AIPMT 2012 Mains 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
89 questions from the AIPMT 2012 Mains paper, conducted by CBSE. 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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- AIPMT 2012 Mains
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- CBSE
- Physics
- 22 questions
- Chemistry
- 21 questions
- Biology
- 46 questions
AIPMT 2012 Mains Physics questions
22 questions. Tap an option to check your answer.
Q1AIPMT 2012 Mains
The dimensions of (μ₀ε₀)^(−1/2) are
Not quite — the answer is C.
Since c = 1/√(μ₀ε₀), the quantity (μ₀ε₀)^(−1/2) is simply the speed of light and has dimensions [LT⁻¹]. The inverse [L⁻¹T] would be the dimensions of √(μ₀ε₀) itself.
Q2AIPMT 2012 Mains
A stone is dropped from a height h. It hits the ground with a certain momentum P. If the same stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground will change by
Not quite — the answer is B.
Momentum on hitting the ground is P = m√(2gh). Doubling the height multiplies P by √2 ≈ 1.414, an increase of about 41%. Doubling the momentum (100%) would need four times the height.
Q3AIPMT 2012 Mains
A car of mass m is moving on a level circular track of radius R. If μs represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by
Not quite — the answer is D.
Static friction supplies the centripetal force: mv²/R ≤ μs·mg, so vmax = √(μs Rg). The mass cancels, so expressions containing m cannot be right.
Q4AIPMT 2012 Mains
A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude P₀. The instantaneous velocity of this car is proportional to
Not quite — the answer is B.
With constant power, the work done equals the kinetic energy gained: P₀t = ½mv², so v = √(2P₀t/m) and v ∝ t^(1/2). The velocity grows ever more slowly, not linearly or as t².
Q5AIPMT 2012 Mains
Three masses are placed on the x-axis : 300 g at origin, 500g at x = 40 cm and 400g at x = 70 cm. The distance of the centre of mass from the origin is :
Not quite — the answer is A.
x_cm = (300 × 0 + 500 × 40 + 400 × 70)/(300 + 500 + 400) = 48000/1200 = 40 cm. Forgetting the 300 g mass at the origin in the denominator gives a larger, wrong value.
Q6AIPMT 2012 Mains
If ve is escape velocity and v₀ is orbital velocity of a satellite for orbit close to the earth's surface, then these are related by
Not quite — the answer is D.
Near the surface, orbital speed v₀ = √(GM/R) and escape speed ve = √(2GM/R), so ve = √2·v₀. The escape speed is the larger one, so v₀ = √2·ve is reversed.
Q7AIPMT 2012 Mains
A slab of stone of area 0.36 m² and thickness 0.1 m is exposed on the lower surface to steam at 100°C. A block of ice at 0°C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is (Given latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹)
Not quite — the answer is A.
Heat rate = mL/t = (4.8 × 3.36 × 10⁵)/3600 = 448 W. Setting KAΔT/d = 448 gives K × 0.36 × 100/0.1 = 360K = 448, so K ≈ 1.24 J m⁻¹ s⁻¹ °C⁻¹.
Q8AIPMT 2012 Mains
The equation of a simple harmonic wave is given by y = 3 sin (π/2)(50t − x), where x and y are in metres and t is in seconds. The ratio of maximum particle velocity to the wave velocity is
Not quite — the answer is B.
y = 3 sin(25πt − πx/2), so A = 3 m, ω = 25π rad/s, k = π/2 m⁻¹. Wave speed = ω/k = 50 m/s and maximum particle speed = Aω = 75π m/s, giving a ratio of 75π/50 = 3π/2.
Q9AIPMT 2012 Mains
A train moving at a speed of 220 m s⁻¹ towards a stationary object emits a sound of frequency 1000 Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is (speed of sound in air is 330 m s⁻¹)
Not quite — the answer is C.
The object receives f₁ = f·v/(v − u) and reflects it; the moving driver then hears f₂ = f₁(v + u)/v = f(v + u)/(v − u) = 1000 × 550/110 = 5000 Hz. Treating only one Doppler shift gives 3000 Hz, which is wrong.
Q10AIPMT 2012 Mains
A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor is
Not quite — the answer is C.
Energy density between the plates is ½ε₀E² and the field occupies volume Ad, so U = ½ε₀E²Ad. The expression ½ε₀E² alone is energy per unit volume, not total energy.
Q11AIPMT 2012 Mains
Two metallic spheres of radii 1 cm and 3 cm are given charges of −1 × 10⁻² C and 5 × 10⁻² C, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
Not quite — the answer is B.
Total charge = −1 × 10⁻² + 5 × 10⁻² = 4 × 10⁻² C. When connected the spheres reach the same potential, so charge divides in the ratio of radii 1 : 3, giving the bigger sphere ¾ × 4 × 10⁻² = 3 × 10⁻² C.
