NEET 2017 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
129 questions from the NEET 2017 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 2017
- Conducted by
- CBSE
- Physics
- 20 questions
- Chemistry
- 37 questions
- Biology
- 72 questions
NEET 2017 Physics questions
20 questions. Tap an option to check your answer.
Q1NEET 2017 · Q136
A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be:
Not quite — the answer is D.
By the Stefan–Boltzmann law P = σ(4πr²)T⁴, so P ∝ r²T⁴. Halving r multiplies P by 1/4 and doubling T multiplies it by 16, a net factor of 4: 450 × 4 = 1800 W. Keeping only the radius change would give the trap value of about 112 W, and assuming the effects cancel gives 450 W.
Q2NEET 2017 · Q138
The ratio of resolving powers of an optical microscope for two wavelengths λ₁ = 4000 Å and λ₂ = 6000 Å is:
Not quite — the answer is C.
Resolving power of a microscope is 2n sin θ/λ, i.e. inversely proportional to wavelength. RP₁ : RP₂ = λ₂ : λ₁ = 6000 : 4000 = 3 : 2. Squaring or cubing the ratio (9 : 4, 8 : 27) has no basis.
Q3NEET 2017 · Q141
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N?
Not quite — the answer is C.
For a hollow cylinder about its axis I = MR² = 3 × (0.4)² = 0.48 kg m². Torque τ = FR = 30 × 0.4 = 12 N m, so α = τ/I = 12/0.48 = 25 rad/s². 25 m/s² has the right number but wrong units for an angular acceleration.
Q4NEET 2017 · Q142
The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be:
Not quite — the answer is C.
Melting and redrawing keeps the volume constant, so if length becomes nl the area becomes A/n. R = ρl/A then becomes ρ(nl)/(A/n) = n²R. Taking only the length change gives nR.
Q5NEET 2017 · Q143
The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is:
Not quite — the answer is C.
The last (series limit) line has n₁ → ∞. Balmer: 1/λ = R(1/4), so λ = 4/R; Lyman: 1/λ = R(1/1), so λ = 1/R. Their ratio is 4. Taking the ratio the other way round gives 0.25, not an option.
Q6NEET 2017 · Q144
A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle θ, the spot of the light is found to move through a distance y on the scale. The angle θ is given by:
Not quite — the answer is A.
Rotating a mirror by θ turns the reflected ray by 2θ. For a small angle the spot moves y = x·2θ on the scale at distance x, so θ = y/2x. Writing θ = y/x forgets the doubling of the reflected beam's rotation.
Q7NEET 2017 · Q146
Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s]:
Not quite — the answer is D.
Source and observer approach each other, so f′ = f(v + v₀)/(v − vₛ) = 400 × (340 + 16.5)/(340 − 22) = 400 × 356.5/318 ≈ 448 Hz. The heard frequency must be higher than 400 Hz, which rules out 350 Hz and 361 Hz.
Q8NEET 2017 · Q147
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:
Not quite — the answer is C.
Speed v = ω√(A² − x²) = ω√(9 − 4) = √5 ω and acceleration a = ω²x = 2ω² (in cm units). Setting them equal gives ω = √5/2 rad/s, so T = 2π/ω = 4π/√5 s. √5/2π is the frequency, not the period.
Q9NEET 2017 · Q148
A carnot engine having an efficiency of 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 B.
For a Carnot cycle the coefficient of performance as a refrigerator is β = (1 − η)/η = (9/10)/(1/10) = 9. Heat drawn from the cold reservoir Q₂ = βW = 9 × 10 = 90 J. 100 J is the heat rejected to the hot reservoir (Q₂ + W).
Q10NEET 2017 · Q151
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t₁. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t₂. The time taken by her to walk up on the moving escalator will be:
Not quite — the answer is C.
With escalator length L her speed is L/t₁ and the escalator's is L/t₂. Moving together the speeds add, so t = L/(L/t₁ + L/t₂) = t₁t₂/(t₁ + t₂), which is shorter than either time. The average (t₁ + t₂)/2 wrongly averages times instead of adding speeds.
Q11NEET 2017 · Q152
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system:
Not quite — the answer is B.
With the battery removed, charge Q is conserved and is shared over capacitance 2C. Energy goes from Q²/2C to Q²/(2 × 2C), i.e. it halves; the lost half is dissipated during charge flow. Energy cannot stay the same even though charge is conserved.
Q12NEET 2017 · Q154
A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves:
Not quite — the answer is C.
At the balance point no current flows through the galvanometer, so no current is drawn from the cell being measured and its terminal voltage equals its emf. This null method is what makes the potentiometer accurate; a voltmeter always draws some current.
Q13NEET 2017 · Q157
The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then:
Not quite — the answer is D.
For small heights g_h = g(1 − 2h/R) and at depth g_d = g(1 − d/R). Equating gives d = 2h = 2 km. Taking d = h ignores the factor 2 in the height formula.
Q14NEET 2017 · Q158
A gas mixture consists of 2 moles of O₂ and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:
Not quite — the answer is D.
Diatomic O₂ (no vibration) has 5 degrees of freedom: U = 2 × (5/2)RT = 5RT. Monatomic Ar has 3: U = 4 × (3/2)RT = 6RT. Total U = 11RT. Treating O₂ as having 3 degrees of freedom would give 9RT.
Q15NEET 2017 · Q160
A thin prism having refracting angle 10° is made of glass of refractive index 1.42. This prism is combined with another thin prism of glass of refractive index 1.7. This combination produces dispersion without deviation. The refracting angle of second prism should be:
Not quite — the answer is B.
For zero net deviation (μ₁ − 1)A₁ = (μ₂ − 1)A₂, so 0.42 × 10° = 0.7 × A₂ and A₂ = 6°. A smaller-angle prism of higher index cancels the deviation of the first.
Q16NEET 2017 · Q161
The bulk modulus of a spherical object is 'B'. If it is subjected to uniform pressure 'p', the fractional decrease in radius is:
Not quite — the answer is D.
Volumetric strain ΔV/V = p/B, and for a sphere V ∝ r³ so ΔV/V = 3Δr/r. Hence Δr/r = p/3B. p/B is the fractional change in volume, not radius.
Q17NEET 2017 · Q163
A physical quantity of the dimensions of length can be formed out of c, G and e²/4πε₀ is [c is velocity of light, G is universal constant of gravitation, e is charge]:
Not quite — the answer is A.
e²/4πε₀ = F r² has dimensions [M L³ T⁻²] and G has [M⁻¹ L³ T⁻²], so G e²/4πε₀ has [L⁶ T⁻⁴]; its square root is [L³ T⁻²]. Dividing by c² [L² T⁻²] leaves [L]. Multiplying by c² instead gives L⁵T⁻⁴.
