AIPMT 2009 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
189 questions from the AIPMT 2009 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 2009 Prelims Physics questions
45 questions. Tap an option to check your answer.
Q1AIPMT 2009 Prelims · Q1
In the nuclear decay given below: (A, Z)X → (A, Z+1)Y → (A−4, Z−1)B* → (A−4, Z−1)B the particles emitted in the sequence are: [notation: (mass number, atomic number)]
Not quite — the answer is D.
X to Y keeps the mass number and raises the atomic number by 1, which is β⁻ emission. Y to B* lowers A by 4 and Z by 2, which is α emission. B* to B changes neither A nor Z, so the excited nucleus de-excites by γ emission, giving the order β, α, γ.
Q2AIPMT 2009 Prelims · Q2
A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity ω. If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity:
Not quite — the answer is A.
No external torque acts, so angular momentum is conserved: MR²ω = (MR² + 2mR²)ω′. Hence ω′ = ωM/(M + 2m). The added masses increase the moment of inertia, so the angular velocity must fall, which rules out the expression larger than ω.
Q3AIPMT 2009 Prelims · Q3
In thermodynamic processes which of the following statements is not true?
Not quite — the answer is A.
In an isochoric process the volume is held constant; constant pressure describes an isobaric process. The other statements (constant temperature in isothermal, PV^γ = constant and thermal insulation in adiabatic) are all correct.
Q4AIPMT 2009 Prelims · Q4
The number of photo electrons emitted for light of a frequency ν (higher than the threshold frequency ν₀) is proportional to:
Not quite — the answer is B.
Above threshold, each photon ejects at most one electron, so the number of photoelectrons depends on the number of photons arriving per second, i.e. the intensity. The frequency (and ν − ν₀) decides the kinetic energy of the electrons, not their number.
Q5AIPMT 2009 Prelims · Q5
A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a/2 will be:
Not quite — the answer is C.
v = ω√(a² − x²) with ω = 2π/T. At x = a/2, v = (2π/T)√(3a²/4) = (2π/T)(√3a/2) = πa√3/T. The value with 2T in the denominator forgets that ω already carries the factor 2π.
Q6AIPMT 2009 Prelims · Q7
A body, under the action of a force F⃗ = 6î − 8ĵ + 10k̂, acquires an acceleration of 1 m/s². The mass of this body must be:
Not quite — the answer is C.
|F| = √(6² + 8² + 10²) = √200 = 10√2 N. With a = 1 m/s², m = F/a = 10√2 kg. Simply adding the components or taking 2√10 is a mistake in the magnitude.
Q7AIPMT 2009 Prelims · Q9
If F⃗ is the force acting on a particle having position vector r⃗ and τ⃗ be the torque of this force about the origin, then:
Not quite — the answer is B.
τ⃗ = r⃗ × F⃗ is perpendicular to the plane containing r⃗ and F⃗. A vector perpendicular to both has zero dot product with each, so r⃗ · τ⃗ = 0 and F⃗ · τ⃗ = 0.
Q8AIPMT 2009 Prelims · Q10
The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T₁ and T₂ (T₁ > T₂). The rate of heat transfer, dQ/dt, through the rod in a steady state is given by:
Not quite — the answer is C.
Fourier's law for steady conduction gives dQ/dt = kA(T₁ − T₂)/L: the heat current grows with area and temperature difference and falls with length. Forms with L in the numerator or A in the denominator have the wrong dependence.
Q9AIPMT 2009 Prelims · Q11
A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength:
Not quite — the answer is C.
The longest detectable wavelength is λ = hc/Eg = 1240 eV nm/2.5 eV ≈ 496 nm ≈ 5000 Å. Only photons of shorter wavelength have enough energy, so 4000 Å (400 nm) is detected. 6000 Å and the nm values of 4000 and 6000 are all longer than the cut-off.
Q10AIPMT 2009 Prelims · Q12
If the dimensions of a physical quantity are given by Mᵃ Lᵇ Tᶜ, then the physical quantity will be:
Not quite — the answer is D.
Pressure = force/area = MLT⁻²/L² = ML⁻¹T⁻², so a = 1, b = −1, c = −2. Velocity has no mass dimension (a = 0), acceleration has no mass dimension, and force is MLT⁻², so the other pairings are wrong.
Q11AIPMT 2009 Prelims · Q13
A transistor is operated in common-emitter configuration at Vc = 2 V such that a change in the base current from 100 µA to 200 µA produces a change in the collector current from 5 mA to 10 mA. The current gain is:
Not quite — the answer is C.
Current gain β = ΔIc/ΔIb = (10 − 5) mA/(200 − 100) μA = 5 × 10⁻³ A/10⁻⁴ A = 50. A value of 100 or more comes from mishandling the milli and micro prefixes.
Q12AIPMT 2009 Prelims · Q14
The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is:
Not quite — the answer is A.
T − mg = ma gives 28000 − 2000 × 10 = 2000a, so a = 4 m s⁻² upwards. The tension exceeds the weight, so the acceleration must be upward; 14 m s⁻² ignores gravity.
Q13AIPMT 2009 Prelims · Q15
Four identical thin rods each of mass M and length l, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is :
Not quite — the answer is D.
Each rod's centre is l/2 from the axis: I = Ml²/12 + M(l/2)² = Ml²/3 (parallel-axis theorem). Four rods give 4Ml²/3. Forgetting the parallel-axis term would give only Ml²/3 for the frame.
Q14AIPMT 2009 Prelims · Q16
Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats is:
Not quite — the answer is A.
Wave speed v = √(T/μ) = √(20/10⁻³) ≈ 141.4 m/s. Fundamentals: f = v/2l gives 141.4/1.032 ≈ 137.0 Hz and 141.4/0.982 ≈ 144.0 Hz. Beat frequency = 144 − 137 = 7 per second.
Q15AIPMT 2009 Prelims · Q17
The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an:
Not quite — the answer is C.
One α emission lowers Z by 2 and A by 4; two β⁻ emissions raise Z by 2 and leave A unchanged. Net: same Z, A reduced by 4, so the daughter is an isotope of the parent, not an isobar (same mass number).
Q16AIPMT 2009 Prelims · Q19
The Figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If t₁ is the time for the planet to move from C to D and t₂ is the time to move from A to B then:
Not quite — the answer is B.
Kepler's second law: the radius vector sweeps equal areas in equal times, so time is proportional to area swept. Area SCD = 2 × area SAB gives t₁ = 2t₂. The fact that the planet moves faster near the sun is already accounted for by the equal-area law.
Q17AIPMT 2009 Prelims · Q20
A black body at 227°C radiates heat at the rate of 7 cal/cm² s. At a temperature of 727°C, the rate of heat radiated in the same units will be:
Not quite — the answer is B.
Stefan's law: E ∝ T⁴ with absolute temperatures. 227°C = 500 K and 727°C = 1000 K, so E rises by (1000/500)⁴ = 16: 7 × 16 = 112 cal/cm² s. Using Celsius temperatures directly gives wrong ratios.
Q18AIPMT 2009 Prelims · Q21
The driver of a car travelling with speed 30 m/sec towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 m/s, the frequency of reflected sound as heard by driver is:
Not quite — the answer is B.
The hill receives f₁ = 600 × 330/(330 − 30) from the approaching car and reflects it; the driver, moving towards the hill, hears f = f₁ × (330 + 30)/330 = 600 × 360/300 = 720 Hz. The reflected sound is heard at a higher, not lower, frequency.
Q19AIPMT 2009 Prelims · Q23
A rectangular, a square, a circular and an elliptical loop, all in the (x−y) plane, are moving out of a uniform magnetic field with a constant velocity, v⃗ = v î. The magnetic field is directed along the negative z axis direction. The induced emf, during the passage of these loops, out of the field region, will not remain constant for:
Not quite — the answer is A.
Induced emf = Bvl, where l is the length of the loop lying on the field boundary as it leaves. For a rectangle or square moving perpendicular to a side, l is a constant side length, so the emf is constant; for a circle or ellipse the chord at the boundary changes continuously, so the emf changes. A shrinking flux alone does not make the emf vary; it is the rate of change of flux that matters.
Q20AIPMT 2009 Prelims · Q24
A galvanometer having a coil resistance of 60 Ω shows full scale deflection when a current of 1.0 amp passes through it. It can be converted into an ammeter to read currents upto 5.0 amp by:
Not quite — the answer is C.
Shunt S = IgG/(I − Ig) = (1 × 60)/(5 − 1) = 15 Ω connected in parallel, so the extra 4 A bypasses the coil. A series resistance would make a voltmeter, not an ammeter.
Q21AIPMT 2009 Prelims · Q25
Power dissipated in an LCR series circuit connected to an a.c. source of emf ε is:
Not quite — the answer is D.
Average power P = εrms Irms cos φ with I = ε/Z and cos φ = R/Z, so P = ε²R/Z² = ε²R/[R² + (Lω − 1/Cω)²]. Using only one factor of Z (the square-root form) gives the wrong power.
Q22AIPMT 2009 Prelims · Q26
Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ, −σ and σ respectively. If VA, VB and VC denote the potentials of the three shells, then for c = a + b, we have:
Not quite — the answer is D.
VA = (σ/ε₀)(a − b + c) = 2aσ/ε₀ for c = a + b. VC = (σ/ε₀)(a² − b² + c²)/c = (σ/ε₀)(2a² + 2ab)/(a + b) = 2aσ/ε₀. VB = (σ/ε₀)(a²/b − b + c) = (σ/ε₀)(a²/b + a), which differs from 2aσ/ε₀ unless a = b, so VC = VA ≠ VB.
Q23AIPMT 2009 Prelims · Q27
An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?
Not quite — the answer is D.
Mass leaving per second = (mass per length) × speed = mv. KE delivered per second = ½(mv)v² = ½mv³. Taking ½mv² treats m as the mass per second instead of per unit length.
