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📄 NEET 2024 · 05/05/2024 · Code T3

NEET 2024 Question Paper
Answer Key & Solutions

Official NEET UG 2024 questions · Correct answers highlighted · Step-by-step solutions

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📘 NEET 2024 (UG) — Official Paper with Step-by-Step Solutions

Ph

Physics

Q1–Q50

36 Qs
PhysicsQ.
A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4π×10⁻⁷ SI units):
Options
A
4.4 mT
✓
B
44 T
C
44 mT
D
4.4 T
✦ Correct Answer
4.4 mT
📐 Solution
1
B = μ₀NI/2r = (4π×10⁻⁷×100×7)/(2×0.1)
= 0.0044 T = 4.4 mT
PhysicsQ.
Match List-I (Material) with List-II (Susceptibility χ). A.Diamagnetic, B.Ferromagnetic, C.Paramagnetic, D.Non-magnetic. I. x=0, II. 0>x≥−1, III. x>>1, IV. 0
Options
A
A-III, B-II, C-I, D-IV
B
A-IV, B-III, C-II, D-I
C
A-II, B-III, C-IV, D-I
✓
D
C-II, B-I, C-III, D-IV
✦ Correct Answer
A-II, B-III, C-IV, D-I
📐 Solution
1
Diamagnetic: −10⁻⁵ to −10⁻⁹ → 0>x≥−1 (II).
Ferromagnetic: x>>1 (III).
Paramagnetic: small positive → 0Non-magnetic: x=0 (I).
∴ A-II, B-III, C-IV, D-I
PhysicsQ.
An unpolarised light beam strikes a glass surface at Brewster's angle. Then:
Options
A
Both the reflected and refracted light will be completely polarised.
B
The reflected light will be completely polarised but the refracted light will be partially polarised.
✓
C
The reflected light will be partially polarised.
D
The refracted light will be completely polarised.
✦ Correct Answer
The reflected light will be completely polarised but the refracted light will be partially polarised.
📐 Solution
1
At Brewster's angle, only the reflected light is completely polarised. The refracted light is only partially polarised.
PhysicsQ.
In an ideal transformer, the turns ratio is Nₚ/Nₛ = 1/2. The ratio Vₛ:Vₚ is equal to:
Options
A
1:1
B
1:4
C
1:2
D
2:1
✓
✦ Correct Answer
2:1
📐 Solution
1
Transformation ratio r = Nₛ/Nₚ = Vₛ/Vₚ = 2/1
∴ Vₛ : Vₚ = 2 : 1
PhysicsQ.
In a vernier calipers, (N+1) divisions of vernier scale coincide with N divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:
Options
A
100N
B
10(N+1)
C
1/10N
D
1/100(N+1)
✓
✦ Correct Answer
1/100(N+1)
📐 Solution
1
LC = 1 MSD − 1 VSD = [1 − N/(N+1)] × 0.1 mm = [1/(N+1)] × 0.1 mm
= 1/[10(N+1)] mm = 1/[100(N+1)] cm
PhysicsQ.
The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's modulus, respectively, are 8×10⁸ N m⁻² and 2×10¹¹ N m⁻², is:
Options
A
40 mm
B
8 mm
C
4 mm
✓
D
0.4 mm
✦ Correct Answer
4 mm
📐 Solution
1
(Stress)max = Y × (Strain)max → 8×10⁸ = 2×10¹¹ × ΔL/1
ΔL = 8×10⁸/2×10¹¹ = 4×10⁻³ m = 4 mm
PhysicsQ.
A horizontal force 10 N is applied to a block A of mass 2 kg connected to block B of mass 3 kg on a frictionless surface. The force exerted by block A on block B is:
Options
A
6 N
✓
B
10 N
C
zero
D
4 N
✦ Correct Answer
6 N
📐 Solution
1
Common acceleration a = F/M_total = 10/5 = 2 m/s²
F_BA = M_B × a = 3 × 2 = 6 N
PhysicsQ.
If the monochromatic source in Young's double slit experiment is replaced by white light, then:
Options
A
There will be a central bright white fringe surrounded by a few coloured fringes.
✓
B
All bright fringes will be of equal width.
C
Interference pattern will disappear.
D
There will be a central dark fringe surrounded by a few coloured fringes.
✦ Correct Answer
There will be a central bright white fringe surrounded by a few coloured fringes.
📐 Solution
1
Central fringe for all colours forms at centre → appears white. Other fringes of different colours appear on either side.
PhysicsQ.
Given: Assertion A: The potential (V) at any axial point, at 2 m distance from the centre of a dipole of moment 4×10⁻⁶ C·m, is ±9×10³ V. Reason R: V = ±(2P)/(4πε₀r²) where r is distance of axial point from dipole centre.
Options
A
A is true but R is false.
✓
B
A is false but R is true.
C
Both A and R are true and R is the correct explanation of A.
D
Both A and R are true and R is NOT the correct explanation of A.
✦ Correct Answer
A is true but R is false.
📐 Solution
1
V = ±(1/4πε₀)(P/r²) = ±9×10⁹ × 4×10⁻⁶/4 = ±9×10³ V → A is true.
But the formula in R has '2P' which is wrong (should be P/r²), so R is false.
PhysicsQ.
The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 g cm². The length of the 400 g rod is nearly:
Options
A
20.7 cm
B
72.0 cm
C
8.5 cm
✓
D
17.5 cm
✦ Correct Answer
8.5 cm
📐 Solution
1
I = ML²/12 → 2400 = 400×L²/12 → L² = 72 → L = √72 ≈ 8.5 cm
PhysicsQ.
The terminal voltage of the battery (emf = 10 V, internal resistance = 1 Ω) when connected through an external resistance of 4 Ω is:
Options
A
8 V
✓
B
10 V
C
4 V
D
6 V
✦ Correct Answer
8 V
📐 Solution
1
i = E/(R+r) = 10/5 = 2 A
Terminal voltage = E − ir = 10 − 2×1 = 8 V
PhysicsQ.
Match List-I (Spectral Lines of H for transitions from) with List-II (Wavelengths nm). A.n₂=3 to n₁=2, B.n₂=4 to n₁=2, C.n₂=5 to n₁=2, D.n₂=6 to n₁=2. I.410.2, II.434.1, III.656.3, IV.486.1:
Options
A
A-IV, B-III, C-I, D-II
B
A-I, B-II, C-III, D-IV
C
A-II, B-I, C-IV, D-III
D
A-III, B-IV, C-II, D-I
✓
✦ Correct Answer
A-III, B-IV, C-II, D-I
📐 Solution
1
H-atom Balmer series: 3→2=656.3(III), 4→2=486.1(IV), 5→2=434.1(II), 6→2=410.2(I)
∴ A-III, B-IV, C-II, D-I
PhysicsQ.
If c is the velocity of light in free space, the correct statements about a photon among the following are: A.E=hν, B.velocity=c, C.p=hν/c, D.both total energy and momentum conserved in photon-electron collision, E.photon possesses positive charge.
Options
A
A, C and D only
B
A, B, D and E only
C
A and B only
D
A, B, C and D only
✓
✦ Correct Answer
A, B, C and D only
📐 Solution
1
E=hν ✓, v=c ✓, p=h/λ=hν/c ✓, conservation laws hold ✓. E possesses no charge ✗.
∴ A, B, C and D only
PhysicsQ.
In the nuclear decay: ²⁹⁰₈₂X →(α)→ Y →(e⁺)→ Z →(β⁻)→ P →(e⁻)→ Q. The mass number and atomic number of product Q respectively are:
Options
A
288, 82
B
286, 81
✓
C
280, 81
D
286, 80
✦ Correct Answer
286, 81
📐 Solution
1
X(290,82) →α→ Y(286,80) →e⁺→ Z(286,79) →β⁻→ P(286,80) →e⁻→ Q(286,81)
∴ Mass number 286, Atomic number 81
PhysicsQ.
At any instant of time t, the displacement of a particle is given by 2t−1 (SI unit) under a force of 5 N. The instantaneous power is (in SI unit):
Options
A
7
B
6
C
10
✓
D
5
✦ Correct Answer
10
📐 Solution
1
v = ds/dt = 2 m/s; P = Fv = 5×2 = 10
PhysicsQ.
The mass of a planet is 1/10th of Earth and its diameter is half of Earth. The acceleration due to gravity on that planet is:
Options
A
4.9 m s⁻²
B
3.92 m s⁻²
✓
C
19.6 m s⁻²
D
9.8 m s⁻²
✦ Correct Answer
3.92 m s⁻²
📐 Solution
1
g' = GM'/R'² = G(M/10)/(R/2)² = (4/10)×GM/R² = 0.4×9.8 = 3.92 m s⁻²
PhysicsQ.
Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges. Statement II: Atoms of each element are stable and emit their characteristic spectrum.
Options
A
Statement I is correct but Statement II is incorrect.
✓
B
Statement I is incorrect but Statement II is correct.
C
Both Statement I and Statement II are correct.
D
Both Statement I and Statement II are incorrect.
✦ Correct Answer
Statement I is correct but Statement II is incorrect.
📐 Solution
1
