AIPMT 2010 Prelims Question PaperPhysics, Chemistry and Biology, with checked answers and solutions
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177 questions from the AIPMT 2010 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 2010 Prelims Physics questions
42 questions. Tap an option to check your answer.
Q1AIPMT 2010 Prelims · Q1
The dimension of (1/2)ε₀E², where ε₀ is permittivity of free space and E is electric field, is
Not quite — the answer is C.
(1/2)ε₀E² is the energy density of the electric field, i.e. energy per unit volume. Its dimensions are ML²T⁻²/L³ = ML⁻¹T⁻². ML²T⁻² is the dimension of energy itself, which forgets the division by volume.
Q2AIPMT 2010 Prelims · Q2
A particle moves a distance x in time t according to equation x = (t + 5)⁻¹. The acceleration of particle is proportional to
Not quite — the answer is B.
With x = (t + 5)⁻¹, v = −(t + 5)⁻² and a = 2(t + 5)⁻³. Since |v| = (t + 5)⁻², (t + 5)⁻³ = |v|^(3/2), so a ∝ (velocity)^(3/2). In terms of distance a = 2x³, so neither (distance)² nor (distance)⁻² fits.
Q3AIPMT 2010 Prelims · Q4
A particle has initial velocity (3î + 4ĵ) and has acceleration (0.4î + 0.3ĵ). Its speed after 10 s is
Not quite — the answer is C.
v = u + at = (3 + 0.4×10)î + (4 + 0.3×10)ĵ = 7î + 7ĵ. Speed = √(49 + 49) = 7√2 ≈ 9.9 units. Adding the initial speed 5 and the gain 5 to get 10 units wrongly adds magnitudes of non-parallel vectors.
Q4AIPMT 2010 Prelims · Q6
A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 m/s. When the stone reaches the floor, the distance of the man above the floor will be
Not quite — the answer is C.
With no external force, momentum is conserved: 0.5 × 2 = 50 × v, so the man moves up at 0.02 m/s. The stone takes 10/2 = 5 s to reach the floor, in which the man rises 0.02 × 5 = 0.1 m, so he is at 10.1 m. Answering 10 m ignores the recoil; 9.9 m takes the recoil in the wrong direction.
Q5AIPMT 2010 Prelims · Q8
A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5 then their velocities (in m/s) after collision will be
Not quite — the answer is B.
Momentum: m×2 = m v₁ + 2m v₂, so v₁ + 2v₂ = 2. Restitution: v₂ − v₁ = 0.5 × 2 = 1. Solving gives v₂ = 1 m/s and v₁ = 0, i.e. velocities 0 and 1 m/s. The pair 1 and 1 would need perfectly inelastic sticking, which conserves momentum only if v = 2/3.
Q6AIPMT 2010 Prelims · Q9
A gramophone record is revolving with an angular velocity ω. A coin is placed at a distance r from the centre of the record. The static coefficient of friction is μ. The coin will revolve with the record if
Not quite — the answer is D.
Static friction supplies the centripetal force: mrω² ≤ μmg, so the coin stays on if r ≤ μg/ω². The reversed inequality r ≥ μg/ω² describes when the coin slips off.
Q7AIPMT 2010 Prelims · Q11
Two particles which are initially at rest, move towards each other under the action of their internal attraction. If their speeds are v and 2v at any instant, then the speed of centre of mass of the system will be
Not quite — the answer is C.
Only internal forces act, so total momentum stays zero (it was zero initially). Hence the centre of mass stays at rest and its speed is zero, whatever the individual speeds v and 2v are.
Q8AIPMT 2010 Prelims · Q12
The radii of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of satellite A is 3V, then the speed of satellite B will be
Not quite — the answer is C.
Orbital speed v = √(GM/r), so v ∝ 1/√r. v_B/v_A = √(4R/R) = 2, so v_B = 2 × 3V = 6V. Using v ∝ 1/r would give 12V.
Q9AIPMT 2010 Prelims · Q13
A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The gravitational potential at a point situated at a/2 distance from the centre, will be
Not quite — the answer is B.
Inside the shell its potential is constant at −GM/a. The point mass at the centre contributes −GM/(a/2) = −2GM/a. Total V = −2GM/a − GM/a = −3GM/a. Forgetting the shell gives −2GM/a.
Q10AIPMT 2010 Prelims · Q14
A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v? (Take g = 10 m/s²)
Not quite — the answer is B.
At t = 18 s the first ball has fallen ½ × 10 × 18² = 1620 m. The second ball moves for 12 s: 12v + ½ × 10 × 12² = 12v + 720. Equating gives 12v = 900, v = 75 m/s.
Q11AIPMT 2010 Prelims · Q15
A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time t?
Not quite — the answer is C.
Volume is fixed: halving the radius makes the area A/4, so the length becomes 4L. Heat flow ∝ A/L, so it becomes ¼ ÷ 4 = 1/16 of the original: Q/16. Taking only the area change gives Q/4.
Q12AIPMT 2010 Prelims · Q16
The total radiant energy per unit area, normal to the direction of incidence, received at a distance R from the centre of a star of radius r, whose outer surface radiates as a black body at a temperature T K is given by (where σ is Stefan's constant)
Not quite — the answer is B.
The star emits σT⁴ × 4πr² in total. At distance R this spreads over a sphere of area 4πR², so energy per unit area per second = σT⁴·4πr²/(4πR²) = σr²T⁴/R². Keeping the 4π in the numerator forgets to divide by the full sphere area.
Q13AIPMT 2010 Prelims · Q17
If ΔU and ΔW represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true?
Not quite — the answer is B.
First law: Q = ΔU + ΔW with ΔW the work done by the system. In an adiabatic process Q = 0, so ΔU = −ΔW. In an isothermal process of an ideal gas ΔU = 0, so neither isothermal relation holds.
Q14AIPMT 2010 Prelims · Q18
The displacement of a particle along the x-axis is given by x = a sin²ωt. The motion of the particle corresponds to
Not quite — the answer is B.
x = a sin²ωt = (a/2)(1 − cos 2ωt). This is a cosine oscillation of amplitude a/2 about the shifted mean position x = a/2 with angular frequency 2ω, so d²(x − a/2)/dt² = −4ω²(x − a/2): it is SHM of frequency 2ω/2π = ω/π. Calling it non-SHM ignores the shift of the mean position; ω/2π uses the wrong angular frequency.
Q15AIPMT 2010 Prelims · Q19
The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be
Not quite — the answer is A.
T = 2π√(M/k), so T ∝ √M. Doubling the mass gives T′ = √2 T. 2T would need four times the mass.
Q16AIPMT 2010 Prelims · Q20
A transverse wave is represented by y = A sin(ωt − kx). For what value of the wavelength is the wave velocity equal to the maximum particle velocity?
Not quite — the answer is D.
Wave speed ω/k equals the maximum particle speed Aω when 1/k = A, i.e. λ/(2π) = A, so λ = 2πA. Choosing πA drops the factor 2 in k = 2π/λ.
Q17AIPMT 2010 Prelims · Q21
A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per sec when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
Not quite — the answer is A.
The string frequency is 512 ± 4, i.e. 508 or 516 Hz. Increasing tension raises the string's frequency; the beats fall to 2, so the frequency moved towards 512, meaning it was below: 508 Hz. If it were 516 Hz, raising it would increase the beats to more than 4.
Q18AIPMT 2010 Prelims · Q22
Which of the following statement is false for the properties of electromagnetic waves?
Not quite — the answer is D.
In an electromagnetic wave E and B are perpendicular to each other and both perpendicular to the direction of propagation, so saying they are parallel to each other is false. The other statements (no medium needed, in-phase maxima, equal energy sharing) are true.
Q19AIPMT 2010 Prelims · Q23
A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d/2 in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively
Not quite — the answer is D.
Covering part of a lens does not change its focal length, which depends only on curvature and refractive index. The light gathered is ∝ area: the covered central part has diameter d/2, i.e. 1/4 of the area, so intensity becomes I − I/4 = 3I/4. I/4 would be the share of the covered part, not what remains.
Q20AIPMT 2010 Prelims · Q24
A ray of light travelling in a transparent medium of refractive index μ, falls on a surface separating the medium from air at an angle of incidence of 45º. For which of the following value of μ the ray can undergo total internal reflection?
Not quite — the answer is D.
TIR needs the incidence angle to exceed the critical angle: sin 45° > 1/μ, so μ > √2 ≈ 1.414. Only μ = 1.50 exceeds this; 1.40 is just below.
Q21AIPMT 2010 Prelims · Q25
Two positive ions, each carrying a charge q, are separated by a distance d. If F is the force of repulsion between the ions, the number of electrons missing from each ion will be (e being the charge on an electron)
Not quite — the answer is D.
Each ion has charge q = ne. F = (ne)²/(4πε₀d²), so n² = 4πε₀Fd²/e² and n = √(4πε₀Fd²/e²). The same expression without the square root gives n², not n.
Q22AIPMT 2010 Prelims · Q26
A square surface of side L meter in the plane of the paper is placed in a uniform electric field E (volt/m) acting along the same plane at an angle θ with the horizontal side of the square. The electric flux linked to the surface, in units of volt-m, is
Not quite — the answer is A.
Flux = E·A cos(angle between E and the area vector). The area vector is normal to the paper, while E lies in the plane of the paper, so the angle is 90° and the flux is zero for any θ. EL² cos θ wrongly treats θ as the angle with the normal.
