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Free mock 1 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, Thermodynamics, Current Electricity, Equilibrium, Molecular Basis of Inheritance, Animal Kingdom, Biotechnology: Principles and Processes.
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Physics
- Q1Two discs of the same moment of inertia I rotate about their regular axis (through centre, perpendicular to plane) with angular velocities ω₁ and ω₂. They are brought into contact face to face, coinciding their axes of rotation. The loss of energy in this process is:
- (1)/(4)I(ω₁ – ω₂)²
- (1)/(2)I(ω₁ + ω₂)²
- I(ω₁ – ω₂)²
- (1)/(8)I(ω₁ – ω₂)²
Answer: (A) (1)/(4)I(ω₁ – ω₂)²
Common final speed ω = (Iω₁ + Iω₂)/(2I) = (ω₁ + ω₂)/(2). Loss = (1)/(2)Iω₁² + (1)/(2)Iω₂² – (1)/(2)(2I)ω² = (1)/(4)I(ω₁ – ω₂)².
- Q2If a flywheel makes 120 rev/min, its angular speed is:
- 6π rad/s
- 4π rad/s
- 8π rad/s
- 2π rad/s
Answer: (B) 4π rad/s
ω = 2πν = 2π × (120)/(60) = 2π × 2 = 4π rad/s.
- Q3The frequency of the sinusoidal wave y=0.40cos(2000t+0.80) would be:
- (1000)/(π) Hz
- 2000 Hz
- 20 Hz
- 1000π Hz
Answer: (A) (1000)/(π) Hz
For y=acos(2π f t+φ), the angular frequency is 2π f=2000, so f=(2000)/(2π)=(1000)/(π) Hz.
- Q4Time intervals measured by a clock give the following readings: 1.25 s, 1.24 s, 1.27 s, 1.21 s and 1.28 s. What is the percentage relative error of the observations?
- 4%
- 1.6%
- 16%
- 2%
Answer: (B) 1.6%
Mean = 1.25 s, mean absolute error = 0.02 s, so percentage relative error = (0.02/1.25)×100 = 1.6%.
- Q5A mass of 0.5 kg on a spring of constant 200 N/m executes SHM. Its time period is approximately:
- 0.2π s
- 0.4π s
- 0.1π s
- 0.05π s
Answer: (C) 0.1π s
T = 2π√(m/k) = 2π√(0.5/200) = 2π√(0.0025) = 2π × 0.05 = 0.1π s ≈ 0.314 s.
- Q6An air bubble in a glass slab (refractive index 1.5, near-normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is:
- 10
- 16
- 12
- 8
Answer: (C) 12
For the two faces: (x)/(μ)+(t-x)/(μ)=3+5, i.e. (t)/(μ)=8. So t=8μ=8×1.5=12 cm.
- Q7A player takes 0.1 s in catching a ball of mass 150 g moving with velocity of 20 m/s. The force imparted by the ball on the hands of the player is
- 3 N
- 0.3 N
- 300 N
- 30 N
Answer: (D) 30 N
F = Δp/Δt = m(v₁ − v₂)/Δt = 0.150×(20−0)/0.1 = 30 N.
- Q8A stone is tied to a string of length l and is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is
- √(2gl)
- √(2(u² − gl))
- √(u² − gl)
- u − √(u² − 2gl)
Answer: (B) √(2(u² − gl))
At horizontal, speed u’ = √(u² − 2gl) and is perpendicular to u. |Δv| = √(u’² + u²) = √(2u² − 2gl) = √(2(u² − gl)).
- Q9In forward bias, the width of the depletion layer in a p-n junction diode:
- first increases then decreases
- increases
- decreases
- remains constant
Answer: (C) decreases
Forward bias applies a field opposing the built-in field, pushing majority carriers toward the junction and neutralising some of the immobile ions. The depletion layer narrows.
- Q10A heat engine has an efficiency of 25% and delivers 100 J of useful work per cycle. The heat it absorbs from the hot reservoir per cycle is:
- 300 J
- 400 J
- 500 J
- 250 J
Answer: (B) 400 J
η = (W)/(Q₁) ⇒ Q₁ = (W)/(η) = (100)/(0.25) = 400 J.
- Q11A battery of EMF 2 V and internal resistance 0.1 Ω is being charged by an external source that pushes 5 A through it. The terminal voltage across the battery during charging is:
- 1.5 V
- 2.5 V
- 2.0 V
- 3.0 V
Answer: (B) 2.5 V
During charging the current is forced into the cell, so V = ε + Ir = 2 + (5)(0.1) = 2.5 V. (During discharge it would be ε − Ir = 1.5 V – note the sign flip.)
- Q12If half life of an element is 69.3 h, then how much of its percent will decay in 10th to 11th h. Initial activity = 50 μCi
- 1%
- 3%
- 2%
- 4%
Answer: (A) 1%
A. Tests: converting a half-life into a decay constant and finding the fraction lost inside one specific hour. Why A: λ = (0.693)/(69.3) = 0.01 per hour. The surviving fraction at time t is e^(-λ t), so the fraction that disappears between t = 10 h and t = 11 h is e^(-0.10) – e^(-0.11) = 0.9048 – 0.8958 = 0.0090, that is about 1 percent of the original. The quoted 50 μCi never enters; it only fixes the scale. Why not B: 2 percent is what the shortcut loss ≈ λ × 1 h returns after doubling the decay constant to λ = 0.02 per hour, i.e. halving the half-life to (0.693)/(0.02) = 34.65 h; the stated 69.3 h fixes λ = 0.01 per hour. [misconception B: “λ = 2 × 0.693/t_(1/2)“] Why not C: no consistent substitution of 69.3 h into e^(-λ t) gives a 3 percent loss over a single hour. Even across the whole first 11 hours the sample sheds only 1 – e^(-0.11) = 10.4 percent, so one hour out of those eleven is worth about 1 percent, not 3. [misconception C: “the fraction lost in an hour grows as the sample ages”] Why not D: no consistent substitution gives 4 percent either; with λ = 0.01 per hour the loss in a single hour is e^(-λ t)(1 – e^(-λ)) = 0.905 × 0.00995 = 0.9 percent, and that shrinks, never grows, as t increases. [misconception D: “a decaying sample sheds a few percent every hour”] Remember: when λ t ≪ 1, the fraction lost in one hour is close to λ × 1 h, so here it is about 1 percent.
- Q13According to law of equipartition of energy the molar specific heat of a non rigid diatomic gas at constant volume is
- (3)/(2)R
- (9)/(2)R
- (5)/(2)R
- (7)/(2)R
Answer: (D) (7)/(2)R
D. Tests: counting degrees of freedom for a diatomic molecule that is allowed to vibrate. Why D: a non-rigid (vibrating) diatomic molecule has 3 translational + 2 rotational + 2 vibrational degrees of freedom, the last pair being the kinetic and potential energy of the single vibrational mode, so f = 7. Equipartition gives U = (f)/(2)RT per mole, and C_v = (dU)/(dT) = (7)/(2)R. Why not A: (9)/(2)R is Cₚ for this same vibrating diatomic gas, since Cₚ = C_v + R = (7)/(2)R + R; the question asks for constant volume [misconception A: ‘pick the largest capacity’]. Why not B: (5)/(2)R is the rigid diatomic C_v, ignoring the vibrational mode the word non-rigid switches on [misconception B: ‘diatomic always means f = 5‘]. Why not C: (3)/(2)R is the monatomic value, counting translation only [misconception C: ‘only translation contributes to internal energy’]. Remember: one vibrational mode adds 2 to f, not 1, because it stores both kinetic and potential energy.
- Q14A planet moves in an elliptical orbit about the Sun. If T, U, E and L are its kinetic energy, gravitational potential energy, total energy and the magnitude of angular momentum about the centre of force, which statement is correct?
- T is conserved
- U is always positive
- L is conserved but its direction changes continuously
- E is always negative
Answer: (D) E is always negative
For a bound elliptical orbit the total energy E = T + U < 0 (always negative). T varies (fastest at perihelion), U = -GMm/r is always negative, and L is conserved both in magnitude and direction (the orbit lies in a fixed plane). Only (C) is correct.
- Q15Which observation of the photoelectric effect could NOT be explained by the classical wave theory of light?
- Light carries energy
- Photocurrent increases with intensity
- Light can be reflected
- Existence of a threshold frequency
Answer: (D) Existence of a threshold frequency
Wave theory predicts that enough intensity should always cause emission, so the existence of a sharp threshold frequency (and instantaneous emission) could not be explained classically.
- Q16A gun recoils backward when a bullet is fired forward. This is most directly explained by the conservation of:
- Linear momentum
- Energy
- Charge
- Angular momentum
Answer: (A) Linear momentum
The gun-bullet system is isolated during firing, so total linear momentum is conserved; the forward momentum of the bullet is balanced by the backward (recoil) momentum of the gun.
- Q17Two long parallel wires are at a distance of 1 m. Both carry 1 A of current. The force of attraction per unit length between the two wires is:
- 2×10⁻⁷ N/m
- 2×10⁻⁸ N/m
- 5×10⁻⁸ N/m
- 10⁻⁷ N/m
Answer: (A) 2×10⁻⁷ N/m
(F)/(L) = (μ₀ I₁ I₂)/(2π d) = ((2×10⁻⁷)(1)(1))/(1) = 2×10⁻⁷ N/m. Same-direction currents attract.
- Q18Two samples A and B contain equal amount of radioactive substances. If ((1)/(8))ᵗʰ of sample A and ((1)/(128))ᵗʰ of sample B, remain after 9 hours, then the ratio of half life period of B to that of A is
- 7:3
- 3:7
- 9:7
- 3:1
Answer: (B) 3:7
C. Tests: reading half-lives off two different survival fractions over the same elapsed time. Why C: (1)/(8) = ((1)/(2))³, so sample A passed through 3 half-lives in 9 hours and T_A = (9)/(3) = 3 h. (1)/(128) = ((1)/(2))⁷, so sample B passed through 7 half-lives in 9 hours and T_B = (9)/(7) h. Hence (T_B)/(T_A) = (9/7)/(3) = (3)/(7), i.e. 3:7. Why not A: 9:7 is T_B = (9)/(7) h read straight off as if it were the answer, without dividing by T_A = 3 h. [misconception A: “the fraction you just computed is already the ratio”] Why not B: 7:3 is the same numbers the other way round, T_A : T_B; the question asks for B to A. [misconception B: “the order of a ratio does not matter”] Why not D: 3:1 would need T_B = 3T_A = 9 h, so sample B would pass through exactly one half-life in 9 hours and half of it would survive, not the stated (1)/(128). [misconception D: “the sample that decayed further has the longer half-life”] Remember: the sample that has decayed further has the shorter half-life, so B being down to (1)/(128) forces the B to A ratio below 1.
- Q19A charged wire is bent in the form of a semicircular arc of radius a. If the charge per unit length is λ C/m, the electric field at the centre O is:
- (λ)/(2πε₀ a)
- (λ)/(4π²ε₀ a)
- (λ)/(2π²ε₀ a)
- zero
Answer: (A) (λ)/(2πε₀ a)
By symmetry only the components along the axis of symmetry survive. An element dl=a dθ at angle θ gives a charge dq=λ a dθ, so dE=(1)/(4πε₀)(λ a dθ)/(a²)=(1)/(4πε₀)(λ dθ)/(a), with surviving component dEcosθ. Integrating θ from -(π)/(2) to +(π)/(2): E=(λ)/(4πε₀ a)[sinθ]_(-π/2)^(π/2)=(λ)/(4πε₀ a)·2=(λ)/(2πε₀ a).
- Q20The electric field due to an infinite charged plane sheet of surface charge density σ is:
- σ/ε₀, depends on distance
- Zero
- σ/(2ε₀r), depends on distance
- σ/(2ε₀), independent of distance
Answer: (D) σ/(2ε₀), independent of distance
Using Gauss’s law, an infinite sheet gives E = σ/(2ε₀), a uniform field independent of distance from the sheet.
- Q21Assertion (A): The intensity of secondary maxima in single-slit diffraction is much smaller than the central maximum. Reason (R): In a single slit, all interfering paths come from the same slit, so most light energy concentrates in the central maximum.
- A is true but R is false
- A is false but R is true
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
Answer: (C) Both A and R are true and R is the correct explanation of A
Secondary maxima are indeed far weaker (the first is only ~4.5% of the central intensity). This is because the bulk of the diffracted energy is concentrated in the central maximum, with side lobes carrying little energy. R correctly explains A.
- Q22A potential difference of 5 V is applied across a uniform wire of length L. The wire is now replaced by another of the same material but length 2L, and the same 5 V is applied. Compared with the first, the electron drift speed in the second wire is:
- unchanged
- doubled
- one-fourth
- halved
Answer: (D) halved
v_d = eEτ/m and E = V/L. With V fixed but L doubled, E halves, so v_d halves. (Drift speed tracks the field, which depends on V/L, not on area or current directly.)
