NEET 2016 Phase 2 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
139 questions from the NEET 2016 Phase 2 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
- NEET 2016 Phase 2
- Conducted by
- CBSE
- Physics
- 36 questions
- Chemistry
- 33 questions
- Biology
- 70 questions
NEET 2016 Phase 2 Physics questions
35 questions. Tap an option to check your answer.
Q1NEET 2016 Phase 2 · Q1
Planck's constant (h), speed of light in vacuum (c) and Newton's gravitational constant (G) are three fundamental constants. Which of the following combinations of these has the dimension of length?
Not quite — the answer is A.
Using [h] = ML²T⁻¹, [G] = M⁻¹L³T⁻², [c] = LT⁻¹: hG has dimensions L⁵T⁻³ and c³ has L³T⁻³, so hG/c³ = L² and √(hG)/c^(3/2) = L (the Planck length). With c^(5/2) in the denominator the result would be T, not L.
Q2NEET 2016 Phase 2 · Q2
Two cars P and Q start from a point at the same time in a straight line and their positions are represented by x_P(t) = at + bt² and x_Q(t) = ft − t². At what time do the cars have the same velocity?
Not quite — the answer is D.
Differentiating, v_P = a + 2bt and v_Q = f − 2t. Setting them equal gives 2t(1 + b) = f − a, so t = (f − a)/(2(1 + b)). Writing a + f in the numerator comes from a sign slip when differentiating −t².
Q3NEET 2016 Phase 2 · Q5
A bullet of mass 10 g moving horizontally with a velocity of 400 m s⁻¹ strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be
Not quite — the answer is C.
The block's speed just after impact is √(2gh) = √(2 × 10 × 0.1) ≈ 1.41 m/s. Momentum conservation: 0.01 × 400 = 0.01v + 2 × 1.41, so 0.01v = 4 − 2.83 = 1.17 and v ≈ 117 ≈ 120 m/s. Energy is not conserved in the bullet–block collision, so equating kinetic energies is the trap.
Q4NEET 2016 Phase 2 · Q6
Two identical balls A and B having velocities of 0.5 m/s and −0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be
Not quite — the answer is B.
In a one-dimensional elastic collision between equal masses the velocities are exchanged, so B moves with 0.5 m/s and A with −0.3 m/s. The question asks for B first, so the order is 0.5 m/s and −0.3 m/s; reversing the order is the common slip.
Q5NEET 2016 Phase 2 · Q7
A particle moves from a point (−2î + 5ĵ) to (4ĵ + 3k̂) when a force of (4î + 3ĵ) N is applied. How much work has been done by the force?
Not quite — the answer is C.
Displacement d = (4ĵ + 3k̂) − (−2î + 5ĵ) = 2î − ĵ + 3k̂. Work W = F·d = 4 × 2 + 3 × (−1) + 0 × 3 = 8 − 3 = 5 J. Forgetting the negative y-component gives 11 J.
Q6NEET 2016 Phase 2 · Q8
Two rotating bodies A and B of masses m and 2m with moments of inertia I_A and I_B (I_B > I_A) have equal kinetic energy of rotation. If L_A and L_B be their angular momenta respectively, then
Not quite — the answer is C.
Rotational KE = L²/(2I), so L = √(2IK). With equal K, the body with the larger moment of inertia has the larger angular momentum, so L_B > L_A. The masses alone do not fix an exact ratio such as 2, since only I enters.
Q7NEET 2016 Phase 2 · Q9
A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation (E_sphere/E_cylinder) will be
Not quite — the answer is B.
E_sphere = ½(2/5 mR²)ω² = mR²ω²/5 and E_cylinder = ½(½mR²)(2ω)² = mR²ω². The ratio is 1 : 5. Ignoring the doubled angular speed of the cylinder would give 4 : 5.
Q8NEET 2016 Phase 2 · Q10
A light rod of length l has two masses m₁ and m₂ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is
Not quite — the answer is A.
The distances from the centre of mass are r₁ = m₂l/(m₁ + m₂) and r₂ = m₁l/(m₁ + m₂). Then I = m₁r₁² + m₂r₂² = m₁m₂l²/(m₁ + m₂), the reduced mass times l². (m₁ + m₂)l² would be the value if both masses were at distance l.
Q9NEET 2016 Phase 2 · Q12
A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g₀, the value of acceleration due to gravity at the earth's surface, is
Not quite — the answer is B.
Total energy of a circular orbit is −GMm/(2r) with r = R + h. Using GM = g₀R² gives E = −mg₀R²/(2(R + h)). The positive expression is the kinetic energy, not the total energy, which must be negative for a bound orbit.
Q10NEET 2016 Phase 2 · Q13
A rectangular film of liquid is extended from (4 cm × 2 cm) to (5 cm × 4 cm). If the work done is 3 × 10⁻⁴ J, the value of the surface tension of the liquid is
Not quite — the answer is B.
Area increases from 8 cm² to 20 cm², i.e. by 12 × 10⁻⁴ m², and a film has two surfaces, so the total new surface is 24 × 10⁻⁴ m². T = W/ΔA = 3 × 10⁻⁴/24 × 10⁻⁴ = 0.125 N/m. Forgetting the two surfaces gives 0.250 N/m.
Q11NEET 2016 Phase 2 · Q14
Three liquids of densities ρ₁, ρ₂ and ρ₃ (with ρ₁ > ρ₂ > ρ₃), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact θ₁, θ₂ and θ₃ obey
Not quite — the answer is B.
From h = 2T cosθ/(rρg), equal h, T and r means cosθ ∝ ρ. The densest liquid has the largest cosθ and hence the smallest angle, so θ₁ < θ₂ < θ₃; since the liquids rise, all angles are below π/2. Angles above π/2 would mean capillary depression.
Q12NEET 2016 Phase 2 · Q15
Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100 °C, while the other one is at 0 °C. If the two bodies are brought into contact, then, assuming no heat loss, the final common temperature is
Not quite — the answer is B.
The hot body has the larger heat capacity, so it gives out more heat per degree of cooling than the cold body absorbs per degree of warming. To balance heat lost and gained, the hot body cools through fewer degrees than the cold one warms, so the final temperature lies above 50 °C. Exactly 50 °C would need a constant heat capacity.
Q13NEET 2016 Phase 2 · Q16
A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be
Not quite — the answer is B.
With the average form: (3T − 2T)/10 = k(2.5T − T) gives k = 1/15 per minute; then (2T − θ)/10 = k((2T + θ)/2 − T) gives 6T − 3θ = θ, so θ = 3T/2. The exponential form agrees: the excess over the room halves from 2T to T, so it becomes T/2 and θ = 1.5T. Assuming equal fall per interval would wrongly give T.
Q14NEET 2016 Phase 2 · Q17
One mole of an ideal monatomic gas undergoes a process described by the equation PV³ = constant. The heat capacity of the gas during this process is
Not quite — the answer is D.
For a polytropic process PVⁿ = constant, C = C_v + R/(1 − n). With C_v = 3R/2 and n = 3, C = 3R/2 − R/2 = R. 3R/2 is just C_v, which applies only at constant volume.
Q15NEET 2016 Phase 2 · Q18
The temperature inside a refrigerator is t₂ °C and the room temperature is t₁ °C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be
Not quite — the answer is B.
For an ideal (Carnot) refrigerator, heat delivered to the hot reservoir per unit work is Q₁/W = T₁/(T₁ − T₂), with T₁ = t₁ + 273 K and T₁ − T₂ = t₁ − t₂. The expression with t₂ + 273 in the numerator is the heat extracted from inside (coefficient of performance), not the heat delivered to the room.
