NEET 2017 (Gujarati paper) 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
147 questions from the NEET 2017 (Gujarati paper) paper, conducted by CBSE. Tap an option to check your answer and read why it is right. Each subject has a free PDF with the answer key and explanations.
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NEET 2017 (Gujarati paper) Physics questions
25 questions. Tap an option to check your answer.
Q1NEET 2017 (Gujarati paper) · Q136
The energy liberated per nuclear fission is 200 MeV. If 10²⁰ fissions occur per second the amount of power produced will be:
Not quite — the answer is A.
200 MeV = 200 × 10⁶ × 1.6 × 10⁻¹⁹ J = 3.2 × 10⁻¹¹ J. Power = 10²⁰ s⁻¹ × 3.2 × 10⁻¹¹ J = 3.2 × 10⁹ W = 32 × 10⁸ W. Forgetting to convert MeV to joules gives a value like 2 × 10²² W.
Q2NEET 2017 (Gujarati paper) · Q137
If the angle of a prism is 60° and angle of minimum deviation is 40°, then the angle of refraction will be:
Not quite — the answer is A.
At minimum deviation the ray passes symmetrically, so r₁ = r₂ and r₁ + r₂ = A. Hence r = A/2 = 60°/2 = 30°, whatever the deviation. 40° belongs to the angle of incidence calculation, i = (A + δ)/2 = 50°, not to r.
Q3NEET 2017 (Gujarati paper) · Q138
In an experiment of photoelectric effect the stopping potential was measured to be V₁ and V₂ with incident light of wavelength λ and λ/2, respectively. The relation between V₁ and V₂ is:
Not quite — the answer is D.
eV₁ = hc/λ − φ and eV₂ = 2hc/λ − φ = 2(eV₁ + φ) − φ = 2eV₁ + φ. Since the work function φ > 0, V₂ > 2V₁. V₂ = 2V₁ would hold only if the work function were zero.
Q4NEET 2017 (Gujarati paper) · Q140
A body initially at rest, breaks up into two pieces of masses 2M and 3M respectively, together having a total kinetic energy E. The piece of mass 2M, after breaking up, has a kinetic energy:
Not quite — the answer is C.
Momentum conservation gives equal and opposite momenta p, and KE = p²/2m, so kinetic energy is inversely proportional to mass. KE(2M) : KE(3M) = 3 : 2, so the 2M piece gets 3E/5. 2E/5 is what you get by wrongly sharing the energy in proportion to mass.
Q5NEET 2017 (Gujarati paper) · Q141
A satellite of mass m is in circular orbit of radius 3R_E about earth (mass of earth M_E, radius of earth R_E). How much additional energy is required to transfer the satellite to an orbit of radius 9R_E?
Not quite — the answer is C.
Total energy in a circular orbit is −GMm/2r. ΔE = GM_E m/2 × (1/3R_E − 1/9R_E) = GM_E m/2 × 2/(9R_E) = GM_E m/(9R_E). Using only potential energies would double this to 2GM_E m/(9R_E).
Q6NEET 2017 (Gujarati paper) · Q142
A thin uniform rod of mass 'M' and length 'L' is rotating about a perpendicular axis passing through its centre with a constant angular velocity 'ω'. Two objects each of mass M/3 are attached gently to the two ends of the rod. The rod will now rotate with an angular velocity of:
Not quite — the answer is D.
I₁ = ML²/12. Each added mass contributes (M/3)(L/2)² = ML²/12, so I₂ = ML²/12 + 2 × ML²/12 = 3ML²/12. With no external torque, I₁ω = I₂ω′, so ω′ = ω/3. Taking each mass at the full length L instead of L/2 leads to the wrong value ω/9.
Q7NEET 2017 (Gujarati paper) · Q144
Imagine earth to be a solid sphere of mass M and radius R. If the value of acceleration due to gravity at a depth 'd' below earth's surface is same as its value at a height 'h' above its surface and equal to g/4 (where g is the value of acceleration due to gravity on the surface of earth), the ratio of h/d will be:
Not quite — the answer is A.
Below the surface g(1 − d/R) = g/4 gives d = 3R/4. Above the surface g/(1 + h/R)² = g/4 gives 1 + h/R = 2, so h = R. Hence h/d = R/(3R/4) = 4/3. Using the small-height formula g(1 − 2h/R) would be wrong here because h is not small compared with R.
Q8NEET 2017 (Gujarati paper) · Q145
A parallel-plate capacitor is to be designed, using a dielectric of dielectric constant 5, so as to have a dielectric strength of 10⁹ V m⁻¹. If the voltage rating of the capacitor is 12 kV, the minimum area of each plate required to have a capacitance of 80 pF is:
Not quite — the answer is A.
The minimum plate separation is d = V/E_max = 12 × 10³ / 10⁹ = 1.2 × 10⁻⁵ m. Then A = Cd/(Kε₀) = (80 × 10⁻¹² × 1.2 × 10⁻⁵)/(5 × 8.85 × 10⁻¹²) ≈ 2.17 × 10⁻⁵ m² = 21.7 × 10⁻⁶ m². Leaving out the dielectric constant would make the area five times larger.
Q9NEET 2017 (Gujarati paper) · Q146
Inside a parallel plate capacitor the electric field E varies with time as t². The variation of induced magnetic field with time is given by:
Not quite — the answer is C.
The induced magnetic field comes from the displacement current, I_d = ε₀ dΦ_E/dt, which is proportional to dE/dt. If E ∝ t², then dE/dt ∝ t, so B ∝ t. B ∝ t² would wrongly assume the field depends on E itself rather than on its rate of change.
Q10NEET 2017 (Gujarati paper) · Q147
A ball of mass 1 kg is thrown vertically upwards and returns to the ground after 3 seconds. Another ball, thrown at 60° with vertical also stays in air for the same time before it touches the ground. The ratio of the two heights are:
Not quite — the answer is B.
Time of flight T = 2u_y/g depends only on the vertical component of velocity, and maximum height H = u_y²/2g = gT²/8. Equal flight times mean equal vertical components, so both balls rise to the same height, 1 : 1 (each H = 10 × 9/8 ≈ 11.25 m with g = 10 m s⁻²). The 60° angle changes only the horizontal motion.
Q11NEET 2017 (Gujarati paper) · Q149
A light beam is incident on a denser medium whose refractive index is 1.414 at an angle of incidence 45°. Find the ratio of width of refracted beam in a medium to the width of the incident beam in air.
Not quite — the answer is D.
sin r = sin 45°/1.414 = 0.5, so r = 30°. A beam crossing a boundary strip of length x has width x cos i in air and x cos r in the medium, so the ratio is cos 30°/cos 45° = (√3/2)/(1/√2) = √3 : √2. The inverse ratio √2 : √3 comes from swapping the two angles.
Q12NEET 2017 (Gujarati paper) · Q152
A cyclist on a level road takes a sharp circular turn of radius 3 m (g = 10 m s⁻²). If the coefficient of static friction between the cycle tyres and the road is 0.2, at which of the following speeds will the cyclist not skid while taking the turn?
Not quite — the answer is A.
Friction must supply the centripetal force, so v_max = √(μgr) = √(0.2 × 10 × 3) = √6 ≈ 2.45 m s⁻¹ ≈ 8.8 km h⁻¹. Only 7.2 km h⁻¹ (2 m s⁻¹) is below this limit; 9 km h⁻¹ (2.5 m s⁻¹) is just above it, so the cyclist would skid.
Q13NEET 2017 (Gujarati paper) · Q154
Two coherent sources of intensity ratio α interfere. The value of (I_max − I_min)/(I_max + I_min) is:
Not quite — the answer is A.
Let I₁ = αI₂. I_max = (√I₁ + √I₂)² and I_min = (√I₁ − √I₂)², so I_max − I_min = 4√(I₁I₂) and I_max + I_min = 2(I₁ + I₂). The ratio is 2√(I₁I₂)/(I₁ + I₂) = 2√α/(1 + α). The inverse form (1 + α)/(2√α) is always at least 1, which a ratio of this kind cannot be.
Q14NEET 2017 (Gujarati paper) · Q155
In a certain planetary system, it is observed that one of the celestial bodies having a surface temperature of 200 K, emits radiation of maximum intensity near the wavelength 12 μm. The surface temperature of a nearby star which emits light of maximum intensity at a wavelength λ = 4800 Å, is:
Not quite — the answer is A.
