Free AI question bank · NEET UG · Physics, Chemistry, Biology
NEET Question Bank, chapter by chapter
Pick an option to see the answer. 5 free answers, then continue in the Lumi app.
780 questions
- BiologyAnatomy of Flowering PlantsDicot and monocot rootsEasy
- BiologyAnatomy of Flowering PlantsEpidermis and stomataEasy
- BiologyAnatomy of Flowering PlantsHairs and ground tissueMediumQ3Match each tissue or feature with the plant organ where it is characteristically found.Column IColumn IIP. Hypodermis of collenchyma1. Monocot stemQ. Hypodermis of sclerenchyma2. Dicot stemR. Bulliform cells3. Grass leafS. Casparian strips4. Root endodermis
- BiologyAnatomy of Flowering PlantsVascular bundlesMediumQ4Vascular bundles differ in their structure. Evaluate the two statements.
- Statement I: Open vascular bundles have a strip of cambium between xylem and phloem, and they are present in dicot stems.
- Statement II: The vascular bundles of a monocot stem are conjoint and closed, and a well-developed phloem parenchyma is present in them.
- BiologyAnatomy of Flowering PlantsDicot and monocot rootsMedium
- BiologyAnatomy of Flowering PlantsDicot and monocot rootsMedium
- BiologyAnatomy of Flowering PlantsMonocot stemMediumQ7Monocot stems typically do not increase in girth by secondary growth. Evaluate the Assertion and Reason.
- Assertion (A): Monocot stems do not show secondary growth.
- Reason (R): The vascular bundles of the monocot stem are closed, with no cambium between the xylem and phloem.
- BiologyAnatomy of Flowering PlantsEpidermis and stomataHard
- BiologyAnatomy of Flowering PlantsIsobilateral leafEasy
- BiologyAnatomy of Flowering PlantsDicot stemHardQ10Which sequence correctly gives the tissues met in a dicot stem, going from the outside inwards up to the vascular bundles?
- AEpidermis, collenchymatous hypodermis, parenchymatous cortical layers, endodermis (starch sheath), pericycle, vascular bundles
- BEpidermis, cortex, hypodermis, pericycle, endodermis, vascular bundles
- CEpidermis, sclerenchymatous hypodermis, cortex, endodermis, pericycle, vascular bundles
- DEpidermis, collenchymatous hypodermis, endodermis, cortex, pericycle, vascular bundles
- BiologyAnimal KingdomLevels of organisationEasy
- BiologyAnimal KingdomCoelom, segmentation and notochordEasy
- BiologyAnimal KingdomFlatworms, roundworms and annelidsMediumQ13Match each phylum with its excretory structure.Column IColumn IIP. Platyhelminthes1. Malpighian tubulesQ. Annelida2. Flame cellsR. Arthropoda3. NephridiaS. Hemichordata4. Proboscis gland
- BiologyAnimal KingdomCoelom, segmentation and notochordMediumQ14Body cavities and gut completeness vary among worms. Evaluate the two statements.
- Statement I: In aschelminthes the body cavity is not lined by mesoderm, so they are pseudocoelomates.
- Statement II: Flatworms have a true coelom and a complete digestive tract with both mouth and anus.
- BiologyAnimal KingdomEchinoderms and hemichordatesMedium
- BiologyAnimal KingdomJawless and jawed fishesMedium
- BiologyAnimal KingdomSponges, cnidarians and comb jelliesMediumQ17Sponges are the simplest multicellular animals. Evaluate the Assertion and Reason.
- Assertion (A): Sponges show the cellular level of organisation.
- Reason (R): In sponges the cells are loosely arranged and show some division of labour, but they are not organised into tissues.
- BiologyAnimal KingdomChordates and non-chordatesHard
- BiologyAnimal KingdomSponges, cnidarians and comb jelliesEasy
- BiologyAnimal KingdomAmphibians and reptilesHard
- BiologyBiological ClassificationThe five kingdom classificationEasy
- BiologyBiological ClassificationArchaebacteria, eubacteria and mycoplasmaEasy
- BiologyBiological ClassificationKingdom ProtistaMediumQ23Match each protistan group with its characteristic feature.Column IColumn IIP. Gonyaulax1. Rapid multiplication causes red tidesQ. Diatoms2. Silica walls that leave diatomaceous earthR. Euglena3. Flexible protein-rich pellicleS. Trypanosoma4. Flagellated parasite causing sleeping sickness
- BiologyBiological ClassificationArchaebacteria, eubacteria and mycoplasmaMediumQ24Blue-green algae are often a source of confusion in classification. Examine the two statements below.
- Statement I: Cyanobacteria are placed in Kingdom Protista because they are algae and photosynthetic.
- Statement II: Nostoc and Anabaena are filamentous forms that fix atmospheric nitrogen in specialised cells called heterocysts.
- BiologyBiological ClassificationKingdom Fungi: body, nutrition and reproductionMedium
- BiologyBiological ClassificationThe four classes of fungiMedium
- BiologyBiological ClassificationThe four classes of fungiHard
- BiologyBiological ClassificationViroids, prions and lichensMediumQ28Viroids are agents of disease in plants. Evaluate the following Assertion and Reason.
- Assertion (A): Viroids, the agents of diseases such as potato spindle tuber disease, are smaller than viruses.
- Reason (R): A viroid is a free RNA of low molecular weight that lacks the protein coat found in viruses.
- BiologyBiological ClassificationViroids, prions and lichensEasy
- BiologyBiological ClassificationKingdom ProtistaHard
- BiologyBiomoleculesLipids, nucleotides and sugarsEasy
- BiologyBiomoleculesLipids, nucleotides and sugarsEasy
- BiologyBiomoleculesClassification of enzymes and co-factorsMediumQ33Match each class of enzymes with the type of reaction it catalyses.Column IColumn IIP. Oxidoreductases1. Hydrolysis of ester, ether, peptide or glycosidic bondsQ. Transferases2. Removal of groups from substrates by mechanisms other than hydrolysis, leaving double bondsR. Hydrolases3. Transfer of a group other than hydrogen between a pair of substratesS. Lyases4. Oxidation-reduction between two substrates
- BiologyBiomoleculesFactors affecting enzyme activityMediumQ34Enzymes and their inhibitors are often misunderstood. Evaluate the two statements.
- Statement I: Malonate inhibits succinic dehydrogenase competitively because it closely resembles the substrate succinate in structure.
- Statement II: An enzyme speeds up a reaction by changing the energy content of the substrate and of the product, not by lowering the activation energy barrier between them.
- BiologyBiomoleculesStructure of proteinsMedium
- BiologyBiomoleculesPolysaccharidesMedium
- BiologyBiomoleculesFactors affecting enzyme activityMediumQ37Enzymes respond differently to cold and heat. Evaluate the Assertion and Reason.
- Assertion (A): An enzyme kept at a low temperature can usually regain its activity when warmed, but one exposed to a very high temperature loses its activity.
- Reason (R): Low temperature only keeps an enzyme in a temporarily inactive state, whereas high temperature denatures its protein structure.