Q16NEET 2016 Phase 2 · Q19
A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule of the gas is m. Which of the following gives the density of the gas?
Not quite — the answer is B.
PV = NkT gives the number density N/V = P/(kT). Multiplying by the mass of each molecule gives density ρ = Pm/(kT). P/(kT) alone is the number of molecules per unit volume, not mass per unit volume.
Q17NEET 2016 Phase 2 · Q20
A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3 s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of m in kg is
Not quite — the answer is D.
T ∝ √m, so 3/5 = √(m/(m + 1)) and 9/25 = m/(m + 1). Then 9m + 9 = 25m, giving m = 9/16 kg. Inverting the ratio leads to 16/9 kg.
Q18NEET 2016 Phase 2 · Q21
The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe L metre long. The length of the open pipe will be
Not quite — the answer is B.
The first overtone of a closed pipe is 3v/(4L). The second overtone of an open pipe of length L′ is the third harmonic 3v/(2L′). Equating, 2L′ = 4L, so L′ = 2L. Treating the second overtone as the second harmonic would give a different length.
Q19NEET 2016 Phase 2 · Q23
An electric dipole is placed at an angle of 30° with an electric field intensity 2 × 10⁵ N/C. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is
Not quite — the answer is B.
τ = pE sinθ = q(2a)E sinθ, so 4 = q × 0.02 × 2 × 10⁵ × 0.5 = 2000q, giving q = 2 × 10⁻³ C = 2 mC. Using sin30° as 1 by mistake gives 1 mC, and forgetting to convert cm to m gives a far smaller charge.
Q20NEET 2016 Phase 2 · Q26
A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is
Not quite — the answer is C.
Rated current I = P/V = 500/100 = 5 A and bulb resistance = 100²/500 = 20 Ω. To draw 5 A from 230 V the total resistance must be 230/5 = 46 Ω, so R = 46 − 20 = 26 Ω. 46 Ω is the total resistance, not the series resistor.
Q21NEET 2016 Phase 2 · Q27
A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be
Not quite — the answer is B.
With the same wire, n turns means each turn has radius r/n. B′ = μ₀nI/(2(r/n)) = n² × μ₀I/(2r) = n²B. nB forgets that the radius also shrinks by a factor of n.
Q22NEET 2016 Phase 2 · Q28
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is
Not quite — the answer is B.
Work to rotate from 0° to 60° is W = MB(1 − cos60°) = MB/2, so MB = 2W. Torque at 60° is MB sin60° = 2W × √3/2 = √3W. Using sin60° with MB = W gives √3W/2.
Q23NEET 2016 Phase 2 · Q29
An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57 × 10⁻² T. If the value of e/m is 1.76 × 10¹¹ C/kg, the frequency of revolution of the electron is
Not quite — the answer is A.
f = (e/m)B/(2π) = (1.76 × 10¹¹ × 3.57 × 10⁻²)/(2π) = 6.28 × 10⁹/6.28 ≈ 1 × 10⁹ Hz = 1 GHz. 6.28 × 10⁹ is the angular frequency ω, so dividing by 2π is essential; 100 MHz is off by a factor of 10.
Q24NEET 2016 Phase 2 · Q30
Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?
Not quite — the answer is C.
Better tuning means a sharper resonance, i.e. a larger quality factor Q = (1/R)√(L/C). The values are about 10.4 for 20 Ω/1.5 H/35 μF, 9.4 for 25 Ω/2.5 H/45 μF, 22.8 for 15 Ω/3.5 H/30 μF and 7.3 for 25 Ω/1.5 H/45 μF. The 15 Ω, 3.5 H, 30 μF combination has the highest Q.
Q25NEET 2016 Phase 2 · Q32
The potential differences across the resistance, capacitance and inductance are 80 V, 40 V and 100 V respectively in an L-C-R circuit. The power factor of this circuit is
Not quite — the answer is C.
Net voltage V = √(V_R² + (V_L − V_C)²) = √(80² + 60²) = 100 V. Power factor cosφ = V_R/V = 80/100 = 0.8. Taking the ratio of V_C to V_L or assuming resonance gives the other values.
Q26NEET 2016 Phase 2 · Q33
A 100 Ω resistance and a capacitor of 100 Ω reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is
Not quite — the answer is A.
The displacement current between the plates equals the conduction current in the circuit. Impedance Z = √(100² + 100²) = 100√2 Ω, and the peak voltage of a 220 V (rms) supply is 220√2 V, so the peak current is 220√2/(100√2) = 2.2 A. The '50% charged' detail does not change the peak value.
Q27NEET 2016 Phase 2 · Q34
Two identical glass (μ_g = 3/2) equiconvex lenses of focal length f each are kept in contact. The space between the two lenses is filled with water (μ_w = 4/3). The focal length of the combination is
Not quite — the answer is D.
For each glass lens 1/f = (0.5)(2/R), so R = f. The water forms a concave lens with 1/f_w = (1/3)(−2/R) = −2/(3f). Combined: 1/F = 2/f − 2/(3f) = 4/(3f), so F = 3f/4. Ignoring the water lens gives f/2, and treating it as converging gives a shorter focal length.
Q28NEET 2016 Phase 2 · Q35
An air bubble in a glass slab with refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is
Not quite — the answer is C.
Real depth = μ × apparent depth, so the bubble is 1.5 × 5 = 7.5 cm from one face and 1.5 × 3 = 4.5 cm from the other. Thickness = 7.5 + 4.5 = 12 cm. Just adding the apparent depths gives 8 cm.
Q29NEET 2016 Phase 2 · Q36
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio (I_max − I_min)/(I_max + I_min) will be
Not quite — the answer is B.
With I₁ = nI₂, I_max = (√I₁ + √I₂)² and I_min = (√I₁ − √I₂)². Their difference is 4√(I₁I₂) and sum is 2(I₁ + I₂), so the ratio is 2√(I₁I₂)/(I₁ + I₂) = 2√n/(n + 1). Dropping the factor 2 gives √n/(n + 1).
Q30NEET 2016 Phase 2 · Q37
A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use, will be
Not quite — the answer is B.
To see distant objects the lens must form the image of an object at infinity at the far point, 400 cm, so f = −400 cm = −4 m. Power P = 1/f = −0.25 D, a concave lens. A convex lens would correct the near point, not the far point.
Q31NEET 2016 Phase 2 · Q38
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10⁻⁵ cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is
Not quite — the answer is D.
For single-slit diffraction the first minimum is at sinθ = λ/a, and on the focal plane y = fλ/a = 60 × 5 × 10⁻⁵/0.02 = 0.15 cm. Using 2λ/a would give the width of the central maximum (0.30 cm), not the position of the first dark band.
Q32NEET 2016 Phase 2 · Q39
Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cutoff wavelength (λ₀) of the emitted X-ray is
Not quite — the answer is A.
The electron's kinetic energy is K = p²/(2m) = h²/(2mλ²). The shortest X-ray wavelength carries all of this energy: hc/λ₀ = h²/(2mλ²), so λ₀ = 2mcλ²/h. The cutoff wavelength is not equal to the electron's de Broglie wavelength.
Q33NEET 2016 Phase 2 · Q40
Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is
Not quite — the answer is D.
Work function φ = 5 − 2 = 3 eV. For 6 eV photons the maximum kinetic energy is 6 − 3 = 3 eV, so the anode must be at least 3 V negative with respect to the cathode: −3 V. A positive potential would attract the electrons rather than stop them.
Q34NEET 2016 Phase 2 · Q41
If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be
Not quite — the answer is C.
