AIPMT 2011 Prelims Question PaperPhysics, Chemistry and Biology, with checked answers and solutions
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171 questions from the AIPMT 2011 Prelims 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 Prelims
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AIPMT 2011 Prelims Physics questions
38 questions. Tap an option to check your answer.
Q1AIPMT 2011 Prelims
The dimensions of (μ₀ε₀)^(−1/2) are
Not quite — the answer is D.
Since c = 1/√(μ₀ε₀), the quantity (μ₀ε₀)^(−1/2) is simply the speed of light and has dimensions [LT⁻¹]. The inverse [L⁻¹T] would be the dimensions of √(μ₀ε₀) itself, not its inverse root.
Q2AIPMT 2011 Prelims
The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is I₀. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is
Not quite — the answer is C.
By the parallel axis theorem, I = I₀ + Md², and the end of the rod is d = L/2 from the centre, so I = I₀ + ML²/4. Using the full length L as the shift distance (giving I₀ + ML²) is the usual mistake.
Q3AIPMT 2011 Prelims
A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 m s⁻², the velocity with which it hits the ground is
Not quite — the answer is C.
For a stone dropped from rest, v² = 2gh = 2 × 10 × 20 = 400, so v = 20 m/s. Forgetting the factor 2 under the root gives about 14 m/s, and v = gh would give an unphysical 200 m/s.
Q4AIPMT 2011 Prelims
A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on the control panel. The lift starts moving upwards with an acceleration 1.0 m/s². If g = 10 m s⁻², the tension in the supporting cable is
Not quite — the answer is D.
The cable accelerates the lift and the person together (total mass 1000 kg) upward at 1 m/s², so T − (M + m)g = (M + m)a, giving T = 1000 × (10 + 1) = 11000 N. Using only the lift's mass or ignoring the acceleration gives the smaller values.
Q5AIPMT 2011 Prelims
The instantaneous angular position of a point on a rotating wheel is given by the equation θ(t) = 2t³ − 6t². The torque on the wheel becomes zero at
Not quite — the answer is B.
Torque is zero when angular acceleration is zero: α = d²θ/dt² = 12t − 12 = 0, so t = 1 s. Setting the angular velocity 6t² − 12t to zero instead gives t = 2 s, which is the trap.
Q6AIPMT 2011 Prelims
A particle moves in a circle of radius 5 cm with constant speed and time period 0.2π s. The acceleration of the particle is
Not quite — the answer is A.
ω = 2π/T = 2π/(0.2π) = 10 rad/s, and the centripetal acceleration is a = ω²r = 100 × 0.05 = 5 m/s². Forgetting to convert 5 cm to metres gives a value 100 times too large.
Q7AIPMT 2011 Prelims
A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is
Not quite — the answer is D.
Impulse equals change in momentum: taking the initial direction as positive, Δp = (−MV) − (MV) = −2MV, so its magnitude is 2MV. Zero would only be the change in speed, not in momentum.
Q8AIPMT 2011 Prelims
A planet moving along an elliptical orbit is closest to the sun at a distance r₁ and farthest away at a distance of r₂. If v₁ and v₂ are the linear velocities at these points respectively, then the ratio v₁/v₂ is
Not quite — the answer is C.
Gravity exerts no torque about the sun, so angular momentum is conserved; at perihelion and aphelion the velocity is perpendicular to the radius, so mv₁r₁ = mv₂r₂ and v₁/v₂ = r₂/r₁. The planet moves faster when closer, so the ratio must exceed 1, which rules out r₁/r₂.
Q9AIPMT 2011 Prelims
A radioactive nucleus of mass M emits a photon of frequency ν and the nucleus recoils. The recoil energy will be
Not quite — the answer is C.
The photon carries momentum p = hν/c, and the nucleus recoils with equal and opposite momentum. Its kinetic energy is p²/2M = h²ν²/(2Mc²). It is not hν, which is the photon's own energy.
Q10AIPMT 2011 Prelims
The potential energy of a system increases if work is done
Not quite — the answer is C.
Potential energy is defined only for conservative forces: ΔU = −W(conservative). When the system does work against a conservative force (the conservative force does negative work), its potential energy increases. Work by non-conservative forces does not change potential energy.
Q11AIPMT 2011 Prelims
A body is moving with velocity 30 m/s towards east. After 10 seconds its velocity becomes 40 m/s towards north. The average acceleration of the body is
Not quite — the answer is A.
The velocity changes from 30 m/s east to 40 m/s north; the change Δv has magnitude √(30² + 40²) = 50 m/s. Average acceleration = 50/10 = 5 m/s². Taking the difference of speeds (1 m/s²) ignores the change in direction.
Q12AIPMT 2011 Prelims
A missile is fired for maximum range with an initial velocity of 20 m/s. If g = 10 m/s², the range of the missile is
Not quite — the answer is B.
Maximum range occurs at 45°, R = u²/g = 20²/10 = 40 m. The value 20 m is the range for u²/2g, which is the maximum height formula at 90°, not the range.
Q13AIPMT 2011 Prelims
A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will
Not quite — the answer is D.
By Gauss's law, the total flux through any closed surface is q(enclosed)/ε₀. The enclosed charge is still Q when the radius doubles, so the flux remains the same; the field falls as 1/r² but the area grows as r².
Q14AIPMT 2011 Prelims
A parallel plate condenser has a uniform electric field E (V/m) in the space between the plates. If the distance between the plates is d (m) and area of each plate is A (m²) the energy (joules) stored in the condenser is
Not quite — the answer is A.
The energy density in the field is (1/2)ε₀E² and the field fills a volume Ad, so U = (1/2)ε₀E²Ad. The expression (1/2)ε₀E² alone is just the energy per unit volume, not the total energy.
Q15AIPMT 2011 Prelims
A current of 2 A flows through a 2 Ω resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 Ω resistor. The internal resistance of the battery is
Not quite — the answer is C.
The emf is the same in both cases: ε = 2(2 + r) = 0.5(9 + r). So 4 + 2r = 4.5 + 0.5r, giving 1.5r = 0.5 and r = 1/3 Ω (ε = 4.67 V). Checking 1/3 Ω: 2 × 2.333 = 4.67 and 0.5 × 9.333 = 4.67, consistent.
Q16AIPMT 2011 Prelims
A current carrying closed loop in the form of a right angle isosceles triangle ABC (right angle at B) is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F⃗, the force on the arm AC is
Not quite — the answer is C.
The field is along AB, so the force on arm AB is zero. The net magnetic force on any closed current loop in a uniform field is zero, so F(BC) + F(AC) = 0 and the force on AC is −F⃗. A factor of √2 is the trap from the longer length of AC, but only the component of AC perpendicular to B (equal to BC) matters.
Q17AIPMT 2011 Prelims
The power obtained in a reactor using U-235 disintegration is 1000 kW. The mass decay of U-235 per hour is
Not quite — the answer is D.
Energy produced in one hour = 10⁶ W × 3600 s = 3.6 × 10⁹ J. Mass converted = E/c² = 3.6 × 10⁹/(9 × 10¹⁶) = 4 × 10⁻⁸ kg = 40 μg. Forgetting the factor 3600 or using kW as W gives the smaller options.
Q18AIPMT 2011 Prelims
There are four light-weight-rod samples, A, B, C, D separately suspended by threads. A bar magnet is slowly brought near each sample and the following observations are noted (i) A is feebly repelled (ii) B is feebly attracted (iii) C is strongly attracted (iv) D remains unaffected Which one of the following is true?
Not quite — the answer is B.
Feeble attraction towards a magnet is the signature of a paramagnetic material. Feeble repulsion indicates diamagnetic (so the first sample is diamagnetic, not non-magnetic), strong attraction indicates ferromagnetic, and the unaffected sample is non-magnetic, not ferromagnetic.
Q19AIPMT 2011 Prelims
The electric and the magnetic field, associated with an e.m. wave, propagating along the +z-axis, can be represented by
Not quite — the answer is B.
An EM wave travels along E⃗ × B⃗, with E and B perpendicular to each other and to the propagation direction. With E along î and B along ĵ, î × ĵ = k̂, i.e. +z. E along ĵ with B along î gives ĵ × î = −k̂, the reverse direction.
Q20AIPMT 2011 Prelims
A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron
Not quite — the answer is C.
Since the velocity is parallel to B, the magnetic force qv × B is zero and there is no sideways turning. The electron's charge is negative, so the electric force is opposite to E, i.e. opposite to its velocity, and it decelerates, so its speed decreases.
Q21AIPMT 2011 Prelims
In an ac circuit an alternating voltage e = 200√2 sin 100t volts is connected to a capacitor of capacity 1 µF. The r.m.s. value of the current in the circuit is
Not quite — the answer is A.
