NEET 2026 · Code 50

Question 1

A particle of mass M moves along a horizontal x-axis from x = 0 to x = L. The coefficient of kinetic friction varies as a function of x as μ_k(x) = μ₀ − αx, with μ_k(L) = 0. If the total work done by the frictional force during the motion is nμ₀MgL (where g is the acceleration due to gravity), the v

Accepted Answer

(D) 1/2

The frictional force is f_k(x) = μ_k(x)N = (μ₀ − αx)Mg (N = Mg on a horizontal surface). From μ_k(L) = 0 we get α = μ₀/L, so μ_k(x) = μ₀(1 − x/L). Work done by friction = ∫₀ᴸ (μ₀ − αx)Mg dx = μ₀Mg[x − x²/2L]₀ᴸ = μ₀Mg(L − L/2) = ½μ₀MgL. Comparing with nμ₀MgL gives n = 1/2 — option (4).

Question 2

The mean free path of molecules in an ideal gas A is half that of another ideal gas B. The diameter of the molecules of gas A is twice the diameter of the molecules of B. If number densities of the gases A and B are n_A and n_B respectively, then the correct option is :

Accepted Answer

(D) n_A = (1/2) n_B

λ = 1/(√2 π d² n); λ_A/λ_B = (d_B² n_B)/(d_A² n_A) = 1/2 with d_A = 2d_B ⇒ n_A = n_B/2.

Question 3

Two identical inductors are connected in two different configurations P and Q, where a time varying current I(t) is flowing, as shown in the figure. The induced emf between points a and b for configuration P is E_P and that for configuration Q is E_Q. The ratio E_P/E_Q is : [Neglect the effect of mu

Accepted Answer

(D) 2

The emf induced by an inductor is |E| = L·|dI/dt|. In configuration P (series) the current I(t) flows through both coils, but points a and b are across only the first coil, so E_P = L·(dI/dt). In configuration Q (parallel) the two identical coils are in parallel between a and b, so L_eq = L/2 and E_Q = (L/2)·(dI/dt). Therefore E_P/E_Q = L/(L/2) = 2 — option (4).

Question 4

For sound waves, if the number of nodes for the 5th harmonic of an open-ended pipe is a and that for the 5th harmonic of the same pipe with one of its ends closed is m, then a/m is :

Accepted Answer

(C) 5/3

Open pipe 5th harmonic: 5 nodes (a=5). Closed pipe 5th harmonic: 3 nodes (m=3). a/m = 5/3. ⚠ Verify options/convention.

Question 5

Consider a long solenoid of length l and radius r. If n is the number of turns per unit length and μ₀ is the permeability of free space, the inductance of the solenoid is :

Accepted Answer

(A) μ₀πn²r²l

L = μ₀n²(πr²)l = μ₀πn²r²l.

Question 6

Consider a particle moving along a straight line, whose position as a function of time is given by s(t) = αt² − βt + γ, where α = 1 m s⁻², β = 6 m s⁻¹ and γ = 5 m. The average speed of the particle, in m s⁻¹, from t = 0 to t = 6 s is :

Accepted Answer

(C) 3

v = 2t − 6 = 0 at t = 3 s. Distance = 9 + 9 = 18 m; average speed = 18/6 = 3 m s⁻¹.

Question 7

Consider the following nuclear reaction : ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He. Take masses of ²³⁸U, ²³⁴Th and ⁴He as 238.050 u, 234.043 u and 4.003 u, respectively. The Q value for the reaction, in keV, is : (Given : 1 u = 931.5 MeV c⁻²)

Accepted Answer

(A) 3726

Δm = 238.050 − 234.043 − 4.003 = 0.004 u; Q = 0.004 × 931.5 = 3.726 MeV = 3726 keV.

Question 8

A beam of light falls on a metal surface such that photo-electrons are generated. If power of the light source starts to decrease linearly with time t, then variation of the photocurrent I and magnitude of the stopping potential |V| with time is best represented by :

Accepted Answer

(A) (1)

Photocurrent ∝ intensity → I falls linearly; stopping potential depends on frequency, so |V| stays constant. ⚠ Figure-based — verify the numbered graph.

Question 9

Three identical capacitors P, Q and S, each of the capacitance C, are connected to a battery of voltage V, as shown in the figure. If the energy stored in the capacitor P and total energy stored in the system are U_P and U_T, respectively, then the ratio U_P/U_T is :

Accepted Answer

(D) 1/6

P and Q are in series: 1/C_PQ = 1/C + 1/C = 2/C, so C_PQ = C/2. S is in parallel with this combination, so C_eq = C_PQ + C = C/2 + C = 3C/2. Total energy U_T = ½C_eq V² = ½(3C/2)V² = (3/4)CV². As P and Q in series share the battery voltage, V_P = V/2, so U_P = ½C V_P² = ½C(V/2)² = (1/8)CV². Therefore U_P/U_T = (1/8)/(3/4) = 1/6 — option (4).

