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Free mock 2 of 3, on the standard NEET blueprint: Physics, Chemistry, Botany and Zoology, 45 questions each, weighted chapter-wise like recent NEET papers. Includes Biomolecules, Principles of Inheritance and Variation, Current Electricity, Equilibrium, Molecular Basis of Inheritance, Biotechnology: Principles and Processes, Animal Kingdom, System of Particles and Rotational Motion.
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Physics
- Q1A torque of 20 N·m is applied to a wheel of moment of inertia 4 kg·m². The angular acceleration produced is:
- 2 rad/s²
- 80 rad/s²
- 8 rad/s²
- 5 rad/s²
Answer: (D) 5 rad/s²
Using τ = Iα → α = τ/I = 20/4 = 5 rad/s².
- Q2The intensity at the maximum in a Young’s double-slit experiment is I₀. When one of the two slits is covered, the intensity at a point on the screen where the path difference is λ/4 from that slit’s contribution becomes (treat the geometry so that the path difference there is λ/4):
- I₀/2
- 3I₀/4
- I₀/4
- I₀
Answer: (A) I₀/2
A path difference of λ/4 corresponds to a phase difference φ = (2π)/(λ)×(λ)/(4) = (π)/(2). Using I = I₀cos²(φ/2) = I₀cos²(π/4) = I₀×(1)/(2) = (I₀)/(2).
- Q3An object of mass m is suspended at the end of a massless wire of length L and area of cross-section A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is:
- f = (1)/(2π)√((YL)/(mA))
- f = (1)/(2π)√((mL)/(YA))
- f = (1)/(2π)√((YA)/(mL))
- f = (1)/(2π)√((mA)/(YL))
Answer: (C) f = (1)/(2π)√((YA)/(mL))
A. Tests: reading a stretched wire as a spring of stiffness k = (YA)/(L) and feeding that into simple harmonic motion. Why A: for an extra extension x the restoring force is F = (YA)/(L)x, so the wire is a spring with k = (YA)/(L). For a mass on a spring, f = (1)/(2π)√((k)/(m)) = (1)/(2π)√((YA)/(mL)). Why not B: this is (1)/(2π)√((m)/(k)), which is proportional to the time period, not to the frequency [misconception: “frequency and period expressions are interchangeable”]. Why not C: A and L are swapped, but stiffness rises with area and falls with length, so A must be upstairs and L downstairs [misconception: “a longer wire is stiffer”]. Why not D: this both swaps A with L and inverts (k)/(m), so it fails on the same two counts at once [misconception: “any arrangement of the symbols under the root is acceptable”]. Remember: a wire is just a spring with k = (YA)/(L), so f = (1)/(2π)√((YA)/(mL)).
- Q4Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen. Reason R: Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell. In the light of the above statements, choose the correct answer from the options given below.
- A is true but R is false
- Both A and R are true but R is NOT the correct explanation of A
- A is false but R is true
- Both A and R are true and R is the correct explanation of A
Answer: (D) Both A and R are true and R is the correct explanation of A
B. Tests: linking path independent zero work inside a shell to the constancy of the potential there. Why B: a Gaussian sphere drawn inside the shell encloses no charge, so E = 0 everywhere inside. Since E = -(dV)/(dr), V cannot vary inside, and matching at the surface fixes it at V = (kQ)/(R), the surface value. R is therefore true. The work done in moving a test charge is W = q(Vᵢ – V_f), and with Vᵢ = V_f for any two interior points, W = 0 on every path. So A is true, and R is precisely the reason A holds. Why not A: this option declares R false, but the constant interior potential equal to the surface value is the standard shell result that follows from E = 0 inside; R is a true statement. Why not C: R is not an unrelated true fact; equal potential at the two endpoints is exactly what makes W = qΔ V vanish, so R does explain A rather than merely accompanying it. Why not D: this declares A false, yet zero work on every path follows immediately from the constant potential, and the electrostatic force is conservative so no path can smuggle in extra work. Remember: inside a charged shell the field is zero but the potential is not; it is frozen at its surface value.
- Q5The length of the string of a musical instrument is 90 cm and it has a fundamental frequency of 120 Hz. Where should it be pressed to produce a fundamental frequency of 180 Hz?
- 60 cm
- 45 cm
- 75 cm
- 80 cm
Answer: (A) 60 cm
For a fixed string under constant tension, f∝(1)/(l). So l₂=l₁(f₁)/(f₂)=90×(120)/(180)=60 cm.
- Q6Which condition is essential for nuclear fusion to occur?
- Presence of slow neutrons
- A moderator like heavy water
- Low temperature and high pressure only
- Extremely high temperature
Answer: (D) Extremely high temperature
Fusion needs very high temperature (~10⁷ K) to give nuclei enough kinetic energy to overcome Coulomb repulsion and get close enough to fuse.
- Q7A parallel-plate capacitor of capacitance 40 μ F is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K = 2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are
- 2 mC and 0.2 J
- 4 mC and 0.2 J
- 8 mC and 2.0 J
- 2 mC and 0.4 J
Answer: (B) 4 mC and 0.2 J
B. Tests: noticing the supply stays connected, so V is held fixed and both Q and U scale with K. Why B: initially Q₀ = C V = (40×10⁻⁶)(100) = 4×10⁻³ C = 4 mC, and U₀ = (1)/(2)CV² = (1)/(2)(40×10⁻⁶)(100)² = 0.2 J. With K = 2 the capacitance becomes 80 μ F while the battery pins V at 100 V, so Q = 8 mC and U = (1)/(2)(80×10⁻⁶)(10⁴) = 0.4 J. The extra charge is 8 – 4 = 4 mC and the change in energy is 0.4 – 0.2 = 0.2 J. Why not A: 8 mC is the final charge, not the extra charge drawn from the supply, so it double counts the original 4 mC; and 2.0 J is five times the final energy of 0.4 J and ten times the correct change of 0.2 J, a value no substitution of C, V or K produces. Why not C: 2 mC needs C = 20 μ F, i.e. the capacitance halved rather than doubled, which is what you get by dividing by K instead of multiplying [misconception C: ‘a dielectric reduces capacitance’]. Why not D: 2 mC is wrong for the same halving reason, and 0.4 J is the final energy rather than the change; the question asks for the increase, U – U₀ [misconception D: ‘the final value is the change’]. Remember: battery connected means V is fixed, so C, Q and U each go up by the factor K.
- Q8A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2°C. The work done by the gas is: Given, R = 8.3 J K⁻¹ mol⁻¹.
- 72.9 J
- 48 J
- 24.9 J
- 23.1 J
Answer: (D) 23.1 J
A. Tests: the first law of thermodynamics combined with C_v = (3)/(2)R for a monoatomic gas. Why A: Helium is monoatomic, so Δ U = nC_vΔ T = 1×(3)/(2)× 8.3× 2 = 24.9 J (a rise of 2°C is a rise of 2 K). The first law then gives W = Q – Δ U = 48 – 24.9 = 23.1 J. Why not B: 48 J is the entire heat input, which would all become work only if Δ U were zero, that is at constant temperature [misconception B: “heat supplied all turns into work”]. Why not C: 24.9 J is Δ U itself, the share of the heat stored inside the gas rather than the share done as work. Why not D: 72.9 J is 48 + 24.9, adding the internal energy change instead of subtracting it. Remember: Q = Δ U + W, and for helium Δ U = (3)/(2)nRΔ T every single time.
- Q9The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without a filter is 10 V. The DC component of the output voltage is:
- (10)/(√(2)) V
- (20)/(π) V
- 10 V
- (10)/(π) V
Answer: (D) (10)/(π) V
For a half-wave rectifier the DC (average) component is V_(dc) = Vₚₑₐₖ/π = 10/π V. (A full-wave rectifier would give 2Vₚₑₐₖ/π = 20/π V.)
- Q10If x = 5sin(π t + (π)/(3)) m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
- 5 m, 1 s
- 5 cm, 2 s
- 5 m, 2 s
- 5 cm, 1 s
Answer: (C) 5 m, 2 s
B. Tests: reading amplitude and period straight off a written SHM equation, units included. Why B: compare with the standard form x = Asin(ω t + φ). The coefficient in front of the sine is the amplitude, A = 5, and the equation is stated in metres, so it is 5 m. The coefficient of t is the angular frequency, ω = π rad/s, so T = (2π)/(ω) = (2π)/(π) = 2 s. Why not A: the period is right, but the equation gives the displacement in metres, so 5 means 5 m and not 5 cm [misconception A: “SHM amplitudes are quoted in centimetres by default”]. Why not C: both halves fail: the wrong unit on the amplitude, and a period of 1 s would require ω = 2π rad/s, whereas the coefficient of t here is only π. Why not D: the amplitude is right, but T = 1 s again needs ω = 2π; the (π)/(3) inside the bracket is a phase constant and has no effect on the period. Remember: the period is 2π divided by the coefficient of t, and the phase constant never changes it.
- Q11An object of mass 3 kg is at rest. If a force F = (6t²î + 4tĵ) N is applied on the object, then the velocity of the object at t = 3 s is
- 18î + 4ĵ
- 18î + 3ĵ
- 18î + 6ĵ
- 3î + 18ĵ
Answer: (C) 18î + 6ĵ
v = (1/3)∫₀³(6t²î + 4tĵ)dt = (1/3)[2t³î + 2t²ĵ]₀³ = (1/3)[54î + 18ĵ] = 18î + 6ĵ.
- Q12A ray is incident at an angle of incidence i on one surface of a small angle prism (with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is μ, then the angle of incidence is nearly equal to:
- (A)/(2μ)
- (2A)/(μ)
- (μ A)/(2)
- μ A
Answer: (D) μ A
B. Tests: prism geometry with normal emergence plus the small-angle form of Snell’s law. Why B: Normal emergence means the refraction angle at the second face is r₂=0. Since r₁+r₂=A, we get r₁=A. Snell’s law at the first face: sin i=μsin r₁; for a small prism the angles are small, so i≈μ r₁=μ A. Why not A: (2A)/(μ) divides by μ where entering a denser medium demands multiplying, and the 2 has no source here [misconception A: “dividing by mu when light enters glass”] Why not C: (μ A)/(2) halves r₁ as if the prism angle splits equally between the faces; that is the minimum-deviation case, but normal emergence loads the whole A onto the first face [misconception C: “r1 = A/2 in every prism problem”] Why not D: (A)/(2μ) makes both errors together [misconception D: “applying Snell’s law upside down and halving”] Remember: normal emergence forces r₂=0, r₁=A, so i=μ A for a thin prism.
- Q13In a particular system, the unit of length, mass and time are chosen to be 10 cm, 10 g and 0.1 s respectively. The unit of force in this system will be equivalent to
- 0.1 N
- 100 N
- 10 N
- 1 N
Answer: (A) 0.1 N
Force F = [MLT⁻²] = (10 g)(10 cm)(0.1 s)⁻² = 10⁻¹ N = 0.1 N.
- Q14Two metal spheres, one of radius R and the other of radius 2R, both have the same surface charge density σ. They are brought into contact and then separated. The new surface charge densities on them are:
- σ₁=(5)/(2)σ, σ₂=(5)/(3)σ
- σ₁=(5)/(3)σ, σ₂=(5)/(6)σ
- σ₁=(5)/(2)σ, σ₂=(5)/(6)σ
- σ₁=(5)/(6)σ, σ₂=(5)/(2)σ
Answer: (B) σ₁=(5)/(3)σ, σ₂=(5)/(6)σ
Initial charges: Q₁=σ 4π R², Q₂=σ 4π(2R)²=16π R²σ, total =20π R²σ. On contact the spheres reach a common potential, so charges share as Q₁’:Q₂’=R:2R=1:2. Then Q₁’=(1)/(3)(20π R²σ), giving σ₁’=(Q₁’)/(4π R²)=(5)/(3)σ; and Q₂’=(2)/(3)(20π R²σ), giving σ₂’=(Q₂’)/(4π(2R)²)=(5)/(6)σ.
- Q15A convex lens made of glass behaves as a diverging lens when placed in a medium that is:
- denser (higher n) than the glass
- of the same n as glass
- vacuum
- less dense (lower n) than the glass
Answer: (A) denser (higher n) than the glass
If the surrounding medium has a higher index than the lens, (nₗₑₙₛ/n_medium – 1) becomes negative, reversing the sign of f, so a convex shape acts as a diverging lens.
- Q16If the magnification produced by a mirror is +0.5, the image is:
- virtual and inverted
- real and inverted
- real and magnified
- virtual and erect, diminished
Answer: (D) virtual and erect, diminished
Positive magnification means erect/virtual; magnitude 0.5 < 1 means diminished. This corresponds to a convex mirror image.
- Q17The ratio of de-Broglie wavelength of an α-particle and proton accelerated from rest by the same potential is (1)/(√(m)). The value of m is
- 8
- 2
- 5
- 4
Answer: (A) 8
B. Tests: applying λ = h/√(2mqV) when the two particles differ in both mass and charge. Why B: λ = h/√(2mqV), so at the same V, (λ_α)/(λₚ) = √((mₚ qₚ)/(m_α q_α)). With m_α = 4mₚ and q_α = 2e against qₚ = e, this is √((mₚ e)/(4mₚ · 2e)) = √((1)/(8)) = (1)/(√(8)), so m = 8. Why not A: m = 2 uses only the charge factor (q_α = 2e) and ignores that the alpha particle is also 4 times heavier [misconception A: “charge alone sets the ratio”]. Why not C: m = 4 uses only the mass factor and forgets the doubled charge, which is the electron-versus-proton habit carried over wrongly [misconception C: “both particles carry the same charge”]. Why not D: m = 5 would need m_α q_α/(mₚ qₚ) = 5, but the mass factor 4 and the charge factor 2 give 8 multiplied and 6 added, so no consistent substitution reaches 5 [misconception D: “guessing a number between the mass factor and the full product instead of computing it”]. Remember: at the same accelerating potential λ ∝ 1/√(mq), so the alpha particle loses a factor √(4 × 2) = √(8) against a proton.
