AIPMT 2010 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
93 questions from the AIPMT 2010 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 2010 Mains
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- CBSE
- Physics
- 24 questions
- Chemistry
- 23 questions
- Biology
- 46 questions
AIPMT 2010 Mains Physics questions
23 questions. Tap an option to check your answer.
Q1AIPMT 2010 Mains · Q61
A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the other in x-z plane. If the current in the loop is i. The resultant magnetic field due to the two semicircular parts at their common centre is
Not quite — the answer is A.
Each semicircle gives μ₀i/(4R) at the centre, half the field of a full loop. Because the two semicircles lie in perpendicular planes, their fields are perpendicular, so the resultant is √2 × μ₀i/(4R) = μ₀i/(2√2 R). Simply adding the two fields gives μ₀i/(2R), which wrongly treats them as parallel.
Q2AIPMT 2010 Mains · Q63
Two parallel metal plates having charges +Q and –Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will
Not quite — the answer is C.
With fixed charges ±Q the field between the plates is σ/(Kε₀) in a dielectric instead of σ/ε₀ in air. Kerosene has K > 1, so the field decreases. The charge on isolated plates does not change, so the field cannot remain the same.
Q3AIPMT 2010 Mains · Q64
The electric field at a distance 3R/2 from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance R/2 from the centre of the sphere is
Not quite — the answer is A.
All the charge of a conducting shell sits on its surface, and by Gauss's law the enclosed charge for any point inside (r = R/2) is zero, so the field there is zero. The value E at 3R/2 only describes points outside the shell and does not scale down inside.
Q4AIPMT 2010 Mains · Q65
A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are e1 and e2 respectively, the percentage error in the estimation of g is
Not quite — the answer is B.
For free fall from rest h = ½gt², so g = 2h/t². Maximum relative errors add with their powers: Δg/g = Δh/h + 2Δt/t, giving e₁ + 2e₂. Subtracting errors (e₁ − 2e₂) is wrong because maximum errors always add.
Q5AIPMT 2010 Mains · Q66
When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectron and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 2I, the number of emitted electrons and their maximum kinetic energy are respectively
Not quite — the answer is B.
Doubling intensity doubles the number of photons per second, so the number of photoelectrons doubles to 2N. The maximum kinetic energy depends only on frequency and work function (hν − φ), so it stays T. Doubling the kinetic energy would need a change in frequency, not intensity.
Q6AIPMT 2010 Mains · Q67
The electric field of an electromagnetic wave in free space is given by E⃗ = 10 cos(10⁷t + kx) ĵ V/m, where t and x are in seconds and metres respectively. It can be inferred that (a) The wavelength λ is 188.4 m (b) The wave number k is 0.33 rad/m (c) The wave amplitude is 10 V/m (d) The wave is propagating along +x direction Which one of the following pairs of statements is correct?
Not quite — the answer is D.
Here ω = 10⁷ rad/s, so k = ω/c = 10⁷/(3 × 10⁸) = 0.033 rad/m and λ = 2π/k ≈ 188.4 m; the amplitude is 10 V/m. So the wavelength and amplitude statements are correct. The wave number is 0.033, not 0.33 rad/m, and the phase (ωt + kx) means the wave travels along −x, not +x.
Q7AIPMT 2010 Mains · Q68
The speed of light in media M₁ and M₂ is 1.5 × 10⁸ m/s and 2.0 × 10⁸ m/s respectively. A ray of light enters from medium M₁ to M₂ at an incidence angle i. If the ray suffers total internal reflection, the value of i is
Not quite — the answer is C.
Refractive indices are μ₁ = c/v₁ = 2 and μ₂ = c/v₂ = 1.5. The critical angle satisfies sin C = μ₂/μ₁ = 3/4, so total internal reflection needs i equal to or greater than sin⁻¹(3/4). The value sin⁻¹(2/3) is obtained by wrongly using 1/μ₁ ratios; TIR never occurs for angles smaller than the critical angle.
Q8AIPMT 2010 Mains · Q69
A ray of light is incident on a 60º prism at the minimum deviation position. The angle of refraction at the first face (i.e., incident face) of the prism is
Not quite — the answer is B.
At minimum deviation the ray passes symmetrically, so r₁ = r₂ and r₁ + r₂ = A. Hence r₁ = A/2 = 60°/2 = 30°. Taking 60° confuses the refraction angle with the prism angle itself.
Q9AIPMT 2010 Mains · Q70
For transistor action (a) Base, emitter and collector regions should have similar size and doping concentrations. (b) The base region must be very thin and lightly doped. (c) The emitter-base junction is forward biased and base-collector junction is reverse biased. (d) Both the emitter-base junction as well as the base collector junction are forward biased. Which one of the following pairs of statements is correct?
Not quite — the answer is C.
A transistor works because the base is very thin and lightly doped, and in active mode the emitter-base junction is forward biased while the collector-base junction is reverse biased. The regions are deliberately unequal in size and doping, and forward-biasing both junctions puts the transistor in saturation, not normal transistor action.
Q10AIPMT 2010 Mains · Q72
The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is
Not quite — the answer is A.
At the highest point only the horizontal component u cos θ remains. Setting u cos θ = u/2 gives cos θ = 1/2, so θ = 60°. An angle of 30° would leave a speed of (√3/2)u at the top.
Q11AIPMT 2010 Mains · Q73
From a circular disc of radius R and mass 9M, a small disc of mass M and radius R/3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is
Not quite — the answer is A.
Full disc: I₁ = (9M)R²/2. Removed disc: I₂ = M(R/3)²/2 = MR²/18. Remaining I = 81MR²/18 − MR²/18 = 80MR²/18 = (40/9)MR². Answers like 4MR² come from subtracting MR²/2 instead of using the small disc's own radius R/3.