Q18NEET 2017 · Q166
A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is k′. Then they are connected in parallel and force constant is k″. Then k′ : k″ is:
Not quite — the answer is C.
Spring constant is inversely proportional to length, so the pieces (l/6, 2l/6, 3l/6) have constants 6k, 3k and 2k. In series they rebuild the original spring, k′ = k; in parallel k″ = 6k + 3k + 2k = 11k. Hence k′ : k″ = 1 : 11.
Q19NEET 2017 · Q169
The x and y coordinates of the particle at any time are x = 5t − 2t² and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2s is:
Not quite — the answer is C.
aₓ = d²x/dt² = −4 m/s² (constant) and a_y = d²y/dt² = 0 since y grows uniformly. The acceleration is therefore −4 m/s² along x at every instant, including t = 2 s. −8 m/s² comes from substituting t = 2 into −4t, which is not the acceleration.
Q20NEET 2017 · Q174
In a common emitter transistor amplifier the audio signal voltage across the collector is 3 V. The resistance of collector is 3 kΩ. If current gain is 100 and the base resistance is 2 kΩ, the voltage and power gain of the amplifier is:
Not quite — the answer is C.
Voltage gain A_V = β × R_C/R_B = 100 × 3/2 = 150, and power gain = β × A_V = 100 × 150 = 15000. The 3 V signal value is not needed.
Chapter-wise split of NEET 2017 Physics
- Ray Optics and Optical Instruments3 Q
- Current Electricity2 Q
- Oscillations2 Q
- Thermal Properties of Matter1 Q
- System of Particles and Rotational Motion1 Q
- Atoms1 Q
- Waves1 Q
- Thermodynamics1 Q
- Motion in a Straight Line1 Q
- Electrostatic Potential and Capacitance1 Q
- Gravitation1 Q
- Kinetic Theory1 Q
- Mechanical Properties of Solids1 Q
- Units and Measurements1 Q
- Motion in a Plane1 Q
- Semiconductor Electronics: Materials, Devices and Simple Circuits1 Q
NEET 2017 Chemistry questions
33 questions. Tap an option to check your answer.
Q1NEET 2017 · Q1
HgCl₂ and I₂ both when dissolved in water containing I⁻ ions the pair of species formed is:
Not quite — the answer is C.
In excess iodide, HgCl₂ first gives HgI₂, which dissolves in more I⁻ as the complex HgI₄²⁻ (as in Nessler's reagent), while I₂ combines with I⁻ to form the triiodide ion I₃⁻. HgI₂ cannot remain as such because excess I⁻ is present.
Q2NEET 2017 · Q2
Predict the correct intermediate and product in the following reaction: H₃C–C≡CH —(H₂O, H₂SO₄, HgSO₄)→ intermediate (A) → product (B)
Not quite — the answer is D.
Mercuric-ion catalysed hydration of propyne adds water by Markovnikov's rule, giving the enol CH₃–C(OH)=CH₂. The enol is unstable and tautomerises to the ketone, acetone (CH₃COCH₃). No sulphate-containing product is formed.
Q3NEET 2017 · Q3
The correct statement regarding electrophile is:
Not quite — the answer is D.
An electrophile is an electron-deficient species: it may be positively charged (H⁺, NO₂⁺) or neutral (BF₃, AlCl₃, carbonyl carbon), and it bonds by accepting an electron pair from a nucleophile. Saying electrophiles are 'generally neutral' ignores the many cationic electrophiles, and negatively charged species are nucleophiles.
Q4NEET 2017 · Q4
Which of the following pairs of compounds is isoelectronic and isostructural?
Not quite — the answer is C.
XeF₂ and IBr₂⁻ both have 22 valence electrons (8 + 14 and 7 + 14 + 1) and both have three lone pairs on the central atom with two bonds, giving a linear shape. BeCl₂ is linear but has only 16 valence electrons, and TeI₂ is bent.
Q5NEET 2017 · Q5
The species, having bond angles of 120° is:
Not quite — the answer is D.
BCl₃ has sp² boron with no lone pair, so it is trigonal planar with 120° angles. PH₃ and NCl₃ are pyramidal (angles below 109.5°) and ClF₃ is T-shaped (angles near 90°).
Q6NEET 2017 · Q7
Which one of the following pairs of species have the same bond order?
Not quite — the answer is C.
CN⁻ and CO are both 14-electron species, isoelectronic with N₂, so each has bond order 3. NO has 2.5 against 3 for CO, O₂ has 2 against 3 for NO⁺, and O₂⁻ has 1.5 against 3 for N₂.
Q7NEET 2017 · Q9
Name the gas that can readily decolourise acidified KMnO₄ solution:
Not quite — the answer is B.
SO₂ is a strong reducing agent: it reduces purple MnO₄⁻ to nearly colourless Mn²⁺ while being oxidised to sulphate, and NCERT gives this decolourisation as a test for SO₂. CO₂ and P₂O₅ are not reducing, and NO₂ is itself an oxidising agent rather than the gas known for this test.
Q8NEET 2017 · Q10
Which one is the wrong statement?
Not quite — the answer is D.
In hydrogen-like (one-electron) species the energy depends only on n, so 2s and 2p are degenerate; the 2s-below-2p order holds only for multi-electron atoms. The other statements are correct: λ = h/mv, ΔE·Δt ≥ h/4π, and half/fully filled subshells are extra stable.
Q9NEET 2017 · Q11
Correct increasing order for the wavelengths of absorption in the visible region for the complexes of Co³⁺ is:
Not quite — the answer is A.
Field strength follows the spectrochemical series en > NH₃ > H₂O, so Δₒ is largest for the en complex. Absorbed wavelength is inversely related to Δₒ (λ = hc/Δₒ), so the order of increasing wavelength is [Co(en)₃]³⁺ < [Co(NH₃)₆]³⁺ < [Co(H₂O)₆]³⁺. Writing the reverse order confuses increasing Δₒ with increasing wavelength.
Q10NEET 2017 · Q12
The correct order of the stoichiometries of AgCl formed when AgNO₃ in excess is treated with the complexes: CoCl₃.6NH₃, CoCl₃.5NH₃, CoCl₃.4NH₃ respectively is:
Not quite — the answer is C.
The complexes are [Co(NH₃)₆]Cl₃, [Co(NH₃)₅Cl]Cl₂ and [Co(NH₃)₄Cl₂]Cl. Only chloride outside the coordination sphere is precipitated by Ag⁺, giving 3, 2 and 1 mol AgCl respectively.