By Wien's law λ_max T is constant: 12 × 10⁻⁶ m × 200 K = 2.4 × 10⁻³ m K. For the star T = 2.4 × 10⁻³ / 4800 × 10⁻¹⁰ = 5000 K. Getting 10000 K means 4800 Å was taken as 2.4 × 10⁻⁷ m, a unit slip.
Q15NEET 2017 (Gujarati paper) · Q157
A bulb connected in series with an air-cored solenoid is lit by an a.c. source. If a soft iron core is introduced in the solenoid:
Not quite — the answer is C.
Soft iron raises the permeability and hence the inductance L, so the inductive reactance ωL and the circuit impedance increase. The rms current falls and the bulb glows dimmer. It does not go out completely, because the impedance is still finite.
Q16NEET 2017 (Gujarati paper) · Q159
Two open organ pipes of fundamental frequencies n₁ and n₂ are joined in series. The fundamental frequency of the new pipe so obtained will be:
Not quite — the answer is C.
For an open pipe n = v/2L, so L₁ = v/2n₁ and L₂ = v/2n₂. The joined pipe has length L₁ + L₂, so n = v/2(L₁ + L₂) and 1/n = 1/n₁ + 1/n₂, giving n = n₁n₂/(n₁ + n₂). A longer pipe must have a lower frequency, so n₁ + n₂ cannot be right.
Q17NEET 2017 (Gujarati paper) · Q161
Two reasons for using soft iron as the material for electromagnets.
Not quite — the answer is A.
An electromagnet should become strongly magnetised by a small current (high permeability) and lose its magnetism when the current is switched off (low retentivity and coercivity). High retentivity suits permanent magnets such as steel, not electromagnets.
Q18NEET 2017 (Gujarati paper) · Q163
One mole of a gas obeying the equation of state P(V − b) = RT is made to expand from a state with coordinates (P₁, V₁) to a state with (P₂, V₂) along a process that is depicted by a straight line on a P–V diagram. Then, the work done is given by:
Not quite — the answer is A.
Work done is the area under the P–V path, whatever the equation of state. For a straight line from (V₁, P₁) to (V₂, P₂) the area is a trapezium: W = ½(P₁ + P₂)(V₂ − V₁). The constant b does not enter, because the work uses the actual volume change.
Q19NEET 2017 (Gujarati paper) · Q164
A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. The induced e.m.f. in the coil is: (Take H_E = 3.0 × 10⁻⁵ T)
Not quite — the answer is C.
A 180° turn reverses the flux, so ΔΦ = 2NBA = 2 × 500 × 3.0 × 10⁻⁵ × π(0.1)² ≈ 9.42 × 10⁻⁴ Wb. Average emf = ΔΦ/Δt = 9.42 × 10⁻⁴ / 0.25 ≈ 3.8 × 10⁻³ V. The 2 Ω resistance affects the current, not the emf; dropping the factor 2 would give about 1.9 × 10⁻³ V.
Q20NEET 2017 (Gujarati paper) · Q166
Due to Doppler effect, the shift in wavelength observed is 0.1 Å, for a star producing a wavelength 6000 Å. The velocity of recession of the star will be:
Not quite — the answer is C.
For v ≪ c, Δλ/λ = v/c, so v = c × Δλ/λ = 3 × 10⁸ × 0.1/6000 = 5 × 10³ m s⁻¹ = 5 km s⁻¹. Doubling the shift by mistake would give 10 km s⁻¹.
Q21NEET 2017 (Gujarati paper) · Q169
The angle between A⃗ − B⃗ and A⃗ × B⃗ is (A⃗ ≠ B⃗):
Not quite — the answer is A.
A⃗ × B⃗ is perpendicular to the plane containing A⃗ and B⃗, and A⃗ − B⃗ lies in that plane. So (A⃗ − B⃗)·(A⃗ × B⃗) = 0 and the angle is 90° for any non-parallel A⃗ and B⃗.
Q22NEET 2017 (Gujarati paper) · Q172
If the longest wavelength in the ultraviolet region of hydrogen spectrum is λ₀ then the shortest wavelength in its infrared region is:
Not quite — the answer is C.
The longest Lyman (UV) line is n = 2 → 1: 1/λ₀ = R(1 − 1/4) = 3R/4. The shortest infrared wavelength is the Paschen series limit, ∞ → 3: 1/λ = R/9, so λ = 9/R = 9 × (3λ₀/4) = 27λ₀/4. 36/5 λ₀ is the longest Balmer line (3 → 2), which is in the visible region.
Q23NEET 2017 (Gujarati paper) · Q173
A parallel beam of light of wavelength λ is incident normally on a single slit of width d. Diffraction bands are obtained on a screen placed at a distance D from the slit. The second dark band from the central bright band will be at a distance given by:
Not quite — the answer is D.
Single-slit minima occur at d sin θ = nλ. For n = 2 and small angles, y = D tan θ ≈ 2λD/d. The form 2λd/D has the slit width and screen distance swapped and does not even have units of length.
Q24NEET 2017 (Gujarati paper) · Q176
The rotational kinetic energy of a solid sphere of mass 3 kg and radius 0.2 m rolling down an inclined plane of height 7 m is:
Not quite — the answer is A.
Total KE at the bottom = mgh = 3 × 10 × 7 = 210 J (taking g = 10 m s⁻²). For a solid sphere, K_rot/K_total = (2/5)/(1 + 2/5) = 2/7, so K_rot = 210 × 2/7 = 60 J. 70 J would be one third of the energy, the fraction for a different shape.
Q25NEET 2017 (Gujarati paper) · Q180
A student performs an experiment of measuring the thickness of a slab with a vernier calliper whose 50 divisions of the vernier scale are equal to 49 divisions of the main scale. He noted that zero of the vernier scale is between 7.00 cm and 7.05 cm mark of the main scale and 23rd division of the vernier scale exactly coincides with the main scale. The measured value of the thickness of the given slab using the calliper will be:
Not quite — the answer is B.
One main scale division is 0.05 cm (the marks are at 7.00 and 7.05 cm), so the least count = 1 MSD − 1 VSD = 0.05/50 = 0.001 cm. Reading = 7.00 + 23 × 0.001 = 7.023 cm. 7.23 cm would wrongly take the least count as 0.01 cm.
Chapter-wise split of NEET 2017 (Gujarati paper) Physics
- Wave Optics3 Q
- Ray Optics and Optical Instruments2 Q
- Laws of Motion2 Q
- Gravitation2 Q
- System of Particles and Rotational Motion2 Q
- Motion in a Plane2 Q
- Nuclei1 Q
- Dual Nature of Radiation and Matter1 Q
- Electrostatic Potential and Capacitance1 Q
- Electromagnetic Waves1 Q
- Thermal Properties of Matter1 Q
- Alternating Current1 Q
- Waves1 Q
- Magnetism and Matter1 Q
- Thermodynamics1 Q
- Electromagnetic Induction1 Q
- Atoms1 Q
- Units and Measurements1 Q
NEET 2017 (Gujarati paper) Chemistry questions
31 questions. Tap an option to check your answer.
Q1NEET 2017 (Gujarati paper) · Q1
The electron distribution in dⁿ coordination complexes depends on magnitude of crystal field splitting, (Δₒ) and pairing energy (P). The condition which favours formation of high spin complexes is:
Not quite — the answer is B.
When the crystal field splitting is smaller than the pairing energy, it costs less energy to put electrons in the higher eg set than to pair them in t₂g, so electrons stay unpaired and a high spin complex forms. When Δₒ exceeds P, pairing is preferred and a low spin complex results.
Q2NEET 2017 (Gujarati paper) · Q2
Toluene in the vapour phase is in equilibrium with a solution of benzene and toluene having mole fraction of toluene 0.50. If vapour pressure of pure benzene is 119 torr and that of toluene is 37.0 torr at the same temperature, mole fraction of toluene in vapour phase will be:
Not quite — the answer is B.
By Raoult's law, p(benzene) = 0.5 × 119 = 59.5 torr and p(toluene) = 0.5 × 37.0 = 18.5 torr, so total pressure = 78.0 torr. Mole fraction of toluene in vapour = 18.5/78.0 = 0.237. The vapour is richer in the more volatile benzene, so the toluene fraction must be below 0.50, which rules out 0.506.
Q3NEET 2017 (Gujarati paper) · Q4
Which one of the following ions is not tetrahedral in shape?
Not quite — the answer is C.
[Cu(NH₃)₄]²⁺ is a d⁹ dsp² complex and is square planar. NH₄⁺ and BF₄⁻ are sp³ tetrahedral, and [NiCl₄]²⁻ with the weak-field Cl⁻ ligand is sp³ tetrahedral.
