CUET PG Nanoscience Question Paper 2024: Download Question paper with Answers PDF

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Sahaj Anand

Content Writer | Journalism Graduate | Updated 3+ months ago

CUET PG Nanoscience Question Paper 2024 is available here for download. NTA conducted CUET PG Nanoscience paper 2024 on from March 28 in Shift 2. CUET PG Question Paper 2024 is based on objective-type questions (MCQs). According to latest exam pattern, candidates get 105 minutes to solve 75 MCQs in CUET PG 2024 Nanoscience question paper.

CUET PG Nanoscience Question Paper 2024 PDF Download

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CUET PG 2024 Nanoscience Questions with Solutions

Question 1:

The First Law of Thermodynamics represents:

  1. Law of conservation of mass
  2. Law of conservation of energy
  3. Law of conservation of entropy
  4. Law of conservation of momentum
Correct Answer: (2) Law of conservation of energy.
View Solution

The First Law of Thermodynamics is a statement of energy conservation, asserting that energy cannot be created or destroyed in an isolated system. The law is mathematically expressed as:

∆U = Q − W

  • ∆U: Change in internal energy of the system.
  • Q: Heat added to the system.
  • W: Work done by the system.

This principle is applicable in various thermodynamic processes, including isochoric, isobaric, isothermal, and adiabatic processes. It also explains energy transfer in heat engines, refrigerators, and closed systems.

Question 2:

A Carnot engine is a reversible engine. It can be proved from the:

  1. Zeroth Law of Thermodynamics
  2. First Law of Thermodynamics
  3. Second Law of Thermodynamics
  4. Third Law of Thermodynamics
Correct Answer: (3) Second Law of Thermodynamics.
View Solution

The Second Law of Thermodynamics introduces the concept of entropy, which governs the direction of energy transfer and the efficiency of heat engines. A Carnot engine is a theoretical reversible engine that operates between two reservoirs and achieves maximum efficiency:

η = 1 − Tc / Th

  • Tc: Absolute temperature of the cold reservoir.
  • Th: Absolute temperature of the hot reservoir.

The Carnot engine is reversible because it undergoes no net entropy change (∆S = 0), making it an ideal standard for evaluating real-world engines.

Question 3:

From the Hall Effect experiment, one can measure:

  1. Carrier concentration
  2. Nature of charge carriers
  3. Carrier concentration and Mobility of charge carriers
  4. Nature of charge carriers, concentration of charge carriers as well as their mobility
Correct Answer: (4) Nature of charge carriers, concentration of charge carriers as well as their mobility.
View Solution

The Hall Effect occurs when a current-carrying conductor is placed in a perpendicular magnetic field, generating a transverse voltage (Hall voltage):

VH = (IB) / (qnd)

  • I: Current.
  • B: Magnetic field strength.
  • q: Charge of carriers.
  • n: Carrier concentration.
  • d: Thickness of the conductor.

This experiment provides:

  • Nature of charge carriers: Positive (holes) or negative (electrons).
  • Carrier concentration (n): Determined from the Hall coefficient.
  • Mobility of charge carriers: Can also be derived from the Hall Effect measurements.

Question 4:

With the reduction in particle size, the optical absorption spectrum shows:

  1. Blue Shift
  2. Red Shift
  3. Green Shift
  4. Yellow Shift
Correct Answer: (1) Blue Shift.
View Solution

When particles are reduced to the nanoscale, quantum confinement effects dominate. The energy gap (Eg) increases as the particle size decreases, causing higher energy (shorter wavelength) absorption and emission. This phenomenon, termed "blue shift," is observed in materials like quantum dots.

Question 5:

Bragg’s law of diffraction is used to determine the:

  1. Crystallite structure
  2. Crystal structure
  3. Particle size
  4. Crystalline structure as well as Particle size
Correct Answer: (2) Crystal structure.
View Solution

Bragg’s law describes the diffraction of X-rays by crystal planes:

nλ = 2d sin θ

  • n: Order of diffraction.
  • λ: Wavelength of X-rays.
  • d: Interplanar spacing.
  • θ: Diffraction angle.

This law is fundamental for determining the arrangement of atoms in crystals.

Question 6:

A bipolar junction transistor is said to be in active mode if:

  1. Base-emitter junction is forward biased and collector-emitter junction is reverse biased
  2. Base-emitter junction is reverse biased and collector-emitter junction is forward biased
  3. Both junctions are forward biased
  4. Both junctions are reverse biased
Correct Answer: (1) Base-emitter junction is forward biased and collector-emitter junction is reverse biased.
View Solution

In active mode:

  • The base-emitter junction is forward biased, facilitating charge flow from emitter to base.
  • The collector-emitter junction is reverse biased, enabling charge collection at the collector.

This mode is essential for amplification in bipolar junction transistors (BJTs).

Question 7:

In a Quantum dot, the degree of confinement is:

  1. Zero
  2. One
  3. Two
  4. Three
Correct Answer: (4) Three.
View Solution

Quantum dots are nanoscale particles in which electrons are confined in all three spatial dimensions, leading to discrete quantized energy levels. Unlike bulk materials with continuous energy bands, quantum dots behave like artificial atoms, showing unique optical and electronic properties. This three-dimensional confinement results in size-dependent behavior, such as tunable fluorescence, making quantum dots valuable in applications like medical imaging, photovoltaics, and quantum computing.

Question 8:

The failure of Michelson-Morley experiment indicated:

  1. Existence of other medium
  2. Speed of light is directional
  3. Speed of light is invariant
  4. Light has dual nature
Correct Answer: (3) Speed of light is invariant.
View Solution

The Michelson-Morley experiment was conducted to detect the presence of a luminiferous ether, a supposed medium for the propagation of light. Its null result demonstrated that the speed of light is constant in all inertial reference frames, regardless of the motion of the source or observer. This pivotal discovery refuted the ether hypothesis and laid the foundation for Einstein’s special theory of relativity, where the speed of light (c) is a fundamental constant.

Question 9:

A rigid body is dropped from a tower top. Midway it splits in two unequal parts. The centre of mass of the system:

  1. Shifts horizontally towards the heavier part
  2. Shifts horizontally towards the lighter part
  3. Does not shift horizontally
  4. Does shift depending on velocity at the time of split
Correct Answer: (3) Does not shift horizontally.
View Solution

When the rigid body is dropped, gravity is the only external force acting on it, and gravity acts vertically. The centre of mass (COM) of the system is initially in a vertical line. The key principle here is that the COM position is determined by the distribution of mass and the external forces acting on the system. Internal forces, such as the splitting of the body into two unequal parts, do not affect the position of the COM horizontally.

Let's break it down:

  • The force of gravity acts vertically on the system, causing the body to fall vertically.
  • When the body splits into two parts, the splitting is an internal event, and the internal forces are equal and opposite (Newton’s Third Law). These forces do not affect the motion of the COM in the horizontal direction.
  • Because no external horizontal forces are applied, there is no reason for the horizontal position of the COM to shift. The COM continues to fall in the same vertical path as before the split.

Therefore, the horizontal position of the COM remains unchanged during the splitting process, and it continues its downward motion along the same vertical line as the body before it split.

