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Physics
- Q1A circular current loop of radius R is placed inside a square loop of side length L (L ≫ R) such that they are co-planar and their centres coincide. The permeability of free space is μ₀. The mutual inductance between the circular loop and the square loop is:
- 2√(2) (μ₀ R²)/(L)
- 2√(2) (μ₀ L²)/(R)
- √(2) (μ₀ R²)/(L)
- √(2) (μ₀ L²)/(R)
Answer: (A) 2√(2) (μ₀ R²)/(L)
D. Tests: computing mutual inductance via the field of the easier loop (reciprocity). Why D: Let current I flow in the square. Each side (a finite wire at distance L/2) gives B = (μ₀ I)/(4π(L/2))(sin 45° + sin 45°) = (√(2)μ₀ I)/(2π L) at the centre; four sides give B = (2√(2)μ₀ I)/(π L). Since L ≫ R, this field is uniform over the small circle: Φ = Bπ R² = (2√(2)μ₀ I R²)/(L) (the π cancels). M = Φ/I = 2√(2) μ₀ R²/L. Why not A: M must vanish as R → 0, but L²/R blows up [misconception A: “using the square’s area for the flux instead of the small circle’s”] Why not B: same wrong L²/R dependence [misconception B: “swapping which loop’s area collects the flux”] Why not C: right dependence, missing the factor 2 [misconception C: “summing only two sides of the square (or dropping one sin 45°)”] Remember: by reciprocity M₁₂ = M₂₁: always drive the loop whose field is easy, and let the SMALL loop’s area collect the flux.
- Q2Young’s double-slit experiment is first performed in air and then in a medium other than air. It is found that the 8th bright fringe in the medium lies where the 5th dark fringe lies in air. The refractive index of the medium is nearly:
- 1.59
- 1.25
- 1.78
- 1.69
Answer: (C) 1.78
Position of 5th dark fringe in air: y = (2×5-1)(λ D)/(2d) = (9λ D)/(2d). Position of 8th bright fringe in the medium (wavelength λ/n): y = 8((λ/n) D)/(d). Equating: (9)/(2) = (8)/(n), so n = (16)/(9) ≈ 1.78.
- Q3Sound travels faster in hydrogen than in oxygen at the same temperature mainly because hydrogen has:
- Lower molar mass
- Higher pressure
- Greater bulk modulus
- Higher density
Answer: (A) Lower molar mass
v = √(γRT/M), so v ∝ 1/√M. Hydrogen (M = 2) is much lighter than oxygen (M = 32), giving it a higher speed of sound at the same temperature.
- Q4A biconvex lens has a radius of curvature of magnitude 20 cm. For an object of height 2 cm placed 30 cm from the lens (take μ=1.5), the image formed is best described as:
- real, inverted, height = 1 cm
- virtual, upright, height = 0.5 cm
- real, inverted, height = 4 cm
- virtual, upright, height = 1 cm
Answer: (C) real, inverted, height = 4 cm
(1)/(f)=(1.5-1)·(2)/(20)=(1)/(20), so f=20 cm. With u=-30: (1)/(v)=(1)/(20)-(1)/(30)=(1)/(60), v=60 cm. Magnification m=(v)/(u)=(60)/(-30)=-2, so the image is real, inverted, of height 2×2=4 cm.
- Q5There are two inclined surfaces of equal length L and the same angle of inclination 45° with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down the rough surface as it takes on the smooth surface. The coefficient of kinetic friction μₖ between the object and the rough surface is close to:
- 0.40
- 0.75
- 0.5
- 0.25
Answer: (B) 0.75
B. Tests: incline kinematics with and without kinetic friction, using L = (1)/(2)at². Why B: Same length, starting from rest: L = (1)/(2)a t², so a ∝ (1)/(t²). With t_(rough) = 2tₛₘₒₒₜₕ, a_(rough) = (aₛₘₒₒₜₕ)/(4). Now aₛₘₒₒₜₕ = gsin 45° and a_(rough) = g(sin 45° – μₖ cos 45°) = gsin 45°(1 – μₖ) since sin 45° = cos 45°. Hence 1 – μₖ = (1)/(4), giving μₖ = 0.75. Why not A: 0.5 comes from setting a_(rough) = (aₛₘₒₒₜₕ)/(2), using the time ratio instead of its square [misconception A: “n times the time means n times smaller acceleration, forgetting L = (1)/(2)at²“]. Why not C: 0.25 is the value of (1 – μₖ), i.e. stopping one step before solving for μₖ [misconception C: “reporting an intermediate quantity as the answer”]. Why not D: 0.40 follows from no consistent route; it rewards guessing near the middle [misconception D: “estimating instead of deriving”]. Remember: On a 45° incline, μₖ = 1 – (1)/(n²) when the rough descent takes n times as long.
- Q6Two charges of 5Q and -2Q are situated at the points (3a, 0) and (-5a, 0) respectively. The electric flux through a sphere of radius ‘4a‘ having center at origin is :
- (5Q)/(ε₀)
- (3Q)/(ε₀)
- (7Q)/(ε₀)
- (2Q)/(ε₀)
Answer: (A) (5Q)/(ε₀)
D. Tests: deciding which charges are actually enclosed before applying Gauss’s law. Why D: The sphere is centred at the origin with radius 4a. The charge 5Q sits at distance 3a, and 3a < 4a, so it is inside. The charge -2Q sits at distance 5a, and 5a > 4a, so it is outside and its net contribution to the flux is zero. Hence Φ = (q_(enc))/(ε₀) = (5Q)/(ε₀). Why not A: (2Q)/(ε₀) uses only the outside charge, and drops its minus sign as well. Why not B: (7Q)/(ε₀) adds the two magnitudes, 5Q + 2Q, treating both charges as enclosed and both as positive. Why not C: (3Q)/(ε₀) is the algebraic sum 5Q – 2Q, which would be correct only if the -2Q were inside the sphere too [misconception C: “every charge named in the question counts towards q_(enc)“]. Remember: compare each charge’s distance with the radius first; an outside charge bends field lines through the surface but its net flux is exactly zero.
- Q7A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness (4/5)d, where d is the separation of the plates. The capacitance in the presence and absence of dielectric are C and C₀ respectively. The ratio (C/C₀) is
- (K + 4)/(5K)
- (5K)/(K + 4)
- (4K)/(K + 5)
- (K + 5)/(4K)
Answer: (B) (5K)/(K + 4)
C. Tests: the effective gap of a partly filled capacitor, written as a ratio. Why C: the slab covers (4d)/(5) and leaves (d)/(5) empty, so the effective gap is (d)/(5) + (4d)/(5K) = (d(K+4))/(5K). Since capacitance goes inversely with the gap, (C)/(C₀) = (d)/(d(K+4)/(5K)) = (5K)/(K+4). It checks out: at K = 1 the ratio is 1, and for very large K it tends to 5, the value for a gap reduced to (d)/(5). Why not A: (K+4)/(5K) is the reciprocal of the answer, so it is (C₀)/(C); for K greater than 1 it is less than 1, and a dielectric can only raise the capacitance [misconception: “the order of a ratio does not matter”]. Why not B: at K = 1 this gives (4)/(6) instead of 1, so it fails the no dielectric check; the 4 belongs with the empty fraction inside the denominator, not on top [misconception: “the slab thickness fraction multiplies the whole ratio”]. Why not D: at K = 1 this gives (6)/(4) = 1.5, again failing the check, and for large K it falls towards (1)/(4) when (C)/(C₀) must climb towards 5 [misconception: “invert the expression and reshuffle the constants”]. Remember: effective gap = uncovered part + covered part divided by K, and then (C)/(C₀) = (d)/(d_(eff)).
- Q8A body of mass 1 kg begins to move under the action of a time dependent force F = (2t î + 3t² ĵ) N, where î and ĵ are unit vectors along X and Y axes. What power will be developed by the force at the time t?
- (2t² + 4t⁴) W
- (2t³ + 3t⁵) W
- (2t³ + 3t⁴) W
- (2t + 3t³) W
Answer: (B) (2t³ + 3t⁵) W
a = F/m = (2t î + 3t² ĵ). v = ∫a dt = (t² î + t³ ĵ). P = F·v = (2t î + 3t² ĵ)·(t² î + t³ ĵ) = 2t³ + 3t⁵ W.
- Q9A current of 2 A flows through a 2 Ω resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 Ω resistor. The internal resistance of the battery is:
- (1)/(4) Ω
- 1 Ω
- 0.5 Ω
- (1)/(3) Ω
Answer: (D) (1)/(3) Ω
ε = I(R + r). Case 1: ε = 2(2 + r) = 4 + 2r. Case 2: ε = 0.5(9 + r) = 4.5 + 0.5r. Equating: 4 + 2r = 4.5 + 0.5r → 1.5r = 0.5 → r = (1)/(3) Ω.
- Q10A 5 A fuse wire can withstand a maximum power of 1 W in a circuit. The resistance of the fuse wire is:
- 0.4 Ω
- 0.04 Ω
- 0.2 Ω
- 5 Ω
Answer: (B) 0.04 Ω
P = I²R → R = (P)/(I²) = (1)/(5²) = (1)/(25) = 0.04 Ω.
- Q11A convex mirror used as a rear-view mirror always forms an image that is:
- real, inverted, magnified
- virtual, inverted, magnified
- virtual, erect, diminished
- real, erect, diminished
Answer: (C) virtual, erect, diminished
A convex mirror forms a virtual, erect, diminished image for all real object positions, giving a wide field of view.
- Q12The half-life of a radioactive material is 3 h. If the initial amount is 300 g, then after 18 h it will remain:
- 93.75 g
- 4.68 g
- 9.375 g
- 46.8 g
Answer: (B) 4.68 g
n = 18/3 = 6 half-lives. M = 300×((1)/(2))⁶ = 300/64 ≈ 4.68 g.
- Q13A satellite revolving around a planet in a stationary orbit has a time period of 6 hours. The mass of the planet is one-fourth the mass of earth. The radius of the orbit of the planet’s satellite is: (Given: radius of geo-stationary orbit for earth is 4.2×10⁴ km)
- 1.05×10⁴ km
- 1.68×10⁵ km
- 8.4×10⁴ km
- 1.4×10⁴ km
Answer: (A) 1.05×10⁴ km
D. Tests: the scaling r ∝ (MT²)^(1/3) that follows from T² = (4π² r³)/(GM). Why D: T² = (4π² r³)/(GM) gives r = ((GMT²)/(4π²))^(1/3). Comparing the planet with the earth’s geostationary case (Tₑ = 24 h, rₑ = 4.2×10⁴ km): (rₚ)/(rₑ) = ((Mₚ)/(Mₑ)((Tₚ)/(Tₑ))²)^(1/3) = ((1)/(4)×((6)/(24))²)^(1/3) = ((1)/(4)×(1)/(16))^(1/3) = ((1)/(64))^(1/3) = (1)/(4). So rₚ = (4.2×10⁴)/(4) = 1.05×10⁴ km. Why not A: 1.68×10⁵ km is 4.2×10⁴×4, the exact reciprocal factor, from writing the ratio upside down as ((Mₑ)/(Mₚ)((Tₑ)/(Tₚ))²)^(1/3) = (4×16)^(1/3) = 4 [misconception: ‘a smaller planet needs a bigger orbit’]. Why not B: no consistent substitution gives this; ignoring the mass entirely gives 4.2×10⁴((1)/(4))^(2/3) = 1.67×10⁴ km and using the mass correctly gives 1.05×10⁴ km, so 1.4×10⁴ km matches neither route. Why not C: no consistent substitution gives this; 8.4×10⁴ km is simply twice the earth’s geostationary radius, and no combination of a quarter mass with a quarter period yields a factor 2. Remember: r ∝ (MT²)^(1/3), so quartering both the mass and the period quarters the orbital radius.
- Q14In a p-n junction diode, the change due to heating (rise in temperature):
- affects the overall V-I characteristics of the p-n junction
- does not affect the resistance of the p-n junction
- affects only the forward resistance
- affects only the reverse resistance
Answer: (A) affects the overall V-I characteristics of the p-n junction
Heating generates more thermal electron-hole pairs, changing carrier concentrations on both sides. This alters both forward and reverse behaviour, so the entire V-I characteristic (and hence both forward and reverse resistance) is affected.
- Q15A bar magnet of magnetic moment M is placed in a magnetic field of induction B. The torque exerted on it is:
- MtimesB
- -McdotB
- -MtimesB
- McdotB
Answer: (A) MtimesB
The torque on a magnetic dipole is τ = MtimesB, with magnitude MBsinθ where θ is the angle between M and B. It is a cross product (a vector), not the scalar dot product.
- Q16A 100 W, 220 V bulb has a resistance of:
- 220 Ω
- 100 Ω
- 484 Ω
- 48.4 Ω
Answer: (C) 484 Ω
R = V²/P = (220)²/100 = 48400/100 = 484 Ω.
- Q17To make an n-type semiconductor, silicon should be doped with a:
- divalent impurity like zinc
- pentavalent impurity like arsenic
- trivalent impurity like boron
- tetravalent impurity like germanium
Answer: (B) pentavalent impurity like arsenic
Pentavalent atoms (As, P, Sb) have five valence electrons; four bond and the fifth is loosely bound, donating a free electron. This makes electrons the majority carriers → n-type.