- Q18The amplitude of the charge oscillating in a circuit decreases exponentially as Q = Q₀ e^(-Rt/2L), where Q₀ is the charge at t = 0 s. The time at which charge amplitude decreases to 0.50 Q₀ is nearly: [Given that R = 1.5 Ω, L = 12 mH, ln(2) = 0.693]
- 11.09 s
- 19.01 s
- 19.01 ms
- 11.09 ms
Answer: (D) 11.09 ms
B. Tests: solving an exponential decay for the half-amplitude time in a damped LC circuit. Why B: 0.5 Q₀ = Q₀ e^(-Rt/2L) gives e^(-Rt/2L) = 0.5, so Rt/2L = ln 2 and t = 2L ln 2/R = (2 × 12 × 10⁻³ × 0.693)/1.5 = 0.016632/1.5 = 0.01109 s, that is 11.09 ms. Why not A: 19.01 ms would need 2L ln 2/R = 0.01901 s; no consistent substitution of 12 mH, 1.5 Ω and ln 2 = 0.693 produces that number. Why not C: 19.01 s carries the same unsupported number and is inflated by a further factor of 10³ in the unit. Why not D: 11.09 s is the right number with L treated as 12 H instead of 12 mH, a pure unit slip of 10³ [misconception: “millihenry can be dropped into the formula as henry”]. Remember: half-amplitude time is t = 2L ln 2/R; convert mH to 10⁻³ H and the answer lands in milliseconds.
- Q19A metal has conductivity σ = 6×10⁷ S/m and free-electron density n = 8.5×10²⁸ m⁻³. The electron mobility (μ = σ/ne) is approximately:
- 4.4×10⁻¹ m²V⁻¹s⁻¹
- 1.4×10⁻² m²V⁻¹s⁻¹
- 7.0×10⁻⁵ m²V⁻¹s⁻¹
- 4.4×10⁻³ m²V⁻¹s⁻¹
Answer: (D) 4.4×10⁻³ m²V⁻¹s⁻¹
μ = σ/(ne) = (6×10⁷)/[(8.5×10²⁸)(1.6×10⁻¹⁹)] = 6×10⁷/(1.36×10¹⁰) ≈ 4.41×10⁻³ m²V⁻¹s⁻¹.
- Q20The gravitational potential energy of a 100 kg satellite in a circular orbit of radius r is -2× 10⁹ J. Its kinetic energy and total energy are respectively:
- 2× 10⁹ J and -4× 10⁹ J
- 1× 10⁹ J and 1× 10⁹ J
- 2× 10⁹ J and 0
- 1× 10⁹ J and -1× 10⁹ J
Answer: (D) 1× 10⁹ J and -1× 10⁹ J
For a circular orbit, KE = -(1)/(2)U = (1)/(2)(2× 10⁹) = 1× 10⁹ J, and total energy E = KE + U = 1× 10⁹ + (-2× 10⁹) = -1× 10⁹ J.
- Q21The working of a laser is based on which branch/principle of physics?
- Bernoulli’s principle
- Classical thermodynamics
- Quantum theory of radiation (stimulated emission)
- Newtonian mechanics
Answer: (C) Quantum theory of radiation (stimulated emission)
A laser works on the quantum theory of radiation, specifically stimulated emission and population inversion – a microscopic-domain principle.
- Q22A rod of length 3 m has a mass per unit length directly proportional to the distance x from one end. The centre of gravity of the rod from that end is at:
- 3 m
- 2 m
- 1.5 m
- 2.5 m
Answer: (B) 2 m
With λ = kx, x_(cg) = (∫₀^L x(kx) dx)/(∫₀^L (kx) dx) = (∫₀^L x² dx)/(∫₀^L x dx) = (L³/3)/(L²/2) = (2L)/(3). For L = 3 m, x_(cg) = 2 m. The COM is pulled toward the denser (far) end.
- Q23A particle moves along a straight line such that its displacement at any time t is given by s = 3t³ + 7t² + 14t + 5. The acceleration of the particle at t = 1 s is
- 18 m/s²
- 29 m/s²
- 32 m/s²
- 24 m/s²
Answer: (C) 32 m/s²
v = 9t² + 14t + 14, a = 18t + 14; at t = 1 s, a = 18 + 14 = 32 m/s².
- Q24Time period of a simple pendulum in a stationary lift is ‘T’. If the lift accelerates with (g)/(6) vertically upwards then the time period will be: (Where g = acceleration due to gravity)
- √((6)/(5))T
- √((5)/(6))T
- √((6)/(7))T
- √((7)/(6))T
Answer: (C) √((6)/(7))T
C. Tests: effective gravity in an accelerating lift and its effect on the pendulum period. Why C: for upward acceleration a, g_(eff) = g + a = g + (g)/(6) = (7g)/(6). Since T ∝ (1)/(√(g_(eff))), (T’)/(T) = √((g)/(g_(eff))) = √((6)/(7)), so T’ = √((6)/(7))T. Why not A: this uses g_(eff) = g – (g)/(6) = (5g)/(6), treating an upward acceleration as weakening gravity; that is the downward-acceleration case [misconception A: “accelerating up makes the bob lighter”]. Why not B: same (5g)/(6) slip, with the ratio inverted on top of it. Why not D: g_(eff) is right but the ratio is flipped; stronger effective gravity must shorten the period, so the factor has to be less than 1. Remember: lift accelerating up adds to g, the bob feels heavier, and the period drops by √(6/7).
- Q25If two vectors are perpendicular to each other, their scalar (dot) product is:
- Zero
- Maximum
- Equal to AB sinθ
- Equal to AB
Answer: (A) Zero
Dot product = AB cosθ. For θ = 90°, cos90° = 0, so the dot product is zero.
- Q26In an experiment with photoelectric effect, the stopping potential,
- is ((1)/(e)) times the maximum kinetic energy of the emitted photoelectrons
- increases with increase in the wavelength of the incident light
- decreases with increase in the intensity of the incident light
- increases with increase in the intensity of the incident light
Answer: (A) is ((1)/(e)) times the maximum kinetic energy of the emitted photoelectrons
A. Tests: the definition of stopping potential as the retarding voltage that just stops the fastest electron. Why A: The stopping potential V₀ is the reverse bias at which even the most energetic photoelectron fails to reach the collector. Its kinetic energy is exactly spent climbing the retarding potential: eV₀ = Kₘₐₓ, so V₀ = (Kₘₐₓ)/(e), that is (1)/(e) times Kₘₐₓ. Why not B: eV₀ = hν – φ₀ contains no intensity term, so a brighter beam raises the saturation photocurrent but leaves V₀ exactly where it was [misconception B: “brighter light gives faster electrons”]. Why not C: the same equation rules out a decrease as well; intensity is simply not a variable in the stopping-potential relation, in either direction. Why not D: eV₀ = (hc)/(λ) – φ₀ falls as λ grows, so a longer wavelength LOWERS the stopping potential and eventually kills emission altogether. Remember: V₀ = (Kₘₐₓ)/(e) = (hν – φ₀)/(e), set by frequency and metal only.
- Q27Monochromatic radiation emitted when an electron in a hydrogen atom jumps from the first excited state to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the material is:
- 4×10¹⁵ Hz
- 5×10¹⁵ Hz
- 2.5×10¹⁵ Hz
- 1.6×10¹⁵ Hz
Answer: (D) 1.6×10¹⁵ Hz
Photon energy = E₂ – E₁ = -3.4 -(-13.6) = 10.2 eV. Work function φ = E – eV₀ = 10.2 – 3.57 = 6.63 eV. ν₀ = φ/h = (6.63×1.6×10⁻¹⁹)/(6.63×10⁻³⁴) = 1.6×10¹⁵ Hz.
- Q28If the voltage across a bulb rated ‘220 V, 100 W’ drops by 2.5% of its rated value, the percentage by which the power decreases is (assume constant resistance):
- 5%
- 2.5%
- 10%
- 20%
Answer: (A) 5%
Since P = V²/R with R constant, (Δ P)/(P) = 2(Δ V)/(V). For a 2.5% drop in V, the power decreases by 2×2.5% = 5%.
- Q29In a meter bridge, with a known resistance of 6 Ω in the right gap, the balance point is found at 40 cm from the left end (where the unknown X is). The value of X is:
- 6 Ω
- 12 Ω
- 4 Ω
- 9 Ω
Answer: (C) 4 Ω
X/R = l/(100−l) with l = 40 cm: X = 6×40/(100−40) = 6×40/60 = 4 Ω.
- Q30A balloon is filled at 27°C and 1 atmospheric pressure by volume 500 m³ helium gas. At -3°C and 0.5 atmospheric pressure, the volume of helium gas will be
- 700 m³
- 500 m³
- 1000 m³
- 900 m³
Answer: (D) 900 m³
C. Tests: the combined gas law (P₁V₁)/(T₁) = (P₂V₂)/(T₂) with a Celsius to kelvin conversion on both readings. Why C: T₁ = 27+273 = 300 K and T₂ = -3+273 = 270 K. Then V₂ = V₁·(P₁)/(P₂)·(T₂)/(T₁) = 500×(1)/(0.5)×(270)/(300) = 500× 2× 0.9 = 900 m³. Why not A: 500 m³ is the starting volume, which would only hold if the balloon could not change size, contradicting the data supplied. Why not B: no consistent substitution of the pressure and temperature ratios gives 700 m³. Why not D: 1000 m³ is 500× 2, the pressure halving applied on its own with the cooling ignored [misconception D: “a 30 degree drop is too small to matter”]. Remember: halving the pressure doubles the volume, and cooling from 300 K to 270 K then trims it by the factor 0.9.
- Q31A box of mass 5 kg is pulled by a cord up along a frictionless plane inclined at 30° with the horizontal. The tension in the cord is 30 N. The acceleration of the box is (take g = 10 m s⁻²):
- 0.1 m s⁻²
- Zero
- 2 m s⁻²
- 1 m s⁻²
Answer: (D) 1 m s⁻²
D. Tests: writing Newton’s second law along an incline with tension and the gravity component. Why D: Along the incline: ma = T – mgsin 30° = 30 – 5 × 10 × (1)/(2) = 30 – 25 = 5 N. So a = (5)/(5) = 1 m s⁻² up the plane. Why not A: 2 m s⁻² needs a gravity component of 20 N, an arithmetic slip in mgsin 30° = 25 N [misconception A: “sloppy evaluation of the mgsinθ term”]. Why not B: zero assumes the 30 N tension exactly balances the gravity component, but mgsin 30° = 25 N, not 30 N [misconception B: “assuming equilibrium whenever a rope pulls a block”]. Why not C: 0.1 comes from dividing the 5 N net force by the weight 50 N instead of the mass 5 kg [misconception C: “using a = F/W instead of a = F/m“]. Remember: Along an incline, ma = T – mgsinθ; always divide net force by mass, never by weight.
- Q32The radiation emitted by the human body lies mainly in the:
- microwave region during sleep only
- ultraviolet region and is hence not visible
- infrared region
- visible region
Answer: (C) infrared region
At body temperature (~310 K), Wien’s law gives a peak wavelength of about 9 mum, well within the infrared region – which is why the body’s thermal radiation is invisible to the eye.
- Q33The specific resistance of a conductor increases with:
- decrease in length
- decrease in cross-sectional area
- increase in cross-sectional area
- increase in temperature
Answer: (D) increase in temperature
Specific resistance (resistivity) is independent of the dimensions (length, area) of the sample. For a metallic conductor it rises with temperature because more frequent lattice collisions reduce the relaxation time τ (ρ = m/ne²τ).
- Q34The ratio of the radii of gyration of a circular disc to that of a circular ring, each of the same mass and radius, about their respective central perpendicular axes is:
- √(2) : √(3)
- √(2) : 1
- 1 : √(2)
- √(3) : √(2)
Answer: (C) 1 : √(2)
k = √(I/M). Disc: I = (1)/(2)MR² ⇒ k_d = R/√(2). Ring: I = MR² ⇒ kᵣ = R. So k_d : kᵣ = (R)/(√(2)) : R = 1 : √(2).
- Q35A straight wire of length 0.5 m carrying a current of 1.2 A is placed in a uniform magnetic field of induction 2 T. The magnetic field is perpendicular to the length of the wire. The force on the wire is:
- 2.0 N
- 2.4 N
- 3.0 N
- 1.2 N
Answer: (D) 1.2 N
F = BILsin90° = 2×1.2×0.5 = 1.2 N.
- Q36An alpha particle of kinetic energy K has a head-on distance of closest approach r₀ to a nucleus. For an alpha particle of kinetic energy 2K, the distance of closest approach to the same nucleus is:
- r₀/4
- 2r₀
- 4r₀
- r₀/2
Answer: (D) r₀/2
At closest approach K = (1)/(4πε₀)(2Ze²)/(r₀), so r₀ ∝ (1)/(K). Doubling K halves the distance: r₀/2.
- Q37A rectangular wire loop of sides 8 cm and 3 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is 2 cm s⁻¹ in a direction normal to the shorter side of the loop, will be:
- 4.8 × 10⁻⁴ volt
- 1.2 × 10⁻⁴ volt
- 1.3 × 10⁻⁴ volt
- 1.8 × 10⁻⁴ volt
Answer: (D) 1.8 × 10⁻⁴ volt
D. Tests: identifying which side of a moving loop generates the motional emf Blv. Why D: Velocity is normal to the shorter side, so the loop moves along its longer side; the conductor that sweeps field lines is the shorter side (3 cm), since it moves perpendicular to its own length. ε = Blv = 0.3 × 0.03 × 0.02 = 1.8 × 10⁻⁴ V. Why not A: 0.3 × 0.08 × 0.02 = 4.8 × 10⁻⁴ V uses the 8 cm side [misconception A: “using the side parallel to the motion as the emf-generating length”] Why not B: comes from an arithmetic slip such as 0.3 × 0.02 × 0.02 [misconception B: “substituting the speed in place of the rod length”] Why not C: a near-miss distractor; no correct substitution of Blv gives 1.3 [misconception C: “rounding/decimal slip in Blv“] Remember: only the side moving perpendicular to its own length cuts field lines; emf = B × (that side) × v.
