NEET 2024 Re-exam 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
175 questions from the NEET 2024 Re-exam 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 2024 Re-exam
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- 45 questions
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- 92 questions
NEET 2024 Re-exam Physics questions
37 questions. Tap an option to check your answer.
Q1NEET 2024 Re-exam
The magnetic potential energy, when a magnetic bar of magnetic moment m⃗ is placed perpendicular to the magnetic field B⃗ is
Not quite — the answer is B.
The potential energy of a magnetic dipole is U = −m⃗·B⃗ = −mB cos θ. For θ = 90°, cos θ = 0, so U = 0. The value −mB belongs to the aligned position (θ = 0°), the minimum-energy orientation.
Q2NEET 2024 Re-exam
A bob is whirled in a horizontal circle by means of a string at an initial speed of 10 rpm. If the tension in the string is quadrupled while keeping the radius constant, the new speed is:
Not quite — the answer is A.
The tension supplies the centripetal force, T = mω²r, so at fixed r, ω ∝ √T. Quadrupling T doubles ω: 10 rpm × √4 = 20 rpm. Taking ω proportional to T itself would give 40 rpm.
Q3NEET 2024 Re-exam
A metal cube of side 5 cm is charged with 6 μC. The surface charge density on the cube is
Not quite — the answer is D.
Charge on a conductor spreads over its whole outer surface, here six faces: A = 6 × (0.05 m)² = 0.015 m². σ = q/A = 6 × 10⁻⁶ / 0.015 = 4 × 10⁻⁴ = 0.4 × 10⁻³ C m⁻². Using one face only (0.0025 m²) would give a value six times larger.
Q4NEET 2024 Re-exam
The incorrect relation for a diamagnetic material (all the symbols carry their usual meaning and ε is a small positive number) is
Not quite — the answer is D.
For a diamagnetic material the susceptibility is negative, −1 ≤ χ < 0, so μᵣ = 1 + χ lies in 0 ≤ μᵣ < 1 and μ < μ₀ (a superconductor is the limit χ = −1). A relative permeability slightly greater than 1, 1 < μᵣ < 1 + ε, describes a paramagnetic material, so that relation is the incorrect one.
Q5NEET 2024 Re-exam
The escape velocity for earth is v. A planet having 9 times mass that of earth and radius, 16 times that of earth, has the escape velocity of:
Not quite — the answer is C.
vₑ = √(2GM/R) ∝ √(M/R). For the planet, v' = v√(9/16) = 3v/4. Forgetting the square root gives 9v/16, which is not offered; 9v/4 would come from using the masses' ratio over the radius' square root.
Q6NEET 2024 Re-exam
An electron and an alpha particle are accelerated by the same potential difference. Let λₑ and λα denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:
Not quite — the answer is A.
After acceleration through V, p = √(2mqV), so λ = h/√(2mqV) ∝ 1/√(mq). The alpha particle has about 7300 times the electron mass and twice its charge, so mq is far larger and its wavelength far shorter; hence λₑ > λα (by a factor of about √14 600 ≈ 120, not 4).
Q7NEET 2024 Re-exam
An object moving along horizontal x-direction with kinetic energy 10 J is displaced through x = (3î) m by the force F⃗ = (−2î + 3ĵ) N. The kinetic energy of the object at the end of the displacement x is
Not quite — the answer is C.
Work done W = F⃗·x⃗ = (−2)(3) + (3)(0) = −6 J; the y-component of force does no work along x. By the work-energy theorem, K_final = 10 − 6 = 4 J. Adding |W| instead gives 16 J.
Q8NEET 2024 Re-exam
An object falls from a height of 10 m above the ground. After striking the ground it loses 50% of its kinetic energy. The height upto which the object can rebound from the ground is:
Not quite — the answer is D.
Just before impact KE = mg(10 m). After losing half, KE = mg(5 m), and all of it turns back into potential energy on the way up, so the rebound height is 5 m. Halving the speed instead of the energy would give 2.5 m.
Q9NEET 2024 Re-exam
A 12 pF capacitor is connected to a 50 V battery, the electrostatic energy stored in the capacitor in nJ is
Not quite — the answer is A.
U = ½CV² = ½ × 12 × 10⁻¹² × (50)² = 6 × 10⁻¹² × 2500 = 1.5 × 10⁻⁸ J = 15 nJ. Omitting the ½ would give 30 nJ, and using V instead of V² gives 0.3 nJ.
Q10NEET 2024 Re-exam
A uniform wire of diameter d carries a current of 100 mA when the mean drift velocity of electrons in the wire is v. For a wire of diameter d/2 of the same material to carry a current of 200 mA, the mean drift velocity of electrons in the wire is
Not quite — the answer is B.
I = neAv_d with A = πd²/4, so v_d ∝ I/d² for the same material (same n). Doubling I multiplies v_d by 2, and halving d multiplies it by 4, giving v_d = 8v. Treating the area as proportional to d instead of d² would give 4v.
Q11NEET 2024 Re-exam
In an electrical circuit, the voltage is measured as V = (200 ± 4) volt and the current is measured as I = (20 ± 0.2) A. The value of the resistance is:
Not quite — the answer is B.
R = V/I = 200/20 = 10 Ω. For a quotient the relative errors add: ΔR/R = 4/200 + 0.2/20 = 0.02 + 0.01 = 0.03, so ΔR = 0.03 × 10 = 0.3 Ω. Adding the absolute errors (4 + 0.2) gives the wrong 4.2 Ω, and subtracting relative errors gives 0.1 Ω.
Q12NEET 2024 Re-exam
A step up transformer is connected to an ac mains supply of 220 V to operate at 11000 V, 88 watt. The current in the secondary circuit, ignoring the power loss in the transformer, is
Not quite — the answer is A.
For an ideal transformer the secondary delivers the stated power at its own voltage: I_s = P/V_s = 88/11000 = 0.008 A = 8 mA. Using the primary voltage, 88/220 = 0.4 A, gives the primary current instead.
Q13NEET 2024 Re-exam
A particle is moving along x-axis with its position (x) varying with time (t) as x = αt⁴ + βt² + γt + δ. The ratio of its initial velocity to its initial acceleration, respectively, is:
Not quite — the answer is D.
v = dx/dt = 4αt³ + 2βt + γ, so v(0) = γ. a = dv/dt = 12αt² + 2β, so a(0) = 2β. The ratio is γ : 2β. The constant δ is the initial position and does not appear in either derivative.
Q14NEET 2024 Re-exam
The capacitance of a capacitor with charge q and a potential difference V depends on
Not quite — the answer is B.
C = q/V, but q and V change in the same proportion, so the ratio is fixed by the size, shape and separation of the conductors and by the dielectric between them, for example C = ε₀A/d for parallel plates. The formula C = q/V defines capacitance; it does not make C depend on q or V.
Q15NEET 2024 Re-exam
Given below are two statements: Statement I : Image formation needs regular reflection and/or refraction. Statement II : The variety in colour of objects we see around us is due to the constituent colours of the light incident on them. In the light of the above statements, choose the most appropriate answer from the options given below :
Not quite — the answer is C.
A sharp image needs light from each point of the object to be brought together in an orderly way, which is done by regular (specular) reflection or refraction; diffuse scattering cannot form an image. Objects look coloured because they reflect or transmit some of the colours in the light falling on them, so the colours we see depend on the constituent colours of the incident light. Both statements are correct, as stated in NCERT.
Q16NEET 2024 Re-exam
A uniform metal wire of length l has 10 Ω resistance. Now this wire is stretched to a length 2l and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is
Not quite — the answer is A.
Stretching at constant volume to twice the length halves the area, so R' = 2² × 10 = 40 Ω. Across a diameter the circle splits into two 20 Ω halves in parallel: (20 × 20)/(20 + 20) = 10 Ω. Forgetting that the area also shrinks gives R' = 20 Ω and an answer of 5 Ω.
Q17NEET 2024 Re-exam
The spectral series which corresponds to the electronic transition from the levels n₂ = 5, 6, ... to the level n₁ = 4 is
Not quite — the answer is B.
Transitions ending on n = 4 form the Brackett series. Lyman ends on n = 1, Balmer on n = 2, Paschen on n = 3 and Pfund on n = 5.
Q18NEET 2024 Re-exam
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Houses made of concrete roofs overlaid with foam keep the room hotter during summer. Reason R: The layer of foam insulation prohibits heat transfer, as it contains air pockets. In the light of the above statements, choose the correct answer from the options given below.
Not quite — the answer is B.
Trapped air is a poor conductor, so a foam layer cuts heat flow through the roof (the reason is true). In summer this keeps the outside heat out, so the room stays cooler, not hotter; the assertion is false.
Q19NEET 2024 Re-exam
A particle executing simple harmonic motion with amplitude A has the same potential and kinetic energies at the displacement
Not quite — the answer is C.
PE = ½kx² and KE = ½k(A² − x²). Setting them equal: x² = A² − x², so x² = A²/2 and x = A/√2. At x = A/2 the potential energy is only a quarter of the total.
Q20NEET 2024 Re-exam
Two slits in Young's double slit experiment are 1.5 mm apart and the screen is placed at a distance of 1 m from the slits. If the wavelength of light used is 600 × 10⁻⁹ m then the fringe separation is
Not quite — the answer is D.
Fringe width β = λD/d = (600 × 10⁻⁹ m × 1 m)/(1.5 × 10⁻³ m) = 4 × 10⁻⁴ m (0.4 mm). Leaving d in mm or mixing units gives the wrong powers of ten in the other options.
Q21NEET 2024 Re-exam
Water is used as a coolant in a nuclear reactor because of its
Not quite — the answer is B.
A coolant must carry away a lot of heat for a small rise in its own temperature. Water's high specific heat capacity (about 4200 J kg⁻¹ K⁻¹) lets it absorb large amounts of heat. A low boiling point would be a drawback, since the coolant would boil off easily.
Q22NEET 2024 Re-exam
The pitch of an error free screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a thick wire, the pitch scale reads 1 mm and 63rd division on the circular scale coincides with the reference line. The diameter of the wire is:
Not quite — the answer is B.
Least count = pitch/number of divisions = 1 mm/100 = 0.01 mm. Reading = main scale + (circular division × LC) = 1 mm + 63 × 0.01 mm = 1.63 mm = 0.163 cm. The value 1.63 cm makes the mm-to-cm conversion wrong by a factor of 10.
Q23NEET 2024 Re-exam
Let us consider two solenoids A and B, made from same magnetic material of relative permeability μᵣ and equal area of cross-section. Length of A is twice that of B and the number of turns per unit length in A is half that of B. The ratio of self inductances of the two solenoids, L_A : L_B is
Not quite — the answer is A.
For a long solenoid L = μ₀μᵣn²Al. With the same μᵣ and A, L_A/L_B = (n_A/n_B)² × (l_A/l_B) = (1/2)² × 2 = 1/2, so L_A : L_B = 1 : 2. Using n instead of n² would give 1 : 1, and inverting the ratio gives 2 : 1.
Q24NEET 2024 Re-exam
When the output of an OR gate is applied as input to a NOT gate, then the combination acts as a
Not quite — the answer is B.
The OR gate gives A + B and the NOT gate inverts it to (A + B)‾, which is the NOR function. A NAND gate is AND followed by NOT, (A·B)‾.
Q25NEET 2024 Re-exam
Interference pattern can be observed due to superposition of the following waves: A. y = a sin ωt B. y = a sin 2ωt C. y = a sin(ωt − φ) D. y = a sin 3ωt Choose the correct answer from the options given below.
Not quite — the answer is C.
A steady interference pattern needs coherent waves: the same frequency and a constant phase difference. Waves A and C both have angular frequency ω and differ by a fixed phase φ. Waves of frequency ω, 2ω and 3ω differ in frequency, so their phase difference changes with time and no steady pattern forms.
Q26NEET 2024 Re-exam
The electromagnetic radiation which has the smallest wavelength are
Not quite — the answer is B.
In the electromagnetic spectrum wavelength increases from gamma rays (below about 10⁻¹² m) through X-rays, ultraviolet, visible, infrared and microwaves to radio waves. So gamma rays have the smallest wavelength; X-rays come next.