- Q18The equation of a particle executing simple harmonic motion is given by x = sinπ(t + (1)/(3)) m. At t = 1 s, the speed of particle will be (Given: π = 3.14)
- 272 cm s⁻¹
- 0 cm s⁻¹
- 314 cm s⁻¹
- 157 cm s⁻¹
Answer: (D) 157 cm s⁻¹
B. Tests: differentiating an SHM equation and evaluating the phase at a given instant. Why B: v = (dx)/(dt) = πcosπ(t + (1)/(3)). At t = 1 s the phase is π(1 + (1)/(3)) = (4π)/(3), and cos(4π)/(3) = -(1)/(2). So v = π × (-(1)/(2)) = -1.57 m/s and the speed is 1.57 m/s = 157 cm s⁻¹. Why not A: zero speed needs the cosine of the phase to vanish, which happens at the extremes; at (4π)/(3) the cosine is -(1)/(2), not 0. Why not C: 272 cm/s uses sin(4π)/(3) = -(√(3))/(2) in place of the cosine, giving π × 0.866 = 2.72 m/s. That is differentiating a sine into a sine [misconception C: “the derivative of sine is sine”]. Why not D: 314 cm/s is vₘₐₓ = Aω = 1 × π = 3.14 m/s, the speed at the mean position, obtained by taking the cosine factor as 1. Remember: differentiate first, substitute the time second; here the phase (4π)/(3) leaves exactly half the maximum speed.
- Q19An electron falls from rest through a vertical distance h in a uniform, vertically upward electric field E. The direction of the field is then reversed (magnitude kept the same) and a proton is allowed to fall from rest through the same distance h. The time of fall of the electron, compared with the time of fall of the proton, is:
- smaller
- 10 times greater
- equal
- 5 times greater
Answer: (A) smaller
For each particle a=(qE)/(m) and h=(1)/(2)at², so t=√((2hm)/(qE))∝√(m) (same q, E, h). Since the electron’s mass is far smaller than the proton’s, the electron’s fall time is smaller.
- Q20A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will:
- be reduced to half
- remain the same
- be doubled
- increase four times
Answer: (B) remain the same
Gauss’s law: flux depends only on the net enclosed charge, not on the size or shape of the surface. Enclosing the same Q, the flux (Q)/(ε₀) is unchanged.
- Q21The r.m.s. speed of oxygen molecules at 47 °C is equal to that of the hydrogen molecules kept at _________ °C. (Mass of oxygen molecule/mass of hydrogen molecule = 32/2)
- -253
- -100
- -235
- -20
Answer: (A) -253
B. Tests: equating vᵣₘₛ of two gases, which fixes T/M, then converting back to Celsius. Why B: vᵣₘₛ = √(3RT/M), so equal speeds mean (T_O)/(M_O) = (T_H)/(M_H). With T_O = 47 + 273 = 320 K, T_H = 320 × (2)/(32) = 20 K = 20 – 273 = -253°C. Why not A: -100°C is 173 K, which would demand a mass ratio 320/173 = 1.85 instead of 16 [misconception A: ‘lighter gas needs a moderately lower temperature’]. Why not C: -20 takes the kelvin answer 20 K and simply attaches a minus sign instead of subtracting 273 [misconception C: ‘converting to Celsius means flipping the sign’]. Why not D: -235 is -253 with the last two digits transposed; no substitution produces 38 K [misconception D: ‘close digits, close physics’]. Remember: equal vᵣₘₛ means equal T/M, so a 16 times lighter gas needs a 16 times lower kelvin temperature.
- Q22In an adiabatic process, which quantity is zero?
- Change in internal energy
- Change in temperature
- Heat exchanged
- Work done
Answer: (C) Heat exchanged
An adiabatic process is defined by no heat exchange with the surroundings, Q = 0. Then ΔU = −W, so temperature and internal energy do change.
- Q23The power factor of a series LCR circuit at resonance is:
- 0.5
- 1
- infinite
- zero
Answer: (B) 1
At resonance Z = R, so cos φ = R/Z = R/R = 1. The circuit is purely resistive and power transfer is maximum.
- Q24A solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of the magnetic field is:
- (B)/(2)
- B
- 2B
- 4B
Answer: (B) B
B = μ₀ n I. New n’ = n/2 and I’ = 2I, so B’ = μ₀(n/2)(2I) = μ₀ n I = B. Unchanged.
- Q25The frequency of oscillation of a mass m suspended by a spring is v₁. If the length of the spring is cut to half, the same mass oscillates with frequency v₂. The value of v₂ / v₁ is ____ .
- 1
- √(3)
- 2
- √(2)
Answer: (D) √(2)
C. Tests: how the force constant of a spring scales when the spring is shortened, and how frequency follows that force constant. Why C: for a uniform spring the force constant is inversely proportional to its length, so halving the length doubles it, k₂ = 2k. The frequency of a spring-mass oscillator is v = (1)/(2π)√(k/m), and the mass is unchanged, so v₂/v₁ = √(k₂/k) = √(2). Why not A: 1 assumes the force constant belongs to the spring’s material only and does not change when the spring is cut [misconception A: “k is a property of the wire, not of the length”]. Why not B: 2 comes from writing v as proportional to k instead of to √(k), so the doubling of k is carried straight into the frequency [misconception B: “frequency is proportional to k”]. Why not D: √(3) would need k₂ = 3k, which is what you get by cutting the spring to one third of its length, not to one half. Remember: cut a spring shorter and it gets stiffer in exact inverse proportion, and frequency follows the square root of that.
- Q26A speed of 72 km/h is equal to:
- 20 m/s
- 25 m/s
- 36 m/s
- 15 m/s
Answer: (A) 20 m/s
Multiply by 5/18 to convert km/h to m/s: 72 × 5/18 = 20 m/s.
- Q27Two wires of the same material and same length carry the same current. Their cross-sectional areas are in the ratio A₁:A₂ = 1:2. The ratio of the drift speeds of electrons v₁:v₂ is:
- 1:2
- 1:1
- 2:1
- 4:1
Answer: (C) 2:1
From I = neAv_d with I and n equal, v_d ∝ 1/A. So v₁/v₂ = A₂/A₁ = 2/1, i.e. v₁:v₂ = 2:1. The thinner wire (smaller area) has the larger drift speed for the same current.
- Q28The perpendicular axis theorem (I_z = Iₓ + I_y) is applicable to:
- Only solid spheres
- Any three-dimensional rigid body
- Only bodies of uniform density
- Only planar (laminar) bodies
Answer: (D) Only planar (laminar) bodies
The perpendicular axis theorem applies only to plane laminae, where the z-axis is perpendicular to the plane containing the x and y axes. It is not valid for 3D bodies.
- Q29A beam of cathode rays is subjected to crossed electric (E) and magnetic (B) fields. The fields are adjusted so that the beam is not deflected. The specific charge (e/m) of the cathode rays, accelerated through a potential difference V, is given by:
- (E²)/(2VB²)
- (2VE²)/(B²)
- (B²)/(2VE²)
- (2VB²)/(E²)
Answer: (A) (E²)/(2VB²)
No deflection means electric and magnetic forces balance: eE = evB ⇒ v = (E)/(B). Acceleration through V gives (1)/(2)mv² = eV ⇒ (e)/(m) = (v²)/(2V) = ((E/B)²)/(2V) = (E²)/(2VB²).
- Q30Assertion (A): In a uniform plane EM wave in vacuum, the E and B fields are in phase. Reason (R): The energy in the wave is carried equally by the E and B fields.
- A is false but R is true
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- Both A and R are true and R is the correct explanation of A
Answer: (B) Both A and R are true but R is not the correct explanation of A
Both statements are true: E and B oscillate in phase, and average energy is shared equally. But the equal energy sharing is not the reason the fields are in phase (that comes from Maxwell’s equations), so R does not explain A.
- Q31Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. The number of beats heard per second is:
- 516
- 2
- 8
- 4
Answer: (D) 4
Beat frequency = |f₁ − f₂| = |260 − 256| = 4 beats per second.
- Q32A particle moving in a straight line covers half the distance with speed 6 m/s. The other half is covered in two equal time intervals with speeds 9 m/s and 15 m/s respectively. The average speed of the particle during the motion is :
- 10 m/s
- 9.2 m/s
- 8 m/s
- 8.8 m/s
Answer: (C) 8 m/s
D. Tests: mixing the two averaging rules in one problem, arithmetic for equal times and harmonic for equal distances. Why D: let the total distance be 2L. First half: t₁=(L)/(6). The second half is covered in two equal times τ, so L=9τ+15τ=24τ, giving τ=(L)/(24) and t₂=2τ=(L)/(12), i.e. the second half averages (9+15)/(2)=12 m/s. Average speed =(2L)/((L)/(6)+(L)/(12))=(2)/((1)/(6)+(1)/(12))=(2)/(0.25)=8 m/s, the harmonic mean of 6 and 12. Why not C: 10 m/s is the plain average (6+9+15)/(3), which would only be valid if all three speeds lasted for equal times [misconception C: “average the given speeds”]. Why not A: no consistent substitution gives 9.2 m/s. Why not B: no consistent substitution gives 8.8 m/s; the harmonic mean of 6, 9 and 15 taken over equal distances is 8.7 m/s, and that is the wrong model anyway. Remember: equal times means average the speeds, equal distances means average the reciprocals, and this question needs both in turn.
- Q33In a carbon monoxide molecule the carbon and oxygen atoms are separated by 1.12 × 10⁻¹⁰ m. The distance of the centre of mass from the carbon atom is (take C = 12 amu, O = 16 amu):
- 0.64 × 10⁻¹⁰ m
- 0.56 × 10⁻¹⁰ m
- 0.51 × 10⁻¹⁰ m
- 0.48 × 10⁻¹⁰ m
Answer: (A) 0.64 × 10⁻¹⁰ m
Take carbon at the origin and oxygen at r = 1.12 × 10⁻¹⁰ m. x_(cm) = (12(0) + 16(r))/(12 + 16) = (16)/(28)r = (4)/(7)(1.12 × 10⁻¹⁰) = 0.64 × 10⁻¹⁰ m from the carbon atom.
- Q34If the ionisation energy for the hydrogen atom is 13.6 eV, then the energy required to excite it from the ground state to the next higher state is nearly
- 13.6 eV
- 10.2 eV
- -10.2 eV
- -3.4 eV
Answer: (B) 10.2 eV
C. Tests: turning the ionisation energy into the level ladder and taking a difference, with the correct sign for absorbed energy. Why C: ionisation energy 13.6 eV means the ground level is at E₁ = -13.6 eV. The next higher state is n = 2, at E₂ = (-13.6)/(4) = -3.4 eV. Energy required = E₂ – E₁ = -3.4 – (-13.6) = +10.2 eV, positive because the atom absorbs it. Why not A: the magnitude is right but the sign is wrong; -10.2 eV would mean the atom gives energy out while climbing to a higher level [misconception A: “every atomic energy carries a minus sign”]. Why not B: -3.4 eV is the energy of the n = 2 level itself, not the gap from n = 1 to n = 2. Why not D: 13.6 eV is the cost of taking the electron all the way to n = ∞, that is full ionisation, not excitation to the next level. Remember: first excitation of hydrogen costs 10.2 eV, ionisation costs 13.6 eV.
- Q35The elastic potential energy stored per unit volume of a stretched wire is:
- stress / strain
- 2 × stress × strain
- ½ × stress × strain
- stress × strain
Answer: (C) ½ × stress × strain
Energy density u = ½ × stress × strain (the ½ comes from the average force during stretching). This also equals ½ (stress)²/Y.
- Q36Three identical coils C₁, C₂ and C₃ are placed close together sharing a common axis, with C₂ exactly midway. C₁ carries current I in the anticlockwise direction while C₃ carries current I in the clockwise direction (both viewed from the same side). An induced current flows through C₂ in the clockwise direction when
- C₁ and C₃ move with equal speeds towards C₂
- C₁ moves away from C₂ and C₃ moves towards C₂
- C₁ moves towards C₂ and C₃ moves away from C₂
- C₁ and C₃ move with equal speeds away from C₂
Answer: (C) C₁ moves towards C₂ and C₃ moves away from C₂
B. Tests: applying Lenz’s law with superposed fluxes from two sources. Why B: Viewed from the common side, C₁‘s anticlockwise current sends flux through C₂ towards the viewer (+n), while C₃‘s clockwise current sends flux through C₂ away from the viewer (-n). Net flux along +n is Φ₁ – Φ₃. A clockwise induced current in C₂ (magnetic moment along -n) requires the +n flux to be increasing. That happens when Φ₁ grows (C₁ approaches) AND Φ₃ shrinks (C₃ recedes) – both changes add up. Why not A: both fluxes decrease in magnitude; with equal speeds the changes cancel by symmetry, giving no net induced current [misconception A: “adding flux magnitudes without tracking their opposite signs”] Why not C: mirror of A – the two changes again cancel [misconception C: “assuming approaching coils always induce a current”] Why not D: this decreases the +n flux, inducing an anticlockwise current [misconception D: “applying Lenz’s law to the wrong sign of flux change”] Remember: superpose the fluxes with signs first, then apply Lenz: induced current opposes the CHANGE of the net flux.
- Q37The wavelength of the first line of the Lyman series for hydrogen is equal to that of the second line of the Balmer series for a hydrogen-like ion. The atomic number Z of this hydrogen-like ion is:
- 4
- 1
- 3
- 2
Answer: (D) 2
Lyman 1st line (H): (1)/(λ) = R(1-(1)/(4)). Balmer 2nd line (4→2) for ion: (1)/(λ) = Z²R((1)/(4)-(1)/(16)). Equating: (3)/(4) = Z²·(3)/(16) ⇒ Z² = 4 ⇒ Z = 2.