Question 10:

A system in thermodynamic equilibrium satisfies the following requirements:

  1. A system is in mechanical equilibrium
  2. A system is in mechanical equilibrium and chemical equilibrium
  3. A system is in mechanical equilibrium and thermal equilibrium
  4. A system is in mechanical equilibrium, chemical equilibrium and thermal equilibrium
Correct Answer: (4) A system is in mechanical equilibrium, chemical equilibrium and thermal equilibrium.
View Solution

For a system to be in thermodynamic equilibrium:

  • Mechanical equilibrium: There are no unbalanced forces; pressure is uniform throughout.
  • Thermal equilibrium: Temperature is uniform, with no heat flow within the system or with its surroundings.
  • Chemical equilibrium: No net chemical reactions occur, and there is no transfer of species between phases.

Only when all three conditions are met can the system be considered in thermodynamic equilibrium, maintaining stability over time.

Question 11:

Which of the following orbitals will have a radial node?

  1. 1s orbital
  2. 2s orbital
  3. 2p orbital
  4. 3d orbital
Correct Answer: (2) 2s orbital.
View Solution

Radial nodes are spherical regions in an orbital where the probability of finding an electron is zero. They are located at specific radii from the nucleus.

Formula for radial nodes: The number of radial nodes is given by n − l − 1, where:

  • n is the principal quantum number
  • l is the azimuthal quantum number (0 for s, 1 for p, 2 for d, etc.

Calculation:

  • (a) 1s orbital: n = 1, l = 0, nodes = 1 − 0 − 1 = 0 (no radial node).
  • (b) 2s orbital: n = 2, l = 0, nodes = 2 − 0 − 1 = 1 (one radial node).
  • (c) 2p orbital: n = 2, l = 1, nodes = 2 − 1 − 1 = 0 (no radial node).
  • (d) 3d orbital: n = 3, l = 2, nodes = 3 − 2 − 1 = 0 (no radial node).

Only the 2s orbital has a radial node among the given options.

Question 12:

Primitive cell has lattice point per unit cell:

  1. One
  2. Two
  3. Three
  4. (4) Four
Correct Answer: (1) One.
View Solution

Definition of Primitive Cell: A primitive cell is the smallest repeating unit in a crystal lattice that can generate the entire lattice through translations.

Lattice Points: Lattice points are locations within the unit cell where atoms, ions, or molecules reside.

Corner Atoms Contribution: Each atom at a corner contributes 1/8 to a single cell since it is shared among eight adjacent cells.

Total Lattice Points: A primitive cell has 8 corner atoms × 1/8 contribution = 1 lattice point.

Question 13:

Filling of electrons in degenerate orbitals is governed by:

  1. Hund’s rule
  2. Planck’s Law
  3. de-Broglie’s wave-particle duality
  4. Photoelectric effect
Correct Answer: (1) Hund’s rule.
View Solution

Degenerate Orbitals: These are orbitals with the same energy, such as the three p orbitals in a given shell.

Hund’s Rule: This rule states that electrons fill degenerate orbitals singly with parallel spins before pairing. This minimizes electron repulsion and maximizes stability.

Illustration: For example, in the p orbitals, electrons will occupy px, py, and pz singly before any pairing occurs.

Relevance of Other Laws:

  • Planck’s Law relates energy to the frequency of light.
  • de-Broglie’s theory addresses the wave-particle duality of matter.
  • The Photoelectric effect explains electron ejection by photons.

Hund’s Rule is the only principle governing electron filling in degenerate orbitals.

Question 14:

Arrange the following radiations in terms of increasing order of their ionizing power:

  1. X-rays
  2. Beta-particles
  3. Gamma-rays
  4. Alpha-particles

Choose the correct answer from the options given below:

  1. (A), (B), (C), (D)
  2. (A), (B), (D), (C)
  3. (B), (A), (D), (C)
  4. (C), (B), (A), (D)
Correct Answer: 2. (A), (B), (D), (C)
View Solution

Ionizing Power: Ionizing power refers to the ability of radiation to ionize atoms or molecules by removing electrons. It depends on the charge, mass, and energy of the radiation.

Explanation of Options:

  • X-rays: These are high-energy photons with zero charge and low ionizing power, but they penetrate deeply into matter.
  • Beta-particles: These are fast-moving electrons or positrons with moderate ionizing power and intermediate penetration ability.
  • Gamma-rays: High-energy photons with more ionizing power than X-rays, but less than alpha particles.
  • Alpha-particles: Consist of two protons and two neutrons, making them highly ionizing but with very low penetration depth.

Order of Ionizing Power: X-rays < Beta-particles < Gamma-rays < Alpha-particles.

Question 15:

No two particles can occupy the same state is the consequence of:

  1. Maxwell-Boltzmann distribution
  2. Bose-Einstein Distribution
  3. Fermi-Dirac Distribution
  4. Classical Distribution
Correct Answer: (3) Fermi-Dirac Distribution.
View Solution

Pauli Exclusion Principle: This principle states that no two fermions (particles with half-integer spin) can occupy the same quantum state simultaneously.

Fermi-Dirac Statistics: This statistical model describes the distribution of fermions in a system and incorporates the Pauli Exclusion Principle. It is used to understand the behavior of electrons in atoms and solids.

Comparison with Other Models:

  • Maxwell-Boltzmann and Classical distributions apply to distinguishable particles without quantum restrictions.
  • Bose-Einstein Distribution applies to bosons (particles with integer spin) and allows multiple particles in the same state.

Fermi-Dirac distribution is unique in restricting identical particles (fermions) from sharing a quantum state.

Question 16:

Which of the following metals possesses the highest melting point?

  1. Sodium (Na)
  2. Lithium (Li)
  3. Potassium (K)
  4. Cesium (Cs)
Correct Answer: (2) Lithium (Li).
View Solution

Melting Point Trends in Alkali Metals: In Group 1 of the periodic table, melting points decrease as atomic size increases. This occurs because:

  • Larger atomic sizes lead to weaker metallic bonding.
  • Lithium, being the smallest alkali metal, has the strongest metallic bonds and the highest melting point.

Order of Melting Points: Li > Na > K > Cs.

Question 17:

The factors that influence the ionization energy are:

  1. Size of the atom
  2. Charge of the atom
  3. Size and charge of the atom
  4. Size, charge, and type of electron involved (s, p, d, or f)
Correct Answer: (4) Size, charge, and type of electron involved (s, p, d, or f).
View Solution

Ionization Energy: The energy required to remove an electron from a gaseous atom or ion.

Factors Influencing Ionization Energy:

  • Atomic Size: Smaller atoms have higher ionization energy due to stronger attraction between the nucleus and valence electrons.
  • Nuclear Charge: Higher nuclear charge increases ionization energy.
  • Type of Electron: Electrons in different orbitals (s, p, d, f) experience different shielding and attraction to the nucleus. For example, s-electrons are harder to remove than p-electrons.