Question 10

Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between P and Q is 15 N m⁻². The area of cross-section at P and Q are 40 cm² and 20 cm², respectively. The rate of flow of water through the pipe, in cm³ s⁻

Accepted Answer

(D) 400

v_Q = 2v_P; ΔP = (3/2)ρv_P² = 15 ⇒ v_P = 0.1 m s⁻¹; rate = 40 × 10 = 400 cm³ s⁻¹.

Question 11

A current I₀ flows through a metallic circular loop of radius r as shown in the figure. Resistance of the segment ABC is half that of ADC. Magnitude of the magnetic field at the center O of the loop is :

Accepted Answer

(A) μ₀I₀/12r

I_ABC = 2I₀/3, I_ADC = I₀/3; each semicircle gives μ₀I/4r, opposing: net = μ₀I₀/12r.

Question 12

In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used. The variation of the terminal velocity (v) with the ratio of density of spherical ball (σ) to density of the liquid (ρ), is best represented by :

Accepted Answer

(C) (3)

Weight F_W = (4/3)πR³σ·g; buoyancy F_B = (4/3)πR³ρ·g; viscous drag F_D = 6πηRv (Stokes' law). At terminal velocity F_W = F_B + F_D, giving (4/3)πR³(σ − ρ)g = 6πηRv, so v = (2R²g/9η)(σ − ρ) = (2R²gρ/9η)(σ/ρ − 1). With K = 2R²gρ/9η (a positive constant), v = K(σ/ρ) − K: a straight line of positive slope that is 0 at σ/ρ = 1 and equals −K at σ/ρ = 0. This is option (3).

Question 13

Two planets P₁ and P₂ with equal mass have radii R₁ and R₂, respectively, where R₂ = R₁/2. The escape speeds of P₁ and P₂ are v₁ and v₂, respectively. Then v₂/v₁ is :

Accepted Answer

(C) √2

v = √(2GM/R) ∝ 1/√R; v₂/v₁ = √(R₁/R₂) = √2.

Question 14

In a solar system, the time-period of revolution of a planet during a circular orbit of radius R is proportional to :

Accepted Answer

(B) R^(3/2)

Kepler's third law T² ∝ R³, so T ∝ R^(3/2).

Question 15

Two infinitely long parallel conducting wires A and B carry currents I and 2I, respectively, in the same direction. The wire A has uniform mass per unit length λ and lies on an insulated floor. The wire B is kept fixed at a height h above the floor. The minimum magnitude of h so that the wire A does

Accepted Answer

(B) μ₀I²/πλg

Wire A experiences a downward gravitational force per unit length F_g = λg. Currents in the same direction attract, so wire B (at height h) exerts an upward magnetic force per unit length F_m = μ₀ I_A I_B/(2πh) = μ₀(I)(2I)/(2πh) = μ₀I²/(πh). For wire A not to rise, F_m ≤ F_g: μ₀I²/(πh) ≤ λg ⇒ h ≥ μ₀I²/(πλg). Hence h_min = μ₀I²/(πλg) — option (2).

Question 16

An ideal Zener diode with breakdown voltage of −3 V is reverse biased with a negative input voltage V_i = −5 V. The magnitude of voltage difference between points B and A is :

Accepted Answer

(B) 2 V

The reverse-bias input magnitude |V_i| = 5 V exceeds the Zener breakdown voltage of 3 V, so the Zener is in breakdown and the voltage across it stays constant at 3 V. By KVL the input is shared between the Zener and the series resistor R: V_i = V_Zener + V_R. With points B and A across the resistor, 5 = 3 + V_BA, so V_BA = 5 − 3 = 2 V — option (2).

Question 17

In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas (γ = 5/3) decreases from 60 K to 50 K. The work done by the gas in the process is : (Take the universal gas constant as R = 8.3 J mol⁻¹ K⁻¹)

Accepted Answer

(C) 124.5 J

W = −nC_VΔT = −(3/2 × 8.3)(50 − 60) = 124.5 J.