Q23AIPMT 2010 Prelims · Q27
A series combination of n₁ capacitors, each of value C₁, is charged by a source of potential difference 4V. When another parallel combination of n₂ capacitors, each of value C₂, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of C₂, in terms of C₁, is then
Not quite — the answer is A.
Series: C_eq = C₁/n₁, energy = ½(C₁/n₁)(4V)² = 8C₁V²/n₁. Parallel: C_eq = n₂C₂, energy = ½n₂C₂V². Equating gives C₂ = 16C₁/(n₁n₂).
Q24AIPMT 2010 Prelims · Q29
A galvanometer has a coil of resistance 100 ohm and gives a full scale deflection for 30 mA current. If it is to work as a voltmeter of 30 volt range, the resistance required to be added will be
Not quite — the answer is B.
Total resistance needed = V/I_g = 30/0.030 = 1000 Ω. Series resistance = 1000 − 100 = 900 Ω. 1000 Ω forgets to subtract the coil's own resistance.
Q25AIPMT 2010 Prelims · Q30
Consider the following two statements: (A) Kirchhoff's junction law follows from the conservation of charge. (B) Kirchhoff's loop law follows from the conservation of energy. Which of the following is correct?
Not quite — the answer is A.
The junction rule says charge does not pile up at a node: conservation of charge. The loop rule says the net potential change round a closed loop is zero, i.e. work per unit charge round a loop is zero: conservation of energy. Both statements are correct.
Q26AIPMT 2010 Prelims · Q31
In producing chlorine by electrolysis 100 kW power at 125 V is being consumed. How much chlorine per minute is liberated? (E.C.E. of chlorine is 0.367 × 10⁻⁶ kg/C)
Not quite — the answer is D.
I = P/V = 100000/125 = 800 A. In one minute m = ZIt = 0.367 × 10⁻⁶ × 800 × 60 = 17.6 × 10⁻³ kg. 3.67 × 10⁻³ kg would need a far smaller charge.
Q27AIPMT 2010 Prelims · Q32
A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is F⃗, the net force on the remaining three arms of the loop is
Not quite — the answer is C.
The net magnetic force on a closed current loop in a uniform field is zero. So the other three arms together must give −F⃗, cancelling the force on the first arm.
Q28AIPMT 2010 Prelims · Q33
A thin ring of radius R meter has charge q coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of f revolutions/s. The value of magnetic induction in Wb/m² at the centre of the ring is
Not quite — the answer is A.
The rotating charge acts as a current I = qf. Field at the centre of a ring is B = μ₀I/(2R) = μ₀qf/(2R). Putting 2πR in the denominator confuses it with the field of a long straight wire.
Q29AIPMT 2010 Prelims · Q34
Electromagnets are made of soft iron because soft iron has
Not quite — the answer is D.
An electromagnet must lose its magnetism as soon as the current is switched off, so it needs low retentivity, and it must be easy to demagnetise, so low coercivity. High retentivity and high coercivity suit permanent magnets.
Q30AIPMT 2010 Prelims · Q35
A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 s in earth's horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be
Not quite — the answer is A.
T ∝ 1/√B. The opposing field reduces the net horizontal field to 24 − 18 = 6 μT. T′ = 2 × √(24/6) = 4 s. Adding the fields (42 μT) would give a shorter period.
Q31AIPMT 2010 Prelims · Q36
A conducting circular loop is placed in a uniform magnetic field, B = 0.025 T with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1 mm s⁻¹. The induced emf when the radius is 2 cm is
Not quite — the answer is C.
φ = Bπr², so |ε| = Bπ·2r·(dr/dt) = 0.025 × π × 2 × 0.02 × 10⁻³ = π × 10⁻⁶ V = π μV. Forgetting the factor 2 from d(r²)/dt gives (π/2) μV.
Q32AIPMT 2010 Prelims · Q38
A 220 volt input is supplied to a transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is
Not quite — the answer is A.
Output power = 440 × 2 = 880 W. Input = 880/0.8 = 1100 W, so primary current = 1100/220 = 5.0 A. Multiplying by 0.8 instead of dividing gives 3.2 A; ignoring efficiency gives 4 A.
Q33AIPMT 2010 Prelims · Q39
A source S₁ is producing 10¹⁵ photons per second of wavelength 5000 Å. Another source S₂ is producing 1.02 × 10¹⁵ photons per second of wavelength 5100 Å. Then (power of S₂)/(power of S₁) is equal to
Not quite — the answer is B.
Power = n·hc/λ. P₂/P₁ = (n₂/n₁)(λ₁/λ₂) = 1.02 × 5000/5100 = 1.00. 1.02 counts only the photon numbers and ignores the lower energy of longer-wavelength photons.
Q34AIPMT 2010 Prelims · Q40
A beam of cathode rays is subjected to crossed electric (E) and magnetic (B) fields. The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by (where V is the potential difference between cathode and anode)
Not quite — the answer is A.
No deflection means eE = evB, so v = E/B. Accelerated through V: eV = ½mv², so e/m = v²/(2V) = E²/(2VB²). The inverted forms come from writing v = B/E.
Q35AIPMT 2010 Prelims · Q42
The activity of a radioactive sample is measured as N₀ counts per minute at t = 0 and N₀/e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is
Not quite — the answer is A.
Activity falls to 1/e in 5 min, so λ = 1/5 per minute (mean life 5 min). T½ = ln 2/λ = 5 logₑ2 min ≈ 3.47 min. 5/logₑ2 inverts the relation.
Q36AIPMT 2010 Prelims · Q43
The energy of a hydrogen atom in the ground state is −13.6 eV. The energy of a He⁺ ion in the first excited state will be
Not quite — the answer is B.
Eₙ = −13.6 Z²/n² eV. For He⁺, Z = 2 and first excited state n = 2: E = −13.6 × 4/4 = −13.6 eV. −54.4 eV is the He⁺ ground state.
Q37AIPMT 2010 Prelims · Q44
The mass of a ⁷₃Li nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of ⁷₃Li nucleus is nearly
Not quite — the answer is C.
Binding energy = 0.042 × 931 ≈ 39.1 MeV. Per nucleon: 39.1/7 ≈ 5.6 MeV. 39 MeV would be the total, not per nucleon.
Q38AIPMT 2010 Prelims · Q45
A alpha nucleus of energy (1/2)mv² bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
Not quite — the answer is D.
At closest approach all kinetic energy becomes potential energy: ½mv² = (1/4πε₀)(Ze)(2e)/r₀, so r₀ = 4Ze²/(4πε₀mv²). Thus r₀ ∝ 1/m, ∝ Ze and ∝ 1/v², so 1/v⁴ and v² are wrong powers and 1/(Ze) has the wrong dependence.
Q39AIPMT 2010 Prelims · Q46
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain of the amplifier is
Not quite — the answer is D.
Voltage gain A_V = β × (R_out/R_in), so β = 50 × 100/200 = 25. Power gain = A_V × β = 50 × 25 = 1250. Taking the power gain equal to the voltage gain (50) ignores the current gain.
Q40AIPMT 2010 Prelims · Q47
Which one of the following bonds produces a solid that reflects light in the visible region and whose electrical conductivity decreases with temperature and has high melting point?
Not quite — the answer is B.
Metallic solids reflect visible light (lustre), have high melting points and their conductivity falls as temperature rises because lattice vibrations scatter the free electrons. Covalent and ionic solids are poor conductors whose conductivity, if any, increases with temperature.
Q41AIPMT 2010 Prelims · Q48
The device that can act as a complete electronic circuit is
Not quite — the answer is C.
An integrated circuit contains many transistors, diodes, resistors and capacitors on one chip, so it can work as a complete circuit. A diode or a single transistor is just one component.
Q42AIPMT 2010 Prelims · Q49
Which of the following statements is false?
Not quite — the answer is C.
In n-type semiconductors the majority carriers are electrons, not holes, so that statement is false. Intrinsic resistance does fall with temperature, trivalent doping does give p-type, and electrons are minority carriers in p-type.
Chapter-wise split of AIPMT 2010 Prelims Physics
- Moving Charges and Magnetism4 Q
- Semiconductor Electronics: Materials, Devices and Simple Circuits4 Q
- Motion in a Straight Line2 Q
- System of Particles and Rotational Motion2 Q
- Gravitation2 Q
- Thermal Properties of Matter2 Q
- Oscillations2 Q
- Waves2 Q
- Ray Optics and Optical Instruments2 Q
- Electric Charges and Fields2 Q
- Current Electricity2 Q
- Magnetism and Matter2 Q
- Nuclei2 Q
- Atoms2 Q
- Units and Measurements1 Q
- Motion in a Plane1 Q
- Work, Energy and Power1 Q
- Laws of Motion1 Q
- Thermodynamics1 Q
- Electromagnetic Waves1 Q
- Electrostatic Potential and Capacitance1 Q
- Electromagnetic Induction1 Q
- Alternating Current1 Q
- Dual Nature of Radiation and Matter1 Q
AIPMT 2010 Prelims Chemistry questions
37 questions. Tap an option to check your answer.
Q1AIPMT 2010 Prelims · Q51
An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to
Not quite — the answer is B.
A strong electrolyte is already fully ionised, so dilution does not add ions. Dilution weakens interionic attractions, so the ions move faster: the increase is due to greater ionic mobility. More ions is the explanation for weak electrolytes.
