NEET 2026 (3 May, cancelled) Question PaperPhysics, Chemistry and Biology, with checked answers and solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
164 questions from the NEET 2026 (3 May, cancelled) paper, conducted by NTA. Tap an option to check your answer and read why it is right. Each subject has a free PDF with the answer key and explanations.
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- NEET 2026 (3 May, cancelled)
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- NTA
- Physics
- 35 questions
- Chemistry
- 41 questions
- Biology
- 88 questions
NEET 2026 (3 May, cancelled) Physics questions
35 questions. Tap an option to check your answer.
Q1NEET 2026 (3 May, cancelled)
The speed of light in vacuum is taken as unity. If light takes 6 min 40 s to reach the Earth from the Sun, the distance between the Sun and the Earth in new unit is:
Not quite — the answer is D.
Time = 6 × 60 + 40 = 400 s. With c = 1 (one unit of length per second), distance = c × t = 400 units (light-seconds). Multiplying by 3 × 10⁸ gives metres, not the new unit.
Q2NEET 2026 (3 May, cancelled)
Match List I with List II: List I — List II A. Young's Modulus — I. (Δd/ΔL)(L/d) B. Compressibility — II. FL/(A ΔL) C. Bulk Modulus — III. −(1/ΔP)(ΔV/V) D. Poisson's Ratio — IV. −P(V/ΔV) Choose the correct answer from the options given below:
Not quite — the answer is D.
Young's modulus = stress/strain = (F/A)/(ΔL/L) = FL/(AΔL). Bulk modulus = −P/(ΔV/V) = −P V/ΔV, and compressibility is its reciprocal, −(1/ΔP)(ΔV/V). Poisson's ratio is lateral strain over longitudinal strain, (Δd/d)/(ΔL/L). Swapping bulk modulus with compressibility is the main trap.
Q3NEET 2026 (3 May, cancelled)
The angular speed of a flywheel is increased from 600 rpm to 1200 rpm in 10 s. The number of revolutions completed by the flywheel during this time is:
Not quite — the answer is C.
With uniform angular acceleration, average speed = (600 + 1200)/2 = 900 rpm = 15 rev/s, so in 10 s the flywheel makes 15 × 10 = 150 revolutions. Answers like 900 come from treating rpm as rev/s or forgetting to convert minutes to seconds.
Q4NEET 2026 (3 May, cancelled)
A resistor is connected to a battery of 12 V emf and internal resistance 2 Ω. If the current in the circuit is 0.6 A, the terminal voltage of the battery is:
Not quite — the answer is D.
Terminal voltage V = E − Ir = 12 − 0.6 × 2 = 12 − 1.2 = 10.8 V. The 1.2 V is only the drop inside the battery, and 12 V ignores the internal resistance.
Q5NEET 2026 (3 May, cancelled)
A flask contains argon and chlorine in the ratio of 2 : 1 by mass. The temperature of the mixture is 27°C. The ratio of root mean square speed of the molecules of the two gases (V_rms(Ar)/V_rms(Cl)) is: (Atomic mass of argon = 40.0 u and molecular mass of chlorine = 70.0 u)
Not quite — the answer is A.
v_rms = √(3RT/M) depends only on temperature and molar mass, not on the mass ratio in the flask. At the same T, v(Ar)/v(Cl₂) = √(70/40) = √(7/4) = √7/2. Inverting the ratio gives 2/√7, and forgetting the square root gives 7/4.
Q6NEET 2026 (3 May, cancelled)
Match List I with List II. List-I — List-II A. E = hν — I. de Broglie wavelength B. Diffraction and Interference — II. Particle nature of light C. λ = h/p — III. Wave nature of light D. Compton effect — IV. Energy of photon
Not quite — the answer is D.
E = hν gives the energy of a photon; diffraction and interference show the wave nature of light; λ = h/p is the de Broglie wavelength; and the Compton effect (photon–electron collisions) shows the particle nature of light.
Q7NEET 2026 (3 May, cancelled)
In the first excited state of hydrogen atom, the energy of its electron is −3.4 eV. The radial distance of the electron from the hydrogen nucleus in this case is approximately: (Take 1 eV = 1.6 × 10⁻¹⁹ J, e = 1.6 × 10⁻¹⁹ C and 1/4πε₀ = 9 × 10⁹ N m²/C²)
Not quite — the answer is C.
Total energy E = −ke²/(2r), so r = ke²/(2|E|) = (9 × 10⁹ × (1.6 × 10⁻¹⁹)²)/(2 × 3.4 × 1.6 × 10⁻¹⁹) = (9 × 10⁹ × 1.6 × 10⁻¹⁹)/6.8 = 2.1 × 10⁻¹⁰ m. This matches 4a₀ = 4 × 0.53 Å for n = 2; dropping the factor 2 would give double.
Q8NEET 2026 (3 May, cancelled)
A box of mass 15 kg is kept on the floor of a stationary trolley. The coefficient of static friction between the box and the trolley is 0.12. Keeping the box in stationary state over the trolley, the maximum acceleration with which the trolley can be moved horizontally in m s⁻² is: (g = 10 m/s²)
Not quite — the answer is D.
Static friction must supply the box's acceleration: ma ≤ μmg, so a(max) = μg = 0.12 × 10 = 1.2 m s⁻². The mass cancels, so the 15 kg value is not needed.
Q9NEET 2026 (3 May, cancelled)
The amount of work done to raise a mass 'm' from the surface of the Earth to a height equal to the radius of the Earth 'R' will be
Not quite — the answer is D.
W = ΔU = −GMm/(2R) − (−GMm/R) = GMm/(2R) = mgR/2, using GM = gR². Taking mgh = mgR assumes g stays constant over a height as large as R.
Q10NEET 2026 (3 May, cancelled)
Each side of a metallic cube of mass 5.580 kg is measured to the 9.0 cm. Keeping the significant figures in view, the density of the material of the cube can be best expressed as X × 10³ kg m⁻³ where the value of X is:
Not quite — the answer is D.
Volume = (0.090 m)³ = 7.29 × 10⁻⁴ m³, so density = 5.580/7.29 × 10⁻⁴ = 7.654 × 10³ kg m⁻³. The side 9.0 cm has only two significant figures, so the result is rounded to two: 7.7 × 10³ kg m⁻³. Keeping 7.654 or 7.65 claims more precision than the data allow, and 7.6 is a wrong rounding.
Q11NEET 2026 (3 May, cancelled)
The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately: (Consider mass of the bob = 20 g)
Not quite — the answer is B.
At the mean position all the mechanical energy is kinetic (taking PE = 0 there): ½ × 0.020 × v² = 0.02, so v² = 2 and v = 1.41 m/s. Using 20 kg instead of 0.020 kg, or forgetting the ½, gives the other values.
Q12NEET 2026 (3 May, cancelled)
In Young's double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where the path difference is λ, is K units. The intensity of light at a point where the path difference is λ/3 will be
Not quite — the answer is A.
I = I₀ cos²(φ/2) with φ = 2π × (path difference)/λ. At path difference λ, φ = 2π and I = I₀ = K. At λ/3, φ = 2π/3, so I = K cos²(π/3) = K × 1/4 = K/4. Using cos(π/3) without squaring gives K/2.
Q13NEET 2026 (3 May, cancelled)
An ac circuit contains a resistance of 1 kΩ, a capacitor of 0.1 μF and an inductor of 1 mH connected in series. The resonance frequency of the circuit is approximately:
Not quite — the answer is D.
f₀ = 1/(2π√(LC)) = 1/(2π√(10⁻³ × 10⁻⁷)) = 1/(2π × 10⁻⁵) = 10⁵/6.283 ≈ 15.9 kHz. The resistance does not affect the resonance frequency; forgetting the 2π would give the angular frequency 100 krad/s.
Q14NEET 2026 (3 May, cancelled)
In interference and diffraction, the light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. A. As there is no gain or loss of energy, these phenomena are consistent with the principle of conservation of energy. B. Diffraction and interference are characteristics exhibited only by light waves. Choose the correct answer from the options given below:
Not quite — the answer is C.
Energy is only redistributed between bright and dark regions, so the total is conserved and the first statement holds. Interference and diffraction are shown by all waves, including sound, water waves and matter waves, so the claim that only light shows them is false.
Q15NEET 2026 (3 May, cancelled)
For a travelling harmonic wave y(x, t) = 2.0 cos 2π(10t − 0.0080x + 0.35), where x and y are in cm and t in s. The phase difference between oscillatory motion of two points separated by a distance of 0.5 m is:
Not quite — the answer is B.
The wave number is k = 2π × 0.0080 rad/cm. For Δx = 0.5 m = 50 cm, Δφ = kΔx = 2π × 0.0080 × 50 = 0.8π rad. Using 0.5 without converting to centimetres gives 0.008π rad.
Q16NEET 2026 (3 May, cancelled)
The magnitude and direction of the acceleration produced in a body of mass 5 kg when two mutually perpendicular forces 8 N and 6 N act on it, are respectively:
Not quite — the answer is D.
Resultant force = √(8² + 6²) = 10 N, so a = 10/5 = 2 m s⁻² along the resultant. Its angle θ with the 8 N force satisfies tan θ = 6/8 = 3/4. The angle tan⁻¹(4/3) is measured from the 6 N force, not the 8 N force.
Q17NEET 2026 (3 May, cancelled)
Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:
Not quite — the answer is A.
Initial energy = ½CV² = ½ × 200 × 10⁻¹² × 100² = 1.0 × 10⁻⁶ J. After sharing, the common voltage is 50 V and the total energy is 2 × ½ × 200 × 10⁻¹² × 50² = 0.5 × 10⁻⁶ J, so 0.5 × 10⁻⁶ J is lost. 1.0 × 10⁻⁶ J is the initial energy, not the loss, and the joule-sized values forget the pico prefix.
Q18NEET 2026 (3 May, cancelled)
The power of a crane, which lifts a mass of 1000 kg to a height of 20 m in 10 s is: (g = 9.8 m/s²)
Not quite — the answer is C.
Work = mgh = 1000 × 9.8 × 20 = 1.96 × 10⁵ J; power = 1.96 × 10⁵ / 10 = 1.96 × 10⁴ W = 19.6 kW. 19.6 W misses the factor of 1000 in the unit.
Q19NEET 2026 (3 May, cancelled)
In a vernier calliper, 20 VSD coincide with 16 MSD (each division of length 1 mm). The least count of the vernier callipers is:
Not quite — the answer is C.
1 VSD = 16/20 MSD = 0.8 mm, so least count = 1 MSD − 1 VSD = 1 − 0.8 = 0.2 mm = 0.02 cm. Writing 0.2 cm forgets to convert millimetres to centimetres.
Q20NEET 2026 (3 May, cancelled)
When a ruler falls vertically, 5 different persons catch it with different reaction times. (g = 9.8 m s⁻²) A. Person A has reaction time of 0.20 s. B. Person B has reaction time of 0.22 s. C. Person C has reaction time of 0.18 s. D. Person D has reaction time of 0.19 s. E. Person E has reaction time of 0.21 s. What is the correct order of the distance travelled by the ruler for each person?
Not quite — the answer is C.
The ruler falls freely from rest, so d = ½gt² grows with reaction time. Ordering the times 0.22 > 0.21 > 0.20 > 0.19 > 0.18 s gives B > E > A > D > C. Reversing the order, or misplacing the 0.18 s and 0.19 s persons, gives the other sequences.
Q21NEET 2026 (3 May, cancelled)
A room heater is rated 400 W, 220 V. If the supply voltage drops to 200 V, what will be the power consumed (approximately)?
Not quite — the answer is C.
R = V²/P = 220²/400 = 121 Ω. At 200 V, P = 200²/121 = 40000/121 ≈ 331 W. 121 is the resistance in ohms, not the power, and the power does not stay at 400 W because it scales as V².
Q22NEET 2026 (3 May, cancelled)
A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14 × 10⁻³ T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively : (Take μ₀ = 4π × 10⁻⁷ T m/A)
Not quite — the answer is D.