Q24AIPMT 2009 Prelims · Q28
A bar magnet having a magnetic moment of 2 × 10⁴ J T⁻¹ is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 × 10⁻⁴ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is:
Not quite — the answer is B.
Work = MB(cos 0° − cos 60°) = 2 × 10⁴ × 6 × 10⁻⁴ × ½ = 6 J. The value 12 J is MB itself, which would be the work only for a 90° turn.
Q25AIPMT 2009 Prelims · Q29
In a Rutherford scattering experiment when a projectile of charge z₁ and mass M₁ approaches a target nucleus of charge z₂ and mass M₂, the distance of closest approach is r₀. The energy of the projectile is:
Not quite — the answer is A.
At the distance of closest approach all kinetic energy becomes electrostatic potential energy: E = kz₁z₂e²/r₀. So, for a given r₀, the energy is directly proportional to z₁z₂ and does not depend on the masses.
Q26AIPMT 2009 Prelims · Q30
Monochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is 9 mW. The number of photons arriving per sec on the average at a target irradiated by this beam is:
Not quite — the answer is A.
Energy per photon = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(667 × 10⁻⁹) ≈ 2.98 × 10⁻¹⁹ J. Photons per second = 9 × 10⁻³/2.98 × 10⁻¹⁹ ≈ 3 × 10¹⁶. Forgetting the milli prefix gives 3 × 10¹⁹.
Q27AIPMT 2009 Prelims · Q31
A wave in a string has an amplitude of 2 cm. The wave travels in the + ve direction of x axis with a speed of 128 m/sec. and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the wave is:
Not quite — the answer is C.
λ = 4/5 = 0.8 m, so k = 2π/0.8 ≈ 7.85 m⁻¹ and ω = kv = 7.85 × 128 ≈ 1005 s⁻¹. A wave moving towards +x has the form sin(kx − ωt), so y = 0.02 sin(7.85x − 1005t); the '+' form travels towards −x and 15.7 uses λ = 0.4 m.
Q28AIPMT 2009 Prelims · Q32
Which one of the following equations of motion represents simple harmonic motion? Where k, k₀, k₁ and a are all positive.
Not quite — the answer is A.
Put X = x + a; then acceleration = −kX, which is SHM about x = −a with ω² = k. Acceleration = +kx or k(x + a) points away from equilibrium, and the x² term makes the motion non-linear.
Q29AIPMT 2009 Prelims · Q33
A student measures the terminal potential difference (V) of a cell (of emf ε and internal resistance r) as a function of the current (I) flowing through it. The slope, and intercept, of the graph between V and I, then, respectively, equal:
Not quite — the answer is A.
Terminal voltage V = ε − Ir, a straight line in I. Its slope is −r and its intercept on the V axis (I = 0) is ε. A positive slope r would mean the voltage rises as more current is drawn.
Q30AIPMT 2009 Prelims · Q34
If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is:
Not quite — the answer is D.
Diamagnetic materials get a weak induced moment opposite to the applied field, so they are pushed towards weaker field regions. This happens regardless of which pole is used, so they are repelled by both poles.
Q31AIPMT 2009 Prelims · Q35
A bus is moving with a speed of 10 m s⁻¹ on a straight road. A scooterist wishes to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist chase the bus?
Not quite — the answer is D.
The scooter must close 1000 m in 100 s, so its speed relative to the bus is 10 m s⁻¹. Its ground speed = 10 + 10 = 20 m s⁻¹. Answering 10 m s⁻¹ gives only the relative speed.
Q32AIPMT 2009 Prelims · Q37
The internal energy change in a system that has absorbed 2 kcal of heat and done 500 J of work is:
Not quite — the answer is C.
First law: ΔU = Q − W = 2 × 4200 − 500 = 7900 J (1 cal = 4.2 J). Adding the work instead of subtracting gives 8900 J.
Q33AIPMT 2009 Prelims · Q38
Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be:
Not quite — the answer is B.
Capacitors in series: 1/C_eq = 3/C, so C_eq = C/3. The same charge sits on each, so the applied voltage divides equally and each can take V before breakdown: total 3V.
Q34AIPMT 2009 Prelims · Q39
An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, 1 kg first part moving with a velocity of 12 m s⁻¹ and 2 kg second part moving with a velocity of 8 m s⁻¹. If the third part flies off with a velocity of 4 m s⁻¹, its mass would be:
Not quite — the answer is D.
Momenta of the two pieces are 1 × 12 = 12 and 2 × 8 = 16 kg m s⁻¹ at right angles, with resultant √(12² + 16²) = 20 kg m s⁻¹. The third piece must carry an equal and opposite 20 kg m s⁻¹, so m = 20/4 = 5 kg.
Q35AIPMT 2009 Prelims · Q40
A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is S₁ and that covered in the first 20 seconds is S₂, then:
Not quite — the answer is B.
From rest under constant force, S = ½at², so S ∝ t². Doubling the time from 10 s to 20 s makes S₂ = 4S₁. S₂ = 3S₁ is the distance in the second 10 s, not the total.
Q36AIPMT 2009 Prelims · Q41
A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (g = 10 m/s²)
Not quite — the answer is D.
Initial KE = ½ × 1 × 20² = 200 J. PE at 18 m = 1 × 10 × 18 = 180 J. Energy lost to air friction = 200 − 180 = 20 J.
Q37AIPMT 2009 Prelims · Q42
A conducting circular loop is placed in a uniform magnetic field 0.04 T with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 2 mm/s. The induced emf in the loop when the radius is 2 cm is:
Not quite — the answer is D.
Emf = B dA/dt = B × 2πr × dr/dt = 0.04 × 2π × 0.02 × 2 × 10⁻³ = 3.2π × 10⁻⁶ V = 3.2π μV. Missing the factor 2 from d(πr²)/dt gives 1.6π μV.
Q38AIPMT 2009 Prelims · Q43
The magnetic force acting on a charged particle of charge −2 μC in a magnetic field of 2 T acting in y direction, when the particle velocity is (2î + 3ĵ) × 10⁶ m s⁻¹, is:
Not quite — the answer is D.
F⃗ = q(v⃗ × B⃗). Only the x component of velocity matters: (2 × 10⁶ î) × (2ĵ) = 4 × 10⁶ k̂. Multiplying by q = −2 × 10⁻⁶ C gives −8k̂ N, i.e. 8 N in −z direction. Ignoring the negative charge gives +z.
Q39AIPMT 2009 Prelims · Q44
Two bodies of mass 1 kg and 3 kg have position vectors î + 2ĵ + k̂ and −3î − 2ĵ + k̂, respectively. The centre of mass of this system has a position vector:
Not quite — the answer is A.
r_cm = (m₁r₁ + m₂r₂)/(m₁ + m₂) = [(î + 2ĵ + k̂) + 3(−3î − 2ĵ + k̂)]/4 = (−8î − 4ĵ + 4k̂)/4 = −2î − ĵ + k̂.
Q40AIPMT 2009 Prelims · Q45
The electric potential at a point (x, y, z) is given by V = −x²y − xz³ + 4. The electric field E⃗ at that point is:
Not quite — the answer is D.
E⃗ = −∇V. −∂V/∂x = 2xy + z³, −∂V/∂y = x², −∂V/∂z = 3xz². So E⃗ = î(2xy + z³) + ĵx² + k̂3xz². A sign slip on the z³ term gives the (2xy − z³) form.
Q41AIPMT 2009 Prelims · Q46
The mean free path of electrons in a metal is 4 × 10⁻⁸ m. The electric field which can give on an average 2 eV energy to an electron in the metal will be in units of V/m:
Not quite — the answer is C.
An electron gains energy eEλ while travelling one mean free path. Setting eEλ = 2 eV gives E = 2 V/(4 × 10⁻⁸ m) = 5 × 10⁷ V/m. The very small values (10⁻¹¹) would come from multiplying by λ instead of dividing.
Q42AIPMT 2009 Prelims · Q47
The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between:
Not quite — the answer is C.
Number of lines from level n = n(n − 1)/2 = 6, so n = 4. The longest wavelength needs the smallest energy gap, which is between n = 4 and n = 3. The 3 → 1 or 2 → 1 lines have larger gaps and shorter wavelengths.
Q43AIPMT 2009 Prelims · Q48
Under the influence of a uniform magnetic field, a charged particle moves with constant speed V in a circle of radius R. The time period of rotation of the particle:
Not quite — the answer is B.
qVB = mV²/R gives R = mV/qB, and T = 2πR/V = 2πm/qB. Since R grows in proportion to V, T depends on neither speed nor radius.
Q44AIPMT 2009 Prelims · Q49
The electric field part of an electromagnetic wave in a medium is represented by Ex = 0; Ey = 2.5 (N/C) cos[(2π × 10⁶ rad/s) t − (π × 10⁻² rad/m) x]; Ez = 0. The wave is:
Not quite — the answer is B.
Comparing with E₀cos(ωt − kx): ω = 2π × 10⁶ rad/s gives f = 10⁶ Hz, and k = π × 10⁻² rad/m gives λ = 2π/k = 200 m. The (ωt − kx) form means propagation along +x. Taking f = 2π × 10⁶ Hz confuses ω with f.
Q45AIPMT 2009 Prelims · Q50
A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be:
Not quite — the answer is A.
Released from rest with the spring unstretched, the block moves until it is momentarily at rest again. Energy conservation: Mgx = ½kx², so the maximum extension is x = 2Mg/k. Mg/k is only the equilibrium extension (where the block moves fastest), not the maximum.
