AIPMT 2008 Prelims 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
180 questions from the AIPMT 2008 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.
- Exam
- AIPMT 2008 Prelims
- Conducted by
- CBSE
- Physics
- 40 questions
- Chemistry
- 45 questions
- Biology
- 95 questions
AIPMT 2008 Prelims Physics questions
39 questions. Tap an option to check your answer.
Q1AIPMT 2008 Prelims · Q1
Which two of the following five physical parameters have the same dimensions? (a) energy density (b) refractive index (c) dielectric constant (d) Young's modulus (e) magnetic field
Not quite — the answer is D.
Energy density is energy per unit volume, [ML²T⁻²]/[L³] = [ML⁻¹T⁻²], and Young's modulus is stress/strain, also [ML⁻¹T⁻²]. Refractive index and dielectric constant are both dimensionless, but the magnetic field [MT⁻²A⁻¹] matches neither of them.
Q2AIPMT 2008 Prelims · Q2
If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be
Not quite — the answer is C.
V = (4/3)πr³, so ΔV/V = 3(Δr/r) = 3 × 2% = 6%. Taking 8% would wrongly cube the error itself instead of multiplying it by the power 3.
Q3AIPMT 2008 Prelims · Q3
The distance travelled by a particle starting from rest and moving with an acceleration (4/3) m s⁻², in the third second is
Not quite — the answer is D.
Distance in the nth second from rest = (a/2)(2n − 1) = (2/3)(2 × 3 − 1) = 10/3 m. 6 m is the total distance covered in the first three seconds, (1/2)(4/3)(3²), not the distance in the third second alone.
Q4AIPMT 2008 Prelims · Q4
A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 m s⁻¹ to 20 m s⁻¹ while passing through a distance 135 m in t second. The value of t is
Not quite — the answer is A.
For uniform acceleration s = (u + v)t/2, so 135 = (10 + 20)t/2 and t = 9 s. Check: a = (20² − 10²)/(2 × 135) = 10/9 m s⁻², and (20 − 10)/(10/9) = 9 s.
Q5AIPMT 2008 Prelims · Q6
A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be
Not quite — the answer is D.
The horizontal momentum mv cos 45° is unchanged, while the vertical component mv sin 45° reverses. The change is 2mv sin 45° = √2 mv. The answer is not zero, because momentum is a vector and its direction has changed even though the speed is the same.
Q6AIPMT 2008 Prelims · Q7
Sand is being dropped on a conveyor belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of v m/s will be
Not quite — the answer is B.
Each second a mass M of sand must be brought from zero to speed v, so F = v(dm/dt) = Mv newton. The force is not zero: even at constant belt speed, the momentum of the system keeps increasing as sand is added.
Q7AIPMT 2008 Prelims · Q9
Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional forces are 10% of energy. How much power is generated by the turbine? (g = 10 m/s²)
Not quite — the answer is B.
Input power = (dm/dt)gh = 15 × 10 × 60 = 9000 W. With 10% lost, the output is 0.9 × 9000 = 8100 W = 8.1 kW. 9 kW without the loss is not offered; adding the loss instead of subtracting it would give about 9.9 kW.
Q8AIPMT 2008 Prelims · Q10
A shell of mass 200 g is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is
Not quite — the answer is B.
Momentum conservation: 0.2v = 4V, so the gun recoils with V = v/20. Energy: (1/2) × 0.2 × v² + (1/2) × 4 × (v/20)² = 0.1v² + 0.005v² = 0.105v² = 1050, so v² = 10⁴ and v = 100 m s⁻¹. Ignoring the gun's recoil energy would give about 102 m s⁻¹, so the recoil share must be included.
Q9AIPMT 2008 Prelims · Q11
The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axes is
Not quite — the answer is C.
About the central axis, I(disc) = MR²/2 gives k = R/√2, and I(ring) = MR² gives k = R. So k(disc) : k(ring) = 1 : √2. √2 : 1 is the inverted ratio (ring to disc).
Q10AIPMT 2008 Prelims · Q12
A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is 90°. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is
Not quite — the answer is C.
Each half has mass M/2 and length L/2 and rotates about its own end: I = 2 × (1/3)(M/2)(L/2)² = ML²/12. The bend angle does not matter, because each element stays at the same distance from the axis, so the √2 factor is a trap.
Q11AIPMT 2008 Prelims · Q14
If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cycle process, then
Not quite — the answer is D.
Internal energy is a state function, so after a complete cycle its change E is zero. From Q = E + W this gives Q = W, but neither heat nor work need be zero: a heat engine running in a cycle does net work.
Q12AIPMT 2008 Prelims · Q15
On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39° W and 239° W respectively. What will be the temperature on the new scale, corresponding to a temperature of 39° C on the Celsius scale?
Not quite — the answer is C.
For a linear scale, (W − 39)/(239 − 39) = (C − 0)/(100 − 0). With C = 39: W = 39 + 200 × 0.39 = 39 + 78 = 117° W. 78° W forgets to add the offset of the freezing point.
Q13AIPMT 2008 Prelims · Q16
At 10°C the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At 110°C this ratio is
Not quite — the answer is A.
From PV = (m/M)RT, ρ/P = M/(RT), so the ratio is inversely proportional to absolute temperature. At 383 K it becomes x × 283/383. Using Celsius temperatures (10/110) is wrong because gas laws need kelvin.
Q14AIPMT 2008 Prelims · Q17
Two simple harmonic motions of angular frequency 100 and 1000 rad s⁻¹ have the same displacement amplitude. The ratio of their maximum accelerations is
Not quite — the answer is C.
Maximum acceleration = ω²A. For equal amplitudes the ratio is (100/1000)² = 1 : 10². 1 : 10 is the ratio of maximum velocities (ωA), not accelerations.
Q15AIPMT 2008 Prelims · Q18
The wave described by y = 0.25 sin(10πx − 2πt), where x and y are in metres and t in seconds, is a wave travelling along the
Not quite — the answer is D.
Compare with y = A sin(kx − ωt): k = 10π gives λ = 2π/k = 0.2 m, and ω = 2π gives f = 1 Hz. The minus sign between the kx and ωt terms means the wave moves along +x. Reading ω = 2π as 'frequency π Hz' confuses ω with 2πf.
Q16AIPMT 2008 Prelims · Q19
A point performs simple harmonic oscillation of period T and the equation of motion is given by x = a sin(ωt + π/6). After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?
Not quite — the answer is A.
v = aω cos(ωt + π/6). Setting v = aω/2 gives ωt + π/6 = π/3, so ωt = π/6 and t = (π/6)(T/2π) = T/12. T/6 would be the answer if the initial phase π/6 were ignored.
Q17AIPMT 2008 Prelims · Q20
Two points are located at a distance of 10 m and 15 m from the source of oscillation. The period of oscillation is 0.05 sec and the velocity of the wave is 300 m/sec. What is the phase difference between the oscillations of two points?
Not quite — the answer is C.
Wavelength λ = vT = 300 × 0.05 = 15 m. Path difference = 15 − 10 = 5 m, so phase difference = (2π/λ) × 5 = 2π/3. Using λ = 5 m by mistake would give 2π, not one of the options.
Q18AIPMT 2008 Prelims · Q21
The velocity of electromagnetic radiation in a medium of permittivity ε₀ and permeability μ₀ is given by
Not quite — the answer is D.
Maxwell's equations give the wave speed v = 1/√(με); with ε₀ and μ₀ this is c = 1/√(μ₀ε₀) ≈ 3 × 10⁸ m/s. √(μ₀ε₀) is its inverse, with dimensions of time per length, and √(μ₀/ε₀) is the impedance of free space (in ohms), not a speed.
Q19AIPMT 2008 Prelims · Q22
Two periodic waves of intensities I₁ and I₂ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
Not quite — the answer is A.
On superposition, Imax = (√I₁ + √I₂)² and Imin = (√I₁ − √I₂)². Adding them, the cross terms ±2√(I₁I₂) cancel, leaving 2(I₁ + I₂). (√I₁ + √I₂)² is only the maximum, not the sum.
Q20AIPMT 2008 Prelims · Q23
Two thin lenses of focal lengths f₁ and f₂ are in contact and coaxial. The power of the combination is
Not quite — the answer is A.
For thin lenses in contact, powers add: P = 1/f₁ + 1/f₂ = (f₁ + f₂)/(f₁f₂). (f₁ + f₂)/2 has dimensions of length, so it cannot be a power.
Q21AIPMT 2008 Prelims · Q24
A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the Sun is 1.39 × 10⁹ m and its mean distance from the earth is 1.5 × 10¹¹ m. What is the diameter of the Sun's image on the paper?
Not quite — the answer is B.
The image forms in the focal plane, and the Sun and its image subtend the same angle at the lens: d = f × D/u = 0.10 × 1.39 × 10⁹/1.5 × 10¹¹ ≈ 9.3 × 10⁻⁴ m, closest to 9.2 × 10⁻⁴ m. 6.5 × 10⁻⁴ m would come from wrongly halving or mis-scaling the angular size.
Q22AIPMT 2008 Prelims · Q25
The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is
Not quite — the answer is A.
Energy = (1/2)CV² with C = ε₀A/d and V = Ed, giving (1/2)(ε₀A/d)(Ed)² = (1/2)ε₀E²Ad, i.e. energy density (1/2)ε₀E² times the volume Ad. ε₀E²Ad misses the factor 1/2, and the forms divided by Ad have the wrong dimensions.
Q23AIPMT 2008 Prelims · Q27
The electric potential at a point in free space due to a charge Q coulomb is Q × 10¹¹ volts. The electric field at that point is
Not quite — the answer is B.
V = Q/(4πε₀r) = Q × 10¹¹ gives r = 1/(4πε₀ × 10¹¹). Then E = V/r = Q × 10¹¹ × 4πε₀ × 10¹¹ = 4πε₀Q × 10²² V/m. The 10²⁰ forms come from mishandling the power of ten.
Q24AIPMT 2008 Prelims · Q29
A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively
Not quite — the answer is C.
The volume stays constant, so R = ρl/A = ρl²/V is proportional to l². A 10% stretch gives R' = (1.1)²R = 1.21R. Specific resistance (resistivity) depends only on the material and temperature, so it stays the same.