B = μ₀NI/(2r) gives I = 2rB/(μ₀N) = (2 × 0.05 × 3.14 × 10⁻³)/(4π × 10⁻⁷ × 100) = 3.14 × 10⁻⁴/1.257 × 10⁻⁴ = 2.5 A. Magnetic moment m = NIA = 100 × 2.5 × π(0.05)² = 1.96 ≈ 2 A m². Forgetting the 100 turns, or using diameter for radius, gives the other pairs.
Q23NEET 2026 (3 May, cancelled)
A rectangular wire loop of sides 8 cm and 3 cm with a small cut, is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is 2 cm s⁻¹, in a direction normal to the shorter side of the loop, will be :
Not quite — the answer is D.
Moving normal to the shorter side, the flux changes as the 3 cm side crosses the boundary: emf = Blv = 0.3 × 0.03 × 0.02 = 1.8 × 10⁻⁴ V. Using the 8 cm side (motion normal to the longer side) gives 4.8 × 10⁻⁴ V.
Q24NEET 2026 (3 May, cancelled)
Four statements are given (A is mass number): A. The volume of a nucleus is proportional to A^(1/3). B. The volume of a nucleus is proportional to A. C. The difference in mass of an atom and its nucleus is called the mass defect. D. The difference in mass of a nucleus and its constituents is called the mass defect. Choose the correct answer from the options given below:
Not quite — the answer is D.
Nuclear radius R = R₀A^(1/3), so volume ∝ R³ ∝ A; it is the radius, not the volume, that goes as A^(1/3). Mass defect is the difference between the total mass of the separated nucleons and the mass of the nucleus, not the electrons' mass (atom minus nucleus). So only the second and fourth statements hold.
Q25NEET 2026 (3 May, cancelled)
An unknown nucleus has a nuclear density of 2.29 × 10¹⁷ kg/m³ and mass of 19.926 × 10⁻²⁷ kg. Its mass number A is approximately: (Take R₀ = 1.2 × 10⁻¹⁵ m, 4π = 12.56)
Not quite — the answer is A.
Volume = mass/density = 19.926 × 10⁻²⁷ / 2.29 × 10¹⁷ = 8.70 × 10⁻⁴⁴ m³. Also V = (4/3)πR₀³A = (12.56/3) × 1.728 × 10⁻⁴⁵ × A = 7.23 × 10⁻⁴⁵ A, so A = 8.70 × 10⁻⁴⁴ / 7.23 × 10⁻⁴⁵ ≈ 12 (carbon-12). Dividing the mass by the nucleon mass would also give 12, so 19 or 20 are mistaken readings of 19.926.
Q26NEET 2026 (3 May, cancelled)
Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum L, notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as: (Take π² = 9.8, and g = 9.8 m/s²)
Not quite — the answer is D.
T = 60/30 = 2 s. From T = 2π√(L/g), L = gT²/(4π²) = 9.8 × 4/(4 × 9.8) = 1 m. Taking T = 2 s as the length, or forgetting to square T, gives the other values.
Q27NEET 2026 (3 May, cancelled)
An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 J/s, then the rate at which internal energy increases will be:
Not quite — the answer is D.
By the first law, dU/dt = dQ/dt − dW/dt = 100 − 75 = 25 W. Adding the work rate (175 W) or taking 125 W treats work done by the system as energy gained.
Q28NEET 2026 (3 May, cancelled)
A galvanometer of resistance 100 Ω gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range 0 − 10 A. The shunt required is:
Not quite — the answer is A.
Shunt S = I_g G/(I − I_g) = (10⁻³ × 100)/(10 − 0.001) ≈ 0.1/10 = 0.01 Ω. Dropping a factor of ten in the current gives 0.1 Ω or 0.001 Ω.
Q29NEET 2026 (3 May, cancelled)
The peak value of an alternating current is 5 A and frequency is 60 Hz. How long will the current, starting from zero, take to reach the peak value?
Not quite — the answer is D.
i = i₀ sin ωt rises from zero to the peak in a quarter period: t = T/4 = 1/(4 × 60) = 1/240 s. 1/120 s is half a period, which brings the current back to zero.
Q30NEET 2026 (3 May, cancelled)
Two statements are given below: A. When the forward bias voltage across a p-n junction diode increases above a certain threshold voltage, the diode current increases significantly. B. This current is called reverse saturation current. Choose the correct answer from the options given below:
Not quite — the answer is B.
Beyond the threshold (cut-in) voltage, the forward current rises sharply, so the first statement holds. That large current is the forward current due to majority carriers; the reverse saturation current is the tiny minority-carrier current under reverse bias, so the second statement is false.
Q31NEET 2026 (3 May, cancelled)
Which of the following statements are correct? A. Inside a conductor, the electrostatic field is zero. B. Electric field at the surface of a charged conductor does not depend on its surface charge density. C. The interior of a charged conductor can have no excess charge in the static situation. D. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point. E. The electrostatic potential is zero everywhere inside a charged conductor. Choose the correct answer from the options given below:
Not quite — the answer is C.
In electrostatics the field inside a conductor is zero, any excess charge resides on the surface, and the surface field is normal to the surface, so those three statements hold. The surface field is E = σ/ε₀, so it does depend on surface charge density, and the potential inside is constant (equal to the surface value), not necessarily zero.
Q32NEET 2026 (3 May, cancelled)
For a metal of work function 6.6 eV, which of the following wavelengths of incident radiation does not give rise to the photoelectric effect? (Take Planck's constant as 6.6 × 10⁻³⁴ J s)
Not quite — the answer is C.
Threshold wavelength λ₀ = hc/φ = (6.6 × 10⁻³⁴ × 3 × 10⁸)/(6.6 × 1.6 × 10⁻¹⁹) = 1.875 × 10⁻⁷ m ≈ 188 nm. Only radiation of wavelength shorter than this ejects electrons, so 200 nm fails while 50, 100 and 150 nm all work.
Q33NEET 2026 (3 May, cancelled)
In a concave lens, a ray of light emanating from the object parallel to the principal axis of the lens after refraction:
Not quite — the answer is B.
A concave lens is diverging: a ray parallel to the axis bends away and appears to come from the principal focus on the object side (the first principal focus). Passing through the second focus is the behaviour of a convex lens, and a ray through the optical centre, not one parallel to the axis, goes straight on.
Q34NEET 2026 (3 May, cancelled)
A submarine is designed to withstand an absolute pressure of 100 atm. How deep can it go below the water surface? (Consider the density of water = 1000 kg m⁻³, 1 atm = 1 × 10⁵ Pa and gravitational acceleration g = 10 m/s²)
Not quite — the answer is A.
Absolute pressure = P_atm + ρgh, so ρgh = 100 − 1 = 99 atm = 99 × 10⁵ Pa and h = 99 × 10⁵/(1000 × 10) = 990 m. Ignoring the atmospheric pressure would give 1000 m, which is not offered; the other values are slips of a factor of ten.
Q35NEET 2026 (3 May, cancelled)
Match List I with List II: List-I (Electromagnetic wave) — List-II (Production) A. Microwave — I. Electrons in atoms emit light when they move from a higher energy level to a lower energy level B. Visible light — II. Radioactive decay of nucleus C. Gamma rays — III. Vibration of atoms and molecules D. Infra-red rays — IV. Klystron valve or magnetron valve Choose the correct answer from the options given below:
Not quite — the answer is C.
Microwaves are produced by special vacuum tubes such as klystrons and magnetrons; visible light by electronic transitions in atoms; gamma rays by radioactive decay of nuclei; and infrared by vibrations of atoms and molecules (hot bodies). Assigning molecular vibrations to microwaves is the main trap.
Chapter-wise split of NEET 2026 (3 May, cancelled) Physics
- Units and Measurements3 Q
- Current Electricity2 Q
- Dual Nature of Radiation and Matter2 Q
- Laws of Motion2 Q
- Oscillations2 Q
- Wave Optics2 Q
- Alternating Current2 Q
- Electrostatic Potential and Capacitance2 Q
- Moving Charges and Magnetism2 Q
- Nuclei2 Q
- Mechanical Properties of Solids1 Q
- System of Particles and Rotational Motion1 Q
- Kinetic Theory1 Q
- Atoms1 Q
- Gravitation1 Q
- Waves1 Q
- Work, Energy and Power1 Q
- Motion in a Straight Line1 Q
- Electromagnetic Induction1 Q
- Thermodynamics1 Q
- Semiconductor Electronics: Materials, Devices and Simple Circuits1 Q
- Ray Optics and Optical Instruments1 Q
- Mechanical Properties of Fluids1 Q
- Electromagnetic Waves1 Q
NEET 2026 (3 May, cancelled) Chemistry questions
40 questions. Tap an option to check your answer.
Q1NEET 2026 (3 May, cancelled)
Select the reagents that reduce nitriles to primary amines. A. (i) LiAlH₄; (ii) H₂O B. Sn + HCl C. H₂/Ni D. Na(Hg)/C₂H₅OH E. Br₂/aq. NaOH Choose the correct answer from the options given below.
Not quite — the answer is B.
Nitriles are reduced to primary amines by LiAlH₄ followed by hydrolysis, by catalytic hydrogenation with H₂/Ni, and by sodium amalgam in ethanol (Mendius reduction). Sn + HCl is the reagent for reducing nitro compounds, and Br₂/aq. NaOH brings about the Hofmann degradation of amides, so neither of these belongs in the set.
Q2NEET 2026 (3 May, cancelled)
Match List I with List II : List I (Transition metal/compound/complex) — List II (Catalytic Role) A. V₂O₅ — I. Preparation of ammonia from N₂/H₂ mixture B. Fe — II. Polymerisation of alkynes C. PdCl₂ — III. Preparation of H₂SO₄ and SO₂ D. Ni complex — IV. Oxidation of ethyne to ethanal Choose the correct answer from the options given below.
Not quite — the answer is D.
V₂O₅ is the catalyst of the contact process for sulphuric acid, finely divided iron catalyses the Haber synthesis of ammonia, PdCl₂ is the Wacker-process catalyst for oxidation to ethanal, and nickel complexes catalyse the polymerisation of alkynes. Pairing iron with the ethanal oxidation is the common slip.
Q3NEET 2026 (3 May, cancelled)
Consider the following reaction : 2A(g) + B(g) → 2D(g) ΔU° = −10 kJ mol⁻¹ and ΔS° = −44 J K⁻¹ at 298 K. Identify the correct option with ΔG° for the reaction and spontaneity of the reaction at 298 K. (Given : R = 8.31 J mol⁻¹ K⁻¹)
Not quite — the answer is C.
Δn(g) = 2 − 3 = −1, so ΔH° = ΔU° + Δn(g)RT = −10000 − 8.31 × 298 = −12476 J. Then ΔG° = ΔH° − TΔS° = −12476 − 298 × (−44) = −12476 + 13112 = +636 J ≈ +0.636 kJ mol⁻¹; a positive ΔG° means non-spontaneous. Using ΔU° directly in place of ΔH° gives the wrong value of about +3.1 kJ, and forgetting the sign of TΔS° gives the negative values.
Q4NEET 2026 (3 May, cancelled)
Match List I with List II : List I (Quantum Numbers 'n', 'l') — List II (Orbital) A. n = 2, l = 1 — I. 3d B. n = 4, l = 0 — II. 2p C. n = 5, l = 3 — III. 4s D. n = 3, l = 2 — IV. 5f Choose the correct answer from the options given below.
Not quite — the answer is C.
l = 0, 1, 2, 3 correspond to s, p, d, f, and n gives the shell number. So (2, 1) is 2p, (4, 0) is 4s, (5, 3) is 5f and (3, 2) is 3d. The other combinations mismatch either the shell or the subshell letter.
Q5NEET 2026 (3 May, cancelled)
In a qualitative analysis, Bi³⁺ is detected by appearance of precipitate of BiO(OH)(s). Calculate pH when the following equilibrium exists at 298 K. BiO(OH)(s) ⇌ BiO⁺(aq) + OH⁻(aq), K = 4 × 10⁻¹⁰ (Given : log 2 = 0.3010)
Not quite — the answer is D.
