NEET 2027 · Full Syllabus Practice · Aamirah Fathima
Answer Key & Solutions — Set 3
Quick-reference grid first, then a one-line reason for every question with its topic tag. Score the paper, then re-work every wrong answer before moving on.
Questions
180
Max marks
720
Correct
+4
Wrong
−1
Set
3
Quick key
All 180 answers
001 B
002 C
003 C
004 B
005 B
006 C
007 B
008 B
009 B
010 B
011 C
012 B
013 D
014 B
015 C
016 C
017 B
018 B
019 B
020 B
021 B
022 B
023 B
024 A
025 A
026 B
027 B
028 B
029 B
030 C
031 B
032 C
033 B
034 C
035 B
036 B
037 C
038 B
039 B
040 B
041 A
042 C
043 B
044 C
045 B
046 B
047 C
048 A
049 C
050 B
051 C
052 B
053 B
054 B
055 B
056 C
057 B
058 C
059 B
060 B
061 C
062 C
063 B
064 B
065 B
066 B
067 B
068 B
069 B
070 C
071 B
072 A
073 B
074 B
075 B
076 B
077 D
078 B
079 B
080 B
081 B
082 C
083 B
084 B
085 B
086 B
087 B
088 B
089 C
090 B
091 A
092 B
093 B
094 B
095 B
096 B
097 C
098 B
099 B
100 B
101 A
102 B
103 A
104 B
105 B
106 B
107 B
108 B
109 B
110 B
111 A
112 B
113 B
114 A
115 B
116 A
117 B
118 B
119 C
120 B
121 B
122 B
123 B
124 B
125 B
126 C
127 C
128 B
129 B
130 B
131 C
132 B
133 B
134 C
135 B
136 B
137 B
138 B
139 B
140 A
141 B
142 B
143 B
144 B
145 C
146 C
147 C
148 B
149 B
150 B
151 B
152 B
153 B
154 B
155 B
156 B
157 B
158 B
159 B
160 B
161 B
162 B
163 B
164 D
165 C
166 B
167 B
168 C
169 C
170 C
171 B
172 B
173 C
174 B
175 A
176 B
177 B
178 B
179 B
180 B
Physics
Section A · worked reasoning
Q001 (B) [M⁻¹L³T⁻²]
From F = Gm₁m₂/r², G = Fr²/m² ⇒ [MLT⁻²][L²]/[M²] = [M⁻¹L³T⁻²].
Units & Dimensions
Q002 (C) 3.1
The least precise factor has two significant figures, so the answer is rounded to two.
Errors & Significant Figures
Q003 (C) 18 m s⁻¹
v = dx/dt = 8t + 2 = 18 m s⁻¹ at t = 2 s.
Kinematics
Q004 (B) 30° and 60°
R is unchanged for complementary angles θ and (90° − θ).
Kinematics
Q005 (B) m(g − a)
The normal reaction N = m(g − a) when the lift accelerates downwards.
Laws of Motion
Q006 (C) 2 m s⁻¹
Momentum conservation: 4v = 0.02 × 400 ⇒ v = 2 m s⁻¹.
Laws of Motion
Q007 (B) 1 : 2
KE ∝ distance fallen for free fall, so the ratio is 1 : 2.
Work, Energy & Power
Q008 (B) MR²/2
By the perpendicular axis theorem, 2I_d = MR² ⇒ I_d = MR²/2.
Rotational Motion
Q009 (B) √(10gh/7)
v = √(2gh/(1 + k²/R²)) with k²/R² = 2/5 gives √(10gh/7).
Rotational Motion
Q010 (B) g/4
g_h = g R²/(R + h)² = g/4 at h = R.
Gravitation
Q011 (C) 24 hours
Its period must match the earth's rotation so that it appears stationary.
Gravitation
Q012 (B) W
Breaking stress depends on material and area, not on length.
Elasticity
Q013 (D) Four times
Equation of continuity: A₁v₁ = A₂v₂ ⇒ v ∝ 1/A.
Fluids
Q014 (B) Inversely proportional to the radius
h = 2T cosθ/rρg, so a narrower tube gives a greater rise.
Fluids
Q015 (C) γ = 3α
For an isotropic solid, γ = 3α and the areal coefficient is 2α.