Q25AIPMT 2008 Prelims · Q31
An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20°C? The temperature of boiling water is 100°C.
Not quite — the answer is B.
Heat needed = mcΔθ = 1 × 4200 × 80 = 3.36 × 10⁵ J; power = VI = 220 × 4 = 880 W. t = 3.36 × 10⁵/880 ≈ 382 s ≈ 6.4 min, closest to 6.3 min. Heating from 0°C to 100°C instead of 20°C would give about 8 min.
Q26AIPMT 2008 Prelims · Q33
A particle of mass m, charge Q and kinetic energy T enters a transverse uniform magnetic field of induction B⃗. After 3 seconds the kinetic energy of the particle will be
Not quite — the answer is D.
The magnetic force qv × B is always perpendicular to the velocity, so it does no work. The particle moves in a circle at constant speed and its kinetic energy stays T. The 3 seconds is a distractor: the elapsed time does not matter.
Q27AIPMT 2008 Prelims · Q35
A circular disc of radius 0.2 metre is placed in a uniform magnetic field of induction (1/π) Wb/m² in such a way that its axis makes an angle of 60° with B⃗. The magnetic flux linked with the disc is
Not quite — the answer is B.
The disc's area vector lies along its axis, so φ = BA cos 60° = (1/π)(π × 0.2²)(1/2) = 0.02 Wb. Using sin 60° (treating the angle as being with the plane of the disc) would give about 0.035 Wb.
Q28AIPMT 2008 Prelims · Q36
A galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be
Not quite — the answer is A.
Full-scale current = 3/(50 + 2950) = 1 mA for 30 divisions. For 20 divisions I = (2/3) mA, so total resistance = 3/(2/3 × 10⁻³) = 4500 Ω and the series resistance = 4500 − 50 = 4450 Ω. 4500 Ω would forget to subtract the galvanometer's own resistance.
Q29AIPMT 2008 Prelims · Q37
Curie temperature is the temperature above which
Not quite — the answer is B.
Above the Curie temperature, thermal agitation destroys the alignment of the domains, and a ferromagnet behaves as a paramagnet. It does not become diamagnetic, because the atoms still carry permanent magnetic moments.
Q30AIPMT 2008 Prelims · Q38
A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 × 10⁻³ Wb. The self-inductance of the solenoid is
Not quite — the answer is D.
Total flux linkage Nφ = LI, so L = 500 × 4 × 10⁻³/2 = 1.0 H. 2.0 H would forget to divide by the current.
Q31AIPMT 2008 Prelims · Q39
In an a.c. circuit the e.m.f. (e) and the current (i) at any instant are given respectively by e = E₀ sin ωt i = I₀ sin(ωt − φ) The average power in the circuit over one cycle of a.c. is
Not quite — the answer is D.
Average power = Erms × Irms × cos φ = (E₀/√2)(I₀/√2) cos φ = (E₀I₀/2) cos φ. E₀I₀/2 holds only when current and emf are in phase (φ = 0), and the sin φ form gives the wattless (reactive) component.
Q32AIPMT 2008 Prelims · Q41
The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the
Not quite — the answer is B.
Photon energy = φ + eV₀ = 6.2 + 5 = 11.2 eV, so λ = 12400/11.2 eV·Å ≈ 1100 Å (about 110 nm). This lies in the ultraviolet; visible light is only about 1.8 to 3.1 eV, and X-rays need wavelengths below about 100 Å.
Q33AIPMT 2008 Prelims · Q42
A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of 3 × 10⁶ m s⁻¹. The velocity of the particle is (mass of electron = 9.1 × 10⁻³¹ kg)
Not quite — the answer is B.
Equal de Broglie wavelengths (λ = h/mv) mean equal momenta. 1 mg = 10⁻⁶ kg, so v = (9.1 × 10⁻³¹ × 3 × 10⁶)/10⁻⁶ ≈ 2.7 × 10⁻¹⁸ m s⁻¹. Taking 1 mg as 10⁻³ kg gives 2.7 × 10⁻²¹ m s⁻¹.
Q34AIPMT 2008 Prelims · Q43
The ground state energy of hydrogen atom is −13.6 eV. When its electron is in the first excited state, its excitation energy is
Not quite — the answer is D.
The first excited state is n = 2 with energy −13.6/4 = −3.4 eV. Excitation energy = −3.4 − (−13.6) = 10.2 eV. 3.4 eV is the energy needed to ionise from n = 2, not the excitation energy from the ground state.
Q35AIPMT 2008 Prelims · Q44
Two radioactive materials X₁ and X₂ have decay constants 5λ and λ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X₁ to that of X₂ will be 1/e after a time
Not quite — the answer is D.
N₁/N₂ = e^(−5λt)/e^(−λt) = e^(−4λt). Setting this to 1/e gives 4λt = 1, so t = 1/(4λ). λ and λ/2 cannot be times, since λ has units of inverse time.
Q36AIPMT 2008 Prelims · Q45
Two nuclei have their mass numbers in the ratio of 1 : 3. The ratio of their nuclear densities would be
Not quite — the answer is A.
Nuclear radius R = R₀A^(1/3), so volume is proportional to A, just as mass is. Their ratio, the density, is the same for all nuclei, 1 : 1. The 3^(1/3) form is the ratio of radii, not of densities.
Q37AIPMT 2008 Prelims · Q46
If M(A; Z), Mp and Mn denote the masses of the nucleus ᴬ_ZX, proton and neutron respectively in units of u (1 u = 931.5 MeV/c²) and BE represents its binding energy in MeV, then
Not quite — the answer is B.
A bound nucleus is lighter than its free nucleons by the mass defect, and the mass defect equals BE/c². So M(A, Z) = ZMp + (A − Z)Mn − BE/c². The plus sign would make the nucleus heavier than its parts, and subtracting BE without c² mixes energy with mass units.
Q38AIPMT 2008 Prelims · Q50
A p-n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly
Not quite — the answer is C.
The photon must at least bridge the band gap: ν = Eg/h = (2.0 × 1.6 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) ≈ 4.8 × 10¹⁴ Hz ≈ 5 × 10¹⁴ Hz. 10 × 10¹⁴ Hz would correspond to a 4 eV gap.
Q39AIPMT 2008 Prelims · Q79
With which one of the following elements silicon should be doped so as to give p-type of semiconductor?
Not quite — the answer is A.
A trivalent dopant such as boron has one electron too few to complete the four bonds with silicon, which creates holes, the majority carriers in a p-type semiconductor. Arsenic is pentavalent and gives n-type, while germanium is tetravalent like silicon and creates no extra carriers.
Physics: not in the current NEET syllabus
These 1 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
1 questions. Tap an option to check your answer.
Q1AIPMT 2008 Prelims · Q47
The voltage gain of an amplifier with 9% negative feedback is 10. The voltage gain without feedback will be
Not quite — the answer is A.
With negative feedback, A_f = A/(1 + βA) with β = 0.09. Then 10 = A/(1 + 0.09A) gives 10 + 0.9A = A, so 0.1A = 10 and A = 100. A gain of 10 without feedback is impossible, because negative feedback always lowers the gain.
Chapter-wise split of AIPMT 2008 Prelims Physics
- Laws of Motion3 Q
- Current Electricity3 Q
- Nuclei3 Q
- Units and Measurements2 Q
- Motion in a Straight Line2 Q
- System of Particles and Rotational Motion2 Q
- Oscillations2 Q
- Waves2 Q
- Ray Optics and Optical Instruments2 Q
- Electrostatic Potential and Capacitance2 Q
- Electromagnetic Induction2 Q
- Dual Nature of Radiation and Matter2 Q
- Semiconductor Electronics: Materials, Devices and Simple Circuits2 Q
- Work, Energy and Power1 Q
- Thermodynamics1 Q
- Thermal Properties of Matter1 Q
- Kinetic Theory1 Q
- Electromagnetic Waves1 Q
- Wave Optics1 Q
- Moving Charges and Magnetism1 Q
- Magnetism and Matter1 Q
- Alternating Current1 Q
- Atoms1 Q
AIPMT 2008 Prelims Chemistry questions
37 questions. Tap an option to check your answer.
Q1AIPMT 2008 Prelims · Q51
If uncertainty in position and momentum are equal, then uncertainty in velocity is
Not quite — the answer is B.
Δx·Δp = h/4π with Δx = Δp gives Δp² = h/4π, so Δp = (1/2)√(h/π). Then Δv = Δp/m = (1/2m)√(h/π). The forms without m are momenta, not velocities.
Q2AIPMT 2008 Prelims · Q53
The value of equilibrium constant of the reaction HI(g) ⇌ ½H₂(g) + ½I₂(g) is 8.0. The equilibrium constant of the reaction H₂(g) + I₂(g) ⇌ 2HI(g) will be
Not quite — the answer is C.
Doubling the given reaction gives 2HI ⇌ H₂ + I₂ with K = 8² = 64. Reversing it gives H₂ + I₂ ⇌ 2HI with K = 1/64. 1/8 only reverses the reaction and forgets that doubling the equation squares K.
Q3AIPMT 2008 Prelims · Q55
Kohlrausch's law states that at
Not quite — the answer is A.
Kohlrausch's law of independent migration says that at infinite dilution each ion contributes a fixed amount to the limiting equivalent (or molar) conductivity, whatever the other ion is. It does not hold at finite dilution, where interionic attractions matter, and it is stated for equivalent/molar conductance, not plain conductance.
Q4AIPMT 2008 Prelims · Q56
The measurement of the electron position is associated with an uncertainty in momentum, which is equal to 1 × 10⁻¹⁸ g cm s⁻¹. The uncertainty in electron velocity is, (mass of an electron is 9 × 10⁻²⁸ g)
Not quite — the answer is B.
Δv = Δp/m = (1 × 10⁻¹⁸ g cm s⁻¹)/(9 × 10⁻²⁸ g) ≈ 1.1 × 10⁹ cm s⁻¹. Both quantities are already in CGS units, so no conversion is needed, and converting only one of them gives the wrong power of ten.
Q5AIPMT 2008 Prelims · Q57
Which of the following are not state functions? (I) q + w (II) q (III) w (IV) H − TS
Not quite — the answer is A.