Q11NEET 2017 · Q13
Which one is the most acidic compound?
Not quite — the answer is D.
Electron-withdrawing –NO₂ groups at ortho and para positions stabilise the phenoxide ion by resonance, so acidity rises with the number of nitro groups; 2,4,6-trinitrophenol (pKa about 0.4) is the strongest acid. The methyl group of p-cresol is electron-donating and makes it weaker than phenol.
Q12NEET 2017 · Q14
The correct increasing order of basic strength for the following compounds is: (I) aniline, C₆H₅NH₂ (II) p-nitroaniline, p-O₂N–C₆H₄–NH₂ (III) p-toluidine, p-CH₃–C₆H₄–NH₂
Not quite — the answer is D.
The electron-donating –CH₃ at the para position increases electron density on nitrogen, so p-toluidine is more basic than aniline, while the –NO₂ group withdraws the lone pair by resonance and makes p-nitroaniline the weakest base. Hence II < I < III.
Q13NEET 2017 · Q15
In which pair of ions both the species contain S–S bond?
Not quite — the answer is B.
Thiosulphate S₂O₃²⁻ has one sulphur bonded directly to the central S, and tetrathionate S₄O₆²⁻ has an –S–S– chain between two SO₃ groups. Disulphate S₂O₇²⁻ has an S–O–S bridge and peroxodisulphate S₂O₈²⁻ an S–O–O–S bridge, with no S–S bond.
Q14NEET 2017 · Q17
Which one is the correct order of acidity?
Not quite — the answer is B.
Acidity of C–H rises with s-character of the carbon: sp (50%) > sp² (33%) > sp³ (25%), so alkynes > alkenes > alkanes. Between the alkynes, the electron-donating methyl group makes propyne less acidic than ethyne, which is why ethene cannot be placed above propyne.
Q15NEET 2017 · Q18
The heating of phenyl-methyl ethers with HI produces.
Not quite — the answer is C.
HI protonates anisole and I⁻ attacks the methyl carbon by SN2, giving phenol and CH₃I. Iodobenzene does not form because the aryl C–O bond has partial double-bond character and SN2 attack on an sp² ring carbon does not occur.
Q16NEET 2017 · Q24
For a given reaction, ΔH = 35.5 kJ mol⁻¹ and ΔS = 83.6 J K⁻¹ mol⁻¹. The reaction is spontaneous at: (Assume that ΔH and ΔS do not vary with temperature)
Not quite — the answer is B.
ΔG = ΔH − TΔS is negative when T > ΔH/ΔS = 35500 / 83.6 ≈ 425 K. Both ΔH and ΔS are positive, so the reaction is non-spontaneous at low temperature and becomes spontaneous only above 425 K; it is not spontaneous at all temperatures.
Q17NEET 2017 · Q25
In the electrochemical cell: Zn|ZnSO₄ (0.01 M)||CuSO₄ (1.0 M)|Cu, the emf of this Daniel cell is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ changed to 0.01 M, the emf changes to E₂. From the followings, which one is the relationship between E₁ and E₂? (Given, RT/F = 0.059)
Not quite — the answer is C.
E = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺]). For E₁ the ratio is 0.01/1 so E₁ = E° + 0.059 V; for E₂ the ratio is 1/0.01 so E₂ = E° − 0.059 V. Hence E₁ > E₂; reversing the concentrations does not make the emf zero.
Q18NEET 2017 · Q26
An example of a sigma bonded organometallic compound is:
Not quite — the answer is B.
In a Grignard reagent RMgX the alkyl carbon is joined to magnesium by a single σ bond. Ferrocene, ruthenocene and cobaltocene are sandwich compounds in which the metal is π-bonded to cyclopentadienyl rings.
Q19NEET 2017 · Q27
The equilibrium constants of the following are: N₂ + 3H₂ ⇌ 2NH₃ K₁; N₂ + O₂ ⇌ 2NO K₂; H₂ + ½O₂ → H₂O K₃. The equilibrium constant (K) of the reaction: 2NH₃ + 5/2 O₂ ⇌ 2NO + 3H₂O, will be:
Not quite — the answer is B.
The required reaction = (reverse of the NH₃ formation) + (NO formation) + 3 × (H₂O formation): 2NH₃ → N₂ + 3H₂ (1/K₁), N₂ + O₂ → 2NO (K₂), 3H₂ + 3/2 O₂ → 3H₂O (K₃³). Adding reactions multiplies constants, so K = K₂K₃³/K₁.
Q20NEET 2017 · Q28
The element Z = 114 has been discovered recently. It will belong to which of the following family/group and electronic configuration?
Not quite — the answer is B.
After radon (Z = 86), 32 more electrons fill 7s² 5f¹⁴ 6d¹⁰ 7p², reaching Z = 86 + 2 + 14 + 10 + 2 = 114. A p² valence configuration places it in group 14, the carbon family (flerovium).
Q21NEET 2017 · Q29
Pick out the correct statement with respect to [Mn(CN)₆]³⁻:
Not quite — the answer is C.
Mn³⁺ is 3d⁴. CN⁻ is a strong-field ligand, so the four electrons pair into three 3d orbitals, leaving two inner 3d orbitals free; with 4s and 4p they give d²sp³ hybridisation and an octahedral (inner-orbital) complex. sp³d² would apply to a weak-field outer-orbital complex.
Q22NEET 2017 · Q31
It is because of inability of ns² electrons of the valence shell to participate in bonding that:
Not quite — the answer is A.
The inert pair effect makes the +2 state more stable down group 14. For tin the +4 state is still the more stable, so Sn²⁺ tends to lose electrons and acts as a reducing agent; for lead the +2 state is more stable, so Pb⁴⁺ gains electrons and acts as an oxidising agent.
Q23NEET 2017 · Q32
Which of the following statements is not correct?
Not quite — the answer is D.
Denaturation destroys the secondary and tertiary structure of a protein, so it loses its biological activity rather than gaining it. The other statements are correct: insulin regulates blood glucose, ovalbumin is the storage protein of egg white, and thrombin and fibrinogen take part in clotting.
Q24NEET 2017 · Q35
The reason for greater range of oxidation states in actinoids is attributed to:
Not quite — the answer is C.
In actinoids the 5f, 6d and 7s orbitals are close in energy, so varying numbers of electrons from all three can be used in bonding, giving many oxidation states. 4f and 5d levels belong to the lanthanoids, and radioactivity or contraction does not create extra oxidation states.