Q2AIPMT 2010 Prelims · Q52
25.3 g of sodium carbonate, Na₂CO₃ is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, Na⁺ and carbonate ions, CO₃²⁻ are respectively (Molar mass of Na₂CO₃ = 106 g mol⁻¹)
Not quite — the answer is C.
Moles = 25.3/106 = 0.2387 mol in 0.250 L, so [Na₂CO₃] = 0.955 M. Each formula unit gives 2 Na⁺ and 1 CO₃²⁻, so [Na⁺] = 1.910 M and [CO₃²⁻] = 0.955 M. The reversed pair swaps the ions.
Q3AIPMT 2010 Prelims · Q54
Which of the following pairs has the same size?
Not quite — the answer is D.
Because of the lanthanoid contraction, Hf (5d) is almost the same size as Zr (4d), so Zr⁴⁺ and Hf⁴⁺ have nearly identical radii. Ti⁴⁺ is a 3d ion and clearly smaller than Zr⁴⁺.
Q4AIPMT 2010 Prelims · Q55
In a buffer solution containing equal concentration of B⁻ and HB, the Kb for B⁻ is 10⁻¹⁰. The pH of buffer solution is
Not quite — the answer is A.
For the buffer B⁻/HB, pOH = pK_b + log([HB]/[B⁻]). With equal concentrations pOH = pK_b = 10, so pH = 14 − 10 = 4. Answering 10 confuses pOH with pH.
Q5AIPMT 2010 Prelims · Q56
An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?
Not quite — the answer is A.
Vapour pressure lowering depends on solute concentration. Adding water dilutes the solution, so its vapour pressure rises. Adding NaCl or Na₂SO₄ adds solute and lowers it; adding more 1.00 molal KI leaves the concentration unchanged.
Q6AIPMT 2010 Prelims · Q57
What is [H⁺] in mol/L of a solution that is 0.20 M in CH₃COONa and 0.10 M in CH₃COOH? Ka for CH₃COOH = 1.8 × 10⁻⁵.
Not quite — the answer is A.
Buffer: [H⁺] = K_a[CH₃COOH]/[CH₃COO⁻] = 1.8 × 10⁻⁵ × 0.10/0.20 = 9.0 × 10⁻⁶ M. 1.8 × 10⁻⁵ would hold only if the acid and salt concentrations were equal.
Q7AIPMT 2010 Prelims · Q58
For the reaction N₂O₅(g) → 2NO₂(g) + (1/2)O₂(g) the value of rate of disappearance of N₂O₅ is given as 6.25 × 10⁻³ mol L⁻¹ s⁻¹. The rate of formation of NO₂ and O₂ is given respectively as:
Not quite — the answer is C.
Rate = −d[N₂O₅]/dt = ½ d[NO₂]/dt = 2 d[O₂]/dt. So d[NO₂]/dt = 2 × 6.25 × 10⁻³ = 1.25 × 10⁻² and d[O₂]/dt = 6.25 × 10⁻³/2 = 3.125 × 10⁻³ mol L⁻¹ s⁻¹.
Q8AIPMT 2010 Prelims · Q59
Standard entropies of X₂, Y₂ and XY₃ are 60, 40 and 50 J K⁻¹ mol⁻¹ respectively. For the reaction (1/2)X₂ + (3/2)Y₂ ⇌ XY₃, ΔH = −30 kJ to be at equilibrium, the temperature should be
Not quite — the answer is B.
ΔS° = 50 − (½ × 60 + 3/2 × 40) = 50 − 90 = −40 J K⁻¹ mol⁻¹. At equilibrium ΔG = 0, so T = ΔH/ΔS = (−30000)/(−40) = 750 K. Forgetting the stoichiometric halves gives a different ΔS and a wrong T.
Q9AIPMT 2010 Prelims · Q60
During the kinetic study of the reaction 2A + B → C + D, following results were obtained: Run I: [A] = 0.1, [B] = 0.1, rate = 6.0 × 10⁻³ Run II: [A] = 0.3, [B] = 0.2, rate = 7.2 × 10⁻² Run III: [A] = 0.3, [B] = 0.4, rate = 2.88 × 10⁻¹ Run IV: [A] = 0.4, [B] = 0.1, rate = 2.40 × 10⁻² ([A], [B] in mol L⁻¹; initial rate of formation of D in mol L⁻¹ min⁻¹) Based on the above data which one of the following is correct?
Not quite — the answer is A.
Runs II and III: [A] fixed, [B] doubled, rate × 4, so order in B = 2. Runs I and IV: [B] fixed, [A] × 4, rate × 4, so order in A = 1. Rate = k[A][B]²; check: Run II/Run I = (3 × 4) = 12 = 7.2 × 10⁻²/6.0 × 10⁻³.
Q10AIPMT 2010 Prelims · Q61
For the reduction of silver ions with copper metal, the standard cell potential was found to be +0.46 V at 25°C. The value of standard Gibbs energy, ΔG° will be (F = 96500 C mol⁻¹)
Not quite — the answer is B.
Cu + 2Ag⁺ → Cu²⁺ + 2Ag involves n = 2. ΔG° = −nFE° = −2 × 96500 × 0.46 = −88780 J ≈ −89.0 kJ. Taking n = 1 gives −44.5 kJ; −89.0 J has the wrong unit.
Q11AIPMT 2010 Prelims · Q62
Which one of the following species does not exist under normal conditions?
Not quite — the answer is C.
Be₂ has 8 electrons: σ1s² σ*1s² σ2s² σ*2s², bond order (4 − 4)/2 = 0, so it does not exist. Li₂ (bond order 1), Be₂⁺ (bond order ½) and B₂ (bond order 1) have positive bond orders.
Q12AIPMT 2010 Prelims · Q65
Which one of the following ions has electronic configuration [Ar]3d⁶? (At. nos. Mn = 25, Fe = 26, Co = 27, Ni = 28)
Not quite — the answer is A.
Co (Z = 27) is [Ar]3d⁷4s²; removing 3 electrons gives Co³⁺ = [Ar]3d⁶. Fe³⁺ is 3d⁵, Mn³⁺ is 3d⁴ and Ni³⁺ is 3d⁷.
Q13AIPMT 2010 Prelims · Q66
In which of the following equilibrium Kc and Kp are not equal?
Not quite — the answer is A.
K_p = K_c(RT)^Δn_g. For 2C(s) + O₂(g) ⇌ 2CO(g), only gases count: Δn_g = 2 − 1 = 1, so K_p ≠ K_c. In the other three reactions the moles of gas are the same on both sides (Δn_g = 0).
Q14AIPMT 2010 Prelims · Q67
If pH of a saturated solution of Ba(OH)₂ is 12, the value of its Ksp is
Not quite — the answer is A.
pH 12 → pOH 2 → [OH⁻] = 10⁻² M. Ba(OH)₂ gives s and 2s, so s = 5 × 10⁻³ M. K_sp = s(2s)² = 5 × 10⁻³ × 10⁻⁴ = 5.00 × 10⁻⁷ M³. Using 4s³ with s = 10⁻² gives 4 × 10⁻⁶.
Q15AIPMT 2010 Prelims · Q68
Which of the following ions will exhibit colour in aqueous solutions?
Not quite — the answer is C.
Ti³⁺ is 3d¹ with one unpaired electron, so a d-d transition gives colour (violet). Sc³⁺ (3d⁰), La³⁺ (4f⁰) and Lu³⁺ (4f¹⁴) have no unpaired d/f electrons and are colourless.
Q16AIPMT 2010 Prelims · Q71
In which of the following pairs of molecules/ions, the central atoms have sp² hybridization?
Not quite — the answer is C.
BF₃ has 3 bond pairs and no lone pair (sp²); NO₂⁻ has 2 bond pairs and 1 lone pair on N (sp²). NH₂⁻, NH₃ and H₂O each have 4 electron pairs on the central atom and are sp³.
Q17AIPMT 2010 Prelims · Q72
The correct order of the decreasing ionic radii among the following isoelectronic species is
Not quite — the answer is D.
All four ions have 18 electrons. Size falls as nuclear charge rises: S²⁻ (Z = 16) > Cl⁻ (17) > K⁺ (19) > Ca²⁺ (20). The orders putting cations first reverse this trend.
Q18AIPMT 2010 Prelims · Q73
The number of atoms in 0.1 mol of a triatomic gas is (NA = 6.02 × 10²³ mol⁻¹)
Not quite — the answer is C.
0.1 mol of molecules × 3 atoms × 6.02 × 10²³ = 1.806 × 10²³ atoms. 6.02 × 10²² is the number of molecules, not atoms.
Q19AIPMT 2010 Prelims · Q76
The existence of two different coloured complexes with the composition of [Co(NH₃)₄Cl₂]⁺ is due to
Not quite — the answer is C.
[Co(NH₃)₄Cl₂]⁺ is octahedral MA₄B₂, which exists as cis (violet) and trans (green) isomers: geometrical isomerism. Ionisation and linkage isomerism need different ions or an ambidentate ligand, which are absent here.
Q20AIPMT 2010 Prelims · Q77
Oxidation states of P in H₄P₂O₅, H₄P₂O₆, H₄P₂O₇, are respectively
Not quite — the answer is A.
With H = +1 and O = −2: H₄P₂O₅: 4 + 2x − 10 = 0, x = +3; H₄P₂O₆: x = +4; H₄P₂O₇: x = +5. So +3, +4, +5.