V(rms) = 200√2/√2 = 200 V and ω = 100 rad/s, so X_C = 1/(ωC) = 1/(100 × 10⁻⁶) = 10⁴ Ω. I(rms) = 200/10⁴ = 0.02 A = 20 mA. Using the peak voltage 200√2 V would give about 28 mA, which is not the rms value.
Q22AIPMT 2011 Prelims
An ac voltage is applied to a resistance R and inductor L in series. If R and the inductive reactance are both equal to 3 Ω, the phase difference between the applied voltage and the current in the circuit is
Not quite — the answer is C.
For a series LR circuit tan φ = X_L/R = 3/3 = 1, so φ = π/4 with voltage leading the current. π/2 would apply only for a pure inductor.
Q23AIPMT 2011 Prelims
When 1 kg of ice at 0°C melts to water at 0°C, the resulting change in its entropy, taking latent heat of ice to be 80 cal/g, is
Not quite — the answer is A.
Heat absorbed Q = mL = 1000 g × 80 cal/g = 8 × 10⁴ cal at constant T = 273 K. ΔS = Q/T = 80000/273 ≈ 293 cal/K. Giving 8 × 10⁴ cal/K forgets to divide by the temperature.
Q24AIPMT 2011 Prelims
Two waves are represented by the equations y₁ = a sin(ωt + kx + 0.57) m and y₂ = a cos(ωt + kx) m, where x is in metre and t in s. The phase difference between them is
Not quite — the answer is B.
Write y₂ = a cos(ωt + kx) = a sin(ωt + kx + π/2). The phase difference is π/2 − 0.57 = 1.57 − 0.57 = 1.0 radian. 1.57 rad would be the phase of the cosine wave alone, ignoring the 0.57 in the first wave.
Q25AIPMT 2011 Prelims
Out of the following functions representing motion of a particle which represents SHM? (A) y = sin ωt − cos ωt (B) y = sin³ωt (C) y = 5 cos(3π/4 − 3ωt) (D) y = 1 + ωt + ω²t²
Not quite — the answer is D.
sin ωt − cos ωt = √2 sin(ωt − π/4) and 5 cos(3π/4 − 3ωt) are single sinusoids, so both are SHM. sin³ωt = (3 sin ωt − sin 3ωt)/4 is periodic but a mix of two frequencies, so not SHM, and 1 + ωt + ω²t² is not even periodic. So only the first and third functions are SHM; the claim that only the last function fails is wrong because sin³ωt also fails.
Q26AIPMT 2011 Prelims
Sound waves travel at 350 m/s through warm air and at 3500 m/s through brass. The wavelength of a 700 Hz acoustic wave as it enters brass from warm air
Not quite — the answer is D.
Frequency is unchanged when a wave enters a new medium, so λ = v/f is proportional to speed. λ goes from 350/700 = 0.5 m to 3500/700 = 5 m, an increase by a factor of 10. It increases, not decreases, because sound is faster in brass.
Q27AIPMT 2011 Prelims
The decreasing order of wavelength of infrared, microwave, ultraviolet and gamma rays is
Not quite — the answer is B.
In the EM spectrum wavelength falls from microwaves (mm–cm) to infrared (μm) to ultraviolet (~100 nm) to gamma rays (< 0.01 nm). The order starting with gamma rays is increasing, not decreasing, wavelength.
Q28AIPMT 2011 Prelims
The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion is
Not quite — the answer is A.
Lyman first line: 1/λ = R(1 − 1/4) = 3R/4. Balmer second line (n = 4 → 2) for the ion: 1/λ = Z²R(1/4 − 1/16) = 3Z²R/16. Equating gives Z² = 4, Z = 2 (He⁺). Using n = 3 → 2 as the second Balmer line gives a non-integer Z.
Q29AIPMT 2011 Prelims
Which of the following is not due to total internal reflection?
Not quite — the answer is C.
The pond looking shallower than it is comes from ordinary refraction at the water surface, not total internal reflection. Sparkle of diamond, light guidance in optical fibres and mirages all rely on total internal reflection.
Q30AIPMT 2011 Prelims
In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is
Not quite — the answer is D.
The stopping potential satisfies eV₀ = K(max) = 0.5 eV, so V₀ = 0.5 V. The work function is irrelevant once K(max) is given; adding it (2.3 V) gives the photon energy, not the stopping potential.
Q31AIPMT 2011 Prelims
Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 kV then the de-Broglie wavelength associated with the electrons would
Not quite — the answer is C.
For electrons accelerated through V, λ = h/√(2meV), so λ ∝ 1/√V. Increasing V from 25 kV to 100 kV (4 times) reduces λ by √4 = 2 times. 'Decrease by 4 times' wrongly takes λ ∝ 1/V.
Q32AIPMT 2011 Prelims
Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be
Not quite — the answer is C.
K(max) = hν − φ gives 1 − 0.5 = 0.5 eV and 2.5 − 0.5 = 2.0 eV. Since v ∝ √K, v₁ : v₂ = √0.5 : √2 = 1 : 2. Taking the ratio of photon energies (1 : 2.5) or of kinetic energies (1 : 4) are the traps.
Q33AIPMT 2011 Prelims
The half life of a radioactive isotope X is 50 years. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio of 1 : 15 in a sample of a given rock. The age of the rock was estimated to be
Not quite — the answer is C.
If X : Y = 1 : 15, the fraction of X remaining is 1/16 = (1/2)⁴, so four half-lives have passed: 4 × 50 = 200 years. Taking 1/15 as the remaining fraction or counting 3 half-lives gives the wrong ages.
Q34AIPMT 2011 Prelims
Photoelectric emission occurs only when the incident light has more than a certain minimum
Not quite — the answer is A.
Photoelectric emission requires the frequency to exceed the threshold frequency of the metal, whatever the intensity. Intensity or power only changes the number of electrons once emission occurs; a minimum wavelength is the wrong sense since there is a maximum wavelength.
Q35AIPMT 2011 Prelims
Fusion reaction takes place at high temperature because
Not quite — the answer is D.
Fusing nuclei are positively charged and repel each other; only at very high temperatures (~10⁷ K) do they have enough kinetic energy to overcome this Coulomb barrier. Ionisation of atoms or breaking of molecules is not the reason fusion needs high temperature.
Q36AIPMT 2011 Prelims
A nucleus ᵐₙX emits one α-particle and two β⁻-particles. The resulting nucleus is
Not quite — the answer is D.
Alpha emission lowers mass number by 4 and atomic number by 2; each β⁻ raises atomic number by 1 with no mass change. Net: mass number m − 4, atomic number n − 2 + 2 = n, so the product is an isotope of X, ᵐ⁻⁴ₙX.
Q37AIPMT 2011 Prelims
If a small amount of antimony is added to germanium crystal
Not quite — the answer is D.
Antimony is pentavalent; in germanium it donates an extra electron, making an n-type semiconductor with free electrons as majority carriers. It is a donor, not an acceptor, and doping lowers the resistance rather than increasing it.
Q38AIPMT 2011 Prelims
In forward biasing of the p-n junction
Not quite — the answer is A.
In forward bias, the p-side is connected to the positive terminal; this opposes the built-in field and narrows the depletion region, letting current flow. Connecting the positive terminal to the n-side is reverse bias, which widens the depletion layer.
Physics: not in the current NEET syllabus
These 3 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
3 questions. Tap an option to check your answer.
Q1AIPMT 2011 Prelims
The rate of increase of thermo e.m.f. with temperature at the neutral temperature of a thermocouple
Not quite — the answer is C.
Thermo emf rises with temperature, reaches a maximum at the neutral temperature, and then falls. At a maximum the slope dE/dθ is zero. It does not depend on the materials at that point, although the neutral temperature itself does.
Q2AIPMT 2011 Prelims
In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by
Not quite — the answer is B.
The electrons gain kinetic energy eV from the accelerating potential difference between filament and anode, so increasing that potential difference increases their speed. Filament current only changes how many electrons are emitted, not their speed.
Q3AIPMT 2011 Prelims
A transistor is operated in common emitter configuration at VC = 2 V such that a change in the base current from 100 µA to 300 µA produces a change in the collector current from 10 mA to 20 mA. The current gain is
Not quite — the answer is B.
Common-emitter current gain β = ΔI_C/ΔI_B = (20 − 10) mA/(300 − 100) μA = 10 mA/0.2 mA = 50. Using the final values I_C/I_B = 20 mA/300 μA ≈ 67 or initial values 100 is the trap.