Q12AIPMT 2012 Mains
A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an α-particle to describe a circle of same radius in the same field?
Not quite — the answer is B.
r = √(2mK)/(qB), so for the same r and B, K ∝ q²/m. For an α-particle q²/m = (2e)²/(4m_p) = e²/m_p, the same as the proton, so it needs the same 1 MeV.
Q13AIPMT 2012 Mains
A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be
Not quite — the answer is B.
Work to turn from 0° to 60° is W = MB(1 − cos 60°) = MB/2 = √3 J, so MB = 2√3 J. The torque to hold it there is τ = MB sin 60° = 2√3 × √3/2 = 3 (in N m).
Q14AIPMT 2012 Mains
The instantaneous values of alternating current and voltage in a circuit are given as i = (1/√2) sin(100πt) ampere e = (1/√2) sin(100πt + π/3) volt The average power in watts consumed in the circuit is
Not quite — the answer is D.
Peak values are 1/√2, so Irms = Vrms = 1/2. The phase difference is π/3, so P = Vrms·Irms·cos φ = ½ × ½ × ½ = 1/8 W. Using peak values instead of rms values gives a larger, wrong answer.
Q15AIPMT 2012 Mains
The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to :
Not quite — the answer is B.
In vacuum E₀ = cB₀, so B₀/E₀ = 1/c, the reciprocal of the speed of light. The speed of light itself is the ratio E₀/B₀, the reverse.
Q16AIPMT 2012 Mains
For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index :
Not quite — the answer is B.
μ = sin((A + δm)/2)/sin(A/2); with δm = A this becomes μ = sin A/sin(A/2) = 2cos(A/2). For a real emergent ray A/2 ranges between 0 and 45°, so μ lies between √2 and 2.
Q17AIPMT 2012 Mains
A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is :
Not quite — the answer is D.
The near end at 20 cm = 2f images at 20 cm. For the far end, u = −30 cm: 1/v = 1/f − 1/u = −1/10 + 1/30, so v = −15 cm. The image extends from 15 cm to 20 cm, a length of 5 cm.
Q18AIPMT 2012 Mains
If the momentum of electron is changed by P, then the de Broglie wavelength associated with it changes by 0.5%. The initial momentum of electron will be
Not quite — the answer is A.
λ = h/p gives |Δλ/λ| = Δp/p for small changes. So 0.005 = P/p₀ and p₀ = P/0.005 = 200P. P/200 inverts the relation.
Q19AIPMT 2012 Mains
Two radiations of photons energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is :
Not quite — the answer is B.
Maximum kinetic energies are 1 − 0.5 = 0.5 eV and 2.5 − 0.5 = 2 eV, a ratio of 1 : 4. Speed goes as √KE, so the speeds are in the ratio 1 : 2; 1 : 4 is the energy ratio, not the speed ratio.
Q20AIPMT 2012 Mains
The transition from the state n = 3 to n = 1 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from :
Not quite — the answer is D.
Infrared needs a much smaller energy gap than the ultraviolet 3 → 1 line. The energy factors (1/n₁² − 1/n₂²) are 3/4 for 2 → 1, 5/36 for 3 → 2, 3/16 for 4 → 2 and only 7/144 for 4 → 3, so 4 → 3 (Paschen-type) is the infrared one; 3 → 2 and 4 → 2 are visible for hydrogen.
Q21AIPMT 2012 Mains
The half life of a radioactive nucleus is 50 days. The time interval (t₂ − t₁) between the time t₂ when 2/3 of it has decayed and the time t₁ when 1/3 of it had decayed is
Not quite — the answer is B.
When 1/3 has decayed, 2/3 remains; when 2/3 has decayed, 1/3 remains. The undecayed amount halves between these instants, which takes exactly one half-life, 50 days.
Q22AIPMT 2012 Mains
The input resistance of a silicon transistor is 100 Ω. Base current is changed by 40 µA which results in a change in collector current by 2 mA. This transistor is used as a common emitter amplifier with a load resistance of 4 kΩ. The voltage gain of the amplifier is
Not quite — the answer is A.
Current gain β = ΔIc/ΔIb = 2 mA/40 µA = 50. Voltage gain = β × RL/Rin = 50 × 4000/100 = 2000.