1/λ ∝ (1/2² − 1/3²) = 5/36 and 1/λ′ ∝ (1/3² − 1/4²) = 7/144. So λ′/λ = (5/36)/(7/144) = 20/7, giving λ′ = 20λ/7. The 4→3 photon has less energy, so its wavelength must be longer than λ, which rules out the fractions smaller than 1.
Q35NEET 2016 Phase 2 · Q42
The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is
Not quite — the answer is D.
After 40% decay 60% remains; after 85% decay 15% remains. 60% → 15% is a factor of 4 = 2², i.e. two half-lives = 2 × 30 = 60 minutes. Subtracting the percentages (45%) and treating decay as linear is the trap.
Physics: not in the current NEET syllabus
These 1 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
1 questions. Tap an option to check your answer.
Q1NEET 2016 Phase 2 · Q43
For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kΩ, then the input signal voltage is
Not quite — the answer is B.
Collector current i_c = 4 V/2 kΩ = 2 mA, so base current i_b = i_c/β = 20 μA. Input voltage = i_b × R_B = 20 × 10⁻⁶ × 1000 = 20 mV. Equivalently voltage gain = β R_C/R_B = 200, and 4 V/200 = 20 mV.
Chapter-wise split of NEET 2016 Phase 2 Physics
- Work, Energy and Power3 Q
- System of Particles and Rotational Motion3 Q
- Ray Optics and Optical Instruments3 Q
- Mechanical Properties of Fluids2 Q
- Thermal Properties of Matter2 Q
- Thermodynamics2 Q
- Moving Charges and Magnetism2 Q
- Alternating Current2 Q
- Wave Optics2 Q
- Dual Nature of Radiation and Matter2 Q
- Units and Measurements1 Q
- Motion in a Straight Line1 Q
- Gravitation1 Q
- Kinetic Theory1 Q
- Oscillations1 Q
- Waves1 Q
- Electric Charges and Fields1 Q
- Current Electricity1 Q
- Magnetism and Matter1 Q
- Electromagnetic Waves1 Q
- Atoms1 Q
- Nuclei1 Q
NEET 2016 Phase 2 Chemistry questions
28 questions. Tap an option to check your answer.
Q1NEET 2016 Phase 2 · Q47
The molar conductivity of a 0.5 mol/dm³ solution of AgNO₃ with electrolytic conductivity of 5.76 × 10⁻³ S cm⁻¹ at 298 K is
Not quite — the answer is B.
Λm = κ × 1000/M = 5.76 × 10⁻³ × 1000/0.5 = 11.52 S cm² mol⁻¹. Multiplying by the concentration instead of dividing gives 2.88 S cm²/mol.
Q2NEET 2016 Phase 2 · Q48
The decomposition of phosphine (PH₃) on tungsten at low pressure is a first-order reaction. It is because the
Not quite — the answer is A.
At low pressure only a small part of the metal surface is covered, and the coverage is proportional to the pressure of PH₃. The rate depends on the coverage, so it is proportional to pressure: first order. At high pressure the surface is saturated and the rate becomes independent of coverage (zero order).
Q3NEET 2016 Phase 2 · Q52
For a sample of perfect gas when its pressure is changed isothermally from pᵢ to p_f, the entropy change is given by
Not quite — the answer is B.
For an isothermal reversible change ΔS = q_rev/T = nR ln(V_f/Vᵢ), and at constant T, V_f/Vᵢ = pᵢ/p_f, so ΔS = nR ln(pᵢ/p_f). Expressions containing T have units of energy, not entropy, and the ratio p_f/pᵢ has the wrong sign.
Q4NEET 2016 Phase 2 · Q53
The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
Not quite — the answer is D.
Ba(OH)₂ dissociates completely in dilute solution: Ba(OH)₂ → Ba²⁺ + 2OH⁻, giving three ions per formula unit, so i = 3. i = 2 would apply to a 1:1 electrolyte like NaCl.
Q5NEET 2016 Phase 2 · Q54
The percentage of pyridine (C₅H₅N) that forms pyridinium ion (C₅H₅N⁺H) in a 0.10 M aqueous pyridine solution (K_b for C₅H₅N = 1.7 × 10⁻⁹) is
Not quite — the answer is B.
For a weak base α = √(K_b/C) = √(1.7 × 10⁻⁹/0.10) = √(1.7 × 10⁻⁸) ≈ 1.3 × 10⁻⁴. As a percentage this is 0.013%. Forgetting to convert the fraction to a percentage, or not taking the square root, gives the other values.
Q6NEET 2016 Phase 2 · Q56
If the E°cell for a given reaction has a negative value, which of the following gives the correct relationships for the values of ΔG° and K_eq?
Not quite — the answer is A.
ΔG° = −nFE°cell, so a negative E° gives a positive ΔG°. Also ΔG° = −RT ln K, so a positive ΔG° means ln K < 0, i.e. K < 1. A negative ΔG° and K > 1 belong to a positive E°.
Q7NEET 2016 Phase 2 · Q57
Which one of the following is incorrect for ideal solution?
Not quite — the answer is D.
For an ideal solution ΔH_mix and ΔU_mix are zero and the vapour pressure follows Raoult's law exactly. Mixing is still spontaneous because ΔS_mix > 0, so ΔG_mix = −TΔS_mix is negative, not zero.
Q8NEET 2016 Phase 2 · Q58
The solubility of AgCl(s) with solubility product 1.6 × 10⁻¹⁰ in 0.1 M NaCl solution would be
Not quite — the answer is B.
Because of the common ion, [Cl⁻] ≈ 0.1 M, so s = K_sp/[Cl⁻] = 1.6 × 10⁻¹⁰/0.1 = 1.6 × 10⁻⁹ M. 1.26 × 10⁻⁵ M is the solubility in pure water, which ignores the common-ion effect.
Q9NEET 2016 Phase 2 · Q59
Suppose the elements X and Y combine to form two compounds XY₂ and X₃Y₂. When 0.1 mole of XY₂ weighs 10 g and 0.05 mole of X₃Y₂ weighs 9 g, the atomic weights of X and Y are
Not quite — the answer is A.
Molar masses: XY₂ = 10/0.1 = 100 and X₃Y₂ = 9/0.05 = 180. So X + 2Y = 100 and 3X + 2Y = 180; subtracting gives 2X = 80, X = 40 and Y = 30. The pair 30, 20 fails the first equation (30 + 40 = 70).
Q10NEET 2016 Phase 2 · Q60
The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = 1.60 × 10⁻¹⁹ C)
Not quite — the answer is C.
Charge passed Q = It = 1 × 60 = 60 C. Number of electrons = 60/(1.60 × 10⁻¹⁹) = 3.75 × 10²⁰. 6 × 10²³ is close to Avogadro's number and would need about one faraday of charge.
Q11NEET 2016 Phase 2 · Q61
Boric acid is an acid because its molecule
Not quite — the answer is C.
B(OH)₃ is not a protonic acid; the electron-deficient boron accepts OH⁻ from water: B(OH)₃ + 2H₂O → [B(OH)₄]⁻ + H₃O⁺. So it behaves as a Lewis acid, and the proton released comes from water, not from boric acid itself.
Q12NEET 2016 Phase 2 · Q62
AlF₃ is soluble in HF only in presence of KF. It is due to the formation of
Not quite — the answer is B.
AlF₃ is insoluble in anhydrous HF because HF is too weakly ionised to supply F⁻. KF provides fluoride ions that form the soluble complex: AlF₃ + 3KF → K₃[AlF₆]. Hydride species such as AlH₃ are not formed in this medium.
Q13NEET 2016 Phase 2 · Q63
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
Not quite — the answer is D.