Q4NEET 2017 (Gujarati paper) · Q7
Amongst the following compounds the one which is most easily sulphonated is:
Not quite — the answer is C.
Sulphonation is electrophilic substitution, so the ring with the most electron-donating group reacts fastest. The methyl group of toluene activates the ring, while –NO₂ and –Cl deactivate it compared with benzene.
Q5NEET 2017 (Gujarati paper) · Q8
For the reaction, XA + YB → ZC, if −d[A]/dt = −d[B]/dt = 1.5 d[C]/dt, then the correct statement among the following is:
Not quite — the answer is B.
Rate = −(1/X)d[A]/dt = −(1/Y)d[B]/dt = (1/Z)d[C]/dt. Equal rates of loss of A and B give X = Y, and −d[A]/dt = 1.5 d[C]/dt gives X = 1.5Z, so the smallest whole numbers are X = Y = 3 and Z = 2. Z cannot also be 3, because then the rates would be equal, not in the ratio 1.5.
Q6NEET 2017 (Gujarati paper) · Q9
If the rate of the reaction: R–CO–Z + Nu⁻ → R–CO–Nu + Z⁻ is fastest, then Z is:
Not quite — the answer is A.
Nucleophilic acyl substitution is fastest when Z⁻ is the best leaving group, i.e. the weakest base. Cl⁻ (conjugate base of the strong acid HCl) leaves most easily, so acid chlorides are most reactive; acetate is next, while ethoxide and especially NH₂⁻ are poor leaving groups.
Q7NEET 2017 (Gujarati paper) · Q11
Of the following alcohols, the one that would react fastest with conc. HCl and anhydrous ZnCl₂ is:
Not quite — the answer is C.
The Lucas reagent reacts through a carbocation, so the tertiary alcohol 2-methylpropan-2-ol, which forms the most stable tertiary carbocation, reacts at once. Secondary butan-2-ol reacts in minutes, and the primary alcohols barely react at room temperature.
Q8NEET 2017 (Gujarati paper) · Q12
Of the following, the largest value of entropy at 25°C and 1 atm is that of:
Not quite — the answer is B.
Standard molar entropy rises with molar mass and the number of atoms (more vibrational and rotational freedom). C₂H₆ (about 229 J K⁻¹ mol⁻¹) exceeds C₂H₂ (about 201), CH₄ (about 186) and H₂ (about 131).
Q9NEET 2017 (Gujarati paper) · Q13
Which of the following hydrides has the largest bond angle?
Not quite — the answer is C.
Oxygen is the smallest and most electronegative central atom, so the bonding pairs sit close to it and repel strongly, keeping the angle near 104.5°. Down the group the bonding pairs move away and the angle falls towards 90° (H₂S about 92°, H₂Te about 90°).
Q10NEET 2017 (Gujarati paper) · Q14
The [Co(H₂O)₆]²⁺ ion has three unpaired electrons. The hybridization of Co in [Co(H₂O)₆]²⁺ is:
Not quite — the answer is C.
Co²⁺ is 3d⁷. Three unpaired electrons means the weak-field water ligands do not force pairing, so the inner 3d orbitals are not free and the six ligands use 4s, 4p and 4d orbitals: sp³d² (outer orbital, octahedral). d²sp³ would need pairing down to one unpaired electron.
Q11NEET 2017 (Gujarati paper) · Q15
A hydrocarbon contains 85.7% C. If 42 mg of the compound contains 3.01 × 10²⁰ molecules, the molecular formula of the compound will be:
Not quite — the answer is B.
Moles = 3.01 × 10²⁰ / 6.02 × 10²³ = 5 × 10⁻⁴ mol, so molar mass = 0.042 g / 5 × 10⁻⁴ mol = 84 g mol⁻¹. 85.7% C corresponds to the empirical formula CH₂ (14 g mol⁻¹), and 84/14 = 6 gives C₆H₁₂. All options have 85.7% C, so only the molar mass separates them.
Q12NEET 2017 (Gujarati paper) · Q16
For the tetrahedral complex [MnBr₄]²⁻, the spin only magnetic moment value is: [At. no. of Mn = 25]
Not quite — the answer is B.
Mn²⁺ is 3d⁵, and in a tetrahedral field with weak Br⁻ ligands all five electrons stay unpaired. μ = √(n(n + 2)) = √35 ≈ 5.9 BM.
Q13NEET 2017 (Gujarati paper) · Q21
Which of the following lanthanoids shows +4 oxidation state to acquire noble gas configuration? (At. nos.: La = 57, Ce = 58, Eu = 63 and Yb = 70)
Not quite — the answer is A.
Ce is [Xe]4f¹5d¹6s²; losing four electrons gives Ce⁴⁺ with the noble gas configuration [Xe]. Eu and Yb show +2 to reach stable 4f⁷ and 4f¹⁴, and La shows only +3, which already gives [Xe].
Q14NEET 2017 (Gujarati paper) · Q22
Consider the following sequence of reactions: Benzene —(CO, HCl, anhydrous AlCl₃/CuCl)→ A —('B', OH⁻, 293 K)→ C₆H₅–CH=CH–CO–C₆H₅. The substance 'B' is:
Not quite — the answer is B.
The Gattermann–Koch reaction converts benzene into benzaldehyde, the intermediate A. A base-catalysed cross aldol (Claisen–Schmidt) condensation of benzaldehyde with acetophenone, C₆H₅COCH₃, then gives benzalacetophenone, C₆H₅CH=CHCOC₆H₅. Acetone would give C₆H₅CH=CHCOCH₃, not the diphenyl product.
Q15NEET 2017 (Gujarati paper) · Q24
Which of the following complex ions is not diamagnetic?
Not quite — the answer is B.
Cr³⁺ is 3d³, with three unpaired electrons in t₂g, so [Cr(NH₃)₆]³⁺ is paramagnetic. Ti⁴⁺ and Sc³⁺ are d⁰ and Zn²⁺ is d¹⁰, so those complexes have no unpaired electrons and are diamagnetic.
Q16NEET 2017 (Gujarati paper) · Q26
Among halogens, the one which can oxidise water to oxygen is:
Not quite — the answer is C.
Fluorine is the strongest oxidising agent (E° = +2.87 V) and oxidises water: 2F₂ + 2H₂O → 4HF + O₂. Chlorine and bromine react with water only to form hypohalous and hydrohalic acids, and iodine is too weak.
Q17NEET 2017 (Gujarati paper) · Q27
The total number of orbitals present for principle quantum number, n = 4 is:
Not quite — the answer is C.
The number of orbitals in a shell is n², so n = 4 has 16 orbitals (one 4s, three 4p, five 4d, seven 4f). 32 would be the number of electrons, 2n².
Q18NEET 2017 (Gujarati paper) · Q28
Given that Λ°ₘ = 133.4 (AgNO₃); Λ°ₘ = 149.9 (KCl); Λ°ₘ = 144.9 S cm² mol⁻¹ (KNO₃), the molar conductivity at infinite dilution for AgCl is:
Not quite — the answer is B.
By Kohlrausch's law, Λ°(AgCl) = Λ°(AgNO₃) + Λ°(KCl) − Λ°(KNO₃) = 133.4 + 149.9 − 144.9 = 138.4 ≈ 138 S cm² mol⁻¹. Adding all three, or subtracting the wrong salt, gives none of the ion combinations needed.
Q19NEET 2017 (Gujarati paper) · Q29
Consider the reaction between chlorine and nitric oxide Cl₂(g) + 2NO(g) → 2NOCl(g). On doubling the concentration of both reactants, the rate of the reaction increases by a factor of 8. However, if only the concentration of Cl₂ is doubled, the rate increases by a factor of 2. The order of this reaction with respect to NO is:
Not quite — the answer is C.
Let rate = k[Cl₂]ˣ[NO]ʸ. Doubling Cl₂ alone doubles the rate, so x = 1. Doubling both gives 2¹ × 2ʸ = 8, so 2ʸ = 4 and y = 2. The overall order is 3, which is the trap.
Q20NEET 2017 (Gujarati paper) · Q30
Strong reducing behaviour of H₃PO₂ is due to:
Not quite — the answer is B.
Hypophosphorous acid, H₃PO₂, has one P–OH and two P–H bonds; the P–H bonds are what make phosphorus oxyacids reducing, and with two of them H₃PO₂ is a strong reducing agent. Two –OH groups and one P–H bond describe H₃PO₃, a weaker reductant.