Chapter-wise split of AIPMT 2009 Prelims Physics
- System of Particles and Rotational Motion4 Q
- Laws of Motion3 Q
- Waves3 Q
- Moving Charges and Magnetism3 Q
- Electrostatic Potential and Capacitance3 Q
- Work, Energy and Power3 Q
- Nuclei2 Q
- Thermodynamics2 Q
- Dual Nature of Radiation and Matter2 Q
- Oscillations2 Q
- Thermal Properties of Matter2 Q
- Semiconductor Electronics: Materials, Devices and Simple Circuits2 Q
- Electromagnetic Induction2 Q
- Magnetism and Matter2 Q
- Atoms2 Q
- Current Electricity2 Q
- Motion in a Straight Line2 Q
- Units and Measurements1 Q
- Gravitation1 Q
- Alternating Current1 Q
- Electromagnetic Waves1 Q
AIPMT 2009 Prelims Chemistry questions
42 questions. Tap an option to check your answer.
Q1AIPMT 2009 Prelims · Q151
The state of hybridization of C2, C3, C5 and C6 of the hydrocarbon CH₃–C(CH₃)₂–CH=CH–CH(CH₃)–C≡CH (numbered from the right: C≡CH carbons are C1 and C2, the CH(CH₃) carbon is C3, CH=CH are C4 and C5, the C(CH₃)₂ carbon is C6, end CH₃ is C7) is in the following sequence:
Not quite — the answer is D.
C2 is a triple-bonded carbon (sp). C3 is the CH(CH₃) carbon with only single bonds (sp³). C5 is a double-bonded carbon (sp²). C6 is the quaternary C(CH₃)₂ carbon (sp³). So the sequence is sp, sp³, sp², sp³.
Q2AIPMT 2009 Prelims · Q152
Oxidation numbers of P in PO₄³⁻, of S in SO₄²⁻ and that of Cr in Cr₂O₇²⁻, are respectively :
Not quite — the answer is D.
PO₄³⁻: x − 8 = −3, x = +5. SO₄²⁻: x − 8 = −2, x = +6. Cr₂O₇²⁻: 2x − 14 = −2, x = +6. So +5, +6 and +6.
Q3AIPMT 2009 Prelims · Q154
Which of the following reactions is an example of nucleophilic substitution reaction?
Not quite — the answer is D.
In RX + KOH(aq) → ROH + KX the nucleophile OH⁻ replaces the halide ion, a nucleophilic substitution. Reaction with Na is the Wurtz coupling, with H₂ is a reduction, and with Mg forms a Grignard reagent.
Q4AIPMT 2009 Prelims · Q156
Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states?
Not quite — the answer is B.
3d⁵4s² (manganese) has seven electrons that can take part in bonding and shows oxidation states from +2 up to +7, the widest range here. 3d⁵4s¹ (Cr) goes up to +6, and the others to +4 and +5.
Q5AIPMT 2009 Prelims · Q157
The stability of +1 oxidation state increases in the sequence:
Not quite — the answer is D.
Because of the inert pair effect, the ns² pair becomes less available down group 13, so the +1 state becomes more stable from Al to Tl: Al < Ga < In < Tl. Tl⁺ is the most stable +1 ion of the group.
Q6AIPMT 2009 Prelims · Q158
Given: (i) Cu²⁺ + 2e⁻ → Cu, E° = 0.337 V (ii) Cu²⁺ + e⁻ → Cu⁺, E° = 0.153 V Electrode potential, E° for the reaction, Cu⁺ + e⁻ → Cu, will be:
Not quite — the answer is D.
Combine free energies, not potentials: ΔG° for Cu⁺ + e⁻ → Cu = ΔG°(i) − ΔG°(ii) = (−2 × 0.337F) − (−1 × 0.153F) = −0.521F. With n = 1, E° = +0.52 V. Simply subtracting the potentials gives 0.18 V and adding them gives 0.49 V, both wrong.
Q7AIPMT 2009 Prelims · Q159
For the reaction N₂ + 3H₂ → 2NH₃, if d[NH₃]/dt = 2 × 10⁻⁴ mol L⁻¹ s⁻¹, the value of −d[H₂]/dt would be :
Not quite — the answer is D.
Rate = −(1/3)d[H₂]/dt = (1/2)d[NH₃]/dt. So −d[H₂]/dt = (3/2) × 2 × 10⁻⁴ = 3 × 10⁻⁴ mol L⁻¹ s⁻¹. Multiplying by 3 alone gives 6 × 10⁻⁴.
Q8AIPMT 2009 Prelims · Q160
Consider the following reaction: ethanol —(PBr₃)→ X —(alc. KOH)→ Y —((i) H₂SO₄, room temperature; (ii) H₂O, heat)→ Z The product Z is:
Not quite — the answer is C.
PBr₃ converts ethanol to bromoethane (X); alcoholic KOH eliminates HBr to give ethene (Y). Cold conc. H₂SO₄ adds across the double bond giving ethyl hydrogen sulphate, which on boiling with water is hydrolysed back to ethanol (Z). Ethyl hydrogen sulphate is only the intermediate.
Q9AIPMT 2009 Prelims · Q161
The energy absorbed by each molecule (A₂) of a substance is 4.4 × 10⁻¹⁹ J and bond energy per molecule is 4.0 × 10⁻¹⁹ J. The kinetic energy of the molecule per atom will be:
Not quite — the answer is D.
Energy left after breaking the bond = 4.4 × 10⁻¹⁹ − 4.0 × 10⁻¹⁹ = 0.4 × 10⁻¹⁹ J per molecule. It is shared by the two atoms: 2.0 × 10⁻²⁰ J per atom. 4.0 × 10⁻²⁰ J forgets to divide between the two atoms.
Q10AIPMT 2009 Prelims · Q162
Amongst the elements with following electronic configurations, which one of them may have the highest ionization energy?
Not quite — the answer is D.
[Ne]3s²3p³ is phosphorus: a small period-3 atom with a stable half-filled 3p subshell, so it has the highest ionization energy here. [Ar]3d¹⁰4s²4p³ (arsenic) is also half-filled but larger, and the 3p¹ and 3p² atoms (Al, Si) lose electrons more easily.
Q11AIPMT 2009 Prelims · Q163
In the reaction BrO₃⁻(aq) + 5Br⁻(aq) + 6H⁺(aq) → 3Br₂(l) + 3H₂O(l), the rate of appearance of bromine (Br₂) is related to rate of disappearance of bromide ions as following:
Not quite — the answer is D.
Rate = (1/3)d[Br₂]/dt = −(1/5)d[Br⁻]/dt, so d[Br₂]/dt = −(3/5)d[Br⁻]/dt. The minus sign is needed because [Br⁻] decreases while [Br₂] increases, and 5/3 inverts the stoichiometric ratio.
Q12AIPMT 2009 Prelims · Q164
A 0.0020 m aqueous solution of an ionic compound Co(NH₃)₅(NO₂)Cl freezes at −0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = −1.86 °C/m)
Not quite — the answer is D.
ΔTf = iKf m, so i = 0.00732/(1.86 × 0.0020) ≈ 1.97 ≈ 2. The compound is [Co(NH₃)₅(NO₂)]Cl, giving one complex cation and one Cl⁻ ion. A value of 3 would need two ionisable chlorides.
Q13AIPMT 2009 Prelims · Q165
What is the dominant intermolecular force or bond that must be overcome in converting liquid CH₃OH to a gas?
Not quite — the answer is D.
Methanol molecules are linked by O–H···O hydrogen bonds, which are the dominant forces to be overcome on boiling. Covalent bonds are broken only in a chemical change, and dipole–dipole and London forces are present but weaker.
Q14AIPMT 2009 Prelims · Q167
The segment of DNA which acts as the instrumental manual for the synthesis of the protein is:
Not quite — the answer is B.
A gene is the segment of DNA that carries the coded information for making a particular protein. Ribose is a sugar, and nucleosides and nucleotides are only the building units of nucleic acids.
Q15AIPMT 2009 Prelims · Q168
Maximum number of electrons in a subshell of an atom is determined by the following:
Not quite — the answer is D.
A subshell with quantum number l has (2l + 1) orbitals, each holding 2 electrons, so the maximum is 2(2l + 1) = 4l + 2. 2l + 1 counts orbitals, not electrons, and 2n² is for a whole shell.
Q16AIPMT 2009 Prelims · Q169
Half life period of a first-order reaction is 1386 seconds. The specific rate constant of the reaction is:
Not quite — the answer is B.
For a first-order reaction k = 0.693/t½ = 0.693/1386 = 5.0 × 10⁻⁴ s⁻¹ = 0.5 × 10⁻³ s⁻¹. 5.0 × 10⁻³ s⁻¹ is ten times too large.
Q17AIPMT 2009 Prelims · Q171
Al₂O₃ is reduced by electrolysis at low potentials and high currents. If 4.0 × 10⁴ amperes of current is passed through molten Al₂O₃ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency. At. mass of Al = 27 g mol⁻¹)
Not quite — the answer is A.
Charge Q = 4.0 × 10⁴ × 6 × 3600 = 8.64 × 10⁸ C. Al³⁺ + 3e⁻ → Al, so moles of Al = 8.64 × 10⁸/(3 × 96500) ≈ 2984 mol, mass ≈ 2984 × 27 ≈ 8.1 × 10⁴ g. Forgetting the factor 3 gives 2.4 × 10⁵ g.
Q18AIPMT 2009 Prelims · Q172
Benzene reacts with CH₃Cl in the presence of anhydrous AlCl₃ to form:
Not quite — the answer is D.
This is Friedel–Crafts alkylation: AlCl₃ generates CH₃⁺ which substitutes on the ring, giving C₆H₅CH₃ (toluene). Chlorobenzene would need Cl₂, benzyl chloride needs side-chain chlorination of toluene, and xylene needs a second methylation.
Q19AIPMT 2009 Prelims · Q173
Which of the following is not a permissible arrangement of electrons in an atom?
Not quite — the answer is B.