Q27NEET 2024 Re-exam
The equilibrium state of a thermodynamic system is described by A. Pressure B. Total heat C. Temperature D. Volume E. Work done Choose the most appropriate answer from the options given below.
Not quite — the answer is D.
An equilibrium state is specified by state variables such as pressure, volume and temperature, which have definite values in that state. Heat and work are not properties of a state; they depend on the process (path) by which the system moves between states. So only pressure, temperature and volume describe the state.
Q28NEET 2024 Re-exam
A box of mass 5 kg is pulled by a cord, up along a frictionless plane inclined at 30° with the horizontal. The tension in the cord is 30 N. The acceleration of the box is (Take g = 10 m s⁻²)
Not quite — the answer is D.
Along the incline (cord parallel to it): T − mg sin 30° = ma, so 30 − 5 × 10 × 0.5 = 5a, giving 30 − 25 = 5a and a = 1 m s⁻² up the plane. Ignoring the gravity component would give 6 m s⁻², and using mg instead of mg sin 30° would make the box slide down.
Q29NEET 2024 Re-exam
If the ratio of relative permeability and relative permittivity of a uniform medium is 1 : 4. The ratio of the magnitudes of electric field intensity (E) to the magnetic field intensity (H) of an EM wave propagating in that medium is (Given that √(μ₀/ε₀) = 120π):
Not quite — the answer is C.
In a medium E/H = √(μ/ε) = √(μᵣμ₀/εᵣε₀) = √(μᵣ/εᵣ) × √(μ₀/ε₀) = √(1/4) × 120π = 60π. So E : H = 60π : 1. The value 120π : 1 is for free space, and 30π comes from using 1/4 without the square root.
Q30NEET 2024 Re-exam
The value of electric potential at a distance of 9 cm from the point charge 4 × 10⁻⁷ C is [Given 1/(4πε₀) = 9 × 10⁹ N m² C⁻²] :
Not quite — the answer is D.
V = kq/r = (9 × 10⁹ × 4 × 10⁻⁷)/0.09 = 3600/0.09 = 4 × 10⁴ V. Using r = 9 (cm left unconverted) gives 400 V, and dividing by r² mixes up potential with field.
Q31NEET 2024 Re-exam
The displacement of a travelling wave y = C sin (2π/λ)(at − x) where t is time, x is distance and λ is the wavelength, all in S.I. units. Then the frequency of the wave is
Not quite — the answer is D.
Writing y = C sin(2πat/λ − 2πx/λ) and comparing with sin(ωt − kx) gives ω = 2πa/λ, so the wave speed is a. Frequency f = ω/2π = a/λ. The expression 2πa/λ is the angular frequency, not the frequency.
Q32NEET 2024 Re-exam
The potential energy of a particle moving along x-direction varies as V = Ax²/(√x + B). The dimensions of A²/B are:
Not quite — the answer is C.
B is added to √x, so [B] = [L^(1/2)]. Then V ≈ Ax²/√x gives [A] = [ML²T⁻²]/[L^(3/2)] = [M L^(1/2) T⁻²]. So [A²/B] = [M² L T⁻⁴]/[L^(1/2)] = [M² L^(1/2) T⁻⁴]. Treating B as dimensionless would give [M² L T⁻⁴].
Q33NEET 2024 Re-exam
The two-dimensional motion of a particle, described by r⃗ = (î + 2ĵ)A cos ωt is a/an: A. parabolic path B. elliptical path C. periodic motion D. simple harmonic motion Choose the correct answer from the options given below:
Not quite — the answer is D.
x = A cos ωt and y = 2A cos ωt, so y = 2x: the particle moves to and fro along a straight line. Along that line its displacement varies as cos ωt, so the motion is simple harmonic and therefore periodic. The path is neither a parabola nor an ellipse; an ellipse would need a phase difference between the x and y motions.
Q34NEET 2024 Re-exam
Let ω₁, ω₂ and ω₃ be the angular speed of the second hand, minute hand and hour hand of a smoothly running analog clock, respectively. If x₁, x₂ and x₃ are their respective angular distances in 1 minute then the factor which remains constant (k) is
Not quite — the answer is A.
Each hand turns uniformly, so its angular distance in time t is x = ωt. For the same t = 1 minute, ω/x = 1/t is the same for all three hands. The products ωx, ωx² or ω²x grow with ω and differ from hand to hand.
Q35NEET 2024 Re-exam
The magnetic moment of an iron bar is M. It is now bent in such a way that it forms an arc section of a circle subtending an angle of 60° at the centre. The magnetic moment of this arc section is
Not quite — the answer is A.
M = m × l, where m is the pole strength. Bending into an arc of angle π/3 gives l = r(π/3), so r = 3l/π. The new pole separation is the chord, 2r sin 30° = r = 3l/π. So M' = m × 3l/π = 3M/π. Using the arc length instead of the chord leaves the moment wrongly unchanged.
Q36NEET 2024 Re-exam
According to the law of equipartition of energy, the number of vibrational modes of a polyatomic gas of constant γ = C_P/C_V is (C_P and C_V are the specific heat capacities of the gas at constant pressure and constant volume, respectively):
Not quite — the answer is C.
A non-linear polyatomic molecule has 3 translational and 3 rotational degrees of freedom, and each vibrational mode contributes kT (½kT kinetic plus ½kT potential). With f vibrational modes, C_V = (3 + f)R and C_P = (4 + f)R, so γ = 1 + 1/(3 + f). Then 3 + f = 1/(γ − 1) and f = (1 − 3γ + 3)/(γ − 1) = (4 − 3γ)/(γ − 1). Counting only ½kT per vibrational mode would give a different expression.
Q37NEET 2024 Re-exam
The amplitude of the charge oscillating in a circuit decreases exponentially as Q = Q₀e^(−Rt/2L), where Q₀ is the charge at t = 0 s. The time at which charge amplitude decreases to 0.50Q₀ is nearly: [Given that R = 1.5 Ω, L = 12 mH, ln(2) = 0.693]
Not quite — the answer is B.
0.5 = e^(−Rt/2L), so Rt/2L = ln 2 and t = 2L ln 2/R = (2 × 0.012 × 0.693)/1.5 = 0.016632/1.5 = 0.01109 s = 11.09 ms. Taking L as 12 H instead of 12 mH gives 11.09 s.
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 Re-exam
Select the correct statements among the following : A. Slow neutrons can cause fission in ²³⁵₉₂U than fast neutrons. B. α-rays are Helium nuclei. C. β-rays are fast moving electrons or positrons. D. γ-rays are electromagnetic radiations of wavelengths larger than X-rays. Choose the most appropriate answer from the options given below :
Not quite — the answer is A.
Slow (thermal) neutrons cause fission of U-235 much more readily than fast ones; α-particles are helium nuclei; β-rays are fast electrons (β⁻) or positrons (β⁺). γ-rays are electromagnetic radiation of wavelength shorter than X-rays, not larger, so that statement is false.
Chapter-wise split of NEET 2024 Re-exam Physics
- Magnetism and Matter3 Q
- Electrostatic Potential and Capacitance3 Q
- Units and Measurements3 Q
- Laws of Motion2 Q
- Work, Energy and Power2 Q
- Current Electricity2 Q
- Alternating Current2 Q
- Thermal Properties of Matter2 Q
- Oscillations2 Q
- Wave Optics2 Q
- Electromagnetic Waves2 Q
- Electric Charges and Fields1 Q
- Gravitation1 Q
- Dual Nature of Radiation and Matter1 Q
- Motion in a Straight Line1 Q
- Ray Optics and Optical Instruments1 Q
- Atoms1 Q
- Electromagnetic Induction1 Q
- Semiconductor Electronics: Materials, Devices and Simple Circuits1 Q
- Thermodynamics1 Q
- Waves1 Q
- Motion in a Plane1 Q
- Kinetic Theory1 Q
NEET 2024 Re-exam Chemistry questions
45 questions. Tap an option to check your answer.
Q1NEET 2024 Re-exam
The correct decreasing order of atomic radii (pm) of Li, Be, B and C is
Not quite — the answer is B.
Li, Be, B and C lie in the same period, and atomic radius falls across a period as nuclear charge rises while electrons enter the same shell. So the order is Li (152 pm) > Be (111 pm) > B (88 pm) > C (77 pm). Orders that put carbon or beryllium first reverse this trend.
Q2NEET 2024 Re-exam
Following data is for a reaction between reactants A and B : Rate (mol L⁻¹ s⁻¹) | [A] | [B] 2 × 10⁻³ | 0.1 M | 0.1 M 4 × 10⁻³ | 0.2 M | 0.1 M 1.6 × 10⁻² | 0.2 M | 0.2 M The order of the reaction with respect to A and B, respectively, are
Not quite — the answer is C.
Runs 1 and 2: [A] doubles at fixed [B] and the rate goes from 2 × 10⁻³ to 4 × 10⁻³ (×2 = 2¹), so the order in A is 1. Runs 2 and 3: [B] doubles at fixed [A] and the rate goes from 4 × 10⁻³ to 1.6 × 10⁻² (×4 = 2²), so the order in B is 2. Reading the second change as a simple doubling would wrongly give order 1 in B.
Q3NEET 2024 Re-exam
Baeyer's reagent is :
Not quite — the answer is C.
Baeyer's reagent is cold, dilute, aqueous (slightly alkaline) KMnO₄; its purple colour disappears as it converts an alkene into a vicinal diol, which is the test for unsaturation. Acidic or hot concentrated permanganate cleaves the C=C bond instead, so those are not Baeyer's reagent.
Q4NEET 2024 Re-exam
Which of the following molecules has "NON ZERO" dipole moment value?
Not quite — the answer is B.
HI is a heteronuclear diatomic molecule with an electronegativity difference between H and I, so it has a net dipole moment. CCl₄ (tetrahedral), CO₂ (linear) and BF₃ (trigonal planar) have polar bonds, but their bond dipoles cancel by symmetry, giving zero net dipole moment.
Q5NEET 2024 Re-exam
Which indicator is used in the titration of sodium hydroxide against oxalic acid and what is the colour change at the end point?
Not quite — the answer is C.
Oxalic acid is a weak acid titrated with a strong base, so the end point is in the basic range and phenolphthalein is used. With the acid in the flask and NaOH from the burette, the solution goes from colourless to pink at the end point. Methyl orange changes in the acidic range and is unsuitable for a weak acid-strong base titration.
Q6NEET 2024 Re-exam
Match List I with List II : List - I (Atom/Molecule): (A) Nitrogen atom (B) Fluorine molecule (C) Oxygen molecule (D) Xenon atom List - II (Property): (I) Paramagnetic (II) Most reactive element in group 18 (III) Element with highest ionisation enthalpy in group 15 (IV) Strongest oxidising agent Identify the correct answer from the options given below :
Not quite — the answer is D.
Nitrogen, the smallest atom of group 15 with a stable half-filled 2p shell, has the highest ionisation enthalpy in that group. F₂ is the strongest oxidising agent. O₂ has two unpaired electrons in π* orbitals (MO theory), so it is paramagnetic. Among the stable noble gases xenon, with the lowest ionisation enthalpy, forms the most compounds and is taken as the most reactive. Matching O₂ with 'strongest oxidising agent' confuses it with F₂.
Q7NEET 2024 Re-exam
From the following select the one which is not an example of corrosion.
Not quite — the answer is B.
Corrosion is the slow, unwanted attack of a metal by its surroundings (air, moisture, gases), as in rusting of iron, tarnishing of silver by sulphur compounds and the green basic carbonate coating on copper and bronze. Electrolysis of water to make hydrogen is a deliberate, current-driven decomposition of water, not attack on a metal, so it is not corrosion.
Q8NEET 2024 Re-exam
Which of the following pairs of ions will have same spin only magnetic moment values within the pair? A. Zn²⁺, Ti²⁺ B. Cr²⁺, Fe²⁺ C. Ti³⁺, Cu²⁺ D. V²⁺, Cu⁺ Choose the correct answer from the options given below :
Not quite — the answer is D.