Heat q and work w depend on the path, so they are not state functions. Their sum q + w equals ΔU, and H − TS is the Gibbs energy G; both depend only on the state. So only q and w are not state functions.
Q6AIPMT 2008 Prelims · Q58
The bromination of acetone that occurs in acid solution is represented by this equation CH₃COCH₃(aq) + Br₂(aq) → CH₃COCH₂Br(aq) + H⁺(aq) + Br⁻(aq) These kinetic data were obtained for given reaction concentrations. Initial concentrations (M) [CH₃COCH₃], [Br₂], [H⁺] → initial rate of disappearance of Br₂ (M s⁻¹): 0.30, 0.05, 0.05 → 5.7 × 10⁻⁵ 0.30, 0.10, 0.05 → 5.7 × 10⁻⁵ 0.30, 0.10, 0.10 → 1.2 × 10⁻⁴ 0.40, 0.05, 0.20 → 3.1 × 10⁻⁴ Based on these data, the rate equation is
Not quite — the answer is B.
Runs 1 and 2: doubling [Br₂] leaves the rate unchanged, so the reaction is zero order in Br₂. Runs 2 and 3: doubling [H⁺] doubles the rate, so it is first order in H⁺. Run 4 checks acetone: 5.7 × 10⁻⁵ × (0.40/0.30) × (0.20/0.05) ≈ 3.0 × 10⁻⁴, matching 3.1 × 10⁻⁴, so it is first order in acetone. Rate = k[CH₃COCH₃][H⁺]; any form containing [Br₂] contradicts runs 1 and 2.
Q7AIPMT 2008 Prelims · Q59
What volume of oxygen gas (O₂) measured at 0°C and 1 atm, is needed to burn completely 1 L of propane gas (C₃H₈) measured under the same conditions?
Not quite — the answer is D.
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. At the same temperature and pressure, volumes are in the same ratio as moles, so 1 L of propane needs 5 L of O₂. 7 L comes from adding the volumes of the products, which is not what the question asks.
Q8AIPMT 2008 Prelims · Q60
Bond dissociation enthalpy of H₂, Cl₂ and HCl are 434, 242 and 431 kJ mol⁻¹ respectively. Enthalpy of formation of HCl is
Not quite — the answer is D.
For ½H₂ + ½Cl₂ → HCl, ΔHf = ½(434) + ½(242) − 431 = 338 − 431 = −93 kJ mol⁻¹. The ±245 values come from using the full bond enthalpies of H₂ and Cl₂ with only one H–Cl bond (434 + 242 − 431 = 245); that equation actually forms 2 mol HCl.
Q9AIPMT 2008 Prelims · Q62
Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H⁺ ion concentration in the mixture?
Not quite — the answer is C.
With equal volumes V, total H⁺ = (10⁻³ + 10⁻⁴ + 10⁻⁵)V = 1.11 × 10⁻³ V mol in a total volume 3V. [H⁺] = 1.11 × 10⁻³/3 = 3.7 × 10⁻⁴ M. 1.11 × 10⁻³ M forgets the threefold dilution.
Q10AIPMT 2008 Prelims · Q63
The values of Kp₁ and Kp₂ for the reactions X ⇌ Y + Z ....(1) and A ⇌ 2B ....(2) are in ratio of 9 : 1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (1) and (2) are in the ratio
Not quite — the answer is D.
With degree of dissociation α: for the first reaction, moles 1 − α, α, α (total 1 + α) give Kp₁ = α²P₁/(1 − α²). For the second reaction, moles 1 − α, 2α (total 1 + α) give Kp₂ = 4α²P₂/(1 − α²). Then Kp₁/Kp₂ = P₁/(4P₂) = 9, so P₁ : P₂ = 36 : 1. 9 : 1 would forget the factor 4 from (2α)².
Q11AIPMT 2008 Prelims · Q64
If the concentration of OH⁻ ions in the reaction Fe(OH)₃(s) ⇌ Fe³⁺(aq) + 3OH⁻(aq) is decreased by 1/4 times, then equilibrium concentration of Fe³⁺ will increase by
Not quite — the answer is D.
K = [Fe³⁺][OH⁻]³ is constant (the solid does not appear). If [OH⁻] becomes one quarter, [OH⁻]³ falls by 4³ = 64, so [Fe³⁺] must increase 64 times. 4 times ignores the cube from the coefficient 3.
Q12AIPMT 2008 Prelims · Q65
For the gas phase reaction, PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), which of the following conditions is correct?
Not quite — the answer is C.
Dissociation of PCl₅ breaks P–Cl bonds, so it absorbs heat (∆H > 0). One mole of gas gives two moles of gas, so disorder increases (∆S > 0). A negative ∆S would require fewer gas molecules on the product side.
Q13AIPMT 2008 Prelims · Q66
The rate constants k₁ and k₂ for two different reactions are 10¹⁶·e^(−2000/T) and 10¹⁵·e^(−1000/T) respectively. The temperature at which k₁ = k₂ is
Not quite — the answer is A.
Setting 10¹⁶e^(−2000/T) = 10¹⁵e^(−1000/T) gives 10 = e^(1000/T), so 1000/T = ln 10 = 2.303 and T = 1000/2.303 K ≈ 434 K. 1000 K would come from wrongly taking ln 10 as 1.
Q14AIPMT 2008 Prelims · Q67
Standard free energies of formation (in kJ/mol) at 298 K are −237.2, −394.4 and −8.2 for H₂O(l), CO₂(g) and pentane(g) respectively. The value of E°cell for the pentane-oxygen fuel cell is
Not quite — the answer is D.
C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O. ∆G° = 5(−394.4) + 6(−237.2) − (−8.2) = −3387.0 kJ. Eight O₂ molecules take 4 electrons each, so n = 32, and E° = −∆G°/nF = 3387000/(32 × 96500) ≈ 1.0968 V. Using a smaller n (for example 16) would double the voltage to about 2.19 V, which is not offered.
Q15AIPMT 2008 Prelims · Q68
The dissociation equilibrium of a gas AB₂ can be represented as, 2AB₂(g) ⇌ 2AB(g) + B₂(g) The degree of dissociation is 'x' and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure P is
Not quite — the answer is C.
Starting with 1 mol AB₂: AB₂ = 1 − x, AB = x, B₂ = x/2, total = 1 + x/2 ≈ 1. Kp = (xP)²(xP/2)/(P)² = x³P/2, so x = (2Kp/P)^(1/3). The square-root form would hold for a reaction where x enters only as x², e.g. AB₂ ⇌ AB + B.
Q16AIPMT 2008 Prelims · Q69
The sequence of ionic mobility in aqueous solution is
Not quite — the answer is C.
Smaller ions have higher charge density and hold a larger shell of water molecules, so the hydrated Na⁺ ion is the bulkiest and moves slowest. Mobility therefore increases down the group: Cs⁺ > Rb⁺ > K⁺ > Na⁺. Na⁺ > ... > Cs⁺ follows bare ionic size and ignores hydration.
Q17AIPMT 2008 Prelims · Q71
The correct order of decreasing second ionisation enthalpy of Ti (22), V (23), Cr (24) and Mn (25) is
Not quite — the answer is B.
The second electron is removed from M⁺. Cr⁺ is 3d⁵, a stable half-filled shell, so removing a 3d electron costs the most. Mn⁺ (3d⁵4s¹) loses its 4s electron more easily, and V⁺ and Ti⁺ follow in order of nuclear charge. Observed values (≈1590, 1510, 1410, 1310 kJ mol⁻¹) give Cr > Mn > V > Ti. Mn > Cr would hold for the third, not the second, ionisation enthalpy.
Q18AIPMT 2008 Prelims · Q72
How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl?
Not quite — the answer is A.
PbO + 2HCl → PbCl₂ + H₂O. Moles of PbO = 6.5/223 ≈ 0.029, and of HCl = 3.2/36.5 ≈ 0.088, more than the 0.058 needed. PbO is the limiting reagent, so 0.029 mol PbCl₂ forms. 0.044 would wrongly treat HCl as limiting (0.088/2).
Q19AIPMT 2008 Prelims · Q74
Volume occupied by one molecule of water (density = 1 g cm⁻³) is
Not quite — the answer is D.
One mole of water (18 g) occupies 18 cm³ at density 1 g cm⁻³. Volume per molecule = 18/(6.022 × 10²³) ≈ 3.0 × 10⁻²³ cm³. 6.023 × 10⁻²³ cm³ just reuses the digits of Avogadro's number.
Q20AIPMT 2008 Prelims · Q75
Number of moles of MnO₄⁻ required to oxidize one mole of ferrous oxalate completely in acidic medium will be
Not quite — the answer is B.
In FeC₂O₄ both parts are oxidised: Fe²⁺ → Fe³⁺ (1 e⁻) and C₂O₄²⁻ → 2CO₂ (2 e⁻), so 3 electrons per mole. Each MnO₄⁻ → Mn²⁺ takes 5 electrons, so 3/5 = 0.6 mol MnO₄⁻ is needed. 0.4 mol counts only the oxalate electrons.
Q21AIPMT 2008 Prelims · Q76
On the basis of the following E° values, the strongest oxidizing agent is [Fe(CN)₆]⁴⁻ → [Fe(CN)₆]³⁻ + e⁻; E° = −0.35 V Fe²⁺ → Fe³⁺ + e⁻; E° = −0.77 V
Not quite — the answer is D.
The values given are oxidation potentials, so the reduction potentials are +0.35 V for [Fe(CN)₆]³⁻/[Fe(CN)₆]⁴⁻ and +0.77 V for Fe³⁺/Fe²⁺. The species with the higher reduction potential is the stronger oxidising agent, so Fe³⁺. [Fe(CN)₆]⁴⁻ and Fe²⁺ are the reduced forms and act as reducing agents.
Q22AIPMT 2008 Prelims · Q80
In which of the following coordination entities the magnitude of Δo (CFSE in octahedral field) will be maximum? (at. no. Co = 27)
Not quite — the answer is D.