- Q23A cell is balanced against 110 cm and 100 cm of a potentiometer wire respectively, without and with being short-circuited through a 10 Ω resistance. Its internal resistance is:
- zero
- 1.0 Ω
- 0.5 Ω
- 2.0 Ω
Answer: (B) 1.0 Ω
The open-circuit balance (110 cm) reads the EMF; the balance with the 10 Ω across the cell (100 cm) reads the terminal voltage. r = R((l₁ – l₂)/(l₂)) = 10×(110 – 100)/(100) = 10×0.1 = 1.0 Ω.
- Q24A simple pendulum of string length 30 cm performs 20 oscillations in 10 s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is ____ cm. [Assume that the mass of the pendulum remains same.]
- 120
- 15
- 0.75
- 7.5
Answer: (D) 7.5
B. Tests: the pendulum period varying as the square root of length, applied to a doubling of the oscillation count. Why B: 20 oscillations in 10 s gives T₁ = 0.5 s, and 40 oscillations in the same 10 s needs T₂ = 0.25 s, so T₂ = (T₁)/(2). From T = 2π√(L/g), L varies as T², so (L₂)/(L₁) = ((1)/(2))² = (1)/(4) and L₂ = (30)/(4) = 7.5 cm. Why not A: 0.75 cm is the correct answer with a decimal place lost; the required ratio L₂/L₁ is (1)/(4), which turns 30 cm into 7.5 cm, not into 0.75 cm. Why not C: 15 cm is (30)/(2), obtained by treating L as proportional to T instead of to T² [misconception C: “halve the period, halve the length”]. Why not D: 120 cm is 30 × 4, the right factor applied in the wrong direction; more oscillations in the same time means a faster, and therefore shorter, pendulum. Remember: double the frequency and the length must drop to a quarter, because T goes as √(L).
- Q25According to Curie’s law, the susceptibility of a paramagnetic material varies with absolute temperature T as:
- χ ∝ T²
- χ ∝ T
- χ ∝ 1/T
- χ is independent of T
Answer: (C) χ ∝ 1/T
Curie’s law: χ = C/T, so susceptibility is inversely proportional to absolute temperature for paramagnets.
- Q26A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of 30° with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 s, the average force acting on the wall is
- 12 N
- 48 N
- 24 N
- 96 N
Answer: (C) 24 N
Only the component perpendicular to the wall reverses: Δp = 2mv sin30°. F = Δp/t = (2×0.5×12×sin30°)/0.25 = 24 N.
- Q27An infinitely long straight wire has linear charge density λ. The electric field at a perpendicular distance r from the wire is proportional to:
- 1/r²
- r
- 1/r
- r²
Answer: (C) 1/r
For an infinite line charge, E = λ/(2πε₀r), so the field is inversely proportional to the distance r.
- Q28In an ac circuit, the instantaneous current is zero, when the instantaneous voltage is maximum. In this case, the source may be connected to : A. pure inductor. B. pure capacitor. C. pure resistor. D. combination of an inductor and capacitor. Choose the correct answer from the options given below :
- B, C and D only
- A and B only
- A, B and D only
- A, B and C only
Answer: (C) A, B and D only
C. Tests: recognising which elements give a π/2 phase difference between voltage and current. Why C: current zero exactly when voltage is maximum means the two waveforms are a quarter cycle apart, a phase difference of π/2. A pure inductor gives current lagging by π/2, a pure capacitor gives current leading by π/2, and a resistanceless L plus C combination still has a purely reactive impedance, so its phase difference is again ±π/2. Statements A, B and D therefore hold, which is option C. Why not A: the listed set includes the pure resistor, whose voltage and current are in phase, so its current is maximum, not zero, when the voltage peaks [misconception A: “every ac element shifts the phase”]. Why not B: it drops the inductor plus capacitor combination, which has zero resistance and so is also fully wattless with a π/2 shift. Why not D: it keeps the pure resistor, which is in phase, and drops the pure inductor, which is not. Remember: zero current at peak voltage means a purely reactive circuit, so no resistance anywhere in it.
- Q29For angles of projection of a projectile at angles (45° − θ) and (45° + θ), the horizontal ranges described by the projectile are in the ratio of
- 2 : 3
- 1 : 1
- 1 : 2
- 2 : 1
Answer: (B) 1 : 1
(45° − θ) and (45° + θ) are complementary angles (they add to 90°), and complementary angles give equal horizontal ranges, so the ratio is 1 : 1.
- Q30A ball is dropped from a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t=18 s. What is the value of v? (Take g=10 m/s²)
- 60 m/s
- 75 m/s
- 55 m/s
- 40 m/s
Answer: (B) 75 m/s
A. Tests: equating the displacements of two bodies released at different times under the same gravity. Why A: the first ball falls for the full 18 s, covering (1)/(2)(10)(18)²=1620 m. The second ball is thrown at t=6 s, so it falls for 12 s, covering 12v+(1)/(2)(10)(12)²=12v+720. They meet, so 12v+720=1620, giving 12v=900 and v=75 m/s. Relative-velocity check: at t=6 s the first ball is (1)/(2)(10)(6)²=180 m down and moving at 10(6)=60 m/s. Both then share the same g, so the gap closes at the constant rate v-60, and (180)/(v-60)=12 gives v=75 m/s. Why not B: at 55 m/s the second ball starts slower than the 60 m/s the first ball already has, so the gap grows at 5 m/s and the balls never meet. Why not C: at 40 m/s the gap grows at 20 m/s; the second ball falls further behind at every instant. Why not D: 60 m/s is the first ball’s speed at t=6 s, the moment of the throw. Launching at exactly that speed makes the closing rate zero, freezing the 180 m gap forever. [misconception D: “match the leading body’s speed and you catch it”] Remember: under equal gravity the closing rate is just the difference of the launch speeds, so meeting time is gap over that difference.
- Q31The energy of He⁺ ion in its first excited state is, (The ground state energy for the Hydrogen atom is -13.6 eV) :
- -13.6 eV
- -27.2 eV
- -3.4 eV
- -54.4 eV
Answer: (A) -13.6 eV
A. Tests: applying Eₙ = -(13.6 Z²)/(n²) eV to a hydrogen like ion in a named excited state. Why A: He⁺ has one electron and Z = 2, and the first excited state is n = 2. So E = -(13.6 × Z²)/(n²) = -(13.6 × 4)/(4) = -13.6 eV. The fourfold Z² deepening and the fourfold n² shallowing cancel exactly. Why not B: -27.2 eV is -13.6 × 2, which comes from scaling with Z instead of Z² while also dropping the n²; equivalently it is the He⁺ ground state -54.4 eV divided by n = 2 rather than by n² = 4 [misconception B: “energy scales as Z, and dividing by n instead of n²“]. Why not C: -3.4 eV is hydrogen’s own first excited state, -(13.6)/(4), obtained by forgetting the Z² = 4 factor entirely [misconception C: “treating He⁺ exactly like hydrogen”]. Why not D: -54.4 eV is the He⁺ ground state, -(13.6 × 4)/(1); the first excited state is n = 2, not n = 1 [misconception D: “first excited state means n = 1”]. Remember: First excited state is n = 2, and E = -(13.6Z²)/(n²), so He⁺ at n = 2 lands exactly on hydrogen’s ground state value.
- Q32Which experiment provided direct experimental confirmation of the wave nature of electrons?
- Davisson-Germer experiment
- Young’s double-slit experiment with light
- Rutherford’s scattering experiment
- Millikan’s oil-drop experiment
Answer: (A) Davisson-Germer experiment
The Davisson-Germer experiment showed diffraction of electrons by a nickel crystal, confirming that electrons (matter) behave as waves with λ = h/p.
- Q33An ideal p-n junction diode in series with a 1 kΩ resistor is operated such that the p-side is at +4 V and the n-side is at −6 V. The current through the diode is:
- 10⁻¹ A
- 10⁻³ A
- 0 A
- 10⁻² A
Answer: (D) 10⁻² A
The p-side (+4 V) is at higher potential than the n-side (−6 V), so the ideal diode is forward biased with zero drop. The full 4-(-6) = 10 V appears across the 1 kΩ resistor: I = 10/1000 = 10⁻² A.
- Q34In a parallel plate capacitor, the distance between the plates is d and the potential difference across the plates is V. The energy stored per unit volume between the plates is:
- (1)/(2)(ε₀ V²)/(d²)
- (Q²)/(2V)
- (1)/(2)(V²)/(ε₀ d²)
- (1)/(2)(ε₀ V)/(d)
Answer: (A) (1)/(2)(ε₀ V²)/(d²)
Energy density u=(1)/(2)ε₀ E² with E=(V)/(d), so u=(1)/(2)ε₀((V)/(d))²=(1)/(2)(ε₀ V²)/(d²).
- Q35In an AC circuit Vᵣₘₛ = 200 V, Iᵣₘₛ = 5 A, and the phase difference is 60°. The average power is:
- 866 W
- 250 W
- 500 W
- 1000 W
Answer: (C) 500 W
P = Vᵣₘₛ Iᵣₘₛ cos φ = 200 × 5 × cos 60° = 1000 × 0.5 = 500 W.
- Q36For high magnification, a compound microscope should have:
- small fₒ and small fₑ
- small fₒ and large fₑ
- large fₒ and small fₑ
- large fₒ and large fₑ
Answer: (A) small fₒ and small fₑ
Magnification m = (L/fₒ)(1 + D/fₑ); both small fₒ and small fₑ increase m. (Contrast: a telescope needs a large fₒ.)
- Q37The energy needed for breaking a liquid drop of radius ‘R‘ into ‘n‘ droplets each of radius ‘r‘ is [T = surface tension of the liquid]
- 4π T[1 + (R³)/(r)]
- 4π T[(R²)/(r²) – 1]
- 4π T R²[(R)/(r) – 1]
- 4π T R[(R)/(r) – 1]
Answer: (C) 4π T R²[(R)/(r) – 1]
A. Tests: volume conservation plus the increase in total surface area on fragmentation. Why A: volume conservation gives (4)/(3)π R³ = n(4)/(3)π r³, so n = (R³)/(r³). The energy required is T times the area increase: T(n · 4π r² – 4π R²) = 4π T((R³)/(r³)r² – R²) = 4π T((R³)/(r) – R²) = 4π T R²[(R)/(r) – 1]. Why not B: this carries only one power of R outside the bracket, so it has the units of surface tension times length, which is a force and not an energy [misconception B: ‘dropping a power of R while factorising’]. Why not C: R²/r² is a pure number, so the whole expression still carries the units of surface tension, force per unit length, and not of energy [misconception C: ‘the ratio of areas is itself the energy’]. Why not D: it adds the two surface energies instead of subtracting them, and the bracket is not even self-consistent, since the 1 is a pure number while (R³)/(r) is an area, so the two terms cannot be added [misconception D: ‘total energy is the sum of the before and after surface energies’]. Remember: work = TΔ A = 4π T R²(n^(1/3) – 1), and n^(1/3) = R/r.
- Q38An electric kettle is rated ‘1000 W, 250 V’. The current it draws at rated operation, and hence the minimum fuse rating suitable for it, is:
- 2.5 A
- 0.25 A
- 4 A
- 8 A
Answer: (C) 4 A
I = P/V = 1000/250 = 4 A. The fuse must carry at least this, so a 4 A (or slightly higher) fuse is required.
- Q39Water stands at a height H in a tank. A small hole is made at depth h below the free surface. The speed of efflux of water from the hole is:
- √(2gH)
- √(g h / 2)
- 2gh
- √(2gh)
Answer: (D) √(2gh)
By Torricelli’s theorem the efflux speed is v = √(2gh), where h is the depth of the hole below the free surface.
- Q40A flywheel rotating about a fixed axis has a kinetic energy of 360 J when its angular speed is 30 rad/s. The moment of inertia of the wheel about the axis of rotation is:
- 0.75 kg·m²
- 0.6 kg·m²
- 0.8 kg·m²
- 0.15 kg·m²
Answer: (C) 0.8 kg·m²
KE = (1)/(2)Iω² ⇒ I = (2 × KE)/(ω²) = (2 × 360)/(30²) = (720)/(900) = 0.8 kg·m².
- Q41A hospital uses an ultrasonic scanner of operating frequency 4.2 MHz to locate tumours in a tissue. If the speed of sound in the tissue is 1.7 km s⁻¹, the wavelength of sound in the tissue is close to:
- 4×10⁻³ m
- 8×10⁻³ m
- 8×10⁻⁴ m
- 4×10⁻⁴ m
Answer: (D) 4×10⁻⁴ m
λ=(v)/(f)=(1.7×10³)/(4.2×10⁶)≈4.05×10⁻⁴≈4×10⁻⁴ m.
- Q42In dispersion through a prism, the colour that deviates the most is:
- green
- yellow
- violet
- red
Answer: (C) violet
Violet has the highest refractive index (shortest wavelength), so it bends the most; red bends the least.