K = [BiO⁺][OH⁻] = x² = 4 × 10⁻¹⁰, so [OH⁻] = 2 × 10⁻⁵ M. pOH = 5 − log 2 = 4.699 and pH = 14 − 4.699 = 9.301. Reporting 4.699 mistakes pOH for pH.
Q6NEET 2026 (3 May, cancelled)
The correct statement with regard to the secondary structure of DNA/RNA is
Not quite — the answer is B.
DNA is a double helix whose bases are adenine, guanine, cytosine and thymine. RNA is single-stranded and has uracil in place of thymine, so the statements giving RNA thymine or a double strand, or DNA uracil or a single strand, are wrong.
Q7NEET 2026 (3 May, cancelled)
The pair of molecules that are metamers among the following is :
Not quite — the answer is D.
Metamers have the same functional group but different alkyl groups on either side of it. Methyl propyl ether and diethyl ether are both C₄H₁₀O ethers differing only in the groups on oxygen. The two propanols are position isomers, the pentanes are chain isomers, and propanone/propanal are functional isomers.
Q8NEET 2026 (3 May, cancelled)
Match List I with List II : List-I (Complex) — List-II (Type of isomerism) A. [Pt(NH₃)₂Cl₂] — I. Optical B. [Co(en)₃]³⁺ — II. Solvate C. [Co(NH₃)₅NO₂]Cl₂ — III. Geometrical D. [Cr(H₂O)₆]Cl₃ — IV. Linkage Choose the correct answer from the options given below :
Not quite — the answer is D.
Square planar [Pt(NH₃)₂Cl₂] exists as cis and trans forms (geometrical); [Co(en)₃]³⁺ has non-superimposable mirror images (optical); NO₂⁻ can bind through N or O (linkage); and [Cr(H₂O)₆]Cl₃ has hydrate (solvate) isomers such as [Cr(H₂O)₅Cl]Cl₂·H₂O. Swapping the linkage and solvate pairings is the main trap.
Q9NEET 2026 (3 May, cancelled)
Match List I with List II : List-I (Order of reaction) — List-II (Unit of rate constant) A. Zero order — I. mol⁻¹ L s⁻¹ B. First order — II. mol⁻² L² s⁻¹ C. Second order — III. s⁻¹ D. Third order — IV. mol L⁻¹ s⁻¹ Choose the correct answer from the options given below :
Not quite — the answer is B.
The unit of k is (mol L⁻¹)¹⁻ⁿ s⁻¹. For n = 0 this is mol L⁻¹ s⁻¹, for n = 1 it is s⁻¹, for n = 2 it is mol⁻¹ L s⁻¹ and for n = 3 it is mol⁻² L² s⁻¹. Interchanging the second- and third-order units is the usual error.
Q10NEET 2026 (3 May, cancelled)
The correct IUPAC name of the following compound is : CH₃–CH₂–CH(CH₂CH₃)–CH₂–CH(CH₃)–CH₂–CH₃
Not quite — the answer is A.
The longest chain has seven carbons, with an ethyl group and a methyl group on C-3 and C-5 whichever end is chosen. With a tie in locants, the substituent first in alphabetical order (ethyl) gets the lower number, giving 3-ethyl-5-methylheptane. The hexane names use a chain that is not the longest, and '3-methyl-5-ethyl' breaks the alphabetical rule.
Q11NEET 2026 (3 May, cancelled)
A bulb is rated at 150 watt, converting 8% energy into light. If energy of one photon is 4.42 × 10⁻¹⁹ J, how many photons are emitted by the bulb per second?
Not quite — the answer is A.
Light power = 8% of 150 W = 12 J s⁻¹. Photons per second = 12 / (4.42 × 10⁻¹⁹) = 2.71 × 10¹⁹. Using the full 150 W would give about 3.4 × 10²⁰, ten times larger than any sensible choice.
Q12NEET 2026 (3 May, cancelled)
Match List I with List II : List-I — List-II A. C₂H₄ — I. 3σ bonds, 2π bonds B. C₂H₂ — II. 3σ bonds, one lone pair C. CH₄ — III. 4σ bonds D. NH₃ — IV. 5σ bonds, 1π bond Choose the correct answer from the options given below :
Not quite — the answer is B.
Ethene has four C–H and one C–C σ bond plus one π bond (5σ, 1π); ethyne has two C–H and one C–C σ bond plus two π bonds (3σ, 2π); methane has 4σ bonds; ammonia has 3σ bonds and one lone pair on N. Swapping the ethene and ethyne counts is the main trap.
Q13NEET 2026 (3 May, cancelled)
The following two reactions give the same foul smelling product Z. C₂H₅Cl —X→ Z C₂H₅CONH₂ —Br₂, NaOH→ Y —CHCl₃/ethanolic KOH, Δ→ Z X and Z, respectively, are :
Not quite — the answer is A.
Hofmann bromamide degradation of propanamide gives ethylamine (Y), and the carbylamine reaction converts it into ethyl isocyanide, C₂H₅NC, the foul-smelling product. Ethyl chloride gives the isocyanide with the covalent AgCN (attack through N), whereas ionic KCN gives the nitrile C₂H₅CN.
Q14NEET 2026 (3 May, cancelled)
The number of hydrogen atoms present in 5.4 g of urea is: (Given: Molar mass of urea : 60 g mol⁻¹ N_A : 6.022 × 10²³ particles mol⁻¹)
Not quite — the answer is D.
Moles of urea = 5.4/60 = 0.09 mol; each NH₂CONH₂ has 4 H atoms, so H = 0.36 mol = 0.36 × 6.022 × 10²³ = 2.168 × 10²³ atoms. Counting only 2 H per molecule gives 1.084 × 10²³.
Q15NEET 2026 (3 May, cancelled)
Identify the incorrect statement from the following:
Not quite — the answer is D.
Nitrogen has no d orbitals in its valence shell, so it cannot form dπ–pπ bonds; its multiple bonds are pπ–pπ, as in N≡N. The heavier group 15 elements can use d orbitals (dπ–dπ bonding of phosphines/arsines with metals) and do show some catenation, so those statements are correct.
Q16NEET 2026 (3 May, cancelled)
Which one of the following is an ambidentate ligand?
Not quite — the answer is C.
An ambidentate ligand has two different donor atoms but binds through only one at a time; SCN⁻ can bind through S (thiocyanato) or N (isothiocyanato). Ethane-1,2-diamine and oxalate are bidentate and EDTA is hexadentate, all binding through several atoms at once.
Q17NEET 2026 (3 May, cancelled)
The correct order of increasing metallic character of Na, Be, P, Mg and Si is
Not quite — the answer is A.
Metallic character decreases across a period and increases down a group. In period 3, P < Si < Mg < Na, and Be lies above Mg so it is less metallic than Mg but more metallic than the metalloid Si: P < Si < Be < Mg < Na. Placing Be as the most metallic reverses the group trend.
Q18NEET 2026 (3 May, cancelled)
Match List I with List II: List I (Conversion) — List II (Reagents) A. Cumene, C₆H₅CH(CH₃)₂ → Phenol — I. (i) Oleum; (ii) NaOH, Δ; (iii) H⁺ B. CH₃COOH → CH₃CH₂OH — II. (i) O₂; (ii) H₃O⁺/H⁺ C. CH₃CH₂CH₂OH → CH₃–CH(OH)–CH₃ — III. (i) CH₃OH, H⁺; (ii) H₂, catalyst D. Benzene → Phenol — IV. (i) conc. H₂SO₄, Δ; (ii) H⁺/H₂O Choose the correct answer from the options given below :
Not quite — the answer is B.
Cumene is oxidised by air to its hydroperoxide, which acid cleaves to phenol. Acetic acid is esterified with methanol and the ester is hydrogenated to ethanol. Propan-1-ol is dehydrated by hot conc. H₂SO₄ to propene, and acid-catalysed hydration (Markovnikov) gives propan-2-ol. Benzene is sulphonated with oleum and the sulphonate fused with NaOH, then acidified, to give phenol.
Q19NEET 2026 (3 May, cancelled)
Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because:
Not quite — the answer is D.
Ce³⁺ is 4f¹; losing one more electron gives Ce⁴⁺ with the empty 4f⁰ configuration (noble gas core of xenon), which is especially stable. Cerium's atomic number is 58, not 61, and its noble-gas core is xenon, not radon; 4f¹⁴ applies to Yb²⁺/Lu³⁺.
Q20NEET 2026 (3 May, cancelled)
In the following reaction sequence, X and Z respectively are : CH₃CH₂CH₂–OH + PCl₅ → CH₃CH₂CH₂Cl + X + HCl CH₃CH₂CH₂Cl —alc. KOH, Δ→ Y Y —HBr, (C₆H₅CO)₂O₂→ Z
Not quite — the answer is B.
PCl₅ converts an alcohol to the chloride with POCl₃ and HCl as by-products (H₃PO₃ comes from PCl₃). Alcoholic KOH dehydrohalogenates 1-chloropropane to propene, and HBr in the presence of benzoyl peroxide adds anti-Markovnikov (peroxide effect), giving 1-bromopropane rather than 2-bromopropane.
Q21NEET 2026 (3 May, cancelled)
Match List I with List II : List I (Complex/ion) — List II (Shape/geometry) A. [Pt(Cl₂)(NH₃)₂] — I. Octahedral B. [Co(NH₃)₆]Cl₃ — II. Trigonal bipyramidal C. [NiCl₄]²⁻ — III. Square planar D. [Fe(CO)₅] — IV. Tetrahedral Choose the correct answer from the options given below :
Not quite — the answer is B.
Pt(II) (d⁸) complexes are dsp² square planar; [Co(NH₃)₆]³⁺ is d²sp³ octahedral; with the weak-field Cl⁻ ligand, [NiCl₄]²⁻ is sp³ tetrahedral; Fe(CO)₅ is dsp³ trigonal bipyramidal. Making [NiCl₄]²⁻ square planar confuses it with [Ni(CN)₄]²⁻.
Q22NEET 2026 (3 May, cancelled)
The functional group that can be identified through phthalein dye test is :
Not quite — the answer is B.
In the phthalein dye test the compound is heated with phthalic anhydride and conc. H₂SO₄ and the melt is poured into dilute NaOH; phenols condense to give coloured phthalein dyes (phenol itself gives phenolphthalein, pink in alkali). Alcohols, aldehydes and acids do not form such dyes.
Q23NEET 2026 (3 May, cancelled)
Two products X and Y are formed in the following reaction sequence. Benzene + CH₃Cl —Anhydr. AlCl₃→ W —dil. HNO₃ + dil. H₂SO₄, warm→ X + Y The suitable method that can be used for the separation of products X and Y is :
Not quite — the answer is A.
Friedel–Crafts methylation gives toluene (W), and nitration gives the ortho- and para-nitrotoluenes, liquids/low-melting solids whose boiling points differ (about 222 °C and 238 °C) but are fairly close, so fractional distillation separates them. Sublimation and the extraction methods rely on properties these isomers do not differ in usefully.
Q24NEET 2026 (3 May, cancelled)
Identify the correct statements : (A) The molality of 2.5 g of ethanoic acid (Molar mass : 60 g mol⁻¹) in 75 g of benzene solution is 0.556 m. (B) The molarity of a solution containing 5 g of NaOH (molar mass : 40 g mol⁻¹) in 450 mL of solution is 0.278 M at 298 K. (C) Aquatic species are more comfortable in cold water. (D) The solubility of gas increases with decrease in pressure. (E) For a binary mixture of A and B, the number of moles of A and B are n_A and n_B respectively. The mole fraction of B will be x_B = n_A/(n_A + n_B). Choose the correct answer from the options given below :
Not quite — the answer is A.
Molality = (2.5/60) mol / 0.075 kg = 0.0417/0.075 = 0.556 m, and molarity = (5/40) mol / 0.450 L = 0.278 M, so the first two statements hold. Gases such as O₂ are more soluble in cold water, so aquatic life is more comfortable there. Gas solubility rises with increasing pressure (Henry's law), not decreasing, and the mole fraction of B is n_B/(n_A + n_B), so the last two statements are false.
