Re-NEET 2026 question paper with answers and solutions (code 60)
This is the full Re-NEET 2026 question paper (NEET-UG re-exam held on 21 June 2026), booklet code 60, in English. It has all 180 questions with the answer from NTA's final answer key and a worked solution for each.
- 180 questions, 720 marks. Physics 45, Chemistry 45, Botany 45 and Zoology 45. All questions are compulsory.
- Marking: +4 for a correct answer, −1 for a wrong one, 0 if left blank.
- Dropped question: NTA dropped Q43 (Physics) in the final key, so every candidate got 4 marks for it.
- Solutions are written by the Testline NEET team. The answers follow NTA's final key.
The paper was held on pen and paper. You can attempt the same paper on an NTA-style computer screen with a 180-minute timer, and get your score and an estimated rank at the end.
Physics questions (Q1–45)
Q1. Alternating Current
An ac voltage is applied to a series LCR circuit. Then the current amplitude in this circuit is :
(Given : , , )
Show answer and solution
Answer: option (3)
, so and . The circuit is at resonance, , and .
Q2. Kinetic Theory
The mean free path of molecules in an ideal gas A is half that of another ideal gas B. The diameter of the spherical molecules of gas A is twice the diameter of the molecules of B. If number densities of the gases A and B are and , respectively, then the correct option is :
Show answer and solution
Answer: option (2)
Mean free path . With and : , so .
Q3. Oscillations
A cylindrical cork of uniform density floats in a liquid of density . If the cork is depressed slightly and released, it oscillates harmonically with time period T. If the same cork floats in another liquid of density , then the similar oscillation has time period 2T. The value of is :
Show answer and solution
Answer: option (2)
For a floating cylinder the restoring force is , so . Doubling needs .
Q4. Oscillations
Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is and the spring constant is . At a given instant, the extension of the spring is meter and the speed of the particle is . On the plane, if the graph of as a function of is a circle, then the correct option is :
Show answer and solution
Answer: option (4)
Energy conservation gives , an ellipse in the - plane. It becomes a circle only when the coefficients are equal, i.e. (numerically).
Q5. Thermodynamics
In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas () decreases from to . The work done by the gas in the process is :
(Take the universal gas constant as )
Show answer and solution
Answer: option (1)
Adiabatic work by the gas .
Q6. Electromagnetic Waves
The following table presents the part of the electromagnetic spectrum and their corresponding major applications.
| Part of the electromagnetic spectrum | Applications | ||
|---|---|---|---|
| P. | Microwave | I. | For purifying the water |
| Q. | UV rays | II. | For warming the food |
| R. | Gamma rays | III. | For AM and FM communication systems |
| S. | Radio wave | IV. | For treating the Cancer cells |
The correct option is :
- P-II, Q-I, R-IV, S-III
- P-II, Q-IV, R-III, S-I
- P-I, Q-II, R-III, S-IV
- P-I, Q-IV, R-II, S-III
Show answer and solution
Answer: option (1)
Microwaves heat food (P-II), UV rays sterilise/purify water (Q-I), gamma rays are used to destroy cancer cells (R-IV) and radio waves carry AM/FM broadcasts (S-III).
Q7. Electrostatic Potential and Capacitance
A unit positive point charge is taken slowly through an infinitesimally thin tube that is inside a charged dielectric sphere of radius , having uniform positive charge density , as shown in the figure. The initial and final positions of the charge are marked by and at distances and respectively, from the centre of the sphere. In this process, the magnitude of the total work done on the point charge is . The value of is :
( is the permittivity of vacuum)

- 9
- 18
- 2
- 6
Show answer and solution
Answer: option (2)
The field is conservative, so the work equals for a unit charge, whatever the path. Outside the sphere , giving , so .
Q8. Dual Nature of Radiation and Matter
A beam of light falls on a metal surface such that photo-electrons are generated. If power of the light source starts to decrease linearly with time , then variation of the photocurrent and magnitude of the stopping potential with time is best represented by :

The four options are in the figure above.
Show answer and solution
Answer: option (3)
Photocurrent is proportional to intensity, so it falls linearly as the power falls. The stopping potential depends only on frequency, so stays constant: graph (3).
Q9. Work, Energy and Power
Bob of mass at rest is hanging vertically from the ceiling via a massless string of length , as shown in the figure. Point mass of mass travelling horizontally with speed hits bob elastically. The bob rises meter after the collision. Taking the acceleration due to gravity and neglecting the size of the bob, the value of is :

- 5
- 2.5
- 8
- 7
Show answer and solution
Answer: option (1)
In an elastic head-on collision between equal masses the velocities are exchanged, so B moves off at . It rises (less than the string length, so the string stays taut).
Q10. Kinetic Theory
An ideal gas is made of polyatomic molecules. Each of the molecules has three translational, three rotational and number of vibrational modes. If the ratio of heat capacities of the gas is , then the value of is :
- 2
- 1
- 4
- 3
Show answer and solution
Answer: option (3)
Each vibrational mode contributes to , so and , giving , .
Q11. Mechanical Properties of Fluids
In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used. The variation of the terminal velocity () with the ratio of density of spherical ball () to density of the liquid (), is best represented by :

The four options are in the figure above.
Show answer and solution
Answer: option (3)
Terminal velocity : a straight line that is zero at and rises linearly after it, as in graph (3).
Q12. Gravitation
In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius is proportional to :
Show answer and solution
Answer: option (4)
By Kepler's third law , so .
Q13. System of Particles and Rotational Motion
A solid sphere of radius and mass is attached at a point to a smaller solid sphere of radius and mass . Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of is and that calculated about a vertical axis passing through the centre of is . The difference is :

Show answer and solution
Answer: option (4)
The centres are apart. and , so .
Q14. Nuclei
Consider the following nuclear reaction :
Take masses of , and as , and , respectively. The value for the reaction, in , is :
[Given : ]
Show answer and solution
Answer: option (3)
Mass defect , so .
Q15. Electrostatic Potential and Capacitance
Consider a fixed uniformly charged insulating sphere with radius and total charge . A point charge () with mass is released from rest at a distance of from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is :
( is the permittivity of vacuum, neglect gravitational forces).
Show answer and solution
Answer: option (1)
Gain in KE = loss in PE: , so .
Q16. Current Electricity
Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, , while the other one has a negative temperature of coefficient, , as shown in the figure. The current through these circuits are denoted by and . At initial temperature, the resistance of the two resistors is . As the temperature is increased, the correct option that describes the variation of current in these circuits is :

- increases while decreases
- both and remain constant
- remains constant while increases
- decreases while increases
Show answer and solution
Answer: option (3)
In series (A) the total resistance is unchanged, so is constant. In parallel (B) decreases, so increases.
Q17. Electromagnetic Induction
A conducting loop of finite resistance lies on the plane. There is a constant magnetic field in the direction. The area of the loop varies with time , as in appropriate units. The figure that correctly indicates the qualitative behaviour of the power dissipated in the loop as a function of time is :

The four options are in the figure above.
Show answer and solution
Answer: option (4)
EMF , so : maximum at , falling to zero and rising again periodically, as in figure (4).
Q18. Dual Nature of Radiation and Matter
A photon and an electron, each of energy, move in free space. The ratio of linear momentum of electron to that of photon is :
(Take speed of light , charge of electron and mass of electron )
Show answer and solution
Answer: option (1)
Electron: SI. Photon: SI. Ratio .
Q19. Units and Measurement
Consider that , , represent Stefan-Boltzmann constant, Boltzmann constant and Wien's displacement law constant, respectively. The dimension of is :
Show answer and solution
Answer: option (3)
, , . So .
Q20. Moving Charges and Magnetism
Two infinitely long parallel conducting wires and carry currents and , respectively, in the same direction. The wire has uniform mass per unit length and lies on an insulated floor. The wire is kept fixed at a height above the floor. The minimum magnitude of so that the wire does not rise from the floor is :
[ is the acceleration due to gravity and is the permeability of free space.]
Show answer and solution
Answer: option (4)
Parallel currents attract, so B pulls A upward with force per length . A stays down if this is at most , i.e. .
Q21. Ray Optics and Optical Instruments
Which of the following measurements require 'index correction'?
- Measurement of focal length of lenses using optical bench
- Measurement of speed of sound using resonance tube
- Measurement of resistance of a wire using meter bridge
- Measurement of gravitational acceleration using simple pendulum
Show answer and solution
Answer: option (1)
On an optical bench the scale distance between the uprights differs from the actual distance between the lens and the needle tips; this difference (index error) must be corrected. The other experiments need no such correction.
Q22. Electromagnetic Induction
Two identical inductors are connected in two different configurations and , where a time varying current is flowing, as shown in the figure. The induced emf between points and for configuration is and that for configuration is . The ratio is :
[Neglect the effect of mutual inductance.]

Show answer and solution
Answer: option (2)
In P, a and b span only one inductor, so . In Q, the inductors are in parallel and each carries , so . Hence .
Q23. Semiconductor Electronics: Materials, Devices and Simple Circuits
Three identical junction diodes , and are connected across a battery as shown in the figure. If the width of the depletion regions of , and are , and , respectively, then the correct option is :

Show answer and solution
Answer: option (1)
D1 is forward biased (narrowest depletion region), D2 is in an open branch so it is unbiased (normal width), and D3 is reverse biased (widest). So .
Q24. Electrostatic Potential and Capacitance
Three identical capacitors , and , each of the capacitance , are connected to a battery of voltage as shown in the figure. If the energy stored in the capacitor and total energy stored in the system are and , respectively, then the ratio is :

Show answer and solution
Answer: option (2)
P and Q are in series (each gets ) and this pair is in parallel with S (gets ). , , so .
Q25. System of Particles and Rotational Motion
A thin horizontal disc is rotating about a vertical axis passing through its fixed centre . Its angular momentum is and computed about points and , respectively, with . The value of is :

Show answer and solution
Answer: option (1)
The disc's centre of mass is at rest, so its total linear momentum is zero and its angular momentum is the same about every point (). Hence .
Q26. Semiconductor Electronics: Materials, Devices and Simple Circuits
An ideal Zener diode with breakdown voltage of is reverse biased with a negative input voltage . The magnitude of voltage difference between point and is :