Q12AIPMT 2010 Mains · Q74
A particle moves in x-y plane according to rule x = a sin ωt and y = a cosωt. The particle follows
Not quite — the answer is B.
Squaring and adding x = a sin ωt and y = a cos ωt gives x² + y² = a²(sin²ωt + cos²ωt) = a², the equation of a circle of radius a. An ellipse would need unequal amplitudes, and a straight line would need the two motions to be in phase.
Q13AIPMT 2010 Mains · Q75
A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10⁻⁴ m² carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 × 10⁻² tesla making an angle of 30° with the axis of the solenoid. The torque on the solenoid will be
Not quite — the answer is C.
Magnetic moment m = NIA = 2000 × 2.0 × 1.5 × 10⁻⁴ = 0.6 A m². Torque τ = mB sin 30° = 0.6 × 5 × 10⁻² × 0.5 = 1.5 × 10⁻² N m. Using cos 30° or dropping the factor N leads to the other values.
Q14AIPMT 2010 Mains · Q76
The decay constant of a radio isotope is λ. If A₁ and A₂ are its activities at times t₁ and t₂ respectively, the number of nuclei which have decayed during the time (t₁ − t₂) is
Not quite — the answer is C.
Activity A = λN, so N₁ = A₁/λ and N₂ = A₂/λ. The number decayed between the two instants is N₁ − N₂ = (A₁ − A₂)/λ. The difference A₁ − A₂ alone is a difference of decay rates, not a number of nuclei.
Q15AIPMT 2010 Mains · Q78
The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is
Not quite — the answer is A.
Binding energy of two deuterons = 2 × (2 × 1.1) = 4.4 MeV; of helium-4 = 4 × 7.0 = 28.0 MeV. Energy released = 28.0 − 4.4 = 23.6 MeV. 28.0 MeV is just the helium binding energy without subtracting that of the reactants.
Q16AIPMT 2010 Mains · Q79
The electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping potential is estimated to be (the energy of the electron in nth state En = −13.6/n² eV)
Not quite — the answer is D.
Photon energy for n = 3 → 1 is 13.6(1 − 1/9) = 12.09 eV. Maximum kinetic energy = 12.09 − 5.1 ≈ 7.0 eV, so the stopping potential is about 7 V. 12.1 V forgets to subtract the work function.
Q17AIPMT 2010 Mains · Q80
If cp and cv denote the specific heats (per unit mass) of an ideal gas of molecular weight M, then (where R is the molar gas constant)
Not quite — the answer is C.
For one mole, Cp − Cv = R (Mayer's relation). Specific heats per unit mass are molar heats divided by M, so cp − cv = R/M. 'cp − cv = R' applies to molar specific heats, not per unit mass.
Q18AIPMT 2010 Mains · Q81
A condenser of capacity C is charged to a potential difference of V₁. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V₂ is
Not quite — the answer is D.
In the LC circuit energy is conserved: ½CV₁² = ½CV₂² + ½LI². So I = [C(V₁² − V₂²)/L]^(1/2). Using (V₁ − V₂)² instead of V₁² − V₂² treats the voltage drop wrongly, and omitting the square root gives I² rather than I.
Q19AIPMT 2010 Mains · Q83
A solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from the same height at the same time on a inclined plane. Both roll down without slipping. Which one will reach the bottom first?
Not quite — the answer is D.
For rolling without slipping a = g sin θ/(1 + k²/R²). The solid cylinder has k²/R² = 1/2 and the hollow one 1, so the solid cylinder has the larger acceleration and reaches first at any angle. Masses being equal does not make them arrive together because the mass distribution differs.
Q20AIPMT 2010 Mains · Q85
(a) Centre of gravity (C.G.) of a body is the point at which the weight of the body acts (b) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius (c) To evaluate the gravitational field intensity due to any body at an external point, the entire mass of the body can be considered to be concentrated at its C.G. (d) The radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the C.G. of the body to the axis Which one of the following pairs of statements is correct?
Not quite — the answer is B.
The weight of a body acts at its centre of gravity, and if g is uniform over the body (earth of infinite radius) the centre of gravity coincides with the centre of mass, so the first two statements are correct. The gravitational field of a body at an external point can be found by concentrating its mass at one point only for spherically symmetric bodies, and the radius of gyration is defined by I = Mk², not as the distance from the C.G. to the axis.
Q21AIPMT 2010 Mains · Q86
The magnetic moment of a diamagnetic atom is
Not quite — the answer is D.
In a diamagnetic atom the orbital and spin magnetic moments of the electrons cancel completely, so the net (permanent) magnetic moment is zero. A moment between zero and one or greater than one would describe paramagnetic or ferromagnetic atoms.
Q22AIPMT 2010 Mains · Q89
A thin circular ring of mass M and radius r is rotating about its axis with constant angular velocity ω. Two objects each of mass m are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by
Not quite — the answer is D.
No external torque acts, so angular momentum is conserved: Mr²ω = (M + 2m)r²ω′, giving ω′ = Mω/(M + 2m). Adding mass increases the moment of inertia, so the angular velocity must fall; expressions bigger than ω such as (M + 2m)ω/M are wrong.
Q23AIPMT 2010 Mains · Q90
A monoatomic gas at pressure P₁ and volume V₁ is compressed adiabatically to 1/8th of its original volume. What is the final pressure of the gas?
Not quite — the answer is D.
For an adiabatic process PV^γ is constant with γ = 5/3 for a monoatomic gas. P₂ = P₁(V₁/V₂)^γ = P₁ × 8^(5/3) = P₁ × 2⁵ = 32P₁. 64P₁ would correspond to γ = 2, and 16P₁ to γ = 4/3.
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.
Q1AIPMT 2010 Mains · Q84
The thermo e.m.f. E in volts of a certain thermo-couple is found to vary with temperature difference θ in °C between the two junctions according to the relation E = 30θ − θ²/15. The neutral temperature for the thermo-couple will be
Not quite — the answer is C.
