AIPMT 2015 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
153 questions from the AIPMT 2015 paper, conducted by CBSE. Tap an option to check your answer and read why it is right. Each subject has a free PDF with the answer key and explanations.
- Exam
- AIPMT 2015
- Conducted by
- CBSE
- Physics
- 34 questions
- Chemistry
- 37 questions
- Biology
- 82 questions
AIPMT 2015 Physics questions
34 questions. Tap an option to check your answer.
Q1AIPMT 2015
A radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = Velocity of light):
Not quite — the answer is C.
Radiation of energy E carries momentum E/c. A perfect reflector sends it back with the same magnitude, so the change in momentum is E/c − (−E/c) = 2E/c. E/c would be the answer only for a perfectly absorbing surface.
Q2AIPMT 2015
A ship A is moving Westwards with a speed of 10 km h⁻¹ and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h⁻¹. The time after which the distance between them becomes shortest, is:
Not quite — the answer is C.
Take A as origin: B starts at (0, −100) km and its velocity relative to A is (10, 10) km/h, i.e. 10√2 km/h at 45°. Time of closest approach t = −(r·v)/v² = (100 × 10)/200 = 5 h. The minimum separation is then 50√2 km, which is the trap behind 5√2 h.
Q3AIPMT 2015
The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius 'a' centred at the origin of the field, will be given by:
Not quite — the answer is D.
At r = a the field is Aa, radial, so the flux through the sphere is Aa × 4πa² = 4πAa³. By Gauss's law q = ε₀ × flux = 4πε₀Aa³. Forgetting the 4π gives ε₀Aa³.
Q4AIPMT 2015
In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?
Not quite — the answer is B.
Fringe width β = λD/d = 5 × 10⁻⁷ × 1/10⁻³ = 0.5 mm. Ten fringes fill the central diffraction maximum: 2λD/a = 10β = 5 mm, so a = 2 × 5 × 10⁻⁷ × 1/(5 × 10⁻³) = 2 × 10⁻⁴ m = 0.2 mm.
Q5AIPMT 2015
A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is:
Not quite — the answer is A.
Take torques about knife edge B: N_A × d = W × (d − x), so N_A = W(d − x)/d. Wx/d is the reaction at B, the usual mix-up.
Q6AIPMT 2015
Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r between sun and planet, i.e. T² = Kr³, here K is constant. If the masses of sun and planet are M and m respectively then as per Newton's law of gravitation force of attraction between them is F = GMm/r², here G is gravitational constant. The relation between G and K is described as:
Not quite — the answer is C.
Gravity provides the centripetal force: GMm/r² = m(2π/T)²r, so T² = (4π²/GM)r³. Hence K = 4π²/GM, i.e. GMK = 4π². GK = 4π² drops the Sun's mass.
Q7AIPMT 2015
Two particles of masses m₁, m₂ move with initial velocities u₁ and u₂. On collision, one of the particles get excited to higher level, after absorbing energy ε. If final velocities of particles be v₁ and v₂ then we must have:
Not quite — the answer is D.
Total energy is conserved: the initial kinetic energy equals the final kinetic energy plus the energy ε absorbed in exciting the particle, so ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂² + ε. Writing −ε on the final side would mean the system gained energy, which is wrong.
Q8AIPMT 2015
The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 × 10⁻¹¹ Pa⁻¹ and density of water is 10³ kg/m³. What fractional compression of water will be obtained at the bottom of the ocean?
Not quite — the answer is D.
Pressure at the bottom ≈ hρg = 2700 × 10³ × 10 = 2.7 × 10⁷ Pa. Fractional compression ΔV/V = compressibility × pressure = 45.4 × 10⁻¹¹ × 2.7 × 10⁷ ≈ 1.2 × 10⁻² (1.20 × 10⁻² with g = 9.8).
Q9AIPMT 2015
The two ends of a metal rod are maintained at temperatures 100°C and 110°C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow will be:
Not quite — the answer is A.
The conduction rate is kAΔT/L and depends on the temperature difference, not the actual temperatures. ΔT is 10 °C in both cases, so the rate stays 4.0 J/s (ignoring change of k with temperature).
Q10AIPMT 2015
A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = βx⁻²ⁿ, where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by:
Not quite — the answer is C.
a = v dv/dx. With v = βx⁻²ⁿ, dv/dx = −2nβx⁻²ⁿ⁻¹, so a = βx⁻²ⁿ × (−2nβx⁻²ⁿ⁻¹) = −2nβ²x⁻⁴ⁿ⁻¹. Using only dv/dx gives the −2n−1 power, which is not an acceleration.
Q11AIPMT 2015
The refracting angle of a prism is A, and refractive index of the material of the prism is cot (A/2). The angle of minimum deviation is:
Not quite — the answer is C.
μ = sin((A + δm)/2)/sin(A/2) = cot(A/2) = cos(A/2)/sin(A/2). So sin((A + δm)/2) = cos(A/2) = sin(90° − A/2), giving (A + δm)/2 = 90° − A/2 and δm = 180° − 2A.
Q12AIPMT 2015
A particle is executing SHM along a straight line. Its velocities at distances x₁ and x₂ from the mean position are V₁ and V₂, respectively. Its time period is:
Not quite — the answer is C.
V₁² = ω²(a² − x₁²) and V₂² = ω²(a² − x₂²). Subtracting, V₁² − V₂² = ω²(x₂² − x₁²), so ω = √((V₁² − V₂²)/(x₂² − x₁²)) and T = 2π√((x₂² − x₁²)/(V₁² − V₂²)). The inverted ratio gives 1/T, not T.
Q13AIPMT 2015
Two similar springs P and Q have spring constants K_P and K_Q, such that K_P > K_Q. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs W_P and W_Q are related as, in case (a) and case (b), respectively:
Not quite — the answer is D.
Same extension: W = ½kx², so the stiffer spring P does more work (W_P > W_Q). Same force: x = F/k and W = F²/2k, so the softer spring Q does more work (W_Q > W_P).
Q14AIPMT 2015
Consider 3rd orbit of He⁺ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K = 9 × 10⁹ constant, Z = 2 and h (Planck's Constant) = 6.6 × 10⁻³⁴ J s]
Not quite — the answer is C.
Bohr speed v = 2πKZe²/(nh) = 2π × 9 × 10⁹ × 2 × (1.6 × 10⁻¹⁹)²/(3 × 6.6 × 10⁻³⁴) ≈ 1.46 × 10⁶ m/s (equivalently 2.19 × 10⁶ × Z/n). 2.92 × 10⁶ m/s would be the n = 1.5 value, i.e. a mistake in n.
Q15AIPMT 2015
A particle of mass m is driven by a machine that delivers a constant power k watts. If the particle starts from rest the force on the particle at time t is:
Not quite — the answer is B.
Constant power: ½mv² = kt, so v = √(2kt/m). Then F = m dv/dt = m × √(2k/m) × ½t^(−1/2) = √(mk/2) t^(−1/2). Forgetting the ½ from differentiation gives √(2mk) t^(−1/2).
Q16AIPMT 2015
The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is:
Not quite — the answer is D.
Closed pipe fundamental = v/(4 × 20 cm) = v/80. Second overtone of an open pipe is the third harmonic, 3v/(2L). Equating, 3v/(2L) = v/80 gives L = 120 cm. Taking second overtone as the second harmonic gives 80 cm.