Show answer and solution
Answer: option (4)
In breakdown the ideal Zener holds across itself, so the rest, , drops across the resistor between B and A.
Q27. Mechanical Properties of Fluids
Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between and is . The area of cross-section at and are and , respectively. The rate of flow of water through the pipe, in , is :
[Take density of water ]

Show answer and solution
Answer: option (2)
Continuity: . Bernoulli: , so and flow rate .
Q28. Work, Energy and Power
A particle of mass moves along a horizontal axis from to . The coefficient of kinetic friction varies as a function of as , where are constants of appropriate dimensions, so that . The total work done by the frictional force during the motion is , where is the acceleration due to gravity. The value of is :
Show answer and solution
Answer: option (2)
gives . , so .
Q29. Dual Nature of Radiation and Matter
A ray of light with wavelength is incident on three different photoelectric cells namely 1, 2 and 3. The threshold wavelength of these photoelectric cells are and , respectively and the magnitude of stopping potentials of these cells are and , respectively. The relation between and threshold wavelengths are , and . The correct option is :
Show answer and solution
Answer: option (3)
Cell 1 has , so no emission and . For cells 2 and 3, grows with threshold wavelength; since , .
Q30. Atoms
Consider that an electron is revolving in an excited state of Hydrogen atom with velocity . The radius of the orbit is . The value of is :
[Take the mass of electron to be , charge of electron and ]
Show answer and solution
Answer: option (2)
Coulomb force supplies centripetal force: , so .
Q31. Laws of Motion
A car travels on a circular racetrack of radius , which is banked at an angle . If the car travels at a speed , then the wear and tear on its tyres is minimum. Taking the acceleration due to gravity to be , the value of is :
Show answer and solution
Answer: option (3)
Tyre wear is minimum when no friction is needed, i.e. at the design speed: .
Q32. Thermodynamics
One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. The total heat supplied to the gas is :

Show answer and solution
Answer: option (1)
For a cycle , so net heat = net work = area of the rectangle (clockwise, so positive).
Q33. System of Particles and Rotational Motion
A frictionless circular wire of unit radius is fixed on the horizontal plane. Two-point particles of unit mass start moving simultaneously from point with identical uniform angular speeds in opposite directions, and meet again at point . During this time, which of the following figures schematically represent the magnitude of the total linear momentum of the system, as a function of ?

The four options are in the figure above.
Show answer and solution
Answer: option (1)
By symmetry the horizontal velocity components cancel and the vertical ones add: . It is zero at and maximum at , a single hump as in figure (1).
Q34. Ray Optics and Optical Instruments
The lens combination as shown in the figure, consists of two lenses, and , of the focal lengths and , respectively. The position of the image formed is :

- to the right of the concave lens
- to the right of the concave lens
- to the left of the concave lens
- to the left of the concave lens
Show answer and solution
Answer: option (4)
: , gives , i.e. 12 cm beyond (virtual object). : , so : 60 cm to the left of the concave lens.
Q35. Gravitation
Two planets and with equal mass have radii and , respectively, where . The escape speeds of and are and , respectively. Then is :
Show answer and solution
Answer: option (1)
for equal masses, so .
Q36. Electromagnetic Waves
An electromagnetic wave travelling in a lossless dielectric medium having a dielectric constant, , has the electric field, where is the amplitude and is the wave vector. Among the following options, the incorrect choice is :
- The magnetic field is given by the relation where is the speed of the electromagnetic wave inside the medium
- The direction of propagation of the electromagnetic wave is along
- The speed of the electromagnetic wave inside the medium is
- The wavelength of the electromagnetic wave inside the medium is
Show answer and solution
Answer: option (4)
and , so , not 300 m. The wave travels along (phase ), and with along the field is along .
Q37. Motion in a Straight Line
Consider a particle moving along a straight line, whose position as a function of time is given by , where , and . The average speed of the particle, in , from to is :
- 3
- 0
- 12
- 6
Show answer and solution
Answer: option (1)
changes sign at . , , , so distance in 6 s: average speed .
Q38. Atoms
In Geiger-Marsden experiment, the number of scattered -particles is plotted as a function of scattering angle . Which of the following options represents the correct plot ?

The four options are in the figure above.
Show answer and solution
Answer: option (1)
Rutherford scattering gives : very large at small angles and falling steeply as increases, as in plot (1).
Q39. Waves
For sound waves, if the number of nodes for the harmonic of an open-ended pipe is and that for the harmonic of the same pipe with one of its ends closed is , the ratio is :
Show answer and solution
Answer: option (1)
An open pipe in its 5th harmonic has 5 nodes. A closed pipe's 9th harmonic () also has 5 nodes (counting the closed end). So .
Q40. Electromagnetic Induction
Consider a long solenoid of length and radius . If is the number of turns per unit length and is the permeability of free space, the inductance of the solenoid is :
Show answer and solution
Answer: option (3)
, flux linkage , so .
Q41. Moving Charges and Magnetism
A current flows through a metallic circular loop of radius as shown in the figure. Resistance of the segment is half that of . Magnitude of magnetic field at the center of the loop is :

Show answer and solution
Answer: option (3)
Currents divide inversely as resistance: through ABC and through ADC. Each semicircle gives in opposite directions, so .
Q42. Thermal Properties of Matter
The temperature of a metallic sphere of radius R is increased by a small amount . If the linear coefficient of thermal expansion of the metal is , the approximate increase in the volume of the sphere is :
Show answer and solution
Answer: option (1)
, so .
Q43. Units and Measurement
One main scale division of a Vernier calliper is equal to and the number of divisions on the Vernier scale is 10. When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of a wire to be , the actual length of the wire is :
Show answer and solution
Dropped by NTA. Every candidate got 4 marks for this question.
NTA dropped this question in the final answer key; every candidate gets +4.
Q44. Ray Optics and Optical Instruments
Consider three media P, Q and R with refractive indices , , and , respectively. The medium Q having a thickness of is placed between extended media P and R as shown in the figure. An object O is placed at the center of medium Q. If viewed from medium P near the normal direction, the apparent depth of O is . For similar observation from medium R, the apparent depth is . The value of , in , is :

- 2
- 3
- 0
- 1
Show answer and solution
Answer: option (4)
O is 2.5 cm from each face. From P: . From R: . So .
Q45. Electric Charges and Fields
A point charge Q is placed inside a cavity within a solid isolated conducting sphere. Consider points A, B and C as shown in the figure, where the magnitudes of the electric fields are , and , respectively. The points B and C are at the same distance from the center of the solid sphere. The correct option is :

Show answer and solution
Answer: option (4)
Inside the cavity the field of Q is present, so . A charge appears uniformly on the outer surface regardless of where Q sits in the cavity, so outside and .
Chemistry questions (Q46–90)
Q46. The d- and f-Block Elements
The lanthanide ion having four unpaired electrons is
(Given : Atomic numbers of , , and )
Show answer and solution
Answer: option (2)
is : seven f orbitals hold 10 electrons, leaving 4 unpaired. () has 6, () has 3 and () has 1.
Q47. Electrochemistry
The standard electrode potential () for the half-cell reaction at is
(Given : and at )
Show answer and solution
Answer: option (4)
Potentials are not additive, but is. : ; : . Sum , so .
Q48. The p-Block Elements
Given below are two statements :
Statement-I : Heating with concentrated and results in oxidation of .
Statement-II : Heating with concentrated and results in reduction of .
In light of the above statements, choose the most appropriate answer from the options given below.
- Statement-I is correct but Statement-II is incorrect
- Statement-I is incorrect but Statement-II is correct
- Both Statement-I and Statement-II are correct
- Both Statement-I and Statement-II are incorrect
Show answer and solution
Answer: option (2)
In both cases acts as the oxidising agent: halide is oxidised to or while Mn goes from in to in /. So Mn is reduced in both; Statement-I is wrong and Statement-II is right.
Q49. Coordination Compounds
The complex which has facial and meridional isomers is
(Given : and )
Show answer and solution
Answer: option (3)
fac/mer isomerism arises in octahedral complexes. Only has this type; and show cis/trans isomerism instead, and has none.
Q50. Organic Chemistry – Some Basic Principles and Techniques
Among the following, the compound having conjugated double bonds is
- hepta-1,5-diene
- hepta-1,6-diene
- hepta-1,3-diene
- hepta-1,4-diene
Show answer and solution
Answer: option (3)
Conjugated double bonds alternate with a single bond, i.e. C=C–C=C. In hepta-1,3-diene the double bonds start at C1 and C3, separated by one single bond; the others are isolated dienes.
Q51. Chemical Bonding and Molecular Structure
Match the species in List-I with their geometry in List-II.
| List-I | List-II |
|---|---|
| A. | I. Tetrahedral |
| B. | II. Square Planar |
| C. | III. Trigonal bipyramidal |
| D. | IV. Square pyramidal |
Choose the correct answer from the options given below:
- A-III, B-I, C-II, D-IV
- A-III, B-II, C-I, D-IV
- A-IV, B-III, C-I, D-II
- A-III, B-IV, C-I, D-II
Show answer and solution
Answer: option (4)
(, 5 bond pairs) is trigonal bipyramidal; (5 bond pairs + 1 lone pair) is square pyramidal; () is tetrahedral; (, strong-field ) is square planar.
Q52. Organic Chemistry – Some Basic Principles and Techniques
The amino acid that gives a red-blood colour on treating its sodium fusion extract with sodium nitroprusside is
- methionine
- serine
- leucine
- threonine
Show answer and solution
Answer: option (1)
A colour test with the sodium fusion extract that indicates sulphur (with N, giving the blood-red thiocyanate test) requires an S-containing amino acid. Of the four, only methionine contains sulphur (and nitrogen); serine, leucine and threonine have no S.
Q53. Thermodynamics
A protein undergoes reversible thermal denaturation from its initial state to denatured state according to . At , the concentrations of both and are equal at equilibrium, and the standard enthalpy change of denaturation is . The standard entropy change ( in ) of the protein upon denaturation at is closest to
- 333.0
- 11.1
- 2.0
- 2000.0
Show answer and solution
Answer: option (3)
Equal concentrations give , so and . With : .
Q54. Chemical Kinetics
is a zero-order reaction, where . If the initial concentration of A is , then the time taken to complete of the reaction will be
- 1.0 min
- 2.0 min
- 1.5 min
- 0.75 min
Show answer and solution
Answer: option (4)
For , rate , so A is consumed at . 75% of 2 M is 1.5 M, which takes .
Q55. Coordination Compounds
Given below are two statements:
Statement-I : is chiral.
Statement-II : trans- is chiral.
(Given : )
In light of the above statements, choose the most appropriate answer from the options given below:
- Statement-I is correct but Statement-II is incorrect
- Statement-I is incorrect but Statement-II is correct
- Both Statement-I and Statement-II are correct
- Both Statement-I and Statement-II are incorrect
Show answer and solution
Answer: option (1)
complexes such as tris(oxalato)ferrate(III) have no mirror plane and exist as enantiomers, so I is correct. In the trans bis(oxalato) complex the two oxalates lie in one plane, which is a mirror plane, so it is achiral and II is wrong.
Q56. The d- and f-Block Elements
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Generally, transition metals have high melting points.
Reason R: Involvement of -electrons in addition to -electrons in the interatomic metallic bonding.
In light of the above statements, choose the most appropriate answer from the options given below:
- A is correct but R is not correct
- A is not correct but R is correct
- Both A and R are correct and R is the correct explanation of A
- Both A and R are correct and R is NOT the correct explanation of A
Show answer and solution
Answer: option (3)
Transition metals have high melting points because, besides the electrons, unpaired electrons also take part in metallic bonding, making it stronger. So both are true and R explains A.
Q57. Coordination Compounds
Among the species given below, the spin-only magnetic moment is highest for
(Given: Atomic number of , , and )
Show answer and solution
Answer: option (3)
Unpaired electrons: low spin ; ; low spin () ; low spin . The Mn complex with has the highest .
Q58. Coordination Compounds
According to crystal field theory, the correct order of ligands with respect to their decreasing order of field strength is
Show answer and solution
Answer: option (3)
In the spectrochemical series , so in decreasing order: .
Q59. The p-Block Elements
In potash alum, the ratio of and ions is
- 2 : 3
- 3 : 2
- 1 : 2
- 2 : 1
Show answer and solution
Answer: option (3)
Potash alum is (i.e. ), so .
Q60. Classification of Elements and Periodicity in Properties
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The first ionization enthalpy of O is lower than that of N and F.
Reason R: The loss of an electron from O leads to stable half-filled orbital.
In light of the above statements, choose the most appropriate answer from the options given below:
- A is correct but R is not correct
- A is not correct but R is correct
- Both A and R are correct and R is the correct explanation of A
- Both A and R are correct and R is NOT the correct explanation of A
Show answer and solution
Answer: option (3)
O () has one doubly occupied orbital; removing that paired electron relieves repulsion and gives the stable half-filled configuration, so O's first IE is lower than N's, and F is higher still. Per the official key, R is taken as the correct explanation of A.
Q61. Amines
Given below are two statements:
Statement-I : Oxidation of -nitrotoluene with acidic gives an acid that is stronger than benzoic acid.
Statement-II : Reduction of -nitrotoluene with followed by neutralization gives an amine that is more basic than aniline.
In light of the above statements, choose the most appropriate answer from the options given below.
- Statement-I is correct but Statement-II is incorrect
- Statement-I is incorrect but Statement-II is correct
- Both Statement-I and Statement-II are correct
- Both Statement-I and Statement-II are incorrect
Show answer and solution
Answer: option (3)
Oxidation gives -nitrobenzoic acid; the electron-withdrawing stabilises the carboxylate, so it is stronger than benzoic acid. Reduction gives -toluidine, whose electron-donating makes it more basic than aniline. Both statements are correct.
Q62. Organic Chemistry – Some Basic Principles and Techniques
The following carbocation is stabilized by the interaction of the empty orbital with