Neutral temperature is where E is maximum: dE/dθ = 30 − 2θ/15 = 0, so θ = 225°C. 450°C is the temperature of inversion (twice the neutral temperature), where E becomes zero again.
Chapter-wise split of AIPMT 2010 Mains Physics
- System of Particles and Rotational Motion4 Q
- Dual Nature of Radiation and Matter2 Q
- Ray Optics and Optical Instruments2 Q
- Motion in a Plane2 Q
- Magnetism and Matter2 Q
- Nuclei2 Q
- Thermodynamics2 Q
- Moving Charges and Magnetism1 Q
- Electrostatic Potential and Capacitance1 Q
- Electric Charges and Fields1 Q
- Units and Measurements1 Q
- Electromagnetic Waves1 Q
- Semiconductor Electronics: Materials, Devices and Simple Circuits1 Q
- Alternating Current1 Q
AIPMT 2010 Mains Chemistry questions
21 questions. Tap an option to check your answer.
Q1AIPMT 2010 Mains · Q91
Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is
Not quite — the answer is B.
Across period 3 radius decreases (Mg > P > Cl), and down group 2 it increases (Ca > Mg). So the increasing order is Cl < P < Mg < Ca. Orders starting with Ca or Mg reverse the periodic trend.
Q2AIPMT 2010 Mains · Q92
The reaction 2A(g) + B(g) ⇌ 3C(g) + D(g) is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression
Not quite — the answer is B.
Forming 0.25 M of D consumes 0.25 M B and 0.50 M A and forms 0.75 M C. At equilibrium [A] = 0.50, [B] = 0.75, [C] = 0.75, [D] = 0.25, so K = (0.75)³(0.25)/[(0.50)²(0.75)]. Using the initial 1.00 M values in the denominator ignores the amounts consumed.
Q3AIPMT 2010 Mains · Q96
Among the following four compounds a. Phenol b. Methyl phenol c. Metanitrophenol d. Paranitrophenol The acidity order is
Not quite — the answer is A.
Electron-withdrawing –NO₂ increases acidity of phenol most from the para position (−R and −I), less from meta (−I only), while –CH₃ donates electrons and lowers acidity. So para-nitrophenol > meta-nitrophenol > phenol > methylphenol. Putting meta above para ignores the resonance effect.
Q4AIPMT 2010 Mains · Q97
Among the following which one has the highest cation to anion size ratio?
Not quite — the answer is B.
The ratio is largest when the cation is largest and the anion smallest: Cs⁺ is the biggest cation and F⁻ the smallest anion, so CsF (about 167/133 ≈ 1.26). CsI has a big cation but a much bigger anion, and LiF has the smallest cation.
Q5AIPMT 2010 Mains · Q98
Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be
Not quite — the answer is D.
Work in expansion is w = −p(ext)ΔV, and for free expansion into vacuum p(ext) = 0, so no work is done whatever the number of moles. Values like 3 J or 9 J wrongly link the work to the amount of gas.
Q6AIPMT 2010 Mains · Q99
Which of the following species is not electrophilic in nature?
Not quite — the answer is C.
Electrophiles are electron-deficient: Cl⁺ and NO₂⁺ are positive and electron-poor, and BH₃ has an empty orbital on boron. In H₃O⁺ the oxygen has a complete octet and still a lone pair, so it acts as an acid (proton donor) but not as an electrophile.
Q7AIPMT 2010 Mains · Q100
A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be (h = 6.6 × 10⁻³⁴ J s)
Not quite — the answer is C.
λ = h/(mv) = 6.6 × 10⁻³⁴/(0.66 × 100) = 6.6 × 10⁻³⁴/66 = 1.0 × 10⁻³⁵ m. 6.6 × 10⁻³⁴ m comes from forgetting to divide by the momentum.
Q8AIPMT 2010 Mains · Q101
Consider the following relations for emf of an electrochemical cell: (a) emf of cell = (Oxidation potential of anode) − (Reduction potential of cathode) (b) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode) (c) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode) (d) emf of cell = (Oxidation potential of anode) − (Oxidation potential of cathode) Which of the above relations are correct?
Not quite — the answer is D.
emf = E(red, cathode) − E(red, anode). Since oxidation potential = −reduction potential, this equals (oxidation potential of anode) + (reduction potential of cathode), and also (oxidation potential of anode) − (oxidation potential of cathode). The relation subtracting the reduction potential of the cathode from the anode's oxidation potential gets the sign wrong.
Q9AIPMT 2010 Mains · Q102
In which of the following molecules the central atom does not have sp3 hybridization?
Not quite — the answer is B.
SF₄ has 4 bond pairs and 1 lone pair on S (5 electron pairs), so S is sp³d with a see-saw shape. CH₄, BF₄⁻ and NH₄⁺ all have four electron pairs and are sp³.
Q10AIPMT 2010 Mains · Q103
For vaporization of water at 1 atmospheric pressure, the values of ΔH and ΔS are 40.63 kJ mol⁻¹ and 108.8 J K⁻¹ mol⁻¹, respectively. The temperature when Gibbs energy change (ΔG) for this transformation will be zero, is
Not quite — the answer is C.
ΔG = ΔH − TΔS = 0 gives T = ΔH/ΔS = 40630/108.8 = 373.4 K. Forgetting to convert kJ to J (or mixing units) gives nonsensical values; 373 K is also the normal boiling point of water, as expected.