Question 18:

The number of gram molecules of a substance present in unit volume is termed as:

  1. Activity
  2. Normal solution
  3. Molar concentration
  4. Active mass
Correct Answer: (3) Molar concentration.
View Solution

Molar Concentration (Molarity): The number of moles (gram molecules) of a solute present in 1 liter of solution, expressed in moles per liter (mol/L).

Other Terms:

  • Activity: Describes the effective concentration of a species in a solution.
  • Normal Solution: Concentration expressed in equivalents per liter.
  • Active Mass: An older term for molar concentration, mostly used in chemical kinetics.

Question 19:

Among the following molecules, the one with the highest dipole moment is:

  1. CH3Cl
  2. CH2Cl2
  3. CHCl3
  4. CCl4
Correct Answer: (1) CH3Cl.
View Solution

Dipole Moment: A measure of the separation of positive and negative charges in a molecule. It depends on bond polarity and molecular geometry.

Analysis of Molecules:

  • CH3Cl: The carbon-chlorine bond is polar, and the tetrahedral geometry results in a significant net dipole moment pointing towards chlorine.
  • CH2Cl2: Contains two C-Cl bonds; the molecular geometry partially cancels some of the dipole, leading to a lower dipole moment than CH3Cl.
  • CHCl3: Contains three C-Cl bonds, but the dipole moments partially cancel due to the 3D tetrahedral geometry, reducing the net dipole moment further.
  • CCl4: Perfectly symmetrical, with all bond dipoles canceling out, resulting in a net dipole moment of zero.

Conclusion: CH3Cl has the highest dipole moment because it has only one polar C-Cl bond, and its geometry does not allow significant cancellation of the dipole.

Question 20:

Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para compounds. The reaction is:

  1. Electrophilic addition reaction
  2. Electrophilic substitution reaction
  3. Free radical addition reaction
  4. Nucleophilic substitution reaction
Correct Answer: (2) Electrophilic substitution reaction.
View Solution

Electrophilic Substitution: This reaction involves the replacement of a hydrogen atom in an aromatic ring with an electrophile.

Mechanism of Reaction:

  • Toluene, an aromatic compound with a methyl group, undergoes halogenation in the presence of FeCl3.
  • FeCl3 acts as a Lewis acid catalyst, generating a halogen electrophile (X+) from the halogen molecule.
  • The electrophile attacks the benzene ring, replacing a hydrogen atom.

Ortho-Para Directing Effect: The methyl group (-CH3) is an electron-donating group, which increases the electron density at the ortho and para positions of the benzene ring. As a result, the halogen substitutes preferentially at these positions.

Conclusion: The reaction is classified as an electrophilic substitution reaction because it involves the substitution of hydrogen by an electrophile in an aromatic system.

Question 21:

Scanning tunneling microscope works on the principle of:

  1. Quantum mechanical tunneling
  2. Photoelectric effect
  3. Stefan’s Law
  4. Meissner effect
Correct Answer: (1) Quantum mechanical tunneling.
View Solution

The scanning tunneling microscope (STM) operates using the principle of quantum mechanical tunneling. When a sharp conducting tip is brought very close to a conducting or semiconducting surface, electrons can tunnel through the vacuum gap between the tip and the sample. This tunneling current depends on the distance between the tip and the sample and can be measured to produce atomic-resolution images of the surface.

STM allows researchers to study surface properties at the atomic level, including individual atoms and molecules.

By scanning the tip across the surface and maintaining a constant tunneling current, the STM maps the topography of the surface.

Question 22:

Strongest acid among the following is:

  1. Option 1
  2. Option 2
  3. Option 3
  4. Option 4
Correct Answer: Option 2.
View Solution

The structure shown as Option 2 contains a nitro group (-NO2), which is a strong electron-withdrawing group. This group stabilizes the conjugate base of the acid through resonance and inductive effects, significantly increasing the acid’s strength.

1. Acidity is enhanced by electron-withdrawing groups as they lower the energy of the conjugate base.

2. Resonance stabilization further contributes to the acidity by delocalizing the negative charge of the conjugate base.

Question 23:

The number of isomers for the compound with molecular formula C2BrClFI is:

  1. 3
  2. 4
  3. 5
  4. 6
Correct Answer: (4) 6.
View Solution

Isomers are compounds with the same molecular formula but different structures. For C2BrClFI, the possible isomers include structural and stereoisomers:

  • Structural isomers: Differ in the connectivity of atoms.
  • Stereoisomers: Differ in the spatial arrangement of atoms.

The molecule C2BrClFI can have 6 possible arrangements based on the different ways the substituents can be positioned around the carbon atoms.

Question 24:

Which of the following statements is incorrect about sucrose?

  1. Ordinary table sugar is called sucrose
  2. Sucrose is a disaccharide
  3. Sucrose has a molecular formula of C12H22O11
  4. Sucrose gives positive test with Benedict’s and Tollens’ solution
Correct Answer: (4) Sucrose gives positive test with Benedict’s and Tollens’ solution.
View Solution

Sucrose is a non-reducing sugar because it does not have a free aldehyde or ketone group. In sucrose, the glucose and fructose units are linked via their anomeric carbons, preventing the molecule from participating in redox reactions with Benedict’s or Tollens’ reagent:

  • Reducing sugars: React with Benedict’s or Tollens’ solutions to form a precipitate due to the reduction of Cu2+ or Ag+ ions.
  • Sucrose: Lacks this reactivity because its glycosidic bond prevents the opening of the sugar ring to form an aldehyde or ketone.

Question 25:

Which of the following are called artificial atoms?

  1. Electrons
  2. Molecules
  3. Quantum dots
  4. H-atom
Correct Answer: (3) Quantum dots.
View Solution

Quantum dots are nanoscale semiconductor particles that exhibit discrete energy levels due to quantum confinement. This property makes their electronic behavior similar to that of isolated atoms, earning them the nickname "artificial atoms."

1. The size of quantum dots controls their electronic and optical properties.

2. Smaller quantum dots have higher energy gaps, resulting in unique fluorescence colors.

Question 26:

Which scientist is often referred to as the Father of Nanotechnology?

  1. Albert Einstein
  2. Max Planck
  3. Niels Bohr
  4. Richard Feynman
Correct Answer: (4) Richard Feynman.
View Solution

Richard Feynman is considered the "Father of Nanotechnology" due to his groundbreaking ideas presented in his famous 1959 lecture, "There’s Plenty of Room at the Bottom." In this lecture, he envisioned the manipulation of individual atoms and molecules to build devices at an atomic scale, laying the foundation for nanotechnology.

  • He suggested writing the entire Encyclopedia Britannica on the head of a pin by controlling matter at the nanoscale.
  • He introduced the concept of nanoscale machines, which are now realized as nanorobots and molecular assemblers.

Question 27:

X-rays are used for the determination of crystal structure because:

  1. They can travel in vacuum
  2. They are mechanical waves
  3. Their wavelength is of the same order as interplanar spacing in crystals
  4. Their speed changes with the change of medium
Correct Answer: (3) Their wavelength is of the same order as interplanar spacing in crystals.
View Solution

X-rays are ideal for determining crystal structures because their wavelengths (approximately 0.1–10 nm) are comparable to the distances between atomic planes in a crystal lattice. This allows X-rays to diffract when they interact with the crystal, following Bragg’s law:

nλ = 2d sin θ

  • n: Order of diffraction
  • λ: Wavelength of X-rays
  • d: Interplanar spacing
  • θ: Angle of incidence

The diffraction pattern can be analyzed to deduce the arrangement of atoms in the crystal.