- Q43A metal rod of length L rotates about one end at the origin with a uniform angular velocity ω. The magnetic field falls off radially as B(r) = B₀ e^(-λ r), λ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is:
- B₀ω[(3)/(λ²) – e^(-3λ L)((3)/(λ²) + (L)/(λ))]
- B₀ω[(4)/(λ²) – e^(-2λ L)((1)/(λ²) + (2L)/(λ))]
- B₀ω[(1)/(λ²) + e^(-λ L)((1)/(λ²) + (L)/(λ))]
- B₀ω[(1)/(λ²) – e^(-λ L)((1)/(λ²) + (L)/(λ))]
Answer: (D) B₀ω[(1)/(λ²) – e^(-λ L)((1)/(λ²) + (L)/(λ))]
A. Tests: integrating the motional emf of a rotating rod in a non-uniform field. Why A: An element dr at radius r moves with speed v = ω r, contributing dε = B(r) ω r dr. So ε = B₀ω ∫₀^L r e^(-λ r) dr. Integration by parts: ∫₀^L r e^(-λ r) dr = [-(r)/(λ)e^(-λ r)]₀^L + (1)/(λ)∫₀^L e^(-λ r) dr = (1)/(λ²) – e^(-λ L)((1)/(λ²) + (L)/(λ)), giving option A. Check: at L = 0 the emf is zero, as required. Why not B: fails the limit check (L → 0 gives 2B₀ω/λ² ≠ 0) [misconception B: “sign slip in the integration-by-parts boundary term”] Why not C: wrong coefficients and a spurious e^(-2λ L) [misconception C: “squaring the exponential while combining terms”] Why not D: invented coefficients; the correct antiderivative has unit coefficients [misconception D: “pattern-matching the algebra without doing the integral”] Remember: rotating rod in field B(r): ε = ω∫₀^L r B(r) dr; for uniform B this reduces to (1)/(2)Bω L².
- Q44Wire A carries 10 A and wire B carries 5 A, parallel and antiparallel respectively at distance d. The magnitude of force per length on A versus B:
- Force on A is twice that on B
- Force on B is twice that on A
- Force on A is zero
- Forces are equal in magnitude
Answer: (D) Forces are equal in magnitude
By Newton’s third law (and the symmetric formula μ₀I₁I₂/2πd), the forces on the two wires are equal and opposite in magnitude regardless of the unequal currents.
- Q45A monochromatic EM wave has a frequency of 5 × 10¹⁴ Hz. Its wavelength in vacuum is approximately, and to which region does it belong?
- 6000 nm, infrared
- 60 nm, ultraviolet
- 600 nm, visible
- 6 nm, X-ray
Answer: (C) 600 nm, visible
λ = c/f = 3×10⁸ / 5×10¹⁴ = 6×10⁻⁷ m = 600 nm, which lies in the visible region.
Chemistry
- Q46In the manufacture of bromine from sea water, the mother liquor containing bromide is treated with:
- carbon dioxide
- iodine
- sulphur dioxide
- chlorine
Answer: (D) chlorine
Chlorine oxidises bromide to bromine: Cl₂ + 2Br⁻ → Br₂ + 2Cl⁻. So the bromide mother liquor is treated with chlorine.
- Q47The [H⁺] of a 0.1 M weak acid solution with Ka = 1.0 × 10⁻⁵ is:
- 1.0 × 10⁻⁵ M
- 3.2 × 10⁻⁴ M
- 1.0 × 10⁻³ M
- 1.0 × 10⁻⁶ M
Answer: (C) 1.0 × 10⁻³ M
[H⁺] = √(Ka·C) = √(10⁻⁵ × 0.1) = √(10⁻⁶) = 10⁻³ M.
- Q48Given below are four compounds: (a) n-propyl chloride (b) iso-propyl chloride (c) sec-butyl chloride (d) neo-pentyl chloride Percentage of carbon in the one which exhibits optical isomerism is:
- 40
- 56
- 46
- 52
Answer: (D) 52
C. Tests: identifying the chiral compound first, then converting its formula into a mass percentage. Why C: Step 1, find the stereocentre. n-propyl chloride CH₃CH₂CH₂Cl has its Cl on a CH₂, iso-propyl chloride (CH₃)₂CHCl has two identical methyls on the C-Cl carbon, and neo-pentyl chloride (CH₃)₃CCH₂Cl again has Cl on a CH₂, so none of these three is chiral. sec-butyl chloride is 2-chlorobutane, CH₃-CHCl-CH₂CH₃, and its C2 carries four different groups, H, Cl, CH₃ and C₂H₅, so it is optically active. Step 2, compute the carbon percentage of C₄H₉Cl: molar mass = 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 g/mol, so %C = (48/92.5) x 100 = 51.9, which rounds to 52. Why not A: 46 is the carbon percentage of a propyl chloride, C₃H₇Cl, molar mass 36 + 7 + 35.5 = 78.5, giving (36/78.5) x 100 = 45.9, so it is what you get by picking (a) or (b) as the chiral one. Why not B: no consistent substitution gives this, none of the four chlorides has a carbon percentage near 40. Why not D: 56 is the carbon percentage of neo-pentyl chloride, C₅H₁₁Cl, molar mass 60 + 11 + 35.5 = 106.5, giving (60/106.5) x 100 = 56.3, but its C-Cl carbon is a CH₂ with two identical hydrogens, so it has no stereocentre [misconception: “a bulky branched halide must be chiral”]. Remember: optical activity needs four DIFFERENT groups on one carbon, and sec-butyl is the smallest simple alkyl chloride that manages it.
- Q49IUPAC name of glycerol is:
- Propene-1,2,3-triol
- Propane-1,2-diol
- Propane-1,2,3-triol
- Propane-1,3-diol
Answer: (C) Propane-1,2,3-triol
C. Tests: recall of the IUPAC name behind a common trivial name of a polyhydric alcohol. Why C: Glycerol is CH₂(OH)-CH(OH)-CH₂(OH), a saturated three carbon chain carrying one -OH on every carbon, so the parent is propane, three hydroxyls give the suffix triol, and the locants are 1,2,3: propane-1,2,3-triol. Why not A: Propane-1,2-diol is propylene glycol and has only two -OH groups, so it misses the third hydroxyl of glycerol [misconception A: “glycerol and glycol are the same compound”]. Why not B: Propane-1,3-diol is again a diol, and it also leaves the middle carbon C-2 without a hydroxyl. Why not D: Propene-1,2,3-triol would need a C=C inside a three carbon chain that already carries three -OH groups, but glycerol is fully saturated and an -OH on a doubly bonded carbon is an unstable enol. Remember: glycerol equals glycerine equals propane-1,2,3-triol, three carbons and three -OH groups.
- Q50Glycolysis is:
- Conversion of glucose to haem
- Oxidation of glucose to glutamate
- Oxidation of glucose to pyruvate
- Conversion of pyruvate to citrate
Answer: (C) Oxidation of glucose to pyruvate
Glycolysis is the first, anaerobic stage of glucose breakdown: one glucose molecule is oxidised to two molecules of pyruvate, generating ATP. It does not produce haem, glutamate, or citrate (citrate forms later, in the Krebs cycle). Hence oxidation of glucose to pyruvate.
- Q51The extra stability of a complex due to ring formation by a polydentate ligand is called the:
- Trans effect
- Jahn-Teller effect
- Chelate effect
- Inert pair effect
Answer: (C) Chelate effect
The chelate effect is the enhanced stability when a bidentate or polydentate ligand forms a ring with the metal (e.g., en, EDTA). More rings give greater stability.
- Q52The rate constants for the forward and backward reactions of ester hydrolysis are 1.1 × 10⁻² and 1.5 × 10⁻³ per minute respectively. The equilibrium constant for the reaction CH₃COOC₂H₅ + H⁺ leftharpoons CH₃COOH + C₂H₅OH is:
- 7.33
- 6.33
- 5.33
- 4.33
Answer: (A) 7.33
At equilibrium K = (k_f)/(k_b) = (1.1 × 10⁻²)/(1.5 × 10⁻³) = (11)/(1.5) = 7.33.
- Q53The oxidation state of nitrogen is NOT +3 in which of the following?
- NH₂OH
- NaNO₂
- N₂O₃
- HNO₂
Answer: (A) NH₂OH
In NH₂OH: N + 2(+1) + (−2) + (+1) = 0 → N = −1. In HNO₂, NaNO₂ and N₂O₃ nitrogen is +3. So NH₂OH is the exception.
- Q54A solution is prepared by dissolving 18 g of glucose (molar mass 180) in 178.2 g of water. The molality of the solution is:
- 0.18 m
- 1.0 m
- 0.10 m
- 0.56 m
Answer: (D) 0.56 m
Moles glucose = 18/180 = 0.1 mol. Mass of solvent = 178.2 g = 0.1782 kg. Molality = 0.1/0.1782 = 0.561 m ≈ 0.56 m.
- Q55The reaction A₂ + B₂ arrow 2AB follows the mechanism: Step 1 (fast equilibrium): A₂ arrow A + A with forward rate constant k₁ and reverse rate constant k₋₁. Step 2 (slow): A + B₂ arrow AB + B with rate constant k₂. Step 3 (fast): A + B arrow AB. The overall order of the reaction is:
- 2.5
- 3
- 2
- 1.5
Answer: (D) 1.5
C. Tests: deriving a rate law from a mechanism whose slow step is preceded by a fast dissociation equilibrium. Why C: the slow step sets the rate, so rate = k₂[A][B₂]. The intermediate A is fixed by the fast equilibrium, k₁[A₂] = k₋₁[A]², which gives [A] = ((k₁)/(k₋₁))^(1/2)[A₂]^(1/2). Substituting, rate = k₂((k₁)/(k₋₁))^(1/2)[A₂]^(1/2)[B₂]. The overall order is (1)/(2) + 1 = 1.5. Why not A: 3 is what you get by adding the molecularity of the fast dissociation step, 1, to the molecularity of the slow step, 2. Molecularities of separate steps never add; the order comes from the derived rate law, whose exponents are (1)/(2) on [A₂] and 1 on [B₂], totalling 1.5 [misconception: “overall order is the sum of the molecularities of the steps”]. Why not B: 2.5 would need [A₂]^(3/2)[B₂] or [A₂]^(1/2)[B₂]², but the slow step is first order in B₂ and the equilibrium supplies only the half power of A₂ [misconception: “the fast equilibrium adds a full order on top of the half order”]. Why not D: 2 is the order you read straight off the slow step, rate = k₂[A][B₂], counting A as an ordinary reactant; but A is an intermediate and must be replaced by [A₂]^(1/2) before the order can be quoted [misconception: “leave the intermediate in the final rate law”]. Remember: a pre-equilibrium that splits a molecule into two identical fragments always injects a square root, and that is where fractional orders come from.
- Q56For a given reaction, Δ H = 35.5 kJ mol⁻¹ and Δ S = 83.6 J K⁻¹ mol⁻¹. The reaction is spontaneous at (assume Δ H and Δ S do not vary with temperature):
- T > 425 K
- all temperatures
- T < 425 K
- T > 298 K
Answer: (A) T > 425 K
Both Δ H and Δ S are positive, so spontaneity needs TΔ S > Δ H ⇒ T > Δ H/Δ S = 35500/83.6 ≈ 425 K.
- Q57Arrange the following compounds according to increasing order of boiling points: n-C₄H₉OH (A), n-C₄H₉NH₂ (B), n-C₄H₁₀ (C) and C₂H₅NHC₂H₅ (D).
- D < B < A < C
- D < C < B < A
- C < B < A < D
- C < D < B < A
Answer: (D) C < D < B < A
C. Tests: ranking boiling points by the number and the strength of the intermolecular hydrogen bonds available. Why C: butane (C) has only weak dispersion forces and boils near 0° C, so it must be lowest. Among the hydrogen bonders, diethylamine (D) is a secondary amine with a single N-H, so it forms the fewest hydrogen bonds and boils near 56° C. n-Butylamine (B) is primary with two N-H bonds and boils near 78° C. n-Butanol (A) hydrogen bonds through O-H, and oxygen is more electronegative than nitrogen, so its hydrogen bonds are the strongest and it boils near 118° C. The increasing order is C < D < B < A. Why not A: it puts B below A correctly but then places D, also an amine, above the alcohol, which contradicts O-H hydrogen bonding being stronger than N-H [misconception A: “the bigger molecule always boils higher, so diethylamine tops butanol”]. Why not B: it places diethylamine below butane, yet D still has an N-H and a large permanent dipole while butane has neither [misconception B: “a dialkylated amine behaves like an alkane”]. Why not D: it ends with butane as the highest boiling compound, reversing the entire hydrogen bonding argument [misconception D: “alkanes boil high because they are heavy and non-polar”]. Remember: no H-bond < one N-H < two N-H < one O-H, so alkane < secondary amine < primary amine < alcohol.
- Q58Among CH₃⁺ (methyl), (CH₃)₂CH⁺ (isopropyl), (CH₃)₃C⁺ (tert-butyl) and CH₃CH₂⁺ (ethyl), the most stable carbocation is:
- (CH₃)₃C⁺ (tert-butyl)
- CH₃⁺ (methyl)
- (CH₃)₂CH⁺ (isopropyl)
- CH₃CH₂⁺ (ethyl)
Answer: (A) (CH₃)₃C⁺ (tert-butyl)
Carbocation stability increases with electron donation by +I and hyperconjugation. The tert-butyl cation has three methyl groups (nine alpha C-H bonds) feeding the positive carbon, so it is the most stable: 3° > 2° > 1° > CH₃⁺.