For l = 2 the magnetic quantum number can only take values from −2 to +2, so m = −3 is not allowed. The other sets obey all the rules (l < n, |m| ≤ l, s = ±1/2).
Q20AIPMT 2009 Prelims · Q175
Propionic acid with Br₂/P yields a dibromo product. Its structure would be:
Not quite — the answer is C.
In the Hell–Volhard–Zelinsky reaction, Br₂/P replaces the α-hydrogens (on the carbon next to –COOH). Propionic acid has two α-H, so the dibromo product is CH₃–CBr₂–COOH. Substitution at the β-carbon or on both carbons does not occur.
Q21AIPMT 2009 Prelims · Q176
The values of ΔH and ΔS for the reaction, C(graphite) + CO₂(g) → 2CO(g) are 170 kJ and 170 J K⁻¹, respectively. This reaction will be spontaneous at:
Not quite — the answer is B.
ΔG = ΔH − TΔS becomes negative when T > ΔH/ΔS = 170000 J/170 J K⁻¹ = 1000 K. Only 1110 K is above 1000 K. Remember to convert kJ to J before dividing.
Q22AIPMT 2009 Prelims · Q178
Predict the product : C₆H₅–NHCH₃ + NaNO₂ + HCl → Product
Not quite — the answer is D.
N-methylaniline is a secondary amine; with HNO₂ (from NaNO₂ + HCl) the N–H hydrogen is replaced by –N=O, giving N-nitroso-N-methylaniline, C₆H₅N(CH₃)–N=O. Ring nitrosation happens with tertiary aromatic amines such as N,N-dimethylaniline.
Q23AIPMT 2009 Prelims · Q179
HOH₂C–CH₂OH on heating with periodic acid gives:
Not quite — the answer is C.
Periodic acid cleaves the C–C bond between the two –OH bearing carbons of a 1,2-diol. Each CH₂OH becomes H₂C=O, so ethylene glycol gives two molecules of formaldehyde. Glyoxal would need oxidation without C–C cleavage.
Q24AIPMT 2009 Prelims · Q180
According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order?
Not quite — the answer is A.
N₂ (14 electrons) has bond order 3. Each extra electron enters an antibonding π* orbital and lowers the bond order by ½: N₂⁻ = 2.5, N₂²⁻ = 2. Increasing order: N₂²⁻ < N₂⁻ < N₂.
Q25AIPMT 2009 Prelims · Q181
Out of TiF₆²⁻, CoF₆³⁻, Cu₂Cl₂ and NiCl₄²⁻ (Z of Ti = 22, Co = 27, Cu = 29, Ni = 28) the colourless species are:
Not quite — the answer is B.
Ti⁴⁺ in TiF₆²⁻ is 3d⁰ and Cu⁺ in Cu₂Cl₂ is 3d¹⁰, so no d–d transition is possible and both are colourless. Co³⁺ (3d⁶) and Ni²⁺ (3d⁸) have partly filled d orbitals and are coloured.
Q26AIPMT 2009 Prelims · Q182
Which of the following molecules acts as a Lewis acid?
Not quite — the answer is D.
In (CH₃)₃B boron has only six valence electrons and an empty p orbital, so it accepts an electron pair: a Lewis acid. The O, P and N compounds each carry a lone pair and act as Lewis bases.
Q27AIPMT 2009 Prelims · Q183
The IUPAC name of the compound having the formula CH≡C–CH=CH₂ is:
Not quite — the answer is C.
CH₂=CH–C≡CH: numbering from the double-bond end gives the double bond at C1 and the triple bond at C3, and when locants tie the double bond gets the lower number. The name is but-1-en-3-yne, written here as 1-butene-3-yne. Names with 'yne' cited first are wrong since 'ene' precedes 'yne'.
Q28AIPMT 2009 Prelims · Q185
Which of the following does not show optical isomerism?
Not quite — the answer is A.
[Co(NH₃)₃Cl₃] is an MA₃B₃ octahedral complex; both its fac and mer forms have a mirror plane, so it is not optically active. [Co(en)₃]³⁺, cis-[Co(en)₂Cl₂]⁺ and one isomer of [Co(en)Cl₂(NH₃)₂]⁺ are chiral.
Q29AIPMT 2009 Prelims · Q187
The ionization constant of ammonium hydroxide is 1.77 × 10⁻⁵ at 298 K. Hydrolysis constant of ammonium chloride is:
Not quite — the answer is D.
For the salt of a weak base and strong acid, Kh = Kw/Kb = 1.0 × 10⁻¹⁴/1.77 × 10⁻⁵ = 5.65 × 10⁻¹⁰. Values with 10⁻¹² or 10⁻¹³ come from slips in the exponent.
Q30AIPMT 2009 Prelims · Q188
Consider the following reaction: Phenol —(Zn dust)→ X —(CH₃Cl, anhydrous AlCl₃)→ Y —(alkaline KMnO₄)→ Z The product Z is:
Not quite — the answer is B.
Distilling phenol with zinc dust removes the –OH, giving benzene (X). Friedel–Crafts methylation gives toluene (Y). Alkaline KMnO₄ oxidises the whole side chain to –COOH, giving benzoic acid (Z), not just benzaldehyde.
Q31AIPMT 2009 Prelims · Q189
The equivalent conductance of M/32 solution of a weak monobasic acid is 8.0 mhos cm² and at infinite dilution is 400 mhos cm². The dissociation constant of this acid is:
Not quite — the answer is D.
Degree of dissociation α = 8.0/400 = 0.02. Ka = Cα²/(1 − α) ≈ Cα² = (1/32)(0.02)² = 1.25 × 10⁻⁵. Forgetting to square α gives 6.25 × 10⁻⁴.
Q32AIPMT 2009 Prelims · Q190
The straight chain polymer is formed by:
Not quite — the answer is C.
(CH₃)₂SiCl₂ hydrolyses to (CH₃)₂Si(OH)₂, which has two –OH groups and condenses into long straight-chain silicones. CH₃SiCl₃ gives three –OH groups and a cross-linked polymer, while (CH₃)₃SiCl has only one and just caps chains.
Q33AIPMT 2009 Prelims · Q191
From the following bond energies: H–H bond energy: 431.37 kJ mol⁻¹ C=C bond energy: 606.10 kJ mol⁻¹ C–C bond energy: 336.49 kJ mol⁻¹ C–H bond energy: 410.50 kJ mol⁻¹ Enthalpy for the reaction H₂C=CH₂ + H–H → H₃C–CH₃ will be:
Not quite — the answer is B.
ΔH = bonds broken − bonds formed = [BE(C=C) + BE(H–H)] − [BE(C–C) + 2BE(C–H)] = (606.10 + 431.37) − (336.49 + 821.00) = −120.0 kJ mol⁻¹. The four C–H bonds common to both sides cancel out.
Q34AIPMT 2009 Prelims · Q192
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be:
Not quite — the answer is B.
2H₂ + O₂ → 2H₂O. 10 g H₂ = 5 mol and 64 g O₂ = 2 mol; 2 mol O₂ needs only 4 mol H₂, so O₂ is limiting and gives 4 mol of water. Taking H₂ as limiting would wrongly give 5 mol.
Q35AIPMT 2009 Prelims · Q193
Among the following which is the strongest oxidising agent?
Not quite — the answer is D.
F₂ has the highest standard reduction potential among the halogens (about +2.87 V), due to low F–F bond energy and high hydration energy of F⁻. So it is the strongest oxidising agent, and oxidising power falls F₂ > Cl₂ > Br₂ > I₂.
Q36AIPMT 2009 Prelims · Q194
In which of the following molecules / ions BF₃, NO₂⁻, NH₂⁻ and H₂O, the central atom is sp² hybridized?
Not quite — the answer is C.
BF₃ has three bond pairs and no lone pair, and NO₂⁻ has two σ bonds and one lone pair on N; both have three electron domains, so sp². NH₂⁻ and H₂O each have two bond pairs and two lone pairs (four domains), so they are sp³.
Q37AIPMT 2009 Prelims · Q195
Nitrobenzene can be prepared from benzene by using a mixture of conc. HNO₃ and conc. H₂SO₄. In the mixture, nitric acid acts as a/an:
Not quite — the answer is B.
H₂SO₄ is the stronger acid and protonates HNO₃: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻. Accepting a proton means nitric acid acts as a base here, producing the nitronium electrophile.
Q38AIPMT 2009 Prelims · Q196
Which of the following complex ions is expected to absorb visible light? (At. no. Zn = 30, Sc = 21, Ti = 22, Cr = 24)
Not quite — the answer is B.
Cr³⁺ is 3d³, so d–d transitions are possible and the complex absorbs visible light. Ti⁴⁺ and Sc³⁺ are 3d⁰ and Zn²⁺ is 3d¹⁰, so they have no d–d transitions and are colourless.
Q39AIPMT 2009 Prelims · Q197
What is the [OH⁻] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)₂?
Not quite — the answer is D.
HCl gives 20 × 0.050 = 1.0 mmol H⁺; Ba(OH)₂ gives 30 × 0.10 × 2 = 6.0 mmol OH⁻. After neutralisation 5.0 mmol OH⁻ remain in 50 mL: [OH⁻] = 0.10 M. Forgetting that Ba(OH)₂ gives two OH⁻ per formula unit leads to a wrong value.
Q40AIPMT 2009 Prelims · Q198
Trichloroacetaldehyde, CCl₃CHO reacts with chlorobenzene in presence of sulphuric acid and produces:
Not quite — the answer is C.
In conc. H₂SO₄, chloral condenses with two molecules of chlorobenzene at their para positions, losing water, to give DDT: (p-Cl–C₆H₄)₂CH–CCl₃. The CCl₃ group stays intact; it is not replaced by a third ring.