Count unpaired electrons, since μ = √(n(n+2)) BM. Zn²⁺ 3d¹⁰ (0) and Ti²⁺ 3d² (2): different. Cr²⁺ 3d⁴ (4) and Fe²⁺ 3d⁶ (4): both 4.90 BM. Ti³⁺ 3d¹ (1) and Cu²⁺ 3d⁹ (1): both 1.73 BM. V²⁺ 3d³ (3) and Cu⁺ 3d¹⁰ (0): different. So only the Cr²⁺/Fe²⁺ and Ti³⁺/Cu²⁺ pairs match.
Q9NEET 2024 Re-exam
At a given temperature and pressure, the equilibrium constant values for the equilibria are given below: 3A₂ + B₂ ⇌ 2A₃B, K₁ A₃B ⇌ (3/2)A₂ + (1/2)B₂, K₂ The relation between K₁ and K₂ is :
Not quite — the answer is D.
The second reaction is the first one reversed and multiplied by 1/2. Reversing gives the constant 1/K₁, and halving the coefficients takes its square root, so K₂ = (1/K₁)^(1/2) = 1/√K₁. Writing K₁ = 1/√K₂ applies the square root to the wrong constant (the correct inverse is K₁ = 1/K₂²).
Q10NEET 2024 Re-exam
Identify the incorrect statement about PCl₅.
Not quite — the answer is B.
PCl₅ is trigonal bipyramidal (sp³d) with five P–Cl σ bonds. The two axial bonds suffer more repulsion from the three equatorial bond pairs and are longer (about 240 pm) than the equatorial bonds (about 202 pm), so the five bonds are not all the same length. The bond angles are 120° (equatorial–equatorial) and 90° (axial–equatorial), so 'two different angles' is correct.
Q11NEET 2024 Re-exam
Choose the correct statement for the work done in the expansion and heat absorbed or released when 5 litres of an ideal gas at 10 atmospheric pressure isothermally expands into vacuum until volume is 15 litres :
Not quite — the answer is C.
Expansion into vacuum is free expansion: the external pressure is zero, so w = −p_ext ΔV = 0. For an ideal gas at constant temperature ΔU = 0, so q = ΔU − w = 0 too. The initial 10 atm and the 5 L to 15 L change are red herrings; no work is done against a vacuum.
Q12NEET 2024 Re-exam
Which of the following set of ions act as oxidising agents?
Not quite — the answer is A.
Lanthanoids are most stable in the +3 state. Ce⁴⁺ (4f⁰) and Tb⁴⁺ (4f⁷) readily gain an electron to return to +3, so both are oxidising agents. Eu²⁺ and Yb²⁺ readily lose an electron to reach +3, so they are reducing agents, and La³⁺/Lu³⁺ are already in the stable +3 state.
Q13NEET 2024 Re-exam
Select the incorrect reaction among the following :
Not quite — the answer is B.
LiAlH₄ reduces an amide C=O to CH₂, so benzamide gives benzylamine (C₆H₅CH₂NH₂), not benzoic acid. Acid chlorides hydrolyse to carboxylic acids, and primary alcohols are oxidised to the acids with the same carbon count by alkaline KMnO₄ or by Jones reagent (CrO₃–H₂SO₄), so those reactions are correct.
Q14NEET 2024 Re-exam
The UV-visible absorption bands in the spectra of lanthanoid ions are 'X', probably because of the excitation of electrons involving 'Y'. The 'X' and 'Y', respectively, are :
Not quite — the answer is B.
Lanthanoid ions are coloured because of f–f transitions. The 4f orbitals are deeply buried and shielded from the ligands by the 5s and 5p electrons, so the absorption bands are very narrow, almost line-like. Broad bands are typical of d–d transitions in transition-metal ions, whose d orbitals interact strongly with the ligands.
Q15NEET 2024 Re-exam
Ethylene diaminetetraacetate ion is a/an:
Not quite — the answer is A.
EDTA⁴⁻ binds a metal through two nitrogen atoms and four carboxylate oxygen atoms, six donor sites in all, so it is hexadentate. An ambidentate ligand has two different donor atoms but uses only one at a time (like SCN⁻), which is not how EDTA behaves.
Q16NEET 2024 Re-exam
The amount of glucose required to prepare 250 mL of M/20 aqueous solution is : (Molar mass of glucose : 180 g mol⁻¹)
Not quite — the answer is A.
Moles needed = M × V = (1/20 mol L⁻¹) × 0.250 L = 0.0125 mol. Mass = 0.0125 × 180 = 2.25 g. Using 500 mL or 125 mL by mistake would give 4.5 g or 1.125 g.
Q17NEET 2024 Re-exam
Identify the incorrect statement from the following :
Not quite — the answer is A.
The acid strength of the hydrogen halides increases down the group, HF < HCl < HBr < HI, because the H–X bond becomes longer and weaker. So the order given with HF strongest is reversed and incorrect. The F–F bond is weaker than Cl–Cl because of lone-pair repulsion in the small F₂ molecule, fluorine shows only −1, and F₂ is the stronger oxidant, so those statements are correct.
Q18NEET 2024 Re-exam
For the reaction in equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −Q Reaction is favoured in forward direction by:
Not quite — the answer is D.
By Le Chatelier's principle: the forward reaction lowers the number of gas moles (4 → 2), so high pressure helps; it is exothermic, so low temperature helps; and adding the reactant H₂ shifts it forward. A catalyst only speeds up reaching equilibrium without shifting it, and removing N₂ or adding NH₃ pushes the reaction backward.
Q19NEET 2024 Re-exam
The major product D formed in the following reaction sequence is : CH₃OH —SOCl₂→ A —KCN, aq. EtOH→ B —Na(Hg), EtOH→ C —(i) NaNO₂ + HCl, (ii) H₂O→ D (major)
Not quite — the answer is C.
SOCl₂ converts methanol to CH₃Cl, which is A; KCN gives CH₃CN, which is B; and Na(Hg)/ethanol reduces the nitrile to ethylamine CH₃CH₂NH₂, which is C. Nitrous acid turns the primary aliphatic amine into an unstable diazonium salt that loses N₂ and reacts with water to give ethanol, which is D. Reduction adds one CH₂ to the chain, so it is not stopped at acetaldehyde or acetamide.
Q20NEET 2024 Re-exam
Match List I with List II: List - I (Block/group in periodic table): (A) Lanthanoid (B) d-block element (C) p-block element (D) s-block element List - II (Element): (I) Ce (II) As (III) Cs (IV) Mn Choose the correct answer from the options given below:
Not quite — the answer is C.
Cerium (Z = 58) is a lanthanoid, manganese (Z = 25) is a 3d transition element, arsenic (group 15) is a p-block element and caesium (group 1) is an s-block alkali metal. The usual trap is confusing Cs (caesium, s-block) with Ce (cerium, lanthanoid).
Q21NEET 2024 Re-exam
Which of the following is not an ambidentate ligand?
Not quite — the answer is A.
An ambidentate ligand can bind through either of two different donor atoms: SCN⁻ (S or N), NO₂⁻ (N or O) and CN⁻ (C or N). Oxalate binds through two oxygen atoms at the same time, so it is a bidentate chelating ligand, not ambidentate.
Q22NEET 2024 Re-exam
The quantum numbers of four electrons are given below : I. n = 4; l = 2; mₗ = −2; s = −1/2 II. n = 3; l = 2; mₗ = 1; s = +1/2 III. n = 4; l = 1; mₗ = 0; s = +1/2 IV. n = 3; l = 1; mₗ = −1; s = +1/2 The correct decreasing order of energy of these electrons is
Not quite — the answer is B.
Use the (n + l) rule: I is 4d (n + l = 6), II is 3d (5), III is 4p (5) and IV is 3p (4). For II and III with equal n + l, the one with larger n (4p) is higher in energy. So the order is 4d > 4p > 3d > 3p, i.e. I > III > II > IV. Putting 3d above 4p ignores the tie-break rule for equal n + l.
Q23NEET 2024 Re-exam
The major product C in the below mentioned reaction is: CH₃CH₂CH₂Br —alc. KOH, Δ→ A —HBr→ B —aq. KOH, Δ→ C
Not quite — the answer is B.
Alcoholic KOH dehydrohalogenates 1-bromopropane to propene, which is A. HBr adds by Markovnikov's rule (no peroxide) to give 2-bromopropane, which is B, and aqueous KOH substitutes Br by OH to give propan-2-ol, which is C. Propan-1-ol would need anti-Markovnikov addition, which is not the case here.
Q24NEET 2024 Re-exam
The compound that does not undergo Friedel-Crafts alkylation reaction but gives a positive carbylamine test is :
Not quite — the answer is A.
The carbylamine (isocyanide) test is given only by primary amines, and aniline is a primary aromatic amine. Aniline does not undergo Friedel-Crafts reactions because its basic NH₂ group forms a salt with the AlCl₃ catalyst, which deactivates the ring. N-methylaniline is secondary, and pyridine and triethylamine are tertiary bases, so they give no carbylamine test.
Q25NEET 2024 Re-exam
For an endothermic reaction: (A) q_p is negative. (B) Δ_rH is positive. (C) Δ_rH is negative. (D) q_p is positive. Choose the correct answer from the options given below:
Not quite — the answer is A.
An endothermic reaction absorbs heat from the surroundings. At constant pressure the heat absorbed is q_p = Δ_rH, so both are positive. A negative Δ_rH or q_p describes an exothermic reaction.
Q26NEET 2024 Re-exam
1.0 g of H₂ has same number of molecules as in:
Not quite — the answer is A.
1.0 g H₂ = 1.0/2 = 0.5 mol. 14 g N₂ = 14/28 = 0.5 mol, so it has the same number of molecules (0.5 Nₐ). 18 g H₂O and 28 g N₂ are 1 mol each, and 16 g CO is 16/28 ≈ 0.57 mol.
Q27NEET 2024 Re-exam
A steam volatile organic compound which is immiscible with water has a boiling point of 250°C. During steam distillation, a mixture of this organic compound and water will boil :
Not quite — the answer is D.
For immiscible liquids each component exerts its own vapour pressure, and the mixture boils when p(water) + p(organic) equals atmospheric pressure. This happens below the boiling point of the more volatile component, water, and since the high-boiling organic compound adds only a small vapour pressure, the mixture boils just below 100°C. That is why steam distillation can carry over high-boiling compounds without heating them to their boiling points.
Q28NEET 2024 Re-exam
Given below are two statements : Statement I : Glycogen is similar to amylose in its structure. Statement II : Glycogen is found in yeast and fungi also. In the light of the above statements, choose the correct answer from the options given below :
Not quite — the answer is B.
Glycogen, the animal storage polysaccharide, is highly branched like amylopectin (and even more so), whereas amylose is an unbranched chain, so Statement I is false. Glycogen is stored in liver, muscle and brain, and is also found in yeast and fungi, so Statement II is true.
Q29NEET 2024 Re-exam
The oxidation states not shown by Mn in given reaction are : 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O A. +6 B. +2 C. +4 D. +7 E. +3 Choose the most appropriate answer from the options given below :
Not quite — the answer is D.
In this disproportionation Mn is +6 in MnO₄²⁻, +7 in MnO₄⁻ and +4 in MnO₂. So +2 and +3 do not appear. Any choice that includes +6 or +7 lists a state that is present.
Q30NEET 2024 Re-exam
Given below are two statements: Statement I: The Balmer spectral line for H atom with lowest energy is located at (5/36)R_H cm⁻¹. (R_H = Rydberg constant) Statement II: When the temperature of blackbody increases, the maxima of the curve (intensity and wavelength) shifts to shorter wavelength. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is C.
The lowest-energy Balmer line is the n = 3 → 2 transition: wavenumber = R_H(1/2² − 1/3²) = R_H(1/4 − 1/9) = 5R_H/36, so Statement I is true. For black-body radiation the peak of the intensity-wavelength curve shifts to shorter wavelength as temperature rises (Wien's displacement), so Statement II is true. 5R_H/36 is not the series limit (R_H/4), which is the highest-energy Balmer line.
Q31NEET 2024 Re-exam
Identify D in the following sequence of reactions: CH₃CH₂OH —P + I₂→ A —Mg, dry ether→ B —HCHO→ C —H₂O→ D (major)
Not quite — the answer is A.