Chapter-wise split of AIPMT 2011 Prelims Physics
- Dual Nature of Radiation and Matter5 Q
- Nuclei4 Q
- Electromagnetic Waves3 Q
- Motion in a Plane3 Q
- System of Particles and Rotational Motion2 Q
- Laws of Motion2 Q
- Moving Charges and Magnetism2 Q
- Alternating Current2 Q
- Waves2 Q
- Semiconductor Electronics: Materials, Devices and Simple Circuits2 Q
- Motion in a Straight Line1 Q
- Gravitation1 Q
- Work, Energy and Power1 Q
- Electric Charges and Fields1 Q
- Electrostatic Potential and Capacitance1 Q
- Current Electricity1 Q
- Magnetism and Matter1 Q
- Thermodynamics1 Q
- Oscillations1 Q
- Atoms1 Q
- Ray Optics and Optical Instruments1 Q
AIPMT 2011 Prelims Chemistry questions
35 questions. Tap an option to check your answer.
Q1AIPMT 2011 Prelims
The total number of atomic orbitals in fourth energy level of an atom is
Not quite — the answer is C.
The number of orbitals in shell n is n². For n = 4 (4s, 4p, 4d, 4f: 1 + 3 + 5 + 7) this is 16 orbitals. 32 is the maximum number of electrons (2n²), not orbitals.
Q2AIPMT 2011 Prelims
The electrode potentials for Cu²⁺(aq) + e⁻ → Cu⁺(aq) and Cu⁺(aq) + e⁻ → Cu(s) are +0.15 V and +0.50 V respectively. The value of E°(Cu²⁺/Cu) will be
Not quite — the answer is C.
Potentials are not additive but free energies are: ΔG° = −nFE°. −2F·E° = −1F(0.15) − 1F(0.50), so E°(Cu²⁺/Cu) = 0.65/2 = 0.325 V. Simply adding the potentials gives 0.65 V, the trap.
Q3AIPMT 2011 Prelims
Mole fraction of the solute in a 1.00 molal aqueous solution is
Not quite — the answer is C.
1.00 molal means 1 mol solute per 1000 g water, i.e. 1000/18 = 55.56 mol water. Mole fraction of solute = 1/(1 + 55.56) = 0.0177. Note this is the mole fraction, not the molality, so it must be much smaller than 1; 0.177 is a decimal slip.
Q4AIPMT 2011 Prelims
A buffer solution is prepared in which the concentration of NH₃ is 0.30 M and the concentration of NH₄⁺ is 0.20 M. If the equilibrium constant, Kb for NH₃ equals 1.8 × 10⁻⁵, what is the pH of this solution?
Not quite — the answer is C.
pKb = −log(1.8 × 10⁻⁵) = 4.74. pOH = pKb + log([NH₄⁺]/[NH₃]) = 4.74 + log(0.20/0.30) = 4.74 − 0.18 = 4.57, so pH = 14 − 4.57 = 9.43. Swapping the ratio gives pOH 4.92 and pH 9.08, the trap.
Q5AIPMT 2011 Prelims
Which of the following is correct option for free expansion of an ideal gas under adiabatic condition?
Not quite — the answer is D.
In free expansion into vacuum, no work is done (w = 0) and in adiabatic conditions q = 0. Thus ΔU = 0, and for an ideal gas U depends only on T, so ΔT = 0 as well. A temperature drop would occur only for a real gas or reversible adiabatic expansion.
Q6AIPMT 2011 Prelims
For the reaction N₂(g) + O₂(g) ⇌ 2NO(g), the equilibrium constant is K₁. The equilibrium constant is K₂ for the reaction 2NO(g) + O₂(g) ⇌ 2NO₂(g). What is K for the reaction NO₂(g) ⇌ (1/2)N₂(g) + O₂(g)?
Not quite — the answer is D.
Adding the two given reactions gives N₂ + 2O₂ ⇌ 2NO₂ with K = K₁K₂. The required reaction is the reverse of this halved, so K = (K₁K₂)^(−1/2) = [1/(K₁K₂)]^(1/2). Halving a reaction takes the square root of K; it does not divide K by 2.
Q7AIPMT 2011 Prelims
If the enthalpy change for the transition of liquid water to steam is 30 kJ mol⁻¹ at 27°C, the entropy change for the process would be
Not quite — the answer is A.
At the transition temperature ΔS = ΔH/T = 30000 J mol⁻¹/300 K = 100 J mol⁻¹ K⁻¹. Forgetting to convert kJ to J gives 0.1, the trap.
Q8AIPMT 2011 Prelims
The van't Hoff factor i for a compound which undergoes dissociation in one solvent and association in other solvent is respectively
Not quite — the answer is D.
Dissociation increases the number of solute particles, so i > 1; association decreases it, so i < 1. Reversing these is the trap.
Q9AIPMT 2011 Prelims
Standard electrode potential for Sn⁴⁺/Sn²⁺ couple is +0.15 V and that for the Cr³⁺/Cr couple is −0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be
Not quite — the answer is C.
The couple with higher reduction potential, Sn⁴⁺/Sn²⁺, is the cathode and Cr is the anode. E°cell = 0.15 − (−0.74) = +0.89 V. Multiplying by stoichiometric coefficients (to get 1.83 V) is wrong since E° is intensive.
Q10AIPMT 2011 Prelims
If the E°cell for a given reaction has a negative value, then which of the following gives the correct relationships for the values of ΔG° and Keq?
Not quite — the answer is A.
ΔG° = −nFE°cell, so a negative E°cell makes ΔG° positive. Since ΔG° = −RT ln K, a positive ΔG° means ln K < 0, i.e. K < 1: the reaction is non-spontaneous as written.
Q11AIPMT 2011 Prelims
The freezing point depression constant for water is −1.86°C m⁻¹. If 5.00 g Na₂SO₄ is dissolved in 45.0 g H₂O, the freezing point is changed by −3.82°C. Calculate the van't Hoff factor for Na₂SO₄
Not quite — the answer is C.
Moles of Na₂SO₄ = 5.00/142 = 0.0352 mol; molality = 0.0352/0.0450 kg = 0.782 m. i = ΔTf/(Kf·m) = 3.82/(1.86 × 0.782) = 3.82/1.455 = 2.63. Ideal complete dissociation would give 3, so 3.11 is too high.
Q12AIPMT 2011 Prelims
The energies E₁ and E₂ of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e. λ₁ and λ₂ will be
Not quite — the answer is C.
E = hc/λ, so λ ∝ 1/E. E₂ is twice E₁, so λ₂ = λ₁/2, i.e. λ₁ = 2λ₂. The higher-energy radiation has the shorter wavelength, so λ₁ = λ₂/2 is the reverse.
Q13AIPMT 2011 Prelims
Standard electrode potential of three metals X, Y and Z are –1.2 V, +0.5 V and –3.0 V respectively. The reducing power of these metals will be
Not quite — the answer is D.
The more negative the standard reduction potential, the more readily the metal is oxidised, so the stronger the reducing agent. Order: Z (−3.0 V) > X (−1.2 V) > Y (+0.5 V).
Q14AIPMT 2011 Prelims
Which one of the following statements for the order of a reaction is incorrect?
Not quite — the answer is A.
Order is an experimental quantity and can be zero, fractional or even negative, so the claim that it is always a whole number is incorrect. It is determined experimentally, does not have to follow stoichiometric coefficients, and equals the sum of the powers of concentration terms in the rate law.
Q15AIPMT 2011 Prelims
Enthalpy change for the reaction, 4H(g) → 2H₂(g) is −869.6 kJ. The dissociation energy of H–H bond is
Not quite — the answer is D.
4H(g) → 2H₂(g) forms two H–H bonds and releases 869.6 kJ. Breaking one mole of H–H bonds therefore needs 869.6/2 = +434.8 kJ. Dissociation energy is always positive, so the negative values are wrong.
Q16AIPMT 2011 Prelims
If n = 6, the correct sequence of filling of electrons will be
Not quite — the answer is B.
For n = 6 the Aufbau order is 6s → 4f → 5d → 6p (as in Cs, Ba, La/Ce onwards, Hf, then Tl), i.e. ns → (n−2)f → (n−1)d → np. Filling np before the inner d and f orbitals is the trap.
Q17AIPMT 2011 Prelims
The correct order of increasing bond length of C–H, C–O, C–C and C=C is
Not quite — the answer is B.
Typical bond lengths: C–H ≈ 109 pm, C=C ≈ 134 pm, C–O ≈ 143 pm, C–C ≈ 154 pm. So the increasing order is C–H < C=C < C–O < C–C. A double bond is shorter than any of the carbon single bonds to C or O.
Q18AIPMT 2011 Prelims
Acidified K₂Cr₂O₇ solution turns green when Na₂SO₃ is added to it. This is due to the formation of
Not quite — the answer is B.