- Q38A pulse of a wave train travels along a stretched string and reaches the fixed end of the string. It will be reflected back with:
- the same phase as the incident pulse but with velocity reversed
- a phase change of 180° with velocity reversed
- a phase change of 180° with no reversal of velocity
- the same phase as the incident pulse with no reversal of velocity
Answer: (B) a phase change of 180° with velocity reversed
Reflection at a rigid (fixed) boundary introduces a phase change of 180° (π), and the direction of propagation reverses, so the velocity is reversed. (A free end would reflect with no phase change.)
- Q39An electromagnetic wave travelling in the x-direction is described by the field equation E_y = 300 sin ω(t – (x)/(c)) V/m. If an electron is restricted to move in the y-direction only, with a speed of 1.5 × 10⁶ m/s, then the ratio of the maximum electric force to the maximum magnetic force acting on the electron is:
- 400
- 300
- 150
- 200
Answer: (D) 200
A. Tests: comparing electric and magnetic forces exerted by an EM wave on a moving charge. Why A: maximum electric force F_E = eE₀. The wave’s magnetic field (B₀ = E₀/c, along k) acts on the electron moving along y with F_B = e v B₀ (since v ⊥ B). The ratio is (F_E)/(F_B) = (eE₀)/(e v E₀/c) = (c)/(v) = (3 × 10⁸)/(1.5 × 10⁶) = 200. Why not D: 300 is just the field amplitude E₀ mistaken for the answer [misconception D: “the ratio equals the electric field value”]. Why not B: 150 echoes the 1.5 of the electron speed (1.5 × 10⁶) promoted by a factor of 100; c/v with these numbers is exactly 200 [misconception B: “recycling a given digit string as the answer”]. Why not C: 400 doubles the result, an arithmetic slip in 3 × 10⁸ / 1.5 × 10⁶ [misconception C: “3/1.5 = 4“]. Remember: for a charge in an EM wave, (F_E)/(F_B) = (c)/(v); magnetic force only rivals electric force at relativistic speeds.
- Q40A Zener diode is used for:
- stabilisation (voltage regulation)
- rectification
- amplification
- producing oscillations in an oscillator
Answer: (A) stabilisation (voltage regulation)
A Zener diode is a silicon diode designed to operate in reverse breakdown, where its voltage stays nearly constant. This makes it ideal for voltage stabilisation/regulation.
- Q41The torque on a magnetic dipole of moment m in a uniform field B is maximum when the angle between m and B is:
- 90°
- 180°
- 0°
- 45°
Answer: (A) 90°
τ = mB sinθ is maximum when sinθ = 1, i.e. θ = 90°. At 0° and 180° the torque is zero.
- Q42A body tied to a string is whirled in a vertical circle of radius L. The minimum speed it must have at the lowest point to just complete the circle is:
- √(gL)
- √(3gL)
- √(5gL)
- √(2gL)
Answer: (C) √(5gL)
At the top the minimum speed is √(gL). Using energy conservation between bottom and top (height difference 2L): ½v_b² = ½vₜ² + g(2L). With vₜ² = gL, v_b² = gL + 4gL = 5gL, so v_b = √(5gL).
- Q43A coil of 800 turns and effective area 0.05 m² is kept perpendicular to a magnetic field 5×10⁻⁵ T. When the plane of the coil is rotated by 90° about any of its coplanar axes in 0.1 s, the EMF induced in the coil is:
- 0.2 V
- 2 V
- 0.02 V
- 2×10⁻³ V
Answer: (C) 0.02 V
ε = N(ΔΦ)/(Δ t) = N B A(cos0° – cos90°)/(Δ t) = (800×5×10⁻⁵×0.05×(1-0))/(0.1) = (2×10⁻³)/(0.1) = 0.02 V.
- Q44A solenoid is 0.4 m long with 800 turns and carries 2 A. The field inside is: (μ₀ = 4π×10⁻⁷)
- 2π×10⁻³ T
- 8π×10⁻⁴ T
- 4π×10⁻³ T
- 1.6π×10⁻³ T
Answer: (D) 1.6π×10⁻³ T
n = 800/0.4 = 2000 turns per metre. B = μ₀nI = (4π×10⁻⁷)(2000)(2) = 16π×10⁻⁴ T = 1.6π×10⁻³ T.
- Q45A square surface of side L metre in the plane of the paper is placed in a uniform electric field E (V/m) acting along the same plane at an angle θ with the horizontal side of the square. The electric flux linked to the surface is:
- EL²cosθ
- EL²
- 0
- EL²sinθ
Answer: (C) 0
The field lies in the plane of the square, so it is parallel to the surface and perpendicular to the area vector. Flux =EcdotA=EAcos90°=0, regardless of θ.
Chemistry
- Q46Decomposition of A is a first order reaction at T(K) and is given by A(g) arrow B(g) + C(g). In a closed 1 L vessel, 1 bar A(g) is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in min⁻¹) of the reaction? (log 2 = 0.3)
- 6.9 × 10⁻¹
- 6.9 × 10⁻⁴
- 6.9 × 10⁻²
- 6.9 × 10⁻³
Answer: (D) 6.9 × 10⁻³
B. Tests: using total pressure to follow a gas phase first order decomposition. Why B: Let x bar of A decompose in 100 minutes. Total pressure = (1-x) + x + x = 1 + x = 1.5, so x = 0.5 bar and the pressure of A left is p_A = 1 – 0.5 = 0.5 bar. Then k = (2.303)/(t)log(p₀)/(p_A) = (2.303)/(100) × log 2 = (2.303 × 0.3)/(100) = 6.9 × 10⁻³ min⁻¹. Why not A: (2.303 × 0.3)/(1000) = 6.9 × 10⁻⁴ min⁻¹, which is the answer for t = 1000 minutes [misconception A: “the power of ten can be fixed at the end by eye”]. Why not C: (2.303 × 0.3)/(1) = 6.9 × 10⁻¹ min⁻¹, the answer for t = 1 minute. Why not D: (2.303 × 0.3)/(10) = 6.9 × 10⁻² min⁻¹, the answer for t = 10 minutes. Remember: for A arrow B + C started from pure A, the reactant left is p_A = 2p₀ – pₜₒₜₐₗ.
- Q47Benzaldehyde CANNOT be prepared by which of the following reactions?
- Benzoyl chloride + H₂ over Pd-BaSO₄
- Benzene + CO + HCl with anhydrous AlCl₃/CuCl
- Benzoic acid + Zn/Hg and conc. HCl
- Toluene + CrO₂Cl₂ (then H₃O⁺)
Answer: (C) Benzoic acid + Zn/Hg and conc. HCl
(A) is the Etard reaction, (B) Rosenmund, (C) Gattermann-Koch – all give benzaldehyde. Benzoic acid + Zn/Hg/HCl (Clemmensen-type conditions) would tend to reduce past the aldehyde; carboxylic acids are not converted to aldehydes this way, so benzaldehyde cannot be obtained from option D.
- Q48Which of the following gives a white precipitate with ammoniacal silver nitrate?
- But-2-ene
- But-1-yne
- But-2-yne
- n-Butane
Answer: (B) But-1-yne
Only terminal alkynes (with ≡C-H) form metal acetylides. But-1-yne is terminal and gives a white silver acetylide precipitate; but-2-yne is internal and does not react.
- Q49Which from following compounds is classified as oligosaccharide?
- Ribose
- Sucrose
- Fructose
- Glucose
Answer: (B) Sucrose
B. Tests: classifying a sugar by how many monosaccharide units it releases on hydrolysis. Why B: an oligosaccharide gives two to ten monosaccharide units on hydrolysis. Sucrose hydrolyses into one glucose and one fructose, so it is a disaccharide and therefore falls inside the oligosaccharide class. Why not A: ribose is an aldopentose, a single sugar unit that cannot be hydrolysed further, so it is a monosaccharide. Why not C: fructose is a ketohexose, again a single unhydrolysable unit, a monosaccharide. Why not D: glucose is an aldohexose and is the standard example of a monosaccharide, not an oligosaccharide [misconception D: “the most common sugar must be the compound sugar”]. Remember: count the units released on hydrolysis, one is mono, two to ten is oligo, many is poly.
- Q50The frequency of radiation emitted when the electron falls from n=4 to n=1 in a hydrogen atom is (ionisation energy of H = 2.18×10⁻¹⁸ J atom⁻¹, h = 6.625×10⁻³⁴ J s):
- 2.00×10¹⁵ s⁻¹
- 1.03×10¹⁵ s⁻¹
- 3.08×10¹⁵ s⁻¹
- 1.54×10¹⁵ s⁻¹
Answer: (C) 3.08×10¹⁵ s⁻¹
Δ E = 2.18×10⁻¹⁸((1)/(1²)-(1)/(4²)) = 2.18×10⁻¹⁸×(15)/(16) = 2.044×10⁻¹⁸ J. Then ν = (Δ E)/(h) = (2.044×10⁻¹⁸)/(6.625×10⁻³⁴) = 3.08×10¹⁵ s⁻¹.
- Q51The general molecular formula, which represents the homologous series of alkanols is
- CₙH₂ₙO₂
- CₙH₂ₙ₊₁O
- CₙH₂ₙ₊₂O
- CₙH₂ₙO
Answer: (C) CₙH₂ₙ₊₂O
A. Tests: deriving a homologous series formula from the parent alkane formula. Why A: an alkanol is an alkane with one H replaced by OH. The alkane is CₙH₂ₙ₊₂, so replacing one H by OH gives CₙH₂ₙ₊₁OH, which collected as a molecular formula is CₙH₂ₙ₊₂O. Check with ethanol, n = 2: C₂H₆O, and with methanol, n = 1: CH₄O. Why not B: CₙH₂ₙO₂ carries two oxygens and one degree of unsaturation, which is the carboxylic acid series, for example ethanoic acid C₂H₄O₂. Why not C: CₙH₂ₙO is two hydrogens short of saturation, so it must contain a ring or a C=O; that is the aldehyde and ketone series, for example propanone C₃H₆O. Why not D: CₙH₂ₙ₊₁O has an odd total valence count and cannot be a neutral closed-shell molecule; it is the alkoxy fragment, not a compound [misconception D: “take the alkyl part CₙH₂ₙ₊₁ and just bolt an O on”]. Remember: alcohol = alkane plus one oxygen, so CₙH₂ₙ₊₂O; two hydrogens missing means a C=O, an extra oxygen with no extra hydrogen means an acid.
- Q52For the reversible reaction N₂(g) + 3H₂(g) leftharpoons 2NH₃(g) + heat, the equilibrium shifts in the forward direction by:
- increasing the concentration of NH₃
- decreasing the concentrations of N₂ and H₂
- decreasing the pressure
- increasing pressure and decreasing temperature
Answer: (D) increasing pressure and decreasing temperature
Forward goes from 4 gas moles to 2 and releases heat (exothermic). Increasing pressure favours fewer moles (forward); decreasing temperature favours the exothermic (forward) direction. So both together shift it forward.
- Q53Which from the following amines when heated with ethanolic KOH and chloroform does NOT produce foul smell?
- Propan-2-amine
- N,N-Dimethyl aniline
- Phenylmethanamine
- Benzenamine
Answer: (B) N,N-Dimethyl aniline
B. Tests: using the carbylamine test as a diagnostic that fires only for primary amines. Why B: chloroform with ethanolic KOH generates dichlorocarbene, CCl₂. A primary amine attacks that carbene with its lone pair and then loses two molecules of HCl to build the isocyanide R-NC, whose appalling smell is the positive test. Building the isocyanide consumes two N-H bonds. In N,N-dimethylaniline both hydrogens on nitrogen have been replaced by methyl groups, making it a tertiary amine with zero N-H, so the sequence stops at the first step and no foul smelling isocyanide is formed. Why not A: benzenamine is aniline, a primary aromatic amine with two N-H bonds, and it gives phenyl isocyanide, the classic positive carbylamine result [misconception A: “aromatic amines do not respond to the carbylamine test”]. Why not C: propan-2-amine is (CH₃)₂CH-NH₂, primary at nitrogen even though the carbon is secondary, and the test counts hydrogens on nitrogen, so it does give isopropyl isocyanide. Why not D: phenylmethanamine is benzylamine, C₆H₅CH₂NH₂, still a primary amine with two N-H bonds, so it gives benzyl isocyanide and the same stench. Remember: the carbylamine test is a 1° amine test, aromatic or aliphatic alike, and it needs two N-H bonds, so 2° and 3° amines stay silent.
- Q54An organic compound has 42.1 % carbon, 6.4 % hydrogen and remainder is oxygen. If its molecular weight is 342, then its molecular formula is :
- C₁₄H₂₀O₁₀
- C₁₂H₂₂O₁₁
- C₁₂H₂₀O₁₂
- C₁₁H₁₈O₁₂
Answer: (B) C₁₂H₂₂O₁₁
B. Tests: getting a molecular formula directly from percentage composition plus the molecular weight. Why B: Oxygen % = 100 – 42.1 – 6.4 = 51.5 %. Apply each percentage to the 342 g mole: carbon = 0.421 x 342 = 144 g, which is 144/12 = 12 atoms; hydrogen = 0.064 x 342 = 21.9 g, which is about 22 atoms; oxygen = 0.515 x 342 = 176 g, which is 176/16 = 11 atoms. So the formula is C₁₂H₂₂O₁₁, and its mass checks out: 144 + 22 + 176 = 342. Why not A: C₁₄H₂₀O₁₀ weighs 168 + 20 + 160 = 348, not 342, and its carbon share is 168/348 = 48.3 %, not 42.1 %. Why not C: C₁₂H₂₀O₁₂ weighs 144 + 20 + 192 = 356, again not 342. Why not D: C₁₁H₁₈O₁₂ does weigh 132 + 18 + 192 = 342, so it passes the mass check, but its carbon share is 132/342 = 38.6 %, not the given 42.1 % [misconception D: “matching the molecular weight is enough”]. Remember: multiply each percentage by the molar mass to read the atom counts straight off, then test BOTH the total mass and the composition.