Q17AIPMT 2015
An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude:
Not quite — the answer is A.
The orbiting electron is a current loop with I = charge passing per second = ne. The field at the centre of a loop is μ₀I/(2r) = μ₀ne/(2r). μ₀ne/(2πr) is the formula for a long straight wire.
Q18AIPMT 2015
Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is:
Not quite — the answer is D.
Each plano-convex glass lens: 1/f = (1.5 − 1)/20 = 1/40 cm⁻¹. The oil forms a biconcave lens: 1/f = (1.7 − 1)(−1/20 − 1/20) = −0.07 cm⁻¹. Total 1/F = 2/40 − 0.07 = −0.02 cm⁻¹, so F = −50 cm.
Q19AIPMT 2015
On observing light from three different stars P, Q and R, it was found that intensity of violet colour is maximum in the spectrum of P, the intensity of green colour is maximum in the spectrum of R and the intensity of red colour is maximum in the spectrum of Q. If T_P, T_Q and T_R are the respective absolute temperatures of P, Q and R, then it can be concluded from the above observations that:
Not quite — the answer is C.
By Wien's displacement law λ_max T = constant, so a shorter peak wavelength means a hotter star. λ_violet < λ_green < λ_red, hence T_P > T_R > T_Q.
Q20AIPMT 2015
If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be:
Not quite — the answer is D.
Surface tension has dimensions [MT⁻²]. With E = [ML²T⁻²] and V = [LT⁻¹], M = E V⁻², so [MT⁻²] = [E V⁻² T⁻²].
Q21AIPMT 2015
A Carnot engine, having an efficiency of η = 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is:
Not quite — the answer is D.
For the Carnot cycle η = W/Q_H = 1/10, so with W = 10 J, Q_H = 100 J. The heat taken from the cold reservoir is Q_L = Q_H − W = 90 J. 100 J is the heat rejected to the hot reservoir, not that absorbed from the cold one.
Q22AIPMT 2015
For a parallel beam of monochromatic light of wavelength 'λ', diffraction is produced by a single slit whose width 'a' is of the order of the wavelength of the light. If 'D' is the distance of the screen from the slit, the width of the central maxima will be:
Not quite — the answer is B.
The first minima are at sin θ ≈ λ/a on either side, so the central maximum spans 2Dλ/a on the screen. Dλ/a is only the half-width.
Q23AIPMT 2015
A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m². Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be: (ρ_air = 1.2 kg/m³)
Not quite — the answer is D.
By Bernoulli the pressure above the roof is lower by ½ρv² = 0.5 × 1.2 × 40² = 960 Pa. Force = 960 × 250 = 2.4 × 10⁵ N, acting upwards because the higher pressure is inside the house.
Q24AIPMT 2015
The ratio of the specific heats Cp/Cv = γ in terms of degrees of freedom (n) is given by:
Not quite — the answer is D.
With n degrees of freedom C_v = (n/2)R and C_p = (n/2 + 1)R, so γ = C_p/C_v = 1 + 2/n. For a monatomic gas (n = 3) this gives 5/3 as expected.
Q25AIPMT 2015
If radius of the ²⁷₁₃Al nucleus is taken to be R_Al, then the radius of ¹²⁵₅₃Te nucleus is nearly:
Not quite — the answer is C.
Nuclear radius R = R₀A^(1/3), so R_Te/R_Al = (125/27)^(1/3) = 5/3. Using atomic numbers instead of mass numbers gives the (53/13)^(1/3) trap.
Q26AIPMT 2015
A block of mass 10 kg, moving in x direction with a constant speed of 10 ms⁻¹, is subjected to a retarding force F = 0.1x J/m during its travel from x = 20 m to 30 m. Its final KE will be:
Not quite — the answer is B.
Initial KE = ½ × 10 × 10² = 500 J. Work by the retarding force = −∫₂₀³⁰ 0.1x dx = −0.05(900 − 400) = −25 J. Final KE = 500 − 25 = 475 J.
Q27AIPMT 2015
A block A of mass m₁ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m₂ is suspended. The coefficient of kinetic friction between the block and the table is μₖ. When the block A is sliding on the table, the tension in the string is:
Not quite — the answer is D.
a = (m₂ − μₖm₁)g/(m₁ + m₂). For block A: T − μₖm₁g = m₁a, so T = m₁[(m₂ − μₖm₁)g/(m₁ + m₂) + μₖg] = m₁m₂(1 + μₖ)g/(m₁ + m₂). Friction increases the tension compared with the frictionless value.
Q28AIPMT 2015
A certain metallic surface is illuminated with monochromatic light of wavelength λ. The stopping potential for photo-electric current for this light is 3V₀. If the same surface is illuminated with light of wavelength 2λ, the stopping potential is V₀. The threshold wavelength for this surface for photo-electric effect is:
Not quite — the answer is C.
hc/λ = φ + 3eV₀ and hc/2λ = φ + eV₀. Subtracting, hc/2λ = 2eV₀, so eV₀ = hc/4λ and φ = hc/2λ − hc/4λ = hc/4λ. Threshold wavelength = hc/φ = 4λ.
Q29AIPMT 2015
When two displacements represented by y₁ = a sin(ωt) and y₂ = b cos(ωt) are superimposed the motion is:
Not quite — the answer is D.
a sin ωt + b cos ωt = √(a² + b²) sin(ωt + φ) with tan φ = b/a: a single SHM of the same ω. The two are 90° out of phase, so amplitudes add as √(a² + b²), not (a + b)/2.
Q30AIPMT 2015
A potentiometer wire has length 4 m and resistance 8 Ω. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2 V, so as to get a potential gradient 1 mV per cm on the wire is:
Not quite — the answer is B.
1 mV/cm = 0.1 V/m, so 0.4 V must drop across the 4 m, 8 Ω wire: I = 0.4/8 = 0.05 A. Total resistance = 2/0.05 = 40 Ω, so the series resistance is 40 − 8 = 32 Ω. 40 Ω is the total, not the added resistance.
Q31AIPMT 2015
Two spherical bodies of mass M and 5 M and radii R and 2 R are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is:
Not quite — the answer is D.
Only internal forces act, so the centre of mass stays fixed. It is 5M × 12R/6M = 10R from the small body initially. At contact the centres are R + 2R = 3R apart and the small body is 2.5R from the centre of mass, so it moves 10R − 2.5R = 7.5R.
Q32AIPMT 2015
A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z', the power drawn will be:
Not quite — the answer is B.
Without L: P = V²/R. With L, I = V/Z and power = I²R = V²R/Z² = (V²/R)(R/Z)² = P(R/Z)². P(R/Z) misses that both current and power factor fall by R/Z.
Q33AIPMT 2015
Across a metallic conductor of non-uniform cross section a constant potential difference is applied. The quantity which remains constant along the conductor is:
Not quite — the answer is C.
The conductor is a single series path, so charge conservation makes the current the same at every cross-section. Current density, field and drift velocity all vary inversely with the local area.
Q34AIPMT 2015
A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?
Not quite — the answer is A.
After disconnection the plates are isolated, so the charge stays constant; the statement that charge is not conserved is the incorrect one. With Q fixed, V = Q/KC falls K times and U = Q²/2KC falls K times, so the change is ½CV²(1/K − 1).
