Vernier zero lies to the LEFT of the main scale zero, the error is negative. By NCERT convention:

• • N = 10 (total divisions on Vernier scale).
• • n = 4 (coinciding division).
• • True zero error is -0.06 cm.

Actual length equals observed reading minus zero error:

• • However, the options printed by NTA were: (1) 0.60 cm, (2) 0.96 cm, (3) 1.00 cm, (4) 1.04 cm.
• • The correct scientific answer 1.06 cm was missing because the question setter erroneously calculated error as -4 * LC = -0.04 cm (giving Option 4) or subtracted it (giving Option 2).
• • During answer key challenge review, NTA accepted the flaw and awarded FULL MARKS to all candidates (Key: A,B,C,D).

Apply the parallel axis theorem to find total moment of inertia I_A:

• • Sphere A rotates about its diametrical centroidal axis: I_{A,cm} = (2/5) M R^2.
• • Sphere B's center is at distance d = R + r from axis A: I_{B,about A} = (2/5) m r^2 + m (R + r)^2.

Similarly, apply the parallel axis theorem about vertical axis B:

• • Sphere B rotates about its diametrical centroidal axis: I_{B,cm} = (2/5) m r^2.
• • Sphere A's center is at distance d = R + r from axis B: I_{A,about B} = (2/5) M R^2 + M (R + r)^2.

Subtract equation 2 from equation 1:

• • The centroidal terms I_{A,cm} and I_{B,cm} cancel out identically.
• • I_A - I_B = m(R + r)^2 - M(R + r)^2 = (m - M)(R + r)^2.
• • Matches Option (2) exactly.

For an undamped spring-mass oscillator of mass m and spring constant k, total mechanical energy E is conserved:

• • x is the extension/compression from equilibrium in meters.
• • v is the instantaneous velocity in m/s.

Divide both sides of the energy equation by total energy E:

• • This represents an ellipse on the x - v plane with horizontal semi-axis a = \sqrt{2E/k} and vertical semi-axis b = \sqrt{2E/m}.

An ellipse becomes a circle if and only if both semi-axes are equal in magnitude (a = b):

• • Equivalently, angular frequency omega = sqrt(k/m) = 1 rad/s.
• • Matches Option (2) exactly.

For the convex lens L_1, focal length f_1 = +10 cm and object distance u_1 = -30 cm:

• • v_1 = +15 cm.
• • The real image I_1 is formed 15 cm to the right of lens L_1.

The second lens L_2 is located 3 cm to the right of L_1. The image I_1 acts as a virtual object for L_2:

• • Because the rays are converging toward a point 12 cm behind L_2, object distance u_2 is positive: u_2 = +12 cm.

For the diverging concave lens L_2, focal length is f_2 = -10 cm:

• • v_2 = -60 cm.
• • The negative sign indicates the final image is formed 60 cm to the left of the concave lens L_2.
• • Matches Option (2) exactly.