NEET 2024 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
170 questions from the NEET 2024 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.
- Exam
- NEET 2024
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- NTA
- Physics
- 33 questions
- Chemistry
- 48 questions
- Biology
- 89 questions
NEET 2024 Physics questions
32 questions. Tap an option to check your answer.
Q1NEET 2024
In a vernier callipers, (N + 1) divisions of vernier scale coincide with N divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:
Not quite — the answer is B.
(N + 1) VSD = N MSD, so 1 VSD = N/(N + 1) MSD. Vernier constant = 1 MSD − 1 VSD = MSD/(N + 1) = 0.1 mm/(N + 1) = 0.01 cm/(N + 1) = 1/(100(N + 1)) cm. Using N instead of N + 1 in the denominator is the usual slip.
Q2NEET 2024
If the monochromatic source in Young's double slit experiment is replaced by white light, then
Not quite — the answer is C.
At the centre the path difference is zero for every wavelength, so all colours interfere constructively and the central fringe is bright white. Away from the centre, fringe widths differ for different wavelengths, so only a few coloured fringes are seen before they overlap into uniform illumination. The pattern does not disappear and fringe widths are not equal.
Q3NEET 2024
A logic circuit provides the output Y as per the following truth table : A B Y 0 0 1 0 1 0 1 0 1 1 1 0 The expression of the output Y is :
Not quite — the answer is C.
Y is 1 exactly when B = 0 and 0 when B = 1, whatever A is, so Y = B̄. Check A·B + Ā: for A = 1, B = 0 it gives 0, but the table shows 1. A·B̄ + Ā gives 1 for A = 0, B = 1, but the table shows 0.
Q4NEET 2024
A wire of length 'l' and resistance 100 Ω is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
Not quite — the answer is B.
Each part is 10 Ω. Five in series give 50 Ω; five in parallel give 10/5 = 2 Ω. In series the total is 50 + 2 = 52 Ω.
Q5NEET 2024
A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4π × 10⁻⁷ SI units):
Not quite — the answer is C.
B = μ₀NI/(2R) = 4π × 10⁻⁷ × 100 × 7/(2 × 0.10) = 8.796 × 10⁻⁴/0.2 ≈ 4.4 × 10⁻³ T = 4.4 mT. Forgetting to convert the radius to metres or the factor 2 gives the larger values.
Q6NEET 2024
In an ideal transformer, the turns ratio is N_P/N_S = 1/2. The ratio V_S : V_P is equal to (the symbols carry their usual meaning) :
Not quite — the answer is B.
For an ideal transformer V_S/V_P = N_S/N_P = 2/1, so V_S : V_P = 2 : 1 (a step-up transformer). Writing the turns ratio as given (1 : 2) inverts the answer.
Q7NEET 2024
A bob is whirled in a horizontal plane by means of a string with an initial speed of ω rpm. The tension in the string is T. If speed becomes 2ω while keeping the same radius, the tension in the string becomes:
Not quite — the answer is B.
The tension provides the centripetal force: T = mω²r. Doubling ω at the same radius multiplies the tension by 2² = 4, giving 4T. A linear dependence would wrongly give 2T.
Q8NEET 2024
The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 g cm². The length of the 400 g rod is nearly:
Not quite — the answer is A.
I = ML²/12, so L² = 12I/M = 12 × 2400/400 = 72 cm², L = √72 ≈ 8.5 cm. The value 72.0 cm is L² mistaken for L; 17.5 cm and 20.7 cm come from using the end-axis formula or other errors.
Q9NEET 2024
A particle moving with uniform speed in a circular path maintains:
Not quite — the answer is D.
In uniform circular motion the speed is constant but the direction of velocity changes continuously, so velocity varies. The centripetal acceleration has constant magnitude v²/r but always points towards the centre, so its direction (and hence the acceleration vector) also varies.
Q10NEET 2024
If c is the velocity of light in free space, the correct statements about photon among the following are: A. The energy of a photon is E = hν. B. The velocity of a photon is c. C. The momentum of a photon, p = hν/c. D. In a photon-electron collision, both total energy and total momentum are conserved. E. Photon possesses positive charge. Choose the correct answer from the options given below:
Not quite — the answer is B.
A photon has energy hν, travels at c in free space, has momentum hν/c, and in photon–electron collisions (Compton-type) total energy and momentum are conserved. A photon is electrically neutral, so the statement about positive charge is false; the correct set is the first four statements.
Q11NEET 2024
At any instant of time t, the displacement of any particle is given by 2t − 1 (SI unit) under the influence of force of 5 N. The value of instantaneous power is (in SI unit):
Not quite — the answer is A.
Velocity v = d(2t − 1)/dt = 2 m s⁻¹. Power P = F·v = 5 × 2 = 10 W. Using the displacement value instead of its derivative gives a time-dependent wrong answer.
Q12NEET 2024
The quantities which have the same dimensions as those of solid angle are:
Not quite — the answer is A.
Solid angle (area/r²) is dimensionless. Strain (change in length/length) and plane angle (arc/radius) are also dimensionless. Stress has dimensions [ML⁻¹T⁻²], arc has [L] and angular speed has [T⁻¹], so the other pairings fail.
Q13NEET 2024
The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's modulus, respectively, are 8 × 10⁸ N m⁻² and 2 × 10¹¹ N m⁻², is:
Not quite — the answer is A.
Maximum strain = stress/Y = 8 × 10⁸/2 × 10¹¹ = 4 × 10⁻³. Maximum elongation = strain × L = 4 × 10⁻³ × 1 m = 4 × 10⁻³ m = 4 mm.
Q14NEET 2024
The mass of a planet is 1/10th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:
Not quite — the answer is D.
g = GM/R², so g_p/g_e = (M_p/M_e)(R_e/R_p)² = (1/10) × 2² = 0.4. g_p = 0.4 × 9.8 = 3.92 m s⁻². Using the diameter ratio without squaring gives 1.96 m s⁻², and ignoring the mass gives 39.2 m s⁻².
Q15NEET 2024
Match List I with List II: List I (Spectral Lines of Hydrogen for transitions from): A. n₂ = 3 to n₁ = 2 B. n₂ = 4 to n₁ = 2 C. n₂ = 5 to n₁ = 2 D. n₂ = 6 to n₁ = 2 List II (Wavelengths (nm)): I. 410.2 II. 434.1 III. 656.3 IV. 486.1 Choose the correct answer from the options given below:
Not quite — the answer is B.
These are Balmer lines; a larger jump to n = 2 releases more energy and gives a shorter wavelength. So 3→2 is the longest, 656.3 nm (Hα); 4→2 is 486.1 nm; 5→2 is 434.1 nm; 6→2 is 410.2 nm. (Check: 1/λ = R(1/4 − 1/9) gives 656 nm for 3→2.)
Q16NEET 2024
An unpolarised light beam strikes a glass surface at Brewster's angle. Then
Not quite — the answer is D.
At Brewster's angle the reflected and refracted rays are perpendicular, and the component of the electric field in the plane of incidence is not reflected at all, so the reflected light is completely plane polarised. The refracted beam still contains both components (with less of the perpendicular one), so it is only partially polarised.
Q17NEET 2024
Match List I with List II List I (Material): A. Diamagnetic B. Ferromagnetic C. Paramagnetic D. Non-magnetic List II (Susceptibility (χ)): I. χ = 0 II. 0 > χ ≥ −1 III. χ >> 1 IV. 0 < χ < ε (a small positive number) Choose the correct answer from the options given below.
Not quite — the answer is A.
Diamagnetic materials have small negative susceptibility (−1 ≤ χ < 0, with −1 for a superconductor). Ferromagnetic materials have very large positive χ. Paramagnetic materials have small positive χ, and a non-magnetic material has χ = 0.
Q18NEET 2024
Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity v₁ while body B is at rest before collision. The velocity of the system after collision is v₂. The ratio v₁ : v₂ is
Not quite — the answer is B.
Momentum is conserved: m·v₁ + 0 = 2m·v₂, so v₂ = v₁/2 and v₁ : v₂ = 2 : 1. The ratio 1 : 2 is inverted, and 4 : 1 would come from wrongly conserving kinetic energy in a form that is not valid for an inelastic collision.
Q19NEET 2024
If x = 5 sin(πt + π/3) m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
Not quite — the answer is B.
Comparing with x = A sin(ωt + φ): A = 5 m and ω = π rad/s, so T = 2π/ω = 2 s. The displacement is in metres, so 5 cm is wrong, and T = 1 s would need ω = 2π.
Q20NEET 2024
A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 N m⁻¹, then the excess force required to take it away from the surface is
Not quite — the answer is A.
Surface tension acts along the circumference of the disc where it touches water: F = S × 2πr = 0.07 × 2π × 0.045 = 0.0198 N ≈ 19.8 mN. Forgetting to convert cm to m gives the large newton values.
Q21NEET 2024
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The potential (V) at any axial point, at 2 m distance (r) from the centre of the dipole of dipole moment vector P⃗ of magnitude, 4 × 10⁻⁶ C m, is ±9 × 10³ V. (Take 1/(4πε₀) = 9 × 10⁹ SI units) Reason R: V = ±2P/(4πε₀r²), where r is the distance of any axial point, situated at 2 m from the centre of the dipole. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is C.
On the axis (r much larger than the dipole length) V = ±P/(4πε₀r²) = ±9 × 10⁹ × 4 × 10⁻⁶/4 = ±9 × 10³ V, so the assertion is true. The reason puts a factor 2 in the potential; 2P/(4πε₀r³) is the axial field form, and with the factor 2 the potential would be ±1.8 × 10⁴ V, so the reason is false.
Q22NEET 2024
A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates:
Not quite — the answer is B.
While charging, the changing electric field between the plates gives a displacement current I_d = ε₀ dΦ_E/dt = dq/dt, equal to the conduction current I and in the same direction, so total current is continuous. There is no conduction current in the gap, but saying there is no current at all ignores the displacement current.
Q23NEET 2024
The property which is not of an electromagnetic wave travelling in free space is that:
Not quite — the answer is D.
Electromagnetic waves are produced by accelerated charges; a charge moving with uniform velocity produces steady fields and does not radiate. The other statements are true: EM waves are transverse, the electric and magnetic energy densities are equal, and their speed in free space is 1/√(μ₀ε₀).
Q24NEET 2024
A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object is:
Not quite — the answer is B.
For normal adjustment (final image at infinity), m = f_o/f_e = 140/5.0 = 28. The value 32 would apply for final image at the near point, m = (f_o/f_e)(1 + f_e/D) = 28 × 1.2 = 33.6, which is not intended for a distant object in normal adjustment.
Q25NEET 2024
Two heaters A and B have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:
Not quite — the answer is B.