- empty and filled orbitals
- empty and empty orbitals
- filled and filled orbitals
- empty and empty orbitals
Show answer and solution
Answer: option (3)
The 1-phenylethyl cation is stabilised by hyperconjugation (overlap of filled C–H orbitals of with the empty p orbital) and by resonance with the filled orbitals of the benzene ring. An empty orbital can only be stabilised by filled ones.
Q63. Structure of Atom
Consider the following schematic plots of orbital wavefunction against distance from the nucleus.
The figure representing two radial nodes in the orbital is

- C
- D
- A
- B
Show answer and solution
Answer: option (1)
A radial node is a point (other than and ) where crosses zero. Plot A has no crossing, B one, C two and D three, so C has two radial nodes (e.g. a orbital).
Q64. Equilibrium
The correct order of solubility of the given salts in water at is
| Salt | at |
|---|---|
Show answer and solution
Answer: option (2)
AgBr: M. and (as ) both follow : M and M. Hence .
Q65. Coordination Compounds
A electrolyte in an aqueous solution is
Show answer and solution
Answer: option (1)
ionises into one complex cation and three , i.e. a 1 : 3 electrolyte. The others give 1 : 0, 1 : 1 and 1 : 2.
Q66. Biomolecules
The correct statement about peptides and proteins is
- In -pleated sheet structures, peptide chains are held together by intermolecular hydrogen bonds
- In -helices, the polypeptide chain is twisted into a left-handed screw (helix) through intramolecular hydrogen bonds
- Tertiary structure of proteins has two or more polypeptide subunits
- Only the proteins having a quaternary structure are biologically active
Show answer and solution
Answer: option (1)
In a -pleated sheet, neighbouring chains lie side by side and are linked by hydrogen bonds between chains (intermolecular). The -helix is right-handed, subunits define quaternary (not tertiary) structure, and many single-chain proteins are active.
Q67. Haloalkanes and Haloarenes
One of the products formed in the following reaction is

The four options are in the figure above.
Show answer and solution
Answer: option (2)
Cyclohexylamine has acidic N–H protons, so the Grignard reagent simply acts as a strong base: . The product shown is cyclohexane (option 2).
Q68. Redox Reactions
In an acidic medium, of oxalic acid is titrated with solution. If the volume of solution required to reach end point is , the strength of the solution is
Show answer and solution
Answer: option (3)
. Oxalic acid mmol, so mmol in 10 mL, i.e. .
Q69. Hydrocarbons
Consider the following reaction sequences and choose the correct option.

- M and N are geometrical isomers
- M and N are stereoisomers
- K and L are geometrical isomers
- K and L are enantiomers
Show answer and solution
Answer: option (3)
Lindlar reduction of gives cis-1-phenylpropene (K) and Na/liq. gives the trans alkene (L), so K and L are geometrical isomers. HBr adds to give (M, via benzylic cation) while HBr/peroxide gives (N, via benzylic radical); M and N are position isomers.
Q70. Alcohols, Phenols and Ethers
Arrange the following compounds in the increasing order of polarity
A.
B.
C.
D.
Choose the correct answer from the options given below.
- C < A < B < D
- A < C < B < D
- A < B < C < D
- C < A < D < B
Show answer and solution
Answer: option (2)
Diethyl ether is the least polar (weakly polar C–O–C, no H-bonding), acetone is more polar due to its C=O group, ethanol is a polar protic H-bonding solvent, and acetic acid with both C=O and O–H is the most polar: A < C < B < D.
Q71. Thermodynamics
Consider the reversible processes for of an ideal gas as shown in the figure.
Processes 2 and 4 are adiabatic.
, , and represent work done (in calories) in the processes 1, 2, 3 and 4, respectively; and are changes in the internal energy for the processes 2 and 4, respectively.
[use ]
The correct option is

Show answer and solution
Answer: option (3)
Processes 1 and 3 are reversible isotherms, so and ; adding gives option 3. For the adiabats and , so , and the cycle's net work is not zero.
Q72. The d- and f-Block Elements
The green paramagnetic species formed by heating at is
Show answer and solution
Answer: option (3)
. Potassium manganate contains (, one unpaired electron), so it is green and paramagnetic.
Q73. Some Basic Concepts of Chemistry
The numbers and were rounded to three figures after the decimal point. The resulting numbers, respectively, are
- and
- and
- and
- and
Show answer and solution
Answer: option (1)
When the digit dropped is exactly 5, the preceding digit is left unchanged if even and raised by one if odd. In 17.0145 the preceding 4 is even, giving 17.014; in 21.0235 the preceding 3 is odd, giving 21.024.
Q74. Hydrocarbons
Given below are two statements :
Statement I : trans-But-2-ene upon treatment with in gives the following product.
Statement II : cis-But-2-ene upon treatment with alkaline gives the following product.
In the light of the above statements, choose the most appropriate answer from the options given below.

- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
Show answer and solution
Answer: option (2)
Anti addition of to trans-but-2-ene gives meso-2,3-dibromobutane, but the Newman projection in Statement I (Br anti, the two Me gauche) is the chiral (±) form, so I is wrong. Syn dihydroxylation of cis-but-2-ene by cold alkaline gives meso-butane-2,3-diol, which is the structure drawn in Statement II, so II is correct.
Q75. Aldehydes, Ketones and Carboxylic Acids
Consider the following reaction, and choose the correct option.
- On treatment with bromine water, compound P gives a white precipitate
- Compound P is obtained by the hydrogenation of benzoyl chloride with Pd on
- On treating compound P with saturated solution, brisk effervescence is observed
- Compound P can be prepared by treating benzene with anhydrous and
Show answer and solution
Answer: option (2)
The Étard reaction converts toluene into benzaldehyde (P). Benzaldehyde is also made by Rosenmund reduction, i.e. hydrogenation of benzoyl chloride over Pd/. It is not a phenol or acid, and Friedel–Crafts acetylation would give acetophenone.
Q76. Biomolecules
Match the vitamins in List I with their sources in List II
| List I | List II |
|---|---|
| A. vitamin A | I. meat |
| B. vitamin B | II. sunflower oil |
| C. vitamin E | III. green leafy vegetables |
| D. vitamin K | IV. carrots |
Choose the correct answer from the options given below.
- A-IV, B-II, C-I, D-III
- A-III, B-I, C-IV, D-II
- A-II, B-III, C-IV, D-I
- A-IV, B-I, C-II, D-III
Show answer and solution
Answer: option (4)
Vitamin A comes from carrots (-carotene), from animal foods such as meat, vitamin E from vegetable oils like sunflower oil, and vitamin K from green leafy vegetables: A-IV, B-I, C-II, D-III.
Q77. Some Basic Concepts of Chemistry
The amount of carbon dioxide evolved upon complete combustion of of -butane is
(Given: atomic mass in amu , and )
Show answer and solution
Answer: option (3)
has , so 116 g is 2 mol. Each mole gives 4 mol , i.e. 8 mol .
Q78. Aldehydes, Ketones and Carboxylic Acids
The compound that CANNOT be obtained from the aldol condensation reaction shown below, is