Chapter-wise split of AIPMT 2015 Physics
- Work, Energy and Power4 Q
- Wave Optics2 Q
- Gravitation2 Q
- Thermal Properties of Matter2 Q
- Ray Optics and Optical Instruments2 Q
- Oscillations2 Q
- Current Electricity2 Q
- Electromagnetic Waves1 Q
- Motion in a Plane1 Q
- Electric Charges and Fields1 Q
- System of Particles and Rotational Motion1 Q
- Mechanical Properties of Solids1 Q
- Motion in a Straight Line1 Q
- Atoms1 Q
- Waves1 Q
- Moving Charges and Magnetism1 Q
- Units and Measurements1 Q
- Thermodynamics1 Q
- Mechanical Properties of Fluids1 Q
- Kinetic Theory1 Q
- Nuclei1 Q
- Laws of Motion1 Q
- Dual Nature of Radiation and Matter1 Q
- Alternating Current1 Q
- Electrostatic Potential and Capacitance1 Q
AIPMT 2015 Chemistry questions
32 questions. Tap an option to check your answer.
Q1AIPMT 2015
The enolic form of ethyl acetoacetate, CH₃–C(OH)=CH–COOC₂H₅ (in equilibrium with the keto form CH₃–CO–CH₂–COOC₂H₅), has:
Not quite — the answer is B.
CH₃–C(OH)=CH–COOC₂H₅ has 9 C–H, 4 C–C, 4 C–O and 1 O–H σ bonds = 18 σ bonds, and two π bonds (one C=C, one C=O). 9 σ bonds counts only the skeleton without hydrogens.
Q2AIPMT 2015
Which of the following species contain equal number of σ- and π-bonds?
Not quite — the answer is C.
XeO₄ has four Xe–O σ bonds and four π bonds (each Xe=O is one σ + one π). CH₂(CN)₂ has 6 σ and 4 π, (CN)₂ has 3 σ and 4 π, and HCO₃⁻ has 4 σ and 1 π.
Q3AIPMT 2015
The correct bond order in the following species is:
Not quite — the answer is A.
By MO theory O₂ has bond order 2; removing antibonding electrons raises it: O₂⁺ = 2.5, O₂²⁺ = 3, while adding one gives O₂⁻ = 1.5. So O₂⁻ < O₂⁺ < O₂²⁺.
Q4AIPMT 2015
Which of these statements about [Co(CN)₆]³⁻ is true?
Not quite — the answer is B.
Co³⁺ is 3d⁶. CN⁻ is a strong-field ligand, so the electrons pair up as t₂g⁶e_g⁰ (d²sp³, inner orbital): low spin with no unpaired electrons. Four unpaired electrons would occur only with a weak-field ligand such as F⁻.
Q5AIPMT 2015
The activation energy of a reaction can be determined from the slope of which of the following graphs?
Not quite — the answer is D.
From the Arrhenius equation ln k = ln A − Ea/RT, a plot of ln k against 1/T is a straight line of slope −Ea/R, from which Ea is found. ln k against T is curved.
Q6AIPMT 2015
The reaction (CH₃)₃C–ONa + CH₃CH₂Cl → (CH₃)₃C–O–CH₂CH₃ + NaCl is called:
Not quite — the answer is B.
An alkoxide reacting with a primary alkyl halide by SN2 to give an ether is the Williamson synthesis. Gattermann–Koch and Etard reactions make aromatic aldehydes, not ethers.
Q7AIPMT 2015
Which one is not equal to zero for an ideal solution?
Not quite — the answer is C.
For an ideal solution ΔH_mix = 0, ΔV_mix = 0 and the vapour pressure obeys Raoult's law exactly, so ΔP = 0. Mixing still increases randomness, so ΔS_mix > 0 (this is what makes ΔG_mix negative).
Q8AIPMT 2015
An organic compound 'X' having molecular formula C₅H₁₀O yields phenyl hydrazone and gives negative response to the Iodoform test and Tollens' test. It produces n-pentane on reduction. 'X' could be:
Not quite — the answer is D.
Forming a phenylhydrazone shows a C=O group; a negative Tollens' test rules out pentanal; a negative iodoform test rules out the CH₃CO– group of 2-pentanone. Reduction to n-pentane needs an unbranched chain, so X is 3-pentanone (CH₃CH₂COCH₂CH₃).
Q9AIPMT 2015
Cobalt(III) chloride forms several octahedral complexes with ammonia. Which of the following will not give test for chloride ions with silver nitrate at 25°C?
Not quite — the answer is B.
Octahedral Co(III) needs six ligands. With only three NH₃ the complex is [Co(NH₃)₃Cl₃], all three Cl⁻ inside the coordination sphere, so no Cl⁻ ions are free to precipitate with AgNO₃. The 6, 5 and 4 NH₃ complexes have 3, 2 and 1 ionisable Cl⁻.
Q10AIPMT 2015
A mixture of gases contains H₂ and O₂ gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the two gases in the mixture?
Not quite — the answer is C.
Take 1 g H₂ and 4 g O₂: moles = 1/2 = 0.5 and 4/32 = 0.125. Mole ratio 0.5 : 0.125 = 4 : 1. 1 : 4 is just the mass ratio.
Q11AIPMT 2015
The electrolytic reduction of nitrobenzene in strongly acidic medium produces:
Not quite — the answer is B.
In strongly acidic medium nitrobenzene is reduced electrolytically to phenylhydroxylamine, which rearranges in acid to p-aminophenol. Aniline is the product of complete reduction in mildly acidic conditions (e.g. Sn/HCl), and azo/azoxy compounds form in alkaline medium.
Q12AIPMT 2015
Which of the following statements is correct for a reversible process in a state of equilibrium?
Not quite — the answer is D.
ΔG = ΔG° + 2.303RT log Q, and at equilibrium ΔG = 0 with Q = K. So ΔG° = −2.303RT log K. ΔG = −2.303RT log K is wrong because ΔG itself is zero at equilibrium.
Q13AIPMT 2015
Which of the following pairs of ions are isoelectronic and isostructural?
Not quite — the answer is A.
ClO₃⁻ (17 + 24 + 1 = 42 electrons) and SO₃²⁻ (16 + 24 + 2 = 42) are isoelectronic, and both have three bond pairs and one lone pair, so both are pyramidal. CO₃²⁻ and NO₃⁻ are trigonal planar with 32 electrons.
Q14AIPMT 2015
The angular momentum of electron in 'd' orbital is equal to:
Not quite — the answer is B.
Orbital angular momentum = √(l(l + 1)) ħ. For a d orbital l = 2, giving √6 ħ. √2 ħ is the value for a p orbital.
Q15AIPMT 2015
Which of the following options represents the correct bond order?
Not quite — the answer is C.
MO bond orders: O₂⁻ = 1.5, O₂ = 2, O₂⁺ = 2.5, because electrons are added to or removed from the π* antibonding orbitals. Hence O₂⁻ < O₂ < O₂⁺.
Q16AIPMT 2015
Magnetic moment 2.84 B.M. is given by : (At. Nos., Ni = 28, Ti = 22, Cr = 24, Co = 27)
Not quite — the answer is B.
μ = √(n(n + 2)) = 2.84 BM gives n = 2. Ni²⁺ (3d⁸) has 2 unpaired electrons, whereas Ti³⁺ (3d¹) has 1, Co²⁺ (3d⁷) has 3 and Cr²⁺ (3d⁴) has 4.