Q41AIPMT 2009 Prelims · Q199
For the reaction A + B → products, it is observed that: (a) on doubling the initial concentration of A only, the rate of reaction is also doubled and (b) on doubling the initial concentrations of both A and B, there is a change by a factor of 8 in the rate of the reaction. The rate of this reaction is given by:
Not quite — the answer is A.
Doubling [A] alone doubles the rate, so the order in A is 1. Doubling both raises the rate 8 times: 2¹ × 2ⁿ = 8, so n = 2 for B. Rate = k[A][B]².
Q42AIPMT 2009 Prelims · Q200
Which of the following hormones contains iodine?
Not quite — the answer is C.
Thyroxine, secreted by the thyroid gland, is an iodinated derivative of tyrosine. Testosterone is a steroid, adrenaline a catecholamine and insulin a polypeptide; none contain iodine.
Chemistry: not in the current NEET syllabus
These 7 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
7 questions. Tap an option to check your answer.
Q1AIPMT 2009 Prelims · Q36
Sodium has body centred packing. Distance between two nearest atoms is 3.7 Å. The lattice parameter is:
Not quite — the answer is A.
In a bcc lattice the nearest neighbours lie along the body diagonal, so 2r = d = √3a/2. a = 2 × 3.7/√3 ≈ 4.27 ≈ 4.3 Å. Doubling 3.7 Å treats the structure as touching along the cube edge.
Q2AIPMT 2009 Prelims · Q153
Lithium metal crystallises in a body centred cubic crystal. If the length of the side of the unit cell of lithium is 351 pm, the atomic radius of the lithium will be:
Not quite — the answer is A.
In bcc the atoms touch along the body diagonal: 4r = √3a. r = 1.732 × 351/4 ≈ 152 pm, matching 151.8 pm. 75.5 pm and 300.5 pm come from wrong geometric relations.
Q3AIPMT 2009 Prelims · Q155
In the case of alkali metals, the covalent character decreases in the order:
Not quite — the answer is C.
By Fajans' rules, larger anions are more easily polarised by the cation, so covalent character is highest for iodides and lowest for fluorides: MI > MBr > MCl > MF. The reverse order describes ionic character.
Q4AIPMT 2009 Prelims · Q166
Which of the following oxides is not expected to react with sodium hydroxide?
Not quite — the answer is A.
CaO is a basic oxide and does not react with the base NaOH. SiO₂ and B₂O₃ are acidic oxides and BeO is amphoteric, so all three dissolve in NaOH.
Q5AIPMT 2009 Prelims · Q170
Which one of the following is employed as a tranquilizer?
Not quite — the answer is D.
Equanil (meprobamate) is a tranquilizer used to relieve anxiety. Naproxen is an analgesic, tetracycline an antibiotic and chlorpheniramine an antihistamine.
Q6AIPMT 2009 Prelims · Q177
Copper crystallises in a face-centred cubic lattice with a unit cell length of 361 pm. What is the radius of copper atom in pm?
Not quite — the answer is D.
In fcc atoms touch along the face diagonal: 4r = √2a. r = 1.414 × 361/4 ≈ 128 pm. Using the bcc relation (√3a/4) gives 157 pm.
Q7AIPMT 2009 Prelims · Q186
Structures of some common polymers are given. Which one is not correctly presented?
Not quite — the answer is A.
Neoprene is polychloroprene, [–CH₂–C(Cl)=CH–CH₂–]ₙ, made from chloroprene (four carbons per unit). The given formula has an extra CH₂, so it is incorrect. The terylene, nylon 66 and teflon repeat units are correctly shown.
Chapter-wise split of AIPMT 2009 Prelims Chemistry
- Chemical Kinetics4 Q
- Electrochemistry3 Q
- Alcohols, Phenols and Ethers3 Q
- Structure of Atom3 Q
- Chemical Bonding and Molecular Structure3 Q
- Coordination Compounds3 Q
- Equilibrium3 Q
- Organic Chemistry: Some Basic Principles and Techniques2 Q
- Haloalkanes and Haloarenes2 Q
- Biomolecules2 Q
- Hydrocarbons2 Q
- Thermodynamics2 Q
- Redox Reactions1 Q
- The d- and f-Block Elements1 Q
- Classification of Elements and Periodicity in Properties1 Q
- Solutions1 Q
- Aldehydes, Ketones and Carboxylic Acids1 Q
- Amines1 Q
- Some Basic Concepts of Chemistry1 Q
AIPMT 2009 Prelims Biology questions
77 questions. Tap an option to check your answer.
Q1AIPMT 2009 Prelims · Q51
Which one of the following is correct pairing of a body part and the kind of muscle tissue that moves it?
Not quite — the answer is C.
The muscles of the iris that change pupil size are smooth, involuntary muscles. The abdominal wall is moved by skeletal (striated, voluntary) muscles, so pairing it with smooth muscle is wrong; the biceps is also skeletal and the heart wall is striated cardiac muscle.
Q2AIPMT 2009 Prelims · Q52
The epithelial tissue present on the inner surface of bronchioles and fallopian tubes is:
Not quite — the answer is B.
NCERT states that ciliated epithelium lines hollow organs such as bronchioles and fallopian tubes, where the beating cilia move mucus or the ovum in one direction. Squamous epithelium lines blood vessels and air sacs, not these tubes.
Q3AIPMT 2009 Prelims · Q55
Polyethylene glycol method is used for:
Not quite — the answer is D.
Polyethylene glycol (PEG) makes plant protoplast membranes permeable or fuse, so naked DNA can be taken up directly; it is a vectorless (direct) gene transfer method. It has nothing to do with biodiesel or seedless fruits.
Q4AIPMT 2009 Prelims · Q56
The floral formula ⊕ ⚥ K₍₅₎ C₍₅₎ A₅ G₍₂₎ is that of:
Not quite — the answer is C.
The formula shows an actinomorphic bisexual flower with five fused sepals, five fused petals, five epipetalous stamens and a bicarpellary syncarpous superior ovary, which is the Solanaceae formula (tobacco). Soybean and sunnhemp (Fabaceae) are zygomorphic with a monocarpellary pistil, and tulip (Liliaceae) has a trimerous perianth.
Q5AIPMT 2009 Prelims · Q57
Which one of the following groups of animals is bilaterally symmetrical and triploblastic?
Not quite — the answer is A.
Roundworms (Aschelminthes) are bilaterally symmetrical, triploblastic pseudocoelomates. Ctenophores and cnidarians are radially symmetrical and diploblastic, and sponges have no true tissue layers.
Q6AIPMT 2009 Prelims · Q58
Which one of the following is commonly used in transfer of foreign DNA into crop plants?
Not quite — the answer is B.
Agrobacterium tumefaciens naturally transfers a piece of its Ti plasmid (T-DNA) into dicot plant cells, and the disarmed Ti plasmid is used as a vector to deliver genes into crop plants. Meloidogyne incognita is a nematode pest and Trichoderma is a biocontrol fungus, not a vector.
Q7AIPMT 2009 Prelims · Q59
Which one of the following is the correct matching of the events occurring during menstrual cycle?
Not quite — the answer is B.
After ovulation the ruptured follicle becomes the corpus luteum, which secretes large amounts of progesterone during the luteal (secretory) phase. The proliferative phase and menstruation involve the endometrium, not the myometrium, so those pairings are wrong.
Q8AIPMT 2009 Prelims · Q60
Which one is the wrong pairing for the disease and its causal organism?
Not quite — the answer is D.
Late blight of potato is caused by the oomycete Phytophthora infestans; Alternaria solani causes early blight of potato. The other three pairings (rust by Puccinia, smut by Ustilago, root-knot by the nematode Meloidogyne) are correct.
Q9AIPMT 2009 Prelims · Q62
What is true about Bt toxin?
Not quite — the answer is D.
Bacillus thuringiensis makes Bt toxin as an inactive protoxin crystal; the alkaline pH of the insect gut solubilises and activates it, and it binds the midgut epithelium, creating pores that kill the insect. It is not active inside the bacterium, which is why the bacterium is unharmed, and it does not act on the ovaries.
Q10AIPMT 2009 Prelims · Q63
Peripatus is a connecting link between:
Not quite — the answer is B.
Peripatus (Onychophora) has annelid features such as nephridia and a soft segmented body along with arthropod features such as tracheae, claws and a haemocoel, so it links Annelida and Arthropoda.
Q11AIPMT 2009 Prelims · Q64
T.O. Diener discovered a:
Not quite — the answer is D.
In 1971 T.O. Diener discovered the viroid, the agent of potato spindle tuber disease, which is a free low-molecular-weight RNA without a protein coat. An infectious protein is a prion, not a viroid.
Q12AIPMT 2009 Prelims · Q65
Seminal plasma in humans is rich in:
Not quite — the answer is D.
NCERT states that secretions of the seminal vesicles, prostate and bulbourethral glands form seminal plasma, which is rich in fructose, calcium and certain enzymes. So the option saying it is poor in calcium is wrong.
Q13AIPMT 2009 Prelims · Q66
A fruit developed from hypanthodium inflorescence is called:
Not quite — the answer is B.
A hypanthodium (as in fig) is a hollow fleshy receptacle with tiny flowers inside, and the composite fruit formed from it is a syconus. Sorosis develops from a spike or spadix (mulberry, pineapple), while caryopsis and hesperidium are simple fruits.
Q14AIPMT 2009 Prelims · Q67
The cell junctions called tight, adhering and gap junctions are found in:
Not quite — the answer is B.
NCERT describes tight, adhering and gap junctions as the special junctions that hold epithelial cells together and link them structurally and functionally. Connective tissue cells are scattered in a matrix and do not form such sheets.
Q15AIPMT 2009 Prelims · Q68
What will happen if the stretch receptors of the urinary bladder wall are totally removed?
Not quite — the answer is C.