The four options are in the figure above.
Show answer and solution
Answer: option (3)
2,2-Dimethylcyclopentanone has α-H only at C5. Crossed condensation with PhCHO gives 5-benzylidene products (E and Z, options 2 and 4); self-condensation followed by dehydration toward the α-H gives the conjugated exocyclic enone (option 1). Option 3 has the C=C inside the second ring, not conjugated with C=O, which base-catalysed aldol dehydration does not form.
Q79. The p-Block Elements
The correct statement is
- Magnesium has a maximum covalency of four.
- Aluminium has five valence orbitals.
- Boron has a maximum covalency of four.
- Beryllium has three valence orbitals.
Show answer and solution
Answer: option (3)
Boron has only and valence orbitals (four in total, no d orbitals), so its maximum covalency is four, as in . Be also has four valence orbitals, Al has nine () and Mg can exceed four.
Q80. Coordination Compounds
The formula of tetraammineaquachloridocobalt(III) chloride is
Show answer and solution
Answer: option (2)
The coordination sphere has 4 , 1 and 1 around , giving a charge, balanced by two chloride counter ions: .
Q81. Solutions
Assertion A: For an ideal solution formed by mixing liquids P and Q, and
Reason R: No interactions occur between P and Q
In the light of the above statements, choose the most appropriate answer from the options given below.
- A is correct but R is not correct
- A is not correct but R is correct
- Both A and R are correct and R is the correct explanation of A
- Both A and R are correct but R is NOT the correct explanation of A
Show answer and solution
Answer: option (1)
An ideal solution does have and , but this is because P–Q interactions are equal in strength to P–P and Q–Q interactions, not because there are no interactions. So A is true and R is false.
Q82. Amines
Identify the reactions which give aniline as the major product.
Choose the correct answer from the options given below.

- A and C only
- C and D only
- A and B only
- B and D only
Show answer and solution
Answer: option (4)
B: Hofmann bromamide degradation of benzamide (/KOH) gives aniline. D: acid hydrolysis of acetanilide gives aniline. A: reduces benzonitrile to benzylamine, and C: does not reduce the nitro group.
Q83. Redox Reactions
The correct decreasing order of oxidation state of the <u>underlined</u> atom in each molecule is
Show answer and solution
Answer: option (4)
Oxidation states: (1) ; (2) ; (3) ; (4) N > Al > S . Only option 4 is in decreasing order.
Q84. Aldehydes, Ketones and Carboxylic Acids
For the following reaction sequence, choose the correct option
- If P gives a carboxylic acid on acidification, Q gives a poisonous gas on exposure to air and light
- Both P and Q are carbonyl compounds
- If P is the sodium salt of a carboxylic acid, Q is a primary alcohol
- P and Q are aromatic compounds
Show answer and solution
Answer: option (1)
Friedel–Crafts acylation gives acetophenone; the haloform reaction with NaOCl then gives sodium benzoate (P, gives benzoic acid on acidification) and chloroform (Q). is oxidised by air in light to poisonous phosgene, .
Q85. Thermodynamics
Two moles of an ideal gas undergo free expansion from to at . The values of and are
(R is universal gas constant)
- ;
- ;
- ;
- ;
Show answer and solution
Answer: option (2)
Entropy is a state function, so . In free expansion no heat is exchanged with the surroundings, so .
Q86. Chemical Kinetics
For an elementary chemical reaction, the Arrhenius plot is given below.
If the energy of activation is and , the temperature at which the rate constant becomes , is