Q17AIPMT 2015
A single compound of the structure OHC–CH₂–CH(CH₃)–CH₂–CO–CH₃ is obtainable from ozonolysis of which of the following cyclic compounds?
Not quite — the answer is B.
In 1,4-dimethylcyclopentene the ring C1(CH₃)=C2 bond is cleaved: C1 becomes a methyl ketone and C2 an aldehyde, still joined through C5–C4(CH₃)–C3. This gives the single product CH₃CO–CH₂–CH(CH₃)–CH₂–CHO. 1,2-Dimethylcyclopentene would give a diketone.
Q18AIPMT 2015
The total number of π-bond electrons in the following structure is: (CH₃)₂C=CH–C(=CH₂)–CH=CH–CH=C(CH₃)₂
Not quite — the answer is C.
The structure contains four C=C double bonds (including the exocyclic =CH₂), i.e. four π bonds. Each π bond has 2 electrons, so there are 8 π electrons; 4 is the number of π bonds, not electrons.
Q19AIPMT 2015
The Ksp of Ag₂CrO₄, AgCl, AgBr and AgI are respectively, 1.1 × 10⁻¹², 1.8 × 10⁻¹⁰, 5.0 × 10⁻¹³, 8.3 × 10⁻¹⁷. Which one of the following salts will precipitate last if AgNO₃ solution is added to the solution containing equal moles of NaCl, NaBr, NaI and Na₂CrO₄?
Not quite — the answer is A.
[Ag⁺] needed to start precipitation (for 1 M anion): AgI 8.3 × 10⁻¹⁷, AgBr 5.0 × 10⁻¹³, AgCl 1.8 × 10⁻¹⁰ M, but Ag₂CrO₄ needs √(1.1 × 10⁻¹²) ≈ 1.05 × 10⁻⁶ M. Ag₂CrO₄ needs the highest Ag⁺, so it precipitates last even though its Ksp is not the largest.
Q20AIPMT 2015
When initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction is:
Not quite — the answer is C.
t½ ∝ [A]₀^(1 − n). It is independent of the initial concentration only when n = 1, as in t½ = 0.693/k. For zero order t½ doubles when [A]₀ doubles.
Q21AIPMT 2015
Which of the following processes does not involve oxidation of iron?
Not quite — the answer is D.
In Fe(CO)₅ iron is in the zero oxidation state, as in the free metal, so no oxidation occurs. Iron is oxidised in rusting, in liberating H₂ from steam (to Fe₃O₄) and in displacing Cu from CuSO₄ (to Fe²⁺).
Q22AIPMT 2015
The boiling point of 0.2 mol kg⁻¹ solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case?
Not quite — the answer is B.
ΔT_b = iK_bm. At equal molality, a higher boiling point means a larger van't Hoff factor for X, so X dissociates in water. Molecular mass does not matter once molality is fixed.
Q23AIPMT 2015
In Dumas' method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is:
Not quite — the answer is D.
Dry gas pressure = 725 − 25 = 700 mm. V(STP) = 40 × 700 × 273/(300 × 760) = 33.5 mL; mass of N₂ = 28 × 33.5/22400 = 0.0419 g. %N = 0.0419/0.25 × 100 = 16.76. Not subtracting aqueous tension gives a higher value.
Q24AIPMT 2015
The species Ar, K⁺ and Ca²⁺ contain the same number of electrons. In which order do their radii increase?
Not quite — the answer is D.
All three have 18 electrons; the greater the nuclear charge (Ca 20 > K 19 > Ar 18), the more strongly the electrons are pulled in. So the radii increase Ca²⁺ < K⁺ < Ar.
Q25AIPMT 2015
Because of lanthanoid contraction, which of the following pairs of elements have nearly same atomic radii ? (Numbers in the parenthesis are atomic numbers).
Not quite — the answer is D.
The 14 lanthanoids between La and Hf cause a contraction that cancels the expected size increase from the 4d to the 5d series. So Zr (160 pm) and Hf (159 pm) have almost the same radius. Ti and Zr differ normally because no f-block intervenes.
Q26AIPMT 2015
The number of d-electrons in Fe²⁺ (Z = 26) is not equal to the number of electrons in which one of the following?
Not quite — the answer is C.
Fe²⁺ is [Ar]3d⁶, so it has 6 d-electrons. Ne has 6 p-electrons, Mg has 6 s-electrons and Fe has 6 d-electrons, but Cl (1s²2s²2p⁶3s²3p⁵) has 11 p-electrons.
Q27AIPMT 2015
Which one of the following electrolytes has the same value of van't Hoff's factor (i) as that of Al₂(SO₄)₃ (if all are 100% ionised)?
Not quite — the answer is A.
Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻ gives i = 5. K₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]⁴⁻ also gives 5 ions. K₃[Fe(CN)₆] and Al(NO₃)₃ give 4 and K₂SO₄ gives 3.
Q28AIPMT 2015 · Q153
Nitrogen dioxide and sulphur dioxide have some properties in common. Which property is shown by one of these compounds, but not by the other?
Not quite — the answer is A.
SO₂ (as sulphite/metabisulphite) is used as a food preservative, while NO₂ is not. Both are water-soluble acidic oxides that cause acid rain, and both can act as reducing agents.
Q29AIPMT 2015
Maximum bond angle at nitrogen is present in which of the following?
Not quite — the answer is D.
NO₂⁺ is isoelectronic with CO₂ and is linear (180°). NO₃⁻ is trigonal planar (120°), NO₂ is bent (~134°) and NO₂⁻ is bent (~115°) due to the lone pair.
Q30AIPMT 2015
The reaction of C₆H₅CH=CHCH₃ with HBr produces:
Not quite — the answer is B.
H⁺ adds to the carbon away from the ring so that the positive charge sits on the benzylic carbon, where it is stabilised by resonance with the ring. Br⁻ then attacks there, giving C₆H₅CH(Br)CH₂CH₃. Ring bromination does not occur with HBr.
Q31AIPMT 2015
A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known as:
Not quite — the answer is B.
A fuel cell converts the free energy of combustion of fuels like H₂ or CH₄ directly into electricity. A Ni–Cd cell is a rechargeable storage cell, and a dynamo converts mechanical energy, not chemical.
Q32AIPMT 2015
If the value of an equilibrium constant for a particular reaction is 1.6 × 10¹², then at equilibrium the system will contain:
Not quite — the answer is D.
K = [products]/[reactants] = 1.6 × 10¹² is very large, so at equilibrium products far exceed reactants. Similar amounts would correspond to K near 1.
Chemistry: not in the current NEET syllabus
These 5 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
5 questions. Tap an option to check your answer.
Q1AIPMT 2015 · Q138
Biodegradable polymer which can be produced from glycine and aminocaproic acid is:
Not quite — the answer is B.
Nylon-2–nylon-6 is an alternating polyamide copolymer of glycine (H₂N–CH₂–COOH) and ε-aminocaproic acid, and it is biodegradable. PHBV is also biodegradable but is made from 3-hydroxybutanoic and 3-hydroxypentanoic acids.
Q2AIPMT 2015 · Q165
Bithional is generally added to the soaps as an additive to function as a/an:
Not quite — the answer is A.
Bithional is added to medicated soaps as an antiseptic to reduce skin bacteria and odour. It does not soften, dry or buffer the soap.