For the same metal ion (Co³⁺), Δo depends on the ligand's place in the spectrochemical series, where CN⁻ is the strongest-field ligand here, well above NH₃, H₂O and oxalate. So [Co(CN)₆]³⁻ has the largest splitting. [Co(NH₃)₆]³⁺ is a strong-field complex too, but ammonia splits less than cyanide.
Q23AIPMT 2008 Prelims · Q81
The angular shape of ozone molecule (O₃) consists of
Not quite — the answer is A.
Ozone has two O–O σ bonds forming the bent skeleton, and one π bond that is delocalised over the three atoms by resonance (each O–O has bond order 1.5). Two π bonds would require one O=O on each side, which is not the structure.
Q24AIPMT 2008 Prelims · Q82
The correct order of increasing bond angles in the following triatomic species is
Not quite — the answer is C.
NO₂⁺ has no non-bonding electrons on N and is linear (180°). NO₂ has a single unpaired electron on N (about 134°), and NO₂⁻ has a full lone pair that repels more strongly (about 115°). So the increasing order is NO₂⁻ < NO₂ < NO₂⁺. The reverse order wrongly treats the linear cation as the most bent.
Q25AIPMT 2008 Prelims · Q83
Four diatomic species are listed below in different sequences. Which of these presents the correct order of their increasing bond order?
Not quite — the answer is A.
From MO theory: He₂⁺ (3 e⁻) has bond order 0.5, O₂⁻ (17 e⁻) 1.5, NO (15 e⁻) 2.5, and C₂²⁻ (14 e⁻, like N₂) 3. The increasing order is He₂⁺ < O₂⁻ < NO < C₂²⁻. Placing He₂⁺ last mistakes its weak half bond for a strong one.
Q26AIPMT 2008 Prelims · Q85
In the hydrocarbon CH₃(6)–CH(5)=CH(4)–CH₂(3)–C(2)≡CH(1) The state of hybridization of carbons 1, 3 and 5 are in the following sequence
Not quite — the answer is C.
C-1 is part of the C≡C triple bond (sp), C-3 is a CH₂ with only single bonds (sp³), and C-5 is part of the C=C double bond (sp²). The sp, sp², sp³ sequence wrongly gives C-3 a double bond.
Q27AIPMT 2008 Prelims · Q87
A strong base can abstract an α-hydrogen from
Not quite — the answer is D.
The α-hydrogen of a ketone is acidic (pKa about 19–20), because the carbanion formed is stabilised as a resonance enolate by the C=O group. Alkanes, alkenes and amines have no such electron-withdrawing group to stabilise the carbanion.
Q28AIPMT 2008 Prelims · Q88
How many stereoisomers does the molecule have? CH₃CH=CHCH₂CHBrCH₃
Not quite — the answer is B.
The C=C bond has two different groups on each carbon, so it shows cis–trans isomerism (2 forms). The CHBr carbon carries H, Br, CH₃ and CH₂CH=CHCH₃, so it is a stereocentre (2 forms). These are independent, giving 2 × 2 = 4 stereoisomers. Counting only one source of isomerism would give 2.
Q29AIPMT 2008 Prelims · Q90
The relative reactivities of acyl compounds towards nucleophilic substitution are in the order of
Not quite — the answer is B.
Reactivity in nucleophilic acyl substitution follows how easily the leaving group departs: Cl⁻ > RCOO⁻ > RO⁻ > NH₂⁻, the weaker the base the better. So acyl chloride > acid anhydride > ester > amide. Placing ester above anhydride ignores that carboxylate is a better leaving group than alkoxide.
Q30AIPMT 2008 Prelims · Q91
Base strength of (a) H₃C–CH₂⁻ (b) H₂C=CH⁻ (c) H–C≡C⁻ is in the order of
Not quite — the answer is A.
The conjugate acids get more acidic from ethane (sp³) to ethene (sp²) to ethyne (sp), because more s-character holds the lone pair more tightly. The stronger the conjugate acid, the weaker the base, so the basicity order is ethyl > vinyl > acetylide anion. The reverse order (acetylide strongest) is the stability order, not the base strength.
Q31AIPMT 2008 Prelims · Q93
In DNA, the complementary bases are
Not quite — the answer is B.
In DNA, adenine pairs with thymine (two hydrogen bonds) and guanine pairs with cytosine (three hydrogen bonds). Uracil occurs in RNA, not DNA, and adenine with guanine would pair two purines, which does not fit the helix.
Q32AIPMT 2008 Prelims · Q94
Which one of the following is most reactive towards electrophilic attack?
Not quite — the answer is D.
The –OH group of phenol donates its lone pair into the ring (+R effect), which strongly activates the ring towards electrophiles. –CH₂OH is only weakly activating, –Cl deactivates by its –I effect, and –NO₂ strongly deactivates.
Q33AIPMT 2008 Prelims · Q95
An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be
Not quite — the answer is B.
O = 100 − 38.71 − 9.67 = 51.62%. Moles: C = 38.71/12 = 3.23, H = 9.67/1 = 9.67, O = 51.62/16 = 3.23, a ratio of 1 : 3 : 1, so the empirical formula is CH₃O. CH₄O (methanol) would need about 12.5% H.
Q34AIPMT 2008 Prelims · Q96
In a SN2 substitution reaction of the type R–Br + Cl⁻ —(DMF)→ R–Cl + Br⁻ Which one of the following has the highest relative rate?
Not quite — the answer is A.
SN2 needs backside attack at the carbon bearing Br, so it slows as branching near that carbon increases steric hindrance. Ethyl bromide is the least hindered, so it reacts fastest. The neopentyl-type (CH₃)₃C–CH₂Br, though primary, is extremely hindered and reacts slowest.
Q35AIPMT 2008 Prelims · Q97
Acetophenone when reacted with a base, C₂H₅ONa, yields a stable compound which has the structure?
Not quite — the answer is B.
Ethoxide removes an α-hydrogen from the CH₃ group, and the enolate attacks a second acetophenone (aldol addition). The β-hydroxy ketone then dehydrates to the conjugated α,β-unsaturated ketone C₆H₅C(CH₃)=CH–CO–C₆H₅, which is stabilised by conjugation. The diols are pinacol-type reduction products, not products of base-catalysed condensation.
Q36AIPMT 2008 Prelims · Q98
In a reaction of aniline a coloured product C was obtained. C₆H₅NH₂ (A) —(NaNO₂, HCl)→ B —(C₆H₅N(CH₃)₂, cold)→ C The structure of C would be
Not quite — the answer is A.
Diazotisation of aniline gives benzenediazonium chloride, the intermediate B. In the cold it couples with N,N-dimethylaniline at the activated para position, giving the coloured azo dye p-(dimethylamino)azobenzene. The –NH–NH– (hydrazo) compound has no N=N chromophore, so it would not be the coloured coupling product.
Q37AIPMT 2008 Prelims · Q100
Which one of the following is an amine hormone?
Not quite — the answer is B.
Thyroxine is made from the amino acid tyrosine and is classed as an amino acid-derived (amine) hormone. Progesterone is a steroid hormone, insulin is a polypeptide, and oxypurin is a purine derivative, not a hormone.
Chemistry: not in the current NEET syllabus
These 8 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
8 questions. Tap an option to check your answer.
Q1AIPMT 2008 Prelims · Q48
If the lattice parameter for a crystalline structure is 3.6 Å, then the atomic radius in fcc crystal is
Not quite — the answer is A.
In fcc, atoms touch along the face diagonal: 4r = √2a, so r = a/(2√2) = 3.6/2.828 ≈ 1.27 Å. 1.81 Å is just a/2, which applies to simple cubic.
Q2AIPMT 2008 Prelims · Q52
If a gas expands at constant temperature, it indicates that
Not quite — the answer is D.
The average kinetic energy of gas molecules is (3/2)kT and depends only on temperature, so at constant temperature it stays the same. In an isothermal expansion the pressure actually falls (Boyle's law), and the number of molecules is fixed.
Q3AIPMT 2008 Prelims · Q54
If 'a' stands for the edge length of the cubic systems: simple cubic, body centred cubic and face centred cubic, then the ratio of radii of the spheres in these systems will be respectively
Not quite — the answer is B.
Simple cubic: atoms touch along the edge, 2r = a, so r = a/2. bcc: they touch along the body diagonal, 4r = √3a, so r = (√3/4)a. fcc: they touch along the face diagonal, 4r = √2a, so r = a/(2√2). The (√3/2)a and (√2/2)a forms forget that the diagonal spans four radii.
Q4AIPMT 2008 Prelims · Q70
Percentage of free space in a body centred cubic unit cell is
Not quite — the answer is C.
In bcc, 2 atoms of radius r = √3a/4 occupy 2 × (4/3)πr³ = (√3π/8)a³ ≈ 0.68a³, a packing efficiency of 68%. The free space is 100 − 68 = 32%. 26% would be the value for fcc/hcp.
Q5AIPMT 2008 Prelims · Q77
The alkali metals form salt-like hydrides by the direct synthesis at elevated temperature. The thermal stability of these hydrides decreases in which of the following orders?
Not quite — the answer is A.
Down the group the cation gets larger, the lattice energy of the ionic hydride falls and the M–H interaction weakens, so the hydrides decompose more easily. Thermal stability decreases LiH > NaH > KH > RbH > CsH. The reverse order would follow increasing metallic character, not the stability of the hydride.
Q6AIPMT 2008 Prelims · Q84
Equimolar solutions of the following were prepared in water separately. Which one of the solutions will record the highest pH?
Not quite — the answer is C.
Small, highly polarising cations such as Mg²⁺ hydrolyse water slightly and make the solution acidic. This tendency falls down the group as the cation grows, and Ba²⁺ hardly hydrolyses, so BaCl₂ is nearly neutral and has the highest pH. MgCl₂ gives the lowest pH.
Q7AIPMT 2008 Prelims · Q86
Green chemistry means such reactions which
Not quite — the answer is C.
Green chemistry is the design of chemical products and processes that reduce or eliminate the use and generation of hazardous substances. The word 'green' does not refer to plants, colour, or ozone depletion.
Q8AIPMT 2008 Prelims · Q99
Which of the following statements is not true?
Not quite — the answer is C.