Show answer and solution
Answer: option (1)
From the plot, the intercept . With and : , so .
Q87. Solutions
Consider the following statements about the solutions formed by mixing two liquids.
A. An ideal solution thus formed obeys Raoult's law throughout the composition range.
B. Mixture of chloroform and acetone shows negative deviation from Raoult's law.
C. Mixture of aniline and phenol shows positive deviation from Raoult's law.
- A only
- A and C only
- A and B only
- B and C only
Show answer and solution
Answer: option (3)
A is the definition of an ideal solution. Chloroform–acetone and phenol–aniline both form hydrogen bonds between the two components, stronger than the original interactions, so both show negative deviation; C is false.
Q88. Chemical Bonding and Molecular Structure
The highest occupied molecular orbital for is
Show answer and solution
Answer: option (2)
has 20 electrons, which fill all the n = 2 MOs, including and finally . The last filled (highest occupied) orbital is therefore (bond order 0).
Q89. Electrochemistry
For a salt , which is a strong electrolyte, the plot of versus has a slope of at . At concentration of , the value of is . The limiting molar conductivity of ion (, in ) at will be
(Given : )
- 90.0
- 76.0
- 80.0
- 100.0
Show answer and solution
Answer: option (3)
Kohlrausch: , so , giving . Then .
Q90. Classification of Elements and Periodicity in Properties
Among the following options, the correct trend in the electron gain enthalpy is
Show answer and solution
Answer: option (1)
Electron gain enthalpy (magnitude, most negative first) is Cl > F > Br > I. F is less negative than Cl because the incoming electron feels strong repulsion in the small, compact subshell.
Botany questions (Q91–135)
Q91. Cell Cycle and Cell Division
Given below are two statements :
Statement I : Chromosomes are fully condensed at the end of prophase I.
Statement II : Meiosis I resembles mitosis.
In the light of the above statements, choose the most appropriate answer from the options given below :
- Statement I is correct, but Statement II is false
- Statement I is incorrect, but Statement II is true
- Both Statement I and Statement II are true
- Both Statement I and Statement II are false
Show answer and solution
Answer: option (1)
By diakinesis at the end of prophase I the chromosomes are fully condensed, so Statement I is true. It is meiosis II, not meiosis I, that resembles mitosis, because meiosis I separates homologous chromosomes and halves the chromosome number.
Q92. Animal Kingdom
Which of the following is not a characteristic of chordates ?
- Absence of gills
- Presence of post anal part (tail)
- Presence of notochord
- Central nervous system is dorsal
Show answer and solution
Answer: option (1)
Chordates characteristically have a notochord, a dorsal hollow nerve cord, paired pharyngeal gill slits and a post-anal tail. Absence of gills is therefore not a chordate feature.
Q93. Morphology of Flowering Plants
Phyllotaxy is the pattern of arrangement of ________
- fruits
- sepals
- leaves
- flowers
Show answer and solution
Answer: option (3)
Phyllotaxy describes how leaves are arranged on the stem or branch: alternate, opposite or whorled.
Q94. Chemical Coordination and Integration
Which one of the following statements is incorrect ?
- Glucagon stimulates glycogenolysis
- β-cells of pancreas secrete insulin
- α-cells of pancreas secrete glucagon
- α-cells of pancreas secrete insulin
Show answer and solution
Answer: option (4)
In the islets of Langerhans, -cells secrete glucagon and -cells secrete insulin. So the statement that -cells secrete insulin is wrong.
Q95. Cell: The Unit of Life
The plastid that stores xanthophyll is known as ________ .
- aleuroplast
- amyloplast
- chloroplast
- chromoplast
Show answer and solution
Answer: option (4)
Chromoplasts contain fat-soluble carotenoid pigments such as carotene and xanthophyll, giving yellow, orange or red colours.
Q96. Biological Classification
Match List-I with List-II.
| List-I | List-II |
|---|---|
| A. Fusion of protoplasms between gametes | I. Meiosis |
| B. Fusion of two nuclei | II. Plasmogamy |
| C. Generation of haploid spores | III. Karyogamy |
Choose the correct answer from the options given below :
- A-III, B-II, C-I
- A-I, B-III, C-II
- A-II, B-III, C-I
- A-II, B-I, C-III
Show answer and solution
Answer: option (3)
Fusion of the protoplasms of two gametes is plasmogamy, fusion of their nuclei is karyogamy, and haploid spores are produced by meiosis. So A-II, B-III, C-I.
Q97. Plant Kingdom
Given below are two statements :
Statement I : In gymnosperms, the male and female gametophytes remain within the sporangia.
Statement II : In gymnosperms, seeds are not covered.
In the light of the above statements, choose the most appropriate answer from the options given below :
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
Show answer and solution
Answer: option (3)
In gymnosperms the gametophytes are not free-living; they stay within the sporangia on the sporophyte. The ovules are not enclosed in an ovary, so the seeds are naked. Both statements are correct.
Q98. Structural Organisation in Animals
In water, frogs respire using ________.
- lungs
- trachea
- skin
- buccal cavity
Show answer and solution
Answer: option (3)
In water a frog breathes through its moist skin (cutaneous respiration), where dissolved oxygen diffuses in. Lungs and the buccal cavity are used on land.
Q99. Biological Classification
Mad cow disease is caused by ________.
- Aspergillus sp.
- Mycoplasma sp.
- prions
- viroids
Show answer and solution
Answer: option (3)
Mad cow disease (bovine spongiform encephalopathy) is caused by prions, which are abnormally folded infectious proteins.
Q100. Photosynthesis in Higher Plants
Which of the following statements regarding photorespiration are correct ?
(a) Do not occur in C3 plants
(b) is consumed and is generated
(c) Phosphoglycolate is formed
(d) No synthesis of ATP and NADPH
Choose the correct answer from the options given below :
- (b) and (d) only
- (a) and (b) only
- (a) and (d) only
- (c) and (d) only
Show answer and solution
Answer: option (4)
Photorespiration happens in C3 plants, uses and releases , so (a) and (b) are wrong. RuBisCO acting as an oxygenase forms phosphoglycolate, and the pathway makes no ATP or NADPH, so (c) and (d) are correct.
Q101. Photosynthesis in Higher Plants
Select the correct sequence of experiments that led to a gradual understanding of photosynthesis in green plants.
- Release of oxygen production of glucose absorption spectra of chlorophyll a and b role of air
- Production of glucose role of air release of oxygen absorption spectra of chlorophyll a and b
- Absorption spectra of chlorophyll a and b production of glucose release of oxygen role of air
- Role of air release of oxygen production of glucose absorption spectra of chlorophyll a and b
Show answer and solution
Answer: option (4)
Priestley first showed the role of air (1770s), Ingenhousz then showed oxygen release by green parts in light, Sachs showed glucose (starch) production, and later work by Engelmann and others linked the absorption spectra of chlorophyll a and b to photosynthesis.
Q102. Photosynthesis in Higher Plants
How many turns of Calvin cycle are required for the formation of three molecules of glucose ?
- 1
- 18
- 6
- 3
Show answer and solution
Answer: option (2)
Each turn of the Calvin cycle fixes one , so one glucose (6 carbons) needs 6 turns. Three glucose molecules need turns.
Q103. Cell: The Unit of Life
Mitochondrial inner membrane encloses ____________.
- mucus
- aqueous humor
- matrix
- cytosol
Show answer and solution
Answer: option (3)
The inner mitochondrial membrane surrounds the inner compartment called the matrix, which holds enzymes of the Krebs cycle, DNA and ribosomes.
Q104. Body Fluids and Circulation
Which of the following statements is incorrect ?
- Fibrin is produced from fibrinogen
- Fibrinogen is produced from fibrin
- Blood coagulates in response to an injury
- Blood clot consists of fibrins
Show answer and solution
Answer: option (2)
During clotting, thrombin converts the soluble plasma protein fibrinogen into insoluble fibrin threads. Fibrinogen is not made from fibrin, so statement (2) is incorrect.
Q105. Plant Kingdom
Sphenopsida class belongs to ____________.
- gymnosperms
- pteridophytes
- bryophytes
- angiosperms
Show answer and solution
Answer: option (2)
Pteridophytes are divided into Psilopsida, Lycopsida, Sphenopsida (e.g. Equisetum) and Pteropsida. So Sphenopsida is a pteridophyte class.
Q106. Biological Classification
Match List-I with List-II.
| List-I | List-II |
|---|---|
| A. Spherical | I. Vibrio |
| B. Rod | II. Cocci |
| C. Comma | III. Spirilla |
| D. Spirillum | IV. Bacilli |
Choose the correct answer from the options given below :
- A-II, B-I, C-IV, D-III
- A-II, B-IV, C-I, D-III
- A-I, B-III, C-II, D-IV
- A-III, B-II, C-I, D-IV
Show answer and solution
Answer: option (2)
Spherical bacteria are cocci, rod-shaped are bacilli, comma-shaped are vibrio and spiral ones are spirilla. So A-II, B-IV, C-I, D-III.
Q107. Cell: The Unit of Life
Which of the following are characteristics of prokaryotic cells ?
(a) Ribosomes are made of 50S and 30S subunits
(b) They can have plasmids
(c) They contain mesosome
(d) They have peroxisomes
Choose the correct answer from the options given below :
- (a), (c) and (d) only
- (a), (b) and (c) only
- (b) and (c) only
- (a) and (c) only
Show answer and solution
Answer: option (2)
Prokaryotes have 70S ribosomes (50S + 30S subunits), may carry plasmids, and have mesosomes formed by infolding of the cell membrane. They lack membrane-bound organelles such as peroxisomes, so (a), (b) and (c) are correct.
Q108. Cell: The Unit of Life
Smooth endoplasmic reticulum ___________ .
- is actively involved in protein synthesis
- is a site for the synthesis of carbohydrates
- has ribosomes attached to its surface
- is the major site for the synthesis of lipids
Show answer and solution
Answer: option (4)
Smooth ER has no ribosomes and is the main site of lipid synthesis, including steroid hormones in animal cells. Protein synthesis takes place on rough ER.
Q109. Animal Kingdom
Given below are two statements :
Statement I : The class name Reptilia refers to creeping or crawling mode of locomotion.
Statement II : All organisms belonging to Reptilia have three chambered heart.
In the light of the above statements, choose the most appropriate answer from the options given below :
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
Show answer and solution
Answer: option (1)
Reptilia comes from the creeping or crawling movement of these animals, so Statement I is correct. Most reptiles have a three-chambered heart but crocodiles have four chambers, so Statement II is false.
Q110. Structural Organisation in Animals
In frogs, the number of pairs of cranial nerves arising from the brain are _______ .
- 10
- 12
- 6
- 9
Show answer and solution
Answer: option (1)
The frog brain gives off 10 pairs of cranial nerves.
Q111. Cell: The Unit of Life
Cell theory was formulated by _______ .
- Singer and Nicolson
- Antonie Von Leeuwenhoek
- Schleiden and Schwann
- Robert Brown
Show answer and solution
Answer: option (3)
Schleiden (plants, 1838) and Schwann (animals, 1839) together formulated the cell theory. Virchow later added that new cells arise from pre-existing cells.
Q112. Chemical Coordination and Integration
Which of the following statements related to pituitary gland are correct ?
(a) It is divided anatomically into adenohypophysis and neurohypophysis
(b) It secretes follicle stimulating hormone
(c) It secretes melanocyte stimulating hormone
(d) It does not secrete prolactin
Choose the correct answer from the options given below :
- (c) and (d) only
- (b) and (c) only
- (a) and (b) only
- (a), (b) and (c) only
Show answer and solution
Answer: option (4)
The pituitary is divided anatomically into the adenohypophysis and neurohypophysis. The anterior part secretes FSH, MSH and prolactin, so (d) is false and (a), (b) and (c) are correct.