- BiologyBiomoleculesClassification of enzymes and co-factorsHard
- BiologyBiomoleculesProteinsEasy
- BiologyBiomoleculesAmino acidsHard
- BiologyBody Fluids and CirculationBlood groupsEasy
- BiologyBody Fluids and CirculationThe cardiac cycleEasy
- BiologyBody Fluids and CirculationFormed elements: RBCs, WBCs and plateletsMediumQ43Match each type of leucocyte with its feature.Column IColumn IIP. Neutrophils1. Most numerous WBC, 60-65 per centQ. Eosinophils2. Resist infections; associated with allergic reactionsR. Basophils3. Secrete histamine, serotonin and heparinS. Lymphocytes4. B and T forms responsible for immune responses
- BiologyBody Fluids and CirculationElectrocardiogram (ECG)MediumQ44Examine the following statements about cardiac activity and the ECG.
- Statement I: The sino-atrial node (SAN) can generate the maximum number of action potentials per minute and therefore acts as the pacemaker.
- Statement II: The QRS complex of an ECG represents the repolarisation of the ventricles.
- BiologyBody Fluids and CirculationBlood groupsMedium
- BiologyBody Fluids and CirculationThe cardiac cycleMedium
- BiologyBody Fluids and CirculationBlood groupsMediumQ47Blood group AB individuals can receive blood of any ABO group. Evaluate the Assertion and Reason.
- Assertion (A): Persons with blood group AB are called universal recipients.
- Reason (R): The plasma of group AB persons contains neither anti-A nor anti-B antibodies.
- BiologyBody Fluids and CirculationCoagulation of bloodHard
- BiologyBody Fluids and CirculationRegulation of cardiac activity and disordersEasy
- BiologyBody Fluids and CirculationBlood groupsHard
- BiologyBreathing and Exchange of GasesRespiratory organs — human respiratory systemEasyQ51Which one of the following correctly gives the sequence of passages through which inhaled air travels, from the external nostrils to the alveoli, in humans?
- AExternal nostrils → larynx → pharynx → trachea → bronchioles → primary bronchi → alveoli
- BExternal nostrils → pharynx → trachea → larynx → primary bronchi → bronchioles → alveoli
- CExternal nostrils → pharynx → larynx → trachea → primary bronchi → bronchioles → alveoli
- DExternal nostrils → larynx → trachea → pharynx → bronchioles → alveoli
- BiologyBreathing and Exchange of GasesRespiratory organs — larynxEasyQ52The larynx, commonly called the sound box, lies between the pharynx and the trachea. Which of the following correctly describes a key feature of the larynx and its role during swallowing?
- AIt lies below the diaphragm and its main job is to trap dust particles in mucus
- BIt is the widest part of the respiratory tract, positioned between the two bronchi, with no role during swallowing
- CIt is a cartilaginous box that helps in sound production, and during swallowing its opening (the glottis) can be covered by a thin, elastic cartilaginous flap, the epiglottis, to prevent food from entering the larynx
- DIt is a spongy, elastic organ made of bronchioles and alveoli where gas exchange occurs
- BiologyBreathing and Exchange of GasesMechanism of breathingMediumQ53During a normal, quiet inspiration in humans, contraction of the diaphragm and the external intercostal muscles brings about a pressure change that draws air into the lungs. Which of the following correctly explains this sequence of events?
- AThe diaphragm relaxes and arches upward while the external intercostals contract, decreasing thoracic volume and pressure, which draws air in
- BBoth the diaphragm and the intercostal muscles contract to decrease thoracic volume, increasing intra-pulmonary pressure and pushing air out
- CThe diaphragm contracts and flattens while the external intercostals lift the ribs and sternum outward, increasing thoracic volume; intra-pulmonary pressure falls below atmospheric pressure, so air flows in
- DAir enters the lungs because the alveoli actively pump air in using ciliary movement, independent of thoracic volume changes
- BiologyBreathing and Exchange of GasesRespiratory volumes and capacitiesMedium
- BiologyBreathing and Exchange of GasesExchange of gasesMediumQ55Efficient gas exchange at the alveoli depends on the properties of the gases themselves as well as on the structure of the membrane they must cross. Evaluate the two statements below on this exchange.
- Statement I: The solubility of CO₂ is 20–25 times higher than that of O₂, so the amount of CO₂ that can diffuse through the diffusion membrane per unit difference in partial pressure is much higher than that of O₂.
- Statement II: The diffusion membrane is made up of three major layers — the thin squamous epithelium of the alveoli, the endothelium of the alveolar capillaries, and the basement substance in between them — and its total thickness is much less than a millimetre.
- BiologyBreathing and Exchange of GasesTransport of gases — oxygenMediumQ56Read the following two statements about the transport of oxygen in human blood and choose the option that correctly evaluates them.
- Statement I: About 97% of the oxygen carried in blood is bound to haemoglobin as oxyhaemoglobin, and each haemoglobin molecule can bind a maximum of four O₂ molecules.
- Statement II: A higher partial pressure of CO₂, a higher hydrogen ion concentration, and a higher temperature, as found in the tissues, all favour dissociation of oxygen from oxyhaemoglobin, helping O₂ unload where it is needed.
- BiologyBreathing and Exchange of GasesTransport of gases — carbon dioxideHardQ57Only a small fraction of the CO₂ produced by tissues travels dissolved in plasma. Which of the following correctly describes how most CO₂ is carried in the blood, and which enzyme catalyses the key step of this route?
- AAbout 70% of CO₂ is carried as carbamino-haemoglobin, formed by CO₂ binding directly to the haem group of haemoglobin, a reaction catalysed by carbonic anhydrase
- BNearly all CO₂ diffuses freely and travels dissolved in plasma, since CO₂, unlike O₂, needs no carrier
- CAbout 70% of CO₂ is transported as bicarbonate ions, formed after CO₂ combines with water inside RBCs to give carbonic acid, which dissociates into H⁺ and HCO₃⁻; carbonic anhydrase, abundant in RBCs, catalyses the CO₂ + H₂O step
- DAbout 70% of CO₂ combines irreversibly with plasma proteins to form stable carbaminohaemoglobin, catalysed by RuBisCO present in RBCs
- BiologyBreathing and Exchange of GasesRegulation of respirationHardQ58Breathing rate is adjusted moment to moment by chemical signals reaching the medulla, not by mechanical stretch alone. Evaluate the following Assertion and Reason on how CO₂ levels drive this regulation.
- Assertion (A): A rise in the CO₂ concentration of blood changes the rate and depth of breathing far more strongly than a comparable fall in O₂ concentration.
- Reason (R): A chemosensitive area adjacent to the rhythm centre in the medulla, together with receptors associated with the aortic arch and carotid artery, is highly sensitive to CO₂ and H⁺ concentration and signals the rhythm centre to make the necessary adjustments.
- BiologyBreathing and Exchange of GasesDisorders of the respiratory systemEasyQ59Emphysema is a chronic respiratory disorder commonly caused by long-term cigarette smoking. Which of the following correctly describes what happens to the lungs in this disease and its immediate functional consequence?
- AThe bronchi and bronchioles become inflamed, causing difficulty in breathing with wheezing
- BFine dust particles accumulate in the lungs and stimulate fibrous tissue growth, reducing lung flexibility
- CThe walls of the alveoli are damaged and adjacent alveoli merge, decreasing the total surface area available for gas exchange
- DThe alveolar surface area increases as new alveoli are formed to compensate for reduced oxygen availability
- BiologyBreathing and Exchange of GasesDisorders of the respiratory systemMediumQ60Workers in industries involving grinding or stone-breaking are exposed over years to excessive dust in their workplace. What kind of respiratory damage does chronic exposure of this sort typically cause, and what is this category of disorders called?