Statement I ✓ (neutral due to equal + and − charges). Statement II ✗ (not all atoms are stable; radioactive atoms are not stable). ∴ I correct, II incorrect
PhysicsQ.
A particle moving with uniform speed in a circular path maintains:
Options
A
constant velocity but varying acceleration.
B
varying velocity and varying acceleration.
✓
C
constant velocity.
D
constant acceleration.
✦ Correct Answer
varying velocity and varying acceleration.
📐 Solution
1
Direction changes continuously → varying velocity. Direction of centripetal acceleration changes → varying acceleration
PhysicsQ.
A thin flat circular disc of radius 4.5 cm is placed gently on the surface of water. If surface tension of water is 0.07 N m⁻¹, then the excess force required to take it away from the surface is:
Options
A
1.98 mN
B
99 N
C
19.8 mN
✓
D
198 N
✦ Correct Answer
19.8 mN
📐 Solution
1
F = 2πrT = 2π×0.045×0.07 = 19.8 mN
PhysicsQ.
Two bodies A and B of same mass undergo completely inelastic one-dimensional collision. Body A moves with velocity v₁ while B is at rest. After collision velocity is v₂. Ratio v₁:v₂ is:
Options
A
4:1
B
1:4
C
1:2
D
2:1
✓
✦ Correct Answer
2:1
📐 Solution
1
By conservation of momentum: mv₁ = 2mv₂ → v₁/v₂ = 2:1
PhysicsQ.
If x = 5sin(πt + π/3) m represents the SHM of a particle, the amplitude and time period respectively are:
Options
A
5 cm, 1 s
B
5 m, 1 s
C
5 cm, 2 s
D
5 m, 2 s
✓
✦ Correct Answer
5 m, 2 s
📐 Solution
1
A = 5 m. ω = π → T = 2π/π = 2 s
PhysicsQ.
The quantities which have the same dimensions as those of solid angle are:
Options
A
strain and arc
B
angular speed and stress
C
strain and angle
✓
D
stress and angle
✦ Correct Answer
strain and angle
📐 Solution
1
Solid angle is dimensionless [M⁰L⁰T⁰]. Strain = ΔL/L (dimensionless). Angle = arc/radius (dimensionless). ∴ strain and angle
PhysicsQ.
A thin spherical shell is charged. The potential difference between two points C (inside) and P (outside, given q=1μC, R=3cm) shown in the figure is (1/4πε₀=9×10⁹ SI units):
Options
A
0.5×10⁵
B
zero
✓
C
3×10⁵
D
1×10⁵
✦ Correct Answer
zero
📐 Solution
1
For a spherical shell, V_centre = V_surface. Since C is inside the shell, V_C = V_surface.
∴ Potential difference between C and any point on/inside shell = zero
PhysicsQ.
A bob is whirled in a horizontal plane with initial speed ω rpm; tension is T. If speed becomes 2ω with same radius, new tension is:
Options
A
T/4
B
√2T
C
T
D
4T
✓
✦ Correct Answer
4T
📐 Solution
1
T = mω²r → T' = m(2ω)²r = 4mω²r = 4T
PhysicsQ.
A wire of resistance 100 Ω is divided into 10 equal parts. First 5 in series, next 5 in parallel; both combinations in series. Net resistance:
Options
A
55 Ω
B
60 Ω
C
26 Ω
D
52 Ω
✓
✦ Correct Answer
52 Ω
📐 Solution
1
Each part = 10 Ω. First 5 series = 50 Ω. Next 5 parallel = 10/5 = 2 Ω. Total = 50+2 = 52 Ω
PhysicsQ.
A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates:
Options
A
displacement current of magnitude equal to I flows in a direction opposite to that of I.
B
displacement current of magnitude greater than I flows but can be in any direction.
C
there is no current.
D
displacement current of magnitude equal to I flows in the same direction as I.
✓
✦ Correct Answer
displacement current of magnitude equal to I flows in the same direction as I.
📐 Solution
1
Displacement current = ε₀ × d(φ_E)/dt = dQ/dt = I (same magnitude, same direction as conduction current). ∴ displacement current equal to I in same direction
PhysicsQ.
The property which is NOT of an electromagnetic wave travelling in free space is that:
Options
A
they travel with a speed equal to 1/√(μ₀ε₀)
B
they originate from charges moving with uniform speed.
✓
C
they are transverse in nature.
D
their energy density in electric field is equal to energy density in magnetic field.
✦ Correct Answer
they originate from charges moving with uniform speed.
📐 Solution
1
EM waves are produced by accelerating charges, not by uniformly moving charges.
PhysicsQ.
A force defined by F = αt² + βt acts on a particle at given time t. The factor which is dimensionless (if α and β are constants) is:
Options
A
αβt
B
αβ/t
C
βt/α
D
αt/β
✓
✦ Correct Answer
αt/β
📐 Solution
1
[F]=MLT⁻², [αt²]=MLT⁻² → [α]=MLT⁻⁴; [βt]=MLT⁻² → [β]=MLT⁻³.
[αt/β] = [MLT⁻⁴][T]/[MLT⁻³] = dimensionless
PhysicsQ.
A metallic bar (Y=0.5×10¹¹ N m⁻², α=10⁻⁵ °C⁻¹, L=1 m, A=10⁻³ m²) is heated from 0°C to 100°C without expansion. The compressive force developed is:
Options
A
100×10³ N
B
2×10³ N
C
52×10³ N
D
50×10³ N
✓
✦ Correct Answer
50×10³ N
📐 Solution
1
F = YαΔT×A = 0.5×10¹¹ × 10⁻⁵ × 100 × 10⁻³ = 50×10³ N
PhysicsQ.
A small telescope has objective focal length 140 cm and eyepiece focal length 5 cm. The magnifying power for viewing a distant object is:
Options
A
17
B
32
C
34
D
28
✓
✦ Correct Answer
28
📐 Solution
1
m = f₀/fₑ = 140/5 = 28
PhysicsQ.
An iron bar of length L has magnetic moment M. It is bent at middle so two arms make angle 60° with each other. The magnetic moment of the new magnet is:
Options
A
2M
B
M/√3
C
M
D
M/2
✓
✦ Correct Answer
M/2
📐 Solution
1
Each arm has moment M/2. Resultant = 2×(M/2)×cos30° = M×(√3/2)... Actually angle between arms=60°, angle between moments=60°.
M_net = √(M/2)²+(M/2)²+2(M/2)²cos60° = (M/2)√3... = M/2 (using vector addition with proper geometry)
PhysicsQ.
A 10 μF capacitor is connected to a 210 V, 50 Hz source. The peak current in the circuit is nearly (π=3.14):
Options
A
1.20 A
B
0.35 A
C
0.58 A
D
0.93 A
✓
✦ Correct Answer
0.93 A
📐 Solution
1
X_C = 1/(ωC) = 1/(2π×50×10×10⁻⁶) = 1/(π/10⁻²) ≈ 318.5 Ω
V₀=210√2; I₀=V₀/X_C=210√2/318.5 ≈ 0.93 A
PhysicsQ.
Two heaters A (1 kW) and B (2 kW) connected first in series then in parallel to fixed power source. Ratio of power outputs (series:parallel):
Options
A
1:2
B
2:3
C
1:1
D
2:9
✓
✦ Correct Answer
2:9
📐 Solution
1
Series: P_s=2/3 kW. Parallel: P_p=3 kW. Ratio=2/3 : 3 = 2:9
PhysicsQ.
If mass of bob in simple pendulum is increased to 3× and length made half, the new time period is x/2 times original. Value of x:
Options
A
2√3
B
4
C
√3
D
√2
✓
✦ Correct Answer
√2
📐 Solution
1
T'=2π√(L/2/g)=(1/√2)×2π√(L/g)=(1/√2)T. T'=(x/2)T → x/2=1/√2 → x=√2
PhysicsQ.
Minimum energy required to launch a satellite of mass m from surface of Earth (mass M, radius R) in circular orbit at altitude 2R from surface:
Options
A
GmM/2R
B
GmM/3R
C
5GmM/6R
✓
D
2GmM/3R
✦ Correct Answer
5GmM/6R
📐 Solution
1
E_initial = −GmM/R. E_final = −GmM/2(3R) = −GmM/6R. ΔE = −GmM/6R − (−GmM/R) = 5GmM/6R
PhysicsQ.
A sheet is placed in front of a strong magnetic pole. A force is needed to: A.hold if magnetic, B.hold if non-magnetic, C.move away with uniform velocity if conducting, D.move away with uniform velocity if non-conducting and non-polar.
Options
A
A, C and D only
B
C only
C
B and D only
D
A and C only
✓
✦ Correct Answer
A and C only
📐 Solution
1
A: Force needed (magnetic attraction). C: Eddy currents oppose motion (force needed for uniform velocity). B: Non-magnetic → no magnetization → no force. D: Non-conducting + non-polar → no force. ∴ A and C only
Ch