Q11AIPMT 2010 Mains · Q108
Match the compounds given in List-I with their characteristic reactions given in List-II. Select the correct option. List-I: a. CH₃CH₂CH₂CH₂NH₂; b. CH₃C≡CH; c. CH₃CH₂COOCH₃; d. CH₃CH(OH)CH₃ List-II: (i) Alkaline hydrolysis; (ii) With KOH (alcohol) and CHCl₃ produces bad smell; (iii) Gives white ppt. with ammoniacal AgNO₃; (iv) With Lucas reagent cloudiness appears after 5 minutes
Not quite — the answer is C.
n-Butylamine, a primary amine, gives the foul-smelling isocyanide with CHCl₃/alcoholic KOH; propyne, a terminal alkyne, gives a white precipitate with ammoniacal AgNO₃; the ester undergoes alkaline hydrolysis; 2-propanol, a secondary alcohol, turns cloudy with Lucas reagent in about 5 minutes. So a(ii), b(iii), c(i), d(iv). Assigning the silver test to the amine misses that only terminal alkynes have an acidic H.
Q12AIPMT 2010 Mains · Q110
The rate of the reaction 2NO + Cl₂ → 2NOCl is given by the rate equation rate = k[NO]²[Cl₂]. The value of the rate constant can be increased by
Not quite — the answer is A.
The rate constant depends on temperature (Arrhenius, k = A e^(−Ea/RT)), not on concentrations. Raising concentrations of NO or Cl₂ raises the rate but leaves k unchanged.
Q13AIPMT 2010 Mains · Q111
Which one of the following complexes is not expected to exhibit isomerism?
Not quite — the answer is C.
[Ni(NH₃)₂Cl₂] is tetrahedral, and an MA₂B₂ tetrahedral complex shows no geometrical or optical isomers. [Ni(NH₃)₄(H₂O)₂]²⁺ (octahedral MA₄B₂) and square planar [Pt(NH₃)₂Cl₂] show cis-trans isomerism, and [Ni(en)₃]²⁺ is optically active.
Q14AIPMT 2010 Mains · Q113
The IUPAC name of the compound CH₃CH=CHC≡CH is
Not quite — the answer is B.
Number the chain so the multiple bonds get the lowest locants: from the C≡CH end the triple bond is at C-1 and the double bond at C-3, giving pent-3-en-1-yne. Numbering from the CH₃ end gives the higher set {2,4}, so pent-2-en-4-yne is wrong.
Q15AIPMT 2010 Mains · Q114
Which of the following oxidation states is the most common among the lanthanoids?
Not quite — the answer is D.
All lanthanoids show the +3 oxidation state, which is the characteristic and most common one. +2 (Eu, Yb) and +4 (Ce) occur only for a few elements with stable f⁰, f⁷ or f¹⁴ configurations.
Q16AIPMT 2010 Mains · Q115
How many bridging oxygen atoms are present in P₄O₁₀?
Not quite — the answer is A.
In P₄O₁₀ each of the six P–P edges of the P₄ tetrahedron is bridged by an oxygen (P–O–P), and each P also carries one terminal P=O. So there are 6 bridging and 4 terminal oxygens. 4 is the number of terminal oxygens.
Q17AIPMT 2010 Mains · Q116
Some of the properties of the two species, NO₃⁻ and H₃O⁺ are described below. Which one of them is correct?
Not quite — the answer is A.
In NO₃⁻ nitrogen has three σ-bonds and no lone pair, so it is sp² and trigonal planar. In H₃O⁺ oxygen has three bonds and one lone pair, so it is sp³ and pyramidal. Thus both the hybridization and the shapes differ; they are not isostructural.
Q18AIPMT 2010 Mains · Q117
The following two reactions are known: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g); ΔH = −26.8 kJ FeO(s) + CO(g) → Fe(s) + CO₂(g); ΔH = −16.5 kJ The value of ΔH for the following reaction Fe₂O₃(s) + CO(g) → 2FeO(s) + CO₂(g) is
Not quite — the answer is D.
Target = (first equation) − 2 × (second equation): Fe₂O₃ + 3CO − 2FeO − 2CO → 2Fe + 3CO₂ − 2Fe − 2CO₂, i.e. Fe₂O₃ + CO → 2FeO + CO₂. ΔH = −26.8 − 2(−16.5) = −26.8 + 33.0 = +6.2 kJ. −10.3 kJ comes from subtracting the second equation only once.
Q19AIPMT 2010 Mains · Q118
Following compounds are given a. CH₃CH₂OH b. CH₃COCH₃ c. CH₃CH(OH)CH₃ d. CH₃OH Which of the above compound(s), on being warmed with iodine solution and NaOH, will give iodoform?
Not quite — the answer is C.
The iodoform test is given by compounds with CH₃CO– or CH₃CH(OH)– groups: ethanol, acetone and 2-propanol all qualify. Methanol has no such group and gives no iodoform.
Q20AIPMT 2010 Mains · Q119
Fructose reduces Tollens' reagent due to
Not quite — the answer is D.
Fructose is a ketose, but in alkaline Tollens' reagent it undergoes enolisation (via an enediol) and isomerises to aldoses, which are then oxidised. The alcoholic groups or asymmetric carbons do not reduce Tollens' reagent.
Q21AIPMT 2010 Mains · Q120
In the following reaction: C₆H₅CH₂Br —(1. Mg, ether; 2. H₃O⁺)→ X the product 'X' is
Not quite — the answer is C.
Benzyl bromide with Mg in dry ether forms the Grignard reagent C₆H₅CH₂MgBr, and hydrolysis with H₃O⁺ replaces MgBr by H, giving toluene C₆H₅CH₃. Benzyl alcohol would need reaction with O₂ or formaldehyde, not plain hydrolysis; bibenzyl is a Wurtz-type side product.
Chemistry: not in the current NEET syllabus
These 2 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
2 questions. Tap an option to check your answer.
Q1AIPMT 2010 Mains · Q94
The pressure exerted by 6.0 g of methane gas in a 0.03 m³ vessel at 129°C is (Atomic masses: C = 12.01, H = 1.01 and R = 8.314 J K⁻¹ mol⁻¹)
Not quite — the answer is C.
n = 6.0/16.05 = 0.374 mol and T = 129 + 273 = 402 K. P = nRT/V = 0.374 × 8.314 × 402/0.03 ≈ 4.16 × 10⁴ Pa = 41648 Pa. Forgetting to convert °C to K gives a much smaller pressure (13409 Pa).