- Q38The pair of physical quantities not having the same dimensions is:
- Surface tension and impulse
- Pressure and Young’s modulus
- Angular momentum and Planck’s constant
- Torque and energy
Answer: (A) Surface tension and impulse
C. Tests: spotting the odd pair by writing out both dimensional formulae rather than trusting intuition. Why C: surface tension is force per unit length, ([MLT⁻²])/([L]) = [MT⁻²]. Impulse is force times time, [MLT⁻²][T] = [MLT⁻¹]. They disagree in both the length and the time exponent, so this is the mismatched pair. Why not A: angular momentum is [ML²T⁻¹], and Planck’s constant is (E)/(ν) = [ML²T⁻²][T] = [ML²T⁻¹], so this pair does match. Why not B: torque and energy are both a force times a length, [ML²T⁻²], so they match despite being physically different [misconception: ‘different physics means different dimensions’]. Why not D: pressure and Young’s modulus are both [ML⁻¹T⁻²], since strain is dimensionless. Remember: surface tension sits alone at [MT⁻²], the same slot as a spring constant, and never matches an impulse.
- Q39A ring of mass 2 kg and radius 0.5 m rolls without slipping with a centre-of-mass speed of 3 m/s. Its total kinetic energy is:
- 9 J
- 18 J
- 27 J
- 13.5 J
Answer: (B) 18 J
For a ring k²/R² = 1, so KE = ½Mv²(1 + 1) = Mv² = 2 × 3² = 18 J.
- Q40Assertion (A): The same physics that powers a nuclear reactor can also power a nuclear weapon. Reason (R): Both rely on energy released in nuclear fission.
- A is false but R is true
- Both A and R are true and R is the correct explanation of A
- A is true but R is false
- Both A and R are true but R is NOT the correct explanation of A
Answer: (B) Both A and R are true and R is the correct explanation of A
A is true and R is true: a reactor releases fission energy in a controlled way while a bomb releases it uncontrolled, but both rest on nuclear fission. R is the correct explanation of A and captures the ethical dilemma physics raises for society.
- Q41A water spray gun is attached to a hose of cross-sectional area 30 cm². The gun comprises 10 perforations, each of cross-sectional area 15 mm². If the water flows in the hose with a speed of 50 cm/s, calculate the speed at which the water flows out from each perforation. (Neglect any edge effects)
- 100 m/s
- 10 m/s
- 15 × 10² m/s
- 1000 m/s
Answer: (B) 10 m/s
B. Tests: applying continuity across one inlet and many outlets, with two unit conversions. Why B: the volume flow rate in the hose is A v = (30 × 10⁻⁴)(0.5) = 1.5 × 10⁻³ m³/s. The perforations share that flow, and their total area is 10 × 15 = 150 mm² = 1.5 × 10⁻⁴ m². So the exit speed is (1.5 × 10⁻³)/(1.5 × 10⁻⁴) = 10 m/s. Why not A: 100 m/s uses the area of a single perforation, 15 mm² = 1.5 × 10⁻⁵ m², so it forces the whole discharge through one hole [misconception A: “each hole carries the full flow rate”]. Why not C: 1000 m/s follows from converting 150 mm² with a factor 10⁻⁸ instead of 10⁻⁶, giving 1.5 × 10⁻⁶ m² and hence (1.5 × 10⁻³)/(1.5 × 10⁻⁶) = 1000. Why not D: no consistent substitution gives 15 × 10² m/s. Remember: continuity balances total areas, so add up every outlet before you divide.
- Q42Which of the following statements is true for a photoelectric experiment?
- The photocurrent increases with increasing intensity of light
- The stopping potential increases with increasing intensity of incident light
- The current in a photocell increases with increasing frequency of light
- The photocurrent is proportional to the applied voltage
Answer: (A) The photocurrent increases with increasing intensity of light
Photocurrent (saturation) is proportional to intensity, so B is true. Stopping potential depends on frequency, not intensity (A false). Saturation current does not depend on frequency (C false), and photocurrent saturates rather than rising with applied voltage (D false).
- Q43The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of V volts is proportional to:
- V²
- (1)/(√(V))
- (1)/(V)
- √(V)
Answer: (C) (1)/(V)
A. Tests: the Duane-Hunt cut-off, where one electron dumps all of its kinetic energy into a single photon. Why A: The most energetic X-ray photon possible is made when the entire kinetic energy eV goes into one photon, so (hc)/(λₘᵢₙ) = eV, giving λₘᵢₙ = (hc)/(eV) = (12400)/(V) angstrom. That is an inverse first power of V. Why not B: (1)/(√(V)) is the de Broglie wavelength of the accelerated electron itself, λ = (h)/(√(2meV)), a different quantity from the X-ray it emits [misconception B: “the X-ray cut-off follows the same √(V) law as the matter wave”]. Why not C: λₘᵢₙ falls as V rises, so it cannot be proportional to any positive power of V; V² has the dependence backwards and exaggerated. Why not D: √(V) also grows with V while the cut-off wavelength shrinks, so the direction of the dependence is wrong here too. Remember: eV = (hc)/(λₘᵢₙ), so λₘᵢₙ = (12400)/(V) angstrom, strictly (1)/(V).
- Q44A particle of mass m is moving with a uniform velocity v₁. It is given an impulse such that its velocity becomes v₂. The impulse is equal to
- m(v₁ + v₂)
- m(v₂ − v₁)
- m[|v₂| − |v₁|]
- ½m(v₂² − v₁²)
Answer: (B) m(v₂ − v₁)
Impulse equals change in linear momentum: I = mv₂ − mv₁ = m(v₂ − v₁).
- Q45An object is placed 15 cm in front of a convex lens of focal length 10 cm. The image distance is:
- +15 cm
- +6 cm
- +30 cm
- -30 cm
Answer: (C) +30 cm
Using 1/v – 1/u = 1/f with u = -15, f = +10: 1/v = 1/10 + 1/(-15) = 3/30 – 2/30 = 1/30 ⇒ v = +30 cm (real image).
Chemistry
- Q46Acetaldehyde reacts with:
- Both electrophiles and nucleophiles
- Only electrophiles
- Only nucleophiles
- Only free radicals
Answer: (A) Both electrophiles and nucleophiles
The polar carbonyl group has two reactive sites: the δ+ carbon is attacked by nucleophiles, while the δ- oxygen (with lone pairs) can be attacked/protonated by electrophiles. Hence the carbonyl of acetaldehyde reacts with both nucleophiles and electrophiles.
- Q47Among CO2, NH3, H2 and He, which gas is expected to be liquefied most easily (highest critical temperature)?
- He
- CO2
- H2
- NH3
Answer: (D) NH3
Ease of liquefaction increases with the van der Waals constant ‘a’ (intermolecular attraction) and critical temperature. NH3 has strong hydrogen bonding giving the highest Tc among these (about 405 K, versus 304 K for CO2), so it liquefies most easily.
- Q48The mole fraction of ethanol (C₂H₅OH) in a solution containing 46 g ethanol and 90 g water is (C₂H₅OH=46, H₂O=18):
- 0.10
- 0.83
- 0.50
- 0.17
Answer: (D) 0.17
Moles ethanol = 46/46 = 1; moles water = 90/18 = 5. Mole fraction of ethanol = 1/(1+5) = 1/6 ≈ 0.17.
- Q49Which one of the following binary liquid mixtures shows positive deviation from Raoult’s law?
- Benzene + toluene
- Acetone + chloroform
- Ethanol + acetone
- Chloroethane + bromoethane
Answer: (C) Ethanol + acetone
Ethanol + acetone shows positive deviation: ethanol’s hydrogen-bonded network is broken by acetone, so A-B forces are weaker than A-A/B-B, raising the vapour pressure. Benzene+toluene and chloroethane+bromoethane are ideal; acetone+chloroform shows negative deviation (H-bonding).
- Q50The rate constant of the reaction A → B is 0.6×10⁻³ mol L⁻¹ s⁻¹. If the concentration of A is 5 M, then the concentration of B after 20 minutes is:
- 0.36 M
- 1.08 M
- 3.60 M
- 0.72 M
Answer: (D) 0.72 M
The unit of k (mol L⁻¹ s⁻¹) shows the reaction is zero order, so x = kt. x = 0.6×10⁻³ × (20×60) = 0.6×10⁻³ × 1200 = 0.72. So [B] formed = 0.72 M.
- Q51Repeated use of which one of the following fertilizers would increase the acidity of the soil?
- Ammonium sulphate
- Superphosphate of lime
- Urea
- Potassium nitrate
Answer: (A) Ammonium sulphate
Ammonium sulphate is a salt of a weak base (NH₃) and a strong acid (H₂SO₄). Its NH₄⁺ ion hydrolyses to give an acidic solution, so repeated use lowers soil pH.
- Q52For the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the correct expression for Kc is:
- [SO₃]/([SO₂][O₂])
- [SO₂]²[O₂]/[SO₃]²
- 2[SO₃]/(2[SO₂]+[O₂])
- [SO₃]²/([SO₂]²[O₂])
Answer: (D) [SO₃]²/([SO₂]²[O₂])
Kc = products over reactants, each raised to its stoichiometric coefficient: [SO₃]²/([SO₂]²[O₂]).
- Q53The correct statement regarding electrophile is
- electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile
- electrophiles are generally neutral species and can form a bond by accepting a pair of electrons from a nucleophile
- electrophile is a negatively charged species and form a bond by accepting a pair of electrons from another electrophile
- electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from a nucleophile
Answer: (A) electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile
C. Tests: the definition of an electrophile in terms of both its charge and the partner it takes electrons from. Why C: An electrophile is an electron pair acceptor, that is a Lewis acid. It may be positively charged, as in NO₂⁺, CH₃⁺ or H⁺, or neutral with an incomplete octet or an electron poor site, as in BF₃, AlCl₃ or SO₃. In either case it forms the new bond by accepting a lone pair supplied by a nucleophile. Option C states both the charge possibilities and the correct partner, so it is the only complete statement. Why not A: It calls the electrophile negatively charged, which describes a nucleophile, and then has it accepting electrons from another electrophile, which is impossible because two electron acceptors have no pair to share [misconception A: “electro in the name implies a negative charge”]. Why not B: The partner is right but the charge description is too narrow; restricting electrophiles to neutral species excludes the very common cationic ones such as NO₂⁺ in nitration [misconception B: “all electrophiles are neutral electron deficient molecules”]. Why not D: The partner is right but it limits the electrophile to negatively charged species, which is exactly backwards; an anion carries surplus electron density to donate, not a vacancy to fill [misconception D: “the charge sign does not matter so long as the species accepts electrons”]. Remember: electrophile means electron loving, so it is an electron pair acceptor that is positive or neutral, never negative.
- Q54Which of the following is a secondary alcohol?
- Propan-2-ol
- Propan-1-ol
- 2-methylpropan-2-ol
- 2-methylpropan-1-ol
Answer: (A) Propan-2-ol
In propan-2-ol, CH3-CH(OH)-CH3, the carbon bearing -OH is attached to two other carbons, making it a secondary (2deg) alcohol. Propan-1-ol and 2-methylpropan-1-ol are primary; 2-methylpropan-2-ol is tertiary.
- Q55For a reaction A arrow product, k = 2 × 10⁻² s⁻¹. If the initial concentration of A is 1.0 mol dm⁻³ find the value of log (1)/([A]ₜ) after 100 second?
- 0.868
- 0.135
- 0.270
- 0.430
Answer: (A) 0.868
D. Tests: rearranging the first order integrated law into base-10 log form and spotting the shortcut hidden in [A]₀ = 1. Why D: The unit s⁻¹ marks the reaction as first order, so log([A]₀)/([A]ₜ) = (kt)/(2.303) = (2 × 10⁻² × 100)/(2.303) = (2)/(2.303) = 0.868. Since [A]₀ = 1.0 mol dm⁻³, log([A]₀)/([A]ₜ) is the same as log(1)/([A]ₜ), so the answer is 0.868. Why not A: 0.135 is [A]ₜ itself, since [A]ₜ = [A]₀e⁻ᵏᵗ = e⁻² = 0.135 mol dm⁻³; that is the concentration, not the logarithm asked for [misconception A: “the exponential result is the final answer”]. Why not B: 0.270 is twice 0.135, that is the leftover concentration e⁻² doubled because kt = 2; back-solving it needs (kt)/(2.303) = 0.270, that is kt = 0.622, which k = 2 × 10⁻² s⁻¹ and t = 100 s never give. Why not C: 0.430 is essentially (1)/(2.303) = 0.434, the value when kt = 1, which follows from reading k as 1 × 10⁻² s⁻¹ instead of 2 × 10⁻² s⁻¹. Remember: log([A]₀)/([A]ₜ) = (kt)/(2.303), so when [A]₀ = 1 the answer is just (kt)/(2.303).
- Q56What is the SI unit of luminous intensity?
- Coulomb
- Candela
- Ampere
- Ohm
Answer: (B) Candela
C. Tests: recall of the seven SI base units and the quantity each one measures. Why C: luminous intensity is one of the seven SI base quantities and its base unit is the candela, symbol cd. Why not A: the ampere is the SI base unit of electric current, not of light. Why not B: the coulomb is a derived unit, that of electric charge, equal to one ampere second. Why not D: the ohm is a derived unit, that of electrical resistance [misconception: “any familiar electrical unit will do for a light quantity”]. Remember: the seven base units are metre, kilogram, second, ampere, kelvin, mole and candela, and candela is the only one about light.