Question 28:

Carbon nanotubes are nanostructures:

  1. Zero-dimensional
  2. One-dimensional
  3. Two-dimensional
  4. Three-dimensional
Correct Answer: (2) One-dimensional.
View Solution

Carbon nanotubes (CNTs) are classified as one-dimensional (1D) nanostructures because their length is much greater than their diameter, typically in the range of a few nanometers.

  • CNTs can be single-walled (SWCNT) or multi-walled (MWCNT), depending on the number of graphene layers rolled into tubes.
  • Their unique 1D structure imparts exceptional electrical, mechanical, and thermal properties, such as high conductivity and tensile strength.

Question 29:

Tendency of an atom to attract electrons to itself when combined in a group is called:

  1. Workfunction
  2. Electronegativity
  3. Polarization
  4. Catenation
Correct Answer: (2) Electronegativity.
View Solution

Electronegativity refers to the ability of an atom to attract shared electrons in a chemical bond. This property is influenced by:

  • Atomic size: Smaller atoms have higher electronegativity due to stronger attraction between the nucleus and valence electrons.
  • Nuclear charge: Atoms with more protons in the nucleus have higher electronegativity.

For example, fluorine is the most electronegative element, with an electronegativity of 3.98 on the Pauling scale.

Question 30:

Which of the following diode is used in voltage-regulation circuits?

  1. Photodiode
  2. Light Emitting Diode (LED)
  3. Tunnel diode
  4. Zener diode
Correct Answer: (4) Zener diode.
View Solution

Zener diodes are designed to operate in the reverse breakdown region. They maintain a constant output voltage across a load by allowing current to flow in reverse when the applied voltage reaches the Zener breakdown voltage. This makes them ideal for voltage regulation in power supply circuits.

  • Zener diodes protect sensitive electronic components by limiting voltage surges.
  • They stabilize voltage, crucial in DC power supplies and voltage reference circuits.

Question 31:

In a bipolar junction transistor:

  1. Collector Current IC = Emitter Current IE + Base Current IB
  2. Emitter Current IE = Collector Current IC + Base Current IB
  3. Base Current IB = Emitter Current IE + Collector Current IC
  4. Base, emitter, and collector currents are equal.
Correct Answer: (2) Emitter Current IE = Collector Current IC + Base Current IB.
View Solution

In a bipolar junction transistor (BJT), the emitter current (IE) is the sum of the collector current (IC) and the base current (IB), derived from Kirchhoff’s Current Law (KCL):

IE = IC + IB

  • The base current is typically very small compared to the emitter and collector currents.

Question 32:

Photons are classified as:

  1. Fermions
  2. Bosons
  3. Classical Particles
  4. Nucleons
Correct Answer: (2) Bosons.
View Solution

Photons are classified as bosons because they have an integer spin (s = 1) and follow Bose-Einstein statistics. Unlike fermions, bosons do not obey the Pauli exclusion principle, meaning multiple photons can occupy the same quantum state.

  • Photons are the force carriers of the electromagnetic interaction.
  • Other bosons include W and Z bosons, gluons, and the Higgs boson.

Question 33:

Which of the following statement is incorrect about Silicon?

  1. Its atomic number is 14
  2. It has semiconducting nature
  3. It is a direct band gap semiconductor
  4. Silicon is the backbone of semiconductor industry
Correct Answer: (3) It is a direct band gap semiconductor.
View Solution

Silicon is an indirect band gap semiconductor, meaning the conduction band minimum and valence band maximum occur at different momentum values. This makes radiative recombination less efficient compared to direct band gap semiconductors like gallium arsenide (GaAs).

  • Silicon has an atomic number of 14 and forms the basis of the modern semiconductor industry.
  • Its properties as a semiconductor make it ideal for use in electronic devices like transistors and integrated circuits.

Question 34:

Meissner effect is a phenomenon of:

  1. Perfect diamagnetism
  2. Perfect paramagnetism
  3. Perfect ferromagnetism
  4. Perfect antiferromagnetism
Correct Answer: (1) Perfect diamagnetism.
View Solution

The Meissner effect is the expulsion of magnetic fields from a superconductor when it is cooled below its critical temperature. This results in perfect diamagnetism, where the magnetic susceptibility (χ) is −1.

  • It demonstrates the transition of a material to the superconducting state.
  • This effect is a defining property of superconductors, alongside zero electrical resistance.

Question 35:

Which of the following process is carried out at constant pressure?

  1. Isothermal Process
  2. Isobaric Process
  3. Adiabatic Process
  4. Isochoric Process
Correct Answer: (2) Isobaric Process.
View Solution

An isobaric process occurs when the pressure of the system remains constant while the volume and temperature may change. Examples include heating or cooling a gas in a piston-cylinder setup at constant external pressure.

  • In an isothermal process, temperature remains constant.
  • In an adiabatic process, no heat exchange occurs.
  • In an isochoric process, volume remains constant.

Question 36:

Which among the following would be a linear species?

  1. ICl2-
  2. IF-
  3. I3-
  4. H2O
Correct Answer: (3) I3-.
View Solution

The I3- ion has a linear geometry because of the arrangement of electron pairs around the central iodine atom. According to the Valence Shell Electron Pair Repulsion (VSEPR) theory:

  • I3- has three iodine atoms, with the central iodine having three lone pairs of electrons.
  • The linear shape arises as the lone pairs arrange themselves to minimize repulsion, leaving the bond angle between the two bonded iodine atoms at 180°.

Question 37:

In gram staining, cell walls of Gram-positive bacteria become dehydrated during treatment with alcohol due to:

  1. Lower Lipid Content
  2. Higher Lipid Content
  3. Lower Protein Content
  4. Higher Protein Content
Correct Answer: (1) Lower Lipid Content.
View Solution

Gram-positive bacteria have a thick peptidoglycan layer in their cell walls, which contains very low lipid content. When treated with alcohol:

  • The alcohol dehydrates the peptidoglycan layer, causing it to shrink and trap the crystal violet-iodine complex.
  • This makes Gram-positive bacteria retain the purple stain.
  • Gram-negative bacteria, having a higher lipid content, lose the stain as the alcohol dissolves their outer membrane.

Question 38:

Which is the polymer of Fructose?

  1. Starch
  2. Glycogen
  3. Inulin
  4. Cellulose
Correct Answer: (3) Inulin.
View Solution

Inulin is a polysaccharide made up of fructose units linked by β-(2→1) glycosidic bonds. It serves as a storage carbohydrate in many plants, including chicory, Jerusalem artichokes, and onions.

  • Unlike starch (polymer of glucose), inulin cannot be digested by human enzymes but serves as dietary fiber.
  • It is also used as a prebiotic, promoting the growth of beneficial gut bacteria.

Question 39:

Which of the following is a thiol-containing amino acid?