- Q59A compound decolourises both cold dilute alkaline KMnO₄ (Baeyer’s reagent) and bromine water. The compound is most likely a(n):
- Saturated alcohol
- Unsaturated hydrocarbon (alkene/alkyne)
- Aromatic hydrocarbon
- Alkane
Answer: (B) Unsaturated hydrocarbon (alkene/alkyne)
Baeyer’s reagent and bromine water are decolourised by C=C/C≡C unsaturation. Alkanes and benzene (under mild conditions) do not react, so decolourisation of both indicates an alkene or alkyne.
- Q60In the fluorite (CaF₂) structure, Ca²⁺ ions form an FCC arrangement and F⁻ ions occupy:
- All tetrahedral voids
- Half the tetrahedral voids
- Half the octahedral voids
- All octahedral voids
Answer: (A) All tetrahedral voids
In CaF₂, Ca²⁺ is FCC (4 per cell) and the 8 F⁻ fill all tetrahedral voids, giving the 1:2 ratio and 8:4 coordination.
- Q61Which one of the following orders is NOT in accordance with the property stated against it?
- F₂ > Cl₂ > Br₂ > I₂ : bond dissociation energy
- HI > HBr > HCl > HF : acidic property in water
- F₂ > Cl₂ > Br₂ > I₂ : electronegativity
- F₂ > Cl₂ > Br₂ > I₂ : oxidising power
Answer: (A) F₂ > Cl₂ > Br₂ > I₂ : bond dissociation energy
Bond dissociation energy is NOT F₂ > Cl₂ > Br₂ > I₂; the actual order is Cl₂ > Br₂ > F₂ > I₂ (F₂ anomalously low). The other three orders are correct.
- Q62Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to a gaseous hydrocarbon containing fewer than four carbon atoms. (A) is
- CH₃-CH₃
- HC≡CH
- CH₄
- CH₂=CH₂
Answer: (C) CH₄
D. Tests: combining reaction-type logic (substitution vs addition) with the carbon-doubling arithmetic of the Wurtz reaction. Why D: Only a saturated hydrocarbon reacts with Br₂ by substitution. CH₄ gives CH₃Br, and Wurtz coupling joins two alkyl groups: 2 CH₃Br + 2 Na arrow CH₃-CH₃. Ethane has 2 carbons (< 4) and is a gas. All conditions fit. Why not A: Ethyne has a C≡C, so it ADDS bromine rather than substituting; it fails the first condition [misconception A: “all hydrocarbons react with Br2 the same way”] Why not B: Ethene likewise undergoes addition across the C=C, not substitution [misconception B: “substitution and addition are interchangeable words”] Why not C: Ethane does substitute (C₂H₅Br), but Wurtz then gives butane, which has exactly 4 carbons, not fewer than four [misconception C: “Wurtz adds one carbon instead of doubling”] Remember: Wurtz doubles the alkyl carbon count: R-X (n C) gives R-R (2n C).
- Q63Deficiency of which vitamin causes osteomalacia?
- Vitamin D
- Vitamin A
- Vitamin E
- Vitamin K
Answer: (A) Vitamin D
Vitamin D deficiency causes rickets in children and osteomalacia (softening of bones) in adults. Vitamin A deficiency arrow xerophthalmia/night blindness, vitamin E arrow fragility of RBCs/muscle weakness, vitamin K arrow increased clotting time. Hence vitamin D.
- Q64A one-litre flask is full of brown bromine vapours. The intensity of the brown colour will NOT decrease appreciably on adding to the flask some:
- animal charcoal powder
- carbon tetrachloride
- pieces of marble
- carbon disulphide
Answer: (C) pieces of marble
Animal charcoal adsorbs bromine; CCl₄ and CS₂ dissolve bromine – all three remove Br₂ vapour and lighten the colour. Marble (CaCO₃) neither adsorbs nor dissolves bromine, so the brown colour does not decrease.
- Q65The most appropriate reagent for conversion of C₂H₅CN into CH₃CH₂CH₂NH₂ is
- Na(CN)BH₃
- CaH₂
- LiAlH₄
- NaBH₄
Answer: (C) LiAlH₄
D. Tests: choosing a hydride source strong enough to reduce a nitrile all the way to a primary amine. Why D: Going from C₂H₅CN to CH₃CH₂CH₂NH₂ keeps all three carbons and adds four hydrogens across the C-N triple bond. LiAlH₄ is a powerful, non-selective hydride donor that reduces nitriles cleanly to primary amines, exactly the transformation asked for (catalytic H₂/Ni or Na in ethanol would do the same job). Why not A: NaBH₄ is a mild hydride that reduces aldehydes and ketones but leaves the far less electrophilic nitrile carbon untouched [misconception A: “any borohydride reduces anything with a multiple bond to carbon”]. Why not B: CaH₂ is used as a drying agent; it reacts violently with water rather than delivering hydride to an organic substrate. Why not C: Na(CN)BH₃ is milder and more selective still, used for reductive amination of iminium ions at controlled pH, and it does not touch nitriles. Remember: nitrile to primary amine needs LiAlH₄, H₂/Ni or Na in ethanol; NaBH₄ is too weak.
- Q66The correct order of second ionization enthalpy for Li, Be, B and C is:
- Li > C > B > Be
- Be > Li > B > C
- C > B > Be > Li
- Li > Be > B > C
Answer: (A) Li > C > B > Be
After losing the first electron, Li⁺ has a stable noble-gas (He) configuration, so removing a second electron is extremely difficult, giving Li the highest second IE. Be⁺ readily loses its second electron to reach the stable 1s² configuration, giving the lowest. Hence order: Li > C > B > Be.
- Q67The iodoform test is NOT given by:
- Ethanal
- 2-pentanone (pentan-2-one)
- Ethanol
- 3-pentanone (pentan-3-one)
Answer: (D) 3-pentanone (pentan-3-one)
A positive iodoform needs a CH₃-CO- group or a CH₃-CH(OH)- group (oxidised to CH₃CO- in situ). Pentan-2-one (CH₃COCH₂CH₂CH₃) has CH₃CO-; ethanol → ethanal → CH₃CO-; ethanal is CH₃CHO. Pentan-3-one (CH₃CH₂-CO-CH₂CH₃) has the carbonyl flanked by ethyl groups, no CH₃CO-, so it gives no iodoform.
- Q68According to VSEPR, which repulsion is the strongest?
- all are equal
- lone pair - bond pair
- bond pair - bond pair
- lone pair - lone pair
Answer: (D) lone pair - lone pair
Lone pairs are held only by one nucleus and occupy more space, so lone pair-lone pair repulsion is greatest: lp-lp > lp-bp > bp-bp.
- Q69Which common name is correctly matched with its formula?
- Formic acid - CH₃COOH
- Formaldehyde - HCHO
- Acetic acid - HCOOH
- Acetone - CH₃CHO
Answer: (B) Formaldehyde - HCHO
HCHO is formaldehyde. CH₃CHO is acetaldehyde, CH₃COCH₃ is acetone, HCOOH is formic acid and CH₃COOH is acetic acid.
- Q70The hybridisation involved in the complex [Ni(CN)₄]²⁻ is (atomic number of Ni = 28):
- d²sp³
- dsp²
- d²sp²
- sp³
Answer: (B) dsp²
Ni²⁺ is 3d⁸. CN⁻ is a strong field ligand, so the d-electrons pair up, vacating one 3d orbital. With one (n−1)d + s + two p orbitals the hybridisation is dsp² → square planar, diamagnetic.
- Q71[Cr(H₂O)₆]Cl₃ (at. no. of Cr = 24) has a magnetic moment of 3.83 BM. The number of unpaired electrons in the chromium ion is:
- 3
- 2
- 1
- 4
Answer: (A) 3
μ = √[n(n+2)] = 3.83 ⟹ n(n+2) ≈ 14.67 ⟹ n = 3. Indeed Cr in this complex is +3 (Cr³⁺ = 3d³), which has 3 unpaired electrons, consistent with 3.83 BM.
- Q72Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Benzene is more stable than hypothetical cyclohexatriene Reason R : The delocalised π electron cloud is attracted more strongly by nuclei of carbon atoms. In the light of the above statements, choose the correct answer from the options given below:
- Both A and R are correct but R is NOT the correct explanation of A
- A is true but R is false
- Both A and R are correct and R is the correct explanation of A
- A is false but R is true
Answer: (C) Both A and R are correct and R is the correct explanation of A
B. Tests: judging whether delocalisation is genuinely the cause of benzene’s extra stability, not merely a true fact about it. Why B: The assertion is true. Cyclohexene releases about 120 kJ per mol on hydrogenation, so a hypothetical cyclohexatriene with three isolated double bonds should release about 3 × 120 = 360 kJ per mol. Benzene actually releases only about 208 kJ per mol, so it starts 360 – 208 = 152 kJ per mol lower than the hypothetical triene; that gap is its resonance energy. The reason is true as well, and it is the cause of that gap: the six π electrons are not locked into three separate double bonds but spread over all six carbons, so each electron is attracted by more than one carbon nucleus, which lowers the energy of the system. Since delocalisation is exactly why benzene lies below cyclohexatriene, R correctly explains A. Why not A: R is not false; delocalisation of the π cloud over all six carbons is the accepted textbook explanation of benzene’s resonance energy [misconception A: “benzene is stable only because a six membered ring is strain free”]. Why not C: R is not a true but unrelated fact; the extra nuclear attraction felt by the delocalised electrons is the direct source of the stabilisation stated in A, so it does explain it [misconception C: “stability and delocalisation are two separate facts about benzene”]. Why not D: A is not false; benzene is measurably more stable than cyclohexatriene, which is precisely why cyclohexatriene remains hypothetical [misconception D: “benzene really is cyclohexatriene with alternating single and double bonds”]. Remember: resonance energy is measured as predicted minus observed heat of hydrogenation, and for benzene that gap of about 152 kJ per mol comes from the delocalised π cloud being held by all six nuclei at once.
- Q73For the cell reaction 2Fe³⁺(aq) + 2I⁻(aq) arrow 2Fe²⁺(aq) + I₂(aq) E^Θ_(cell) = 0.24 V at 298 K. The standard Gibbs energy (Δᵣ Gᵒ) of the cell reaction is : [Given that Faraday constant F = 96500 C mol⁻¹]
- 46.32 kJ mol⁻¹
- -23.16 kJ mol⁻¹
- -46.32 kJ mol⁻¹
- 23.16 kJ mol⁻¹
Answer: (C) -46.32 kJ mol⁻¹
C. Tests: counting n from a balanced cell reaction and applying Δᵣ Gᵒ = -nFE^Θ_(cell) with the right sign. Why C: two Fe³⁺ ions each gain one electron, so n = 2. Then Δᵣ Gᵒ = -nFE^Θ = -2 × 96500 × 0.24 = -46320 J mol⁻¹ = -46.32 kJ mol⁻¹. The negative sign is forced by the positive EMF, which marks the reaction as spontaneous. Why not A: the magnitude is right but the sign is positive, which would make a cell with positive EMF non-spontaneous, a contradiction [misconception A: “drop the minus sign in -nFE because energies are quoted as positive numbers”]. Why not B: +23.16 uses n = 1 and also loses the minus sign, so it carries both errors at once. Why not D: -23.16 has the correct sign but n = 1; the balanced equation shows two electrons transferred, so the magnitude must double [misconception D: “n is the electrons in one ion’s half change, not in the balanced reaction”]. Remember: Δᵣ Gᵒ = -nFE^Θ_(cell), n comes from the balanced equation, and positive EMF always means negative Δ Gᵒ.
- Q74When phenol is treated with excess of bromine water, it gives:
- 2,4,6-tribromophenol
- m-bromophenol
- 2,4-dibromophenol
- a mixture of o- and p-bromophenols
Answer: (A) 2,4,6-tribromophenol
Bromine water is polar and the -OH group strongly activates the ring, so all three free ortho/para positions are brominated, precipitating 2,4,6-tribromophenol (white solid). Mono-bromination (mainly para) requires a non-polar solvent like CS2 at low temperature.
- Q75Which statement about molecularity is INCORRECT?
- It can be zero
- It is the number of species colliding
- It is always a whole number
- It refers to an elementary reaction
Answer: (A) It can be zero
Molecularity is the number of reacting species in an elementary step; it is a positive integer (1, 2, or 3) and can never be zero or fractional. Order, not molecularity, can be zero or fractional.
- Q76A first order reaction is 99% complete in time t. In terms of the half-life t₁/₂, the time t for 99% completion of a first-order reaction is closest to:
- about 6.6 t₁/₂
- about 2 t₁/₂
- about 4.6 t₁/₂
- about 10 t₁/₂
Answer: (A) about 6.6 t₁/₂
For 99% completion, t = (2.303/k)log 100 = (2.303/k)(2) = 4.606/k. Since t₁/₂ = 0.693/k, t/t₁/₂ = 4.606/0.693 = 6.65. So 99% completion takes about 6.6 half-lives.