Q21AIPMT 2010 Prelims · Q78
The tendency of BF₃, BCl₃ and BBr₃ to behave as Lewis acid decreases in the sequence
Not quite — the answer is C.
Lewis acidity follows how electron-deficient boron remains. Back-donation from the halogen p orbital into boron's empty 2p is strongest for F (similar size 2p–2p) and weakest for Br, so BF₃ is the weakest and BBr₃ the strongest acid: BBr₃ > BCl₃ > BF₃. Going by electronegativity alone wrongly gives BF₃ as strongest.
Q22AIPMT 2010 Prelims · Q79
Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements O, S, F and Cl?
Not quite — the answer is C.
Negative electron gain enthalpies: O ≈ 141, S ≈ 200, F ≈ 328, Cl ≈ 349 kJ/mol. O and F are less negative than S and Cl because their small size causes electron repulsion. So O < S < F < Cl.
Q23AIPMT 2010 Prelims · Q80
Crystal field stabilization energy for high spin d⁴ octahedral complex is
Not quite — the answer is A.
High-spin d⁴ in an octahedral field is t₂g³ e_g¹. CFSE = 3(−0.4Δ₀) + 1(+0.6Δ₀) = −0.6Δ₀. −1.6Δ₀ + P is the low-spin t₂g⁴ value.
Q24AIPMT 2010 Prelims · Q81
In which one of the following species the central atom has the type of hybridisation which is not the same as that present in the other three?
Not quite — the answer is D.
PCl₅ (5 bond pairs), SF₄ (4 bp + 1 lp) and I₃⁻ (2 bp + 3 lp) all have 5 electron pairs: sp³d. SbCl₅²⁻ has Sb(III) with 5 bond pairs and 1 lone pair, 6 pairs: sp³d², so it is the odd one out.
Q25AIPMT 2010 Prelims · Q82
Which of the following molecular hydrides acts as a Lewis acid?
Not quite — the answer is D.
B₂H₆ is electron-deficient (each B has an incomplete octet) and accepts electron pairs, so it acts as a Lewis acid. CH₄ is electron-precise, while NH₃ and H₂O have lone pairs and act as Lewis bases.
Q26AIPMT 2010 Prelims · Q83
Aniline in a set of the following reactions yielded a coloured product Y: C₆H₅NH₂ —(NaNO₂/HCl, 273–278 K)→ X —(N,N-dimethylaniline)→ Y The structure of Y would be
Not quite — the answer is B.
Aniline with NaNO₂/HCl at 273–278 K gives benzenediazonium chloride (X). This couples with N,N-dimethylaniline at the para position to give the yellow azo dye p-(dimethylamino)azobenzene, C₆H₅–N=N–C₆H₄–N(CH₃)₂. The other structures do not contain the unchanged phenyl ring from aniline joined to the dimethylamino ring.
Q27AIPMT 2010 Prelims · Q84
The reaction of toluene with Cl₂ in presence of FeCl₃ gives X and reaction in presence of light gives Y. Thus, X and Y are
Not quite — the answer is D.
With FeCl₃ (Lewis acid) chlorine substitutes the ring at the o- and p-positions activated by CH₃, giving o- and p-chlorotoluene (X). In light, free-radical substitution occurs on the side chain; with excess Cl₂ this goes to C₆H₅CCl₃, trichloromethylbenzene (Y). Benzyl chloride is a side-chain product, so it cannot be X.
Q28AIPMT 2010 Prelims · Q85
Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by
Not quite — the answer is C.
Cracking (pyrolysis) breaks large hydrocarbon molecules into smaller ones at high temperature, turning liquid hydrocarbons into gaseous ones. Distillation under reduced pressure only separates components without changing them.
Q29AIPMT 2010 Prelims · Q87
Which one of the following does not exhibit the phenomenon of mutarotation?
Not quite — the answer is B.
Mutarotation needs a free hemiacetal/hemiketal group that can open and re-close. In sucrose the anomeric carbons of glucose and fructose are joined in the glycosidic bond, so sucrose is non-reducing and shows no mutarotation. Fructose, lactose and maltose have a free anomeric carbon.
Q30AIPMT 2010 Prelims · Q89
In a set of reactions, ethyl benzene yielded a product D: C₆H₅CH₂CH₃ —(KMnO₄, KOH)→ B —(Br₂, FeCl₃)→ C —(C₂H₅OH, H⁺)→ D 'D' would be
Not quite — the answer is A.
Alkaline KMnO₄ oxidises the whole ethyl side chain to –COOH, so B is benzoic acid. –COOH is meta-directing, so Br₂/FeCl₃ gives m-bromobenzoic acid as C. Esterification with ethanol/H⁺ then gives D, ethyl m-bromobenzoate. Products that keep a CH₂ side chain ignore the side-chain oxidation.
Q31AIPMT 2010 Prelims · Q90
Which is most reactive towards SN1 reaction?
Not quite — the answer is D.
SN1 rate depends on carbocation stability. C₆H₅C(CH₃)(C₆H₅)Br gives a tertiary carbocation stabilised by two phenyl rings and a methyl group, the most stable of the four. Benzyl bromide gives only a primary benzylic cation.
Q32AIPMT 2010 Prelims · Q92
Which one of the following compounds has the most acidic nature?
Not quite — the answer is C.
Phenol is far more acidic (pKa ≈ 10) than any alcohol because the phenoxide ion is resonance-stabilised by the ring. Benzyl alcohol, cyclohexanol and dicyclohexylmethanol give alkoxide ions with no such delocalisation.
Q33AIPMT 2010 Prelims · Q93
Given are cyclohexanol (I), acetic acid (II), 2, 4, 6-trinitrophenol (III) and phenol (IV). In these the order of decreasing acidic character will be
Not quite — the answer is B.
Picric acid (2,4,6-trinitrophenol, pKa ≈ 0.4) is the strongest because three –NO₂ groups stabilise the phenoxide; acetic acid (pKa ≈ 4.8) is next, then phenol (≈ 10), then cyclohexanol (≈ 16). Putting phenol above acetic acid is the common trap.
Q34AIPMT 2010 Prelims · Q95
Acetamide is treated with the following reagents separately. Which one of these would yield methyl amine?
Not quite — the answer is B.
Hofmann bromamide degradation: CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂, losing one carbon. PCl₅ or hot conc. H₂SO₄ dehydrate acetamide to acetonitrile, and soda lime does not give methylamine.
Q35AIPMT 2010 Prelims · Q96
Which of the following reactions will not result in the formation of carbon-carbon bonds?
Not quite — the answer is C.
The Cannizzaro reaction is a disproportionation of an aldehyde into an alcohol and a carboxylate by hydride transfer; no new C–C bond is formed. Friedel–Crafts acylation, Reimer–Tiemann (ring –CHO) and Wurtz coupling all make new C–C bonds.
Q36AIPMT 2010 Prelims · Q99
Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is
Not quite — the answer is A.
Acyl chlorides are the most reactive acid derivatives: Cl is strongly electron-withdrawing and gives almost no resonance donation to C=O, and Cl⁻ is an excellent leaving group. Anhydrides, esters and amides (least) are progressively less reactive.
Q37AIPMT 2010 Prelims · Q100
A solution of sucrose (molar mass = 342 g mol⁻¹) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol⁻¹)
Not quite — the answer is B.
Molality = (68.5/342)/1 kg = 0.200 mol/kg. ΔT_f = 1.86 × 0.200 = 0.372 K, so the freezing point is −0.372 °C. +0.372 °C has the wrong sign; freezing point is lowered.
Chemistry: not in the current NEET syllabus
These 6 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
6 questions. Tap an option to check your answer.
Q1AIPMT 2010 Prelims · Q53
Property of the alkaline earth metals that increases with their atomic number
Not quite — the answer is B.
Down group 2 the lattice enthalpy of hydroxides falls faster than hydration enthalpy, so hydroxide solubility increases (Be(OH)₂ is insoluble, Ba(OH)₂ is fairly soluble). Sulphate solubility does the opposite and decreases; electronegativity and ionisation energy also decrease.
Q2AIPMT 2010 Prelims · Q63
AB crystallizes in a body centred cubic lattice with edge length 'a' equal to 387 pm. The distance between two oppositely charged ions in the lattice is
Not quite — the answer is B.
In a body-centred (CsCl-type) AB lattice the oppositely charged ions touch along the body diagonal: 2(r⁺ + r⁻) = √3a. Distance = √3 × 387/2 ≈ 335 pm. Using a/2 or a/√2 gives the wrong distances.
Q3AIPMT 2010 Prelims · Q70
Which one of the following compounds is a peroxide?
Not quite — the answer is C.
BaO₂ contains the peroxide ion O₂²⁻ (O–O single bond). KO₂ is a superoxide (O₂⁻), and NO₂ and MnO₂ are ordinary oxides with no O–O bond.
Q4AIPMT 2010 Prelims · Q75
Which of the following alkaline earth metal sulphates has hydration enthalpy higher than the lattice enthalpy?
Not quite — the answer is C.
The small Be²⁺ ion has a very high hydration enthalpy that exceeds the lattice enthalpy of BeSO₄, so BeSO₄ is soluble. For Ca, Sr and Ba sulphates hydration enthalpy falls below lattice enthalpy, making them sparingly soluble.