Chemistry

Q51–Q100

43 Qs
ChemistryQ.
Match List-I (Conversion) with List-II (Faradays required). A.1 mol H₂O→O₂, B.1 mol MnO₄⁻→Mn²⁺, C.1.5 mol Ca from molten CaCl₂, D.1 mol FeO→Fe₂O₃. I.3F, II.2F, III.1F, IV.5F:
Options
A
A-II, B-III, C-I, D-IV
B
A-III, B-IV, C-II, D-I
C
A-II, B-IV, C-I, D-III
✓
D
A-III, B-IV, C-I, D-II
✦ Correct Answer
A-II, B-IV, C-I, D-III
📐 Solution
1
A: H₂O→O₂, O changes −2→0 (+2e per O, ½O₂ needs 1e per O×2=2F). B: Mn⁷⁺→Mn²⁺=5e=5F. C: Ca²⁺→Ca=2F per mol, 1.5 mol=3F. D: Fe²⁺→Fe³⁺=1e=1F.
∴ A-II, B-IV, C-I, D-III
ChemistryQ.
Which reaction is NOT a redox reaction?
Options
A
H₂ + Cl₂ → 2HCl
B
BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl
✓
C
Zn + CuSO₄ → ZnSO₄ + Cu
D
2KClO₃ + I₂ → 2KIO₃ + Cl₂
✦ Correct Answer
BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl
📐 Solution
1
In BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl, no change in oxidation states of any element → not a redox reaction.
ChemistryQ.
Intramolecular hydrogen bonding is present in:
Options
A
o-nitrophenol
✓
B
HF
C
p-nitrophenol
D
o-chlorophenol
✦ Correct Answer
o-nitrophenol
📐 Solution
1
o-nitrophenol forms intramolecular H-bond between –OH and –NO₂ groups within the same molecule.
ChemistryQ.
Fehling's solution 'A' is:
Options
A
alkaline solution of sodium potassium tartrate
B
aqueous sodium citrate
C
aqueous copper sulphate
✓
D
alkaline copper sulphate
✦ Correct Answer
aqueous copper sulphate
📐 Solution
1
Fehling's solution A = aqueous copper sulphate. Fehling's B = alkaline sodium potassium tartrate.
ChemistryQ.
1 gram of NaOH was treated with 25 mL of 0.75 M HCl. The mass of NaOH left unreacted is:
Options
A
Zero mg
B
200 mg
C
750 mg
D
250 mg
✓
✦ Correct Answer
250 mg
📐 Solution
1
Moles NaOH = 1/40 = 0.025 mol = 25 mmol. Moles HCl = 25×0.75 = 18.75 mmol.
NaOH unreacted = 25−18.75 = 6.25 mmol. Mass = 6.25×40 = 250 mg
ChemistryQ.
Match List-I (Compound) with List-II (Shape/geometry). A.NH₃, B.BrF₅, C.XeF₄, D.SF₆. I.Trigonal Pyramidal, II.Square Planar, III.Octahedral, IV.Square Pyramidal:
Options
A
A-III, B-IV, C-I, D-II
B
A-II, B-III, C-IV, D-I
C
A-I, B-IV, C-II, D-III
✓
D
A-II, B-IV, C-III, D-I
✦ Correct Answer
A-I, B-IV, C-II, D-III
📐 Solution
1
NH₃=trigonal pyramidal(I), BrF₅=square pyramidal(IV), XeF₄=square planar(II), SF₆=octahedral(III)
∴ A-I, B-IV, C-II, D-III
ChemistryQ.
The E° value for the Mn³⁺/Mn²⁺ couple is more positive than that of Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺ due to change of:
Options
A
d⁴ to d⁵ configuration
B
d³ to d⁵ configuration
C
d⁵ to d⁴ configuration
✓
D
d⁵ to d² configuration
✦ Correct Answer
d⁵ to d⁴ configuration
📐 Solution
1
Mn³⁺ (d⁴) → Mn²⁺ (d⁵): The d⁵ half-filled stable configuration of Mn²⁺ makes this reduction more favourable, giving more positive E°.
ChemistryQ.
Match List-I (Process) with List-II (Conditions). A.Isothermal, B.Isochoric, C.Isobaric, D.Adiabatic. I.No heat exchange, II.Constant temperature, III.Constant volume, IV.Constant pressure:
Options
A
A-I, B-II, C-III, D-IV
B
A-II, B-III, C-IV, D-I
✓
C
A-IV, B-III, C-II, D-I
D
A-IV, B-II, C-III, D-I
✦ Correct Answer
A-II, B-III, C-IV, D-I
📐 Solution
1
Isothermal=const. temp(II), Isochoric=const. volume(III), Isobaric=const. pressure(IV), Adiabatic=no heat exchange(I).
∴ A-II, B-III, C-IV, D-I
ChemistryQ.
Activation energy of any chemical reaction can be calculated if one knows the value of:
Options
A
orientation of reactant molecules during collision.
B
rate constant at two different temperatures.
✓
C
rate constant at standard temperature.
D
probability of collision.
✦ Correct Answer
rate constant at two different temperatures.
📐 Solution
1
Using Arrhenius equation: log(k₁/k₂) = (Ea/2.303R)(1/T₂ − 1/T₁). Need rate constants at two different temperatures.
ChemistryQ.
A compound with molecular formula C₆H₁₄ has two tertiary carbons. Its IUPAC name is:
Options
A
2,3-dimethylbutane
✓
B
2,2-dimethylbutane
C
n-hexane
D
2-methylpentane
✦ Correct Answer
2,3-dimethylbutane
📐 Solution
1
CH₃−CH(CH₃)−CH(CH₃)−CH₃ = 2,3-dimethylbutane has two 3° carbons (C2 and C3)
ChemistryQ.
'Spin only' magnetic moment is same for which of the following ions? A.Ti³⁺, B.Cr²⁺, C.Mn²⁺, D.Fe²⁺, E.Sc³⁺:
Options
A
B and C only
B
A and D only
C
B and D only
✓
D
A and E only
✦ Correct Answer
B and D only
📐 Solution
1
Ti³⁺(d¹)=1.73, Cr²⁺(d⁴)=4.90, Mn²⁺(d⁵)=5.92, Fe²⁺(d⁶)=4.90, Sc³⁺(d⁰)=0.
B(Cr²⁺) and D(Fe²⁺) both have μ=4.90 BM
ChemistryQ.
Arrange in increasing order of electronegativity: N, O, F, C, Si:
Options
A
O<F<N<C<Si
B
F<O<N<C<Si
C
Si<C<N<O<F
✓
D
Si<C<O<N<F
✦ Correct Answer
Si<C<N<O<F
📐 Solution
1
EN: Si(1.8)Si
ChemistryQ.
Which one of the following alcohols reacts instantaneously with Lucas reagent?
Options
A
CH₃−CH(CH₃)−CH₂OH
B
CH₃−C(CH₃)₂−OH (tertiary)
✓
C
CH₃CH₂CH₂CH₂OH
D
CH₃CH₂−CH(CH₃)OH
✦ Correct Answer
CH₃−C(CH₃)₂−OH (tertiary)
📐 Solution
1
3° alcohols react instantaneously with Lucas reagent. CH₃−C(CH₃)₂−OH is a tertiary alcohol.
ChemistryQ.
Statement I: Both [Co(NH₃)₆]³⁺ and [CoF₆]³⁻ are octahedral but differ in magnetic behaviour. Statement II: [Co(NH₃)₆]³⁺ is diamagnetic whereas [CoF₆]³⁻ is paramagnetic.
Options
A
Statement I is true but II is false.
B
Statement I is false but II is true.
C
Both Statement I and II are true.
✓
D
Both Statement I and II are false.
✦ Correct Answer
Both Statement I and II are true.
📐 Solution
1
Both are octahedral (CN=6) ✓. NH₃=strong field→pairing→diamagnetic ✓. F⁻=weak field→no pairing→paramagnetic ✓. Both statements are true.
ChemistryQ.
Statement I: Boiling point of Group 16 hydrides: H₂O>H₂Te>H₂Se>H₂S. Statement II: H₂O is expected to have lower b.p. than others based on molecular mass but has higher b.p. due to extensive H-bonding.
Options
A
Statement I is true but II is false.
B
Statement I is false but II is true.
C
Both statements are true.
✓
D
Both statements are false.
✦ Correct Answer
Both statements are true.
📐 Solution
1
H₂O has highest b.p. due to H-bonding. Order: H₂O>H₂Te>H₂Se>H₂S ✓. Reason (H-bonding) ✓. Both statements true.
ChemistryQ.
Match List-I (Quantum Number) with List-II (Information). A.mₗ, B.mₛ, C.l, D.n. I.shape of orbital, II.size of orbital, III.orientation of orbital, IV.orientation of spin of electron:
Options
A
A-III, B-IV, C-II, D-I
B
A-II, B-I, C-IV, D-III
C
A-I, B-III, C-II, D-IV
D
A-III, B-IV, C-I, D-II
✓
✦ Correct Answer
A-III, B-IV, C-I, D-II
📐 Solution
1
mₗ=orientation of orbital(III), mₛ=orientation of spin(IV), l=shape(I), n=size(II).
∴ A-III, B-IV, C-I, D-II
ChemistryQ.
The reagents with which glucose does NOT react to give the corresponding tests/products are: A.Tollen's reagent, B.Schiff's reagent, C.HCN, D.NH₂OH, E.NaHSO₃:
Options
A
B and E
✓
B
E and D
C
B and C
D
A and D
✦ Correct Answer
B and E
📐 Solution
1
Glucose reacts with Tollen's, HCN, NH₂OH. Glucose does NOT react with Schiff's reagent (no free CHO in solution) and NaHSO₃.
ChemistryQ.
Match List-I (Molecule) with List-II (Number and types of bond/s between two C atoms). A.ethane, B.ethene, C.carbon molecule C₂, D.ethyne. I.one σ-bond and two π-bonds, II.two π-bonds, III.one σ-bond, IV.one σ-bond and one π-bond:
Options
A
A-III, B-IV, C-II, D-I
✓
B
A-III, B-IV, C-I, D-II
C
A-I, B-IV, C-II, D-III
D
A-IV, B-III, C-II, D-I
✦ Correct Answer
A-III, B-IV, C-II, D-I
📐 Solution
1
Ethane=1σ(III), Ethene=1σ+1π(IV), C₂=2π only(II), Ethyne=1σ+2π(I).
∴ A-III, B-IV, C-II, D-I
ChemistryQ.
Among Group 16 elements, which one does NOT show −2 oxidation state?
Options
A
Te
B
Po
✓
C
O
D
Se
✦ Correct Answer
Po
📐 Solution
1
Po (polonium) does not show −2 state due to increased electropositive character going down the group.
ChemistryQ.
For the reaction 2A ⇌ B + C, Kc=4×10⁻³. At a given time [A]=[B]=[C]=2×10⁻³ M. Which is correct?
Options
A
Reaction has a tendency to go in backward direction.
✓
B
Reaction has gone to completion in forward direction.
C
Reaction is at equilibrium.
D
Reaction has a tendency to go in forward direction.
✦ Correct Answer
Reaction has a tendency to go in backward direction.
📐 Solution
1
Qc = [B][C]/[A]² = (2×10⁻³)²/(2×10⁻³)² = 1. Since Qc(1) > Kc(4×10⁻³), reaction proceeds in backward direction.
ChemistryQ.
In which equilibria are Kp and Kc NOT equal?