P/I₂ converts ethanol to ethyl iodide, which is A; with Mg in dry ether it gives CH₃CH₂MgI, which is B. This adds to formaldehyde to give CH₃CH₂CH₂OMgI, which is C, and hydrolysis gives the primary alcohol CH₃CH₂CH₂OH, n-propyl alcohol, which is D. Formaldehyde always gives a primary alcohol with one more carbon; isopropyl alcohol would need acetaldehyde with CH₃MgX.
Q32NEET 2024 Re-exam
Identify the incorrect statement.
Not quite — the answer is B.
The N–N single bond is weaker than the P–P single bond: the small N–N bond length brings the non-bonding lone pairs close and their repulsion weakens the bond. Nitrogen forms pπ–pπ multiple bonds because of its small size and has no d orbitals for dπ–pπ bonding, while P and As in ligands like P(C₂H₅)₃ and As(C₆H₅)₃ can form dπ–dπ bonds with transition metals. So those statements are correct.
Q33NEET 2024 Re-exam
Match List-I with List-II : List - I (Test/reagent): (A) Lake Test (B) Nessler's Reagent (C) Potassium sulphocyanide (D) Brown Ring Test List - II (Radical identified): (I) NO₃⁻ (II) Fe³⁺ (III) Al³⁺ (IV) NH₄⁺ Choose the correct answer from the options given below :
Not quite — the answer is D.
The lake test (blue litmus adsorbed on Al(OH)₃) identifies Al³⁺. Nessler's reagent gives a brown precipitate with NH₄⁺ (as ammonia). Potassium sulphocyanide (thiocyanate) gives a blood-red colour with Fe³⁺, and the brown ring test detects NO₃⁻. Swapping the Al³⁺ and Fe³⁺ tests is the common error.
Q34NEET 2024 Re-exam
Match List-I with List-II: List - I (Molecule): (A) HCl (B) N₂ (C) H₂ (D) O₂ List - II (Bond enthalpy (kJ mol⁻¹)): (I) 435.8 (II) 498 (III) 946.0 (IV) 431.0 Choose the correct answer from the options given below:
Not quite — the answer is D.
N₂ has a triple bond and the highest bond enthalpy, 946 kJ mol⁻¹; O₂ with a double bond has 498 kJ mol⁻¹. H–H is 435.8 kJ mol⁻¹ and H–Cl is 431 kJ mol⁻¹, the two single bonds with close values. Swapping H₂ and HCl, or giving N₂ a value below O₂'s, are the traps.
Q35NEET 2024 Re-exam
The standard cell potential of the following cell Zn | Zn²⁺(aq) | Fe²⁺(aq) | Fe is 0.32 V. Calculate the standard Gibbs energy change for the reaction: Zn(s) + Fe²⁺(aq) → Zn²⁺(aq) + Fe(s) (Given : 1 F = 96487 C)
Not quite — the answer is A.
ΔG° = −nFE°cell with n = 2: ΔG° = −2 × 96487 × 0.32 = −61 752 J mol⁻¹ ≈ −61.75 kJ mol⁻¹. It is negative because E°cell is positive (spontaneous cell). The positive value has the wrong sign, and ±5.006 kJ does not follow from −nFE° at all.
Q36NEET 2024 Re-exam
Match List I with List II: List - I (Solid salt treated with dil. H₂SO₄): (A) effervescence of colourless gas (B) gas with smell of rotten egg (C) gas with pungent smell (D) brown fumes List - II (Anion detected): (I) NO₂⁻ (II) CO₃²⁻ (III) S²⁻ (IV) SO₃²⁻ Choose the correct answer from the options given below:
Not quite — the answer is A.
With dilute H₂SO₄, carbonate gives brisk effervescence of colourless, odourless CO₂; sulphide gives H₂S with its rotten-egg smell; sulphite gives SO₂ with a pungent (burning sulphur) smell; and nitrite gives brown fumes of NO₂. Matching sulphite with the colourless effervescence confuses SO₂'s sharp smell with odourless CO₂.
Q37NEET 2024 Re-exam
The ratio of solubility of AgCl in 0.1 M KCl solution to the solubility of AgCl in water is: (Given : Solubility product of AgCl = 10⁻¹⁰)
Not quite — the answer is A.
In water s₀ = √Ksp = √10⁻¹⁰ = 10⁻⁵ M. In 0.1 M KCl the common Cl⁻ ion gives s = Ksp/[Cl⁻] = 10⁻¹⁰/0.1 = 10⁻⁹ M. The ratio s/s₀ = 10⁻⁹/10⁻⁵ = 10⁻⁴. The value 10⁻⁹ is the solubility in KCl itself, not the ratio.
Q38NEET 2024 Re-exam
On complete combustion, 0.3 g of an organic compound gave 0.2 g of CO₂ and 0.1 g of H₂O. The percentage composition of carbon and hydrogen in the compound, respectively is:
Not quite — the answer is B.
%C = (12/44) × (0.2/0.3) × 100 = 0.0545/0.3 × 100 = 18.18%. %H = (2/18) × (0.1/0.3) × 100 = 0.0111/0.3 × 100 = 3.70%. The option with the same numbers reversed swaps carbon and hydrogen.
Q39NEET 2024 Re-exam
The following reaction method C₆H₅CH₃ + X₂ —Fe, dark→ o-XC₆H₄CH₃ + p-XC₆H₄CH₃ is not suitable for the preparation of the corresponding haloarene products, due to high reactivity of halogen, when X is :
Not quite — the answer is A.
Fluorine is so reactive that it reacts violently and uncontrollably with the arene, so aryl fluorides cannot be made by direct electrophilic fluorination; they are made via diazonium salts instead. Iodination is also unsuitable, but for the opposite reason: it is reversible and needs an oxidising agent to remove HI, so it is not about high reactivity.
Q40NEET 2024 Re-exam
The alkane that can be oxidized to the corresponding alcohol by KMnO₄ as per the equation R₁R₂R₃C–H —KMnO₄→ R₁R₂R₃C–OH is, when :
Not quite — the answer is B.
KMnO₄ oxidises an alkane that has a tertiary C–H bond to the tertiary alcohol: (CH₃)₃CH (2-methylpropane) gives (CH₃)₃COH (2-methylpropan-2-ol). With all three groups CH₃ the carbon is tertiary. Methane, ethane and propane (the other combinations) have only primary or secondary C–H bonds and are not converted this way.
Q41NEET 2024 Re-exam
For the following reaction at 300 K A₂(g) + 3B₂(g) → 2AB₃(g) the enthalpy change is +15 kJ, then the internal energy change is :
Not quite — the answer is A.
Δn_g = 2 − (1 + 3) = −2. ΔU = ΔH − Δn_g RT = 15 000 J − (−2)(8.314)(300) J = 15 000 + 4988.4 = 19 988.4 J. Using Δn_g = +2 would give 10 011.6 J; the values around 2 kJ come from misplacing the decimal.
Q42NEET 2024 Re-exam
Rate constants of a reaction at 500 K and 700 K are 0.04 s⁻¹ and 0.14 s⁻¹, respectively; then, activation energy of the reaction is : (Given: log 3.5 = 0.5441, R = 8.31 J K⁻¹ mol⁻¹)
Not quite — the answer is C.
log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂). k₂/k₁ = 0.14/0.04 = 3.5, and 1/500 − 1/700 = 200/350 000 = 1/1750 K⁻¹. So Ea = 0.5441 × 2.303 × 8.31 × 1750 ≈ 18 220 J, i.e. 18219 J (the small difference is rounding of 2.303). 182310 J is ten times too large.
Q43NEET 2024 Re-exam
Mass of glucose (C₆H₁₂O₆) required to be dissolved to prepare one litre of its solution which is isotonic with 15 g L⁻¹ solution of urea (NH₂CONH₂) is (Given: Molar mass in g mol⁻¹ C : 12, H : 1, O : 16, N : 14)
Not quite — the answer is D.
Isotonic solutions of non-electrolytes have equal molar concentrations. Urea (M = 60 g mol⁻¹): 15/60 = 0.25 mol L⁻¹. Glucose (M = 180 g mol⁻¹) at 0.25 mol L⁻¹ needs 0.25 × 180 = 45 g in one litre. Taking the same mass, 15 g, ignores the different molar masses.
Q44NEET 2024 Re-exam
[Mn₂(CO)₁₀] and [Co₂(CO)₈] structures have A. Metal-Metal linkage B. Terminal CO groups C. Bridging CO groups D. Metal in zero oxidation state Choose the correct answer from the options given below
Not quite — the answer is D.
Both carbonyls have a metal-metal bond, terminal CO groups and the metal in the zero oxidation state (CO is neutral). [Mn₂(CO)₁₀] has only terminal CO ligands, with five on each Mn joined by a Mn–Mn bond, whereas [Co₂(CO)₈] also has two bridging CO groups. Bridging CO is therefore not common to both.
Q45NEET 2024 Re-exam
Methyl group attached to a positively charged carbon atom stabilizes the carbocation due to
Not quite — the answer is C.
The σ electrons of the C–H bonds of the methyl group overlap with the empty p orbital of the carbocation carbon (σ–p conjugation, called hyperconjugation), which spreads out the positive charge. A methyl group is electron-releasing (+I), not −I, and it has no π system for a mesomeric effect.
Chapter-wise split of NEET 2024 Re-exam Chemistry
- Chemical Bonding and Molecular Structure4 Q
- Equilibrium4 Q
- The d- and f-Block Elements3 Q
- Thermodynamics3 Q
- Coordination Compounds3 Q
- Organic Chemistry: Some Basic Principles and Techniques3 Q
- Classification of Elements and Periodicity in Properties2 Q
- Chemical Kinetics2 Q
- Hydrocarbons2 Q
- Electrochemistry2 Q
- Solutions2 Q
- Amines2 Q
- Structure of Atom2 Q
- Haloalkanes and Haloarenes2 Q
- Aldehydes, Ketones and Carboxylic Acids1 Q
- Some Basic Concepts of Chemistry1 Q
- Biomolecules1 Q
- Redox Reactions1 Q
- Alcohols, Phenols and Ethers1 Q
NEET 2024 Re-exam Biology questions
92 questions. Tap an option to check your answer.
Q1NEET 2024 Re-exam
The regions with high level of species richness, high degree of endemism and a loss of 70% of the species and habitat are identified as:
Not quite — the answer is C.
Biodiversity hotspots are regions with very high species richness and endemism that have already lost a large share (about 70%) of their original habitat, which is why they are given priority for conservation. Sacred groves are small community-protected forest patches, not regions identified by these criteria.
Q2NEET 2024 Re-exam
Which of the following simple tissues are commonly found in the fruit walls of nuts and pulp of pear?
Not quite — the answer is A.
Sclereids are short, irregular, highly lignified sclerenchyma cells and are typical of the hard shells of nuts, the gritty pulp of pear and guava, and tea leaves. Fibres are long, pointed cells found mainly in bundles in stems and leaves, not as stone cells in fruit pulp.
Q3NEET 2024 Re-exam
In a chromosome, there is a specific DNA sequence, responsible for initiating replication. It is called as:
Not quite — the answer is D.
Replication of DNA starts at a specific sequence called the origin of replication (ori); any DNA linked to an ori can be replicated, which is why vectors carry one. Recognition and restriction sites are sequences read and cut by restriction enzymes, and a cloning site is where foreign DNA is inserted.
Q4NEET 2024 Re-exam
Given below are two statements: Statement I: When many alleles of a single gene govern a character, it is called polygenic inheritance. Statement II: In Polygenic inheritance, the effect of each allele is additive. In the light of the above statements, choose the correct answer from the options given below.
Not quite — the answer is B.
When many alleles of one gene control a character it is multiple allelism (as in ABO blood groups), so the first statement is false; polygenic inheritance involves several genes. In polygenic traits such as human skin colour each contributing allele adds a small effect, so the second statement is true.
Q5NEET 2024 Re-exam
Which of the following are required for the light reaction of Photosynthesis? A. CO₂ B. O₂ C. H₂O D. Chlorophyll E. Light Choose the correct answer from the options given below:
Not quite — the answer is B.
