NEET UG 2025 Question Paper with Solutions (4 May, Code 45) — Solved PYQ with Answer Key

Physics — Solved Questions

1. Q1: Consider a water tank shown in the figure. It has one wall at x=L and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density ρ, the liquid surface
   Answer: B — d^2y/dx^2=ρ g/S\,y
   The correct answer is Option B. Key Points - At the free surface, the excess pressure due to curvature balances the hydrostatic pressure: δ P = S×(curvature). - For a nearly flat surface the curvature d^2y/dx^2 (since dy/dx=tanθ 0). - Hydro
2. Q2: A microscope has an objective of focal length 2 cm, an eyepiece of focal length 4 cm and a tube length of 40 cm. If the distance of distinct vision of the eye is 25 cm, the magnification of the micros
   Answer: B — 125
   The correct answer is 125. Key Points - Magnification of a compound microscope: m=L/fo×D/fe. - Given L=40 cm, fo=2 cm, fe=4 cm, D=25 cm. - m=40/2×25/4=20× 6.25=125.
3. Q3: An electron (mass 9×10^-31 kg and charge 1.6×10^-19 C) moving with speed c/100 (c= speed of light) is injected into a magnetic field B of magnitude 9×10^-4 T perpendicular to its direction of motion.
   Answer: B — E is perpendicular to B and its magnitude is 27×10^2 V m^-1
   The correct answer is Option B. Key Points - For zero deflection the electric force must balance the magnetic force, which requires E=v×B, so EB. - Magnitude: E=vB=c/100× B=3×10^8100× 9×10^-4. - E = 3×10^6× 9×10^-4=27×10^2 V m^-1.
4. Q4: There are two inclined surfaces of equal length L and the same angle of inclination 45^ with the horizontal. One is rough and the other is perfectly smooth. A given body takes 2 times as much time to
   Answer: D — 0.75
   The correct answer is 0.75. Key Points - Smooth incline: asmooth=gsinθ. Rough incline: arough=gsinθ-k gcosθ. - Since t 1/√a, we have troughtsmooth=asmootharough=2. - So sinθ/sinθ-kcosθ=4. At θ=45^ (sinθ=cosθ): 1/1-k=4 k=0.75.
5. Q5: The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying brakes, car A stops after 1000 m and car B stops after 1500 m. If FA and FB are the forces applied by the
   Answer: B — 2/3
   The correct answer is 2/3. Key Points - By the work-energy theorem the braking force does work equal to the car's kinetic energy: F\,S=δ KE=KE (final KE is zero). - So F=KE/S, giving FA/FB=KEA/KEB×SB/SA. - FA/FB=100/225×1500/1000=150/225=2/
6. Q6: The current passing through the battery in the given circuit is:
   Answer: B — 0.5 A
   The correct answer is 0.5 A. Key Points - The bridge in the circuit is a balanced Wheatstone bridge, so the 6 ω arm carries no current and is removed. - The bridge then reduces to two parallel branches (4 ω and 8 ω): R'=4× 8/4+8=8/3\,ω. - T
7. Q7: A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v0 as shown in the figure. If the string gets slack at some point P making an angle θ from the
   Answer: D — (sinθ/2+3sinθ)^1/2
   The correct answer is Option D. Key Points - At P the string goes slack, so tension =0 and the component of gravity along the string supplies the centripetal force: mgsinθ=mv^2/l ... (i) - Energy conservation from the lowest point to P (ris
8. Q8: The output (Y) of the given logic implementation is similar to the output of a/an gate.
   Answer: D — NOR
   The correct answer is NOR. Key Points - Top gate: Y1=A+B= A\, B (NOR of A and B). - Bottom gate: Y2=A+A= A (NOR of A with itself = NOT A). - Final AND gate: Y=Y1· Y2= A\, B· A= A\, B=A+B. - This is exactly the output of a NOR gate.
9. Q9: The electric field in a plane electromagnetic wave is given by Ez = 60cos(5x + 1.5×10^9t) V/m. Then the expression for the corresponding magnetic field is (here subscripts denote the direction of the
   Answer: A — By = 2×10^-7cos(5x + 1.5×10^9t) T
   The correct answer is Option A. Key Points - In an EM wave, E and B are in the same phase with B0=E0/c; E, B and the propagation direction are mutually perpendicular. - B0=60/c=603×10^8=2×10^-7 T. - E is along z and propagation is along x,
10. Q10: A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take g=9.8 m/s
    Answer: A — 21 NS
    The correct answer is 21 NS. Key Points - Speed just before impact: v1=√2gh1=√2×9.8×40=√784=28 m s^-1 (downward). - Speed just after impact: v2=√2gh2=√2×9.8×10=√196=14 m s^-1 (upward). - Impulse =δ p=m(v2-v1)=0.5×(14-(-28))=0.5×42=21 N s.
11. Q11: AB is a part of an electrical circuit (see figure). The potential difference "VA - VB", at the instant when current i=2 A and is increasing at a rate of 1 amp/second, is:
    Answer: D — 10 volt
    The correct answer is 10 volt. Key Points - Traversing A to B across the inductor (1 H), the 5 V cell and the 2 ω resistor: VA - Ldi/dt - 5 - iR = VB. - Substituting L=1 H, di/dt=1 A/s, i=2 A, R=2\,ω: VA-(1)(1)-5-(2)(2)=VB. - VA-VB = 1+5+4
12. Q12: A 2 A current is flowing through two different small circular copper coils having radii in the ratio 1 : 2. The ratio of their respective magnetic moments will be:
    Answer: A — 1 : 4
    The correct answer is 1 : 4. Key Points - Magnetic moment of a current-carrying circular loop: M=IA=Iπ r^2. - With the same current, M r^2. - M1/M2=r1^2/r2^2=(1/2)^2=1/4.
13. Q13: In a certain camera, a combination of four similar thin convex lenses is arranged axially in contact. Then the power of the combination and the total magnification in comparison to the power (p) and m
    Answer: C — 4p and m^4
    The correct answer is 4p and m^4. Key Points - For thin lenses in contact, powers add: peff=p1+p2+p3+p4=4p. - Magnifications multiply: meff=m1× m2× m3× m4=m^4.
14. Q14: An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature 27°C. The mass of
    Answer: C — 0.116 kg
    The correct answer is 0.116 kg. Key Points - Absolute pressure after withdrawal =11+1=12 atm. Moles left: n=PV/RT=12×1.01×10^5×30×10^-3100/12×300=14.54 mol. - Moles withdrawn =18.2-14.54=3.656 mol. - Mass withdrawn =3.656×32116.99 g 0.116 k
15. Q15: In some appropriate units, the time (t) and position (x) relation of a moving particle is given by t=x^2+x. The acceleration of the particle is:
    Answer: B — -2/(2x+1)^3
    The correct answer is Option B. Key Points - t=x^2+x dt/dx=2x+1, so velocity v=dx/dt=1/2x+1. - dv/dx=-2/(2x+1)^2. - Acceleration a=vdv/dx=1/2x+1·-2/(2x+1)^2=-2/(2x+1)^3.
16. Q16: To an AC power supply of 220 V at 50 Hz, a resistor of 20 ω, a capacitor of reactance 25 ω and an inductor of reactance 45 ω are connected in series. The corresponding current in the circuit and the p
    Answer: B — 7.8 A and 45°
    The correct answer is 7.8 A and 45°. Key Points - Impedance: Z=√(XL-XC)^2+R^2=√(45-25)^2+20^2=√20^2+20^2=202\,ω. - Current: I=V/Z=220/202=11/2 7.8 A. - Phase angle: tanφ=XL-XC/R=45-25/20=1 φ=45^.
17. Q17: The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sp
    Answer: D — 108 days
    The correct answer is 108 days. Key Points - With no external torque, angular momentum is conserved: I'ω'=Iω. - For a solid sphere I=2/5mR^2, so I R^2 and ω=2π/T. - 2/5m(2R)^22π/T'=2/5mR^22π/T\;\;T'=4T. - T'=4× 27=108 days.