Sulphite reduces dichromate in acid: Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O. The green colour is due to Cr³⁺, i.e. Cr₂(SO₄)₃. Chromate is yellow and Cr would not be reduced to +2.
Q19AIPMT 2011 Prelims
Which of the following is least likely to behave as Lewis base?
Not quite — the answer is D.
A Lewis base donates an electron pair. OH⁻, H₂O and NH₃ all have lone pairs, whereas BF₃ has an incomplete octet on boron and accepts a pair, so it is a Lewis acid and least likely to act as a base.
Q20AIPMT 2011 Prelims
The complex [Pt(Py)(NH₃)BrCl] will have how many geometrical isomers?
Not quite — the answer is B.
[Pt(py)(NH₃)BrCl] is square planar of type [Mabcd]. Fixing one ligand, any of the other three can be trans to it, giving 3 geometrical isomers. Two isomers is the answer only for [Ma₂b₂]-type complexes.
Q21AIPMT 2011 Prelims
The complexes [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆] are the examples of which type of isomerism?
Not quite — the answer is D.
Both compounds have the same formula, but the ligands are exchanged between the cobalt and chromium centres of the cationic and anionic complexes. This is coordination isomerism. Ionisation isomerism would involve exchange of an ion between the coordination sphere and the counter-ion.
Q22AIPMT 2011 Prelims
The d-electron configurations of Cr²⁺, Mn²⁺, Fe²⁺ and Co²⁺ are d⁴, d⁵, d⁶ and d⁷ respectively. Which one of the following will exhibit minimum paramagnetic behaviour? (At. nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)
Not quite — the answer is D.
H₂O is a weak-field ligand, so all are high-spin: d⁴ (Cr²⁺) has 4 unpaired electrons, d⁵ (Mn²⁺) 5, d⁶ (Fe²⁺) 4 and d⁷ (Co²⁺) 3. Fewest unpaired electrons means least paramagnetism, so the cobalt complex. The manganese complex is the most paramagnetic.
Q23AIPMT 2011 Prelims
Of the following complex ions, which is diamagnetic in nature?
Not quite — the answer is C.
Ni²⁺ is d⁸; strong-field CN⁻ pairs all electrons and gives a dsp² square planar complex with no unpaired electrons, so [Ni(CN)₄]²⁻ is diamagnetic. [NiCl₄]²⁻ is tetrahedral sp³ with 2 unpaired electrons, [CoF₆]³⁻ has 4 and [CuCl₄]²⁻ (d⁹) has 1.
Q24AIPMT 2011 Prelims
Which of the following has the minimum bond length?
Not quite — the answer is B.
MO bond orders: O₂⁺ 2.5, O₂ 2, O₂⁻ 1.5, O₂²⁻ 1. Bond length decreases as bond order increases, so O₂⁺ has the shortest bond. Adding electrons fills π* orbitals and lengthens the bond.
Q25AIPMT 2011 Prelims
The value of ΔH for the reaction X₂(g) + 4Y₂(g) ⇌ 2XY₄(g) is less than zero. Formation of XY₄(g) will be favoured at
Not quite — the answer is A.
The forward reaction reduces gas moles from 5 to 2, so high pressure shifts it forward (Le Chatelier). It is exothermic (ΔH < 0), so low temperature favours products. Hence high pressure and low temperature.
Q26AIPMT 2011 Prelims
What is the product obtained in the following reaction? C₆H₅NO₂ —(Zn, NH₄Cl)→ ?
Not quite — the answer is B.
Nitrobenzene in neutral medium (Zn dust with NH₄Cl) is reduced only as far as N-phenylhydroxylamine, C₆H₅NHOH. Aniline forms in acidic medium (Sn/HCl or Fe/HCl), and azo/azoxy compounds in alkaline reductions.
Q27AIPMT 2011 Prelims
In a set of reactions m-bromobenzoic acid (A) gave a product D: A —(SOCl₂)→ B —(NH₃)→ C —(NaOH, Br₂)→ D Identify the product D.
Not quite — the answer is D.
SOCl₂ converts the acid to m-bromobenzoyl chloride, NH₃ converts it to m-bromobenzamide, and Br₂/NaOH (Hoffmann bromamide degradation) removes the carbonyl carbon to give m-bromoaniline. The amide is only the intermediate C, not the final product.
Q28AIPMT 2011 Prelims
In Dumas' method of estimation of nitrogen 0.35 g of an organic compound gave 55 mL of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be (Aqueous tension at 300 K = 15 mm)
Not quite — the answer is C.
Pressure of dry N₂ = 715 − 15 = 700 mm. Volume at STP = 55 × (700/760) × (273/300) = 46.1 mL. Mass of N₂ = 28 × 46.1/22400 = 0.0576 g, so %N = 0.0576/0.35 × 100 ≈ 16.5 (16.45). Forgetting the aqueous tension correction gives a higher value.
Q29AIPMT 2011 Prelims
Which one is a nucleophilic substitution reaction among the following?
Not quite — the answer is D.
In the reaction of the primary alkyl bromide with NH₃, the nucleophile NH₃ replaces Br⁻, which is nucleophilic substitution. The cyanohydrin and Grignard reactions are nucleophilic additions to C=O, and acid-catalysed hydration of propene is electrophilic addition.
Q30AIPMT 2011 Prelims
Considering the state of hybridization of carbon atoms, find out the molecule among the following which is linear?
Not quite — the answer is C.
In CH₃–C≡C–CH₃ both triple-bonded carbons are sp hybridised (180°), so all four carbons lie on a straight line. Butane and but-2-ene contain sp³ or sp² carbons that bend the chain, and pent-1-en-4-yne has sp³ and sp² centres.
Q31AIPMT 2011 Prelims
In the following reactions, (a) CH₃–CH(CH₃)–CH(OH)–CH₃ —(H⁺/heat)→ A (major product) + B (minor product) (b) A —(HBr, dark, in absence of peroxide)→ C (major product) + D (minor product) The major products (A) and (C) are respectively
Not quite — the answer is C.
Protonation and loss of water from 3-methylbutan-2-ol gives a secondary carbocation, which undergoes a 1,2-hydride shift to the tertiary cation; loss of H⁺ gives the more substituted alkene 2-methylbut-2-ene (Saytzeff). HBr adds by Markovnikov's rule without peroxide, giving 2-bromo-2-methylbutane. The 2-methylbut-1-ene or anti-Markovnikov products are the minor/trap options.
Q32AIPMT 2011 Prelims
The Lassaigne's extract is boiled with conc. HNO₃ while testing for halogens. By doing so it
Not quite — the answer is B.
If N or S is present, Na₂S and NaCN are in the Lassaigne extract and would give black Ag₂S or white AgCN with AgNO₃, interfering with the halide test. Boiling with HNO₃ decomposes them (releasing H₂S and HCN). It does not change the solubility product of AgCl.
Q33AIPMT 2011 Prelims · Q81
Name the type of the structure of silicate in which one oxygen atom of [SiO₄]⁴⁻ is shared?
Not quite — the answer is D.
Sharing one oxygen atom between two SiO₄⁴⁻ tetrahedra gives the Si₂O₇⁶⁻ unit, the pyrosilicate. Chains need two shared oxygens per tetrahedron, sheets three and three-dimensional networks all four.
Q34AIPMT 2011 Prelims
Clemmensen reduction of a ketone is carried out in the presence of which of the following?
Not quite — the answer is C.
Clemmensen reduction converts C=O to CH₂ using zinc amalgam and concentrated HCl. Hydrazine with KOH in ethylene glycol is the Wolff–Kishner reduction, and LiAlH₄ reduces ketones only to alcohols.
Q35AIPMT 2011 Prelims
Which one of the following statements is not true regarding (+) Lactose?
Not quite — the answer is D.
Lactose has a free hemiacetal group on its glucose unit, so it is a reducing sugar and does show mutarotation; the statement that it does not mutarotate is false. It has 8 –OH groups, is a β-glycoside of D-galactose and D-glucose, and hydrolyses to equal amounts of the two.
Chemistry: not in the current NEET syllabus
These 9 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
9 questions. Tap an option to check your answer.
Q1AIPMT 2011 Prelims · Q54
By what factors does the average velocity of a gaseous molecule increase when the temperature (in Kelvin) is doubled?
Not quite — the answer is A.
Average speed is proportional to √T (v = √(8RT/πM)). Doubling the Kelvin temperature increases it by √2 ≈ 1.4 times. A factor 2 would apply to kinetic energy, not speed.
Q2AIPMT 2011 Prelims · Q56
Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. Molecular mass of B will be
Not quite — the answer is C.
Time to diffuse equal volumes is proportional to √M (Graham's law). tA/tB = 20/10 = 2 = √(49/MB), so MB = 49/4 = 12.25 u. Taking the time ratio directly instead of its square gives 24.5 u.