  1. Cysteine
  2. Glycine
  3. Serine
  4. Tryptophan
Correct Answer: (1) Cysteine.
View Solution

Cysteine is a sulfur-containing amino acid with a thiol (-SH) group in its side chain. This group:

  • Enables cysteine to form disulfide bonds (-S-S-) when oxidized, which are crucial for protein structure stabilization.
  • Plays a role in enzyme active sites due to its ability to form reactive thiol intermediates.

Question 40:

The antigen specificity of a particular B cell:

  1. Is induced by interaction with antigen
  2. Is due to light chain of antibody
  3. Is due to heavy chain of antibody
  4. Is determined by heavy and light chain variable region sequence
Correct Answer: (4) Is determined by heavy and light chain variable region sequence.
View Solution

The antigen-binding specificity of a B cell is determined by the unique sequence in the variable regions of the heavy and light chains of its antibodies. These regions:

  • Form the antigen-binding site and are highly diverse due to somatic recombination.
  • Allow B cells to recognize a vast array of antigens.
  • Are critical for adaptive immunity, enabling specific responses to pathogens.

Question 41:

Which of the following properties is not exhibited by TH cells?

  1. Stimulate division of B cells
  2. Are cytotoxic for other cells
  3. Stimulate division of cytotoxic T cells
  4. Stimulate migration of macrophages
Correct Answer: (2) Are cytotoxic for other cells.
View Solution

Helper T cells (TH cells) regulate immune responses through cytokines. They:

  • Stimulate B cell division and antibody production.
  • Activate cytotoxic T cells indirectly.
  • Stimulate macrophage activity.

However, TH cells are not cytotoxic; this function is exclusive to cytotoxic T cells (CD8+ cells).

Question 42:

In germ line gene therapy, healthy gene is inserted into:

  1. Hepatic cells
  2. Fibroblast cells
  3. Endothelium
  4. Sperms or Eggs
Correct Answer: (4) Sperms or Eggs.
View Solution

Germ line gene therapy involves introducing a functional gene into germ cells (sperms or eggs) or embryos. This ensures that the genetic modification is inheritable and passed on to future generations, potentially eliminating genetic disorders permanently.

Question 43:

Restriction Endonuclease type I is involved in:

  1. Endonuclease and Methylase
  2. Cleaving and Methylation
  3. Cut DNA within palindromic sequences
  4. Cut DNA within non-palindromic sequences
Correct Answer: (1) Endonuclease and Methylase.
View Solution

Type I restriction enzymes have both endonuclease and methyltransferase activities. They:

  • Recognize specific DNA sequences.
  • Cleaves DNA at random locations far from the recognition site.
  • Methylate specific bases to protect host DNA from cleavage.

Question 44:

The Ti plasmid of Agrobacterium tumefaciens contains a transposon called:

  1. T DNA
  2. R DNA
  3. cDNA
  4. A DNA
Correct Answer: (1) T DNA.
View Solution

The Ti (tumor-inducing) plasmid in Agrobacterium tumefaciens contains a segment of DNA called T DNA, which is transferred to the plant genome during infection. This causes tumor formation in plants and is widely used in genetic engineering to introduce foreign genes into plants.

Question 45:

Nitrogenase enzyme involved in nitrogen fixation works best in:

  1. Aerobic conditions
  2. Anaerobic conditions
  3. Insensitive to oxygen
  4. Both aerobic and anaerobic conditions
Correct Answer: (2) Anaerobic conditions.
View Solution

Nitrogenase is highly sensitive to oxygen, as it is irreversibly inactivated by it. Anaerobic conditions or protective mechanisms like leghemoglobin in nitrogen-fixing nodules create an oxygen-free environment, enabling the enzyme to function effectively.

Question 46:

Arrange the following electromagnetic radiations in the order of decreasing wavelength:

A. Microwaves

B. Gamma rays

C. X-rays

D. Visible rays

  1. (A), (D), (C), (B)
  2. (B), (C), (D), (A)
  3. (B), (A), (D), (C)
  4. (C), (B), (D), (A)
Correct Answer: (1) (A), (D), (C), (B).
View Solution

The order of electromagnetic radiations in terms of decreasing wavelength is:

  • Microwaves: Longest wavelength, low energy.
  • Visible rays: Intermediate wavelength, detectable by the human eye.
  • X-rays: Shorter wavelength, higher energy.
  • Gamma rays: Shortest wavelength, highest energy.

Question 47:

Which of the following is correct about electromagnetic waves?

A. Sound waves are electromagnetic in nature
B. They travel in vacuum
C. Electromagnetic waves are longitudinal (formation of compression and rarefaction) in nature
D. X-rays are electromagnetic in nature

  1. (A), (D), and (D) only
  2. (A), (B), and (C) only
  3. (B) only
  4. (B) and (D) only
Correct Answer: (4) (B) and (D) only.
View Solution

Electromagnetic waves:

  • Travel through vacuum, unlike sound waves, which require a medium.
  • Are transverse waves, characterized by oscillating electric and magnetic fields perpendicular to the direction of wave propagation.
  • Include X-rays, which are high-energy electromagnetic radiation.

Therefore, only statements (B) and (D) are correct. Statement (A) is incorrect because sound waves are mechanical, not electromagnetic, and statement (C) is incorrect because electromagnetic waves are transverse, not longitudinal.

Question 48:

Which of the following statement is correct?

A. The stronger the acid, the weaker will be its conjugate base
B. The stronger the acid, the stronger will be its conjugate base
C. The larger the pKa of the conjugate acid, the weaker is its base
D. The larger the pKa of the conjugate acid, the stronger is its base

  1. D only
  2. B and C only
  3. A and D only
  4. B only
Correct Answer: (3) A and D only.
View Solution

1. A strong acid dissociates completely, producing a weak conjugate base with little tendency to accept protons.

2. The pKa value indicates the strength of an acid; a larger pKa corresponds to a weaker acid, making its conjugate base stronger.

Thus, statements (A) and (D) are correct.

Question 49:

All the metal halides have negative enthalpies of formation. Thermodynamically feasible from them. Arrange the following halides from most to least negative enthalpies of formation for any given metal:

A. Iodide
B. Chloride
C. Bromide
D. Fluoride

  1. (A), (C), (B), (D)
  2. (A), (B), (C), (D)
  3. (D), (B), (C), (A)
  4. (D), (C), (B), (A)
Correct Answer: (3) (D), (B), (C), (A).
View Solution

The enthalpy of formation of metal halides depends on several factors, including lattice energy and the size of the halide ion. Lattice energy is the energy released when oppositely charged ions combine to form a lattice:

  • Fluoride (D): Fluoride ions are the smallest halide ions, leading to a very high lattice energy, resulting in the most negative enthalpy of formation.
  • Chloride (B): Larger than fluoride ions but still relatively high lattice energy, leading to the next most negative enthalpy.
  • Bromide (C): Larger than chloride, so the lattice energy decreases further, leading to a less negative enthalpy.
  • Iodide (A): Largest halide ions, resulting in the lowest lattice energy and least negative enthalpy.

Thus, the order is Fluoride > Chloride > Bromide > Iodide.