Q25NEET 2017 · Q37
Ionic mobility of which of the following alkali metal ions is lowest when aqueous solution of their salts are put under an electric field?
Not quite — the answer is D.
Li⁺ is the smallest alkali metal ion, so it has the highest charge density and is the most heavily hydrated. Its large hydrated radius makes it move most slowly in solution, so Li⁺ has the lowest ionic mobility even though the bare ion is the smallest.
Q26NEET 2017 · Q38
Which of the following is dependent on temperature?
Not quite — the answer is B.
Molarity is moles per litre of solution, and the volume of a solution changes with temperature, so molarity changes too. Molality, mole fraction and weight percentage involve only masses or mole numbers, which do not depend on temperature.
Q27NEET 2017 · Q39
If molality of the dilute solution is doubled, the value of molal depression constant (K_f) will be:
Not quite — the answer is D.
K_f is a property of the solvent alone (K_f = R·M·T_f²/1000ΔH_fus), so it does not change with concentration. Doubling the molality doubles the depression ΔT_f = K_f·m, not K_f itself.
Q28NEET 2017 · Q40
Mechanism of a hypothetical reaction X₂ + Y₂ → 2XY is given below: (i) X₂ → X + X (fast) (ii) X + Y₂ ⇌ XY + Y (slow) (iii) X + Y → XY (fast). The overall order of the reaction will be:
Not quite — the answer is D.
The slow step gives rate = k[X][Y₂]. X comes from the fast pre-equilibrium X₂ ⇌ 2X, so [X] = √(K[X₂]) and rate = k′[X₂]^½[Y₂]. The overall order is ½ + 1 = 1.5; taking the slow step's molecularity of 2 as the order ignores that X is an intermediate.
Q29NEET 2017 · Q41
Concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.2 × 10⁻⁴ mol L⁻¹. Solubility product of Ag₂C₂O₄ is:
Not quite — the answer is D.
Ag₂C₂O₄ ⇌ 2Ag⁺ + C₂O₄²⁻, so [C₂O₄²⁻] = ½ × 2.2 × 10⁻⁴ = 1.1 × 10⁻⁴ M. Ksp = [Ag⁺]²[C₂O₄²⁻] = (2.2 × 10⁻⁴)² × 1.1 × 10⁻⁴ = 4.84 × 10⁻⁸ × 1.1 × 10⁻⁴ ≈ 5.3 × 10⁻¹². Forgetting to square [Ag⁺] gives a wrong value.
Q30NEET 2017 · Q42
Match the interhalogen compounds of column I with the geometry in column II and assign the correct code. Column I: (a) XX′ (b) XX′₃ (c) XX′₅ (d) XX′₇. Column II: (i) T - shape (ii) Pentagonal bipyramidal (iii) Linear (iv) Square - pyramidal (v) Tetrahedral
Not quite — the answer is B.
XX′ is diatomic and linear. XX′₃ has 3 bond pairs and 2 lone pairs (sp³d), giving a T shape; XX′₅ has 5 bond pairs and 1 lone pair (sp³d²), giving a square pyramid; XX′₇ has 7 bond pairs (sp³d³), giving a pentagonal bipyramid.
Q31NEET 2017 · Q43
Which one of the following statements is not correct?
Not quite — the answer is B.
A catalyst speeds up the forward and backward reactions equally, so equilibrium is reached sooner but the equilibrium constant stays the same. The other statements are correct: a catalyst cannot start a reaction that is not feasible, enzymes are biological catalysts, and coenzymes enhance enzyme activity.
Q32NEET 2017 · Q44
Consider the reactions: X (C₂H₆O) —(Cu/573 K)→ A; A —([Ag(NH₃)₂]⁺, ⁻OH, Δ)→ Silver mirror observed; A —(⁻OH, Δ)→ Y; A —(NH₂–NH–CO–NH₂)→ Z. Identify A, X, Y and Z
Not quite — the answer is C.
C₂H₆O that is dehydrogenated by Cu at 573 K must be ethanol, giving ethanal as A, which gives the silver mirror with Tollens' reagent. With dilute alkali and heat ethanal undergoes aldol condensation to but-2-enal as Y, and with semicarbazide it forms the semicarbazone Z. Methoxymethane is also C₂H₆O but cannot be dehydrogenated to an aldehyde.
Q33NEET 2017 · Q45
Which of the following reactions is appropriate for converting acetamide to methanamine?
Not quite — the answer is B.
CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O: the Hofmann bromamide (hypobromite) degradation gives an amine with one carbon fewer, exactly acetamide to methanamine. Gabriel synthesis starts from an alkyl halide, Stephens reduction gives aldehydes from nitriles, and the carbylamine reaction is a test for primary amines.
Chemistry: not in the current NEET syllabus
These 4 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
4 questions. Tap an option to check your answer.
Q1NEET 2017 · Q6
Which of the following is a sink for CO?
Not quite — the answer is B.
Soil microorganisms remove carbon monoxide from the atmosphere by oxidising it, so the soil acts as the natural sink for CO. Haemoglobin binds CO in the body but is not an environmental sink, and oceans are a major sink for CO₂, not CO.
Q2NEET 2017 · Q16
Mixture of chloroxylenol and terpineol acts as:
Not quite — the answer is B.
A mixture of chloroxylenol and terpineol is the antiseptic Dettol, applied to living tissue to stop microbial growth. It does not relieve pain or fever and is not a microbially derived antibiotic.
Q3NEET 2017 · Q33
Which is the incorrect statement?
Not quite — the answer is D.
Frenkel defects occur when the cation is much smaller than the anion (AgCl, ZnS), so the small cation can move into an interstitial site; similar-sized ions favour Schottky defects instead. FeO₀.₉₈ is metal deficient, and Schottky defects lower density because ions are missing.
Q4NEET 2017 · Q36
Extraction of gold and silver involves leaching with CN⁻ ion. Silver is later recovered by:
Not quite — the answer is D.
Leaching with cyanide forms the soluble complex [Ag(CN)₂]⁻. Zinc, being more electropositive, displaces silver: 2[Ag(CN)₂]⁻ + Zn → [Zn(CN)₄]²⁻ + 2Ag. Liquation, distillation and zone refining are refining methods, not recovery from the cyanide solution.
Chapter-wise split of NEET 2017 Chemistry
- Coordination Compounds4 Q
- Chemical Bonding and Molecular Structure3 Q
- Equilibrium3 Q
- The d- and f-Block Elements2 Q
- Hydrocarbons2 Q
- Alcohols, Phenols and Ethers2 Q
- Amines2 Q
- Electrochemistry2 Q
- Solutions2 Q
- Organic Chemistry: Some Basic Principles and Techniques1 Q
- Structure of Atom1 Q
- Thermodynamics1 Q
- Classification of Elements and Periodicity in Properties1 Q
- Biomolecules1 Q
- Chemical Kinetics1 Q
- Aldehydes, Ketones and Carboxylic Acids1 Q
NEET 2017 Biology questions
64 questions. Tap an option to check your answer.