The light reaction needs light, chlorophyll (pigments of the photosystems) and water, whose photolysis supplies electrons and releases O₂. CO₂ is used only in the Calvin cycle (dark reaction), and O₂ is a product of the light reaction, not a requirement, so only water, chlorophyll and light are needed.
Q6NEET 2024 Re-exam
Match List I with List II : List-I – List-II (A) Flemming – (I) Disc shaped sacs or cisternae near cell nucleus (B) Robert Brown – (II) Chromatin (C) George Palade – (III) Ribosomes (D) Camillo Golgi – (IV) Nucleus Choose the correct answer from the options given below:
Not quite — the answer is A.
Flemming named chromatin, Robert Brown discovered the nucleus, George Palade observed ribosomes under the electron microscope, and Camillo Golgi described the disc-shaped cisternae near the nucleus (Golgi apparatus). The common trap is pairing Palade with the Golgi body because both are endomembrane discoveries.
Q7NEET 2024 Re-exam
Match List-I with List-II: List-I (Type of Inheritance) – List-II (Example) (A) Incomplete dominance – (I) Blood groups in human (B) Co-dominance – (II) Flower colour in Antirrhinum (C) Pleiotropy – (III) Skin colour in human (D) Polygenic inheritance – (IV) Phenylketonuria Choose the correct answer from the options given below:
Not quite — the answer is B.
Flower colour in Antirrhinum (snapdragon) shows incomplete dominance, human ABO blood groups show co-dominance, phenylketonuria is a pleiotropic single-gene disorder, and human skin colour is polygenic. Skin colour is not incomplete dominance because it is controlled by several genes.
Q8NEET 2024 Re-exam
Which part of the ovule stores reserve food materials?
Not quite — the answer is A.
The nucellus is the mass of cells inside the integuments that contains reserve food materials and encloses the embryo sac. Integuments only form protective layers, and the funicle merely attaches the ovule to the placenta.
Q9NEET 2024 Re-exam
Which one of the following is not found in Gymnosperms?
Not quite — the answer is D.
Gymnosperm xylem consists of tracheids and lacks vessels, which are characteristic of angiosperms (Gnetales being an exception). Sieve cells and albuminous cells are the conducting and companion-like cells of gymnosperm phloem, so they are present.
Q10NEET 2024 Re-exam
Which one of the following is not included under in-situ conservation?
Not quite — the answer is B.
A botanical garden conserves plants away from their natural habitat, so it is ex-situ conservation. Wildlife sanctuaries, national parks and biosphere reserves protect species within their natural ecosystems, i.e. in-situ.
Q11NEET 2024 Re-exam
Given below are two statements: Statement I : The Indian Government has set up GEAC, which will make decisions regarding the validity of GM research. Statement II : Biopiracy is the term used to refer to the use of bio-resources by native people. In the light of the above statements, choose the correct answer from the options given below :
Not quite — the answer is A.
The Genetic Engineering Approval Committee (GEAC) was set up by the Indian Government to decide on the validity of GM research and the safety of introducing GM organisms, so the first statement is true. Biopiracy is the use of bio-resources and traditional knowledge by multinational companies without proper authorisation or compensation to the native people, so the second statement is false.
Q12NEET 2024 Re-exam
Pollen grains remain preserved as fossils due to the presence of :
Not quite — the answer is C.
The exine of the pollen wall is made of sporopollenin, one of the most resistant organic materials, which withstands high temperature, strong acids and alkalis and enzymatic degradation, so pollen is well preserved as fossils. The intine is a thin pectocellulosic layer that decays easily.
Q13NEET 2024 Re-exam
Identify the incorrect pair :
Not quite — the answer is A.
Adiantum is a fern and belongs to Pteropsida; Sphenopsida is represented by Equisetum, so that pair is incorrect. Dryopteris (Pteropsida), Psilotum (Psilopsida) and Selaginella (Lycopsida) are correctly placed.
Q14NEET 2024 Re-exam
Which of the following is the correct match?
Not quite — the answer is A.
Cedrus, Pinus and Sequoia are all gymnosperms, so that set is correct. Sequoia is wrongly listed under angiosperms, Polysiphonia is a red alga rather than a bryophyte, and Ginkgo is a gymnosperm rather than a pteridophyte.
Q15NEET 2024 Re-exam
Given below are two statements regarding RNA polymerase in prokaryotes. Statement I : In prokaryotes, RNA polymerase is capable of catalysing the process of elongation during transcription. Statement II : RNA polymerase associate transiently with ‘Rho’ factor to initiate transcription. In the light of the above statements, choose the correct answer from the options given below :
Not quite — the answer is A.
Bacterial RNA polymerase alone catalyses chain elongation during transcription, so the first statement is true. It associates transiently with the sigma factor to initiate transcription and with the rho factor to terminate it, so the second statement is false.
Q16NEET 2024 Re-exam
Which of the following is a nucleotide?
Not quite — the answer is B.
A nucleotide is a nitrogen base + sugar + phosphate; adenylic acid (adenosine monophosphate) has all three. Uridine and guanosine are nucleosides (base + sugar only), and guanine is just a nitrogenous base.
Q17NEET 2024 Re-exam
Match List-I with List-II : List-I – List-II (A) Vexillary aestivation – (I) Brinjal (B) Epipetalous stamens – (II) Peach (C) Epiphyllous stamens – (III) Pea (D) Perigynous flower – (IV) Lily Choose the correct answer from the options given below :
Not quite — the answer is A.
Vexillary aestivation is seen in pea, epipetalous stamens (attached to petals) in brinjal, epiphyllous stamens (attached to perianth) in lily, and perigynous flowers in peach (also plum and rose). Matching lily with epipetalous is the trap, but lily has tepals, not separate petals.
Q18NEET 2024 Re-exam
Match List-I with List-II : List-I – List-II (A) China rose – (I) Free central (B) Mustard – (II) Basal (C) Primrose – (III) Axile (D) Marigold – (IV) Parietal Choose the correct answer from the options given below :
Not quite — the answer is C.
China rose has axile placentation, mustard parietal (with a false septum), primrose free central, and marigold (sunflower family) basal placentation. Mustard is often mistaken for axile because of its false septum, but the ovules are borne on the ovary wall.
Q19NEET 2024 Re-exam
Which of the following helps in maintenance of the pressure gradient in sieve tubes?
Not quite — the answer is D.
Companion cells are connected to sieve tube elements by pit fields and their nuclei control the sieve tubes; they help maintain the pressure gradient in the sieve tubes. Albuminous and sieve cells are found in gymnosperms, and phloem parenchyma only stores food and other substances.
Q20NEET 2024 Re-exam
Mesosome in a cell is a :
Not quite — the answer is C.
A mesosome is a special membranous structure formed by infoldings (extensions) of the plasma membrane of prokaryotes, helping in wall formation, DNA replication and respiration. A chain of ribosomes on one mRNA is a polysome, and mesosomes are not membrane-bound vesicles.
Q21NEET 2024 Re-exam
Match List-I with List-II : List-I – List-II (A) Abscisic acid – (I) Promotes female flowers in cucumber (B) Ethylene – (II) Helps seeds to withstand desiccation (C) Gibberellin – (III) Helps in nutrient mobilisation (D) Cytokinin – (IV) Promotes bolting in beet, cabbage etc. Choose the correct answer from the options given below:
Not quite — the answer is C.
Abscisic acid helps seeds withstand desiccation, ethylene promotes female flowers in cucumber, gibberellins cause bolting in beet and cabbage, and cytokinins promote nutrient mobilisation and delay senescence. Assigning nutrient mobilisation to ethylene is the trap.
Q22NEET 2024 Re-exam
Match List-I with List - II : List-I – List-II (A) Genetically engineered Human Insulin – (I) Gene therapy (B) GM Cotton – (II) E. coli (C) ADA Deficiency – (III) Antigen-antibody interaction (D) ELISA – (IV) Bacillus thuringiensis Choose the correct answer from the options given below:
Not quite — the answer is D.
Genetically engineered human insulin was produced in E. coli, GM cotton carries the cry gene of Bacillus thuringiensis, ADA deficiency was the first disease treated by gene therapy, and ELISA works on antigen–antibody interaction. These pairings follow directly from each technology's basis.
Q23NEET 2024 Re-exam
Match List-I with List-II: List-I – List-II (A) ETS Complex I – (I) NADH Dehydrogenase (B) ETS Complex II – (II) Cytochrome bC₁ (C) ETS Complex III – (III) Cytochrome C oxidase (D) ETS Complex IV – (IV) Succinate Dehydrogenase Choose the correct answer from the options given below :
Not quite — the answer is B.
Complex I of the electron transport system is NADH dehydrogenase, complex II is succinate dehydrogenase, complex III is the cytochrome bc₁ complex, and complex IV is cytochrome c oxidase. Mixing up complexes I and II is the common trap, since both are dehydrogenases.
Q24NEET 2024 Re-exam
Cryopreservation technique is used for :
Not quite — the answer is C.
Cryopreservation stores gametes, embryos and other tissues at very low temperature (in liquid nitrogen at −196 °C) so they remain viable and fertile for long periods; it is an ex-situ method. It is therefore not an in-situ technique and does not protect hotspots or the environment as such.
Q25NEET 2024 Re-exam
Which of the following are correct about cellular respiration? A. Cellular respiration is the breaking of C-C bonds of complex organic molecules by oxidation. B. The entire cellular respiration takes place in Mitochondria. C. Fermentation takes place under anaerobic condition in germinating seeds. D. The fate of pyruvate formed during glycolysis depends on the type of organism also. E. Water is formed during respiration as a result of O₂ accepting electrons and getting reduced. Choose the correct answer from the options given below :
Not quite — the answer is A.
Respiration is the oxidative breakdown of C–C bonds of food, the fate of pyruvate depends on the organism and conditions, fermentation can occur under anaerobic conditions (as in germinating seeds deprived of oxygen), and water forms when O₂ accepts electrons at the end of the ETS. The statement that the entire process occurs in mitochondria is false because glycolysis takes place in the cytoplasm.
Q26NEET 2024 Re-exam
Given below are two statements: Statement I: In eukaryotes there are three RNA polymerases in the nucleus in addition to the RNA polymerase found in the organelles. Statement II: All the three RNA polymerases in eukaryotic nucleus have different roles. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is C.
Eukaryotic nuclei contain RNA polymerase I, II and III, in addition to the separate RNA polymerases of mitochondria and chloroplasts, so the first statement is true. They divide the work: polymerase I makes rRNAs, II makes hnRNA (mRNA precursor) and III makes tRNA, 5S rRNA and snRNAs, so the second statement is also true.
Q27NEET 2024 Re-exam
Match List-I with List-II: List-I – List-II (A) Histones – (I) Loosely packed chromatin (B) Nucleosome – (II) Densely packed chromatin (C) Euchromatin – (III) Positively charged basic proteins (D) Heterochromatin – (IV) DNA wrapped around histone octamer Choose the correct answer from the options given below:
Not quite — the answer is D.
Histones are positively charged basic proteins rich in lysine and arginine, a nucleosome is DNA wrapped around a histone octamer, euchromatin is loosely packed (transcriptionally active) chromatin and heterochromatin is densely packed chromatin. Confusing histones with nucleosomes is the main trap.
Q28NEET 2024 Re-exam
Given below are two statements: Statement I: Failure of segregation of chromatids during cell cycle resulting in the gain or loss of whole set of chromosome in an organism is known as aneuploidy. Statement II: Failure of cytokinesis after anaphase stage of cell division results in the gain or loss of a chromosome is called polyploidy. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is D.
Failure of chromatid segregation causes gain or loss of one or a few chromosomes (aneuploidy), not of a whole set, so the first statement is false. Failure of cytokinesis after telophase leads to an increase in a whole set of chromosomes (polyploidy), not gain or loss of one chromosome, so the second statement is also false.
Q29NEET 2024 Re-exam
Recombination between homologous chromosomes is completed by the end of
Not quite — the answer is D.
Crossing over between non-sister chromatids of homologous chromosomes occurs during pachytene, and recombination is completed by the end of this stage. Chiasmata only become visible in diplotene, which is why diplotene is a tempting but wrong choice.