Question 18

A ray of light with wavelength λ is incident on three different photo-electric cells namely 1, 2 and 3. The threshold wavelengths of these cells are λ₁, λ₂ and λ₃ respectively and the magnitude of stopping potentials are V₁, V₂ and V₃ respectively. The relation between λ and threshold wavelengths ar

Accepted Answer

(A) V₁ = 0, V₂ < V₃ By Einstein's photoelectric equation eV₀ = hc(1/λ − 1/λ₀), and emission needs λ ≤ λ₀ (otherwise V₀ = 0). Cell 1: λ₁ < λ means λ > λ₁, so no emission and V₁ = 0. Cells 2 and 3: λ < λ₂ and λ ≪ λ₃, so both emit. eV₂ = hc(1/λ − 1/λ₂) and eV₃ = hc(1/λ − 1/λ₃); since λ₃ > λ₂, 1/λ₃ < 1/λ₂, so (1/λ − 1/λ₃) > (1/λ − 1/λ₂), giving V₃ > V₂, i.e. V₂ < V₃. Hence V₁ = 0 and V₂ < V₃ — option (1).

Question 19

A photon and an electron, each of 20 eV energy, move in free space. The ratio of the momentum of the electron p_e to that of photon p_ph, i.e. p_e/p_ph, is : (Take speed of light = 3 × 10⁸ m s⁻¹, charge of electron = 1.6 × 10⁻¹⁹ C and mass of electron = 9 × 10⁻³¹ kg)

Accepted Answer

(C) 225

For a photon, momentum p_Ph = E/c. For a non-relativistic electron of kinetic energy E, p_e = √(2mE). Hence p_e/p_Ph = √(2mE)/(E/c) = c·√(2m/E). Substituting E = 20 eV = 20 × 1.6 × 10⁻¹⁹ J, m = 9 × 10⁻³¹ kg and c = 3 × 10⁸ m/s: the term inside the root is 2 × 9 × 10⁻³¹ / (32 × 10⁻¹⁹) = 9/16 × 10⁻¹², whose square root is 3/4 × 10⁻⁶. Therefore p_e/p_Ph = (3 × 10⁸)(3/4 × 10⁻⁶) = 9/4 × 10² = 225 — option (3).

Question 20

Which of the following measurements require 'index correction' ?

Accepted Answer

(C) Measurement of focal length of lenses using optical bench

Index correction accounts for the gap between the index pointer and the actual position of the lens/object on an optical bench, so it applies to the optical-bench focal-length measurement.

Question 21

A unit positive point charge is taken slowly through an infinitesimally thin tube that is inside a charged dielectric sphere of radius R, having uniform positive charge density ρ, as shown in the figure. The initial and final positions of the charge are marked by A and B at distances 2R and 3R respe

Accepted Answer

(D) 18

A and B are outside the sphere, so it acts as a point charge Q = ρ(4/3)πR³. |W| = (Q/4πε₀)(1/2R − 1/3R) = (ρR³/3ε₀)(1/6R) = ρR²/(18ε₀), so n = 18.

Question 22

Consider three media P, Q and R with refractive indices 1, 1.25 and 1.5, respectively. The medium Q having a thickness of 5 cm is placed between extended media P and R as shown in the figure. An object is placed at the centre of medium Q. If viewed from medium P near the normal direction, the appare

Accepted Answer

(B) 1

The object lies at the centre of the 5 cm medium Q, so its real depth from each interface is 5/2 = 2.5 cm. Apparent depth d_app = d_real × (n_observer / n_object). Viewed from P: h₁ = 2.5 × (1/1.25) = 2.5 × 4/5 = 2.0 cm. Viewed from R: h₂ = 2.5 × (1.5/1.25) = 2.5 × 6/5 = 3.0 cm. Therefore |h₁ − h₂| = |2.0 − 3.0| = 1 cm — option (2).

Question 23

A frictionless circular wire of unit radius is fixed on the horizontal plane. Two point particles of unit mass start moving simultaneously from point A (θ = π/2) with identical uniform angular speeds in opposite directions, and meet again at point B (θ = −π/2). During this time, which of the followi

Accepted Answer

(C) (3)

Both unit-mass particles start at A (θ = π/2) and move with equal angular speed ω in opposite directions. Taking θ as Particle 1's angular position, Particle 2 is at (π − θ). Their velocities are v₁ = (ω sinθ, −ω cosθ) and v₂ = (−ω sinθ, −ω cosθ). Since m = 1, the total momentum is P = v₁ + v₂ = (0, −2ω cosθ), so its magnitude is |P| = 2ω|cosθ|. This is 0 at θ = ±π/2 (points A and B), rises to a maximum of 2ω at θ = 0, and is symmetric — an inverted-U |cosθ| curve. This matches Figure (3).

Question 24

The temperature of a metallic sphere of radius R is raised by a small amount ΔT. If the linear coefficient of thermal expansion of the metal is α, the approximate increase in the volume of the sphere is :

Accepted Answer

(C) 4πR³αΔT

ΔV = γVΔT = 3α·(4/3)πR³·ΔT = 4πR³αΔT.