Q2AIPMT 2010 Mains · Q104
Match List-I (substances) with List-II (process) employed in the manufacture of the substances and select the correct option. List-I: a. Sulphuric acid; b. Steel; c. Sodium hydroxide; d. Ammonia List-II: (i) Haber's Process; (ii) Bessemer's Process; (iii) Leblanc Process; (iv) Contact Process
Not quite — the answer is D.
Sulphuric acid is made by the Contact process, steel by the Bessemer process, ammonia by Haber's process, and the Leblanc process (soda) is the one left for sodium hydroxide. This gives a(iv), b(ii), c(iii), d(i). Matching sulphuric acid with Haber's process confuses it with ammonia.
Chapter-wise split of AIPMT 2010 Mains Chemistry
- Thermodynamics3 Q
- Classification of Elements and Periodicity in Properties2 Q
- Alcohols, Phenols and Ethers2 Q
- Organic Chemistry: Some Basic Principles and Techniques2 Q
- Chemical Bonding and Molecular Structure2 Q
- Equilibrium1 Q
- Structure of Atom1 Q
- Electrochemistry1 Q
- Amines1 Q
- Chemical Kinetics1 Q
- Coordination Compounds1 Q
- The d- and f-Block Elements1 Q
- Biomolecules1 Q
- Haloalkanes and Haloarenes1 Q
AIPMT 2010 Mains Biology questions
41 questions. Tap an option to check your answer.
Q1AIPMT 2010 Mains · Q1
Which one of the following pairs of structures is correctly matched with their correct description? Structures – Description
Not quite — the answer is B.
Cartilage and the cornea are avascular tissues: they have no blood vessels, yet their living cells still respire and obtain oxygen by diffusion (from perichondrium, aqueous humour and air). The knee joint is formed by the femur, tibia and patella; the fibula does not take part. The elbow is a hinge joint, not ball and socket, and premolars plus molars (8 + 12 = 20) are not all three-rooted.
Q2AIPMT 2010 Mains · Q3
Fastest distribution of some injectible material/medicine and with no risk of any kind can be achieved by injecting it into the
Not quite — the answer is C.
An intravenous injection puts the drug straight into venous blood, which returns to the heart and is pumped all over the body within seconds. Arteries carry blood at high pressure, so puncturing them is risky, and intramuscular injection only releases the drug slowly from the muscle. Lymph flow is very slow.
Q3AIPMT 2010 Mains · Q4
Which one of the following statements about the particular entity is true?
Not quite — the answer is B.
Every somatic cell carries the full genome, so the insulin gene is present in all body cells even though it is expressed only in pancreatic beta cells. Asters are formed by centrioles (centrosome), not by the centromere. A nucleosome is DNA wrapped around a histone octamer, so it is the nucleosome (not DNA) that has a core of eight histones.
Q4AIPMT 2010 Mains · Q7
Which one of the following is a xerophytic plant in which the stem is modified into the flat green and succulent structure?
Not quite — the answer is A.
Opuntia is a desert plant whose stem is flattened, green, fleshy and photosynthetic (a phylloclade), while its leaves are reduced to spines. Hydrilla is an aquatic plant, and Acacia and Casuarina do not have succulent flattened stems.
Q5AIPMT 2010 Mains · Q9
Which of the following are used in gene cloning?
Not quite — the answer is D.
Plasmids are small, autonomously replicating, extrachromosomal circular DNA molecules of bacteria and are the commonest cloning vectors (e.g. pBR322). Mesosomes and nucleoids are bacterial cell structures and lomasomes are fungal membrane bodies; none of them carry foreign DNA.
Q6AIPMT 2010 Mains · Q13
In which one of the following organisms its excretory organs are correctly stated?
Not quite — the answer is B.
Earthworm excretes through three types of nephridia: septal, integumentary and pharyngeal. Tear glands are not excretory in humans, enteric caeca in cockroach are digestive (its excretory organs are Malpighian tubules plus fat body and uricose glands), and buccal epithelium is not an excretory organ in frog.
Q7AIPMT 2010 Mains · Q15
The most apparent change during the evolutionary history of Homo sapiens is traced in
Not quite — the answer is D.
Across the human lineage the most marked trend is the steady rise in cranial capacity, from about 650-800 cc in Homo habilis, to about 900 cc in Homo erectus, to about 1400 cc in Homo sapiens. Upright walking appeared early and changed little afterwards, so it is not the most apparent trend through the lineage.
Q8AIPMT 2010 Mains · Q16
Which one of the following is now being commercially produced by biotechnological procedures?
Not quite — the answer is D.
Human insulin is now made by recombinant DNA technology: Eli Lilly produced the A and B chains separately in E. coli and then joined them. Nicotine, morphine and quinine are plant alkaloids still obtained mainly from plants.
Q9AIPMT 2010 Mains · Q19
Which one of the following is most appropriately defined?
Not quite — the answer is C.
A predator catches and kills its prey for food, which is the standard definition. One species benefiting while the other is unaffected is commensalism, not amensalism (one harmed, other unaffected). Parasites may be external (lice, ticks) and generally do not kill the host, and 'host' refers to the organism a parasite lives on, not any food provider.
Q10AIPMT 2010 Mains · Q20
Read the following four statements, A, B, C and D and select the right option having both correct statements. STATEMENTS: (A) Z scheme of light reaction takes place in presence of PSI only. (B) Only PS I is functional in cyclic photophosphorylation. (C) Cyclic photophosphorylation results into synthesis of ATP and NADPH₂ (D) Stroma lamellae lack PS II as well as NADP.
Not quite — the answer is A.