- AThe dust particles are trapped entirely by nasal hair and mucus, so no lasting lung damage occurs even after years of exposure
- BThe dust causes reversible bronchoconstriction that resolves fully within hours of leaving the workplace, similar to asthma
- CChronic inhalation of fine dust particles causes inflammation leading to fibrosis of the lungs, reducing their functioning; such conditions are called occupational respiratory disorders
- DThe dust particles are absorbed into the blood through the alveoli and cause damage mainly to the liver rather than the lungs
- BiologyCell Cycle and Cell DivisionInterphase: G1, S and G2Easy
- BiologyCell Cycle and Cell DivisionProphase IEasy
- BiologyCell Cycle and Cell DivisionProphase IMediumQ63Match each sub-stage of prophase I of meiosis with its characteristic event.Column IColumn IIP. Zygotene1. Terminalisation of chiasmataQ. Pachytene2. X-shaped chiasmata become visibleR. Diplotene3. Pairing of homologous chromosomesS. Diakinesis4. Crossing over at recombination nodules
- BiologyCell Cycle and Cell DivisionQuiescent stage and which cells divideMediumQ64Cells vary in their division behaviour. Evaluate the following statements.
- Statement I: In the S phase the DNA content of a cell doubles from 2C to 4C, but the number of chromosomes stays the same.
- Statement II: Cells that leave G1 and enter the quiescent stage (G0) become metabolically inactive.
- BiologyCell Cycle and Cell DivisionProphase and metaphaseMedium
- BiologyCell Cycle and Cell DivisionMetaphase I to interkinesisMedium
- BiologyCell Cycle and Cell DivisionSignificance of mitosisMediumQ67Mitosis is called an equational division. Evaluate the Assertion and Reason.
- Assertion (A): Mitosis is described as an equational division.
- Reason (R): The number of chromosomes in the daughter cells is the same as that in the parent cell.
- BiologyCell Cycle and Cell DivisionMeiosis and its key featuresHard
- BiologyCell Cycle and Cell DivisionCytokinesisEasy
- BiologyCell Cycle and Cell DivisionAnaphase and telophaseHard
- BiologyCell: The Unit of LifeCell TheoryEasy
- BiologyCell: The Unit of LifeRibosomesEasy
- BiologyCell: The Unit of LifePlasma membraneMediumQ73The fluid mosaic model of the plasma membrane, proposed by Singer and Nicolson in 1972, describes how membrane lipids and proteins are organised. Which statement correctly describes this arrangement?
- AThe membrane is a rigid, static lipid bilayer, with all proteins fixed permanently on its outer surface only
- BProteins form a continuous layer completely coating both surfaces of the lipid bilayer, like bread around a filling
- CThe lipid bilayer is quasi-fluid, and most proteins are embedded to varying depths, giving the membrane a mosaic look; proteins and lipids can move laterally
- DThe membrane is composed almost entirely of protein, with only a few scattered lipid molecules trapped between them
- BiologyCell: The Unit of LifeEndomembrane system — Golgi apparatusMediumQ74The Golgi apparatus is an important part of the endomembrane system that packages materials received from the endoplasmic reticulum. Evaluate the two statements below on its structure.
- Statement I: The Golgi apparatus consists of many flat, disc-shaped sacs called cisternae, which are stacked parallel to each other and concentrically arranged near the nucleus.
- Statement II: The Golgi cisternae have a convex cis face, called the forming face, and a concave trans face, called the maturing face; the two faces are entirely different but interconnected.
- BiologyCell: The Unit of LifeMitochondriaMediumQ75Mitochondria are unusual among cytoplasmic organelles in carrying a genome of their own, in addition to their distinctive double-membrane, cristae-folded architecture. Evaluate the two statements below on mitochondrial structure and genetics.
- Statement I: Mitochondria are bound by a double membrane, and the inner membrane is thrown into infoldings called cristae that increase the surface area for ATP synthesis.
- Statement II: The mitochondrial matrix contains single circular DNA, a few RNA molecules, ribosomes, and the enzymes required for the synthesis of some of the organelle's own proteins.
- BiologyCell: The Unit of LifeEndomembrane system — LysosomesMediumQ76Lysosomes, popularly nicknamed the 'suicide bags' of the cell because of the powerful digestive enzymes they hold, belong to the endomembrane system. Which statement correctly describes how lysosomes are formed and what they contain?
- AThey are formed by the packaging activity of the Golgi apparatus and contain hydrolytic enzymes that work best at an alkaline pH
- BThey are bound by a membrane called the tonoplast and store cell sap that can occupy up to 90% of a plant cell's volume
- CThey are formed by the packaging activity of the Golgi apparatus and are rich in hydrolytic enzymes (lipases, proteases, carbohydrases) that are optimally active at an acidic pH
- DThey are membrane-less particles of RNA and protein that synthesise the digestive enzymes they carry
- BiologyCell: The Unit of LifeMitochondria and plastidsHardQ77Mitochondria and chloroplasts are unlike most other organelles in retaining a trace of genetic independence from the nucleus. Evaluate the following Assertion and Reason on why they are termed semi-autonomous.
- Assertion (A): Mitochondria and plastids are described as semi-autonomous organelles.
- Reason (R): Both organelles possess their own double-stranded circular DNA, 70S ribosomes, and the machinery to synthesise some of their own proteins, but still depend on nuclear genes for many other proteins.
- BiologyCell: The Unit of LifeCell wallMedium
- BiologyCell: The Unit of LifeCilia and flagellaHardQ79A transverse section through the shaft of a cilium or flagellum, viewed under the electron microscope, shows a characteristic microtubule arrangement called the axoneme. Which of the following correctly describes this arrangement and the structure that anchors it in the cell?
- ANine central microtubules surrounded by two peripheral doublets, anchored by the nucleolus
- BNine peripheral triplets and no central microtubules, anchored by a centrosome located inside the nucleus
- CNine peripheral doublets of microtubules and one central pair (the '9 + 2' array), anchored at the base by a basal body structurally similar to a centriole
- DNine peripheral doublets with no central microtubules, anchored directly to the Golgi apparatus
- BiologyCell: The Unit of LifeNucleusEasy
- ChemistryChemical Bonding and Molecular StructureHybridisationEasy
- ChemistryChemical Bonding and Molecular StructureVSEPR theoryEasy
- ChemistryChemical Bonding and Molecular StructureHydrogen bondingEasy
- ChemistryChemical Bonding and Molecular StructureMolecular orbital theory / bond orderMedium
- ChemistryChemical Bonding and Molecular StructureDipole momentMediumQ85Boron trifluoride (BF₃) contains three polar B–F bonds, yet its measured net dipole moment is zero. What is the correct explanation?
- AThe B–F bonds are actually non-polar because B and F have similar electronegativity
- BBF₃ is trigonal planar (sp² boron, 120° F–B–F angles), so the three equal bond dipoles cancel exactly by symmetry
- CThe lone pair on boron cancels the bond dipoles
- DFluorine's very high electronegativity prevents any dipole from forming in the first place
- ChemistryChemical Bonding and Molecular StructureHybridisation and VSEPR (expanded octet)Medium
- ChemistryChemical Bonding and Molecular StructureHybridisation and molecular shapeMediumQ87The Assertion (A) and Reason (R) below concern the shape of a xenon fluoride. Read them and choose the option that correctly describes their relationship.
- Assertion (A): Xenon tetrafluoride (XeF₄) has a square planar molecular shape.
- Reason (R): The Xe atom in XeF₄ is sp³d² hybridised, and its two lone pairs occupy the axial positions of the octahedral electron-pair geometry, leaving the four F atoms and Xe in one plane.
