AIPMT 2011 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
100 questions from the AIPMT 2011 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 2011 Mains
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- 27 questions
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- 50 questions
AIPMT 2011 Mains Physics questions
23 questions. Tap an option to check your answer.
Q1AIPMT 2011 Mains
A galvanometer of resistance G is shunted by a resistance S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is
Not quite — the answer is A.
For the main current to stay unchanged, the shunted galvanometer plus the series resistance must still equal G. The shunted combination is GS/(G + S), so R = G − GS/(G + S) = G²/(G + S). The answer GS/(G + S) is only the shunted combination itself, not the extra resistance needed.
Q2AIPMT 2011 Mains
A particle covers half of its total distance with speed v₁ and the rest half distance with speed v₂. Its average speed during the complete journey is
Not quite — the answer is D.
Each half takes d/v₁ and d/v₂, so average speed = 2d/(d/v₁ + d/v₂) = 2v₁v₂/(v₁ + v₂), the harmonic mean. The arithmetic mean (v₁ + v₂)/2 would apply only to equal time intervals, not equal distances.
Q3AIPMT 2011 Mains
A thermocouple of negligible resistance produces an e.m.f. of 40 µV/°C in the linear range of temperature. A galvanometer of resistance 10 ohm whose sensitivity is 1 µA/div, is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system will be
Not quite — the answer is B.
The smallest detectable current is 1 µA (one division), so the smallest detectable emf is 1 µA × 10 Ω = 10 µV. With 40 µV per °C, ΔT = 10/40 = 0.25°C. Answers like 0.1°C or 0.5°C come from mixing up the ratio.
Q4AIPMT 2011 Mains
A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to a mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is
Not quite — the answer is C.
Momentum is conserved in this perfectly inelastic collision: m v î + 3m(2v) ĵ = 4m V⃗. So V⃗ = (v/4) î + (6v/4) ĵ = (1/4)v î + (3/2)v ĵ. Swapping the components, (3/2)v î + (1/4)v ĵ, is the common trap.
Q5AIPMT 2011 Mains
A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is
Not quite — the answer is A.
Without the lens the rays meet 10 cm behind the lens position, so that point is a virtual object at u = +10 cm; with the lens the image is at v = +15 cm. 1/f = 1/v − 1/u = 1/15 − 1/10 = −1/30, so f = −30 cm. The negative sign confirms a diverging lens, ruling out the positive values.
Q6AIPMT 2011 Mains
The threshold frequency for a photosensitive metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly
Not quite — the answer is C.
eV₀ = h(ν − ν₀) = 6.63 × 10⁻³⁴ × (8.2 − 3.3) × 10¹⁴ = 3.25 × 10⁻¹⁹ J, so V₀ = 3.25 × 10⁻¹⁹/1.6 × 10⁻¹⁹ ≈ 2 V. Using the incident frequency alone (ignoring the threshold) would give about 3.4 V, which is not the cut-off voltage.
Q7AIPMT 2011 Mains
An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, then the value of n is
Not quite — the answer is D.
Photon energy = work function + eV₀ = 2.75 + 10 = 12.75 eV. For hydrogen, 13.6(1 − 1/n²) = 12.75 gives 1/n² = 0.0625, so n = 4. The n = 3 → 1 transition gives only 12.09 eV, too little.
Q8AIPMT 2011 Mains
A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The magnitude of the gravitational potential at a point situated at a/2 distance from the centre, will be
Not quite — the answer is D.
Potential due to the central particle at a/2 is −GM/(a/2) = −2GM/a; inside a shell the potential is constant at −GM/a. Total = −3GM/a, magnitude 3GM/a. Taking only the particle's contribution gives 2GM/a, forgetting that the shell's potential inside is not zero.
Q9AIPMT 2011 Mains
Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is
Not quite — the answer is D.
x = A sin φ = A/2 gives φ = π/6 for one particle and 5π/6 for the other, since they move in opposite directions at that displacement. Phase difference = 5π/6 − π/6 = 2π/3. Zero would mean they move together, and π would mean they cross only at the mean position.
Q10AIPMT 2011 Mains
A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between them is μ = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g = 10 m s⁻², is
Not quite — the answer is B.
Relative to the belt the box starts at 2 m/s and friction gives it a relative deceleration μg = 5 m/s². Relative distance = u²/2a = 4/10 = 0.4 m. The answer zero ignores that the box slips until it reaches the belt speed.
Q11AIPMT 2011 Mains
A mass of diatomic gas (γ = 1.4) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from 27°C to 927°C. The pressure of the gas in the final state is
Not quite — the answer is A.
For an adiabatic process P^(1−γ)T^γ is constant, so P ∝ T^(γ/(γ−1)) = T^3.5. T rises from 300 K to 1200 K, a factor of 4, so P₂ = 2 × 4^3.5 = 2 × 128 = 256 atm. Using Celsius temperatures or P ∝ T (isochoric) gives the wrong values.
Q12AIPMT 2011 Mains
Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the center of the ring is
Not quite — the answer is C.
The rotating ring is equivalent to a current I = q/T = qf. The field at the centre of a circular loop is μ₀I/(2R) = μ₀qf/(2R). The expression with 2πR in the denominator is the formula for a long straight wire, not a loop.
Q13AIPMT 2011 Mains
Two radioactive nuclei P and Q in a given sample decay into a stable nucleus R. At time t = 0, number of P species are 4N₀ and that of Q are N₀. Half-life of P (for conversion to R) is 1 minute whereas that of Q is 2 minutes. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be
Not quite — the answer is D.