Question 50:

Arrange the following oxoacids in the order of their increasing acidic strength:

A. HOCl
B. HOClO
C. HOClO2
D. HOClO3

  1. (D), (C), (A), (B)
  2. (B), (A), (C), (D)
  3. (A), (B), (C), (D)
  4. (C), (A), (B), (D)
Correct Answer: (3) (A), (B), (C), (D).
View Solution

The acidic strength of oxoacids increases with the number of oxygen atoms attached to the central atom, as these increase the stability of the conjugate base through resonance:

HOCl < HOClO < HOClO2 < HOClO3

Question 51:

Arrange the following in the order of increasing basic strength:

A. NH3
B. PH3
C. AsH3
D. BiH3

  1. (A), (C), (B), (D)
  2. (A), (B), (C), (D)
  3. (B), (D), (C), (A)
  4. (D), (C), (B), (A)
Correct Answer: (4) (D), (C), (B), (A).
View Solution

The basic strength of group 15 hydrides decreases down the group as the atomic size increases, and the lone pair of electrons becomes less available for donation:

Order: BiH3 < AsH3 < PH3 < NH3.

Ammonia (NH3) is the strongest base due to its small size, making the lone pair highly accessible for bonding. As we move down the group, the atoms (P, As, Bi) get larger, and the lone pair becomes more diffuse, reducing basicity.

Question 52:

Arrange the following in decreasing order of nucleophilicity:

A. H2O
B. NH3
C. OH-
D. NH2-

  1. (A), (B), (C), (D)
  2. (D), (C), (B), (A)
  3. (C), (D), (B), (A)
  4. (D), (B), (C), (A)
Correct Answer: (2) (D), (C), (B), (A).
View Solution

Nucleophilicity refers to the ability of a species to donate a lone pair to form a bond. Factors affecting nucleophilicity include charge, electronegativity, and the solvent:

  • Anionic species like NH2- and OH- are stronger nucleophiles than neutral species like NH3 and H2O because the negative charge increases their electron density.
  • NH2- is the strongest nucleophile as nitrogen is less electronegative than oxygen, making its lone pair more available for bonding.

Order: NH2- > OH- > NH3 > H2O.

Question 53:

The correct order of steps in PCR is:

A. Annealing
B. Denaturation
C. Extension

  1. (A), (B), (C)
  2. (B), (A), (C)
  3. (B), (C), (A)
  4. (C), (B), (A)
Correct Answer: (2) (B), (A), (C).
View Solution

Polymerase Chain Reaction (PCR) involves three main steps:

  1. Denaturation: Heating (~95°C) to separate the double-stranded DNA into single strands.
  2. Annealing: Cooling (50-65°C) to allow primers to bind to complementary sequences on the single-stranded DNA.
  3. Extension: Raising the temperature (~72°C) to enable Taq DNA polymerase to synthesize new DNA strands by adding nucleotides to the primers.

These steps are repeated in cycles to exponentially amplify the DNA sequence of interest.

Question 54:

Which of the following enzymes of Krebs’ cycle is also involved in the mitochondrial electron transport system?

  1. Citrate synthase
  2. Succinate dehydrogenase
  3. Malate dehydrogenase
  4. Isocitrate dehydrogenase
Correct Answer: (2) Succinate dehydrogenase.
View Solution

Succinate dehydrogenase is unique as it functions in both the Krebs’ cycle and the mitochondrial electron transport chain:

  • In the Krebs’ cycle, it catalyzes the oxidation of succinate to fumarate.
  • In the electron transport chain, it transfers electrons from FADH2 to ubiquinone (Coenzyme Q), contributing to ATP production.

Question 55:

Sequential steps of Prophase of Meiosis I:

A. Zygotene
B. Pachytene
C. Leptotene
D. Diakinesis
E. Diplotene

  1. (A), (B), (C), (E), (D)
  2. (C), (A), (B), (E), (D)
  3. (B), (E), (D), (C), (A)
  4. (C), (B), (D), (E), (A)
Correct Answer: (2) (C), (A), (B), (E), (D).
View Solution

Prophase I of meiosis is divided into five stages:

  1. Leptotene: Chromosomes condense and become visible under a microscope.
  2. Zygotene: Homologous chromosomes pair up and form synaptonemal complexes (synapsis).
  3. Pachytene: Crossing-over occurs between homologous chromosomes, exchanging genetic material.
  4. Diplotene: Synaptonemal complexes disassemble, and chiasmata (cross-over points) become visible.
  5. Diakinesis: Chromosomes condense further, and chiasmata move to terminal positions, preparing for separation.

Question 56:

The conditions for a reversible isothermal process are:

A. Infinitesimally slow process
B. Infinitely fast process
C. Constant Pressure
D. Constant Temperature

  1. (A) and (D) only
  2. (A), (C), and (D) only
  3. (B), (C), and (D) only
  4. (D) only
Correct Answer: (1) (A) and (D) only.
View Solution

A reversible isothermal process occurs when:

  • Temperature remains constant throughout the process (T = constant).
  • The process is carried out infinitesimally slowly to maintain equilibrium with the surroundings, allowing maximum energy exchange without dissipation.

Question 57:

Crystalline materials possess:

A. Sharp melting point and boiling point
B. Anisotropic properties
C. Periodic arrangement of atoms
D. Isotropic properties

  1. (A), (C), and (D) only
  2. (A) and (C) only
  3. (A), (B), and (D) only
  4. (C) and (D) only
Correct Answer: (2) (A) and (C) only.
View Solution

Crystalline materials have:

  • Sharp melting and boiling points: Due to uniform bonding throughout the material.
  • Periodic arrangement of atoms: Forming a lattice structure.

They are anisotropic
(properties vary with direction), whereas isotropic properties are characteristic of amorphous materials.

Question 58:

Which of the following are the correct postulates of the quantum theory of light?

A. Each photon of frequency ν has the energy hν, the same as Planck’s energy.
B. The higher ν, the greater the photon energy hν.
C. Blue color has lower energy than green color.
D. Changing the intensity of the incident light beam will change the frequency ν of the photon.

  1. (A), (B), and (C) only
  2. (A), (B), and (D) only
  3. (A) and (B) only
  4. (A) and (D) only
Correct Answer: (3) (A) and (B) only.
View Solution

The quantum theory of light states that:

  • Each photon carries energy E = hν, where h is Planck’s constant and ν is the frequency.
  • Photon energy increases with frequency (higher ν → higher energy).

Statement C is incorrect because blue light has higher energy than green light. Statement D is also incorrect because changing the intensity affects the number of photons, not their frequency.

Question 59:

The following species have an equal number of electrons. Arrange them in increasing order of their size:

A. Ne
B. Na+
C. O2-
D. F-

  1. (A), (B), (C), (D)
  2. (B), (C), (A), (D)
  3. (B), (A), (D), (C)
  4. (D), (C), (A), (B)
Correct Answer: (3) (B), (A), (D), (C).
View Solution

The size of isoelectronic species depends on nuclear charge:

  • Na+: Highest nuclear charge, smallest size.
  • Ne: Neutral atom, slightly larger than Na+.
  • F-: Extra electron increases electron repulsion, making it larger than Ne.
  • O2-: Most electron repulsion, largest size.