- ChemistryChemical Bonding and Molecular StructureMolecular orbital theory / bond order trendsHardQ88Two statements about the bond orders of O₂, O₂⁺ (dioxygenyl cation), and O₂⁻ (superoxide ion) are given below. Study them and choose the correct option.
- Statement I: The bond order of O₂⁺ is higher than the bond order of O₂.
- Statement II: The bond order of O₂⁻ is lower than the bond order of O₂.
- ChemistryChemical Bonding and Molecular StructureVSEPR theory (expanded octet)Hard
- BiologyChemical Coordination and IntegrationThe pituitary glandEasy
- BiologyChemical Coordination and IntegrationPineal and thyroid glandsEasy
- BiologyChemical Coordination and IntegrationThe pituitary glandMediumQ92Match each disorder with its hormonal cause.Column IColumn IIP. Acromegaly1. Excess growth hormone in adultsQ. Diabetes insipidus2. Impaired ADH functionR. Pituitary dwarfism3. Deficiency of growth hormone in childhoodS. Cretinism4. Thyroid hormone deficiency during pregnancy
- BiologyChemical Coordination and IntegrationPancreasMediumQ93Evaluate the following statements about the endocrine pancreas.
- Statement I: Glucagon, secreted by the α-cells of the islets of Langerhans, acts mainly on hepatocytes and stimulates glycogenolysis, causing hyperglycemia.
- Statement II: Insulin, secreted by the β-cells, is a hyperglycemic hormone because it increases the release of glucose from the liver.
- BiologyChemical Coordination and IntegrationParathyroid gland and thymusMedium
- BiologyChemical Coordination and IntegrationAdrenal glandMedium
- BiologyChemical Coordination and IntegrationMechanism of hormone actionHardQ96Hormones act through receptors located in different places. Evaluate the Assertion and Reason.
- Assertion (A): Steroid hormones such as cortisol usually act through intracellular receptors and regulate gene expression.
- Reason (R): Steroid hormones are lipid-soluble, so they can cross the cell membrane to reach the intracellular receptors, whereas protein hormones bind to membrane-bound receptors.
- BiologyChemical Coordination and IntegrationTestis and ovaryHard
- BiologyChemical Coordination and IntegrationHormones of heart, kidney and gutEasy
- BiologyChemical Coordination and IntegrationThe pituitary glandMedium
- ChemistryClassification of Elements and Periodicity in PropertiesNomenclature of elements with Z > 100Easy
- ChemistryClassification of Elements and Periodicity in PropertiesElectronic configuration and position in the tableEasy
- ChemistryClassification of Elements and Periodicity in PropertiesIonic radiusEasy
- ChemistryClassification of Elements and Periodicity in PropertiesIonization enthalpyMedium
- ChemistryClassification of Elements and Periodicity in PropertiesElectron gain enthalpyMedium
- ChemistryClassification of Elements and Periodicity in PropertiesIonization enthalpyMediumQ105Magnesium and aluminium sit next to each other in period 3. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): The first ionization enthalpy of magnesium is higher than that of aluminium.
- Reason (R): The electron removed from aluminium is a 3p electron, which is shielded by the 3s electrons and lies higher in energy than a 3s electron.
- ChemistryClassification of Elements and Periodicity in PropertiesAnomalous properties of second period elementsMediumQ106Lithium, beryllium and boron often behave differently from the heavier members of their groups. Which of the statements below are correct?
- Statement I: The maximum covalency of a second-period element is 4, because it has only four valence orbitals.
- Statement II: Boron forms the ion [BF₆]³⁻ just as aluminium forms [AlF₆]³⁻.
- Statement III: Lithium shows several similarities to magnesium, an example of a diagonal relationship.
- Statement IV: Sodium oxide, Na₂O, is amphoteric.
- ChemistryClassification of Elements and Periodicity in PropertiesPeriodic trends and chemical reactivityMediumQ107Oxides change character across the periodic table. Match each oxide in List I with its acid–base nature in List II.Column IColumn IIP. Na₂O1. AmphotericQ. Cl₂O₇2. NeutralR. As₂O₃3. BasicS. N₂O4. Acidic
- ChemistryClassification of Elements and Periodicity in PropertiesIonization enthalpyHard
- ChemistryClassification of Elements and Periodicity in PropertiesIonization enthalpyHard
- ChemistryEquilibriumRelationship between Kp and KcEasy
- ChemistryEquilibriumFactors affecting equilibria: Le Chatelier's principleEasy
- ChemistryEquilibriumCalculating equilibrium constantsMedium
- ChemistryEquilibriumIonization constants of weak acidsMedium
- ChemistryEquilibriumCommon ion effect in ionization of acidsMediumQ114Read the Assertion (A) and Reason (R) and choose the correct option.
- Assertion (A): Dissolving solid sodium acetate in an aqueous solution of acetic acid raises the pH of the solution.
- Reason (R): The added acetate ions shift the equilibrium CH₃COOH ⇌ CH₃COO⁻ + H⁺ to the left, lowering [H⁺].
- ChemistryEquilibriumSolubility product constantMedium
- ChemistryEquilibriumThe pH scale: strong basesEasy
- ChemistryEquilibriumBuffer solutionsHard
- ChemistryEquilibriumRelationship between equilibrium constant K and ΔG°Hard
- BiologyExcretory Products and Their EliminationNitrogenous wastesEasy
- BiologyExcretory Products and Their EliminationGlomerular filtrationEasy
- BiologyExcretory Products and Their EliminationExcretory structures in animalsMediumQ121Excretory organs differ across animal groups. Pair each animal in the left list with the organ that removes its wastes.Column IColumn IIP. Planaria1. Protonephridia (flame cells)Q. Earthworm2. NephridiaR. Cockroach3. Malpighian tubulesS. Prawn4. Antennal or green glands
- BiologyExcretory Products and Their EliminationFunction of the tubulesMediumQ122The loop of Henle has two limbs with different permeabilities. Evaluate the statements.
- Statement I: The descending limb of the loop of Henle is permeable to water but almost impermeable to electrolytes.
- Statement II: The ascending limb is permeable to water and impermeable to electrolytes.
- BiologyExcretory Products and Their EliminationRegulation of kidney functionMedium
- BiologyExcretory Products and Their EliminationRegulation of kidney functionMedium
- BiologyExcretory Products and Their EliminationCounter current mechanismMediumQ125Human kidneys can produce urine about four times as concentrated as the initial filtrate. Evaluate the Assertion and Reason.
- Assertion (A): Human kidneys can produce urine that is about four times more concentrated than the glomerular filtrate.
- Reason (R): The counter current arrangement of the loop of Henle and vasa recta maintains an osmolarity gradient in the medullary interstitium, from about 300 mOsmol L⁻¹ in the cortex to about 1200 mOsmol L⁻¹ in the inner medulla.
- BiologyExcretory Products and Their EliminationReabsorption and secretionHard
- BiologyExcretory Products and Their EliminationThe nephronEasy
- BiologyExcretory Products and Their EliminationRole of other organs in excretionHard
- PhysicsGravitationKepler's lawsEasy
- PhysicsGravitationg above and below the surfaceEasy
- PhysicsGravitationEscape speedEasyQ131Consider statements I to IV about the escape speed from the surface of a planet (atmosphere ignored). Select the option that names every correct statement and no incorrect one.