Q113. Biological Classification
Which of the following is not a prokaryote ?
- Mycoplasma
- Fungi
- Bacteria
- Blue green algae
Show answer and solution
Answer: option (2)
Mycoplasma, bacteria and blue-green algae (cyanobacteria) are prokaryotes in Kingdom Monera. Fungi have a true nucleus and are eukaryotes.
Q114. Plant Growth and Development
Length of the stem at time 0 is 20 cm. The arithmetic growth rate is 30 cm per day. What is the length of the stem at the end of the day ?
- 230 cm
- 460 cm
- 50 cm
- 170 cm
Show answer and solution
Answer: option (1)
Arithmetic growth follows .
Q115. Neural Control and Coordination
Which of the following is not a part of human central neural system ?
- Pia mater
- Pericardium
- Arachnoid
- Dura mater
Show answer and solution
Answer: option (2)
The brain is covered by three meninges: dura mater, arachnoid and pia mater. The pericardium is the membrane around the heart.
Q116. Plant Growth and Development
Which of the following plant growth regulators promotes internode elongation prior to flowering in cabbage ?
- Indole butyric acid
- Ethephon
- Abscisic acid
- Gibberellin
Show answer and solution
Answer: option (4)
Gibberellins cause bolting, which is internode elongation just before flowering in rosette plants such as cabbage and beet.
Q117. Morphology of Flowering Plants
Match List-I with List-II.
| List-I | List-II |
|---|---|
| A. Marginal placentation | I. _Argemone_ |
| B. Axile placentation | II. Tomato |
| C. Parietal placentation | III. _Primrose_ |
| D. Free central placentation | IV. Pea |
Choose the correct answer from the options given below :
- A-IV, B-III, C-I, D-II
- A-IV, B-II, C-I, D-III
- A-II, B-IV, C-I, D-III
- A-IV, B-II, C-III, D-I
Show answer and solution
Answer: option (2)
Pea has marginal placentation, tomato axile, Argemone parietal and Primrose free central. So A-IV, B-II, C-I, D-III.
Q118. Plant Growth and Development
Which of the following plant growth regulators is used as herbicide ?
- Abscisic acid
- Gibberellin
- 2,4-D
- Kinetin
Show answer and solution
Answer: option (3)
2,4-D is a synthetic auxin used as a weedicide to kill broad-leaved dicot weeds. It does not affect mature monocots.
Q119. Animal Kingdom
Given below are two statements :
Statement I : When any plane passing through the central axis of the body divides the organism into two identical halves, it is called radial symmetry.
Statement II : In phylum Echinodermata, both adults and larvae are radially symmetrical.
In the light of the above statements, choose the most appropriate answer from the options given below :
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
Show answer and solution
Answer: option (1)
Statement I is the correct definition of radial symmetry. Adult echinoderms are radially symmetrical but their larvae are bilaterally symmetrical, so Statement II is false.
Q120. Biomolecules
Match List-I with List-II.
| List-I | List-II |
|---|---|
| A. Starch | I. Fights infection |
| B. Antibody | II. Energy storage |
| C. Concanavalin A | III. Glucose transport |
| D. Glut-4 | IV. Lectin |
Choose the correct answer from the options given below :
- A-II, B-I, C-III, D-IV
- A-I, B-II, C-III, D-IV
- A-I, B-II, C-IV, D-III
- A-II, B-I, C-IV, D-III
Show answer and solution
Answer: option (4)
Starch is a storage polysaccharide, antibodies fight infection, concanavalin A is a lectin, and GLUT-4 carries glucose into cells. So A-II, B-I, C-IV, D-III.
Q121. Locomotion and Movement
The number of vertebrae in a human is ________.
- 26
- 206
- 7
- 12
Show answer and solution
Answer: option (1)
The adult human vertebral column has 26 vertebrae: 7 cervical, 12 thoracic, 5 lumbar, 1 sacrum and 1 coccyx.
Q122. Photosynthesis in Higher Plants
Which pigment has absorption peak at 700 nm in the photosynthetic reaction centre PS I (P700) ?
- Xanthophylls
- Carotenoids
- Chlorophyll b
- Chlorophyll a
Show answer and solution
Answer: option (4)
The reaction centre of PS I is a chlorophyll a molecule with an absorption peak at 700 nm, hence the name P700.
Q123. The Living World
Arrange the following taxonomic categories in ascending order.
(a) Genus
(b) Class
(c) Order
(d) Phylum
(e) Family
(f) Kingdom
(g) Species
Choose the correct answer from the options given below :
- (g), (c), (d), (b), (e), (a), (f)
- (f), (c), (b), (g), (d), (e), (a)
- (g), (a), (e), (c), (b), (d), (f)
- (a), (c), (d), (g), (f), (b), (e)
Show answer and solution
Answer: option (3)
From the lowest rank upward the order is species, genus, family, order, class, phylum, kingdom, which is (g), (a), (e), (c), (b), (d), (f).
Q124. The Living World
Match List-I with List-II.
| List-I | List-II |
|---|---|
| A. Family | I. Sapindales |
| B. Genus | II. Dicotyledonae |
| C. Class | III. Anacardiaceae |
| D. Phylum | IV. Angiospermae |
| E. Order | V. _Mangifera_ |
Choose the correct answer from the options given below :
- A-II, B-III, C-V, D-I, E-IV
- A-III, B-V, C-II, D-IV, E-I
- A-I, B-V, C-II, D-IV, E-III
- A-II, B-I, C-III, D-IV, E-V
Show answer and solution
Answer: option (2)
For mango, the family is Anacardiaceae, the genus is Mangifera, the class is Dicotyledonae, the phylum/division is Angiospermae and the order is Sapindales. So A-III, B-V, C-II, D-IV, E-I.
Q125. Biomolecules
Arrange the following elements in descending order of their contribution to percentage weight of the human body.
(a) Oxygen
(b) Carbon
(c) Hydrogen
(d) Nitrogen
Choose the correct answer from the options given below :
- (b), (c), (d), (a)
- (b), (a), (c), (d)
- (a), (b), (c), (d)
- (c), (a), (b), (d)
Show answer and solution
Answer: option (3)
By percentage weight the human body is about 65% oxygen, 18.5% carbon, 9.5% hydrogen and 3.3% nitrogen, so the order is O > C > H > N.
Q126. Cell: The Unit of Life
Match List-I with List-II.
| List-I | List-II |
|---|---|
| A. Cristae | I. Flat membrane sacs in stroma of chloroplast |
| B. Cisternae | II. Infoldings in mitochondria |
| C. Thylakoids | III. Cell membrane |
| D. Phospholipid | IV. Disc shaped sacs in the Golgi apparatus |
Choose the correct answer from the options given below :
- A-II, B-IV, C-III, D-I
- A-IV, B-III, C-I, D-II
- A-III, B-IV, C-I, D-II
- A-II, B-IV, C-I, D-III
Show answer and solution
Answer: option (4)
Cristae are infoldings of the inner mitochondrial membrane, cisternae are the disc-shaped sacs of the Golgi apparatus, thylakoids are the flat membrane sacs in the chloroplast stroma, and phospholipids make up the cell membrane. So A-II, B-IV, C-I, D-III.
Q127. Body Fluids and Circulation
The number of action potentials generated by sino-atrial node (SAN) in a healthy human is __________ per minute.
- 100 – 110
- 120 – 140
- 28 – 30
- 70 – 75
Show answer and solution
Answer: option (4)
The SA node is the pacemaker. In a healthy adult it fires about 70–75 action potentials per minute, which sets the normal heart rate.
Q128. Biological Classification
Symbiotic association between fungi and algae are called __________.
- mycorrhiza
- chrysophytes
- lichens
- sponges
Show answer and solution
Answer: option (3)
Lichens are mutualistic associations between algae and fungi. Mycorrhiza is a fungus living with plant roots.
Q129. Respiration in Plants
How many molecules of pyruvic acid are produced at the end of glycolysis from 206 molecules of glucose ?
- 103
- 412
- 206
- 309
Show answer and solution
Answer: option (2)
Glycolysis splits each glucose into 2 pyruvic acid molecules, so 206 glucose give pyruvic acid.
Q130. Locomotion and Movement
Which of the following represents the correct sequence of arrangement of bones in the lower limb of humans ?
- Femur-patella-tibia-tarsal
- Femur-tarsal-patella-tibia
- Femur-tibia-patella-tarsal
- Patella-femur-tibia-tarsal
Show answer and solution
Answer: option (1)
In the lower limb the femur is the thigh bone. The patella covers the knee joint, the tibia (with the fibula) forms the leg, and the tarsals form the ankle.
Q131. The Living World
Genus represents ___________ .
- a group of closely related species
- a group of closely related families
- an individual plant or animal
- a population of plants and animals
Show answer and solution
Answer: option (1)
A genus is a group of closely related species that share more characters with one another than with species of other genera.
Q132. Cell Cycle and Cell Division
The correct sequence of adult cell cycle phases is ___________ .
- G1-S-G2-M
- S-M-G2-G1
- G1-G2-S-M
- G1-M-G2-S
Show answer and solution
Answer: option (1)
Interphase runs G1 → S → G2 and is followed by the M phase, so the cycle is G1-S-G2-M.
Q133. Cell: The Unit of Life
Endomembrane system includes ___________
- mitochondria, chloroplast, peroxisomes and vacuole
- Golgi complex, chloroplast, peroxisomes and vacuole
- endoplasmic reticulum, Golgi complex, lysosomes and vacuole
- endoplasmic reticulum, chloroplast, peroxisomes and vacuole
Show answer and solution
Answer: option (3)
The endomembrane system is made up of the ER, the Golgi complex, lysosomes and vacuoles, whose functions are coordinated. Mitochondria, chloroplasts and peroxisomes are not included.
Q134. Photosynthesis in Higher Plants
Photorespiration reaction catalyzed by RuBisCo is shown below :
Identify "X" from the given options :
- Oxaloacetate
- Malate
- Phosphoenolpyruvate
- 2-Phosphoglycolate
Show answer and solution
Answer: option (4)
When RuBisCO acts as an oxygenase, RuBP + gives one 3-phosphoglycerate and one 2-phosphoglycolate.
Q135. Morphology of Flowering Plants
Which of the following are characteristic features of Solanaceae family ?
(a) Flowers are bisexual and actinomorphic
(b) Calyx have five sepals and are united
(c) Androecium have five stamens and are epipetalous
(d) Ovary is inferior
Choose the correct answer from the options given below :
- (a) and (b) only
- (b), (c) and (d) only
- (a), (b) and (c) only
- (d) only
Show answer and solution
Answer: option (3)
Solanaceae flowers are bisexual and actinomorphic, with 5 united sepals and 5 epipetalous stamens. The bicarpellary syncarpous ovary is superior, so (d) is false.
Zoology questions (Q136–180)
Q136. Human Reproduction
Sperm motility is due to ________.
- amoeboid movement
- muscular movement
- flagellar movement
- ciliary movement
Show answer and solution
Answer: option (3)
A sperm swims by the whip-like beating of its tail, a flagellum powered by mitochondria in the middle piece. So its motility is flagellar movement.
Q137. Biodiversity and Conservation
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Abingdon tortoise in Galapagos islands became extinct within a decade after goats were introduced.
Reason R : Goats were more efficient at browsing than Abingdon tortoise.
In the light of the above statements, choose the most appropriate answer from the options given below :
- A is correct but R is not correct
- A is not correct but R is correct
- Both A and R are correct and R is the correct explanation of A
- Both A and R are correct but R is not the correct explanation of A
Show answer and solution
Answer: option (3)
Goats were introduced to the Galapagos and, being far better browsers, outcompeted the Abingdon tortoise for food, driving it extinct within about ten years. Both statements are true and the browsing advantage of goats explains the extinction.
Q138. Organisms and Populations
Match List-I with List-II.
| List-I | List-II |
|---|---|
| A. Both species are harmed | I. Predation |
| B. One species is harmed and the other is benefited | II. Mutualism |
| C. Both species are benefited | III. Competition |
| D. One is benefited while the other has no effect | IV. Commensalism |
Choose the correct answer from the options given below :
- A-II, B-I, C-IV, D-III
- A-III, B-I, C-II, D-IV
- A-III, B-IV, C-II, D-I
- A-I, B-II, C-III, D-IV
Show answer and solution
Answer: option (2)
Competition harms both species (−/−), predation benefits one and harms the other (+/−), mutualism benefits both (+/+), and commensalism benefits one with no effect on the other (+/0). So A-III, B-I, C-II, D-IV.
Q139. Human Reproduction