Q3AIPMT 2011 Prelims · Q59
If x is the amount of adsorbate and m is the amount of adsorbent, which of the following relations is not related to adsorption process?
Not quite — the answer is A.
Adsorption is described by isotherms (x/m as a function of p at constant T), isobars (x/m against T at constant p) and isosteres (p against T at constant x/m). A simple product x/m = p × T is not an adsorption relation.
Q4AIPMT 2011 Prelims · Q63
A gaseous mixture was prepared by taking equal mole of CO and N₂. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N₂) in the mixture is
Not quite — the answer is B.
By Dalton's law, partial pressure = mole fraction × total pressure. Equal moles of CO and N₂ make the mole fraction of N₂ 0.5, so p(N₂) = 0.5 × 1 = 0.5 atm. Their equal molar masses do not change this.
Q5AIPMT 2011 Prelims · Q72
Which of the following pairs of metals is purified by van Arkel method?
Not quite — the answer is C.
In the van Arkel method, the impure metal forms a volatile iodide (e.g. ZrI₄, TiI₄) which is decomposed on a hot tungsten filament to give very pure metal. It is used for zirconium and titanium; nickel is refined by the Mond process instead.
Q6AIPMT 2011 Prelims · Q77
Which of the following elements is present as the impurity to the maximum extent in the pig iron?
Not quite — the answer is C.
Pig iron from the blast furnace contains about 4% carbon, far more than its silicon, manganese or phosphorus impurities. That high carbon content is what makes it hard and brittle.
Q7AIPMT 2011 Prelims · Q87
Of the following which one is classified as polyester polymer?
Not quite — the answer is B.
Terylene (Dacron) is made from ethylene glycol and terephthalic acid, linked by ester bonds, so it is a polyester. Nylon-66 is a polyamide, while Bakelite and melamine are formaldehyde resins.
Q8AIPMT 2011 Prelims · Q98
Which one of the following is employed as Antihistamine?
Not quite — the answer is C.
Diphenhydramine (Benadryl) blocks histamine receptors and is an antihistamine. Omeprazole is an antacid (proton pump inhibitor), chloramphenicol is an antibiotic and norethindrone is an antifertility drug.
Q9AIPMT 2011 Prelims · Q100
Which one of the following statements is not true?
Not quite — the answer is C.
Fish need dissolved oxygen of about 6 ppm or more; DO below 6 ppm suppresses their growth, so calling it good is false. Clean water has BOD below 5 ppm, and oxides of S, N and C are the most widespread air pollutants.
Chapter-wise split of AIPMT 2011 Prelims Chemistry
- Electrochemistry4 Q
- Coordination Compounds4 Q
- Structure of Atom3 Q
- Solutions3 Q
- Equilibrium3 Q
- Thermodynamics3 Q
- Chemical Bonding and Molecular Structure3 Q
- Amines2 Q
- Organic Chemistry: Some Basic Principles and Techniques2 Q
- Hydrocarbons2 Q
- Chemical Kinetics1 Q
- The d- and f-Block Elements1 Q
- Haloalkanes and Haloarenes1 Q
- Aldehydes, Ketones and Carboxylic Acids1 Q
- Biomolecules1 Q
AIPMT 2011 Prelims Biology questions
71 questions. Tap an option to check your answer.
Q1AIPMT 2011 Prelims
The "eyes" of the potato tuber are
Not quite — the answer is A.
The potato tuber is an underground stem, and each 'eye' is a node carrying an axillary bud in the axil of a scale leaf. These buds sprout into new shoots, which is why potato pieces with eyes are used for propagation. They are not root buds, since roots do not form tubers with nodes.
Q2AIPMT 2011 Prelims
Organisms called Methanogens are most abundant in a
Not quite — the answer is C.
Methanogens are strictly anaerobic archaebacteria found in marshy areas and in the rumen of cattle, where they produce methane from cellulose breakdown products. Cattle dung is therefore rich in them, which is why it is used for biogas. Hot springs harbour thermoacidophiles, a different archaebacterial group.
Q3AIPMT 2011 Prelims
Which one of the following have the highest number of species in nature?
Not quite — the answer is C.
Insects account for more than 70 percent of all recorded animal species, far more than any other group. Angiosperms and fungi have far fewer described species, and birds number only around ten thousand species.
Q4AIPMT 2011 Prelims
Archegoniophore is present in
Not quite — the answer is B.
Marchantia, a liverwort, bears its sex organs on stalked branches: archegonia on archegoniophores and antheridia on antheridiophores. Funaria is a moss whose sex organs sit at branch tips without such stalks, Chara is an alga and Adiantum is a fern.
Q5AIPMT 2011 Prelims
Compared with the gametophytes of the bryophytes the gametophytes of vascular plants tend to be
Not quite — the answer is A.
Moving from bryophytes to vascular plants, the gametophyte becomes progressively reduced while the sporophyte becomes dominant. The fern prothallus is small and short-lived, and in seed plants the gametophyte is just a few cells, with sex organs reduced or absent. So both the gametophyte and its sex organs become smaller.
Q6AIPMT 2011 Prelims
The gametophyte is not an independent, free-living generation in
Not quite — the answer is A.
In gymnosperms such as Pinus, the male and female gametophytes develop within the pollen grain and ovule and stay dependent on the sporophyte. Mosses like Polytrichum, liverworts like Marchantia and ferns like Adiantum all have free-living gametophytes.
Q7AIPMT 2011 Prelims
Important site for formation of glycoproteins and glycolipids is
Not quite — the answer is C.
The Golgi apparatus packages and modifies proteins and lipids coming from the ER, and it is the main site where glycoproteins and glycolipids are formed. Lysosomes are digestive vesicles budded from the Golgi, not the site of glycosylation.
Q8AIPMT 2011 Prelims
Peptide synthesis inside a cell takes place in
Not quite — the answer is A.
Peptide bonds are formed on ribosomes during translation. Chloroplasts and mitochondria do contain their own ribosomes, but the synthesis there still happens on those ribosomes, and chromoplasts are pigment-storing plastids.
Q9AIPMT 2011 Prelims
In eubacteria, a cellular component that resembles eukaryotic cell is
Not quite — the answer is B.
The bacterial plasma membrane is a lipid bilayer with proteins, structurally similar to the eukaryotic membrane. The bacterial cell wall is made of peptidoglycan, there is no true nucleus (only a nucleoid), and bacterial ribosomes are 70S rather than the 80S of the eukaryotic cytoplasm.
Q10AIPMT 2011 Prelims
Which one of the following also acts as a catalyst in a bacterial cell?
Not quite — the answer is A.
In bacteria, the 23S rRNA of the large ribosomal subunit acts as a ribozyme, catalysing peptide bond formation (peptidyl transferase activity). snRNA and hnRNA are eukaryotic nuclear RNAs, and 5S rRNA is structural.
Q11AIPMT 2011 Prelims
Which one of the following statements is correct?
Not quite — the answer is A.
A flower is a modified shoot in which the axis is condensed and the floral parts are modified leaves, and the tulip flower is no exception. Tomato fruit is a berry, orchid seeds are minute and lack endosperm, and primrose shows free central placentation, not basal.
Q12AIPMT 2011 Prelims
A drupe develops in
Not quite — the answer is B.
A drupe develops from a monocarpellary superior ovary and has a stony endocarp, as in mango and coconut. Tomato is a berry, wheat is a caryopsis and pea is a legume.
Q13AIPMT 2011 Prelims
Ground tissue includes
Not quite — the answer is C.
The ground tissue system comprises all tissues except the epidermis and the vascular bundles, so it includes cortex, pericycle, pith and medullary rays in a dicot stem, and mesophyll in leaves. It is not limited to tissues on one side of the endodermis.
Q14AIPMT 2011 Prelims
In land plants the guard cells differ from other epidermal cells in having
Not quite — the answer is A.
Guard cells contain chloroplasts, whereas ordinary epidermal cells generally lack them. Mitochondria, ER and the cytoskeleton are present in all living epidermal cells, so they do not distinguish guard cells.
Q15AIPMT 2011 Prelims
The ovary is half inferior in flowers of
Not quite — the answer is B.
In perigynous flowers such as peach, plum and rose, the thalamus forms a cup around the ovary at about its level, so the ovary is described as half inferior. Guava and cucumber have epigynous flowers with an inferior ovary, and cotton has a superior ovary.
Q16AIPMT 2011 Prelims
The cork cambium, cork and secondary cortex are collectively called
Not quite — the answer is D.
During secondary growth the phellogen (cork cambium), phellem (cork) and phelloderm (secondary cortex) together make up the periderm. Phellem, phelloderm and phellogen are each just one component of it.