Cyclic photophosphorylation uses only PS I, and the stroma lamellae lack PS II and NADP reductase, so cyclic flow there makes only ATP. The Z scheme (non-cyclic flow) needs both PS II and PS I, and cyclic photophosphorylation does not produce NADPH.
Q11AIPMT 2010 Mains · Q21
Which one of the following techniques is safest for the detection of cancers?
Not quite — the answer is A.
MRI uses strong magnetic fields and non-ionising radiation, so it does not damage DNA the way X-rays do. Radiography and CT both use ionising X-rays, which can themselves cause mutations. Histopathology is accurate but needs a biopsy, which is an invasive procedure.
Q12AIPMT 2010 Mains · Q22
Signals from fully developed foetus and placenta ultimately lead to parturition which requires the release of
Not quite — the answer is B.
Signals from the mature foetus and placenta set up the foetal ejection reflex, which triggers release of oxytocin from the mother's posterior pituitary; oxytocin causes strong uterine contractions that expel the baby. The oxytocin acts on the mother's uterus, so it is the maternal, not foetal, pituitary that releases it.
Q13AIPMT 2010 Mains · Q23
Select the correct matching of a hormone, its source and function. Hormone – Source – Function
Not quite — the answer is B.
Norepinephrine (with epinephrine) from the adrenal medulla is an emergency hormone that raises heart rate, breathing rate and alertness. Vasopressin (ADH) reduces water loss in urine, glucagon comes from alpha cells (beta cells make insulin), and prolactin is secreted by the anterior pituitary.
Q14AIPMT 2010 Mains · Q24
In eukaryotic cell transcription, RNA splicing and RNA capping take place inside the
Not quite — the answer is B.
In eukaryotes the DNA is in the nucleus, so the primary transcript (hnRNA) is made there and is also capped, tailed and spliced there before the mature mRNA leaves for the cytoplasm. Ribosomes are where translation happens, after these steps.
Q15AIPMT 2010 Mains · Q25
Given below are four statements (a-d) regarding human blood circulatory system (a) Arteries are thick-walled and have narrow lumen as compared to veins (b) Angina is acute chest pain when the blood circulation to the brain is reduced (c) Persons with blood group AB can donate blood to any person with any blood group under ABO system (d) Calcium ions play a very important role in blood clotting Which two of the above statements are correct?
Not quite — the answer is A.
Arteries have thick muscular walls and a narrow lumen compared with veins, and Ca²⁺ ions are needed at several steps of the clotting cascade. Angina is chest pain caused by too little oxygen reaching the heart muscle, not the brain, and AB individuals are universal recipients, while O is the universal donor.
Q16AIPMT 2010 Mains · Q26
In human female the blastocyst
Not quite — the answer is D.
The outer trophoblast layer of the blastocyst attaches to the endometrium and the uterine cells then cover it, which is implantation. The placenta forms only after implantation, implantation happens about 7 days after fertilisation (not 3), and the blastocyst is already nourished by uterine secretions before it implants.
Q17AIPMT 2010 Mains · Q27
The haemoglobin content per 100 ml of blood of a normal healthy human adult is
Not quite — the answer is D.
NCERT states that a healthy adult has 12-16 g of haemoglobin in every 100 ml of blood. Values of 5-11 g would indicate anaemia, and 17-20 g or more are above the normal range.
Q18AIPMT 2010 Mains · Q28
An example of endomycorrhiza is
Not quite — the answer is B.
Glomus is a fungus that forms vesicular-arbuscular mycorrhiza, growing inside root cortical cells (endomycorrhiza) and helping the plant absorb phosphorus. Nostoc is a cyanobacterium, Rhizobium is a nitrogen-fixing bacterium, and Agaricus is a mushroom, none of which form endomycorrhiza.
Q19AIPMT 2010 Mains · Q29
One of the commonly used plant growth hormone in tea plantations is
Not quite — the answer is D.
Auxins such as IAA are used in tea plantations, where shoot tips are removed to break auxin-driven apical dominance and get bushy growth, and auxins are applied to make stem cuttings root for propagation. Ethylene and ABA are not used commercially there in this way.
Q20AIPMT 2010 Mains · Q32
Kranz anatomy is one of the characteristics of the leaves of
Not quite — the answer is C.
Kranz anatomy, with large chloroplast-rich bundle sheath cells ringing the vascular bundles, is typical of C4 plants such as sugarcane and maize. Wheat, potato and mustard are C3 plants and lack this arrangement.
Q21AIPMT 2010 Mains · Q33
In Antirrhinum two plants with pink flowers were hybridized. The F₁ plants produced red, pink and white flowers in the proportion of 1 red, 2 pink and 1 white. What could be the genotype of the two plants used for hybridization? Red flower colour is determined by RR, and white by rr genes.
Not quite — the answer is C.
Flower colour in Antirrhinum (dog flower) shows incomplete dominance, so the heterozygote Rr is pink. Crossing Rr × Rr gives RR : Rr : rr in a 1 : 2 : 1 ratio, i.e. red : pink : white. RR or rr parents would breed true for red or white.
Q22AIPMT 2010 Mains · Q34
Transport of food material in higher plants takes place through
Not quite — the answer is D.
Food is carried in phloem, through the sieve tube elements (sieve cells in gymnosperms) joined end to end by sieve plates. Companion cells only support the sieve tubes, tracheids conduct water, and transfusion tissue is a gymnosperm leaf tissue.
Q23AIPMT 2010 Mains · Q35
Which one of the following is monoecious?
Not quite — the answer is C.
In Pinus both male and female cones are borne on the same tree, so it is monoecious. Cycas has separate male and female plants, Marchantia has separate male and female thalli, and date palm is also dioecious.
Q24AIPMT 2010 Mains · Q36
A cross in which an organism showing a dominant phenotype is crossed with the recessive parent in order to know its genotype is called :
Not quite — the answer is C.