Which of the following structure is not a part of the male reproductive system?
- Vasa efferentia
- Infundibulum
- Rete testis
- Epididymis
Show answer and solution
Answer: option (2)
The infundibulum is the funnel-shaped end of the fallopian tube in the female tract. Vasa efferentia, rete testis and epididymis are all ducts of the male system.
Q140. Sexual Reproduction in Flowering Plants
Which of the following plant produces non-albuminous seeds?
- Barley
- Pea
- Wheat
- Maize
Show answer and solution
Answer: option (2)
In pea the endosperm is used up by the developing embryo, so the mature seed has no endosperm (non-albuminous). Barley, wheat and maize are cereals with persistent endosperm (albuminous).
Q141. Molecular Basis of Inheritance
Which of the following statements about lac-operon is correct?
- Genes , , and share single common promoter
- Galactose can act as an inducer of lac operon
- Gene is constitutively expressed
- Lactose activates repressor to bind to the operator
Show answer and solution
Answer: option (3)
The regulator gene has its own promoter and is expressed constitutively, making repressor all the time. Only , , share the lac promoter; the inducer is lactose (allolactose), which inactivates the repressor rather than activating it.
Q142. Human Health and Disease
Which of the following is used as an effective sedative and painkiller for treating post-surgery patients?
- Morphine
- Anti-retroviral drugs
- Interferon
- Antibiotics
Show answer and solution
Answer: option (1)
Morphine, an opioid from the poppy latex, is a strong sedative and painkiller given to patients after surgery. The other options do not relieve pain.
Q143. Locomotion and Movement
Which of the following is the correct order of arrangement of vertebrate column from the head to toe?
- Cervical vertebra, lumbar vertebra, thoracic vertebra, sacrum
- Cervical vertebra, thoracic vertebra, lumbar vertebra, sacrum
- Cervical vertebra, thoracic vertebra, sacrum, lumbar vertebra
- Sacrum, lumbar vertebra, thoracic vertebra, cervical vertebra
Show answer and solution
Answer: option (2)
The vertebral column runs cervical (7) → thoracic (12) → lumbar (5) → sacral (fused sacrum) → coccygeal from head downward.
Q144. Locomotion and Movement
Muscle contraction is initiated by a signal sent by the central nervous system by the release of _______ .
- cyclic guanine monophosphate
- cyclic adenine monophosphate
- acetyl choline
- acetyl coenzyme A
Show answer and solution
Answer: option (3)
A motor neuron signal reaches the neuromuscular junction, where acetylcholine is released and generates an action potential in the sarcolemma, starting contraction.
Q145. Evolution
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Forelimbs of human and bats are homologous.
Reason R : Forelimbs of humans and bats have similar anatomical structure.
In the light of the above statements, choose the most appropriate answer from the options given below.
- A is true but R is false
- A is false but R is true
- Both A and R are correct and R is the correct explanation of A
- Both A and R are true, but R is not the correct explanation of A
Show answer and solution
Answer: option (3)
Human and bat forelimbs share the same basic bone plan (humerus, radius-ulna, carpals, metacarpals, phalanges) despite different functions, which is exactly what makes them homologous. So both are true and R explains A.
Q146. Principles of Inheritance and Variation
Given below are two statements :
Statement I : Down’s syndrome is caused by the absence of one of the X-chromosomes.
Statement II : Turner’s syndrome is caused by the presence of an additional copy of the chromosomes.
In the light of the above statements, choose the correct answer from the options given below :
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
Show answer and solution
Answer: option (4)
Down's syndrome is trisomy 21 (an extra chromosome 21), and Turner's syndrome is 45,XO (loss of one X). Each statement describes the other's cause, so both are incorrect.
Q147. Excretory Products and their Elimination
Which of the following statements about the reabsorption process in Henle's loop are correct?
(a) The descending limb of Henle's loop is permeable to water but almost impermeable to electrolytes.
(b) Urine gets concentrated in Henle's loop.
(c) Reabsorption of and water takes place in Henle's loop.
(d) Active or passive transport of electrolytes occurs in the ascending limb of Henle's loop.
Choose the correct answer from the options given below :
- (a), (b) and (c) only
- (a), (b) and (d) only
- (a) and (b) only
- (b), (c) and (d) only
Show answer and solution
Answer: option (2)
The descending limb lets water out but not salts, the ascending limb moves electrolytes out actively or passively but not water, and this countercurrent arrangement concentrates the filtrate in the loop. NCERT notes that reabsorption in Henle's loop is minimal, so (c) is excluded and (a), (b), (d) are correct.
Q148. Sexual Reproduction in Flowering Plants
If the diploid chromosome number of typical angiosperm is 36, what would be the chromosome number in its endosperm?
- 54
- 72
- 18
- 36
Show answer and solution
Answer: option (1)
Here , so . Endosperm is triploid (), giving chromosomes.
Q149. Biodiversity and Conservation
Arrange the following in descending order of number of species in the Amazonian rain forest.
(a) Plants
(b) Birds
(c) Fishes
(d) Invertebrates
(e) Mammals
Choose the correct answer from the options given below :
- (e) > (b) > (a) > (c) > (d)
- (b) > (a) > (d) > (c) > (e)
- (c) > (b) > (d) > (e) > (a)
- (d) > (a) > (c) > (b) > (e)
Show answer and solution
Answer: option (4)
NCERT figures for the Amazon: about 1,25,000 invertebrates, 40,000 plants, 3,000 fishes, 1,300 birds and 427 mammals. So invertebrates > plants > fishes > birds > mammals.
Q150. Evolution
A population of diploid organisms is at Hardy-Weinberg equilibrium. If the frequency of allele A is 0.1, the frequency of AA is ________.
- 0.10
- 0.99
- 0.01
- 0.02
Show answer and solution
Answer: option (3)
At Hardy-Weinberg equilibrium the frequency of AA is .
Q151. Body Fluids and Circulation
The opening between the right atrium and the right ventricle is guarded by ________.
- semilunar valve
- sino-atrial node
- bicuspid valve
- tricuspid valve
Show answer and solution
Answer: option (4)
The right atrioventricular opening is guarded by the three-cusped tricuspid valve; the bicuspid (mitral) valve guards the left side and semilunar valves guard the arterial exits.
Q152. Molecular Basis of Inheritance
Which of the following enzymes synthesizes precursor mRNA?
- RNA polymerase III
- DNA polymerase
- RNA polymerase I
- RNA polymerase II
Show answer and solution
Answer: option (4)
In eukaryotes RNA polymerase II transcribes hnRNA, the precursor of mRNA. Pol I makes rRNAs and Pol III makes tRNA, 5S rRNA and snRNAs.
Q153. Biotechnology: Principles and Processes
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : In recombinant DNA technology, lysozyme is used for disrupting bacterial cells while cellulase is for plant cells.
Reason R : Isolation of genetic material needs disruption of cells.
In the light of the above statements, choose the most appropriate answer from the options given below :
- A is correct but R is not correct
- A is not correct but R is correct
- Both A and R are correct and R is the correct explanation of A
- Both A and R are correct but R is not the correct explanation of A
Show answer and solution
Answer: option (3)
To get DNA out, the cell must first be broken open, so bacterial walls are digested with lysozyme and plant walls with cellulase. Both are true and the need for cell disruption is why these enzymes are used.
Q154. Human Reproduction
Which of the following hormone is not secreted by human placenta?
- Progesterone
- LH
- hCG
- Estrogen
Show answer and solution
Answer: option (2)
The placenta acts as an endocrine tissue making hCG, hPL, estrogens and progesterone. LH is made by the anterior pituitary, not the placenta.
Q155. Human Health and Disease
Colostrum, secreted by mother during initial days of lactation, is abundant in ________.
- IgA
- IgD
- IgG
- IgM
Show answer and solution
Answer: option (1)
Colostrum is rich in IgA antibodies, which give the newborn passive immunity.
Q156. Organisms and Populations
Consider a population of 10 million cells. Given the per-capita birth rate of 0.002 (per unit time) and the per-capita death rate of 0.002 (per unit time), the expected number of cells after 10 generations is ________.
- 10 million
- 100 million
- 1 million
- 5 million
Show answer and solution
Answer: option (1)
Net growth rate , so the population neither grows nor declines and stays at 10 million.
Q157. Biotechnology: Principles and Processes
Given below are two statements :
Statement I : Plasmids are autonomously replicating DNA.
Statement II : Plasmids are extrachromosomal DNA.
In the light of the above statements, choose the most appropriate answer from the options given below :
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
Show answer and solution
Answer: option (3)
Plasmids are small circular DNA molecules that lie outside the bacterial chromosome and replicate independently of it. Both statements are true.
Q158. Biotechnology: Principles and Processes
During PCR, primers bind to the DNA strands in the ________ step.
- annealing
- ligation
- denaturation
- extension
Show answer and solution
Answer: option (1)
PCR cycles are denaturation (strands separate), annealing (primers bind to complementary sites) and extension (Taq polymerase builds new strands). Primer binding is annealing.
Q159. Biotechnology: Principles and Processes
Match List-I with List-II.
| List-I | List-II |
|---|---|
| A. Transformation | I. Restriction enzyme |
| B. Cloning site | II. Transfer DNA to host bacteria |
| C. Selection | III. Replication |
| D. Ori | IV. Antibiotic |
Choose the correct answer from the options given below :
- A-III, B-IV, C-II, D-I
- A-IV, B-I, C-III, D-II
- A-II, B-I, C-IV, D-III
- A-I, B-II, C-IV, D-III
Show answer and solution
Answer: option (3)
Transformation is getting DNA into host bacteria, a cloning site is a recognition site for a restriction enzyme, selection uses antibiotic-resistance markers, and ori is where replication starts. So A-II, B-I, C-IV, D-III.
Q160. Evolution
Adaptive radiation in placental mammals and Australian Marsupials leading to similarity between distant species is an example of ________.
- founder effect
- genetic drift
- divergent evolution
- convergent evolution
Show answer and solution
Answer: option (4)
Placental mammals and Australian marsupials evolved separately yet produced look-alike forms (e.g. wolf and Tasmanian wolf) because they adapted to similar habitats. Similarity in unrelated lines from similar pressures is convergent evolution.
Q161. Human Reproduction
The covering of ovum at ovulation is ________.
- zona pellucida
- chorion
- endometrium
- zona radiata
Show answer and solution
Answer: option (1)
The secondary oocyte released at ovulation is surrounded by the zona pellucida, a glycoprotein layer, with corona radiata cells outside it.
Q162. Ecosystem
Which of the following statements are Correct?
(a) Energy flow from producers to consumers is unidirectional
(b) Energy pyramid can never be inverted
(c) Transfer of energy follows the 1% law
Choose the correct answer from the options given below :
- (a) and (c) only
- (b) and (c) only
- (a), (b) and (c)
- (a) and (b) only
Show answer and solution
Answer: option (4)
Energy moves one way from producers to consumers and is lost as heat at each step, so an energy pyramid is always upright. Transfer follows the 10% law, not 1%, so only (a) and (b) are correct.
Q163. Sexual Reproduction in Flowering Plants
How many theca are present in each lobe of a typical bilobed angiosperm anther?
- 8
- 12
- 2
- 6
Show answer and solution
Answer: option (3)
A typical anther is bilobed and each lobe has two thecae (dithecous), giving four microsporangia in all. So each lobe holds 2 thecae.
Q164. Human Health and Disease