- Q771 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to
- Zero mg
- 250 mg
- 750 mg
- 200 mg
Answer: (B) 250 mg
B. Tests: identifying the limiting reagent in a 1:1 neutralisation and reporting the leftover mass. Why B: moles of NaOH = 1/40 = 0.025 mol. Moles of HCl = 0.025 L × 0.75 mol L⁻¹ = 0.01875 mol. NaOH and HCl react 1:1, so HCl is limiting and only 0.01875 mol of NaOH is consumed. Left over = 0.025 – 0.01875 = 0.00625 mol, mass = 0.00625 × 40 = 0.25 g = 250 mg. Why not A: 750 mg is the NaOH that reacted, 0.01875 × 40 = 0.75 g, not the part left behind [misconception A: “answering with the consumed amount instead of the excess”]. Why not C: zero would need at least 0.025 mol of HCl, that is 33.3 mL of the 0.75 M acid; only 25 mL was added, so acid runs out first. Why not D: 200 mg is 0.005 mol of NaOH; no consistent substitution of 1 g, 25 mL and 0.75 M leaves that amount. Remember: leftover = initial moles minus moles consumed, converted back to grams only at the very end.
- Q78The correct order of atomic radii in group 13 elements is:
- B < Ga < Al < In < Tl
- B < Ga < Al < Tl < In
- B < Al < Ga < In < Tl
- B < Al < In < Ga < Tl
Answer: (A) B < Ga < Al < In < Tl
Radii increase down the group, but gallium shows an anomaly: the poorly shielding 3d¹⁰ electrons make Ga smaller than Al. Thus B (85 pm) < Ga (135 pm) < Al (143 pm) < In (167 pm) < Tl (170 pm).
- Q79A reaction is spontaneous at all temperatures when:
- ΔH > 0 and ΔS > 0
- ΔH < 0 and ΔS < 0
- ΔH > 0 and ΔS < 0
- ΔH < 0 and ΔS > 0
Answer: (D) ΔH < 0 and ΔS > 0
ΔG = ΔH – TΔS. If ΔH is negative and ΔS is positive, then -TΔS is also negative, so ΔG < 0 at every temperature – always spontaneous.
- Q80For an endothermic reaction at equilibrium, increasing the temperature will:
- Decrease K and shift backward
- Not change K
- Increase K and shift forward
- Shift toward fewer gas moles only
Answer: (C) Increase K and shift forward
For endothermic reactions, heat acts as a reactant; raising temperature shifts the equilibrium forward and increases K.
- Q81In which compound will C-Cl bond ionisation give the MOST stable carbonium ion (carbocation)?
- O₂N-CH₂-CH₂Cl
- CH₃CH(CH₃)Cl (isopropyl chloride)
- C₆H₅CH₂Cl (benzyl chloride)
- (CH₃)₃CCl (tert-butyl chloride)
Answer: (C) C₆H₅CH₂Cl (benzyl chloride)
Benzyl chloride ionises to the benzyl cation, which is stabilised by resonance with the aromatic ring (charge delocalised over the ring). This resonance stabilisation outweighs the hyperconjugative stabilisation of the tert-butyl (3°) cation, so the benzyl cation is the most stable. The 2° isopropyl and the -NO₂-destabilised cation are far less stable.
- Q82If α is the fraction of HI dissociated at equilibrium in the reaction 2HI(g) leftharpoons H₂(g) + I₂(g), starting with 2 moles of HI, then the total number of moles of reactants and products at equilibrium are:
- 2 + 2α
- 2
- 1 + α
- 2 – α
Answer: (B) 2
Start: 2 mol HI. At equilibrium HI = 2 – 2α, H₂ = α, I₂ = α. Total = (2 – 2α) + α + α = 2. Since Δ n_(gas) = 0 for this reaction, total moles stay constant at 2.
- Q83Find the mass of sodium carbonate in grams to form 100 ml 0.1 M solution in water? ( Molar mass of Na₂CO₃ = 106 g/mol)
- 3.18 g
- 1.06 g
- 2.12 g
- 4.16 g
Answer: (B) 1.06 g
A. Tests: preparing a solution of stated molarity, that is molarity x volume x molar mass. Why A: Moles required = M x V(in litres) = 0.1 x 0.100 = 0.01 mol. Mass = 0.01 x 106 = 1.06 g. Why not B: 3.18 g is 0.03 mol x 106, three times the moles actually needed, which would suit 300 mL of 0.1 M. Why not C: 2.12 g is 0.02 mol x 106, twice the moles needed, which would suit 100 mL of 0.2 M [misconception C: “the 2 in Na₂CO₃ doubles the mass required”]. Why not D: 4.16 g is 4.16/106 = 0.0392 mol, which is not any product of the given 0.1 M and 0.100 L; no consistent substitution gives it. Remember: grams needed = molarity x litres x molar mass, so 100 mL of a 0.1 M solution is always one hundredth of the gram molar mass.
- Q84The freezing point depression of a 0.5 molal aqueous solution of a non-electrolyte (K_f of water = 1.86 K kg/mol) is:
- 0.5 K
- 1.86 K
- 0.93 K
- 3.72 K
Answer: (C) 0.93 K
ΔT_f = K_f × m = 1.86 × 0.5 = 0.93 K.
- Q85The number of unpaired electrons in the ground state of Fe (Z=26) is:
- 4
- 2
- 6
- 3
Answer: (A) 4
Fe is [Ar]3d⁶4s². The 3d⁶ subshell, by Hund’s rule, has 5 orbitals filled as one paired and four singly occupied → 4 unpaired electrons; 4s² is fully paired.
- Q86Which of the following has the maximum bond angle?
- NH₄⁺
- PCl₃
- NH₃
- SCl₂
Answer: (A) NH₄⁺
NH₄⁺ has 4 bond pairs and no lone pair (regular tetrahedron, 109.5°). NH₃ (107°), PCl₃ (≈100°) and SCl₂ (≈103°) all have lone pairs that compress the angle below 109.5°, so NH₄⁺ has the maximum bond angle.
- Q87The dipole moment of NH₃ (1.5 D) is larger than that of NF₃ (0.2 D). This is because:
- in NH₃ as well as in NF₃ the atomic dipole and bond dipole are in the same direction
- in NH₃ the dipoles are opposite whereas in NF₃ they are in the same direction
- in NH₃ the atomic (lone-pair) dipole and the bond dipole are in the same direction, whereas in NF₃ they are in opposite directions
- in NH₃ as well as in NF₃ the atomic dipole and bond dipole are in opposite directions
Answer: (C) in NH₃ the atomic (lone-pair) dipole and the bond dipole are in the same direction, whereas in NF₃ they are in opposite directions
N is more electronegative than H, so each N-H bond dipole points toward N, the same direction as the lone-pair (atomic) dipole – they add. In NF₃, F is more electronegative than N, so each N-F dipole points away from N, opposing the lone-pair dipole – they nearly cancel, leaving a small net μ.
- Q88Among the following series of transition-metal ions, the one in which all the metal ions have a 3d² electronic configuration is (At. no. Ti = 22, V = 23, Cr = 24, Mn = 25):
- Ti³⁺, V²⁺, Cr³⁺, Mn⁴⁺
- Ti²⁺, V³⁺, Cr⁴⁺, Mn⁵⁺
- Ti⁴⁺, V³⁺, Cr²⁺, Mn³⁺
- Ti⁺, V⁴⁺, Cr⁶⁺, Mn⁷⁺
Answer: (B) Ti²⁺, V³⁺, Cr⁴⁺, Mn⁵⁺
3d² means 2 d-electrons. Ti²⁺ (22−2=20e core+3d²), V³⁺ (23−3), Cr⁴⁺ (24−4), Mn⁵⁺ (25−5) all leave [Ar]3d². The other sets do not.
- Q89The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0°C is:
- 10.52 cal/mol K
- 5.260 cal/mol K
- 21.04 cal/mol K
- 0.526 cal/mol K
Answer: (B) 5.260 cal/mol K
Δ S = (Δ H_(fus))/(T) = (1435 cal/mol)/(273 K) = 5.26 cal/mol K.
- Q90Assertion (A): The standard electrode potential of SHE is taken as zero. Reason (R): The absolute electrode potential of a single electrode cannot be measured.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer: (A) Both A and R are true and R is the correct explanation of A
Only potential differences between two electrodes can be measured, never a single absolute electrode potential. So a reference (SHE) is assigned 0 V; R correctly explains why a zero reference is needed.
Botany
- Q91A bicarpellary ovary with an obliquely placed septum is characteristically seen in:
- Sesbania
- Brassica
- Aloe
- Solanum
Answer: (D) Solanum
An obliquely placed septum dividing a bicarpellary, syncarpous ovary is a diagnostic feature of Solanaceae (Solanum). Brassica has a bicarpellary ovary with a false septum (replum), Sesbania (Fabaceae) is monocarpellary, and Aloe (Liliaceae) is tricarpellary.
- Q92The product of conjugation in Spirogyra or fertilisation in Chlamydomonas is the:
- Zygospore
- Carpospore
- Zoospore
- Oospore
Answer: (A) Zygospore
Fusion of the two gametes produces a diploid zygote that secretes a thick wall to become a resting zygospore (the only diploid stage in these haplontic algae). It later germinates by meiosis.
- Q93In a short-day plant, flowering can be prevented by:
- Removing the roots
- Extending the dark period uninterrupted
- Interrupting the long dark period with a brief flash of light
- Lowering the temperature
Answer: (C) Interrupting the long dark period with a brief flash of light
For short-day plants it is really the long uninterrupted dark period that is critical. A brief flash of light during the night breaks the dark period and prevents flowering.
- Q94Vegetative propagation in mint (Mentha) occurs by:
- Sucker
- Runner
- Offset
- Rhizome
Answer: (A) Sucker
Mint spreads by suckers – underground lateral branches from the stem base that grow obliquely and emerge as new shoots. Runners (grass) creep above ground, offsets occur in aquatic Pistia/Eichhornia, and a rhizome (ginger) is a thick horizontal stem, none of which is mint’s mode.
- Q95Assertion (A): Co-extinction is a cause of biodiversity loss. Reason (R): When a host or partner species becomes extinct, the animals and plants obligately associated with it (e.g. its parasites or mutualists) also become extinct. Choose the correct option:
- 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
- A is true but R is false
Answer: (C) Both A and R are true and R is the correct explanation of A
Co-extinction is indeed one of the Evil Quartet causes, and the reason correctly explains it: obligately associated species (specific parasites, obligate mutualists like plant-pollinator pairs) die out when their partner vanishes.
- Q96The Respiratory Quotient (RQ) value of tripalmitin (a fat) is approximately:
- 0.09
- 0.07
- 0.7
- 0.9
Answer: (C) 0.7
For the fat tripalmitin the balanced equation 2 C51H98O6 + 145 O2 -> 102 CO2 + 98 H2O gives RQ = 102/145 = approximately 0.7. Fats are H-rich and O-poor, so they consume more O2 than the CO2 they release, making RQ less than 1.
- Q97An association of individuals of different species living in the same habitat and having functional interactions is called a:
- Ecological niche
- Ecosystem
- Population
- Biotic community
Answer: (D) Biotic community
A biotic community is an assemblage of populations of different species living together in a defined area with functional interactions. An ecosystem (C) additionally includes the abiotic environment; a population (D) is a single species.
- Q98Which of the following is a man-made ecosystem?
- Forest
- Pond
- Grassland
- Crop field
Answer: (D) Crop field
A crop field (and an aquarium) is an artificial, man-made ecosystem. Pond, grassland and forest are natural ecosystems.
- Q99A bivalent consists of:
- Four chromatids and two centromeres
- Two chromatids and two centromeres
- Two chromatids and one centromere
- Four chromatids and four centromeres
Answer: (A) Four chromatids and two centromeres
A bivalent (tetrad) is a pair of synapsed homologous chromosomes. Each homologue has two sister chromatids and one centromere, so the bivalent has four chromatids and two centromeres in total.
- Q100Which statement about transcription in bacteria is correct?
- A poly-A tail is added before translation
- It requires three different RNA polymerases
- Transcription and translation can occur simultaneously (coupled)
- The mRNA undergoes splicing before translation
Answer: (C) Transcription and translation can occur simultaneously (coupled)
Bacteria lack a nucleus, so transcription and translation are coupled. They use a single RNA polymerase, and the mRNA needs no splicing/capping/tailing.
- Q101Mendel studied inheritance of seven pairs of traits in pea, which can have 21 possible combinations. If you are told that in one of these combinations independent assortment is NOT observed in later studies, your reaction would be that:
- later studies may be wrong
- Mendel might not have studied all the combinations
- the independent assortment principle may be wrong
- it is impossible
Answer: (B) Mendel might not have studied all the combinations
Mendel’s seven genes lie on only four chromosomes, but the gene pairs he actually scored together happened to be unlinked, so he never saw linkage. If some pair shows non-independent assortment, the sensible inference is that Mendel did not study that particular (linked) combination, so B is correct.