Chapter-wise split of AIPMT 2012 Mains Physics
- Electromagnetic Waves2 Q
- Work, Energy and Power2 Q
- Waves2 Q
- Electrostatic Potential and Capacitance2 Q
- Ray Optics and Optical Instruments2 Q
- Dual Nature of Radiation and Matter2 Q
- Laws of Motion1 Q
- System of Particles and Rotational Motion1 Q
- Gravitation1 Q
- Thermal Properties of Matter1 Q
- Moving Charges and Magnetism1 Q
- Magnetism and Matter1 Q
- Alternating Current1 Q
- Atoms1 Q
- Nuclei1 Q
- Semiconductor Electronics: Materials, Devices and Simple Circuits1 Q
AIPMT 2012 Mains Chemistry questions
16 questions. Tap an option to check your answer.
Q1AIPMT 2012 Mains
Given that the equilibrium constant for the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature? SO₃(g) ⇌ SO₂(g) + (1/2)O₂(g)
Not quite — the answer is C.
The required reaction is the given one reversed and halved, so K′ = (1/278)^(1/2) = 1/16.7 ≈ 6.0 × 10⁻². Reversing without taking the square root gives 3.6 × 10⁻³.
Q2AIPMT 2012 Mains
Given the reaction between 2 gases represented by A₂ and B₂ to give the compound AB(g): A₂(g) + B₂(g) ⇌ 2AB(g). At equilibrium, the concentration of A₂ = 3.0 × 10⁻³ M, of B₂ = 4.2 × 10⁻³ M and of AB = 2.8 × 10⁻³ M. If the reaction takes place in a sealed vessel at 527°C, then the value of Kc will be
Not quite — the answer is C.
Kc = [AB]²/([A₂][B₂]) = (2.8 × 10⁻³)²/(3.0 × 10⁻³ × 4.2 × 10⁻³) = 7.84 × 10⁻⁶/1.26 × 10⁻⁵ ≈ 0.62. The temperature is not needed, since Kc is computed directly from concentrations.
Q3AIPMT 2012 Mains
During change of O₂ to O₂⁻ ion, the electron adds on which one of the following orbitals?
Not quite — the answer is A.
In O₂ the two π*2p orbitals each hold one electron, and they are the lowest vacant-capacity orbitals, so the extra electron of O₂⁻ enters a π* orbital. The bonding π and σ orbitals are already full.
Q4AIPMT 2012 Mains
Standard reduction potentials of the half reactions are given below: F₂(g) + 2e⁻ → 2F⁻(aq); E° = +2.85 V Cl₂(g) + 2e⁻ → 2Cl⁻(aq); E° = +1.36 V Br₂(l) + 2e⁻ → 2Br⁻(aq); E° = +1.06 V I₂(s) + 2e⁻ → 2I⁻(aq); E° = +0.53 V The strongest oxidising and reducing agents respectively are
Not quite — the answer is A.
The highest reduction potential (F₂, +2.85 V) marks the strongest oxidising agent. The lowest reduction potential (I₂/I⁻, +0.53 V) means I⁻ is most easily oxidised, so I⁻ is the strongest reducing agent; I₂ itself is an oxidant, not a reductant.
Q5AIPMT 2012 Mains
The orbital angular momentum of a p-electron is given as :
Not quite — the answer is A.
Orbital angular momentum = √(l(l + 1))·h/2π. For a p-electron l = 1, giving √2·h/2π = h/(√2 π). √6·h/2π would be for a d-electron (l = 2).
Q6AIPMT 2012 Mains
Molar conductivities (Λ°m) at infinite dilution of NaCl, HCl and CH₃COONa are 126.4, 425.9 and 91.0 S cm² mol⁻¹ respectively. Λ°m for CH₃COOH will be
Not quite — the answer is D.
By Kohlrausch's law, Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl) = 91.0 + 425.9 − 126.4 = 390.5 S cm² mol⁻¹.
Q7AIPMT 2012 Mains
Equal volumes of two monoatomic gases, A and B, at same temperature and pressure are mixed. The ratio of specific heats (Cp/Cv) of the mixture will be
Not quite — the answer is D.
For any monoatomic ideal gas Cv = (3/2)R and Cp = (5/2)R, so a mixture of two monoatomic gases still has Cp/Cv = 5/3 ≈ 1.67. 1.50 is not the value for any simple ideal gas, and values below 1 are impossible.
Q8AIPMT 2012 Mains
Which one of the following does not correctly represent the correct order of the property indicated against it?
Not quite — the answer is C.
Manganese has a much lower melting point (about 1246 °C) than chromium (about 1907 °C) because its stable 3d⁵4s² configuration weakens metallic bonding, so Ti < V < Cr < Mn for melting point is wrong. The other orders are right: Cr has the highest 2nd ionisation enthalpy (removing an electron from stable 3d⁵), and unpaired electrons rise from d¹ to d⁴ across Ti³⁺ to Mn³⁺.
Q9AIPMT 2012 Mains
Four successive members of the first series of the transition metals are listed below. For which one of them the standard potential E°(M²⁺/M) value has a positive sign?