Q17AIPMT 2011 Prelims
Which one of the following is wrongly matched?
Not quite — the answer is C.
Puccinia causes rust disease of wheat; smuts are caused by Ustilago, so this pair is wrongly matched. Cassia does show imbricate aestivation, guttation is caused by root pressure, and roots have exarch protoxylem.
Q18AIPMT 2011 Prelims
Flowers are zygomorphic in
Not quite — the answer is C.
Gulmohur (and Cassia) have zygomorphic flowers, which can be divided into two equal halves in only one plane. Datura, mustard and tomato have actinomorphic flowers that can be divided into equal halves in any radial plane.
Q19AIPMT 2011 Prelims
CAM helps the plants in
Not quite — the answer is B.
CAM plants open their stomata at night to take up CO₂, which is stored as organic acids and used for photosynthesis in the day while the stomata stay closed. This cuts transpiration loss sharply, so CAM is an adaptation for conserving water in dry habitats.
Q20AIPMT 2011 Prelims
Of the total incident solar radiation the proportion of PAR is
Not quite — the answer is D.
Of the total incident solar radiation, less than 50 percent is photosynthetically active radiation (PAR, 400–700 nm). Plants capture only about 2–10 percent of this PAR.
Q21AIPMT 2011 Prelims
A prokaryotic autotrophic nitrogen fixing symbiont is found in
Not quite — the answer is C.
The coralloid roots of Cycas harbour cyanobacteria such as Anabaena, which are prokaryotic, photoautotrophic and fix nitrogen. Pisum, Cicer and Alnus have nitrogen-fixing symbionts too (Rhizobium, Frankia), but these are not autotrophic.
Q22AIPMT 2011 Prelims
Nucellar polyembryony is reported in species of
Not quite — the answer is B.
In many Citrus species, nucellar cells surrounding the embryo sac divide and grow into extra embryos, so seeds often contain several embryos. This is nucellar polyembryony, a form of apomixis.
Q23AIPMT 2011 Prelims
Filiform apparatus is a characteristic feature of
Not quite — the answer is D.
The filiform apparatus is a set of finger-like wall thickenings at the micropylar tip of the synergids, which guides the pollen tube into the embryo sac. It is not found in the egg, zygote or suspensor.
Q24AIPMT 2011 Prelims
What would be the number of chromosomes of the aleurone cells of a plant with 42 chromosomes in its root tip cells?
Not quite — the answer is C.
Root tip cells are diploid, so 2n = 42 and n = 21. The aleurone layer is the outer layer of the endosperm, which is triploid: 3n = 3 × 21 = 63. 42 is the diploid count and would be the trap.
Q25AIPMT 2011 Prelims
Wind pollination is common in
Not quite — the answer is D.
Wind pollination is common in grasses, which have light non-sticky pollen, well-exposed stamens and large feathery stigmas. Orchids, legumes and lilies have showy flowers pollinated by insects.
Q26AIPMT 2011 Prelims
In which one of the following pollination is autogamous?
Not quite — the answer is A.
Cleistogamous flowers never open, so pollen from the anthers can only reach the stigma of the same flower: pollination is always autogamous. Geitonogamy and xenogamy involve transfer between flowers, and chasmogamous flowers are open and may be cross-pollinated.
Q27AIPMT 2011 Prelims
Mass of living matter at a trophic level in an area at any time is called
Not quite — the answer is B.
Each trophic level holds a certain mass of living material at a given time, called the standing crop, measured as biomass or number per unit area. Standing state refers to the amount of nutrients in the soil, and detritus and humus are dead organic matter.
Q28AIPMT 2011 Prelims
Secondary sewage treatment is mainly a
Not quite — the answer is A.
Secondary treatment of sewage is a biological process: the primary effluent is aerated so that aerobic microbes form flocs and consume the organic matter, reducing the BOD. Primary treatment, by contrast, is a physical removal of debris by filtration and sedimentation.
Q29AIPMT 2011 Prelims
Which one of the following is not a biofertilizer?
Not quite — the answer is B.
Agrobacterium tumefaciens is a plant pathogen that causes crown gall and is used as a vector in genetic engineering, not as a biofertiliser. Rhizobium and Nostoc fix nitrogen, and mycorrhiza helps phosphorus uptake.
Q30AIPMT 2011 Prelims
Which of the following acts as a physiological barrier to the entry of microorganisms in human body?
Not quite — the answer is C.
Physiological barriers to infection include acid in the stomach, saliva in the mouth and tears from the eyes, which prevent microbial growth. Skin and the mucus-coated epithelium of the urogenital tract are physical barriers, and monocytes are cellular barriers.
Q31AIPMT 2011 Prelims
Which one of the following helps in absorption of phosphorus from soil by plants?
Not quite — the answer is B.
Fungi of the genus Glomus form mycorrhiza with plant roots and help absorb phosphorus from the soil. Anabaena, Rhizobium and Frankia are nitrogen fixers.
Q32AIPMT 2011 Prelims
Which of the following is mainly produced by the activity of anaerobic bacteria on sewage?
Not quite — the answer is A.
In sludge digesters, anaerobic bacteria (including methanogens) break down sewage and give off a gas mixture rich in methane, also called marsh gas or biogas. Laughing gas is nitrous oxide, and mustard gas is a chemical weapon.
Q33AIPMT 2011 Prelims
Agarose extracted from sea weeds finds use in
Not quite — the answer is A.
Agarose, a polysaccharide extracted from seaweeds, forms the gel matrix used in gel electrophoresis to separate DNA fragments by size. It plays no matrix role in spectrophotometry or PCR.
Q34AIPMT 2011 Prelims
Maximum number of existing transgenic animals is of
Not quite — the answer is C.
Although transgenic rats, rabbits, pigs, sheep, cows and fish have been produced, more than 95 percent of all existing transgenic animals are mice.
Q35AIPMT 2011 Prelims
The process of RNA interference has been used in the development of plants resistant to
Not quite — the answer is B.
Using Agrobacterium vectors, nematode-specific genes were introduced into tobacco so that the plant makes dsRNA that silences a vital mRNA of the nematode Meloidogyne incognita. The parasite then cannot survive in the transgenic host, giving nematode-resistant plants.
Q36AIPMT 2011 Prelims
Which one of the following organisms is not an example of eukaryotic cells
Not quite — the answer is C.
Escherichia coli is a bacterium and so has a prokaryotic cell without a nuclear membrane. Amoeba, Paramecium and Euglena are protists with true nuclei, so they are eukaryotes.
Q37AIPMT 2011 Prelims
Which one of the following animals is correctly matched with its particular named taxonomic category?
Not quite — the answer is B.
In Panthera tigris, the tiger, 'tigris' is the specific epithet that denotes the species. Musca is the genus of the housefly, Mollusca is a phylum, and Primata is an order, not a family.
Q38AIPMT 2011 Prelims
Which of the following groups of animals is correctly matched with its one characteristic feature without even a single exception?
Not quite — the answer is D.
Every member of Chondrichthyes (sharks, rays) has a cartilaginous endoskeleton, with no exceptions. Not all mammals give birth (platypus lays eggs), crocodiles have a four-chambered heart, and jawless cyclostomes are chordates without jaws.
Q39AIPMT 2011 Prelims
What will you look for to identify the sex of the following?
Not quite — the answer is A.
Male sharks bear claspers on their pelvic fins, used to transfer sperm. In Ascaris it is the male whose posterior end is sharply curved, the copulatory pad of male frogs is on the first digit of the fore limb, and anal cerci are present in both sexes of cockroach (anal styles are male-only).
Q40AIPMT 2011 Prelims
The ciliated columnar epithelial cells in humans are known to occur in
Not quite — the answer is C.
Ciliated columnar epithelium lines hollow organs such as the bronchioles and the Fallopian tubes, where the cilia move mucus or the ovum along. The stomach is lined by non-ciliated columnar epithelium and the oesophagus by stratified squamous epithelium.
Q41AIPMT 2011 Prelims
Select the correct option with respect to mitosis
Not quite — the answer is A.
In metaphase the chromosomes move to the spindle equator and line up along the metaphase plate. Chromatids separate and move to opposite poles in anaphase, not telophase, and the Golgi complex and ER are no longer visible at the end of prophase.
Q42AIPMT 2011 Prelims
What was the most significant trend in the evolution of modern man (Homo sapiens) from his ancestors?
Not quite — the answer is A.
The most significant trend in the evolution of modern humans was the steady increase in brain capacity, from about 650–800 cc in Homo habilis to around 1,400 cc in Homo sapiens. Upright posture appeared much earlier, in the australopithecines.
Q43AIPMT 2011 Prelims
Which one of the following conditions correctly describes the manner of determining the sex in the given example?