Q21NEET 2017 (Gujarati paper) · Q31
Under isothermal and reversible conditions, the term "free energy" in thermodynamics signifies:
Not quite — the answer is A.
At constant temperature and pressure for a reversible change, −ΔG equals the maximum useful (non-expansion) work the system can do, such as electrical work in a cell. Expansion (PV) work is already accounted for in the enthalpy term and is not what ΔG measures.
Q22NEET 2017 (Gujarati paper) · Q32
The standard equilibrium constant, Kp at 298 K for the reaction, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 5.8 × 10⁵. The value of standard equilibrium constant, if the concentration of gases is expressed in terms of mol/L, will be: [Given: R = 0.08314 L bar K⁻¹ mol⁻¹]
Not quite — the answer is C.
Kp = Kc(RT)^Δn with Δn = 2 − 4 = −2, so Kc = Kp(RT)². RT = 0.08314 × 298 = 24.78, (RT)² = 613.9, and Kc = 5.8 × 10⁵ × 613.9 ≈ 3.56 × 10⁸. Dividing by (RT)² instead would give a much smaller value.
Q23NEET 2017 (Gujarati paper) · Q33
Among the following acids, the strongest acid is:
Not quite — the answer is C.
Electron-withdrawing groups on the α-carbon stabilise the carboxylate ion. Three fluorine atoms, the most electronegative, withdraw electrons most strongly, making trifluoroacetic acid the strongest (pKa about 0.2), ahead of trichloroacetic acid and the single-substituent acids.
Q24NEET 2017 (Gujarati paper) · Q34
The letter 'D' in D-glucose signifies:
Not quite — the answer is D.
'D' refers to the relative configuration of the lowest (penultimate) chiral carbon, C-5 in glucose, which matches D-glyceraldehyde (–OH on the right). It says nothing about the sign of rotation; that is shown by (+) or (−), and D-fructose, for example, is laevorotatory.
Q25NEET 2017 (Gujarati paper) · Q37
The tendency to form monovalent compounds among the Group 13 elements is correctly exhibited in:
Not quite — the answer is A.
Down group 13 the ns² electrons become increasingly reluctant to take part in bonding (inert pair effect), so the +1 state becomes more stable from B to Tl. Thallium mainly forms Tl(I) compounds, while boron hardly shows +1 at all.
Q26NEET 2017 (Gujarati paper) · Q38
Consider the following reaction for which the change in enthalpy is positive. 2A(g) + B(g) ⇌ C(g) + D(g). Which of the following will not affect the equilibrium?
Not quite — the answer is D.
A catalyst speeds up the forward and backward reactions equally, so equilibrium is reached faster but its position and K do not change. Since Δn = −1, pressure shifts the equilibrium; changing temperature changes K for this endothermic reaction; and changing reactant concentration shifts the position.
Q27NEET 2017 (Gujarati paper) · Q40
Which of the following will react faster through Sɴ1 mechanism?
Not quite — the answer is A.
Sɴ1 rate depends on carbocation stability. Allyl chloride ionises to the resonance-stabilised allyl cation, so it reacts fastest. Ethyl chloride would give an unstable primary cation, and vinyl and aryl chlorides have a C–Cl bond with partial double-bond character and do not ionise easily.
Q28NEET 2017 (Gujarati paper) · Q41
For the reaction CO(g) + Cl₂(g) ⇌ COCl₂(g), Kp/Kc is equal to:
Not quite — the answer is A.
Kp = Kc(RT)^Δn, and here Δn = 1 − 2 = −1. So Kp/Kc = (RT)⁻¹ = 1/RT. RT would be the answer if the number of gas moles increased by one.
Q29NEET 2017 (Gujarati paper) · Q42
Which of the following pair of species is not iso-structural?
Not quite — the answer is B.
ClO₃⁻ has three bond pairs and one lone pair on Cl (sp³), so it is trigonal pyramidal, whereas CO₃²⁻ is sp² trigonal planar. ICl₄⁻ and XeF₄ are both square planar, IBr₂⁻ and XeF₂ both linear, and BrO₃⁻ and XeO₃ both pyramidal.
Q30NEET 2017 (Gujarati paper) · Q43
Which of the following amino acid is not optically active?
Not quite — the answer is D.
Glycine, H₂N–CH₂–COOH, has two hydrogens on its α-carbon, so it has no chiral centre and is optically inactive. Proline, serine and leucine all have an α-carbon with four different groups.
Q31NEET 2017 (Gujarati paper) · Q44
The correct statement regarding ethane conformation is:
Not quite — the answer is C.
Ethane has only a C–C σ bond, and a σ bond is cylindrically symmetric about the bond axis, so its overlap does not change as the groups rotate; this free rotation produces the staggered and eclipsed conformations. Ethane has no π bond, so the statements saying rotation is blocked by a π bond are false.
Chemistry: not in the current NEET syllabus
These 6 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
6 questions. Tap an option to check your answer.
Q1NEET 2017 (Gujarati paper) · Q17
Which of the following statements is incorrect?
Not quite — the answer is B.
By the Hardy–Schulze rule, coagulating power rises with the charge of the oppositely charged ion, so for a positive sol the correct order is [Fe(CN)₆]⁴⁻ > PO₄³⁻ > SO₄²⁻ > Cl⁻, the reverse of the statement. The order for cations acting on a negative sol is correctly stated.
Q2NEET 2017 (Gujarati paper) · Q18
Which of these artificial sweetener is unstable at cooking temperature?
Not quite — the answer is A.
Aspartame, a methyl ester of a dipeptide, breaks down on heating and is used only in cold foods and drinks. Sucralose, saccharin and alitame are heat-stable.
Q3NEET 2017 (Gujarati paper) · Q20
Match the polymer in Column-I to the monomer from Column-II and assign the correct code: Column-I (Polymer): (a) Nylon-6, (b) Dacron, (c) Glyptal, (d) Novolac. Column-II (Monomer): (i) ethylene glycol, terephthalic acid; (ii) urea, formaldehyde; (iii) ethylene glycol, phthalic acid; (iv) phenol, formaldehyde; (v) caprolactam.
Not quite — the answer is A.
Nylon-6 is made from caprolactam, Dacron from ethylene glycol and terephthalic acid, Glyptal from ethylene glycol and phthalic acid, and Novolac from phenol and formaldehyde. Urea–formaldehyde gives a different resin, so any code pairing Novolac with it is wrong.
Q4NEET 2017 (Gujarati paper) · Q25
Which of the following absorbs carbon dioxide and releases oxygen?
Not quite — the answer is B.
Potassium superoxide reacts as 4KO₂ + 2CO₂ → 2K₂CO₃ + 3O₂, so it removes CO₂ and supplies O₂ (used in breathing apparatus). CaO, KOH and K₂O absorb CO₂ but release no oxygen.
Q5NEET 2017 (Gujarati paper) · Q35
For dry cleaning of clothes instead of tetrachloroethane which is carcinogen in nature, which of the following solvents can be used?
Not quite — the answer is A.
Liquid (supercritical) CO₂ with a suitable detergent is the green-chemistry replacement for chlorinated dry-cleaning solvents. It is non-toxic and non-flammable, unlike petrol, and does not bleach fabrics the way H₂O₂ does.
Q6NEET 2017 (Gujarati paper) · Q36
Depressant used in the concentration of an ore containing ZnS and PbS is:
Not quite — the answer is C.
In froth flotation of a mixed ZnS/PbS ore, NaCN forms a complex on the ZnS surface and stops it from frothing, so PbS floats off first. Na₂CO₃, NaCl and Na₂SO₄ do not act selectively in this way.
Chapter-wise split of NEET 2017 (Gujarati paper) Chemistry
- Coordination Compounds5 Q
- Aldehydes, Ketones and Carboxylic Acids3 Q
- Equilibrium3 Q
- Hydrocarbons2 Q
- Chemical Kinetics2 Q
- Thermodynamics2 Q
- Chemical Bonding and Molecular Structure2 Q
- Biomolecules2 Q
- Solutions1 Q
- Alcohols, Phenols and Ethers1 Q
- Some Basic Concepts of Chemistry1 Q
- The d- and f-Block Elements1 Q
- Structure of Atom1 Q
- Electrochemistry1 Q
- Haloalkanes and Haloarenes1 Q
NEET 2017 (Gujarati paper) Biology questions
75 questions. Tap an option to check your answer.