P: 4N₀(1/2)^t and Q: N₀(1/2)^(t/2) with t in minutes. Setting them equal gives 2^(2 − t) = 2^(−t/2), so t = 4 min, when P = Q = N₀/4. R = 5N₀ − N₀/2 = 9N₀/2. 5N₀/2 would count only the decayed P nuclei incorrectly.
Q14AIPMT 2011 Mains
Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5 × 10¹⁶ m⁻³. Doping by indium increases nh to 4.5 × 10²² m⁻³. The doped semiconductor is of
Not quite — the answer is B.
Indium is trivalent, so the doped crystal is p-type. By the mass-action law ne = ni²/nh = (1.5 × 10¹⁶)²/(4.5 × 10²²) = 2.25 × 10³²/4.5 × 10²² = 5 × 10⁹ m⁻³. The n-type choices are wrong because an acceptor dopant was added.
Q15AIPMT 2011 Mains
A thin prism of angle 15° made of glass of refractive index μ₁ = 1.5 is combined with another prism of glass of refractive index μ₂ = 1.75. The combination of the prisms produces dispersion without deviation. The angle of the second prism should be
Not quite — the answer is D.
For no net deviation with thin prisms, (μ₁ − 1)A₁ = (μ₂ − 1)A₂, so 0.5 × 15° = 0.75 × A₂ and A₂ = 10°. Using the ratio of refractive indices directly instead of (μ − 1) leads to wrong angles.
Q16AIPMT 2011 Mains
Two identical piano wires, kept under the same tension T, have a fundamental frequency of 600 Hz. The fractional increase in the tension of one of the wires which will lead to occurrence of 6 beats/s when both the wires oscillate together would be
Not quite — the answer is C.
f ∝ √T, so Δf/f = ½ ΔT/T. Six beats means Δf = 6 Hz, so 6/600 = ½ ΔT/T and ΔT/T = 0.02. Forgetting the square-root dependence gives 0.01.
Q17AIPMT 2011 Mains
A small mass attached to a string rotates on a frictionless table top (the string passes through a hole at the centre of the table). If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will
Not quite — the answer is A.
The string tension acts towards the hole, so angular momentum L = mvr is conserved. KE = L²/(2mr²), so halving r makes KE four times larger. Saying it stays constant ignores the work done by the pulling force.
Q18AIPMT 2011 Mains
The electric potential V at any point (x, y, z), in meters in space is given by V = 4x² volt. The electric field at the point (1, 0, 2) in volt/meter, is
Not quite — the answer is B.
E = −dV/dx = −8x, which at x = 1 m is −8 V/m, i.e. 8 V/m along the negative x-axis. The value 16 V/m comes from wrong differentiation, and the positive direction ignores the minus sign in E = −∇V.
Q19AIPMT 2011 Mains
A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth, is
Not quite — the answer is C.
To not return, the particle must reach infinity: ½mu² = GMm/R, so u = √(2GM/R) (= √(2gR)). √(2GM/R²) and √(2gR²) are dimensionally wrong as speeds.
Q20AIPMT 2011 Mains
A short bar magnet of magnetic moment 0.4 J T⁻¹ is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium when the potential energy is
Not quite — the answer is C.
U = −MB cos θ; stable equilibrium is θ = 0, so U = −MB = −0.4 × 0.16 = −0.064 J. +0.064 J is the unstable (antiparallel) equilibrium.
Q21AIPMT 2011 Mains
A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is
Not quite — the answer is D.
At the top the projectile is at height H and horizontal distance R/2. tan α = H/(R/2) = (u² sin²θ/2g)/(u² sinθ cosθ/g) = tanθ/2 = 1/2 for θ = 45°. So α = tan⁻¹(1/2); 45° would be the launch angle, not the elevation of the highest point.
Q22AIPMT 2011 Mains
A coil has resistance 30 ohm and inductive reactance 20 ohm at 50 Hz frequency. If an ac source of 200 volt, 100 Hz is connected across the coil, the current in the coil will be
Not quite — the answer is C.
Inductive reactance doubles with frequency: at 100 Hz XL = 40 Ω. Z = √(30² + 40²) = 50 Ω, so I = 200/50 = 4 A. Keeping XL = 20 Ω gives 20/√13 A, which is the trap.
Q23AIPMT 2011 Mains
The density of a material in CGS system of units is 4 g/cm³. In a system of units in which unit of length is 10 cm and unit of mass is 100 g, the value of density of material will be
Not quite — the answer is D.
New unit of density = 100 g/(10 cm)³ = 0.1 g/cm³. So 4 g/cm³ = 4/0.1 = 40 new units. Dividing the wrong way gives 0.4.
Chapter-wise split of AIPMT 2011 Mains Physics
- Current Electricity2 Q
- Ray Optics and Optical Instruments2 Q
- Gravitation2 Q
- Motion in a Straight Line1 Q
- Work, Energy and Power1 Q
- Dual Nature of Radiation and Matter1 Q
- Atoms1 Q
- Oscillations1 Q
- Laws of Motion1 Q
- Thermodynamics1 Q
- Moving Charges and Magnetism1 Q
- Nuclei1 Q
- Semiconductor Electronics: Materials, Devices and Simple Circuits1 Q
- Waves1 Q
- System of Particles and Rotational Motion1 Q
- Electrostatic Potential and Capacitance1 Q
- Magnetism and Matter1 Q
- Motion in a Plane1 Q
- Alternating Current1 Q
- Units and Measurements1 Q
AIPMT 2011 Mains Chemistry questions
22 questions. Tap an option to check your answer.
Q1AIPMT 2011 Mains
Which of the statements about "Denaturation" given below are correct? (a) Denaturation of proteins causes loss of secondary and tertiary structures of the protein (b) Denaturation leads to the conversion of double strand of DNA into single strand (c) Denaturation affects primary structure which gets distorted
Not quite — the answer is A.