Not quite — the answer is C.
Copper is the only one of these with a positive E°(M²⁺/M), about +0.34 V, because its high atomisation enthalpy and low hydration enthalpy are not compensated. Fe, Co and Ni all have negative values.
Q10AIPMT 2012 Mains
Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them:
Not quite — the answer is D.
Bond orders from MO theory: He₂⁺ = 0.5, O₂⁻ = 1.5, NO = 2.5 and C₂²⁻ = 3 (isoelectronic with N₂). So the increasing order is He₂⁺ < O₂⁻ < NO < C₂²⁻.
Q11AIPMT 2012 Mains
Which of the following exhibits only +3 oxidation state?
Not quite — the answer is C.
Actinium (5f⁰) shows only the +3 oxidation state. Thorium is mainly +4, protactinium up to +5 and uranium shows +3 to +6.
Q12AIPMT 2012 Mains
Consider the following reaction: C₆H₅COCl —(H₂, Pd–BaSO₄)→ 'A' The product 'A' is
Not quite — the answer is A.
This is the Rosenmund reduction: hydrogen over poisoned Pd–BaSO₄ reduces an acid chloride only as far as the aldehyde, so benzoyl chloride gives benzaldehyde, C₆H₅CHO. The poisoned catalyst prevents further reduction to the alcohol.
Q13AIPMT 2012 Mains
An organic compound C₃H₉N (A), when treated with nitrous acid, gave an alcohol and N₂ gas was evolved. (A) on warming with CHCl₃ and caustic potash gave (C) which on reduction gave isopropylmethylamine. Predict the structure of (A).
Not quite — the answer is A.
Evolving N₂ with nitrous acid and giving a carbylamine (isocyanide) with CHCl₃/KOH both show A is a primary amine. Reducing (CH₃)₂CH–NC gives (CH₃)₂CH–NH–CH₃ (isopropylmethylamine), so A is isopropylamine, (CH₃)₂CH–NH₂. n-Propylamine would give n-propylmethylamine instead.
Q14AIPMT 2012 Mains
Which of the following reagents will be able to distinguish between 1-butyne and 2-butyne?
Not quite — the answer is A.
1-Butyne is a terminal alkyne with an acidic ≡C–H and reacts with NaNH₂ to form a sodium acetylide; 2-butyne has no such hydrogen. Br₂ and HCl add to both alkynes, so they cannot tell them apart.
Q15AIPMT 2012 Mains
Consider the reaction: RCHO + NH₂NH₂ → RCH=N–NH₂. What sort of reaction is it?
Not quite — the answer is D.
The nucleophilic nitrogen of hydrazine adds to the electrophilic carbonyl carbon, and the intermediate then loses water to form the hydrazone C=N bond. This is nucleophilic addition followed by elimination, not an electrophilic or free-radical process.
Q16AIPMT 2012 Mains
Which of the following compounds can be used as antifreeze in automobile radiators ?
Not quite — the answer is B.
Ethylene glycol (ethane-1,2-diol) is the standard radiator antifreeze: it lowers the freezing point of water and, unlike methyl or ethyl alcohol, has a high boiling point so it does not evaporate away.
Chemistry: not in the current NEET syllabus
These 5 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
5 questions. Tap an option to check your answer.
Q1AIPMT 2012 Mains · Q32
Structure of a mixed oxide is cubic close-packed (c.c.p.). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is
Not quite — the answer is D.
A ccp unit cell has 4 oxide ions, 8 tetrahedral voids and 4 octahedral voids. A occupies 1/4 of the tetrahedral voids (2 per cell) and B all octahedral voids (4 per cell), giving A₂B₄O₄ = AB₂O₂; the charges balance (2 × 2 + 4 × 1 = 8 = 4 × 2).
Q2AIPMT 2012 Mains · Q37
A certain gas takes three times as long to effuse out as helium. Its molecular mass will be :
Not quite — the answer is B.
By Graham's law, effusion time ∝ √M, so t/t_He = √(M/4) = 3, giving M = 36 u. Assuming time is proportional to M itself would give 12 u, and squaring wrongly gives 9 u.
Q3AIPMT 2012 Mains · Q52
The Gibbs energy for the decomposition of Al₂O₃ at 500°C is as follows: (2/3)Al₂O₃ → (4/3)Al + O₂; ΔrG = +960 kJ mol⁻¹ The potential difference needed for the electrolytic reduction of aluminium oxide (Al₂O₃) at 500°C is at least
Not quite — the answer is C.
Making 4/3 mol Al from Al³⁺ needs (4/3) × 3 = 4 mol electrons per mol O₂. E = ΔG/(nF) = 960000/(4 × 96500) ≈ 2.5 V. Taking n = 3 instead would give about 3.3 V.