Question 25

A cylindrical cork of uniform density floats in a liquid of density ρ. If the cork is depressed slightly and released, it oscillates harmonically with time period T. If the same cork floats in another liquid of density ρ₀, then another oscillation has time period 2T. The value of ρ₀/ρ is :

Accepted Answer

(D) 1/4

For a floating cylinder T = 2π√(Lρ_cork/(ρ_liquid g)) ∝ 1/√ρ_liquid. So T₀/T = √(ρ/ρ₀) = 2 ⇒ ρ/ρ₀ = 4 ⇒ ρ₀/ρ = 1/4.

Question 26

One main scale division of a Vernier callipers is equal to 1 mm and the number of divisions on the Vernier scale is 10. When both the jaws touch each other, the zero of the Vernier scale shifts to the left of the zero of the main scale in such a way that 4th Vernier division coincides with a divisio

Accepted Answer

(D) 1.04 cm

Least count = 1 mm/10 = 0.1 mm. The vernier zero to the left of the main zero is a negative zero error of magnitude 4 × 0.1 = 0.4 mm = 0.04 cm, which is added: actual = 1.00 + 0.04 = 1.04 cm. ⚠ Zero-error convention can vary — verify against the official key.

Question 27

A solid sphere A of radius R and mass M is attached at a point to a smaller solid sphere B of radius r < R and mass m < M, so that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of A is I_A, an

Accepted Answer

(B) (m − M)(R + r)²

The spheres touch externally, so the centre-to-centre distance is d = R + r. About a vertical axis through A's centre, sphere A contributes (2/5)MR² and sphere B (by the parallel-axis theorem) contributes (2/5)mr² + m(R + r)²: I_A = (2/5)MR² + (2/5)mr² + m(R + r)². About a vertical axis through B's centre: I_B = (2/5)MR² + (2/5)mr² + M(R + r)². Subtracting, the central terms cancel: I_A − I_B = m(R + r)² − M(R + r)² = (m − M)(R + r)² — option (2).

Question 28

Consider a spring-mass system executing simple harmonic motion, where the spring constant is k and the mass is m. At any instant the displacement is x and the speed of the particle is v. On the v−x plane, if the graph of v as a function of x is a circle, then the correct option is :

Accepted Answer

(B) k = m

v = ω√(A² − x²) traces an ellipse on the v–x plane; it is a circle only when ω = 1, i.e. √(k/m) = 1 ⇒ k = m.

Question 29

The lens combination as shown in the figure consists of two lenses, L₁ and L₂, of focal lengths +10 cm and −10 cm, respectively. The position of the image formed is :

Accepted Answer

(B) 60 cm to the left of the concave lens

First lens L₁ (convex, f₁ = +10 cm) with object at u₁ = −30 cm: 1/v₁ = 1/f₁ + 1/u₁ = 1/10 − 1/30 = 2/30, so v₁ = +15 cm — a real image 15 cm to the right of L₁. This image acts as a virtual object for L₂ (concave, f₂ = −10 cm), which is 3 cm away, so u₂ = 15 − 3 = +12 cm. Then 1/v₂ = 1/f₂ + 1/u₂ = −1/10 + 1/12 = (−6 + 5)/60 = −1/60, giving v₂ = −60 cm. The negative sign means the final image is virtual, 60 cm to the left of the concave lens — option (2).

Question 30

An ac voltage V = 220 sin(2π × 10⁴ t) Volt is applied to a series LCR circuit. Then the current amplitude in this circuit is : (Given : L = 10 mH, C = 25 μF, R = 100 Ω)

Accepted Answer

(A) 2.2 A

At series resonance Z = R = 100 Ω, so I₀ = 220/100 = 2.2 A. ⚠ The printed L, C, ω do not actually satisfy resonance (give ≈0.35 A) — verify against the official key.

Question 31

A thin horizontal disc is rotating about a vertical axis passing through its fixed centre O. Its angular momentum is L_A and L_B computed about points A and B, respectively, with OB = 2 × OA. The value of L_A/L_B is :

Accepted Answer

(C) 1

The disc spins about its fixed centre O, so its total linear momentum is zero — for every mass element moving with velocity v there is an identical one on the opposite side moving with −v. The angular momentum about any point P is L_P = L_O + r_{O/P} × P_total; since P_total = 0, the second term vanishes, so L_P = L_O for every point. Hence L_A = L_B = L_O regardless of OA or OB, and L_A/L_B = 1 — option (3).