Q30NEET 2024 Re-exam
Match List-I with List-II : List-I – List-II (A) Metacentric chromosome – (I) Chromosome has a terminal centromere (B) Sub-metacentric chromosome – (II) Middle centromere forming two equal arms of chromosome (C) Acrocentric chromosome – (III) Centromere is slightly away from the middle of chromosome resulting into two unequal arms (D) Telocentric chromosome – (IV) Centromere is situated close to its end forming one extremely short and one very long arm Choose the correct answer from the options given below:
Not quite — the answer is D.
A metacentric chromosome has a middle centromere and two equal arms, a sub-metacentric one has the centromere slightly away from the middle giving unequal arms, an acrocentric one has it close to one end giving one very short and one long arm, and a telocentric one has a terminal centromere. Swapping acrocentric and telocentric is the usual mistake.
Q31NEET 2024 Re-exam
Ligases is a class of enzymes responsible for catalysing the linking together of two compounds. Which of the following bonds is not catalysed by it?
Not quite — the answer is A.
Ligases are described as catalysing the linking of two compounds by forming C–O, C–S, C–N and P–O bonds; C–C is not in this list, so it is the answer expected. P–O bonds (for example joining DNA fragments) are the classic ligase reaction.
Q32NEET 2024 Re-exam
F. Skoog observed that callus proliferated from the internodal segments of tobacco stem when auxin was supplied with one of the following except :
Not quite — the answer is C.
Skoog and Miller found that callus from tobacco internodes proliferated when auxin was supplied together with extracts of vascular tissue, yeast extract, coconut milk or DNA, which led to the discovery of cytokinins. Abscisic acid is a growth inhibitor and was not among these substances.
Q33NEET 2024 Re-exam
Given below are some statements about plant growth regulators. A. All GAs are acidic in nature. B. Auxins are antagonists to GAs. C. Zeatin was isolated from coconut milk. D. Ethylene induces flowering in Mango. E. Abscisic acid induces parthenocarpy. Choose the correct set of statements from the options given below :
Not quite — the answer is A.
All gibberellins are acidic, zeatin was isolated from corn kernels and coconut milk, and ethylene induces flowering in mango, so these three are correct. It is abscisic acid, not auxin, that acts as the antagonist of gibberellins, and parthenocarpy is induced by auxins and gibberellins, not by abscisic acid.
Q34NEET 2024 Re-exam
Identify the incorrect statement related to gel electrophoresis.
Not quite — the answer is A.
DNA fragments in an agarose gel cannot be seen directly; they become visible as orange bands only after staining with ethidium bromide and exposure to UV light, so that statement is incorrect. DNA is negatively charged and moves towards the anode, separated by the sieving effect of the gel, and bands can be cut out and eluted.
Q35NEET 2024 Re-exam
Which of the following examples show monocarpellary, unilocular ovary with many ovules? A. Sesbania B. Brinjal C. Indigofera D. Tobacco E. Asparagus Choose the correct answer from the options given below :
Not quite — the answer is D.
Sesbania and Indigofera belong to Fabaceae, whose pistil is monocarpellary with a unilocular ovary bearing many ovules (marginal placentation). Brinjal and tobacco (Solanaceae) have a bicarpellary ovary, and Asparagus (Liliaceae) has a tricarpellary ovary.
Q36NEET 2024 Re-exam
Given below are two statements : Statement I: In the lac operon, the z gene codes for beta-galactosidase which is primarily responsible for the hydrolysis of lactose into galactose and glucose. Statement II: In addition to lactose, glucose or galactose can also induce lac operon. In the light of the above statements, choose the correct answer from the options given below :
Not quite — the answer is A.
The z gene codes for β-galactosidase, which hydrolyses lactose into galactose and glucose, so the first statement is true. The lac operon inducer is lactose (or its analogues); glucose or galactose cannot induce it, so the second statement is false.
Q37NEET 2024 Re-exam
Given below are two statements: Statement I: In a dicotyledonous leaf, the adaxial epidermis generally bears more stomata than the abaxial epidermis. Statement II: In a dicotyledonous leaf, the adaxially placed palisade parenchyma is made up of elongated cells, which are arranged vertically and parallel to each other. In the light of the above statements, choose the correct answer from the options given below:
Not quite — the answer is B.
In a dorsiventral (dicot) leaf the abaxial (lower) epidermis generally bears more stomata than the adaxial one, so the first statement is false. The adaxially placed palisade parenchyma has elongated cells arranged vertically and parallel to each other, so the second statement is true.
Q38NEET 2024 Re-exam
Which of the following are not fatty acids? A. Glutamic acid B. Arachidonic acid C. Palmitic acid D. Lecithin E. Aspartic acid Choose the correct answer from the options given below :
Not quite — the answer is C.
Glutamic acid and aspartic acid are amino acids, and lecithin is a phospholipid, so these are not fatty acids. Arachidonic acid (20 C) and palmitic acid (16 C) are fatty acids with a carboxyl group attached to an R group.
Q39NEET 2024 Re-exam
Which one of the following products diffuses out of the chloroplast during photosynthesis?
Not quite — the answer is C.
O₂ released by photolysis of water diffuses out of the chloroplast. ATP and NADPH formed in the light reaction are used up inside the chloroplast stroma in the Calvin cycle, and ADP is regenerated there too.
Q40NEET 2024 Re-exam
Recombinant DNA molecule can be created normally by cutting the vector DNA and source DNA respectively with:
Not quite — the answer is A.
To join vector and source DNA, both are normally cut with the same restriction enzyme so that their ends are compatible; using Hind II on both satisfies this. Using two different enzymes generally produces ends that cannot be ligated together.
Q41NEET 2024 Re-exam
Which one of the following is not a limitation of ecological pyramids?
Not quite — the answer is C.
Ecological pyramids are limited because they assume a simple food chain, ignore saprophytes and cannot place one species in two or more trophic levels; in fact they do not accommodate a food web. So the statement that they accommodate a food web is not a limitation (it is untrue).
Q42NEET 2024 Re-exam
The Bt toxin in genetically engineered Bt cotton kills the pest by:
Not quite — the answer is A.
The Bt protoxin becomes active in the alkaline insect gut, binds to the surface of midgut epithelial cells and creates pores, causing swelling, lysis and death of the insect. It does not act on the respiratory or nervous system.
Q43NEET 2024 Re-exam
Match List-I with List-II: List-I (Organisms) – List-II (Mode of Nutrition) (A) Euglenoid – (I) Parasitic (B) Dinoflagellate – (II) Saprophytic (C) Slime mould – (III) Photosynthetic (D) Plasmodium – (IV) Switching between photosynthetic and heterotrophic mode Choose the correct answer from the options given below:
Not quite — the answer is C.
Euglenoids are photosynthetic in light but heterotrophic in its absence, dinoflagellates are mostly photosynthetic, slime moulds are saprophytic, and Plasmodium is a parasite. Labelling dinoflagellates as saprophytic is the trap.
Q44NEET 2024 Re-exam
When will the population density increase, under special conditions? When the number of :
Not quite — the answer is D.
Population density increases when additions exceed losses, i.e. births plus immigration is greater than deaths plus emigration (Nt+1 = Nt + [(B + I) − (D + E)]). Equal additions and losses keep it constant, and more deaths than births reduce it.
Q45NEET 2024 Re-exam
When a tall pea plant with round seeds was selfed, it produced the progeny of : (a) Tall plants with round seeds and (b) Tall plants with wrinkled seeds. Identify the genotype of the parent plant.
Not quite — the answer is D.
All progeny are tall, so the parent is homozygous TT; both round and wrinkled seeds appear, so it must be heterozygous Rr, giving TTRr. TtRr would also produce dwarf plants, and TTRR or TtRR could not give wrinkled seeds.
Q46NEET 2024 Re-exam
Match List-I with List-II: List-I – List-II (A) Biodiversity hotspot – (I) Khasi and Jaintia hills in Meghalaya (B) Sacred groves – (II) World Summit on Sustainable Development 2002 (C) Johannesburg, South Africa – (III) Parthenium (D) Alien species invasion – (IV) Western Ghats Choose the correct answer from the options given below:
Not quite — the answer is A.
The Western Ghats is a biodiversity hotspot, the Khasi and Jaintia hills of Meghalaya have sacred groves, the 2002 World Summit on Sustainable Development was held in Johannesburg, and Parthenium is an alien invasive weed. These pairings follow directly from the conservation examples.
Q47NEET 2024 Re-exam
Match List-I with List-II: List-I – List-II (A) Predator – (I) Ophrys (B) Mutualism – (II) Pisaster (C) Parasitism – (III) Female wasp and fig (D) Sexual deceit – (IV) Plasmodium Choose the correct answer from the options given below:
Not quite — the answer is D.
Pisaster (a starfish) is a predator, the fig and its pollinating female wasp are mutualists, Plasmodium is a parasite, and the orchid Ophrys uses sexual deceit to get pollinated by bees. Wasp–fig is the classic mutualism, not parasitism.
Q48NEET 2024 Re-exam
Match List-I with List-II: List-I (Location of Joint) – List-II (Type of Joint) (A) Joint between humerus and pectoral girdle – (I) Gliding joint (B) Knee joint – (II) Ball and Socket joint (C) Joint between atlas and axis – (III) Hinge joint (D) Joint between carpals – (IV) Pivot joint Choose the correct answer from the options given below:
Not quite — the answer is A.
The humerus and pectoral girdle form a ball-and-socket joint, the knee is a hinge joint, the atlas–axis joint is a pivot joint, and joints between carpals are gliding joints. Swapping the knee with the shoulder type is the usual error.
Q49NEET 2024 Re-exam
Following are the steps involved in action of toxin in Bt. Cotton A. The inactive toxin converted into active form due to alkaline pH of gut of insect. B. Bacillus thuringiensis produces crystals with toxic insecticidal proteins. C. The alkaline pH solubilises the crystals. D. The activated toxin binds to the surface of midgut cells, creates pores and causes death of the insect. E. The toxin proteins exist as inactive protoxins in bacteria. Choose the correct sequence of steps from the options given below:
Not quite — the answer is D.
Bacillus thuringiensis forms protein crystals, the toxin exists in them as inactive protoxin, the alkaline insect gut pH solubilises the crystals, the protoxin is converted into active toxin, and the active toxin binds to midgut cells, creates pores and kills the insect. Activation cannot precede solubilisation, which rules out the other sequences.
Q50NEET 2024 Re-exam
Match List-I with List-II : List-I – List-II (A) Gene pool – (I) Stable within a generation (B) Genetic drift – (II) Change in gene frequency by chance (C) Gene flow – (III) Transfer of genes into or out of population (D) Gene frequency – (IV) Total number of genes and their alleles Choose the correct answer from the options given below :
Not quite — the answer is B.
The gene pool is the total number of genes and their alleles in a population, genetic drift is change in gene frequency by chance, gene flow is transfer of genes into or out of a population, and gene frequency stays stable within a generation in Hardy–Weinberg equilibrium. Confusing gene flow with drift is the trap.
Q51NEET 2024 Re-exam
A person with blood group ARh– can receive the blood transfusion from which of the following types? A. BRh⁻ B. ABRh⁻ C. ORh⁻ D. ARh⁻ E. ARh⁺ Choose the correct answer from the options given below :
Not quite — the answer is D.
An A Rh⁻ person has anti-B antibodies and lacks Rh antigen, so the donor must have no B antigen and be Rh negative: O Rh⁻ and A Rh⁻ are safe. B or AB blood carries B antigen, and A Rh⁺ blood would sensitise the Rh negative recipient.
Q52NEET 2024 Re-exam
Enzymes that catalyse the removal of groups from substrates by mechanisms other than hydrolysis leaving double bonds, are known as :
Not quite — the answer is D.
Lyases remove groups from substrates by mechanisms other than hydrolysis, leaving double bonds. Transferases move groups between substrates, and oxidoreductases/dehydrogenases catalyse oxidation–reduction.
Q53NEET 2024 Re-exam
Match List-I with List-II. List-I (Event) – List-II (Stage of Prophase-I (Meiosis-I)) (A) Chiasmata formation – (I) Pachytene (B) Crossing over – (II) Diakinesis (C) Synaptonemal complex formation – (III) Diplotene (D) Terminalisation of chiasmata – (IV) Zygotene Choose the correct answer from the options given below :
Not quite — the answer is A.