Q3AIPMT 2015 · Q167
A given metal crystallizes out with a cubic structure having edge length of 361 pm. If there are four metal atoms in one unit cell, what is the radius of one atom?
Not quite — the answer is C.
Four atoms per cubic unit cell means fcc, where atoms touch along the face diagonal: 4r = √2 a. So r = √2 × 361/4 = 0.3536 × 361 ≈ 127.6 pm ≈ 127 pm. Using the bcc relation √3a/4 would give about 156 pm.
Q4AIPMT 2015 · Q177
Which property of colloidal solution is independent of charge on the colloidal particles?
Not quite — the answer is A.
The Tyndall effect is scattering of light by colloidal particles and depends on their size, not on their charge. Coagulation, electrophoresis and electro-osmosis all depend on the particle charge.
Q5AIPMT 2015 · Q178
Solubility of the alkaline earth metal sulphates in water decreases in the sequence:
Not quite — the answer is B.
Down group 2 the lattice energy of the sulphates changes little (large SO₄²⁻), but hydration energy falls sharply as the cation grows. So solubility decreases Mg > Ca > Sr > Ba; BaSO₄ is practically insoluble.
Chapter-wise split of AIPMT 2015 Chemistry
- Chemical Bonding and Molecular Structure6 Q
- Solutions3 Q
- Equilibrium3 Q
- Coordination Compounds2 Q
- Chemical Kinetics2 Q
- Structure of Atom2 Q
- The d- and f-Block Elements2 Q
- Hydrocarbons2 Q
- Organic Chemistry: Some Basic Principles and Techniques2 Q
- Alcohols, Phenols and Ethers1 Q
- Aldehydes, Ketones and Carboxylic Acids1 Q
- Some Basic Concepts of Chemistry1 Q
- Amines1 Q
- Redox Reactions1 Q
- Classification of Elements and Periodicity in Properties1 Q
- Electrochemistry1 Q
AIPMT 2015 Biology questions
72 questions. Tap an option to check your answer.
Q1AIPMT 2015
Leaves become modified into spines in:
Not quite — the answer is B.
In Opuntia the leaves are reduced to spines that cut water loss, while the flattened green stem (phylloclade) does photosynthesis. Pea has leaflets modified into tendrils, onion has fleshy storage leaves and silk cotton has normal leaves with prickles on the stem.
Q2AIPMT 2015
Vertical distribution of different species occupying different levels in a biotic community is known as:
Not quite — the answer is C.
Stratification is the vertical layering of species in a community, e.g. trees at the top, then shrubs, then herbs and grasses in a forest. Zonation is a horizontal change of communities along a gradient, not vertical layering within one community.
Q3AIPMT 2015
Gene regulation governing lactose operon of E.coli that involves the lac I gene product is:
Not quite — the answer is C.
The lac I gene codes for a repressor that binds the operator and blocks transcription, so control is negative. Lactose (allolactose) inactivates the repressor and switches the operon on, so it is inducible, not repressible.
Q4AIPMT 2015
Which one of the following matches is correct?
Not quite — the answer is C.
Alternaria is a deuteromycete (imperfect fungus) in which only the asexual phase is known. Agaricus is a saprophytic basidiomycete, Phytophthora and Mucor have aseptate mycelium but belong to Phycomycetes, not Basidiomycetes or Ascomycetes.
Q5AIPMT 2015
Which of these is not an important component of initiation of parturition in humans?
Not quite — the answer is A.
Parturition is triggered by signals from the fully developed foetus and placenta: a rising estrogen to progesterone ratio, prostaglandins and oxytocin cause strong uterine contractions. Prolactin controls milk production after birth and has no role in starting labour.
Q6AIPMT 2015
A chemical signal that has both endocrine and neural roles is:
Not quite — the answer is D.
Epinephrine (adrenaline) is released into the blood by the adrenal medulla as a hormone and is also a neurotransmitter of sympathetic nerves. Cortisol, melatonin and calcitonin act only as hormones.
Q7AIPMT 2015
Match each disease with its correct type of vaccine: (a) tuberculosis (b) Whooping cough (c) diphtheria (d) polio — (i) harmless virus (ii) inactivated toxin (iii) killed bacteria (iv) harmless bacteria
Not quite — the answer is D.
BCG for tuberculosis uses harmless (attenuated) bacteria, the whooping cough vaccine uses killed bacteria, the diphtheria vaccine is a toxoid (inactivated toxin) and oral polio vaccine uses harmless (attenuated) virus. The other matchings pair diphtheria or polio with bacteria, which is wrong.
Q8AIPMT 2015
Nuclear envelope is a derivative of:
Not quite — the answer is A.
The outer membrane of the nuclear envelope is continuous with the rough endoplasmic reticulum and also carries ribosomes, so it is regarded as derived from rough ER. Microtubules are protein structures, not membranes.
Q9AIPMT 2015
The crops engineered for glyphosate are resistant/tolerant to:
Not quite — the answer is A.
Glyphosate is a broad-spectrum herbicide, and glyphosate-tolerant (e.g. Roundup Ready) crops survive spraying that kills weeds. Insect resistance is a different trait, given by Bt genes.
Q10AIPMT 2015
Vascular bundles in monocotyledons are considered closed because:
Not quite — the answer is C.
A vascular bundle is closed when there is no cambium between xylem and phloem, so it cannot form secondary tissues. Monocot bundles lack this cambium; having a bundle sheath does not make a bundle closed.
Q11AIPMT 2015
True nucleus is absent in:
Not quite — the answer is B.
Anabaena is a cyanobacterium, a prokaryote whose DNA lies free in the cytoplasm without a nuclear membrane. Volvox and Vaucheria are algae and Mucor is a fungus, all eukaryotes with a true nucleus.
Q12AIPMT 2015
Which one of the following statements is not true?
Not quite — the answer is A.
Bees make honey from the nectar they collect, not by digesting pollen, so that statement is false. Pollen is used as a nutritional supplement, causes allergies in some people, and flowers pollinated by flies and bats often give out foul odours.
Q13AIPMT 2015
A gymnast is able to balance his body upside down even in the total darkness because of:
Not quite — the answer is C.
The vestibular apparatus (semicircular canals and otolith organ) senses body position and movement and keeps balance without needing vision. The organ of Corti, cochlea and tectorial membrane are parts of the hearing mechanism.
Q14AIPMT 2015
The hilum is a scar on the:
Not quite — the answer is B.
The hilum is the scar on the seed coat where the funicle (stalk) attached the ovule to the placenta. The micropyle is a separate small pore next to the hilum, not the scar itself.
Q15AIPMT 2015
Which one of the following is correct?
Not quite — the answer is A.
Blood is plasma plus the formed elements: RBCs, WBCs and platelets. Serum is plasma without clotting factors, and lymph has no RBCs, so the other equations are wrong.
Q16AIPMT 2015
The guts of cow and buffalo possess:
Not quite — the answer is D.
The rumen of cattle and buffalo contains methanogens such as Methanobacterium, which help break down cellulose and release methane. Cyanobacteria, Chlorella and Fucus are photosynthetic and do not live in the gut.
Q17AIPMT 2015
Which one of the following may require pollinators, but is genetically similar to autogamy?
Not quite — the answer is B.
In geitonogamy pollen goes to another flower of the same plant, so a pollinating agent is needed, but the pollen and ovule are genetically the same, as in self-pollination. Xenogamy brings genetically different pollen, and cleistogamy needs no pollinator at all.