R = V²/P, so R_A = V²/1000 and R_B = V²/2000. In series P_s = V²/(R_A + R_B) = 1/(1/1000 + 1/2000) = 2000/3 W ≈ 0.667 kW. In parallel P_p = 1 + 2 = 3 kW. Ratio P_s : P_p = (2/3) : 3 = 2 : 9.
Q26NEET 2024
If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is x/2 times its original time period. Then the value of x is:
Not quite — the answer is B.
T = 2π√(L/g) does not depend on the bob's mass. Halving L gives T′ = T/√2 = (√2/2)T. Comparing with T′ = (x/2)T gives x = √2. Including the mass (factor √3) is the trap.
Q27NEET 2024
The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2R from the surface of the earth is:
Not quite — the answer is A.
At altitude 2R the orbit radius is 3R, and the total energy in orbit is −GMm/(2 × 3R) = −GMm/(6R). On the surface the energy is −GMm/R. Energy required = −GMm/(6R) + GMm/R = 5GMm/(6R). Counting only the potential energy change gives 2GMm/(3R), which ignores the orbital kinetic energy.
Q28NEET 2024
A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to: A. hold the sheet there if it is magnetic. B. hold the sheet there if it is non-magnetic. C. move the sheet away from the pole with uniform velocity if it is conducting. D. move the sheet away from the pole with uniform velocity if it is both, non-conducting and non-polar. Choose the correct statement(s) from the options given below:
Not quite — the answer is B.
A magnetic sheet is attracted by the pole, so a force is needed to hold it in place. A conducting sheet moved in the non-uniform field has eddy currents induced that oppose the motion, so a force is needed to keep it moving at uniform velocity. A non-magnetic sheet at rest, or a non-conducting, non-polar sheet, experiences no magnetic force, so no extra force is needed in those cases.
Q29NEET 2024
A metallic bar of Young's modulus, 0.5 × 10¹¹ N m⁻² and coefficient of linear thermal expansion 10⁻⁵ °C⁻¹, length 1 m and area of cross-section 10⁻³ m² is heated from 0°C to 100°C without expansion or bending. The compressive force developed in it is :
Not quite — the answer is B.
Thermal stress = Y α ΔT = 0.5 × 10¹¹ × 10⁻⁵ × 100 = 5 × 10⁷ N m⁻². Force = stress × area = 5 × 10⁷ × 10⁻³ = 5 × 10⁴ N = 50 × 10³ N. The length of the bar does not enter the result.
Q30NEET 2024
An iron bar of length L has magnetic moment M. It is bent at the middle of its length such that the two arms make an angle 60° with each other. The magnetic moment of this new magnet is :
Not quite — the answer is B.
M = m × L, where m is pole strength. After bending at the middle with a 60° angle, the two arms (each L/2) and the line joining the poles form an equilateral triangle, so the pole separation becomes L/2. New moment = m × L/2 = M/2.
Q31NEET 2024
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then A. the charge stored in it, increases. B. the energy stored in it, decreases. C. its capacitance increases. D. the ratio of charge to its potential remains the same. E. the product of charge and voltage increases. Choose the most appropriate answer from the options given below:
Not quite — the answer is B.
With the battery connected, V stays fixed. Reducing d increases C = ε₀A/d, so Q = CV increases and QV increases. Energy ½CV² increases (not decreases), and Q/V = C changes. So the correct statements are that the charge increases, the capacitance increases and the product QV increases.
Q32NEET 2024
A force defined by F = αt² + βt acts on a particle at a given time t. The factor which is dimensionless, if α and β are constants, is:
Not quite — the answer is B.
Each term must have the dimension of force: [α][T²] = [β][T] = [F]. So [α] = [F]/T² and [β] = [F]/T, giving [αt/β] = ([F]/T²)(T)/([F]/T) = 1, which is dimensionless. βt/α has dimensions T², not dimensionless.
Physics: 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 2024
²⁹⁰₈₂X —α→ Y —e⁺→ Z —β⁻→ P —e⁻→ Q In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are
Not quite — the answer is D.
α emission: A − 4, Z − 2 → ²⁸⁶₈₀Y. Positron emission: Z − 1 → ²⁸⁶₇₉Z. β⁻ emission: Z + 1 → ²⁸⁶₈₀P. Emission of another electron: Z + 1 → ²⁸⁶₈₁Q. Mass number stays 286 after the α step since e⁺ and e⁻ carry no nucleons, so Q has A = 286 and Z = 81.
Chapter-wise split of NEET 2024 Physics
- Units and Measurements3 Q
- Wave Optics2 Q
- Current Electricity2 Q
- Work, Energy and Power2 Q
- Gravitation2 Q
- Magnetism and Matter2 Q
- Oscillations2 Q
- Electrostatic Potential and Capacitance2 Q
- Electromagnetic Waves2 Q
- Semiconductor Electronics: Materials, Devices and Simple Circuits1 Q
- Moving Charges and Magnetism1 Q
- Alternating Current1 Q
- Laws of Motion1 Q
- System of Particles and Rotational Motion1 Q
- Motion in a Plane1 Q
- Dual Nature of Radiation and Matter1 Q
- Mechanical Properties of Solids1 Q
- Atoms1 Q
- Mechanical Properties of Fluids1 Q
- Ray Optics and Optical Instruments1 Q
- Electromagnetic Induction1 Q
- Thermal Properties of Matter1 Q
NEET 2024 Chemistry questions
48 questions. Tap an option to check your answer.
Q1NEET 2024
Match List I with List II. List I (Molecule): A. ethane B. ethene C. carbon molecule, C₂ D. ethyne List II (Number and types of bond/s between two carbon atoms): I. one σ-bond and two π-bonds II. two π-bonds III. one σ-bond IV. one σ-bond and one π-bond Choose the correct answer from the options given below :
Not quite — the answer is C.
Ethane has a C–C single bond (one σ), ethene a double bond (one σ + one π) and ethyne a triple bond (one σ + two π). In the C₂ molecule MO theory gives bond order 2 with both bonds π (the σ2p orbital stays empty), so C₂ has two π-bonds. The trap is assigning C₂ a σ + π double bond like ethene.
Q2NEET 2024
The Henry's law constant (K_H) values of three gases (A, B, C) in water are 145, 2 × 10⁻⁵ and 35 kbar respectively. The solubility of these gases in water follows the order:
Not quite — the answer is B.
From p = K_H·x, at a given partial pressure the mole fraction dissolved is inversely proportional to K_H. Smallest K_H (2 × 10⁻⁵ kbar, gas B) means highest solubility, then C (35 kbar), then A (145 kbar): B > C > A. Ordering by increasing K_H value directly would reverse the trend.
Q3NEET 2024
Given below are two statements: Statement I: The boiling point of hydrides of Group 16 elements follow the order H₂O > H₂Te > H₂Se > H₂S. Statement II: On the basis of molecular mass, H₂O is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H₂O, it has higher boiling point. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is A.
Observed boiling points are H₂O (373 K) > H₂Te (271 K) > H₂Se (232 K) > H₂S (213 K), so Statement I is true. Among H₂S, H₂Se, H₂Te boiling point rises with molar mass, and H₂O would be the lowest on mass alone; extensive intermolecular hydrogen bonding makes it the highest, so Statement II is also true.
Q4NEET 2024
Intramolecular hydrogen bonding is present in
Not quite — the answer is A.
In o-nitrophenol the –OH and –NO₂ groups are on adjacent carbons, so the O–H hydrogen bonds to an oxygen of the nitro group within the same molecule (a six-membered chelate ring). In the meta and para isomers the groups are too far apart and hydrogen bonding is intermolecular, and HF shows only intermolecular hydrogen bonding.
Q5NEET 2024
Given below are two statements: Statement I : The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane Statement II : When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point. In the light of the above statements, choose the most appropriate answer from the options given below:
Not quite — the answer is A.
Boiling points are n-pentane 309 K > isopentane 301 K > neopentane 282.5 K, so Statement I is correct. Branching makes the molecule more compact and nearly spherical, reducing contact area and van der Waals forces, which is exactly the reason given in Statement II.
Q6NEET 2024
The compound that will undergo Sɴ1 reaction with the fastest rate is
Not quite — the answer is D.
Sɴ1 rate depends on carbocation stability. (1-Bromoethyl)benzene ionises to a secondary benzylic cation stabilised by resonance with the ring, far more stable than the primary or secondary alkyl cations from the cyclohexyl bromides. Bromobenzene does not ionise at all because the C–Br bond has partial double-bond character and a phenyl cation is very unstable.
Q7NEET 2024
Which one of the following alcohols reacts instantaneously with Lucas reagent?
Not quite — the answer is D.
Lucas reagent (conc. HCl + anhydrous ZnCl₂) reacts by an Sɴ1-type path, so the rate follows carbocation stability: 3° > 2° > 1°. tert-Butyl alcohol, (CH₃)₃C–OH, gives turbidity immediately; the secondary butan-2-ol takes about five minutes, and the primary alcohols do not react at room temperature.
Q8NEET 2024
1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to
Not quite — the answer is B.
Moles of NaOH = 1/40 = 0.025 mol (25 mmol). Moles of HCl = 0.025 L × 0.75 M = 0.01875 mol (18.75 mmol), which neutralises 18.75 mmol NaOH in a 1:1 reaction. NaOH left = 6.25 mmol × 40 mg/mmol = 250 mg; 750 mg is the mass that reacted, not the mass left.
Q9NEET 2024
Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N Choose the correct answer from the options given below:
Not quite — the answer is B.
Ionization enthalpy generally rises across period 2, but boron (≈801 kJ/mol) is below beryllium (≈899 kJ/mol) because boron's outermost electron is in a 2p orbital, which is shielded by 2s electrons and easier to remove than beryllium's 2s electron. So Li (520) < B (801) < Be (899) < C (1086) < N (1402); the simple steady increase misses this inversion.
Q10NEET 2024
The most stable carbocation among the following is :
Not quite — the answer is D.
The 1-methylcyclohexyl cation is tertiary: the positive carbon carries three alkyl groups and has the most hyperconjugating C–H bonds and the strongest +I effect. The two open-chain cations are only secondary and the cyclopentylmethyl cation is primary, so they are less stable.
Q11NEET 2024
Activation energy of any chemical reaction can be calculated if one knows the value of
Not quite — the answer is D.
From the Arrhenius equation, log(k₂/k₁) = (Eₐ/2.303R)(1/T₁ − 1/T₂), so knowing k at two temperatures gives Eₐ directly. A single rate constant is not enough because the pre-exponential factor A is also unknown; collision probability and orientation alone do not give Eₐ.
Q12NEET 2024
Given below are two statements: Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II : Aniline cannot be prepared through Gabriel synthesis. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is A.
Aniline is a Lewis base and forms a salt with the AlCl₃ catalyst, which puts a positive charge on nitrogen and deactivates the ring, so Friedel–Crafts alkylation fails. Gabriel synthesis needs Sɴ2 attack of phthalimide anion on an alkyl halide; aryl halides do not undergo this substitution, so aniline cannot be made this way. Both statements are true.