Q25NEET 2026 (3 May, cancelled)
During Lassaigne's test, the elements present in an organic compound are converted from :
Not quite — the answer is B.
Fusion with sodium converts N, S and halogens, which are covalently bonded in the organic compound, into ionic NaCN, Na₂S and NaX, which can then be detected by ionic tests in the sodium fusion extract. The elements are not ionic to begin with, so the other choices fail.
Q26NEET 2026 (3 May, cancelled)
A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is : (Given : Molar mass of Cu = 63 g mol⁻¹; 1 F = 96487 C mol⁻¹)
Not quite — the answer is B.
Q = It = 1.5 × 600 = 900 C. Cu²⁺ + 2e⁻ → Cu, so mass = (63 × 900)/(2 × 96487) = 56700/192974 = 0.2938 g. Taking one electron per Cu gives double, 0.5876 g.
Q27NEET 2026 (3 May, cancelled)
At a certain temperature, T(K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is :
Not quite — the answer is B.
ΔU = q + w with q = +500 J (absorbed) and w = −200 J (work done by the system), so ΔU = +300 J. Adding the two magnitudes to get 700 J ignores the sign of work done by the system.
Q28NEET 2026 (3 May, cancelled)
Identify the correct statement about ClF₃ from the following options :
Not quite — the answer is A.
Cl has 7 valence electrons; three are used in bonds to F, leaving 4 electrons = 2 lone pairs, so 5 electron pairs (sp³d, trigonal bipyramidal arrangement). The two lone pairs occupy equatorial positions, leaving a T-shaped molecule. Three lone pairs would make it linear, and the trigonal shapes ignore the lone pairs' positions.
Q29NEET 2026 (3 May, cancelled)
In a test tube containing a salt, a few drops of dilute H₂SO₄ was added, which gave colourless vapours having the smell of vinegar. The vapours turned the blue litmus paper red. Identify the correct anion from the following :
Not quite — the answer is C.
Dilute H₂SO₄ liberates acetic acid from an acetate; its vapour smells of vinegar and is acidic, turning blue litmus red. A sulphide gives H₂S with a rotten-egg smell, a carbonate gives odourless CO₂ with effervescence, and a sulphate gives no gas with dilute acid.
Q30NEET 2026 (3 May, cancelled)
At 298 K, a certain buffer solution contains equal concentrations of X⁻ and HX. K_b for X⁻ is 10⁻¹⁰. What is the pH of this buffer solution?
Not quite — the answer is B.
Ka of HX = Kw/Kb = 10⁻¹⁴/10⁻¹⁰ = 10⁻⁴, so pKa = 4. With [X⁻] = [HX], Henderson's equation gives pH = pKa + log 1 = 4. Taking pH = pKb = 10 confuses the base constant with the acid constant.
Q31NEET 2026 (3 May, cancelled)
The calculated 'spin-only' magnetic moment of Ti²⁺ (3d²) is :
Not quite — the answer is C.
3d² has n = 2 unpaired electrons, so μ = √(n(n + 2)) = √8 = 2.83 ≈ 2.84 BM. The values 3.87, 4.90 and 5.92 BM correspond to 3, 4 and 5 unpaired electrons.
Q32NEET 2026 (3 May, cancelled)
Identify the incorrect statement from the following:
Not quite — the answer is D.
Oxygen shows −2 in most compounds but also −1 in peroxides, −½ in superoxides and +2 in OF₂, so 'only −2' is false. Carbon's small size allows strong pπ–pπ multiple bonds, BCl₃ is monomeric while AlCl₃ dimerises to Al₂Cl₆, and catenation falls in the order C ≫ Si > Ge ≈ Sn, so those statements are correct.
Q33NEET 2026 (3 May, cancelled)
Phenolphthalein is used as an indicator for the titration of sodium hydroxide solution against a standard solution of oxalic acid. The colour change that is observed at an alkaline pH close to the equivalence point during this titration is:
Not quite — the answer is D.
Phenolphthalein is colourless in acid and turns pink above about pH 8.3. NaOH is added from the burette to the oxalic acid, so at the end point the colourless solution just turns permanently pink. Yellow/red changes belong to methyl orange, and pink to colourless would occur only if acid were added to alkali.
Q34NEET 2026 (3 May, cancelled)
When 1 dm³ of CO₂ gas is passed over hot coke the volume of gaseous mixture after complete reaction at STP becomes 1.4 dm³. The composition of the gaseous mixture at STP is :
Not quite — the answer is B.
CO₂ + C → 2CO. If x dm³ of CO₂ reacts, the mixture contains (1 − x) CO₂ and 2x CO, total 1 + x = 1.4, so x = 0.4: CO = 0.8 dm³ and CO₂ = 0.6 dm³. The other pairs do not add up to 1.4 dm³ or do not conserve carbon.
Q35NEET 2026 (3 May, cancelled)
The major product Z formed in the following sequence of reactions is C₂H₆ —Cl₂, UV light→ X (monochlorinated product) —NH₃→ Y —(i) NaNO₂, HCl; (ii) H₂O→ Z
Not quite — the answer is C.
Free-radical chlorination gives chloroethane (X), ammonolysis gives ethylamine (Y), and nitrous acid converts a primary aliphatic amine into an unstable diazonium salt that loses N₂ and reacts with water to give ethanol. Aliphatic diazonium salts do not survive, so the azo-type product is not obtained, and ethylamine is only the intermediate.
Q36NEET 2026 (3 May, cancelled)
Given below is an expression for the rate constant of a first-order reaction occurring at a certain temperature, T(K). ln k = 14.34 − (1.25 × 10⁴)/T The energy of activation in kcal mol⁻¹ for the reaction is : (Given: k in s⁻¹, R = 1.987 cal mol⁻¹ K⁻¹)
Not quite — the answer is A.
Comparing with ln k = ln A − Ea/(RT): Ea/R = 1.25 × 10⁴ K, so Ea = 1.25 × 10⁴ × 1.987 = 24 838 cal ≈ 24.84 kcal mol⁻¹. 14.34 is ln A, not the activation energy, and using log (÷2.303) wrongly gives other values.
Q37NEET 2026 (3 May, cancelled)
Given below are certain reactions. Identify the reaction for which K_P ≠ K_C.
Not quite — the answer is B.
K_P = K_C(RT)^Δn, so they differ only when Δn(gas) ≠ 0. For ammonia synthesis Δn = 2 − 4 = −2; the other three reactions have equal moles of gas on both sides (Δn = 0), so K_P = K_C for them.
Q38NEET 2026 (3 May, cancelled)
Identify the incorrect statement from the following :
Not quite — the answer is A.
Size falls across a period and cations are smaller than their atoms, and Al³⁺ (higher charge, same electron count as Mg²⁺) is smaller than Mg²⁺. So the largest is Mg and the smallest is Al³⁺, making the 'Al and Mg²⁺' statement wrong. Element 107 is unnilseptium, Li–Mg is a diagonal pair, and Al in [AlCl(H₂O)₅]²⁺ is +3 with six ligands, so the rest are correct.
Q39NEET 2026 (3 May, cancelled)
Mixture of chloroform and acetone forms a solution with negative deviation from Raoult's law due to :
Not quite — the answer is B.
The C–H of chloroform (made acidic by three Cl atoms) hydrogen-bonds to the carbonyl oxygen of acetone, so A–B attractions are stronger than A–A and B–B. This lowers the escaping tendency and the vapour pressure, a negative deviation. Increased escaping tendency or weaker A–B forces would give a positive deviation.
Q40NEET 2026 (3 May, cancelled)
The number of chlorine atoms present in the organic products X and Y of the following reactions, respectively, are : Benzene + 6Cl₂ —Anhydr. AlCl₃, dark, cold→ X Benzene + 3Cl₂ —UV, 500 K→ Y
Not quite — the answer is C.
With a Lewis acid in the dark and excess chlorine, electrophilic substitution replaces all six H atoms to give hexachlorobenzene, C₆Cl₆ (6 Cl). Under UV light at 500 K, three Cl₂ molecules add across the ring to give benzene hexachloride, C₆H₆Cl₆ (also 6 Cl). Counting the 3 Cl₂ molecules as 3 atoms is the trap.
Chemistry: not in the current NEET syllabus
These 1 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
1 questions. Tap an option to check your answer.
Q1NEET 2026 (3 May, cancelled)
Methane reacts with steam at 1273 K in the presence of nickel catalyst to form :
Not quite — the answer is C.
Steam reforming: CH₄ + H₂O → CO + 3H₂ over Ni at about 1270 K, giving water gas (syngas). CO₂ and H₂ arise only in the later water-gas shift step with more steam, and CO₂ + H₂O are combustion products.
Chapter-wise split of NEET 2026 (3 May, cancelled) Chemistry
- Organic Chemistry: Some Basic Principles and Techniques4 Q
- Amines3 Q
- The d- and f-Block Elements3 Q
- Equilibrium3 Q
- Coordination Compounds3 Q
- Thermodynamics2 Q
- Structure of Atom2 Q
- Chemical Kinetics2 Q
- Chemical Bonding and Molecular Structure2 Q
- Some Basic Concepts of Chemistry2 Q
- Classification of Elements and Periodicity in Properties2 Q
- Alcohols, Phenols and Ethers2 Q
- Solutions2 Q
- Biomolecules1 Q
- Haloalkanes and Haloarenes1 Q
- Electrochemistry1 Q
- Hydrocarbons1 Q
NEET 2026 (3 May, cancelled) Biology questions
87 questions. Tap an option to check your answer.
Q1NEET 2026 (3 May, cancelled)
In angiosperms, root hairs arise from which one of the following regions of the root?
Not quite — the answer is D.
Root hairs develop from epidermal cells in the region of maturation, which lies just above the region of elongation. Cells of the elongation zone are still enlarging and do not yet bear root hairs, while the meristematic zone and the root cap only produce and protect new cells.
Q2NEET 2026 (3 May, cancelled)
In which one of the following, the ovules are not enclosed by an ovary wall and remain exposed?
Not quite — the answer is B.
Pinus is a gymnosperm, and gymnosperms bear naked ovules that are not enclosed by an ovary wall. Funaria (a moss) and Selaginella (a pteridophyte) do not form ovules at all, while Wolffia is an angiosperm whose ovules lie inside an ovary.
Q3NEET 2026 (3 May, cancelled)
In the lac operon, the z gene codes for
Not quite — the answer is C.
In the lac operon the structural gene z codes for beta-galactosidase, which hydrolyses lactose into glucose and galactose. Permease is coded by the y gene, transacetylase by the a gene, and the repressor by the separate regulatory i gene.
Q4NEET 2026 (3 May, cancelled)
Exploring molecular, genetic and species-level diversity for products of economic importance is called
Not quite — the answer is C.
Bioprospecting is the search through molecular, genetic and species diversity for products of economic value such as drugs. Biofortification is breeding crops for higher nutrient content, bioremediation is using organisms to clean pollutants, and biomagnification is the build-up of toxins along a food chain.
Q5NEET 2026 (3 May, cancelled)
Match List I with List II: List I: A. Genetically modified organism B. Thermostable DNA polymerase C. Ti plasmid D. pBR322 List II: I. Agrobacterium tumefaciens II. Bt cotton III. Thermus aquaticus IV. Escherichia coli Choose the correct answer from the options given below:
Not quite — the answer is A.
Bt cotton is a genetically modified crop, the thermostable Taq DNA polymerase comes from Thermus aquaticus, the Ti plasmid belongs to Agrobacterium tumefaciens, and pBR322 is an Escherichia coli cloning vector. Pairing the Ti plasmid with Bt cotton is the common trap; Agrobacterium is only the source of the plasmid.
Q6NEET 2026 (3 May, cancelled)
Match List I with List II: List I: A. Productivity B. Net primary productivity C. Gross primary productivity D. Secondary productivity List II: I. Gross primary productivity minus respiration losses II. Rate of formation of new organic matter by consumers III. Rate of biomass production IV. Rate of production of organic matter during photosynthesis Choose the correct answer from the options given below:
Not quite — the answer is A.