Q5AIPMT 2010 Prelims · Q86
Which of the following is employed as a tranquilizer drug?
Not quite — the answer is C.
Valium (diazepam) is a tranquilizer. Mifepristone is an anti-fertility drug, promethazine an antihistamine and naproxen an analgesic/anti-inflammatory.
Q6AIPMT 2010 Prelims · Q88
Which of the following structures represents Neoprene polymer?
Not quite — the answer is B.
Neoprene is polychloroprene, made from chloroprene (2-chloro-1,3-butadiene): [–CH₂–C(Cl)=CH–CH₂–]ₙ. The –CH(C₆H₅)– polymer is polystyrene, –CH(CN)– is polyacrylonitrile and –CH(Cl)– is PVC.
Chapter-wise split of AIPMT 2010 Prelims Chemistry
- Equilibrium4 Q
- Chemical Bonding and Molecular Structure4 Q
- The d- and f-Block Elements3 Q
- Aldehydes, Ketones and Carboxylic Acids3 Q
- Electrochemistry2 Q
- Some Basic Concepts of Chemistry2 Q
- Solutions2 Q
- Chemical Kinetics2 Q
- Classification of Elements and Periodicity in Properties2 Q
- Coordination Compounds2 Q
- Amines2 Q
- Hydrocarbons2 Q
- Alcohols, Phenols and Ethers2 Q
- Thermodynamics1 Q
- Redox Reactions1 Q
- Biomolecules1 Q
- Haloalkanes and Haloarenes1 Q
AIPMT 2010 Prelims Biology questions
82 questions. Tap an option to check your answer.
Q1AIPMT 2010 Prelims · Q102
Some hyperthermophilic organisms that grow in highly acidic (pH2) habitats belong to the two groups
Not quite — the answer is B.
Only prokaryotes tolerate the combination of very high temperature and very low pH; archaebacteria such as thermoacidophiles live in hot acidic springs, and some eubacteria also survive there. Eukaryotic groups like mosses, liverworts, yeasts and diatoms cannot survive such extreme conditions.
Q2AIPMT 2010 Prelims · Q103
Infectious proteins are present in
Not quite — the answer is C.
Prions are abnormally folded infectious proteins that cause diseases such as BSE (mad cow disease) and CJD in humans. Viroids are the opposite case: they are free infectious RNA with no protein coat.
Q3AIPMT 2010 Prelims · Q104
Male and female gametophytes are independent and free-living in
Not quite — the answer is A.
In bryophytes such as the moss Sphagnum, the gametophyte is the dominant, free-living, photosynthetic generation. In gymnosperms (Pinus) and angiosperms (mustard, castor), the gametophytes are greatly reduced and stay inside the sporophyte's sporangia.
Q4AIPMT 2010 Prelims · Q105
Single-celled eukaryotes are included in
Not quite — the answer is B.
In Whittaker's five-kingdom system, Protista holds all single-celled eukaryotes. Monera and archaea are prokaryotes, and most fungi are multicellular.
Q5AIPMT 2010 Prelims · Q106
One example of animals having a single opening to the outside that serves both as mouth as well as anus is
Not quite — the answer is A.
Platyhelminthes such as the liver fluke Fasciola have an incomplete digestive system, with one opening that works as both mouth and anus. Molluscs (Octopus), echinoderms (Asterias) and urochordates (Ascidia) have a complete gut with a separate mouth and anus.
Q6AIPMT 2010 Prelims · Q107
Which one of the following statements about all the four of Spongilla, Leech, Dolphin and Penguin is correct ?
Not quite — the answer is D.
Collar cells (choanocytes) lining the canal system are unique to sponges such as Spongilla. The homoiothermy statement fails because the dolphin, a mammal, is also warm-blooded. Spongilla is a freshwater sponge and is asymmetrical, so the other two statements fail too.
Q7AIPMT 2010 Prelims · Q108
Which one of the following kinds of animals are triploblastic?
Not quite — the answer is B.
Triploblasty, with a mesoderm between ectoderm and endoderm, begins with the Platyhelminthes (flatworms). Corals (cnidarians) and ctenophores are diploblastic, and sponges have a cellular level of organisation without true germ layers.
Q8AIPMT 2010 Prelims · Q109
Which one of the following statements about certain given animals is correct ?
Not quite — the answer is B.
Aschelminthes (roundworms) have a body cavity that is not lined by mesoderm all round (mesoderm occurs only as scattered pouches), so they are pseudocoelomates. Flatworms are acoelomate, while molluscs and arthropods such as insects are true coelomates.
Q9AIPMT 2010 Prelims · Q110
The plasma membrane consists mainly of
Not quite — the answer is C.
According to Singer and Nicolson's fluid mosaic model, the membrane is a quasi-fluid lipid bilayer with proteins embedded in it as integral and peripheral proteins. The bilayer is made of phospholipids, not proteins or carbohydrates.
Q10AIPMT 2010 Prelims · Q111
Which one of the following structures between two adjacent cells is an effective transport pathway?
Not quite — the answer is B.
Plasmodesmata are cytoplasmic strands that cross the cell walls of neighbouring plant cells, connecting their cytoplasm into a continuous symplast for transport. The plasmalemma surrounds a single cell, and plastoquinone is an electron carrier in photosynthesis.
Q11AIPMT 2010 Prelims · Q112
Which one of the following has its own DNA?
Not quite — the answer is B.
Mitochondria have their own circular double-stranded DNA and 70S ribosomes, so they are semi-autonomous. Peroxisomes, dictyosomes (Golgi) and lysosomes have no DNA of their own.
Q12AIPMT 2010 Prelims · Q113
The main arena of various types of activities of a cell is
Not quite — the answer is D.
NCERT describes the cytoplasm as the main arena of cellular activities, since most metabolic reactions take place there. The nucleus controls the cell, but most of the cell's chemical work happens in the cytoplasm.
Q13AIPMT 2010 Prelims · Q115
Algae have cell wall made up of
Not quite — the answer is B.
NCERT states that the algal cell wall is made of cellulose, galactans, mannans and minerals like calcium carbonate. Cellulose, hemicellulose and pectins together describe the wall of other (higher) plants.
Q14AIPMT 2010 Prelims · Q116
Membrane-bound organelles are absent in
Not quite — the answer is C.
Streptococcus is a bacterium, a prokaryote, so it has no membrane-bound organelles. Plasmodium (protist), Saccharomyces (yeast) and Chlamydomonas (alga) are eukaryotes with nuclei, mitochondria and other organelles.
Q15AIPMT 2010 Prelims · Q117
The kind of epithelium which forms the inner walls of blood vessels is
Not quite — the answer is A.
Blood vessels are lined by a single layer of thin, flat cells (simple squamous epithelium, the endothelium), which suits diffusion and filtration. Cuboidal epithelium lines ducts and kidney tubules, not vessels.
Q16AIPMT 2010 Prelims · Q119
Which one of the following cannot be explained on the basis of Mendel’s Law of Dominance?
Not quite — the answer is D.
The return of both parental traits unchanged in the F₂, without blending, is explained by the Law of Segregation (purity of gametes), not by the Law of Dominance. Paired factors and one dominant factor masking a recessive one are the ideas behind dominance.
Q17AIPMT 2010 Prelims · Q120
The genotype of a plant showing the dominant phenotype can be determined by
Not quite — the answer is B.
In a test cross, the dominant-phenotype plant is crossed with a homozygous recessive. All dominant offspring mean it is homozygous, and a 1:1 ratio means it is heterozygous. A back cross can be made to either parent, so it does not necessarily reveal the genotype.
Q18AIPMT 2010 Prelims · Q121
The one aspect which is not a salient feature of genetic code, is its being
Not quite — the answer is C.
The genetic code is unambiguous and specific: each codon codes for only one amino acid. It is degenerate (several codons per amino acid) and nearly universal, so 'ambiguous' is the one property it lacks.
Q19AIPMT 2010 Prelims · Q122
Satellite DNA is useful tool in
Not quite — the answer is D.
Satellite DNA repeats such as VNTRs are highly polymorphic between individuals, so they form the basis of DNA fingerprinting used in forensic identification and paternity testing.
Q20AIPMT 2010 Prelims · Q123
Which one of the following does not follow the central dogma of molecular biology?
Not quite — the answer is A.
HIV is a retrovirus: it copies its RNA genome into DNA with reverse transcriptase, which reverses the usual DNA → RNA direction of the central dogma. Pea, Mucor and Chlamydomonas are cellular organisms with DNA genomes that follow DNA → RNA → protein.
Q21AIPMT 2010 Prelims · Q124
ABO blood groups in humans are controlled by the gene I. It has three alleles–Iᴬ, Iᴮ and i. Since there are three different alleles, six different genotypes are possible. How many phenotypes can occur?
Not quite — the answer is D.
The six genotypes give four blood groups: A (IᴬIᴬ, Iᴬi), B (IᴮIᴮ, Iᴮi), AB (IᴬIᴮ, codominance) and O (ii). The answer 'three' misses the AB group, where both alleles are expressed.
Q22AIPMT 2010 Prelims · Q125
Select the correct statement from the ones given below with respect to dihybrid cross
Not quite — the answer is A.
Morgan found that genes lying close together (tightly linked) rarely separate by crossing over, so they show few recombinants, as with the white-eye and yellow-body genes in Drosophila. Genes far apart or loosely linked show more recombination, not less.