Micturition is a reflex: stretch receptors in the filling bladder send signals to the CNS, which then sends motor messages to contract the bladder muscle and relax the sphincter. Without stretch receptors, the reflex cannot start, so there will be no micturition. Urine would still reach the bladder from the ureters, but voluntary/reflex emptying would fail.
Q16AIPMT 2009 Prelims · Q70
The most popularly known blood grouping is the ABO grouping. It is named ABO and not ABC, because O in it refers to having:
Not quite — the answer is C.
Group O red cells carry neither antigen A nor antigen B on their surface, so 'O' stands for the absence (zero) of these antigens. The O allele is recessive, not overdominant, and antibodies are in the plasma, not on the RBCs.
Q17AIPMT 2009 Prelims · Q71
One of the synthetic auxin is:
Not quite — the answer is D.
NCERT states that IAA and IBA have been isolated from plants, whereas NAA (naphthalene acetic acid) and 2,4-D are synthetic auxins. GA is a gibberellin, not an auxin.
Q18AIPMT 2009 Prelims · Q72
A person likely to develop tetanus is immunised by administering:
Not quite — the answer is A.
When a quick immune response is needed, as in a person at risk of tetanus after an injury, NCERT says preformed antibodies or antitoxin are injected directly; this is passive immunisation. Weakened or dead germs give active immunity, which takes too long to protect someone already exposed.
Q19AIPMT 2009 Prelims · Q74
Biochemical Oxygen Demand (BOD) in a river water:
Not quite — the answer is C.
BOD is the amount of oxygen consumed when microbes oxidise the organic matter in a litre of water; sewage adds a lot of organic matter, so BOD rises. BOD is about oxygen use by microbes in general, not a count of Salmonella.
Q20AIPMT 2009 Prelims · Q75
The genetic defect – adenosine deaminase (ADA) deficiency may be cured permanently by:
Not quite — the answer is B.
NCERT notes that if the gene from bone marrow cells producing ADA is introduced into cells at early embryonic stages, it could be a permanent cure. Infusing engineered lymphocytes is not permanent because lymphocytes are not immortal and the infusion must be repeated; enzyme replacement is also only a treatment.
Q21AIPMT 2009 Prelims · Q76
Compared to blood our lymph has:
Not quite — the answer is B.
Lymph is tissue fluid drained into lymphatic vessels; it has no RBCs and is rich in lymphocytes (WBCs). It does contain plasma with fewer proteins, so 'plasma without proteins' and 'no plasma' are wrong.
Q22AIPMT 2009 Prelims · Q77
Sickle cell anemia is:
Not quite — the answer is B.
Sickle cell anaemia results from a single base substitution (GAG to GUG) in the β-globin gene, a point mutation, so glutamic acid at the sixth position is replaced by valine, not the other way round. The trait is autosomal recessive and mature RBCs have no nucleus.
Q23AIPMT 2009 Prelims · Q81
Which one of the following has maximum genetic diversity in India?
Not quite — the answer is A.
NCERT gives genetic diversity examples for India: more than 50,000 strains of rice and about 1,000 varieties of mango. Of the listed plants, mango is the one cited for very high genetic diversity in India; wheat is not given as an example.
Q24AIPMT 2009 Prelims · Q82
Oxygenic photosynthesis occurs in:
Not quite — the answer is A.
Oscillatoria is a cyanobacterium with chlorophyll a and photosystem II, so it splits water and releases oxygen like plants. Rhodospirillum, Chlorobium and Chromatium are purple or green photosynthetic bacteria that use H₂S or organic donors and do not release oxygen.
Q25AIPMT 2009 Prelims · Q83
There is no DNA in:
Not quite — the answer is A.
Mature human (mammalian) RBCs lose their nucleus and have no mitochondria, so they contain no DNA. An enucleated ovum still has mitochondrial DNA, and a sperm and hair root cells have nuclei.
Q26AIPMT 2009 Prelims · Q85
Tiger is not a resident in which one of the following national park?
Not quite — the answer is B.
Gir National Park in Gujarat is the home of the Asiatic lion and has no resident tigers. Sunderbans, Jim Corbett and Ranthambhore are tiger reserves.
Q27AIPMT 2009 Prelims · Q87
Synapsis occurs between:
Not quite — the answer is C.
Synapsis is the pairing of homologous chromosomes during zygotene of prophase I of meiosis, aided by the synaptonemal complex. Spindle fibres attach to kinetochores, which is a different event.
Q28AIPMT 2009 Prelims · Q89
What is not true for genetic code?
Not quite — the answer is D.
The genetic code is read in a contiguous way, codon after codon without punctuation, so a 'non-contiguous' reading is false. It is indeed nearly universal, degenerate (several codons per amino acid) and unambiguous (one codon codes for one amino acid).
Q29AIPMT 2009 Prelims · Q90
Which one of the following plants is monoecious?
Not quite — the answer is A.
In Pinus both male and female cones are borne on the same tree, so it is monoecious. Cycas, papaya and Marchantia have separate male and female plants (dioecious).
Q30AIPMT 2009 Prelims · Q91
Cyclic photophosphorylation results in the formation of
Not quite — the answer is C.
In cyclic photophosphorylation only PS I works and electrons return to the same reaction centre, so only ATP is made. NADPH and O₂ come from the non-cyclic pathway, which involves PS II and water splitting.
Q31AIPMT 2009 Prelims · Q92
The letter T in T-lymphocyte refers to:
Not quite — the answer is C.
T-lymphocytes arise in the bone marrow but mature and become immunocompetent in the thymus, hence the letter T. Tonsils and thyroid are not sites of T-cell maturation.
Q32AIPMT 2009 Prelims · Q93
Foetal ejection reflex in human female is induced by:
Not quite — the answer is B.
NCERT states that the signals for parturition originate from the fully developed foetus and the placenta, which induce mild uterine contractions called the foetal ejection reflex. This reflex then triggers oxytocin release, so oxytocin is a result, not the initial inducer.
Q33AIPMT 2009 Prelims · Q94
Anatomically fairly old dicotyledonous root is distinguished from the dicotyledonous stem by
Not quite — the answer is C.
Even after secondary growth, a dicot root keeps exarch protoxylem (towards the periphery), whereas the stem has endarch protoxylem (towards the centre). Both old root and stem have secondary xylem and phloem, so their presence or absence cannot distinguish them.
Q34AIPMT 2009 Prelims · Q95
Plasmodesmata are:
Not quite — the answer is C.
Plasmodesmata are cytoplasmic strands passing through the cell walls that connect the cytoplasm of neighbouring plant cells. The cementing layer between cells is the middle lamella, made of calcium pectate, not lignin.
Q35AIPMT 2009 Prelims · Q96
Removal of introns and joining the exons in a defined order in a transcription unit is called:
Not quite — the answer is D.
In eukaryotes the primary transcript (hnRNA) has introns, which are removed and the exons joined in order by splicing. Capping adds methyl guanosine at the 5′ end and tailing adds a poly-A tail at the 3′ end; neither removes introns.
Q36AIPMT 2009 Prelims · Q97
Phylogenetic system of classification is based on:
Not quite — the answer is D.
Phylogenetic classification groups organisms according to their evolutionary relationships, assuming members of a taxon share a common ancestor. Systems based mainly on morphology or floral characters are artificial or natural systems.
Q37AIPMT 2009 Prelims · Q98
Which part of human brain is concerned with the regulation of body temperature?
Not quite — the answer is C.
The hypothalamus contains centres that control body temperature, hunger and thirst. The cerebellum coordinates movement and balance, and the medulla controls breathing and heart rate.
Q38AIPMT 2009 Prelims · Q99
Semiconservative replication of DNA was first demonstrated in:
Not quite — the answer is A.
Meselson and Stahl (1958) grew E. coli in ¹⁵N and then ¹⁴N medium and showed by CsCl density gradient centrifugation that DNA replicates semiconservatively. Streptococcus pneumoniae was used by Griffith for transformation, not replication.
Q39AIPMT 2009 Prelims · Q100
Which one of the following pairs of animals comprises jawless fishes?
Not quite — the answer is B.
Lampreys (Petromyzon) and hagfishes (Myxine) are cyclostomes, the jawless vertebrates (Agnatha). Eels, guppies, mackerel and rohu are bony fishes with jaws.
Q40AIPMT 2009 Prelims · Q101
Which of the following is a pair of viral diseases?
Not quite — the answer is A.
Common cold is caused by rhinoviruses and AIDS by HIV, so both are viral. Typhoid and tuberculosis are bacterial, dysentery is bacterial or protozoan, and ringworm is fungal.
Q41AIPMT 2009 Prelims · Q102
Aerobic respiratory pathway is appropriately termed:
Not quite — the answer is B.
NCERT calls the respiratory pathway amphibolic because it breaks down substrates (catabolism) and also supplies intermediates for synthesis of fatty acids, amino acids etc. (anabolism). Calling it only catabolic misses this dual role.
Q42AIPMT 2009 Prelims · Q104
Cytoskeleton is made up of:
Not quite — the answer is C.
The cytoskeleton is a network of protein filaments — microtubules (tubulin), microfilaments (actin) and intermediate filaments — in the cytoplasm. Cellulose microfibrils are part of the plant cell wall, outside the cytoplasm.
Q43AIPMT 2009 Prelims · Q105
An example of axile placentation is:
Not quite — the answer is B.
In lemon (like china rose and tomato) the ovules are attached to a central axis in a multilocular ovary, which is axile placentation. Dianthus has free central, Argemone parietal and marigold (sunflower family) basal placentation.
Q44AIPMT 2009 Prelims · Q106
Which one of the following has haplontic life cycle?
Not quite — the answer is B.
In fungi like Ustilago the dominant phase is haploid and the only diploid stage is the zygote nucleus, which undergoes meiosis, giving a haplontic pattern. Mosses like Funaria and Polytrichum are haplo-diplontic, and wheat is diplontic.