Q1NEET 2017 (Gujarati paper) · Q47
The correct sequence of involvement of cell organelles in secretion of proteins from the cell is:
Not quite — the answer is A.
mRNA made in the nucleus is translated on ribosomes attached to the ER; the protein enters the ER lumen, is passed to the Golgi for modification and packaging, and leaves in secretory vesicles that fuse with the plasma membrane. Lysosomes are not on the secretory route, and ribosomes come before (not after) the ER in the sequence.
Q2NEET 2017 (Gujarati paper) · Q48
Flippers of Penguins and Dolphins are the example of:
Not quite — the answer is B.
Penguin flippers are modified bird wings while dolphin flippers are modified mammalian forelimbs; they look alike and do the same job (swimming) because of similar habitat, which is convergent evolution producing analogous structures. They are not homologous in the sense intended, since their resemblance is due to function rather than shared ancestry of the flipper form.
Q3NEET 2017 (Gujarati paper) · Q49
Signals for the onset of parturition originate from:
Not quite — the answer is A.
Parturition is triggered by a neuroendocrine mechanism whose signals come from the fully developed foetus and the placenta; these induce mild uterine contractions (the foetal ejection reflex). The maternal pituitary only releases oxytocin in response, so it is not where the signal originates.
Q4NEET 2017 (Gujarati paper) · Q50
Which of the following is not true of organisms in the kingdom Monera?
Not quite — the answer is D.
Monerans are prokaryotes without a nucleus or spindle apparatus, so they divide by binary fission, not mitosis. They are indeed the oldest life forms (about 3.5 billion years) and include both autotrophs (cyanobacteria, chemosynthetic bacteria) and heterotrophs.
Q5NEET 2017 (Gujarati paper) · Q51
Which of the following is not true for callus culture?
Not quite — the answer is C.
Callus is an undifferentiated mass of parenchyma that grows only by repeated mitosis; meiosis does not occur in callus culture. Somaclonal variation does arise in callus and callus can be used to regenerate many plantlets, so those statements are true.
Q6NEET 2017 (Gujarati paper) · Q52
Which of the following statements regarding enzyme inhibition is correct?
Not quite — the answer is B.
In competitive inhibition a substrate-like inhibitor and the substrate compete for the same active site. It is competitive (not non-competitive) inhibition that can be overcome by raising substrate concentration, and non-competitive inhibitors typically bind reversibly at an allosteric site.
Q7NEET 2017 (Gujarati paper) · Q53
Which of the following ossicles is adhered to tympanic membrane of middle ear?
Not quite — the answer is D.
The malleus is attached to the tympanic membrane, the incus links it to the stapes, and the stapes sits on the oval window of the cochlea. The utricle is part of the vestibular apparatus of the inner ear, not an ossicle.
Q8NEET 2017 (Gujarati paper) · Q54
Trace the correct path of sperm from seminiferous tubules:
Not quite — the answer is D.
Seminiferous tubules open into the rete testis, which drains through the vasa efferentia into the epididymis; from there sperms pass into the vas deferens. Putting the epididymis before the vasa efferentia reverses the actual order.
Q9NEET 2017 (Gujarati paper) · Q55
A large amount of about 180 L of fluid is filtered by the human kidneys per day. Which one of the following renal corpuscle features does not justify this?
Not quite — the answer is D.
High filtration needs high glomerular pressure, which comes from the efferent arteriole being narrower than the afferent arteriole; the statement that it is wider is false and would lower pressure. Large surface area, fenestrations and high capillary pressure all genuinely favour the large filtration volume.
Q10NEET 2017 (Gujarati paper) · Q57
A fat molecule is formed from:
Not quite — the answer is C.
A fat (triglyceride) is formed by esterification of the three hydroxyl groups of one glycerol with three fatty acids. One glycerol with one fatty acid gives only a monoglyceride.
Q11NEET 2017 (Gujarati paper) · Q58
The recessive genes located on X-chromosome in humans are always:
Not quite — the answer is B.
Males are hemizygous (XY), so a single recessive allele on their only X chromosome is always expressed, as in haemophilia and colour blindness. Females express it only when homozygous, and most such genes are neither lethal nor sub-lethal.
Q12NEET 2017 (Gujarati paper) · Q60
Select the incorrect option with respect to features present in three animals. Characters | Cockroach | Earthworm | Frog
Not quite — the answer is D.
Cockroach development is indirect (paurometabolous, through nymphs) and earthworm development is direct, so the development row is wrong for both. The circulation row (open, closed, closed), body surface and eye rows correctly describe the three animals.
Q13NEET 2017 (Gujarati paper) · Q62
In roots, absorption of water and minerals mostly occurs in the:
Not quite — the answer is B.
The region of maturation bears root hairs, which enormously increase the absorbing surface, so most water and mineral uptake happens there. The region of elongation is mainly concerned with growth in length, and the meristematic region and root cap do not absorb significantly.
Q14NEET 2017 (Gujarati paper) · Q63
The type of ribosomes is same in:
Not quite — the answer is C.
Prokaryotes, mitochondria and chloroplasts all have 70S ribosomes, which supports the endosymbiotic origin of these organelles. The eukaryotic cytoplasm and rough ER carry 80S ribosomes, so any grouping that mixes cytoplasm with mitochondria or chloroplasts is wrong.
Q15NEET 2017 (Gujarati paper) · Q64
Select the correct option:
Not quite — the answer is D.
Klinefelter's syndrome (47, XXY) results from an extra X chromosome and gives a sterile male. Phenylketonuria is autosomal recessive, Down's syndrome is trisomy of chromosome 21 (not triploidy), and Turner's syndrome is a monosomy (45, XO).
Q16NEET 2017 (Gujarati paper) · Q65
Which of the following values will depict correct respiratory quotient when tripalmitin (a fatty acid) is used as a respiratory substrate?
Not quite — the answer is B.
For tripalmitin, 2C₅₁H₉₈O₆ + 145 O₂ → 102 CO₂ + 98 H₂O, so RQ = 102/145 ≈ 0.7. Fats are poor in oxygen and need more O₂ than the CO₂ they release; an RQ of 1 applies to carbohydrates.
Q17NEET 2017 (Gujarati paper) · Q66
Which of the following cannot be measured by spirometry?
Not quite — the answer is C.
Residual volume is the air that always remains in the lungs after forced expiration, so it never passes through the spirometer and cannot be measured by it. Tidal volume, inspiratory reserve volume and vital capacity are all exhaled or inhaled volumes that a spirometer records.
Q18NEET 2017 (Gujarati paper) · Q67
Select the sac fungus:
Not quite — the answer is B.
Sac fungi are Ascomycetes, which produce ascospores in sac-like asci; Neurospora is an ascomycete. Agaricus is a basidiomycete (club fungus), while Mucor and Albugo belong to the Phycomycetes.
Q19NEET 2017 (Gujarati paper) · Q68
Consider the following statements and choose the correct option: (a) Six codons do not code for any amino-acid. (b) Codon is read in m-RNA in a contiguous fashion. (c) Three codons function as stop codons. (d) The initiator codon AUG codes for methionine.
Not quite — the answer is D.
Only three codons (UAA, UAG, UGA) do not code for any amino acid, so the claim of six is the only false statement. The code is read contiguously without punctuation, the three stop codons terminate translation, and AUG both initiates and codes for methionine.
Q20NEET 2017 (Gujarati paper) · Q69
Air pollution can result in Emphysema, which is:
Not quite — the answer is A.
Emphysema is a chronic disorder in which alveolar walls are damaged, so the respiratory surface area decreases; smoking and air pollution are major causes. Persistent inflammation of the bronchi describes bronchitis, allergic bronchial spasms describe asthma, and emphysema reduces rather than increases lung elasticity.
Q21NEET 2017 (Gujarati paper) · Q70
Which of the following ecologists has tried to put price-tags on nature's life support services?
Not quite — the answer is B.
Robert Constanza and his colleagues estimated the value of global ecosystem services at about US $33 trillion a year. David Tilman worked on productivity-diversity experiments, Paul Ehrlich proposed the rivet popper hypothesis, and Robert May estimated global species diversity.
Q22NEET 2017 (Gujarati paper) · Q71
In Lactational amenorrhoea, ovulation does not occur during the period of intense lactation because of:
Not quite — the answer is C.
During intense breastfeeding, high prolactin suppresses GnRH and hence FSH and LH release, so follicles do not mature and ovulation does not occur. A surge of estrogen or high FSH and LH would instead promote ovulation.
Q23NEET 2017 (Gujarati paper) · Q72
Which of the following type of muscle fibres will be the first one to undergo fatigue?
Not quite — the answer is C.