Q13NEET 2024
Arrange the following elements in increasing order of electronegativity: N, O, F, C, Si Choose the correct answer from the options given below :
Not quite — the answer is A.
Electronegativity increases across a period and decreases down a group. Si (1.8) is below C (2.5) in group 14, and across period 2 the values rise C (2.5) < N (3.0) < O (3.5) < F (4.0). Placing O below N is the main trap.
Q14NEET 2024
Match List I with List II. List I (Conversion): A. 1 mole of H₂O to O₂ B. 1 mol of MnO₄⁻ to Mn²⁺ C. 1.5 mol of Ca from molten CaCl₂ D. 1 mol of FeO to Fe₂O₃ List II (Number of Faraday required): I. 3F II. 2F III. 1F IV. 5F Choose the correct answer from the options given below :
Not quite — the answer is A.
H₂O → ½O₂ + 2H⁺ + 2e⁻, so 1 mol H₂O needs 2F. MnO₄⁻ (Mn +7) → Mn²⁺ needs 5 e⁻, so 5F. Ca²⁺ + 2e⁻ → Ca: 1.5 mol × 2 = 3F. FeO (Fe +2) → Fe₂O₃ (Fe +3) loses 1 e⁻ per Fe, so 1F. Thus 2F, 5F, 3F, 1F in order.
Q15NEET 2024
Match List I with List II. List I (Complex): A. [Co(NH₃)₅(NO₂)]Cl₂ B. [Co(NH₃)₅(SO₄)]Br C. [Co(NH₃)₆][Cr(CN)₆] D. [Co(H₂O)₆]Cl₃ List II (Type of isomerism): I. Solvate isomerism II. Linkage isomerism III. Ionization isomerism IV. Coordination isomerism Choose the correct answer from the options given below:
Not quite — the answer is A.
NO₂⁻ is ambidentate (–NO₂ or –ONO), giving linkage isomers. [Co(NH₃)₅(SO₄)]Br and [Co(NH₃)₅Br]SO₄ are ionization isomers. Exchange of ligands between the complex cation and anion in [Co(NH₃)₆][Cr(CN)₆] gives coordination isomerism, and [Co(H₂O)₆]Cl₃ can exchange water for chloride inside the sphere, giving solvate (hydrate) isomers.
Q16NEET 2024
Match List I with List II. List I (Compound): A. NH₃ B. BrF₅ C. XeF₄ D. SF₆ List II (Shape/geometry): I. Trigonal Pyramidal II. Square Planar III. Octahedral IV. Square Pyramidal Choose the correct answer from the options given below:
Not quite — the answer is A.
NH₃ has 3 bond pairs + 1 lone pair: trigonal pyramidal. BrF₅ has 5 bond pairs + 1 lone pair: square pyramidal. XeF₄ has 4 bond pairs + 2 lone pairs placed trans: square planar. SF₆ has 6 bond pairs: octahedral.
Q17NEET 2024
The energy of an electron in the ground state (n = 1) for He⁺ ion is −x J, then that for an electron in (n = 2) state for Be³⁺ ion in J is
Not quite — the answer is A.
For hydrogen-like species Eₙ = −k·Z²/n². He⁺ (Z = 2), n = 1: E = −4k = −x, so k = x/4. Be³⁺ (Z = 4), n = 2: E = −k·16/4 = −4k = −x. The Z² and n² factors cancel exactly; using Z instead of Z² gives a wrong value.
Q18NEET 2024
In which of the following processes entropy increases? A. A liquid evaporates to vapour. B. Temperature of a crystalline solid lowered from 130 K to 0 K. C. 2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g) D. Cl₂(g) → 2Cl(g) Choose the correct answer from the options given below:
Not quite — the answer is C.
Evaporation increases disorder; decomposition of solid NaHCO₃ produces two gases from a solid; dissociation of Cl₂ doubles the number of gas particles, so entropy rises in all three. Cooling a crystal from 130 K to 0 K reduces molecular motion and entropy, so the choice including the cooling process is wrong.
Q19NEET 2024
Which reaction is NOT a redox reaction?
Not quite — the answer is D.
In BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl every element keeps its oxidation state (Ba +2, Cl −1, Na +1, S +6, O −2); it is a double displacement (precipitation) reaction. In the others Zn is oxidised and Cu²⁺ reduced, I₂ is oxidised to +5 while Cl goes from +5 to 0, and H₂ is oxidised while Cl₂ is reduced.
Q20NEET 2024
Match List I with List II. List I (Quantum Number): A. mₗ B. mₛ C. l D. n List II (Information provided): I. Shape of orbital II. Size of orbital III. Orientation of orbital IV. Orientation of spin of electron Choose the correct answer from the options given below :
Not quite — the answer is B.
The magnetic quantum number mₗ gives the orientation of the orbital, the spin quantum number mₛ the orientation of electron spin, the azimuthal quantum number l the shape of the orbital, and the principal quantum number n its size (and energy). Swapping the roles of l and n is the common trap.
Q21NEET 2024
'Spin only' magnetic moment is same for which of the following ions? A. Ti³⁺ B. Cr²⁺ C. Mn²⁺ D. Fe²⁺ E. Sc³⁺ Choose the most appropriate answer from the options given below.
Not quite — the answer is A.
Unpaired electrons: Ti³⁺ 3d¹ has 1, Cr²⁺ 3d⁴ has 4, Mn²⁺ 3d⁵ has 5, Fe²⁺ 3d⁶ has 4 and Sc³⁺ 3d⁰ has none. Cr²⁺ and Fe²⁺ both have 4 unpaired electrons, so μ = √(4×6) = √24 ≈ 4.90 BM for each. Ti³⁺ and Sc³⁺ differ (1.73 BM vs zero).
Q22NEET 2024
The highest number of helium atoms is in
Not quite — the answer is A.
4 mol He = 4 × 6.022 × 10²³ atoms. 4 u is the mass of a single He atom; 4 g is 1 mol (6.022 × 10²³ atoms); 2.271098 L at STP is 2.271098/22.71098 = 0.1 mol. So 4 mol contains the most atoms.
Q23NEET 2024
Among Group 16 elements, which one does NOT show -2 oxidation state?
Not quite — the answer is D.
Stability of the −2 state falls down group 16 as electronegativity decreases and metallic character increases. O, Se and Te all form compounds in the −2 state (oxides, selenides, tellurides), while polonium, a metal, hardly shows −2 and mainly shows +2 and +4.
Q24NEET 2024
The E° value for the Mn³⁺/Mn²⁺ couple is more positive than that of Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺ due to change of
Not quite — the answer is C.
Reduction of Mn³⁺ (3d⁴) to Mn²⁺ (3d⁵) gives the extra-stable half-filled d⁵ configuration, so Mn³⁺ is strongly reduced and E° (+1.57 V) is high. The reverse direction (d⁵ to d⁴) describes oxidation of Mn²⁺, which is unfavourable, not the reason for the positive E°.
Q25NEET 2024
The reagents with which glucose does not react to give the corresponding tests/products are A. Tollen's reagent B. Schiff's reagent C. HCN D. NH₂OH E. NaHSO₃ Choose the correct options from the given below:
Not quite — the answer is C.
Glucose reduces Tollens' reagent and forms a cyanohydrin with HCN and an oxime with NH₂OH, but it does not give Schiff's test and does not form the hydrogensulphite addition product with NaHSO₃. This failure is evidence that the aldehyde group is mostly tied up in the cyclic hemiacetal form.
Q26NEET 2024
Identify the correct reagents that would bring about the following transformation. C₆H₁₁–CH₂–CH=CH₂ → C₆H₁₁–CH₂–CH₂–CHO (C₆H₁₁ = cyclohexyl)
Not quite — the answer is B.
Hydroboration–oxidation (BH₃ then H₂O₂/OH⁻) adds water anti-Markovnikov, giving the primary alcohol C₆H₁₁–CH₂CH₂CH₂OH, and PCC oxidises a primary alcohol only to the aldehyde. Acid-catalysed hydration gives the Markovnikov secondary alcohol, which oxidises to a ketone, and alkaline KMnO₄ would take the alcohol on to the carboxylic acid.
Q27NEET 2024
In which of the following equilibria, Kp and Kc are NOT equal?
Not quite — the answer is A.
Kp = Kc(RT)^Δn, with Δn = moles of gaseous products − reactants. For PCl₅ ⇌ PCl₃ + Cl₂, Δn = 2 − 1 = 1, so Kp ≠ Kc. In the other three equilibria Δn = 0, so Kp = Kc.
Q28NEET 2024
A compound with a molecular formula of C₆H₁₄ has two tertiary carbons. Its IUPAC name is :
Not quite — the answer is C.
A tertiary carbon is bonded to three other carbons. In 2,3-dimethylbutane, (CH₃)₂CH–CH(CH₃)₂, both C-2 and C-3 are tertiary. 2-Methylpentane has only one tertiary carbon, 2,2-dimethylbutane has a quaternary carbon and no tertiary one, and n-hexane has none.
Q29NEET 2024
Fehling's solution 'A' is
Not quite — the answer is A.
Fehling's solution A is aqueous copper(II) sulphate, and Fehling's solution B is alkaline sodium potassium tartrate (Rochelle salt); they are mixed just before the test. Alkaline Rochelle salt is solution B, and sodium citrate is used in Benedict's reagent.
Q30NEET 2024
Match List I with List II. List I (Process): A. Isothermal process B. Isochoric process C. Isobaric process D. Adiabatic process List II (Conditions): I. No heat exchange II. Carried out at constant temperature III. Carried out at constant volume IV. Carried out at constant pressure Choose the correct answer from the options given below:
Not quite — the answer is D.
Isothermal means constant temperature, isochoric constant volume, isobaric constant pressure, and adiabatic means no heat exchange with the surroundings (q = 0).
Q31NEET 2024
Given below are two statements : Statement I: Both [Co(NH₃)₆]³⁺ and [CoF₆]³⁻ complexes are octahedral but differ in their magnetic behaviour. Statement II: [Co(NH₃)₆]³⁺ is diamagnetic whereas [CoF₆]³⁻ is paramagnetic. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is A.
Both are six-coordinate octahedral Co³⁺ (3d⁶) complexes. NH₃ is a strong-field ligand, so [Co(NH₃)₆]³⁺ is low-spin t₂g⁶ with no unpaired electrons (diamagnetic, d²sp³), while F⁻ is weak-field, so [CoF₆]³⁻ is high-spin with four unpaired electrons (paramagnetic, sp³d²). Both statements are true.
Q32NEET 2024
On heating, some solid substances change from solid to vapour state without passing through liquid state. The technique used for the purification of such solid substances based on the above principle is known as
Not quite — the answer is B.
The direct change from solid to vapour is sublimation, and it is used to separate sublimable solids (such as camphor, naphthalene) from non-sublimable impurities. Distillation involves a liquid boiling, and crystallization depends on solubility differences.