- I. The escape speed does not depend on the mass of the body being launched.
- II. The escape speed from the moon's surface is less than that from the earth's surface.
- III. The escape speed from a planet's surface is √2 times the speed of a satellite orbiting just above that surface.
- IV. The escape speed from the earth's surface is about 7.9 km s⁻¹.
- PhysicsGravitationg above and below the surfaceMedium
- PhysicsGravitationEnergy of an orbiting satelliteHard
- PhysicsGravitationEarth satellitesMedium
- PhysicsGravitationEscape speedMedium
- PhysicsGravitationEnergy of an orbiting satelliteMediumQ136Why is a satellite bound to the earth? Weigh the Assertion (A) on the sign of its total mechanical energy against the Reason (R) comparing its kinetic and potential energies.
- Assertion (A): The total mechanical energy of a satellite in a circular orbit round the earth is negative.
- Reason (R): The gravitational potential energy of the satellite is negative and its magnitude is greater than the satellite's kinetic energy.
- PhysicsGravitationEarth satellitesMediumQ137A satellite of mass m moves in a circular orbit of radius r round a planet of mass M. Match each quantity in List I with its expression in List II.Column IColumn IIP. Orbital speed1. −GMm/2rQ. Kinetic energy2. √(GM/r)R. Potential energy3. −GMm/rS. Total energy4. GMm/2r
- PhysicsGravitationg above and below the surfaceHard
- ChemistryHydrocarbonsAddition of hydrogen halides to alkenesEasy
- ChemistryHydrocarbonsAcidic character of alkynesEasy
- ChemistryHydrocarbonsConformations of ethaneEasyQ141Which statement about the conformations of ethane is correct?
- AThe staggered form is more stable because the hydrogen atoms on the two carbons are as far apart as possible.
- BBond angles and C–H bond lengths are different in the staggered and eclipsed forms.
- CThe staggered and eclipsed forms can be separated at room temperature.
- DThe eclipsed form is more stable because the C–H bonds are aligned for maximum overlap.
- ChemistryHydrocarbonsOzonolysis of alkenesMedium
- ChemistryHydrocarbonsPeroxide effectMediumQ143Propene is treated with HCl in the presence of benzoyl peroxide. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): The product is 2-chloropropane, the Markovnikov product, even though peroxide is present.
- Reason (R): The H–Cl bond is too strong to be split by the free radicals formed from the peroxide, so no radical chain starts.
- ChemistryHydrocarbonsPreparation of alkanesMedium
- ChemistryHydrocarbonsBenzene structure and aromaticityMediumQ145Benzene was once drawn with alternating single and double bonds. Which of the statements below about its structure and aromaticity are correct?
- Statement I: Benzene is a planar molecule in which every carbon atom is sp² hybridised.
- Statement II: Cyclooctatetraene is aromatic because it is cyclic and fully conjugated.
- Statement III: All six carbon–carbon bonds in benzene have the same length, 139 pm.
- Statement IV: Benzene readily undergoes addition reactions in preference to substitution.
- ChemistryHydrocarbonsDirective influence of functional groupsMediumQ146A substituent already on a benzene ring decides where the next electrophile goes and how fast. Match each substituent in List I with its behaviour in List II.Column IColumn IIP. –CH₃1. ortho/para-directing, deactivatingQ. –NO₂2. meta-directing, deactivatingR. –Cl3. ortho/para-directing, strongly activating by resonanceS. –OH4. ortho/para-directing, activating by +I effect and hyperconjugation
- ChemistryHydrocarbonsElectrophilic substitution in benzeneHard
- ChemistryHydrocarbonsPreparation of alkenes from alkynesHard
- PhysicsKinetic TheoryKinetic interpretation of temperatureEasy
- PhysicsKinetic TheoryKinetic interpretation of temperatureEasy
- PhysicsKinetic TheoryPressure of an ideal gas from molecular impactsEasyQ151A bicycle pump's outlet is blocked and its piston is pushed in slowly, so the trapped air stays at room temperature. Evaluate the Assertion (A) about the air's pressure and the molecular Reason (R) offered for it.
- Assertion (A): When the trapped air is compressed at constant temperature, its pressure increases.
- Reason (R): The number of molecules per unit volume increases, so molecules strike each unit area of the walls more often.
- PhysicsKinetic TheoryKinetic interpretation of temperatureMedium
- PhysicsKinetic TheoryKinetic interpretation of temperatureMedium
- PhysicsKinetic TheorySpecific heats of gasesHard
- PhysicsKinetic TheoryDalton's law of partial pressuresMedium
- PhysicsKinetic TheoryDegrees of freedom and equipartitionMediumQ156Using the law of equipartition of energy, match each substance in List I with its molar specific heat at constant volume in List II.Column IColumn IIP. Monatomic gas1. 3RQ. Rigid diatomic gas2. 7R/2R. Diatomic gas with one vibrational mode3. 3R/2S. Solid (equipartition)4. 5R/2
- PhysicsKinetic TheoryMean free pathMediumQ157Gas molecules move fast, yet a smell spreads slowly across a room because each molecule travels only a short distance between collisions. Which of the claims I to IV about this mean free path are correct?
- I. At a fixed temperature, doubling the pressure of the gas halves the mean free path.
- II. The mean free path is directly proportional to the square of the molecular diameter.
- III. At a fixed pressure, raising the temperature increases the mean free path.
- IV. The mean free path increases when the number of molecules per unit volume increases.
- PhysicsKinetic TheoryPressure of an ideal gas from molecular impactsHard
- PhysicsLaws of MotionImpulseEasy
- PhysicsLaws of MotionNewton's second law of motionEasy
- PhysicsLaws of MotionNewton's third law of motionEasyQ161Four claims about action-reaction force pairs (Newton's third law) are listed below as I to IV. Select the option that names every correct claim and no incorrect one.
- I. The action and reaction forces of a pair always act on two different bodies.
- II. Action and reaction cancel each other, so the net force on any body is always zero.
- III. For a book resting on a table, the book's weight and the normal force from the table on the book form an action-reaction pair.
- IV. Action and reaction are equal in magnitude even when the interacting bodies are accelerating.
- PhysicsLaws of MotionSolving problems in mechanics (connected bodies)Medium
- PhysicsLaws of MotionFrictionMedium
- PhysicsLaws of MotionApparent weight (Newton's second law)Medium
- PhysicsLaws of MotionCircular motion (car on a level road)Medium
- PhysicsLaws of MotionSolving problems in mechanics (friction with connected bodies)Hard
- PhysicsLaws of MotionConservation of momentumHard
- BiologyLocomotion and MovementAxial skeletonEasy
- BiologyLocomotion and MovementTypes of muscleEasy
- BiologyLocomotion and MovementJointsMediumQ170Match each type of synovial joint with an example.Column IColumn IIP. Ball and socket1. Between humerus and pectoral girdleQ. Hinge2. Knee jointR. Pivot3. Between atlas and axisS. Gliding4. Between the carpals
- BiologyLocomotion and MovementMechanism of muscle contractionMediumQ171Examine the statements about muscle contraction.
- Statement I: During contraction of a sarcomere, the I band and the H zone become narrower, while the length of the A band remains the same.
- Statement II: Ca²⁺ ions released from the sarcoplasmic reticulum bind to tropomyosin and thereby expose the active sites on actin.
- BiologyLocomotion and MovementAxial skeletonMedium
- BiologyLocomotion and MovementMechanism of muscle contractionMedium
- BiologyLocomotion and MovementRed and white fibresMediumQ174Muscle fibres are classified as red or white. Evaluate the Assertion and Reason.