Question 32

A conducting loop of finite resistance lies on the x−y plane. There is a constant magnetic field in the z direction. The area of the loop varies with time as A = A₀(1 + sin t). The figure that correctly indicates the qualitative behaviour of the power P dissipated in the loop as a function of time i

Accepted Answer

(B) (2)

EMF = −B·dA/dt = −BA₀ cos t, so P = EMF²/R ∝ cos²t — identical humps touching zero, period π, never negative. ⚠ Figure-based — verify the numbered graph.

Question 33

A point charge Q is placed inside a cavity within a solid isolated conducting sphere. Consider points A, B and C as shown in the figure, where the magnitudes of the electric fields are E_A, E_B and E_C, respectively. The points B and C are at the same distance from the centre of the solid sphere. Th

Accepted Answer

(B) E_A = 0, E_B = E_C

Inside the conductor material E = 0 (E_A = 0). Outside, the induced charge on the outer surface makes the field spherically symmetric, so points at equal distance have equal fields (E_B = E_C). ⚠ Depends on the positions of A, B, C in the figure.

Question 34

Consider a fixed uniformly charged insulating sphere with radius R and total charge +Q. A point charge −q (q < Q) with mass m is released from rest at a distance of 3R from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is : (ε₀ is the permittivi

Accepted Answer

(C) √(Qq/3πε₀mR)

Only the conservative electrostatic force acts, so mechanical energy is conserved: K_i + U_i = K_f + U_f. Outside the sphere it behaves as a point charge Q at the centre. At r = 3R: V_i = Q/(4πε₀·3R), so U_i = (−q)V_i = −Qq/(12πε₀R); released from rest, K_i = 0. At the surface r = R: V_f = Q/(4πε₀R), so U_f = −Qq/(4πε₀R) and K_f = ½mv². Then 0 − Qq/(12πε₀R) = ½mv² − Qq/(4πε₀R), giving ½mv² = Qq/(4πε₀R) − Qq/(12πε₀R) = (3 − 1)Qq/(12πε₀R) = Qq/(6πε₀R). Hence v² = Qq/(3πε₀mR) and v = √(Qq/(3πε₀mR)) — option (3).

Question 35

In Geiger-Marsden's experiment, the number of scattered α-particles N(θ) is plotted as a function of the scattering angle θ. Which of the following options represents the correct plot ?

Accepted Answer

(C) (3)

By Rutherford's scattering formula, N(θ) ∝ 1/sin⁴(θ/2). At small angles sin(θ/2) is tiny, so N(θ) is very large (≈10⁷ — most α-particles pass nearly straight through). As θ increases toward 180°, sin(θ/2) → 1 and N(θ) falls to its smallest value (≈10¹ — only about 1 in 8000 are scattered through large angles). So the plot starts very high near 0°, drops sharply, then flattens out — option (3).

Question 36

Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, +α, while the other one has a negative temperature coefficient of resistance, −α, as shown in the figure. The current through these circuits are denoted by I_A and I_B. At

Accepted Answer

(A) I_A remains constant while I_B increases

In the series circuit R₀(1+αT)+R₀(1−αT)=2R₀ is constant (current constant); in the parallel circuit the resistance decreases as T rises (current increases). ⚠ Which of A/B is series vs parallel depends on the figure.

Question 37

Consider that α, k_B and b represent the Stefan-Boltzmann constant, Boltzmann constant and Wien's displacement law constant, respectively. The dimension of α·k_B³·b is :

Accepted Answer

(A) [M⁴L⁷T⁻⁹K⁻⁶]

α = [MT⁻³K⁻⁴], k_B = [ML²T⁻²K⁻¹], b = [LK]; α·k_B³·b = [M⁴L⁷T⁻⁹K⁻⁶]. ⚠ The printed options/expression were hard to read on the scan — verify against the official key.

Question 38

An electromagnetic wave travelling in a lossless dielectric medium having dielectric constant ε_r = 9 has the electric field E_x = E₀ sin(kz − 2π×10⁶ t) V m⁻¹, where E₀ is the amplitude and k is the wave vector. Among the following options, the incorrect choice is :

Accepted Answer

(B) The wavelength of the electromagnetic wave inside the medium is 300 m

The medium has ε_r = 9 and μ_r = 1. From E_x = E₀ sin(kz − 2π×10⁶ t), ω = 2π×10⁶ rad s⁻¹, so f = 10⁶ Hz. The speed inside the medium is v = c/√(ε_r μ_r) = (3×10⁸)/√9 = 10⁸ m s⁻¹ — so statement (1) is correct, and the kz − ωt phase gives propagation along +z, making statement (4) correct. The wavelength is λ = v/f = 10⁸/10⁶ = 100 m, not 300 m, so statement (2) is the incorrect one. Hence the incorrect choice is option (2). (This question asks for the wrong statement, so option (2) is the marked answer.)