Natural rubber is cis-1,4-polyisoprene, so saying it has the trans configuration at every double bond is false; the all-trans polymer is gutta-percha. The other statements are true: vulcanization cross-links chains with sulphur, and Buna-S is a butadiene–styrene copolymer.
Chapter-wise split of AIPMT 2008 Prelims Chemistry
- Equilibrium5 Q
- Electrochemistry4 Q
- Some Basic Concepts of Chemistry4 Q
- Thermodynamics3 Q
- Chemical Bonding and Molecular Structure3 Q
- Organic Chemistry: Some Basic Principles and Techniques3 Q
- Aldehydes, Ketones and Carboxylic Acids3 Q
- Structure of Atom2 Q
- Chemical Kinetics2 Q
- Biomolecules2 Q
- The d- and f-Block Elements1 Q
- Redox Reactions1 Q
- Coordination Compounds1 Q
- Alcohols, Phenols and Ethers1 Q
- Haloalkanes and Haloarenes1 Q
- Amines1 Q
AIPMT 2008 Prelims Biology questions
76 questions. Tap an option to check your answer.
Q1AIPMT 2008 Prelims · Q101
Select one of the following of important features distinguishing Gnetum from Cycas and Pinus and showing affinities with angiosperms
Not quite — the answer is C.
Gnetum stands apart from Cycas and Pinus because its xylem contains true vessels and its female gametophyte has no archegonia, two features otherwise typical of flowering plants. Apical meristems and embryo development are common to all seed plants, so they do not separate Gnetum from the other gymnosperms.
Q2AIPMT 2008 Prelims · Q102
Thermococcus, Methanococcus and Methanobacterium exemplify
Not quite — the answer is B.
These genera are methanogenic and hyperthermophilic archaebacteria, and many euryarchaea such as these carry small histone proteins clearly homologous to the eukaryotic core histones (H3/H4 fold) that wrap their DNA. The statement that they lack any eukaryote-like histones is therefore wrong; hyperthermophiles like Thermococcus also tend to have relaxed or positively supercoiled DNA owing to reverse gyrase.
Q3AIPMT 2008 Prelims · Q103
Which one of the following is heterosporous?
Not quite — the answer is C.
Salvinia, a water fern, produces two kinds of spores (microspores and megaspores), as NCERT notes for Selaginella and Salvinia. Equisetum, Dryopteris and Adiantum all produce a single kind of spore and are homosporous.
Q4AIPMT 2008 Prelims · Q104
In which one of the following male and female gametophytes do not have free living independent existence?
Not quite — the answer is A.
Cedrus is a gymnosperm, in which the male gametophyte (pollen) and female gametophyte (inside the ovule) are highly reduced and remain dependent on the sporophyte. In Pteris the prothallus is free-living, and in the mosses Funaria and Polytrichum the gametophyte is the dominant independent plant.
Q5AIPMT 2008 Prelims · Q105
Which one of the following groups of three animals each is correctly matched with their one characteristic morphological feature? Animals - Morphological feature
Not quite — the answer is D.
Scorpion, spider and cockroach are all arthropods, and non-chordates like these have a ventral, solid nerve cord. Taenia is not metamerically segmented, sea anemone is radially symmetrical, and sea urchin (an echinoderm) has no jointed appendages, so the other groupings each contain a misfit.
Q6AIPMT 2008 Prelims · Q106
Which one of the following phyla is correctly matched with its two general characteristics?
Not quite — the answer is A.
Molluscs are usually oviparous with indirect development, typically via trochophore and veliger larvae. Arthropods breathe by gills, book lungs or tracheae and are not all divided into head, thorax and abdomen; echinoderms have external fertilisation; and many chordates have a common cloacal opening.
Q7AIPMT 2008 Prelims · Q107
Which one of the following in birds, indicates their reptilian ancestry?
Not quite — the answer is B.
NCERT notes that the hind limbs of birds generally bear scales, a retained reptilian character. A four-chambered heart is a feature birds share with mammals rather than with reptiles (which mostly have three chambers), so it does not indicate reptilian ancestry.
Q8AIPMT 2008 Prelims · Q108
Ascaris is characterized by
Not quite — the answer is C.
Ascaris is an aschelminth (roundworm): it is pseudocoelomate and its body is not segmented. True coelom together with metamerism is the annelid condition, not that of nematodes.
Q9AIPMT 2008 Prelims · Q109
Which one of the following is not a characteristic of phylum Annelida?
Not quite — the answer is D.
Annelids are true coelomates (schizocoelous); a pseudocoelom is the feature of Aschelminthes. Annelids do have a ventral nerve cord, closed circulation and metameric segmentation.
Q10AIPMT 2008 Prelims · Q110
Cellulose is the major component of cell walls of
Not quite — the answer is B.
Pythium is an oomycete (traditionally a phycomycete) whose wall is made mainly of cellulose. Saccharomyces, a true fungus, has a chitin/glucan wall, while Xanthomonas and Pseudomonas are bacteria with peptidoglycan walls.
Q11AIPMT 2008 Prelims · Q111
Vacuole in a plant cell
Not quite — the answer is C.
The plant vacuole is bounded by a single membrane, the tonoplast, and holds water, sap, excretory products and other materials not useful to the cell. Lipids are stored in oil bodies, not the vacuole, and the vacuole is not membrane-less.
Q12AIPMT 2008 Prelims · Q112
A competitive inhibitor of succinic dehydrogenase is
Not quite — the answer is B.
Malonate closely resembles succinate in structure, so it binds the active site of succinic dehydrogenase and competes with the substrate, the NCERT example of competitive inhibition. Malate and oxaloacetate are downstream Krebs cycle intermediates, not the classic inhibitor.
Q13AIPMT 2008 Prelims · Q113
Polysome is formed by
Not quite — the answer is B.
A polysome (polyribosome) is a chain of several ribosomes translating the same mRNA at the same time. The ribosomes are linked through the mRNA, not directly to one another, and ribosomes bound to ER form rough ER rather than a polysome.
Q14AIPMT 2008 Prelims · Q114
Carbohydrates are commonly found as starch in plant storage organs. Which of the following five properties of starch (a - e) make it useful as a storage material? a. Easily translocated b. Chemically non-reactive c. Easily digested by animals d. Osmotically inactive e. Synthesized during photosynthesis The useful properties are
Not quite — the answer is C.
Starch is a large, insoluble polymer, so it is chemically unreactive and exerts no osmotic effect, letting cells store a lot of it without drawing in water. It is not easily translocated (sugars like sucrose are), and being digestible by animals is irrelevant to the plant's storage needs.
Q15AIPMT 2008 Prelims · Q115
In the light of recent classification of living organisms into three domains of life (bacteria, archaea and eukarya), which one of the following statements is true about archaea?
Not quite — the answer is C.
Archaea are prokaryotes that share some traits with bacteria and some with eukaryotes, and also have unique features such as ether-linked membrane lipids and distinctive cell walls. They neither completely differ from nor fully resemble the other groups, which rules out the absolute options.
Q16AIPMT 2008 Prelims · Q116
Keeping in view the 'fluid mosaic model' for the structure of cell membrane, which one of the following statements is correct w.r.t. the movement of lipids and proteins from one lipid monolayer to the other (described as flip-flop movement)?
Not quite — the answer is C.
Membrane lipids occasionally move from one leaflet to the other (transverse diffusion), though this is slow and rare. Proteins, with their large hydrophilic domains, cannot cross the hydrophobic core, so they move only laterally.
Q17AIPMT 2008 Prelims · Q117
In germinating seeds fatty acids are degraded exclusively in the
Not quite — the answer is C.
In fat-storing germinating seeds, beta-oxidation of fatty acids and the glyoxylate cycle take place in glyoxysomes, specialised microbodies that convert fat to sugar. In plants mitochondria do not carry out fatty acid beta-oxidation for this purpose.
Q18AIPMT 2008 Prelims · Q118
The two sub-units of ribosome remain united at a critical ion level of
Not quite — the answer is D.
Mg²⁺ ions hold the large and small ribosomal subunits together; when Mg²⁺ falls below a critical level the subunits dissociate. Calcium, copper and manganese do not play this role.
Q19AIPMT 2008 Prelims · Q119
Thorn of Bougainvillea and tendril of Cucurbita are example of
Not quite — the answer is C.
Both the thorn of Bougainvillea and the tendril of Cucurbita are modified axillary branches (stem origin) that now do different jobs, which makes them homologous, the NCERT example of divergent evolution. Analogous organs would share function but differ in origin.
Q20AIPMT 2008 Prelims · Q120
Haploids are more suitable for mutation studies than the diploids. This is because
Not quite — the answer is A.
A haploid has only one copy of each gene, so a recessive mutation cannot be masked by a dominant allele and shows up directly in the phenotype. Mutagen penetration does not depend on ploidy, so that option gives a wrong reason.
Q21AIPMT 2008 Prelims · Q121
Which one of the following pairs of nitrogenous bases of nucleic acids, is wrongly matched with the category mentioned against it?
Not quite — the answer is A.
Adenine is a purine but thymine is a pyrimidine, so grouping them together as purines is wrong. Adenine and guanine are purines, and cytosine, thymine and uracil are pyrimidines, so the other pairs are correct.
Q22AIPMT 2008 Prelims · Q122
Which one of the following conditions in humans is correctly matched with its chromosomal abnormality/ linkage?
Not quite — the answer is B.
Klinefelter syndrome results from an extra X chromosome in a male, giving 44 autosomes + XXY (47 chromosomes). Down syndrome is trisomy 21 (44 + XO is Turner syndrome), colour blindness is X-linked, and erythroblastosis foetalis comes from Rh incompatibility, not sex linkage.
Q23AIPMT 2008 Prelims · Q123
In the DNA molecules
Not quite — the answer is A.
The two polynucleotide chains of the double helix run antiparallel, one 5'→3' and the other 3'→5'. By Chargaff's rule, purines always equal pyrimidines and A always equals T in double-stranded DNA, so those options are false.
Q24AIPMT 2008 Prelims · Q124
What is true about the isolated small tribal populations?
Not quite — the answer is B.
A small isolated tribe has a small gene pool and no gene flow, so marriage within the tribe leads to inbreeding depression and can contribute to population decline. Inbreeding actually helps recessive hereditary disorders like colour blindness spread, and acquired traits such as a wrestler's muscles are not inherited.