18. Q18: A model for quantized motion of an electron in a uniform magnetic field B states that the flux passing through the orbit of the electron is n(h/e) where n is an integer, h is Planck's constant and e i
    Answer: B — he/2π m
    The correct answer is he/2π m. Key Points - The magnetic force provides the centripetal force: evB=mv^2/r v=eBr/m. - Quantized flux: φ=BA=Bπ r^2=nh/e Br^2=nh/eπ. - Magnetic moment μ=IA=e/Tπ r^2=evr/2=1/2e·eBr/m· r=e^2Br^2/2m. - Substituting
19. Q19: Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity 2K while that in the middle has thermal conductivity K. The left en
    Answer: C — 5/3
    The correct answer is 5/3. Key Points - Thermal resistance R=l/KA. The two side rods have R=l/2KA each and the middle rod R=l/KA, so Req=2l/KA. - In steady state the heat current is the same through each rod. Across the first rod: 3T-T1/l/2
20. Q20: The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K1 and K2 with thickness 3/8d and d/2 respectively are inserted in the capacitor. Due to this, t
    Answer: A — 2.66
    The correct answer is 2.66. Key Points - The dielectrics fill 3d/8 and d/2, leaving an air gap of d/8 (t3=d/8, K3=1). - Series combination: Ceq=0 At1/K1+t2/K2+t3/K3 with K2=K1/1.25. - Setting Ceq=2C0=20 A/d gives 2=13/8K1+5/8K1+1/8. - Solvi
21. Q21: Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every T min. A girl is driving a scooty with a speed of 60 km/h in the direction X to Y notices that a
    Answer: D — 15 min, 120 km/h
    The correct answer is 15 min, 120 km/h. Key Points - Let the bus speed be v. Spacing between consecutive buses =vT. - Buses in the direction of motion (relative speed v-60) pass every 30 min: (v-60)· 30=vT. - Buses in the opposite direction
22. Q22: A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60^ with it. The other end rests on a rough horizontal floor. The friction force that the floor exert
    Answer: B — 100√3 N
    The correct answer is 100√3 N. Key Points - The rod makes 60^ with the wall, i.e. 30^ with the floor. Let N1 be the normal from the floor, N2 the normal from the smooth wall, and f the friction. - Translational equilibrium: N1=Mg=200 N and
23. Q23: In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency ω(t) and average
    Answer: B — ω(t) increases, A(t) decreases
    The correct answer is ω(t) increases and A(t) decreases (Option 2 in the paper). Key Points - Time period T=2πm/k. As sand leaks out, m decreases, so T decreases and ω=2π/T increases. - At equilibrium the spring stretch is x0=mg/k. As m dec
24. Q24: A balloon is made of a material of surface tension S and its inflation outlet (from where gas is filled in it) has small area A. It is filled with a gas of density ρ and takes a spherical shape of rad
    Answer: C — a=-12,\ α=-12,\ β=-1,\ γ=12,\ δ=72
    The correct answer is Option C. Key Points - Dimensional analysis of T S^αA^βρ^γR^δ: writing [S]=MT^-2, [A]=L^2, [ρ]=ML^-3, [R]=L and matching M^0L^0T^1. - Time exponent: -2α=1α=-12; mass exponent: α+γ=0γ=12. - Length exponent: 2β-3γ+δ=0; w
25. Q25: Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The l
    Answer: C — 4.98 cm
    The correct answer is 4.98 cm. Key Points - Least count =1\,MSD-1\,VSD=1\,MSD-9/10\,MSD=1/10×0.1=0.01 cm. - Zero error =+0.1 cm (zero of vernier at x=0.1 cm when jaws closed), so a positive correction of -0.1 cm is applied. - Observed readi
26. Q26: A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to dis
    Answer: C — Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates
    The correct answer is Option C. Key Points - Surface charge density σ=q/A with dσ/dt= constant, so the displacement current id=Adσ/dt is constant. - The system behaves like a uniform cylindrical current distribution of radius R (the plate r
27. Q27: An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Then:
    Answer: A — Reflected light is completely polarized and the angle of reflection is close to 60^
    The correct answer is Option A. Key Points - Brewster's law: μ=tanB, so tanB=1.73=√3B=60^. - At the Brewster (polarising) angle the reflected light is completely (perfectly) polarized. - The angle of reflection equals the angle of incidence
28. Q28: Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharge
    Answer: D — 3F/8
    The correct answer is 3F/8. Key Points - Initially F=kq^2/r^2. - Third sphere touches A: charge shares equally, so A and the third sphere each become q/2. - The third sphere (q/2) then touches B (q): they share equally to q/2+q/2=3q/4 each;
29. Q29: A container has two chambers of volumes V1=2 litres and V2=3 litres separated by a partition made of a thermal insulator. The chambers contain n1=5 and n2=4 moles of ideal gas at pressures p1=1 atm an
    Answer: B — 1.6 atm
    The correct answer is 1.6 atm. Key Points - The internal energy (and hence pV) is conserved on mixing at the same temperature: p1V1+p2V2=p\,(V1+V2). - 1(2)+2(3)=p(2+3) 8=5p. - p=8/5=1.6 atm.
30. Q30: A particle of mass m is moving around the origin with a constant force F pulling it towards the origin. If Bohr model is used to describe its motion, the radius of the n^th orbit and the particle's sp
    Answer: C — r n^2/3;\ v n^1/3
    The correct answer is r n^2/3 and v n^1/3. Key Points - The constant force provides the centripetal force: F=mv^2/r v^2=Fr/m. - Bohr quantization: mvr=nh/2π v=nh/2π mr. - Combining: Fr/m=n^2h^2/4π^2m^2r^2 r^3 n^2 r n^2/3. - Then v=nh/2π mrn
31. Q31: The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?
    Answer: A — 88 earth days
    The correct answer is 88 earth days. Key Points - Kepler's third law: T^2 R^3. - With RMars=4RMercury: (TMarsTMercury)^2=(4R/R)^3=64TMarsTMercury=8. - TMercury=687/8 85.9 days, closest to 88 earth days.
32. Q32: A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:
    Answer: B — 27 N
    The correct answer is 27 N. Key Points - At height h, gh=GM/(R+h)^2, so Wh/W=R^2/(R+h)^2. - With h=R/3: Wh/W=R^2(4R/3)^2=9/16. - Wh=9/16× 48=27 N.
33. Q33: A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net
    Answer: C — R/16
    The correct answer is R/16. Key Points - Each of the 8 pieces has resistance R/8. - Four pieces in parallel: R/8/4=R/32 (one set); the second set is likewise R/32. - The two sets in series: Req=R/32+R/32=R/16.
34. Q34: De-Broglie wavelength of an electron orbiting in the n=2 state of hydrogen atom is close to (Given Bohr radius =0.052 nm):
    Answer: B — 0.67 nm
    The correct answer is 0.67 nm. Key Points - Bohr radius of the n-th orbit: r=0.052\,n^2 nm. For n=2: r=0.052×4=0.208 nm. - Angular momentum quantization mvr=nh/2π gives λ=h/mv=2π r/n. - For n=2: λ=2π×0.208/2=π×0.208 0.65 nm 0.67 nm.
35. Q35: An electric dipole with dipole moment 5×10^-6 C m is aligned with the direction of a uniform electric field of magnitude 4×10^5 N/C. The dipole is then rotated through an angle of 60^ with respect to
    Answer: B — 1.0 J
    The correct answer is 1.0 J. Key Points - Potential energy of a dipole: U=-PEcosθ, so δ U=PE(cosi-cosf). - Here i=0^, f=60^: δ U=PE(1-1/2)=PE/2. - δ U=1/2× 5×10^-6× 4×10^5=1.0 J.
36. Q36: A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is:
    Answer: B — 2.0 A
    The correct answer is 2.0 A. Key Points - The network is two parallel branches between A and B: branch through C has 1\,ω then 2\,ω; branch through D has 3\,ω then 4\,ω. - RAB=3×1/3+1+2×4/2+4=3/4+8/6=25/12\,ω, so total current I=50/25/12=24
37. Q37: A photon and an electron (mass m) have the same energy E. The ratio (photonelectron) of their de Broglie wavelengths is: (c is the speed of light)
    Answer: C — c2m/E
    The correct answer is c2m/E. Key Points - Photon: E=hcphph=hc/E. - Electron: E=p^2/2m p=√2mE, so e=h/√2mE. - Ratio: phe=hc/E/h/√2mE=c√2mE/E=c2m/E.
38. Q38: Which of the following options represent the variation of photoelectric current with the property of light shown on the x-axis? (A) current vs intensity — straight line through origin; (B) current vs
    Answer: A — A only
    The correct answer is A only. Key Points - The photoelectric current is directly proportional to the intensity of the incident light (more photons per second more photoelectrons). - So the current-vs-intensity graph is a straight line passi
39. Q39: A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-
    Answer: C — 7/57
    The correct answer is 7/57. Key Points - Let the big sphere (2R) have mass M: Iwhole=2/5M(2R)^2=8/5MR^2. - By uniform density, the cut-out sphere of radius R has mass M'=M/8 (volume ratio 1:8). - Its centre is at distance R from the Y-axis,
40. Q40: A full wave rectifier circuit with diodes (D1) and (D2) is shown in the figure. If input supply voltage Vin=220sin(100π t) volt, then at t=15 msec:
    Answer: B — D1 is reverse biased, D2 is forward biased
    The correct answer is D1 reverse biased, D2 forward biased. Key Points - At t=15 ms, phase =100π×0.015=1.5π rad, so Vin=220sin(1.5π)=-220 V (negative half-cycle). - In a centre-tapped full-wave rectifier, during the negative half-cycle the
41. Q41: Two gases A and B are filled at the same pressure in separate cylinders with movable pistons of radius rA and rB, respectively. On supplying an equal amount of heat to both the systems reversibly unde
    Answer: B — 3/4
    The correct answer is 3/4. Key Points - First law: δ Q=δ U+Pδ V. Since δ Q and δ U are the same for both, the work Pδ V is equal, and P is the same, so δ VA=δ VB. - δ V=A\,d=π r^2 d, hence rA^2 dA=rB^2 dB. - rA/rB=dB/dA=9/16=3/4.
42. Q42: A physical quantity P is related to four observations a, b, c and d as follows: P=a^3 b^2/c√d. The percentage errors of measurement in a, b, c and d are 1%, 3%, 2% and 4% respectively. The percentage
    Answer: C — 13%
    The correct answer is 13%. Key Points - For P=a^3 b^2/c√d, the maximum percentage error adds up each contribution weighted by its power. - δ P/P×100=3(1\%)+2(3\%)+1(2\%)+1/2(4\%). - =3+6+2+2=13\%.
43. Q43: The intensity of transmitted light when a polaroid sheet, placed between two crossed polaroids at 22.5^ from the polarization axis of one of the polaroids, is (I0 is the intensity of polarised light a
    Answer: C — I0/8
    The correct answer is I0/8. Key Points - After the middle polaroid (at 22.5^): I1=I0cos^2(22.5^). - The last polaroid is crossed with the first, so it is at 90^-22.5^ from the middle one: I2=I1cos^2(90^-22.5^)=I0cos^2(22.5^)sin^2(22.5^). -
44. Q44: Two identical point masses P and Q, suspended from two separate massless springs of spring constants k1 and k2, respectively, oscillate vertically. If their maximum speeds are the same, the ratio (AQ/
    Answer: D — k1/k2
    The correct answer is k1/k2. Key Points - Maximum speed vmax=Aω, and it is equal for both: APP=AQQ. - With ω=k/m and equal masses: AQ/AP=P/Q=kP/kQ=k1/k2.
45. Q45: A pipe open at both ends has a fundamental frequency f in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to:
    Answer: B — f
    The correct answer is f. Key Points - Open pipe (both ends open) of length L: f=v/2L. - When dipped to half its length in water, the effective air column length is L/2 and it now behaves as a closed pipe (one end closed): f'=v/4(L/2)=v/2L.