- Q55A 0.5 M aqueous solution is diluted from 100 mL to 500 mL by adding water. The molarity of the diluted solution is:
- 0.25 M
- 0.5 M
- 2.5 M
- 0.1 M
Answer: (D) 0.1 M
On dilution the moles of solute are conserved: M₁ V₁ = M₂ V₂. So 0.5 × 100 = M₂ × 500 → M₂ = 50/500 = 0.1 M.
- Q56The solubility of a saturated solution of calcium fluoride is 2 × 10⁻⁴ mol/L. Its solubility product is:
- 32 × 10⁻¹²
- 12 × 10⁻²
- 14 × 10⁻⁴
- 22 × 10⁻²
Answer: (A) 32 × 10⁻¹²
CaF₂ leftharpoons Ca²⁺ + 2F⁻, Kₛₚ = (S)(2S)² = 4S³ = 4(2 × 10⁻⁴)³ = 4 × 8 × 10⁻¹² = 32 × 10⁻¹².
- Q57The standard enthalpy of formation of an element in its most stable (reference) state is:
- Always positive
- Equal to its enthalpy of combustion
- Zero
- Always negative
Answer: (C) Zero
By definition, the standard enthalpy of formation of an element in its standard reference state (e.g. O₂ gas, C as graphite) is taken as zero.
- Q58The boiling point of p-nitrophenol is higher than that of o-nitrophenol because:
- the NO₂ group at the para position behaves differently from that at the ortho position
- p-nitrophenol has a higher molecular weight
- there is intermolecular hydrogen bonding in p-nitrophenol
- intramolecular hydrogen bonding exists in p-nitrophenol
Answer: (C) there is intermolecular hydrogen bonding in p-nitrophenol
In p-nitrophenol the -OH and -NO₂ groups are far apart, so they form intermolecular hydrogen bonds that link molecules and raise the boiling point. In o-nitrophenol the groups are adjacent and form an intramolecular (chelate) H-bond, reducing intermolecular association and lowering the boiling point.
- Q59The molar heat capacity C of water at constant pressure is 75 J K⁻¹ mol⁻¹. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of water is:
- 4.8 K
- 1.2 K
- 6.6 K
- 2.4 K
Answer: (D) 2.4 K
Moles of water = 100/18 = 5.55 mol. q = nCΔT ⇒ 1000 = 5.55 × 75 × ΔT ⇒ ΔT = 1000/(416.6) ≈ 2.4 K.
- Q60Assertion (A): Ionic compounds conduct electricity in the molten state. Reason (R): In the molten state, ions become free to move and carry charge.
- 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
- A is false but R is true
Answer: (C) Both A and R are true and R is the correct explanation of A
In the solid lattice ions are fixed; on melting, the lattice breaks and ions move freely, conducting electricity. R correctly explains A.
- Q61At 25°C, the dissociation constant of a base BOH is 1.0 × 10⁻¹². The concentration of hydroxyl ions in a 0.01 M aqueous solution of the base would be:
- 1.0 × 10⁻⁵ mol L⁻¹
- 1.0 × 10⁻⁶ mol L⁻¹
- 1.0 × 10⁻⁷ mol L⁻¹
- 2.0 × 10⁻⁶ mol L⁻¹
Answer: (C) 1.0 × 10⁻⁷ mol L⁻¹
[OH⁻] = √(K_b · C) = √((1.0 × 10⁻¹²)(0.01)) = √(10⁻¹⁴) = 1.0 × 10⁻⁷ mol L⁻¹.
- Q62In Lassaigne’s test, the appearance of a Prussian blue colour confirms the presence of:
- Sulphur
- Phosphorus
- Nitrogen
- A halogen
Answer: (C) Nitrogen
Nitrogen forms sodium cyanide on fusion with Na; with Fe²⁺/Fe³⁺ it gives ferric ferrocyanide, the Prussian blue colour, confirming nitrogen.
- Q63An oxide has oxide ions in CCP with 2/3 of the octahedral voids occupied by metal M. The formula of the oxide is:
- M₂O₃
- MO₂
- MO
- M₃O
Answer: (A) M₂O₃
CCP gives N oxide ions and N octahedral voids. M occupies 2/3 of N = 2N/3. Ratio M:O = (2/3):1 = 2:3 → M₂O₃.
- Q64Using anhydrous AlCl₃ as catalyst, which reaction produces ethylbenzene?
- H₂C=CH-OH + C₆H₆
- H₃C-CH₃ + C₆H₆
- CH₃-CH=CH₂ + C₆H₆
- H₂C=CH₂ + C₆H₆
Answer: (D) H₂C=CH₂ + C₆H₆
Friedel-Crafts alkylation of benzene with ethene (ethylene) over anhydrous AlCl₃ adds a -CH₂CH₃ group to the ring, giving ethylbenzene (C₆H₅CH₂CH₃). Propene would give cumene, and ethane (no pi bond/leaving group) does not react.
- Q65A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is: (Given: Molar mass of Cu = 63 g mol⁻¹; 1 F = 96487 C mol⁻¹)
- 2.4036 g
- 1.7018 g
- 0.2938 g
- 0.5876 g
Answer: (C) 0.2938 g
B. Tests: Faraday’s first law with the correct electron count for a divalent ion. Why B: Q = It = 1.5 × (10 × 60) = 900 C. Moles of electrons = (900)/(96487) = 9.328 × 10⁻³. The cathode reaction Cu²⁺ + 2e⁻ arrow Cu needs 2 electrons per atom, so moles of Cu = (9.328 × 10⁻³)/(2) = 4.664 × 10⁻³ and mass = 4.664 × 10⁻³ × 63 = 0.2938 g. Why not A: 1.7018 g follows from no consistent substitution of 1.5 A, 600 s, 63 g mol⁻¹ and 96487 C into m = (ItM)/(nF). A magnitude check kills it on sight: 900 C is under a hundredth of a Faraday, so the deposit has to be a fraction of a gram [misconception A: “a few hundred coulombs can plate out gram quantities of metal”]. Why not C: 2.4036 g likewise cannot be reached by any valid combination of the given numbers, with or without the factor 2, and it fails the same magnitude check as A [misconception C: “the charge passed sets the answer loosely, so the size of the number does not need checking”]. Why not D: 0.5876 g is exactly twice the answer, from forgetting that Cu²⁺ needs 2 electrons and using n = 1: (900 × 63)/(96487) = 0.5876 [misconception D: “one electron deposits one metal atom”]. Remember: m = (ItM)/(nF), and for Cu²⁺ the n in the denominator is 2, not 1.
- Q66Adding a catalyst to a system at equilibrium:
- Increases the value of K
- Shifts equilibrium forward
- Has no effect on the equilibrium position
- Shifts equilibrium toward fewer moles
Answer: (C) Has no effect on the equilibrium position
A catalyst speeds up forward and backward reactions equally, so equilibrium is reached faster but the position and K are unchanged.
- Q67What is the term used for the minimum kinetic energy required for the reactant molecules to undergo reaction?
- Potential energy
- Activation energy
- Bond energy
- Thermal energy
Answer: (B) Activation energy
D. Tests: the definition of activation energy within collision theory. Why D: collision theory says a collision only leads to reaction if the colliding molecules bring at least a threshold amount of energy, enough to reach the activated complex at the top of the energy barrier. That minimum extra energy, measured above the average energy of the reactants, is exactly what activation energy Eₐ means. Collisions below it simply bounce apart unchanged. Why not A: potential energy is stored energy of position or configuration; it describes where a species sits on the energy diagram, not the threshold a collision has to clear [misconception: “activation energy is just the potential energy of the reactants”]. Why not B: bond energy is the energy needed to break one specified bond homolytically, a property of that bond alone, and a reaction barrier is generally lower than the full bond energy because new bonds form as old ones break [misconception: “the barrier equals the bond dissociation energy”]. Why not C: thermal energy is the average kinetic energy the molecules already possess at a given temperature, which is what they bring to the collision, not the threshold they must exceed [misconception: “RT itself is the activation energy”]. Remember: activation energy is the hill every reacting collision must climb, measured up from the reactant energy level.
- Q68The method by which aniline cannot be prepared is:
- hydrolysis of phenylisocyanide with acidic solution
- reduction of nitrobenzene with H₂/Pd in ethanol
- potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution
- degradation of benzamide with bromine in alkaline solution
Answer: (C) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution
C. Tests: valid versus invalid laboratory routes to aniline. Why C: Option C is the Gabriel route, which requires potassium phthalimide to displace the halide by SN2 attack. Chlorobenzene cannot undergo SN2: the C-Cl bond has partial double-bond character and the ring blocks backside attack, so no N-aryl phthalimide forms and aniline is never obtained. Why not A: Br₂ in alkali on benzamide is Hoffmann bromamide degradation, C₆H₅CONH₂ arrow C₆H₅NH₂, a standard aniline preparation [misconception A: “Hoffmann degradation fails for aromatic amides”]. Why not B: catalytic hydrogenation reduces the -NO₂ group of nitrobenzene cleanly to -NH₂ [misconception B: “H₂/Pd attacks the aromatic ring instead of the nitro group”]. Why not D: acidic hydrolysis of phenyl isocyanide gives aniline and formic acid, reversing the carbylamine reaction [misconception D: “hydrolysis of an isocyanide destroys the C-N bond to the ring”]. Remember: Gabriel synthesis makes aliphatic 1° amines only, because aryl halides refuse SN2.
- Q69The experimental rate data for A + B → 2AB are: ([A],[B],Rate) = (0.50,0.50,1.6×10⁻⁴); (0.50,1.00,3.2×10⁻⁴); (1.00,1.00,3.2×10⁻⁴), with rate in M s⁻¹. The rate equation for the data is:
- rate = k[A][B]
- rate = k[A]²[B]²
- rate = k[B]²
- rate = k[B]
Answer: (D) rate = k[B]
Compare rows 2 and 3 ([B] fixed at 1.00): [A] doubles (0.50→1.00) but rate is unchanged (3.2×10⁻⁴) ⇒ order in A = 0. Compare rows 1 and 2 ([A] fixed): [B] doubles, rate doubles (1.6→3.2×10⁻⁴) ⇒ order in B = 1. So rate = k[A]⁰[B] = k[B].
- Q70A gas expands against a constant external pressure of 2 atm from 1 L to 5 L. The work done by the gas (1 L·atm ≈ 101.3 J) is approximately:
- +810 J
- +405 J
- -405 J
- -810 J
Answer: (D) -810 J
w = -pₑₓₜ ΔV = -(2 atm)(5-1 L) = -8 L·atm = -8 × 101.3 ≈ -810 J. Negative because the gas does work on the surroundings.
- Q71Match the following and identify the correct option. (A) CO(g) + H₂(g) (B) Temporary hardness of water (C) B₂H₆ (D) H₂O₂ (i) Mg(HCO₃)₂ + Ca(HCO₃)₂ (ii) An electron deficient hydride (iii) Synthesis gas (iv) Non-planar structure
- (A)-(iii), (B)-(ii), (C)-(i), (D)-(iv)
- (A)-(iii), (B)-(iv), (C)-(ii), (D)-(i)
- (A)-(iii), (B)-(i), (C)-(ii), (D)-(iv)
- (A)-(i), (B)-(iii), (C)-(ii), (D)-(iv)
Answer: (C) (A)-(iii), (B)-(i), (C)-(ii), (D)-(iv)
D. Tests: matching four unrelated species to their defining descriptions, anchored on the pairs you are surest of. Why D: (A) an equimolar mixture of CO and H₂ is water gas, industrially called synthesis gas or syngas, so (A)-(iii). (B) temporary hardness is caused by dissolved bicarbonates of calcium and magnesium, removable by boiling, so (B)-(i). (C) B₂H₆ has 12 valence electrons for 8 bonds and must use three-centre two-electron bridges, the definition of an electron deficient hydride, so (C)-(ii). (D) H₂O₂ has an open-book shape with the two O-H bonds in different planes and a dihedral angle of about 111 degrees in the gas phase, so it is non-planar, giving (D)-(iv). The set is (A)-(iii), (B)-(i), (C)-(ii), (D)-(iv). Why not A: it sends B₂H₆ to the calcium and magnesium bicarbonates and temporary hardness to the electron deficient hydride, two chemically impossible pairings. Why not B: it calls temporary hardness a non-planar structure and assigns the bicarbonate mixture to H₂O₂. Why not C: it pairs CO + H₂ with the bicarbonates instead of synthesis gas, and then calls temporary hardness synthesis gas. Remember: CO + H₂ is synthesis gas, temporary hardness is Ca and Mg bicarbonates, B₂H₆ is the electron deficient hydride, H₂O₂ is the non-planar open book.
- Q72Identify the incorrect statement among the following:
- Lanthanide contraction is the accumulation of successive shrinkages
- As a result of lanthanide contraction, the properties of the 4d series of transition elements have no similarities with the 5d series
- There is a decrease in the radii of atoms or ions as one proceeds from La to Lu
- The shielding power of 4f electrons is quite weak
Answer: (B) As a result of lanthanide contraction, the properties of the 4d series of transition elements have no similarities with the 5d series
Because of the lanthanide contraction, the 4d and 5d series actually become very SIMILAR in size and properties (e.g., Zr/Hf), so the statement that they have no similarities is incorrect.
- Q73Which of the following orders is WRONG?