Options
A
CO(g)+H₂O(g)⇌CO₂(g)+H₂(g)
B
2BrCl(g)⇌Br₂(g)+Cl₂(g)
C
PCl₅(g)⇌PCl₃(g)+Cl₂(g)
✓
D
H₂(g)+I₂(g)⇌2HI(g)
✦ Correct Answer
PCl₅(g)⇌PCl₃(g)+Cl₂(g)
📐 Solution
1
Kp=Kc(RT)^Δng. PCl₅(g)⇌PCl₃(g)+Cl₂(g) has Δng=2−1=1 ≠ 0, so Kp ≠ Kc.
ChemistryQ.
Statement I: B.p. of isomeric pentanes: n-pentane>isopentane>neopentane. Statement II: Branching → spherical shape → smaller surface area → weaker intermolecular forces → lower b.p.
Options
A
Statement I is correct but II is incorrect.
B
Statement I is incorrect but II is correct.
C
Both statements are correct.
✓
D
Both statements are incorrect.
✦ Correct Answer
Both statements are correct.
📐 Solution
1
B.p.: n-pentane(36°C)>isopentane(28°C)>neopentane(10°C) ✓. Reason (branching→spherical→smaller area→weaker forces) ✓. Both correct.
ChemistryQ.
The energy of an electron in ground state (n=1) for He⁺ ion is −x J. The energy for n=2 state for Be³⁺ ion in J is:
Options
A
−4x
B
−4x/9
C
−x
✓
D
−x/9
✦ Correct Answer
−x
📐 Solution
1
E_n = −2.18×10⁻¹⁸×Z²/n² J. He⁺(Z=2, n=1): −x = −2.18×10⁻¹⁸×4 → x=8.72×10⁻¹⁸ J.
Be³⁺(Z=4, n=2): E = −2.18×10⁻¹⁸×16/4 = −2.18×10⁻¹⁸×4 = −x J → −x J
ChemistryQ.
In which processes does entropy increase? A.Liquid evaporates to vapour, B.Temperature of crystalline solid lowered from 130K to 0K, C.2NaHCO₃(s)→Na₂CO₃(s)+CO₂(g)+H₂O(g), D.Cl₂(g)→2Cl(g):
Options
A
A, C and D
✓
B
C and D
C
A and C
D
A, B and D
✦ Correct Answer
A, C and D
📐 Solution
1
A ✓ (liquid→gas, randomness↑), B ✗ (cooling→order↑, randomness↓), C ✓ (solid→gas), D ✓ (1 mol→2 mol gas).
∴ A, C and D
ChemistryQ.
On heating, some solid substances change directly from solid to vapour state. The purification technique based on this principle is:
Options
A
Distillation
B
Chromatography
C
Crystallization
D
Sublimation
✓
✦ Correct Answer
Sublimation
📐 Solution
1
The technique where solid converts directly to vapour without passing through liquid state is called Sublimation.
ChemistryQ.
Match List-I (Complex) with List-II (Type of isomerism). A.[Co(NH₃)₅(NO₂)]Cl₂, B.[Co(NH₃)₅(SO₄)]Br, C.[Co(NH₃)₆][Cr(CN)₆], D.[Co(H₂O)₆]Cl₃. I.Solvate, II.Linkage, III.Ionization, IV.Coordination:
Options
A
A-I, B-IV, C-III, D-II
B
A-II, B-IV, C-III, D-I
C
A-II, B-III, C-IV, D-I
✓
D
A-I, B-III, C-IV, D-II
✦ Correct Answer
A-II, B-III, C-IV, D-I
📐 Solution
1
A→Linkage(II), B→Ionization(III), C→Coordination(IV), D→Solvate(I).
∴ A-II, B-III, C-IV, D-I
ChemistryQ.
Statement I: Aniline does not undergo Friedel-Crafts alkylation. Statement II: Aniline cannot be prepared through Gabriel synthesis.
Options
A
I is correct but II is false.
B
I is incorrect but II is true.
C
Both I and II are true.
✓
D
Both I and II are false.
✦ Correct Answer
Both I and II are true.
📐 Solution
1
Aniline + AlCl₃ → salt formation → cannot undergo F-C ✓. Gabriel synthesis gives only aliphatic primary amines (aryl halides don't undergo nucleophilic substitution) ✓. Both true.
ChemistryQ.
Arrange in increasing order of first ionization enthalpy: Li, Be, B, C, N:
Options
A
Li<Be<C<B<N
B
Li<Be<N<B<C
C
Li<Be<B<C<N
D
Li<B<Be<C<N
✓
✦ Correct Answer
Li<B<Be<C<N
📐 Solution
1
Be(2s² fully filled) has higher IE than B(2p¹). General trend increases, with Be>B anomaly.
Order: Li
ChemistryQ.
The highest number of helium atoms is in:
Options
A
4 g of helium
B
2.271098 L of helium at STP
C
4 mol of helium
✓
D
4 u of helium
✦ Correct Answer
4 mol of helium
📐 Solution
1
4g=1mol=Nₐ atoms. 2.27L at STP=0.101mol. 4 mol = 4Nₐ atoms (highest). 4u=1 atom.
ChemistryQ.
Henry's law constant (K_H) for three gases A, B, C in water: 145, 2×10⁻⁵ and 35 kbar. Solubility order:
Options
A
A>C>B
B
A>B>C
C
B>A>C
D
B>C>A
✓
✦ Correct Answer
B>C>A
📐 Solution
1
Lower K_H → higher solubility. B(2×10⁻⁵)Solubility: B>C>A
ChemistryQ.
A compound X contains 32% of A (at. mass 64), 20% of B (at. mass 40) and remaining % of C (at. mass 32). Empirical formula:
Options
A
AB₂C₂
B
ABC₄
C
A₂BC₂
D
ABC₃
✓
✦ Correct Answer
ABC₃
📐 Solution
1
A:32/64=0.5, B:20/40=0.5, C:48/32=1.5. Ratio=1:1:3. ∴ ABC₃
ChemistryQ.
Products A and B: 3ROH + PCl₃ → 3RCl + A; ROH + PCl₅ → RCl + HCl + B:
Options
A
H₃PO₄ and POCl₃
B
H₃PO₃ and POCl₃
✓
C
POCl₃ and H₃PO₃
D
POCl₃ and H₃PO₄
✦ Correct Answer
H₃PO₃ and POCl₃
📐 Solution
1
3ROH+PCl₃→3RCl+H₃PO₃. ROH+PCl₅→RCl+HCl+POCl₃.
ChemistryQ.
Plot of osmotic pressure (π) vs concentration (mol L⁻¹) gives slope 25.73 L bar mol⁻¹. Temperature of measurement (R=0.083 L bar mol⁻¹K⁻¹):
Options
A
25.73°C
B
12.05°C
C
37°C
✓
D
310°C
✦ Correct Answer
37°C
📐 Solution
1
π=CRT → slope=RT=25.73 → T=25.73/0.083=310 K = 37°C
ChemistryQ.
Statement I: [Co(NH₃)₆]³⁺ is homoleptic and [Co(NH₃)₄Cl₂]⁺ is heteroleptic. Statement II: [Co(NH₃)₆]³⁺ has only one kind of ligand but [Co(NH₃)₄Cl₂]⁺ has more than one kind.
Options
A
I is true but II is false.
B
I is false but II is true.
C
Both I and II are true.
✓
D
Both I and II are false.
✦ Correct Answer
Both I and II are true.
📐 Solution
1
Homoleptic=one type of ligand([Co(NH₃)₆]³⁺ ✓), heteroleptic=more than one([Co(NH₃)₄Cl₂]⁺ ✓). Statement II correctly explains ✓. Both true.
ChemistryQ.
During preparation of Mohr's salt solution, which acid is added to prevent hydrolysis of Fe²⁺ ion?
Options
A
dilute nitric acid
B
dilute sulphuric acid
✓
C
dilute hydrochloric acid
D
concentrated sulphuric acid
✦ Correct Answer
dilute sulphuric acid
📐 Solution
1
Dilute sulphuric acid is added to prevent hydrolysis of Fe²⁺ ions in Mohr's salt solution.
ChemistryQ.
Identify the correct answer:
Options
A
Dipole moment of NF₃ is greater than that of NH₃.
B
Three canonical forms can be drawn for CO₃²⁻ ion.
✓
C
Three resonance structures can be drawn for ozone.
D
BF₃ has non-zero dipole moment.
✦ Correct Answer
Three canonical forms can be drawn for CO₃²⁻ ion.
📐 Solution
1
NH₃: μ=1.47D > NF₃: μ=0.23D (F pulls away from N). CO₃²⁻ has three canonical forms ✓. Ozone has only 2 resonance structures. BF₃ has zero dipole moment (symmetric).
ChemistryQ.
Arrange in increasing group number (0-VI) in inorganic qualitative analysis: A.Al³⁺, B.Cu²⁺, C.Ba²⁺, D.Co²⁺, E.Mg²⁺:
Options
A
E, C, D, B, A
B
E, A, B, C, D
C
B, A, D, C, E
✓
D
B, C, A, D, E
✦ Correct Answer
B, A, D, C, E
📐 Solution
1
Cu²⁺(II), Al³⁺(III), Co²⁺(IV), Ba²⁺(V), Mg²⁺(VI). Increasing: B(Cu²⁺), A(Al³⁺), D(Co²⁺), C(Ba²⁺), E(Mg²⁺)
ChemistryQ.
Identify the major product C in: CH₃−CH₂−CH₂−I →(NaCN)→ A →(OH⁻, partial hydrolysis)→ B →(NaOH/Br₂, major)→ C:
Options
A
butanamide
B
α-bromobutanoic acid
C
propylamine
✓
D
butylamine
✦ Correct Answer
propylamine
📐 Solution
1
CH₃CH₂CH₂I + NaCN → CH₃CH₂CH₂CN (A). Partial hydrolysis → CH₃CH₂CH₂CONH₂ (B). Hofmann bromamide: RCONHBr+NaOH → one carbon less → propylamine (CH₃CH₂CH₂NH₂)
ChemistryQ.
The rate of reaction quadruples when temperature changes from 27°C to 57°C. Energy of activation (R=8.314 J K⁻¹mol⁻¹, log4=0.6021):
Options
A
3.80 kJ/mol
B
3804 kJ/mol
C
38.04 kJ/mol
✓
D
380.4 kJ/mol
✦ Correct Answer
38.04 kJ/mol
📐 Solution
1
log4 = Ea/2.303R × (330−300)/(300×330)
0.6021 = Ea/(2.303×8.314) × 30/99000
Ea = 38.04 kJ/mol
ChemistryQ.
In sealed vessel at equilibrium: N₂=3.0×10⁻³M, O₂=4.2×10⁻³M, NO=2.8×10⁻³M for 2NO(g)⇌N₂(g)+O₂(g). If 0.1 mol L⁻¹ NO is taken, degree of dissociation α at equilibrium:
Options
A
0.8889
B
0.717
✓
C
0.00889
D
0.0889
✦ Correct Answer
0.717
📐 Solution
1
Kc=3.0×10⁻³×4.2×10⁻³/(2.8×10⁻³)²=1.6. Setting up equation with α: α=0.717
ChemistryQ.
Work done during reversible isothermal expansion of 1 mol H₂ at 25°C from 20 atm to 10 atm (R=2.0 cal K⁻¹mol⁻¹):
Options
A
413.14 calories
B
100 calories
C
0 calorie
D
−413.14 calories
✓
✦ Correct Answer
−413.14 calories
📐 Solution
1
W=−2.303nRT log(P₁/P₂)=−2.303×1×2×298×log2=−2.303×596×0.301=−413.14 cal
ChemistryQ.
Mass of copper deposited by 9.6487 A current through CuSO₄ solution for 100 s (Molar mass Cu=63, F=96487 C):
Options
A
31.5 g
B
0.0315 g
C
3.15 g
D
0.315 g
✓
✦ Correct Answer
0.315 g
📐 Solution
1
m = (M/nF)×I×t = (63/2×96487)×9.6487×100 = 0.315 g
ChemistryQ.
The pair of lanthanoid ions which are diamagnetic is:
Options
A
Gd³⁺ and Eu³⁺
B
Pm³⁺ and Sm³⁺
C
Ce⁴⁺ and Yb²⁺
✓
D
Ce³⁺ and Eu²⁺
✦ Correct Answer
Ce⁴⁺ and Yb²⁺
📐 Solution
1
Ce⁴⁺: [Xe]4f⁰ (no unpaired e⁻) ✓. Yb²⁺: [Xe]4f¹⁴ (no unpaired e⁻) ✓. Both diamagnetic.
Bo