Q25AIPMT 2008 Prelims · Q125
Which one of the following scientist's name is correctly matched with the theory put forth by him?
Not quite — the answer is B.
August Weismann proposed the continuity of germplasm, holding that only changes in germ cells are inherited. Pangenesis was Darwin's idea, inheritance of acquired characters was Lamarck's, and de Vries gave the mutation theory.
Q26AIPMT 2008 Prelims · Q126
Which one of the following is incorrect about the characteristics of protobionts (coacervates and microspheres) as envisaged in the abiogenic origin of life?
Not quite — the answer is B.
Protobionts such as coacervates and microspheres were membrane-bounded droplets that kept an internal environment partly separate from their surroundings, but they lacked genetic material and so could not truly reproduce. Maintaining an internal environment is a correct property, not the incorrect one.
Q27AIPMT 2008 Prelims · Q127
Darwin's Finches are an excellent example of
Not quite — the answer is B.
From a seed-eating ancestor, Galapagos finches diversified into many species with different beaks suited to different diets, which NCERT presents as the classic case of adaptive radiation. They are not intermediates between two groups, so 'connecting links' does not apply.
Q28AIPMT 2008 Prelims · Q128
Which one of the following pairs of items correctly belongs to the category of organs mentioned against it?
Not quite — the answer is D.
Nephridia in earthworm and Malpighian tubules in cockroach are both excretory organs. Wings of bee and crow are analogous (not homologous), thorn and tendril are homologous (not analogous), and the blind spot is just where the optic nerve leaves the eye, not a vestigial organ.
Q29AIPMT 2008 Prelims · Q129
The fruit is chambered, developed from inferior ovary and has seeds with succulent testa in
Not quite — the answer is B.
Pomegranate (balausta) develops from an inferior ovary, is divided into chambers, and its edible part is the juicy outer seed coat (testa). In orange the edible part is juicy hairs from the endocarp, and orange develops from a superior ovary.
Q30AIPMT 2008 Prelims · Q130
The C₄ plants are photosynthetically more efficient than C₃ plants because
Not quite — the answer is C.
C₄ plants use PEP carboxylase in mesophyll cells to capture CO₂ and concentrate it in bundle sheath cells, so RuBisCO works near saturation and photorespiration is minimal; any CO₂ that leaks out can be refixed by PEP carboxylase. Their efficiency comes from this CO₂-concentrating mechanism, not from simply having more chloroplasts, and their CO₂ compensation point is lower, not higher.
Q31AIPMT 2008 Prelims · Q131
The chemiosmotic coupling hypothesis of oxidative phosphorylation proposes that adenosine triphosphate (ATP) is formed because
Not quite — the answer is D.
According to the chemiosmotic hypothesis, the electron transport chain pumps protons into the intermembrane space, building a proton gradient across the inner mitochondrial membrane. Protons flowing back through F₀-F₁ ATP synthase drive ATP formation; high-energy protein intermediates were part of an older, rejected hypothesis.
Q32AIPMT 2008 Prelims · Q132
Dry indehiscent single-seeded fruit formed from bicarpellary syncarpous inferior ovary is
Not quite — the answer is C.
A cypsela, as in sunflower and other Asteraceae, is a dry, one-seeded indehiscent fruit from a bicarpellary, syncarpous, inferior ovary. A cremocarp splits into two mericarps, and a caryopsis comes from a monocarpellary superior ovary.
Q33AIPMT 2008 Prelims · Q135
Vascular tissues in flowering plants develop from
Not quite — the answer is C.
In Hanstein's histogen theory the plerome, the central histogen of the apex, gives rise to the stele including the vascular tissues. Dermatogen forms the epidermis, periblem forms the cortex, and phellogen is the cork cambium of secondary growth.
Q34AIPMT 2008 Prelims · Q136
In leaves of C₄ plants malic acid synthesis during CO₂ fixation occurs in
Not quite — the answer is C.
In C₄ plants PEP carboxylase in the mesophyll cells fixes CO₂ into oxaloacetate, which is converted to malic acid there. The malate then moves to the bundle sheath, where it is decarboxylated, so the bundle sheath is where malate breaks down, not where it is made.
Q35AIPMT 2008 Prelims · Q138
Endosperm is consumed by developing embryo in the seed of
Not quite — the answer is D.
Pea seeds are non-albuminous: the endosperm is fully used up by the developing embryo, and food ends up stored in the cotyledons. Maize, coconut and castor seeds keep endosperm in the mature seed (albuminous seeds).
Q36AIPMT 2008 Prelims · Q140
The energy releasing process in which the substrate is oxidised without an external electron acceptor is called
Not quite — the answer is B.
In fermentation the NADH produced during glycolysis is reoxidised by passing electrons to an organic product of the substrate itself (pyruvate or acetaldehyde), so no external acceptor such as O₂ is needed. Aerobic respiration depends on oxygen as the terminal external acceptor.
Q37AIPMT 2008 Prelims · Q141
Replum is present in the ovary of flower of
Not quite — the answer is C.
In mustard (Brassicaceae) the ovary is bicarpellary with parietal placentation and becomes two-chambered by a false septum called the replum. Pea has a monocarpellary ovary with marginal placentation and no replum.
Q38AIPMT 2008 Prelims · Q142
The fleshy receptacle of syconus of fig encloses a number of
Not quite — the answer is B.
The fig syconus is a composite fruit: a hollow fleshy receptacle encloses many tiny achene-like fruitlets formed from the female flowers. Mericarps are parts of a schizocarpic fruit like a cremocarp, not the fruitlets of a fig.
Q39AIPMT 2008 Prelims · Q143
Electrons from excited chlorophyll molecule of photosystem II are accepted first by
Not quite — the answer is D.
Electrons leaving the PS II reaction centre are passed to a quinone acceptor (via pheophytin to plastoquinone) before reaching the cytochrome b₆f complex. Ferredoxin accepts electrons on the PS I side, much later in the chain.
Q40AIPMT 2008 Prelims · Q144
Which type of white blood cells are concerned with the release of histamine and the natural anticoagulant heparin?
Not quite — the answer is C.
Basophils secrete histamine, serotonin and heparin and take part in inflammatory reactions, as NCERT states. Eosinophils resist infections and are linked to allergies but are not the source of heparin.
Q41AIPMT 2008 Prelims · Q145
Which one of the following is the true description about an animal concerned?
Not quite — the answer is A.
The cockroach has 10 pairs of spiracles, 2 on the thorax and 8 on the abdomen. In earthworm the gizzard comes before the stomach, the frog has no neck (only head and trunk), and in rat the right kidney lies slightly higher than the left.
Q42AIPMT 2008 Prelims · Q147
What is vital capacity of our lungs?
Not quite — the answer is A.
Vital capacity is ERV + TV + IRV, the most air that can be breathed out after a forced inspiration. Total lung capacity is VC + RV, so VC = TLC − RV. IRV + ERV leaves out the tidal volume, and IRV + TV is the inspiratory capacity.
Q43AIPMT 2008 Prelims · Q149
Which one of the following items gives its correct total number?
Not quite — the answer is B.
Humans have 12 pairs of ribs, and the last 2 pairs (11th and 12th), which are not joined to the sternum, are the floating ribs, giving 4 in total. There are 7 cervical vertebrae, and proteins are built from 20 amino acids.
Q44AIPMT 2008 Prelims · Q151
Given below are four methods (i - iv) and their modes of action (a - d) in achieving contraception. Select their correct matching from the four options that follow Method: (i) The pill (ii) Condom (iii) Vasectomy (iv) Copper T Mode of Action: (a) Prevents sperms reaching cervix (b) Prevents implantation (c) Prevents ovulation (d) Semen contains no sperms
Not quite — the answer is B.
Pills block ovulation (and implantation), a condom is a barrier that stops sperm reaching the cervix, vasectomy cuts the vas deferens so semen has no sperms, and Copper-T acts in the uterus against fertilisation and implantation. Matching a condom to 'semen contains no sperms' confuses a barrier with a surgical method.
Q45AIPMT 2008 Prelims · Q153
During the propagation of a nerve impulse, the action potential results from the movement of
Not quite — the answer is A.
At the stimulated site voltage-gated Na⁺ channels open and Na⁺ rushes in from the extracellular fluid, reversing the polarity and producing the action potential. K⁺ moving outward causes repolarisation afterwards, not the action potential itself.
Q46AIPMT 2008 Prelims · Q154
The blood calcium level is lowered by the deficiency of
Not quite — the answer is B.
Parathormone raises blood Ca²⁺ by mobilising it from bone and increasing reabsorption, so its deficiency causes hypocalcaemia. Calcitonin lowers blood calcium, so a lack of it would tend to raise the level.
Q47AIPMT 2008 Prelims · Q155
The most active phagocytic white blood cells are
Not quite — the answer is A.
NCERT states that neutrophils and monocytes (which become macrophages) are phagocytic cells that destroy foreign organisms. Lymphocytes are responsible for immune responses, not phagocytosis.
Q48AIPMT 2008 Prelims · Q157
In humans, blood passes from the post caval to the diastolic right atrium of heart due to
Not quite — the answer is A.
During atrial diastole the pressure in the right atrium falls below that in the venae cavae, so blood flows down this pressure gradient into the atrium. SA node stimulation starts contraction (systole), not the filling of a relaxed atrium.
Q49AIPMT 2008 Prelims · Q158
In humans, at the end of the first meiotic division, the male germ cells differentiate into the
Not quite — the answer is C.
A primary spermatocyte finishes meiosis I to form two haploid secondary spermatocytes, which then complete meiosis II to form spermatids. Spermatids come only after the second meiotic division.
Q50AIPMT 2008 Prelims · Q159
Which one of the following is resistant to enzyme action?
Not quite — the answer is D.
Pollen exine is made of sporopollenin, among the most resistant organic materials known, which no known enzyme can degrade; this is why pollen survives as fossils. Cuticle, cork and wood are eventually broken down by microbial enzymes.
Q51AIPMT 2008 Prelims · Q160
The length of different internodes in a culm of sugarcane is variable because of
Not quite — the answer is A.