- NH₃ < PH₃ < AsH₃ : acidic character
- Al₂O₃ < MgO < Na₂O < K₂O : basic character
- Li < Be < B < C : first ionisation potential
- Li⁺ < Na⁺ < K⁺ < Cs⁺ : ionic radius
Answer: (C) Li < Be < B < C : first ionisation potential
In period 2 the first ionisation potential does not rise monotonically: boron dips below beryllium (Be has a stable filled 2s²). So Li < Be < B < C is wrong; the correct order is Li < B < Be < C. The other three orders are correct.
- Q74In context with beryllium, which one of the following statements is INCORRECT?
- It is rendered passive by nitric acid
- Its salts rarely hydrolyse
- Its hydride is electron-deficient and polymeric
- It forms Be₂C
Answer: (B) Its salts rarely hydrolyse
Beryllium salts are largely covalent (small, highly polarising Be²⁺) and therefore hydrolyse readily in water, so ‘its salts rarely hydrolyse’ is incorrect. Be is passivated by conc. HNO₃, forms Be₂C, and BeH₂ is electron-deficient and polymeric – all true.
- Q75According to VSEPR theory, the shape of the SF₄ molecule is:
- trigonal bipyramidal
- see-saw (distorted tetrahedron)
- square planar
- tetrahedral
Answer: (B) see-saw (distorted tetrahedron)
SF₄ has 4 bond pairs and 1 lone pair (steric number 5). The lone pair occupies an equatorial position of the trigonal bipyramid, giving a see-saw shape.
- Q76The hybridisation of the carbonyl carbon and approximate C=O bond angle region in an aldehyde are:
- sp, 180°
- sp³, 109.5°
- sp², 90°
- sp², 120°
Answer: (D) sp², 120°
The carbonyl carbon forms three sigma bonds and is sp² hybridised, giving a trigonal planar geometry with bond angles close to 120°.
- Q77Crystal field stabilisation energy for a high spin d⁴ octahedral complex is:
- −0.6 Δ₀
- −1.8 Δ₀
- −1.2 Δ₀
- −1.6 Δ₀ + P
Answer: (A) −0.6 Δ₀
High spin d⁴ has configuration t₂g³ eg¹ (no extra pairing). CFSE = [3(−0.4) + 1(+0.6)]Δ₀ = (−1.2 + 0.6)Δ₀ = −0.6 Δ₀.
- Q78Identify correct statements from the following: A. Propanal and propanone are functional isomers. B. Ethoxyethane and methoxypropane are metamers. C. But-2-ene shows optical isomerism. D. But-1-ene and but-2-ene are functional isomers. E. Pentane and 2, 2-dimethyl propane are chain isomers. Choose the correct answer from the options given below:
- A, B and C only
- C, D and E only
- A, B and E only
- B, C and D only
Answer: (C) A, B and E only
A. Tests: applying five different isomerism labels precisely, including the metamer and optical cases. Why A: Statement A is true, propanal CH₃CH₂CHO and propanone CH₃COCH₃ are both C₃H₆O with different functional groups, aldehyde versus ketone. Statement B is true, ethoxyethane C₂H₅-O-C₂H₅ and methoxypropane CH₃-O-C₃H₇ are both C₄H₁₀O ethers that differ only in how the carbons are split on either side of the same functional group, which is the definition of metamerism. Statement C is false, but-2-ene CH₃-CH=CH-CH₃ has no carbon bearing four different groups, so it is not optically active, it shows cis-trans isomerism instead. Statement D is false, but-1-ene and but-2-ene are both alkenes with the same skeleton and differ only in the position of the double bond, making them position isomers. Statement E is true, pentane and 2,2-dimethylpropane are both C₅H₁₂ with different carbon skeletons. So A, B and E only. Why not B: it adds statement C, but optical activity needs a stereocentre and neither sp² carbon of but-2-ene has four different groups. Why not C: it keeps both false statements C and D and drops the two clearly true ones, A and B. Why not D: statement B is right, but C and D are both wrong, and calling but-1-ene and but-2-ene functional isomers ignores that they belong to the same alkene family [misconception: “moving a double bond changes the functional group”]. Remember: metamers share one functional group with the carbons split differently around it, and a moved double bond is position isomerism, never functional.
- Q79Pure water can be obtained from sea water by:
- plasmolysis
- reverse osmosis
- centrifugation
- sedimentation
Answer: (B) reverse osmosis
Applying a pressure greater than the osmotic pressure of sea water to its surface forces pure solvent (water) out through a semipermeable membrane against the natural osmotic flow. This is reverse osmosis, the basis of desalination.
- Q80Which statement about Werner’s theory is INCORRECT?
- Primary valencies are ionisable
- Secondary valency equals the coordination number
- Primary valency equals the oxidation state of the metal
- Secondary valencies are non-directional and variable
Answer: (D) Secondary valencies are non-directional and variable
Secondary valencies are DIRECTIONAL and FIXED (they determine geometry), not non-directional or variable. All other statements are correct postulates of Werner’s theory.
- Q81In the sequence CH₃CH₂Cl →[NaCN] X →[Ni/H₂] Y xrightarrowacetic anhydride Z, the final product Z is:
- CH₃CH₂CH₂CONHCOCH₃
- CH₃CH₂CH₂CONHCH₃
- CH₃CH₂CH₂NHCOCH₃
- CH₃CH₂CH₂NH₂
Answer: (C) CH₃CH₂CH₂NHCOCH₃
CH₃CH₂Cl + NaCN → CH₃CH₂CN (propanenitrile, X). Catalytic reduction (Ni/H₂) of the nitrile adds one carbon’s amine: CH₃CH₂CH₂NH₂ (propan-1-amine, Y). The primary amine is then acetylated by acetic anhydride to the amide CH₃CH₂CH₂NHCOCH₃ (Z).
- Q82Which of the following is most acidic?
- Cyclohexanol
- Phenol
- Benzyl alcohol
- m-Chlorophenol
Answer: (D) m-Chlorophenol
D. Tests: ranking acidity by conjugate-base stability, taking resonance first and then the inductive effect as tie-breaker. Why D: benzyl alcohol and cyclohexanol are ordinary alcohols whose alkoxide gets no delocalisation, so their pKₐ sits near 16. Phenol is far more acidic (pKₐ about 10) because the phenoxide charge is spread over the ring by resonance. Putting a chlorine on that ring adds a -I pull that drains more electron density away from the oxygen, stabilising the phenoxide further, so m-chlorophenol (pKₐ about 9) is the strongest acid here. Why not A: in benzyl alcohol the OH sits on an sp³ carbon outside the ring, so the ring cannot delocalise the alkoxide charge; it behaves as a plain alcohol [misconception A: “a benzene ring anywhere in the molecule makes it phenol-like”]. Why not B: cyclohexanol is a simple secondary alcohol with no delocalisation at all, so it is the weakest acid in the set. Why not C: phenol is acidic, but m-chlorophenol is phenol plus an electron-withdrawing chlorine, so phenol cannot be the most acidic once that option is on the list [misconception C: “phenol is automatically the strongest acid on any list”]. Remember: resonance beats induction to separate phenols from alcohols, then induction breaks the tie, so an electron-withdrawing group on the ring means a stronger acid.
- Q83Oxidation of glucose with concentrated nitric acid (HNO3) gives:
- Saccharic (glucaric) acid
- Sorbitol
- Gluconic acid
- n-Hexane
Answer: (A) Saccharic (glucaric) acid
Strong oxidation by HNO3 oxidises BOTH the terminal -CHO and the primary -CH2OH to -COOH, giving the dicarboxylic saccharic (glucaric) acid. Mild oxidation by bromine water gives only the monocarboxylic gluconic acid. Sorbitol is the reduction product; n-hexane comes from HI reduction.
- Q84Based on the data given below: Eᵒ_(Cr₂O₇²⁻/Cr³⁺) = 1.33 V Eᵒ_(Cl₂/Cl⁻) = 1.36 V Eᵒ_(MnO₄⁻/Mn²⁺) = 1.51 V Eᵒ_(Cr³⁺/Cr) = -0.74 V the strongest reducing agent is:
- Cl⁻
- MnO₄⁻
- Cr
- Mn²⁺
Answer: (C) Cr
C. Tests: picking the strongest reducing agent out of a table of standard reduction potentials. Why C: a strong reducing agent is a species that is itself easily oxidised, which means the couple it belongs to has the most negative standard reduction potential. Of the four couples listed the lowest value is Eᵒ_(Cr³⁺/Cr) = -0.74 V, and the reduced form of that couple is chromium metal. So Cr is the strongest reducing agent. Why not A: Cl⁻ belongs to Cl₂/Cl⁻ at +1.36 V, a high positive value, so chloride is very reluctant to give up an electron [misconception A: “an anion already carries extra electrons, so it must be a good reducing agent”]. Why not B: MnO₄⁻ is the oxidised form of its couple and at +1.51 V it is the strongest oxidising agent in the list, so it cannot be the strongest reducing agent. Why not D: Mn²⁺ is the reduced form of that same highest potential couple (+1.51 V), which makes it the most reluctant of all these reduced species to be oxidised. Remember: scan the table for the smallest Eᵒ and take the reduced species of that couple, that is your best electron donor.
- Q85A solution containing components A and B follows Raoult’s law when:
- volume of solution is different from sum of volumes of solute and solvent
- A-B attraction force remains same as A-A and B-B
- A-B attraction force is greater than A-A and B-B
- A-B attraction force is less than A-A and B-B
Answer: (B) A-B attraction force remains same as A-A and B-B
Raoult’s law is strictly obeyed (ideal solution) only when the A-B attractive force equals the A-A and B-B forces, so mixing causes no enthalpy or volume change. Stronger or weaker A-B forces give negative or positive deviations.
- Q86The straight-chain silicone polymer is formed by:
- hydrolysis of (CH₃)₂SiCl₂ followed by condensation polymerisation
- hydrolysis of (CH₃)₄Si by addition polymerisation
- hydrolysis of CH₃SiCl₃ followed by condensation polymerisation
- hydrolysis of (CH₃)₃SiCl followed by condensation polymerisation
Answer: (A) hydrolysis of (CH₃)₂SiCl₂ followed by condensation polymerisation
Dimethyldichlorosilane (CH₃)₂SiCl₂ has two hydrolysable Cl atoms, giving (CH₃)₂Si(OH)₂, which condenses (losing water) into a linear -Si-O-Si- (R₂SiO)ₙ chain. The two reactive ends are exactly what give a straight-chain silicone.
- Q87Given below are two statements: Statement I: Vapours of the liquid with higher boiling point condense before vapours of the liquid with lower boiling points in fractional distillation. Statement II: The vapours rising up in the fractionating column become richer in high boiling component of the mixture. In the light of the above statements, choose the correct answer from the options given below:
- Both Statement I and Statement II are false
- Statement I is false but Statement II is true
- Statement I is true but Statement II is false
- Both Statement I and Statement II are true
Answer: (C) Statement I is true but Statement II is false
C. Tests: tracking which component enriches the vapour as it climbs a fractionating column. Why C: A fractionating column is a stack of repeated vaporisation and condensation cycles. At every plate the vapour is cooled a little, and the component that condenses first is the one with the HIGHER boiling point, because it has the weaker tendency to stay in the vapour phase. Statement I is therefore true. Since that high boiling component keeps dropping back as liquid, the vapour that survives to the next plate up is progressively enriched in the LOW boiling component, which is exactly why the low boiling liquid distils over first. Statement II reverses this, so it is false. Why not A: it accepts Statement II, but if the rising vapour got richer in the high boiling component, the high boiling liquid would come over first and the column could never separate the mixture [misconception: “the heavier component travels up”]. Why not B: Statement I is a correct description of what happens at each plate, so calling both statements false throws away a true statement. Why not D: it inverts both verdicts, treating the lower boiling vapour as the one that condenses first [misconception: “low boiling point means condenses easily”]. Remember: going up a fractionating column the vapour gets richer in the LOWER boiling liquid, and the higher boiling one rains back down.
- Q88Noble gases are named because of their inertness towards reactivity. Identify the INCORRECT statement about them:
- Noble gases are sparingly soluble in water
- Noble gases have very high melting and boiling points
- Noble gases have weak dispersion forces
- Noble gases have large positive values of electron gain enthalpy
Answer: (B) Noble gases have very high melting and boiling points
Noble gases are held only by weak London dispersion forces, so they have very LOW melting and boiling points – statement (b) is incorrect. The other statements are correct.
- Q89An ‘inner orbital’ (low spin) octahedral complex uses which set of d-orbitals for hybridisation?
- Inner 3d orbitals (d²sp³)
- Outer 4d orbitals (sp³d²)
- f-orbitals
- Only 4s and 4p
Answer: (A) Inner 3d orbitals (d²sp³)
An inner orbital complex uses the inner (n−1)d orbitals, giving d²sp³ hybridisation, which occurs with strong field ligands that cause pairing.
- Q90H₂O₂ acts as a reducing agent in its reaction with:
- Acidified KMnO₄
- SO₂
- KI in acidic medium
- PbS
Answer: (A) Acidified KMnO₄
With acidified KMnO₄, H₂O₂ is oxidised (O: −1 → 0, releasing O₂) while Mn is reduced; thus H₂O₂ acts as a reducing agent. With KI and PbS it acts as an oxidising agent.
Botany
- Q91The phenomenon of heterospory in pteridophytes is biologically significant because it:
- Allows the gametophyte to become dominant
- Is a precursor to the seed habit seen in higher plants
- Removes the requirement of vascular tissue
- Eliminates the need for a sporophyte stage
Answer: (B) Is a precursor to the seed habit seen in higher plants
In heterosporous pteridophytes the megaspores germinate and are retained on the parent sporophyte; the development of the female gametophyte within is regarded as a precursor to the seed habit of gymnosperms and angiosperms.
- Q92The following ratio is generally constant for a given species:
- (A + C)/(T + G)
- (G + C)/(A + T)
- (T + C)/(G + A)
- (A + G)/(C + T)
Answer: (B) (G + C)/(A + T)
By Chargaff’s rule A = T and G = C, so (A + G)/(C + T) = 1 always. The base composition ratio (G + C)/(A + T) varies from species to species but is constant for a given species (it is rarely equal to 1, ranging ~0.4-1.9).