- Q102African sleeping sickness is due to:
- Trypanosoma lewisii transmitted by bed bug
- Entamoeba gingivalis spread by house fly
- Trypanosoma gambiense transmitted by Glossina palpalis
- Plasmodium vivax transmitted by tse-tse fly
Answer: (C) Trypanosoma gambiense transmitted by Glossina palpalis
African sleeping sickness is caused by Trypanosoma gambiense, transmitted by the tse-tse fly Glossina palpalis. Plasmodium causes malaria (via Anopheles), so the other pairings are incorrect.
- Q103Many ribosomes may associate with a single mRNA to form multiple copies of a polypeptide simultaneously. Such strings of ribosomes are termed:
- nucleosome
- polysome
- polyhedral bodies
- plastidome
Answer: (B) polysome
A polysome (polyribosome) is a string of ribosomes attached to a single mRNA, allowing several copies of the same polypeptide to be made at once. A nucleosome is DNA wound on histones, unrelated to this.
- Q104A ‘polyarch’ xylem condition (many xylem bundles) and a large central pith are typical of a:
- Monocot leaf
- Dicot root
- Dicot stem
- Monocot root
Answer: (D) Monocot root
Monocot roots usually have many (more than six) xylem and phloem patches (polyarch) and a well-developed large pith, with no secondary growth. Dicot roots are di- to tetrarch with little/no pith.
- Q105Oxygen is NOT produced during photosynthesis by:
- Green sulphur bacteria
- Cycas
- Nostoc
- Chara
Answer: (A) Green sulphur bacteria
Green (and purple) sulphur bacteria carry out anoxygenic photosynthesis: they use H₂S, not water, as the electron donor, so no O₂ is released. Cycas (gymnosperm), Nostoc (cyanobacterium) and Chara (green alga) all do oxygenic photosynthesis and release O₂.
- Q106The stomata of CAM (Crassulacean Acid Metabolism) plants are:
- Open during the night and closed during the day
- Never open
- Always open
- Open during the day and closed at night
Answer: (A) Open during the night and closed during the day
CAM plants (e.g. Agave, Opuntia) open stomata at night when temperature is low and humidity high, fixing CO2 into malic acid, and keep stomata closed in the day to minimise water loss. This is an adaptation to arid conditions.
- Q107In which one of the following habitats does the diurnal (day-to-night) temperature of the soil surface vary the most?
- Shrubland
- Grassland
- Forest
- Desert
Answer: (D) Desert
Deserts have bare, exposed soil with sparse plant cover, so the surface heats strongly by day and loses heat rapidly at night, giving the largest diurnal temperature swing. Vegetation in forests and grasslands buffers this variation.
- Q108As we go from species to kingdom in a taxonomic hierarchy, the number of common characteristics
- Will decrease
- Will increase
- Remain same
- May increase or decrease
Answer: (A) Will decrease
A. Tests: the inverse relation between the rank of a taxon and the number of characters its members share. Why A: a higher category has to accommodate more and more dissimilar organisms, so the definition holding them together must be built on fewer features. Members of one species share almost every character including the ability to interbreed, members of a family share a handful of floral or structural features, and members of a kingdom such as Plantae share only very general ones like being eukaryotic, having a cell wall and being autotrophic. So going up from species to kingdom the number of common characteristics falls. Why not B: what increases going up is the number of organisms included, not the number of shared characters, and these two trends run in opposite directions [misconception B: “more members must mean more in common”]. Why not C: it cannot stay the same, because if every rank were defined by the same set of characters there would be no basis for splitting a kingdom into phyla and a phylum into classes. Why not D: the trend is strictly one-way; there is no rank at which climbing the hierarchy adds shared characters back. Remember: climb the ladder and the group gets bigger while the shared character list gets shorter.
- Q109The Respiratory Quotient (RQ) of a fat being used as a respiratory substrate is approximately:
- 0.7
- 0.9
- More than 1
- 1.0
Answer: (A) 0.7
Fats are rich in H and poor in O, so more O2 is consumed than CO2 released, giving RQ less than 1, approximately 0.7.
- Q110Which is the basis of genetic mapping of the human genome as well as DNA fingerprinting?
- Polymorphism in RNA sequence
- Polymorphism in DNA sequence
- Single nucleotide polymorphism only
- Polymorphism in hnRNA sequence
Answer: (B) Polymorphism in DNA sequence
Polymorphism (variation) in DNA sequence is the basis of both genetic mapping of the human genome and DNA fingerprinting; these variations arise due to mutations.
- Q111Which one of the following expanded forms of the acronym is correct?
- UNEP – United Nations Environmental Policy
- IUCN – International Union for Conservation of Nature and Natural Resources
- EPA – Environmental Pollution Agency
- IPCC – International Panel for Climate Change
Answer: (B) IUCN – International Union for Conservation of Nature and Natural Resources
IUCN stands for the International Union for Conservation of Nature and Natural Resources. UNEP = United Nations Environment Programme (not ‘Policy’); EPA = Environmental Protection Agency (not ‘Pollution’); IPCC = Intergovernmental (not ‘International’) Panel on Climate Change.
- Q112Pollination in water hyacinth and water lily is brought about by the agency of
- insects or wind
- bats
- water
- birds
Answer: (A) insects or wind
B. Tests: recognising that living in water does not make a plant water pollinated. Why B: In water hyacinth and water lily the flowers are held above the water surface on emergent stalks, and they are showy and coloured. Once the flower opens in air, the agents available to it are the same as for a land plant, so these species are pollinated by insects or by wind. True hydrophily is rare and is limited to plants whose flowers stay submerged or reach the surface for pollination, such as Vallisneria, Hydrilla and the marine Zostera. Why not A: Water pollination needs pollen to travel in or on water, but the flowers of these two plants are emergent and never release pollen into the water; the aquatic habitat is the trap in the question [misconception A: “an aquatic plant must be hydrophilous”]. Why not C: Bird pollination (ornithophily) fits large, tubular, brightly coloured flowers with copious nectar such as Bombax and Callistemon, not these. Why not D: Bat pollination (chiropterophily) fits big, sturdy, night-opening flowers with a strong odour and plenty of nectar and pollen, such as Adansonia and Kigelia. Remember: Only truly submerged flowers are water pollinated; emergent aquatic flowers use insects or wind.
- Q113The hormone primarily connected with cell division is:
- Cytokinin (zeatin)
- IAA
- Gibberellic acid
- NAA
Answer: (A) Cytokinin (zeatin)
Cell division (cytokinesis) is the characteristic property associated with cytokinins (e.g. zeatin). Cytokinin acts with auxin to stimulate cell division even in non-meristematic tissues.
- Q114An element playing an important role in nitrogen fixation is:
- zinc
- copper
- manganese
- molybdenum
Answer: (D) molybdenum
Molybdenum is absorbed as molybdate and is a component of the enzyme nitrogenase (and nitrate reductase), so it is directly required for biological nitrogen fixation. Its deficiency causes whiptail of cauliflower.
- Q115Down’s syndrome is due to:
- crossing over
- linkage
- non-disjunction of chromosomes
- sex-linked inheritance
Answer: (C) non-disjunction of chromosomes
Down’s syndrome (trisomy 21) arises from non-disjunction of chromosome 21 during meiosis (usually oogenesis), giving an extra chromosome 21. It is not caused by crossing over or linkage.
- Q116Stroma lamellae differ from grana thylakoids in that stroma lamellae:
- Lack PS II and NADP reductase
- Contain PS II but lack PS I
- Contain no pigments
- Are the site of carbon fixation
Answer: (A) Lack PS II and NADP reductase
Stroma lamellae lack PS II and NADP reductase, so they carry out only cyclic photophosphorylation (PS I only), whereas grana thylakoids have both photosystems for non-cyclic flow.
- Q117If the egg is fertilised by a male gamete carrying a mutation, the zygote ploidy and the endosperm ploidy will respectively be:
- 2n and 2n
- 3n and 3n
- 2n and 3n
- n and 2n
Answer: (C) 2n and 3n
Syngamy gives a diploid (2n) zygote (n egg + n sperm). Triple fusion gives a triploid (3n) endosperm (n sperm + n + n polar nuclei). Hence 2n and 3n.
- Q118A dihybrid test cross (RrYy × rryy) is expected to give offspring in the ratio:
- 9:3:3:1
- 3:1
- 1:2:1
- 1:1:1:1
Answer: (D) 1:1:1:1
The double heterozygote produces four equal gamete types (RY, Ry, rY, ry); the recessive parent contributes only ry. Hence four phenotypes in a 1:1:1:1 ratio.
- Q119How does the desert kangaroo rat meet its water requirement in the absence of drinking water?
- By storing water in a special bladder filled during rains
- By concentrating urine and using metabolic (internal fat oxidation) water
- By reducing all metabolic activity to zero
- By absorbing dew through its skin
Answer: (B) By concentrating urine and using metabolic (internal fat oxidation) water
The kangaroo rat in North American deserts has the ability to concentrate its urine so little water is used in excretion, and it derives water internally by oxidising fat, allowing survival with no drinking water.
- Q120Which one of the following is a characteristic feature of a cropland ecosystem?
- The absence of weeds
- Ecological succession
- Least genetic diversity
- The absence of soil organisms
Answer: (C) Least genetic diversity
A cropland is a man-made (anthropogenic) ecosystem dominated by a single crop species (monoculture), giving it the least genetic diversity but high productivity. Weeds and soil organisms are still present, and a managed crop field is maintained against natural succession.
- Q121The primary significance of meiosis in sexually reproducing organisms is:
- Maintaining constant chromosome number across generations and generating variation
- Increasing the chromosome number each generation
- Growth and repair of body tissues
- Producing genetically identical cells
Answer: (A) Maintaining constant chromosome number across generations and generating variation
Meiosis halves the chromosome number so that fertilisation restores the diploid number, keeping it constant across generations, and crossing over plus independent assortment generate variation for evolution.
- Q122An allele is said to be dominant if it is expressed in:
- the second generation only
- both homozygous and heterozygous states
- the homozygous condition only
- the heterozygous condition only
Answer: (B) both homozygous and heterozygous states
A dominant allele expresses its phenotype whether present in homozygous (TT) or heterozygous (Tt) condition. A recessive allele expresses only when homozygous (tt).
- Q123Some enzymes that convert fats into carbohydrates are present in:
- glyoxysomes
- liposomes
- Golgi bodies
- microsomes
Answer: (A) glyoxysomes
Glyoxysomes contain the enzymes of the glyoxylate cycle, through which fats are converted into carbohydrates (gluconeogenesis from lipids) in germinating fatty seeds. (Catalase and beta-oxidation enzymes also occur there.)
- Q124In eukaryotes, the precursor of mRNA (hnRNA) is transcribed by:
- RNA polymerase II
- RNA polymerase III
- DNA polymerase
- RNA polymerase I
Answer: (A) RNA polymerase II
RNA Pol II transcribes the mRNA precursor (hnRNA). Pol I makes rRNA and Pol III makes tRNA and 5S rRNA.
- Q125According to the chemiosmotic hypothesis, ATP synthesis in chloroplasts is driven by the movement of protons:
- From stroma into the cytoplasm
- Across the outer chloroplast membrane
- From lumen into the stroma through F0-F1 ATP synthase
- From stroma into the lumen through ATP synthase
Answer: (C) From lumen into the stroma through F0-F1 ATP synthase
Protons accumulate in the thylakoid lumen creating a gradient; they flow back into the stroma through the F0-F1 (CF0-CF1) ATP synthase, driving ATP synthesis. ATP is released into the stroma.
- Q126Which one of the following precedes the reformation of the nuclear envelope during the M-phase of the cell cycle?
- Transcription from chromosomes and reassembly of the nuclear lamina
- Formation of the contractile ring and transcription from chromosomes
- Decondensation from chromosomes and reassembly of the nuclear lamina
- Formation of the contractile ring and formation of the phragmoplast
Answer: (C) Decondensation from chromosomes and reassembly of the nuclear lamina
At telophase, the chromosomes decondense and the nuclear lamina is reassembled; only then do nuclear-membrane vesicles coalesce to reform the nuclear envelope. Transcription resumes only after the nucleus reforms, and the contractile ring/phragmoplast belong to cytokinesis, not nuclear-envelope reformation. So A is correct.
- Q127An onion bulb stores food mainly in its:
- Terminal bud
- Fleshy scale leaves
- Reduced disc-like stem
- Adventitious roots
Answer: (B) Fleshy scale leaves
In an onion bulb the stem is a small reduced disc; food is stored in concentric fleshy scale leaves attached to this disc.
- Q128The ovule of an angiosperm is technically equivalent to:
- Megaspore mother cell
- Megaspore
- Megasporophyll
- Megasporangium
Answer: (D) Megasporangium
The ovule is the megasporangium: it is the structure within which the megaspore mother cell forms and undergoes meiosis to give megaspores. A megasporophyll (carpel) bears ovules, while the megaspore and megaspore mother cell are cells inside the ovule.