Order: Na+ < Ne < F- < O2-.

Question 60:

Identify the correct statements with reference to VSEPR theory:

A. Lone pairs cause more repulsion than double bonds.
B. Double bonds cause more repulsion than a single bond.
C. Triple bonds cause more repulsion than a double bond.
D. Single bonds cause more repulsion than a lone pair.

  1. (A), (B), and (C) only
  2. (B) and (C) only
  3. (A) and (B) only
  4. (B), (C), and (D) only
Correct Answer: (2) (B) and (C) only.
View Solution

According to VSEPR theory:

  • Lone pairs exert the strongest repulsion, followed by triple bonds, double bonds, and single bonds.
  • Double bonds cause more repulsion than single bonds, and triple bonds cause more repulsion than double bonds.

Thus, statements B and C are correct, while A and D are incorrect.

Question 61:

Which of the following arrangements does not represent the correct order of properties stated against it?

A. V2+ < Cr2+ < Mn2+ < Fe2+: Paramagnetic behavior
B. Zn2+ < Ni2+ < Cr2+ < Fe2+: Number of unpaired electrons
C. Sc3+ < Ti3+ < V3+ < Cr3+ < Mn3+: Number of oxidation states
D. Co2+ < Cr2+ < Mn2+ < Ni2+: Stability in aqueous solution

  1. (A), (B), and (C) only
  2. (A) and (C) only
  3. (A), (B), and (D)
  4. (B), (C), and (D) only
Correct Answer: (3) (A), (B), and (D).
View Solution

The paramagnetic behavior in option (A) is incorrect because Fe2+ has fewer unpaired electrons than Mn2+. In option (B), Cr2+ has more unpaired electrons than Ni2+, contradicting the sequence. The stability order in aqueous solution for option (D) does not align with experimental observations.

Question 62:

The process of vulcanization makes rubber:

A. Soluble in water
B. Elastic
C. Hard
D. Soft

  1. (A), (B), and (D) only
  2. (B) and (C) only
  3. (B) and (D)
  4. (B) only
Correct Answer: (2) (B) and (C) only.
View Solution

Vulcanization introduces sulfur cross-links between the polymer chains of rubber, enhancing its elasticity and hardness. This makes it suitable for industrial applications such as tires and seals.

Question 63:

Which of the following will give benzoic acid on acidic hydrolysis?

A. Phenyl cyanide
B. Benzimidazole
C. Benzoic anhydride
D. Methyl benzoate

  1. (A) and (D) only
  2. (A), (C), and (D) only
  3. (A), (B), (C), and (D)
  4. (C) and (D) only
Correct Answer: (2) (A), (C), and (D) only.
View Solution

Phenyl cyanide undergoes hydrolysis in acidic conditions to produce benzoic acid. Similarly, benzoic anhydride and methyl benzoate yield benzoic acid upon hydrolysis. Benzimidazole does not produce benzoic acid.

Question 64:

Match List-I with List-II:

List-I List-II
(A) CCl4 (ii) Insecticide
(B) C6F6Cl6 (i) Refrigerant
(C) CHCl3 (iv) Antiseptic for wounds
(D) CH2Cl2 (iii) Propellant for aerosols
  1. (A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)
  2. (A) - (ii), (B) - (i), (C) - (iv), (D) - (iii)
  3. (A) - (i), (B) - (ii), (C) - (iv), (D) - (iii)
  4. (A) - (iii), (B) - (iv), (C) - (i), (D) - (ii)
Correct Answer: (2) (A) - (ii), (B) - (i), (C) - (iv), (D) - (iii).
View Solution

Each compound in List-I is matched with its respective application in List-II based on its chemical properties:

  • CCl4: Used as an insecticide.
  • C6F6Cl6: Used as a refrigerant.
  • CHCl3: Used as an antiseptic for wounds.
  • CH2Cl2: Used as a propellant for aerosols.

Question 65:

Two metals are completely miscible with each other to form a continuous range of solid solutions only if:

A. Two metals must be similar in size—their metallic radii must not differ by more than 15%.
B. Both metals must have the same crystal structure.
C. The chemical properties must be similar—particularly the same electronegativity.
D. Both metals must have the same electrical conductivity.

  1. (A) and (D) only
  2. (A), (B), and (C) only
  3. (A), (B), (C), and (D)
  4. (B) and (C) only
Correct Answer: (2) (A), (B), and (C) only.
View Solution

According to Hume-Rothery rules, metals must have:

  • Similar atomic radii (within 15%).
  • The same crystal structure.
  • Comparable electronegativity.

These conditions ensure the formation of a continuous solid solution.

Question 66:

Arrange the following structures in terms of increasing order of confinement:

A. Quantum Well
B. Bulk Material
C. Quantum Dot
D. Carbon Nanotube

  1. (A), (B), (C), (D)
  2. (B), (A), (D), (C)
  3. (C), (B), (A), (D)
  4. (D), (B), (C), (A)
Correct Answer: (2) (B), (A), (D), (C).
View Solution

Confinement refers to the restriction of particle movement:

  • Bulk Material: No confinement, free movement in all dimensions.
  • Quantum Well: Restricts motion in one dimension.
  • Carbon Nanotube: Restricts motion in two dimensions.
  • Quantum Dot: Restricts motion in all three dimensions.

Question 67:

Match List-I with List-II:

List-I (System) List-II (Axial lengths and angles)
(A) Cubic (I) a = b = c, α = β = γ = 90°
(B) Tetragonal (III) a ≠ b = c, α = β = γ = 90°
(C) Orthorhombic (II) a ≠ b ≠ c, α = β = γ = 90°
(D) Hexagonal (IV) a = b ≠ c, α = β = 90°, γ = 120°
  1. (A) - (IV), (B) - (II), (C) - (III), (D) - (I)
  2. (A) - (I), (B) - (III), (C) - (II), (D) - (IV)
  3. (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
  4. (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
Correct Answer: (2) (A) - (I), (B) - (III), (C) - (II), (D) - (IV).
View Solution

The classification of crystal systems is based on axial lengths and interaxial angles:

  • (A) Cubic: a = b = c, α = β = γ = 90° (I).
  • (B) Tetragonal: a ≠ b = c, α = β = γ = 90° (III).
  • (C) Orthorhombic: a ≠ b ≠ c, α = β = γ = 90° (II).
  • (D) Hexagonal: a = b ≠ c, α = β = 90°, γ = 120° (IV).

Question 68:

Match List-I with List-II:

List-I (Type of Crystalline Solid) List-II (Example)
(A) Ionic (III) Sodium Chloride, NaCl
(B) Covalent (IV) Diamond, C
(C) Molecular (I) Methane, CH4
(D) Metallic (II) Copper, Cu
  1. (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
  2. (A) - (III), (B) - (I), (C) - (II), (D) - (IV)
  3. (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
  4. (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Correct Answer: (4) (A) - (III), (B) - (IV), (C) - (I), (D) - (II).
View Solution

The classification of crystalline solids is based on bonding and structure:

  • (A) Ionic: NaCl, held by electrostatic forces.
  • (B) Covalent: Diamond (C), network of covalent bonds.
  • (C) Molecular: CH4, weak van der Waals forces.
  • (D) Metallic: Cu, sea of delocalized electrons.