Productivity is the rate of biomass production; gross primary productivity is the rate of production of organic matter during photosynthesis; net primary productivity is GPP minus respiration losses; and secondary productivity is the rate of formation of new organic matter by consumers. Swapping the GPP and NPP definitions is the usual mistake.
Q7NEET 2026 (3 May, cancelled)
Since the origin and diversification of life on Earth, there have been five episodes of mass extinction of species. How is the sixth extinction, which is in progress, different from the previous episodes?
Not quite — the answer is C.
The ongoing sixth extinction is driven largely by human activities, and species are being lost about 100 to 1000 times faster than in pre-human times. Saying it is only 10 times faster understates the rate, and the rate is neither zero nor lower than before.
Q8NEET 2026 (3 May, cancelled)
Alpha-helix is found in which level of protein structure?
Not quite — the answer is A.
The alpha-helix is a secondary structure, formed when the polypeptide chain coils through hydrogen bonds between nearby peptide groups. The primary structure is only the amino acid sequence, the tertiary structure is the three-dimensional folding of the whole chain, and the quaternary structure is the assembly of several chains.
Q9NEET 2026 (3 May, cancelled)
The main function of bulliform cells in grasses is :
Not quite — the answer is D.
Bulliform cells are large, empty, colourless cells in the upper epidermis of grass leaves. When they lose turgor during water stress the leaf rolls inward, which reduces the exposed surface and cuts water loss. They are not photosynthetic and play no role in water transport.
Q10NEET 2026 (3 May, cancelled)
Identify the correct sequence of steps in each cycle of Polymerase Chain Reaction :
Not quite — the answer is D.
Each PCR cycle first separates the DNA strands by heat (denaturation), then lets the primers bind to the single strands (annealing), and finally Taq polymerase extends the primers (extension). Primers cannot anneal until the strands have separated, and extension needs annealed primers.
Q11NEET 2026 (3 May, cancelled)
Match List I with List II : List I (Phase of cell cycle): A. G₁ phase B. S phase C. G₂ Phase D. M phase List II (Activity): I. Actual cell division occurs II. Cell is metabolically active and continuously grows but does not replicate its DNA III. Synthesis of DNA occurs and the amount of DNA per cell doubles IV. Proteins are synthesized while cell growth continues Choose the correct answer from the options given below:
Not quite — the answer is D.
In G₁ the cell is metabolically active and grows without replicating DNA; in S phase DNA is synthesised and its amount doubles; in G₂ proteins are made while growth continues; and actual division happens in M phase. G₁ and G₂ are both growth phases, but only G₂ is described as preparing proteins after replication.
Q12NEET 2026 (3 May, cancelled)
Which of the following statements are correct? A. The Amazon rainforest being cut and cleared for cultivation of soyabeans is an example of habitat loss. B. Steller's sea cow and passenger pigeon became extinct due to over-exploitation by humans. C. The Nile perch introduced into Lake Victoria in East Africa helped in population growth of cichlid fish in the lake. D. Water hyacinth is an invasive species. E. When a species becomes extinct, the plant and animal species associated with it are not affected. Choose the correct answer from the options given below:
Not quite — the answer is B.
Clearing the Amazon for soyabeans is habitat loss, Steller's sea cow and the passenger pigeon were wiped out by over-exploitation, and water hyacinth is an invasive alien weed, so these three are correct. The Nile perch actually drove more than 200 cichlid species of Lake Victoria to extinction, and extinction of one species can cause co-extinction of species tied to it.
Q13NEET 2026 (3 May, cancelled)
Which of the following statements are correct with reference to a transcription unit? A. A transcription unit in DNA is defined primarily by three regions : promoter, structural gene and terminator. B. The promoter is said to be located towards the 5'-end of the structural gene. C. The promoter is a DNA sequence that provides binding site for RNA polymerase. D. The promoter defines the template and coding strands. E. The terminator is located towards the 3'-end of the coding strand and it defines the end of the process of transcription. Choose the correct answer from the options given below:
Not quite — the answer is D.
All five statements are correct. Every transcription unit consists of a promoter, the structural gene and a terminator. The promoter sits upstream, towards the 5'-end of the structural gene (with respect to the coding strand), provides the binding site for RNA polymerase and so fixes which strand is the template. The terminator lies downstream, towards the 3'-end of the coding strand, and marks where transcription stops.
Q14NEET 2026 (3 May, cancelled)
Which one of the following statements is not true about the universal rules of binomial nomenclature?
Not quite — the answer is D.
In binomial nomenclature the first word is the genus and the second is the specific epithet, as in Mangifera indica, so the statement that reverses them is false. The other rules hold: names are Latin or Latinised, each word is underlined separately when handwritten or printed in italics, and the specific epithet starts with a small letter.
Q15NEET 2026 (3 May, cancelled)
Match List I with List II: List I: A. Decomposition B. Detritus C. Mineralisation D. Humification List II: I. Accumulation of dark coloured amorphous colloidal substance II. Release of inorganic nutrients by the activity of microbes in soil III. Breaking down of complex organic matter into inorganic substances. IV. Dead remains of plants and animals including fecal matter Choose the correct answer from the options given below:
Not quite — the answer is B.
Decomposition is the breakdown of complex organic matter into inorganic substances, detritus is the dead remains of plants and animals including faecal matter, mineralisation is the release of inorganic nutrients by soil microbes, and humification is the build-up of the dark, amorphous, colloidal humus. Humification and mineralisation are the two steps most often swapped.
Q16NEET 2026 (3 May, cancelled)
Which one of the following is the site for active ribosomal RNA synthesis?
Not quite — the answer is C.
The nucleolus is the site of active ribosomal RNA synthesis, which is why it is larger in cells making a lot of protein. Chromatin carries the genes in general, while the centrosome organises spindle fibres and the kinetochore attaches chromosomes to the spindle.
Q17NEET 2026 (3 May, cancelled)
2(C₅₁H₉₈O₆) + 145O₂ → 102CO₂ + 98H₂O + energy The Respiratory Quotient (RQ) of a biomolecule used for respiration, as per the above equation would be :
Not quite — the answer is A.
RQ = volume of CO₂ released / volume of O₂ consumed = 102/145 ≈ 0.70, which falls between 0.5 and 0.95. The equation is the respiration of tripalmitin, a fat. An RQ of 1.0 applies to carbohydrates, and values above 1 apply to organic acids.
Q18NEET 2026 (3 May, cancelled)
Match List I with List II : List I: A. Incomplete dominance B. Co-dominance C. Pleiotropy D. Polygenic inheritance List II: I. Human skin colour II. Inheritance of flower colour in Antirrhinum sp. III. Phenylketonuria disease in humans IV. ABO blood groups Choose the correct answer from the options given below:
Not quite — the answer is A.
Flower colour in Antirrhinum (snapdragon) shows incomplete dominance, the ABO blood groups show co-dominance of the two dominant blood-group alleles, phenylketonuria is a classic case of pleiotropy since one gene affects several traits, and human skin colour is polygenic.
Q19NEET 2026 (3 May, cancelled)
Arrange the following steps of DNA fingerprinting in a correct sequence. A. Isolation of DNA and its digestion by restriction endonucleases. B. Hybridisation using a labelled VNTR probe. C. Transferring of separated DNA fragments to synthetic membranes. D. Detection of hybridised DNA fragments by autoradiography. E. Separation of DNA fragments by electrophoresis. Choose the correct answer from the options given below:
Not quite — the answer is C.
DNA is first isolated and cut with restriction enzymes, the fragments are separated by electrophoresis, blotted onto a synthetic membrane, hybridised with a labelled VNTR probe and finally detected by autoradiography. Hybridisation can only be done after the fragments have been transferred to the membrane, which rules out sequences placing hybridisation before blotting.
Q20NEET 2026 (3 May, cancelled)
Which of the following statements are correct with reference to packaging of DNA helix ? A. Histones are organized to form a unit of eight molecules called histone octamer. B. Histones are negatively charged basic proteins. C. Histones are rich in the basic amino acid residues - lysine and arginine. D. The positively charged DNA is wrapped around the histone octamer to form nucleosome. E. The packaging of chromatin at higher levels requires an additional set of proteins called non-histone chromosomal proteins. Choose the correct answer from the options given below:
Not quite — the answer is A.
Eight histone molecules form the octamer, histones are rich in lysine and arginine, and higher-order packaging needs non-histone chromosomal proteins, so these three statements are correct. The charges are reversed in the other two: histones are positively charged basic proteins, and the DNA wrapped around them is negatively charged.
Q21NEET 2026 (3 May, cancelled)
Find the incorrect statement(s) about photosynthesis from the following: A. The water splitting complex is associated with PSI. B. C₄ plants use the C₃ pathway of CO₂ fixation as the main biosynthetic pathway. C. In C₄ plants, photorespiration does not occur. D. C₃ plants exhibit 'Kranz' anatomy. E. ATP synthesis in chloroplast occurs through chemiosmosis. Choose the answer from the options given below:
Not quite — the answer is D.
The water-splitting complex is linked to PS II, not PS I, and Kranz anatomy is a feature of C₄ plants, not C₃ plants, so these two statements are incorrect. C₄ plants do use the Calvin (C₃) cycle in bundle-sheath cells as their main biosynthetic pathway, they avoid photorespiration by concentrating CO₂, and chloroplast ATP is made by chemiosmosis.
Q22NEET 2026 (3 May, cancelled)
Match List-I with List-II : List-I: A. Conjunctive tissue B. Casparian strips C. Subsidiary cells D. Starch sheath List-II: I. Specialised cells in the vicinity of guard cells II. Endodermal cells rich in starch III. Tissue between xylem and phloem IV. Endodermal cells with suberin deposition Choose the correct answer from the options given below:
Not quite — the answer is C.
Conjunctive tissue is the parenchyma between xylem and phloem bundles, Casparian strips are suberin deposits on endodermal cell walls, subsidiary cells are the specialised epidermal cells near the guard cells, and the starch sheath is the starch-rich endodermis of the dicot stem. Casparian strips and the starch sheath both concern the endodermis, so they are easy to swap.
Q23NEET 2026 (3 May, cancelled)
Which one of the following is not a characteristic of plant cells in the phase of elongation?
Not quite — the answer is D.
Large conspicuous nuclei, dense protoplasm and thin walls are features of meristematic cells. Cells in the elongation phase show more vacuolation, cell enlargement and new cell wall deposition.
Q24NEET 2026 (3 May, cancelled)
Match List-I with List-II List-I (Growth Regulator): A. 2,4-D B. GA₃ C. Kinetin D. ABA List-II (Function/Effect): I. Brewing industry II. Stimulation of stomatal closure III. Herbicide IV. Nutrient mobilisation Choose the correct answer from the options given below:
Not quite — the answer is A.
2,4-D is an auxin used as a weedicide/herbicide, GA₃ speeds up malting and is used in the brewing industry, the cytokinin kinetin promotes nutrient mobilisation and delays senescence, and ABA stimulates stomatal closure.
Q25NEET 2026 (3 May, cancelled)
The enzyme required for carboxylation in the Calvin cycle is
Not quite — the answer is C.
In the Calvin cycle CO₂ is fixed onto RuBP by RuBisCO (RuBP carboxylase-oxygenase), giving two molecules of 3-PGA. PEP carboxylase does the first fixation in the mesophyll of C₄ plants, not in the Calvin cycle; hexokinase works in glycolysis and carboxypeptidase digests proteins.
Q26NEET 2026 (3 May, cancelled)
How many ATP and NADPH molecules are required to make one molecule of glucose through the Calvin pathway?
Not quite — the answer is A.
Fixing one CO₂ takes 3 ATP and 2 NADPH, and a glucose needs six turns of the cycle, so 6 × 3 = 18 ATP and 6 × 2 = 12 NADPH. The option with 12 ATP and 18 NADPH reverses the two numbers.
Q27NEET 2026 (3 May, cancelled)
Which of the following is an in situ conservation method?
Not quite — the answer is A.