Q1NEET 2017 · Q46
Which of the following in sewage treatment removes suspended solids?
Not quite — the answer is C.
Primary treatment is the physical stage: floating debris is removed by sequential filtration and grit and suspended solids settle out by sedimentation as primary sludge. Secondary treatment is biological and removes dissolved organic matter (BOD) through microbes, not suspended solids.
Q2NEET 2017 · Q47
Which one of the following is related to Ex-situ conservation of threatened animals and plants?
Not quite — the answer is A.
Ex-situ conservation protects threatened species outside their natural habitat, in places such as zoos, botanical gardens and wildlife safari parks. Biodiversity hotspots, the Amazon rainforest and the Himalayan region are natural habitats, so protecting them is in-situ conservation.
Q3NEET 2017 · Q48
Phosphoenol pyruvate (PEP) is the primary CO₂ acceptor in:
Not quite — the answer is B.
In C₄ plants PEP in mesophyll cells accepts CO₂ (catalysed by PEP carboxylase) to form the 4-carbon oxaloacetic acid. C₃ plants use RuBP as the primary CO₂ acceptor, so PEP is not the acceptor in both groups.
Q4NEET 2017 · Q50
In case of poriferans, the spongocoel is lined with flagellated cells called:
Not quite — the answer is C.
The spongocoel and canals of sponges are lined by flagellated collar cells called choanocytes, which drive the water current. Ostia are the pores through which water enters and the osculum is the opening through which it leaves; they are openings, not cells.
Q5NEET 2017 · Q52
Lungs are made up of air-filled sacs, the alveoli. They do not collapse even after forceful expiration, because of:
Not quite — the answer is A.
Residual volume (about 1100–1200 mL) is the air that remains in the lungs even after the most forceful expiration, and it keeps the alveoli from collapsing. Expiratory reserve volume is the extra air that can be breathed out forcibly, so it is exactly the air that is removed.
Q6NEET 2017 · Q53
Viroids differ from viruses in having:
Not quite — the answer is D.
Viroids, discovered by T.O. Diener as the cause of potato spindle tuber disease, are free low-molecular-weight RNA without any protein coat. Viruses always have a protein capsid around their nucleic acid, so a coated RNA describes a virus, not a viroid.
Q7NEET 2017 · Q54
Which of the following are not polymeric?
Not quite — the answer is D.
Lipids are small molecules (molecular weight below about 800 Da) built from fatty acids and glycerol, not repeating monomers, so they are not true polymers; they only appear in the acid-insoluble fraction because they form membrane vesicles. Nucleic acids, proteins and polysaccharides are polymers of nucleotides, amino acids and monosaccharides.
Q8NEET 2017 · Q55
Select the mismatch:
Not quite — the answer is A.
Pinus is monoecious, bearing both male and female cones on the same plant, so pairing it with dioecious is the mismatch. Cycas is truly dioecious, Salvinia is a heterosporous water fern and Equisetum is homosporous.
Q9NEET 2017 · Q56
A gene whose expression helps to identify transformed cell is known as:
Not quite — the answer is A.
A selectable marker, such as a gene for ampicillin or tetracycline resistance, lets transformants be picked out because only they survive on the selective medium. A vector or plasmid is the carrier of the DNA, not the identifying gene.
Q10NEET 2017 · Q57
A decrease in blood pressure/volume will not cause the release of:
Not quite — the answer is B.
Atrial natriuretic factor is released by the atrial walls when blood pressure or volume rises, and it causes vasodilation to lower pressure. A fall in pressure or volume instead triggers renin (and through it aldosterone) and ADH, which raise it.
Q11NEET 2017 · Q58
In Bougainvillea thorns are the modifications of:
Not quite — the answer is C.
Thorns of Bougainvillea arise from axillary buds, which marks them as modified stems. Spines of cacti are modified leaves, and stipules can also form spines, but neither applies to Bougainvillea.
Q12NEET 2017 · Q59
An important characteristic that Hemichordates share with Chordates is:
Not quite — the answer is C.
Hemichordates have pharyngeal gill slits, one of the basic chordate features. Chordates have a notochord and a dorsal (not ventral) hollow nerve cord, so the absence of a notochord and a ventral nerve cord are not shared features.
Q13NEET 2017 · Q61
Which of the following statements is correct?
Not quite — the answer is A.
The ascending limb of Henle's loop is impermeable to water but transports electrolytes, which helps build the medullary concentration gradient. The descending limb is the opposite: permeable to water and almost impermeable to electrolytes.
Q14NEET 2017 · Q62
Which of the following are found in extreme saline conditions?
Not quite — the answer is A.
Halophiles, a group of archaebacteria, live in extremely salty habitats; their distinctive cell wall helps them survive such conditions. Eubacteria, cyanobacteria and mycobacteria are not characteristic of extreme salinity.
Q15NEET 2017 · Q63
The morphological nature of the edible part of coconut is:
Not quite — the answer is C.
The edible white kernel and the coconut water are the endosperm (cellular and free-nuclear respectively). The pericarp of the drupe forms the outer fibrous husk and hard shell, which is not eaten.
Q16NEET 2017 · Q64
Identify the wrong statement in context of heartwood:
Not quite — the answer is C.
Heartwood is the dark central wood whose dead, highly lignified elements are filled with tannins, resins, oils and gums, which make it durable but non-conducting. Water and minerals are conducted by the outer sapwood, so that statement is wrong.
Q17NEET 2017 · Q65
If there are 999 bases in an RNA that codes for a protein with 333 amino acids, and the base at position 901 is deleted such that the length of the RNA becomes 998 bases, how many codons will be altered?
Not quite — the answer is C.
Base 901 is the first base of codon 301 (bases 901–903). Deleting it shifts the reading frame from codon 301 to the end, so codons 301 to 333, that is 333 − 300 = 33 codons, are altered while the first 300 stay the same. Choosing 333 wrongly assumes the whole message is misread.
Q18NEET 2017 · Q67
A dioecious flowering plant prevents both:
Not quite — the answer is B.
In a dioecious plant male and female flowers are on different plants, so pollen cannot reach a stigma of the same flower (autogamy) or of another flower on the same plant (geitonogamy). Xenogamy, pollination between different plants, is the only route left, so it is not prevented.