In grasses like sugarcane, intercalary meristems at the base of internodes cause each internode to elongate on its own, so internodes end up with different lengths. The shoot apical meristem adds new nodes and internodes but does not set their individual lengths.
Q52AIPMT 2008 Prelims · Q161
Which one of the following pairs of plant structures has haploid number of chromosomes?
Not quite — the answer is C.
The egg and the antipodal cells both belong to the haploid female gametophyte (embryo sac). The secondary nucleus is diploid (two polar nuclei fused), and the megaspore mother cell and nucellus are diploid sporophytic tissue.
Q53AIPMT 2008 Prelims · Q162
What does the filiform apparatus do at the entrance into ovule?
Not quite — the answer is B.
The filiform apparatus, the finger-like wall ingrowths at the micropylar tip of the synergids, guides the pollen tube into a synergid. It does not lead the tube from the synergid to the egg; the male gametes are released inside the synergid.
Q54AIPMT 2008 Prelims · Q163
Unisexuality of flowers prevents
Not quite — the answer is B.
A unisexual flower carries either anthers or stigma, so pollination within the same flower (autogamy) is impossible. In monoecious plants, pollen can still pass between male and female flowers of the same plant, so geitonogamy is not prevented; only dioecy blocks both.
Q55AIPMT 2008 Prelims · Q164
Which extraembryonic membrane in humans prevents desiccation of the embryo inside the uterus?
Not quite — the answer is A.
The amnion surrounds the embryo with amniotic fluid, keeping it moist and cushioned. The chorion mainly helps form the placenta, and the allantois and yolk sac do not keep the embryo hydrated.
Q56AIPMT 2008 Prelims · Q165
Which one of the following statements is incorrect about menstruation?
Not quite — the answer is C.
Menstrual blood normally does not clot, because fibrinolytic activity in the endometrium dissolves the clots. The first menstruation is called menarche, and about 40 ml of blood is lost per cycle. At menopause gonadotropins rise because the ovaries stop responding.
Q57AIPMT 2008 Prelims · Q166
The haemoglobin of a human foetus
Not quite — the answer is A.
Foetal haemoglobin (α₂γ₂) binds O₂ more strongly than adult HbA, so it can take oxygen from maternal blood across the placenta. It is still a tetramer of four globin chains, not two.
Q58AIPMT 2008 Prelims · Q167
Consider the statements given below regarding contraception and answer as directed thereafter (a) Medical Termination of Pregnancy (MTP) during first trimester is generally safe (b) Generally chances of conception are nil until mother breast-feeds the infant upto two years (c) Intrauterine devices like copper-T are effective contraceptives (d) Contraception pills may be taken upto one week after coitus to prevent conception Which two of the above statements are correct?
Not quite — the answer is D.
NCERT notes that MTP in the first trimester (up to 12 weeks) is relatively safe, and that IUDs like Copper-T are effective contraceptives. Lactational amenorrhoea is reliable only up to about six months after birth, and emergency pills must be taken within 72 hours, not a week.
Q59AIPMT 2008 Prelims · Q168
In human adult females oxytocin
Not quite — the answer is A.
Oxytocin causes strong contractions of the uterine smooth muscle during birth and triggers milk ejection. It is made in the hypothalamus and released from the posterior pituitary, not the anterior. Mammary gland growth is driven by other hormones such as prolactin, oestrogen and progesterone.
Q60AIPMT 2008 Prelims · Q172
Which one of the following is not observed in biodiversity hotspots?
Not quite — the answer is D.
Hotspots are defined by very high species richness, high endemism and severe threat of habitat and species loss. With so many species packed together, interspecific competition is high, not low.
Q61AIPMT 2008 Prelims · Q175
Consider the following statements concerning food chains (a) Removal of 80% tigers from an area resulted in greatly increased growth of vegetation (b) Removal of most of the carnivores resulted in an increased population of deers (c) The length of food chains is generally limited to 3-4 trophic levels due to energy loss (d) The length of food chains may vary from 2 to 8 trophic levels. Which two of the above statements are correct?
Not quite — the answer is B.
Removing predators lets herbivores like deer multiply, and because only about 10% of energy passes to each higher level, food chains are usually limited to 3-4 trophic levels. Removing tigers would reduce vegetation through overgrazing rather than increase it, and 8-level chains do not occur.
Q62AIPMT 2008 Prelims · Q180
A transgenic food crop which may help in solving the problem of night blindness in developing countries is
Not quite — the answer is A.
Golden rice is engineered to make β-carotene, the precursor of vitamin A, whose deficiency causes night blindness. Flavr Savr tomato was modified for delayed ripening and Bt crops for insect resistance, neither of which targets vitamin A.
Q63AIPMT 2008 Prelims · Q184
Trichoderma harzianum has proved a useful microorganism for
Not quite — the answer is A.
Trichoderma species are free-living fungi in the root ecosystem, and NCERT lists them as biocontrol agents against several plant pathogens. Gene transfer in plants is done with Agrobacterium, not Trichoderma.
Q64AIPMT 2008 Prelims · Q185
Gel electrophoresis is used for
Not quite — the answer is C.
In agarose gel electrophoresis negatively charged DNA fragments move toward the anode, and the gel's sieving effect separates them by size, smaller ones travelling farther. Cutting DNA is done by restriction enzymes and joining by ligase.
Q65AIPMT 2008 Prelims · Q186
To which type of barriers under innate immunity, do the saliva in the mouth and the tears from the eyes, belong?
Not quite — the answer is D.
NCERT classes acid in the stomach, saliva in the mouth and tears from the eyes as physiological barriers that stop microbial growth. Physical barriers are the skin and the mucus lining of the tracts.
Q66AIPMT 2008 Prelims · Q187
Match the disease in Column I with the appropriate items (pathogen/prevention/treatment) in Column II Column I: (a) Amoebiasis (b) Diphtheria (c) Cholera (d) Syphilis Column II: (i) Treponema pallidum (ii) Use only sterilized food and water (iii) DPT Vaccine (iv) Use oral rehydration therapy
Not quite — the answer is A.
Amoebiasis spreads through contaminated food and water, so sterilised food and water prevent it; diphtheria is prevented by the DPT vaccine; cholera dehydration is treated with oral rehydration; and syphilis is caused by Treponema pallidum. Treponema is a bacterium, not the cause of amoebiasis.
Q67AIPMT 2008 Prelims · Q189
Which one of the following pairs of codons is correctly matched with their function or the signal for the particular amino acid?
Not quite — the answer is C.
UAA, UAG and UGA are the three stop codons. In the other options only one codon of each pair fits: UCA codes serine, GUU codes valine and ACG codes threonine.
Q68AIPMT 2008 Prelims · Q191
Cry 1 endotoxins obtained from Bacillus thuringiensis are effective against
Not quite — the answer is A.
NCERT notes that the proteins coded by cryIAc and cryIIAb control cotton bollworms, and cryIAb controls the corn borer. Nematodes are controlled by RNA interference in the NCERT example, not by Cry toxins.
Q69AIPMT 2008 Prelims · Q192
Modern detergents contain enzyme preparations of
Not quite — the answer is C.
Laundry detergents work in alkaline wash water, so the proteases and lipases added to them come from alkaliphilic microbes (for example alkaliphilic Bacillus species) whose enzymes stay active at high pH. Enzymes from acidophiles would lose activity in an alkaline detergent solution.
Q70AIPMT 2008 Prelims · Q193
The linking of antibiotic resistance gene with the plasmid vector became possible with
Not quite — the answer is B.
Cohen and Boyer joined the antibiotic resistance gene to the plasmid using DNA ligase, which seals the sugar-phosphate backbone between the cut ends. Restriction endonucleases only cut DNA; they do not join fragments.
Q71AIPMT 2008 Prelims · Q194
Which one of the following proved effective for biological control of nematodal disease in plants?
Not quite — the answer is A.
Paecilomyces lilacinus is a nematophagous fungus that parasitises the eggs of root-knot and cyst nematodes and is sold as a bionematicide. Pisolithus tinctorius is an ectomycorrhizal fungus, and Gliocladium virens and Pseudomonas cepacia are mainly used against fungal pathogens.
Q72AIPMT 2008 Prelims · Q195
Main objective of production/use of herbicide resistant GM crops is to
Not quite — the answer is B.
Herbicide-tolerant crops survive a broad-spectrum herbicide that kills the weeds around them, so fields can be weeded chemically instead of by manual labour. Herbicides are still sprayed, so 'without the use of herbicides' is wrong.
Q73AIPMT 2008 Prelims · Q197
Human insulin is being commercially produced from a transgenic species of
Not quite — the answer is B.
Eli Lilly produced human insulin (1983) by inserting DNA for the A and B chains into plasmids of E. coli and then joining the chains. Saccharomyces has also been used, but the standard NCERT answer is Escherichia coli.
Q74AIPMT 2008 Prelims · Q198
Cornea transplant in humans is almost never rejected. This is because
Not quite — the answer is C.
The cornea is avascular, so immune cells and antibodies carried in the blood rarely reach the graft and rejection seldom occurs. The cornea is living tissue with nucleated cells, so the 'non-living' and 'enucleated' options are false.
Q75AIPMT 2008 Prelims · Q199
Which of the following pairs of organs includes only the endocrine glands?
Not quite — the answer is B.
Parathyroid and adrenal glands are purely endocrine. The pancreas also has an exocrine part, and the ovary and testis also produce gametes, so each of the other pairs includes an organ with non-endocrine functions.
Q76AIPMT 2008 Prelims · Q200
What is antisense technology?
Not quite — the answer is D.
Antisense technology uses an RNA (or DNA) strand complementary to a target mRNA. It pairs with the mRNA and blocks its translation, silencing the gene, the same principle as RNA interference in NCERT. Making DNA from RNA is reverse transcription, not antisense technology.
Biology: not in the current NEET syllabus
These 19 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
19 questions. Tap an option to check your answer.
Q1AIPMT 2008 Prelims · Q133
The rupture and fractionation do not usually occur in the water column in vessel/tracheids during the ascent of sap because of
Not quite — the answer is C.
Water molecules cohere strongly to one another and adhere to the walls of the xylem conduits, so the column under tension stays continuous instead of breaking. Transpiration pull creates the tension but does not by itself stop the column from breaking.