Q4AIPMT 2012 Mains · Q53
Chloramphenicol is an
Not quite — the answer is D.
Chloramphenicol is a broad-spectrum antibiotic, effective against a wide range of Gram-positive and Gram-negative bacteria (used, for example, in typhoid). It is not an antiseptic applied to skin or an antihistamine.
Q5AIPMT 2012 Mains · Q55
Which one of the following sets forms the biodegradable polymer?
Not quite — the answer is B.
Glycine (H₂N–CH₂–COOH) and 6-aminohexanoic acid (H₂N–(CH₂)₅–COOH) condense to form nylon-2-nylon-6, a biodegradable polyamide. Ethylene glycol with terephthalic acid gives non-biodegradable polyester (Dacron/PET), and the other pairs give synthetic rubbers.
Chapter-wise split of AIPMT 2012 Mains Chemistry
AIPMT 2012 Mains Biology questions
39 questions. Tap an option to check your answer.
Q1AIPMT 2012 Mains
How many organisms in the list given below are autotrophs ? Lactobacillus, Nostoc, Chara, Nitrosomonas, Nitrobacter, Streptomyces, Sacharomyces, Trypanosoma, Porphyra, Wolfia
Not quite — the answer is C.
Autotrophs make their own food: Nostoc, Chara, Porphyra and Wolffia are photosynthetic, while Nitrosomonas and Nitrobacter are chemoautotrophs, giving six. Lactobacillus, Streptomyces, Saccharomyces and Trypanosoma are heterotrophs, so counting only the photosynthetic ones (four) is the trap.
Q2AIPMT 2012 Mains
For its activity, carboxypeptidase requires
Not quite — the answer is A.
Carboxypeptidase is a metalloenzyme that needs zinc as its cofactor. Niacin is a vitamin forming the coenzymes NAD/NADP, not the cofactor of this enzyme.
Q3AIPMT 2012 Mains
Which one of the following structures is an organelle within an organelle ?
Not quite — the answer is A.
Mitochondria and chloroplasts contain their own 70S ribosomes, so ribosomes occur as an organelle within an organelle. Peroxisomes and ER lie free in the cytoplasm, and mesosomes are infoldings of the bacterial plasma membrane.
Q4AIPMT 2012 Mains
Which one of the following is a wrong statement regarding mutations ?
Not quite — the answer is D.
A change in a single base pair is itself a mutation (a point mutation, as in sickle-cell anaemia), so that statement is wrong. Insertions and deletions do shift the reading frame, cancer cells commonly show chromosomal aberrations, and UV and gamma rays are mutagens.
Q5AIPMT 2012 Mains
A test cross is carried out to
Not quite — the answer is A.
A test cross crosses a dominant-phenotype plant (such as one from the F₂) with the homozygous recessive parent; the offspring ratio reveals whether that plant is homozygous or heterozygous. It is not a tool for testing linkage or counting alleles.
Q6AIPMT 2012 Mains
How many plants in the list given below have marginal placentation? Mustard, Gram, Tulip, Asparagus, Arhar, Sun hemp, Chilli, Colchicum, Onion, Moong, Pea, Tobacco, Lupin
Not quite — the answer is C.
Marginal placentation is typical of the monocarpellary ovary of Fabaceae: gram, arhar, sun hemp, moong, pea and lupin, which makes six. Mustard has parietal placentation and tulip, asparagus, colchicum, onion, chilli and tobacco have axile placentation.
Q7AIPMT 2012 Mains
Which one of the following generally acts as an antagonist to gibberellins ?
Not quite — the answer is C.
ABA promotes dormancy and inhibits growth, so it generally opposes gibberellins, which break dormancy and promote germination and elongation. Zeatin (a cytokinin) and IAA (an auxin) are growth promoters.
Q8AIPMT 2012 Mains
As compared to a dicot root, a monocot root has :
Not quite — the answer is B.
Monocot roots are polyarch, with many xylem bundles, whereas dicot roots usually have two to six. Monocot roots show no secondary growth, so they have no secondary xylem, annual rings or periderm to compare.
Q9AIPMT 2012 Mains
What is the function of germ pore ?
Not quite — the answer is C.
Germ pores are thin areas of the exine where sporopollenin is absent, and the pollen tube grows out through them. The male gametes leave only later, from the tip of the tube inside the synergid, and the radicle comes out of the seed, not the pollen.
Q10AIPMT 2012 Mains
Which one of the following statements is wrong ?
Not quite — the answer is A.
Pollen shed at the two-celled stage still achieves double fertilisation, because the generative cell divides into two male gametes inside the pollen tube. The other statements are true: the vegetative cell is larger, some pollen (e.g. of Solanaceae) stays viable for months, and the intine is cellulose and pectin.