Crossing a dominant-phenotype individual of unknown genotype with a homozygous recessive is a test cross: all dominant offspring means the individual was homozygous, and a 1 : 1 ratio means it was heterozygous. A back cross is any cross of F₁ with either parent, so only the cross with the recessive parent is the test cross.
Q25AIPMT 2010 Mains · Q37
The Indian Rhinoceros is a natural inhabitant of which one of the Indian states?
Not quite — the answer is D.
The one-horned Indian rhinoceros lives naturally in the floodplain grasslands of Assam, especially Kaziranga National Park, which holds most of the world's population. The other states are not its natural range.
Q26AIPMT 2010 Mains · Q39
Black (stem) rust of wheat is caused by :
Not quite — the answer is C.
Black stem rust of wheat is caused by the basidiomycete rust fungus Puccinia graminis (f. sp. tritici). Ustilago nuda causes loose smut of wheat, Alternaria solani causes early blight of potato, and Xanthomonas oryzae is the bacterium of rice blight.
Q27AIPMT 2010 Mains · Q40
Secretions from which one of the following are rich in fructose, calcium and some enzymes?
Not quite — the answer is A.
The male accessory glands (seminal vesicles, prostate and bulbourethral glands) produce seminal plasma, which NCERT describes as rich in fructose, calcium and certain enzymes. Fructose is the energy source for sperm; the liver, pancreas and salivary glands make digestive secretions instead.
Q28AIPMT 2010 Mains · Q41
A person suffering from a disease caused by Plasmodium, experiences recurring chill and fever at the time when?
Not quite — the answer is C.
Plasmodium multiplies asexually inside RBCs and then bursts them, releasing new merozoites to infect fresh RBCs along with the toxin haemozoin. This bursting happens in cycles (every 48 hours in vivax), which is when the chill and high fever recur. Sporozoites enter through the mosquito bite and go to the liver, not RBCs.
Q29AIPMT 2010 Mains · Q42
ABO blood grouping is controlled by gene I which has three alleles and show co-dominance. There are six genotypes. How many phenotypes in all are possible?
Not quite — the answer is C.
The three alleles Iᴬ, Iᴮ and i give six genotypes, but Iᴬ and Iᴮ are dominant over i and co-dominant with each other, so only four phenotypes appear: A, B, AB and O. Six would be correct only if every genotype looked different.
Q30AIPMT 2010 Mains · Q43
Three of the following statements about enzymes are correct and one is wrong. Which one is wrong?
Not quite — the answer is D.
Almost all enzymes are proteins, and the non-protein exceptions are RNA molecules (ribozymes), not lipids. Enzymes of thermophilic organisms in hot springs do stay active at 80-90°C, and enzymes do have an optimum pH and high specificity.
Q31AIPMT 2010 Mains · Q44
An elaborate network of filamentous proteinaceous structures present in the cytoplasm which helps in the maintenance of cell shape is called :
Not quite — the answer is D.
The cytoskeleton is a network of protein filaments (microtubules, microfilaments and intermediate filaments) in the cytoplasm that gives mechanical support, maintains cell shape and helps in motility. The ER is a membrane system, not a protein filament network.
Q32AIPMT 2010 Mains · Q47
Consider the following four statements A, B, C and D and select the right option for two correct statements. Statements (A) In vexillary aestivation, the large posterior petal is called standard, two lateral ones are wings and two small anterior petals are termed keel (B) The floral formula for Liliaceae is ⊕ ⚥ P₃₊₃ A₃₊₃ + G̲₃ (C) In pea flower the stamens are monadelphous (D) The floral formula for Solanaceae is ⊕ ⚥ K₍₃₎ C₍₃₎ A₍₄₎ + G̲₍₂₎ The correct statements are
Not quite — the answer is B.
Statement A describes vexillary aestivation correctly, and B gives the Liliaceae formula (actinomorphic, bisexual, 3+3 tepals, 3+3 stamens, tricarpellary superior ovary). Pea stamens are diadelphous (9 + 1), not monadelphous, and Solanaceae has five fused sepals, five fused petals and five stamens, not 3, 3 and 4.
Q33AIPMT 2010 Mains · Q49
In genetic engineering, a DNA segment (gene) of interest, is transferred to the host cell through a vector. Consider the following four agents (A-D) in this regard and select the correct option about which one or more of these can be used as a vector/vectors Statements (A) A bacterium (B) Plasmid (C) Plasmodium (D) Bacteriophage
Not quite — the answer is D.
The standard cloning vectors are plasmids and bacteriophages, because both replicate inside bacteria and can carry inserted DNA. Plasmodium is a protozoan parasite and is never used as a vector.
Q34AIPMT 2010 Mains · Q51
The fruit fly Drosophila melanogaster was found to be very suitable for experimental verification of chromosomal theory of inheritance by Morgan and his colleagues because :
Not quite — the answer is D.
Drosophila is useful for genetics because it completes its life cycle in about two weeks, a single mating gives a large number of progeny, and the sexes are easy to tell apart. The female is larger than the male, not smaller, and flies reproduce sexually, not parthenogenetically.
Q35AIPMT 2010 Mains · Q52
The lac operon consists of
Not quite — the answer is B.
The lac operon has one regulatory gene (i, coding the repressor) and three structural genes: z (β-galactosidase), y (permease) and a (transacetylase), along with promoter and operator sites. The promoter and operator are DNA sites, not genes, so it does not have two or three regulatory genes.
Q36AIPMT 2010 Mains · Q53
Crocodile and penguin are similar to Whale and Dogfish in which one of the following features?
Not quite — the answer is D.
All four are chordates, and every chordate has pharyngeal gill slits at some stage of its life, at least in the embryo. Chordates have a hollow dorsal nerve cord, not a solid one; the dogfish has a cartilaginous skeleton and the whale is viviparous, so the other features are not common to all.