- Assertion (A): Red muscle fibres depend mainly on aerobic respiration for energy.
- Reason (R): Red fibres have a high amount of myoglobin and plenty of mitochondria, and they utilise large amounts of oxygen.
- BiologyLocomotion and MovementAxial skeletonHard
- BiologyLocomotion and MovementJointsEasy
- BiologyLocomotion and MovementStructure of contractile proteinsHardQ177Four descriptions of the protein filaments of a sarcomere are given below. Which description is accurate?
- AA thin filament is made of two F actins, two tropomyosin filaments and troponin at regular intervals
- BA thick filament is made of polymerised actin
- CThe head of a myosin molecule is called light meromyosin and has no binding sites
- DTroponin runs continuously along the entire length of the actin filament
- PhysicsMechanical Properties of FluidsAtmospheric and gauge pressureEasy
- PhysicsMechanical Properties of FluidsHydraulic machinesEasy
- PhysicsMechanical Properties of FluidsAngle of contact, drops and bubblesEasyQ180A soap bubble and a liquid drop of equal radius are made from the same soap solution. Decide whether the Assertion (A) about their excess pressures is true, and whether the Reason (R) accounts for it.
- Assertion (A): The excess pressure inside a soap bubble is twice the excess pressure inside a liquid drop of the same radius, made from the same solution.
- Reason (R): A soap bubble film has two surfaces in contact with air, while a liquid drop has only one.
- PhysicsMechanical Properties of FluidsBernoulli's principleMedium
- PhysicsMechanical Properties of FluidsBernoulli's principleMedium
- PhysicsMechanical Properties of FluidsViscosity and Stokes' lawMedium
- PhysicsMechanical Properties of FluidsCapillary riseMedium
- PhysicsMechanical Properties of FluidsViscosity and Stokes' lawMediumQ185Honey, air and glycerine all resist flow. Which of the claims I to IV about viscosity, its temperature dependence and Stokes' drag are correct?
- I. The viscosity of a liquid decreases as its temperature rises.
- II. The viscosity of a gas increases as its temperature rises.
- III. The SI unit of the coefficient of viscosity is the pascal second (Pa s).
- IV. In Stokes' law, the viscous drag on a sphere is proportional to the square of its speed.
- PhysicsMechanical Properties of FluidsDynamic liftHard
- PhysicsMechanical Properties of FluidsAngle of contact, drops and bubblesHard
- PhysicsMechanical Properties of SolidsYoung's modulusEasy
- PhysicsMechanical Properties of SolidsYoung's modulusEasyQ189Rubber stretches far more than steel under the same load, yet physics calls steel the more elastic material. Judge the Assertion (A) and Reason (R) on this point.
- Assertion (A): Steel is more elastic than rubber.
- Reason (R): For the same stress, steel shows a much smaller strain than rubber, so its Young's modulus is much larger.
- PhysicsMechanical Properties of SolidsBulk modulus and compressibilityEasy
- PhysicsMechanical Properties of SolidsYoung's modulusMedium
- PhysicsMechanical Properties of SolidsBulk modulus and compressibilityMedium
- PhysicsMechanical Properties of SolidsPoisson's ratio and elastic energyMedium
- PhysicsMechanical Properties of SolidsHooke's law and the stress-strain curveMediumQ194A metal wire is loaded steadily until it breaks. Match each feature of its stress-strain curve in List I with its description in List II.Column IColumn IIP. Proportional limit1. Strain that remains after the load is removed from beyond the yield pointQ. Yield point (elastic limit)2. Largest stress up to which stress is proportional to strainR. Permanent set3. Stress beyond which the wire does not regain its original length on unloadingS. Fracture point4. Stress at which the wire breaks
- PhysicsMechanical Properties of SolidsShear modulusMediumQ195Young's, shear and bulk moduli and Poisson's ratio describe how a solid deforms. Pick the option listing exactly the true claims among I to IV on their signs, sizes and dimensions.
- I. Poisson's ratio has neither unit nor dimensions.
- II. For most materials the shear modulus is about one-third of Young's modulus.
- III. The bulk modulus of a material is negative because its volume decreases when pressure increases.
- IV. Stress has the same dimensional formula as pressure.
- PhysicsMechanical Properties of SolidsShear modulusHard
- PhysicsMechanical Properties of SolidsPoisson's ratio and elastic energyHard
- BiologyMorphology of Flowering PlantsRoot systems and the root tipEasy
- BiologyMorphology of Flowering PlantsCompound leaves and phyllotaxyEasy
- BiologyMorphology of Flowering PlantsGynoecium and placentationHardQ200Match each type of placentation with a plant that shows it.Column IColumn IIP. Marginal1. SunflowerQ. Axile2. MustardR. Parietal3. LemonS. Basal4. Pea
- BiologyMorphology of Flowering PlantsRacemose and cymose inflorescenceMediumQ201Flowers on an inflorescence axis may open in different orders. Evaluate the following statements.
- Statement I: In a racemose inflorescence the main axis continues to grow, and the flowers are borne in acropetal succession, with younger flowers towards the tip.
- Statement II: In a cymose inflorescence the main axis terminates in a flower, and the flowers are borne in acropetal order.
- BiologyMorphology of Flowering PlantsOvary position on the thalamusMedium
- BiologyMorphology of Flowering PlantsCalyx, corolla and aestivationMedium
- BiologyMorphology of Flowering PlantsAndroeciumMediumQ204Stamens may be attached to different floral parts. Evaluate the Assertion and Reason.
- Assertion (A): The stamens of lily are described as epiphyllous.
- Reason (R): In lily the calyx and corolla are not distinct and together form a perianth, and the stamens are attached to this perianth.
- BiologyMorphology of Flowering PlantsFruit and seedMedium
- BiologyMorphology of Flowering PlantsFruit and seedEasy
- BiologyMorphology of Flowering PlantsFloral formula and SolanaceaeHard
- PhysicsMotion in a PlaneAdding vectors analyticallyEasy
- PhysicsMotion in a PlaneProjectile motionEasy
- PhysicsMotion in a PlaneUniform circular motionEasy
- PhysicsMotion in a PlaneResolving vectors into componentsMedium
- PhysicsMotion in a PlaneProjectile motionMedium
- PhysicsMotion in a PlaneProjectile motionMedium
- PhysicsMotion in a PlaneUniform circular motionMediumQ214A stone whirls round a horizontal circle at constant speed. Is its acceleration a constant vector? Evaluate the Assertion (A) and the Reason (R) given for it.
- Assertion (A): The acceleration of a body in uniform circular motion is constant.
- Reason (R): The speed of a body in uniform circular motion is constant.
- PhysicsMotion in a PlaneProjectile motionMediumQ215A projectile is launched from level ground and lands back on the same level; air resistance is negligible. Consider statements I to IV and select the option that names every correct statement and no incorrect one.
- I. The horizontal component of the projectile's velocity stays constant.
- II. The time of flight depends only on the vertical component of the launch velocity.
- III. The speed of the projectile at the highest point is zero.
- IV. For a given launch speed, the range is maximum at a launch angle of 45°.