Q23AIPMT 2010 Prelims · Q126
Select the two correct statements out of the four (a–d) given below about lac operon. (a) Glucose or galactose may bind with the repressor and inactivate it (b) In the absence of lactose the repressor binds with the operator region (c) The z-gene codes for permease (d) This was elucidated by Francois Jacob and Jacque Monod The correct statements are
Not quite — the answer is D.
Without lactose, the repressor binds the operator and blocks transcription, and the lac operon was described by Jacob and Monod. The inducer that inactivates the repressor is lactose (allolactose), not glucose or galactose. The z gene codes for β-galactosidase, while permease comes from the y gene.
Q24AIPMT 2010 Prelims · Q127
Which one of the following symbols and its representation, used in human pedigree analysis is correct?
Not quite — the answer is B.
In pedigree charts, a double line joining a male (square) and a female (circle) shows a consanguineous mating, that is, mating between relatives. A shaded diamond is an affected person of unspecified sex, a plain circle is an unaffected female, and a plain square is an unaffected male.
Q25AIPMT 2010 Prelims · Q128
Darwin’s finches are a good example of
Not quite — the answer is D.
From an ancestral seed-eating finch, many species with different beak shapes evolved on the Galapagos Islands to use different foods. Many forms branching from one ancestor into different habitats is adaptive radiation. Convergent evolution is the reverse: unrelated lineages becoming similar.
Q26AIPMT 2010 Prelims · Q129
The scutellum observed in a grain of wheat or maize is comparable to which part of the seed in other monocotyledons?
Not quite — the answer is B.
In grasses, the single large shield-shaped cotyledon of the embryo is called the scutellum. It sits next to the endosperm and absorbs food from it, but it is not part of the endosperm or the aleurone layer.
Q27AIPMT 2010 Prelims · Q132
Which one of the following is not a lateral meristem?
Not quite — the answer is A.
Intercalary meristem lies between mature tissues at the bases of internodes (as in grasses) and is a primary meristem, not a lateral one. Fascicular and interfascicular cambium and cork cambium (phellogen) are lateral meristems that cause growth in thickness.
Q28AIPMT 2010 Prelims · Q133
C₄ plants are more efficient in photosynthesis than C₃ plants due to
Not quite — the answer is A.
In C₄ plants, CO₂ is concentrated around RuBisCO in the bundle sheath cells, so its oxygenase activity, and with it photorespiration, is almost absent. This is why their productivity is higher. Having more chloroplasts is not the reason, and the mesophyll cells of C₄ plants do not carry out the Calvin cycle.
Q29AIPMT 2010 Prelims · Q134
In unilocular ovary with a single ovule the placentation is
Not quite — the answer is C.
In basal placentation, the placenta is at the base of a one-chambered ovary and carries a single ovule, as in sunflower and marigold. Free central placentation also has a unilocular ovary, but its ovules are many and sit on a central axis.
Q30AIPMT 2010 Prelims · Q135
The chief water conducting elements of xylem in gymnosperms are
Not quite — the answer is A.
Gymnosperms lack vessels in their xylem, so tracheids are their main water-conducting elements. Vessels are characteristic of angiosperms, and fibres provide support, not conduction.
Q31AIPMT 2010 Prelims · Q136
The technical term used for the androecium in a flower of China rose (Hibiscus rosasinensis) is
Not quite — the answer is B.
In China rose, the filaments of all stamens fuse into one bundle, the staminal tube, around the style. This is the monadelphous condition. Diadelphous (two bundles) is seen in pea, and polyadelphous (many bundles) in citrus.
Q32AIPMT 2010 Prelims · Q137
Ovary is half-inferior in the flowers of
Not quite — the answer is C.
In plum and rose, the flower is perigynous: the thalamus forms a cup at the level of the ovary, so the ovary is called half-inferior. Cucumber and guava are epigynous with inferior ovaries, and brinjal is hypogynous with a superior ovary.
Q33AIPMT 2010 Prelims · Q138
Heartwood differs from sapwood in
Not quite — the answer is D.
Heartwood is the dark central secondary xylem. Its elements are dead and filled with tannins, resins, gums and similar deposits, so it no longer conducts water. These deposits make it more resistant to microbes and insects, not more susceptible.
Q34AIPMT 2010 Prelims · Q139
Keel is characteristic of the flowers of
Not quite — the answer is A.
Bean belongs to Fabaceae, which has papilionaceous corolla with a large standard, two wings and two anterior petals fused into a keel. Gulmohur and Cassia have imbricate ascending (non-papilionaceous) corolla, so they lack a true keel.
Q35AIPMT 2010 Prelims · Q141
PGA as the first CO₂ fixation product was discovered in photosynthesis of
Not quite — the answer is A.
Melvin Calvin traced radioactive ¹⁴C in the green alga Chlorella and found that 3-phosphoglyceric acid (PGA) was the first stable product of CO₂ fixation. The C₄ pathway (OAA first) was found later in angiosperms such as sugarcane and maize.
Q36AIPMT 2010 Prelims · Q142
The energy releasing metabolic process in which substrate is oxidised without an external electron acceptor is called
Not quite — the answer is C.
In fermentation, NADH from glycolysis is reoxidised by reducing an internal organic molecule (pyruvate or acetaldehyde), so no external electron acceptor such as O₂ is needed. Aerobic respiration uses O₂ as the final external acceptor. Glycolysis is only one stage and needs NAD⁺ to be regenerated.
Q37AIPMT 2010 Prelims · Q144
Listed below are four respiratory capacities (a - c) and four jumbled respiratory volumes of a normal human adult Respiratory capacities — Respiratory volumes (a) Residual volume — 2500 mL (b) Vital capacity — 3500 mL (c) Inspiratory reserve volume — 1200 mL (d) Inspiratory capacity — 4500 mL Which one of the following is the correct matching of two capacities and volumes?
Not quite — the answer is D.
Inspiratory capacity = tidal volume (about 500 mL) + IRV (2500–3000 mL), which is about 3500 mL. Residual volume is about 1100–1200 mL. Vital capacity (about 4500 mL) and IRV (about 2500 mL) do not match the other options as given.
Q38AIPMT 2010 Prelims · Q145
What is true about RBCs in humans?
Not quite — the answer is B.
About 20–25% of CO₂ is carried by RBCs as carbamino-haemoglobin, about 70% as bicarbonate, and about 7% dissolved in plasma. For oxygen, about 97% travels bound to haemoglobin in RBCs and 3% dissolved, so the 99.5% and 80% figures are wrong.
Q39AIPMT 2010 Prelims · Q146
If due to some injury the chordae tendinae of the tricuspid valve of the human heart is partially non-functional, what will be the immediate effect?
Not quite — the answer is A.
Chordae tendineae stop the tricuspid valve from turning inside out during ventricular systole. If they fail, part of the right ventricle's blood leaks back into the right atrium, so less is pumped into the pulmonary artery. The left atrium is on the other side of the heart (the bicuspid valve), so backflow there is not the effect.
Q40AIPMT 2010 Prelims · Q147
Which one of the following statements in regard to the excretion by the human kidneys is correct?
Not quite — the answer is D.
About 180 L of filtrate is formed each day but only about 1.5 L of urine is passed, so nearly 99% is reabsorbed. The ascending limb is impermeable to water (not electrolytes), the descending limb is permeable to water, and the DCT does reabsorb HCO₃⁻.
Q41AIPMT 2010 Prelims · Q148
Low Ca⁺⁺ in the body fluid may be the cause of
Not quite — the answer is B.
Tetany is rapid spasms (wild contractions) of muscles caused by low Ca²⁺ in body fluids, for example when parathyroid hormone is deficient. Gout is caused by uric acid crystals building up in joints.
Q42AIPMT 2010 Prelims · Q149
If for some reason our goblet cells are non-functional this will adversely affect
Not quite — the answer is A.
Goblet cells are unicellular glands in the columnar epithelium of the alimentary canal, and they secrete mucus. Mucus lubricates the gut lining so food moves smoothly. Somatostatin comes from delta cells, and sebum from sebaceous glands, so goblet cells play no part in either.
Q43AIPMT 2010 Prelims · Q150
The nerve centres which control the body temperature and the urge for eating are contained in
Not quite — the answer is B.
The hypothalamus contains centres that control body temperature, hunger and thirst. The thalamus is mainly a relay centre for sensory signals, and the cerebellum coordinates movement and balance.
Q44AIPMT 2010 Prelims · Q152
The first movements of the foetus and appearance of hair on its head are usually observed during which month of pregnancy?
Not quite — the answer is C.
NCERT states that the first movements of the foetus and the appearance of hair on the head are usually observed during the fifth month. Limbs and digits form by the end of the second month, and most major organ systems by the end of the first trimester (third month).
Q45AIPMT 2010 Prelims · Q153
Cu ions released from copper- releasing Intra Uterine Devices (IUDs)
Not quite — the answer is D.
Copper ions released by copper IUDs (CuT, Cu7, Multiload 375) suppress sperm motility and fertilising capacity. Making the uterus unsuitable for implantation is attributed to hormone-releasing IUDs, and increased phagocytosis of sperm is a general effect of IUDs, not of Cu ions specifically.
Q46AIPMT 2010 Prelims · Q155
Which one of the following pairs is incorrectly matched?
Not quite — the answer is B.