Denaturation disrupts the secondary and tertiary structure of a protein, and in DNA it separates the double helix into single strands. Peptide bonds are not broken, so the primary structure is unchanged; any combination including that statement is wrong.
Q2AIPMT 2011 Mains
The IUPAC name of the compound ClC(CH₃)=C(I)CH₂CH₃, in which Cl and I lie on opposite sides of the double bond, is
Not quite — the answer is C.
The longest chain containing the double bond has five carbons, numbered from the end nearer the double bond: C2 carries Cl and C3 carries I, giving 2-chloro-3-iodopent-2-ene. Cl and I, the heavier substituents on each alkene carbon, are on opposite sides, so it is trans. Names like 3-iodo-4-chloro-3-pentene number from the wrong end.
Q3AIPMT 2011 Mains
Which of the following compounds undergoes nucleophilic substitution reaction most easily?
Not quite — the answer is C.
Aryl halides undergo nucleophilic substitution only when electron-withdrawing groups at ortho/para positions stabilise the negatively charged intermediate. A para –NO₂ does this strongly, so p-nitrochlorobenzene reacts most easily. –OCH₃ and –CH₃ donate electrons and make substitution harder than in chlorobenzene.
Q4AIPMT 2011 Mains
An organic compound 'A' on treatment with NH₃ gives 'B' which on heating gives 'C'. 'C' when treated with Br₂ in the presence of KOH produces ethylamine. Compound 'A' is
Not quite — the answer is B.
Hoffmann bromamide degradation removes one carbon, so ethylamine must come from propanamide (compound 'C'), which forms on heating ammonium propanoate ('B'), made from propanoic acid ('A') and NH₃. Acetic acid would give methylamine and butanoic acid would give propylamine.
Q5AIPMT 2011 Mains
Consider the following processes: (1/2)A → B; ΔH = +150 kJ/mol 3B → 2C + D; ΔH = −125 kJ/mol E + A → 2D; ΔH = +350 kJ/mol For B + D → E + 2C, ΔH will be
Not quite — the answer is D.
Double the first equation (A → 2B, +300), add the second (3B → 2C + D, −125) and reverse the third (2D → E + A, −350). Adding gives B + D → E + 2C with ΔH = 300 − 125 − 350 = −175 kJ/mol. Forgetting to double the first equation leads to −325 kJ/mol.
Q6AIPMT 2011 Mains
A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be
Not quite — the answer is B.
HA ⇌ H⁺ + A⁻ with α = 0.3 gives i = 1 + α = 1.3. ΔTf = iKf m = 1.3 × 1.86 × 0.1 = 0.24°C, so the freezing point is −0.24°C. Ignoring ionisation gives −0.18°C, and taking full dissociation gives −0.36°C.
Q7AIPMT 2011 Mains
In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag⁺ and Pb²⁺ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl⁻ concentration is 0.10 M. What will the concentrations of Ag⁺ and Pb²⁺ be at equilibrium? (Ksp for AgCl = 1.8 × 10⁻¹⁰, Ksp for PbCl₂ = 1.7 × 10⁻⁵)
Not quite — the answer is A.
At [Cl⁻] = 0.10 M: [Ag⁺] = Ksp/[Cl⁻] = 1.8 × 10⁻¹⁰/0.1 = 1.8 × 10⁻⁹ M and [Pb²⁺] = Ksp/[Cl⁻]² = 1.7 × 10⁻⁵/0.01 = 1.7 × 10⁻³ M. Choices that use [Cl⁻] instead of [Cl⁻]² for PbCl₂ give 1.7 × 10⁻⁴ M, which is wrong.
Q8AIPMT 2011 Mains
The unit of rate constant for a zero order reaction is
Not quite — the answer is C.
For a zero-order reaction rate = k[A]⁰ = k, so k has the units of rate, mol L⁻¹ s⁻¹. s⁻¹ is the first-order unit and L mol⁻¹ s⁻¹ is the second-order unit.
Q9AIPMT 2011 Mains
200 mL of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of this solution at 300 K is found to be 2.57 × 10⁻³ bar. The molar mass of protein will be (R = 0.083 L bar mol⁻¹ K⁻¹)
Not quite — the answer is B.
π = (w/M)RT/V, so M = wRT/(πV) = 1.26 × 0.083 × 300/(2.57 × 10⁻³ × 0.200) = 31.37/5.14 × 10⁻⁴ ≈ 61038 g mol⁻¹. Forgetting to convert 200 mL to 0.200 L or halving/doubling the volume gives the other values.
Q10AIPMT 2011 Mains
The rate of the reaction 2N₂O₅ → 4NO₂ + O₂ can be written in three ways: −d[N₂O₅]/dt = k[N₂O₅] d[NO₂]/dt = k′[N₂O₅] d[O₂]/dt = k″[N₂O₅] The relationship between k and k′ and between k and k″ are
Not quite — the answer is D.
Rate = −½ d[N₂O₅]/dt = ¼ d[NO₂]/dt = d[O₂]/dt. So d[NO₂]/dt = 2 × (−d[N₂O₅]/dt), giving k′ = 2k, and d[O₂]/dt = ½ × (−d[N₂O₅]/dt), giving k″ = k/2. Setting all three constants equal ignores the stoichiometric coefficients.
Q11AIPMT 2011 Mains
The half life of a substance in a certain enzyme-catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L⁻¹ to 0.04 mg L⁻¹ is
Not quite — the answer is A.
A constant half-life means first-order kinetics. 1.28/0.04 = 32 = 2⁵, so five half-lives are needed: 5 × 138 = 690 s. 552 s corresponds to only four half-lives.