- PhysicsMotion in a PlaneConstant acceleration in a planeHard
- PhysicsMotion in a PlaneAdding vectors analyticallyHard
- PhysicsMotion in a Straight LineAverage velocity and average speedEasy
- PhysicsMotion in a Straight LineKinematic equations for uniformly accelerated motionEasy
- PhysicsMotion in a Straight LinePosition-time and velocity-time graphsEasyQ220A student reviewing kinematics graphs for a body moving along a straight line writes four claims, labelled I to IV below. Identify which of the student's claims are physically correct.
- I. The area under a velocity-time graph over an interval equals the displacement in that interval.
- II. The slope of a position-time graph at an instant equals the acceleration at that instant.
- III. A body can have zero velocity and a non-zero acceleration at the same instant.
- IV. If the velocity and acceleration of a body have opposite signs, the body is speeding up.
- PhysicsMotion in a Straight LineFree fallMedium
- PhysicsMotion in a Straight LineVertical motion under gravityMedium
- PhysicsMotion in a Straight LineUniform motion: position as a function of timeMedium
- PhysicsMotion in a Straight LineAccelerationMediumQ224Read the Assertion (A) and Reason (R) about a ball thrown vertically upward, with air resistance neglected, and choose the correct option.
- Assertion (A): At the highest point of its flight, the acceleration of the ball is zero.
- Reason (R): At the highest point of its flight, the velocity of the ball is zero.
- PhysicsMotion in a Straight LineVelocity-time graph and uniformly accelerated motionHard
- PhysicsMotion in a Straight LineInstantaneous velocity; distance vs displacementHard
- BiologyNeural Control and CoordinationResting potentialEasy
- BiologyNeural Control and CoordinationThe brainEasy
- BiologyNeural Control and CoordinationThe brainMediumQ229Match each part of the brain with its function or description.Column IColumn IIP. Hypothalamus1. Contains centres that control body temperature and the urge for eating and drinkingQ. Corpus callosum2. Tract of nerve fibres connecting the two cerebral hemispheresR. Cerebellum3. Very convoluted surface that provides space for many more neuronsS. Medulla oblongata4. Contains centres that control respiration, cardiovascular reflexes and gastric secretions
- BiologyNeural Control and CoordinationNerve impulseMediumQ230Consider the changes in a nerve fibre when a stimulus is applied.
- Statement I: When a stimulus is applied, the membrane becomes freely permeable to Na⁺, so Na⁺ rushes in and the polarity of the membrane is reversed.
- Statement II: Resting potential is restored by the rapid inflow of K⁺ ions into the axon.
- BiologyNeural Control and CoordinationTransmission of impulsesMediumQ231At a chemical synapse, an arriving action potential triggers a sequence of events. Which sequence is correct?
- AImpulse arrives at pre-synaptic knob, synaptic vesicles fuse with the membrane, neurotransmitter released into the cleft, binds receptors on the post-synaptic membrane, ion channels open
- BImpulse crosses the cleft as an electric current
- CNeurotransmitter released first, then impulse arrives at pre-synaptic knob
- DImpulse arrives, neurotransmitter binds to the pre-synaptic membrane, vesicles fuse
- BiologyNeural Control and CoordinationForebrainMedium
- BiologyNeural Control and CoordinationStructure of a neuronMediumQ233Axons may be myelinated or non-myelinated. Evaluate the Assertion and Reason.
- Assertion (A): Myelinated nerve fibres are found in spinal and cranial nerves.
- Reason (R): The gaps between two adjacent myelin sheaths along an axon are called nodes of Ranvier.
- BiologyNeural Control and CoordinationTransmission of impulsesHardQ234Which statement correctly compares electrical and chemical synapses?
- AChemical synapses lack a synaptic cleft and have a direct current
- BElectrical synapses are the commonest type in humans
- CImpulse transmission across an electrical synapse is always faster than across a chemical synapse, and electrical synapses are rare in humans
- DImpulse transmission across a chemical synapse is always faster than across an electrical synapse
- BiologyNeural Control and CoordinationStructure of a neuronEasy
- BiologyNeural Control and CoordinationForebrainHardQ236Which statement about the cerebral cortex is correct?
- AThe outer cortex is white matter because of myelin, and the inner part is grey matter
- BThe outer cortex is grey matter because it contains neuron cell bodies, and the inner part is white matter made of myelinated nerve fibres
- CThe cortex is made only of nerve fibres with no cell bodies
- DBoth outer cortex and inner layer are grey matter
- ChemistryOrganic Chemistry – Some Basic Principles and TechniquesIUPAC nomenclatureEasy
- ChemistryOrganic Chemistry – Some Basic Principles and TechniquesPurification techniquesEasy
- ChemistryOrganic Chemistry – Some Basic Principles and TechniquesInductive effectEasy
- ChemistryOrganic Chemistry – Some Basic Principles and TechniquesCarbocation stabilityMedium
- ChemistryOrganic Chemistry – Some Basic Principles and TechniquesResonanceMediumQ241In the acetate ion, CH₃COO⁻, both carbon–oxygen bonds are experimentally found to be equal in length, intermediate between a typical C–O single bond and a C=O double bond. What best explains this observation?
- AThe ion rapidly interconverts between two structures, with the negative charge hopping back and forth between the two oxygens
- BThe two C–O bonds are best described by a resonance hybrid, in which the negative charge and π-electron density are delocalised equally over both oxygens
- COne C–O bond is truly single and the other truly double, and they only appear equal when measured because of experimental averaging
- DBoth oxygens carry a full unit negative charge simultaneously
- ChemistryOrganic Chemistry – Some Basic Principles and TechniquesIsomerismMedium
- ChemistryOrganic Chemistry – Some Basic Principles and TechniquesInductive effectMediumQ243The Assertion (A) and Reason (R) below concern how far the inductive effect reaches along a carbon chain. Read them and choose the option that correctly describes their relationship.
- Assertion (A): The inductive effect of a substituent becomes negligible after about three bonds along a carbon chain.
- Reason (R): The inductive effect is a permanent polarisation transmitted through σ bonds, and the displacement of electron density falls off rapidly with each additional bond from the substituent.
- ChemistryOrganic Chemistry – Some Basic Principles and TechniquesResonance / carbocation stabilityHardQ244Allyl chloride (CH₂=CH–CH₂–Cl) ionises to give the allyl cation, represented by the two resonance structures CH₂=CH–CH₂⁺ ↔ ⁺CH₂–CH=CH₂. Which statement about this cation is correct?
- AThe two resonance structures interconvert rapidly, and the cation exists as one structure at a time
- BThe two structures are of identical energy and contribute equally, so the actual cation is a single delocalised hybrid more stable than either individual contributing structure
- COnly the first structure is correct, because the positive charge must remain on the carbon nearest the original C–Cl bond
- DThe cation is destabilised relative to a simple (non-allylic) primary carbocation because the positive charge is spread over two carbons
- ChemistryOrganic Chemistry – Some Basic Principles and TechniquesIUPAC nomenclatureHard
- PhysicsOscillationsForce law for SHM: F = −kxEasy
- PhysicsOscillationsThe simple pendulumEasy
- PhysicsOscillationsVelocity and acceleration in SHMEasyQ248A particle performs SHM along a straight line about its mean position O. Which of the claims I to IV about how its displacement, velocity and acceleration are related are correct?
- I. The speed of the particle is maximum at O.
- II. The magnitude of its acceleration is maximum at the extreme positions.
- III. Its velocity and displacement are in phase.
- IV. Its acceleration and displacement are opposite in phase.