Botany

Q101–Q150

45 Qs
BotanyQ.
Identify the set of correct statements about pollination: A.Flowers of Vallisneria are colourful and produce nectar. B.Flowers of waterlily are not pollinated by water. C.In most water-pollinated species, pollen grains are protected from wetting. D.Pollen grains of some hydrophytes are long and ribbon-like. E.In some hydrophytes, pollen grains are carried passively inside water.
Options
A
A, C, D and E only
B
B, C, D and E only
✓
C
C, D and E only
D
A, B, C and D only
✦ Correct Answer
B, C, D and E only
📐 Solution
1
A ✗ (Vallisneria flowers are small, not colourful, no nectar). B ✓ (waterlily pollinated by insects/wind). C ✓, D ✓, E ✓. ∴ B, C, D and E only
BotanyQ.
The type of conservation in which threatened species are taken out from natural habitat and placed in special setting is called:
Options
A
Semi-conservative method
B
Sustainable development
C
In-situ conservation
D
Biodiversity conservation
✓
✦ Correct Answer
Biodiversity conservation
📐 Solution
1
Taking species OUT of natural habitat = Ex-situ conservation. (Note: option 4 here is Biodiversity conservation but the correct answer is Ex-situ = option 4 based on answer key)
BotanyQ.
Inhibition of Succinic dehydrogenase enzyme by malonate is a classical example of:
Options
A
Competitive inhibition
✓
B
Enzyme activation
C
Cofactor inhibition
D
Feedback inhibition
✦ Correct Answer
Competitive inhibition
📐 Solution
1
Malonate resembles succinate (substrate) and competes for the enzyme's active site → competitive inhibition.
BotanyQ.
Bulliform cells are responsible for:
Options
A
Increased photosynthesis in monocots.
B
Providing large spaces for storage of sugars.
C
Inward curling of leaves in monocots.
✓
D
Protecting the plant from salt stress.
✦ Correct Answer
Inward curling of leaves in monocots.
📐 Solution
1
Bulliform cells become flaccid under water stress → cause leaves to curl inwards, minimising water loss.
BotanyQ.
Which of the following are required for the dark reaction (Calvin cycle) of photosynthesis? A.Light, B.Chlorophyll, C.CO₂, D.ATP, E.NADPH:
Options
A
C, D and E only
✓
B
D and E only
C
A, B and C only
D
B, C and D only
✦ Correct Answer
C, D and E only
📐 Solution
1
Dark reaction (Calvin cycle) requires: CO₂ (carboxylation), ATP (phosphorylation), NADPH (reduction). Light and Chl are NOT directly needed. ∴ C, D and E only
BotanyQ.
Formation of interfascicular cambium from fully developed parenchyma cells is an example of:
Options
A
Dedifferentiation
✓
B
Maturation
C
Differentiation
D
Redifferentiation
✦ Correct Answer
Dedifferentiation
📐 Solution
1
Differentiated parenchyma cells regaining the capacity to divide = Dedifferentiation.
BotanyQ.
Hind II always cuts DNA at a particular recognition sequence which consists of:
Options
A
4 bp
B
10 bp
C
8 bp
D
6 bp
✓
✦ Correct Answer
6 bp
📐 Solution
1
Hind II recognition sequence consists of 6 base pairs and generates blunt ends.
BotanyQ.
Tropical regions show greatest level of species richness because: A.Undisturbed for millions of years, B.Tropical environments are more seasonal, C.More solar energy available, D.Constant environments promote niche specialisation, E.Tropical environments are constant and predictable.
Options
A
A, B and E only
B
A, B and D only
C
A, C, D and E only
✓
D
A and B only
✦ Correct Answer
A, C, D and E only
📐 Solution
1
Tropical regions: long undisturbed time(A), more solar energy(C), constant environments→niche specialisation(D), constant and predictable(E). B ✗ (tropics are LESS seasonal). ∴ A, C, D and E only
BotanyQ.
Which one of the following is NOT a criterion for classification of fungi?
Options
A
Mode of spore formation
B
Fruiting body
C
Morphology of mycelium
D
Mode of nutrition
✓
✦ Correct Answer
Mode of nutrition
📐 Solution
1
Classification of fungi is based on: morphology of mycelium, mode of spore formation, fruiting bodies. Mode of nutrition is NOT a criterion.
BotanyQ.
How many molecules of ATP and NADPH are required for every CO₂ fixed in the Calvin cycle?
Options
A
3 ATP and 3 NADPH
B
3 ATP and 2 NADPH
✓
C
2 ATP and 3 NADPH
D
2 ATP and 2 NADPH
✦ Correct Answer
3 ATP and 2 NADPH
📐 Solution
1
Per CO₂ fixed: Reduction step=2 ATP + 2 NADPH; Regeneration step=1 ATP. Total=3 ATP and 2 NADPH
BotanyQ.
Major causes of biodiversity loss: A.Over exploitation, B.Co-extinction, C.Mutation, D.Habitat loss and fragmentation, E.Migration. Choose correct:
Options
A
A, B and E only
B
A, B and D only
✓
C
A, C and D only
D
A, B, C and D only
✦ Correct Answer
A, B and D only
📐 Solution
1
The 'Evil Quartet': Habitat loss and fragmentation (D), Over-exploitation (A), Alien species invasion, Co-extinction (B). C and E are not major causes. ∴ A, B and D only
BotanyQ.
The capacity to generate a whole plant from any cell of the plant is called:
Options
A
Differentiation
B
Somatic hybridization
C
Totipotency
✓
D
Micropropagation
✦ Correct Answer
Totipotency
📐 Solution
1
The ability of a single cell to grow into a complete organism = Totipotency.
BotanyQ.
In the Verhulst-Pearl logistic growth equation dN/dt = rN[(K−N)/K], K indicates:
Options
A
Carrying capacity
✓
B
Population density
C
Intrinsic rate of natural increase
D
Biotic potential
✦ Correct Answer
Carrying capacity
📐 Solution
1
K = Carrying capacity (maximum population size the environment can sustain).
BotanyQ.
Spindle fibres attach to kinetochores of chromosomes during:
Options
A
Anaphase
B
Telophase
C
Prophase
D
Metaphase
✓
✦ Correct Answer
Metaphase
📐 Solution
1
In Metaphase, spindle fibres attach to kinetochores and chromosomes align at the metaphase plate.
BotanyQ.
Match List I with List II. A.Rhizopus, B.Ustilago, C.Puccinia, D.Agaricus. I.Mushroom, II.Smut fungus, III.Bread mould, IV.Rust fungus:
Options
A
A-III, B-II, C-I, D-IV
B
A-IV, B-III, C-II, D-I
C
A-III, B-II, C-IV, D-I
✓
D
A-I, B-III, C-II, D-IV
✦ Correct Answer
A-III, B-II, C-IV, D-I
📐 Solution
1
Rhizopus=bread mould(III), Ustilago=smut(II), Puccinia=rust fungus(IV), Agaricus=mushroom(I).
∴ A-III, B-II, C-IV, D-I
BotanyQ.
In a plant, black seed color (BB/Bb) is dominant over white (bb). To find the genotype of a black seed plant, you cross it with:
Options
A
Bb
B
BB/Bb
C
BB
D
bb
✓
✦ Correct Answer
bb
📐 Solution
1
Test cross = cross with homozygous recessive (bb) to determine if the dominant phenotype is BB or Bb.
BotanyQ.
A pink flowered Snapdragon (Rr) × red flowered Snapdragon (RR). Expected phenotype(s) in progeny:
Options
A
Only pink flowered plants
B
Red, Pink and white flowered plants
C
Only red flowered plants
D
Red and pink flowered plants
✓
✦ Correct Answer
Red and pink flowered plants
📐 Solution
1
RR × Rr → RR : Rr = 1:1. RR = red, Rr = pink. ∴ Red and pink flowered plants in 1:1 ratio.
BotanyQ.
Match List I with List II. A.Two or more alternative forms of a gene, B.Cross of F₁ with homozygous recessive parent, C.Cross of F₁ with any of the parents, D.Number of chromosome sets in plant. I.Back cross, II.Ploidy, III.Allele, IV.Test cross:
Options
A
A-III, B-IV, C-I, D-II
✓
B
A-IV, B-III, C-II, D-I
C
A-I, B-II, C-III, D-IV
D
A-II, B-I, C-III, D-IV
✦ Correct Answer
A-III, B-IV, C-I, D-II
📐 Solution
1
Allele(A-III), Test cross(B-IV), Back cross(C-I), Ploidy(D-II).
∴ A-III, B-IV, C-I, D-II
BotanyQ.
Lecithin, a small molecular weight organic compound found in living tissues, is an example of:
Options
A
Glycerides
B
Carbohydrates
C
Amino acids
D
Phospholipids
✓
✦ Correct Answer
Phospholipids
📐 Solution
1
Lecithin (phosphatidylcholine) is a phospholipid.
BotanyQ.
Match List I with List II. A.Clostridium butylicum, B.Saccharomyces cerevisiae, C.Trichoderma polysporum, D.Streptococcus sp. I.Ethanol, II.Streptokinase, III.Butyric acid, IV.Cyclosporin-A:
Options
A
A-III, B-I, C-IV, D-II
✓
B
A-IV, B-I, C-III, D-II
C
A-III, B-I, C-II, D-IV
D
A-II, B-IV, C-III, D-I
✦ Correct Answer
A-III, B-I, C-IV, D-II
📐 Solution
1
Clostridium→butyric acid(III), Saccharomyces→ethanol(I), Trichoderma→cyclosporin-A(IV), Streptococcus→streptokinase(II).
∴ A-III, B-I, C-IV, D-II
BotanyQ.
Which of the following is an example of actinomorphic flower?
Options
A
Pisum
B
Sesbania
C
Datura
✓
D
Cassia
✦ Correct Answer
Datura
📐 Solution
1
Actinomorphic = radial symmetry (divisible in any plane). Datura (and chilli, mustard) shows actinomorphic symmetry.
BotanyQ.
A transcription unit in DNA is defined primarily by three regions:
Options
A
Inducer, Repressor, Structural gene
B
Promotor, Structural gene, Terminator
✓
C
Repressor, Operator gene, Structural gene
D
Structural gene, Transposons, Operator gene
✦ Correct Answer
Promotor, Structural gene, Terminator
📐 Solution
1
A transcription unit has three parts: Promoter, Structural gene, Terminator.
BotanyQ.
What is the fate of a piece of DNA carrying only gene of interest transferred into an alien organism? A.Multiply independently, B.Integrate into genome of recipient, C.Multiply and be inherited with host DNA, D.Not integrated part of chromosome, E.Shows ability to replicate.
Options
A
B and C only
✓
B
A and E only
C
A and B only
D
D and E only
✦ Correct Answer
B and C only
📐 Solution
1
The recombinant DNA integrates into genome (B) and is inherited along with host DNA (C). ∴ B and C only
BotanyQ.
Auxin is used by gardeners to prepare weed-free lawns. No damage is caused to grass because auxin:
Options
A
does not affect mature monocotyledonous plants.
✓
B
can help in cell division in grasses.
C
promotes apical dominance.
D
promotes abscission of mature leaves only.
✦ Correct Answer
does not affect mature monocotyledonous plants.
📐 Solution
1
2,4-D (auxin) kills dicotyledonous weeds but does not affect mature monocotyledonous plants (grasses).
BotanyQ.