Q45AIPMT 2009 Prelims · Q108
Which one of the following is considered important in the development of seed habit?
Not quite — the answer is A.
NCERT notes that heterospory, as in Selaginella and Salvinia, with the female gametophyte retained on the parent sporophyte, is a precursor to the seed habit. A free-living gametophyte is the opposite of what happens in seed plants.
Q46AIPMT 2009 Prelims · Q109
The annular and spirally thickened conducting elements generally develop in the protoxylem when the root or stem is:
Not quite — the answer is A.
Protoxylem forms in regions that are still elongating, and its annular and spiral wall thickenings allow the vessels to stretch with the growing organ without breaking. Metaxylem, with reticulate and pitted thickenings, forms later when the organ has stopped elongating and is maturing.
Q47AIPMT 2009 Prelims · Q112
A health disorder that results from the deficiency of thyroxine in adults and characterised by (i) a low metabolic rate, (ii) increase in body weight and (iii) tendency to retain water in tissues is:
Not quite — the answer is B.
Myxoedema is the specific adult form of thyroxine deficiency, with low BMR, weight gain and puffy water-retaining tissues. Cretinism is thyroxine deficiency in children, and simple goitre is thyroid enlargement due to iodine shortage.
Q48AIPMT 2009 Prelims · Q113
Mannitol is the stored food in:
Not quite — the answer is B.
Brown algae (Phaeophyceae) such as Fucus store food as laminarin or mannitol. Porphyra and Gracilaria are red algae storing floridean starch, and Chara is a green alga storing starch.
Q49AIPMT 2009 Prelims · Q114
Which one of the following pairs is wrongly matched?
Not quite — the answer is A.
Alcohol is produced by yeast fermentation using zymase enzymes; nitrogenase is the nitrogen-fixing enzyme and has no role in alcohol production. Pectinases clarify fruit juices, lipases are used in detergents and amylases are used for desizing textiles.
Q50AIPMT 2009 Prelims · Q115
Which of the following is not used as a biopesticide?
Not quite — the answer is C.
Xanthomonas campestris is a plant pathogen (black rot of crucifers) and is used commercially for xanthan gum, not as a biopesticide. Trichoderma, nucleopolyhedroviruses and Bacillus thuringiensis are all NCERT examples of biocontrol agents.
Q51AIPMT 2009 Prelims · Q116
Which one of the following is a vascular cryptogam?
Not quite — the answer is D.
Pteridophytes like Equisetum have xylem and phloem but no seeds or flowers, so they are vascular cryptogams. Marchantia is a bryophyte without vascular tissue, and Ginkgo and Cedrus are seed-bearing gymnosperms.
Q52AIPMT 2009 Prelims · Q117
In a standard ECG which one of the following alphabets is the correct representation of the respective activity of the human heart?
Not quite — the answer is C.
In an ECG the P-wave represents electrical excitation (depolarisation) of the atria, leading to atrial contraction. The QRS complex is ventricular depolarisation and the T-wave is ventricular repolarisation, so R is not repolarisation.
Q53AIPMT 2009 Prelims · Q118
Uric acid is the chief nitrogenous component of the excretory products of:
Not quite — the answer is B.
Insects such as the cockroach are uricotelic: they excrete nitrogen mainly as uric acid to save water. Frog and man are ureotelic, and earthworm excretes mostly urea and ammonia.
Q54AIPMT 2009 Prelims · Q119
Guard cells help in:
Not quite — the answer is A.
Guard cells open and close the stomatal pore and so regulate transpiration and gaseous exchange. Guttation happens through hydathodes, not stomata.
Q55AIPMT 2009 Prelims · Q122
Vegetative propagation in mint occurs by:
Not quite — the answer is C.
Mint propagates by suckers: lateral branches arise from the underground base of the stem, grow horizontally under the soil and then turn upward to form new leafy shoots. Runners are found in grasses and Oxalis, and offsets in aquatic plants like Pistia and Eichhornia.
Q56AIPMT 2009 Prelims · Q123
Select the incorrect statement from the following:
Not quite — the answer is C.
Baldness is a sex-influenced trait (expressed more often in males because of hormones), not a sex-limited trait, which is expressed in only one sex. The other statements are correct: galactosemia is an inborn error, genetic drift is strong in small populations, and linked genes do not assort independently.
Q57AIPMT 2009 Prelims · Q124
Cotyledons and testa respectively are edible parts in:
Not quite — the answer is D.
In groundnut we eat the fleshy cotyledons of the seed, and in pomegranate the juicy outer seed coat (testa). In litchi the edible part is the aril and in coconut it is the endosperm, so those pairs fail.
Q58AIPMT 2009 Prelims · Q125
Which one of the following statements is correct?
Not quite — the answer is C.
Malignant tumours shed cells that travel through blood to form secondary tumours elsewhere; this is metastasis. Benign tumours stay confined to their site, and heroin is a depressant that slows body functions.
Q59AIPMT 2009 Prelims · Q126
The correct sequence of spermatogenetic stages leading to the formation of sperms in a mature human testis is:
Not quite — the answer is A.
Spermatogonia divide by mitosis and grow into primary spermatocytes, which undergo meiosis to form secondary spermatocytes and then spermatids; spermatids change into sperms by spermiogenesis. Spermatids come after spermatocytes, so any sequence placing spermatid earlier is wrong.
Q60AIPMT 2009 Prelims · Q127
Use of anti-histamines and steroids give a quick relief from:
Not quite — the answer is D.
In allergy, mast cells release histamine and serotonin, so antihistamines, adrenalin and steroids quickly reduce the symptoms. They are not the standard treatment for nausea or headache.
Q61AIPMT 2009 Prelims · Q129
Which one of the following is the most likely root cause why menstruation is not taking place in regularly cycling human female?
Not quite — the answer is D.
Absence of menstruation in a regularly cycling woman is most likely due to pregnancy, which starts with fertilisation of the ovum. The retained corpus luteum, high hormone levels and thick endometrium are consequences of fertilisation, not its root cause.
Q62AIPMT 2009 Prelims · Q130
Globulins contained in human blood plasma are primarily involved in:
Not quite — the answer is D.
NCERT states that globulins are primarily involved in the defence mechanisms of the body (antibodies are gamma globulins). Albumins maintain osmotic balance and fibrinogen helps in clotting.
Q63AIPMT 2009 Prelims · Q131
Palisade parenchyma is absent in leaves of:
Not quite — the answer is D.
Sorghum is a monocot with an isobilateral leaf whose mesophyll is not differentiated into palisade and spongy layers. Mustard, soybean and gram are dicots with dorsiventral leaves that have palisade parenchyma.
Q64AIPMT 2009 Prelims · Q132
In barley stem vascular bundles are:
Not quite — the answer is A.
Barley is a monocot, and monocot stems have many conjoint, closed (no cambium) vascular bundles scattered in the ground tissue. Open bundles arranged in a ring are typical of dicot stems.
Q65AIPMT 2009 Prelims · Q133
Which one of the following is the correct matching of three items and their grouping category?
Not quite — the answer is A.
Each coxal (hip) bone of the pelvic girdle is formed by fusion of the ilium, ischium and pubis. Rhodopsin is a visual pigment, thiamine is a vitamin (thymine is the pyrimidine) and the cochlea is part of the inner ear, not an ossicle.
Q66AIPMT 2009 Prelims · Q135
In the case of peppered moth (Biston betularia) the black-coloured form became dominant over the light-coloured form in England during industrial revolution. This is an example of:
Not quite — the answer is D.
On soot-darkened tree trunks the dark moths were less visible to predatory birds and survived and reproduced more, so their frequency increased; NCERT uses this industrial melanism as an example of natural selection. It is camouflage, not mimicry of another species, and not inheritance of acquired characters.
Q67AIPMT 2009 Prelims · Q136
Transgenic plants are the ones:
Not quite — the answer is A.
Transgenic (GM) plants carry a foreign gene that was introduced into a cell, from which a whole plant is regenerated. Protoplast fusion produces somatic hybrids, not transgenic plants.
Q68AIPMT 2009 Prelims · Q139
Middle lamella is composed mainly of:
Not quite — the answer is B.
NCERT states that the middle lamella is a layer mainly of calcium pectate that holds neighbouring plant cells together. Muramic acid is found in bacterial cell walls, and hemicellulose is part of the primary wall.
Q69AIPMT 2009 Prelims · Q140
Elbow joint is an example of:
Not quite — the answer is A.
The elbow (like the knee) allows movement in one plane only, so it is a hinge joint. The shoulder is a ball and socket joint, and the joint between atlas and axis is a pivot joint.
Q70AIPMT 2009 Prelims · Q141
Which of the following is a symbiotic nitrogen fixer?
Not quite — the answer is B.
Frankia is an actinomycete that fixes nitrogen in root nodules of non-leguminous plants such as Alnus, a symbiotic association. Azotobacter is free-living, Azolla is a fern that hosts the nitrogen-fixer Anabaena, and Glomus is a mycorrhizal fungus that helps phosphorus uptake.
Q71AIPMT 2009 Prelims · Q142
Whose experiments cracked the DNA and discovered unequivocally that a genetic code is a triplet?
Not quite — the answer is D.
Nirenberg and Matthaei used cell-free protein synthesis with synthetic RNA (poly-U giving polyphenylalanine), which helped decode the codons and confirm the triplet code. Hershey and Chase proved DNA is the genetic material, not the nature of the code.
Q72AIPMT 2009 Prelims · Q143
Which one of the following types of organisms occupy more than one trophic level in a pond ecosystem?
Not quite — the answer is A.
Fish in a pond may feed on plants or algae (primary consumer) as well as on zooplankton or smaller fish (secondary or higher consumer), so they occupy several trophic levels. Phytoplankton are only producers and zooplankton are mainly primary consumers.