- Q57When ethanol is heated with conc. H₂SO₄, a gas is produced. The compound formed, when this gas is treated with cold dilute aqueous solution of Baeyer’s reagent, is
- glycol
- formic acid
- ethanoic acid
- formaldehyde
Answer: (A) glycol
C. Tests: chaining the acid dehydration of ethanol with the syn-dihydroxylation done by Baeyer’s reagent. Why C: ethanol heated with conc. H₂SO₄ at about 443 K loses a molecule of water to give ethene, CH₂=CH₂. Baeyer’s reagent is cold dilute alkaline KMnO₄, which adds two OH groups across the double bond in one syn step, so ethene becomes HOCH₂-CH₂OH, ethane-1,2-diol, that is glycol. The purple colour is discharged, which is the visible test. Why not A: formaldehyde is what you get when the C=C is cleaved, as in ozonolysis of ethene; cold dilute Baeyer’s reagent adds across the bond and does not cut it [misconception A: “any KMnO₄ breaks the double bond”]. Why not B: formic acid is one oxidation step beyond formaldehyde, so it needs both cleavage and further oxidation, which only hot or acidic KMnO₄ would do [misconception B: “oxidation always runs on to the acid”]. Why not D: ethanoic acid is what hot acidified KMnO₄ makes from ethanol itself; the question asks about the evolved gas, not the leftover alcohol [misconception D: “the reagent acts on the ethanol, not on the gas”]. Remember: cold dilute Baeyer’s reagent adds two OH across C=C to give a glycol; hot or acidic permanganate cuts the C=C.
- Q58What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?
- 2.0
- 12.65
- 1.04
- 7.0
Answer: (B) 12.65
NaOH is in excess. After mixing equal volumes, net [OH⁻] = (0.1 – 0.01)/(2) = 0.045 M. pOH = -log(0.045) = 1.35, so pH = 14 – 1.35 = 12.65.
- Q59For a sample of perfect gas when its pressure is changed isothermally from pᵢ to p_f, the entropy change is given by:
- Δ S = RT ln(pᵢ)/(p_f)
- Δ S = nR ln(pᵢ)/(p_f)
- Δ S = nR ln(p_f)/(pᵢ)
- Δ S = nRT ln(p_f)/(pᵢ)
Answer: (B) Δ S = nR ln(pᵢ)/(p_f)
For an isothermal change, Δ S = nRln(V_f/Vᵢ). Since at constant T, V ∝ 1/p, V_f/Vᵢ = pᵢ/p_f. Thus Δ S = nRln(pᵢ/p_f).
- Q60For the reaction 2Cl(g) arrow Cl₂(g), the correct option is:
- Δᵣ H > 0 and Δᵣ S > 0
- Δᵣ H < 0 and Δᵣ S > 0
- Δᵣ H < 0 and Δᵣ S < 0
- Δᵣ H > 0 and Δᵣ S < 0
Answer: (C) Δᵣ H < 0 and Δᵣ S < 0
A Cl−Cl bond forms, releasing energy ⇒ Δᵣ H < 0. Gaseous moles fall (2 → 1, Δ n_g = -1) ⇒ disorder decreases ⇒ Δᵣ S < 0.
- Q61Which of the following acids is a vitamin ?
- Aspartic acid
- Saccharic acid
- Ascorbic acid
- Adipic acid
Answer: (C) Ascorbic acid
B. Tests: recognising vitamin C from its chemical name. Why B: every vitamin has a letter name and a chemical name, so the test is whether the acid in front of you appears in the vitamin table at all, not whether it sounds biological. Of these four, only ascorbic acid does: it is the chemical name of vitamin C, a water-soluble vitamin needed for collagen synthesis, and its deficiency causes scurvy. Why not A: saccharic acid is the dicarboxylic acid obtained when glucose is oxidised by concentrated nitric acid, so it is a carbohydrate derivative and not a vitamin. Why not C: aspartic acid is a non-essential amino acid, a building block of proteins rather than a vitamin [misconception C: “aspartic acid and ascorbic acid sound alike, so they are the same substance”]. Why not D: adipic acid is the six-carbon dicarboxylic acid used with hexamethylenediamine to make nylon-6,6, purely an industrial monomer. Remember: on this list only ascorbic acid is a vitamin, and it is vitamin C.
- Q62Which one of the following statements is correct?
- Proteins are composed of only one type of α-amino acid
- Starch is a polymer of α-glucose
- Amylose is a component of cellulose
- In cyclic structure of fructose, there are four carbons and one oxygen in the ring
Answer: (B) Starch is a polymer of α-glucose
Starch (amylose + amylopectin) is a polymer of α-D-glucose – correct. Amylose is a component of starch, not cellulose. Proteins contain about twenty different amino acids, not one. The fructofuranose ring has five carbons total in the molecule but the ring itself has four carbons and one oxygen; the option as a general ‘fructose’ statement is treated as the incorrect distractor, leaving (A) as the correct statement.
- Q63Isotonic solutions have the same:
- freezing temperature
- osmotic pressure
- boiling temperature
- vapour pressure
Answer: (B) osmotic pressure
Isotonic solutions are defined as having the same osmotic pressure at a given temperature (π = CRT), which means they have the same molar concentration of particles.
- Q64Which of the following molecules has a square planar shape?
- PCl₅
- CH₄
- SF₄
- XeF₄
Answer: (D) XeF₄
XeF₄ has 4 bond pairs and 2 lone pairs (steric number 6, octahedral arrangement). The two lone pairs occupy opposite (trans) positions, leaving a square planar shape.
- Q65Which one of the elements with the following outer-orbital configurations may exhibit the largest number of oxidation states?
- 3d⁵ 4s²
- 3d⁵ 4s¹
- 3d³ 4s²
- 3d² 4s²
Answer: (A) 3d⁵ 4s²
The maximum number of oxidation states for a d-block element equals the sum of (n-1)d and ns electrons. (a) 3+2=5; (b) 5+1=6; (c) 5+2=7; (d) 2+2=4. The 3d⁵4s² (manganese) configuration gives 7, the largest.
- Q66The Ksp of Ag₂CrO₄ in terms of its solubility s is:
- 27s⁴
- 4s³
- s²
- s³
Answer: (B) 4s³
Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻; [Ag⁺] = 2s, [CrO₄²⁻] = s, so Ksp = (2s)²(s) = 4s³.
- Q67Which one of the following lanthanoid ions is diamagnetic? (At. no. Ce = 58, Sm = 62, Eu = 63, Yb = 70)
- Eu²⁺
- Sm²⁺
- Yb²⁺
- Ce²⁺
Answer: (C) Yb²⁺
Yb²⁺ is [Xe]4f¹⁴ – a completely filled f-subshell with no unpaired electrons – so it is diamagnetic. Ce²⁺, Sm²⁺ and Eu²⁺ all have unpaired 4f electrons and are paramagnetic.
- Q68Which of the following molecules acts as a Lewis acid?
- (CH₃)₃P
- (CH₃)₃N
- (CH₃)₃B
- (CH₃)₂O
Answer: (C) (CH₃)₃B
A Lewis acid accepts an electron pair. Boron in (CH₃)₃B has only 6 electrons (an incomplete octet) and an empty orbital, so it accepts a lone pair. The O, P and N species each have a lone pair to donate (Lewis bases).
- Q69The pair of species with the same bond order is:
- N₂, O₂
- O₂⁺, NO⁺
- NO, CO
- O₂²⁻, B₂
Answer: (D) O₂²⁻, B₂
O₂²⁻ (18 e⁻) has B.O. = (10−8)/2 = 1 and B₂ (10 e⁻) has B.O. = (6−4)/2 = 1 – same bond order. The other pairs differ (O₂⁺ 2.5 vs NO⁺ 3; NO 2.5 vs CO 3; N₂ 3 vs O₂ 2).
- Q70The electronic configuration of gadolinium (atomic number 64) is:
- [Xe] 4f⁸ 5d⁹ 6s²
- [Xe] 4f⁹ 5s¹ 6s²
- [Xe] 4f⁵ 5d⁵ 6s²
- [Xe] 4f⁷ 5d¹ 6s²
Answer: (D) [Xe] 4f⁷ 5d¹ 6s²
Gd (Z=64) shows the stable half-filled 4f⁷ configuration: [Xe] 4f⁷ 5d¹ 6s². The single 5d electron plus a half-filled 4f gives extra stability over [Xe] 4f⁸ 6s².
- Q71Which structure is linear?
- SO₂
- CO₃²⁻
- CO₂
- SO₄²⁻
Answer: (C) CO₂
CO₂ has the carbon sp hybridised with two σ-bonds and no lone pair → linear. SO₂ is bent, CO₃²⁻ is trigonal planar, and SO₄²⁻ is tetrahedral.
- Q72Which species CANNOT undergo disproportionation?
- H₂O₂
- ClO⁻
- MnO₄²⁻
- F⁻
Answer: (D) F⁻
Disproportionation requires an element in an intermediate oxidation state. In F⁻, fluorine is at its lowest (and only negative) state −1 and cannot be reduced further, so it cannot disproportionate. The others have intermediate states.
- Q73Given below are two statements : Statement I: The second ionization enthalpy of B, Al and Ga is in the order of B > Al > Ga. Statement II: The correct order in terms of first ionization enthalpy is Si < Ge < Pb < Sn. In the light of the above statements, choose the correct answer from the options given below :
- Both Statement I and Statement II are false
- Both Statement I and Statement II are true
- Statement I is true but Statement II is false
- Statement I is false but Statement II is true
Answer: (A) Both Statement I and Statement II are false
B. Tests: recalling the actual ionisation enthalpy values in Groups 13 and 14 rather than assuming a smooth trend. Why B: Statement I. The second ionisation enthalpies are B 2427, Al 1817 and Ga 1979 kJ mol⁻¹, so the real order is B > Ga > Al. Gallium overtakes aluminium because the intervening 3d¹⁰ electrons shield the nuclear charge poorly, so Ga’s remaining electrons are held more tightly. Statement I is therefore false. Statement II. The first ionisation enthalpies are Si 786, Ge 761, Sn 708 and Pb 715 kJ mol⁻¹, giving the increasing order Sn < Pb < Ge < Si. The statement claims Si < Ge < Pb < Sn, which makes silicon the lowest when it is actually the highest. Statement II is false too, so both are false. Why not A: it needs Al > Ga in the second ionisation enthalpy, but Ga (1979) exceeds Al (1817) because of poor 3d shielding. Why not C: Statement I fails for that same Ga versus Al reason [misconception C: “ionisation enthalpy always decreases down a group”]. Why not D: Statement II puts Si lowest, yet Si has the largest first ionisation enthalpy of the four at 786 kJ mol⁻¹. Remember: poor d-shielding pushes Ga’s ionisation enthalpy above Al’s, and in Group 14 the first ionisation enthalpy falls Si > Ge > Pb > Sn.
- Q74Match the complexes with their spin-only magnetic moments: A) [Fe(CN)₆]³⁻, B) [Fe(H₂O)₆]³⁺, C) [Fe(CN)₆]⁴⁻, D) [Fe(H₂O)₆]²⁺ with 1) 5.92 BM, 2) 0 BM, 3) 4.90 BM, 4) 1.73 BM.
- A→2, B→1, C→3, D→4
- A→4, B→1, C→2, D→3
- A→1, B→2, C→3, D→4
- A→3, B→4, C→1, D→2
Answer: (B) A→4, B→1, C→2, D→3
A: Fe³⁺ d⁵ + strong CN⁻ → 1 unpaired → 1.73 BM (4). B: Fe³⁺ d⁵ + weak H₂O → 5 unpaired → 5.92 BM (1). C: Fe²⁺ d⁶ + strong CN⁻ → 0 unpaired → 0 BM (2). D: Fe²⁺ d⁶ + weak H₂O → 4 unpaired → 4.90 BM (3). So A→4, B→1, C→2, D→3.
- Q75What is the order and molecularity of the following elementary reaction? 2NO_(2(g)) arrow 2NO_((g)) + O_(2(g)), rate = k × [NO₂]²
- The reaction is first order and bimolecular.
- The reaction is second order and bimolecular.
- The reaction is second order and unimolecular.
- The reaction is zero order and unimolecular.
Answer: (B) The reaction is second order and bimolecular.
A. Tests: reading order from the rate law and molecularity from the elementary step, and recognising that they coincide for an elementary reaction. Why A: The rate law is rate = k[NO₂]², so the exponents sum to 2 and the order is 2. The reaction is stated to be elementary, and its single step needs two NO₂ molecules to collide, so the molecularity is 2, that is bimolecular. Why not B: first order would require rate = k[NO₂]¹, but the given rate law carries the square [misconception B: “order is the number of different reactant species”]. Why not C: unimolecular means one molecule reacts on its own in the elementary step, whereas here two NO₂ molecules must meet. Why not D: zero order would mean the rate is independent of [NO₂], which contradicts the squared term, and unimolecular again misreads the collision as involving one molecule. Remember: for an elementary step, order and molecularity match, and both are read off the two NO₂ that collide.
- Q76The shortest carbon-carbon bond distance is found in:
- ethane
- benzene
- diamond
- acetylene
Answer: (D) acetylene
Bond length shortens as bond order rises. Acetylene (ethyne) has a C≡C TRIPLE bond (about 120 pm), shorter than the aromatic C-C of benzene (about 139 pm) and the single C-C bonds in ethane and diamond (about 154 pm).