Synaptonemal complex forms during zygotene, crossing over occurs in pachytene, chiasmata become visible in diplotene, and terminalisation of chiasmata occurs in diakinesis. Placing chiasma formation in diakinesis confuses formation with terminalisation.
Q54NEET 2024 Re-exam
Match List-I with List-II. List-I – List-II (A) Primary structure of protein – (I) Human haemoglobin (B) Secondary structure of protein – (II) Disulphide bonds (C) Tertiary structure of protein – (III) Polypeptide chain (D) Quaternary structure of protein – (IV) Alpha helix and β sheet Choose the correct answer from the options given below :
Not quite — the answer is A.
The primary structure is the sequence of amino acids in the polypeptide chain, secondary structure is the alpha helix and β sheet, tertiary structure is stabilised by bonds such as disulphide bonds, and quaternary structure is the assembly of subunits as in human haemoglobin. Haemoglobin is the standard example of quaternary structure.
Q55NEET 2024 Re-exam
Match List-I with List-II. List-I – List-II (A) Epinephrine – (I) Hyperglycemia (B) Thyroxine – (II) Smooth muscle contraction (C) Oxytocin – (III) Basal metabolic rate (D) Glucagon – (IV) Emergency hormone Choose the correct answer from the options given below :
Not quite — the answer is C.
Epinephrine is the emergency (fight-or-flight) hormone, thyroxine regulates the basal metabolic rate, oxytocin causes smooth muscle contraction (uterus during birth, milk ejection), and glucagon raises blood glucose (hyperglycaemia). Linking epinephrine to smooth muscle contraction misses its classic emergency role.
Q56NEET 2024 Re-exam
Which of the following statements is correct about the type of junction and their role in our body?
Not quite — the answer is B.
Tight junctions stop substances from leaking across a tissue. Cementing neighbouring cells together is the role of adhering junctions, and gap junctions allow cells to communicate by connecting their cytoplasm, not to create gaps.
Q57NEET 2024 Re-exam
Select the restriction endonuclease enzymes whose restriction sites are present for the tetracycline resistance (tetᴿ) gene in the pBR322 cloning vector.
Not quite — the answer is A.
In pBR322, BamHI and SalI have their recognition sites inside the tetracycline resistance gene, so inserting DNA there inactivates tetᴿ. PstI and PvuI sites lie within the ampicillin resistance gene.
Q58NEET 2024 Re-exam
Match List-I with List-II. List-I – List-II (A) Chondrichthyes – (I) Clarias (B) Cyclostomata – (II) Carcharodon (C) Osteichthyes – (III) Myxine (D) Amphibia – (IV) Ichthyophis Choose the correct answer from the options given below :
Not quite — the answer is C.
Carcharodon (great white shark) is a cartilaginous fish, Myxine (hagfish) is a cyclostome, Clarias (magur) is a bony fish, and Ichthyophis (limbless caecilian) is an amphibian. Ichthyophis is often mistaken for a fish because of its name.
Q59NEET 2024 Re-exam
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: During menstrual cycle, the ovulation takes place approximately on 14th day. Reason R: Rapid secretion of LH in the middle of menstrual cycle induces rupture of Graafian follicle and thereby the release of ovum. In the light of the above statements, choose the most appropriate answer from the options given below.
Not quite — the answer is C.
In a 28-day cycle ovulation occurs around day 14, and this happens because the LH surge in the middle of the cycle induces rupture of the Graafian follicle and release of the ovum. So both statements are true and the reason explains the assertion.
Q60NEET 2024 Re-exam
Match List-I with List-II with respect to convergent evolution: List-I – List-II (A) Lemur – (I) Flying phalanger (B) Bobcat – (II) Numbat (C) Anteater – (III) Spotted cuscus (D) Flying squirrels – (IV) Tasmanian tiger cat Choose the correct answer from the options given below:
Not quite — the answer is A.
In convergent evolution of Australian marsupials and placental mammals, the lemur matches the spotted cuscus, the bobcat the Tasmanian tiger cat, the anteater the numbat, and the flying squirrel the flying phalanger. Pairing the bobcat with the numbat is the trap; the numbat is the marsupial anteater.
Q61NEET 2024 Re-exam
Match List-I with List-II. List-I – List-II (A) Cells are metabolically active and proliferate – (I) G₂ phase (B) DNA replication takes place – (II) G₁ phase (C) Proteins are synthesised – (III) G₀ phase (D) Quiescent stage with metabolically active cells – (IV) S phase Choose the correct answer from the options given below:
Not quite — the answer is D.
In G₁ cells are metabolically active and grow continuously, in S phase DNA replicates, in G₂ proteins are synthesised in preparation for mitosis, and G₀ is the quiescent stage where cells stay metabolically active but stop proliferating. G₀ cannot be the proliferating stage.
Q62NEET 2024 Re-exam
Match List-I with List-II relating to microbes and their products: List-I (Microbes) – List-II (Products) (A) Streptococcus – (I) Citric acid (B) Trichoderma polysporum – (II) Clot buster (C) Monascus purpureus – (III) Cyclosporin A (D) Aspergillus niger – (IV) Statins Choose the correct answer from the options given below:
Not quite — the answer is A.
Streptococcus produces streptokinase, a clot buster; Trichoderma polysporum produces cyclosporin A, an immunosuppressant; Monascus purpureus produces statins; and Aspergillus niger produces citric acid. Linking Streptococcus to citric acid is the common error.
Q63NEET 2024 Re-exam
Match List-I with List-II. List-I – List-II (A) F Particles – (I) Chromosomes (B) Histones – (II) Cilia (C) Axoneme – (III) Golgi apparatus (D) Cisternae – (IV) Mitochondria Choose the correct answer from the options given below:
Not quite — the answer is B.
F₀–F₁ particles (oxysomes) occur on the inner mitochondrial membrane, histones are part of chromosomes, the axoneme is the core of cilia and flagella, and cisternae make up the Golgi apparatus. Cisternae also occur in ER, but among the given choices they match the Golgi apparatus.
Q64NEET 2024 Re-exam
Match List-I with List-II relating to examples of various kind of IUDs and barrier: List-I – List-II (A) Copper releasing IUD – (I) Vaults (B) Non-medicated IUD – (II) Multiload 375 (C) Contraceptive barrier – (III) LNG-20 (D) Hormone releasing IUD – (IV) Lippes loop Choose the correct answer from the options given below:
Not quite — the answer is D.
Multiload 375 is a copper-releasing IUD, the Lippes loop is a non-medicated IUD, vaults are barrier contraceptives, and LNG-20 is a hormone-releasing IUD. Confusing LNG-20 with a barrier is the trap.
Q65NEET 2024 Re-exam
Given below are two statements: Statement I: Antibiotics are chemicals produced by microbes that kill other microbes. Statement II: Antibodies are chemicals formed in body that eliminate microbes. In the light of the above statements, choose the most appropriate answer from the options given below.
Not quite — the answer is C.
Antibiotics are chemical substances produced by some microbes that kill or retard the growth of other (disease-causing) microbes, and antibodies are proteins formed in the body by B cells that help eliminate pathogens. Both statements are therefore correct.
Q66NEET 2024 Re-exam
Arrange the following parts in human Mammary gland, traversing the route of milk ejection. A. Mammary duct B. Lactiferous duct C. Mammary alveolus D. Ampulla E. Mammary tubule Choose the correct answer from the options given below:
Not quite — the answer is C.
Milk is secreted in the mammary alveoli, passes into mammary tubules, then mammary ducts, which join to form a wider ampulla, and finally drains through the lactiferous duct. Any order that starts at the ampulla or duct reverses the route.
Q67NEET 2024 Re-exam
Which of the following are correct about EcoRI? A. Cut the DNA with blunt end B. Cut the DNA with sticky end C. Recognise a specific palindromic sequence D. Cut the DNA between the base G and A when encounters the DNA sequence 'GAATTC' E. Exonuclease Choose the correct answer from the options given below:
Not quite — the answer is D.
EcoRI recognises the palindromic sequence GAATTC and cuts between G and A on each strand, leaving sticky ends. It is an endonuclease that cuts within DNA and does not produce blunt ends, so the blunt-end and exonuclease statements are wrong.
Q68NEET 2024 Re-exam
Which of the following is/are present in female cockroach? A. Collateral gland B. Mushroom gland C. Spermatheca D. Anal style E. Phallic gland Choose the most appropriate answer from the options given below:
Not quite — the answer is D.
The female cockroach has collateral glands (which form the ootheca) and a spermatheca that stores sperm. Mushroom glands, phallic glands and anal styles are features of the male cockroach.
Q69NEET 2024 Re-exam
Match List-I with List-II: List-I – List-II (A) Malignant tumors – (I) Destroy tumors (B) MALT – (II) AIDS (C) NACO – (III) Metastasis (D) α-Interferons – (IV) Lymphoid tissue Choose the correct answer from the options given below:
Not quite — the answer is A.
Malignant tumours spread by metastasis, MALT is mucosa-associated lymphoid tissue, NACO (National AIDS Control Organisation) deals with AIDS, and α-interferons activate the immune system and help destroy tumours. Linking interferons to lymphoid tissue is the trap.
Q70NEET 2024 Re-exam
Open Circulatory system is present in :
Not quite — the answer is C.
Arthropods (Anopheles, Limulus) and most molluscs (Limax, a slug) have an open circulatory system. Nereis, Pheretima and Hirudinaria are annelids with closed circulation, so any set containing them is wrong.
Q71NEET 2024 Re-exam
In which of the following connective tissues, the cells secrete fibres of collagen or elastin? A. Cartilage B. Bone C. Adipose tissue D. Blood E. Areolar tissue Choose the most appropriate answer from the options given below :
Not quite — the answer is B.
In all connective tissues except blood, the cells secrete fibres of structural proteins such as collagen or elastin; so cartilage, bone, adipose tissue and areolar tissue qualify. Blood is a fluid connective tissue whose cells do not secrete such fibres.
Q72NEET 2024 Re-exam
Match List-I with List-II: List-I – List-II (A) Residual Volume – (I) Maximum volume of air that can be breathed in after forced expiration (B) Vital Capacity – (II) Volume of air inspired or expired during normal respiration (C) Expiratory Capacity – (III) Volume of air remaining in lungs after forcible expiration (D) Tidal Volume – (IV) Total volume of air expired after normal inspiration Choose the correct answer from the options given below:
Not quite — the answer is C.
Residual volume is the air left in the lungs after forcible expiration, vital capacity is the maximum air breathed in after forced expiration, expiratory capacity is the total air expired after a normal inspiration (TV + ERV), and tidal volume is the air inspired or expired in normal breathing. Mixing vital capacity with expiratory capacity is the trap.
Q73NEET 2024 Re-exam
Match List I with List II: List-I – List-II (A) Living Fossil – (I) Elongated canine teeth (B) Connecting Link – (II) Vermiform appendix (C) Vestigial Organ – (III) Echidna (D) Atavism – (IV) Latimeria Choose the correct answer from the options given below:
Not quite — the answer is A.
Latimeria (coelacanth) is a living fossil, Echidna (egg-laying mammal) is a connecting link between reptiles and mammals, the vermiform appendix is a vestigial organ, and elongated canines in humans are an atavism (reappearance of an ancestral trait). Calling the appendix an atavism confuses vestigial with reappearing traits.
Q74NEET 2024 Re-exam
Match List-I with List -II List-I – List-II (A) Schwann cells – (I) Neurotransmitter (B) Synaptic knob – (II) Cerebral cortex (C) Bipolar neurons – (III) Myelin sheath (D) Multipolar neurons – (IV) Retina Choose the correct answer from the options given below:
Not quite — the answer is A.
Schwann cells form the myelin sheath, synaptic knobs contain vesicles of neurotransmitter, bipolar neurons are found in the retina, and multipolar neurons are found in the cerebral cortex. Retina is the textbook location for bipolar neurons.
Q75NEET 2024 Re-exam
Diuresis is prevented by:
Not quite — the answer is D.
Vasopressin (ADH), released from the neurohypophysis, increases water reabsorption from the distal parts of the tubule and so prevents diuresis. ANF promotes loss of water and salt, and aldosterone comes from the adrenal cortex, not the medulla.