Q18AIPMT 2015
In sea urchin DNA, which is double stranded, 17% of the bases were shown to be cytosine. The percentages of the other three bases expected to be present in this DNA are:
Not quite — the answer is D.
By Chargaff's rule G = C = 17%, so G + C = 34% and A + T = 100 - 34 = 66%. Since A = T, each is 33%. Options with unequal A and T break the base-pairing rule.
Q19AIPMT 2015
Capacitation refers to changes in the:
Not quite — the answer is B.
Capacitation is the set of changes in sperm inside the female reproductive tract that make them able to fertilise the ovum, so it happens before fertilisation. It is a change in the sperm, not in the ovum.
Q20AIPMT 2015
Which of the following had the smallest brain capacity?
Not quite — the answer is A.
Homo habilis had a brain of about 650-800 cc, smaller than Homo erectus (about 900 cc), Neanderthal man (about 1400 cc) and modern humans (about 1350-1450 cc).
Q21AIPMT 2015
Which of the following viruses is not transferred through semen of an infected male?
Not quite — the answer is D.
Chikungunya virus spreads through the bite of Aedes mosquitoes and is not passed on through semen. HIV and hepatitis B are sexually transmitted, and Ebola virus can persist in and spread through semen of survivors.
Q22AIPMT 2015
A major characteristic of the monocot root is the presence of:
Not quite — the answer is D.
Monocot roots have no cambium, so they do not undergo secondary growth. Scattered vascular bundles are a feature of the monocot stem, not the root, where xylem and phloem are arranged radially.
Q23AIPMT 2015
Blood pressure in the mammalian aorta is maximum during:
Not quite — the answer is D.
The left ventricle pumps blood into the aorta, so aortic pressure reaches its peak (systolic pressure, about 120 mm Hg) during left ventricular systole. Atrial contraction only tops up the ventricles and does not raise aortic pressure.
Q24AIPMT 2015
In Bt cotton, the Bt toxin present in plant tissue as pro-toxin is converted into active toxin due to:
Not quite — the answer is B.
Bt toxin is made as an inactive protoxin crystal. The alkaline pH of the insect gut dissolves the crystal and activates the toxin, which then binds the midgut cells and makes pores. An acidic gut would not activate it.
Q25AIPMT 2015
In an ecosystem the rate of production of organic matter during photosynthesis is termed as:
Not quite — the answer is C.
Gross primary productivity is the rate of production of organic matter by photosynthesis. Net primary productivity is what remains after the producers' own respiratory loss is subtracted.
Q26AIPMT 2015
Keel is the characteristic feature of flower of:
Not quite — the answer is C.
A keel (carina) of two fused anterior petals is part of the papilionaceous corolla of Fabaceae, to which Indigofera belongs. Tomato is Solanaceae and tulip and Aloe are Liliaceae, which do not have a keel.
Q27AIPMT 2015
In which of the following gametophyte is not independent free living?
Not quite — the answer is A.
In gymnosperms such as Pinus, the gametophytes are much reduced and stay within the ovule or pollen grain, depending on the sporophyte. Funaria and Marchantia (bryophytes) and Pteris (fern) have free-living gametophytes.
Q28AIPMT 2015
The structures that are formed by stacking of organized flattened membranous sacs in the chloroplasts are:
Not quite — the answer is C.
Thylakoids are flattened membranous sacs that are stacked like coins to form grana. Stroma lamellae are the flat tubules joining grana, and cristae are folds of the inner mitochondrial membrane.
Q29AIPMT 2015
Which of the following does not favour the formation of large quantities of dilute urine?
Not quite — the answer is D.
Renin starts the renin-angiotensin-aldosterone pathway, which increases Na⁺ and water reabsorption and so reduces urine volume. Atrial natriuretic factor, alcohol (which inhibits ADH) and caffeine all increase output of dilute urine.
Q30AIPMT 2015
DNA is not present in:
Not quite — the answer is C.
Ribosomes are made of rRNA and proteins and contain no DNA. Mitochondria and chloroplasts have their own circular DNA, and the nucleus holds the chromosomal DNA.
Q31AIPMT 2015
Which of the following are the important flora rewards to the animal pollinators?
Not quite — the answer is C.
Nectar and pollen are the usual food rewards that flowers offer to animal pollinators. Colour, size and fragrance attract pollinators but are signals, not rewards.
Q32AIPMT 2015
Which of the following represents the correct combination without any exception? (Characteristics – Class)
Not quite — the answer is C.
NCERT describes Chondrichthyes as having a ventral mouth, gill slits without an operculum, skin with minute placoid scales and a notochord that persists throughout life, so this set holds for all members. The mammal set has exceptions: whales lack external pinnae and hind limbs. Birds have dry skin without glands, and cyclostomes lack paired appendages.
Q33AIPMT 2015
Alleles are:
Not quite — the answer is D.
Alleles are alternative forms of the same gene at one locus, differing slightly in their DNA sequence. Heterozygotes and true-breeding homozygotes are kinds of individuals, and a phenotype is the visible trait, not the gene form.
Q34AIPMT 2015
Hysterectomy is surgical removal of:
Not quite — the answer is B.
Hysterectomy is surgical removal of the uterus. Removing part of the vas deferens is vasectomy, and removal of the mammary gland is mastectomy.
Q35AIPMT 2015
HIV that causes AIDS, first starts destroying:
Not quite — the answer is D.
After entering the body, HIV multiplies in macrophages and then enters and destroys helper T-lymphocytes (CD4 cells), whose number falls steadily. B-lymphocytes and platelets are not its primary targets.
Q36AIPMT 2015
Which one of the following statements is wrong?
Not quite — the answer is A.
Mannitol (with laminarin) is the stored food of brown algae (Phaeophyceae); red algae store floridean starch, so that statement is wrong. Algin and carrageen come from algae, agar from Gelidium and Gracilaria, and Chlorella and Spirulina are used as space food.
Q37AIPMT 2015
Cryopreservation of gametes of threatened species in viable and fertile condition can be referred to as:
Not quite — the answer is C.
Keeping gametes of threatened species frozen and viable for long periods is cryopreservation, which is done away from the natural habitat, so it is an advanced ex situ method. In situ methods, such as sacred groves, protect species where they live.
Q38AIPMT 2015
Select the correct matching in the following pairs:
Not quite — the answer is C.
Smooth ER is the main site of lipid synthesis, including steroid hormones in animal cells. Rough ER carries ribosomes and makes proteins; fatty acid oxidation happens in mitochondria and peroxisomes.
Q39AIPMT 2015
Secondary Succession takes place on/in:
Not quite — the answer is C.
Secondary succession begins in areas where a natural community has been destroyed but soil remains, such as a degraded, burnt or cut forest, so it is faster. Newly cooled lava, bare rock and a newly created pond have no previous community and undergo primary succession.
Q40AIPMT 2015
Which of the following is not a sexually transmitted disease?
Not quite — the answer is A.
Encephalitis is inflammation of the brain, usually from mosquito-borne viruses, and is not sexually transmitted. Syphilis, AIDS and trichomoniasis are all sexually transmitted infections.
Q41AIPMT 2015
The movement of a gene from one linkage group to another is called:
Not quite — the answer is D.