Sacred groves are forest patches protected in their natural habitat by local communities, which makes them in situ conservation. Wildlife safari parks, botanical gardens and seed banks all keep organisms or their germplasm away from their natural habitat, which is ex situ conservation.
Q28NEET 2026 (3 May, cancelled)
Which of the following statements are not true regarding restriction endonucleases? A. They are called molecular scissors. B. These are the enzymes responsible for restricting the growth of bacteriophages in E. coli. C. They cut the DNA only at the centre of the palindromic sites. D. They remove nucleotides only from the ends of DNA fragments. E. They recognise specific palindromic base-pair sequences. Choose the answer from the options given below :
Not quite — the answer is D.
Restriction endonucleases usually cut a little away from the centre of the palindrome, between the same two bases on both strands, leaving sticky ends; and removing nucleotides from the ends of DNA is what exonucleases do. So these two statements are untrue. The enzymes are called molecular scissors, recognise specific palindromic sequences and were found because they restrict bacteriophage growth in E. coli.
Q29NEET 2026 (3 May, cancelled)
In racemose inflorescence, ____.
Not quite — the answer is D.
In a racemose inflorescence the main axis keeps growing, so flowers are borne laterally in acropetal succession, with the oldest at the base. Having the main axis end in a flower with limited growth describes a cymose inflorescence, and solitary flowers are not an inflorescence at all.
Q30NEET 2026 (3 May, cancelled)
Arrange the following in the correct developmental sequence related to microsporogenesis : A. Microspore tetrads B. Sporogenous tissue C. Pollen grains D. Pollen mother cells Choose the correct answer from the options given below :
Not quite — the answer is B.
Sporogenous tissue in the young anther gives rise to pollen (microspore) mother cells, which divide by meiosis to form microspore tetrads, and these separate and mature into pollen grains. Pollen grains are the end product, so they cannot come before the tetrads.
Q31NEET 2026 (3 May, cancelled)
Identify the correct statements about biomolecules. A. Lipids are generally water soluble. B. Proteins are polypeptides. C. Polysaccharides are long chains of sugars. D. Adenine and guanine are substituted pyrimidines. E. Almost all enzymes are proteins. Choose the correct answer from the options given below :
Not quite — the answer is B.
Proteins are polypeptides, polysaccharides are long chains of sugars, and nearly all enzymes are proteins (ribozymes are the exception), so these are correct. Lipids are generally insoluble in water, and adenine and guanine are substituted purines, not pyrimidines.
Q32NEET 2026 (3 May, cancelled)
Which of the following statements are true with reference to the sex-determination in honeybees? A. An offspring formed from the union of a sperm and an egg, develops as a female (queen or worker). B. An unfertilized egg develops as a male by parthenogenesis. C. A male has half the number of chromosomes than that of a female. D. Males produce sperms by meiosis. E. Honeybees have a haplodiploid sex-determination system. Choose the correct answer from the options given below :
Not quite — the answer is A.
Honeybees are haplodiploid: fertilised eggs (2n = 32) become females (queens or workers) and unfertilised eggs develop parthenogenetically into haploid males (n = 16). Because drones are already haploid, they produce sperms by mitosis, not meiosis, so that statement is the false one.
Q33NEET 2026 (3 May, cancelled)
Heterophyllous development in response to environment is an example of which of the following phenomena?
Not quite — the answer is D.
Heterophylly, as in buttercup leaves that take different shapes in air and in water, shows that a plant can follow different growth pathways in response to its environment. This ability is called plasticity. Dedifferentiation and redifferentiation describe cells regaining and then losing the ability to divide, not changes in organ form.
Q34NEET 2026 (3 May, cancelled)
Which of the following statements are correct regarding amino acids? A. They are substituted methanes. B. Serine is an aromatic amino acid. C. Valine is a neutral amino acid. D. Lysine is an acidic amino acid. Choose the correct answer from the options given below:
Not quite — the answer is C.
Amino acids are substituted methanes, with the α-carbon carrying H, an amino group, a carboxyl group and an R group, and valine is a neutral amino acid. Serine is a hydroxy amino acid, not an aromatic one (tyrosine, phenylalanine and tryptophan are aromatic), and lysine is basic, not acidic.
Q35NEET 2026 (3 May, cancelled)
"The Evil Quartet" of biodiversity loss includes which of the following?
Not quite — the answer is C.
The "Evil Quartet" is habitat loss and fragmentation, over-exploitation, alien species invasions and co-extinctions. Pollution harms biodiversity too, but it is not one of the four named causes.
Q36NEET 2026 (3 May, cancelled)
Match List-I with List-II: List-I (Process): A. Glycolysis B. ETS C. Accumulation of protons D. Krebs' cycle List-II (Location): I. Inner mitochondrial membrane II. Mitochondrial matrix III. Cytoplasm IV. Intermembrane space Choose the correct answer from the options given below:
Not quite — the answer is C.
Glycolysis takes place in the cytoplasm, the electron transport system lies in the inner mitochondrial membrane, protons pumped by the ETS build up in the intermembrane space, and the Krebs cycle runs in the mitochondrial matrix.
Q37NEET 2026 (3 May, cancelled)
Which one of the following is a triploid cell?
Not quite — the answer is D.
The primary endosperm cell forms when a male gamete fuses with the two polar nuclei (triple fusion), so it is triploid (3n). Synergids are haploid, the zygote is diploid, and the central cell holds two separate haploid polar nuclei before fertilisation.
Q38NEET 2026 (3 May, cancelled)
Which one of the following types of pollination brings genetically different types of pollen grains to the stigma?
Not quite — the answer is B.
Xenogamy carries pollen from a flower of a different plant, so the pollen is genetically different from the stigma. Geitonogamy moves pollen between flowers of the same plant, so it is genetically the same as autogamy, and cleistogamous flowers never open and are always self-pollinated.
Q39NEET 2026 (3 May, cancelled)
Match List I with List II : List I (Placentation): A. Marginal B. Axile C. Parietal D. Basal List II (Example): I. Mustard II. Pea III. Marigold IV. Lemon Choose the correct answer from the options given below:
Not quite — the answer is A.
Pea has marginal placentation, lemon has axile placentation, mustard has parietal placentation and marigold (a composite) has basal placentation.
Q40NEET 2026 (3 May, cancelled)
The main criteria used for Five Kingdom Classification proposed by R.H. Whittaker (1969) included : A. Cell structure B. Body organization C. Presence of flagellum D. Reproduction E. Phylogenetic relationships Choose the correct answer from the options given below :
Not quite — the answer is C.
Whittaker's criteria were cell structure, body (thallus) organisation, mode of nutrition, reproduction and phylogenetic relationships. The presence of a flagellum was not one of the criteria.
Q41NEET 2026 (3 May, cancelled)
Match List I with List II : List I: A. Trypsin B. Morphine C. Concanavalin A D. Collagen List II: I. Intercellular ground substance II. Lectin III. Enzyme IV. Alkaloid Choose the correct answer from the options given below:
Not quite — the answer is A.
Trypsin is a digestive enzyme, morphine is an alkaloid, concanavalin A is a lectin, and collagen is the main protein of the intercellular ground substance.
Q42NEET 2026 (3 May, cancelled)
Which of the following statements are correct with respect to DNA separation, isolation and visualization? A. The cutting of DNA is done by molecular scissors. B. The DNA fragments separate according to their size in an agarose gel, upon electrophoresis. C. The separated DNA fragments can be seen without staining when exposed to UV light. D. The separated DNA fragments, when stained with ethidium bromide, can be seen in visible light. Choose the correct answer from the options given below :
Not quite — the answer is B.
DNA is cut by restriction enzymes ("molecular scissors"), and in agarose gel electrophoresis the fragments separate by size, so these two statements are correct. The fragments cannot be seen without staining; after ethidium bromide staining they are seen as orange bands under UV light, not in visible light.
Q43NEET 2026 (3 May, cancelled)
Which one of the following disorders is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain of the haemoglobin molecule?
Not quite — the answer is B.
Sickle-cell anaemia comes from a single base change (GAG to GUG) that puts valine in place of glutamic acid at position six of the β-globin chain. Thalassemia is a quantitative defect where too little globin is made, not an amino acid substitution.
Q44NEET 2026 (3 May, cancelled)
Match List I with List II: List I: A. Cortisol B. Aldosterone C. Cholecystokinin D. Progesterone List II: I. Stimulates the formation of alveoli in mammary glands II. Produces anti-inflammatory reactions III. Stimulates reabsorption of Na⁺ and water from renal tubule IV. Stimulates secretion of pancreatic enzymes and bile juice Choose the correct answer from the options given below:
Not quite — the answer is C.
Cortisol, a glucocorticoid, produces anti-inflammatory reactions; aldosterone stimulates reabsorption of Na⁺ and water from the renal tubules; cholecystokinin stimulates secretion of pancreatic enzymes and bile; and progesterone stimulates formation of alveoli in the mammary glands.
Q45NEET 2026 (3 May, cancelled)
Arrange the following events occuring in Renin-Angiotensin mechanism in the correct order: A. Increase in blood pressure and Glomerular filtration rate B. Reabsorption of Na⁺ and water from distal parts of tubule due to Aldosterone C. Fall in Glomerular filtration rate D. Vasoconstriction by Angiotensin II and release of Aldosterone. E. Renin converts Angiotensinogen into Angiotensin I, followed by Angiotensin II. Choose the correct answer from the options given below:
Not quite — the answer is D.
Falling GFR triggers renin release; renin converts angiotensinogen to angiotensin I, which becomes angiotensin II; angiotensin II causes vasoconstriction and aldosterone release; aldosterone drives Na⁺ and water reabsorption in the distal tubule; and blood pressure and GFR rise again. The rise in GFR is the result of the process, so it must come last.
Q46NEET 2026 (3 May, cancelled)
In humans, respiration occurs in the following steps. Arrange these steps in the correct order. A. Diffusion of O₂ and CO₂ between blood and tissues B. Diffusion of O₂ and CO₂ across alveolar membrane C. Pulmonary ventilation by which atmospheric air is drawn in and CO₂ rich alveolar air is released out D. Cellular respiration E. Transport of gases by the blood Choose the correct answer from the options given below
Not quite — the answer is C.
Respiration starts with pulmonary ventilation, followed by diffusion of gases across the alveolar membrane, transport of gases by blood, diffusion between blood and tissues, and finally use of O₂ in cellular respiration. Gas exchange at the alveoli has to come before transport, so placing tissue diffusion second is wrong.
Q47NEET 2026 (3 May, cancelled)
The following are the stages of life cycle of Plasmodium. Arrange the stages in the proper order. A. The parasites reproduce asexually in RBCs, bursting the cells. B. The parasites reproduce asexually in liver cells, bursting the cells and releasing into blood. C. Gametocytes develop in RBCs. D. Sporozoites reach the liver through the blood. E. Female mosquito injects sporozoites into humans during bite. Choose the correct answer from the options given below:
Not quite — the answer is A.
The female Anopheles injects sporozoites, which travel through blood to the liver, multiply asexually in liver cells and are released into blood. They then multiply asexually in RBCs, bursting them, and finally gametocytes develop in the RBCs. Gametocytes form only after the RBC cycles, so they cannot come right after the bite.
Q48NEET 2026 (3 May, cancelled)
Insertion of a foreign DNA at BamHI site in an E.coli cloning vector pBR322 results in the loss of antibiotic resistance towards:
Not quite — the answer is C.
The BamHI site of pBR322 lies inside the tetracycline-resistance gene (tet^R), so inserting DNA there inactivates it and the recombinants lose tetracycline resistance. Ampicillin resistance stays intact, which is what allows recombinants to be selected; pBR322 carries no gentamycin-resistance gene.
Q49NEET 2026 (3 May, cancelled)
The following reaction depicts the activity of a particular class of enzymes : X–C–C–Y (Substrate) —E→ X–Y (Product) + C=C (Product) Identify the enzymes class ' E ' from the following options :
Not quite — the answer is C.
The enzyme removes the groups X and Y from adjacent carbon atoms without hydrolysis, leaving a C=C double bond. Removing groups to form a double bond is what lyases do. Transferases move a group from one molecule to another, and ligases join two molecules.