Question 39

One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. The total heat supplied to the gas is :

Accepted Answer

(C) 600 J

For a complete cyclic process the internal energy returns to its starting value (ΔU = 0), so by the first law of thermodynamics the net heat equals the net work, which equals the area enclosed by the cycle on the P–V diagram. The cycle is a rectangle with ΔP = 300 − 100 = 200 N/m² and ΔV = 5 − 2 = 3 m³, so Q_net = W_net = ΔP × ΔV = 200 × 3 = 600 J — option (3).

Question 40

Consider that an electron is revolving in an excited state of a Hydrogen atom with velocity √25.6 × 10⁵ m s⁻¹. The radius of the orbit is x × 10⁻⁹ m. The value of x is : (e = 1.6×10⁻¹⁹ C, mₑ = 9×10⁻³¹ kg, 1/4πε₀ = 9×10⁹ N m² C⁻²)

Accepted Answer

(A) 1

In Bohr's model the electrostatic attraction provides the centripetal force: mv²/r = (1/4πε₀)(e²/r²), so r = (1/4πε₀)(e²/mv²). With v = √25.6 × 10⁵ m/s, v² = 25.6 × 10¹⁰ m²/s²; e² = 2.56 × 10⁻³⁸ C²; m = 9 × 10⁻³¹ kg; 1/4πε₀ = 9 × 10⁹. Then r = (9×10⁹)(2.56×10⁻³⁸) / [(9×10⁻³¹)(25.6×10¹⁰)] = (2.56/25.6) × 10^(−29−(−21)) = 0.1 × 10⁻⁸ = 1 × 10⁻⁹ m. Hence x = 1 — option (1).

Question 41

A car travels on a circular racetrack of radius 50 m, which is banked at an angle θ. If the car travels at a speed of 10 m s⁻¹, then the wear and tear on its tyres is minimum. Taking the acceleration due to gravity to be 10 m s⁻², the value of θ is :

Accepted Answer

(A) tan⁻¹(1/√5)

Minimum wear ⇒ no friction ⇒ tan θ = v²/(rg) = 100/(50×10) = 1/5, θ = tan⁻¹(1/5). ⚠ None of the printed options is exactly tan⁻¹(1/5) — verify the option values against the official key.

Question 42

Three identical p-n junction diodes D₁, D₂ and D₃ are connected across a battery as shown in the figure. If the widths of the depletion regions of D₁, D₂ and D₃ are W₁, W₂ and W₃, respectively, then the correct option is :

Accepted Answer

(C) W₃ > W₂ > W₁

The 5 V battery's left terminal is positive and the right is negative. D₁: its p-side (triangle) connects to + and n-side to −, so it is forward biased → depletion width W₁ decreases (minimum). D₂: its branch has an open switch/break, so no voltage is applied across it — it is unbiased → W₂ stays at its standard value (middle). D₃: its p-side connects to − and n-side to +, so it is reverse biased → W₃ increases (maximum). Hence W₃ > W₂ > W₁ — option (3).

Question 43

The following table presents parts of the electromagnetic spectrum (P–S) and their corresponding major applications (I–IV). The correct option is :

Accepted Answer

(C) P-II, Q-I, R-IV, S-III

Microwave → warming food (II); UV rays → purifying water (I); Gamma rays → treating cancer cells (IV); Radio wave → AM and FM communication (III). Hence P-II, Q-I, R-IV, S-III.

Question 44

Bob B of mass m at rest is hanging vertically from the ceiling via a massless string of length 10 m, as shown in the figure. Point mass A of mass m travelling horizontally with speed 10 m s⁻¹ hits bob B elastically. The bob B rises to a height h after the collision. Taking the acceleration due to gr

Accepted Answer

(C) 5

Elastic collision of equal masses: A stops, B moves with 10 m s⁻¹. h = v²/2g = 100/20 = 5 m.

Question 45

An ideal gas is made of polyatomic molecules. Each of the molecules has three translational and three rotational modes. If the number of vibrational modes is f and the ratio of heat capacities C_P/C_V is 8/7, then the value of f is :

Accepted Answer

(A) 4

Degrees of freedom d = 3 + 3 + 2f. γ = (d+2)/d = 8/7 ⇒ d = 14 ⇒ 6 + 2f = 14 ⇒ f = 4.