Chemistry — Solved Questions

1. Q46: The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n=2 n=3 and n=4 n=6 transitions, respectively, is:
   Answer: D — 1/4
   The correct answer is 1/4. Key Points - 1/λδ E=RH(1/n1^2-1/n2^2), so λ1/δ E. - 23: δ E=RH(1/4-1/9)=RH·5/36. - 46: δ E=RH(1/16-1/36)=RH·20/576. - 2346=δ E46δ E23=20/576/5/36=1/4.
2. Q47: Which of the following statements are true? (A) Unlike Ga that has a very high melting point, Cs has a very low melting point. (B) On Pauling scale, the electronegativity values of N and Cl are not th
   Answer: B — C and E only
   The correct answer is C and E only. Key Points - (A) is false: both Ga (m.p. 303 K) and Cs (m.p. 302 K) have low melting points, so the contrast stated is wrong. - (B) is false: on the Pauling scale both N and Cl have electronegativity =3.0
3. Q48: Match List I with List II. List-I (Ion): A. Co^2+, B. Mg^2+, C. Pb^2+, D. Al^3+. List-II (Group Number in Cation Analysis): I. Group-I, II. Group-III, III. Group-IV, IV. Group-VI. Choose the correct a
   Answer: B — A-III, B-IV, C-I, D-II
   The correct answer is A-III, B-IV, C-I, D-II. Key Points - Co^2+ is precipitated as sulphide in Group-IV (III in the list). - Mg^2+ is a Group-VI cation (IV in the list). - Pb^2+ precipitates as chloride in Group-I (I in the list). - Al^3+
4. Q49: Predict the major product 'P' in the following sequence of reactions: (methylenecyclopentane bearing a CH3 group) (i) HBr, benzoyl peroxide(ii) KCN(iii) Na(Hg)/C2H5OH P (Major).
   Answer: A — Cyclopentane bearing a -CH3 and a -CH2NH2 group (on adjacent carbons)
   The correct answer is Option 1 — the cyclopentane carrying -CH3 and -CH2NH2. Key Points - Step (i) HBr with benzoyl peroxide gives anti-Markovnikov (peroxide/Kharasch) addition, placing Br on the terminal (less substituted) carbon of the ex
5. Q50: Energy and radius of the first Bohr orbit of He^+ and Li^2+ are [Given RH=2.18×10^-18 J, a0=52.9 pm]:
   Answer: A — En(Li^2+)=-19.62×10^-18 J, rn(Li^2+)=17.6 pm; En(He^+)=-8.72×10^-18 J, rn(He^+)=26.4 pm
   The correct answer is Option 1. Key Points - En=-2.18×10^-18\,z^2n^2 J and rn=52.9\,n^2/z pm (for n=1). - He^+ (z=2): E=-2.18×10^-18×4=-8.72×10^-18 J; r=52.9/2=26.4 pm. - Li^2+ (z=3): E=-2.18×10^-18×9=-19.62×10^-18 J; r=52.9/3=17.6 pm.
6. Q51: Which of the following are paramagnetic? (A) [NiCl4]^2-, (B) Ni(CO)4, (C) [Ni(CN)4]^2-, (D) [Ni(H2O)6]^2+, (E) Ni(PPh3)4. Choose the correct answer from the options given below:
   Answer: C — A and D only
   The correct answer is A and D only. Key Points - [NiCl4]^2-: Ni^2+ (3d^8), weak-field Cl^- sp^3, tetrahedral, 2 unpaired electrons — paramagnetic. - [Ni(H2O)6]^2+: Ni^2+ (3d^8), sp^3d^2, octahedral, 2 unpaired electrons — paramagnetic. - Ni
7. Q52: Given below are two statements: Statement I: Like nitrogen that can form ammonia, arsenic can form arsine. Statement II: Antimony cannot form antimony pentoxide. In the light of the above statements,
   Answer: C — Statement I is correct but Statement II is incorrect
   The correct answer is Statement I is correct but Statement II is incorrect. Key Points - All group-15 elements form EH3-type hydrides: N forms ammonia (NH3) and As forms arsine (AsH3), so Statement I is correct. - Group-15 elements form bot
8. Q53: Which among the following electronic configurations belong to main group elements? (A) [Ne]3s^1, (B) [Ar]3d^34s^2, (C) [Kr]4d^105s^25p^5, (D) [Ar]3d^104s^1, (E) [Rn]5f^06d^27s^2. Choose the correct an
   Answer: B — A and C only
   The correct answer is A and C only. Key Points - Main group (representative) elements have their outermost electron in an s- or p-orbital. - (A) [Ne]3s^1 = Na (s-block, main group); (C) [Kr]4d^105s^25p^5 = I (p-block, main group). - (B) [Ar
9. Q54: Dalton's Atomic theory could not explain which of the following?
   Answer: D — Law of gaseous volume
   The correct answer is Law of gaseous volume. Key Points - Dalton's atomic theory successfully explained the laws of chemical combination — conservation of mass, constant proportion, and multiple proportion. - It could not explain Gay-Lussac
10. Q55: Consider the following compounds: KO2, H2O2 and H2SO4. The oxidation state of the underlined elements in them are, respectively:
    Answer: A — +1, -1, and +6
    The correct answer is +1, -1, and +6. Key Points - KO2 (potassium superoxide): alkali metal K is always +1 (the O is the superoxide ion, -12). - H2O2: it is a peroxide, so each O is -1. - H2SO4: with H = +1 and O = -2, sulphur is +6.
11. Q56: If the half-life (t1/2) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to:
    Answer: D — 10 minutes
    The correct answer is 10 minutes. Key Points - First order: t=2.303/k0/At, and k=0.693t1/2. - For 99.9% completion, A0/At=1000, so t=2.303/k1000=2.303×3/k. - This equals 10\,t1/2=10×1=10 minutes.
12. Q57: The correct order of the wavelength of light absorbed by the following complexes is: (A) [Co(NH3)6]^3+, (B) [Co(CN)6]^3-, (C) [Cu(H2O)4]^2+, (D) [Ti(H2O)6]^3+. Choose the correct answer from the optio
    Answer: B — B<A<D<C
    The correct answer is B<A<D<C. Key Points - Wavelength absorbed λ1/o (larger crystal-field splitting shorter wavelength). - Measured absorptions: B [Co(CN)6]^3- 310 nm, A [Co(NH3)6]^3+ 475 nm, D [Ti(H2O)6]^3+ 498 nm, C [Cu(H2O)4]^2
13. Q58: Which one of the following compounds can exist as cis-trans isomers?
    Answer: D — 1,2-Dimethylcyclohexane
    The correct answer is 1,2-Dimethylcyclohexane. Key Points - Geometrical (cis-trans) isomerism needs restricted rotation and two different groups on each of the two centres. - 1,2-Dimethylcyclohexane has two methyls on adjacent ring carbons
14. Q59: Phosphoric acid ionizes in three steps with their ionization constant values Ka1, Ka2 and Ka3, respectively, while K is the overall ionization constant. Which of the following statements are true? (A)
    Answer: D — A, B and C only
    The correct answer is A, B and C only. Key Points - The overall K=Ka1· Ka2· Ka3, so log K=log Ka1+log Ka2+log Ka3 — (A) true. - Successive ionization constants fall sharply: Ka1=7.5×10^-3Ka2=6.2×10^-8Ka3=1.7×10^-12 — so (C) true and H3PO4 i
15. Q60: Which one of the following reactions does NOT give benzene as the product? (1) Sodium benzoate (C6H5COONa) with sodalime, δ; (2) n-hexane over Mo2O3 at 773 K, 10-20 atm; (3) ethyne (HCCH) through a re
    Answer: D — Benzene diazonium chloride + H2O, warm
    The correct answer is benzene diazonium chloride + H2O (warm). Key Points - Sodium benzoate + sodalime δ (decarboxylation) benzene. - n-Hexane over Mo2O3 (aromatization) benzene; ethyne through a red-hot iron tube (cyclic trimerization) ben
16. Q61: If the molar conductivity (m) of a 0.050 mol L^-1 solution of a monobasic weak acid is 90 S cm^2 mol^-1, its extent (degree) of dissociation will be [Assume λ^+=349.6 S cm^2 mol^-1 and λ^-=50.4 S cm^2
    Answer: C — 0.225
    The correct answer is 0.225. Key Points - Degree of dissociation α=m/m^. - m^=λ^++λ^-=349.6+50.4=400 S cm^2 mol^-1. - α=90/400=0.225.
17. Q62: Given below are two statements: Statement I: A hypothetical diatomic molecule with bond order zero is quite stable. Statement II: As bond order increases, the bond length increases. In the light of th
    Answer: B — Both Statement I and Statement II are false
    The correct answer is Both Statement I and Statement II are false. Key Points - A molecule with zero (or negative) bond order is unstable and does not exist — so Statement I is false. - Bond length is inversely related to bond order: as bon
18. Q63: Out of the following complex compounds, which of the compound will be having the minimum conductance in solution?
    Answer: A — [Co(NH3)3Cl3] (NTA awarded marks for both option 1 and option 2)
    The correct answer is [Co(NH3)3Cl3] — NTA accepted both options 1 and 2 as correct. Key Points - Conductance depends on the number of ions furnished in solution; more ions higher conductance. - [Co(NH3)3Cl3] (Co^3+, all 3 Cl inside) is a ne
19. Q64: Match List-I with List-II. List-I: A. XeO3, B. XeF2, C. XeOF4, D. XeF6. List-II: (I) sp^3d; linear, (II) sp^3; pyramidal, (III) sp^3d^3; distorted octahedral, (IV) sp^3d^2; square pyramidal. Choose th
    Answer: A — A-II, B-I, C-IV, D-III
    The correct answer is A-II, B-I, C-IV, D-III. Key Points - XeO3: sp^3, pyramidal (II). - XeF2: sp^3d, linear (I). - XeOF4: sp^3d^2, square pyramidal (IV). - XeF6: sp^3d^3, distorted octahedral (III).