Q33NEET 2024
Match List I with List II. List I (Reaction): A. Cyclohexylidenecyclohexane → 2 Cyclohexanone B. Benzene → Benzophenone (C₆H₅COC₆H₅) C. Cyclohexanol → Cyclohexanone D. Ethylbenzene (C₆H₅CH₂CH₃) → C₆H₅COOK List II (Reagents/Condition): I. C₆H₅COCl/Anhyd. AlCl₃ II. CrO₃ III. KMnO₄/KOH, Δ IV. (i) O₃ (ii) Zn-H₂O Choose the correct answer from the options given below:
Not quite — the answer is C.
Ozonolysis with Zn–H₂O cleaves the C=C of cyclohexylidenecyclohexane into two cyclohexanone molecules. Friedel–Crafts acylation of benzene with C₆H₅COCl/AlCl₃ gives benzophenone. CrO₃ oxidises the secondary alcohol cyclohexanol to cyclohexanone. Hot alkaline KMnO₄ oxidises the whole side chain of ethylbenzene to the benzoate, C₆H₅COOK.
Q34NEET 2024
For the reaction 2A ⇌ B + C, Kc = 4 × 10⁻³. At a given time, the composition of reaction mixture is: [A] = [B] = [C] = 2 × 10⁻³ M. Then, which of the following is correct?
Not quite — the answer is C.
Qc = [B][C]/[A]² = (2 × 10⁻³)(2 × 10⁻³)/(2 × 10⁻³)² = 1. Since Qc = 1 is greater than Kc (4 × 10⁻³), there are too many products and the reaction moves in the backward direction. Forgetting to square [A] would still give Qc ≫ Kc, but the system is clearly not at equilibrium.
Q35NEET 2024
The pair of lanthanoid ions which are diamagnetic is
Not quite — the answer is A.
Ce⁴⁺ is 4f⁰ and Yb²⁺ is 4f¹⁴, so both have no unpaired electrons and are diamagnetic. Ce³⁺ (4f¹), Eu²⁺ and Gd³⁺ (4f⁷), Eu³⁺ (4f⁶), Pm³⁺ (4f⁴) and Sm³⁺ (4f⁵) all have unpaired f electrons.
Q36NEET 2024
Given below are two statements : Statement I : [Co(NH₃)₆]³⁺ is a homoleptic complex whereas [Co(NH₃)₄Cl₂]⁺ is a heteroleptic complex. Statement II : Complex [Co(NH₃)₆]³⁺ has only one kind of ligands but [Co(NH₃)₄Cl₂]⁺ has more than one kind of ligands. In the light of the above statements, choose the correct answer from the options given below.
Not quite — the answer is A.
A homoleptic complex has only one kind of donor ligand and a heteroleptic complex has more than one kind. [Co(NH₃)₆]³⁺ has only NH₃, while [Co(NH₃)₄Cl₂]⁺ has NH₃ and Cl⁻, so both statements are true (Statement II is the definition behind Statement I).
Q37NEET 2024
Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu : 63 g mol⁻¹, 1 F = 96487 C)
Not quite — the answer is B.
Charge Q = I·t = 9.6487 A × 100 s = 964.87 C = 0.01 F. Cu²⁺ + 2e⁻ → Cu, so moles of Cu = 0.01/2 = 0.005 mol and mass = 0.005 × 63 = 0.315 g. Forgetting the factor 2 would give 0.63 g, which is not offered, and the other choices are off by powers of ten.
Q38NEET 2024
For the given reaction: C₆H₁₁–CH=CH–C₆H₁₁ (1,2-dicyclohexylethene) —KMnO₄/H⁺→ 'P' (major product) 'P' is
Not quite — the answer is B.
Hot acidified KMnO₄ cleaves the C=C bond; each =CH– carbon carries one H, so it is oxidised first to an aldehyde and then on to a carboxylic acid, giving two molecules of cyclohexanecarboxylic acid. The diol forms only with cold, dilute alkaline KMnO₄, and the aldehyde does not survive acidic permanganate.
Q39NEET 2024
The products A and B obtained in the following reactions, respectively, are 3ROH + PCl₃ → 3RCl + A ROH + PCl₅ → RCl + HCl + B
Not quite — the answer is D.
With PCl₃, three –OH groups replace the three Cl atoms on phosphorus, giving phosphorous acid: 3ROH + PCl₃ → 3RCl + H₃PO₃. With PCl₅ one Cl is exchanged and the P=O compound phosphoryl chloride forms: ROH + PCl₅ → RCl + HCl + POCl₃. H₃PO₄ is not formed in either reaction.
Q40NEET 2024
The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation. Given R = 8.314 J K⁻¹ mol⁻¹, log 4 = 0.6021
Not quite — the answer is A.
T₁ = 300 K, T₂ = 330 K. log(k₂/k₁) = (Eₐ/2.303R)(T₂ − T₁)/(T₁T₂): 0.6021 = Eₐ × 30/(2.303 × 8.314 × 300 × 330). Eₐ = 0.6021 × 2.303 × 8.314 × 99000/30 = 0.6021 × 19.147 × 3300 ≈ 38,044 J/mol ≈ 38.04 kJ/mol. The other choices are this value with a misplaced decimal.
Q41NEET 2024
During the preparation of Mohr's salt solution (Ferrous ammonium sulphate), which of the following acid is added to prevent hydrolysis of Fe²⁺ ion?
Not quite — the answer is D.
A little dilute H₂SO₄ suppresses hydrolysis of Fe²⁺ (common-ion effect on H⁺) without adding a foreign anion, since the salt is already a sulphate. Nitric acid would oxidise Fe²⁺ to Fe³⁺, HCl introduces chloride, and concentrated H₂SO₄ is an oxidising and dehydrating agent.
Q42NEET 2024
Major products A and B formed in the following reaction sequence, are 2-Methylcyclohexan-1-ol —PBr₃→ A (major) —alc. KOH, Δ→ B (major)
Not quite — the answer is A.
PBr₃ replaces the –OH by –Br (without rearrangement), giving 1-bromo-2-methylcyclohexane. Alcoholic KOH then causes dehydrohalogenation; by Saytzeff's rule the more substituted alkene, 1-methylcyclohexene (trisubstituted double bond), is the major product rather than the less substituted 3-methylcyclohexene. PBr₃ does not brominate the ring while keeping the –OH.
Q43NEET 2024
The plot of osmotic pressure (Π) vs concentration (mol L⁻¹) for a solution gives a straight line with slope 25.73 L bar mol⁻¹. The temperature at which the osmotic pressure measurement is done is (Use R = 0.083 L bar mol⁻¹ K⁻¹)
Not quite — the answer is A.
Π = CRT, so the slope of Π vs C equals RT. T = 25.73/0.083 = 310 K = 310 − 273 = 37°C. Taking 310 as °C forgets the Kelvin conversion.
Q44NEET 2024
Identify the correct answer.
Not quite — the answer is D.
The carbonate ion has three equivalent canonical structures with the C=O double bond on each oxygen in turn. Ozone has only two resonance structures, trigonal planar BF₃ has zero dipole moment, and NF₃ (≈0.23 D) has a smaller dipole moment than NH₃ (≈1.47 D) because the lone-pair moment opposes the N–F bond moments.
Q45NEET 2024
Identify the major product C formed in the following reaction sequence : CH₃–CH₂–CH₂–I —NaCN→ A —OH⁻, Partial hydrolysis→ B —NaOH, Br₂→ C (major)
Not quite — the answer is A.
NaCN gives butanenitrile (CH₃CH₂CH₂CN, one carbon added). Partial hydrolysis gives butanamide, CH₃CH₂CH₂CONH₂. Hofmann bromamide degradation (Br₂/NaOH) removes one carbon, giving propylamine, CH₃CH₂CH₂NH₂. Butylamine would need reduction of the nitrile or amide, not Hofmann degradation.
Q46NEET 2024
Given below are certain cations. Using inorganic qualitative analysis, arrange them in increasing group number from 0 to VI. A. Al³⁺ B. Cu²⁺ C. Ba²⁺ D. Co²⁺ E. Mg²⁺ Choose the correct answer from the options given below.
Not quite — the answer is A.
In the qualitative analysis scheme Cu²⁺ is in group II (sulphide in acid medium), Al³⁺ in group III (hydroxide with NH₄OH/NH₄Cl), Co²⁺ in group IV (sulphide in ammoniacal medium), Ba²⁺ in group V (carbonate) and Mg²⁺ in group VI. So the order is Cu²⁺, Al³⁺, Co²⁺, Ba²⁺, Mg²⁺.
Q47NEET 2024
Consider the following reaction in a sealed vessel at equilibrium with concentrations of N₂ = 3.0 × 10⁻³ M, O₂ = 4.2 × 10⁻³ M and NO = 2.8 × 10⁻³ M. 2NO(g) ⇌ N₂(g) + O₂(g) If 0.1 mol L⁻¹ of NO(g) is taken in a closed vessel, what will be degree of dissociation (α) of NO(g) at equilibrium?
Not quite — the answer is D.
Kc = [N₂][O₂]/[NO]² = (3.0 × 10⁻³)(4.2 × 10⁻³)/(2.8 × 10⁻³)² = 12.6/7.84 = 1.607. Starting with 0.1 M NO: [NO] = 0.1(1 − α), [N₂] = [O₂] = 0.05α, so Kc = (0.5α/(1 − α))². Taking square roots, 0.5α/(1 − α) = 1.268, giving α = 1.268/1.768 ≈ 0.717.
Q48NEET 2024
A compound X contains 32% of A, 20% of B and remaining percentage of C. Then, the empirical formula of X is : (Given atomic masses of A = 64 ; B = 40 ; C = 32 u)
Not quite — the answer is B.
C = 100 − 32 − 20 = 48%. Mole ratios: A = 32/64 = 0.5, B = 20/40 = 0.5, C = 48/32 = 1.5. Dividing by 0.5 gives 1 : 1 : 3, so the empirical formula is ABC₃. Using the percentages directly without dividing by atomic masses gives a wrong ratio.
Chapter-wise split of NEET 2024 Chemistry
- Chemical Bonding and Molecular Structure4 Q
- Aldehydes, Ketones and Carboxylic Acids4 Q
- Equilibrium4 Q
- Some Basic Concepts of Chemistry3 Q
- Coordination Compounds3 Q
- The d- and f-Block Elements3 Q
- Solutions2 Q
- Hydrocarbons2 Q
- Haloalkanes and Haloarenes2 Q
- Alcohols, Phenols and Ethers2 Q
- Classification of Elements and Periodicity in Properties2 Q
- Organic Chemistry: Some Basic Principles and Techniques2 Q
- Chemical Kinetics2 Q
- Amines2 Q
- Electrochemistry2 Q
- Structure of Atom2 Q
- Thermodynamics2 Q
- Redox Reactions1 Q
- Biomolecules1 Q
NEET 2024 Biology questions
87 questions. Tap an option to check your answer.