Translocation moves a chromosome segment, with its genes, to a non-homologous chromosome, which is a different linkage group. Crossing over only exchanges segments between homologous chromosomes of the same linkage group.
Q42AIPMT 2015
Typical growth curve in plants is:
Not quite — the answer is B.
Plant growth passes through a slow lag phase, a fast exponential (log) phase and a stationary phase, giving an S-shaped or sigmoid curve when plotted against time.
Q43AIPMT 2015
Cytochromes are found in:
Not quite — the answer is D.
Cytochromes are electron carriers of the electron transport chain, which lies in the inner mitochondrial membrane, folded into cristae. The matrix holds the Krebs cycle enzymes, not the cytochromes.
Q44AIPMT 2015
Which of the following regions of the brain is incorrectly paired with its function?
Not quite — the answer is C.
The cerebellum coordinates movement, posture and balance; language comprehension is a function of the cerebral cortex. The other pairings, including corpus callosum linking the two hemispheres, are correct.
Q45AIPMT 2015
Which of the following characteristics is mainly responsible for diversification of insects on land?
Not quite — the answer is D.
The chitinous exoskeleton prevents water loss and protects the body, which let insects colonise dry land in great variety. Segmentation and bilateral symmetry are shared with many aquatic animals and do not solve the problem of drying out.
Q46AIPMT 2015
Sliding filament theory can be best explained as:
Not quite — the answer is D.
In sliding filament theory, thin actin filaments slide over thick myosin filaments toward the centre of the sarcomere; the I band and H zone shorten but the filaments themselves keep their length. The options in which one or both filaments shorten are wrong.
Q47AIPMT 2015
Which one of the following is not an inclusion body found in prokaryotes?
Not quite — the answer is A.
Inclusion bodies in prokaryotes are stores of reserve material lying free in the cytoplasm, such as phosphate granules, cyanophycean granules and glycogen granules. A polysome is a chain of ribosomes translating one mRNA, not a storage body.
Q48AIPMT 2015
The mass of living material at a trophic level at a particular time is called:
Not quite — the answer is A.
Standing crop is the mass of living material present at a trophic level at a given time. Standing state is the amount of inorganic nutrients in the soil at a given time, and productivity is a rate, not a mass.
Q49AIPMT 2015
Select the correct option: Column I — (a) Synapsis aligns homologous chromosomes (b) Synthesis of RNA and protein (c) Action of enzyme recombinase (d) Centromeres do not separate but chromatids move towards opposite poles; Column II — (i) Anaphase-II (ii) Zygotene (iii) G₂-phase (iv) Anaphase-I (v) Pachytene
Not quite — the answer is C.
Synapsis of homologues occurs in zygotene, RNA and protein synthesis happen in the G₂ phase, recombinase acts during crossing over in pachytene, and in anaphase I centromeres do not split, so the two chromatids of each chromosome move together to one pole.
Q50AIPMT 2015
Multiple alleles are present:
Not quite — the answer is D.
Multiple alleles are more than two alternative forms of one gene, so they all occupy the same locus (e.g. Iᴬ, Iᴮ and i of the ABO gene). Genes at different loci or on different chromosomes are different genes, not alleles.
Q51AIPMT 2015
Which is the most common mechanism of genetic variation in the population of a sexually-reproducing organism?
Not quite — the answer is A.
In sexually reproducing organisms, crossing over and independent assortment in meiosis and random fusion of gametes recombine genes every generation, so recombination is the most common source of variation. Transduction is a bacterial process, and chromosomal aberrations are rare.
Q52AIPMT 2015
Transmission tissue is characteristic feature of
Not quite — the answer is C.
A solid style is filled with transmitting tissue through which the pollen tube grows to the ovary. In a hollow style the tube grows along the canal, and the type of stigma does not decide this.
Q53AIPMT 2015
A man with blood group 'A' marries a woman with blood group 'B'. What are all the possible blood groups of their offsprings?
Not quite — the answer is D.
If both parents are heterozygous (Iᴬi × Iᴮi), the children can be IᴬIᴮ (group AB), Iᴬi (group A), Iᴮi (group B) or ii (group O). So all four groups are possible; 'A, B and AB only' would apply only if one parent were homozygous.
Q54AIPMT 2015
Perigynous flowers are found in:
Not quite — the answer is A.
In perigynous flowers the ovary is half-inferior, with sepals, petals and stamens on the rim of a cup-like thalamus, as in rose, plum and peach. Guava and cucumber are epigynous, and China rose is hypogynous.
Q55AIPMT 2015
What causes a green plant exposed to the light on only one side, to bend toward the source of light as it grows?
Not quite — the answer is A.
Auxin moves away from light to the shaded side, where it makes cells elongate more, so the shoot bends toward the light. The lighted side grows less, not faster, and 'seeking light' only names the response without explaining it.
Q56AIPMT 2015
The chromosomes in which centromere is situated close to one end are:
Not quite — the answer is C.
In an acrocentric chromosome the centromere lies close to one end, giving one very short and one very long arm. A telocentric chromosome has the centromere at the very end, and a sub-metacentric one has it slightly away from the middle.
Q57AIPMT 2015
A somatic cell that has just completed the S phase of its cell cycle, as compared to gamete of the same species, has:
Not quite — the answer is D.
A gamete is haploid with n chromosomes and C amount of DNA. A somatic cell after S phase still has 2n chromosomes but its DNA has doubled from 2C to 4C, so it has twice the chromosomes and four times the DNA of the gamete.
Q58AIPMT 2015
Which of the following animals is not viviparous?
Not quite — the answer is D.
The platypus is a monotreme mammal that lays eggs, so it is oviparous. Whale, bat (flying fox) and elephant give birth to live young.
Q59AIPMT 2015
In which of the following both pairs have correct combination?
Not quite — the answer is B.
A national park protects species in their natural habitat (in situ), while a botanical garden keeps plants away from their habitat (ex situ). Tissue culture, cryopreservation and seed banks are ex situ, and sacred groves, wildlife sanctuaries and national parks are in situ, so the other pairs are swapped.
Q60AIPMT 2015
Which body of the Government of India regulates GM research and safety of introducing GM organisms for public services?
Not quite — the answer is D.
The Genetic Engineering Approval Committee (GEAC) is the Indian body that decides on the validity of GM research and the safety of releasing GM organisms for public use. The Review Committee on Genetic Manipulation and institutional bio-safety committees handle research-level oversight only.
Q61AIPMT 2015
Which of the following endoparasites of humans does show viviparity?
Not quite — the answer is D.
Trichinella spiralis is viviparous: the female releases live larvae that migrate to and encyst in muscle. Ascaris, Ancylostoma and Enterobius lay eggs.
Q62AIPMT 2015
The terga, sterna and pleura of cockroach body are joined by:
Not quite — the answer is D.
The hardened plates of the cockroach exoskeleton (tergites, sternites and pleurites) are joined by a thin, flexible arthrodial membrane that allows movement. Cartilage and muscle do not join these plates.
Q63AIPMT 2015
Most animals are tree dwellers in a:
Not quite — the answer is A.
Tropical rain forests have a tall, dense, many-layered canopy that provides most of the food and shelter, so most animals there are arboreal. Coniferous forests, thorn woodlands and temperate deciduous forests have more open or seasonal canopies and more ground-dwelling fauna.
Q64AIPMT 2015
Glenoid cavity articulates:
Not quite — the answer is A.