Fast glycolytic (white) fibres have little myoglobin and few mitochondria and rely on anaerobic glycolysis, so lactic acid builds up quickly and they fatigue first. Slow oxidative (red, aerobic) fibres are rich in myoglobin and mitochondria and resist fatigue.
Q24NEET 2017 (Gujarati paper) · Q73
Given below are defined the various taxonomic aids used in taxonomy which facilitate identification and classification of organisms. Which one of the following is wrong?
Not quite — the answer is A.
A herbarium stores dried, pressed and preserved plant specimens mounted on sheets, not live plants; living plants are kept in botanical gardens. The statements on museums, botanical gardens and zoos, and keys and floras are correct.
Q25NEET 2017 (Gujarati paper) · Q74
At what phase of meiosis homologous chromosomes are separated?
Not quite — the answer is C.
In anaphase I the homologous chromosomes of each bivalent separate and move to opposite poles while sister chromatids stay together. In anaphase II it is the sister chromatids that separate.
Q26NEET 2017 (Gujarati paper) · Q75
Identify and select the wrong statement out of the following:
Not quite — the answer is C.
Sequoia, the giant redwood and one of the tallest trees, is a gymnosperm (a conifer), not an angiosperm. Pine roots do form mycorrhizae, Cycas coralloid roots harbour nitrogen-fixing cyanobacteria, and needle-like leaves reduce water loss.
Q27NEET 2017 (Gujarati paper) · Q76
Tetradynamous stamens are characteristic of:
Not quite — the answer is D.
Tetradynamous condition, six stamens with four long and two short, is characteristic of Brassicaceae (mustard family). Fabaceae has diadelphous stamens, Solanaceae has five epipetalous stamens and Liliaceae has six stamens in two whorls.
Q28NEET 2017 (Gujarati paper) · Q77
Reserved material in prokaryotic cells is stored as:
Not quite — the answer is A.
Prokaryotes store reserve materials such as phosphate granules, cyanophycean granules and glycogen granules in the cytoplasm as inclusion bodies, which are not bounded by membranes. The mesosome is an infolding of the plasma membrane, a polysome is a chain of ribosomes on mRNA, and the basal body anchors the flagellum.
Q29NEET 2017 (Gujarati paper) · Q79
Which of the following statements is correct with respect to cell cycle?
Not quite — the answer is C.
Adult nerve cells have exited the cycle and stay in the quiescent G0 stage, remaining metabolically active without dividing. DNA doubles in S phase, so a G2 cell has twice the DNA of a G1 cell, and chromosomes have two chromatids only after S phase.
Q30NEET 2017 (Gujarati paper) · Q80
Match Column-I with Column-II and select the correct option using codes given below. Column-I: (a) Cytokinin (b) Ethylene (c) Gibberellin (d) Abscisic acid Column-II: (i) stimulates closure of stomata (ii) increases stem length (iii) promotes lateral shoot growth (iv) found in large amount in tissues undergoing senescence
Not quite — the answer is D.
Cytokinins overcome apical dominance and promote lateral shoot growth; ethylene is produced in large amounts by senescing tissues and ripening fruits; gibberellins cause stem elongation; abscisic acid stimulates stomatal closure. The trap code giving ethylene the stem-elongation role confuses it with gibberellin.
Q31NEET 2017 (Gujarati paper) · Q81
Which of the following is not a ciliary movement?
Not quite — the answer is C.
Macrophages and leucocytes move by pseudopodia, which is amoeboid movement. Cilia lining the trachea and the fallopian tube, and those of Paramecium driving food into the gullet, all show ciliary movement.
Q32NEET 2017 (Gujarati paper) · Q82
Which of the following structures does not open into the genital chamber of female cockroaches?
Not quite — the answer is C.
Anal cerci are jointed filamentous appendages on the 10th abdominal segment of both sexes and are external sense organs, so they do not open into the genital chamber. The median oviduct (vagina), spermatheca and collateral glands all open into the female genital chamber.
Q33NEET 2017 (Gujarati paper) · Q84
Which of the following is a proteinaceous and water soluble photosynthetic pigment?
Not quite — the answer is B.
Phycocyanin is a phycobiliprotein, a water-soluble pigment-protein complex of cyanobacteria and red algae that captures light for photosynthesis. Xanthophylls and chlorophylls are fat-soluble, and anthocyanin is a water-soluble vacuolar pigment that is not proteinaceous and not photosynthetic.
Q34NEET 2017 (Gujarati paper) · Q85
Which of the following statements is correct?
Not quite — the answer is A.
Acquired immunity is specific to the pathogen encountered and shows memory. An exaggerated response to environmental antigens is allergy, the spleen (not bone marrow) filters blood-borne microbes, and AIDS is caused by HIV, a retrovirus.
Q35NEET 2017 (Gujarati paper) · Q86
All the components of the conducting system can generate an action potential for the contraction of heart muscle, but the sino-atrial node acts as the pacemaker because:
Not quite — the answer is C.
The SA node generates the maximum number of action potentials per minute (about 70-75), so it sets the rhythm for the whole heart. Other nodal tissue such as the AV node is also autoexcitable but fires more slowly, so it is not true that only the SA node is autorhythmic.
Q36NEET 2017 (Gujarati paper) · Q87
If the ratio of (T + C)/(A + G) in one strand of DNA is 1.43, same ratio in the complementary strand is:
Not quite — the answer is B.
Because A pairs with T and G with C, (T + C) of the complementary strand equals (A + G) of the given strand, so the ratio inverts: 1/1.43 ≈ 0.70. Keeping the ratio at 1.43 would apply only to (A + T)/(G + C), which stays the same in both strands.
Q37NEET 2017 (Gujarati paper) · Q88
In a hypothetical population of 100 individual having 'r' = 0.5/female/year, what will be the population size in 6 years (with e = 2.72) showing exponential rate of growth?
Not quite — the answer is C.
Exponential growth gives Nₜ = N₀e^(rt) = 100 × e^(0.5 × 6) = 100 × e³ = 100 × (2.72)³ ≈ 100 × 20.12 ≈ 2012. Using rt = 2 instead (100 × 2.72² ≈ 739) is the common slip.
Q38NEET 2017 (Gujarati paper) · Q89
Thymosin is responsible for:
Not quite — the answer is C.
The thymus secretes thymosins, which promote the differentiation of T-lymphocytes for cell-mediated immunity and also aid antibody production. Lowering blood calcium is the job of thyrocalcitonin, not thymosin.
Q39NEET 2017 (Gujarati paper) · Q90
Choose the correct sequence representing the ploidy of Nucellus; Megaspore mother cell; Megaspore; Egg cell; Zygote; A polar nucleus of embryo sac; Secondary nucleus and Primary endosperm nucleus.
Not quite — the answer is B.
Nucellus and the megaspore mother cell are diploid; meiosis gives a haploid megaspore, whose products include the haploid egg and haploid polar nuclei; the zygote is 2n, the fused secondary nucleus is 2n, and its fusion with a male gamete gives the triploid primary endosperm nucleus. Sequences that make the egg 2n or the PEN 2n are wrong.
Q40NEET 2017 (Gujarati paper) · Q91
An example of flagellate protozoan is:
Not quite — the answer is B.
Trypanosoma, the cause of sleeping sickness, is a flagellated protozoan. Paramoecium is ciliated, Entamoeba is amoeboid (moves by pseudopodia) and Plasmodium is a sporozoan.
Q41NEET 2017 (Gujarati paper) · Q92
Which statement is wrong about photorespiration?
Not quite — the answer is D.
When RuBisCO binds O₂, one RuBP gives one molecule of phosphoglycerate and one molecule of phosphoglycolate, not two phosphoglycolates. RuBisCO does have a much greater affinity for CO₂ than O₂, photorespiration produces no ATP or NADPH, and it does not occur in C₄ plants.
Q42NEET 2017 (Gujarati paper) · Q93
Which technique helps to identify a bacterial or viral pathogen in a human body even when its concentration is very low and clinical symptoms are not yet visible?
Not quite — the answer is C.
PCR amplifies the pathogen's nucleic acid many times, so a bacterium or virus can be detected at very low concentration, before symptoms appear. ELISA detects antigens or antibodies and needs a larger pathogen load or an immune response, and leucocyte counts are not pathogen-specific.
Q43NEET 2017 (Gujarati paper) · Q94
The Pacinian corpuscle responds to rapid changes in:
Not quite — the answer is B.
Pacinian corpuscles are lamellated mechanoreceptors deep in the dermis that respond to deep pressure and vibration. Temperature is sensed by thermoreceptors, light by photoreceptors and gravity by the vestibular apparatus.