20. Q65: C(s)+2H2(g)CH4(g); δ H=-74.8 kJ mol^-1. Which of the following diagrams gives an accurate representation of the above reaction? [R reactants; P products]
    Answer: A — Reactants (R) at higher energy, products (P) lower by 74.8 kJ, with an activation-energy hump (exothermic)
    The correct answer is Option 1. Key Points - δ H=-74.8 kJ mol^-1 is negative, so the reaction is exothermic: products have lower energy than reactants. - The energy profile shows reactants (R) rising over an activation-energy barrier and th
21. Q66: Match List-I (Example) with List-II (Type of Solution). List-I: A. Humidity, B. Alloys, C. Amalgams, D. Smoke. List-II: (I) Solid in solid, (II) Liquid in gas, (III) Solid in gas, (IV) Liquid in solid
    Answer: B — A-II, B-I, C-IV, D-III
    The correct answer is A-II, B-I, C-IV, D-III. Key Points - Humidity = water vapour in air = liquid in gas (II). - Alloys = solid in solid (I). - Amalgams = a metal dissolved in mercury = liquid in solid (IV). - Smoke = solid particles in ga
22. Q67: The correct order of decreasing basic strength of the given amines is:
    Answer: C — N-ethylethanamine ethanamine N-methylaniline benzenamine
    The correct answer is N-ethylethanamine ethanamine N-methylaniline benzenamine. Key Points - Lower the pKb, higher the basicity; aliphatic amines are stronger bases than aromatic amines (whose lone pair is delocalised into the ring). - Amon
23. Q68: Among the following, choose the ones with equal number of atoms. A. 212 g of Na2CO3(s) [molar mass = 106 g], B. 248 g of Na2O(s) [molar mass = 62 g], C. 240 g of NaOH(s) [molar mass = 40 g], D. 12 g o
    Answer: B — A, B, and D only
    The correct answer is A, B, and D only. Key Points - Number of atoms =given massmolar mass×(atomicity)× NA. - A: 212/106×6=12NA; B: 248/62×3=12NA; D: 12/2×2=12NA — all equal to 12NA. - C: 240/40×3=18NA and E: 220/44×3=15NA differ. - Hence A
24. Q69: Match List-I (Name of Vitamin) with List-II (Deficiency disease). List-I, List-II; ---, ---; A. Vitamin B₁₂, I. Cheilosis; B. Vitamin D, II. Convulsions; C. Vitamin B₂, III. Rickets; D. Vitamin B₆, IV
    Answer: B — A-IV, B-III, C-I, D-II
    The correct answer is A-IV, B-III, C-I, D-II. Key Points - Vitamin B12 deficiency pernicious anaemia (IV). - Vitamin D deficiency rickets (III). - Vitamin B2 (riboflavin) deficiency cheilosis (I). - Vitamin B6 (pyridoxine) deficiency convul
25. Q70: The correct order of decreasing acidity of the following aliphatic acids is:
    Answer: C — HCOOH CH3COOH (CH3)2CHCOOH (CH3)3CCOOH
    The correct answer is HCOOH CH3COOH (CH3)2CHCOOH (CH3)3CCOOH. Key Points - Electron-donating (+I) alkyl groups destabilise the carboxylate anion and decrease acidity. - HCOOH has no alkyl group, so it is the strongest; adding more/bulkier a
26. Q71: Given below are two statements: Statement I: Ferromagnetism is considered as an extreme form of paramagnetism. Statement II: The number of unpaired electrons in a Cr^2+ ion (Z=24) is the same as that
    Answer: C — Statement I is true but Statement II is false
    The correct answer is Statement I is true but Statement II is false. Key Points - Ferromagnetic substances are attracted very strongly in a magnetic field; ferromagnetism is indeed an extreme form of paramagnetism — Statement I is true. - C
27. Q72: Match List-I (Mixture) with List-II (Method of separation). List-I: A. CHCl3 + C6H5NH2, B. Crude oil in petroleum industry, C. Glycerol from spent-lye, D. Aniline-water. List-II: (I) Distillation unde
    Answer: A — A-IV, B-III, C-I, D-II
    The correct answer is A-IV, B-III, C-I, D-II. Key Points - CHCl3 + aniline (large boiling-point difference) simple distillation (IV). - Crude oil fractional distillation (III). - Glycerol from spent-lye (decomposes on strong heating) distil
28. Q73: For the reaction A(g) 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K. [Given: R = 0.0831 L atm mol^-1 K^-1] KP for the reac
    Answer: C — 0.033
    The correct answer is 0.033. Key Points - KC=kf/kb=1/2500 (backward is 2500 times faster). - KP=KC(RT)^δ ng with δ ng=2-1=1. - KP=1/2500×0.0831×1000=0.033.
29. Q74: Given below are two statements: Statement-I: Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273-278 K. It decomposes easily in the dry state. Statement-II: Insertion
    Answer: A — Both Statement I and Statement II are correct
    The correct answer is Both Statement I and Statement II are correct. Key Points - Benzenediazonium chloride is prepared by diazotisation of aniline with nitrous acid (NaNO2 + HCl) at 273-278 K, and it decomposes readily when dry — Statement
30. Q75: How many products (including stereoisomers) are expected from monochlorination of the following compound? (2-methylbutane, i.e. (CH3)2CH-CH2-CH3)
    Answer: D — 6
    The correct answer is 6. Key Points - 2-Methylbutane has four types of hydrogens, giving four constitutional monochloro products. - Chlorination at the C-2 (giving 2-chloro-3-methylbutane... i.e. (CH3)2CH-CHCl-CH3) and at the terminal CH3 o
31. Q76: Among the given compounds I-III, the correct order of bond dissociation energy of the C-H bond marked with is: (I) the aromatic C-H of benzene (sp^2 carbon); (II) the terminal alkyne C-H of phenylacet
    Answer: A — II I III
    The correct answer is II I III. Key Points - The bond dissociation energy of a C-H bond increases with the s-character of the carbon's hybrid orbital (more s-character shorter, stronger bond). - (II) sp carbon (50% s), (I) sp^2 carbon (33%
32. Q77: Which one of the following compounds does NOT decolourize bromine water? (1) cyclohexane; (2) phenol (C6H5OH); (3) styrene (C6H5-CH=CH2); (4) aniline (C6H5-NH2).
    Answer: A — Cyclohexane
    The correct answer is cyclohexane. Key Points - Bromine water is decolourised either by addition across a C=C (unsaturation) or by electrophilic substitution on an activated ring. - Styrene adds Br2 across its double bond; phenol and anilin
33. Q78: The major product of the following reaction is: PhCO-CH2CH2-CN (a keto-nitrile) treated with (i) CH3MgBr (excess), then (ii) H3O^+.
    Answer: B — Ph-C(CH3)(OH)-CH2CH2-CO-CH3 (tertiary alcohol at the benzylic C and a methyl ketone from the nitrile)
    The correct answer is Option 2. Key Points - Excess CH3MgBr attacks both functional groups. At the ketone (PhC=O) it adds CH3 to give, after workup, a tertiary benzylic alcohol Ph-C(CH3)(OH)-. - At the nitrile (-CN) the Grignard adds once t
34. Q79: Which of the following aqueous solutions will exhibit highest boiling point?
    Answer: C — 0.01M Na2SO4
    The correct answer is 0.01M Na2SO4. Key Points - Elevation of boiling point δ Tb=i\,Kb\,m, so it depends on the product i× m (van't Hoff factor × molality). - Urea: i× m=1×0.01=0.01; KNO3: 2×0.01=0.02; Na2SO4: 3×0.01=0.03; glucose: 1×0.015=
35. Q80: Match List-I with List-II. List-I: A. Haber process, B. Wacker oxidation, C. Wilkinson catalyst, D. Ziegler catalyst. List-II: (I) Fe catalyst, (II) PdCl2, (III) [(PPh3)3RhCl], (IV) TiCl4 with Al(CH3)
    Answer: C — A-I, B-II, C-III, D-IV
    The correct answer is A-I, B-II, C-III, D-IV. Key Points - Haber process (NH3 synthesis) uses an Fe catalyst (I). - Wacker oxidation (ethene ethanal) uses PdCl2 (II). - Wilkinson's catalyst is [(PPh3)3RhCl] (III). - Ziegler-Natta catalyst i
36. Q81: 5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure X and Y are 63 torr and 78 torr respectively. Which of the following is t
    Answer: B — The solution shows negative deviation
    The correct answer is the solution shows negative deviation. Key Points - For an ideal solution (Raoult's law): Ptotal=xXP^X+xYP^Y=5/15(63)+10/15(78)=21+52=73 torr. - The observed vapour pressure (70 torr) is less than the ideal value (73 t
37. Q82: Sugar 'X': A. is found in honey; B. is a keto sugar; C. exists in α and β-anomeric forms; D. is laevorotatory. 'X' is:
    Answer: B — D-Fructose
    The correct answer is D-Fructose. Key Points - D-Fructose is present in honey (with glucose) and is a ketohexose (a keto sugar). - It forms a cyclic (furanose) structure that exists in α- and β-anomeric forms. - Fructose is strongly laevoro
38. Q83: Identify the suitable reagent for the following conversion: methyl benzoate (C6H5COOCH3, an ester) benzaldehyde (C6H5CHO).