Glucagon is secreted by the alpha (α) cells of the islets of Langerhans, while beta (β) cells secrete insulin, so this pair is wrong. Delta cells secrete somatostatin, the corpus luteum secretes relaxin, and insulin deficiency causes diabetes mellitus.
Q47AIPMT 2010 Prelims · Q156
The principal nitrogenous excretory compound in humans is synthesised
Not quite — the answer is A.
Humans are ureotelic: the liver turns ammonia into urea through the ornithine cycle, and the urea travels in blood to the kidneys to be excreted in urine. The kidneys filter urea but do not make it.
Q48AIPMT 2010 Prelims · Q157
Injury to adrenal cortex is not likely to affect the secretion of which one of the following?
Not quite — the answer is D.
Adrenaline (epinephrine) is made by the adrenal medulla, so damage to the cortex leaves it unaffected. The cortex makes glucocorticoids (cortisol), mineralocorticoids (aldosterone) and adrenal androgens such as DHEA and androstenedione.
Q49AIPMT 2010 Prelims · Q158
Which one of the following statements about human sperm is correct?
Not quite — the answer is C.
The acrosome is a cap-like structure that comes from the Golgi apparatus. It contains hydrolytic enzymes (sperm lysins) that dissolve the egg's coverings so the sperm can enter. It works chemically, by enzymes, not by mechanically piercing the egg.
Q50AIPMT 2010 Prelims · Q160
Apomictic embryos in citrus arise from
Not quite — the answer is C.
In Citrus and mango, some nucellar cells around the embryo sac (maternal sporophytic tissue) divide and grow directly into embryos, a form of apomixis called adventive embryony. They do not come from a diploid egg, synergids or antipodals.
Q51AIPMT 2010 Prelims · Q161
Wind pollinated flowers are
Not quite — the answer is C.
Wind-pollinated flowers are small and inconspicuous and make large amounts of light, dry, non-sticky pollen, because most of it is lost in transit. They give no reward to animals, so they lack nectar and bright colours.
Q52AIPMT 2010 Prelims · Q162
Phototropic curvature is the result of uneven distribution of
Not quite — the answer is A.
Under one-sided light, auxin builds up on the shaded side of the shoot tip, and those cells elongate more, so the shoot bends towards the light. This was shown in the coleoptile experiments of Darwin and Went. Phytochrome is a light-sensing pigment linked to flowering and germination, not the growth hormone that causes the curvature.
Q53AIPMT 2010 Prelims · Q163
Transfer of pollen grains from the anther to the stigma of another flower of the same plant is called
Not quite — the answer is C.
Geitonogamy is pollination between different flowers of the same plant. It needs a pollinating agent, as cross-pollination does, but genetically it is the same as self-pollination. Autogamy is within the same flower, and xenogamy is between flowers of different plants.
Q54AIPMT 2010 Prelims · Q164
Seminal plasma in human males is rich in
Not quite — the answer is B.
Seminal plasma, the secretion of the seminal vesicles, prostate and bulbourethral glands, is rich in fructose, calcium and certain enzymes. Fructose is the sperm's energy source, so glucose is not the right sugar.
Q55AIPMT 2010 Prelims · Q165
Sertoli cells are found in
Not quite — the answer is D.
Sertoli cells line the seminiferous tubules together with the male germ cells and nourish them as they develop. Testosterone is made by the Leydig (interstitial) cells, not by Sertoli cells.
Q56AIPMT 2010 Prelims · Q166
The part of Fallopian tube closest to the ovary is
Not quite — the answer is C.
The funnel-shaped infundibulum, with its finger-like fimbriae, lies closest to the ovary and collects the ovum after ovulation. The order towards the uterus is infundibulum, then ampulla, then isthmus, and the cervix belongs to the uterus, not the tube.
Q57AIPMT 2010 Prelims · Q167
In vitro fertilisation is a technique that involves transfer of which one of the following into the fallopian tube?
Not quite — the answer is C.
In ZIFT (zygote intra-fallopian transfer), the zygote or an early embryo of up to 8 blastomeres is placed in the fallopian tube. Embryos with more than 8 blastomeres are placed in the uterus instead (IUT), so the 32-cell option is wrong.
Q58AIPMT 2010 Prelims · Q168
The permissible use of the technique amniocentesis is for
Not quite — the answer is A.
Amniocentesis tests foetal cells in the amniotic fluid for chromosomal or genetic disorders. Using it to find the sex of the foetus is a misuse that leads to female foeticide, and it is banned by law in India.
Q59AIPMT 2010 Prelims · Q170
The biomass available for consumption by the herbivores and the decomposers is called
Not quite — the answer is B.
Net primary productivity is gross primary productivity minus what plants lose in respiration. It is the biomass actually available to herbivores and decomposers. Standing crop is the biomass present at a given moment, not the rate of production available to consumers.
Q60AIPMT 2010 Prelims · Q171
Which one of the following is one of the characteristics of a biological community?
Not quite — the answer is B.
Stratification, the vertical layering of different species such as trees, shrubs and herbs, is a structural feature of a community. Sex ratio, natality and mortality are attributes of a single-species population.
Q61AIPMT 2010 Prelims · Q172
Which one of the following is an example ex-situ conservation?
Not quite — the answer is C.
Ex-situ conservation protects species outside their natural habitat, for example in seed banks, zoos, botanical gardens and cryopreservation. National parks, wildlife sanctuaries and sacred groves protect species where they naturally live, so they are in-situ.
Q62AIPMT 2010 Prelims · Q175
Select the correct statement from the following
Not quite — the answer is A.
Activated sludge is the settled mass of aerobic bacterial flocs from secondary treatment, and part of it is pumped back as inoculum. Biogas is made by anaerobic methanogens such as Methanobacterium, and it is a mixture of mainly methane with CO₂ and H₂, not pure methane.
Q63AIPMT 2010 Prelims · Q178
Study the four statements (a–d) given below and select the two correct ones out of them (a) A lion eating a deer and a sparrow feeding on grain are ecologically similar in being consumers (b) Predator star fish Pisaster helps in maintaining species diversity of some invertebrates (c) Predators ultimately lead to the extinction of prey species (d) Production of chemicals such as nicotine, strychnine by the plants are metabolic disorders The two correct statements are:
Not quite — the answer is A.
The lion and the sparrow both eat other organisms, so both are consumers. When the predatory starfish Pisaster was removed from rocky shores, more than 10 invertebrate species disappeared, which shows it maintains diversity. Predators usually keep prey numbers in check rather than wiping them out, and nicotine and strychnine are defensive chemicals, not metabolic disorders.
Q64AIPMT 2010 Prelims · Q180
Widal test is used for the diagnosis of
Not quite — the answer is A.
The Widal test detects antibodies against Salmonella typhi and is used to confirm typhoid fever. Malaria is diagnosed by examining blood smears for Plasmodium.
Q65AIPMT 2010 Prelims · Q181
Ringworm in humans is caused by
Not quite — the answer is C.
Ringworm is a skin infection caused by fungi such as Microsporum, Trichophyton and Epidermophyton. Despite the name, no worm (nematode) is involved.
Q66AIPMT 2010 Prelims · Q182
Which one of the following is not used in organic farming?
Not quite — the answer is A.
Glomus (a mycorrhizal fungus) and Oscillatoria (a cyanobacterium) are biofertilisers, and earthworms make vermicompost, so all three help organic farming. Snails have no such role and are usually crop pests.
Q67AIPMT 2010 Prelims · Q183
A common biocontrol agent for the control of plant diseases is
Not quite — the answer is A.
The free-living fungus Trichoderma lives in the root ecosystem and controls several plant pathogens, so it is used against plant diseases. Baculoviruses and Bacillus thuringiensis are biocontrol agents against insect pests rather than diseases, and Glomus is a biofertiliser.
Q68AIPMT 2010 Prelims · Q185
Consider the following four statements (a–d) regarding kidney transplant and select the two correct ones out of these. a. Even if a kidney transplant is proper the recipient may need to take immunosuppressants for a long time b. The cell-mediated immune response is responsible for the graft rejection c. The B-lymphocytes are responsible for rejection of the graft d. The acceptance or rejection of a kidney transplant depends on specific interferons The two correct statements are
Not quite — the answer is A.
NCERT states that graft rejection is due to the cell-mediated (T-lymphocyte) immune response, and that even after tissue and blood-group matching the patient has to take immunosuppressants for life. B-lymphocytes give humoral immunity, and graft acceptance depends on matching tissue antigens (HLA), not interferons.
Q69AIPMT 2010 Prelims · Q186
Which one of the following statements is correct with respect to AIDS?
Not quite — the answer is A.
HIV infects helper T-cells (T_H) and multiplies in them, so their number steadily falls and immunity weakens. HIV does not spread through shared food, drug addicts who share needles are at high risk, and antiretroviral therapy only prolongs life without curing AIDS.
Q70AIPMT 2010 Prelims · Q187
Select the correct statement from the ones given below
Not quite — the answer is C.
Morphine is an opioid that works as a strong sedative and painkiller, and it is used for patients who have undergone surgery. Tobacco contains nicotine, which raises blood pressure and heart rate rather than lowering them. Cocaine is not used to aid recovery.
Q71AIPMT 2010 Prelims · Q188
Toxic agents present in food which interfere with thyroxine synthesis lead to the development of
Not quite — the answer is D.
Substances in food that block thyroid hormone synthesis (goitrogens) lower thyroxine levels. The pituitary responds by releasing more TSH, so the gland enlarges into a simple (non-toxic) goitre. Thyrotoxicosis and toxic goitre are conditions of excess thyroxine.