Q12AIPMT 2011 Mains
Which of the following complex compounds will exhibit highest paramagnetic behaviour? (At. No. Ti = 22, Cr = 24, Co = 27, Zn = 30)
Not quite — the answer is D.
Cr³⁺ is d³ with 3 unpaired electrons in octahedral complexes. Co³⁺ (d⁶) with NH₃ is low spin with no unpaired electrons, Zn²⁺ is d¹⁰ with none, and Ti³⁺ is d¹ with one. So [Cr(NH₃)₆]³⁺ is the most paramagnetic.
Q13AIPMT 2011 Mains
Which of the following is not a fat soluble vitamin?
Not quite — the answer is C.
Vitamins A, D, E and K are fat soluble, while the B-complex vitamins and vitamin C are water soluble. So vitamin B complex is not fat soluble.
Q14AIPMT 2011 Mains
Which of the following compounds is most basic?
Not quite — the answer is D.
In benzylamine the –NH₂ is attached to a CH₂ group, so its lone pair is not delocalised into the ring and it behaves like an alkylamine. In aniline the lone pair is delocalised into the ring, a para –NO₂ withdraws it further, and in acetanilide it is also delocalised into the C=O group, so these are much weaker bases.
Q15AIPMT 2011 Mains
Which has the maximum number of molecules among the following?
Not quite — the answer is A.
Moles: 8 g H₂ = 8/2 = 4 mol; 64 g SO₂, 44 g CO₂ and 48 g O₃ are each 1 mol. The largest number of moles means the most molecules, so 8 g of H₂ has the most, even though it has the smallest mass.
Q16AIPMT 2011 Mains
Match the compounds given in List-I with List-II and select the suitable option using the code given below. List-I: a. Benzaldehyde, b. Phthalic anhydride, c. Phenyl benzoate, d. Methyl salicylate List-II: (i) Phenolphthalein, (ii) Benzoin condensation, (iii) Oil of wintergreen, (iv) Fries rearrangement
Not quite — the answer is B.
Benzaldehyde undergoes benzoin condensation; phthalic anhydride condenses with phenol to give phenolphthalein; phenyl benzoate undergoes Fries rearrangement to hydroxybenzophenones; methyl salicylate is oil of wintergreen. Only one code gives all four pairs correctly.
Q17AIPMT 2011 Mains
What is the value of electron gain enthalpy of Na⁺ if IE₁ of Na = 5.1 eV?
Not quite — the answer is C.
Na⁺ + e⁻ → Na is exactly the reverse of Na → Na⁺ + e⁻ (IE₁ = 5.1 eV), so energy is released and ΔegH = −5.1 eV. +2.55 eV or ±10.2 eV come from wrongly halving or doubling the ionisation energy.
Q18AIPMT 2011 Mains · Q95
Which of the following oxide is amphoteric?
Not quite — the answer is C.
SnO₂ reacts with both acids and bases, so it is amphoteric. SiO₂ and CO₂ are acidic oxides and CaO is a basic oxide.
Q19AIPMT 2011 Mains
Which of the following carbonyls will have the strongest C – O bond?
Not quite — the answer is C.
Back-donation from metal d orbitals into CO π* orbitals weakens the C–O bond. The positive charge in Mn(CO)₆⁺ holds the metal electrons tightly, so back-bonding is least and the C–O bond is strongest. The anion V(CO)₆⁻ has the most back-bonding and the weakest C–O bond.
Q20AIPMT 2011 Mains
The order of reactivity of phenyl magnesium bromide (PhMgBr) with the following compounds: (I) CH₃CHO, (II) (CH₃)₂C=O and (III) Ph₂C=O is
Not quite — the answer is B.
Nucleophilic addition of PhMgBr gets slower as electron-donating and bulky groups around the carbonyl carbon increase. Acetaldehyde (one H, one CH₃) is most reactive, acetone (two CH₃) is next, and benzophenone (two bulky phenyls with resonance) is least reactive: I > II > III.
Q21AIPMT 2011 Mains
The pairs of species of oxygen and their magnetic behaviours are noted below. Which of the following presents the correct description?
Not quite — the answer is A.
By MO theory O₂ has 2 unpaired electrons and O₂⁺ has 1, so both are paramagnetic. O₂²⁻ has all electrons paired (diamagnetic), so the pairings with peroxide as paramagnetic are wrong, and O₂⁻ has 1 unpaired electron, so calling it diamagnetic is wrong.
Q22AIPMT 2011 Mains
According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?
Not quite — the answer is A.
ΔE ∝ (1/n₁² − 1/n₂²): for 6 → 5 it is 1/25 − 1/36 = 0.0122, for 5 → 4 it is 1/16 − 1/25 = 0.0225, and the others are larger. So the 6 → 5 transition gives the least energetic photon.
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 2011 Mains · Q91
Match List-I with List-II for the composition of substances and select the correct answer using the code given below the lists. List-I: (A) Plaster of paris (B) Epsomite (C) Kieserite (D) Gypsum List-II: (i) CaSO₄·2H₂O (ii) CaSO₄·½H₂O (iii) MgSO₄·7H₂O (iv) MgSO₄·H₂O (v) CaSO₄
Not quite — the answer is D.
Plaster of Paris is CaSO₄·½H₂O, epsomite is MgSO₄·7H₂O, kieserite is MgSO₄·H₂O and gypsum is CaSO₄·2H₂O. Anhydrous CaSO₄ is a distractor that matches none of them. Mixing up gypsum and plaster of Paris is the usual trap.
Q2AIPMT 2011 Mains · Q92
Which of the following statements is incorrect?
Not quite — the answer is A.