- Q77In the nitration mixture (conc. HNO₃ + conc. H₂SO₄) used to nitrate benzene, nitric acid acts as a/an:
- base
- catalyst
- reducing agent
- acid
Answer: (A) base
Sulphuric acid is the stronger acid and protonates nitric acid: HNO₃ + 2H₂SO₄ arrow NO₂⁺ + H₃O⁺ + 2HSO₄⁻. Because HNO₃ accepts a proton from H₂SO₄ on the way to forming the nitronium ion, it behaves as a BASE in this mixture.
- Q78Which of the following electrolyte can be used to obtain H₂S₂O₈ by the process of electrolysis?
- Dilute solution of sulphuric acid
- Concentrated solution of sulphuric acid
- Acidified dilute solution of sodium sulphate.
- Dilute solution of sodium sulphate.
Answer: (B) Concentrated solution of sulphuric acid
A. Tests: which species wins the competition for discharge at the anode, sulphate or water. Why A: peroxodisulphuric acid, H₂S₂O₈, is made by the anodic oxidation 2HSO₄⁻ arrow S₂O₈²⁻ + 2H⁺ + 2e⁻, which couples two sulphate units through a peroxide bridge. Water is easier to oxidise than HSO₄⁻, so this route only wins when HSO₄⁻ is present in overwhelming concentration and water is scarce. That is exactly concentrated sulphuric acid. Why not B: dilute sodium sulphate solution is mostly water, so the anode simply oxidises water to O₂ and no peroxo bond is ever formed. Why not C: acidifying a dilute sulphate solution still leaves water as the dominant species at the anode, so O₂ is still the product [misconception C: “adding acid is what triggers sulphate discharge, rather than sheer concentration”]. Why not D: dilute sulphuric acid is the classic electrolysis of water set up, giving H₂ at the cathode and O₂ at the anode. Remember: concentrated sulphuric acid at the anode builds the peroxodisulphate bridge, while anything dilute just gives oxygen from water.
- Q79The IUPAC name of the compound CH≡C-CH=CH₂ is:
- 1-butyne-3-ene
- but-1-yn-3-ene
- but-3-en-1-yne
- but-1-en-3-yne
Answer: (D) but-1-en-3-yne
Number the 4-carbon chain to give the lowest locants to the multiple bonds. Both directions give the set {1,3}. The tie-breaking rule gives the LOWER locant to the double bond, so the double bond is at C-1 and the triple bond at C-3: but-1-en-3-yne.
- Q80Which condition makes the heat absorbed by a system equal to its change in internal energy (q = ΔU)?
- Constant pressure
- Constant volume
- Adiabatic process
- Constant temperature
Answer: (B) Constant volume
At constant volume ΔV = 0, so PV-work w = 0. Then ΔU = q + w = q_V. Hence q_V = ΔU. (At constant pressure qₚ = ΔH instead.)
- Q81Propane on chlorination under photochemical conditions gives two di-chloro products, x and y. Amongst x and y, x is an optically active molecule. How many tri-chloro products (consider only structural isomers) will be obtained from x when it is further treated with chlorine under photochemical conditions?
- 2
- 4
- 3
- 5
Answer: (C) 3
C. Tests: identifying the chiral dichloropropane, then counting distinct structural isomers from a further free radical chlorination. Why C: Photochemical chlorination of propane gives 1,1-, 1,2-, 1,3- and 2,2-dichloropropane. Only 1,2-dichloropropane, CH₃-CHCl-CH₂Cl, has a carbon bearing four different groups (the CHCl carbon carries H, Cl, CH₃ and CH₂Cl), so x is 1,2-dichloropropane. Replacing one further H gives three distinct skeletons: at the CH₃ end gives 1,2,3-trichloropropane, at the CHCl carbon gives 1,2,2-trichloropropane, at the CH₂Cl carbon gives 1,1,2-trichloropropane. That is 3 structural isomers. Why not A: 4 is the number of dichloro products of propane itself, carried over from the previous step instead of counting the trichloro products of x [misconception A: “reusing the previous step’s count”]. Why not B: 2 comes from substituting only at the two carbons that already carry Cl and forgetting the free CH₃ end, or from forgetting the lone H on the chiral CHCl carbon [misconception B: “scanning only the carbons that already bear a substituent”]. Why not D: 5 is the total number of trichloropropane structural isomers that can be written from propane at all (1,1,1-, 1,1,2-, 1,1,3-, 1,2,2- and 1,2,3-), but only the three reachable from x are being asked for [misconception D: “listing every isomer of the target formula instead of only those the given reactant can reach”]. Remember: a chiral carbon needs four different groups, and 1,2-dichloropropane is the only chiral dichloropropane.
- Q82Hofmann Bromamide Degradation reaction is shown by __________.
- ArNO₂
- ArCH₂NH₂
- ArNH₂
- ArCONH₂
Answer: (D) ArCONH₂
B. Tests: recognising the one functional group Hofmann bromamide degradation can start from, an unsubstituted amide. Why B: the reaction needs a -CONH₂ group, because Br₂/KOH must first brominate an N-H on the amide nitrogen, then remove the second N-H proton, and only then can the aryl group migrate across from the carbonyl carbon to nitrogen. ArCONH₂ has exactly that arrangement and gives ArNH₂ with one carbon less. Why not A: ArNH₂ is the product of this reaction, not the substrate, and it has no carbonyl carbon for a group to migrate from [misconception A: “the amine is what reacts with Br2/KOH”]. Why not C: ArNO₂ is a nitro compound with no N-H and no carbonyl, it becomes an amine by reduction with Sn/HCl or H₂/Pd, which is a different route entirely. Why not D: ArCH₂NH₂ is already a primary amine, there is no carbonyl carbon in it to lose, so the degradation has nothing to start on. Remember: Hofmann needs R-CONH₂, an amide with two N-H bonds, anything that is already an amine is the product and not the reactant.
- Q83Cobalt(III) chloride forms several octahedral complexes with ammonia. Which of the following will NOT give a test for chloride ions with silver nitrate at 25°C?
- CoCl₃·3NH₃
- CoCl₃·6NH₃
- CoCl₃·4NH₃
- CoCl₃·5NH₃
Answer: (A) CoCl₃·3NH₃
Only chloride OUTSIDE the coordination sphere ionises and precipitates with AgNO₃. CoCl₃·3NH₃ is [Co(NH₃)₃Cl₃] – all three Cl⁻ are coordinated (inside the sphere) and none ionise, so it gives no test for free Cl⁻. (·4NH₃ gives 1, ·5NH₃ gives 2, ·6NH₃ gives 3 AgCl.)
- Q84Which order is correct regarding the -I (electron-withdrawing inductive) effect of the substituents -NR₂, -OR and -F (R = alkyl)?
- -NR₂ > -OR < -F
- -NR₂ < -OR > -F
- -NR₂ < -OR < -F
- -NR₂ > -OR > -F
Answer: (C) -NR₂ < -OR < -F
The -I effect tracks the electronegativity of the key atom: F > O > N. Hence the electron-withdrawing inductive strength rises as -NR₂ < -OR < -F.
- Q85Which one of the following esters cannot undergo Claisen self-condensation?
- C₆H₅CH₂COOC₂H₅
- C₆H₅COOC₂H₅
- CH₃CH₂CH₂COOC₂H₅
- C₆H₅CH₂CH₂COOC₂H₅
Answer: (B) C₆H₅COOC₂H₅
Claisen self-condensation needs an α-hydrogen on the ester. Ethyl benzoate, C₆H₅-COOC₂H₅, has the carbonyl bonded directly to the phenyl ring with no α-hydrogen on the acid side, so it cannot self-condense. The other three have α-CH₂/α-CH groups and can.
- Q86A certain compound X, when treated with copper sulphate solution, yields a brown precipitate. On adding hypo (sodium thiosulphate) solution, the precipitate turns white. The compound X is:
- K₃PO₄
- K₂CO₃
- KBr
- KI
Answer: (D) KI
KI reduces Cu²⁺: 2CuSO₄ + 4KI → Cu₂I₂ (white) + I₂ + 2K₂SO₄. The liberated I₂ colours the precipitate brown. Adding hypo removes the iodine (2Na₂S₂O₃ + I₂ → Na₂S₄O₆ + 2NaI), leaving the white Cu₂I₂. Hence X is KI.
- Q87The correct order of N-compounds in their decreasing order of oxidation states of nitrogen is:
- HNO₃, NO, N₂, NH₄Cl
- HNO₃, NH₄Cl, NO, N₂
- HNO₃, NO, NH₄Cl, N₂
- NH₄Cl, N₂, NO, HNO₃
Answer: (A) HNO₃, NO, N₂, NH₄Cl
Oxidation states of N: HNO₃ = +5, NO = +2, N₂ = 0, NH₄Cl = −3. Decreasing order is therefore HNO₃ (+5) > NO (+2) > N₂ (0) > NH₄Cl (−3).
- Q88Assertion : Tert-butyl amine can be formed by Gabriel phthalimide synthesis. Reason : It follows S_N1 mechanism.
- If both assertion and reason are true, but reason is not the correct explanation of assertion.
- If assertion is true, but reason is false.
- If both assertion and reason are true and reason is the correct explanation of assertion.
- If both assertion and reason are false.
Answer: (D) If both assertion and reason are false.
D. Tests: the mechanistic limit that decides which amines Gabriel phthalimide synthesis can and cannot make. Why D: the phthalimide anion is a bulky nucleophile that displaces halide by backside S_N2 attack. A tertiary halide such as (CH₃)₃C-Br is completely shielded from backside attack, and with so basic an anion it undergoes elimination instead, so tert-butylamine cannot be prepared this way and the assertion is false. The reason is false too, because the alkylation step is S_N2, not S_N1. Both statements false gives option D. Why not A: A requires both statements to be true, but the synthesis fails for tertiary substrates and the mechanism is misnamed [misconception A: “Gabriel synthesis makes any primary amine from any halide”]. Why not B: B still requires a true assertion, and tert-butylamine is precisely the product Gabriel synthesis cannot deliver [misconception B: “only the mechanism label is wrong here”]. Why not C: C requires a true assertion with a false reason; the reason is indeed false, but the assertion fails as well [misconception C: “a stable tertiary carbocation would rescue the reaction”]. Remember: Gabriel synthesis is strictly S_N2, so it works on methyl, primary, benzylic and allylic halides only.
- Q89The complex ion [Co(NH₃)₆]³⁺ is formed by sp³d² hybridisation. Hence the ion should possess:
- octahedral geometry
- square planar geometry
- tetragonal geometry
- tetrahedral geometry
Answer: (A) octahedral geometry
Six sp³d² hybrid orbitals point to the corners of an octahedron, so a complex with sp³d² (or d²sp³) hybridisation and coordination number 6 is octahedral.
- Q90Which of the following expressions indicates the correct relationship between molar conductivity of strong electrolyte and its concentration?
- Λ = Λ₀ – √(c)
- Λ = Λ₀ + a√(c)
- Λ = Λ₀ – a√(c)
- Λ = Λ₀ + √(c)
Answer: (C) Λ = Λ₀ – a√(c)
B. Tests: the Debye-Huckel-Onsager equation for a strong electrolyte. Why B: A strong electrolyte is fully ionised at every concentration, so the number of ions per mole does not change. What changes is how freely they move: as concentration rises, the relaxation and electrophoretic effects from neighbouring ions slow each ion down, and the loss turns out to be proportional to √(c). Hence Λ = Λ₀ – a√(c), a straight line falling from the intercept Λ₀, with a fixed by the solvent and the electrolyte type. Why not A: a plus sign predicts that molar conductivity rises with concentration, the opposite of the observed behaviour, and it also drops the constant a [misconception A: “a more concentrated solution conducts better per mole”]. Why not C: it carries the constant a correctly but keeps the wrong sign, again predicting an increase with concentration [misconception C: “the sign of the slope does not matter”]. Why not D: the sign is right but a is missing, which forces the slope to be exactly 1 for every electrolyte in every solvent [misconception D: “the slope of the plot is always unity”]. Remember: strong electrolyte, a straight line falling with √(c), intercept Λ₀ and slope -a.
Botany
- Q91In the somatic cell cycle, which statement is correct?
- In G1-phase DNA content is double the amount present in the original cell
- DNA replication takes place in S-phase
- A short interphase is followed by a long mitotic phase
- G2-phase follows the mitotic phase
Answer: (B) DNA replication takes place in S-phase
DNA replication occurs in the S-phase, where DNA content doubles relative to the parent cell. In G1 the DNA is still at the original (2C) amount; interphase is long and mitosis short; and G2 precedes (not follows) the M-phase. Hence B is correct.
- Q92Cuscuta growing on a host shrub and deriving its nutrition from it represents which interaction?
- Parasitism (+ −)
- Mutualism (+ +)
- Amensalism (− 0)
- Commensalism (+ 0)
Answer: (A) Parasitism (+ −)
Cuscuta is a leafless stem parasite lacking chlorophyll that draws nutrition from its host, harming it. The parasite benefits while the host is harmed: parasitism (+ −).