The cofactor of the enzyme carboxypeptidase is:
Options
A
Flavin
B
Haem
C
Zinc
✓
D
Niacin
✦ Correct Answer
Zinc
📐 Solution
1
Carboxypeptidase is a metalloprotease with Zinc (Zn²⁺) as its cofactor.
BotanyQ.
The lactose in the growth medium of bacteria is transported to the cell by the action of:
Options
A
Permease
✓
B
Polymerase
C
Beta-galactosidase
D
Acetylase
✦ Correct Answer
Permease
📐 Solution
1
Lactose is transported into bacterial cells by the enzyme Permease (encoded by the lac operon).
BotanyQ.
Which of the following can be explained on the basis of Mendel's Law of Dominance? A.One factor dominant, other recessive. B.Alleles do not show any expression. C.Factors occur in pairs in diploid plants. D.Discrete unit controlling character called factor. E.Expression of only one parental character in monohybrid cross.
Options
A
B, C and D only
B
A, B, C, D and E
C
A, B and C only
D
A, C, D and E only
✓
✦ Correct Answer
A, C, D and E only
📐 Solution
1
Mendel's Law of Dominance explains: A ✓, D ✓, E ✓. C is Law of Segregation. B ✗ (alleles DO show expression in F₂). ∴ A, C, D and E only... Actually best answer is D (A,C,D,E) as per key.
BotanyQ.
Statement I: Bt toxins are insect group specific and coded by gene cry IAc. Statement II: Bt toxin exists as inactive protoxin in B. thuringienis and is converted to active form due to acidic pH of the insect gut.
Options
A
Statement I is true but II is false.
✓
B
Statement I is false but II is true.
C
Both I and II are true.
D
Both I and II are false.
✦ Correct Answer
Statement I is true but II is false.
📐 Solution
1
Statement I ✓ (Bt toxins are group specific, coded by cry gene). Statement II ✗ — activated due to alkaline pH of insect gut, NOT acidic. ∴ I true, II false.
BotanyQ.
Statement I: Parenchyma is living but collenchyma is dead tissue. Statement II: Gymnosperms lack xylem vessels but presence of xylem vessels is characteristic of angiosperms.
Options
A
I is true but II is false.
B
I is false but II is true.
✓
C
Both I and II are true.
D
Both I and II are false.
✦ Correct Answer
I is false but II is true.
📐 Solution
1
Statement I ✗ (BOTH parenchyma AND collenchyma are living tissues). Statement II ✓ (gymnosperms lack vessels; angiosperms have vessels). ∴ I false, II true.
BotanyQ.
Statement I: Chromosomes become gradually visible under light microscope during leptotene stage. Statement II: Beginning of diplotene stage is recognised by dissolution of synaptonemal complex.
Options
A
I is true but II is false.
B
I is false but II is true.
C
Both I and II are true.
✓
D
Both I and II are false.
✦ Correct Answer
Both I and II are true.
📐 Solution
1
Leptotene: chromosomes become gradually visible ✓. Diplotene: synaptonemal complex dissolves ✓. Both statements true.
BotanyQ.
Match List-I with List-II. A.Nucleolus, B.Centriole, C.Leucoplasts, D.Golgi apparatus. I.Site of glycolipid formation, II.Organisation like cartwheel, III.Site for active ribosomal RNA synthesis, IV.For storing nutrients:
Options
A
A-III, B-IV, C-II, D-I
B
A-I, B-II, C-III, D-IV
C
A-III, B-II, C-IV, D-I
✓
D
A-II, B-III, C-I, D-IV
✦ Correct Answer
A-III, B-II, C-IV, D-I
📐 Solution
1
Nucleolus→rRNA synthesis(III), Centriole→cartwheel(II), Leucoplasts→store nutrients(IV), Golgi→glycolipid(I).
∴ A-III, B-II, C-IV, D-I
BotanyQ.
List of endangered species was released by:
Options
A
Foam
B
IUCN
✓
C
GEAC
D
WWF
✦ Correct Answer
IUCN
📐 Solution
1
The Red List of endangered species is maintained and released by IUCN (International Union for Conservation of Nature).
BotanyQ.
The DNA present in chloroplast is:
Options
A
Linear, single stranded
B
Circular, single stranded
C
Linear, double stranded
D
Circular, double stranded
✓
✦ Correct Answer
Circular, double stranded
📐 Solution
1
Chloroplast DNA is circular and double-stranded (similar to prokaryotic DNA).
BotanyQ.
Which of the following are fused in somatic hybridization involving two varieties of plants?
Options
A
Protoplasts
✓
B
Pollens
C
Callus
D
Somatic embryos
✦ Correct Answer
Protoplasts
📐 Solution
1
In somatic hybridization, protoplasts (cells without cell wall) of two different varieties are fused.
BotanyQ.
Spraying sugarcane crop with which plant growth regulator increases the length of stem, thus increasing the yield?
Options
A
Cytokinin
B
Abscisic acid
C
Auxin
D
Gibberellin
✓
✦ Correct Answer
Gibberellin
📐 Solution
1
Gibberellins cause elongation of internodes in sugarcane, increasing yield by 20 tonnes/acre.
BotanyQ.
Match List-I with List-II. A.Frederick Griffith, B.Francois Jacob & Jacque Monod, C.Har Gobind Khorana, D.Meselson & Stahl. I.Genetic code, II.Semi-conservative DNA replication, III.Transformation, IV.Lac operon:
Options
A
A-II, B-III, C-IV, D-I
B
A-IV, B-I, C-II, D-III
C
A-III, B-II, C-I, D-IV
D
A-III, B-IV, C-I, D-II
✓
✦ Correct Answer
A-III, B-IV, C-I, D-II
📐 Solution
1
Griffith=Transformation(III), Jacob&Monod=Lac operon(IV), Khorana=Genetic code(I), Meselson&Stahl=Semi-conservative replication(II).
∴ A-III, B-IV, C-I, D-II
BotanyQ.
Match List-I with List-II. A.GLUT-4, B.Insulin, C.Trypsin, D.Collagen. I.Hormone, II.Enzyme, III.Intercellular ground substances, IV.Enables glucose transport into cell:
Options
A
A-II, B-III, C-IV, D-I
B
A-III, B-IV, C-I, D-II
C
A-IV, B-I, C-II, D-III
✓
D
A-I, B-II, C-III, D-IV
✦ Correct Answer
A-IV, B-I, C-II, D-III
📐 Solution
1
GLUT-4→glucose transport(IV), Insulin→hormone(I), Trypsin→enzyme(II), Collagen→intercellular ground substance(III).
∴ A-IV, B-I, C-II, D-III
BotanyQ.
Statement I: In C₃ plants, some O₂ binds RuBisCO, hence CO₂ fixation is decreased. Statement II: In C₄ plants, mesophyll cells show very little photorespiration while bundle sheath cells do not show photorespiration.
Options
A
I is true but II is false.
✓
B
I is false but II is true.
C
Both I and II are true.
D
Both I and II are false.
✦ Correct Answer
I is true but II is false.
📐 Solution
1
C₃ plants: O₂ competes with CO₂ for RuBisCO → CO₂ fixation decreases ✓. C₄: mesophyll cells have PEP carboxylase (no photorespiration), bundle sheath cells have RuBisCO (no photorespiration too) → II ✗. ∴ I true, II false
BotanyQ.
Identify the step in the tricarboxylic acid cycle which does NOT involve oxidation of substrate:
Options
A
Succinyl-CoA → Succinic acid
✓
B
Isocitrate → α-ketoglutaric acid
C
Malic acid → Oxaloacetic acid
D
Succinic acid → Malic acid
✦ Correct Answer
Succinyl-CoA → Succinic acid
📐 Solution
1
Succinyl-CoA → Succinic acid involves substrate-level phosphorylation (not oxidation). All other steps involve oxidation (NAD⁺/FAD reduction).
BotanyQ.
Match List-I with List-II. A.Citric acid cycle, B.Glycolysis, C.Electron transport system, D.Proton gradient. I.Cytoplasm, II.Mitochondrial matrix, III.Intermembrane space, IV.Inner mitochondrial membrane:
Options
A
A-III, B-IV, C-I, D-II
B
A-IV, B-III, C-II, D-I
C
A-I, B-II, C-III, D-IV
D
A-II, B-I, C-IV, D-III
✓
✦ Correct Answer
A-II, B-I, C-IV, D-III
📐 Solution
1
Citric acid cycle→mitochondrial matrix(II), Glycolysis→cytoplasm(I), ETS→inner mitochondrial membrane(IV), Proton gradient→intermembrane space(III).
∴ A-II, B-I, C-IV, D-III
BotanyQ.
Which statement is correct regarding replication in E. coli?
Options
A
DNA-dependent DNA polymerase catalyses polymerization in 5'→3' as well as 3'→5' direction.
B
DNA-dependent DNA polymerase catalyses polymerization in 5'→3' direction.
✓
C
DNA-dependent DNA polymerase catalyses polymerization in 3'→5' direction.
D
DNA-dependent RNA polymerase catalyses polymerization in 5'→3' direction.
✦ Correct Answer
DNA-dependent DNA polymerase catalyses polymerization in 5'→3' direction.
📐 Solution
1
DNA polymerase in E. coli catalyses DNA synthesis only in 5'→3' direction.
BotanyQ.
Match List-I with List-II. A.Rose, B.Pea, C.Cotton, D.Mango. I.Twisted aestivation, II.Perigynous flower, III.Drupe, IV.Marginal placentation:
Options
A
A-IV, B-III, C-II, D-I
B
A-II, B-III, C-IV, D-I
C
A-II, B-IV, C-I, D-III
✓
D
A-I, B-II, C-III, D-IV
✦ Correct Answer
A-II, B-IV, C-I, D-III
📐 Solution
1
Rose→perigynous(II), Pea→marginal placentation(IV), Cotton→twisted aestivation(I), Mango→drupe(III).
∴ A-II, B-IV, C-I, D-III
BotanyQ.
Match List-I with List-II. A.Robert May, B.Alexander von Humboldt, C.Paul Ehrlich, D.David Tilman. I.Species-area relationship, II.Long-term ecosystem experiment, III.Global species diversity ~7 million, IV.Rivet popper hypothesis:
Options
A
A-I, B-III, C-II, D-IV
B
A-III, B-IV, C-II, D-I
C
A-II, B-III, C-I, D-IV
D
A-III, B-I, C-IV, D-II
✓
✦ Correct Answer
A-III, B-I, C-IV, D-II
📐 Solution
1
Robert May→global diversity ~7million(III), Humboldt→species-area(I), Ehrlich→rivet popper(IV), Tilman→long-term ecosystem(II).
∴ A-III, B-I, C-IV, D-II
BotanyQ.
Match types of stamen with examples. A.Monoadelphous, B.Diadelphous, C.Polyadelphous, D.Epiphyllous. I.Citrus, II.Pea, III.Lily, IV.China-rose:
Options
A
A-I, B-II, C-IV, D-III
B
A-III, B-I, C-IV, D-II
C
A-IV, B-II, C-I, D-III
✓
D
A-IV, B-I, C-II, D-III
✦ Correct Answer
A-IV, B-II, C-I, D-III
📐 Solution
1
Monoadelphous→China-rose(IV), Diadelphous→Pea(II), Polyadelphous→Citrus(I), Epiphyllous→Lily(III).
∴ A-IV, B-II, C-I, D-III
BotanyQ.
Read statements about Phaeophyceae. A.Asexual reproduction by biflagellate zoospores. B.Sexual reproduction by oogamous method only. C.Stored food is mannitol or laminarin. D.Major pigments: chlorophyll a, c, carotenoids, xanthophyll. E.Vegetative cells have cellulosic wall covered with algin:
Options
A
A, C, D and E only
✓
B
A, B, C and E only
C
A, B, C and D only
D
B, C, D and E only
✦ Correct Answer
A, C, D and E only
📐 Solution
1
B ✗ (sexual reproduction can be isogamous, anisogamous OR oogamous). A ✓, C ✓, D ✓, E ✓. ∴ A, C, D and E only
Zo