Aluminium dissolves in excess NaOH to give sodium aluminate, Na[Al(OH)₄], and H₂; any Al(OH)₃ formed redissolves, so that statement is incorrect. NaHCO₃ does decompose to Na₂CO₃ on heating, sodium gives a blue solution in liquid ammonia (ammoniated electrons), and NaOH attacks glass to form sodium silicate.
Q3AIPMT 2011 Mains · Q97
A bubble of air is underwater at temperature 15°C and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25°C and the pressure is 1.0 bar what will happen to the volume of the bubble?
Not quite — the answer is C.
From P₁V₁/T₁ = P₂V₂/T₂: V₂/V₁ = (P₁/P₂)(T₂/T₁) = (1.5/1.0)(298/288) = 1.55 ≈ 1.6. Using Celsius temperatures or ignoring the temperature change gives the other factors, and the volume cannot get smaller since both changes make it expand.
Q4AIPMT 2011 Mains · Q101
The following reactions take place in the blast furnace in the preparation of impure iron. Identify the reaction pertaining to the formation of the slag.
Not quite — the answer is A.
Slag is the fusible calcium silicate made when the flux CaO combines with the silica gangue: CaO + SiO₂ → CaSiO₃. Decomposition of limestone only supplies CaO, the carbon reaction makes the CO reducing agent, and Fe₂O₃ + 3CO is the reduction step.
Q5AIPMT 2011 Mains · Q110
A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y⁻) will be
Not quite — the answer is A.
In the NaCl (octahedral) structure the limiting radius ratio r⁺/r⁻ = 0.414, so r⁻ = 100/0.414 ≈ 241.5 pm. Using 0.732 (the CsCl value) or 0.225 (tetrahedral) gives the other radii.
Chapter-wise split of AIPMT 2011 Mains Chemistry
- Chemical Kinetics3 Q
- Biomolecules2 Q
- Amines2 Q
- Solutions2 Q
- Coordination Compounds2 Q
- Aldehydes, Ketones and Carboxylic Acids2 Q
- Organic Chemistry: Some Basic Principles and Techniques1 Q
- Haloalkanes and Haloarenes1 Q
- Thermodynamics1 Q
- Equilibrium1 Q
- Some Basic Concepts of Chemistry1 Q
- Classification of Elements and Periodicity in Properties1 Q
- Chemical Bonding and Molecular Structure1 Q
- Structure of Atom1 Q
AIPMT 2011 Mains Biology questions
45 questions. Tap an option to check your answer.
Q1AIPMT 2011 Mains
The cells lining the blood vessels belong to the category of
Not quite — the answer is D.
The inner lining of blood vessels (endothelium) is a single layer of flat, thin cells, i.e. simple squamous epithelium, which suits diffusion across the vessel wall. Columnar epithelium lines the stomach and intestine, not blood vessels; smooth muscle lies in the middle layer of the wall, not the lining.
Q2AIPMT 2011 Mains
Select the correct statement with respect to diseases and immunisation
Not quite — the answer is C.
B-lymphocytes produce antibodies and need helper T-cells to do so, so damage to both lymphocyte types stops antibody production against a pathogen. Hepatitis B vaccine is produced in yeast, not protozoans; snake antivenom supplies ready-made antibodies (passive immunity); and dead or inactivated pathogens stimulate the body's own response (active immunity).
Q3AIPMT 2011 Mains
Which one of the following animals may occupy more than one trophic level in the same ecosystem at the same time?
Not quite — the answer is C.
A sparrow eats grains and fruits (primary consumer) as well as insects (secondary consumer), so it occupies two trophic levels at once. A goat is strictly herbivorous and a lion strictly a top carnivore; a frog feeds mainly on insects.
Q4AIPMT 2011 Mains
Which one of the following is essential for photolysis of water ?
Not quite — the answer is C.
Splitting of water in photosystem II is carried out by the oxygen-evolving complex, which needs manganese along with calcium and chloride ions. Copper (in plastocyanin), boron and zinc are not involved in photolysis of water.
Q5AIPMT 2011 Mains
What happens during fertilisation in humans after many sperms reach close to the ovum?
Not quite — the answer is C.
When sperms reach the ovum, the acrosome releases enzymes that help a single sperm penetrate the zona pellucida and enter the ovum; changes in the zona then block other sperms. The corona radiata does not selectively trap sperms, only one (not two) sperm enters, and the other sperms do not lose their tails.
Q6AIPMT 2011 Mains
Bulk of carbon dioxide (CO₂) released from body tissues into the blood is present as
Not quite — the answer is C.
About 70% of CO₂ is carried as bicarbonate, formed mainly in RBCs by carbonic anhydrase and carried largely in plasma; about 20-25% travels as carbamino-haemoglobin and about 7% as dissolved CO₂. The trap reverses the percentages: carbamino-haemoglobin is the minor form, not 70%.
Q7AIPMT 2011 Mains
In mitochondria, protons accumulate in the
Not quite — the answer is A.
During electron transport, complexes in the inner membrane pump protons from the matrix into the intermembrane space, where they build up; their return through ATP synthase drives ATP formation. Protons are pumped out of the matrix, so they do not accumulate there.
Q8AIPMT 2011 Mains
The unequivocal proof of DNA as the genetic material came from the studies on a
Not quite — the answer is B.
Hershey and Chase grew bacteriophages (bacterial viruses) with radioactive phosphorus or sulphur and found that only the labelled DNA entered the bacteria, giving unequivocal proof that DNA is the genetic material. Griffith's and Avery's work on the bacterium Streptococcus was suggestive but not unequivocal.
Q9AIPMT 2011 Mains
Whorled simple leaves with reticulate venation are present in
Not quite — the answer is B.