- Q93The requirement of more ATP than NADPH (ratio 3:2) by the Calvin cycle is partly met by:
- Glycolysis
- Photorespiration
- Photolysis of water
- Cyclic photophosphorylation
Answer: (D) Cyclic photophosphorylation
Non-cyclic photophosphorylation produces ATP and NADPH in roughly equal amounts, but the Calvin cycle needs more ATP (3 ATP : 2 NADPH per CO2). Cyclic photophosphorylation produces extra ATP to balance this demand.
- Q94Scattered vascular bundles in the ground tissue, each surrounded by a sclerenchymatous bundle sheath, indicate a transverse section of:
- Dicot root
- Dicot leaf
- Monocot stem
- Dicot stem
Answer: (C) Monocot stem
Monocot stems show numerous vascular bundles scattered throughout the ground tissue, each ensheathed by sclerenchyma. Dicot stems have bundles arranged in a single ring.
- Q95Which of the following pairs is correctly matched?
- Uricotelism – Aquatic habitat
- Excessive perspiration – Xeric adaptation
- Streamlined body – Aquatic adaptation
- Parasitism – Intra-specific relationship
Answer: (C) Streamlined body – Aquatic adaptation
A streamlined body reduces drag and is a true aquatic adaptation, so (D) is correct. Uricotelism is a terrestrial water-conserving adaptation, parasitism is interspecific, and excessive perspiration would waste water in a xeric habitat.
- Q96Tendrils of Cucurbita (gourd) and thorns of Citrus are both modifications of the stem because they:
- Arise in the axil of leaves (from axillary buds)
- Store food
- Bear root hairs
- Are green in colour
Answer: (A) Arise in the axil of leaves (from axillary buds)
Stem tendrils and thorns develop from axillary (or terminal) buds, marking them as stem modifications, unlike leaf tendrils/spines which replace the leaf or its parts.
- Q97An mRNA contains 99 nucleotides in its coding sequence including the stop codon. The maximum number of amino acids in the polypeptide produced is:
- 32
- 33
- 34
- 99
Answer: (A) 32
99 nucleotides = 99/3 = 33 codons. One of these is the stop codon (no amino acid), so 33 − 1 = 32 amino acids are incorporated.
- Q98Which one of the following events does NOT occur in the rough endoplasmic reticulum?
- Protein folding
- Cleavage of signal peptide
- Phospholipid synthesis
- Protein glycosylation
Answer: (C) Phospholipid synthesis
Phospholipid (lipid) synthesis occurs in the SMOOTH ER, not the rough ER. The RER, bearing ribosomes, handles synthesis of secretory proteins, signal-peptide cleavage, glycosylation and folding, so phospholipid synthesis is the event that does not occur there.
- Q99Which of the following cell organelles is responsible for extracting energy from carbohydrates to form ATP?
- Mitochondrion
- Ribosome
- Lysosome
- Chloroplast
Answer: (A) Mitochondrion
The mitochondrion (power house of the cell) oxidises carbohydrates by aerobic respiration to produce ATP. Chloroplasts make food but do not extract energy from carbohydrates; lysosomes digest and ribosomes synthesise protein.
- Q100Assertion (A): In incomplete dominance the F2 phenotypic and genotypic ratios are identical. Reason (R): Each genotype produces a distinct, distinguishable phenotype.
- Both A and R are true and R is the correct explanation of A
- A is true but R is false
- A is false but R is true
- Both A and R are true but R is not the correct explanation of A
Answer: (A) Both A and R are true and R is the correct explanation of A
In incomplete dominance the heterozygote is phenotypically distinct (e.g. pink), so all three genotypes (RR, Rr, rr) look different. Hence phenotypic ratio (1:2:1) equals genotypic ratio – and R correctly explains A.
- Q101A population grows exponentially with N₀ = 100 and r = 0.0693 per day. Using Nt = N₀eʳᵗ, what is the population after 10 days (take e⁰.693 ≈ 2)?
- 200
- 150
- 100
- 693
Answer: (A) 200
rt = 0.0693 × 10 = 0.693, so e⁰.693 ≈ 2. Nt = 100 × 2 = 200. (0.693 is ln 2, the doubling time exponent.)
- Q102Of the incident solar radiation, the fraction that is photosynthetically active radiation (PAR) is about:
- Exactly 100%
- Only 2-10%
- About 80%
- Less than 50%
Answer: (D) Less than 50%
Less than 50% of incident solar radiation is PAR. Plants then capture only 2-10% of this PAR, which sets the energy entering the ecosystem.
- Q103If the coding (sense) strand of a gene reads 5′-ATGCGA-3′, the mRNA sequence produced will be:
- 5′-TACGCT-3′
- 5′-AUGCGA-3′
- 5′-UACGCU-3′
- 3′-AUGCGA-5′
Answer: (B) 5′-AUGCGA-3′
The mRNA has the same sequence as the coding strand except that T is replaced by U. So 5′-ATGCGA-3′ becomes 5′-AUGCGA-3′.
- Q104The stored food and the type of flagella found in Chlorophyceae (green algae) are respectively:
- Starch; 2-8 equal apical flagella
- Floridean starch; no flagella
- Laminarin; 2 equal apical flagella
- Mannitol; 2 unequal lateral flagella
Answer: (A) Starch; 2-8 equal apical flagella
Green algae (Chlorophyceae) store food as starch and have 2-8 equal, apical (whiplash) flagella. Mannitol/laminarin and lateral unequal flagella belong to brown algae; floridean starch and no flagella to red algae.
- Q105Meiosis II is often described as ‘equational’ because:
- It halves the chromosome number again
- It separates homologous chromosomes
- It involves crossing over
- It keeps the chromosome number the same in resulting cells (like mitosis)
Answer: (D) It keeps the chromosome number the same in resulting cells (like mitosis)
Meiosis II resembles mitosis: it separates sister chromatids without changing the (already halved) chromosome number, so it is the equational division of meiosis. Reduction occurred in meiosis I.
- Q106After fertilisation in a flowering plant, the ovule and the ovary respectively develop into:
- Pericarp and seed coat
- Fruit and seed
- Embryo and endosperm
- Seed and fruit
Answer: (D) Seed and fruit
In post-fertilisation events of angiosperms, the ovule matures into the seed and the ovary develops into the fruit (its wall forms the pericarp). The zygote becomes the embryo.
- Q107The end products of the citric acid (Krebs) cycle, after complete oxidation, are:
- CO2 + H2O
- Citric acid
- Lactic acid
- Pyruvic acid
Answer: (A) CO2 + H2O
In aerobic respiration the carbons of acetyl CoA are fully oxidised, so the ultimate products are CO2 (released in the cycle) and H2O (formed at the end of the ETS when oxygen accepts electrons). Citric and pyruvic acids are intermediates, and lactic acid is an anaerobic product.
- Q108In which of the following are both autogamy and geitonogamy prevented?
- Wheat
- Castor
- Maize
- Papaya
Answer: (D) Papaya
Papaya is dioecious (male and female flowers on separate plants), so both autogamy (within a flower) and geitonogamy (between flowers of the same plant) are impossible. Maize and castor are monoecious (both sexes on one plant) so geitonogamy is still possible.
- Q109Two strands of the DNA double helix are described as antiparallel because:
- both run in the 5’→3′ direction
- one runs 5’→3′ and the other 3’→5′
- they have identical base sequences
- they are held by ionic bonds
Answer: (B) one runs 5’→3′ and the other 3’→5′
Antiparallel means the two strands have opposite polarity: one is 5’→3′ while its complement is 3’→5′. They are complementary (not identical) and held by hydrogen bonds.
- Q110Chiasmata, the X-shaped sites that mark where crossing over occurred, become clearly visible during:
- Diplotene
- Zygotene
- Pachytene
- Leptotene
Answer: (A) Diplotene
In diplotene the synaptonemal complex dissolves and homologues begin to separate but remain joined at chiasmata, the visible X-shaped sites of earlier crossing over.
- Q111An organism with two identical alleles for a gene (e.g. TT or tt) is said to be:
- homozygous
- heterozygous
- hybrid
- dominant
Answer: (A) homozygous
Two identical alleles (TT or tt) make the organism homozygous (pure) for that gene. Two different alleles (Tt) would make it heterozygous (hybrid).
- Q112Assertion (A): Metabolism is a defining feature of all living organisms. Reason (R): No non-living object exhibits metabolism, although isolated metabolic reactions can occur in cell-free systems in vitro.
- A is true but R is false
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is NOT the correct explanation of A
- A is false but R is true
Answer: (B) Both A and R are true and R is the correct explanation of A
All living things show metabolism and no non-living object does, so metabolism is defining (A true). Isolated metabolic reactions in a test tube (in vitro) are neither living nor non-living, yet a complete living metabolism is unique to organisms, which correctly explains why metabolism defines life (R true and explains A).
- Q113Maximum rate of photosynthesis occurs in which regions of the visible spectrum?
- Green and yellow
- Only green
- Orange and green
- Blue and red
Answer: (D) Blue and red
Chlorophyll a absorbs maximally in the blue-violet and red regions; therefore photosynthesis is highest in blue and red light and least in the green region (green is mostly reflected).
- Q114Opening of stomata is most directly associated with:
- Efflux of K⁺ from guard cells and loss of turgor
- Active influx of K⁺ into guard cells, lowering their water potential and making them turgid
- Thickening of the outer wall of guard cells
- Decrease in guard cell water content
Answer: (B) Active influx of K⁺ into guard cells, lowering their water potential and making them turgid
Stomatal opening is driven by active uptake of K⁺ ions into guard cells, which lowers their water potential; water then enters by osmosis, the guard cells become turgid and the differential wall thickening makes the pore open.
- Q115During transcription, if the nucleotide sequence of the DNA strand being coded (template) is ATACG, then the nucleotide sequence in the mRNA would be:
- UATGG
- TATGC
- UAUGC
- TCTGG
Answer: (C) UAUGC
The mRNA is complementary to the template with U for T: A->U, T->A, A->U, C->G, G->C = UAUGC.
- Q116In an area where DDT had been used extensively, the population of birds declined significantly because:
- Many of the birds laid eggs that did not hatch
- Birds stopped laying eggs
- Earthworms in the area got eradicated
- Cobras were feeding exclusively on birds
Answer: (A) Many of the birds laid eggs that did not hatch
DDT biomagnifies up the food chain and accumulates in birds, where it disturbs calcium metabolism and thins eggshells. The fragile eggs break or fail to hatch, so the bird population declines.
- Q117In the three-domain system of Carl Woese, Kingdom Monera of the five-kingdom system is split into:
- Protista and Bacteria
- Archaea and Eukarya
- Bacteria and Eukarya
- Archaea and Bacteria
Answer: (D) Archaea and Bacteria
Woese’s three domains are Archaea, Bacteria and Eukarya. Monera (prokaryotes) is divided into the two prokaryotic domains Archaea and Bacteria; eukaryotes form Eukarya.
- Q118PGA as the first CO2-fixation product of photosynthesis was discovered in:
- Alga
- Angiosperm
- Gymnosperm
- Bryophyte
Answer: (A) Alga
Melvin Calvin used radioactive 14C in the alga Chlorella to trace CO2 fixation and found the first stable product to be the 3-carbon 3-phosphoglyceric acid (PGA). The work was done in an alga, not a bryophyte/gymnosperm/angiosperm.
- Q119The cell organelle involved in the glycosylation of proteins is the:
- peroxisome
- mitochondria
- endoplasmic reticulum
- ribosome
Answer: (C) endoplasmic reticulum
Proteins synthesised on ribosomes bound to the ER are passed into the lumen of the endoplasmic reticulum where oligosaccharides are added (N-linked glycosylation begins here); glycosylation is then completed in the Golgi. Of the options the ER is the keyed organelle.
- Q120The phenomenon in which mature living differentiated cells regain the capacity to divide is called:
- Differentiation
- Redifferentiation
- Determination
- Dedifferentiation
Answer: (D) Dedifferentiation
Dedifferentiation is when already differentiated living cells (e.g. parenchyma) regain the ability to divide, as in formation of interfascicular and cork cambium. Redifferentiation is the subsequent loss of dividing ability.
- Q121Which one of the following is NOT observed in biodiversity hotspots?
- Endemism
- Accelerated species loss
- Lesser interspecific competition
- Species richness
Answer: (C) Lesser interspecific competition
Hotspots are extremely rich in species (high species richness) and endemism and face accelerated species loss (high threat). With so many species packed together, interspecific competition is intense, not lesser – so ‘lesser interspecific competition’ is the feature NOT observed in hotspots.
- Q122Match the element in Column I with its associated function or role in Column II and choose the correct option. Column I: A. Boron, B. Manganese, C. Molybdenum, D. Zinc, E. Iron Column II: i. splitting of H₂O to liberate O₂ during photosynthesis, ii. needed for synthesis of auxins, iii. component of nitrogenase, iv. pollen germination, v. component of ferredoxin
- A-ii, B-iii, C-v, D-i, E-iv
- A-iii, B-ii, C-iv, D-v, E-i
- A-iv, B-i, C-iii, D-ii, E-v
- A-i, B-ii, C-iii, D-iv, E-v
Answer: (C) A-iv, B-i, C-iii, D-ii, E-v
B. Tests: holding five micronutrient roles at once and using one certain pair to eliminate whole option sets. Why B: boron is required for pollen germination and carbohydrate translocation, so A-iv. Manganese is the metal of the oxygen evolving complex that splits water in photosystem II, so B-i. Molybdenum sits in the Mo-Fe cofactor of nitrogenase and in nitrate reductase, so C-iii. Zinc is needed to make tryptophan, the precursor of the auxin IAA, so D-ii. Iron is the metal of ferredoxin and of the cytochromes, so E-v. That combination is exactly option B. Why not A: it pairs boron with water splitting and manganese with auxin synthesis, but water splitting is manganese and auxin synthesis is zinc [misconception A: “the two columns are printed in matching order”]. Why not C: it makes boron a component of nitrogenase and iron the water splitting metal, whereas nitrogenase carries Mo and Fe and the oxygen evolving complex is a manganese cluster [misconception C: “iron, being the classic redox metal, must run photolysis”]. Why not D: it assigns manganese to nitrogenase and zinc to water splitting, but manganese splits water and molybdenum is the nitrogenase metal [misconception D: “any transition metal can substitute inside nitrogenase”]. Remember: Mn splits water, Mo fixes nitrogen, Zn makes auxin, B germinates pollen, Fe builds ferredoxin.