    Answer: B — (i) AlH(iBu)2 (DIBAL-H), (ii) H2O
    The correct answer is (i) AlH(iBu)2 (DIBAL-H), (ii) H2O. Key Points - DIBAL-H (diisobutylaluminium hydride) is a mild reducing agent that reduces esters (and nitriles) to aldehydes, stopping at the aldehyde stage. - LiAlH4 and NaBH4 would o
39. Q84: Given below are two statements: one labelled Assertion (A) and the other Reason (R). Assertion (A): 1-iodobutane (an alkyl iodide) undergoes SN2 reaction faster than 1-chlorobutane (an alkyl chloride)
    Answer: A — Both A and R are true and R is the correct explanation of A
    The correct answer is Both A and R are true and R is the correct explanation of A. Key Points - In an SN2 reaction the rate depends strongly on how good the leaving group is. - Iodide (I^-) is a better leaving group than chloride because th
40. Q85: The standard heat of formation, in kcal/mol, of Ba^2+ is: [Given: standard heat of formation of SO4^2- ion (aq) = -216 kcal/mol, standard heat of crystallisation of BaSO4(s) = -4.5 kcal/mol, standard
    Answer: A — -128.5
    The correct answer is -128.5 kcal/mol. Key Points - Combine: (1) formation of SO4^2- (-216), (2) crystallisation Ba^2+(g) + SO4^2-(g) BaSO4 (-4.5), (3) formation of BaSO4(s) (-349). - δ Hf(Ba^2+)=δ Hf(BaSO4)-δ Hf(SO4^2-)-δ Hcryst. - =-349-(
41. Q86: Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula C4H8O is:
    Answer: C — 10
    The correct answer is 10. Key Points - Cyclic ethers of C4H8O keep the O in the ring; counting each ring skeleton and its stereoisomers: - Oxetane rings (4-membered) and oxolane/methyl-substituted variants that contain a chiral carbon each
42. Q87: Identify the correct orders against the property mentioned. A. H2O NH3 CHCl3 - dipole moment. B. XeF4 XeO3 XeF2 - number of lone pairs on central atom. C. O-H C-H N-O - bond length. D. N2 O2 H2 - bond
    Answer: A — A, D only
    The correct answer is A, D only. Key Points - A (true): dipole moments H2O (1.85 D) NH3 (1.47 D) CHCl3 (1.04 D). - D (true): bond enthalpy follows bond order — N2 (b.o. 3) O2 (b.o. 2) H2 (b.o. 1). - B (false): lone pairs are XeF2 (3) XeF4 (
43. Q88: Higher yield of NO in N2(g) + O2(g) 2NO(g) can be obtained at [δ H of the reaction = +180.7 kJ mol^-1]: A. Higher temperature, B. Lower temperature, C. Higher concentration of N2, D. Higher concentrat
    Answer: D — A, C, D only
    The correct answer is A, C, D only. Key Points - The reaction is endothermic (δ H=+180.7 kJ mol^-1), so increasing temperature shifts equilibrium forward, raising NO yield (A). - Increasing the concentration of a reactant (N2 or O2) also sh
44. Q89: If the rate constant of a reaction is 0.03 s^-1, how much time does it take for 7.2 mol L^-1 concentration of the reactant to get reduced to 0.9 mol L^-1? (Given: log 2 = 0.301)
    Answer: A — 69.3 s
    The correct answer is 69.3 s. Key Points - The unit of k (s^-1) shows it is a first order reaction: t=2.303/k/a-x. - t=2.303/0.037.2/0.9=2.303/0.03log 8=2.303/0.03×32. - =2.303/0.03×3×0.301=69.3 s.
45. Q90: Which one of the following reactions does NOT belong to "Lassaigne's test"? (1) Na + C + N δ NaCN; (2) 2Na + S δ Na2S; (3) Na + X δ NaX; (4) 2CuO + C δ 2Cu + CO2.
    Answer: D — 2CuO + C 2Cu + CO2
    The correct answer is 2CuO + C 2Cu + CO2. Key Points - Lassaigne's (sodium fusion) test detects N, S, halogens and P in an organic compound by fusing it with sodium: Na + C + N NaCN, 2Na + S Na2S, Na + X NaX. - The reaction 2CuO + C 2Cu + C

Botany — Solved Questions

1. Q91: The complex II of mitochondrial electron transport chain is also known as:
   Answer: B — Succinate dehydrogenase
   The correct answer is Succinate dehydrogenase. Key Points - Complex II of the mitochondrial electron transport chain is succinate dehydrogenase. - For reference: Complex I = NADH dehydrogenase, Complex III = cytochrome bc1, Complex IV = cyt
2. Q92: Polymerase chain reaction (PCR) amplifies DNA following the equation:
   Answer: B — 2^n
   The correct answer is 2^n. Key Points - In PCR the DNA amount doubles every cycle, so after n cycles it is amplified by a factor of 2^n. - For example, after 3 cycles: 2^3=8 copies (fragments) are formed.
3. Q95: Which one of the following statements refers to Reductionist Biology?
   Answer: A — Physico-chemical approach to study and understand living organisms
   The correct answer is Physico-chemical approach to study and understand living organisms. Key Points - Reductionist Biology is the physico-chemical approach — explaining living processes in terms of the physics and chemistry of their molecu
4. Q96: Given below are two statements: Statement I: In the RNA world, RNA is considered the first genetic material evolved to carry out essential life processes. RNA acts as a genetic material and also as a
   Answer: A — Both statement I and statement II are correct
   The correct answer is Both statement I and statement II are correct. Key Points - In the RNA world, RNA was the first genetic material and also acted as a catalyst (ribozyme); being reactive, it is unstable — Statement I is correct. - DNA e
5. Q97: Epiphytes that are growing on a mango branch is an example of which of the following?
   Answer: A — Commensalism
   The correct answer is Commensalism. Key Points - Commensalism is an interaction where one species benefits and the other is neither harmed nor benefited (+/0). - An orchid/epiphyte growing on a mango branch gains support and access to light
6. Q98: From the statements given below choose the correct option: A. The eukaryotic ribosomes are 80S and prokaryotic ribosomes are 70S. B. Each ribosome has two sub-units. C. The two sub-units of 80S riboso
   Answer: A — A, B, C are true
   The correct answer is A, B, C are true. Key Points - Eukaryotic ribosomes are 80S, prokaryotic are 70S (A true), and each ribosome has two sub-units (B true). - The 80S ribosome has 60S + 40S sub-units and the 70S has 50S + 30S sub-units (C
7. Q99: Which one of the following is an example of ex-situ conservation?
   Answer: C — Zoos and botanical gardens
   The correct answer is Zoos and botanical gardens. Key Points - Ex-situ conservation protects species outside their natural habitat: zoological parks (zoos), botanical gardens, wildlife safari parks, seed banks, etc. - National parks, wildli
8. Q100: Given below are two statements: Statement I: The primary source of energy in an ecosystem is solar energy. Statement II: The rate of production of organic matter during photosynthesis in an ecosystem
   Answer: C — Statement I is correct but statement II is incorrect
   The correct answer is Statement I is correct but statement II is incorrect. Key Points - The primary source of energy in an ecosystem is solar energy — Statement I is correct. - The rate of production of organic matter during photosynthesis
9. Q102: Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Both wind and water pollinated flowers are not very colourful and do not produc
   Answer: B — Both A and R are true but R is NOT the correct explanation of A
   The correct answer is Both A and R are true but R is NOT the correct explanation of A. Key Points - Wind- and water-pollinated flowers are not colourful and lack nectar because they do not need to attract animal pollinators — A is true. - T
10. Q103: Which of the following is an example of non-distilled alcoholic beverage produced by yeast?
    Answer: C — Beer
    The correct answer is Beer. Key Points - Wine and beer are produced without distillation (directly from fermentation by yeast). - Whisky, brandy and rum are produced by distillation of the fermented broth.
11. Q104: Given below are two statements: Statement I: In a floral formula stands for zygomorphic nature of the flower, and G stands for inferior ovary. Statement II: In a floral formula stands for actinomorphi
    Answer: D — Statement I is incorrect but Statement II is correct
    The correct answer is Statement I is incorrect but Statement II is correct. Key Points - The symbol denotes an actinomorphic (radially symmetric) flower; a zygomorphic flower is shown by the % symbol — so Statement I is wrong on both counts
12. Q105: Streptokinase produced by the bacterium Streptococcus is used for:
    Answer: D — Removing clots from blood vessels
    The correct answer is Removing clots from blood vessels. Key Points - Streptokinase, produced by Streptococcus and modified by genetic engineering, is used as a 'clot buster' to remove clots from the blood vessels of patients who have had a
13. Q106: Which chromosome in the human genome has the highest number of genes?
    Answer: C — Chromosome 1
    The correct answer is Chromosome 1. Key Points - In the human genome, Chromosome 1 carries the highest number of genes (about 2968). - The Y chromosome has the fewest genes (around 231).
14. Q108: Which one of the following phytohormones promotes nutrient mobilization which helps in the delay of leaf senescence in plants?
    Answer: D — Cytokinin
    The correct answer is Cytokinin. Key Points - Cytokinins promote nutrient mobilisation, which helps delay leaf senescence (the Richmond-Lang effect). - They also help overcome apical dominance and promote cell division.
15. Q111: Given below are the stages in the life cycle of pteridophytes. Arrange the following stages in the correct sequence. A. Prothallus stage, B. Meiosis in spore mother cells, C. Fertilisation, D. Formati
    Answer: A — B, A, D, E, C
    The correct answer is B, A, D, E, C. Key Points - The pteridophyte life cycle proceeds: B meiosis in spore mother cells A prothallus (gametophyte) stage D formation of archegonia and antheridia E transfer of antherozoids to archegonia in pr
16. Q113: Which of the following organisms cannot fix nitrogen? A. Azotobacter, B. Oscillatoria, C. Anabaena, D. Volvox, E. Nostoc. Choose the correct answer from the options given below:
    Answer: B — D only
    The correct answer is D only (Volvox). Key Points - Azotobacter (a free-living aerobic bacterium) and the cyanobacteria Oscillatoria, Anabaena and Nostoc are all nitrogen fixers. - Volvox is a colonial green alga and cannot fix nitrogen.
17. Q114: Given below are two statements: Statement I: Transfer RNAs and ribosomal RNA do not interact with mRNA. Statement II: RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cel
    Answer: D — Statement I is incorrect but statement II is correct
    The correct answer is Statement I is incorrect but statement II is correct. Key Points - Both tRNA and rRNA interact with mRNA during translation (tRNA reads codons, rRNA of the ribosome binds mRNA) — so Statement I is incorrect. - RNA inte
18. Q115: In the represented plasmid (with Ori, amp^R and tet^R resistance genes, and the β-galactosidase gene at the EcoRI site), an alien piece of DNA is inserted at the EcoRI site. Which of the following str
    Answer: C — White color colonies will be selected
    The correct answer is White color colonies will be selected. Key Points - The alien DNA is inserted into the β-galactosidase gene at the EcoRI site, causing insertional inactivation of that gene. - Non-recombinants (intact β-galactosidase)
19. Q116: Which of the following genetically engineered organisms was used by Eli Lilly to prepare human insulin?
    Answer: A — Bacterium
    The correct answer is Bacterium. Key Points - In 1983 Eli Lilly prepared the two DNA sequences for the A and B chains of human insulin and introduced them into plasmids of E. coli (a bacterium) to produce the insulin chains.