- Q129Purines found in both DNA and RNA are:
- adenine and thymine
- adenine and guanine
- guanine and cytosine
- cytosine and thymine
Answer: (B) adenine and guanine
Adenine and guanine are the two purines present in both DNA and RNA. Cytosine and thymine are pyrimidines (in DNA); in RNA thymine is replaced by uracil, so the purines stay the same.
- Q130Robert Costanza and colleagues estimated the value of the biosphere’s ecosystem services to be about US$33 trillion per year. The single largest contributor to this value was:
- Pollination
- Recreation
- Soil formation
- Climate regulation
Answer: (C) Soil formation
In Costanza et al.’s estimate, soil formation accounted for about 50% of the total value of ecosystem services, the largest single share.
- Q131Stomata of a plant open due to:
- efflux of potassium ions
- influx of calcium ions
- influx of potassium ions
- influx of hydrogen ions
Answer: (C) influx of potassium ions
Active influx (uptake) of K+ into guard cells raises their solute content, lowering water potential so water enters by osmosis; the cells become turgid and the stoma opens. K+ efflux would close it. Hence (C).
- Q132Two genes show a recombination frequency of 8%. The map distance between them is:
- 80 cM
- 0.8 cM
- 16 cM
- 8 cM
Answer: (D) 8 cM
1% recombination = 1 centimorgan (map unit). A recombination frequency of 8% therefore equals a map distance of 8 cM.
- Q133The centriole has a characteristic cartwheel structure made of:
- Nine peripheral triplet microtubules with no central tubule (9 + 0)
- Nine doublet microtubules and two central ones
- Two central triplets only
- Microfilaments of actin
Answer: (A) Nine peripheral triplet microtubules with no central tubule (9 + 0)
Each centriole has a 9 + 0 arrangement: nine peripheral fibrils, each a triplet of microtubules, with no central microtubule. The centrioles form the spindle apparatus and basal bodies of cilia/flagella.
- Q134Polyblend, a fine powder of recycled modified plastic, has proved to be a good material for:
- Use as a fertiliser
- Making plastic sacks
- Making tubes and pipes
- Construction of roads
Answer: (D) Construction of roads
Polyblend, mixed with bitumen, has proved to be a good material for road construction; the first polyblend road was laid in Bangalore by Ahmed Khan.
- Q135Which one of the following is NOT used for ex-situ plant conservation?
- Botanical gardens
- Shifting cultivation
- Field gene banks
- Seed banks
Answer: (B) Shifting cultivation
Shifting cultivation (jhum) clears and burns forest patches, causing deforestation – it destroys biodiversity rather than conserving it. Field gene banks, seed banks and botanical gardens all preserve plant genetic material off-site, so they are ex-situ conservation methods.
Zoology
- Q136Which respiratory disorder is an occupational disease caused by long-term inhalation of dust leading to fibrosis of the upper part of the lungs?
- Asthma
- Emphysema
- Silicosis / Asbestosis
- Pneumonia
Answer: (C) Silicosis / Asbestosis
Silicosis and asbestosis are occupational lung diseases caused by chronic inhalation of silica or asbestos dust, leading to proliferation of fibrous tissue (fibrosis) in the upper lungs.
- Q137Which immunoglobulin class is most abundant in the colostrum and protects the mucosal surfaces and infant gut?
- IgG
- IgA
- IgE
- IgM
Answer: (B) IgA
Colostrum (the first milk) is rich in antibodies of the IgA class, which protect the infant’s mucosal surfaces and gut, providing passive immunity.
- Q138Select the correct matching of the type of joint with its example in the human skeletal system:
- Pivot joint - Between third and fourth cervical vertebra
- Hinge joint - Between humerus and pectoral girdle
- Cartilaginous joint - Between frontal and parietal
- Gliding joint - Between carpals
Answer: (D) Gliding joint - Between carpals
Gliding joints occur between the carpal bones – correct. Between frontal and parietal is a fibrous (not cartilaginous) suture; between cervical vertebrae is cartilaginous (not pivot); humerus-pectoral girdle is a ball-and-socket (shoulder), not a hinge.
- Q139Which set of conditions favours maximum diffusion of gases at the respiratory surface?
- Large surface area, high solubility, short diffusion distance
- Small surface area, low solubility, large diffusion distance
- High diffusion distance and low pressure gradient
- Low pressure gradient, thick membrane
Answer: (A) Large surface area, high solubility, short diffusion distance
Diffusion is favoured by a large pressure gradient, large surface area, high solubility of the gas, and a short (thin) diffusion distance. Option B captures these conditions.
- Q140Transgenic animals that carry genes making them more sensitive to toxic substances are primarily used for:
- Chemical (toxicity) safety testing
- Studying normal development
- Vaccine production
- Milk production
Answer: (A) Chemical (toxicity) safety testing
Such transgenic animals are used for chemical safety / toxicity testing; their increased sensitivity to toxins gives test results in less time than non-transgenic animals.
- Q141Among the following edible fishes, which one is a MARINE fish rich in omega-3 fatty acids?
- Mrigala
- Mackerel
- Mangur
- Mystus
Answer: (B) Mackerel
Mackerel is a marine fish rich in omega-3 fatty acids. Mystus, Mangur (Clarias) and Mrigala are freshwater fishes. Hence the marine, omega-3-rich choice is Mackerel (D).
- Q142Which one of the following amino acids was NOT found to be synthesised in Miller’s experiment?
- Alanine
- Aspartic acid
- Glycine
- Glutamic acid
Answer: (D) Glutamic acid
Miller and Urey obtained glycine, alanine and aspartic acid in abundance, but glutamic acid was not produced in their spark-discharge experiment. Hence glutamic acid is the exception.
- Q143A common characteristic of all vertebrates is:
- division of the body into head, neck, trunk and tail
- presence of skull (cranium)
- presence of two pairs of functional appendages
- body covered with an exoskeleton
Answer: (B) presence of skull (cranium)
All vertebrates possess a well-developed cranium (skull) protecting the brain, hence they are also called Craniata. Not all vertebrates have a neck, two pairs of limbs (e.g. fishes, snakes) or an exoskeleton, so those features are not universal.
- Q144Which of the following secretes the hormone relaxin during the later phase of pregnancy?
- Corpus luteum
- Uterus
- Foetus
- Graafian follicle
Answer: (A) Corpus luteum
In the later phase of pregnancy the corpus luteum (and placenta) secretes relaxin, which softens/dilates the cervix and pelvic ligaments to aid parturition. The Graafian follicle, foetus and uterus do not secrete relaxin.
- Q145Coca alkaloid (cocaine) is obtained from:
- Erythroxylum coca
- Papaver somniferum
- Datura
- Atropa belladonna
Answer: (A) Erythroxylum coca
Cocaine is obtained from the coca plant Erythroxylum coca and interferes with dopamine transport, acting as a CNS stimulant. Papaver gives opioids, while Atropa and Datura yield hallucinogens.
- Q146Match the enzyme with its product. Which pairing is INCORRECT?
- Pepsin → amino acids
- Sucrase → glucose + fructose
- Maltase → glucose + glucose
- Lactase → glucose + galactose
Answer: (A) Pepsin → amino acids
Pepsin only partially digests proteins into proteoses and peptones, not free amino acids. Final breakdown to amino acids requires intestinal dipeptidases and pancreatic proteases. The carbohydrase pairings are all correct.
- Q147Which of the following are NOT polymeric?
- Nucleic acid
- Proteins
- Lipids
- Polysaccharides
Answer: (C) Lipids
Nucleic acids (polymers of nucleotides), proteins (polymers of amino acids) and polysaccharides (polymers of monosaccharides) are all polymers. Lipids are not polymers: a fat is just a few fatty acids esterified to one glycerol, with no repeating monomer chain.
- Q148Bacterial cells are made ‘competent’ to take up DNA by treatment with a specific cation followed by a brief heat shock. The cation and temperature are:
- Divalent Ca²⁺ and 42°C
- Monovalent Na⁺ and 0°C
- Divalent Mg²⁺ and 4°C
- Trivalent Al³⁺ and 100°C
Answer: (A) Divalent Ca²⁺ and 42°C
Cells are treated with a specific concentration of divalent Ca²⁺ to increase wall permeability, then DNA is forced in by a brief heat shock (~42°C) followed by placing on ice. This is the standard transformation method.
- Q149Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Amniocentesis for sex determination is one of the strategies of the Reproductive and Child Health Care Programme. Reason R: Ban on amniocentesis checks the increasing menace of female foeticide. In the light of the above statements, choose the correct answer from the options given below:
- A is false but R is true.
- Both A and R are true and R is the correct explanation of A.
- A is true but R is false.
- Both A and R are true and R is NOT the correct explanation of A.
Answer: (A) A is false but R is true.
C. Tests: distinguishing RCH strategies from practices the programme actively bans. Why C: A is false: sex-determination by amniocentesis is not a strategy of RCH; the programme’s stand is the opposite, a statutory ban on it. R is true: prohibiting misuse of amniocentesis for knowing the sex of the foetus curbs female foeticide, which is exactly why the ban exists. Why not A: Treats A as true; RCH promotes awareness and care, it never promotes sex determination [misconception A: “any use of a prenatal test is part of government health strategy”]. Why not B: Gets both halves wrong; the ban genuinely reduces female foeticide, so R is true [misconception B: “the ban is symbolic and ineffective, so R is false”]. Why not D: Both cannot be true together when A itself is false [misconception D: “amniocentesis is encouraged under RCH for family planning”]. Remember: RCH bans sex-determination; amniocentesis is legal only for genetic-disorder screening.
- Q150Name the blood cells whose reduction in number can cause a clotting disorder leading to excessive loss of blood from the body.
- Erythrocytes
- Thrombocytes
- Neutrophils
- Leucocytes
Answer: (B) Thrombocytes
Thrombocytes (platelets) release factors that initiate clotting. A fall in their number (thrombocytopenia) impairs clotting and causes excessive bleeding. Erythrocytes carry O2 and leucocytes/neutrophils are for defence.
- Q151The correct route through which the pulse-making (pacemaker) impulse travels in the heart is:
- SA node → AV node → Bundle of His → Purkinje fibres → Heart muscles
- AV node → Bundle of His → SA node → Purkinje fibres → Heart muscles
- AV node → SA node → Purkinje fibres → Bundle of His → Heart muscles
- SA node → Purkinje fibres → AV node → Bundle of His → Heart muscles
Answer: (A) SA node → AV node → Bundle of His → Purkinje fibres → Heart muscles
The impulse originates at the SA node, spreads to the AV node, then down the bundle of His and its branches, into the Purkinje fibres, finally exciting the heart (ventricular) muscle. Hence the route is SA node → AV node → Bundle of His → Purkinje fibres → Heart muscles.
- Q152The water-vascular system is a unique feature of which phylum?
- Echinodermata
- Mollusca
- Arthropoda
- Annelida
Answer: (A) Echinodermata
Echinoderms possess a unique water-vascular system used for locomotion, capture and transport of food, and respiration. It is found in no other phylum.
- Q153Withdrawal of which of the following hormones is the immediate cause of menstruation?
- Progesterone
- Oestrogen
- FSH
- FSH-RH
Answer: (A) Progesterone
The fall (withdrawal) of progesterone (and oestrogen) at the end of the cycle removes endometrial support, causing the lining to break down – menstruation. Progesterone withdrawal is the immediate trigger.
- Q154Match the following columns and select the correct option. Column I (a) Floating ribs (b) Acromion (c) Scapula (d) Glenoid cavity Column II (i) Located between second and seventh ribs (ii) Head of the humerus (iii) Clavicle (iv) Do not connect with the sternum
- (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
- (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
- (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
- (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Answer: (D) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
C. Tests: the anatomy of the last two rib pairs and of the pectoral girdle, matched item by item. Why C: Work each pair. (a) The 11th and 12th rib pairs are called floating ribs precisely because their ventral ends are free and do not connect with the sternum, so (a) goes to (iv). (b) The acromion is the expanded flat process at the top of the scapular spine, and it articulates with the clavicle to form the acromioclavicular joint, so (b) goes to (iii). (c) The scapula is the triangular bone of the dorsal thorax lying between the second and the seventh ribs, so (c) goes to (i). (d) The glenoid cavity is the shallow socket on the scapula that receives the head of the humerus to form the shoulder joint, so (d) goes to (ii). The set (iv), (iii), (i), (ii) is option C. Why not A: It sends floating ribs to the head of the humerus and the glenoid cavity to the clavicle, so both of the shoulder items and the rib item are swapped out of place [misconception A: “the glenoid cavity holds the clavicle”]. Why not B: It pairs floating ribs with the second to seventh rib position, which is the location of the scapula, and pairs the scapula itself with the head of the humerus [misconception B: “floating ribs are the ones behind the shoulder blade”]. Why not D: It pairs the acromion with the head of the humerus, but the humeral head sits in the glenoid cavity, and it leaves floating ribs matched to the clavicle [misconception D: “the acromion is the socket of the shoulder joint”]. Remember: on the scapula, the acromion takes the clavicle and the glenoid cavity takes the humerus.
- Q155Which of the following events is NOT associated with ovulation in the human female?