- Q93Which one of the following pairs is mismatched?
- Solar energy - Greenhouse effect
- Fossil fuel burning - Release of CO₂
- Biomass burning - Release of CO₂
- Nuclear power - Radioactive wastes
Answer: (A) Solar energy - Greenhouse effect
Solar energy is a clean source and is NOT responsible for the greenhouse effect; the greenhouse effect is caused by gases like CO₂. So ‘Solar energy – Greenhouse effect’ is the mismatched pair. The other three are correctly matched.
- Q94Which of the following bacteria reduces nitrate in the soil into nitrogen?
- Nitrococcus
- Thiobacillus
- Nitrosomonas
- Nitrobacter
Answer: (B) Thiobacillus
Thiobacillus carries out denitrification, reducing soil nitrate to gaseous nitrogen (N₂). Nitrosomonas, Nitrobacter and Nitrococcus oxidise ammonia/nitrite (nitrification), the opposite process.
- Q95The process in which water-soluble inorganic nutrients of detritus go down into the soil horizon and get precipitated as unavailable salts is called:
- Humification
- Mineralisation
- Catabolism
- Leaching
Answer: (D) Leaching
Leaching is the percolation of water-soluble inorganic nutrients into deeper soil layers where they may become precipitated as unavailable salts.
- Q96Expressed Sequence Tags (ESTs) refers to:
- polypeptide expression
- novel DNA sequences
- genes expressed as RNA
- DNA polymorphism
Answer: (C) genes expressed as RNA
ESTs are short sub-sequences of expressed genes (i.e. the genes that are transcribed as RNA/expressed as mRNA). The EST approach in the HGP focused on identifying the expressed genes.
- Q97Vessels are found in:
- All pteridophytes
- All angiosperms and few gymnosperms and some pteridophytes
- Most angiosperms and few gymnosperms
- All angiosperms and some gymnosperms
Answer: (C) Most angiosperms and few gymnosperms
Vessels are characteristic of angiosperms but a few angiosperms (e.g. Drimys, Tetracentron, Trochodendron) lack them, while a few gymnosperms (the Gnetales – Gnetum, Ephedra) do have vessels. Hence ‘most angiosperms and few gymnosperms’.
- Q98In a logistic growth curve, the asymptote is obtained when:
- K is greater than N
- K = N
- K is less than N
- The value of r approaches zero
Answer: (B) K = N
In logistic growth (dN)/(dt) = rN(K-N)/(K), when N equals the carrying capacity K the bracket (K-N)/K = 0, so dN/dt = 0 and the curve flattens into a horizontal asymptote at N = K.
- Q99Arrange the following events of meiosis in the correct sequence: I. Crossing over II. Synapsis III. Terminalisation of chiasmata IV. Disappearance of nucleolus
- I, II, III, IV
- II, I, III, IV
- III, II, IV, I
- II, I, IV, III
Answer: (B) II, I, III, IV
The order through prophase I is: synapsis in zygotene (II) → crossing over in pachytene (I) → terminalisation of chiasmata in diplotene/diakinesis (III) → disappearance of nucleolus in diakinesis (IV). Hence II, I, III, IV.
- Q100Attractants and rewards are required for:
- Entomophily
- Cleistogamy
- Anemophily
- Hydrophily
Answer: (A) Entomophily
Entomophily (insect pollination) needs attractants (colour, scent) and rewards (nectar, pollen) to draw insects. Wind and water pollination (anemophily, hydrophily) need no such attractants, and cleistogamous flowers never open.
- Q101Which one of the following is a true fruit?
- Apple
- Pear
- Cashewnut
- Coconut
Answer: (D) Coconut
Coconut develops only from the ripened ovary, so it is a true fruit. Apple and pear are false fruits (the edible part is the thalamus), and in cashew the fleshy edible ‘apple’ is the swollen pedicel/thalamus with the true nut hanging below.
- Q102In which of the following would meiosis NOT typically occur?
- Root tip meristem cells
- Megaspore mother cell in the ovule
- Meiocytes of the anther
- Microspore mother cells
Answer: (A) Root tip meristem cells
Root tip meristem cells divide by mitosis for growth. Meiosis occurs in reproductive (meiocyte/germ) cells such as microspore and megaspore mother cells to produce gametes/spores.
- Q103Which one of the following statements is wrong?
- Eubacteria are also called false bacteria
- Cyanobacteria are also called blue-green algae
- Golden algae are also called desmids
- Phycomycetes are also called algal fungi
Answer: (A) Eubacteria are also called false bacteria
Eubacteria are the ‘true’ bacteria, so calling them ‘false bacteria’ is wrong. Golden algae (desmids) are chrysophytes, Phycomycetes are indeed called algal fungi, and cyanobacteria are correctly called blue-green algae.
- Q104Which of these statements is incorrect?
- Glycolysis occurs in cytosol.
- Enzymes of TCA cycle are present in mitochondrial matrix.
- Oxidative phosphorylation takes place in outer mitochondrial membrane.
- Glycolysis operates as long as it is supplied with NAD that can pick up hydrogen atoms.
Answer: (C) Oxidative phosphorylation takes place in outer mitochondrial membrane.
C. Tests: placing each stage of respiration in its correct cellular compartment. Why C: Oxidative phosphorylation needs the electron transport complexes together with ATP synthase, and both are embedded in the INNER mitochondrial membrane, whose cristae folds enlarge the working area. The outer membrane is porin rich and freely permeable, so it could never hold the proton gradient the process depends on. Naming the outer membrane makes C the incorrect statement. Why not A: it is correct. Glycolysis must keep regenerating NAD⁺ to go on oxidising glyceraldehyde-3-phosphate, so the pathway runs only while NAD is available to pick up hydrogen [misconception A: “glycolysis is independent of coenzyme supply”]. Why not B: it is correct as the textbook states it. The TCA enzymes are matrix enzymes, which is why pyruvate must first be carried into the matrix before the cycle can act on it. The lone exception, succinate dehydrogenase in the inner membrane, does not make this the statement being asked for [misconception B: “TCA enzymes float in the cytosol beside glycolysis”]. Why not D: it is correct. Glycolysis is entirely cytosolic in every cell, which is why organisms without mitochondria still carry it out [misconception D: “glycolysis begins inside the mitochondrion”]. Remember: the outer membrane leaks protons, so oxidative phosphorylation must live on the inner membrane and its cristae.
- Q105Foetal sex can be determined by examining cells from the amniotic fluid by looking for:
- chiasmata
- Barr bodies
- autosomes
- kinetochore
Answer: (B) Barr bodies
A Barr body (a condensed, inactivated X-chromosome) is present in female (XX) cells but absent in male (XY) cells. Examining amniotic-fluid cells for a Barr body reveals the foetal sex.
- Q106Which of the following is incorrect for wind-pollinated (anemophilous) plants?
- Well-exposed stamens and stigma
- Flowers are small and not brightly coloured
- Pollen grains are light and non-sticky
- Many ovules in each ovary
Answer: (D) Many ovules in each ovary
Wind pollination is wasteful and non-directional, so wind-pollinated flowers typically have a single ovule per ovary (e.g. grasses), not many. They do have well-exposed stamens/stigma, small dull flowers and light, dry, non-sticky pollen. Hence ‘many ovules in each ovary’ is the incorrect statement.
- Q107AGGTATCGCAT is a sequence from the coding strand of a gene. The corresponding mRNA sequence is:
- AGGUAUCGCAU
- ACCUAUGCGUA
- AGGUUCGCAU
- UCCAUAGCGUA
Answer: (A) AGGUAUCGCAU
The mRNA has the same sequence as the coding strand with T replaced by U: AGGTATCGCAT -> AGGUAUCGCAU.
- Q108A scrubber installed in an industrial chimney is primarily intended to remove:
- Gaseous pollutants such as SO₂
- Solid particulate matter
- Heat from flue gas
- Radioactive isotopes
Answer: (A) Gaseous pollutants such as SO₂
A scrubber passes exhaust through a spray of water or lime, which absorbs gases like sulphur dioxide. Particulates are removed by an electrostatic precipitator instead.
- Q109The experimental proof for the semiconservative replication of DNA was first shown in a:
- virus
- bacterium
- fungus
- plant
Answer: (B) bacterium
Meselson and Stahl (1958) first demonstrated semiconservative replication in the bacterium Escherichia coli using 15N/14N density-gradient experiments.
- Q110A pleiotropic gene:
- is a gene evolved during the Pliocene
- is expressed only in primitive plants
- controls multiple traits in an individual
- controls a trait only in combination with another gene
Answer: (C) controls multiple traits in an individual
A pleiotropic gene is one that controls (affects) multiple traits in an individual, e.g. the PKU gene affecting both intelligence and pigmentation.
- Q111Which one of the following cells is surrounded by a callose wall?
- Pollen grain
- Egg
- Male gamete
- Microspore mother cell
Answer: (D) Microspore mother cell
As the microspore (pollen) mother cell enters meiosis, a callose (beta-1,3-glucan) wall is deposited around it; this temporary wall isolates the meiotic product and dissolves to release the microspores. The mature pollen grain itself has exine and intine, not callose.
- Q112Number of Barr bodies in an XXXX female is:
- 2
- 1
- 3
- 4
Answer: (C) 3
The number of Barr bodies = (number of X-chromosomes) – 1. For an XXXX female, 4 – 1 = 3 Barr bodies (three X’s are inactivated, one stays active).
- Q113The first stable product of CO2 fixation in Sorghum is:
- Succinic acid
- Phosphoglyceric acid
- Pyruvic acid
- Oxaloacetic acid
Answer: (D) Oxaloacetic acid
Sorghum is a C4 plant. In C4 plants, PEP in mesophyll cells fixes CO2 (via PEPcase) to give the 4-carbon oxaloacetic acid (OAA) as the first stable product. PGA is the first product in C3 plants.
- Q114Plants grown in magnesium-deficient soil that is sprayed with urea would show:
- yellowing of leaves
- loss of pigments in petals
- deep green foliage
- early flowering
Answer: (A) yellowing of leaves
Spraying urea only supplies nitrogen, not magnesium. Because Mg is the central atom of chlorophyll, its deficiency still causes chlorosis (yellowing of leaves) regardless of the added nitrogen.
- Q115The decline of cichlid fishes in Lake Victoria is attributed mainly to:
- Introduction of the alien Nile perch
- Over-exploitation by fishing
- Habitat fragmentation of the lake
- Co-extinction with their parasites
Answer: (A) Introduction of the alien Nile perch
The introduction of the alien predatory Nile perch led to the extinction of over 200 species of native cichlid fishes – a classic example of alien species invasion.
- Q116The ‘eyes’ of a potato tuber represent:
- Lenticels for respiration
- Adventitious roots
- Reduced leaves
- Nodes bearing axillary buds
Answer: (D) Nodes bearing axillary buds
A potato is an underground stem tuber; its eyes are nodes, each with a scaly leaf and axillary bud, confirming its stem nature.
- Q117In a food chain water → zooplankton → small fish → large fish → fish-eating bird, DDT concentration was found to be highest in the bird. This phenomenon is called:
- Biomagnification
- Biodegradation
- Bioremediation
- Eutrophication
Answer: (A) Biomagnification
Biomagnification is the increase in concentration of a non-degradable toxicant (like DDT) at successive trophic levels. The top consumer (bird) accumulates the highest concentration; in birds DDT disturbs calcium metabolism and thins eggshells.
- Q118Which one of the following is heterosporous?
- Adiantum
- Dryopteris
- Equisetum
- Salvinia
Answer: (D) Salvinia
Salvinia is heterosporous, producing small microspores and large megaspores. Dryopteris, Adiantum and Equisetum are homosporous (one kind of spore).
- Q119Vivipary is best described as:
- Seed germination with subterranean cotyledons
- Fruit development without pollination
- Germination of the seed inside the fruit while still attached to the parent plant
- Seed germination with epiterranean cotyledons
Answer: (C) Germination of the seed inside the fruit while still attached to the parent plant
Vivipary is the germination of seeds inside the fruit while it is still attached to the parent (typical of mangroves in salty/marshy coasts); the seedling then drops and roots. It is not a type of cotyledon position nor parthenocarpy (fruit without fertilisation).
- Q120The axoneme of a eukaryotic cilium or flagellum shows which microtubule arrangement?
- 9 + 0
- 11 + 2
- 9 + 2
- 9 + 3
Answer: (C) 9 + 2
Cilia and flagella have a 9 + 2 axoneme: nine peripheral doublet microtubules surrounding two central singlet microtubules. Centrioles and basal bodies have a 9 + 0 arrangement of triplets.
- Q121Calcium is required for the formation of which of the following?
- Chlorophyll
- ATP
- Auxin
- Middle lamella (calcium pectate)
Answer: (D) Middle lamella (calcium pectate)
Calcium is used in synthesis of the middle lamella as calcium pectate, in spindle formation during cell division, and in regulating membrane permeability.
- Q122Assertion (A): RuBisCO functions both as a carboxylase and an oxygenase. Reason (R): RuBisCO has a much greater affinity for CO2 than for O2, but the relative concentrations of CO2 and O2 decide which activity dominates.
- Both A and R are true but R is not the correct explanation of A
- 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
Answer: (B) Both A and R are true and R is the correct explanation of A
RuBisCO indeed has dual activity. It has a greater affinity for CO2 than O2, but because the active site is the same, the ratio of CO2 to O2 determines whether carboxylation (Calvin cycle) or oxygenation (photorespiration) occurs. R correctly explains A.
- Q123In a marriage between a male with blood group A and a female with blood group B, the progeny had either blood group AB or B. What are the possible genotypes of the parents?