E°(Zn²⁺/Zn) = −0.76 V is more negative than E°(Fe²⁺/Fe) = −0.44 V, so zinc is oxidised in preference and protects iron sacrificially. An iron coating on zinc would not protect the zinc, since iron is less easily oxidised. Melting point and density are not the reason.
Q14NEET 2016 Phase 2 · Q65
The hybridizations of atomic orbitals of nitrogen in NO₂⁺, NO₃⁻ and NH₄⁺ respectively are
Not quite — the answer is C.
NO₂⁺ is linear with two σ bonds and no lone pair on N (sp). NO₃⁻ is trigonal planar with three σ bonds (sp²). NH₄⁺ is tetrahedral with four σ bonds (sp³). So the sequence is sp, sp², sp³.
Q15NEET 2016 Phase 2 · Q66
Which of the following fluoro-compounds is most likely to behave as a Lewis base?
Not quite — the answer is B.
PF₃ has a lone pair on phosphorus that it can donate, so it acts as a Lewis base (a ligand). BF₃ is electron deficient (a Lewis acid), SiF₄ can accept electron pairs into vacant orbitals, and CF₄ has neither a lone pair on C nor vacant orbitals.
Q16NEET 2016 Phase 2 · Q69
Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behaviour?
Not quite — the answer is D.
In the reaction with CaF₂ sulphur stays at +6 (H₂SO₄ → CaSO₄); it is only an acid–base (double decomposition) reaction that releases HF. In the reactions with Cu, S and C, sulphuric acid is reduced to SO₂ (S from +6 to +4), which is oxidising behaviour.
Q17NEET 2016 Phase 2 · Q70
Which of the following pairs of d-orbitals will have electron density along the axes?
Not quite — the answer is C.
The lobes of d_x²−y² lie along the x and y axes and the main lobes of d_z² lie along the z axis. d_xy, d_yz and d_xz have their lobes between the axes, so any pair containing them is wrong.
Q18NEET 2016 Phase 2 · Q71
The correct geometry and hybridization for XeF₄ are
Not quite — the answer is D.
Xe in XeF₄ has 4 bond pairs and 2 lone pairs, six electron pairs in all, so it is sp³d² hybridised. The electron-pair arrangement is octahedral, but with the two lone pairs opposite each other the shape of the molecule is square planar.
Q19NEET 2016 Phase 2 · Q72
Among the following, which one is a wrong statement?
Not quite — the answer is C.
SeF₄ has 4 bond pairs and 1 lone pair (sp³d), so it is see-saw shaped, while CH₄ is tetrahedral; they do not have the same shape. PH₅ and BiCl₅ do not exist, SO₂ has pπ–dπ bonding, and I₃⁺ (2 bond pairs, 2 lone pairs) is bent, so those statements are correct.
Q20NEET 2016 Phase 2 · Q75
Jahn-Teller effect is not observed in high spin complexes of
Not quite — the answer is B.
High-spin d⁸ in an octahedral field is t₂g⁶e_g², with symmetrically filled degenerate sets, so there is no Jahn–Teller distortion. High-spin d⁴ (t₂g³e_g¹) and d⁹ (t₂g⁶e_g³) have unevenly filled e_g orbitals and show strong distortion; d⁷ high spin (t₂g⁵e_g²) has an uneven t₂g set.
Q21NEET 2016 Phase 2 · Q76
Which of the following can be used as the halide component for Friedel-Crafts reaction?
Not quite — the answer is D.
Friedel–Crafts alkylation needs a halide that forms a carbocation with AlCl₃, and isopropyl chloride readily gives the isopropyl cation. In chlorobenzene, bromobenzene and chloroethene the C–X bond has partial double-bond character (sp² carbon), so aryl and vinyl cations are not formed.
Q22NEET 2016 Phase 2 · Q80
Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or direct only elimination reaction?
Not quite — the answer is C.
Cyclopropane adds HBr to give 1-bromopropane, which eliminates HBr to give propene; propan-1-ol dehydrates and 1-bromopropane dehydrohalogenates directly to propene. Ketene (H₂C=C=O) has only two carbons, so it cannot give a three-carbon alkene.
Q23NEET 2016 Phase 2 · Q81
Which one of the following nitro-compounds does not react with nitrous acid?
Not quite — the answer is C.
Nitroalkanes react with nitrous acid through the hydrogen on the carbon carrying the nitro group (primary give nitrolic acids, secondary give pseudonitroles). 2-Methyl-2-nitropropane is a tertiary nitro compound with no α-hydrogen, so it does not react. The other compounds all have at least one α-H.
Q24NEET 2016 Phase 2 · Q82
The central dogma of molecular genetics states that the genetic information flows from
Not quite — the answer is C.
Genetic information is copied from DNA into messenger RNA (transcription) and then read to make proteins (translation). Carbohydrates are not products of genetic information transfer, and information does not flow from proteins back to DNA.
Q25NEET 2016 Phase 2 · Q84
In the given reaction, benzene + cyclohexene —(HF, 0 °C)→ P, the product P is
Not quite — the answer is C.
HF protonates cyclohexene to give the cyclohexyl carbocation, which alkylates benzene (Friedel–Crafts type): the product is cyclohexylbenzene. Fluoride is not retained in the product, and the low temperature and 1:1 reactants do not give a diphenyl product.
Q26NEET 2016 Phase 2 · Q85
A given nitrogen-containing aromatic compound A reacts with Sn/HCl, followed by HNO₂ to give an unstable compound B. B, on treatment with phenol, forms a beautiful coloured compound C with the molecular formula C₁₂H₁₀N₂O. The structure of compound A is
Not quite — the answer is B.
Sn/HCl reduces nitrobenzene to aniline; HNO₂ converts aniline to the unstable benzenediazonium salt, which couples with phenol to give p-hydroxyazobenzene, C₆H₅–N=N–C₆H₄OH (C₁₂H₁₀N₂O). Aniline needs no reduction, so Sn/HCl would be pointless as the first step.
Q27NEET 2016 Phase 2 · Q86
Consider the reaction CH₃CH₂CH₂Br + NaCN → CH₃CH₂CH₂CN + NaBr. This reaction will be the fastest in
Not quite — the answer is C.
This is an S_N2 reaction of a primary bromide. Polar aprotic solvents such as DMF solvate Na⁺ but leave CN⁻ free and highly nucleophilic, so the rate is greatest. Protic solvents (water, methanol, ethanol) hydrogen-bond to CN⁻ and slow the reaction.
Q28NEET 2016 Phase 2 · Q87
The correct structure of the product A formed in the reaction cyclohex-2-en-1-one —(H₂ (gas, 1 atmosphere), Pd/carbon, ethanol)→ A is
Not quite — the answer is B.
H₂ with Pd/C at 1 atm hydrogenates the C=C double bond of the α,β-unsaturated ketone but does not reduce the C=O group under these mild conditions, so cyclohexanone is formed. Reducing the carbonyl to an alcohol while keeping the C=C would need a hydride reagent such as NaBH₄ with CeCl₃.
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.
Q1NEET 2016 Phase 2 · Q49
The coagulation values in millimoles per litre of the electrolytes used for the coagulation of As₂S₃ are given below: I. (NaCl) = 52, II. (BaCl₂) = 0.69, III. (MgSO₄) = 0.22. The correct order of their coagulating power is
Not quite — the answer is C.
Coagulating power is inversely proportional to the coagulation value: the smaller the amount needed, the stronger the electrolyte. So MgSO₄ (0.22) > BaCl₂ (0.69) > NaCl (52). Reading the larger number as greater power reverses the order.