Q50NEET 2026 (3 May, cancelled)
The specific receptors for neurotransmitters in a synapse are present on ____.
Not quite — the answer is D.
Neurotransmitter released from the pre-synaptic terminal binds to specific receptors on the post-synaptic membrane, opening ion channels there. The pre-synaptic membrane releases the transmitter; Schwann cells and the myelin sheath only insulate axons.
Q51NEET 2026 (3 May, cancelled)
What is the probability of having children with 'O' blood group, where both mother and father are heterozygous for 'A' and 'B' blood group, respectively?
Not quite — the answer is A.
The mother is IAi and the father is IBi. The cross gives IAIB, IAi, IBi and ii in equal proportions, so the chance of an ii (O group) child is 1/4 = 25%. The answer is not 0% because each heterozygous parent can pass on the recessive i allele.
Q52NEET 2026 (3 May, cancelled)
Match List-I with List-II. List-I (Respiratory Volume): A. ERV (Expiratory Reserve Volume) B. RV (Residual Volume) C. IRV (Inspiratory Reserve Volume) D. TV (Tidal Volume) List-II (Capacity in mL): I. 2500 – 3000 mL II. 500 mL III. 1000 – 1100 mL IV. 1100 – 1200 mL Choose the correct answer from the options given below :
Not quite — the answer is A.
In a healthy adult, tidal volume is about 500 mL, IRV about 2500–3000 mL, ERV about 1000–1100 mL and RV about 1100–1200 mL. ERV and RV have close values, but the residual volume is slightly larger.
Q53NEET 2026 (3 May, cancelled)
Which of the following is not an example of convergent evolution?
Not quite — the answer is C.
The forelimbs of whales and bats share the same basic bone plan inherited from a common ancestor but do different jobs. They are homologous organs, which shows divergent evolution. Penguin and dolphin flippers, octopus and mammal eyes, and butterfly and bird wings are analogous structures from convergent evolution.
Q54NEET 2026 (3 May, cancelled)
Male frogs can be distinguished from female frogs due to the presence of A. Bulging eyes B. Vocal sacs C. Webbed digits in feet D. Copulatory pad on first digit of fore limbs E. Olive green-coloured skin with dark irregular spots Choose the correct answer from the options given below
Not quite — the answer is D.
Male frogs have sound-producing vocal sacs and a copulatory pad on the first digit of each forelimb, and females lack both. Bulging eyes, webbed feet and olive-green spotted skin are found in both sexes, so they cannot tell them apart.
Q55NEET 2026 (3 May, cancelled)
A group of researchers procured some fish like animals and upon investigation the following characters were observed: A. Endoskeleton was made of cartilage. B. Ectoparasitic; as they were found attached on fish skin with their circular sucking mouth. C. Paired fins and scales were absent, but 7 pairs of gill slits were present. Which of the following species of animals did they consider to fit best with these characters?
Not quite — the answer is B.
Having a cartilaginous skeleton, a circular sucking mouth used to attach as an ectoparasite on fish, no scales or paired fins, and seven pairs of gill slits are features of cyclostomes such as Petromyzon (the lamprey). Scoliodon is a shark with paired fins and placoid scales, Exocoetus is a bony fish, and Branchiostoma is a cephalochordate with no cartilaginous skeleton.
Q56NEET 2026 (3 May, cancelled)
Match List I with List II with respect to chronology of evolution of life forms List-I: A. About 65 mya B. About 500 mya C. About 350 mya D. About 320 mya List-II: I. Jawless fish probably evolved II. The dinosaurs suddenly disappeared from the earth III. Seaweeds and few plants probably existed IV. Invertebrates were formed and became active Choose the correct answer from the options given below:
Not quite — the answer is D.
The dinosaurs suddenly disappeared about 65 mya, invertebrates were formed and active by about 500 mya, jawless fish probably evolved around 350 mya, and seaweeds and a few plants probably existed around 320 mya.
Q57NEET 2026 (3 May, cancelled)
Match List I with List II List-I: A. Progestasert B. Multiload 375 C. Diaphragm D. Saheli List-II: I. Barrier made of rubber used by females II. Oral contraceptive III. Hormone releasing IUD IV. Copper releasing IUD Choose the correct answer from the options given below:
Not quite — the answer is A.
Progestasert is a hormone-releasing IUD, Multiload 375 is a copper-releasing IUD, the diaphragm is a rubber barrier used by females, and Saheli is a non-steroidal oral contraceptive pill. Progestasert and Multiload are both IUDs, so they are easy to swap.
Q58NEET 2026 (3 May, cancelled)
The WBC count of a person's blood sample is 8000/cumm. How many eosinophils and lymphocytes would be in the same blood sample approximately?
Not quite — the answer is B.
Eosinophils make up about 2–3% of WBCs, so 2–3% of 8000 = 160–240/cumm. Lymphocytes make up about 20–25%, so 20–25% of 8000 = 1600–2000/cumm. The other ranges do not match these percentages.
Q59NEET 2026 (3 May, cancelled)
Match List I with List II: List I (Drug): A. Nicotine B. Morphine C. Heroin D. Cocaine List II (Effect): I. Causes sense of euphoria and increased energy II. Stimulates adrenal gland to release catecholamines into blood circulation III. Effective sedative and painkiller IV. A depressant; slows down body function Choose the correct answer from the options given below:
Not quite — the answer is C.
Nicotine stimulates the adrenal gland to release adrenaline and noradrenaline, morphine is an effective sedative and painkiller, heroin (smack) is a depressant that slows body functions, and cocaine causes euphoria and increased energy. Heroin and morphine are both opioids; the difference is that morphine is described as the medical sedative and painkiller.
Q60NEET 2026 (3 May, cancelled)
The human protein named α-1-antitrypsin, obtained from transgenic animals, is used for the treatment of ____.
Not quite — the answer is A.
Human α-1-antitrypsin produced by transgenic animals is used to treat emphysema. Cystic fibrosis is another lung disease and a common distractor, but this protein is not its treatment.
Q61NEET 2026 (3 May, cancelled)
Select the set of fishes which belong to the class Osteichthyes:
Not quite — the answer is D.
Flying fish (Exocoetus), angel fish (Pterophyllum) and fighting fish (Betta) are all bony fishes. Saw fish and dog fish are cartilaginous fishes, hagfish is a cyclostome, and cuttlefish, devil fish and starfish are not fishes at all.
Q62NEET 2026 (3 May, cancelled)
Select the incorrect statements from the following: A. Digestive system in Platyhelminthes is incomplete. B. Bilateral symmetry is a characteristic feature of adult Echinoderms. C. Pseudocoelom is possessed by Aschelminthes. D. Notochord is persistent throughout life in the class Chondrichthyes. E. Members of class Reptilia maintain a constant body temperature. Choose the answer from the options given below:
Not quite — the answer is B.
Adult echinoderms are radially symmetrical (only their larvae are bilateral), and reptiles are cold-blooded (poikilothermous), so these two statements are incorrect. Flatworms have an incomplete gut with a single opening, aschelminthes are pseudocoelomates, and cartilaginous fishes keep their notochord throughout life.
Q63NEET 2026 (3 May, cancelled)
Non-membrane bound cell organelles found in both prokaryotic and eukaryotic cells are ____.
Not quite — the answer is A.
Ribosomes have no membrane and occur in both prokaryotes (70S) and eukaryotes (80S). Centrosomes have no membrane but are absent from prokaryotes, and lysosomes and mitochondria are membrane-bound organelles found only in eukaryotes.
Q64NEET 2026 (3 May, cancelled)
Which of the following equations depicts Verhulst-Pearl logistic population growth?
Not quite — the answer is B.
Verhulst-Pearl logistic growth is dN/dt = rN[(K − N)/K], so growth slows as N approaches the carrying capacity K and stops when N = K. The other forms either put N in the denominator or make growth speed up as N nears K, which is not logistic growth.
Q65NEET 2026 (3 May, cancelled)
Select the incorrect statement with reference to Rh grouping. A. Erythroblastosis foetalis is a condition observed having foetus with Rh⁻ᵛᵉ blood and mother with Rh⁺ᵛᵉ blood. B. Rh antigen is observed on RBCs in the majority of human beings. C. Before blood transfusion, Rh group should also be matched. D. Rh incompatibility is observed when a pregnant mother is Rh⁻ᵛᵉ and the foetus is Rh⁺ᵛᵉ. E. Erythroblastosis foetalis can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the second child. Choose the answer from the options given below :
Not quite — the answer is C.
Erythroblastosis foetalis occurs when the mother is Rh-negative and the foetus Rh-positive, not the reverse. It is prevented by giving anti-Rh antibodies to the mother right after the first delivery, not the second. So these two statements are wrong. About 80% of people carry the Rh antigen, and Rh grouping must be matched before transfusion.
Q66NEET 2026 (3 May, cancelled)
Match List I with List II: List I (Bioactive molecules): A. Streptokinase B. Statins C. Lipases D. Cyclosporin A List II (Importance): I. Immunosuppressive agent II. Removal of clots from the blood vessels III. Blood cholesterol-lowering agent IV. Detergent formulations Choose the correct answer from the options given below:
Not quite — the answer is D.
Streptokinase, from Streptococcus, removes clots from blood vessels; statins, from Monascus purpureus, lower blood cholesterol; lipases are used in detergent formulations to remove oily stains; and cyclosporin A, from Trichoderma polysporum, is an immunosuppressant used in organ transplants.
Q67NEET 2026 (3 May, cancelled)
Match List I with List II: List I: A. Molluscs B. Reptiles C. Adult amphibians D. Amoeba List II: I. Pulmonary respiration only II. Branchial respiration III. Cellular respiration IV. Pulmonary and cutaneous respiration Choose the correct answer from the options given below:
Not quite — the answer is A.
Molluscs mostly breathe through gills (branchial respiration), reptiles breathe only through lungs, adult amphibians use both lungs and moist skin, and Amoeba exchanges gases by simple diffusion across its body surface, which this list calls cellular respiration.
Q68NEET 2026 (3 May, cancelled)
The sixth mutant codon of beta globin gene causing polymerization of Haemoglobin and change in RBC shape is ____
Not quite — the answer is A.
In sickle-cell anaemia the normal sixth codon GAG (glutamic acid) is changed to GUG (valine). This makes the mutant haemoglobin polymerise at low oxygen and turns RBCs sickle-shaped. GAG is the normal codon, not the mutant one.
Q69NEET 2026 (3 May, cancelled)
Choose the correct statements regarding muscle contraction. A. A motor neuron carries a signal sent by the Central Nervous System (CNS) to the sarcolemma of the muscle fibre. B. The neural signal generates an action potential which causes the release of Ca²⁺ into sarcoplasm. C. Increase in Ca²⁺ inactivates the actin for breaking cross bridges. D. Actin binds to the myosin head to form a cross bridge. E. Shortening of sarcomere takes place, by pulling actin filaments towards the centre of 'A' band. Choose the correct answer from the options given below :
Not quite — the answer is D.
The motor neuron carries the CNS signal to the sarcolemma, the action potential releases Ca²⁺ into the sarcoplasm, actin and myosin heads link to form cross bridges, and the actin filaments are pulled towards the middle of the dark band, shortening the sarcomere. Ca²⁺ binds troponin and exposes the active sites on actin, so it activates cross-bridge formation rather than inactivating actin.
Q70NEET 2026 (3 May, cancelled)
Which of the following statements are correct with reference to human endoskeleton? A. Human skull is monocondylic. B. The joint between any two adjoining vertebrae is a cartilaginous joint. C. In human beings, the number of cervical vertebrae is seven. D. All ribs except the last 2 pairs are bicephalic. E. The occipital bone of skull is articulated with atlas vertebra. Choose the correct answer from the options given below:
Not quite — the answer is B.
Adjacent vertebrae are joined by cartilaginous joints, humans have seven cervical vertebrae, and the two occipital condyles articulate with the atlas, so these are correct. The human skull is dicondylic, not monocondylic, and every rib, including the floating ribs, is bicephalic, with two articulating surfaces at its dorsal end.