Q76NEET 2024 Re-exam
Following is the list of STDs. Select the diseases which are not completely curable. A. Genital warts B. Genital herpes C. Syphilis D. Hepatitis-B E. Trichomoniasis Choose the correct answer from the options given below:
Not quite — the answer is B.
Except for hepatitis-B, genital herpes and HIV infection, sexually transmitted diseases are completely curable if detected early and treated properly. Syphilis, trichomoniasis and genital warts are curable, so genital herpes and hepatitis-B are the ones not completely curable.
Q77NEET 2024 Re-exam
What is the correct order (old to recent) of periods in Paleozoic era?
Not quite — the answer is B.
From oldest to youngest the later Palaeozoic periods are Silurian, Devonian, Carboniferous and Permian; the Permian closes the Palaeozoic era. Any order placing the Permian before the Carboniferous reverses the sequence.
Q78NEET 2024 Re-exam
'Lub' sound of Heart is caused by the _________.
Not quite — the answer is D.
The first heart sound ‘lub’ is produced by the closure of the tricuspid and bicuspid (atrioventricular) valves at the start of ventricular systole. Closure of the semilunar valves produces the second sound ‘dub’.
Q79NEET 2024 Re-exam
Match List-I with List-II relating to human female external genitalia. List-I – List-II (A) Mons pubis – (I) A fleshy fold of tissue surrounding the vaginal opening (B) Clitoris – (II) Fatty cushion of cells covered by skin and hair (C) Hymen – (III) Tiny finger-like structure above labia minora (D) Labia majora – (IV) A thin membrane-like structure covering vaginal opening Choose the correct answer from the option given below :
Not quite — the answer is A.
The mons pubis is a fatty cushion covered by skin and pubic hair, the clitoris is a tiny finger-like structure above the urethral opening at the junction of the labia minora, the hymen is a thin membrane partially covering the vaginal opening, and the labia majora are fleshy folds surrounding the vaginal opening. Hymen and labia majora are the pair most often swapped.
Q80NEET 2024 Re-exam
Aneuploidy is a chromosomal disorder where chromosome number is not the exact copy of its haploid set of chromosomes, due to : A. Substitution B. Addition C. Deletion D. Translocation E. Inversion Choose the most appropriate answer from the options given below :
Not quite — the answer is D.
Aneuploidy is the gain (addition) or loss (deletion) of one or more chromosomes, so the chromosome number is not an exact multiple of the haploid set. Substitution, translocation and inversion rearrange genes or segments without changing the chromosome number.
Q81NEET 2024 Re-exam
Given below are two statements : Statement I : RNA interference takes place in all Eukaryotic organisms as method of cellular defense. Statement II : RNAi involves the silencing of a specific mRNA due to a complementary single-stranded RNA molecule that binds and prevents translation of mRNA In the light of the above statements, choose the correct answer from the options given below.
Not quite — the answer is A.
RNA interference occurs in all eukaryotic organisms as a method of cellular defence, so the first statement is true. RNAi silences a specific mRNA by a complementary double-stranded RNA molecule that binds and prevents its translation, so describing it as a single-stranded RNA makes the second statement false.
Q82NEET 2024 Re-exam
Identify the wrong statements : A. Erythropoietin is produced by juxtaglomerular cells of the kidney B. Leydig cells produce Androgens C. Atrial Natriuretic factor, a peptide hormone is secreted by the seminiferous tubules of the testes D. Cholecystokinin is produced by gastrointestinal tract E. Gastrin acts on intestinal wall and helps in the production of pepsinogen Choose the most appropriate answer from the options given below :
Not quite — the answer is C.
Atrial natriuretic factor is secreted by the atrial wall of the heart, not by the seminiferous tubules, and gastrin acts on the gastric glands (stomach) to stimulate HCl and pepsinogen secretion, not on the intestinal wall; these two are wrong. Erythropoietin from juxtaglomerular cells, androgens from Leydig cells and cholecystokinin from the gut are correct.
Q83NEET 2024 Re-exam
Following are the steps involved in the process of PCR. A. Annealing B. Amplification (~1 billion times) C. Denaturation D. Treatment with Taq polymerase and deoxynucleotides E. Extension Choose the correct sequence of steps of PCR from the options given below :
Not quite — the answer is A.
PCR begins with denaturation of the double-stranded DNA, followed by annealing of primers, then Taq polymerase with deoxynucleotides carries out extension; repeating these cycles amplifies the DNA about a billion times. Annealing cannot come first because the strands must be separated before primers can bind.
Q84NEET 2024 Re-exam
Given below are two statements: Statement I: Concentrated urine is formed due to counter current mechanism in nephron. Statement II: Counter current mechanism helps to maintain osmotic gradient in the medullary interstitium. In the light of the above statements, choose the most appropriate answer from the options given below.
Not quite — the answer is C.
The counter-current mechanism of the loop of Henle and vasa recta maintains an increasing osmolarity in the medullary interstitium, and this gradient allows water to be drawn out of the collecting duct to form concentrated urine. Both statements are therefore correct.
Q85NEET 2024 Re-exam
Given below are two statements : Statement I : Concentrically arranged cisternae of Golgi complex are arranged near the nucleus with distinct convex cis or maturing and concave trans or forming face. Statement II : A number of proteins are modified in the cisternae of Golgi complex before they are released from cis face. In the light of the above statements, choose the correct answer from the options given below.
Not quite — the answer is D.
The convex cis face of the Golgi is the forming face and the concave trans face is the maturing face; the statement reverses them, so it is false. Proteins are modified in the cisternae before release from the trans face, not the cis face, so the second statement is also false.
Q86NEET 2024 Re-exam
Match List-I with List-II: List-I – List-II (A) Parturition – (I) Several antibodies for new-born babies (B) Placenta – (II) Collection of ovum after ovulation (C) Colostrum – (III) Foetal ejection reflex (D) Fimbriae – (IV) Secretion of the hormone hCG Choose the correct answer from the options given below:
Not quite — the answer is A.
Parturition is triggered by the foetal ejection reflex, the placenta secretes hCG, colostrum supplies several antibodies (IgA) to the newborn, and the fimbriae collect the ovum after ovulation. Linking colostrum to ovum collection is clearly wrong.
Q87NEET 2024 Re-exam
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Members of subphylum vertebrata possess notochord during the embryonic period. The notochord is replaced by a cartilaginous or bony vertebral column in the adult. Reason R: Thus all chordates are vertebrates not all vertebrates are chordates. In the light of the above statements choose the correct answer from the options given below.
Not quite — the answer is A.
Vertebrates have a notochord in the embryo that is replaced by a cartilaginous or bony vertebral column in the adult, so the assertion is true. The correct conclusion is that all vertebrates are chordates but not all chordates are vertebrates; the reason states the reverse, so it is false.
Q88NEET 2024 Re-exam
The mother has A+ blood group the father has B+ and the child is A+. What can be the possibility of genotypes of all three, respectively? A. IᴬIᴬ | Iᴮi | Iᴮi B. IᴬIᴬ | Iᴮi | Iᴬi C. Iᴮi | IᴬIᴬ | IᴬIᴮ D. IᴬIᴬ | IᴮIᴮ | Iᴬi E. Iᴬi | Iᴮi | Iᴬi Choose the correct answer from the options given below:
Not quite — the answer is D.
The group A child must receive Iᴬ from the mother and i from the father (B group father must be Iᴮi), so valid trios are mother IᴬIᴬ or Iᴬi, father Iᴮi, child Iᴬi; these match the trios IᴬIᴬ | Iᴮi | Iᴬi and Iᴬi | Iᴮi | Iᴬi. A trio with an Iᴮi child or an Iᴮi mother contradicts the stated groups, and an IᴮIᴮ father could only give an AB child, not an Iᴬi child.
Q89NEET 2024 Re-exam
Select the correct statements regarding mechanism of muscle contraction. A. It is initiated by a signal sent by CNS via sensory neuron. B. Neurotransmitter generates action potential in the sarcolemma. C. Increased Ca⁺⁺ level leads to the binding of calcium with troponin on actin filaments. D. Masking of active site for actin is activated. E. Utilising the energy from ATP hydrolysis to form cross bridge. Choose the most appropriate answer from the options given below:
Not quite — the answer is A.
The neurotransmitter released at the neuromuscular junction generates an action potential in the sarcolemma, the rise in Ca⁺⁺ lets calcium bind troponin on actin filaments, and myosin uses energy from ATP hydrolysis to form cross bridges. The signal from the CNS travels via a motor neuron, not a sensory neuron, and calcium causes unmasking, not masking, of the active sites on actin.
Q90NEET 2024 Re-exam
Match List I with List II : List-I – List-II (A) Squamous Epithelium – (I) Goblet cells of alimentary canal (B) Ciliated Epithelium – (II) Inner lining of pancreatic ducts (C) Glandular Epithelium – (III) Walls of blood vessels (D) Compound Epithelium – (IV) Inner surface of Fallopian tubes Choose the correct answer from the options given below :
Not quite — the answer is D.
Squamous epithelium lines the walls of blood vessels, ciliated epithelium lines the inner surface of the Fallopian tubes, goblet cells of the alimentary canal are glandular epithelium, and compound epithelium lines the ducts of salivary glands and pancreatic ducts. Pancreatic ducts are not lined by squamous epithelium.
Q91NEET 2024 Re-exam
Match List I with List II : List-I – List-II (A) B-Lymphocytes – (I) Passive immunity (B) Interferons – (II) Cell mediated immunity (C) T-Lymphocytes – (III) Produce an army of proteins in response to pathogens (D) Colostrum – (IV) Innate immunity Choose the correct answer from the options given below :
Not quite — the answer is C.
B-lymphocytes produce an army of antibodies in response to pathogens, interferons are part of the innate (cytokine) barrier, T-lymphocytes mediate cell-mediated immunity, and colostrum gives the newborn passive immunity through IgA. Calling colostrum innate immunity is the trap; the antibodies are received ready-made.
Q92NEET 2024 Re-exam
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: During the transportation of gases, about 20-25 percent of CO₂ is carried by Haemoglobin as carbamino-haemoglobin. Reason R: This binding is related to high pCO₂ and low pO₂ in tissues. In the light of the above statements, choose the correct answer from the options given below.
Not quite — the answer is C.
About 20–25 per cent of CO₂ is carried by haemoglobin as carbamino-haemoglobin, and this binding depends on partial pressures: high pCO₂ and low pO₂ in the tissues favour CO₂ binding, while the reverse in the alveoli releases it. So both statements are true and the reason explains the assertion.
Chapter-wise split of NEET 2024 Re-exam Biology
- Biotechnology: Principles and Processes6 Q
- Principles of Inheritance and Variation6 Q
- Cell: The Unit of Life5 Q
- Biotechnology and Its Applications5 Q
- Biomolecules5 Q
- Biodiversity and Conservation4 Q
- Anatomy of Flowering Plants4 Q
- Molecular Basis of Inheritance4 Q
- Evolution4 Q
- Structural Organisation in Animals4 Q
- Human Reproduction4 Q
- Morphology of Flowering Plants3 Q
- Plant Growth and Development3 Q
- Cell Cycle and Cell Division3 Q
- Animal Kingdom3 Q
- Human Health and Disease3 Q
- Photosynthesis in Higher Plants2 Q
- Sexual Reproduction in Flowering Plants2 Q
- Plant Kingdom2 Q
- Respiration in Plants2 Q
- Organisms and Populations2 Q
- Locomotion and Movement2 Q
- Body Fluids and Circulation2 Q
- Chemical Coordination and Integration2 Q
- Reproductive Health2 Q
- Breathing and Exchange of Gases2 Q
- Excretory Products and Their Elimination2 Q
- Ecosystem1 Q
- Biological Classification1 Q
- Microbes in Human Welfare1 Q
- Neural Control and Coordination1 Q
About this paper
How many questions from NEET 2024 Re-exam are on this page?
175 questions (Physics 38, Chemistry 45, Biology 92). 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 Re-exam 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 Re-exam useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books. 1 questions here are on topics NCERT has since dropped; they are listed separately so you can skip them.
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