Q2NEET 2016 Phase 2 · Q55
In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion (Ca²⁺) and fluoride ion (F⁻) are
Not quite — the answer is C.
In fluorite Ca²⁺ ions form a face-centred cubic array and F⁻ ions occupy all the tetrahedral voids. Each F⁻ is surrounded by 4 Ca²⁺ and each Ca²⁺ by 8 F⁻, so the coordination numbers are 8 and 4 (in the 1:2 ratio). Swapping them gives 4 and 8.
Q3NEET 2016 Phase 2 · Q64
The suspension of slaked lime in water is known as
Not quite — the answer is C.
A suspension of Ca(OH)₂ in water is called milk of lime. Limewater is the clear filtered solution of Ca(OH)₂, and quicklime is CaO.
Q4NEET 2016 Phase 2 · Q68
In context with beryllium, which one of the following statements is incorrect?
Not quite — the answer is C.
Be²⁺ is very small with a high charge density, so beryllium salts are extensively hydrolysed; saying they rarely hydrolyse is wrong. Beryllium is made passive by nitric acid, forms the carbide Be₂C, and BeH₂ is an electron-deficient, polymeric hydride.
Q5NEET 2016 Phase 2 · Q78
Which one of the following structures represents nylon 6,6 polymer?
Not quite — the answer is D.
Nylon 6,6 is a polyamide made by condensing hexamethylenediamine with adipic acid, so each repeat unit is –CO–(CH₂)₄–CO–NH–(CH₂)₆–NH–, with six carbons from each monomer. The other structures are vinyl-type chains with no amide links; the subscript 66 in them is a trap.
Chapter-wise split of NEET 2016 Phase 2 Chemistry
- Electrochemistry4 Q
- Solutions2 Q
- Equilibrium2 Q
- Chemical Bonding and Molecular Structure2 Q
- Haloalkanes and Haloarenes2 Q
- Hydrocarbons2 Q
- Amines2 Q
- Chemical Kinetics1 Q
- Thermodynamics1 Q
- Some Basic Concepts of Chemistry1 Q
- Structure of Atom1 Q
- Coordination Compounds1 Q
- Biomolecules1 Q
- Aldehydes, Ketones and Carboxylic Acids1 Q
NEET 2016 Phase 2 Biology questions
64 questions. Tap an option to check your answer.
Q1NEET 2016 Phase 2 · Q91
Which one of the following is wrong for fungi?
Not quite — the answer is B.
Fungal cell walls are made mainly of chitin and polysaccharides, not purely of cellulose, so the claim of a purely cellulosic wall is wrong. Fungi are eukaryotic heterotrophs and include both unicellular forms such as yeast and multicellular filamentous forms.
Q2NEET 2016 Phase 2 · Q92
Methanogens belong to
Not quite — the answer is B.
Methanogens are archaebacteria that live in marshy areas and the rumen of cattle and produce methane from CO₂. Their cell wall structure differs from true bacteria, so they are not placed in Eubacteria; dinoflagellates and slime moulds are protists.
Q3NEET 2016 Phase 2 · Q93
Select the wrong statement.
Not quite — the answer is A.
Diatom cell walls are impregnated with silica and are practically indestructible, which is why they accumulate as diatomaceous earth over billions of years. So saying the walls are easily destructible is wrong, while the other statements about diatoms being planktonic chief producers are correct.
Q4NEET 2016 Phase 2 · Q94
The label of a herbarium sheet does not carry information on
Not quite — the answer is D.
A herbarium label records the date and place of collection, the collector's name, and the botanical, family and local names of the specimen. The height of the plant is not a standard entry on the label, since the pressed specimen itself is kept on the sheet.
Q5NEET 2016 Phase 2 · Q95
Conifers are adapted to tolerate extreme environmental conditions because of
Not quite — the answer is C.
Conifer needles have a thick cuticle and sunken stomata, which cut water loss and let them survive cold and dry conditions. Their leaves are needle-like rather than broad, the stomata are sunken rather than superficial, and gymnosperms generally lack vessels in their xylem.
Q6NEET 2016 Phase 2 · Q96
Which one of the following statements is wrong?
Not quite — the answer is B.
The sources are reversed in this statement: algin comes from brown algae and carrageenan from red algae. Agar from Gelidium and Gracilaria, the use of Laminaria and Sargassum as food, and the release of oxygen by photosynthetic algae are all correct.
Q7NEET 2016 Phase 2 · Q97
The term 'polyadelphous' is related to
Not quite — the answer is B.
Polyadelphous describes stamens whose filaments are united into more than two bundles, as in citrus, so it is a term for the androecium. Terms for fused carpels (syncarpous) or fused petals and sepals (gamopetalous, gamosepalous) are different.
Q8NEET 2016 Phase 2 · Q99
Radial symmetry is found in the flowers of
Not quite — the answer is A.
Brassica (mustard) has actinomorphic flowers that can be cut into two equal halves along any radial plane. Trifolium and Pisum have papilionaceous zygomorphic flowers, and Cassia flowers are also zygomorphic.
Q9NEET 2016 Phase 2 · Q100
Free-central placentation is found in
Not quite — the answer is A.
In Dianthus and Primrose the ovules are borne on a central axis with no septa, which is free-central placentation. Argemone and Brassica have parietal placentation, and Citrus has axile placentation.
Q10NEET 2016 Phase 2 · Q102
The balloon-shaped structures called tyloses
Not quite — the answer is C.
Tyloses are balloon-like outgrowths of neighbouring xylem parenchyma cells that push into the vessel lumen through pits. They block the vessels and are typical of heartwood, not sapwood, so they stop rather than aid the conduction of sap.
Q11NEET 2016 Phase 2 · Q103
A non-proteinaceous enzyme is
Not quite — the answer is B.
A ribozyme is an RNA molecule with catalytic activity, such as the 23S rRNA that acts as peptidyl transferase, so it is not a protein. Lysozyme, ligase and deoxyribonuclease are all protein enzymes.
Q12NEET 2016 Phase 2 · Q104
Select the mismatch.
Not quite — the answer is B.
A large central vacuole is a feature of mature plant cells; animal cells have small vacuoles or none. Gas vacuoles do occur in blue-green, purple and green photosynthetic bacteria, protists are eukaryotes and methanogens are prokaryotic archaebacteria.
Q13NEET 2016 Phase 2 · Q105
Select the wrong statement.
Not quite — the answer is B.
Bacterial motility comes from flagella; pili and fimbriae are surface outgrowths that help in attachment to surfaces or host tissues (and pili in conjugation), not in movement. The other statements about peptidoglycan walls, non-flagellated cyanobacteria and wall-less Mycoplasma are correct.
Q14NEET 2016 Phase 2 · Q106
A cell organelle containing hydrolytic enzymes is
Not quite — the answer is A.
Lysosomes are membrane-bound vesicles from the Golgi that are rich in acid hydrolases such as lipases, proteases and carbohydrases. Ribosomes make proteins, mesosomes are infoldings of the bacterial membrane, and microsomes are ER fragments formed when cells are homogenised.
Q15NEET 2016 Phase 2 · Q107
During cell growth, DNA synthesis takes place in
Not quite — the answer is A.
DNA replication happens in the S (synthesis) phase of interphase, doubling the DNA content from 2C to 4C. G₁ and G₂ are growth phases in which the cell makes proteins and organelles but does not replicate DNA, and M phase is division.
Q16NEET 2016 Phase 2 · Q108
Which of the following biomolecules is common to respiration-mediated breakdown of fats, carbohydrates and proteins?
Not quite — the answer is D.