Q71NEET 2026 (3 May, cancelled)
Spermatogonia undergo a series of cell divisions statements to produce sperms. Select the correct from the following : A. Spermatogonia always undergo meiotic cell division. B. Primary spermatocytes divide mitotically to produce secondary spermatocytes. C. Secondary spermatocytes, through their second meiotic division, produce haploid spermatids. D. Spermatids produce spermatozoa through mitosis. E. Spermatids transform into spermatozoa by spermiogenesis. Choose the correct answer from the options given below:
Not quite — the answer is B.
Secondary spermatocytes complete the second meiotic division to form haploid spermatids, and spermatids change into spermatozoa by spermiogenesis, which involves no division. Spermatogonia multiply by mitosis (only those that become primary spermatocytes enter meiosis), and primary spermatocytes divide meiotically, not mitotically.
Q72NEET 2026 (3 May, cancelled)
The JGA (Juxta Glomerular Apparatus) is a special sensitive region formed by cellular modifications in ____ related to the same nephron.
Not quite — the answer is D.
The JGA forms where the distal convoluted tubule touches the afferent arteriole of the same nephron. The arteriole wall carries the renin-secreting juxtaglomerular cells and the tubule carries the macula densa. The PCT and the efferent arteriole take no part in it.
Q73NEET 2026 (3 May, cancelled)
Which one of the following is an appropriate example of sexual deceit?
Not quite — the answer is B.
One petal of the Mediterranean orchid Ophrys looks like a female bee, so male bees try to mate with it ("pseudocopulation") and pick up pollen. This is sexual deceit. The fig and wasp relationship is mutualism, the clownfish and sea anemone is commensalism, and the cuckoo laying eggs in a crow's nest is brood parasitism.
Q74NEET 2026 (3 May, cancelled)
Choose the correct statements regarding frog's anatomy: A. Hepatic portal system is the special venous connection between liver and intestine. B. There are twelve pairs of cranial nerves arising from the brain. C. The ureters and oviducts open separately into the cloaca in female frogs. D. Hind-brain consists of cerebellum, medulla oblongata and optic lobes. E. Sinus venosus joins the right atrium of heart. Choose the correct answer from the options given below:
Not quite — the answer is D.
In the frog the hepatic portal system links the intestine and liver, the ureters and oviducts open separately into the cloaca in females, and the sinus venosus joins the right atrium, so these are correct. Frogs have ten pairs of cranial nerves, not twelve, and the optic lobes belong to the midbrain, not the hindbrain.
Q75NEET 2026 (3 May, cancelled)
Match List I with List II related to embryonic development at various months of pregnancy: List I: A. The foetus movement starts and hair appears on the head B. The foetus develops limbs and digits C. The foetus develops external genital organs D. The foetus body is covered with fine hair; eyelids separate and eyelashes are formed List II: I. 24 weeks of pregnancy II. 20 weeks of pregnancy III. 8 weeks of pregnancy IV. 12 weeks of pregnancy Choose the correct answer from the options given below:
Not quite — the answer is B.
Limbs and digits develop by the end of the second month (8 weeks), external genital organs are well developed by 12 weeks, foetal movements and hair on the head appear in the fifth month (about 20 weeks), and by 24 weeks the body is covered with fine hair and the eyelids separate.
Q76NEET 2026 (3 May, cancelled)
In a population of a grasshopper species, the chromosome number of some members is 23 and some other members possess 24 chromosomes. The 23 and 24 chromosome-bearing members in this species are ____.
Not quite — the answer is C.
Grasshoppers have XO sex determination: females are XX and males have a single X with no Y. So the 23-chromosome individuals are males and the 24-chromosome individuals are females; the reverse pairing is the trap.
Q77NEET 2026 (3 May, cancelled)
In which animal do haploid cells divide mitotically to produce gametes?
Not quite — the answer is A.
Male honeybees (drones) develop from unfertilised eggs and are haploid, so they form sperms by mitosis. The males of earthworms, frogs and grasshoppers have diploid germ cells and make gametes by meiosis.
Q78NEET 2026 (3 May, cancelled)
Arrange the following cell layers/structures around the female gamete, from outer to inner side : A. Zona pellucida B. Perivitelline space C. Corona radiata D. Plasma membrane of ovum Choose the correct answer from the options given below :
Not quite — the answer is C.
From the outside in, the ovum is surrounded by the corona radiata of follicle cells, then the zona pellucida, then the perivitelline space, and then its own plasma membrane. The perivitelline space lies between the zona and the plasma membrane, so it cannot come after the membrane.
Q79NEET 2026 (3 May, cancelled)
What is the reason behind production of large holes in 'Swiss Cheese'?
Not quite — the answer is A.
The large holes in Swiss cheese come from the large amount of CO₂ released by the bacterium Propionibacterium sharmanii. Clostridium butylicum produces butyric acid, Lactobacillus forms curd, and Trichoderma polysporum is the fungus that yields the immunosuppressant cyclosporin.
Q80NEET 2026 (3 May, cancelled)
The toxin proteins isolated from Bacillus thuringiensis, coded by which of the following genes would control cotton bollworms and corn borer, respectively?
Not quite — the answer is A.
The proteins coded by cryIAc and cryIIAb control cotton bollworms, while the protein coded by cryIAb controls the corn borer. So the pair "cryIAc then cryIAb" fits. The pair with cryIIAb second gives two cotton-bollworm genes, and cryIIIAb is not a gene named for either pest.
Q81NEET 2026 (3 May, cancelled)
Ecological pyramids represent the relationship between the organisms at different trophic levels and they are generally inverted for:
Not quite — the answer is D.
In the sea the standing biomass of phytoplankton is smaller than that of the fishes feeding on them, so the pyramid of biomass is inverted. Grassland pyramids of number and biomass are upright, and a pyramid of energy can never be inverted.
Q82NEET 2026 (3 May, cancelled)
Choose the correct statement regarding GIFT to overcome infertility.
Not quite — the answer is D.
Gamete intra-fallopian transfer (GIFT) places an ovum from a donor into the fallopian tube of a woman who cannot make ova but can support fertilisation and development. Transferring embryos of up to 8 blastomeres into the fallopian tube is ZIFT, and placing embryos in the uterus is IUT.
Q83NEET 2026 (3 May, cancelled)
Choose the correct statements regarding cell organelles and their inclusions. A. The endomembrane system includes Golgi complex, endoplasmic reticulum and mitochondria. B. Rough endoplasmic reticulum bears ribosomes on its surface. C. Both mitochondria and plastids have circular DNA. D. A network of microtubules, microfilaments and intermediate filaments present in the cytoplasm is called cytoskeleton. E. Mitochondrion is a single membrane-bound structure. Choose the correct answer from the options given below :
Not quite — the answer is D.
RER carries ribosomes, mitochondria and plastids contain circular DNA, and the cytoskeleton is the network of microtubules, microfilaments and intermediate filaments, so these are correct. Mitochondria are not part of the endomembrane system because their functions are not coordinated with it, and they are bounded by a double membrane.
Q84NEET 2026 (3 May, cancelled)
Match List I with List II related to muscular/skeletal system: List I: A. Tetany B. Arthritis C. Myasthenia gravis D. Muscular dystrophy List II: I. Inflammation of joints II. Autoimmune disorder affecting neuromuscular junction III. Wild contraction in muscle due to low Ca²⁺ in body fluid IV. Progressive degeneration of skeletal muscle Choose the correct answer from the options given below:
Not quite — the answer is A.
Tetany is rapid spasms (wild contractions) caused by low Ca²⁺ in body fluid, arthritis is inflammation of joints, myasthenia gravis is an autoimmune disorder of the neuromuscular junction, and muscular dystrophy is progressive degeneration of skeletal muscle, mostly genetic.
Q85NEET 2026 (3 May, cancelled)
Evolution of human appears parallel to the progressive development of brain and language skills. As such, the evolution of individual species in the sequence of their appearance is:
Not quite — the answer is A.
Ramapithecus, a hominid-like primate of about 15 mya, came first; then Homo habilis (about 2 mya), Homo erectus (about 1.5 mya), Neanderthal man (about 1,00,000–40,000 years ago) and finally Homo sapiens. Sequences that put Ramapithecus after Homo, or Homo sapiens at the start, reverse the timeline.
Q86NEET 2026 (3 May, cancelled)
The flightless bird with forelimbs modified as paddle-like structures suited for swimming is known as:
Not quite — the answer is C.
Aptenodytes is the penguin, a flightless bird whose forelimbs are modified into flippers (paddles) for swimming. Struthio, the ostrich, is also flightless but is a running bird, while Neophron (vulture) and Psittacula (parrot) can fly.
Q87NEET 2026 (3 May, cancelled)
Choose the correct statements regarding population interactions between two species. A. In both parasitism and commensalism, only one species benefits and the other species is harmed. B. Both species benefit in mutualism. C. Both species benefit in commensalism. D. In parasitism, only one species benefits and the other species is harmed. E. In amensalism, one species is harmed and the other is unaffected. Choose the correct answer from the options given below:
Not quite — the answer is C.
Both partners benefit in mutualism, parasitism benefits one species and harms the other, and in amensalism one species is harmed while the other is unaffected, so these three are correct. In commensalism one species benefits and the other is neither harmed nor benefited, so both statements claiming harm or mutual benefit in commensalism are wrong.
Biology: not in the current NEET syllabus
These 1 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
1 questions. Tap an option to check your answer.
Q1NEET 2026 (3 May, cancelled)
Arrange the following steps of somatic hybridisation in a correct sequence. A. Digestion of cell walls. B. Isolation of naked protoplasts. C. Fusion of protoplasts to get hybrid protoplast. D. Isolation of single cells from two different varieties of plants. E. Growing of hybrid protoplast to form a new plant. Choose the correct answer from the options given below:
Not quite — the answer is A.
Single cells are first isolated from the two plant varieties, their cell walls are digested to release naked protoplasts, the protoplasts are fused into a hybrid protoplast, and the hybrid is grown into a new plant. Protoplasts cannot be isolated before the walls are digested, and the process cannot begin with growing the plant.
Chapter-wise split of NEET 2026 (3 May, cancelled) Biology
- Principles of Inheritance and Variation7 Q
- Biodiversity and Conservation5 Q
- Biotechnology: Principles and Processes5 Q
- Biomolecules5 Q
- Molecular Basis of Inheritance4 Q
- Animal Kingdom4 Q
- Morphology of Flowering Plants3 Q
- Ecosystem3 Q
- Cell: The Unit of Life3 Q
- Photosynthesis in Higher Plants3 Q
- Plant Growth and Development3 Q
- Sexual Reproduction in Flowering Plants3 Q
- Breathing and Exchange of Gases3 Q
- Evolution3 Q
- Organisms and Populations3 Q
- Locomotion and Movement3 Q
- Human Reproduction3 Q
- Anatomy of Flowering Plants2 Q
- Respiration in Plants2 Q
- Excretory Products and Their Elimination2 Q
- Human Health and Disease2 Q
- Structural Organisation in Animals2 Q
- Reproductive Health2 Q
- Body Fluids and Circulation2 Q
- Biotechnology and Its Applications2 Q
- Microbes in Human Welfare2 Q
- Plant Kingdom1 Q
- Cell Cycle and Cell Division1 Q
- The Living World1 Q
- Biological Classification1 Q
- Chemical Coordination and Integration1 Q
- Neural Control and Coordination1 Q
About this paper
How many questions from NEET 2026 (3 May, cancelled) are on this page?
164 questions (Physics 35, Chemistry 41, Biology 88). Questions that need a diagram we could not reproduce, or whose wording in the source was unclear, are left out.
Are these the original questions?
Yes. The questions and options are as set in the NEET 2026 (3 May, cancelled) paper, conducted by NTA. The answer to every question was checked by the RankUp team, and the explanations are RankUp's own.
Is NEET 2026 (3 May, cancelled) useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books. 2 questions here are on topics NCERT has since dropped; they are listed separately so you can skip them.
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