20. Q117: Name the class of enzyme that usually catalyzes the following reaction: S-G+S^\ S+S^\-G, where G is a group other than hydrogen, S is a substrate and S^\ is another substrate.
    Answer: C — Transferase
    The correct answer is Transferase. Key Points - Enzymes that catalyse the transfer of a group G (other than hydrogen) between two substrates are transferases. - Ligases join two molecules (forming C-O, C-S, C-N bonds); lyases remove groups
21. Q118: Find the statement that is NOT correct with regard to the structure of monocot stem.
    Answer: A — Hypodermis is parenchymatous
    The correct answer (the incorrect statement) is Hypodermis is parenchymatous. Key Points - In a monocot stem the hypodermis is sclerenchymatous (not parenchymatous), so this statement is wrong. - Correct features: vascular bundles are scatt
22. Q119: The correct sequence of events in the life cycle of bryophytes is: A. Fusion of antherozoid with egg, B. Attachment of gametophyte to substratum, C. Reduction division to produce haploid spores, D. Fo
    Answer: C — B, E, A, D, C
    The correct answer is B, E, A, D, C. Key Points - Bryophyte life cycle: B attachment of gametophyte to substratum E release of antherozoids into water A fusion of antherozoid with egg D formation of sporophyte C reduction division producing
23. Q121: Match List-I with List-II. List-I: A. Centromere, B. Cilium, C. Cristae, D. Cell membrane. List-II: (I) Mitochondrion, (II) Cell division, (III) Cell movement, (IV) Phospholipid Bilayer. Choose the co
    Answer: D — A-II, B-III, C-I, D-IV
    The correct answer is A-II, B-III, C-I, D-IV. Key Points - Centromere helps in cell division (II). - Cilium helps in cell movement (III). - Cristae are the finger-like folds of the mitochondrion (I). - Cell membrane is a phospholipid bilaye
24. Q122: Match List-I with List-II. List-I: A. Chlorophyll a, B. Chlorophyll b, C. Xanthophylls, D. Carotenoids. List-II: (I) Yellow-green, (II) Yellow, (III) Blue-green, (IV) Yellow to Yellow-orange. Choose t
    Answer: B — A-III, B-I, C-II, D-IV
    The correct answer is A-III, B-I, C-II, D-IV. Key Points - Chlorophyll a — blue-green (III). - Chlorophyll b — yellow-green (I). - Xanthophylls — yellow (II). - Carotenoids — yellow to yellow-orange (IV).
25. Q124: In the seeds of cereals, the outer covering of endosperm separates the embryo by a protein-rich layer called:
    Answer: D — Aleurone layer
    The correct answer is Aleurone layer. Key Points - In monocot (cereal) seeds, the outer covering of the endosperm is a proteinaceous layer called the aleurone layer, which separates the embryo from the endosperm. - Coleoptile and coleorhiza
26. Q126: Identify the part of a bio-reactor which is used as a foam breaker from the given figure (with parts labelled A: flat-bladed impeller, B: motor, C, D: sterile air outlet at the base).
    Answer: D — C
    The correct answer is C. Key Points - In a stirred-tank bioreactor: B = motor, A = flat-bladed impeller, D = sterile air (sparger inlet), and C = foam breaker. - The foam breaker (C) controls and breaks the foam formed during the fermentati
27. Q127: Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): A typical unfertilised, angiosperm embryo sac at maturity is 8 nucleate and 7-c
    Answer: C — A is true but R is false
    The correct answer is A is true but R is false. Key Points - A mature angiosperm embryo sac is 8-nucleate and 7-celled — so Assertion A is correct. - The egg apparatus (at the micropylar end) consists of two synergids and one egg cell — it
28. Q128: A specialised membranous structure in a prokaryotic cell which helps in cell wall formation, DNA replication and respiration is:
    Answer: A — Mesosome
    The correct answer is Mesosome. Key Points - The mesosome is a membranous in-folding of the plasma membrane in bacteria that helps in cell wall formation, DNA replication and distribution to daughter cells, and it contains enzymes for respi
29. Q129: Which of the following are the post-transcriptional events in a eukaryotic cell? A. Transport of pre-mRNA to cytoplasm prior to splicing. B. Removal of introns and joining of exons. C. Addition of met
    Answer: B — B, C, D only
    The correct answer is B, C, D only. Key Points - Post-transcriptional processing of hnRNA includes splicing (removal of introns and joining of exons, B), 5' capping (addition of a methyl guanosine cap, C) and 3' polyadenylation (addition of
30. Q130: What is the pattern of inheritance for polygenic trait?
    Answer: B — Non-mendelian inheritance pattern
    The correct answer is Non-mendelian inheritance pattern. Key Points - Polygenic inheritance is a trait controlled by two or more genes (e.g. human skin colour, height). - Because the phenotype is the additive effect of several genes (and of
31. Q131: Which one of the following enzymes contains 'Haem' as the prosthetic group?
    Answer: D — Catalase
    The correct answer is Catalase. Key Points - In catalase (and peroxidase), which break down hydrogen peroxide into water and oxygen, haem is the prosthetic group forming part of the active site. - Carbonic anhydrase uses zinc as a cofactor;
32. Q132: Each of the following characteristics represent a Kingdom proposed by Whittaker. Arrange the following in increasing order of complexity of body organization. A. Multicellular heterotrophs with cell w
    Answer: B — C, E, A, D, B
    The correct answer is C, E, A, D, B. Key Points - The five kingdoms in increasing complexity of body organisation: C Monera (prokaryotes) E Protista (eukaryotes, cellular level) A Fungi (multicellular heterotrophs, chitin wall) D Plantae (e
33. Q133: Who is known as the father of Ecology in India?
    Answer: B — Ramdeo Misra
    The correct answer is Ramdeo Misra. Key Points - Ramdeo Misra is regarded as the father of Ecology in India for his pioneering contributions to ecological research and education.
34. Q134: Match List-I with List-II. List-I: A. Alfred Hershey and Martha Chase, B. Euchromatin, C. Frederick Griffith, D. Heterochromatin. List-II: (I) Streptococcus pneumoniae, (II) Densely packed and dark-st
    Answer: C — A-IV, B-III, C-I, D-II
    The correct answer is A-IV, B-III, C-I, D-II. Key Points - Hershey and Chase gave the unequivocal proof that DNA is the genetic material (IV). - Euchromatin is loosely packed and light-stained (III); Heterochromatin is densely packed and da
35. Q136: Given below are two statements: Statement I: The DNA fragments extracted from gel electrophoresis can be used in construction of recombinant DNA. Statement II: Smaller size DNA fragments are observed
    Answer: A — Both statement I and statement II are correct
    The correct answer is Both statement I and statement II are correct. Key Points - DNA bands separated on an agarose gel are cut out and extracted (elution); these purified fragments are joined with cloning vectors to construct recombinant D
36. Q137: Match List-I with List-II. List-I: A. Adenosine, B. Adenylic acid, C. Adenine, D. Alanine. List-II: (I) Nitrogen base, (II) Nucleotide, (III) Nucleoside, (IV) Amino acid. Choose the option with all co
    Answer: C — A-III, B-II, C-I, D-IV
    The correct answer is A-III, B-II, C-I, D-IV. Key Points - Adenosine = a nucleoside (nitrogen base + sugar) — III. - Adenylic acid = a nucleotide (base + sugar + phosphate) — II. - Adenine = a nitrogen base (purine) — I. - Alanine = an amin
37. Q140: Given below are two statements: one is labelled as Assertion (A), and the other is labelled as Reason (R). Assertion (A): The primary function of the Golgi apparatus is to package the materials made b
    Answer: B — Both A and R are true but R is not the correct explanation of A
    The correct answer is Both A and R are true but R is not the correct explanation of A. Key Points - The Golgi apparatus packages materials made by the ER and delivers them to intracellular targets or outside the cell — A is correct. - Vesic
38. Q141: Match List-I with List-II. List-I: A. Scutellum, B. Non-albuminous seed, C. Epiblast, D. Perisperm. List-II: (I) Persistent nucellus, (II) Cotyledon of Monocot seed, (III) Groundnut, (IV) Rudimentary
    Answer: A — A-II, B-III, C-IV, D-I
    The correct answer is A-II, B-III, C-IV, D-I. Key Points - Scutellum = cotyledon of a monocot seed (II). - Non-albuminous seed example = groundnut (III). - Epiblast = rudimentary cotyledon in a monocot seed (IV). - Perisperm = the persisten
39. Q144: Silencing of specific mRNA is possible via RNAi because of:
    Answer: A — Complementary dsRNA
    The correct answer is Complementary dsRNA. Key Points - RNA interference silences a specific mRNA because a complementary double-stranded RNA (dsRNA) binds to the target mRNA and prevents its translation. - This mechanism is used as a cellu
40. Q145: Genes R and Y follow independent assortment. If RRYY produce round yellow seeds and rryy produce wrinkled green seeds, what will be the phenotypic ratio of the F2 generation?
    Answer: C — Phenotypic ratio - 9 : 3 : 3 : 1
    The correct answer is 9 : 3 : 3 : 1. Key Points - This is a dihybrid cross with two independently assorting genes (R/r and Y/y). - Selfing the F1 (RrYy) gives the classic Mendelian F2 phenotypic ratio of 9 (round yellow) : 3 (round green) :
41. Q146: Histones are enriched with:
    Answer: A — Lysine & Arginine
    The correct answer is Lysine & Arginine. Key Points - Histones are positively charged basic proteins that package DNA; they are rich in the basic amino acids lysine and arginine. - The positive charge lets them bind the negatively charg
42. Q149: The protein portion of an enzyme is called:
    Answer: C — Apoenzyme
    The correct answer is Apoenzyme. Key Points - When a non-protein cofactor is bound to an enzyme, the protein portion is called the apoenzyme. - Cofactors are the non-protein parts — they include prosthetic groups (tightly bound), coenzymes
43. Q150: Which of the following is the unit of productivity of an Ecosystem?
    Answer: D — (KCal m^-2)yr^-1
    The correct answer is (KCal m^-2)yr^-1. Key Points - Productivity is the rate of biomass production, so it must include a per-area and per-time unit. - It is expressed as g m^-2\,yr^-1 or (KCal m^-2)\,yr^-1 to compare the productivity of di
44. Q152: With the help of the given pedigree, find out the probability for the birth of a child having no disease and being a carrier (has the disease mutation in one allele of the gene) in the F3 generation.