Q1NEET 2024
Lecithin, a small molecular weight organic compound found in living tissues, is an example of:
Not quite — the answer is B.
Lecithin is a phosphatidylcholine: glycerol esterified with two fatty acids and a phosphate group carrying choline, which makes it a phospholipid found in cell membranes. It is not a simple glyceride because of the phosphorylated head group, and it is neither an amino acid nor a carbohydrate.
Q2NEET 2024
How many molecules of ATP and NADPH are required for every molecule of CO₂ fixed in the Calvin cycle?
Not quite — the answer is D.
Fixing one CO₂ in the Calvin cycle needs 2 ATP and 2 NADPH for reduction of 3-PGA plus 1 more ATP to regenerate RuBP, i.e. 3 ATP and 2 NADPH. Producing one glucose (6 CO₂) therefore needs 18 ATP and 12 NADPH.
Q3NEET 2024
Hind II always cuts DNA molecules at a particular point called recognition sequence and it consists of:
Not quite — the answer is B.
Hind II, the first restriction endonuclease isolated, always cuts DNA at a specific recognition sequence six base pairs long. Many other restriction enzymes recognise palindromes of different lengths, but for Hind II it is 6 bp.
Q4NEET 2024
The cofactor of the enzyme carboxypeptidase is:
Not quite — the answer is A.
Carboxypeptidase is a metalloenzyme whose prosthetic cofactor is the zinc ion. Haem is the prosthetic group of peroxidase and catalase, while niacin and flavin are parts of coenzymes such as NAD and FAD.
Q5NEET 2024
Match List I with List II List - I | List - II A. Rhizopus | I. Mushroom B. Ustilago | II. Smut fungus C. Puccinia | III. Bread mould D. Agaricus | IV. Rust fungus Choose the correct answer from the options given below:
Not quite — the answer is A.
Rhizopus is the bread mould, Ustilago causes smut, Puccinia causes rust and Agaricus is a mushroom. So the correct pairing is Rhizopus-bread mould, Ustilago-smut fungus, Puccinia-rust fungus, Agaricus-mushroom.
Q6NEET 2024
Given below are two statements: Statement I : Bt toxins are insect group specific and coded by a gene cry IAc. Statement II : Bt toxin exists as inactive protoxin in B. thuringiensis. However, after ingestion by the insect the inactive protoxin gets converted into active form due to acidic pH of the insect gut. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is C.
Bt toxins are insect-group specific and the cry IAc gene codes one such toxin that controls cotton bollworms, so Statement I is accepted as true. Statement II is false because the inactive protoxin is converted to the active toxin by the alkaline pH of the insect gut, not an acidic pH.
Q7NEET 2024
Which of the following is an example of actinomorphic flower?
Not quite — the answer is A.
Datura (like mustard and chilli) has radially symmetrical, actinomorphic flowers that can be divided into equal halves through any radial plane. Cassia, Pisum and Sesbania belong to the pea/cassia families and have zygomorphic flowers.
Q8NEET 2024
Identify the set of correct statements: A. The flowers of Vallisneria are colourful and produce nectar. B. The flowers of water lily are not pollinated by water. C. In most of water-pollinated species, the pollen grains are protected from wetting. D. Pollen grains of some hydrophytes are long and ribbon like. E. In some hydrophytes, the pollen grains are carried passively inside water. Choose the correct answer from the options given below.
Not quite — the answer is D.
Water-pollinated flowers such as Vallisneria are not colourful and produce no nectar, so the first statement is wrong. Water lily is insect/wind pollinated even though it is aquatic; pollen of water-pollinated plants is protected by a mucilaginous covering; sea grasses have long ribbon-like pollen; and in some hydrophytes pollen is carried passively inside water. Thus all statements except the one on Vallisneria are correct.
Q9NEET 2024
The lactose present in the growth medium of bacteria is transported to the cell by the action of
Not quite — the answer is C.
In the lac operon the y gene codes permease, which increases the permeability of the cell to β-galactosides and transports lactose into the cell. β-galactosidase (z gene) hydrolyses lactose, and transacetylase (a gene) is the third product.
Q10NEET 2024
Match List I with List II List - I | List - II A. Clostridium butylicum | I. Ethanol B. Saccharomyces cerevisiae | II. Streptokinase C. Trichoderma polysporum | III. Butyric acid D. Streptococcus sp. | IV. Cyclosporin-A Choose the correct answer from the options given below:
Not quite — the answer is C.
Clostridium butylicum produces butyric acid, Saccharomyces cerevisiae is used for ethanol, Trichoderma polysporum yields the immunosuppressant cyclosporin A, and Streptococcus gives streptokinase, the clot buster. Hence butyric acid, ethanol, cyclosporin-A and streptokinase in that order.
Q11NEET 2024
The equation of Verhulst-Pearl logistic growth is dN/dt = rN[(K − N)/K]. From this equation, K indicates:
Not quite — the answer is C.
In the logistic equation dN/dt = rN[(K − N)/K], r is the intrinsic rate of natural increase and K is the carrying capacity, the maximum population the environment can support. As N approaches K the growth rate falls to zero.
Q12NEET 2024
Auxin is used by gardeners to prepare weed-free lawns. But no damage is caused to grass as auxin
Not quite — the answer is C.
Synthetic auxins such as 2,4-D act as herbicides against broad-leaved dicot weeds but do not affect mature monocotyledonous plants, so grasses in a lawn are spared. Promotion of apical dominance or of abscission does not explain why only the weeds are killed.
Q13NEET 2024
Given below are two statements: Statement I : Parenchyma is living but collenchyma is dead tissue. Statement II : Gymnosperms lack xylem vessels but presence of xylem vessels is the characteristic of angiosperms. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is D.
Both parenchyma and collenchyma are living tissues (sclerenchyma is the dead simple tissue), so Statement I is false. Gymnosperms lack vessels in their xylem, whereas vessels are a characteristic feature of angiosperms, so Statement II is true.
Q14NEET 2024
These are regarded as major causes of biodiversity loss: A. Over exploitation B. Co-extinction C. Mutation D. Habitat loss and fragmentation E. Migration Choose the correct option:
Not quite — the answer is D.
The 'Evil Quartet' of causes of biodiversity loss is habitat loss and fragmentation, over-exploitation, alien species invasions and co-extinctions. Mutation and migration are not counted among them, so the set is over-exploitation, co-extinction and habitat loss.
Q15NEET 2024
Which one of the following is not a criterion for classification of fungi?
Not quite — the answer is B.
Fungi are classified on the basis of the morphology of the mycelium, the mode of spore formation and the type of fruiting bodies. The mode of nutrition is not used, because nearly all fungi are heterotrophic (saprophytic or parasitic).
Q16NEET 2024
List of endangered species was released by
Not quite — the answer is D.
The IUCN (International Union for Conservation of Nature) publishes the Red List of threatened and endangered species. GEAC is an Indian body that approves GM organisms, and WWF is a conservation NGO, not the Red List authority.
Q17NEET 2024
What is the fate of a piece of DNA carrying only gene of interest which is transferred into an alien organism? A. The piece of DNA would be able to multiply itself independently in the progeny cells of the organism. B. It may get integrated into the genome of the recipient. C. It may multiply and be inherited along with the host DNA. D. The alien piece of DNA is not an integral part of chromosome. E. It shows ability to replicate. Choose the correct answer from the options given below:
Not quite — the answer is C.
A piece of DNA carrying only the gene of interest has no origin of replication, so it cannot multiply on its own; it can multiply only if it integrates into the recipient genome and is then replicated and inherited along with host DNA. Independent replication and an ability to replicate on its own are therefore wrong.
Q18NEET 2024
Which one of the following can be explained on the basis of Mendel's Law of Dominance? A. Out of one pair of factors one is dominant and the other is recessive. B. Alleles do not show any expression and both the characters appear as such in F₂ generation. C. Factors occur in pairs in normal diploid plants. D. The discrete unit controlling a particular character is called factor. E. The expression of only one of the parental characters is found in a monohybrid cross. Choose the correct answer from the options given below:
Not quite — the answer is B.
The law of dominance explains that characters are controlled by discrete units called factors, that factors occur in pairs, and that in a dissimilar pair one is dominant and the other recessive, so only one parental character is expressed in a monohybrid F₁. The statement that both characters appear unchanged in F₂ without expression is not part of this law.
Q19NEET 2024
Inhibition of Succinic dehydrogenase enzyme by malonate is a classical example of:
Not quite — the answer is C.
Malonate closely resembles succinate in structure and competes with it for the active site of succinic dehydrogenase, which is classical competitive inhibition. Feedback inhibition is inhibition by an end product at a different site, which is not the case here.
Q20NEET 2024
Formation of interfascicular cambium from fully developed parenchyma cells is an example for
Not quite — the answer is C.
Living differentiated cells that have lost the capacity to divide can regain it; this is dedifferentiation. Interfascicular cambium (and cork cambium) arises in this way from mature parenchyma cells, while redifferentiation is the later step in which the products of such meristems mature again.
Q21NEET 2024
Tropical regions show greatest level of species richness because A. Tropical latitudes have remained relatively undisturbed for millions of years, hence more time was available for species diversification. B. Tropical environments are more seasonal. C. More solar energy is available in tropics. D. Constant environments promote niche specialization. E. Tropical environments are constant and predictable. Choose the correct answer from the options given below.
Not quite — the answer is A.
Tropics have been relatively undisturbed for millions of years, have more solar energy for productivity, and are less seasonal, more constant and predictable, which promotes niche specialisation. The statement that tropical environments are more seasonal is false, so the correct set excludes it.
Q22NEET 2024
Given below are two statements: Statement I : Chromosomes become gradually visible under light microscope during leptotene stage. Statement II : The beginning of diplotene stage is recognized by dissolution of synaptonemal complex. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is A.
During leptotene chromosomes become gradually visible under the light microscope, and the beginning of diplotene is recognised by dissolution of the synaptonemal complex while homologues remain joined at chiasmata. Both statements are true.
Q23NEET 2024
Match List I with List II List - I | List - II A. Nucleolus | I. Site of formation of glycolipid B. Centriole | II. Organization like the cartwheel C. Leucoplasts | III. Site of active ribosomal RNA synthesis D. Golgi apparatus | IV. For storing nutrients Choose the correct answer from the options given below:
Not quite — the answer is A.
The nucleolus is the site of active rRNA synthesis, the centriole has a cartwheel organisation of nine peripheral triplet fibrils, leucoplasts store nutrients such as starch, oils and proteins, and the Golgi apparatus is the site of glycoprotein and glycolipid formation.
Q24NEET 2024
Bulliform cells are responsible for
Not quite — the answer is A.