Q72AIPMT 2010 Prelims · Q189
The second maturation division of the mammalian ovum occurs
Not quite — the answer is C.
The secondary oocyte is released at ovulation arrested in meiosis II. It completes the second meiotic division only after a sperm enters it, producing the ovum and the second polar body. This happens before the sperm and egg nuclei fuse, so the nuclear-fusion option is wrong.
Q73AIPMT 2010 Prelims · Q190
Which one of the following statements about morula in humans is correct?
Not quite — the answer is B.
Cleavage divisions are rapid mitoses with no growth in between, so the zygote's cytoplasm is just split among the blastomeres while every new cell receives a full copy of DNA. The morula (8–16 cells) therefore has roughly the same cytoplasm as the zygote but many times more DNA.
Q74AIPMT 2010 Prelims · Q191
Stirred-tank bioreactors have been designed for
Not quite — the answer is A.
A stirred-tank bioreactor uses a stirrer and a sparging system to mix the contents evenly and keep oxygen available throughout the culture. Purification and adding preservatives belong to downstream processing, not to the bioreactor.
Q75AIPMT 2010 Prelims · Q193
DNA or RNA segment tagged with a radioactive molecule is called
Not quite — the answer is C.
A probe is a single-stranded DNA or RNA labelled with a radioactive marker. It finds its complementary sequence by hybridisation, as in DNA fingerprinting and molecular diagnosis. A plasmid or vector carries genes but is not labelled.
Q76AIPMT 2010 Prelims · Q194
Which one of the following is used as vector for cloning genes into higher organisms?
Not quite — the answer is A.
Retroviruses can turn normal animal cells cancerous, but when disarmed they are used as vectors to deliver genes into animal cells. Baculoviruses are used as biocontrol agents against insects, and Salmonella and Rhizopus are not cloning vectors.
Q77AIPMT 2010 Prelims · Q195
The genetically-modified (GM) brinjal in India has been developed for
Not quite — the answer is B.
Bt brinjal carries a Bacillus thuringiensis cry gene whose toxin protects the crop against the fruit and shoot borer, making it insect-resistant. Delayed ripening (longer shelf life) was the trait of the Flavr Savr tomato, not brinjal.
Q78AIPMT 2010 Prelims · Q196
Genetic engineering has been successfully used for producing
Not quite — the answer is B.
NCERT states that transgenic mice are used to test the safety of vaccines, specifically polio vaccine, before they are used in humans. Rosie's milk was enriched with human alpha-lactalbumin protein, not fat. The disease models NCERT lists are cancer, cystic fibrosis, rheumatoid arthritis and Alzheimer's, not cardiac disease.
Q79AIPMT 2010 Prelims · Q197
Restriction endonucleases are enzymes which
Not quite — the answer is B.
Restriction endonucleases recognise a specific palindromic sequence and cut the DNA at specific positions within the molecule. Removing nucleotides from the ends is what exonucleases do.
Q80AIPMT 2010 Prelims · Q198
Some of the characteristics of Bt cotton are
Not quite — the answer is A.
Bt cotton expresses the cryIAc and cryIIAb genes, whose toxins control cotton bollworms (lepidopteran larvae), cutting pest losses and raising yield. The cry genes in Bt cotton act on bollworms, not on dipterans such as flies and mosquitoes, aphids or beetles.
Q81AIPMT 2010 Prelims · Q199
An improved variety of transgenic basmati rice
Not quite — the answer is C.
Transgenic rice has been engineered with genes for β-carotene (the vitamin A precursor), known as Golden Rice, so 'rich in vitamin A' is the intended answer. No crop is completely resistant to all pests, none can do without fertilisers, and losing the aroma would not count as an improvement.
Q82AIPMT 2010 Prelims · Q200
Which one of the following palindromic base sequences in DNA can be easily cut at about the middle by some particular restriction enzyme?
Not quite — the answer is D.
5′-GAATTC-3′ / 3′-CTTAAG-5′ reads the same 5′→3′ on both strands, which makes it a true palindrome. It is the EcoRI recognition site, which is cut between G and A. The other pairs are not even complementary strands, let alone palindromes.
Biology: not in the current NEET syllabus
These 10 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
10 questions. Tap an option to check your answer.
Q1AIPMT 2010 Prelims · Q130
Which one of the following is not a micronutrient?
Not quite — the answer is C.
Magnesium is a macronutrient: it is needed in large amounts and is the central atom of chlorophyll. Boron, molybdenum and zinc are needed in trace amounts, so they are micronutrients.
Q2AIPMT 2010 Prelims · Q131
An element playing important role in nitrogen fixation is
Not quite — the answer is B.
Nitrogenase, the enzyme that reduces N₂ to ammonia, is a molybdenum-iron protein, so molybdenum is essential for nitrogen fixation. Zinc and manganese activate other enzymes (for example auxin synthesis and photolysis of water), not nitrogenase.
Q3AIPMT 2010 Prelims · Q140
One of the free-living anaerobic nitrogen-fixer is
Not quite — the answer is C.
Rhodospirillum is a free-living anaerobic (photosynthetic) nitrogen-fixing bacterium. Azotobacter and Beijerinckia are free-living but aerobic, and Rhizobium is symbiotic.
Q4AIPMT 2010 Prelims · Q143
Photoperiodism was first characterised in
Not quite — the answer is B.
Garner and Allard (1920) discovered photoperiodism while studying the Maryland Mammoth variety of tobacco, which flowered only when days were short.
Q5AIPMT 2010 Prelims · Q159
Coiling of garden pea tendrils around any support is an example of
Not quite — the answer is D.
Tendril coiling is a directional growth response to contact with a support, which makes it a tropism to touch (thigmotropism). Taxis means movement of a whole free-moving organism or cell, and a nastic movement such as Mimosa leaf folding does not depend on the direction of the stimulus.
Q6AIPMT 2010 Prelims · Q173
A renewable exhaustible natural resource is
Not quite — the answer is A.
Forests can regenerate, so they are renewable, but they can still be exhausted if they are cut faster than they regrow. Coal, petroleum and minerals form over geological time and are non-renewable.
Q7AIPMT 2010 Prelims · Q174
The two gases making highest relative contribution to the greenhouse gases are
Not quite — the answer is B.
The old NCERT figure puts the relative contribution to global warming at about 60% for CO₂, 20% for CH₄, 14% for CFCs and 6% for N₂O. So CO₂ and methane together contribute the most.
Q8AIPMT 2010 Prelims · Q176
Which two of the following changes (a–d) usually tend to occur in the plain dwellers when they move to high altitudes (3,500 m or more)? (a) Increase in red blood cell size (b) Increase in red blood cell production (c) Increased breathing rate (d) Increase in thrombocyte count Changes occurring are
Not quite — the answer is B.
At high altitude the low partial pressure of O₂ makes the body compensate by producing more RBCs (through erythropoietin) and by breathing faster. RBC size and platelet count do not rise as part of this acclimatisation.
Q9AIPMT 2010 Prelims · Q177
dB is a standard abbreviation used for the quantitative expression of
Not quite — the answer is C.
The decibel (dB) is the unit used to measure sound level, so it expresses noise, which is a pollutant. Sound above about 150 dB can damage the eardrum. It has nothing to do with pesticides or bacterial counts.
Q10AIPMT 2010 Prelims · Q192
Breeding of crops with high levels of minerals, vitamins and proteins is called
Not quite — the answer is C.
Biofortification is breeding crops for higher levels of vitamins, minerals, proteins or healthier fats, for example Atlas 66 wheat and vitamin-A-rich carrots. Biomagnification is the rise in toxin concentration up a food chain, which is unrelated.
Chapter-wise split of AIPMT 2010 Prelims Biology
- Human Reproduction7 Q
- Cell: The Unit of Life6 Q
- Principles of Inheritance and Variation5 Q
- Morphology of Flowering Plants5 Q
- Human Health and Disease5 Q
- Biotechnology and Its Applications5 Q
- Animal Kingdom4 Q
- Molecular Basis of Inheritance4 Q
- Biotechnology: Principles and Processes4 Q
- Biological Classification3 Q
- Anatomy of Flowering Plants3 Q
- Reproductive Health3 Q
- Chemical Coordination and Integration3 Q
- Sexual Reproduction in Flowering Plants3 Q
- Microbes in Human Welfare3 Q
- Structural Organisation in Animals2 Q
- Photosynthesis in Higher Plants2 Q
- Breathing and Exchange of Gases2 Q
- Excretory Products and Their Elimination2 Q
- Ecosystem2 Q
- Plant Kingdom1 Q
- Evolution1 Q
- Respiration in Plants1 Q
- Body Fluids and Circulation1 Q
- Locomotion and Movement1 Q
- Neural Control and Coordination1 Q
- Plant Growth and Development1 Q
- Biodiversity and Conservation1 Q
- Organisms and Populations1 Q
About this paper
How many questions from AIPMT 2010 Prelims are on this page?
177 questions (Physics 42, Chemistry 43, Biology 92). Questions that need a diagram we could not reproduce, or whose wording in the source was unclear, are left out.
Are these the original questions?
Yes. The questions and options are as set in the AIPMT 2010 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 2010 Prelims useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books. 16 questions here are on topics NCERT has since dropped; they are listed separately so you can skip them.
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