Alstonia has whorled phyllotaxy (more than two leaves at a node) with simple leaves showing reticulate venation. China rose has alternate phyllotaxy, Calotropis opposite phyllotaxy, and neem has pinnately compound leaves.
Q10AIPMT 2011 Mains
Three of the following pairs of the human skeletal parts are correctly matched with their respective inclusive skeletal category and one pair is not matched. Identify the non-matching pair.
Not quite — the answer is D.
The clavicle and the glenoid cavity (on the scapula) belong to the pectoral girdle, not the pelvic girdle, so this pair is mismatched. Humerus and ulna are limb bones of the appendicular skeleton, malleus and stapes are ear ossicles, and sternum and ribs belong to the axial skeleton.
Q11AIPMT 2011 Mains
What is common between vegetative reproduction and Apomixis ?
Not quite — the answer is B.
Neither vegetative reproduction nor apomixis involves meiosis and fusion of gametes, so both give progeny genetically identical to the parent. Apomixis forms seeds and so does not bypass flowering, and neither process is restricted to dicots.
Q12AIPMT 2011 Mains
Read the following statement having two blanks (A and B): "A drug used for ___(A)___ patients is obtained from a species of the organism ___(B)___." The one correct option for the two blanks is
Not quite — the answer is D.
Cyclosporin A, an immunosuppressant given to organ-transplant patients, is obtained from the fungus Trichoderma polysporum. Monascus yields statins (cholesterol-lowering), and Penicillium yields penicillin, an antibiotic rather than a heart drug.
Q13AIPMT 2011 Mains
Which one of the following conditions of the zygotic cell would lead to the birth of a normal human female child?
Not quite — the answer is C.
A normal human female has 44 autosomes plus two X chromosomes (44 + XX). A single X gives Turner's syndrome, X with Y gives a male, and a zygote with only a Y chromosome is not viable.
Q14AIPMT 2011 Mains
The 24 hour (diurnal) rhythm of our body such as the sleep-wake cycle is regulated by the hormone
Not quite — the answer is B.
Melatonin, secreted by the pineal gland, regulates the 24-hour (diurnal) rhythm including the sleep-wake cycle. Adrenaline is the emergency hormone, calcitonin lowers blood calcium and prolactin stimulates milk production.
Q15AIPMT 2011 Mains
Which one of the following pairs is wrongly matched while the remaining three are correct?
Not quite — the answer is D.
Water hyacinth (Eichhornia) propagates vegetatively by offsets, not runners, so this pair is wrongly matched. Bryophyllum forms buds on leaf margins, Agave forms bulbils and Penicillium reproduces by conidia.
Q16AIPMT 2011 Mains
Which one of the following is not considered as a part of the endomembrane system?
Not quite — the answer is D.
The endomembrane system comprises the ER, Golgi complex, lysosomes and vacuoles, whose functions are coordinated. Peroxisomes, like mitochondria and chloroplasts, work independently of this system and are not counted in it.
Q17AIPMT 2011 Mains
In Kranz anatomy, the bundle sheath cells have
Not quite — the answer is D.
In Kranz anatomy of C₄ leaves, the bundle sheath cells are large, have thick walls, lack intercellular spaces and contain a large number of chloroplasts, where the Calvin cycle runs. Thin walls, intercellular spaces or few chloroplasts describe mesophyll-type cells, not the bundle sheath.
Q18AIPMT 2011 Mains
Sweet potato is homologous to
Not quite — the answer is B.
Sweet potato is a modified (adventitious) root swollen with food, and turnip is also a modified root, so the two are homologous. Ginger and potato are modified stems (rhizome and tuber) and Colocasia has a corm, so they are only analogous to sweet potato.
Q19AIPMT 2011 Mains
Ureters act as urinogenital ducts in
Not quite — the answer is B.
In male frogs the testes connect to the kidneys through vasa efferentia, so the ureter carries both sperms and urine and works as a urinogenital duct. In female frogs the ureter carries only urine, and in humans the ureter never carries gametes.
Q20AIPMT 2011 Mains
Which one of the following is a possibility for most of us in regard to breathing, by making a conscious effort ?
Not quite — the answer is A.
With conscious effort one can breathe by moving the diaphragm alone (abdominal breathing) while keeping the ribs still. The lungs can never be emptied completely because the residual volume always remains, and expired air still contains about 16% oxygen.
Q21AIPMT 2011 Mains
Read the following four statements (A–D) about certain mistakes in two of them: (A) The first transgenic buffalo Rosie produced milk which was human alpha-lactalbumin enriched (B) Restriction enzymes are used in isolation of DNA from other macro-molecules (C) Downstream processing is one of the steps of R-DNA technology (D) Disarmed pathogen vectors are also used in transfer of R-DNA into the host Which are the two statements having mistakes?
Not quite — the answer is B.
Rosie was a transgenic cow, not a buffalo, and restriction enzymes cut DNA at specific sites; isolation of DNA from other macromolecules uses enzymes such as lysozyme, cellulase, ribonuclease and protease. Downstream processing is a recognised step of recombinant DNA technology and disarmed pathogens are used as vectors, so those two statements are correct.
Q22AIPMT 2011 Mains
The pathogen Microsporum responsible for ringworm disease in humans belongs to the same Kingdom of organisms as that of
Not quite — the answer is A.
Microsporum, which causes ringworm, is a fungus, so it shares kingdom Fungi with the mould Rhizopus. Ascaris, Taenia and Wuchereria are worms belonging to kingdom Animalia.
Q23AIPMT 2011 Mains
Which one of the following techniques made it possible to genetically engineer living organisms?
Not quite — the answer is C.
Recombinant DNA technology, cutting and joining DNA with restriction enzymes and ligase and inserting it into a host, made it possible to genetically engineer organisms. Isotope labelling and X-ray diffraction were tools for studying DNA, and hybridization is conventional breeding.