Q44NEET 2017 (Gujarati paper) · Q95
The hollow foliar structure in a wheat embryo that encloses the shoot apex and a few leaf primordia is called:
Not quite — the answer is D.
In grass embryos the shoot apex and a few leaf primordia are enclosed in a hollow foliar sheath called the coleoptile. The coleorrhiza is the undifferentiated sheath around the radicle and root cap, while epicotyl and hypocotyl are portions of the embryonal axis.
Q45NEET 2017 (Gujarati paper) · Q97
The chief function of vessels in the plant body is to:
Not quite — the answer is C.
Vessels are long tubes of dead lignified cells in the xylem whose main role is conducting water and minerals from roots to the aerial parts. Transporting food made in leaves is the function of phloem sieve tubes, not vessels.
Q46NEET 2017 (Gujarati paper) · Q98
A complex of ribosomes attached to a single strand of mRNA is known as:
Not quite — the answer is B.
Several ribosomes translating the same mRNA together form a polyribosome (polysome). A polypeptide is the product of translation and an Okazaki fragment is a short piece of lagging-strand DNA made in replication.
Q47NEET 2017 (Gujarati paper) · Q99
The sequential events from initial stage till climax stage in a succession are called:
Not quite — the answer is B.
The entire sequence of communities that replace one another from the pioneer stage to the climax community is called a sere, and each stage is a seral stage. Nudation (formation of a bare area), migration and ecesis (establishment) are only individual steps of succession.
Q48NEET 2017 (Gujarati paper) · Q100
The protein coat around a virus is called:
Not quite — the answer is B.
The nucleic acid of a virus is enclosed in a protein coat called the capsid, made of subunits called capsomeres. A capsule is the slimy outer layer of some bacteria, and trichome refers to plant hairs or filaments of cyanobacteria.
Q49NEET 2017 (Gujarati paper) · Q102
Which of the following represents correct match of feature with the given set of animals? Feature – Animals
Not quite — the answer is C.
Prawn, centipede and grasshopper are all arthropods with jointed appendages. Tapeworm has no respiratory system, Hydra is radially symmetrical and adult sea urchin is radial, and the liver fluke is a flatworm without metameric segmentation.
Q50NEET 2017 (Gujarati paper) · Q103
Cholecystokinin acts on:
Not quite — the answer is D.
Cholecystokinin (CCK), released by the duodenum, acts on the pancreas to secrete pancreatic enzymes and on the gall bladder to release bile. Gastric glands are stimulated by gastrin, not CCK.
Q51NEET 2017 (Gujarati paper) · Q105
Which of the following pathways is involved for packaging of secretory proteins?
Not quite — the answer is B.
Proteins made on the RER reach the receiving cis (forming) face of the Golgi, move through the cisternae and leave from the trans (maturing) face in secretory vesicles. Entering at the trans face reverses the polarity of the Golgi.
Q52NEET 2017 (Gujarati paper) · Q107
Opposite type of phyllotaxy is not present in:
Not quite — the answer is D.
Mango has alternate phyllotaxy, with a single leaf at each node. Guava and Calotropis have opposite leaves (a pair at each node), and mint has opposite decussate leaves.
Q53NEET 2017 (Gujarati paper) · Q108
In gene therapy to treat adenosine deaminase deficiency disorder, which of the patients blood cells are used?
Not quite — the answer is B.
In the first gene therapy (1990) for ADA deficiency, lymphocytes from the patient's blood were grown in culture, given a functional ADA cDNA using a retroviral vector, and returned to the patient. Erythrocytes lack a nucleus, so they cannot carry and express the introduced gene.
Q54NEET 2017 (Gujarati paper) · Q109
Which one of the following is not an IUD?
Not quite — the answer is D.
Vaults are barrier contraceptives placed over the cervix by the female. Cu T and Multiload 375 are copper-releasing IUDs and Progestasert is a hormone-releasing IUD.
Q55NEET 2017 (Gujarati paper) · Q110
Growth hormone Auxin was isolated by F.W. Went from tips of seedling coleoptile of:
Not quite — the answer is C.
F.W. Went isolated auxin from the coleoptile tips of oat (Avena sativa) seedlings using agar blocks. Maize, wheat and rice coleoptiles were not used in his classic experiment.
Q56NEET 2017 (Gujarati paper) · Q111
In the heart, as the action potential reaches the AV node from the SA node, there is a delay of the action potential. This delay is important because:
Not quite — the answer is C.
The short AV-nodal delay lets the atria finish contracting and push their blood into the ventricles before ventricular systole begins. It does not make atrial contraction stronger or serve atrial filling from the venae cavae.
Q57NEET 2017 (Gujarati paper) · Q112
Pollen grains can be stored for years in liquid nitrogen, maintained at temperature:
Not quite — the answer is C.
Liquid nitrogen boils at –196°C, and pollen grains are cryopreserved at this temperature in pollen banks for years. The other temperatures are not the temperature of liquid nitrogen.
Q58NEET 2017 (Gujarati paper) · Q113
Which of the following statements is true for phloem in plants?
Not quite — the answer is B.
Companion cells are connected to sieve tube elements by pit fields and help maintain the pressure gradient in the sieve tubes. Each sieve tube element is a single cell that loses its nucleus, phloem parenchyma is absent in most monocots, and phloem fibres are sclerenchymatous.
Q59NEET 2017 (Gujarati paper) · Q114
Match Column - I with Column - II and select the correct option. Column - I: (a) Plasmodium (b) Wuchereria (c) Entamoeba (d) Microsporum Column - II: (i) Ringworm (ii) Amoebiasis (iii) Elephantiasis (iv) Malaria
Not quite — the answer is D.
Plasmodium causes malaria, Wuchereria (a filarial worm) causes elephantiasis, Entamoeba histolytica causes amoebiasis and the fungus Microsporum causes ringworm. Pairing Wuchereria with ringworm swaps a helminth with a fungal disease.
Q60NEET 2017 (Gujarati paper) · Q117
Which scientist proposed 'Rivet popper hypothesis' related to biodiversity and Ecosystems?
Not quite — the answer is B.
Stanford ecologist Paul Ehrlich proposed the rivet popper hypothesis, comparing species in an ecosystem to rivets in an aeroplane. Humboldt described species-area relationships and David Tilman showed that plots with more species have more stable productivity.
Q61NEET 2017 (Gujarati paper) · Q118
Select the wrong statement:
Not quite — the answer is B.
Linkage and recombination were discovered by T.H. Morgan through his Drosophila experiments, not by Sutton. Sutton (with Boveri) proposed the chromosomal theory of inheritance, and de Vries, Correns and von Tschermak rediscovered Mendel's work in 1900.
Q62NEET 2017 (Gujarati paper) · Q119
Continued self-pollination results in:
Not quite — the answer is B.
Continued self-pollination increases homozygosity, exposing harmful recessive alleles and reducing fertility and vigour, which is inbreeding depression. Heterosis (hybrid vigour) is the opposite outcome produced by outcrossing.
Q63NEET 2017 (Gujarati paper) · Q121
Continental drift led to disappearance of a number of South American mammals because:
Not quite — the answer is C.
When South America joined North America, the native pouched mammals of South America were overridden and outcompeted by North American fauna that migrated in. Disease outbreaks or climatic shifts are not the explanation NCERT gives.
Q64NEET 2017 (Gujarati paper) · Q122
Which one of the following is not true for the experiments of Mendel on pea?
Not quite — the answer is C.
Mendel made his crosses by controlled artificial pollination (emasculation and bagging), not by natural open pollination. He did choose seven characters with two contrasting forms, start from true-breeding lines and use large samples, which gave his data statistical reliability.
Q65NEET 2017 (Gujarati paper) · Q123
Restriction endonucleases are:
Not quite — the answer is B.
Restriction endonucleases are made by bacteria to cut the DNA of invading bacteriophages, restricting their growth, which is a defence mechanism. Joining DNA molecules is done by DNA ligase, and DNA synthesis in vitro needs DNA polymerase.
Q66NEET 2017 (Gujarati paper) · Q124
Which of the following statements is not true?
Not quite — the answer is C.
The descending limb of Henle's loop is permeable to water but almost impermeable to electrolytes, so water leaves it and the filtrate becomes concentrated. It is the ascending limb that is impermeable to water.
Q67NEET 2017 (Gujarati paper) · Q125
Which of the following pairs is not correctly matched? Vegetative Propagules – Example
Not quite — the answer is D.