Question 69:

Match List-I with List-II:

List-I List-II
(A) Principal Quantum Number (n) (ii) 1, 2, 3,...
(B) Azimuthal Quantum Number (l) (iv) 0, 1, 2,... n - 1
(C) Magnetic Quantum Number (ml) (iii) 0, ±1, ±2, ±3,... ±l
(D) Spin Quantum Number (s) (i) −1/2, +1/2
  1. (A) - (ii), (B) - (iv), (C) - (iii), (D) - (i)
  2. (A) - (ii), (B) - (iii), (C) - (iv), (D) - (i)
  3. (A) - (iv), (B) - (ii), (C) - (iii), (D) - (i)
  4. (A) - (iv), (B) - (ii), (C) - (i), (D) - (iii)
Correct Answer: (1) (A) - (ii), (B) - (iv), (C) - (iii), (D) - (i).
View Solution

Quantum numbers describe the quantum state of an electron:

  • (A) Principal Quantum Number (n): Main shell (1, 2, 3,...).
  • (B) Azimuthal Quantum Number (l): Subshell (0 to n - 1).
  • (C) Magnetic Quantum Number (ml): Orbital orientation (−l to +l).
  • (D) Spin Quantum Number (s): Spin (−1/2, +1/2).

Question 70:

Match List-I with List-II:

List-I List-II
(A) XeF2 (iii) Linear
(B) BF3 (i) Trigonal Planar
(C) XeF4 (ii) Distorted Octahedral
(D) SF6 (iv) Regular Octahedral
  1. (A) - (i), (B) - (iii), (C) - (ii), (D) - (iv)
  2. (A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)
  3. (A) - (iv), (B) - (iii), (C) - (ii), (D) - (i)
  4. (A) - (iii), (B) - (iv), (C) - (i), (D) - (ii)
Correct Answer: (2) (A) - (iii), (B) - (i), (C) - (ii), (D) - (iv).
View Solution

The geometries of the molecules are determined by VSEPR theory:

  • (A) XeF2: Linear geometry.
  • (B) BF3: Trigonal planar geometry.
  • (C) XeF4: Distorted octahedral geometry.
  • (D) SF6: Regular octahedral geometry.

Question 71:

Match List-I with List-II:

List-I List-II
(A) Linus Pauling (iii) Electronegativity
(B) Albert Einstein (ii) Photoelectric effect
(C) Ernest Rutherford (i) Alpha particle scattering experiment
(D) Max Planck (iv) Black body radiations
  1. (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
  2. (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
  3. (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
  4. (A) - (III), (B) - (II), (C) - (IV), (D) - (I)
Correct Answer: (1) (A) - (III), (B) - (II), (C) - (I), (D) - (IV).
View Solution

Key scientific achievements are matched as follows:

  • Linus Pauling: Introduced the concept of electronegativity.
  • Albert Einstein: Explained the photoelectric effect.
  • Ernest Rutherford: Conducted the alpha particle scattering experiment.
  • Max Planck: Developed the quantum theory of black body radiation.

Question 72:

Match List-I with List-II:

List-I List-II
(A) Alkaline phosphatase (iv) Removes phosphate group present at 5’ end of DNA
(B) DNA Polymerase (iii) Nick translation
(C) SI Nuclease (ii) Cleaves ssDNA
(D) DNase I (i) Cleaves DNA
  1. (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
  2. (A) - (I), (B) - (III), (C) - (II), (D) - (IV)
  3. (A) - (I), (B) - (II), (C) - (IV), (D) - (III)
  4. (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Correct Answer: (1) (A) - (IV), (B) - (III), (C) - (II), (D) - (I).
View Solution

Key enzymatic functions:

  • Alkaline phosphatase: Removes phosphate groups from the 5’ end of DNA.
  • DNA Polymerase: Replaces damaged DNA regions during nick translation.
  • SI Nuclease: Cleaves single-stranded DNA.
  • DNase I: Randomly cleaves DNA, aiding in digestion.

Question 73:

Match List-I with List-II:

List-I List-II
(A) IgA (ii) Secretory components
(B) IgE (i) Basophils
(C) IgG (iv) Crosses Placenta
(D) IgM (iii) Pentamer
  1. (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
  2. (A) - (I), (B) - (III), (C) - (II), (D) - (IV)
  3. (A) - (I), (B) - (II), (C) - (IV), (D) - (III)
  4. (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Correct Answer: (1) (A) - (II), (B) - (I), (C) - (IV), (D) - (III).
View Solution

Key roles of immunoglobulins:

  • IgA: Found in mucosal secretions; associated with secretory components.
  • IgE: Activates basophils and mast cells during allergic reactions.
  • IgG: Crosses the placenta, providing immunity to the fetus.
  • IgM: Exists as a pentamer and is the first antibody produced in an immune response.

Question 74:

Match List-I with List-II:

List-I List-II
(A) Thiamine (B1) (iv) Co-carboxylase
(B) Riboflavin (B2) (i) Riboflavin adenine nucleotide
(C) Niacin (iii) Nicotinamide adenine dinucleotide
(D) Folic acid (ii) Tetrahydrofolic acid
  1. (A) - (IV), (B) - (I), (C) - (III), (D) - (II)
  2. (A) - (I), (B) - (III), (C) - (II), (D) - (IV)
  3. (A) - (I), (B) - (II), (C) - (IV), (D) - (III)
  4. (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Correct Answer: (1) (A) - (IV), (B) - (I), (C) - (III), (D) - (II).
View Solution

Key metabolic roles of vitamins:

  • Thiamine (B1): Precursor for co-carboxylase, aiding carbohydrate metabolism.
  • Riboflavin (B2): Forms coenzymes like riboflavin adenine nucleotide.
  • Niacin: Converted into NAD/NADP for redox reactions.
  • Folic acid: Metabolized into tetrahydrofolic acid for nucleotide synthesis.

Question 75:

Match List-I with List-II:

List-I List-II
(A) Lyman (i) n = 2, 3, 4,...
(B) Balmer (ii) n = 3, 4, 5,...
(C) Paschen (iii) n = 4, 5, 6,...
(D) Bracket (iv) n = 5, 6, 7,...
  1. (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
  2. (A) - (I), (B) - (III), (C) - (II), (D) - (IV)
  3. (A) - (I), (B) - (II), (C) - (IV), (D) - (III)
  4. (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Correct Answer: (1) (A) - (I), (B) - (II), (C) - (III), (D) - (IV).
View Solution

The spectral series in hydrogen correspond to electron transitions:

  • Lyman: Transitions ending at n = 1 (ultraviolet).
  • Balmer: Transitions ending at n = 2 (visible).
  • Paschen: Transitions ending at n = 3 (infrared).
  • Bracket: Transitions ending at n = 4 (far-infrared).

CUET PG Previous Year Question Paper

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