Not quite — the answer is C.
In grasshoppers, males have only one X chromosome (XO) and females have XX, so the XO condition determines the male sex. In Drosophila XX is female, in birds ZZ is male, and in humans sex is decided by the presence or absence of the Y chromosome.
Q44AIPMT 2011 Prelims
A person with unknown blood group under ABO system, has suffered much blood loss in an accident and needs immediate blood transfusion. His one friend who has a valid certificate of his own blood type, offers for blood donation without delay. What would have been the type of blood group of the donor friend?
Not quite — the answer is D.
Blood group O red cells carry neither the A nor the B antigen, so they are not agglutinated by the recipient's antibodies: O is the universal donor. Since the patient's group is unknown, only O blood can safely be given without delay; AB is the universal recipient, not donor.
Q45AIPMT 2011 Prelims
What are those structures that appear as 'beads-on-string' in the chromosomes when viewed under electron microscope?
Not quite — the answer is D.
DNA wrapped around histone octamers forms nucleosomes, which repeat along the chromatin fibre and look like beads on a string under the electron microscope. Base pairs, nucleotides and genes are far too small or not visible as separate beads.
Q46AIPMT 2011 Prelims
Match the source gland with its respective hormone as well as the function
Not quite — the answer is C.
Vasopressin (ADH) is released from the posterior pituitary and acts on the distal tubule and collecting duct to increase water reabsorption. Calcium is regulated by thyrocalcitonin and parathyroid hormone, oxytocin is released from the posterior pituitary, and the corpus luteum supports pregnancy mainly through progesterone.
Q47AIPMT 2011 Prelims
A large proportion of oxygen is left unused in the human blood even after its uptake by the body tissues. This O₂
Not quite — the answer is B.
Each 100 mL of oxygenated blood delivers only about 5 mL of O₂ to the tissues at rest, so venous blood remains largely saturated. This unused oxygen acts as a reserve that can be released when muscular activity increases the demand.
Q48AIPMT 2011 Prelims
Which one of the following is not a part of a renal pyramid?
Not quite — the answer is C.
The convoluted tubules (PCT and DCT), along with Malpighian corpuscles, lie in the cortex of the kidney. Loops of Henle, collecting ducts and their capillary network dip into the medullary pyramids.
Q49AIPMT 2011 Prelims
Two friends are eating together at a dining table. One of them suddenly starts coughing while swallowing some food. This coughing would have been due to improper movement of
Not quite — the answer is B.
During swallowing, the epiglottis closes the glottis so that food passes into the oesophagus and not the larynx. If it does not close properly, food enters the air passage and triggers a coughing reflex.
Q50AIPMT 2011 Prelims
Arteries are best defined as the vessels which
Not quite — the answer is C.
An artery is defined by the direction of flow: it carries blood away from the heart to the organs. Not all arteries carry oxygenated blood (the pulmonary artery carries deoxygenated blood), so oxygen content does not define them.
Q51AIPMT 2011 Prelims
'Bundle of His' is a part of which one of the following organs in humans?
Not quite — the answer is C.
The bundle of His (AV bundle) is part of the conducting system of the heart: it arises from the AV node, runs through the interventricular septum and divides into right and left bundle branches.
Q52AIPMT 2011 Prelims
Which of the following plasma proteins is involved in the coagulation of blood?
Not quite — the answer is A.
Fibrinogen is the plasma protein converted by thrombin into fibrin threads, which form the network of a clot. Albumins maintain osmotic balance and globulins serve in defence; amylase is an enzyme, not a clotting factor.
Q53AIPMT 2011 Prelims
When a neuron is in resting state i.e. not conducting any impulse, the axonal membrane is
Not quite — the answer is A.
At rest the axonal membrane is far more permeable to K⁺ ions than to Na⁺ ions and nearly impermeable to negatively charged proteins. With the Na⁺/K⁺ pump, this keeps the inside negative, giving the resting potential.
Q54AIPMT 2011 Prelims
Which one of the following correctly explains the function of a specific part of a human nephron?
Not quite — the answer is B.
Podocytes of Bowman's capsule are arranged so that minute filtration slits (slit pores) are left between them, through which blood is filtered into the capsule. The afferent arteriole brings blood to the glomerulus, most reabsorption happens in the PCT, and the DCT secretes K⁺ rather than reabsorbing it.
Q55AIPMT 2011 Prelims
Given below is an incomplete table about certain hormones, their source glands and one major effect of each on the body in humans. Identify the correct option for the three blanks A, B and C. Gland A – Oestrogen – Maintenance of secondary sexual characters Alpha cells of Islets of Langerhans – B – Raises blood sugar level Anterior pituitary – C – Over secretion leads to gigantism
Not quite — the answer is B.
Oestrogen that maintains female secondary sexual characters comes from the ovary, the α-cells of the islets of Langerhans secrete glucagon which raises blood sugar, and over-secretion of growth hormone from the anterior pituitary causes gigantism. Insulin lowers blood sugar, and calcitonin is from the thyroid.
Q56AIPMT 2011 Prelims
Uricotelic mode of passing out nitrogenous wastes is found in
Not quite — the answer is B.
Reptiles, birds, land snails and insects excrete nitrogen mainly as uric acid, which conserves water. Amphibians are ureotelic as adults, and annelids are mostly ammonotelic.
Q57AIPMT 2011 Prelims
Which one of the following statements is correct regarding blood pressure?
Not quite — the answer is A.
Normal blood pressure is about 120/80 mmHg, and repeated readings of 140/90 or higher indicate hypertension. A pressure of 190/110 mmHg is severe and can damage vital organs such as the brain and kidneys. Readings like 100/55 or 105/50 are on the low side, not ideal or energising.
Q58AIPMT 2011 Prelims
Which one of the following statements is correct with respect to kidney function regulation?
Not quite — the answer is B.
Drinking a lot of water dilutes the body fluids, so the hypothalamic osmoreceptors suppress ADH release and dilute urine is passed. Water loss in summer would stimulate ADH, and it is a fall in glomerular blood flow that triggers renin and angiotensin II formation.
Q59AIPMT 2011 Prelims
The testes in humans are situated outside the abdominal cavity inside a pouch called scrotum. The purpose served is for
Not quite — the answer is B.
The scrotum keeps the testes about 2–2.5 °C below the internal body temperature, which is necessary for spermatogenesis. Testes that stay in the abdomen fail to produce sperm.
Q60AIPMT 2011 Prelims
If for some reason, the vasa efferentia in the human reproductive system get blocked, the gametes will not be transported from
Not quite — the answer is B.
Sperms leave the seminiferous tubules through the rete testis and pass via the vasa efferentia into the epididymis. If the vasa efferentia are blocked, sperms cannot move from the testes to the epididymis.
Q61AIPMT 2011 Prelims
Medical Termination of Pregnancy (MTP) is considered safe up to how many weeks of pregnancy?
Not quite — the answer is C.
MTP is considered relatively safe during the first trimester, that is, up to 12 weeks of pregnancy. Second-trimester abortions are much riskier.
Q62AIPMT 2011 Prelims
Which one of the following is categorised as a parasite in true sense?
Not quite — the answer is D.
The head louse lives on the host's scalp, lays its eggs on the hair and feeds on the host throughout its life, a true ectoparasite. The cuckoo is a brood parasite, the female Anopheles only visits to feed, and the foetus is not a separate parasitic organism.
Q63AIPMT 2011 Prelims
Which one of the following statements for pyramid of energy is incorrect, whereas the remaining three are correct?
Not quite — the answer is D.
Because energy is lost at each transfer, the energy at each trophic level is always less than at the level below. The pyramid of energy is therefore always upright with a broad base and can never be inverted.
Q64AIPMT 2011 Prelims
Ethanol is commercially produced through a particular species of
Not quite — the answer is B.
Ethanol is produced commercially by fermentation with the yeast Saccharomyces cerevisiae. Aspergillus is used for citric acid, Clostridium for butyric acid and Trichoderma for cyclosporin A.
Q65AIPMT 2011 Prelims
Where will you look for the sporozoites of the malarial parasite?
Not quite — the answer is B.
Sporozoites, the infective stage of Plasmodium, are stored in the salivary glands of an infected female Anopheles and enter humans with her saliva when she bites. A freshly moulted mosquito has not yet fed on an infected person, and human RBCs contain merozoites and trophozoites.
Q66AIPMT 2011 Prelims
A certain patient is suspected to be suffering from Acquired Immuno Deficiency Syndrome. Which diagnostic technique will you recommend for its detection?
Not quite — the answer is B.
ELISA (enzyme-linked immunosorbent assay) detects antibodies against HIV and is the widely used diagnostic test for AIDS. WIDAL is a test for typhoid, and MRI and ultrasound are imaging techniques.