    Answer: A — 1/4
    The correct answer is 1/4. Key Points - Since a carrier female and a normal (non-carrier) male produced an affected male child, the disorder is X-linked recessive. - Cross of carrier female (X^cX) with normal male (XY): offspring are X^cX^c
45. Q153: Given below are two statements: One is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A): Cells of the tapetum possess dense cytoplasm and generally have more than one nucle
    Answer: C — A is true but R is false
    The correct answer is A is true but R is false. Key Points - The tapetum cells do have dense cytoplasm and usually more than one nucleus — A is true. - The extra nuclei increase the efficiency of nourishing the developing microspores/pollen
46. Q154: How many meiotic and mitotic divisions need to occur for the development of a mature female gametophyte from the megaspore mother cell in an angiosperm plant?
    Answer: C — 1 Meiosis and 3 Mitosis
    The correct answer is 1 Meiosis and 3 Mitosis. Key Points - The megaspore mother cell undergoes 1 meiotic division to produce four megaspores (one functional). - The functional megaspore then undergoes 3 successive mitotic (free-nuclear) di
47. Q155: Which of the following is an example of a zygomorphic flower?
    Answer: C — Pea
    The correct answer is Pea. Key Points - A zygomorphic flower can be divided into two equal halves in only one vertical plane (bilateral symmetry). - Pea has zygomorphic flowers, whereas chilli, petunia and datura have actinomorphic (radiall
48. Q157: Given below are two statements: Statement I: Fig fruit is a non-vegetarian fruit as it has enclosed fig wasps in it. Statement II: Fig wasp and fig tree exhibit mutual relationship as fig wasp complet
    Answer: B — Both statement I and statement II are incorrect
    The correct answer is Both statement I and statement II are incorrect. Key Points - The fig is considered a vegetarian fruit; the fig tree and fig wasp show mutualism (both benefit) — so Statement I is incorrect. - Statement II is also cons
49. Q158: What is the main function of the spindle fibers during mitosis?
    Answer: A — To separate the chromosomes
    The correct answer is To separate the chromosomes. Key Points - During mitosis the spindle fibres attach to the kinetochores of the chromosomes and pull the sister chromatids apart. - Their main function is thus the separation (segregation)
50. Q159: Which one of the following is the characteristic feature of gymnosperms?
    Answer: B — Seeds are naked
    The correct answer is Seeds are naked. Key Points - In gymnosperms (Gymnos = naked, sperma = seed) the ovules are not enclosed by an ovary wall and remain exposed before and after fertilisation. - Hence the seeds that form are naked (not en
51. Q162: Match List-I with List-II. List-I: A. Pteridophyte, B. Bryophyte, C. Angiosperm, D. Gymnosperm. List-II: (I) Salvia, (II) Ginkgo, (III) Polytrichum, (IV) Salvinia. Choose the option with all correct m
    Answer: B — A-IV, B-III, C-I, D-II
    The correct answer is A-IV, B-III, C-I, D-II. Key Points - Pteridophyte — Salvinia (IV). - Bryophyte — Polytrichum (III). - Angiosperm — Salvia (I). - Gymnosperm — Ginkgo (II).
52. Q163: Who proposed that the genetic code for amino acids should be made up of three nucleotides?
    Answer: A — George Gamow
    The correct answer is George Gamow. Key Points - The physicist George Gamow proposed that, since there are 20 amino acids and only 4 bases, the genetic code should be a triplet code (three nucleotides per amino acid).
53. Q164: Match List-I with List-II. List-I: A. The Evil Quartet, B. Ex situ conservation, C. Lantana camara, D. Dodo. List-II: (I) Cryopreservation, (II) Alien species invasion, (III) Causes of biodiversity lo
    Answer: B — A-III, B-I, C-II, D-IV
    The correct answer is A-III, B-I, C-II, D-IV. Key Points - The Evil Quartet — the four major causes of biodiversity loss (III). - Ex situ conservation — includes cryopreservation (I). - Lantana camara — an example of alien species invasion
54. Q168: In bryophytes, the gemmae help in which one of the following?
    Answer: B — Asexual reproduction
    The correct answer is Asexual reproduction. Key Points - Gemmae are green, multicellular, asexual buds that develop in small receptacles called gemma cups. - They help in the asexual (vegetative) reproduction of bryophytes.
55. Q170: Given below are two statements: Statement I: In ecosystem, there is unidirectional flow of energy of sun from producers to consumers. Statement II: Ecosystems are exempted from 2nd law of thermodynami
    Answer: C — Statement I is correct but statement II is incorrect
    The correct answer is Statement I is correct but statement II is incorrect. Key Points - Energy flow in an ecosystem is unidirectional — from the sun to producers to consumers — so Statement I is correct. - Ecosystems are not exempt from th
56. Q171: Which of the following statements about RuBisCO is true?
    Answer: D — It catalyzes the carboxylation of RuBP
    The correct answer is It catalyzes the carboxylation of RuBP. Key Points - RuBisCO (RuBP carboxylase-oxygenase) catalyses the carboxylation of RuBP — the most crucial step of the Calvin cycle where CO2 is fixed. - It has a higher affinity f
57. Q172: Which of the following enzyme(s) are NOT essential for gene cloning? A. Restriction enzymes, B. DNA ligase, C. DNA mutase, D. DNA recombinase, E. DNA polymerase. Choose the correct answer from the opt
    Answer: A — C and D only
    The correct answer is C and D only. Key Points - Gene cloning primarily uses restriction enzymes (cutting), DNA ligase (joining) and DNA polymerase (replication) — A, B and E are essential. - DNA mutase and DNA recombinase are not used in t
58. Q173: Read the following statements on plant growth and development. (A) Parthenocarpy can be induced by auxins. (B) Plant growth regulators can be involved in promotion as well as inhibition of growth. (C)
    Answer: A — A, B, C only
    The correct answer is A, B, C only. Key Points - A, B and C are correct: auxins induce parthenocarpy; PGRs can both promote and inhibit growth; dedifferentiation precedes re-differentiation. - D is wrong — abscisic acid is a growth inhibito
59. Q174: Which factor is important for termination of transcription?
    Answer: C — ρ (rho)
    The correct answer is ρ (rho). Key Points - In prokaryotes, the RNA polymerase catalyses elongation and associates transiently with an initiation factor (σ, sigma) to begin and a termination factor (ρ, rho) to end transcription. - Hence the
60. Q177: Which of the following microbes is NOT involved in the preparation of household products? A. Aspergillus niger, B. Lactobacillus, C. Trichoderma polysporum, D. Saccharomyces cerevisiae, E. Propionibac
    Answer: B — A and C only
    The correct answer is A and C only. Key Points - Household products: Lactobacillus (curd), Saccharomyces cerevisiae (toddy/bread) and Propionibacterium sharmanii (swiss cheese) — B, D, E are used for household products. - Aspergillus niger
61. Q179: The blue and white selectable markers differentiate recombinant colonies from non-recombinant colonies on the basis of their ability to produce colour in the presence of a chromogenic substrate. Given
    Answer: D — Statement I is incorrect but Statement II is correct
    The correct answer is Statement I is incorrect but Statement II is correct. Key Points - The DNA insert goes into the coding sequence of β-galactosidase, causing insertional inactivation of that gene. - Recombinants (with insert) cannot mak
62. Q180: Which one of the following equations represents the Verhulst-Pearl Logistic Growth of population?
    Answer: B — dN/dt=rN(K-N/K)
    The correct answer is dN/dt=rN(K-N/K). Key Points - Logistic (Verhulst-Pearl) growth of a population in a habitat with limited resources is described by dN/dt=rN(K-N/K). - Here N = population size, r = intrinsic rate of natural increase, an

Zoology — Solved Questions

1. Q93: What are the potential drawbacks in adoption of the IVF method? A. High fatality risk to mother, B. Expensive instruments and reagents, C. Husband/wife necessary for being donors, D. Less adoption of
   Answer: A — B, D, F only
   The correct answer is B, D, F only. Key Points - Genuine drawbacks: IVF needs expensive instruments and reagents (B), it can reduce the adoption of orphans (D), and there is a possibility the early embryo does not survive (F). - A, C and E
2. Q94: What is the name of the blood vessel that carries deoxygenated blood from the body to the heart in a frog?
   Answer: D — Vena cava
   The correct answer is Vena cava. Key Points - In a frog, deoxygenated blood from the body parts is returned to the heart through the major veins called the vena cava. - Aorta and pulmonary vein carry oxygenated blood; the pulmonary artery c
3. Q101: Match List-I with List-II. List-I: A. Emphysema, B. Angina Pectoris, C. Glomerulonephritis, D. Tetany. List-II: (I) Rapid spasms in muscle due to low Ca in body fluid, (II) Damaged alveolar walls and
   Answer: D — A-II, B-III, C-IV, D-I
   The correct answer is A-II, B-III, C-IV, D-I. Key Points - Emphysema = damaged alveolar walls and decreased respiratory surface (II). - Angina pectoris = acute chest pain when not enough oxygen reaches the heart muscle (III). - Glomerulonep
4. Q107: Which of the following statement is correct about the location of the male frog copulatory pad?
   Answer: D — First digit of the fore limb
   The correct answer is First digit of the fore limb. Key Points - In male frogs the copulatory pad is present on the first digit of the forelimbs. - These pads are absent in female frogs and help the male grip the female during amplexus.