Fatty acids are broken down by beta-oxidation to acetyl CoA, glucose gives pyruvate which becomes acetyl CoA, and many amino acids also enter as acetyl CoA, so it is the common entry point into the Krebs cycle. Pyruvic acid is not formed from fatty acids, and the sugar phosphates belong only to glycolysis.
Q17NEET 2016 Phase 2 · Q110
You are given a tissue with its potential for differentiation in an artificial culture. Which of the following pairs of hormones would you add to the medium to secure shoots as well as roots?
Not quite — the answer is B.
In tissue culture the auxin to cytokinin ratio controls organ formation: a higher auxin ratio favours roots and a higher cytokinin ratio favours shoots, so both are needed to obtain shoots and roots. Gibberellin and abscisic acid do not drive this organ differentiation.
Q18NEET 2016 Phase 2 · Q113
The process which makes major difference between C₃ and C₄ plants is
Not quite — the answer is C.
C₄ plants concentrate CO₂ around RuBisCO in bundle sheath cells, so photorespiration is almost absent, while C₃ plants lose much fixed carbon to it. Glycolysis, respiration and the Calvin cycle run in both types of plant.
Q19NEET 2016 Phase 2 · Q116
Match Column-I with Column-II and select the correct option using the codes given below: Column-I: a. Pistils fused together; b. Formation of gametes; c. Hyphae of higher Ascomycetes; d. Unisexual female flower Column-II: (i) Gametogenesis; (ii) Pistillate; (iii) Syncarpous; (iv) Dikaryotic
Not quite — the answer is D.
Fused pistils make a syncarpous gynoecium, gamete formation is gametogenesis, the hyphae of higher ascomycetes and basidiomycetes are dikaryotic, and a unisexual female flower is pistillate. Only one code gives all four of these pairs.
Q20NEET 2016 Phase 2 · Q117
In majority of angiosperms
Not quite — the answer is C.
The diploid megaspore mother cell undergoes meiosis to form four haploid megaspores, so reduction division happens there. The filiform apparatus belongs to the synergids, the typical embryo sac has only three antipodal cells, and the central cell is the largest cell of the embryo sac.
Q21NEET 2016 Phase 2 · Q118
Pollination in water hyacinth and water lily is brought about by the agency of
Not quite — the answer is B.
In water hyacinth and water lily the flowers rise above the water surface and are pollinated by insects or wind, as in most land plants. Water pollination is limited to plants like Vallisneria, Hydrilla and Zostera.
Q22NEET 2016 Phase 2 · Q119
The ovule of an angiosperm is technically equivalent to
Not quite — the answer is A.
The ovule is the megasporangium: its nucellus produces the megaspore mother cell, which forms megaspores. The carpel is the megasporophyll that bears the ovules, and the megaspore mother cell and megaspore are cells inside the ovule.
Q23NEET 2016 Phase 2 · Q120
Taylor conducted the experiments to prove semiconservative mode of chromosome replication on
Not quite — the answer is B.
Taylor and colleagues in 1958 used radioactive thymidine on root tips of Vicia faba (faba bean) to show that chromosomes replicate semiconservatively. The E. coli experiment with heavy nitrogen was done by Meselson and Stahl and dealt with DNA, not whole chromosomes.
Q24NEET 2016 Phase 2 · Q121
The mechanism that causes a gene to move from one linkage group to another is called
Not quite — the answer is C.
In translocation a chromosome segment breaks off and joins a non-homologous chromosome, so its genes shift to a different linkage group. Crossing over exchanges segments between homologous chromosomes of the same linkage group, and inversion and duplication change order or copy number within a chromosome.
Q25NEET 2016 Phase 2 · Q122
The equivalent of a structural gene is
Not quite — the answer is B.
A cistron is the segment of DNA coding for one polypeptide, which is what a structural gene is. A muton and a recon are the smallest units of mutation and recombination, and an operon is a group of structural genes with their regulatory elements.
Q26NEET 2016 Phase 2 · Q123
A true breeding plant is
Not quite — the answer is C.
A true-breeding line has been self-pollinated for generations, making it nearly homozygous, so it gives offspring with the same trait. It may be homozygous dominant or recessive, and cross-pollination among unrelated plants increases heterozygosity instead.
Q27NEET 2016 Phase 2 · Q124
Which of the following rRNAs acts as structural RNA as well as ribozyme in bacteria?
Not quite — the answer is C.
The 23S rRNA of the bacterial 50S subunit forms part of the ribosome's structure and also catalyses peptide bond formation as peptidyl transferase. 18S and 5.8S rRNAs are found in eukaryotic ribosomes, and 5S rRNA is structural only.
Q28NEET 2016 Phase 2 · Q125
Stirred-tank bioreactors have been designed for
Not quite — the answer is C.
The stirrer in a stirred-tank bioreactor mixes the contents and spreads sterile air bubbles evenly, so oxygen is available throughout the culture. Purification and preservation belong to downstream processing, and the design supplies air rather than keeping conditions anaerobic.
Q29NEET 2016 Phase 2 · Q126
A foreign DNA and plasmid cut by the same restriction endonuclease can be joined to form a recombinant plasmid using
Not quite — the answer is D.
DNA ligase seals the sugar-phosphate backbone between the matching sticky ends of the vector and insert, forming the recombinant plasmid. EcoRI only cuts DNA, and Taq polymerase and polymerase III synthesise new strands rather than join fragments.
Q30NEET 2016 Phase 2 · Q127
Which of the following is not a component of downstream processing?
Not quite — the answer is D.
Downstream processing starts after the product is made and includes separation, purification and formulation with preservatives. Expression of the gene in the host happens earlier, during the bioreactor stage, so it is not part of downstream processing.
Q31NEET 2016 Phase 2 · Q128
Which of the following restriction enzymes produces blunt ends?
Not quite — the answer is B.
EcoRV cuts both strands at the same point in the middle of its recognition site GATATC, giving blunt ends. SalI, XhoI and HindIII make staggered cuts and leave sticky single-stranded overhangs.
Q32NEET 2016 Phase 2 · Q129
Which kind of therapy was given in 1990 to a four-year-old girl with adenosine deaminase (ADA) deficiency?
Not quite — the answer is A.
In 1990 the first clinical gene therapy was given to a four-year-old girl with ADA deficiency: her lymphocytes were given a working ADA gene with a retroviral vector and returned to her body. Chemotherapy, immunotherapy and radiation therapy do not replace a missing gene.
Q33NEET 2016 Phase 2 · Q131
The primary producers of the deep-sea hydrothermal vent ecosystem are
Not quite — the answer is B.
No sunlight reaches deep-sea vents, so the producers are chemosynthetic bacteria that oxidise hydrogen sulphide and other compounds to fix carbon. Green algae and blue-green algae need light, and coral reefs are shallow-water communities.
Q34NEET 2016 Phase 2 · Q132
Which of the following is correct for r-selected species?
Not quite — the answer is A.
r-selected species such as oysters and many insects produce very many small offspring with little parental care. Producing few, large, well-cared-for offspring is the K-selected strategy.
Q35NEET 2016 Phase 2 · Q133
If '+' sign is assigned to beneficial interaction, '–' sign to detrimental and '0' sign to neutral interaction, then the population interaction represented by '+' '–' refers to
Not quite — the answer is D.
In parasitism the parasite benefits (+) while the host is harmed (–). Mutualism is +/+, commensalism is +/0 and amensalism is –/0; predation is also +/– but is not offered.
Q36NEET 2016 Phase 2 · Q134
Which of the following is correctly matched?
Not quite — the answer is C.
Parthenium (carrot grass) is an invasive alien weed that spreads aggressively and threatens native species. Aerenchyma is a feature of hydrophytes, not the xerophyte Opuntia, age pyramids describe populations and stratification describes the vertical layering of a community.