Q67AIPMT 2011 Prelims
At which stage of HIV infection does one usually show symptoms of AIDS?
Not quite — the answer is D.
After infection, HIV multiplies in macrophages and later in helper T-lymphocytes; AIDS symptoms appear when the virus replicates rapidly and destroys large numbers of these cells, crippling immunity. This usually takes years, not days after contact.
Q68AIPMT 2011 Prelims
Given below is a sample of a portion of DNA strand giving the base sequence on the opposite strands. What is so special shown in it? 5′ ____ GAATTC ____ 3′ 3′ ____ CTTAAG ____ 5′
Not quite — the answer is A.
The sequence 5′-GAATTC-3′ reads the same on the opposite strand when both are read 5′→3′, so it is a palindromic sequence. This is the recognition site of EcoRI. Nothing in it indicates deletion or a start codon (AUG/ATG).
Q69AIPMT 2011 Prelims
The most common substrate used in distilleries for the production of ethanol is
Not quite — the answer is A.
Molasses, the sugar-rich by-product of sugarcane processing, is the most common substrate fermented by yeast in distilleries. Corn meal can be used but needs starch hydrolysis first, and soya meal and gram are protein-rich.
Q70AIPMT 2011 Prelims
An organism used as biofertilizer for raising soyabean crop is
Not quite — the answer is D.
Soyabean is a legume, and Rhizobium forms root nodules on legumes and fixes atmospheric nitrogen symbiotically. Azotobacter and Azospirillum are free-living or associative fixers, and Nostoc is a cyanobacterium used for paddy fields.
Q71AIPMT 2011 Prelims
There is a restriction endonuclease called EcoRI. What does "co" part in it stand for?
Not quite — the answer is A.
Restriction enzymes are named after their source: in EcoRI, 'E' is the genus Escherichia, 'co' is the species coli, 'R' is the strain RY13 and 'I' the order of isolation.
Biology: not in the current NEET syllabus
These 15 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
15 questions. Tap an option to check your answer.
Q1AIPMT 2011 Prelims · Q110
Mutations can be induced with
Not quite — the answer is A.
Mutation breeding uses mutagens such as gamma radiation to create new genetic variation. Infrared radiation is non-ionising and not mutagenic, and IAA and ethylene are plant hormones.
Q2AIPMT 2011 Prelims · Q111
A collection of plants and seeds having diverse alleles of all the genes of a crop is called
Not quite — the answer is C.
The entire collection of plants and seeds carrying the diverse alleles of all the genes of a crop is called its germplasm collection. A herbarium is a store of pressed dried specimens, and a gene library is a set of cloned DNA fragments.
Q3AIPMT 2011 Prelims
Large woody vines are more commonly found in
Not quite — the answer is D.
Large woody climbers (lianas) are a characteristic feature of tropical rainforests, where they climb trees to reach light in the dense, multilayered canopy. They are rare in alpine and temperate forests and in mangroves.
Q4AIPMT 2011 Prelims · Q115
Nitrifying bacteria
Not quite — the answer is B.
Nitrifying bacteria oxidise ammonia: Nitrosomonas converts it to nitrite and Nitrobacter converts nitrite to nitrate. Reducing nitrate to free nitrogen is denitrification, and converting free nitrogen to compounds is nitrogen fixation.
Q5AIPMT 2011 Prelims · Q116
The function of leghaemoglobin in the root nodules of legumes is
Not quite — the answer is C.
Nitrogenase works only without oxygen, and leghaemoglobin in legume root nodules binds oxygen, acting as an oxygen scavenger that protects the enzyme. It does not inhibit nitrogenase or control nif gene expression.
Q6AIPMT 2011 Prelims · Q117
Which one of the following elements in plants is not remobilised?
Not quite — the answer is C.
Calcium is part of the structural middle lamella (calcium pectate) and cannot be moved once laid down, so its deficiency appears first in young tissues. Phosphorus, potassium and sulphur are mobile and are remobilised from older leaves.
Q7AIPMT 2011 Prelims · Q134
Which one of the following statements is wrong in case of Bhopal tragedy?
Not quite — the answer is D.
The Bhopal gas tragedy on the night of 2–3 December 1984 was caused by a leak of methyl isocyanate from a pesticide plant and killed thousands. There was no radioactive fallout; that statement is the wrong one.
Q8AIPMT 2011 Prelims · Q136
Eutrophication is often seen in
Not quite — the answer is C.
Eutrophication is the nutrient enrichment of water, mostly by nitrates and phosphates from sewage and fertiliser run-off, which causes algal blooms. It is seen most commonly in freshwater lakes, which are enclosed bodies that accumulate nutrients.
Q9AIPMT 2011 Prelims
Which of the following is correctly stated as it happens in the common cockroach?
Not quite — the answer is A.
In the cockroach, food is first chewed by the mandibles and then ground further in the gizzard, which has chitinous teeth. Malpighian tubules arise at the junction of the midgut and hindgut, the blood has no haemoglobin, and the excretory product is uric acid.
Q10AIPMT 2011 Prelims
One very special feature in the earthworm Pheretima is that
Not quite — the answer is C.
In Pheretima, the intestine has an internal median fold called the typhlosole, from the 26th segment onward, which greatly increases the absorptive area. The earthworm has no long tubular heart, fertilisation happens in the cocoon outside the body, and the setae help in locomotion, not defence.
Q11AIPMT 2011 Prelims · Q144
'Himgiri' developed by hybridisation and selection for disease resistance against rust pathogens is a variety of
Not quite — the answer is A.
Himgiri is a wheat variety bred by hybridisation and selection for resistance to leaf and stripe rust and to hill bunt.
Q12AIPMT 2011 Prelims
The purplish red pigment rhodopsin contained in the rods type of photoreceptor cells of the human eye, is a derivative of
Not quite — the answer is A.
Rhodopsin consists of the protein opsin bound to retinal, an aldehyde of vitamin A. This is why vitamin A deficiency causes night blindness.
Q13AIPMT 2011 Prelims · Q150
"Jaya" and "Ratna" developed for green revolution in India are the varieties of
Not quite — the answer is C.
Jaya and Ratna are semi-dwarf rice varieties developed in India during the Green Revolution. Sonalika and Kalyan Sona are the corresponding semi-dwarf wheat varieties.
Q14AIPMT 2011 Prelims
Consider the following four conditions (a–d) and select the correct pair of them as adaptation to environment in desert lizards. (a) Burrowing in soil to escape high temperature (b) Losing heat rapidly from the body during high temperature (c) Bask in sun when temperature is low (d) Insulating body due to thick fatty dermis
Not quite — the answer is C.
Desert lizards cannot regulate body temperature physiologically, so they use behaviour: they bask in the sun when it is cold and burrow into the soil to escape midday heat. Rapid heat loss and thick fatty insulation are not their adaptations.
Q15AIPMT 2011 Prelims · Q194
When two unrelated individuals or lines are crossed, the performance of F₁ hybrid is often superior to both its parents. This phenomenon is called
Not quite — the answer is B.
When unrelated individuals or lines are crossed, the F₁ hybrid often outperforms both parents in yield and vigour; this is heterosis or hybrid vigour. Transformation is uptake of foreign DNA and splicing is removal of introns.
Chapter-wise split of AIPMT 2011 Prelims Biology
- Microbes in Human Welfare7 Q
- Morphology of Flowering Plants5 Q
- Sexual Reproduction in Flowering Plants5 Q
- Body Fluids and Circulation5 Q
- Plant Kingdom4 Q
- Human Health and Disease4 Q
- Excretory Products and Their Elimination4 Q
- Cell: The Unit of Life3 Q
- Anatomy of Flowering Plants3 Q
- Ecosystem3 Q
- Biotechnology: Principles and Processes3 Q
- Biological Classification2 Q
- Molecular Basis of Inheritance2 Q
- Biotechnology and Its Applications2 Q
- The Living World2 Q
- Animal Kingdom2 Q
- Chemical Coordination and Integration2 Q
- Breathing and Exchange of Gases2 Q
- Human Reproduction2 Q
- Biodiversity and Conservation1 Q
- Photosynthesis in Higher Plants1 Q
- Structural Organisation in Animals1 Q
- Cell Cycle and Cell Division1 Q
- Evolution1 Q
- Principles of Inheritance and Variation1 Q
- Neural Control and Coordination1 Q
- Reproductive Health1 Q
- Organisms and Populations1 Q
About this paper
How many questions from AIPMT 2011 Prelims are on this page?
171 questions (Physics 41, Chemistry 44, Biology 86). 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 Prelims 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 Prelims useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books. 27 questions here are on topics NCERT has since dropped; they are listed separately so you can skip them.
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