- Release of the secondary oocyte
- LH surge
- Full development of the Graafian follicle
- Decrease in oestradiol
Answer: (D) Decrease in oestradiol
Oestradiol (oestrogen) actually peaks just before ovulation and triggers the LH surge; a decrease in oestradiol occurs after ovulation and before menstruation, so it is NOT an event of ovulation. Full Graafian follicle, secondary oocyte release and the LH surge all accompany ovulation.
- Q156Uricotelism (excretion of uric acid) is found in:
- fishes and freshwater protozoans
- birds, reptiles and insects
- frogs and toads
- mammals and birds
Answer: (B) birds, reptiles and insects
Birds, reptiles and insects conserve water by excreting nitrogen as insoluble uric acid (uricotelic). Mammals, frogs and toads are ureotelic; aquatic fishes/protozoans are ammonotelic. Hence birds, reptiles and insects.
- Q157In a typical reflex arc, the correct order of components is:
- Receptor → motor neuron → CNS → sensory neuron → effector
- Receptor → sensory (afferent) neuron → CNS → motor (efferent) neuron → effector
- Receptor → CNS → sensory neuron → effector → motor neuron
- Effector → sensory neuron → CNS → motor neuron → receptor
Answer: (B) Receptor → sensory (afferent) neuron → CNS → motor (efferent) neuron → effector
The reflex arc runs: receptor → afferent (sensory) neuron → CNS (spinal cord) → efferent (motor) neuron → effector (muscle/gland), producing a rapid involuntary response.
- Q158A phosphoglyceride is always made up of:
- A saturated or unsaturated fatty acid esterified to a phosphate group, which is also attached to a glycerol
- Only a saturated fatty acid esterified to a glycerol to which a phosphate group is also attached
- A saturated or unsaturated fatty acid esterified to a glycerol to which a phosphate group is also attached
- Only an unsaturated fatty acid esterified to a glycerol to which a phosphate group is also attached
Answer: (C) A saturated or unsaturated fatty acid esterified to a glycerol to which a phosphate group is also attached
A phosphoglyceride has a glycerol backbone esterified with fatty acids (which may be saturated OR unsaturated) at positions 1 and 2, with a phosphate at position 3. The fatty acid is attached to glycerol (not directly to the phosphate), and may be either saturated or unsaturated.
- Q159Human population growth in India:
- Tends to follow a sigmoid curve as in many other animal species
- Can be regulated by following the National programme of family planning
- Can be reduced by permitting natural calamities and enforcing birth-control measures
- Tends to reach zero population growth as in some animal species
Answer: (B) Can be regulated by following the National programme of family planning
Although India is crowded and over-populated, its human population can be regulated by following the National family planning / RCH programme (raising marriage age, promoting contraception, small-family incentives). Relying on natural calamities is unethical and incorrect, and human population is not simply following an automatic sigmoid plateau or zero growth.
- Q160In pBR322, foreign DNA is inserted at the BamHI site located within the tetracycline-resistance gene. The recombinant bacteria will be:
- Resistant to both ampicillin and tetracycline
- Sensitive to both ampicillin and tetracycline
- Resistant to tetracycline but sensitive to ampicillin
- Resistant to ampicillin but sensitive to tetracycline
Answer: (D) Resistant to ampicillin but sensitive to tetracycline
Insertion at BamHI inactivates the tetracycline-resistance gene (insertional inactivation), so recombinants become tetracycline-sensitive. The ampicillin-resistance gene stays intact, so they remain ampicillin-resistant.
- Q161A plasmid is best described as:
- an autotrophic fragment
- a fragment which joins two genes
- a fragment of DNA which acts as a vector
- mRNA which acts as a carrier
Answer: (C) a fragment of DNA which acts as a vector
A plasmid is a piece of (mostly bacterial) DNA, not part of the chromosomal DNA, that replicates independently and acts as a vehicle (vector) for gene transfer. It is DNA, not mRNA, and it does not itself join genes (that is ligase).
- Q162Which one of the following chemicals is used for causing defoliation of forest trees?
- 2,4-D
- Malic hydrazide
- Phosphon-D
- Amo-1618
Answer: (A) 2,4-D
2,4-D (2,4-dichlorophenoxyacetic acid) is a synthetic auxin/herbicide that over-stimulates growth and is used as a defoliant and to kill broad-leaved (dicot) weeds. Amo-1618 and Phosphon-D are growth retardants; malic hydrazide is a growth inhibitor. Hence (D).
- Q163Which one of the following groups of animals reflects organ-system level of organisation with bilateral symmetry, true coelom and segmented body? (Annelida, Arthropoda and Chordata)
- Arthropoda, Mollusca and Chordata
- Annelida, Arthropoda and Chordata
- Annelida, Arthropoda and Mollusca
- Annelida, Mollusca and Chordata
Answer: (B) Annelida, Arthropoda and Chordata
Annelida, Arthropoda and Chordata all show organ-system level, bilateral symmetry, a true coelom AND a segmented (metameric) body. Mollusca, though bilateral coelomate organ-system animals, are NOT segmented, which is why every option containing Mollusca is rejected.
- Q164Seminal plasma in human males is rich in:
- Fructose and calcium
- Ribose and potassium
- Glucose and calcium
- DNA and testosterone
Answer: (A) Fructose and calcium
Seminal plasma (the fluid part of semen, mainly from the seminal vesicles and prostate) is rich in fructose (energy source for sperm) and certain ions including calcium. It is not rich in glucose, DNA, or ribose.
- Q165In which animal is the nerve cell present but the brain absent?
- Hydra
- Earthworm
- Sponge
- Cockroach
Answer: (A) Hydra
Hydra possesses nerve cells arranged in a diffuse nerve net but has no brain or even a nerve cord. Sponges lack nerve cells altogether, whereas earthworm and cockroach both have a definite cerebral ganglion (brain).
- Q166The gastric (hepatic) caeca in the cockroach are found at the junction of the:
- Pharynx and oesophagus
- Crop and gizzard
- Midgut and hindgut
- Foregut and midgut
Answer: (D) Foregut and midgut
At the junction of foregut and midgut, a ring of 6-8 blind hepatic/gastric caeca secretes digestive juice. (Malpighian tubules lie at the midgut-hindgut junction.)
- Q167Graves’ disease is caused due to:
- Hypersecretion of adrenal gland
- Hyposecretion of adrenal gland
- Hyposecretion of thyroid gland
- Hypersecretion of thyroid gland
Answer: (D) Hypersecretion of thyroid gland
Graves’ disease (exophthalmic goitre) is a hyperthyroidism – caused by hypersecretion of thyroid hormones – producing raised metabolism and protruding eyes.
- Q168Lipids are insoluble in water because lipid molecules are:
- Zwitter ions
- Hydrophobic
- Neutral
- Hydrophilic
Answer: (B) Hydrophobic
Lipid molecules are largely hydrophobic (water-fearing) because of their long non-polar hydrocarbon chains, which do not interact with polar water. They dissolve instead in non-polar solvents like benzene and chloroform.
- Q169If due to some injury the chordae tendineae of the tricuspid valve of the human heart is partially non-functional, what will be the immediate effect?
- The flow of blood into the pulmonary artery will be reduced
- The blood will tend to flow back into the left atrium
- The flow of blood into the aorta will be slowed down
- The ‘pacemaker’ will stop working
Answer: (A) The flow of blood into the pulmonary artery will be reduced
Chordae tendineae anchor the tricuspid valve (right side) so it allows one-way flow from right atrium to right ventricle. If they fail, blood leaks back into the right atrium during ventricular systole, so less blood is pumped from the right ventricle into the pulmonary artery. (It cannot affect the left atrium, which involves the bicuspid valve.) Hence D.
- Q170What is a sarcomere?
- Part between two Z-lines
- Part between two A-lines
- Part between two H-lines
- Part between two I-bands
Answer: (A) Part between two Z-lines
A sarcomere is the segment of a myofibril between two successive Z-lines and is the structural and functional unit of contraction. ‘H-lines’ and ‘A-lines’ are not the boundary markers, and ‘between two I-bands’ does not define it.
- Q171Given a sample of DNA with the strands 5′-GAATTC-3′ and 3′-CTTAAG-5′, what is special about this sequence?
- It is a palindromic sequence of base pairs
- Deletion mutation
- Replication is completed here
- Start codon at the 5′ end
Answer: (A) It is a palindromic sequence of base pairs
Reading each strand in its own 5’→3′ direction gives GAATTC on both – the sequence reads the same both ways, i.e. it is a palindromic sequence. Such palindromes are the recognition sites for restriction endonucleases (here, the EcoRI site).
- Q172Floridean starch (the storage carbohydrate of red algae) has a structure most similar to:
- Starch and cellulose
- Amylopectin and glycogen
- Mannitol and algin
- Laminarin and cellulose
Answer: (B) Amylopectin and glycogen
Floridean starch resembles amylopectin and glycogen, all branched polymers of alpha-D-glucose. Amylopectin is a soluble form while glycogen is insoluble; floridean starch lies between them.
- Q173During the purification process in recombinant DNA technology, addition of chilled ethanol precipitates out:
- Polysaccharides
- Histones
- DNA
- RNA
Answer: (C) DNA
A. Tests: the role of chilled ethanol in isolating genetic material. Why A: after cells are lysed and treated with ribonuclease (removes RNA) and protease (removes proteins, including histones), the purified DNA is precipitated by adding chilled ethanol; it appears as fine spooling threads. Why not D: RNA has already been degraded by ribonuclease before the ethanol step [misconception D: “ethanol precipitates all nucleic acids present at the end”] Why not B: histones are proteins and are removed earlier by protease treatment [misconception B: “chromatin precipitates as a whole”] Why not C: other macromolecules like polysaccharides are removed by appropriate treatments; the final ethanol step is for DNA [misconception C: “ethanol precipitates whatever is left in bulk”] Remember: RNase and protease clean up first; chilled ethanol then spools out pure DNA threads.
- Q174The correct sequence of events in the killing of an insect larva by a Bt crop is:
- Active toxin ingested -> binds gut cells -> alkaline pH activates -> larva dies
- Toxin enters via roots -> reaches gut -> acidic pH activates -> larva dies
- Protoxin ingested -> alkaline gut pH activates toxin -> binds midgut epithelium -> pores form -> larva dies
- Protoxin binds gut cells -> acidic pH activates -> cells shrink -> larva survives
Answer: (C) Protoxin ingested -> alkaline gut pH activates toxin -> binds midgut epithelium -> pores form -> larva dies
The larva ingests inactive protoxin; the alkaline gut pH solubilises and activates it; the active toxin binds midgut epithelial cells, forms pores, cells swell and lyse, and the larva dies.
- Q175Which one of the following is incorrect about the characteristics of protobionts (coacervates and microspheres) as envisaged in the abiogenic origin of life?
- They could separate combinations of molecules from the surroundings
- They could maintain an internal environment
- They were able to reproduce
- They were partially isolated from the surroundings
Answer: (C) They were able to reproduce
Coacervates/microspheres could selectively concentrate molecules, were partially isolated, and maintained a different internal environment (a primitive form of homeostasis). However, true heritable reproduction (with a genetic system) had not yet evolved, so the statement that they were able to reproduce is incorrect.
- Q176Koch’s postulates are NOT applicable to:
- Leprosy
- Cholera
- Diphtheria
- Tuberculosis
Answer: (A) Leprosy
Koch’s postulates require culturing the pathogen in vitro. Mycobacterium leprae (leprosy) cannot be grown in artificial culture media, so the postulates do not apply. Cholera, TB and diphtheria pathogens can be cultured, so Koch’s postulates hold for them.
- Q177The photosensitive compound in the human eye is made up of:
- Opsin and retinol
- Transducin and retinene
- Opsin and retinal
- Guanosine and retinal
Answer: (C) Opsin and retinal
The visual pigment rhodopsin is made of the protein opsin combined with retinal (the aldehyde form of vitamin-A). Retinol is the alcohol form, and transducin/guanosine are not structural parts of the photopigment.
- Q178During inspiration, the diaphragm:
- Contracts and flattens
- Relaxes and becomes dome-shaped
- Has no role in breathing
- Moves upward into the thorax
Answer: (A) Contracts and flattens
Contraction of the diaphragm flattens it, increasing the vertical volume of the thoracic cavity. This lowers intra-pulmonary pressure below atmospheric, drawing air in.
- Q179The part of the nephron involved in the active reabsorption of sodium is the:
- Bowman’s capsule
- distal convoluted tubule
- descending limb of Henle’s loop
- proximal convoluted tubule
Answer: (D) proximal convoluted tubule
The proximal convoluted tubule actively reabsorbs the bulk of sodium (about 70%) into the epithelial cells, along with glucose, amino acids and water. The DCT reabsorbs Na⁺ too but conditionally (under aldosterone) and in much smaller amount; the descending limb handles water, not active Na⁺. Hence the PCT.
- Q180Genetic drift operates only in:
- island populations
- smaller populations
- larger populations
- Mendelian populations
Answer: (B) smaller populations
Genetic drift – a chance change in allele frequency by random sampling – is of significance only in small populations, where alleles may easily be lost or fixed by chance. In large populations the effect is negligible.