Q19NEET 2017 · Q68
Which statement is wrong for Krebs' cycle?
Not quite — the answer is D.
The Krebs cycle begins when acetyl CoA condenses with oxaloacetic acid and water to form citric acid, not with pyruvic acid; pyruvate is converted to acetyl CoA before the cycle. The cycle does have three NAD⁺ reductions, one FAD reduction and substrate-level GTP formation from succinyl CoA.
Q20NEET 2017 · Q69
Which among these is the correct combination of aquatic mammals?
Not quite — the answer is C.
Whales, dolphins and seals are all aquatic mammals. Sharks and Trygon (sting ray) are cartilaginous fishes, so any group containing them is wrong.
Q21NEET 2017 · Q70
The hepatic portal vein drains blood to liver from:
Not quite — the answer is D.
The hepatic portal vein carries blood from the intestine to the liver before it reaches the systemic circulation, so absorbed nutrients are processed first. Blood from the heart reaches the liver through the hepatic artery, and kidneys drain into the renal veins.
Q22NEET 2017 · Q71
Functional megaspore in an angiosperm develops into:
Not quite — the answer is C.
The single functional megaspore undergoes three free-nuclear mitotic divisions to form the 7-celled, 8-nucleate embryo sac (female gametophyte). The ovule is the structure that contains it, and endosperm and embryo form only after fertilisation.
Q23NEET 2017 · Q72
Mycorrhizae are the example of:
Not quite — the answer is D.
In mycorrhiza the fungus absorbs mineral nutrients such as phosphorus for the plant, and the plant supplies the fungus with carbohydrates, so both partners benefit: mutualism. Amensalism harms one partner without affecting the other.
Q24NEET 2017 · Q73
Transplantation of tissues/organs fails often due to non-acceptance by the patient's body. Which type of immune-response is responsible for such rejections?
Not quite — the answer is B.
Graft rejection is caused by cell-mediated immunity, in which T-lymphocytes recognise the graft as non-self and attack it. An autoimmune response is directed against the body's own cells, not against a transplanted organ.
Q25NEET 2017 · Q74
Adult human RBCs are enucleate. Which of the following statement(s) is/are most appropriate explanation for this feature? (a) They do not need to reproduce (b) They are somatic cells (c) They do not metabolize (d) All their internal space is available for oxygen transport
Not quite — the answer is A.
Losing the nucleus leaves the whole biconcave cell free to be packed with haemoglobin, so it can carry the most oxygen. RBCs do metabolise (they run glycolysis), and being somatic or non-dividing does not explain why the nucleus is lost.
Q26NEET 2017 · Q75
Alexander Von Humbolt described for the first time:
Not quite — the answer is C.
During his explorations of South America Alexander von Humboldt observed that species richness rises with the area explored, the species–area relationship (log S = log C + Z log A). The law of limiting factors comes from Liebig and Blackman, not Humboldt.
Q27NEET 2017 · Q76
Attractants and rewards are required for:
Not quite — the answer is B.
Insect-pollinated (entomophilous) flowers attract pollinators with colour, scent and nectar, and reward them with nectar and pollen. Wind and water pollination need no attractants, and cleistogamous flowers never open.
Q28NEET 2017 · Q78
An example of colonial alga is:
Not quite — the answer is B.
Volvox is a colonial green alga, a hollow sphere of many flagellated cells. Chlorella is unicellular, while Ulothrix and Spirogyra are filamentous.
Q29NEET 2017 · Q80
DNA fragments are:
Not quite — the answer is B.
The phosphate groups of the sugar–phosphate backbone carry negative charges, so all DNA fragments are negatively charged and move towards the anode in electrophoresis. Size changes how fast they move, not the sign of their charge.
Q30NEET 2017 · Q81
The pivot joint between atlas and axis is a type of:
Not quite — the answer is C.
The joint between atlas and axis is a pivot joint, one kind of synovial joint with a fluid-filled cavity that allows rotation of the head. A saddle joint is a different synovial joint, found between the carpal and metacarpal of the thumb.
Q31NEET 2017 · Q82
Asymptote in a logistic growth curve is obtained when:
Not quite — the answer is B.
In logistic growth dN/dt = rN(K − N)/K, so growth stops and the curve flattens into the asymptote when population density N reaches the carrying capacity K. The intrinsic rate r stays the same; it is the (K − N)/K term that falls to zero.
Q32NEET 2017 · Q83
Myelin sheath is produced by:
Not quite — the answer is A.
Myelin is formed by Schwann cells around axons in the peripheral nervous system and by oligodendrocytes in the central nervous system. Astrocytes support neurons but do not make myelin, and osteoclasts are bone-resorbing cells.
Q33NEET 2017 · Q84
The process of separation and purification of expressed protein before marketing is called:
Not quite — the answer is B.
After biosynthesis in the bioreactor, the product is separated, purified, formulated with preservatives and put through quality control; these steps together are downstream processing. Upstream processing is the earlier stage of preparing the organism and growing it.
Q34NEET 2017 · Q85
GnRH, a hypothalamic hormone, needed in reproduction, acts on:
Not quite — the answer is B.
Gonadotropin-releasing hormone from the hypothalamus acts on the anterior pituitary and stimulates release of the gonadotropins LH and FSH. Oxytocin is made in the hypothalamus and released from the posterior pituitary, and relaxin comes from the placenta and ovary.
Q35NEET 2017 · Q86
Hypersecretion of Growth Hormone in adults does not cause further increase in height, because:
Not quite — the answer is B.
Growth in length of long bones happens at the epiphyseal (growth) plates, which ossify and close after adolescence, so excess GH in adults cannot add height. It instead thickens bones of the face, hands and feet (acromegaly), showing the hormone is still active.
Q36NEET 2017 · Q87
Which ecosystem has the maximum biomass?
Not quite — the answer is A.
Forests hold by far the largest standing biomass because of their massive woody trees. Grasslands have small herbaceous producers, and ponds and lakes have tiny phytoplankton with low standing crop.
Q37NEET 2017 · Q88
Fruit and leaf drop at early stages can be prevented by the application of:
Not quite — the answer is C.
Auxins such as NAA prevent premature drop of fruits and leaves, while they promote abscission of older mature leaves and fruits. Ethylene does the opposite and promotes abscission.
Q38NEET 2017 · Q93
Thalassemia and sickle cell anemia are caused due to a problem in globin molecule synthesis. Select the correct statement.
Not quite — the answer is C.
Thalassemia is a quantitative problem in which too few globin chains are made. Sickle cell anaemia is a qualitative problem, an abnormal β-globin with valine in place of glutamic acid, so calling it quantitative is wrong.