Question 46

For the following reaction sequence, choose the correct option :

Accepted Answer

(C) If P gives a carboxylic acid on acidification, Q gives a poisonous gas on exposure to air and light

Benzene undergoes Friedel–Crafts acylation with CH₃COCl/AlCl₃ to give acetophenone (C₆H₅COCH₃). Acetophenone has a methyl ketone group, so with NaOCl it undergoes the haloform reaction, giving P = sodium benzoate (C₆H₅COONa) and Q = chloroform (CHCl₃). On acidification, P gives benzoic acid (C₆H₅COOH); Q (chloroform) on exposure to air and light is slowly oxidised to the poisonous gas phosgene: 2CHCl₃ + O₂ →(light) 2COCl₂ + 2HCl. This matches option (3). (Q is chloroform, not a primary alcohol; chloroform is aliphatic, not aromatic; and neither product is a carbonyl compound — so options 1, 2 and 4 are wrong.)

Question 47

Given below are two statements : Statement-I : [Fe(ox)₃]³⁻ is chiral. Statement-II : trans-[Cr(C₂O₄)₂(H₂O)₂]⁻ is chiral. (Given : ox = HOOC-COOH). In light of the above statements, choose the most appropriate answer from the options given below :

Accepted Answer

(C) Statement-I is correct but Statement-II is incorrect

The tris-chelate [Fe(ox)₃]³⁻ is chiral (Statement-I correct); the trans-bis(oxalato) complex has a plane of symmetry and is achiral (Statement-II incorrect).

Question 48

The following carbocation is stabilized by the interaction of the empty p orbital with :

Accepted Answer

(A) filled σ and filled π orbitals

A carbocation is stabilised by donation from filled orbitals into the empty p orbital — filled σ(C–H) (hyperconjugation) and a filled π system (conjugation). ⚠ Structure-based — verify the figure.

Question 49

In potash alum, the ratio of K⁺ and SO₄²⁻ ions is :

Accepted Answer

(A) 1:2

Potash alum is KAl(SO₄)₂·12H₂O, with one K⁺ and two SO₄²⁻ per formula unit, so K⁺ : SO₄²⁻ = 1:2.

Question 50

The correct statement about peptides and proteins is :

Accepted Answer

(C) In β-pleated sheet structures, peptide chains are held together by intermolecular hydrogen bonds

(1) Wrong — having two or more polypeptide subunits describes the quaternary structure, not the tertiary (tertiary = the overall 3-D folding of a single polypeptide chain). (2) Wrong — many proteins are biologically active at the tertiary level as single chains (e.g. myoglobin, lysozyme). (3) Correct — in a β-pleated sheet the adjacent polypeptide strands are held together by intermolecular hydrogen bonds (between the C=O of one chain and the N–H of an adjacent chain). (4) Wrong — the α-helix is a right-handed (not left-handed) screw, stabilised by intramolecular hydrogen bonds. Hence the correct statement is option (3).

Question 51

The numbers 17.0145 and 21.0235 were rounded to three figures after the decimal point. The resulting numbers, respectively, are :

Accepted Answer

(C) 17.014 and 21.024

Round-half-to-even: 17.0145 → 17.014 (kept digit 4 is even) and 21.0235 → 21.024 (kept digit 3 is odd, rounds up to even 4).

Question 52

The correct order of solubility of the given salts in water at 298 K (K_sp: AgBr = 5.0×10⁻¹³, Zn(OH)₂ = 1.0×10⁻¹⁵, Hg₂Cl₂ = 1.3×10⁻¹⁸) is :

Accepted Answer

(D) Zn(OH)₂ > AgBr > Hg₂Cl₂

Solubilities: AgBr s=√Ksp=7.1×10⁻⁷; Zn(OH)₂ s=(Ksp/4)^⅓=6.3×10⁻⁶; Hg₂Cl₂ s=(Ksp/4)^⅓=6.9×10⁻⁷. So Zn(OH)₂ > AgBr > Hg₂Cl₂. ⚠ AgBr and Hg₂Cl₂ are close — verify option order.

Question 53

Among the following options, the correct trend in the electron gain enthalpy is :

Accepted Answer

(C) Cl > F > Br > I

Electron gain enthalpy generally becomes less negative down a group (the larger atom adds the incoming electron farther from the nucleus): Cl > Br > I. The exception is F vs Cl — fluorine's very small, compact 2p subshell causes strong inter-electronic repulsion on the incoming electron, whereas in chlorine the electron enters the larger 3p subshell with less repulsion, so chlorine has a more negative electron gain enthalpy than fluorine (Cl > F). Combining the two gives Cl > F > Br > I — option (3).