- Q123Annular and spirally thickened conducting elements generally develop in the protoxylem when the root or stem is:
- Maturing
- Elongating
- Widening
- Differentiating
Answer: (B) Elongating
Protoxylem forms while the organ is still elongating, so its vessels lay down only annular (ring-like) and spiral wall thickenings that can stretch as the organ grows. Later-formed metaxylem (with scalariform/reticulate/pitted thickenings) matures after elongation stops.
- Q124In pea plants, yellow seeds are dominant to green. If a heterozygous yellow-seeded plant is crossed with a green-seeded plant, what ratio of yellow and green seeded plants would you expect in the F1 generation?
- 1 : 3
- 3 : 1
- 9 : 1
- 50 : 50
Answer: (D) 50 : 50
Heterozygous yellow (Yy) x green (yy) is effectively a test cross. Yy gives Y or y gametes; yy gives only y. Offspring: Yy (yellow) and yy (green) in 1:1, i.e. 50% yellow : 50% green.
- Q125If the turgidity of a cell surrounded by water increases, the wall pressure will:
- increase
- fluctuate
- remain unchanged
- decrease
Answer: (A) increase
As a cell in pure/hypotonic water takes in water, the protoplast swells and the turgor pressure rises, pushing outward on the wall. By Newton’s third law the rigid wall pushes back with an equal and opposite wall pressure, so wall pressure increases along with turgor pressure: (A).
- Q126The phenotypic ratio obtained in the F2 of a typical dihybrid cross is:
- 1:2:1
- 3:1
- 1:1:1:1
- 9:3:3:1
Answer: (D) 9:3:3:1
Selfing the dihybrid F1 (e.g. RrYy) gives the classic 9:3:3:1 F2 phenotypic ratio, the product of two independent 3:1 ratios.
- Q127The pioneer species in primary succession on a bare rock are usually:
- Mosses
- Shrubs
- Lichens
- Grasses
Answer: (C) Lichens
On bare rock, lichens are the pioneer colonisers; they secrete acids that weather rock and help form soil, paving the way for mosses and higher plants.
- Q128The CO₂ content by volume in normal atmospheric air is about:
- 0.34%
- 4%
- 3.34%
- 0.0314%
Answer: (D) 0.0314%
CO₂ makes up about 0.0314% (~0.03%) of atmospheric air by volume. It is used by producers in photosynthesis; it becomes a pollutant only when it rises well above this baseline.
- Q129The connecting link between glycolysis, the Krebs cycle and the beta-oxidation of fatty acids (carbohydrate and fat metabolism) is:
- Citric acid
- Acetyl Co-A
- Oxaloacetic acid
- Succinic acid
Answer: (B) Acetyl Co-A
Pyruvate from glycolysis and fatty acids from beta-oxidation are both converted to acetyl CoA, which then enters the Krebs cycle. Acetyl CoA is therefore the common link between carbohydrate and fat metabolism. OAA and citric acid are within the cycle, and succinic acid is just an intermediate.
- Q130The region of a biosphere reserve which is legally protected and where no human activity is allowed is known as:
- restoration zone
- transition zone
- buffer zone
- core zone
Answer: (D) core zone
A biosphere reserve has core, buffer and transition zones. The core zone is the legally protected, undisturbed area where no human activity is permitted. The buffer zone allows limited research and use, and the outermost transition (manipulation) zone permits human settlement and cropping.
- Q131A man and a woman who do not show any apparent signs of a certain inherited disease have seven children (2 daughters and 5 sons). Three of the sons suffer from the disease but none of the daughters are affected. Which mode of inheritance do you suggest for this disease?
- Sex-linked recessive
- Autosomal recessive
- Sex-linked dominant
- Autosomal dominant
Answer: (A) Sex-linked recessive
Unaffected parents with affected SONS only (no affected daughters) points to X-linked recessive: the mother is a carrier (X^A Xᵃ) and passes Xᵃ to half her sons. Daughters get a normal X from the unaffected father, so none are affected.
- Q132Assertion (A): The pyramid of energy can never be inverted. Reason (R): When energy flows from one trophic level to the next, some energy is always lost as heat.
- A is true but R is false
- A is false but R is true
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
Answer: (C) Both A and R are true and R is the correct explanation of A
The energy pyramid is always upright because energy is lost as heat at each transfer (only ~10% passes on), so a higher level can never hold more energy than the level below. R correctly explains A.
- Q133The scutellum of a wheat or maize grain is comparable to which part of the seed in other monocotyledons?
- Cotyledon
- Endosperm
- Plumule
- Aleurone layer
Answer: (A) Cotyledon
In grasses the scutellum is regarded as a modified cotyledon (seed leaf). It is not the endosperm, aleurone layer or plumule.
- Q134Which set correctly pairs the plant with its sub-aerial stem modification?
- Eichhornia - sucker
- Grass - stolon
- Mint - offset
- Jasmine - stolon
Answer: (D) Jasmine - stolon
Jasmine produces a stolon (a lateral branch arching to the ground). Eichhornia/Pistia have offsets, mint/Chrysanthemum have suckers, and grass has runners.
- Q135Which of the following pairs of organisms would share the GREATER number of common characteristics?
- Two organisms belonging to the same kingdom
- Two organisms belonging to the same phylum
- Two organisms belonging to the same class
- Two organisms belonging to the same family
Answer: (D) Two organisms belonging to the same family
The lower (more specific) the shared category, the more characters two organisms have in common. Family is lower than class, phylum and kingdom, so two organisms in the same family share the greatest number of common characters. Hence A.
Zoology
- Q136Select the correct statement with respect to Periplaneta americana.
- Nervous system located dorsally, consisting of segmentally arranged ganglia joined by a pair of longitudinal connectives
- Grinding of food is carried out only by the mouth parts
- Males bear a pair of short thread-like anal styles
- There are 16 very long Malpighian tubules present at the junction of midgut and hindgut
Answer: (C) Males bear a pair of short thread-like anal styles
Male cockroaches bear a pair of short thread-like anal styles (on the 9th sternum) – correct. The nervous system is VENTRAL (not dorsal); there are about 100-150 (not 16) Malpighian tubules; and grinding of food is done by the gizzard, not only the mouth parts.
- Q137If fertilisation does NOT occur, menstruation results mainly because:
- The LH surge is repeated
- The corpus luteum degenerates and progesterone falls
- FSH levels rise sharply
- Estrogen rises to a peak
Answer: (B) The corpus luteum degenerates and progesterone falls
Without pregnancy, the corpus luteum regresses; the resulting fall in progesterone (and estrogen) causes the endometrium to break down, producing menstrual flow.
- Q138The pneumotaxic centre influences respiration by:
- Reducing the duration of inspiration, thereby altering breathing rate
- Initiating the basic respiratory rhythm
- Producing surfactant in alveoli
- Sensing oxygen levels in blood
Answer: (A) Reducing the duration of inspiration, thereby altering breathing rate
The pneumotaxic centre in the pons sends signals that can reduce the duration of inspiration and thus alter the respiratory rate. The basic rhythm itself is generated by the medullary rhythm centre.
- Q139Which of the following restriction enzymes produces blunt ends?
- SalI
- EcoRV
- HindIII
- XhoI
Answer: (B) EcoRV
EcoRV cuts its recognition sequence straight across (at the centre, with no stagger), producing blunt ends. SalI, XhoI and HindIII cut at staggered positions, generating single-stranded overhangs (sticky ends).
- Q140Systemic heart refers to:
- The entire heart in lower vertebrates that pumps blood to body parts and not to lungs
- The heart that contracts under stimulation from the nervous system
- The two ventricles together in humans
- The left auricle and left ventricle in higher vertebrates
Answer: (D) The left auricle and left ventricle in higher vertebrates
In higher vertebrates, the left auricle and left ventricle handle the oxygenated blood and pump it to the body (systemic circulation), so together they form the ‘systemic heart’. The right side (pulmonary heart) sends blood to the lungs.
- Q141Which one of the following features is NOT present in the phylum Arthropoda?
- Chitinous exoskeleton
- Metameric segmentation
- Parapodia
- Jointed appendages
Answer: (C) Parapodia
Parapodia are lateral, fleshy locomotory outgrowths of aquatic annelids (e.g. Nereis), NOT a feature of arthropods. Arthropods do show metameric segmentation, jointed appendages and a chitinous exoskeleton.
- Q142Collagen is a:
- Globular protein
- Carbohydrate
- Lipid
- Fibrous protein
Answer: (D) Fibrous protein
Collagen is a fibrous (structural) protein of connective tissue, occurring as white fibres produced by fibroblasts. Globular proteins (enzymes, haemoglobin) are compact and soluble; collagen is elongated and insoluble.
- Q143The function of copper ions in copper-releasing IUDs is that they:
- Make the uterus unsuitable for implantation
- Suppress sperm motility and the fertilising capacity of sperms
- Inhibit ovulation
- Inhibit gametogenesis
Answer: (B) Suppress sperm motility and the fertilising capacity of sperms
Copper ions released by Cu IUDs act as a spermicide: they suppress sperm motility and the fertilising capacity of sperms (and increase phagocytosis of sperms). They do not inhibit gametogenesis or ovulation; making the uterus unsuitable for implantation is the action of hormone-releasing IUDs, not the copper ions.
- Q144Which one does NOT favour the Lamarckian concept of inheritance of acquired characters?
- Absence of limbs in snakes
- Melanisation of peppered moth in industrial areas
- Presence of webbed toes in aquatic birds
- Lack of pigment in cave dwellers
Answer: (B) Melanisation of peppered moth in industrial areas
Cave-dweller depigmentation, snake limblessness and webbed toes were cited as use/disuse effects supporting Lamarck. Industrial melanism in Biston betularia, however, is a result of natural selection on existing variation, not inheritance of an acquired character – so it does NOT favour Lamarckism.
- Q145Assertion (A): Bile salts are essential for the absorption of fats. Reason (R): Bile salts form micelles that transport the otherwise insoluble fatty acids and monoglycerides to the mucosal surface.
- Both A and R are true and R is the correct explanation of A
- A is false but R is true
- Both A and R are true but R is NOT the correct explanation of A
- A is true but R is false
Answer: (A) Both A and R are true and R is the correct explanation of A
Bile salts aid absorption by forming micelles that ferry insoluble fatty acids and monoglycerides to the mucosal cell surface. Both statements are true and R correctly explains A.
- Q146The use of anti-histamines and steroids gives quick relief from:
- Allergy
- Cough
- Headache
- Nausea
Answer: (A) Allergy
Allergic reactions are driven by histamine released from mast cells; anti-histamines (and steroids) counter this and relieve allergy symptoms. They are not the standard quick relief for nausea, cough or headache.
- Q147Which one of the following statements about blood constituents and transport of respiratory gases is most accurate?
- RBCs as well as WBCs transport both oxygen and CO₂
- RBCs transport oxygen whereas plasma transports only CO₂
- RBCs transport oxygen whereas WBCs transport CO₂
- RBCs as well as plasma transport both oxygen and CO₂
Answer: (D) RBCs as well as plasma transport both oxygen and CO₂
Both RBCs and plasma carry both gases. O₂ is ~97% on haemoglobin in RBCs and ~3% dissolved in plasma; CO₂ is ~70% as bicarbonate (largely formed in RBCs and carried in plasma), ~20-25% as carbaminohaemoglobin in RBCs and ~7% dissolved in plasma. So RBCs and plasma transport both O₂ and CO₂ (in chemically bound or dissolved states). WBCs play no role in gas transport.
- Q148Select the correct statement regarding the specific disorder of the muscular or skeletal system:
- Myasthenia gravis - Autoimmune disorder which inhibits sliding of myosin filaments
- Osteoporosis - Decrease in bone mass and higher chances of fractures with advancing age
- Gout - Inflammation of joints due to extra deposition of calcium
- Muscular dystrophy - Age-related shortening of muscle
Answer: (B) Osteoporosis - Decrease in bone mass and higher chances of fractures with advancing age
Osteoporosis is an age-related decrease in bone mass with increased fracture risk – correct. Muscular dystrophy is a genetic progressive degeneration (not ‘age-related shortening’); myasthenia gravis is autoimmune but acts at the neuromuscular junction by blocking ACh receptors (not by inhibiting myosin sliding); gout is due to uric acid crystals, not calcium deposition.
- Q149Roughly what fraction of the world’s livestock is found in India and China combined, even though their farm output is comparatively low?
- About 40%
- About 20%
- About 90%
- About 70%
Answer: (D) About 70%
NCERT states that India and China together have about 70% of the world’s livestock, yet contribute only ~25% to the world farm produce, highlighting the need for better management.