Q39NEET 2017 · Q94
Which of the following is made up of dead cells?
Not quite — the answer is C.
Phellem (cork) cells are dead at maturity, with suberin deposited in their walls. Xylem parenchyma, collenchyma and the sieve tubes and companion cells of phloem are living (phloem fibres are the only dead part of phloem).
Q40NEET 2017 · Q96
Which of the following cell organelles is responsible for extracting energy from carbohydrates to form ATP?
Not quite — the answer is D.
Mitochondria are the sites of aerobic respiration, where the energy of carbohydrates is extracted and stored as ATP; they are called the powerhouses of the cell. Chloroplasts trap light energy to make carbohydrates rather than break them down.
Q41NEET 2017 · Q97
Capacitation occurs in:
Not quite — the answer is D.
Capacitation, the final activation of sperms that makes them able to fertilise the ovum, happens in the female reproductive tract. Sperms mature and are stored in the epididymis, but that is not capacitation.
Q42NEET 2017 · Q102
Select the correct route for the passage of sperms in male frogs:
Not quite — the answer is D.
In male frogs the vasa efferentia leave the testes and enter the kidney, where they open into Bidder's canal, which joins the urinogenital duct (ureter) that opens into the cloaca. Routes that place Bidder's canal before the vasa efferentia or skip the kidney do not match frog anatomy.
Q43NEET 2017 · Q103
Spliceosomes are not found in cells of:
Not quite — the answer is D.
Spliceosomes remove introns from eukaryotic hnRNA. Bacterial genes are not split into exons and introns, so bacteria have no spliceosomes, while plants, fungi and animals, all eukaryotes, do.
Q44NEET 2017 · Q104
Which one from those given below is the period for Mendel's hybridization experiments?
Not quite — the answer is A.
Mendel carried out his hybridisation experiments on garden pea for seven years, from 1856 to 1863, and published them in 1865. The other ranges do not match this period.
Q45NEET 2017 · Q105
The DNA fragments separated on an agarose gel can be visualised after staining with:
Not quite — the answer is D.
DNA in the gel is stained with ethidium bromide and seen as bright orange bands under UV light. Bromophenol blue is only a tracking dye that shows how far the run has gone, and does not stain DNA bands.
Q46NEET 2017 · Q106
The function of copper ions in copper releasing IUD's is:
Not quite — the answer is A.
Copper ions released by IUDs such as CuT suppress sperm motility and the fertilising capacity of sperms. Making the uterus unsuitable for implantation is the action of hormone-releasing IUDs, and stopping ovulation is the action of oral pills.
Q47NEET 2017 · Q107
Presence of plants arranged into well defined vertical layers depending on their height can be seen best in:
Not quite — the answer is B.
Vertical stratification, with emergent trees, canopy, understorey shrubs and ground herbs, is best developed in tropical rain forests. Grasslands and savannas have mostly herbaceous layers, and temperate forests have fewer layers.
Q48NEET 2017 · Q112
Which among the following are the smallest living cells, known without a definite cell wall, pathogenic to plants as well as animals and can survive without oxygen?
Not quite — the answer is C.
Mycoplasmas have no cell wall, are the smallest known living cells (about 0.3 μm), can survive without oxygen, and many cause diseases in plants and animals. Bacillus, Pseudomonas and Nostoc all have cell walls and are much larger.
Q49NEET 2017 · Q113
Root hairs develop from the region of:
Not quite — the answer is A.
Cells of the region of elongation gradually differentiate in the region of maturation, and some epidermal cells there form root hairs. The region of elongation is mainly responsible for growth in length, and the root cap only protects the tip.
Q50NEET 2017 · Q114
Flowers which have single ovule in the ovary and are packed into inflorescence are usually pollinated by:
Not quite — the answer is C.
Wind-pollinated flowers often have a single ovule in each ovary and many flowers packed into an inflorescence, as in corn cobs and grasses. Bee- and bat-pollinated flowers rely on showy flowers, scent and nectar instead.
Q51NEET 2017 · Q115
Receptor sites for neurotransmitters are present on:
Not quite — the answer is D.
Neurotransmitters released into the synaptic cleft bind to specific receptors on the post-synaptic membrane, opening ion channels there. The synaptic vesicles and pre-synaptic axon terminal store and release the transmitter but do not carry its receptors.
Q52NEET 2017 · Q116
Plants which produce characteristic pneumatophores and show vivipary belong to:
Not quite — the answer is B.
Mangroves such as Rhizophora are halophytes of salty swamps; they bear negatively geotropic breathing roots (pneumatophores) and show vivipary, seeds germinating on the parent plant. Hydrophytes are freshwater plants and psammophytes grow on sand.
Q53NEET 2017 · Q117
DNA replication in bacteria occurs:
Not quite — the answer is C.
Bacteria have no defined cell cycle phases or nucleolus; they replicate their circular DNA before the cell divides by binary fission. The S phase belongs to the eukaryotic cell cycle.
Q54NEET 2017 · Q118
The genotypes of a Husband and Wife are IᴬIᴮ and Iᴬi. Among the blood types of their children, how many different genotypes and phenotypes are possible?
Not quite — the answer is C.
Crossing IᴬIᴮ × Iᴬi gives IᴬIᴬ, Iᴬi, IᴬIᴮ and Iᴮi, which is 4 genotypes. IᴬIᴬ and Iᴬi are both blood group A, so the phenotypes are A, AB and B, which is 3. Group O cannot appear because no child can be ii.
Q55NEET 2017 · Q119
Which of the following components provides sticky character to the bacterial cell?
Not quite — the answer is D.
The glycocalyx, the outermost envelope as a loose slime layer or a thick capsule, makes the bacterial cell sticky. The cell wall gives shape and support, and bacteria have no nuclear membrane at all.
Q56NEET 2017 · Q120
Which of the following RNAs should be abundant in animal cell?
Not quite — the answer is A.
rRNA is the most abundant RNA in any cell, about 80% of total RNA, because it is the structural and catalytic part of the very large number of ribosomes. mRNA forms only a few percent, as it is short-lived and varies from gene to gene.
Q57NEET 2017 · Q123
Select the mismatch:
Not quite — the answer is B.
Rhodospirillum is a free-living, anaerobic photosynthetic nitrogen-fixing bacterium, while mycorrhiza is an association of a fungus with plant roots, so that pairing is the mismatch. Frankia forms root nodules on Alnus, Anabaena is a nitrogen-fixing cyanobacterium and Rhizobium nodulates legumes such as alfalfa.