Which of the following statements is correct about Plasmodium?
- Gametocytes develop in mosquito gut
- Fertilization takes place in mosquito gut
- Reproduces sexually in liver cells
- Reproduces sexually in RBCs
Show answer and solution
Answer: option (2)
Male and female gametes of Plasmodium fuse in the stomach (gut) of the female Anopheles mosquito. Gametocytes form in human RBCs, and the liver and RBC stages are asexual.
Q165. Principles of Inheritance and Variation
For a person with blood group ‘O’, which of the following is not a possible combination of parents’ blood group genotypes?
- Father : and Mother :
- Father : and Mother :
- Father : and Mother :
- Father : and Mother :
Show answer and solution
Answer: option (2)
A blood group O child is and needs an allele from each parent. A father who is has no allele, so option 2 cannot give an O child, while every other pair has in both parents.
Q166. Microbes in Human Welfare
Which of the following is used as a clot buster?
- Cyclosporin A
- Statins
- Streptokinase
- Penicillin
Show answer and solution
Answer: option (3)
Streptokinase from Streptococcus dissolves blood clots and is used as a clot buster. Cyclosporin A is an immunosuppressant, statins lower cholesterol and penicillin is an antibiotic.
Q167. Reproductive Health
Which of the following disease is not sexually transmitted?
- Gonorrhoea
- Genital warts
- Syphilis
- Tuberculosis
Show answer and solution
Answer: option (4)
Gonorrhoea, genital warts and syphilis are STIs; tuberculosis spreads through air droplets, not sexual contact.
Q168. Principles of Inheritance and Variation
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : In an experiment, Mendel observed that the F progeny plants are all tall and none are dwarf.
Reason R : Stem height is a contrasting trait, with tall being dominant and dwarf being recessive.
In the light of the above statements, choose the most appropriate answer from the options given below :
- A is correct but R is not correct
- A is not correct but R is correct
- Both A and R are correct and R is the correct explanation of A
- Both A and R are correct but R is not the correct explanation of A
Show answer and solution
Answer: option (3)
Crossing true-breeding tall and dwarf plants gave an all-tall F1 because tall is dominant over dwarf for this contrasting trait. Both statements are true and R explains A.
Q169. Reproductive Health
The method of directly of injecting a sperm into ovum in assisted reproductive technology is called :
- Intra cytoplasmic sperm injection (ICSI)
- Embryo transfer (ET)
- Gamete intra fallopian transfer (GIFT)
- Zygote intra fallopian transfer (ZIFT)
Show answer and solution
Answer: option (1)
In ICSI a single sperm is injected straight into the cytoplasm of the ovum in the lab. GIFT, ZIFT and ET transfer gametes or embryos instead.
Q170. Biotechnology and its Applications
The inactive form of Bt toxin is converted to the active form in the insect gut ________.
- by proteases
- by nucleases
- due to alkaline pH
- due to acidic pH
Show answer and solution
Answer: option (3)
Bt protoxin crystals dissolve in the alkaline pH of the insect midgut, which converts them to the active toxin that binds midgut cells and makes pores. NCERT credits the activation to the alkaline pH.
Q171. Organisms and Populations
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The logistic growth model of populations is considered more realistic than the exponential growth model.
Reason R : Resources are finite.
In the light of the above statements, choose the most appropriate answer from the options given below :
- A is correct but R is not correct
- A is not correct but R is correct
- Both A and R are correct and R is the correct explanation of A
- Both A and R are correct but R is not the correct explanation of A
Show answer and solution
Answer: option (3)
Because resources are finite, no population can grow exponentially forever. The logistic model includes a carrying capacity, which makes it more realistic, so R explains A.
Q172. Human Health and Disease
Which of the following are secondary lymphoid organs?
(a) Bone marrow
(b) Tonsils
(c) Spleen
(d) Thymus
Choose the correct answer from the options given below:
- (b) and (d) only
- (a) and (d) only
- (a) and (b) only
- (b) and (c) only
Show answer and solution
Answer: option (4)
Bone marrow and thymus are primary lymphoid organs where lymphocytes form and mature. Tonsils and spleen are secondary lymphoid organs where lymphocytes meet antigens.
Q173. Sexual Reproduction in Flowering Plants
Which of the following in female gametophyte of an angiosperm helps in guiding the pollen tube for fertilizing the eggs?
- Central cells
- Polar nucleus
- Antipodals
- Synergids
Show answer and solution
Answer: option (4)
The synergids carry a filiform apparatus at the micropylar end that guides the pollen tube into the embryo sac.
Q174. Human Reproduction
Given below are two statements :
Statement I : Ovulation is caused by LH surge leading to rupture of Graafian follicles.
Statement II : Graafian follicle remaining after ovulation transform into corpus luteum and secretes large amount of estrogen.
In the light of the above statements, choose the most appropriate answer from the options given below :
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
Show answer and solution
Answer: option (1)
The LH surge in mid-cycle ruptures the Graafian follicle and causes ovulation. The leftover follicle becomes the corpus luteum, which mainly secretes large amounts of progesterone, not estrogen, so Statement II is wrong.
Q175. Evolution
Given below are two statements :
Statement I : Modern Homo sapiens arose in Australia and moved across continents.
Statement II : Homo sapiens arose around 75000 to 10000 years ago.
In the light of the above statements, choose the most appropriate answer from the options given below :
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
Show answer and solution
Answer: option (2)
Modern Homo sapiens arose in Africa, not Australia, and spread across continents. NCERT places this during the ice age about 75,000–10,000 years ago, so only Statement II is correct.
Q176. Ecosystem
Which of the following are primary consumers in a food chain?
- Herbivores
- Carnivores
- Parasites
- Predators
Show answer and solution
Answer: option (1)
Herbivores eat producers directly and so occupy the second trophic level as primary consumers.
Q177. Evolution
Which of the following is not evidence for evolution?
- Embryological support for evolution as proposed by Ernst Heckel
- Divergent evolution of anatomical structures such as forelimbs
- Convergent evolution of traits like wings of birds and butterflies
- Paleontological evidence from fossil records
Show answer and solution
Answer: option (1)
Haeckel's idea that embryos pass through adult stages of ancestors was rejected by Karl Ernst von Baer, so it is not accepted evidence. Homology, analogy and fossils are all valid evidence.
Q178. Animal Kingdom
Sponges exchange with by ________.
- tracheal tubes
- gills
- simple diffusion over their entire body surfaces
- moist cuticle
Show answer and solution
Answer: option (3)
Sponges have no respiratory organs; water flowing through the canal system lets and diffuse across the body cells.
Q179. Chemical Coordination and Integration
Match List-I with List-II.
| List-I | List-II |
|---|---|
| (A) Excess growth hormone | (I) Reabsorption of water and electrolytes in kidney |
| (B) Luteinizing hormone | (II) Contraction of uterus during child birth |
| (C) Vasopressin | (III) Acromegaly |
| (D) Oxytocin | (IV) Ovulation |
Choose the correct answer from the options given below:
- A-II, B-IV, C-I, D-III
- A-IV, B-III, C-I, D-II
- A-III, B-IV, C-II, D-I
- A-III, B-IV, C-I, D-II
Show answer and solution
Answer: option (4)
Excess GH in adults causes acromegaly, LH triggers ovulation, vasopressin (ADH) increases water and electrolyte reabsorption in the kidney, and oxytocin contracts the uterus during birth. So A-III, B-IV, C-I, D-II.
Q180. Evolution
Natural selection can lead to ________.
(a) stabilisation
(b) genetic drift
(c) directional change
(d) disruption
Choose the correct answer from the options given below :
- (a), (b), (c) and (d)
- (a) and (c) only
- (a) only
- (a), (c) and (d) only
Show answer and solution
Answer: option (4)
Natural selection can be stabilising, directional or disruptive. Genetic drift is a separate, chance-driven change in allele frequency, not an outcome of selection.
Questions by chapter
How the 180 questions were spread across chapters in this paper.
| Physics chapter | Questions |
|---|---|
| Dual Nature of Radiation and Matter | 3 |
| Electromagnetic Induction | 3 |
| Electrostatic Potential and Capacitance | 3 |
| Ray Optics and Optical Instruments | 3 |
| System of Particles and Rotational Motion | 3 |
| Atoms | 2 |
| Electromagnetic Waves | 2 |
| Gravitation | 2 |
| Kinetic Theory | 2 |
| Mechanical Properties of Fluids | 2 |
| Moving Charges and Magnetism | 2 |
| Oscillations | 2 |
| Semiconductor Electronics: Materials, Devices and Simple Circuits | 2 |
| Thermodynamics | 2 |
| Units and Measurement | 2 |
| Work, Energy and Power | 2 |
| Alternating Current | 1 |
| Current Electricity | 1 |
| Electric Charges and Fields | 1 |
| Laws of Motion | 1 |
| Motion in a Straight Line | 1 |
| Nuclei | 1 |
| Thermal Properties of Matter | 1 |
| Waves | 1 |
| Chemistry chapter | Questions |
|---|---|
| Coordination Compounds | 6 |
| Aldehydes, Ketones and Carboxylic Acids | 3 |
| Organic Chemistry – Some Basic Principles and Techniques | 3 |
| The d- and f-Block Elements | 3 |
| The p-Block Elements | 3 |
| Thermodynamics | 3 |
| Amines | 2 |
| Biomolecules | 2 |
| Chemical Bonding and Molecular Structure | 2 |
| Chemical Kinetics | 2 |
| Classification of Elements and Periodicity in Properties | 2 |
| Electrochemistry | 2 |
| Hydrocarbons | 2 |
| Redox Reactions | 2 |
| Solutions | 2 |
| Some Basic Concepts of Chemistry | 2 |
| Alcohols, Phenols and Ethers | 1 |
| Equilibrium | 1 |
| Haloalkanes and Haloarenes | 1 |
| Structure of Atom | 1 |
| Botany chapter | Questions |
|---|---|
| Cell: The Unit of Life | 7 |
| Biological Classification | 5 |
| Photosynthesis in Higher Plants | 5 |
| Animal Kingdom | 3 |
| Morphology of Flowering Plants | 3 |
| Plant Growth and Development | 3 |
| The Living World | 3 |
| Biomolecules | 2 |
| Body Fluids and Circulation | 2 |
| Cell Cycle and Cell Division | 2 |
| Chemical Coordination and Integration | 2 |
| Locomotion and Movement | 2 |
| Plant Kingdom | 2 |
| Structural Organisation in Animals | 2 |
| Neural Control and Coordination | 1 |
| Respiration in Plants | 1 |
| Zoology chapter | Questions |
|---|---|
| Evolution | 6 |
| Human Reproduction | 5 |
| Biotechnology: Principles and Processes | 4 |
| Human Health and Disease | 4 |
| Sexual Reproduction in Flowering Plants | 4 |
| Organisms and Populations | 3 |
| Principles of Inheritance and Variation | 3 |
| Biodiversity and Conservation | 2 |
| Ecosystem | 2 |
| Locomotion and Movement | 2 |
| Molecular Basis of Inheritance | 2 |
| Reproductive Health | 2 |
| Animal Kingdom | 1 |
| Biotechnology and its Applications | 1 |
| Body Fluids and Circulation | 1 |
| Chemical Coordination and Integration | 1 |
| Excretory Products and their Elimination | 1 |
| Microbes in Human Welfare | 1 |
Questions students ask
How many questions were in the Re-NEET 2026 paper?
180 questions for 720 marks: 45 each from Physics, Chemistry, Botany and Zoology. All were compulsory, with +4 for a correct answer and −1 for a wrong one.
Which question was dropped in Re-NEET 2026?
In booklet code 60, NTA dropped Q43 (Physics, units and measurement) in the final answer key. Every candidate got 4 marks for it.
Was Re-NEET 2026 a computer-based test?
No. The 21 June 2026 re-exam was held on pen and paper (OMR). You can still practise this paper on an NTA-style CBT screen on this site.
Who wrote these solutions?
The solutions are written by the Testline NEET team. The answers follow NTA's final answer key.