- I^A I^A (male) and I^B i (female)
- I^A i (male) and I^B I^B (female)
- I^A i (male) and I^B i (female)
- I^A I^B (male) and I^B I^B (female)
Answer: (B) I^A i (male) and I^B I^B (female)
Progeny are only AB and B (no A and no O), so the mother must contribute only I^B (she is I^B I^B) and the father must be able to give i (to make B) and I^A (to make AB), i.e. father is I^A i. I^A i x I^B I^B gives I^A I^B (AB) and I^B i (B) only.
- Q124A glass bottle filled with moistened mustard seeds and sealed burst after some time, throwing glass pieces around. This demonstrates:
- Anaerobic respiration
- Imbibition
- Osmosis
- Diffusion
Answer: (B) Imbibition
Dry seeds are hydrophilic colloids; on absorbing water they swell and develop an enormous imbibition pressure. In a sealed rigid bottle this pressure has nowhere to go and shatters the glass. The phenomenon is imbibition, so (D). (Anaerobic respiration would build gas slowly but cannot generate this mechanical swelling force.)
- Q125The exchange of genetic material between chromatids of paired homologous chromosomes during the first meiotic division is called:
- Chiasmata
- Crossing over
- Synapsis
- Transformation
Answer: (B) Crossing over
The exchange of segments between non-sister chromatids of paired homologues in prophase I is crossing over. A chiasma is the visible site where it occurred, synapsis is the pairing that precedes it, and transformation is a bacterial uptake of DNA (unrelated).
- Q126For a species-area study, log C = 2 and Z = 0.2. If the area sampled is 100 units (log A = 2), the species richness S is:
- 100
- 251
- 398
- 158
Answer: (B) 251
log S = log C + Z log A = 2 + 0.2 × 2 = 2.4. So S = 10².4 ≈ 251. (10⁰.4 ≈ 2.51, times 100 = 251.)
- Q127The National Botanical Research Institute (NBRI) is located at Lucknow. Which type of taxonomical aid is it best classified as?
- Botanical garden
- Herbarium
- Zoological park
- Museum
Answer: (A) Botanical garden
NBRI, Lucknow maintains collections of living plants for reference and study, so it is a botanical garden (like Kew, England and the Indian Botanical Garden, Howrah). Hence B.
- Q128The primary producers of the deep-sea hydrothermal vent ecosystem are:
- green algae
- blue-green algae
- coral reefs
- chemosynthetic bacteria
Answer: (D) chemosynthetic bacteria
No sunlight reaches deep-sea hydrothermal vents, so photosynthesis is impossible. The primary producers are chemosynthetic (chemoautotrophic) bacteria/archaebacteria that oxidise inorganic compounds such as H2S for energy.
- Q129In the logistic growth model, the population growth rate (dN/dt) is maximum when N equals:
- Zero
- K/2
- K
- 2K
Answer: (B) K/2
In logistic growth dN/dt = rN(K−N)/K is a parabola in N, maximised at N = K/2 (the inflection point of the sigmoid curve). At N = K growth is zero.
- Q130What is the net ATP yield per glucose molecule in anaerobic fermentation?
- 38
- 36
- 2
- 8
Answer: (C) 2
All ATP in fermentation comes from glycolysis (net 2 ATP); the fermentation steps themselves yield no ATP.
- Q131Which of the following statements about lysosomes is NOT correct?
- Lysosomes have numerous hydrolytic enzymes
- The hydrolytic enzymes of lysosomes are active under acidic pH
- Lysosomes are formed by the process of packaging in the endoplasmic reticulum
- Lysosomes are membrane-bound structures
Answer: (C) Lysosomes are formed by the process of packaging in the endoplasmic reticulum
Lysosomes are formed by the budding off of the trans-face of the GOLGI bodies, not by packaging in the ER (their enzyme precursors are made in the RER but the lysosome itself is Golgi-derived). The other three statements are correct, so C is the incorrect one.
- Q132Which of the following is considered the most ecologically relevant abiotic factor?
- Atmospheric pressure
- Light intensity
- Soil pH
- Temperature
Answer: (D) Temperature
NCERT states temperature is the most ecologically relevant environmental factor because it affects enzyme kinetics, metabolism and hence the distribution of organisms.
- Q133The osmotic expansion of a cell kept in water is chiefly regulated by:
- plastids
- mitochondria
- ribosomes
- vacuoles
Answer: (D) vacuoles
The vacuole, bounded by the tonoplast, stores water and solutes and its cell sap has a high osmotic pressure; it therefore chiefly regulates the osmotic expansion (turgor) of a cell kept in water.
- Q134A colourblind man marries a woman who is homozygous for normal colour vision. The probability of their son being colourblind is:
- 0.75
- 0.5
- 0
- 1
Answer: (C) 0
Colourblind man (X^C Y) x homozygous normal woman (X X). Sons get Y from father and a normal X from mother, so all sons are X Y (normal). Probability of a colourblind son = 0.
- Q135Which one of the following is a C4 plant?
- Potato
- Papaya
- Maize/Corn
- Pea
Answer: (C) Maize/Corn
Maize (corn) is a classic C4 plant in which the first stable product of CO2 fixation is the 4-carbon OAA and the leaf shows Kranz anatomy. Papaya, pea and potato are C3 plants.
Zoology
- Q136Adaptive radiation refers to:
- migration of members of a species to different geographical areas
- adaptations due to geographical isolation
- evolution of different species from a common ancestor
- power of adaptation in an individual to a variety of environments
Answer: (C) evolution of different species from a common ancestor
Adaptive radiation is the evolution of different species, each adapted to a different niche, from a single common ancestral form (e.g. Darwin’s finches). It is not migration or individual acclimation.
- Q137Match the following columns and select the correct option. (a) Bt cotton – (i) Gene therapy (b) Adenosine deaminase deficiency – (ii) Cellular defence (c) RNAi – (iii) Detection of HIV infection (d) PCR – (iv) Bacillus thuringiensis
- (a)arrow(ii), (b)arrow(iii), (c)arrow(iv), (d)arrow(i)
- (a)arrow(iv), (b)arrow(i), (c)arrow(ii), (d)arrow(iii)
- (a)arrow(iii), (b)arrow(ii), (c)arrow(i), (d)arrow(iv)
- (a)arrow(i), (b)arrow(ii), (c)arrow(iii), (d)arrow(iv)
Answer: (B) (a)arrow(iv), (b)arrow(i), (c)arrow(ii), (d)arrow(iii)
C. Tests: one-line associations across the four flagship applications of biotechnology. Why C: Bt cotton carries cry genes from Bacillus thuringiensis (a-iv); ADA deficiency (SCID) was the first disorder treated by gene therapy (b-i); RNAi is a natural method of cellular defence in all eukaryotes, exploited against nematodes (c-ii); PCR amplifies viral nucleic acid to detect HIV infection early (d-iii). Why not A: it links Bt cotton to gene therapy and PCR to the bacterium, mixing plant and medical tools [misconception A: “match in the listed order”] Why not B: Bt cotton has nothing to do with HIV detection [misconception B: “cotton and diagnostics both involve kits”] Why not D: RNAi is not from Bacillus thuringiensis, and ADA deficiency is linked to gene therapy, not to HIV detection [misconception D: “any biology pairings sound plausible together”] Remember: Bt-bacterium, ADA-gene therapy, RNAi-cell defence, PCR-early HIV detection.
- Q138Genetic engineering is possible because:
- we can see DNA by an electron microscope
- the phenomenon of transduction in bacteria is well understood
- we can cut DNA at specific sites by endonucleases like DNase-I
- restriction endonucleases purified from bacteria can be used in vitro
Answer: (D) restriction endonucleases purified from bacteria can be used in vitro
Genetic engineering depends on cutting DNA at specific sites; restriction endonucleases purified from bacteria can be used in vitro to do exactly that, making site-specific cutting and recombination possible. DNase-I cuts non-specifically (so C is wrong), and merely visualising DNA (B) does not enable manipulation.
- Q139Which of the following features is NOT present in Periplaneta americana?
- Exoskeleton composed of N-acetylglucosamine (chitin)
- Indeterminate and radial cleavage during embryonic development
- Metamerically segmented body
- Schizocoelom as the body cavity
Answer: (B) Indeterminate and radial cleavage during embryonic development
The cockroach (a protostome arthropod) shows DETERMINATE and SPIRAL cleavage, not indeterminate/radial – so A is the feature NOT present. It does have a chitinous (N-acetylglucosamine) exoskeleton, a metamerically segmented body, and a schizocoelom.
- Q140Which feature is NOT desirable in an ideal contraceptive?
- Effective with least side-effects
- User-friendly and reversible
- Interferes with sexual drive and desire
- Easily available
Answer: (C) Interferes with sexual drive and desire
An ideal contraceptive should NOT interfere with the sexual drive, desire or the sexual act of the user. The other features are all desirable.
- Q141Inflammation of joints due to accumulation of uric acid crystals is called:
- Tetany
- Osteoporosis
- Arthritis
- Gout
Answer: (D) Gout
Gout is inflammation of the joints caused by accumulation of uric acid crystals. General joint inflammation is arthritis, but the uric-acid-specific form is gout.
- Q142Concanavalin A is a:
- A pigment
- An essential oil
- An alkaloid
- A lectin
Answer: (D) A lectin
Concanavalin A is a lectin – a carbohydrate-binding protein and T-cell mitogen used to activate lymphocytes. It is not an oil, pigment or alkaloid.
- Q143The ‘Bundle of His’ is a part of which one of the following organs in humans?
- Heart
- Pancreas
- Brain
- Kidney
Answer: (A) Heart
The Bundle of His is specialised conducting muscle fibres of the heart that carry the impulse from the AV node to the ventricles. It is part of the heart’s conduction system.
- Q144Which of the following pairings of a contraceptive approach with its action is INCORRECT?
- Hormonal contraceptives - prevent/retard entry of sperms, prevent ovulation and fertilisation
- Barrier methods - prevent fertilisation
- Vasectomy - prevents spermatogenesis
- Intra-uterine devices - increase phagocytosis of sperms, suppress sperm motility and fertilising capacity
Answer: (C) Vasectomy - prevents spermatogenesis
Vasectomy cuts/ties the vas deferens, blocking the transport of sperms; it does NOT affect spermatogenesis, which continues in the testes. The other three pairings correctly describe the action of IUDs, hormonal contraceptives and barrier methods, so the incorrect one is C.
- Q145Which band of the sarcomere contains ONLY thin (actin) filaments?
- I-band
- A-band
- H-zone
- M-line region
Answer: (A) I-band
The I-band (isotropic, light) contains only thin actin filaments and is bisected by the Z-line. The A-band contains thick filaments; the H-zone has only thick myosin.
- Q146The causative agent of typhoid is:
- Plasmodium vivax
- Streptococcus pneumoniae
- Salmonella typhi
- Entamoeba histolytica
Answer: (C) Salmonella typhi
Typhoid is caused by the bacterium Salmonella typhi, confirmed by the Widal test.
- Q147In a random-mating population in equilibrium, which one of the following brings about a change in gene frequency in a NON-directional manner?
- Random drift
- Mutation
- Migration
- Selection
Answer: (B) Mutation
Mutation introduces new alleles at random and changes gene frequencies in a non-directional way. Natural selection acts directionally (favouring particular alleles); migration and drift act through other mechanisms, but the classic non-directional source of new variation is mutation.
- Q148Menstrual flow occurs due to lack of:
- Vasopressin
- Progesterone
- Oxytocin
- FSH
Answer: (B) Progesterone
Progesterone from the corpus luteum maintains the endometrium. When the corpus luteum regresses and progesterone falls, the endometrium breaks down, producing menstrual flow. FSH, oxytocin and vasopressin do not maintain the endometrium.
- Q149Which cells of the Islets of Langerhans secrete insulin?
- δ-cells
- Acinar cells
- β-cells
- α-cells
Answer: (C) β-cells
β-cells secrete insulin (hypoglycemic). α-cells secrete glucagon (hyperglycemic). Acinar cells are the exocrine part secreting digestive enzymes.
- Q150Plasmid pBR322 has a PstI restriction enzyme site within the gene amp^R that confers ampicillin resistance. If a gene for β-galactosidase production is inserted at this site and the recombinant plasmid is introduced into an E. coli strain, then:
- the transformed cells will resist ampicillin as well as produce β-galactosidase
- it will lead to lysis of the host cell
- it will not be able to confer ampicillin resistance to the host cell
- it will produce a novel protein with dual ability
Answer: (C) it will not be able to confer ampicillin resistance to the host cell
Inserting foreign DNA at the PstI site, which lies within the amp^R gene, disrupts that gene (insertional inactivation). The ampicillin-resistance gene is therefore non-functional, so the host can no longer confer/express ampicillin resistance. (The tet^R gene would be the one used for selection in this case.)
- Q151The human hindbrain comprises three parts, one of which is the:
- Corpus callosum
- Cerebellum
- Spinal cord
- Hypothalamus
Answer: (B) Cerebellum
The hindbrain consists of the pons, cerebellum and medulla oblongata. The corpus callosum and hypothalamus belong to the forebrain, and the spinal cord is not part of the brain.
- Q152Which pair of cry genes is incorporated into Bt cotton to provide resistance against cotton bollworms?
- cryIAc and cryIIAb
- cryIIIA and cryIAc
- cryIAb and cryIIAb
- cryIAb and cryIIAc
Answer: (A) cryIAc and cryIIAb
Bt cotton carries cryIAc and cryIIAb genes, controlling cotton bollworms. The gene cryIAb controls corn borer (a different crop pest).
- Q153The extinct human who lived 100000 to 40000 years ago in Europe, Asia and parts of Africa, with short stature, heavy eyebrows, retreating forehead, large jaws with heavy teeth, stocky body and a lumbering, stooped gait, was:
- Ramapithecus
- Neanderthal human
- Homo habilis
- Cro-Magnon human
Answer: (B) Neanderthal human
These features (heavy brow ridges, retreating forehead, large jaws, stocky stooped body, living 1,00,000-40,000 years ago across Europe/Asia/Africa) describe Neanderthal man (Homo neanderthalensis), brain ~1400 cc, who used hides and buried the dead.