Q24AIPMT 2011 Mains
Biodiversity of a geographical region represents
Not quite — the answer is D.
Biodiversity of a region is the total variety of all organisms living there, at genetic, species and ecosystem levels. Endemic, endangered or dominant species are only parts of that diversity.
Q25AIPMT 2011 Mains
Which one of the following is a wrong matching of a microbe and its industrial product, while the remaining three are correct?
Not quite — the answer is A.
Clostridium butylicum produces butyric acid; lactic acid comes from bacteria such as Lactobacillus, so this pair is wrong. Aspergillus niger produces citric acid, Acetobacter aceti acetic acid, and the yeast Monascus purpureus statins.
Q26AIPMT 2011 Mains
Silencing of mRNA has been used in producing transgenic plants resistant to
Not quite — the answer is D.
RNA interference, in which double-stranded RNA silences the parasite's mRNA, was used to make tobacco plants resistant to the nematode Meloidogyne incognita. Bollworm resistance comes from cry genes (Bt toxin), not mRNA silencing.
Q27AIPMT 2011 Mains
Which one of the following statements is totally wrong about the occurrence of notochord while the other three are correct
Not quite — the answer is A.
All chordates, humans included, have a notochord at some stage; in humans it forms in the embryo and is later replaced by the vertebral column, so saying it is absent from the very beginning is totally wrong. Amphioxus keeps it throughout life, ascidians have it only in the larval tail and in frogs it is replaced by the vertebral column.
Q28AIPMT 2011 Mains
The technique called gamete intrafallopian transfer (GIFT) is recommended for those females
Not quite — the answer is C.
In GIFT an ovum collected from a donor is placed in the fallopian tube of a female who cannot produce ova but can provide the right environment for fertilisation and development. A female who cannot provide a suitable environment for fertilisation would not benefit, because fertilisation in GIFT takes place inside her own tube.
Q29AIPMT 2011 Mains
Common cold is not cured by antibiotics because it is
Not quite — the answer is C.
Common cold is caused by rhinoviruses, and antibiotics act on bacterial structures and processes, so they have no effect on viruses. It is an infectious disease spread by droplets or contact, not a bacterial one.
Q30AIPMT 2011 Mains
Which one of the following options gives the correct matching of a disease with its causative organism and mode of infection?
Not quite — the answer is D.
Pneumonia is caused by Streptococcus pneumoniae (and Haemophilus influenzae) and spreads by droplet infection. Elephantiasis spreads through the bite of the female Culex mosquito, malaria through female (not male) Anopheles, and typhoid through contaminated food and water, not inspired air.
Q31AIPMT 2011 Mains
Frogs differ from humans in possessing
Not quite — the answer is A.
Frog RBCs are nucleated, while mature human RBCs lack a nucleus. Both frogs and humans have thyroid and parathyroid glands, paired cerebral hemispheres and a hepatic portal system.
Q32AIPMT 2011 Mains
Test cross in plants or in Drosophila involves crossing
Not quite — the answer is A.
A test cross crosses an organism showing the dominant phenotype, typically the F₁ hybrid, with the recessive parent; for two genes that is the double recessive genotype. Crossing two F₁ hybrids is a selfing or F₂ cross, not a test cross.
Q33AIPMT 2011 Mains
The logistic population growth is expressed by the equation
Not quite — the answer is D.
Logistic (Verhulst-Pearl) growth is dN/dt = rN[(K − N)/K]: growth slows as N approaches the carrying capacity K. dN/dt = rN is exponential growth, and the other forms invert either the derivative or the term (N − K).
Q34AIPMT 2011 Mains
About which day in a normal human menstrual cycle does rapid secretion of LH (popularly called LH-surge) normally occur?
Not quite — the answer is C.
In a 28-day cycle, LH rises to a peak (the LH surge) around the middle of the cycle, about day 14, and this surge triggers ovulation. Days 5 and 11 fall in the follicular phase and day 20 in the luteal phase.
Q35AIPMT 2011 Mains
In the history of biology, the human genome project led to the development of
Not quite — the answer is A.
The Human Genome Project produced enormous amounts of sequence data, and storing, retrieving and analysing it led to the rapid growth of bioinformatics. Biotechnology existed long before the project and is not its product.
Q36AIPMT 2011 Mains
The type of muscles present in our
Not quite — the answer is A.
Thigh muscles are skeletal muscles: striated and voluntary. Upper arm muscles are also skeletal, not smooth; heart muscle is striated though involuntary; and intestinal muscle is smooth (unstriated) and involuntary.
Q37AIPMT 2011 Mains
Consider the following four statements whether they are correct or wrong: (A) The sporophyte in liverworts is more elaborate than that in mosses (B) Salvinia is heterosporous (C) The life-cycle in all seed bearing plants is diplontic (D) In Pinus male and female cones are borne on different trees The two wrong statements together are
Not quite — the answer is D.
The sporophyte of mosses is more elaborate than that of liverworts, so the liverwort statement is wrong, and Pinus is monoecious with male and female cones on the same tree, so that statement is wrong too. Salvinia is heterosporous and all seed plants have a diplontic life cycle, so those two are correct.
Q38AIPMT 2011 Mains
In angiosperms, functional megaspore develops into
Not quite — the answer is C.
In angiosperms the functional megaspore divides by three free-nuclear mitoses to form the embryo sac, the female gametophyte. The ovule is the structure that contains the megaspore, and endosperm forms only after fertilisation.
Q39AIPMT 2011 Mains
Which one of the following aspects is an exclusive characteristic of living things?
Not quite — the answer is A.