The glenoid cavity is a depression on the scapula that receives the head of the humerus, forming the shoulder joint. The clavicle articulates with the acromion process, not with the glenoid cavity.
Q65AIPMT 2015
A population will not exist in Hardy-Weinberg equilibrium if:
Not quite — the answer is B.
Hardy-Weinberg equilibrium requires a large population, random mating, no mutation, no migration and no selection. Selective (non-random) mating changes genotype frequencies, so the population will not remain in equilibrium.
Q66AIPMT 2015
Male gametes are flagellated in:
Not quite — the answer is D.
Ectocarpus, a brown alga, produces pear-shaped gametes with two laterally attached flagella. Spirogyra and Polysiphonia have non-motile gametes, and Anabaena is a cyanobacterium with no flagellated gametes.
Q67AIPMT 2015
When you hold your breath, which of the following gas changes in blood would first lead to the urge to breathe?
Not quite — the answer is C.
The chemosensitive area of the medulla responds to rising CO₂ and H⁺ in the blood, and this is the strongest drive to breathe. The fall in O₂ during breath-holding is small and has little effect at first, so it is not what first causes the urge.
Q68AIPMT 2015
Which of the following cells during gametogenesis is normally diploid?
Not quite — the answer is D.
Spermatogonia are diploid (2n) germ cells that divide by mitosis. The primary and secondary polar bodies and spermatids are formed by meiosis and are haploid.
Q69AIPMT 2015
In ginger vegetative propagation occurs through:
Not quite — the answer is B.
Ginger has an underground stem, the rhizome, with nodes and buds, and pieces of it are planted to raise new plants. Runners and offsets are creeping stems of plants like grasses and Pistia, and bulbils are seen in Agave.
Q70AIPMT 2015
Which one of the following statements is incorrect?
Not quite — the answer is A.
A competitive inhibitor competes with the substrate for the active site, so more substrate is needed to reach half Vmax, i.e. the apparent Km increases, not decreases. It binds reversibly and is not chemically changed by the enzyme.
Q71AIPMT 2015
The active form of Entamoeba histolytica feeds upon:
Not quite — the answer is B.
The trophozoite of Entamoeba histolytica invades the mucosa and submucosa of the colon, causing ulcers, and also ingests red blood cells. Feeding on blood alone or on gut food alone is incomplete.
Q72AIPMT 2015
How many pairs of contrasting characters in pea plants were studied by Mendel in his experiments?
Not quite — the answer is A.
Mendel studied seven pairs of contrasting characters in pea, such as tall/dwarf stem, round/wrinkled seed and yellow/green seed.
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 2015 · Q3
Transpiration and root pressure cause water to rise in plants by:
Not quite — the answer is C.
Transpiration pull draws water up from the leaves by evaporation and cohesion, while root pressure pushes water up from the roots. So the correct order is pulling and pushing, respectively.
Q2AIPMT 2015 · Q30
In a ring girdled plant:
Not quite — the answer is C.
Ring girdling removes the bark with the phloem, so sugars made in the leaves cannot reach the roots, while xylem still carries water upward. The starved roots die first, and the shoot dies later when water supply fails.
Q3AIPMT 2015 · Q41
The UN Conference of Parties on climate change in the year 2011 was held in:
Not quite — the answer is C.
The 17th Conference of Parties (COP-17) of the UN climate change convention was held in 2011 at Durban, South Africa. Qatar hosted COP-18 (2012), Poland COP-19 (2013) and Peru COP-20 (2014).
Q4AIPMT 2015 · Q51
Which one gives the most valid and recent explanation for stomatal movements?
Not quite — the answer is C.
The accepted explanation of stomatal movement is active K⁺ influx into guard cells, which draws in water and opens the pore, and K⁺ efflux, which closes it. The older starch-sugar hydrolysis theory does not explain the observations as well.
Q5AIPMT 2015 · Q53
Rachel Carson's famous book "Silent Spring" is related to:
Not quite — the answer is B.
Silent Spring (1962) by Rachel Carson described how pesticides such as DDT accumulate in food chains and harm birds and other wildlife. It is about pesticide pollution, not noise or population growth.
Q6AIPMT 2015 · Q61
Which of the following is not one of the prime health risks associated with greater UV radiation through the atmosphere due to depletion of stratospheric ozone?
Not quite — the answer is A.
More UV-B reaching the earth causes skin cancer, damage to the eyes (cataract, snow blindness) and suppression of the immune system. UV does not penetrate to internal organs, so liver cancer is not a recognised risk.
Q7AIPMT 2015 · Q63
Minerals known to be required in large amounts for plant growth include:
Not quite — the answer is B.
Phosphorus, potassium, sulphur and calcium are all macronutrients needed in large amounts. Each other option includes micronutrients such as iron, zinc, manganese, copper or boron, or selenium, which plants do not need.
Q8AIPMT 2015 · Q66
Which of the following statements is not correct?
Not quite — the answer is B.
Brunner's glands lie in the submucosa of the duodenum and secrete alkaline mucus, not pepsinogen; pepsinogen comes from the chief cells of the gastric glands. Pancreatic acini do secrete carboxypeptidase, goblet cells secrete mucus and oxyntic cells secrete HCl.
Q9AIPMT 2015 · Q73
Gastric juice of infants contains:
Not quite — the answer is D.
The gastric juice of infants contains pepsinogen, gastric lipase and rennin, which curdles milk protein. Amylase, maltase and nucleases are not secreted in the stomach.
Q10AIPMT 2015 · Q81
Which of the following enhances or induces fusion of protoplasts?
Not quite — the answer is C.
Polyethylene glycol (PEG) and sodium nitrate are chemical fusogens that make protoplasts stick together and fuse. IAA, gibberellins and kinetin are plant hormones used for growth in culture, not for fusion.
Chapter-wise split of AIPMT 2015 Biology
- Cell: The Unit of Life6 Q
- Morphology of Flowering Plants5 Q
- Principles of Inheritance and Variation5 Q
- Ecosystem4 Q
- Sexual Reproduction in Flowering Plants4 Q
- Animal Kingdom4 Q
- Human Reproduction3 Q
- Human Health and Disease3 Q
- Biotechnology and Its Applications3 Q
- Evolution3 Q
- Reproductive Health3 Q
- Plant Kingdom3 Q
- Biodiversity and Conservation3 Q
- Molecular Basis of Inheritance2 Q
- Biological Classification2 Q
- Anatomy of Flowering Plants2 Q
- Neural Control and Coordination2 Q
- Body Fluids and Circulation2 Q
- Plant Growth and Development2 Q
- Locomotion and Movement2 Q
- Cell Cycle and Cell Division2 Q
- Chemical Coordination and Integration1 Q
- Microbes in Human Welfare1 Q
- Excretory Products and Their Elimination1 Q
- Respiration in Plants1 Q
- Structural Organisation in Animals1 Q
- Breathing and Exchange of Gases1 Q
- Biomolecules1 Q
About this paper
How many questions from AIPMT 2015 are on this page?
153 questions (Physics 34, Chemistry 37, Biology 82). 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 2015 paper, conducted by CBSE. The answer to every question was checked by the RankUp team, and the explanations are RankUp's own.
Is AIPMT 2015 useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books. 15 questions here are on topics NCERT has since dropped; they are listed separately so you can skip them.
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