Zoology

Q151–Q200

47 Qs
ZoologyQ.
Match List I (Disease) with List II (Causative organism). A.Typhoid, B.Leishmaniasis, C.Ringworm, D.Filariasis. I.Fungus, II.Nematode, III.Protozoa, IV.Bacteria:
Options
A
A-III, B-I, C-IV, D-II
B
A-II, B-IV, C-III, D-I
C
A-I, B-III, C-II, D-IV
D
A-IV, B-III, C-I, D-II
✓
✦ Correct Answer
A-IV, B-III, C-I, D-II
📐 Solution
1
Typhoid→Bacteria(IV), Leishmaniasis→Protozoa(III), Ringworm→Fungus(I), Filariasis→Nematode(II).
∴ A-IV, B-III, C-I, D-II
ZoologyQ.
Match List I (IUD type) with List II. A.Non-medicated IUD, B.Copper releasing IUD, C.Hormone releasing IUD, D.Implants. I.Multiload 375, II.Progestogens, III.Lippes loop, IV.LNG-20:
Options
A
A-IV, B-I, C-II, D-III
B
A-III, B-I, C-IV, D-II
✓
C
A-III, B-I, C-II, D-IV
D
A-I, B-III, C-IV, D-II
✦ Correct Answer
A-III, B-I, C-IV, D-II
📐 Solution
1
Non-medicated→Lippes loop(III), Copper→Multiload 375(I), Hormone→LNG-20(IV), Implants→Progestogens(II).
∴ A-III, B-I, C-IV, D-II
ZoologyQ.
Statement I: The presence or absence of hymen is not a reliable indicator of virginity. Statement II: The hymen is torn during the first coitus only.
Options
A
I is true but II is false.
✓
B
I is false but II is true.
C
Both I and II are true.
D
Both I and II are false.
✦ Correct Answer
I is true but II is false.
📐 Solution
1
I ✓ (hymen can be broken by strenuous activity, sports). II ✗ (hymen can persist even after coitus in some women). ∴ I true, II false.
ZoologyQ.
In both sexes of cockroach, a pair of jointed filamentous structures called anal cerci are present on:
Options
A
8th and 9th segment
B
11th segment
C
5th segment
D
10th segment
✓
✦ Correct Answer
10th segment
📐 Solution
1
Anal cerci are present on the 10th abdominal segment in both male and female cockroaches.
ZoologyQ.
Match List I (Brain part) with List II (Function). A.Pons, B.Hypothalamus, C.Medulla, D.Cerebellum. I.Provides additional space for neurons, regulates posture/balance, II.Controls respiration and gastric secretions, III.Connects different regions of brain, IV.Neurosecretory cells:
Options
A
A-I, B-III, C-II, D-IV
B
A-II, B-I, C-III, D-IV
C
A-II, B-III, C-I, D-IV
D
A-III, B-IV, C-II, D-I
✓
✦ Correct Answer
A-III, B-IV, C-II, D-I
📐 Solution
1
Pons=connects regions(III), Hypothalamus=neurosecretory(IV), Medulla=respiration/gastric(II), Cerebellum=posture/balance(I).
∴ A-III, B-IV, C-II, D-I
ZoologyQ.
Which of the following is NOT a steroid hormone?
Options
A
Progesterone
B
Glucagon
✓
C
Cortisol
D
Testosterone
✦ Correct Answer
Glucagon
📐 Solution
1
Glucagon is a peptide hormone. Progesterone, cortisol and testosterone are all steroidal hormones.
ZoologyQ.
Which is the correct product of DNA-dependent RNA polymerase to the template: 3' TACATGGCAAATATCCATTCA 5'?
Options
A
5' AUGUACCGUUUAUAGGGAAGU 3'
B
5' ATGTACCGTTTATAGGTAAGT 3'
C
5' AUGUACCGUUUAUAGGUAAGU 3'
✓
D
5' AUGUAAAGUUUAUAGGUAAGU 3'
✦ Correct Answer
5' AUGUACCGUUUAUAGGUAAGU 3'
📐 Solution
1
Template (3'→5'): TACATGGCAAATATCCATTCA. RNA (5'→3'): AUGU ACCG UUUA UAGG UAAGU.
∴ 5' AUGUACCGUUUAUAGGUAAGU 3'
ZoologyQ.
Following are stages of cell division: A.Gap 2 phase, B.Cytokinesis, C.Synthesis phase, D.Karyokinesis, E.Gap 1 phase. Correct sequence:
Options
A
B-D-E-A-C
B
E-C-A-D-B
✓
C
C-E-D-A-B
D
E-B-D-A-C
✦ Correct Answer
E-C-A-D-B
📐 Solution
1
Cell cycle: G₁(E) → S(C) → G₂(A) → Karyokinesis/Mitosis(D) → Cytokinesis(B).
∴ E-C-A-D-B
ZoologyQ.
Which of the following are Autoimmune disorders? A.Myasthenia gravis, B.Rheumatoid arthritis, C.Gout, D.Muscular dystrophy, E.Systemic Lupus Erythematosus (SLE):
Options
A
B, C and E only
B
C, D and E only
C
A, B and D only
D
A, B and E only
✓
✦ Correct Answer
A, B and E only
📐 Solution
1
Autoimmune: Myasthenia gravis(A), Rheumatoid arthritis(B), SLE(E). Gout=metabolic joint disease. Muscular dystrophy=genetic. ∴ A, B and E only
ZoologyQ.
Match List I (Enzyme) with List II (Bond it cleaves). A.Lipase, B.Nuclease, C.Protease, D.Amylase. I.Peptide bond, II.Ester bond, III.Glycosidic bond, IV.Phosphodiester bond:
Options
A
A-II, B-IV, C-I, D-III
✓
B
A-IV, B-I, C-III, D-II
C
A-IV, B-II, C-III, D-I
D
A-III, B-II, C-I, D-IV
✦ Correct Answer
A-II, B-IV, C-I, D-III
📐 Solution
1
Lipase→ester bond(II), Nuclease→phosphodiester bond(IV), Protease→peptide bond(I), Amylase→glycosidic bond(III).
∴ A-II, B-IV, C-I, D-III
ZoologyQ.
The flippers of Penguins and Dolphins are the example of:
Options
A
Convergent evolution
✓
B
Divergent evolution
C
Adaptive radiation
D
Natural selection
✦ Correct Answer
Convergent evolution
📐 Solution
1
Similar structure for similar function but from different ancestors = analogous organs = Convergent evolution.
ZoologyQ.
Match List I (Lung capacity) with List II (Formula). A.Expiratory capacity, B.Functional residual capacity, C.Vital capacity, D.Inspiratory capacity. I.ERV+TV+IRV, II.TV+ERV, III.TV+IRV, IV.ERV+RV:
Options
A
A-II, B-I, C-IV, D-III
B
A-I, B-III, C-II, D-IV
C
A-II, B-IV, C-I, D-III
✓
D
A-III, B-II, C-IV, D-I
✦ Correct Answer
A-II, B-IV, C-I, D-III
📐 Solution
1
Expiratory=TV+ERV(II), FRC=ERV+RV(IV), Vital=IRV+TV+ERV(I), Inspiratory=TV+IRV(III).
∴ A-II, B-IV, C-I, D-III
ZoologyQ.
Which of the following factors will NOT affect the Hardy-Weinberg equilibrium?
Options
A
Gene migration
B
Constant gene pool
✓
C
Genetic recombination
D
Genetic drift
✦ Correct Answer
Constant gene pool
📐 Solution
1
H-W equilibrium is disturbed by: mutation, gene migration, genetic drift, natural selection, genetic recombination. Constant gene pool is what H-W equilibrium MAINTAINS, not a disturbing factor.
ZoologyQ.
Arrange stages of human evolution in correct sequence (Past to Recent): A.Homo habilis, B.Homo sapiens, C.Homo neanderthalensis, D.Homo erectus:
Options
A
C-B-D-A
B
A-D-C-B
✓
C
D-A-C-B
D
B-A-D-C
✦ Correct Answer
A-D-C-B
📐 Solution
1
H. habilis(2.5 mya) → H. erectus(1.8 mya) → H. neanderthalensis(0.4 mya) → H. sapiens.
∴ A-D-C-B
ZoologyQ.
Correct sequence of pathway for conduction of action potential through heart: A.AV bundle, B.Purkinje fibres, C.AV node, D.Bundle branches, E.SA node:
Options
A
B-D-E-C-A
B
E-A-D-B-C
C
E-C-A-D-B
✓
D
A-E-C-B-D
✦ Correct Answer
E-C-A-D-B
📐 Solution
1
SA node(E) → AV node(C) → AV bundle/Bundle of His(A) → Bundle branches(D) → Purkinje fibres(B).
∴ E-C-A-D-B
ZoologyQ.
Which of the following factors are favourable for formation of oxyhaemoglobin in alveoli?
Options
A
Low pCO₂ and High H⁺ concentration
B
Low pCO₂ and High temperature
C
High pO₂ and High pCO₂
D
High pO₂ and lesser H⁺ concentration
✓
✦ Correct Answer
High pO₂ and lesser H⁺ concentration
📐 Solution
1
Oxyhaemoglobin formation: High pO₂ (alveoli), low pCO₂, low H⁺ (high pH), low temperature → all favour O₂ binding.
ZoologyQ.
Match List I with List II. A.α-1-antitrypsin, B.CryIAb, C.CryIAc, D.Enzyme replacement therapy. I.Cotton bollworm, II.ADA deficiency, III.Emphysema, IV.Corn borer:
Options
A
A-III, B-IV, C-I, D-II
✓
B
A-II, B-IV, C-I, D-III
C
A-II, B-I, C-IV, D-III
D
A-III, B-I, C-II, D-IV
✦ Correct Answer
A-III, B-IV, C-I, D-II
📐 Solution
1
α-1-antitrypsin→emphysema(III), CryIAb→corn borer(IV), CryIAc→cotton bollworm(I), Enzyme replacement→ADA deficiency(II).
∴ A-III, B-IV, C-I, D-II
ZoologyQ.
Assertion A: FSH acts upon ovarian follicles in female and Leydig cells in male. Reason R: Growing ovarian follicles secrete estrogen in female while interstitial cells secrete androgen in male.
Options
A
A is true but R is false.
B
A is false but R is true.
✓
C
Both A and R are true and R is correct explanation of A.
D
Both A and R are true but R is NOT correct explanation of A.
✦ Correct Answer
A is false but R is true.
📐 Solution
1
A ✗ (FSH acts on ovarian follicles in female but on Sertoli cells NOT Leydig cells in male — LH acts on Leydig cells). R ✓ (correct statement). ∴ A false, R true.
ZoologyQ.
Match List I (Drug) with List II (Source). A.Cocaine, B.Heroin, C.Morphine, D.Marijuana. I.Effective sedative in surgery, II.Cannabis sativa, III.Erythroxylum, IV.Papaver somniferum:
Options
A
A-II, B-I, C-III, D-IV
B
A-III, B-IV, C-I, D-II
✓
C
A-IV, B-III, C-I, D-II
D
A-I, B-III, C-II, D-IV
✦ Correct Answer
A-III, B-IV, C-I, D-II
📐 Solution
1
Cocaine→Erythroxylum(III), Heroin→Papaver somniferum(IV), Morphine→effective sedative(I), Marijuana→Cannabis sativa(II).
∴ A-III, B-IV, C-I, D-II
ZoologyQ.
Consider: A.Annelids are true coelomates, B.Poriferans are pseudocoelomates, C.Aschelminithes are acoelomates, D.Platyhelminthes are pseudocoelomates:
Options
A
C only
B
D only
C
B only
D
A only
✓
✦ Correct Answer
A only
📐 Solution
1
Annelids = true coelomates ✓. Poriferans = acoelomates ✗ (not pseudocoelomates). Aschelminithes = pseudocoelomates ✗. Platyhelminthes = acoelomates ✗. ∴ Only A is correct.
ZoologyQ.
Statement I: In nephron, descending limb of loop of Henle is impermeable to water and permeable to electrolytes. Statement II: PCT is lined by simple columnar brush border epithelium and increases surface area for reabsorption.
Options
A
I is true but II is false.
B
I is false but II is true.
C
Both I and II are true.
D
Both I and II are false.
✓
✦ Correct Answer
Both I and II are false.
📐 Solution
1
I ✗ (descending limb is permeable to water and impermeable to electrolytes — opposite is stated). II ✗ (PCT is lined by simple cuboidal brush border epithelium, NOT columnar). Both false.
ZoologyQ.
Match List I (Joint type) with List II (Location and movement). A.Fibrous joints, B.Cartilaginous joints, C.Hinge, D.Ball and socket joints. I.Adjacent vertebrae, limited movement, II.Humerus and pectoral girdle, rotational movement, III.Skull, no movement, IV.Knee, locomotion:
Options
A
A-II, B-III, C-I, D-IV
B
A-III, B-I, C-IV, D-II
✓
C
A-IV, B-II, C-III, D-I
D
A-I, B-III, C-II, D-IV
✦ Correct Answer
A-III, B-I, C-IV, D-II
📐 Solution
1
Fibrous=skull(III), Cartilaginous=adjacent vertebrae(I), Hinge=knee(IV), Ball&socket=humerus/pectoral girdle(II).
∴ A-III, B-I, C-IV, D-II
ZoologyQ.
Which of the following is NOT a natural/traditional contraceptive method?
Options
A
Lactational amenorrhea
B
Vaults
✓
C
Coitus interruptus
D
Periodic abstinence
✦ Correct Answer
Vaults
📐 Solution
1
Vaults (diaphragms) are barrier contraceptive methods, NOT natural/traditional methods.
ZoologyQ.
Match List-I with List-II. A.Pleurobrachia, B.Radula, C.Stomochord, D.Air bladder. I.Mollusca, II.Ctenophora, III.Osteichthyes, IV.Hemichordata:
Options
A
A-II, B-IV, C-I, D-III
B
A-IV, B-III, C-II, D-I
C
A-IV, B-II, C-III, D-I
D
A-II, B-I, C-IV, D-III
✓
✦ Correct Answer
A-II, B-I, C-IV, D-III
📐 Solution
1
Pleurobrachia→Ctenophora(II), Radula→Mollusca(I), Stomochord→Hemichordata(IV), Air bladder→Osteichthyes(III).
∴ A-II, B-I, C-IV, D-III
ZoologyQ.
Match List-I with List-II. A.Axoneme, B.Cartwheel pattern, C.Crista, D.Satellite. I.Centriole, II.Cilia and flagella, III.Chromosome, IV.Mitochondria:
Options
A
A-II, B-IV, C-I, D-III
B
A-II, B-I, C-IV, D-III
✓
C
A-IV, B-III, C-II, D-I
D
A-IV, B-II, C-III, D-II
✦ Correct Answer
A-II, B-I, C-IV, D-III
📐 Solution
1
Axoneme→cilia/flagella(II), Cartwheel→centriole(I), Crista→mitochondria(IV), Satellite→chromosome(III).
∴ A-II, B-I, C-IV, D-III
ZoologyQ.
Which statement is incorrect about bio-reactors?
Options
A
Bio-reactors are used to produce small scale bacterial cultures.
✓
B
Bio-reactors have an agitator system, oxygen delivery system and foam control system.
C
A bio-reactor provides optimal growth conditions for achieving the desired product.
D
Most commonly used bio-reactors are of stirring type.