- Q150The correct sequence of structures through which urine flows after leaving the nephron’s collecting duct is:
- Renal pelvis → ureter → calyx → bladder
- Renal pelvis → calyx → ureter → bladder
- Ureter → renal pelvis → calyx → bladder
- Calyx → renal pelvis → ureter → bladder
Answer: (D) Calyx → renal pelvis → ureter → bladder
Collecting ducts open at the pyramid tips into minor calyces, which join to form the renal pelvis, which leads into the ureter and then the urinary bladder. Order: calyx → renal pelvis → ureter → bladder.
- Q151The splenic artery arises from the:
- Coeliac (celiac) artery
- Anterior mesenteric artery
- Posterior mesenteric artery
- Intestinal artery
Answer: (A) Coeliac (celiac) artery
The splenic artery, which supplies oxygenated blood to the spleen, branches from the coeliac (celiac) artery/trunk and runs superior to the pancreas. Hence the coeliac artery.
- Q152Which is the most abundant ORGANIC biomolecule in the entire biosphere (on Earth)?
- DNA
- Cellulose
- Protein
- Water
Answer: (B) Cellulose
Cellulose, the structural polysaccharide of plant cell walls, is the most abundant organic biomolecule on Earth. Water is most abundant overall but is inorganic; protein is the most abundant organic molecule within a single cell, not the biosphere.
- Q153Vital capacity is equal to:
- TV + IRV + ERV
- ERV + RV
- TV + IRV + ERV + RV
- TV + IRV
Answer: (A) TV + IRV + ERV
Vital Capacity = TV + IRV + ERV, i.e., the maximum air a person can exhale after a maximal inhalation. Adding RV gives Total Lung Capacity, not Vital Capacity.
- Q154Cytochrome is a:
- Fe-containing porphyrin pigment
- Glycoprotein
- Lipid
- Metallo-flavoprotein
Answer: (A) Fe-containing porphyrin pigment
Cytochrome is a respiratory pigment made of protein, iron and a porphyrin ring; the iron changes valency to carry electrons in the respiratory chain. So it is an iron-containing porphyrin pigment, not a glycoprotein or lipid.
- Q155Select the correct statement about endocrine hormones:
- Glucagon is associated with hypoglycemia
- Insulin acts on pancreatic cells and adipocytes
- Insulin is associated with hyperglycemia
- Glucocorticoids stimulate gluconeogenesis
Answer: (D) Glucocorticoids stimulate gluconeogenesis
Glucocorticoids (e.g. cortisol) stimulate gluconeogenesis – correct. Glucagon causes hyperglycemia (not hypoglycemia); insulin acts on hepatocytes/adipocytes and is associated with hypoglycemia (not hyperglycemia).
- Q156Which of the following is NOT a feature of plasmids?
- Circular structure
- Single-stranded
- Transferable
- Independent replication
Answer: (B) Single-stranded
Plasmids are double-stranded (not single-stranded), circular, extra-chromosomal DNA molecules that replicate independently and can be transferred. So ‘single-stranded’ is not a feature of plasmids.
- Q157Eutherians are characterised by:
- ovoviviparity
- hairy skin
- true placentation
- glandular skin
Answer: (C) true placentation
Eutheria (placental mammals) are characterised by true placentation, in which the young are nourished through a placenta inside the mother and born well developed. Hairy skin is common to all mammals (not just eutherians), so it is not the distinguishing eutherian feature.
- Q158The elbow joint is an example of a:
- Gliding joint
- Pivot joint
- Hinge joint
- Ball and socket joint
Answer: (C) Hinge joint
The elbow (between humerus and ulna) is a hinge joint, allowing movement in one plane like a door hinge. A pivot rotates, gliding joints slide, and ball-and-socket (shoulder/hip) move in all planes.
- Q159According to WHO, reproductive health means well-being in all the following aspects EXCEPT:
- Physical
- Economic profit
- Emotional
- Social and behavioural
Answer: (B) Economic profit
WHO defines reproductive health as total well-being in physical, emotional, behavioural and social aspects of reproduction. Economic profit is not part of the definition.
- Q160Which one of the following characteristic features always holds true for the corresponding group of animals?
- Possess a mouth with an upper and a lower jaw - Chordata
- Cartilaginous endoskeleton - Chondrichthyes
- Viviparous - Mammalia
- 3-chambered heart with one incompletely divided ventricle - Reptilia
Answer: (B) Cartilaginous endoskeleton - Chondrichthyes
All members of Chondrichthyes (without exception) have a cartilaginous endoskeleton, so (D) always holds. (A) fails because monotreme mammals lay eggs; (B) fails because protochordates and cyclostomes lack jaws; (C) fails because crocodiles (reptiles) have a four-chambered heart.
- Q161The sensitive pigmented layer of the eye that contains the photoreceptors is the:
- Cornea
- Iris
- Sclerotic
- Retina
Answer: (D) Retina
The retina is the innermost, pigmented and light-sensitive layer bearing the rods and cones. The cornea and sclerotic are protective outer coats and the iris regulates the pupil.
- Q162What is true about T-lymphocytes in mammals?
- There are three main types – cytotoxic T-cells, helper T-cells and suppressor T-cells
- They scavenge damaged cells and cellular debris
- They originate in lymphoid tissues
- They are produced in the thyroid
Answer: (A) There are three main types – cytotoxic T-cells, helper T-cells and suppressor T-cells
T-lymphocytes occur as three main types – cytotoxic, helper and suppressor T-cells. They originate in the bone marrow and mature in the thymus (not the thyroid or lymphoid tissue), and they provide cell-mediated immunity rather than scavenging debris (a macrophage role).
- Q163Flight muscles of a bird are attached to the:
- clavicle
- coracoid
- scapula
- keel of the sternum
Answer: (D) keel of the sternum
The large flight (pectoral) muscles of birds are attached to the keel (carina), a vertical midventral ridge of the sternum, which provides a broad surface for muscle attachment. The clavicle, scapula and coracoid are part of the pectoral girdle but are not the main flight-muscle attachment.
- Q164Which plasma protein is directly involved in blood clotting?
- Haemoglobin
- Albumin
- Globulin
- Fibrinogen
Answer: (D) Fibrinogen
Fibrinogen is converted to insoluble fibrin during clotting. Albumin maintains osmotic balance, globulins act in immunity, and haemoglobin is inside RBCs, not a plasma protein.
- Q165Which one of the following scientists’ name is correctly matched with the theory put forth by him?
- Mendel - Theory of pangenesis
- Pasteur - Inheritance of acquired characters
- Weismann - Theory of continuity of germplasm
- de Vries - Natural selection
Answer: (C) Weismann - Theory of continuity of germplasm
August Weismann proposed the Theory of continuity of germplasm. The other matches are wrong: inheritance of acquired characters is Lamarck’s; natural selection is Darwin’s; pangenesis is Darwin’s; Mendel gave the laws of inheritance.
- Q166During muscle contraction, which of the following remains UNCHANGED in length?
- H-zone
- I-band
- Sarcomere
- A-band
Answer: (D) A-band
As thin filaments slide inward, the I-band and H-zone shorten and the whole sarcomere shortens, but the A-band (length of the thick filaments) stays constant.
- Q167The maximum amount of electrolytes and water (70-80%) from the glomerular filtrate is reabsorbed in which part of the nephron?
- Distal convoluted tubule
- Proximal convoluted tubule
- Ascending limb of loop of Henle
- Descending limb of loop of Henle
Answer: (B) Proximal convoluted tubule
The proximal convoluted tubule reabsorbs nearly all glucose and amino acids and about 70-80% of the electrolytes and water, plus most of the urea and HCO₃⁻. It is the principal site of bulk reabsorption. Hence the PCT.
- Q168In the mammalian eye, the ‘fovea’ is the centre of the visual field, where:
- The optic nerve leaves the eye
- Only rods are present
- High density of cones occur, but has no rods
- More rods than cones are found
Answer: (C) High density of cones occur, but has no rods
The fovea (in the macula lutea) is a thinned pit packed densely with cones and lacking rods, giving the sharpest visual acuity. The point where the optic nerve leaves the eye is the blind spot, which lacks photoreceptors.
- Q169Who discovered the malarial parasite Plasmodium in the red blood cells of human beings?
- Ronald Ross
- Charles Laveran
- Stephen Hales
- Gregor Mendel
Answer: (B) Charles Laveran
In 1880 Charles Laveran discovered Plasmodium in human RBCs. Ronald Ross later (1897) discovered the parasite’s oocysts in the mosquito stomach and established mosquito transmission. Mendel (genetics) and Hales (plant physiology) are unrelated.
- Q170Colostrum, the first milk secreted after childbirth, is important because it is rich in:
- Antibodies (IgA) giving passive immunity
- Iron and calcium salts
- Lactose only
- Fat for energy
Answer: (A) Antibodies (IgA) giving passive immunity
Colostrum is rich in antibodies (notably IgA), providing the newborn with passive immunity in the first few days of life.
- Q171Which vector can clone only a small fragment of DNA?
- Cosmid
- Bacterial artificial chromosome
- Plasmid
- Yeast artificial chromosome
Answer: (C) Plasmid
A plasmid is small (~10 kbp capacity) and can replicate freely within the cell, so it can clone only small DNA fragments. BACs, YACs and cosmids accommodate much larger inserts (tens to thousands of kbp).
- Q172In the nematode-resistant tobacco, the dsRNA is produced in the host because the introduced gene generates:
- Only sense RNA
- Both sense and antisense (complementary) RNA strands
- Only antisense RNA
- A protein that cleaves nematode DNA
Answer: (B) Both sense and antisense (complementary) RNA strands
Genes introduced via Agrobacterium produce both sense and antisense RNA in the host cells. Being complementary, they pair to form dsRNA, which initiates RNAi against the nematode transcript.
- Q173Production of a human protein in bacteria by genetic engineering is possible because:
- Bacterial cells can carry out the RNA splicing reactions
- The mechanism of gene regulation is identical in humans and bacteria
- The human chromosome can replicate in a bacterial cell
- The genetic code is universal
Answer: (D) The genetic code is universal
Because the genetic code is universal – the same codons specify the same amino acids in bacteria and humans – a human gene is translated correctly in a bacterium. Bacteria lack RNA-splicing machinery, cannot replicate human chromosomes as such, and their gene regulation is not identical to ours.
- Q174During cleavage of the zygote, which of the following is TRUE?
- Cells grow larger after each division
- The zona pellucida is shed at the 2-cell stage
- Meiosis occurs at each division
- Cell number increases but overall size does not increase
Answer: (D) Cell number increases but overall size does not increase
Cleavage is a series of rapid mitotic divisions inside the zona pellucida; cell number rises (2,4,8,16) but the total mass stays roughly the same, so individual cells get smaller.
- Q175The molecule that actually triggers the RNA interference process is:
- Single-stranded sense mRNA
- Transfer RNA
- Ribosomal RNA
- Double-stranded RNA (dsRNA)
Answer: (D) Double-stranded RNA (dsRNA)
RNAi is triggered by double-stranded RNA (dsRNA), which is complementary to and silences the target mRNA, preventing its translation.
- Q176One turn of the helix in a B-form DNA is approximately:
- 0.34 nm
- 2 nm
- 3.4 nm
- 20 nm
Answer: (C) 3.4 nm
In B-DNA the rise between adjacent base pairs is 0.34 nm and there are 10 base pairs per turn, so one complete turn (pitch) is 10 x 0.34 = 3.4 nm. (2 nm is the diameter of the helix, not the pitch.)
- Q177Which sequence is a true palindrome recognised by restriction enzymes (reads same 5’→3′ on both strands)?
- 5′-GATATC-3′ / 3′-GATATC-5′
- 5′-AACGTT-3′ / 3′-AACGTT-5′
- 5′-GAATTC-3′ / 3′-CTTAAG-5′
- 5′-GGGCCC-3′ / 3′-GGGCCC-5′
Answer: (C) 5′-GAATTC-3′ / 3′-CTTAAG-5′
A DNA palindrome reads the same 5’→3′ on both strands. For 5′-GAATTC-3′, the complementary strand read 5’→3′ is also GAATTC (the EcoRI site). The other listed pairings are not valid complementary base-paired duplexes.
- Q178One of the major difficulties in the biological control of insect pests is that the:
- method is less effective compared with the use of insecticides
- predator does not always survive when transferred to a new environment
- predator develops a preference to other diets and may itself become a pest
- practical difficulty of introducing the predator to specific areas
Answer: (C) predator develops a preference to other diets and may itself become a pest
A major drawback of biological control is that the introduced predator may shift to other diets (non-target species) and itself become a pest, disrupting the ecosystem. This is the standard cited difficulty of using living biocontrol agents.
- Q179The wings of a butterfly and the wings of a bird are an example of:
- Vestigial organs
- Analogous organs showing convergent evolution
- Homologous organs showing divergent evolution
- Homologous organs showing convergent evolution
Answer: (B) Analogous organs showing convergent evolution
Butterfly and bird wings differ in structure and origin but perform the same function (flight). Such analogous organs result from convergent evolution, where unrelated groups evolve similar features under similar selection pressures.
- Q180Insulin is a/an:
- Enzyme
- Hormone
- Vitamin
- Lipid
Answer: (B) Hormone
Insulin is a peptide hormone (the earliest-known hormone), secreted by the β-cells of the islets of Langerhans; it is hypoglycemic and antidiabetic. It is not a vitamin, lipid or enzyme.