Q37AIPMT 2010 Mains · Q54
Select the answer with correct matching of the structure, its location and function Structure – Location – Function
Not quite — the answer is C.
The hypothalamus lies at the base of the forebrain (diencephalon) and has centres for body temperature, hunger and thirst. The Eustachian tube connects the middle ear (not internal ear) with the pharynx, the cerebellum is part of the hindbrain and controls balance, and the blind spot has no rods or cones at all.
Q38AIPMT 2010 Mains · Q55
Select the correct combination of the statements (a-d) regarding the characteristics of certain organisms (a) Methanogens are Archaebacteria which produce methane in marshy areas (b) Nostoc is a filamentous blue-green alga which fixes atmospheric nitrogen (c) Chemosynthetic autotrophic bacteria synthesize cellulose from glucose (d) Mycoplasma lack a cell wall and can survive without oxygen The correct statement are
Not quite — the answer is D.
Methanogens are archaebacteria that produce methane in marshes, Nostoc is a filamentous cyanobacterium with heterocysts that fix nitrogen, and Mycoplasma has no cell wall and can survive without oxygen. Chemosynthetic bacteria oxidise inorganic substances such as nitrites or ammonia to get energy for making food; making cellulose from glucose is not what defines them.
Q39AIPMT 2010 Mains · Q56
Which one of the following is the correct description of a certain part of a normal human skeleton?
Not quite — the answer is A.
Skull bones such as the parietal and temporal are joined by sutures, which are fibrous, immovable joints. The first vertebra is the atlas (not the axis), the floating ribs are the 11th and 12th pairs, and the glenoid cavity takes the head of the humerus, whereas the femur fits into the acetabulum.
Q40AIPMT 2010 Mains · Q57
Vegetative propagation in Pistia occurs by
Not quite — the answer is B.
In aquatic plants like Pistia and Eichhornia, a short lateral branch with one internode grows out and bears a rosette of leaves and a tuft of roots at its tip; this is an offset, and it separates to form a new plant. Runners and stolons are longer creeping branches of land plants such as grasses and strawberry.
Q41AIPMT 2010 Mains · Q60
The 3′-5′ phosphodiester linkages inside a polynucleotide chain serve to join
Not quite — the answer is C.
In a polynucleotide, the phosphate of one nucleotide links the 3′-carbon of one sugar to the 5′-carbon of the next, so a 3′-5′ phosphodiester bond joins successive nucleotides. The two DNA strands are held together by hydrogen bonds, and a base is joined to its sugar by an N-glycosidic bond.
Biology: 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 2010 Mains · Q6
Leguminous plants are able to fix atmospheric nitrogen through the process of symbiotic nitrogen fixation. Which one of the following statements is not correct during this process of nitrogen fixation?
Not quite — the answer is D.
Nitrogenase is highly sensitive to oxygen, which inactivates it; that is exactly why root nodules contain the pink pigment leghaemoglobin, which binds and removes oxygen. The other statements are correct: nodules are the sites of fixation and nitrogenase reduces N₂ to NH₃.
Q2AIPMT 2010 Mains · Q10
When domestic sewage mixes with river water
Not quite — the answer is C.
Sewage adds a large amount of biodegradable organic matter, and the microbes decomposing it consume dissolved oxygen, raising the BOD and causing oxygen depletion that kills fish. Even though impurities are a small percentage, they include pathogens, so the water is not fit for drinking.
Q3AIPMT 2010 Mains · Q18
If for some reason the parietal cells of the gut epithelium become partially non-functional, what is likely to happen?
Not quite — the answer is D.
Parietal (oxyntic) cells secrete HCl, which converts inactive pepsinogen into pepsin and gives the acidic pH pepsin needs. With less HCl, pepsin is poorly formed and proteins are not properly broken down into proteoses and peptones. The stomach pH would rise, not fall, and pancreatic enzymes work in the intestine, which is unaffected.
Q4AIPMT 2010 Mains · Q31
Jaundice is a disorder of
Not quite — the answer is C.
Jaundice is a disorder of the liver, a digestive gland: when the liver is damaged or the bile duct blocked, bile pigments build up in the blood. The yellowing of skin and eyes is only the symptom, not the organ system affected.
Q5AIPMT 2010 Mains · Q50
Which one of the following can not be used for preparation of vaccines against plague?
Not quite — the answer is C.
Plague vaccines have been made from killed Yersinia pestis (heat-killed or formalin-inactivated) and from live avirulent (attenuated) strains. Capsular polysaccharide vaccines are used against encapsulated bacteria such as pneumococcus, Haemophilus and meningococcus, not against plague.
Chapter-wise split of AIPMT 2010 Mains Biology
- Molecular Basis of Inheritance4 Q
- Principles of Inheritance and Variation4 Q
- Body Fluids and Circulation3 Q
- Morphology of Flowering Plants3 Q
- Human Reproduction3 Q
- Locomotion and Movement2 Q
- Biotechnology: Principles and Processes2 Q
- Photosynthesis in Higher Plants2 Q
- Human Health and Disease2 Q
- Biological Classification2 Q
- Excretory Products and Their Elimination1 Q
- Evolution1 Q
- Biotechnology and Its Applications1 Q
- Organisms and Populations1 Q
- Chemical Coordination and Integration1 Q
- Microbes in Human Welfare1 Q
- Plant Growth and Development1 Q
- Anatomy of Flowering Plants1 Q
- Plant Kingdom1 Q
- Biodiversity and Conservation1 Q
- Biomolecules1 Q
- Cell: The Unit of Life1 Q
- Animal Kingdom1 Q
- Neural Control and Coordination1 Q
About this paper
How many questions from AIPMT 2010 Mains are on this page?
93 questions (Physics 24, Chemistry 23, 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 2010 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 2010 Mains useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books. 8 questions here are on topics NCERT has since dropped; they are listed separately so you can skip them.
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