Q58NEET 2017 · Q124
Double fertilization is exhibited by:
Not quite — the answer is D.
Double fertilisation, syngamy plus triple fusion to form the triploid primary endosperm nucleus, is a unique feature of angiosperms. Gymnosperms have a haploid endosperm formed before fertilisation.
Q59NEET 2017 · Q126
A temporary endocrine gland in the human body is:
Not quite — the answer is C.
The corpus luteum forms from the ruptured Graafian follicle after ovulation, secretes progesterone, and degenerates if pregnancy does not follow, so it is temporary. The pineal gland is permanent, and the corpus cardiacum and corpus allatum are insect glands.
Q60NEET 2017 · Q127
The vascular cambium normally gives rise to:
Not quite — the answer is C.
The vascular cambium cuts off secondary xylem towards the inside and secondary phloem towards the outside. Phelloderm and the rest of the periderm come from the cork cambium, and primary phloem comes from procambium.
Q61NEET 2017 · Q128
During DNA replication, Okazaki fragments are used to elongate:
Not quite — the answer is D.
DNA polymerase works only 5′→3′, so on the lagging strand it makes short Okazaki fragments in the direction away from the moving replication fork, and ligase joins them. The leading strand is made continuously towards the fork and needs no fragments.
Q62NEET 2017 · Q129
Artificial selection to obtain cows yielding higher milk output represents:
Not quite — the answer is B.
Choosing only high-yielding cows for breeding generation after generation moves the mean milk yield of the population steadily towards one extreme, which is directional selection. Stabilising selection favours the average value, and disruptive selection favours both extremes.
Q63NEET 2017 · Q131
Coconut fruit is a:
Not quite — the answer is A.
Coconut develops from a monocarpellary superior ovary and is one-seeded, with a fibrous mesocarp and a hard stony endocarp, which makes it a drupe like mango. A berry has a fleshy pericarp throughout, and a nut has a single hard woody pericarp.
Q64NEET 2017 · Q135
MALT constitutes about _____ percent of the lymphoid tissue in human body.
Not quite — the answer is A.
NCERT states that mucosa-associated lymphoid tissue (MALT), lining the respiratory, digestive and urogenital tracts, makes up about 50 percent of the lymphoid tissue in the human body. The other percentages do not match this figure.
Biology: not in the current NEET syllabus
These 8 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
8 questions. Tap an option to check your answer.
Q1NEET 2017 · Q49
Which one of the following statements is not valid for aerosols?
Not quite — the answer is C.
Aerosols (fine particles and droplets in air) harm human health, alter cloud formation, rainfall and monsoon patterns, and lower crop yields by cutting sunlight and depositing on land. They do not raise agricultural productivity, so that statement is the invalid one.
Q2NEET 2017 · Q51
Which cells of 'Crypts of Lieberkuhn' secrete antibacterial lysozyme?
Not quite — the answer is B.
Paneth cells at the base of the intestinal crypts of Lieberkühn secrete lysozyme, an antibacterial enzyme. Argentaffin cells are enteroendocrine cells, zymogen cells are found in gastric glands, and Kupffer cells are macrophages of the liver.
Q3NEET 2017 · Q60
Which of the following facilitates opening of stomatal aperture?
Not quite — the answer is C.
Cellulose microfibrils in guard cell walls are oriented radially, so when the cells become turgid the outer walls bow out and the inner walls are pulled into a crescent, opening the pore. A loss of turgor closes the stoma.
Q4NEET 2017 · Q95
A baby boy aged two years is admitted to play school and passes through a dental check - up. The dentist observed that the boy had twenty teeth. Which teeth were absent?
Not quite — the answer is C.
The 20 milk (deciduous) teeth have the dental formula 2102/2102, with no premolars; the 12 premolars appear only among the 32 permanent teeth. Incisors, canines and molars are all present in the milk set.
Q5NEET 2017 · Q110
Zygotic meiosis is characteristic of:
Not quite — the answer is D.
Zygotic meiosis means the zygote is the only diploid stage and divides meiotically at once, the haplontic life cycle seen in Chlamydomonas. Fucus is diplontic (gametic meiosis), while Marchantia and Funaria are haplo-diplontic with sporic meiosis.
Q6NEET 2017 · Q111
Life cycle of Ectocarpus and Fucus respectively are:
Not quite — the answer is C.
Ectocarpus is haplo-diplontic, with free-living multicellular haploid and diploid phases, while Fucus is diplontic, the gametes being the only haploid stage. Calling Fucus haplontic or haplo-diplontic is the trap.
Q7NEET 2017 · Q130
Which of the following options best represents the enzyme composition of pancreatic juice?
Not quite — the answer is D.
Pancreatic juice contains inactive trypsinogen, chymotrypsinogen and procarboxypeptidase, along with amylase, lipases and nucleases. Pepsin and rennin are gastric enzymes and maltase comes from intestinal juice, so groups containing them are wrong.
Q8NEET 2017 · Q132
The water potential of pure water is:
Not quite — the answer is A.
By convention, pure water at standard temperature and atmospheric pressure has the highest water potential, set at zero. Adding any solute lowers it, so solutions have negative values.
Chapter-wise split of NEET 2017 Biology
- Morphology of Flowering Plants5 Q
- Molecular Basis of Inheritance5 Q
- Sexual Reproduction in Flowering Plants5 Q
- Biotechnology: Principles and Processes4 Q
- Animal Kingdom3 Q
- Anatomy of Flowering Plants3 Q
- Principles of Inheritance and Variation3 Q
- Cell: The Unit of Life3 Q
- Microbes in Human Welfare2 Q
- Biodiversity and Conservation2 Q
- Biological Classification2 Q
- Plant Kingdom2 Q
- Excretory Products and Their Elimination2 Q
- Body Fluids and Circulation2 Q
- Organisms and Populations2 Q
- Human Health and Disease2 Q
- Neural Control and Coordination2 Q
- Human Reproduction2 Q
- Chemical Coordination and Integration2 Q
- Ecosystem2 Q
- Photosynthesis in Higher Plants1 Q
- Breathing and Exchange of Gases1 Q
- Biomolecules1 Q
- Respiration in Plants1 Q
- Locomotion and Movement1 Q
- Plant Growth and Development1 Q
- Structural Organisation in Animals1 Q
- Reproductive Health1 Q
- Evolution1 Q
About this paper
How many questions from NEET 2017 are on this page?
129 questions (Physics 20, Chemistry 37, Biology 72). 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 2017 paper, conducted by CBSE. The answer to every question was checked by the RankUp team, and the explanations are RankUp's own.
Is NEET 2017 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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