Q2AIPMT 2008 Prelims · Q137
Importance of day length in flowering of plants was first shown in
Not quite — the answer is C.
Garner and Allard (1920) discovered photoperiodism while working with the Maryland Mammoth variety of tobacco, which flowered only when days became short. The other plants were not the organisms in which the effect was first shown.
Q3AIPMT 2008 Prelims · Q139
Nitrogen fixation in root nodules of Alnus is brought about by
Not quite — the answer is D.
Non-leguminous Alnus forms root nodules with Frankia, a nitrogen-fixing actinomycete. Rhizobium-group bacteria such as Bradyrhizobium nodulate legumes, and Clostridium is a free-living anaerobic fixer.
Q4AIPMT 2008 Prelims · Q146
Which one of the following is the correct matching of the site of action on the given substrate, the enzyme acting upon it and the end product?
Not quite — the answer is C.
In the small intestine pancreatic α-amylase breaks starch down to the disaccharide maltose. Trypsin is a protease and does not act on triglycerides, and pepsin works in the stomach, turning proteins into peptones and proteoses, not amino acids.
Q5AIPMT 2008 Prelims · Q148
Which one of the following is the correct difference between Rod Cells and Cone Cells of our retina? (Feature : Rod Cells | Cone Cells)
Not quite — the answer is D.
Rods contain rhodopsin and handle twilight (scotopic) vision, while cones handle colour and sharp daylight (photopic) vision. Cones, not rods, are packed in the fovea and give high acuity, so any option that swaps the pigments or puts rods in the fovea is wrong.
Q6AIPMT 2008 Prelims · Q152
What will happen if the secretion of parietal cells of gastric glands is blocked with an inhibitor?
Not quite — the answer is D.
Parietal (oxyntic) cells secrete HCl and intrinsic factor, and HCl is what converts pepsinogen into active pepsin. Pepsinogen itself is made by chief (peptic) cells, so blocking parietal cells would not reduce pepsinogen output.
Q7AIPMT 2008 Prelims · Q156
Earthworms have no skeleton but during burrowing, the anterior end becomes turgid and acts as a hydraulic skeleton. It is due to
Not quite — the answer is B.
The fluid-filled coelomic compartments of the earthworm, squeezed by body-wall muscles, make the anterior end turgid and act as a hydrostatic skeleton. Setae only grip the soil during locomotion and do not provide the turgidity.
Q8AIPMT 2008 Prelims · Q169
Which one of the following is the correct percentage of the two (out of the total of 4) green house gases that contribute to the total global warming?
Not quite — the answer is B.
The figures in the old NCERT pie chart were CO₂ 60%, CH₄ 20%, CFCs 14% and N₂O 6%. Only the pair CFCs 14% with methane 20% matches; the other options pair one right figure with a wrong one.
Q9AIPMT 2008 Prelims · Q170
Quercus species are the dominant component in
Not quite — the answer is B.
Oaks (Quercus) are the typical dominant trees of temperate broad-leaved deciduous forests, including those of the Himalayan temperate zone. Tropical rain forests are dominated by trees such as Dipterocarpus, and alpine zones by Rhododendron and junipers.
Q10AIPMT 2008 Prelims · Q171
About 70% of total global carbon is found in
Not quite — the answer is D.
The old NCERT carbon cycle section states that about 71% of global carbon is dissolved in the oceans, which act as the main reservoir and regulate atmospheric CO₂. Forests, while important carbon sinks, hold much less.
Q11AIPMT 2008 Prelims · Q173
World summit on Sustainable Development (2002) was held in
Not quite — the answer is A.
The 2002 World Summit on Sustainable Development was held in Johannesburg, South Africa, where 190 countries pledged to reduce the rate of biodiversity loss by 2010. The 1992 Earth Summit was held in Rio de Janeiro, Brazil, which is the usual trap.
Q12AIPMT 2008 Prelims · Q176
According to Central Pollution Control Board (CPCB), which particulate size in diameter (in micrometers) of the air pollutants is responsible for greatest harm to human health?
Not quite — the answer is B.
CPCB identifies particulate matter of 2.5 µm or less (PM2.5) as the most harmful, because these particles are inhaled deep into the lungs and cause breathing problems, inflammation and damage. Larger particles are mostly trapped in the upper respiratory tract.
Q13AIPMT 2008 Prelims · Q178
A lake near a village suffered heavy mortality of fishes within a few days. Consider the following reasons for this (a) Lots of urea and phosphate fertilizer were used in the crops in the vicinity (b) The area was sprayed with DDT by an aircraft (c) The lake water turned green and stinky (d) Phytoplankton populations in the lake declined initially thereby greatly reducing photosynthesis Which two of the above were the main causes of fish mortality in the lake?
Not quite — the answer is D.
Fertiliser runoff rich in nitrates and phosphates triggers algal blooms (the water turns green), and when the algae die their decomposition uses up dissolved oxygen and gives off foul smells, killing fish within days. DDT biomagnifies slowly over time and would not explain sudden mass death alongside these signs, and phytoplankton increase rather than decline.
Q14AIPMT 2008 Prelims · Q179
Consider the following four statements (a - d) about certain desert animals such as kangaroo rat. (a) They have dark colour and high rate of reproduction and excrete solid urine (b) They do not drink water, breathe at a slow rate to conserve water and have their body covered with thick hairs (c) They feed on dry seeds and do not require drinking water (d) They excrete very concentrated urine and do not use water to regulate body temperature Which two of the above statements for such animals are true?
Not quite — the answer is B.
The kangaroo rat lives on dry seeds and meets its water needs from metabolic water produced by oxidising its food. It produces highly concentrated urine and does not sweat to cool itself. Dark colour, solid urine and a thick hair coat are not its adaptations.
Q15AIPMT 2008 Prelims · Q181
Bacterial leaf blight of rice is caused by a species of
Not quite — the answer is B.
Bacterial leaf blight of rice is caused by Xanthomonas oryzae pv. oryzae. Alternaria is a fungus, and Erwinia mostly causes soft rots.
Q16AIPMT 2008 Prelims · Q182
Which one of the following is linked to the discovery of Bordeaux mixture as a popular fungicide?
Not quite — the answer is C.
Millardet in Bordeaux, France, noticed that vines sprayed with a copper sulphate and lime mixture escaped downy mildew (Plasmopara viticola), which led to Bordeaux mixture being used as a fungicide. Bacterial leaf blight is bacterial and is not controlled this way.
Q17AIPMT 2008 Prelims · Q183
Which one of the following is being tried in India as a biofuel substitute for fossil fuels?
Not quite — the answer is B.
Jatropha curcas seeds yield a non-edible oil that can be turned into biodiesel, and it has been promoted in India as a biofuel crop on wastelands. Azadirachta (neem) is used as a biopesticide, not a fuel.
Q18AIPMT 2008 Prelims · Q188
Consider the following statements about biomedical technologies (a) During open heart surgery blood is circulated in the heart-lung machine (b) Blockage in coronary arteries is removed by angiography (c) Computerised Axial Tomography (CAT) shows detailed internal structure as seen in a section of body (d) X-ray provides clear and detailed images of organs like prostate glands and lungs Which two of the above statements are correct?
Not quite — the answer is D.
A heart-lung machine takes over circulation during open-heart surgery, and CAT scanning builds cross-sectional images of internal structures. Angiography only images vessels (angioplasty removes blockages), and plain X-rays do not give detailed soft-tissue images of organs such as the prostate.
Q19AIPMT 2008 Prelims · Q196
Consider the following four measures (a - d) that could be taken to successfully grow chick-pea in an area where bacterial blight disease is common (a) Spray with Bordeaux mixture (b) Control of the insect vector of the disease pathogen (c) Use of only disease-free seeds (d) Use of varieties resistant to the disease Which two of the above measures can control the disease?
Not quite — the answer is D.
Bacterial blight is seed-borne, so sowing disease-free seed and growing resistant varieties are the reliable controls. Bordeaux mixture is mainly a fungicide, and this disease has no insect vector to target.
Chapter-wise split of AIPMT 2008 Prelims Biology
- Evolution6 Q
- Animal Kingdom5 Q
- Sexual Reproduction in Flowering Plants5 Q
- Biotechnology and Its Applications5 Q
- Cell: The Unit of Life4 Q
- Molecular Basis of Inheritance4 Q
- Morphology of Flowering Plants4 Q
- Plant Kingdom3 Q
- Biological Classification3 Q
- Photosynthesis in Higher Plants3 Q
- Body Fluids and Circulation3 Q
- Chemical Coordination and Integration3 Q
- Human Reproduction3 Q
- Microbes in Human Welfare3 Q
- Human Health and Disease3 Q
- Biomolecules2 Q
- Principles of Inheritance and Variation2 Q
- Respiration in Plants2 Q
- Anatomy of Flowering Plants2 Q
- Breathing and Exchange of Gases2 Q
- Reproductive Health2 Q
- Biotechnology: Principles and Processes2 Q
- Structural Organisation in Animals1 Q
- Locomotion and Movement1 Q
- Neural Control and Coordination1 Q
- Biodiversity and Conservation1 Q
- Ecosystem1 Q
About this paper
How many questions from AIPMT 2008 Prelims are on this page?
180 questions (Physics 40, Chemistry 45, Biology 95). 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 2008 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 2008 Prelims useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books. 28 questions here are on topics NCERT has since dropped; they are listed separately so you can skip them.
More NEET and AIPMT papers
Practise thousands more questions chapter-wise in the RankUp app.
- NEET 2021
- NEET 2020
- NEET 2019
- NEET 2018
- NEET 2017
- NEET 2017 (Gujarati paper)
- NEET 2016 Phase 2
- NEET 2016 Phase 1
- AIPMT 2015
- AIPMT 2014
- NEET 2013
- AIPMT 2012 Mains
- AIPMT 2012 Prelims
- AIPMT 2011 Mains
- AIPMT 2011 Prelims
- AIPMT 2010 Mains
- AIPMT 2010 Prelims
- AIPMT 2009 Prelims
- AIPMT 2007 Prelims
- AIPMT 2006 Prelims
- AIPMT 2005 Prelims
- All NEET PYQs