Q37NEET 2016 Phase 2 · Q135
Red List contains data or information on
Not quite — the answer is C.
The IUCN Red List records the conservation status of plant and animal species, mainly those that are threatened with extinction. It is not limited to economically important or traded plants, or to marine vertebrates.
Q38NEET 2016 Phase 2 · Q136
Which of the following sets of diseases is caused by bacteria?
Not quite — the answer is A.
Cholera is caused by Vibrio cholerae and tetanus by Clostridium tetani, so both are bacterial. Smallpox, mumps, herpes and influenza are viral, so each other pair has at least one viral disease.
Q39NEET 2016 Phase 2 · Q137
Match Column-I with Column-II for housefly classification and select the correct option using the codes given below: Column-I: a. Family; b. Order; c. Class; d. Phylum Column-II: (i) Diptera; (ii) Arthropoda; (iii) Muscidae; (iv) Insecta
Not quite — the answer is A.
The housefly Musca domestica belongs to family Muscidae, order Diptera, class Insecta and phylum Arthropoda. Only one code pairs family with Muscidae, order with Diptera, class with Insecta and phylum with Arthropoda.
Q40NEET 2016 Phase 2 · Q138
Choose the correct statement.
Not quite — the answer is B.
Cyclostomes such as lampreys and hagfishes are jawless and lack paired fins. Egg-laying mammals like the platypus are not viviparous, crocodiles have a four-chambered heart, and cartilaginous fishes have gill slits without an operculum.
Q41NEET 2016 Phase 2 · Q139
Study the four statements (A–D) given below and select the two correct ones out of them: A. Definition of biological species was given by Ernst Mayr. B. Photoperiod does not affect reproduction in plants. C. Binomial nomenclature system was given by R. H. Whittaker. D. In unicellular organisms, reproduction is synonymous with growth. The two correct statements are
Not quite — the answer is C.
Ernst Mayr gave the biological species concept, and in unicellular organisms division of the cell is both growth and reproduction, so A and D are correct. Photoperiod does affect flowering and reproduction, and binomial nomenclature was given by Linnaeus, while Whittaker proposed the five-kingdom system.
Q42NEET 2016 Phase 2 · Q141
Smooth muscles are
Not quite — the answer is A.
Smooth muscle fibres are spindle-shaped (fusiform), uninucleate, have no striations and are not under voluntary control, as in the walls of the gut and blood vessels. Voluntary, multinucleate, cylindrical and striated fibres describe skeletal muscle.
Q43NEET 2016 Phase 2 · Q142
Oxidative phosphorylation is
Not quite — the answer is D.
In oxidative phosphorylation electrons removed from substrates (as NADH and FADH₂) pass along the electron transport chain, and the energy released drives ATP synthase to make ATP from ADP and Pi. Direct transfer of a phosphate from a substrate to ADP is substrate-level phosphorylation.
Q44NEET 2016 Phase 2 · Q143
Which of the following is the least likely to be involved in stabilizing the three-dimensional folding of most proteins?
Not quite — the answer is D.
The tertiary structure of proteins is held by hydrogen bonds, ionic (electrostatic) interactions, hydrophobic interactions and disulphide bridges. Ester bonds join fatty acids to glycerol in lipids and are not used to hold protein folds.
Q45NEET 2016 Phase 2 · Q146
Match the stages of meiosis in Column-I to their characteristic features in Column-II and select the correct option using the codes given below: Column-I: a. Pachytene; b. Metaphase I; c. Diakinesis; d. Zygotene Column-II: (i) Pairing of homologous chromosomes; (ii) Terminalization of chiasmata; (iii) Crossing-over takes place; (iv) Chromosomes align at equatorial plate
Not quite — the answer is A.
Crossing over occurs in pachytene, bivalents align at the equator in metaphase I, chiasmata terminalise in diakinesis and homologues pair (synapsis) in zygotene. Only one code gives all four of these pairs.
Q46NEET 2016 Phase 2 · Q147
Which hormones do stimulate the production of pancreatic juice and bicarbonate?
Not quite — the answer is C.
Secretin, from the duodenum, makes the exocrine pancreas release water and bicarbonate, and cholecystokinin stimulates secretion of pancreatic enzymes and bile release. Gastrin acts on the stomach, while insulin and glucagon are endocrine pancreatic hormones that regulate blood glucose.
Q47NEET 2016 Phase 2 · Q148
The partial pressure of oxygen in the alveoli of the lungs is
Not quite — the answer is B.
Alveolar pO₂ is about 104 mm Hg, higher than the 40 mm Hg of deoxygenated blood and the 95 mm Hg of oxygenated blood, so oxygen diffuses from the alveoli into the blood. Alveolar pCO₂ is only about 40 mm Hg, so pO₂ is not less than pCO₂.
Q48NEET 2016 Phase 2 · Q150
Graves' disease is caused due to
Not quite — the answer is B.
Graves' disease (exophthalmic goitre) is a form of hyperthyroidism caused by excess thyroid hormone, with bulging eyes, high metabolic rate and weight loss. Too little thyroid hormone causes cretinism or myxoedema, and adrenal disorders cause Addison's disease or Cushing's syndrome.
Q49NEET 2016 Phase 2 · Q151
Name the ion responsible for unmasking of active sites for myosin for cross-bridge activity during muscle contraction.
Not quite — the answer is A.
Ca²⁺ released from the sarcoplasmic reticulum binds troponin, which moves tropomyosin and uncovers the myosin-binding sites on actin so cross-bridges can form. Sodium and potassium handle the action potential, and magnesium supports ATP binding but does not unmask actin.
Q50NEET 2016 Phase 2 · Q152
Name the blood cells, whose reduction in number can cause clotting disorder, leading to excessive loss of blood from the body.
Not quite — the answer is D.
Thrombocytes (platelets) release factors needed for blood coagulation, so too few of them cause clotting disorders and heavy blood loss. Erythrocytes carry gases, and leucocytes, including neutrophils, are involved in defence.
Q51NEET 2016 Phase 2 · Q153
Name a peptide hormone which acts mainly on hepatocytes, adipocytes and enhances cellular glucose uptake and utilization.
Not quite — the answer is A.
Insulin, a peptide hormone from pancreatic beta cells, acts mainly on liver cells and fat cells to increase glucose uptake and use, lowering blood sugar. Glucagon raises blood glucose, and secretin and gastrin are gastrointestinal hormones.
Q52NEET 2016 Phase 2 · Q154
Osteoporosis, an age-related disease of skeletal system, may occur due to
Not quite — the answer is C.
Osteoporosis is an age-related loss of bone mass with more fractures, commonly caused by a fall in oestrogen after menopause. An immune disorder at the neuromuscular junction describes myasthenia gravis, and uric acid crystals in joints cause gout.
Q53NEET 2016 Phase 2 · Q155
Serum differs from blood in
Not quite — the answer is C.
Serum is plasma from which the clotting factors such as fibrinogen have been removed by clotting. It still has albumins and globulins, including antibodies.
Q54NEET 2016 Phase 2 · Q156
Lungs do not collapse between breaths and some air always remains in the lungs which can never be expelled because
Not quite — the answer is B.
The pleural space is kept below the pressure inside the lungs, and this negative intrapleural pressure pulls on the lung walls and keeps the alveoli from collapsing, so a residual volume always stays. Intrapulmonary pressure itself swings either side of atmospheric during breathing and is not permanently negative or higher.
Q55NEET 2016 Phase 2 · Q157
The posterior pituitary gland is not a 'true' endocrine gland because
Not quite — the answer is B.