Q11AIPMT 2012 Mains
Plants with ovaries having only one or a few ovules are generally pollinated by
Not quite — the answer is D.
Wind-pollinated flowers usually have a single ovule in each ovary and many flowers in an inflorescence. Insect- and bird-pollinated flowers often have many ovules, since one visit can deliver many pollen grains.
Q12AIPMT 2012 Mains
Sacred groves are specially useful in
Not quite — the answer is D.
Sacred groves are forest patches protected by religious tradition, and they act as refuges for many rare and threatened species, a form of in situ conservation. Benefits such as soil protection are side effects, not their special value.
Q13AIPMT 2012 Mains
The rate of formation of new organic matter by rabbit in a grassland is called
Not quite — the answer is B.
Secondary productivity is the rate at which consumers such as a rabbit form new organic matter. Gross and net primary productivity refer only to producers (green plants).
Q14AIPMT 2012 Mains
Cuscuta is an example of
Not quite — the answer is A.
Cuscuta (dodder) grows on the outer surface of its host plant and draws nutrients through haustoria, which makes it an ectoparasite. Brood parasitism is egg-laying in another bird's nest, as the cuckoo does.
Q15AIPMT 2012 Mains
In gobar gas, the maximum amount is that of
Not quite — the answer is B.
Biogas (gobar gas) produced by methanogens in cattle dung is mostly methane, about 50–70%. Carbon dioxide is the second main component but present in a smaller amount.
Q16AIPMT 2012 Mains
Read the following four statements (A–D): (A) Colostrum is recommended for the new born because it is rich in antigens. (B) Chikungunya is caused by a Gram negative bacterium. (C) Tissue culture has proved useful in obtaining virus-free plants. (D) Beer is manufactured by distillation of fermented grape juice. How many of the above statements are wrong?
Not quite — the answer is B.
Three are wrong. Colostrum is valuable because it is rich in antibodies (IgA), not antigens; chikungunya is caused by a virus; and beer is brewed from malted barley without distillation. Only the statement about tissue (meristem) culture giving virus-free plants is right.
Q17AIPMT 2012 Mains
Tobacco plants resistant to a nematode have been developed by the introduction of DNA that produced (in the host cells)
Not quite — the answer is A.
Nematode-resistant tobacco uses RNA interference. The introduced DNA makes both sense and antisense RNA, which pair into double-stranded RNA that silences the nematode's mRNA, so no toxin or hormone is involved.
Q18AIPMT 2012 Mains
Biolistics (gene-gun) is suitable for
Not quite — the answer is B.
In biolistics, gold or tungsten microparticles coated with DNA are shot into cells, a method suited to transforming plant cells. Disarming pathogens and joining DNA to vectors are done with enzymes, not a gene gun.
Q19AIPMT 2012 Mains
In genetic engineering, the antibiotics are used
Not quite — the answer is A.
Antibiotic-resistance genes on a vector act as selectable markers: only transformed cells survive on medium containing the antibiotic. The replication start site is the ori, not an antibiotic sequence.
Q20AIPMT 2012 Mains
Which one of the following pairs of animals are similar to each other pertaining to the feature stated against them?
Not quite — the answer is D.
The sea horse and the flying fish are both bony fishes and therefore cold-blooded (poikilothermous). Ornithorhynchus lays eggs, unlike the viviparous bat Pteropus; the crocodile has a four-chambered heart; and Ascaris and Ancylostoma are unsegmented roundworms.
Q21AIPMT 2012 Mains
Which one of the following categories of animals is correctly described with no single exception in it?
Not quite — the answer is B.
Every bony fish has four pairs of gills covered by an operculum on each side. The other groups have exceptions: crocodiles have a four-chambered heart, some sponges (Spongilla) live in freshwater, and monotremes lay eggs.
Q22AIPMT 2012 Mains
Which one of the following organisms is scientifically correctly named, correctly printed according to the International Rules of Nomenclature and correctly described?
Not quite — the answer is B.
Plasmodium falciparum is correctly named and causes malignant malaria, the most serious kind. Musca domestica is the housefly, the tiger is Panthera tigris and Gir protects lions, and E. coli stands for Escherichia coli, not Entamoeba.
Q23AIPMT 2012 Mains
Which one of the following cellular parts is correctly described ?
Not quite — the answer is A.
Thylakoids are flattened membranous sacs stacked into grana inside chloroplasts. Chloroplast ribosomes are the smaller 70S type and cytoplasmic ones are 80S, lysosomes work best at an acidic pH of about 5, and centrioles organise spindles rather than making RNA.
Q24AIPMT 2012 Mains
Identify the meiotic stage in which the homologous chromosomes separate while the sister chromatids remain associated at their centromeres :
Not quite — the answer is C.