Q73AIPMT 2009 Prelims · Q144
Which one of the following acids is a derivative of carotenoids?
Not quite — the answer is C.
Abscisic acid is a sesquiterpene formed from the breakdown of carotenoids (xanthoxin pathway). IAA and IBA are indole compounds derived from tryptophan, and gibberellic acid is a diterpene not derived from carotenoids.
Q74AIPMT 2009 Prelims · Q145
The bacterium Bacillus thuringiensis is widely used in contemporary biology as:
Not quite — the answer is A.
Bacillus thuringiensis spores and its Bt toxin genes are used to kill insect larvae such as caterpillars, making it a biological insecticide. Dairy products are made with lactic acid bacteria, not Bt.
Q75AIPMT 2009 Prelims · Q146
An example of a seed with endosperm, perisperm, and caruncle is:
Not quite — the answer is C.
Castor seed is endospermic and has a caruncle, a spongy outgrowth near the micropyle; a thin remnant of nucellus (perisperm) is also reported. Cotton seeds bear hairs, not a caruncle.
Q76AIPMT 2009 Prelims · Q148
Point mutation involves:
Not quite — the answer is A.
A point mutation is a change in a single base pair of DNA, as in sickle cell anaemia. Duplication and larger deletions or insertions alter segments of chromosomes or DNA rather than a single base.
Q77AIPMT 2009 Prelims · Q150
The kind of tissue that forms the supportive structure in our pinna (external ears) is also found in:
Not quite — the answer is C.
The pinna is supported by cartilage, which NCERT says is also present in the tip of the nose, the outer ear joints and between adjacent vertebrae. Ear ossicles and vertebrae themselves are bone, and nails are keratin.
Biology: not in the current NEET syllabus
These 18 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
18 questions. Tap an option to check your answer.
Q1AIPMT 2009 Prelims · Q54
Manganese is required in:
Not quite — the answer is B.
Manganese is a component of the water-splitting (oxygen-evolving) complex of photosystem II, so it is needed for photolysis of water and release of oxygen. Calcium (not manganese) is used in the middle lamella of the cell wall, and magnesium/iron are linked to chlorophyll synthesis.
Q2AIPMT 2009 Prelims · Q61
Global agreement in specific control strategies to reduce the release of ozone depleting substances, was adopted by:
Not quite — the answer is A.
The Montreal Protocol (1987) is the international treaty that set specific controls to phase out ozone-depleting substances such as CFCs. The Kyoto Protocol deals with greenhouse gas emissions, and the Vienna Convention was only a framework agreement without specific control measures.
Q3AIPMT 2009 Prelims · Q69
If a live earthworm is pricked with a needle on its outer surface without damaging its gut, the fluid that comes out is:
Not quite — the answer is A.
The earthworm has a true coelom filled with coelomic fluid lying between the body wall and the gut, so a needle prick through the body wall releases coelomic fluid. Haemolymph is found in arthropods with an open circulation, while the earthworm has a closed blood system.
Q4AIPMT 2009 Prelims · Q73
Alzheimer disease in humans is associated with the deficiency of:
Not quite — the answer is B.
Alzheimer disease involves loss of cholinergic neurons, causing a deficiency of the neurotransmitter acetylcholine and memory loss. Dopamine deficiency is linked to Parkinson disease, not Alzheimer.
Q5AIPMT 2009 Prelims · Q78
Which of the following plant species you would select for the production of bioethanol?
Not quite — the answer is A.
Bioethanol is made by fermenting sugars or starch, and maize (Zea mays) starch is a major feedstock for it. Jatropha, Pongamia and Brassica are oil-seed plants whose oils are used for biodiesel, not ethanol.
Q6AIPMT 2009 Prelims · Q79
When breast feeding is replaced by less nutritive food low in proteins and calories; the infants below the age of one year are likely to suffer from:
Not quite — the answer is D.
Marasmus results from a combined deficiency of proteins and calories and is seen in infants below one year when mother's milk is replaced too early by food poor in both. Kwashiorkor is mainly protein deficiency (with adequate calories) and occurs in children over one year.
Q7AIPMT 2009 Prelims · Q80
A young infant may be feeding entirely on mother's milk which is white in colour but the stools which the infant passes out is quite yellowish. What is this yellow colour due to?
Not quite — the answer is A.
Bile carries the pigments bilirubin and biliverdin, formed from haemoglobin breakdown, into the intestine, and their products colour the faeces yellow. Pancreatic and intestinal juices carry enzymes, not pigments.
Q8AIPMT 2009 Prelims · Q86
Which one of the following statements is true regarding digestion and absorption of food in humans?
Not quite — the answer is A.
Old NCERT states that fructose and some amino acids are absorbed with the help of carrier ions like Na⁺ (facilitated transport). Chylomicrons go into the lymph vessels (lacteals), not blood capillaries; salivary amylase hydrolyses about 30% of starch; and oxyntic cells secrete HCl, while chief cells secrete pepsinogen.
Q9AIPMT 2009 Prelims · Q107
Steps taken by the Government of India to control air pollution include:
Not quite — the answer is A.
PUC certification, which checks the emission of carbon monoxide and hydrocarbons from vehicles, is a compulsory measure in India. The other options overstate the rules: CNG was mandated for public transport in Delhi, not all buses and trucks nationwide, sulphur limits were lowered far below 500 ppm, and 20% blending was not compulsory.
Q10AIPMT 2009 Prelims · Q110
The correct sequence of plants in a hydrosere is:
Not quite — the answer is A.
A hydrosere begins with phytoplankton (Volvox), then submerged plants (Hydrilla), floating plants (Pistia), reed-swamp plants (Scirpus), shrubs (Lantana) and finally a forest climax (Oak). Sequences starting with Oak run backwards from the climax.
Q11AIPMT 2009 Prelims · Q120
Montreal Protocol aims at:
Not quite — the answer is D.
The Montreal Protocol (1987) is an international treaty to cut down the emission of ozone-depleting substances such as CFCs. Control of CO₂ emissions is the aim of the Kyoto Protocol.
Q12AIPMT 2009 Prelims · Q121
DDT residues are rapidly passed through food chain causing biomagnification because DDT is:
Not quite — the answer is D.
DDT is fat-soluble and cannot be metabolised or excreted, so it accumulates in fatty tissues and its concentration increases at each trophic level (biomagnification). A water-soluble substance would be excreted rather than accumulate.
Q13AIPMT 2009 Prelims · Q128
Chipko movement was launched for the protection of:
Not quite — the answer is A.
In the Chipko movement (Garhwal Himalayas, 1974) local women hugged trees to stop contractors from felling them, so it aimed to protect forests.
Q14AIPMT 2009 Prelims · Q134
Somaclones are obtained by
Not quite — the answer is D.
Plants regenerated from somatic cells through tissue culture (micropropagation) are genetically identical to the source plant and are called somaclones. Irradiation and genetic engineering are used to create variation, not clones.
Q15AIPMT 2009 Prelims · Q137
Which one of the following pairs of food components in humans reaches the stomach totally undigested?
Not quite — the answer is B.
Saliva contains salivary amylase that starts starch digestion in the mouth but has no fat-digesting enzyme in humans, and cellulose is never digested by human enzymes. So fat and cellulose reach the stomach unchanged, while starch is already partly digested.
Q16AIPMT 2009 Prelims · Q138
A change in the amount of yolk and its distribution in the egg will affect:
Not quite — the answer is A.
Yolk slows down cleavage furrows, so the amount and distribution of yolk decide whether cleavage is holoblastic or meroblastic, equal or unequal. Fertilisation and zygote formation do not depend on yolk distribution.
Q17AIPMT 2009 Prelims · Q147
Reduction in vascular tissue, mechanical tissue and cuticle is characteristic of:
Not quite — the answer is C.
Water plants are supported by water and absorb water over their whole surface, so they have poorly developed xylem, little mechanical tissue and a thin or absent cuticle. Xerophytes show the opposite: thick cuticle and well-developed mechanical tissue.
Q18AIPMT 2009 Prelims · Q149
Which one of the following correctly describes the location of some body parts in the earthworm Pheretima?
Not quite — the answer is C.
Pheretima has two pairs of testes in the 10th and 11th segments. The four pairs of spermathecae lie in segments 6-9, the single pair of ovaries is at the septum between segments 12 and 13, and the two pairs of accessory glands are in segments 17 and 19.
Chapter-wise split of AIPMT 2009 Prelims Biology
- Morphology of Flowering Plants6 Q
- Plant Kingdom6 Q
- Human Reproduction5 Q
- Human Health and Disease5 Q
- Microbes in Human Welfare5 Q
- Anatomy of Flowering Plants5 Q
- Structural Organisation in Animals4 Q
- Body Fluids and Circulation4 Q
- Cell: The Unit of Life4 Q
- Molecular Basis of Inheritance4 Q
- Animal Kingdom3 Q
- Biological Classification3 Q
- Biotechnology and Its Applications3 Q
- Principles of Inheritance and Variation3 Q
- Biotechnology: Principles and Processes2 Q
- Excretory Products and Their Elimination2 Q
- Plant Growth and Development2 Q
- Biodiversity and Conservation2 Q
- Locomotion and Movement2 Q
- Cell Cycle and Cell Division1 Q
- Photosynthesis in Higher Plants1 Q
- Neural Control and Coordination1 Q
- Respiration in Plants1 Q
- Chemical Coordination and Integration1 Q
- Evolution1 Q
- Ecosystem1 Q
About this paper
How many questions from AIPMT 2009 Prelims are on this page?
189 questions (Physics 45, Chemistry 49, 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 2009 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 2009 Prelims useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books. 25 questions here are on topics NCERT has since dropped; they are listed separately so you can skip them.
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