Agave propagates by bulbils, which are modified buds, not by stolons. Water hyacinth spreads by offsets, potato tubers sprout from their eyes (nodes), and ginger is a rhizome.
Q68NEET 2017 (Gujarati paper) · Q126
What is not true for an angiospermic embryo sac?
Not quite — the answer is D.
The pollen tube releases two male gametes into the synergid of the embryo sac: one fuses with the egg and the other with the polar nuclei (double fertilisation). The embryo sac is the female gametophyte inside the ovule, and it develops from a megaspore formed by meiosis.
Q69NEET 2017 (Gujarati paper) · Q127
Which of the following enzymes is not protein?
Not quite — the answer is D.
Ribozymes are catalytic RNA molecules (for example the peptidyl transferase activity of rRNA), so they are not proteins. Polymerases, ligases and lysozyme are all protein enzymes.
Q70NEET 2017 (Gujarati paper) · Q128
The technique of DNA fingerprinting is superior to conventional fingerprinting because it can:
Not quite — the answer is C.
DNA fingerprinting compares repetitive, highly polymorphic DNA sequences (VNTRs), which differ between individuals and are inherited from parents. It does not compare whole genomes, and it is slower and more expensive than taking conventional fingerprints.
Q71NEET 2017 (Gujarati paper) · Q129
Choose the false statement regarding Petromyzon.
Not quite — the answer is C.
Petromyzon (lamprey) is marine but migrates to fresh water for spawning; the adults die soon after and the larvae return to the ocean after metamorphosis, so migrating to the ocean to spawn is false. Cyclostomes are scaleless, have a circular jawless sucking mouth and a closed circulatory system, so the statement about closed circulation is true.
Q72NEET 2017 (Gujarati paper) · Q130
Which of the following symmetry is exhibited by Echinoderm Larvae?
Not quite — the answer is D.
Echinoderm larvae are bilaterally symmetrical, while the adults become radially symmetrical. Radial symmetry describes the adult, not the larva.
Q73NEET 2017 (Gujarati paper) · Q131
What is the meaning of Bt in Bt cotton?
Not quite — the answer is D.
Bt stands for Bacillus thuringiensis; Bt cotton carries its cry gene coding for an insecticidal protein toxin that kills bollworms. Baculoviruses are separate biocontrol agents, and Bt cotton gives insect resistance rather than resistance to microbial infection.
Q74NEET 2017 (Gujarati paper) · Q132
A couple claimed in court that a child belonged to them. Their claim can be true if the DNA fingerprint pattern of the child shows:
Not quite — the answer is D.
A child gets one set of chromosomes from each parent, so about half of the bands in its DNA fingerprint match the father and half match the mother. A complete match with either parent, or with both, is impossible for a biological child.
Q75NEET 2017 (Gujarati paper) · Q134
Which of the following statements is not true?
Not quite — the answer is B.
Detritivores feed on dead organic matter from every trophic level, including dead leaves and wood of producers, so excluding the producer level is false. Primary consumers are by definition herbivores, one organism such as a sparrow can feed at more than one trophic level, and energy decreases at each higher level.
Biology: not in the current NEET syllabus
These 10 questions are from topics NCERT dropped in its 2023 revision. Shown for completeness; you can skip them for NEET 2027.
10 questions. Tap an option to check your answer.
Q1NEET 2017 (Gujarati paper) · Q56
Special feature about Strobilanthus kunthiana (neelakuranji) is its flowering once in:
Not quite — the answer is A.
Neelakuranji (Strobilanthus kunthiana) of the Western Ghats hills of Kerala, Karnataka and Tamil Nadu flowers once in 12 years, turning the hill slopes blue. Bamboo species are the plants that flower once in 50-100 years.
Q2NEET 2017 (Gujarati paper) · Q59
Which of the following plants has association with Frankia?
Not quite — the answer is B.
Frankia is a nitrogen-fixing actinomycete that forms root nodules on non-leguminous plants such as Alnus. Alfalfa, sweet pea and lentils are legumes whose nodules contain Rhizobium.
Q3NEET 2017 (Gujarati paper) · Q61
A genetically engineered bacterium first used for cleaning the oil spills, was a species of:
Not quite — the answer is B.
The first patented genetically engineered 'superbug' that degrades oil hydrocarbons was a strain of Pseudomonas putida developed by Ananda Mohan Chakrabarty. Escherichia, Bacillus and Rhizobium were not used for this purpose.
Q4NEET 2017 (Gujarati paper) · Q78
'MOET' technique is used for super-ovulation in:
Not quite — the answer is B.
Multiple Ovulation Embryo Transfer is used in cattle: a cow is given FSH-like hormones to release 6-8 eggs instead of one, the embryos are collected and transferred to surrogate mothers. It is a technique of animal husbandry for high-yielding cattle, not for fish or poultry.
Q5NEET 2017 (Gujarati paper) · Q96
The increase in concentration of the toxicant at successive trophic levels is referred to as:
Not quite — the answer is D.
Biomagnification is the rise in concentration of a persistent toxicant such as DDT or mercury at each higher trophic level, because it cannot be metabolised or excreted. Eutrophication is nutrient enrichment and ageing of a water body, not toxicant build-up.
Q6NEET 2017 (Gujarati paper) · Q101
Name the element which is the main constituent of the ring structure of chlorophyll and helps to maintain the ribosome structure.
Not quite — the answer is B.
Magnesium is the central atom of the porphyrin ring of chlorophyll and also maintains the structure of ribosomes. Nitrogen is part of the ring too but is not the element that maintains ribosome structure, while calcium is used in the middle lamella.
Q7NEET 2017 (Gujarati paper) · Q104
To protect and improve the quality of environment, the Government of India passed the Environment (Protection) Act in the year:
Not quite — the answer is C.
The Environment (Protection) Act was passed by the Government of India in 1986, after the Bhopal gas tragedy. The other years do not correspond to this legislation.
Q8NEET 2017 (Gujarati paper) · Q115
Tree planting helps reduce global warming as trees:
Not quite — the answer is C.
Trees fix atmospheric CO₂ in photosynthesis and store the carbon in their biomass, lowering the concentration of this greenhouse gas. Giving out O₂ or shading the ground does not reduce the greenhouse effect.
Q9NEET 2017 (Gujarati paper) · Q116
Van Mahotsava is a festival of:
Not quite — the answer is A.
Van Mahotsava, celebrated in the first week of July in India, is a festival of planting trees. Sacred groves are traditional forest patches protected for religious reasons, a different practice.
Q10NEET 2017 (Gujarati paper) · Q135
The difference between Marasmus and Kwashiorkor is that:
Not quite — the answer is C.
Marasmus results from a combined deficiency of proteins and calories, typically in infants under a year, whereas kwashiorkor is caused by protein deficiency when calorie intake is adequate. Calling marasmus purely a calorie deficiency misses the protein component, and neither disease is a vitamin deficiency.
Chapter-wise split of NEET 2017 (Gujarati paper) Biology
- Cell: The Unit of Life5 Q
- Sexual Reproduction in Flowering Plants5 Q
- Biological Classification4 Q
- Biotechnology and Its Applications4 Q
- Principles of Inheritance and Variation4 Q
- Morphology of Flowering Plants4 Q
- Molecular Basis of Inheritance4 Q
- Biomolecules3 Q
- Ecosystem3 Q
- Animal Kingdom3 Q
- Evolution2 Q
- Human Reproduction2 Q
- Neural Control and Coordination2 Q
- Excretory Products and Their Elimination2 Q
- Structural Organisation in Animals2 Q
- Breathing and Exchange of Gases2 Q
- Reproductive Health2 Q
- Locomotion and Movement2 Q
- Cell Cycle and Cell Division2 Q
- Plant Kingdom2 Q
- Plant Growth and Development2 Q
- Human Health and Disease2 Q
- Body Fluids and Circulation2 Q
- Chemical Coordination and Integration2 Q
- Anatomy of Flowering Plants2 Q
- Respiration in Plants1 Q
- The Living World1 Q
- Organisms and Populations1 Q
- Photosynthesis in Higher Plants1 Q
- Biodiversity and Conservation1 Q
- Biotechnology: Principles and Processes1 Q
About this paper
How many questions from NEET 2017 (Gujarati paper) are on this page?
147 questions (Physics 25, Chemistry 37, Biology 85). 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 (Gujarati paper) 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 (Gujarati paper) useful for NEET 2027?
Yes. The NEET syllabus is built on the same NCERT books. 16 questions here are on topics NCERT has since dropped; they are listed separately so you can skip them.
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