Oxytocin and vasopressin are made by neurosecretory cells of the hypothalamus and only stored in and released from the posterior pituitary (neurohypophysis), so it makes no hormones of its own. It is ductless and secretes no enzymes, and being regulated by the hypothalamus applies to the anterior pituitary too.
Q56NEET 2016 Phase 2 · Q159
Which of the following is hormone-releasing IUD?
Not quite — the answer is A.
LNG-20 releases levonorgestrel, a progestogen, so it is a hormone-releasing IUD. Multiload 375 and Cu7 are copper-releasing IUDs, and the Lippes loop is a non-medicated IUD.
Q57NEET 2016 Phase 2 · Q161
Embryo with more than 16 blastomeres formed due to in vitro fertilization is transferred into
Not quite — the answer is A.
In assisted reproduction the zygote or early embryo with up to 8 blastomeres is placed in the fallopian tube (ZIFT), while later embryos with more than 8 blastomeres are transferred into the uterus (IUT). An embryo of more than 16 cells therefore goes into the uterus.
Q58NEET 2016 Phase 2 · Q162
Which of the following depicts the correct pathway of transport of sperms?
Not quite — the answer is A.
Sperms pass from the seminiferous tubules into the rete testis, then through the vasa efferentia into the epididymis, and on into the vas deferens. Any sequence that puts the epididymis before the efferent ductules, or the vas deferens before the epididymis, reverses this path.
Q59NEET 2016 Phase 2 · Q164
Several hormones like hCG, hPL, estrogen, progesterone are produced by
Not quite — the answer is B.
During pregnancy the placenta acts as an endocrine tissue and makes hCG, hPL, oestrogens and progesterone. The ovary makes oestrogen and progesterone but not hCG or hPL, and the pituitary makes neither.
Q60NEET 2016 Phase 2 · Q165
If a colour-blind man marries a woman who is homozygous for normal colour vision, the probability of their son being colour-blind is
Not quite — the answer is A.
Colour blindness is X-linked recessive, and a son gets his only X chromosome from his mother. Here the mother is XX with two normal alleles and the father passes only his Y to sons, so every son is normal and the probability is 0; the daughters will be carriers.
Q61NEET 2016 Phase 2 · Q166
Genetic drift operates in
Not quite — the answer is A.
Genetic drift is a random change in allele frequency by chance, and its effect is strongest in small, isolated populations such as those formed by founder effect or bottlenecks. In large populations chance changes even out, and a non-reproducing population passes on no alleles at all.
Q62NEET 2016 Phase 2 · Q167
In Hardy-Weinberg equation, the frequency of heterozygous individual is represented by
Not quite — the answer is B.
From (p + q)² = p² + 2pq + q² = 1, the heterozygote frequency is 2pq, because the heterozygote can form in two ways. p² and q² are the frequencies of the two homozygotes, and pq on its own is only half the heterozygote frequency.
Q63NEET 2016 Phase 2 · Q168
The chronological order of human evolution from early to the recent is
Not quite — the answer is B.
Ramapithecus lived about 15 million years ago, Australopithecines about 2 million years ago, then Homo habilis, followed by Homo erectus about 1.5 million years ago. Putting Australopithecus before Ramapithecus, or Homo habilis before Australopithecus, breaks this NCERT sequence.
Q64NEET 2016 Phase 2 · Q169
Which of the following is the correct sequence of events in the origin of life? I. Formation of protobionts II. Synthesis of organic monomers III. Synthesis of organic polymers IV. Formation of DNA-based genetic systems
Not quite — the answer is C.
Chemical evolution first formed simple organic monomers, which joined into polymers; these then gathered into protobionts (coacervates), and DNA-based genetic systems arose last after RNA-based ones. Sequences that start with protobionts, or place the DNA genetic system before protobionts, do not fit this order.
Biology: 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.
Q1NEET 2016 Phase 2 · Q109
A few drops of sap were collected by cutting across a plant stem by a suitable method. The sap was tested chemically. Which one of the following test results indicates that it is phloem sap?
Not quite — the answer is B.
Phloem sap is rich in sucrose and has a slightly alkaline pH of about 7.5 to 8, while xylem sap is slightly acidic. The high sugar content gives phloem sap a high refractive index, so low refractive index or absence of sugar would point away from phloem.
Q2NEET 2016 Phase 2 · Q111
Phytochrome is a
Not quite — the answer is D.
Phytochrome is a protein bound to a light-absorbing pigment (a linear tetrapyrrole chromophore) that switches between red and far-red absorbing forms, so it is a chromoprotein. It does not carry a flavin, sugar or lipid as its prosthetic group.
Q3NEET 2016 Phase 2 · Q114
Which one of the following statements is not correct?
Not quite — the answer is C.
In potato tubers, banana suckers and ginger rhizomes the new plantlets arise from buds at the nodes (the eyes of a potato are nodes), not from internodes. Clones, zoospores and the oxygen-draining water hyacinth are all correctly described.
Q4NEET 2016 Phase 2 · Q115
Which one of the following generates new genetic combinations leading to variation?
Not quite — the answer is C.
Sexual reproduction involves meiosis with crossing over and fusion of gametes from two parents, which creates new combinations of genes. Vegetative reproduction, parthenogenesis and nucellar embryos all come from a single parent without fertilisation and give near-identical offspring.
Q5NEET 2016 Phase 2 · Q140
In male cockroaches, sperms are stored in which part of the reproductive system?
Not quite — the answer is A.
In the male cockroach the sperms are stored in the seminal vesicles and are glued together into spermatophores that are discharged during mating. The mushroom glands are accessory glands, the testes produce sperms and the vasa deferentia only carry them.
Q6NEET 2016 Phase 2 · Q149
Choose the correct statement.
Not quite — the answer is C.
Rods and cones keep sodium channels open in the dark and stay depolarised; light closes these channels and hyperpolarises them. Nociceptors detect pain, Meissner's corpuscles are touch receptors, and sensory receptors do produce graded receptor potentials.
Chapter-wise split of NEET 2016 Phase 2 Biology
- Morphology of Flowering Plants4 Q
- Biotechnology: Principles and Processes4 Q
- Chemical Coordination and Integration4 Q
- Evolution4 Q
- Biological Classification3 Q
- The Living World3 Q
- Cell: The Unit of Life3 Q
- Sexual Reproduction in Flowering Plants3 Q
- Molecular Basis of Inheritance3 Q
- Principles of Inheritance and Variation3 Q
- Plant Kingdom2 Q
- Biomolecules2 Q
- Cell Cycle and Cell Division2 Q
- Respiration in Plants2 Q
- Organisms and Populations2 Q
- Biodiversity and Conservation2 Q
- Breathing and Exchange of Gases2 Q
- Locomotion and Movement2 Q
- Body Fluids and Circulation2 Q
- Reproductive Health2 Q
- Human Reproduction2 Q
- Anatomy of Flowering Plants1 Q
- Plant Growth and Development1 Q
- Photosynthesis in Higher Plants1 Q
- Biotechnology and Its Applications1 Q
- Ecosystem1 Q
- Human Health and Disease1 Q
- Animal Kingdom1 Q
- Structural Organisation in Animals1 Q
About this paper
How many questions from NEET 2016 Phase 2 are on this page?
139 questions (Physics 36, Chemistry 33, Biology 70). 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 NEET 2016 Phase 2 paper, conducted by CBSE. The answer to every question was checked by the RankUp team, and the explanations are RankUp's own.
Is NEET 2016 Phase 2 useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books. 12 questions here are on topics NCERT has since dropped; they are listed separately so you can skip them.
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