Thermal Properties
Q016 (C) 800 cal
Q = mL = 10 × 80 = 800 cal; temperature stays constant during melting.
Thermal Properties
Q017 (B) Entirely increases the internal energy
Volume is constant so W = 0 and Q = ΔU.
Thermodynamics
Q018 (B) 5
COP = T₂/(T₁ − T₂) = 250/50 = 5.
Thermodynamics
Q019 (B) P = ⅓ρv²_rms
From kinetic theory, P = (1/3)ρ v²_rms.
Kinetic Theory
Q020 (B) ½mω²A²
Total energy is constant and equals the maximum kinetic energy ½mω²A².
Oscillations
Q021 (B) λ/2
Nodes recur every half wavelength.
Waves
Q022 (B) √(T/μ)
v = √(T/μ).
Waves
Q023 (B) Never intersect one another
Two directions of field at one point are impossible, so lines never cross.
Electrostatics
Q024 (A) Zero
Potentials are scalars and the equal and opposite contributions cancel.
Electrostatics
Q025 (A) 0.01 J
U = ½CV² = ½ × 2 × 10⁻⁶ × 10⁴ = 0.01 J.
Capacitors
Q026 (B) Charge
The total current entering a junction equals the total current leaving it.
Current Electricity
Q027 (B) 2.4 Ω
R = (4 × 6)/(4 + 6) = 2.4 Ω.
Current Electricity
Q028 (B) The 60 W bulb glows brighter
In series the current is common and the 60 W bulb has the greater resistance, so I²R is larger for it.
Current Electricity
Q029 (B) 5 × 10⁻⁵ T
B = μ₀I/2πr = (2 × 10⁻⁷ × 5)/0.02 = 5 × 10⁻⁵ T.
Moving Charges & Magnetism
Q030 (C) Increasing the number of turns and the area of the coil
Sensitivity = NAB/k, so more turns or greater area raises it.
Moving Charges & Magnetism
Q031 (B) Isolated magnetic monopoles do not exist
Magnetic field lines are continuous closed loops, so as many enter as leave.
Magnetism & Matter
Q032 (C) 20 V
ε = ΔΦ/Δt = 4/0.2 = 20 V.
Electromagnetic Induction
Q033 (B) Induction furnaces and electromagnetic braking
They are minimised by laminating cores but exploited in furnaces and braking systems.
Electromagnetic Induction
Q034 (C) 5 Ω
Z = √(R² + (X_L − X_C)²) = √(9 + 16) = 5 Ω.
Alternating Current
Q035 (B) Only a pure inductor or capacitor
The phase difference is 90°, so cosφ = 0 and no power is dissipated.
Alternating Current
Q036 (B) Mutually perpendicular
EM waves are transverse, with E × B along the direction of propagation.
Electromagnetic Waves
Q037 (C) 4 D
P = 1/f (in metres) = 1/0.25 = 4 D.
Ray Optics
Q038 (B) f_o/f_e
The objective must have a long focal length and the eyepiece a short one.
Ray Optics
Q039 (B) nλ
Bright fringes correspond to an integral number of wavelengths.
Wave Optics
Q040 (B) The refractive index of the medium
tan i_B = μ, and at this angle the reflected ray is completely plane polarised.
Wave Optics
Q041 (A) 12.27/√V Å
λ = h/√(2meV) = 12.27/√V Å.
Dual Nature of Matter
Q042 (C) Is independent of mass number
R = R₀A^(1/3), so volume ∝ A and density stays nearly constant.
Nuclei
Q043 (B) Balmer series, visible
Transitions ending at n = 2 form the Balmer series, which lies in the visible region.
Atoms
Q044 (C) 100 Hz
Both halves of the input cycle are used, so the output pulses at twice the input frequency.
Semiconductors
Q045 (B) Electrons
Pentavalent dopants donate free electrons, which become the majority carriers.
Semiconductors
Chemistry
Section B · worked reasoning
Q046 (B) 0.2 M
0.1 mol in 0.5 L gives 0.2 mol L⁻¹.
Some Basic Concepts
Q047 (C) 44 g
1 mol C gives 1 mol CO₂ = 44 g.
Some Basic Concepts
Q048 (A) [Ar]3d⁹4s⁰
Cu is [Ar]3d¹⁰4s¹; losing two electrons (4s first) leaves 3d⁹.