Question 54

Given below are two statements : Assertion A : For an ideal solution formed by mixing liquids P and Q, ΔmixH = 0 and ΔmixV = 0. Reason R : No interactions occur between P and Q. In light of the above statements, choose the most appropriate answer from the options given below :

Accepted Answer

(C) A is correct but R is not correct

Assertion A is correct: an ideal solution obeys Raoult's law over the entire composition range, and for it there is no enthalpy change (ΔmixH = 0) and no volume change (ΔmixV = 0) on mixing. Reason R is incorrect: in a liquid solution intermolecular interactions are always present — for an ideal solution the new P–Q interactions are exactly equal in magnitude to the original P–P and Q–Q interactions, not absent. 'No interactions occur' would describe a mixture of ideal gases, not liquids. Hence A is true but R is false — option (3).

Question 55

The amino acid that gives a red-to-blood-red colour on treating with sodium nitroprusside is :

Accepted Answer

(C) methionine

The sodium nitroprusside test detects sulfur-containing amino acids; of the options, methionine is the only sulfur-containing amino acid. ⚠ The test is classically for a free thiol (cysteine) — verify the intended answer.

Question 56

The standard electrode potential (E°) for the half-cell reaction Fe³⁺ + e⁻ → Fe²⁺ at 298 K is : (Given : E°(Fe³⁺/Fe) = −0.04 V and E°(Fe²⁺/Fe) = −0.44 V at 298 K)

Accepted Answer

(B) +0.76 V

Standard electrode potentials are intensive and cannot be added directly, but their Gibbs free energies (ΔG° = −nFE°) are extensive and add up. Reaction 1: Fe³⁺ + 3e⁻ → Fe, E° = −0.04 V → ΔG₁° = −3F(−0.04) = +0.12F. Reaction 2: Fe²⁺ + 2e⁻ → Fe, E° = −0.44 V → ΔG₂° = −2F(−0.44) = +0.88F. The target Fe³⁺ + e⁻ → Fe²⁺ = Reaction 1 − Reaction 2, so ΔG₃° = ΔG₁° − ΔG₂° = 0.12F − 0.88F = −0.76F. Since ΔG₃° = −1·F·E₃°, E₃° = +0.76 V — option (2).

Question 57

In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO₄ solution. If the volume of KMnO₄ solution required to reach the end point is 10 mL, the strength of the KMnO₄ solution is :

Accepted Answer

(A) 0.10 M

At the end point the gram-equivalents of KMnO₄ equal those of oxalic acid: M₁v₁V₁ = M₂v₂V₂, where v is the n-factor. In acidic medium KMnO₄ (MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O) has n-factor 5; oxalic acid (C₂O₄²⁻ → 2CO₂ + 2e⁻) has n-factor 2. So M₁·5·10 = 0.25·2·10 ⟹ M₁·5 = 0.50 ⟹ M₁ = 0.10 M — option (1).

Question 58

According to crystal field theory, the correct order of ligands with respect to their decreasing order of field strength is :

Accepted Answer

(A) CO > NH₃ > H₂O > Cl⁻

From the spectrochemical series (strong → weak): CO > NH₃ > H₂O > Cl⁻.

Question 59

Two moles of an ideal gas undergo free expansion from 10 L to 100 L at 300 K. The values of ΔS_system and ΔS_surroundings are : (R is the universal gas constant)

Accepted Answer

(D) ΔS_system = 4.606R, ΔS_surroundings = 0

In free expansion the gas expands against a vacuum (P_ext = 0), so the work W = −P_ext·ΔV = 0. The process is isothermal and for an ideal gas ΔU = 0, so by the first law q = ΔU − W = 0 — no heat is exchanged with the surroundings, giving ΔS_surroundings = q_surr/T = 0. Entropy is a state function, so ΔS_system is found via a reversible isothermal path: ΔS_system = nR ln(V₂/V₁) = 2 × R × 2.303 × log₁₀(100/10) = 2 × R × 2.303 × 1 = 4.606 R. Hence ΔS_system = 4.606 R and ΔS_surroundings = 0 — option (4).

Question 60

2A → B is a zero-order reaction, where k = 1.0 mol L⁻¹ min⁻¹. If the initial concentration of A is 2 M, then the time taken to complete 75% of the reaction will be :

Accepted Answer

(B) 0.75 min

For 2A → B (zero order), rate = −(1/2)d[A]/dt = k[A]⁰ = k, so −d[A]/dt = 2k and the integrated rate law is [A]_t = [A]₀ − 2kt, giving t = ([A]₀ − [A]_t)/(2k). 75% complete means 25% of A remains: [A]_t = 0.25 × 2 = 0.5 M, so the amount reacted is [A]₀ − [A]_t = 2 − 0.5 = 1.5 M. Then t = 1.5/(2 × 1.0) = 0.75 min — option (2). (Note: 1.5 M is the amount reacted, not the time.)

NEET 2026 · Code 50 — Answer Key & Solutions

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