In anaphase I the homologous chromosomes move apart while each still has two sister chromatids joined at the centromere. The sister chromatids separate only in anaphase II.
Q25AIPMT 2012 Mains
Which one of the following biomolecules is correctly characterised ?
Not quite — the answer is A.
Lecithin is a phospholipid, a glyceride carrying a phosphate group, and it is a major part of cell membranes. Palmitic acid is a saturated 16-carbon fatty acid, and adenylic acid contains ribose, not glucose.
Q26AIPMT 2012 Mains
The idea of mutations was brought forth by
Not quite — the answer is A.
Hugo de Vries proposed the mutation theory after his work on the evening primrose (Oenothera lamarckiana), describing sudden large changes that he called mutations. Darwin stressed small continuous variations, and Mendel described factors, not mutations.
Q27AIPMT 2012 Mains
What is it that forms the basis of DNA fingerprinting?
Not quite — the answer is D.
DNA fingerprinting compares repetitive DNA, satellite DNA such as VNTRs whose copy number differs between people. The purine to pyrimidine ratio is the same in everyone, and the technique has nothing to do with skin ridges.
Q28AIPMT 2012 Mains
Which one of the following options gives the correct categorisation of six animals according to the type of nitrogenous wastes (A – Ammonotelic, B – Ureotelic, C – Uricotelic) they give out?
Not quite — the answer is C.
Aquatic amphibians excrete ammonia; frogs and humans are ureotelic; and pigeons, lizards and cockroaches conserve water by excreting uric acid. The adult frog is ureotelic, which rules out the options that call it ammonotelic or uricotelic.
Q29AIPMT 2012 Mains
Which one of the following pairs of chemical substances is correctly categorised?
Not quite — the answer is C.
Troponin and myosin are both proteins of the contractile machinery of striated muscle. Thymosin comes from the thymus, not the thyroid; prolactin is a pituitary hormone; and rhodopsin is a visual pigment, not a hormone.
Q30AIPMT 2012 Mains
The supportive skeletal structures in the human external ears and in the nose tip are examples of
Not quite — the answer is D.
The external ear (pinna) and the tip of the nose are supported by flexible cartilage. Bone would be rigid, and ligaments join bone to bone rather than forming such frameworks.
Q31AIPMT 2012 Mains
A fall in glomerular filtration rate (GFR) activates
Not quite — the answer is A.
A fall in GFR is sensed by the juxtaglomerular apparatus, whose JG cells release renin. Renin begins the renin–angiotensin pathway, which only later causes aldosterone release from the adrenal cortex.
Q32AIPMT 2012 Mains
Which one of the following characteristics is common both in humans and adult frogs ?
Not quite — the answer is D.
Both humans and adult frogs excrete urea (ureotelic). Frogs have a three-chambered heart, nucleated RBCs and external fertilisation, whereas humans have a four-chambered heart and enucleate RBCs.
Q33AIPMT 2012 Mains
Which one of the following human organs is often called the "graveyard" of RBCs?
Not quite — the answer is C.
The spleen filters the blood and destroys old, worn-out RBCs, earning the name 'graveyard of RBCs'. The liver breaks down haemoglobin products but is not the organ called the graveyard.
Q34AIPMT 2012 Mains
The secretory phase in the human menstrual cycle is also called
Not quite — the answer is C.
After ovulation the corpus luteum secretes progesterone, so the secretory phase is the luteal phase, lasting from about day 15 to day 28 (roughly 13–14 days). The follicular phase is the proliferative phase before ovulation.
Q35AIPMT 2012 Mains
Select the correct statement about biodiversity :
Not quite — the answer is C.
The Western Ghats are a biodiversity hotspot with very high species richness and many endemic species, especially amphibians and plants. Deserts have low diversity, and biodiversity conservation is a global need, not a fad.
Q36AIPMT 2012 Mains
Which one of the following sets of items in the options given below are correctly categorised with one exception in it?
Not quite — the answer is C.
Plasmodium and Trypanosoma are protozoan parasites, while Cuscuta is a parasitic flowering plant, so it is the true exception. UAG is itself a stop codon, the wombat is a marsupial, and diphtheria is a bacterial disease, so the 'exceptions' in the other sets are not exceptions.
Q37AIPMT 2012 Mains
Which one of the following statements is correct with respect to immunity ?
Not quite — the answer is A.
Snake antivenom supplies preformed antibodies (passive immunisation), which is needed for a quick response to viper venom. Antibodies are made by B-lymphocytes and have two light and two heavy chains, and graft rejection is mediated by T-lymphocytes.
Q38AIPMT 2012 Mains
The first clinical gene therapy was given for treating
Not quite — the answer is D.