Structure of Atom
Q049 (C) 2
Radial nodes = n − l − 1 = 3 − 0 − 1 = 2.
Structure of Atom
Q050 (B) The 2p electron of boron is more easily removed than the paired 2s electron of beryllium
Be has a stable filled 2s² configuration and its electron is more tightly held.
Classification & Periodicity
Q051 (C) Fluorine
Fluorine has the highest electronegativity, 4.0 on the Pauling scale.
Classification & Periodicity
Q052 (B) NH₄⁺
The lone pair of NH₃ is donated to H⁺, forming a coordinate bond.
Chemical Bonding
Q053 (B) The bond moments and the lone pair moment act in the same direction
In NF₃ the bond moments oppose the lone pair moment, reducing the resultant.
Chemical Bonding
Q054 (B) Zero
Two bonding and two antibonding electrons give BO = (2 − 2)/2 = 0.
Chemical Bonding
Q055 (B) Smaller cation with high charge and larger anion
A small, highly charged cation polarises a large anion strongly.
Chemical Bonding
Q056 (C) Positive
ΔS_total > 0 for a spontaneous change in an isolated system.
Thermodynamics
Q057 (B) Volume, giving ΔU
The rigid vessel keeps volume constant, so the heat measured is ΔU.
Thermodynamics
Q058 (C) Basic
The salt of a weak acid and strong base hydrolyses to give OH⁻ ions.
Equilibrium
Q059 (B) Products predominate at equilibrium
K is the ratio of product to reactant activities at equilibrium.
Equilibrium
Q060 (B) Weak electrolytes
It relates the dissociation constant to the degree of dissociation of a weak electrolyte.
Equilibrium
Q061 (C) M/5
Mn goes from +7 to +2, a five-electron change.
Redox Reactions
Q062 (C) CN⁻
Nucleophiles are electron-rich species that donate a lone pair.
Organic Chemistry – Basic Principles
Q063 (B) 3° > 2° > 1°
Hyperconjugation and the inductive effect stabilise the more substituted radical.
Organic Chemistry – Basic Principles
Q064 (B) 2-Chlorobutane
C-2 of 2-chlorobutane carries four different groups, making it a chiral centre.
Organic Chemistry – Basic Principles
Q065 (B) Ethane
Decarboxylation at the anode gives methyl radicals that dimerise to ethane.
Hydrocarbons
Q066 (B) HBr only
Only the H–Br bond has suitable bond energy for the free radical chain mechanism.
Hydrocarbons
Q067 (B) (4n + 2) π electrons in a planar cyclic system
Planarity, cyclic delocalisation and (4n + 2) π electrons are all required.
Hydrocarbons
Q068 (B) Directly proportional to its partial pressure
p = K_H x, so more pressure dissolves more gas.
Solutions
Q069 (B) A maximum boiling azeotrope
Stronger solute–solvent interaction lowers vapour pressure and raises boiling point.
Solutions
Q070 (C) 3 F
Al³⁺ + 3e⁻ → Al requires three faradays per mole.
Electrochemistry
Q071 (B) Aqueous KOH
Concentrated aqueous KOH carries the ions between the porous carbon electrodes.
Electrochemistry
Q072 (A) Zero
Rate ∝ [A]⁰ means the rate is independent of that concentration.
Chemical Kinetics
Q073 (B) [A]₀/2k
For [A] = [A]₀ − kt, half the initial concentration is consumed at t = [A]₀/2k.
Chemical Kinetics
Q074 (B) d–d electronic transitions
Absorption of visible light promotes an electron between split d orbitals.
d- and f-Block Elements
Q075 (B) Ce
Ce⁴⁺ attains the stable xenon core configuration and is a good oxidising agent.
d- and f-Block Elements
Q076 (B) Cr³⁺
Orange Cr₂O₇²⁻ is reduced to green Cr³⁺, a six-electron change.
d- and f-Block Elements
Q077 (D) CO
CO is the strongest field ligand of those listed.
Coordination Compounds
Q078 (B) Ionisation isomers
They give different ions in solution because the ligand and counter ion are exchanged.
Coordination Compounds
Q079 (B) +3 and 6
Oxalate is bidentate, so three of them give a coordination number of six.
Coordination Compounds
Q080 (B) Inversion of configuration
Backside attack by the nucleophile causes Walden inversion.