- Q154Match correctly: (i) Ascariasis (ii) Filariasis (iii) Common cold (iv) Amoebiasis with their agents: (1) Rhinovirus (2) Wuchereria (3) Ascaris (4) Entamoeba histolytica.
- i-2, ii-3, iii-4, iv-1
- i-4, ii-1, iii-2, iv-3
- i-1, ii-4, iii-3, iv-2
- i-3, ii-2, iii-1, iv-4
Answer: (D) i-3, ii-2, iii-1, iv-4
Ascariasis – Ascaris (roundworm); Filariasis – Wuchereria; Common cold – Rhinovirus; Amoebiasis – Entamoeba histolytica.
- Q155A person who is on a long hunger strike and is surviving only on water will have:
- more sodium in his urine
- less amino acids in his urine
- less urea in his urine
- more glucose in his blood
Answer: (C) less urea in his urine
Urea comes from protein/amino-acid (deamination) metabolism. A person fasting on only water takes in no protein, so urea production falls and the urine contains less urea. Blood glucose would tend to fall (not rise) during starvation. Hence less urea in the urine.
- Q156The quaternary structure of a protein is best illustrated by:
- The α-helix of a single chain
- The 3-D folding of one polypeptide into a globule
- Haemoglobin made of four polypeptide subunits
- The linear sequence of amino acids
Answer: (C) Haemoglobin made of four polypeptide subunits
Quaternary structure refers to the spatial arrangement of two or more polypeptide chains. Haemoglobin (2α + 2β subunits) is the classic example. The α-helix is secondary, the sequence is primary, and single-chain 3-D folding is tertiary.
- Q157Which of the following conditions will stimulate the parathyroid gland to release parathyroid hormone?
- Fall in active vitamin-D levels
- Rise in blood Ca²⁺ levels
- Fall in bone Ca²⁺ levels
- Fall in blood Ca²⁺ levels
Answer: (D) Fall in blood Ca²⁺ levels
A fall in blood Ca²⁺ is sensed by the parathyroid glands, which release PTH to mobilise Ca²⁺ from bone and increase renal/intestinal reabsorption, restoring blood calcium.
- Q158If a DNA segment is amplified by PCR and one molecule becomes 8 copies, how many cycles of amplification occurred (assuming ideal doubling)?
- 2 cycles
- 3 cycles
- 4 cycles
- 8 cycles
Answer: (B) 3 cycles
Each cycle doubles the DNA, so after n cycles copies = 2ⁿ. Since 2³ = 8, three cycles are needed to go from 1 to 8 copies.
- Q159If you suspect a major deficiency of antibodies in a person, to which of the following would you look for confirmatory evidence?
- Serum globulins
- Serum albumins
- Fibrinogen in the plasma
- Haemocytes
Answer: (A) Serum globulins
Antibodies (immunoglobulins) constitute the gamma-globulin fraction of blood proteins. So a deficiency of antibodies is confirmed by examining serum globulins. Albumins maintain osmotic balance, fibrinogen is for clotting.
- Q160Which two classes of vertebrates are warm-blooded (homeothermic)?
- Osteichthyes and Aves
- Aves and Mammalia
- Amphibia and Mammalia
- Reptilia and Aves
Answer: (B) Aves and Mammalia
Only Aves (birds) and Mammalia maintain a constant body temperature (warm-blooded/homeothermic). Fishes, amphibians and reptiles are cold-blooded (poikilothermic).
- Q161In the production of recombinant human insulin (Humulin) by Eli Lilly, the strategy used was to:
- Insert the whole proinsulin gene and let E. coli process the C-peptide
- Use yeast to secrete mature insulin with C-peptide intact
- Produce A and B chains separately in E. coli and join them by disulphide bonds
- Extract insulin directly from pig pancreas
Answer: (C) Produce A and B chains separately in E. coli and join them by disulphide bonds
Eli Lilly (1983) produced the two polypeptide chains A and B separately in E. coli and then combined them via disulphide bonds to form functional human insulin.
- Q162Identify the animal with the correct class: Ichthyophis, Pristis, Hippocampus, Chelone.
- Amphibia, Osteichthyes, Chondrichthyes, Reptilia
- Reptilia, Osteichthyes, Osteichthyes, Amphibia
- Reptilia, Chondrichthyes, Osteichthyes, Amphibia
- Amphibia, Chondrichthyes, Osteichthyes, Reptilia
Answer: (D) Amphibia, Chondrichthyes, Osteichthyes, Reptilia
Ichthyophis is a limbless amphibian (caecilian); Pristis (sawfish) is cartilaginous (Chondrichthyes); Hippocampus (seahorse) is a bony fish (Osteichthyes); Chelone (sea turtle) is a reptile. Hence Amphibia, Chondrichthyes, Osteichthyes, Reptilia.
- Q163Assertion (A): The deposition of calcium, fat and fibrous tissue in coronary arteries can lead to heart problems. Reason (R): This deposition narrows the lumen of coronary arteries, reducing blood flow to the heart muscle (coronary artery disease).
- Both A and R are true but R is not the correct explanation of A
- 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
Answer: (B) Both A and R are true and R is the correct explanation of A
Atherosclerosis is the deposition of Ca²⁺, fat and fibrous tissue inside coronary arteries, narrowing them and reducing blood supply to the heart muscle – the basis of coronary artery disease. R correctly explains A.
- Q164Capacitation of sperm occurs in the:
- Vas deferens
- Epididymis
- Female reproductive tract
- Rete testis
Answer: (C) Female reproductive tract
Capacitation is the final functional maturation that makes sperm capable of fertilising; it occurs in the female reproductive tract, not in the male ducts (rete testis, epididymis or vas deferens) where sperm are only stored and partially matured.
- Q165Which one of the following statements regarding enzyme inhibition is correct?
- Competitive inhibition is seen when the substrate and the inhibitor compete for the active site on the enzyme
- Competitive inhibition is seen when a substrate competes with an enzyme for binding to an inhibitor protein
- Non-competitive inhibitors often bind the enzyme irreversibly
- Non-competitive inhibition of an enzyme can be overcome by adding a large amount of substrate
Answer: (A) Competitive inhibition is seen when the substrate and the inhibitor compete for the active site on the enzyme
Competitive inhibition occurs when an inhibitor resembling the substrate competes with it for the active site of the enzyme, so statement B is correct. A states it backwards; C is false (excess substrate overcomes competitive, not non-competitive, inhibition); D is not generally true.
- Q166The counter-current mechanism in the kidney involves the loop of Henle and the:
- Bowman’s capsule
- Peritubular capillaries of the cortex
- Glomerulus
- Vasa recta
Answer: (D) Vasa recta
The vasa recta runs parallel to the loop of Henle in a counter-current arrangement. Together they form a counter-current multiplier/exchanger that builds and maintains the medullary concentration gradient.
- Q167Which one of the following is a fat-soluble vitamin and its correctly related deficiency disease?
- Ascorbic acid - Scurvy
- Calciferol - Pellagra
- Cobalamine - Beri-beri
- Retinol - Xerophthalmia
Answer: (D) Retinol - Xerophthalmia
Retinol (vitamin-A) is fat-soluble and its deficiency causes xerophthalmia, so B is correct. Ascorbic acid (C) is water-soluble; cobalamine (B₁₂) deficiency causes pernicious anaemia, not beri-beri; calciferol (D) deficiency causes rickets, not pellagra.
- Q168What is common among amylase, rennin and trypsin?
- These are produced in the stomach
- These are all proteins
- These are proteolytic enzymes
- These act at a pH lower than 7
Answer: (B) These are all proteins
Amylase, rennin and trypsin are all enzymes, and enzymes are proteins, so ‘all are proteins’ is the common feature. They are not all proteolytic (amylase digests starch), not all gastric (trypsin is pancreatic) and not all acidic (trypsin works at alkaline pH).
- Q169Mutations in plant cells can be induced by:
- Zeatin
- Gamma rays
- Infrared rays
- Kinetin
Answer: (B) Gamma rays
Gamma rays are high-energy ionising radiation that damage DNA and induce mutations, the basis of mutation breeding. Kinetin and zeatin are cytokinins (growth regulators), and infrared is non-ionising and does not induce mutations. Hence (C).
- Q170In a stirred-tank bioreactor, the main function of the sparger is to:
- Stir and mix the contents
- Withdraw samples for testing
- Maintain the pH of the medium
- Supply/deliver oxygen by introducing air bubbles
Answer: (D) Supply/deliver oxygen by introducing air bubbles
The sparger introduces air as bubbles to increase the available surface area for oxygen transfer into the medium. Mixing is done by the agitator, and a separate port is used for sampling.
- Q171Acrosome reaction in sperm is triggered by:
- Release of fertilisin
- Release of lysin
- Influx of Na⁺
- Capacitation
Answer: (A) Release of fertilisin
Fertilisin, a glycoprotein released by the egg, binds the antifertilisin on the sperm surface; this fertilisin-antifertilisin reaction triggers the acrosome reaction (releasing sperm lysins). Capacitation precedes but does not itself trigger the acrosome reaction.
- Q172When CO₂ concentration in blood increases, breathing becomes:
- shallower and slow
- slow and deep
- there is no effect on breathing
- faster and deeper
Answer: (D) faster and deeper
Rising blood CO₂ (and H⁺) is detected by the chemosensitive area near the respiratory rhythm centre and by the aortic/carotid bodies, which drive the medullary centre to increase ventilation – breathing becomes faster and deeper to blow off the excess CO₂. So the answer is faster and deeper.
- Q173An important characteristic that hemichordates share with chordates is:
- pharynx without gill slits
- absence of notochord
- pharynx with gill slits
- ventral tubular nerve cord
Answer: (C) pharynx with gill slits
Hemichordates share a pharynx with gill slits with the chordates. They lack a true notochord (having only a stomochord), and the chordate nerve cord is dorsal, not ventral, so the other options are incorrect.
- Q174Malonate inhibits succinate dehydrogenase by competing with succinate for the active site. This is an example of:
- Non-competitive inhibition
- Allosteric activation
- Feedback inhibition
- Competitive inhibition
Answer: (D) Competitive inhibition
Malonate structurally resembles succinate and competes for the same active site of succinate dehydrogenase, blocking substrate binding. This is the textbook example of competitive inhibition.
- Q175The H-zone in the skeletal muscle fibre is due to:
- The central gap between myosin filaments in the A-band
- The absence of myofibrils in the central portion of A-band
- The central gap between actin filaments extending through myosin filaments in the A-band
- Extension of myosin filaments in the central portion of the A-band
Answer: (C) The central gap between actin filaments extending through myosin filaments in the A-band
The H-zone is the lighter central part of the A-band where the thin (actin) filaments do not reach – i.e. the gap between the two sets of actin filaments, a region of myosin only. It is not an absence of myofibrils, and myosin spans the whole A-band, so the other options are wrong.
- Q176A high Biochemical Oxygen Demand (BOD) of a water sample indicates that:
- The water is rich in dissolved oxygen and clean
- The water is suitable for drinking
- The water has very low microbial activity
- The water contains a large amount of organic (polluting) matter
Answer: (D) The water contains a large amount of organic (polluting) matter
BOD measures the oxygen that bacteria would consume to oxidise organic matter in water. Greater BOD means more organic matter, and hence greater polluting potential of the water.
- Q177Among reptiles, a four-chambered heart is found in:
- Lizards only
- Turtles only
- All snakes
- Crocodiles
Answer: (D) Crocodiles
Most reptiles have a three-chambered heart (two atria, one incompletely divided ventricle), but crocodiles are the exception with a complete four-chambered heart.
- Q178Destruction of the anterior horn cells of the spinal cord would result in loss of:
- Commissural impulses
- Integrating impulses
- Sensory impulses
- Voluntary motor impulses
Answer: (D) Voluntary motor impulses
The anterior (ventral) horn of the spinal cord grey matter contains the cell bodies of motor neurons that drive skeletal muscle. Their destruction abolishes voluntary motor output. Sensory fibres enter through the dorsal (posterior) horn, so sensation is not lost.
- Q179A large proportion of oxygen remains unused in the human blood even after its uptake by the body tissues. This O₂
- helps in releasing more O₂ to the epithelial tissues
- raises the pCO₂ of blood to 75 mm of Hg
- is enough to keep oxyhaemoglobin saturation at 96%
- acts as a reserve during muscular exercise
Answer: (D) acts as a reserve during muscular exercise
A. Tests: interpreting the venous oxygen reserve from the numbers of oxygen transport. Why A: every 100 mL of arterial blood carries about 20 mL of O₂, and resting tissues extract only about 5 mL of it, so venous blood leaves still roughly 75% saturated. That unextracted oxygen is a standing reserve: during exercise tissue pO₂ falls and the Bohr shift lowers affinity, so extraction from the same 20 mL can rise threefold or more before ventilation and cardiac output need to catch up. Why not B: venous pCO₂ is about 45 mm Hg and arterial about 40 mm Hg, and leftover oxygen does not raise pCO₂ at all; no consistent substitution gives 75 mm Hg. Why not C: 96 to 97% is the arterial saturation, set in the alveolar capillary, where an alveolar pO₂ of about 104 mm Hg loads blood to a pO₂ of about 95 mm Hg. It is fixed before the tissues take anything, not maintained by what they leave behind. Why not D: epithelial tissues get no preferential supply; the unextracted O₂ stays bound to haemoglobin and returns to the right heart. Remember: venous blood is still about 75% saturated, and that leftover is your exercise reserve.
- Q180Which lung parameter CANNOT be measured directly by a simple spirometer?
- Inspiratory reserve volume
- Tidal volume
- Residual volume
- Vital capacity
Answer: (C) Residual volume
Residual volume is the air remaining after a forced expiration and never leaves the lungs, so a spirometer cannot measure it (nor capacities containing it, like FRC and TLC). It is measured indirectly.