Bulliform cells are large, empty, colourless cells in the upper epidermis of grass leaves. When they lose turgor under water stress the leaf rolls inward, reducing water loss; they do not store sugars or protect against salt.
Q25NEET 2024
A pink flowered Snapdragon plant was crossed with a red flowered Snapdragon plant. What type of phenotype/s is/are expected in the progeny?
Not quite — the answer is B.
Flower colour in snapdragon shows incomplete dominance: red is RR, pink is Rr. Crossing pink (Rr) with red (RR) gives RR and Rr in a 1:1 ratio, i.e. red and pink flowered plants; white (rr) cannot arise because the red parent gives only R.
Q26NEET 2024
A transcription unit in DNA is defined primarily by the three regions in DNA and these are with respect to upstream and down stream end;
Not quite — the answer is D.
A transcription unit is defined by a promoter at the upstream end, the structural gene in the middle and a terminator at the downstream end. Repressor, operator and inducer are elements of operon regulation, not of the basic transcription unit.
Q27NEET 2024
In a plant, black seed color (BB/Bb) is dominant over white seed color (bb). In order to find out the genotype of the black seed plant, with which of the following genotype will you cross it?
Not quite — the answer is B.
To find whether a dominant phenotype is BB or Bb, it is crossed with the homozygous recessive bb (a test cross). All black progeny means the plant is BB, while a 1:1 black:white ratio means it is Bb.
Q28NEET 2024
Which of the following are required for the dark reaction of photosynthesis? A. Light B. Chlorophyll C. CO₂ D. ATP E. NADPH Choose the correct answer from the options given below:
Not quite — the answer is C.
The biosynthetic (dark) phase uses CO₂ together with ATP and NADPH produced in the light reaction. Light and chlorophyll are needed only for the light reaction, not directly for the Calvin cycle.
Q29NEET 2024
Match List I with List II List - I | List - II A. Two or more alternative forms of a gene | I. Back cross B. Cross of F₁ progeny with homozygous recessive parent | II. Ploidy C. Cross of F₁ progeny with any of the parents | III. Allele D. Number of chromosome sets in plant | IV. Test cross Choose the correct answer from the options given below:
Not quite — the answer is C.
Alternative forms of a gene are alleles; crossing F₁ with the homozygous recessive parent is a test cross; crossing F₁ with either parent is a back cross; and the number of chromosome sets is ploidy.
Q30NEET 2024
The DNA present in chloroplast is:
Not quite — the answer is B.
Chloroplasts, like mitochondria, contain their own small, circular, double-stranded DNA molecule together with 70S ribosomes, a feature similar to prokaryotes. They do not contain linear or single-stranded DNA.
Q31NEET 2024
Match List I with List II List - I | List - II A. Robert May | I. Species-Area relationship B. Alexander von Humboldt | II. Long term ecosystem experiment using out door plots C. Paul Ehrlich | III. Global species diversity at about 7 million D. David Tilman | IV. Rivet popper hypothesis Choose the correct answer from the options given below:
Not quite — the answer is B.
Robert May estimated global species diversity at about 7 million, Alexander von Humboldt described the species-area relationship, Paul Ehrlich proposed the rivet popper hypothesis, and David Tilman ran long-term ecosystem experiments using outdoor plots.
Q32NEET 2024
Match List I with List II List - I | List - II A. Rose | I. Twisted aestivation B. Pea | II. Perigynous flower C. Cotton | III. Drupe D. Mango | IV. Marginal placentation Choose the correct answer from the options given below :
Not quite — the answer is A.
Rose has perigynous flowers, pea has marginal placentation, cotton (like China rose) shows twisted aestivation and mango fruit is a drupe.
Q33NEET 2024
Which of the following statement is correct regarding the process of replication in E.coli?
Not quite — the answer is D.
DNA-dependent DNA polymerases can add nucleotides only to a free 3′-OH, so they catalyse polymerisation only in the 5′ → 3′ direction; this is why one strand is synthesised discontinuously. RNA polymerase is not the replication enzyme, and DNA polymerase does not polymerise 3′ → 5′.
Q34NEET 2024
Match List I with List II List - I | List - II A. Citric acid cycle | I. Cytoplasm B. Glycolysis | II. Mitochondrial matrix C. Electron transport system | III. Intermembrane space of mitochondria D. Proton gradient | IV. Inner mitochondrial membrane Choose the correct answer from the options given below:
Not quite — the answer is B.
The citric acid cycle occurs in the mitochondrial matrix, glycolysis in the cytoplasm, the electron transport system is located in the inner mitochondrial membrane, and the proton gradient is built up in the intermembrane space.
Q35NEET 2024
Read the following statements and choose the set of correct statements: In the members of Phaeophyceae, A. Asexual reproduction occurs usually by biflagellate zoospores. B. Sexual reproduction is by oogamous method only. C. Stored food is in the form of carbohydrates which is either mannitol or laminarin. D. The major pigments found are chlorophyll a, c and carotenoids and xanthophyll. E. Vegetative cells have a cellulosic wall, usually covered on the outside by gelatinous coating of algin. Choose the correct answer from the options given below:
Not quite — the answer is C.
In brown algae asexual reproduction is by biflagellate zoospores, food is stored as mannitol or laminarin, pigments are chlorophyll a and c with carotenoids and fucoxanthin, and the cellulosic wall is covered by algin. Sexual reproduction may be isogamous, anisogamous or oogamous, so 'oogamous only' is wrong.
Q36NEET 2024
Match List I with List II List - I | List - II A. GLUT-4 | I. Hormone B. Insulin | II. Enzyme C. Trypsin | III. Intercellular ground substance D. Collagen | IV. Enables glucose transport into cells Choose the correct answer from the options given below.
Not quite — the answer is A.
GLUT-4 is the transporter that enables glucose transport into cells, insulin is a hormone, trypsin is an enzyme and collagen is an intercellular ground substance protein.
Q37NEET 2024
Identify the step in tricarboxylic acid cycle, which does not involve oxidation of substrate.
Not quite — the answer is C.
Conversion of succinyl-CoA to succinic acid releases energy used for substrate-level phosphorylation (GTP/ATP) without removal of hydrogen, so there is no oxidation. Isocitrate → α-ketoglutarate, succinate → fumarate/malate route and malate → oxaloacetate are all dehydrogenations (oxidations).
Q38NEET 2024
Given below are two statements: Statement I: In C₃ plants, some O₂ binds to RuBisCO, hence CO₂ fixation is decreased. Statement II: In C₄ plants, mesophyll cells show very little photorespiration while bundle sheath cells do not show photorespiration. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is C.
In C₃ plants RuBisCO also binds O₂ (photorespiration), so some RuBP is wasted and CO₂ fixation decreases; Statement I is true. In C₄ plants mesophyll cells lack RuBisCO and so do not photorespire, while bundle sheath cells have RuBisCO but high CO₂ there minimises photorespiration, so Statement II reverses the facts and is false.
Q39NEET 2024
Match List I with List II List - I (Types of Stamens) | List - II (Example) A. Monoadelphous | I. Citrus B. Diadelphous | II. Pea C. Polyadelphous | III. Lily D. Epiphyllous | IV. China-rose Choose the correct answer from the options given below:
Not quite — the answer is A.
Stamens united into one bundle (monoadelphous) occur in China rose, two bundles (diadelphous) in pea, more than two bundles (polyadelphous) in citrus, and stamens attached to the perianth (epiphyllous) in lily.
Q40NEET 2024
Match List I with List II List - I | List - II A. Frederick Griffith | I. Genetic code B. Francois Jacob & Jacque Monod | II. Semi-conservative mode of DNA replication C. Har Gobind Khorana | III. Tranformation D. Meselson & Stahl | IV. Lac operon Choose the correct answer from the options given below:
Not quite — the answer is B.
Griffith discovered transformation, Jacob and Monod proposed the lac operon, Har Gobind Khorana helped decipher the genetic code, and Meselson and Stahl proved semi-conservative DNA replication.
Q41NEET 2024
Spraying sugarcane crop with which of the following plant growth regulators, increases the length of stem, thus, increasing the yield?
Not quite — the answer is B.
Gibberellins cause internode elongation; spraying sugarcane with GA increases stem length and therefore yield by as much as 20 tonnes per acre. Auxin, cytokinin and abscisic acid are not used this way.
Q42NEET 2024
Following are the stages of pathway for conduction of an action potential through the heart A. AV bundle B. Purkinje fibres C. AV node D. Bundle branches E. SA node Choose the correct sequence of pathway from the options given below
Not quite — the answer is A.
The impulse starts at the SA node, passes to the AV node, then along the AV bundle (bundle of His), into the right and left bundle branches and finally the Purkinje fibres. So SA node → AV node → AV bundle → bundle branches → Purkinje fibres.
Q43NEET 2024
In both sexes of cockroach, a pair of jointed filamentous structures called anal cerci are present on
Not quite — the answer is B.
In both sexes of cockroach a pair of jointed filamentous anal cerci arises from the 10th abdominal segment. Anal styles, by contrast, are present only in males on the 9th segment.
Q44NEET 2024
The flippers of the Penguins and Dolphins are the example of the
Not quite — the answer is C.
Flippers of penguins (birds) and dolphins (mammals) are analogous structures: similar form and function evolved independently in unrelated groups adapting to the same aquatic habitat, which is convergent evolution. Divergent evolution produces homologous structures from a common ancestor.
Q45NEET 2024
Which of the following is not a component of Fallopian tube?
Not quite — the answer is A.
The fallopian tube consists of the funnel-shaped infundibulum with fimbriae, the wider ampulla and the narrow isthmus that joins the uterus. The fundus is the dome-shaped upper part of the uterus itself, not of the tube.
Q46NEET 2024
Given below are some stages of human evolution. Arrange them in correct sequence. (Past to Recent) A. Homo habilis B. Homo sapiens C. Homo neanderthalensis D. Homo erectus Choose the correct sequence of human evolution from the options given below:
Not quite — the answer is D.
Homo habilis (about 2 million years ago) came first, followed by Homo erectus (about 1.5 million years ago), then Neanderthal man (about 1,00,000-40,000 years ago) and finally Homo sapiens. So habilis → erectus → neanderthalensis → sapiens.
Q47NEET 2024
Which of the following is not a steroid hormone?
Not quite — the answer is D.
Cortisol, testosterone and progesterone are steroid hormones derived from cholesterol. Glucagon from the α-cells of the pancreas is a peptide hormone.
Q48NEET 2024
Match List I with List II List - I | List - II A. α-1 antitrypsin | I. Cotton bollworm B. Cry IAb | II. ADA deficiency C. Cry IAc | III. Emphysema D. Enzyme replacement therapy | IV. Corn borer Choose the correct answer form the options given below:
Not quite — the answer is C.
Human α-1 antitrypsin made by transgenic animals is used to treat emphysema, cry IAb codes a toxin that controls corn borer, cry IAc controls cotton bollworms, and enzyme replacement therapy is used for ADA deficiency.