The first clinical gene therapy (1990) was given to a four-year-old girl with adenosine deaminase (ADA) deficiency, putting a functional ADA cDNA into her lymphocytes. Diabetes is treated with recombinant insulin, which is not gene therapy.
Q39AIPMT 2012 Mains
Which one of the following represents a palindromic sequence in DNA ?
Not quite — the answer is A.
5′–GAATTC–3′ reads the same 5′→3′ on both strands (the EcoRI site), which makes it a palindrome. In the other options the second strand does not read the same as the first in the 5′→3′ direction.
Biology: not in the current NEET syllabus
These 7 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
7 questions. Tap an option to check your answer.
Q1AIPMT 2012 Mains
Through their effect on plant growth regulators, what do the temperature and light control in the plants?
Not quite — the answer is B.
Light (photoperiodism) and low temperature (vernalisation) both act through growth regulators to decide when a plant flowers. Stomatal closure is driven by ABA in response to water stress, not by photoperiod and temperature.
Q2AIPMT 2012 Mains
Vernalisation stimulates flowering in
Not quite — the answer is C.
Biennials such as carrot need a cold period (vernalisation) before they can flower. Zamikand, turmeric and ginger are propagated vegetatively by underground stems and are not classic vernalisation examples.
Q3AIPMT 2012 Mains
The second stage of hydrosere is occupied by plants like
Not quite — the answer is D.
A hydrosere passes from phytoplankton to submerged plants (Vallisneria, Hydrilla), then floating plants (Azolla), reed swamp (Typha), sedge meadow and finally woodland (Salix). The second stage is therefore the submerged stage, not the floating stage of Azolla.
Q4AIPMT 2012 Mains · Q76
For its action, nitrogenase requires :
Not quite — the answer is A.
Nitrogenase needs a large energy input, about 16 ATP for each N₂ reduced to ammonia, and it works only in the absence of oxygen. It does not need light, and oxygen radicals inactivate it.
Q5AIPMT 2012 Mains · Q85
Green revolution in India occurred during
Not quite — the answer is A.
India's Green Revolution came in the 1960s with high-yielding semi-dwarf wheat and rice varieties. By the 1970s food production had already risen as a result.
Q6AIPMT 2012 Mains · Q102
Where do certain symbiotic microorganisms normally occur in human body?
Not quite — the answer is A.
The caecum, a blind sac at the start of the large intestine, hosts certain symbiotic micro-organisms. The vermiform appendix is a vestigial outgrowth of the caecum and the rectum only stores faeces.
Q7AIPMT 2012 Mains · Q115
Consider the following four statements (a–d) and select the option which includes all the correct ones only. (a) Single cell Spirulina can produce large quantities of food rich in protein, minerals, vitamins etc. (b) Body weight-wise the microorganism Methylophilus methylotrophus may be able to produce several times more proteins than the cows per day (c) Common button mushrooms are a very rich source of vitamin C (d) A rice variety has been developed which is very rich in calcium
Not quite — the answer is D.
Spirulina is a protein-, mineral- and vitamin-rich single-cell food, and 250 kg of Methylophilus methylotrophus can make about 25 tonnes of protein a day, far more than a cow. Button mushrooms are not a rich vitamin C source, and the biofortified rice variety is rich in iron, not calcium.
Chapter-wise split of AIPMT 2012 Mains Biology
- Sexual Reproduction in Flowering Plants3 Q
- Biotechnology: Principles and Processes3 Q
- Excretory Products and Their Elimination3 Q
- Biological Classification2 Q
- Biomolecules2 Q
- Cell: The Unit of Life2 Q
- Principles of Inheritance and Variation2 Q
- Biodiversity and Conservation2 Q
- Human Health and Disease2 Q
- Biotechnology and Its Applications2 Q
- Animal Kingdom2 Q
- Morphology of Flowering Plants1 Q
- Plant Growth and Development1 Q
- Anatomy of Flowering Plants1 Q
- Ecosystem1 Q
- Organisms and Populations1 Q
- Microbes in Human Welfare1 Q
- The Living World1 Q
- Cell Cycle and Cell Division1 Q
- Evolution1 Q
- Molecular Basis of Inheritance1 Q
- Locomotion and Movement1 Q
- Structural Organisation in Animals1 Q
- Body Fluids and Circulation1 Q
- Human Reproduction1 Q
About this paper
How many questions from AIPMT 2012 Mains are on this page?
89 questions (Physics 22, Chemistry 21, Biology 46). 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 AIPMT 2012 Mains paper, conducted by CBSE. The answer to every question was checked by the RankUp team, and the explanations are RankUp's own.
Is AIPMT 2012 Mains useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books. 12 questions here are on topics NCERT has since dropped; they are listed separately so you can skip them.
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