Consciousness, the ability to sense the environment and remember it, is the defining property found only in living things. Increase in mass by surface accumulation also occurs in non-living things such as crystals and mountains, and isolated metabolic reactions in vitro are not living.
Q40AIPMT 2011 Mains
Some vascular bundles are described as open because these
Not quite — the answer is D.
A vascular bundle is open when cambium is present between xylem and phloem, so it can produce secondary xylem and phloem. Conjunctive tissue lies between separate xylem and phloem patches in roots and does not make a bundle open.
Q41AIPMT 2011 Mains
Bacillus thuringiensis forms protein crystals which contain insecticidal protein. This protein:
Not quite — the answer is C.
Bt protoxin is activated by the alkaline pH of the insect's gut; the active toxin binds to midgut epithelial cells, makes pores and kills the insect. The bacterium is not harmed because it keeps the toxin as an inactive protoxin, not because it is resistant, and acid pH does not activate it.
Q42AIPMT 2011 Mains
Which one of the following structures in Pheretima is correctly matched with its function?
Not quite — the answer is C.
The clitellum of Pheretima secretes the cocoon in which the eggs are deposited. Setae help in locomotion, the typhlosole increases the absorptive area of the intestine, and the gizzard grinds food rather than absorbing it.
Q43AIPMT 2011 Mains
Both, hydrarch and xerarch successions lead to
Not quite — the answer is C.
Hydrarch succession moves from water towards drier conditions and xerarch from dry towards wetter conditions, so both end in a mesic (medium water) climax. Neither ends in extreme wet or xeric conditions.
Q44AIPMT 2011 Mains
The breakdown of detritus into smaller particles by earthworm is a process called
Not quite — the answer is D.
Breaking detritus into smaller particles by detritivores such as earthworms is fragmentation. Humification is the formation of humus, and mineralisation is the later release of inorganic nutrients from humus.
Q45AIPMT 2011 Mains
Consider the following statements (A)–(D) each with one or two blanks. (A) Bears go into ___(1)___ during winter to ___(2)___ cold weather (B) A conical age pyramid with a broad base represents ___(3)___ human population (C) A wasp pollinating a fig flower is an example of ___(4)___ (D) An area with high levels of species richness is known as ___(5)___ Which one of the following options gives the correct fill ups for the respective blank numbers from (1) to (5) in the statements?
Not quite — the answer is B.
Bears hibernate to escape the cold winter, a broad-based conical age pyramid indicates an expanding population, and an area with very high species richness is a hotspot. Fig and wasp is mutualism, not commensalism, and a broad base means growth, not stability, so only the choice with hibernation, escape, expanding and hot spot fills its blanks correctly.
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 2011 Mains · Q28
"Good ozone" is found in the
Not quite — the answer is A.
Ozone in the stratosphere absorbs harmful ultraviolet-B radiation from the sun, so it is called 'good ozone'. Ozone in the troposphere, near the ground, is a harmful pollutant ('bad ozone').
Q2AIPMT 2011 Mains · Q29
Guttation is the result of
Not quite — the answer is B.
Guttation, the loss of water droplets from leaf tips and margins through hydathodes at night or early morning, is caused by root pressure when transpiration is low. Transpiration is water loss as vapour, which is the opposite situation.
Q3AIPMT 2011 Mains · Q44
Which one of the following correctly represents the normal adult human dental formula?
Not quite — the answer is A.
Each half of the adult human upper and lower jaw has 2 incisors, 1 canine, 2 premolars and 3 molars, giving 2123/2123 and 32 teeth. Formulas with 3 incisors or 3 premolars per half do not match the human count of 32 teeth.
Q4AIPMT 2011 Mains · Q46
Which one of the following is not an essential mineral element for plants while the remaining three are?
Not quite — the answer is A.
Cadmium is a toxic heavy metal and not an essential element for plants. Phosphorus is a macronutrient, while iron and manganese are essential micronutrients.
Q5AIPMT 2011 Mains · Q54
One of the constituents of the pancreatic juice while poured into the duodenum in humans is
Not quite — the answer is C.
Pancreatic juice enters the duodenum carrying inactive trypsinogen, which is converted to active trypsin by enterokinase secreted by the intestinal mucosa. Trypsin and chymotrypsin are the active forms produced only after this activation, and enterokinase comes from the intestine, not the pancreas.
Chapter-wise split of AIPMT 2011 Mains Biology
- Human Health and Disease3 Q
- Ecosystem3 Q
- Morphology of Flowering Plants3 Q
- Biotechnology and Its Applications3 Q
- Structural Organisation in Animals2 Q
- Photosynthesis in Higher Plants2 Q
- Human Reproduction2 Q
- Breathing and Exchange of Gases2 Q
- Molecular Basis of Inheritance2 Q
- Locomotion and Movement2 Q
- Sexual Reproduction in Flowering Plants2 Q
- Microbes in Human Welfare2 Q
- Principles of Inheritance and Variation2 Q
- Organisms and Populations2 Q
- Respiration in Plants1 Q
- Chemical Coordination and Integration1 Q
- Cell: The Unit of Life1 Q
- Excretory Products and Their Elimination1 Q
- Biological Classification1 Q
- Biotechnology: Principles and Processes1 Q
- Biodiversity and Conservation1 Q
- Animal Kingdom1 Q
- Reproductive Health1 Q
- Body Fluids and Circulation1 Q
- Plant Kingdom1 Q
- The Living World1 Q
- Anatomy of Flowering Plants1 Q
About this paper
How many questions from AIPMT 2011 Mains are on this page?
100 questions (Physics 23, Chemistry 27, Biology 50). 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 2011 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 2011 Mains useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books. 10 questions here are on topics NCERT has since dropped; they are listed separately so you can skip them.
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