Q49NEET 2024
Following are the stages of cell division : A. Gap 2 phase B. Cytokinesis C. Synthesis phase D. Karyokinesis E. Gap 1 phase Choose the correct sequence of stages from the options given below :
Not quite — the answer is D.
Interphase runs G₁ → S → G₂, and is followed by M phase, in which karyokinesis (nuclear division) precedes cytokinesis. So Gap 1 → Synthesis → Gap 2 → Karyokinesis → Cytokinesis.
Q50NEET 2024
Which one of the following factors will not affect the Hardy-Weinberg equilibrium?
Not quite — the answer is D.
Hardy-Weinberg equilibrium is disturbed by gene migration (gene flow), genetic drift, mutation, genetic recombination and natural selection. A constant gene pool is the very condition of equilibrium, so it does not disturb it.
Q51NEET 2024
Which of the following are Autoimmune disorders? A. Myasthenia gravis B. Rheumatoid arthritis C. Gout D. Muscular dystrophy E. Systemic Lupus Erythematosus (SLE) Choose the most appropriate answer from the options given below:
Not quite — the answer is B.
Myasthenia gravis, rheumatoid arthritis and systemic lupus erythematosus are autoimmune diseases in which the body attacks its own tissues. Gout is caused by uric acid crystal deposition, and muscular dystrophy is a genetic degenerative disorder.
Q52NEET 2024
Match List I with List II: List - I | List - II A. Typhoid | I. Fungus B. Leishmaniasis | II. Nematode C. Ringworm | III. Protozoa D. Filariasis | IV. Bacteria Choose the correct answer from the options given below:
Not quite — the answer is B.
Typhoid is caused by the bacterium Salmonella typhi, leishmaniasis by a protozoan (Leishmania), ringworm by fungi (e.g. Microsporum, Trichophyton) and filariasis by the nematode Wuchereria.
Q53NEET 2024
Match List I with List II: List - I | List - II A. Expiratory capacity | I. Expiratory reserve volume + Tidal volume + Inspiratory reserve volume B. Functional residual capacity | II. Tidal volume + Expiratory reserve volume C. Vital capacity | III. Tidal volume + Inspiratory reserve volume D. Inspiratory capacity | IV. Expiratory reserve volume + Residual volume Choose the correct answer from the options given below :
Not quite — the answer is A.
Expiratory capacity = TV + ERV; functional residual capacity = ERV + RV; vital capacity = ERV + TV + IRV; inspiratory capacity = TV + IRV.
Q54NEET 2024
Match List I with List II : List - I | List - II A. Down's syndrome | I. 11th chromosome B. α-thalassemia | II. 'X' chromosome C. β-thalassemia | III. 21st chromosome D. Klinefelter's syndrome | IV. 16th chromosome Choose the correct answer from the options given below :
Not quite — the answer is C.
Down's syndrome is trisomy of chromosome 21, α-thalassemia is controlled by genes on chromosome 16, β-thalassemia by the HBB gene on chromosome 11, and Klinefelter's syndrome involves an extra X chromosome (XXY).
Q55NEET 2024
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R: Assertion A : FSH acts upon ovarian follicles in female and Leydig cells in male. Reason R : Growing ovarian follicles secrete estrogen in female while interstitial cells secrete androgen in male human being. In the light of the above statements, choose the correct answer from the options given below :
Not quite — the answer is D.
In males FSH acts on the Sertoli cells, while LH (ICSH) acts on the Leydig cells, so the Assertion is false. The Reason is true: growing ovarian follicles secrete oestrogen and interstitial (Leydig) cells secrete androgens.
Q56NEET 2024
Match List I with List II: List - I | List - II A. Pleurobrachia | I. Mollusca B. Radula | II. Ctenophora C. Stomochord | III. Osteichthyes D. Air bladder | IV. Hemichordata Choose the correct answer from the options given below :
Not quite — the answer is B.
Pleurobrachia is a ctenophore, the radula is the rasping feeding organ of molluscs, the stomochord is the notochord-like structure of hemichordates, and the air bladder is found in bony fishes (Osteichthyes).
Q57NEET 2024
Match List I with List II : List - I (Sub Phases of Prophase I) | List - II (Specific Characters) A. Diakinesis | I. Synaptonemal complex formation B. Pachytene | II. Completion of terminalisation of chiasmata C. Zygotene | III. Chromosomes look like thin threads D. Leptotene | IV. Appearance of recombination nodules Choose the correct answer from the options given below
Not quite — the answer is C.
Diakinesis is marked by completion of terminalisation of chiasmata, pachytene by appearance of recombination nodules, zygotene by formation of the synaptonemal complex, and in leptotene chromosomes look like thin threads.
Q58NEET 2024
The "Ti plasmid" of Agrobacterium tumefaciens stands for
Not quite — the answer is C.
Ti stands for tumour inducing: this plasmid of Agrobacterium tumefaciens transfers T-DNA into dicot plant cells and causes crown gall tumours. Disarmed Ti plasmids are used as cloning vectors.
Q59NEET 2024
Match List I with List II: List - I | List - II A. Cocaine | I. Effective sedative in surgery B. Heroin | II. Cannabis sativa C. Morphine | III. Erythroxylum D. Marijuana | IV. Papaver somniferum Choose the correct answer from the options given below:
Not quite — the answer is D.
Cocaine comes from Erythroxylum coca, heroin is derived (by acetylation) from morphine obtained from the poppy Papaver somniferum, morphine is an effective sedative and painkiller used in surgery, and marijuana comes from Cannabis sativa.
Q60NEET 2024
Which one is the correct product of DNA dependent RNA polymerase to the given template? 3′TACATGGCAAATATCCATTCA5′
Not quite — the answer is A.
RNA is complementary and antiparallel to the template, with U pairing opposite A: 3′TACATGGCAAATATCCATTCA5′ gives 5′AUGUACCGUUUAUAGGUAAGU3′. The sequence with T in place of U is DNA, and the other two contain mismatched bases.
Q61NEET 2024
Match List I with List II: List - I | List - II A. Pons | I. Provides additional space for Neurons, regulates posture and balance. B. Hypothalamus | II. Controls respiration and gastric secretions. C. Medulla | III. Connects different regions of the brain. D. Cerebellum | IV. Neuro secretory cells Choose the correct answer from the options given below :
Not quite — the answer is B.
Pons consists of fibre tracts that interconnect different regions of the brain, the hypothalamus contains neurosecretory cells, the medulla controls respiration and gastric secretions, and the cerebellum provides additional space for neurons and regulates posture and balance.
Q62NEET 2024
Match List I with List II: List - I | List - II A. Lipase | I. Peptide bond B. Nuclease | II. Ester bond C. Protease | III. Glycosidic bond D. Amylase | IV. Phosphodiester bond Choose the correct answer from the options given below :
Not quite — the answer is C.
Lipases hydrolyse ester bonds of fats, nucleases break phosphodiester bonds of nucleic acids, proteases break peptide bonds, and amylase breaks glycosidic bonds of starch.
Q63NEET 2024
Which of the following is not a natural/traditional contraceptive method?
Not quite — the answer is D.
Coitus interruptus, periodic abstinence and lactational amenorrhoea are natural methods that need no device. Vaults are rubber barrier devices placed over the cervix, so they are not a natural method.
Q64NEET 2024
Which of the following statements is incorrect?
Not quite — the answer is C.
Bioreactors are large vessels (100-1000 litres) used for large-scale production of cultures and products; small-scale cultures are grown in flasks. Bioreactors do provide optimal conditions, are commonly of the stirring type, and include agitator, oxygen delivery and foam control systems.
Q65NEET 2024
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion A : Breast-feeding during initial period of infant growth is recommended by doctors for bringing a healthy baby. Reason R : Colostrum contains several antibodies absolutely essential to develop resistance for the new born baby. In the light of the above statements, choose the most appropriate answer from the options given below:
Not quite — the answer is A.
Breast-feeding in the initial period is recommended because the colostrum secreted in the first few days contains several antibodies (IgA) essential for developing resistance in the newborn. Both statements are correct and the Reason explains the Assertion.
Q66NEET 2024
Which of the following factors are favourable for the formation of oxyhaemoglobin in alveoli?
Not quite — the answer is B.
In the alveoli high pO₂, low pCO₂, lower H⁺ concentration (higher pH) and lower temperature favour binding of O₂ to haemoglobin. High pCO₂, high H⁺ and high temperature favour dissociation of oxyhaemoglobin, as in tissues.
Q67NEET 2024
Match List I with List II : List - I | List - II A. Common cold | I. Plasmodium B. Haemozoin | II. Typhoid C. Widal test | III. Rhinoviruses D. Allergy | IV. Dust mites Choose the correct answer from the options given below :
Not quite — the answer is C.
Common cold is caused by rhinoviruses, haemozoin is the toxin released by Plasmodium in malaria, the Widal test confirms typhoid, and dust mites are common allergens.
Q68NEET 2024
Consider the following statements : A. Annelids are true coelomates B. Poriferans are pseudocoelomates C. Aschelminthes are acoelomates D. Platyhelminthes are pseudocoelomates Choose the correct answer from the options given below :
Not quite — the answer is B.
Annelids are true coelomates (eucoelomates). Poriferans have no coelom at all, aschelminthes are pseudocoelomates and platyhelminthes are acoelomates, so only the annelid statement is correct.
Q69NEET 2024
Match List I with List II List - I | List - II A. Non-medicated IUD | I. Multiload 375 B. Copper releasing IUD | II. Progestogens C. Hormone releasing IUD | III. Lippes loop D. Implants | IV. LNG-20 Choose the correct answer from the options given below:
Not quite — the answer is D.
Lippes loop is a non-medicated IUD, Multiload 375 is a copper-releasing IUD, LNG-20 is a hormone-releasing IUD, and implants contain progestogens (alone or with oestrogen).
Q70NEET 2024
Given below are two statements : Statement I : In the nephron, the descending limb of loop of Henle is impermeable to water and permeable to electrolytes. Statement II : The proximal convoluted tubule is lined by simple columnar brush border epithelium and increases the surface area for reabsorption. In the light of the above statements, choose the correct answer from the options given below :
Not quite — the answer is B.
The descending limb of Henle's loop is permeable to water but almost impermeable to electrolytes, so Statement I is false. The PCT is lined by simple cuboidal (not columnar) brush border epithelium, so Statement II as worded is also false.
Q71NEET 2024
Match List I with List II : List - I | List - II A. Axoneme | I. Centriole B. Cartwheel pattern | II. Cilia and flagella C. Crista | III. Chromosome D. Satellite | IV. Mitochondria Choose the correct answer from the options given below :
Not quite — the answer is D.
The axoneme is the core of cilia and flagella, the cartwheel pattern is seen in centrioles, cristae are infoldings of the inner mitochondrial membrane, and a satellite is the small fragment beyond the secondary constriction of some chromosomes.