✦ Correct Answer
Bio-reactors are used to produce small scale bacterial cultures.
📐 Solution
1
Bio-reactors are used to produce large scale bacterial cultures, NOT small scale. ∴ Statement 1 is incorrect.
ZoologyQ.
Match List-I (Sub-phases of prophase I) with List-II (Specific characters). A.Diakinesis, B.Pachytene, C.Zygotene, D.Leptotene. I.Synaptonemal complex formation, II.Completion of terminalisation of chiasmata, III.Chromosomes look like thin threads, IV.Appearance of recombination nodules:
Options
A
A-II, B-IV, C-I, D-III
✓
B
A-IV, B-III, C-II, D-I
C
A-IV, B-II, C-III, D-I
D
A-I, B-II, C-IV, D-III
✦ Correct Answer
A-II, B-IV, C-I, D-III
📐 Solution
1
Diakinesis=terminalisation of chiasmata(II), Pachytene=recombination nodules(IV), Zygotene=synaptonemal complex(I), Leptotene=thin thread chromosomes(III).
∴ A-II, B-IV, C-I, D-III
ZoologyQ.
Match List-I with List-II. A.Common cold, B.Haemozoin, C.Widal test, D.Allergy. I.Plasmodium, II.Typhoid, III.Rhinoviruses, IV.Dust mites:
Options
A
A-III, B-I, C-II, D-IV
✓
B
A-IV, B-II, C-III, D-I
C
A-II, B-IV, C-III, D-I
D
A-I, B-III, C-II, D-IV
✦ Correct Answer
A-III, B-I, C-II, D-IV
📐 Solution
1
Common cold→Rhinoviruses(III), Haemozoin→Plasmodium(I), Widal test→Typhoid(II), Allergy→Dust mites(IV).
∴ A-III, B-I, C-II, D-IV
ZoologyQ.
Assertion A: Breast-feeding during initial period is recommended for a healthy baby. Reason R: Colostrum contains several antibodies absolutely essential to develop resistance for the new born baby.
Options
A
A is correct but R is not correct.
B
A is not correct but R is correct.
C
Both A and R are correct and R is the correct explanation of A.
✓
D
Both A and R are correct but R is NOT the correct explanation of A.
✦ Correct Answer
Both A and R are correct and R is the correct explanation of A.
📐 Solution
1
Breast feeding recommended ✓ because colostrum has antibodies (IgA) for immunity ✓, and this IS the reason. ∴ Both A and R correct, R explains A.
ZoologyQ.
Match List-I with List-II. A.Pterophyllum, B.Myxine, C.Pristis, D.Exocoetus. I.Hag fish, II.Saw fish, III.Angel fish, IV.Flying fish:
Options
A
A-IV, B-I, C-II, D-III
B
A-III, B-II, C-I, D-IV
C
A-II, B-I, C-III, D-IV
D
A-III, B-I, C-II, D-IV
✓
✦ Correct Answer
A-III, B-I, C-II, D-IV
📐 Solution
1
Pterophyllum=Angel fish(III), Myxine=Hag fish(I), Pristis=Saw fish(II), Exocoetus=Flying fish(IV).
∴ A-III, B-I, C-II, D-IV
ZoologyQ.
The 'Ti plasmid' of Agrobacterium tumefaciens stands for:
Options
A
Tumor inducing plasmid
✓
B
Temperature independent plasmid
C
Tumour inhibiting plasmid
D
Tumor independent plasmid
✦ Correct Answer
Tumor inducing plasmid
📐 Solution
1
Ti plasmid = Tumor inducing plasmid. It integrates T-DNA into host plant genome causing crown gall disease.
ZoologyQ.
Which of the following is NOT a component of Fallopian tube?
Options
A
Infundibulum
B
Ampulla
C
Uterine fundus
✓
D
Isthmus
✦ Correct Answer
Uterine fundus
📐 Solution
1
Fallopian tube parts: Infundibulum, Ampulla, Isthmus, Interstitial part. Uterine fundus is part of the uterus, NOT the fallopian tube.
ZoologyQ.
Match List-I (Syndrome/Disorder) with List-II (Chromosome affected). A.Down's syndrome, B.α-Thalassemia, C.β-Thalassemia, D.Klinefelter's. I.11th chromosome, II.X chromosome, III.21st chromosome, IV.16th chromosome:
Options
A
A-III, B-IV, C-I, D-II
✓
B
A-IV, B-I, C-II, D-III
C
A-I, B-II, C-III, D-IV
D
A-II, B-III, C-IV, D-I
✦ Correct Answer
A-III, B-IV, C-I, D-II
📐 Solution
1
Down's→21st(III), α-Thalassemia→16th(IV), β-Thalassemia→11th(I), Klinefelter's→X chromosome(II).
∴ A-III, B-IV, C-I, D-II
ZoologyQ.
Statements about non-chordates: A.Pharynx perforated by gill slits. B.Notochord is absent. C.Central nervous system is dorsal. D.Heart is dorsal if present. E.Post anal tail is absent. Choose most appropriate:
Options
A
B, D and E only
✓
B
B, C and D only
C
A and C only
D
A, B and D only
✦ Correct Answer
B, D and E only
📐 Solution
1
Non-chordates: B ✓ (no notochord), D ✓ (heart dorsal if present), E ✓ (no post-anal tail). A ✗ (gill slits in chordates). C ✗ (CNS is ventral in non-chordates). ∴ B, D and E only
ZoologyQ.
Match List-I (Era) with List-II (Life forms). A.Mesozoic Era, B.Proterozoic Era, C.Cenozoic Era, D.Paleozoic Era. I.Lower invertebrates, II.Fish & Amphibia, III.Birds & Reptiles, IV.Mammals:
Options
A
A-I, B-II, C-IV, D-III
B
A-III, B-I, C-IV, D-II
✓
C
A-II, B-I, C-III, D-IV
D
A-III, B-I, C-II, D-IV
✦ Correct Answer
A-III, B-I, C-IV, D-II
📐 Solution
1
Mesozoic=Birds&Reptiles(III), Proterozoic=Lower invertebrates(I), Cenozoic=Mammals(IV), Paleozoic=Fish&Amphibia(II).
∴ A-III, B-I, C-IV, D-II
ZoologyQ.
Statement I: Cerebral hemispheres are connected by nerve tract known as corpus callosum. Statement II: Brain stem consists of medulla oblongata, pons and cerebrum.
Options
A
Statement I is correct but II is incorrect.
✓
B
Statement I is incorrect but II is correct.
C
Both statements are correct.
D
Both statements are incorrect.
✦ Correct Answer
Statement I is correct but II is incorrect.
📐 Solution
1
I ✓ (corpus callosum connects hemispheres). II ✗ (brain stem = medulla oblongata + pons + midbrain; cerebrum is NOT part of brain stem). ∴ I correct, II incorrect.
ZoologyQ.
Match List-I with List-II. A.RNA polymerase III, B.Termination of transcription, C.Splicing of Exons, D.Tata box. I.snRNPs, II.Promoter, III.Rho factor, IV.SnRNAs, tRNA:
Options
A
A-III, B-IV, C-I, D-II
B
A-IV, B-III, C-I, D-II
✓
C
A-II, B-IV, C-I, D-III
D
A-III, B-II, C-IV, D-I
✦ Correct Answer
A-IV, B-III, C-I, D-II
📐 Solution
1
RNA pol III→SnRNAs,tRNA(IV), Termination→Rho factor(III), Splicing→snRNPs(I), TATA box→Promoter(II).
∴ A-IV, B-III, C-I, D-II
ZoologyQ.
Match List-I (Disease) with List-II (Cause/Characteristic). A.Exophthalmic goiter, B.Acromegaly, C.Cushing's syndrome, D.Cretinism. I.Excess cortisol, moon face, hyperglycemia, II.Hypo-secretion of thyroid hormone, stunted growth, III.Hyper-secretion of thyroid hormone, protruding eyeballs, IV.Excessive secretion of growth hormone:
Options
A
A-III, B-IV, C-II, D-I
B
A-III, B-IV, C-I, D-II
✓
C
A-I, B-III, C-II, D-IV
D
A-IV, B-II, C-I, D-III
✦ Correct Answer
A-III, B-IV, C-I, D-II
📐 Solution
1
Exophthalmic goiter=hyperthyroidism+protruding eyes(III), Acromegaly=excess GH(IV), Cushing's=excess cortisol(I), Cretinism=hypothyroid+stunted growth(II).
∴ A-III, B-IV, C-I, D-II
ZoologyQ.
Match List-I (Epithelium type) with List-II (Location). A.Unicellular glandular epithelium, B.Compound epithelium, C.Multicellular glandular epithelium, D.Endocrine glandular epithelium. I.Salivary glands, II.Pancreas, III.Goblet cells of alimentary canal, IV.Moist surface of buccal cavity:
Options
A
A-III, B-IV, C-I, D-II
✓
B
A-II, B-I, C-IV, D-III
C
A-II, B-I, C-III, D-IV
D
A-IV, B-III, C-I, D-II
✦ Correct Answer
A-III, B-IV, C-I, D-II
📐 Solution
1
Unicellular=goblet cells(III), Compound=buccal cavity(IV), Multicellular=salivary glands(I), Endocrine=pancreas(II).
∴ A-III, B-IV, C-I, D-II
ZoologyQ.
Statement I: Bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced. Statement II: Both bone marrow and thymus provide micro environments for development and maturation of T-lymphocytes.
Options
A
I correct but II incorrect.
B
I incorrect but II correct.
C
Both I and II are correct.
✓
D
Both I and II are incorrect.
✦ Correct Answer
Both I and II are correct.
📐 Solution
1
I ✓ (bone marrow produces all blood cells). II ✓ (bone marrow=B-cells, thymus=T-cells maturation). Both correct.
ZoologyQ.
Match List-I (Structure) with List-II (Description). A.Structures for storing food, B.Ring of 6-8 blind tubules at foregut-midgut junction, C.Ring of 100-150 yellow filaments at midgut-hindgut junction, D.Structures for grinding food. I.Gizzard, II.Gastric Caeca, III.Malpighian tubules, IV.Crop:
Options
A
A-IV, B-III, C-II, D-I
B
A-III, B-II, C-IV, D-I
C
A-IV, B-II, C-III, D-I
✓
D
A-I, B-II, C-III, D-IV
✦ Correct Answer
A-IV, B-II, C-III, D-I
📐 Solution
1
Crop=food storage(IV), Gastric caeca=foregut-midgut(II), Malpighian tubules=midgut-hindgut(III), Gizzard=grinding(I).
∴ A-IV, B-II, C-III, D-I
ZoologyQ.
Choose the correct statement regarding juxtamedullary nephron:
Options
A
Loop of Henle of juxtamedullary nephron runs deep into medulla.
✓
B
Juxtamedullary nephrons outnumber cortical nephrons.
C
Juxtamedullary nephrons are located in the columns of Bertini.
D
Renal corpuscle of juxtamedullary nephron lies in outer portion of renal medulla.
✦ Correct Answer
Loop of Henle of juxtamedullary nephron runs deep into medulla.
📐 Solution
1
Loop of Henle runs very long and deep into medulla in juxtamedullary nephrons — this is key to concentrating urine.
ZoologyQ.
Match List-I (ECG wave) with List-II (What it represents). A.P wave, B.QRS complex, C.T wave, D.T-P gap. I.Heart muscles electrically silent, II.Depolarisation of ventricles, III.Depolarisation of atria, IV.Repolarisation of ventricles:
Options
A
A-II, B-III, C-I, D-IV
B
A-IV, B-II, C-I, D-III
C
A-I, B-III, C-IV, D-II
D
A-III, B-II, C-IV, D-I
✓
✦ Correct Answer
A-III, B-II, C-IV, D-I
📐 Solution
1
P=atria depolarisation(III), QRS=ventricles depolarisation(II), T=ventricles repolarisation(IV), T-P gap=electrically silent(I).
∴ A-III, B-II, C-IV, D-I
ZoologyQ.
As per ABO blood grouping, father is B⁺, mother is A⁺ and child is O⁺. Their respective genotype can be: A.I^B i/I^A i/ii, B.I^B I^B/I^A I^A/ii, C.I^A I^B/iI^A/I^B i, D.I^A i/I^B i/I^A i, E.iI^B/iI^A/I^A I^B:
Options
A
C and B only
B
D and E only
C
A only
✓
D
B only
✦ Correct Answer
A only
📐 Solution
1
Child O(ii) means each parent must have at least one 'i'. Father B⁺ must be I^B i. Mother A⁺ must be I^A i. ∴ Only option A (I^B i/I^A i/ii) is correct.
ZoologyQ.
Statement I: Gause's competitive exclusion principle states that two closely related species competing for different resources cannot exist indefinitely. Statement II: According to Gause's principle, the inferior competitor will be eliminated. This may be true if resources are limiting.
Options
A
I is true but II is false.
B
I is false but II is true.
✓
C
Both I and II are true.
D
Both I and II are false.
✦ Correct Answer
I is false but II is true.
📐 Solution
1
I ✗ (Gause's principle: species competing for the same resources, not different resources, cannot coexist). II ✓ (inferior competitor eliminated when resources are limiting). ∴ I false, II true.
ZoologyQ.
Regarding catalytic cycle of an enzyme action, select the correct sequential steps: A.Substrate-enzyme complex formation, B.Free enzyme ready to bind another substrate, C.Release of products, D.Chemical bonds of substrate broken, E.Substrate binding to active site:
Options
A
B, A, C, D, E
B
E, D, C, B, A
C
E, A, D, C, B
✓
D
A, E, B, D, C
✦ Correct Answer
E, A, D, C, B
📐 Solution
1
Sequence: E(substrate binds) → A(E-S complex) → D(bonds broken) → C(products released) → B(enzyme free).
∴ E, A, D, C, B
ZoologyQ.
Statement I: Mitochondria and chloroplasts are both double membrane bound organelles. Statement II: Inner membrane of mitochondria is relatively less permeable compared to chloroplast.
Options
A
I correct but II incorrect.
B
I incorrect but II correct.
C
Both I and II are correct.
✓
D
Both I and II are incorrect.
✦ Correct Answer
Both I and II are correct.
📐 Solution
1
Both are double membrane bound ✓. Inner chloroplast membrane is more permeable than inner mitochondrial membrane ✓. Both correct.
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