5. Q109: While trying to find out the characteristic of a newly found animal, a researcher did the histology of an adult animal and observed a cavity with presence of mesodermal tissue towards the body wall bu
   Answer: B — Pseudocoelomate
   The correct answer is Pseudocoelomate. Key Points - In pseudocoelomates, the body cavity is not fully lined by mesoderm — mesodermal tissue is present as scattered pouches only along the body wall, not around the gut. - Acoelomates lack a c
6. Q110: Match List-I with List-II. List-I: A. Head, B. Middle piece, C. Acrosome, D. Tail. List-II: (i) Enzymes, (ii) Sperm motility, (iii) Energy, (iv) Genetic material. Choose the correct answer from the op
   Answer: A — A-IV, B-III, C-I, D-II
   The correct answer is A-IV, B-III, C-I, D-II. Key Points - Head contains the elongated nucleus with the genetic material (iv). - Middle piece has numerous mitochondria that provide energy for movement (iii). - Acrosome is a cap filled with
7. Q112: Cardiac activities of the heart are regulated by: A. Nodal tissue, B. A special neural centre in the medulla oblongata, C. Adrenal medullary hormones, D. Adrenal cortical hormones. Choose the correct
   Answer: A — A, B and C Only
   The correct answer is A, B and C Only. Key Points - The heart is myogenic: its activity is auto-regulated by specialised muscle (nodal tissue) — A. - A neural centre in the medulla oblongata moderates cardiac function through the autonomic
8. Q120: Which are correct: A. Computed tomography and magnetic resonance imaging detect cancers of internal organs. B. Chemotherapeutic drugs are used to kill non-cancerous cells. C. α-interferon activate the
   Answer: D — A and C only
   The correct answer is A and C only. Key Points - A true: CT and MRI are used to detect cancers of internal organs. - C true: α-interferons activate the patient's immune system to help destroy the tumour. - B is wrong (chemotherapeutic drugs
9. Q123: Find the correct statement: (A) In human pregnancy, the major organ systems are formed at the end of 12 weeks. (B) In human pregnancy the major organ systems are formed at the end of 8 weeks. (C) In h
   Answer: D — A, C, D and E only
   The correct answer is A, C, D and E only. Key Points - By the end of 12 weeks (first trimester) most major organ systems are formed (A correct; so B is wrong). - The embryo's heart forms after one month of gestation (C). - Limbs and digits
10. Q125: Which of the following diagrams is correct with regard to the proximal (P) and distal (D) tubule of the Nephron (showing which substances are reabsorbed into and secreted out of the tubule)?
    Answer: B — PCT reabsorbs HCO3^-/NaCl/H2O and secretes H^+/NH3; DCT reabsorbs NaCl/H2O/HCO3^- and secretes K^+/H^+ (Option 2)
    The correct answer is Option 2. Key Points - The PCT reabsorbs Na^+, Cl^-, HCO3^- and water, and selectively secretes H^+, ammonia (NH3) and K^+ into the filtrate to maintain ionic and acid-base balance. - The DCT reabsorbs HCO3^- (and Na^+
11. Q135: Neoplastic characteristics of cells refer to: A. A mass of proliferating cell, B. Rapid growth of cells, C. Invasion and damage to the surrounding tissue, D. Those confined to original location. Choos
    Answer: B — A, B, C only
    The correct answer is A, B, C only. Key Points - Neoplastic (malignant/tumour) cells are a mass of proliferating cells (A) that show rapid growth (B) and invade and damage surrounding tissue (C). - D is wrong — cells confined to their origi
12. Q138: Consider the following: A. The reductive division for the human female gametogenesis starts earlier than that of the male gametogenesis. B. The gap between the first meiotic division and the second me
    Answer: A — A and B are true
    The correct answer is A and B are true. Key Points - A true: in females, meiosis (reductive division) begins during foetal life — much earlier than in males. - B true: in males the two meiotic divisions occur in quick succession, whereas in
13. Q139: All living members of the class Cyclostomata are:
    Answer: D — Ectoparasite
    The correct answer is Ectoparasite. Key Points - All living members of the class Cyclostomata (e.g. Petromyzon, Myxine) are ectoparasites on fishes. - They are jawless vertebrates that attach to the host's body surface. Additional Informati
14. Q142: Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): All vertebrates are chordates but all chordates are not vertebrate. Reason (R):
    Answer: A — Both (A) and (R) are true and (R) is the correct explanation of (A)
    The correct answer is Both (A) and (R) are true and (R) is the correct explanation of (A). Key Points - Members of subphylum Vertebrata possess a notochord in the embryonic period, which is later replaced by a cartilaginous/bony vertebral c
15. Q143: Identify the statement that is NOT correct (about antibody structure).
    Answer: C — Antigen binding site is located at C-terminal region of antibody molecules
    The correct answer (the incorrect statement) is Antigen binding site is located at C-terminal region of antibody molecules. Key Points - An antibody (H2L2) has two light and two heavy chains held together by disulfide bonds — statements A,
16. Q147: The first menstruation is called:
    Answer: B — Menarche
    The correct answer is Menarche. Key Points - The first menstruation at puberty is called menarche. - Menopause is the cessation of menstrual cycles (~50 years); ovulation is the release of the secondary oocyte; diapause is a state of dorman
17. Q148: Match List-I with List-II. List-I: A. Heart, B. Kidney, C. Gastro-intestinal tract, D. Adrenal Cortex. List-II: (I) Erythropoietin, (II) Aldosterone, (III) Atrial natriuretic factor, (IV) Secretin. Ch
    Answer: D — A-III, B-I, C-IV, D-II
    The correct answer is A-III, B-I, C-IV, D-II. Key Points - Heart secretes atrial natriuretic factor (III). - Kidney secretes erythropoietin (I). - Gastro-intestinal tract secretes secretin (IV). - Adrenal cortex secretes aldosterone (II).
18. Q151: Sweet potato and potato represent a certain type of evolution. Select the correct combination of terms to explain the evolution.
    Answer: A — Analogy, convergent
    The correct answer is Analogy, convergent. Key Points - Sweet potato is a root modification while potato is a stem modification, yet both store food and look alike — they are analogous (similar function, different origin). - Analogous struc
19. Q156: After maturation, in primary lymphoid organs, the lymphocytes migrate for interaction with antigens to secondary lymphoid organ(s)/tissue(s) like: A. thymus, B. bone marrow, C. spleen, D. lymph nodes,
    Answer: D — C, D, E only
    The correct answer is C, D, E only. Key Points - Primary lymphoid organs (bone marrow and thymus) are where lymphocytes mature. - Secondary lymphoid organs, where matured lymphocytes interact with antigens, are the spleen, lymph nodes and P
20. Q160: Consider the following statements regarding function of adrenal medullary hormones: (A) It causes pupilary constriction. (B) It is a hyperglycemic hormone. (C) It causes piloerection. (D) It increases
    Answer: B — B, C and D only
    The correct answer is B, C and D only. Key Points - Adrenal medullary hormones (adrenaline/noradrenaline) are hyperglycemic (B), cause piloerection (C) and increase the strength of heart contraction (D). - They cause pupillary dilation, not
21. Q161: Why can't insulin be given orally to diabetic patients?
    Answer: B — It will be digested in Gastro-Intestinal (GI) tract
    The correct answer is It will be digested in Gastro-Intestinal (GI) tract. Key Points - Insulin is a protein (polypeptide) hormone. If taken orally it would be broken down by the proteolytic enzymes in the gastro-intestinal tract before it
22. Q165: Which of the following hormones released from the pituitary is actually synthesized in the hypothalamus?
    Answer: B — Anti-diuretic hormone (ADH)
    The correct answer is Anti-diuretic hormone (ADH). Key Points - The neurohypophysis (posterior pituitary) only stores and releases oxytocin and vasopressin (ADH), which are actually synthesised in the hypothalamus and transported axonally.
23. Q166: Role of the water vascular system in Echinoderms is: A. Respiration and Locomotion, B. Excretion and Locomotion, C. Capture and transport of food, D. Digestion and Respiration, E. Digestion and Excret
    Answer: B — A and C Only
    The correct answer is A and C Only. Key Points - The water vascular system in echinoderms helps in locomotion, capture and transport of food, and respiration — so A (respiration and locomotion) and C (capture and transport of food) are corr
24. Q167: Which of the following type of immunity is present at the time of birth and is a non-specific type of defence in the human body?
    Answer: B — Innate Immunity
    The correct answer is Innate Immunity. Key Points - Innate immunity is a non-specific defence present from birth, providing various barriers to the entry of foreign agents. - Acquired immunity is pathogen-specific and develops with memory;
25. Q169: In frog, the Renal portal system is a special venous connection that acts to link:
    Answer: D — Kidney and lower part of body
    The correct answer is Kidney and lower part of body. Key Points - In frogs the renal portal system is a special venous connection carrying blood from the lower parts of the body to the kidney. - (The hepatic portal system is the analogous c
26. Q175: Frogs respire in water by skin and buccal cavity and on land by skin, buccal cavity and lungs. Choose the correct answer from the following:
    Answer: C — The statement is false for water but true for land
    The correct answer is The statement is false for water but true for land. Key Points - In water, frogs respire only through the skin (cutaneous respiration) — not through the buccal cavity, so the statement is false for water. - On land, th
27. Q176: Twins are born to a family that lives next door to you. The twins are a boy and a girl. Which of the following must be true?
    Answer: B — They are fraternal twins
    The correct answer is They are fraternal twins. Key Points - Monozygotic (identical) twins come from a single fertilised egg and are always the same sex. - A boy-and-girl pair must therefore have come from two separate fertilised eggs — the
28. Q178: Match List-I with List-II. List-I: A. Progesterone, B. Relaxin, C. Melanocyte stimulating hormone, D. Catecholamines. List-II: (I) Pars intermedia, (II) Ovary, (III) Adrenal Medulla, (IV) Corpus luteu
    Answer: A — A-IV, B-II, C-I, D-III
    The correct answer is A-IV, B-II, C-I, D-III. Key Points - Progesterone — secreted by the corpus luteum (IV). - Relaxin — secreted by the ovary in later pregnancy (II). - Melanocyte stimulating hormone — released by the pars intermedia of t

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The NEET UG 2025 Question Paper with Solutions (4 May, Code 45) has 180 questions. Every question is solved here with the correct answer and a detailed explanation.

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