Q72NEET 2024
Match List I with List II : List - I | List - II A. Pterophyllum | I. Hag fish B. Myxine | II. Saw fish C. Pristis | III. Angel fish D. Exocoetus | IV. Flying fish Choose the correct answer from the options given below :
Not quite — the answer is B.
Pterophyllum is the angel fish, Myxine the hag fish, Pristis the saw fish and Exocoetus the flying fish.
Q73NEET 2024
Match List I with List II : List - I | List - II A. Fibrous joints | I. Adjacent vertebrae, limited movement B. Cartilaginous joints | II. Humerus and Pectoral girdle, rotational movement C. Hinge joints | III. Skull, don't allow any movement D. Ball and socket joints | IV. Knee, help in locomotion Choose the correct answer from the options given below :
Not quite — the answer is D.
Fibrous joints, as between skull bones, allow no movement; cartilaginous joints, as between adjacent vertebrae, allow limited movement; the knee is a hinge joint; and the joint between humerus and pectoral girdle is a ball and socket joint allowing rotation.
Q74NEET 2024
Given below are two statements: Statement I: The presence or absence of hymen is not a reliable indicator of virginity. Statement II: The hymen is torn during the first coitus only. In the light of the above statements, choose the correct answer from the options given below :
Not quite — the answer is C.
The hymen can be torn by a sudden fall or jolt, insertion of a tampon, or active sports such as horse riding or cycling, and in some women it persists even after coitus. Hence its presence or absence is not a reliable indicator of virginity, and the claim that it is torn only during first coitus is false.
Q75NEET 2024
Given below are two statements: Statement I: Gause's competitive exclusion principle states that two closely related species competing for different resources cannot exist indefinitely. Statement II: According to Gause's principle, during competition, the inferior will be eliminated. This may be true if resources are limiting. In the light of the above statements, choose the correct answer from the options given below :
Not quite — the answer is D.
Gause's competitive exclusion principle states that two closely related species competing for the same resources cannot coexist indefinitely; saying 'different resources' makes Statement I false. Statement II is true: the competitively inferior species is eliminated, which may hold when resources are limiting.
Q76NEET 2024
Match List I with List II related to digestive system of cockroach. List - I | List - II A. The structures used for storing of food | I. Gizzard B. Ring of 6-8 blind tubules at junction of foregut and midgut. | II. Gastric Caeca C. Ring of 100-150 yellow coloured thin filaments at junction of midgut and hindgut. | III. Malpighian tubules D. The structures used for grinding the food. | IV. Crop Choose the correct answer from the options given below:
Not quite — the answer is A.
In cockroach the crop stores food, a ring of 6-8 blind gastric caeca lies at the junction of foregut and midgut, a ring of 100-150 yellow Malpighian tubules lies at the junction of midgut and hindgut, and the gizzard grinds the food.
Q77NEET 2024
The following are the statements about non-chordates: A. Pharynx is perforated by gill slits. B. Notochord is absent. C. Central nervous system is dorsal. D. Heart is dorsal if present. E. Post anal tail is absent. Choose the most appropriate answer from the options given below:
Not quite — the answer is C.
Non-chordates have no notochord, have a dorsal heart (if present), and lack a post-anal tail. Their CNS is ventral, solid and double, and the pharynx is not perforated by gill slits, which are chordate features.
Q78NEET 2024
Choose the correct statement given below regarding juxta medullary nephron.
Not quite — the answer is C.
In juxta medullary nephrons the loop of Henle is very long and runs deep into the medulla. Their renal corpuscles lie in the inner cortex near the medulla, they are fewer than cortical nephrons, and the columns of Bertini are cortical extensions between medullary pyramids.
Q79NEET 2024
Given below are two statements: Statement I: The cerebral hemispheres are connected by nerve tract known as corpus callosum. Statement II: The brain stem consists of the medulla oblongata, pons and cerebrum. In the light of the above statements, choose the most appropriate answer from the options given below:
Not quite — the answer is C.
The two cerebral hemispheres are connected by the corpus callosum, so Statement I is correct. The brain stem consists of the midbrain, pons and medulla oblongata; the cerebrum is part of the forebrain, so Statement II is incorrect.
Q80NEET 2024
Given below are two statements : Statement I : Bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced. Statement II : Both bone marrow and thymus provide micro environments for the development and maturation of T-lymphocytes. In the light of above statements, choose the most appropriate answer from the options given below
Not quite — the answer is A.
Bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced. Bone marrow and thymus together provide micro-environments for the development and maturation of T-lymphocytes, so both statements are correct.
Q81NEET 2024
Regarding catalytic cycle of an enzyme action, select the correct sequential steps : A. Substrate enzyme complex formation. B. Free enzyme ready to bind with another substrate. C. Release of products. D. Chemical bonds of the substrate broken. E. Substrate binding to active site. Choose the correct answer from the options given below :
Not quite — the answer is A.
The substrate first binds the active site, forming an enzyme-substrate complex; the bonds of the substrate are then broken, the products are released, and the free enzyme is ready to bind another substrate. That gives binding → complex → bond breaking → release → free enzyme.
Q82NEET 2024
Identify the correct option (A), (B), (C), (D) with respect to spermatogenesis. GnRH → LH → (B) → Androgens → Formation of separmatids GnRH → (A) → (C) → Factors → (D)
Not quite — the answer is A.
GnRH stimulates the pituitary to release LH and FSH. LH acts on Leydig cells to stimulate androgen secretion, which drives spermatogenesis, while FSH acts on Sertoli cells, whose factors support spermiogenesis. So the blanks are FSH, Leydig cells, Sertoli cells and spermiogenesis.
Q83NEET 2024
Match List I with List II: List - I | List - II A. Mesozoic Era | I. Lower invertebrates B. Proterozoic Era | II. Fish & Amphibia C. Cenozoic Era | III. Birds & Reptiles D. Paleozoic Era | IV. Mammals Choose the correct answer from the options given below :
Not quite — the answer is D.
The Mesozoic era is the age of reptiles and birds, the Proterozoic era had only lower invertebrates, the Cenozoic is the age of mammals, and the Paleozoic saw the rise of fishes and amphibians.
Q84NEET 2024
Match List I with List II: List - I | List - II A. Unicellular glandular epithelium | I. Salivary glands B. Compound epithelium | II. Pancreas C. Multicellular glandular epithelium | III. Goblet cells of alimentary canal D. Endocrine glandular epithelium | IV. Moist surface of buccal cavity Choose the correct answer from the options given below:
Not quite — the answer is C.
Goblet cells of the alimentary canal are unicellular glandular epithelium, compound epithelium covers the moist surface of the buccal cavity, salivary glands are multicellular glands, and the pancreas has endocrine glandular epithelium (islets).
Q85NEET 2024
Match List I with List II: List - I | List - II A. RNA polymerase III | I. snRNPs B. Termination of transcription | II. Promotor C. Splicing of Exons | III. Rho factor D. TATA box | IV. SnRNAs, tRNA Choose the correct answer from the options given below :
Not quite — the answer is D.
RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs; termination of transcription in bacteria involves the rho factor; splicing of exons is done by snRNPs (spliceosome); and the TATA box lies in the promoter.
Q86NEET 2024
Match List I with List II : List - I | List - II A. P wave | (i) Heart muscles are electrically silent. B. QRS complex | (ii) Depolarisation of ventricles. C. T wave | (iii) Depolarisation of atria. D. T-P gap | (iv) Repolarisation of ventricles. Choose the correct answer from the options given below :
Not quite — the answer is B.
The P wave represents depolarisation of the atria, the QRS complex depolarisation of the ventricles, the T wave repolarisation of the ventricles, and during the T-P gap the heart muscle is electrically silent.
Q87NEET 2024
Match List I with List II : List - I | List - II A. Exophthalmic goiter | I. Excess secretion of cortisol, moon face & hyperglycemia. B. Acromegaly | II. Hypo-secretion of thyroid hormone and stunted growth. C. Cushing's syndrome | III. Hyper secretion of thyroid hormone & protruding eye balls. D. Cretinism | IV. Excessive secretion of growth hormone. Choose the correct answer from the options given below :
Not quite — the answer is D.
Exophthalmic goitre (Graves' disease) results from hyper-secretion of thyroid hormone with protruding eyeballs; acromegaly from excess growth hormone in adults; Cushing's syndrome from excess cortisol with moon face and hyperglycaemia; and cretinism from hypo-secretion of thyroid hormone with stunted growth.
Biology: not in the current NEET syllabus
These 2 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
2 questions. Tap an option to check your answer.
Q1NEET 2024
The capacity to generate a whole plant from any cell of the plant is called:
Not quite — the answer is A.
Totipotency is the capacity of a single plant cell or explant to regenerate an entire plant, which is the basis of plant tissue culture. Micropropagation is the technique that exploits this capacity, not the capacity itself, and differentiation is only maturation of cells.
Q2NEET 2024
Which of the following are fused in somatic hybridization involving two varieties of plants?
Not quite — the answer is C.
In somatic hybridisation the cell walls are digested away and the naked protoplasts of two different varieties or species are fused to give hybrid protoplasts, which are then grown into somatic hybrids (e.g. pomato). Callus, somatic embryos and pollen are not what is fused.
Chapter-wise split of NEET 2024 Biology
- Biomolecules6 Q
- Molecular Basis of Inheritance6 Q
- Principles of Inheritance and Variation5 Q
- Human Reproduction5 Q
- Human Health and Disease5 Q
- Biotechnology: Principles and Processes4 Q
- Biodiversity and Conservation4 Q
- Evolution4 Q
- Animal Kingdom4 Q
- Photosynthesis in Higher Plants3 Q
- Morphology of Flowering Plants3 Q
- Plant Growth and Development3 Q
- Cell Cycle and Cell Division3 Q
- Cell: The Unit of Life3 Q
- Structural Organisation in Animals3 Q
- Biological Classification2 Q
- Biotechnology and Its Applications2 Q
- Organisms and Populations2 Q
- Anatomy of Flowering Plants2 Q
- Respiration in Plants2 Q
- Body Fluids and Circulation2 Q
- Chemical Coordination and Integration2 Q
- Breathing and Exchange of Gases2 Q
- Neural Control and Coordination2 Q
- Reproductive Health2 Q
- Excretory Products and Their Elimination2 Q
- Sexual Reproduction in Flowering Plants1 Q
- Microbes in Human Welfare1 Q
- Plant Kingdom1 Q
- Locomotion and Movement1 Q
About this paper
How many questions from NEET 2024 are on this page?
170 questions (Physics 33, Chemistry 48, Biology 89). 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 2024 paper, conducted by NTA. The answer to every question was checked by the RankUp team, and the explanations are RankUp's own.
Is NEET 2024 useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books. 3 questions here are on topics NCERT has since dropped; they are listed separately so you can skip them.
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