- PhysicsOscillationsVelocity and acceleration in SHMMedium
- PhysicsOscillationsEnergy in SHMMedium
- PhysicsOscillationsEnergy in SHMMediumQ251A block on a spring oscillates with period T. Decide whether the Assertion (A) about how often its kinetic energy repeats is true, and whether the trigonometric Reason (R) explains it.
- Assertion (A): The kinetic energy of a particle in SHM varies periodically with a period of T/2.
- Reason (R): For x = A cos(ωt + φ), the kinetic energy is proportional to sin²(ωt + φ), and sin² repeats every half period of the motion.
- PhysicsOscillationsSimple harmonic motion: amplitude, phase and angular frequencyMedium
- PhysicsOscillationsSimple harmonic motion: amplitude, phase and angular frequencyMedium
- PhysicsOscillationsForce law for SHM: F = −kxHard
- PhysicsOscillationsThe simple pendulumHard
- BiologyPhotosynthesis in Higher PlantsWhere does photosynthesis take placeEasyQ256Within the chloroplast, the light reactions of photosynthesis and the carbon-fixation reactions occur in two distinct sub-compartments. Where do these two sets of reactions respectively take place?
- ALight reactions occur in the stroma, while carbon-fixation reactions occur on the thylakoid membrane
- BBoth the light reactions and the carbon-fixation reactions occur exclusively on the outer chloroplast membrane
- CLight reactions occur on the thylakoid membrane, while carbon-fixation reactions occur in the stroma
- DLight reactions occur in the mitochondrial matrix, while carbon fixation occurs in the chloroplast stroma
- BiologyPhotosynthesis in Higher PlantsPigments involved in photosynthesisEasy
- BiologyPhotosynthesis in Higher PlantsEarly experiments — van NielMediumQ258Cornelius van Niel studied purple and green sulphur bacteria, which photosynthesise using H₂S instead of H₂O as the hydrogen donor and release sulphur or sulphate, not O₂. He summarised photosynthesis as 2H₂A + CO₂ → 2A + CH₂O + H₂O (in light), where H₂A is the hydrogen donor. What did he infer about the O₂ released by green plants?
- AIt comes from CO₂, because CO₂ is the reactant that green plants and sulphur bacteria share
- BIt comes from H₂O, the hydrogen donor of green plants, and not from CO₂ — an inference later proved using radioisotopic techniques
- CIt comes from chlorophyll, which is broken down during illumination to release oxygen
- DIt is released directly from the ATP molecules synthesised during the light reaction
- BiologyPhotosynthesis in Higher PlantsLight reaction — photosystemsMediumQ259Photosystem I and Photosystem II each have a reaction-centre chlorophyll a molecule, named after the wavelength at which it shows its absorption peak. Which of the following correctly pairs each photosystem with its reaction-centre pigment and identifies the photosystem associated with the splitting of water into O₂, H⁺ and electrons?
- APSI has reaction centre P680, PSII has reaction centre P700, and water is split at PSI
- BPSI has reaction centre P700, PSII has reaction centre P680, and water is split at PSII
- CPSI has reaction centre P700, PSII has reaction centre P680, and water is split at PSI
- DBoth PSI and PSII have the same reaction centre, P680, and water is split where PSI and PSII physically overlap
- BiologyPhotosynthesis in Higher PlantsThe electron transport — photophosphorylationMediumQ260The light reaction of photosynthesis can proceed along two alternative electron-flow routes through the photosystems, each yielding a different set of products. Evaluate the two statements below on these two routes.
- Statement I: Non-cyclic photophosphorylation involves both Photosystem I and Photosystem II acting in series and generates ATP, NADPH, and O₂.
- Statement II: Cyclic photophosphorylation involves only Photosystem I, produces only ATP with no NADPH or O₂ formed, and can occur in regions such as the stroma lamellae, which lack Photosystem II.
- BiologyPhotosynthesis in Higher PlantsCalvin cycle (C₃ pathway)MediumQ261The biosynthetic phase of photosynthesis begins when a specific five-carbon acceptor molecule reacts with CO₂ in the chloroplast stroma. Evaluate the two statements below on this carbon-fixation step.
- Statement I: The primary CO₂ acceptor of the Calvin cycle is ribulose-1,5-bisphosphate (RuBP), a five-carbon sugar, and the enzyme catalysing its carboxylation is RuBisCO.
- Statement II: The first stable product formed when CO₂ combines with RuBP is a three-carbon compound, 3-phosphoglyceric acid (3-PGA), which is why this pathway is also called the C₃ pathway.
- BiologyPhotosynthesis in Higher PlantsThe C₄ pathwayHardQ262Plants such as maize and sugarcane show Kranz anatomy. They first fix CO₂ in their mesophyll cells using PEP carboxylase (PEPcase), and the Calvin cycle then operates in their bundle sheath cells. Which of the following correctly explains why this two-step strategy gives C₄ plants better productivity than C₃ plants?
- APEP carboxylase directly reduces CO₂ to glucose in the mesophyll cells, making the Calvin cycle in the bundle sheath unnecessary
- BKranz anatomy allows C₄ plants to run the entire Calvin cycle in the mesophyll cells, while bundle sheath cells only store starch
- CThe C₄ acids formed by PEPcase in the mesophyll are broken down in the bundle sheath cells to release CO₂; this raises the CO₂ concentration at the RuBisCO site, so RuBisCO works as a carboxylase, its oxygenase activity is minimised and photorespiration does not occur
- DRuBisCO in C₄ plants has evolved a higher affinity for CO₂ than PEP carboxylase, which is why photorespiration never occurs in these plants
- BiologyPhotosynthesis in Higher PlantsPhotorespirationHardQ263RuBisCO's active site can bind both CO₂ and O₂, and the relative concentration of the two gases decides which one binds. Evaluate the following Assertion and Reason about what happens in C₃ plants when O₂ binds instead.
- Assertion (A): Photorespiration in C₃ plants is regarded as a wasteful process that lowers the net amount of CO₂ fixed.
- Reason (R): In the photorespiratory pathway there is neither synthesis of sugars nor of ATP; instead it results in the release of CO₂ with the utilisation of ATP.
- BiologyPhotosynthesis in Higher PlantsATP synthesis — chemiosmotic hypothesisMediumQ264According to the chemiosmotic hypothesis of ATP synthesis in chloroplasts, protons (H⁺) accumulate in a particular compartment during the light reaction, creating a gradient that drives ATP synthase. In which compartment do these protons accumulate, and in which direction do they move to synthesise ATP?
- AProtons accumulate in the stroma and move into the thylakoid lumen through ATP synthase to synthesise ATP
- BProtons accumulate inside the thylakoid lumen and move from the lumen, through ATP synthase, into the stroma to synthesise ATP
- CProtons accumulate in the mitochondrial matrix and diffuse into the chloroplast stroma through ATP synthase
- DProtons accumulate equally in the stroma and the lumen, so no net gradient or directional flow is needed for ATP synthesis
- BiologyPhotosynthesis in Higher PlantsFactors affecting photosynthesisEasyQ265A crop is grown under bright sunlight with adequate water, but the surrounding air has an unusually low CO₂ concentration. According to Blackman's Law of Limiting Factors, what would you expect to happen to the rate of photosynthesis if light intensity is increased further while CO₂ availability stays low?
- AThe rate of photosynthesis would keep increasing in proportion to light intensity, regardless of CO₂ availability, since light is always the master factor
- BThe rate of photosynthesis would show little to no further increase, because CO₂, being the factor nearest its minimal level, is limiting the rate, not light
- CThe rate of photosynthesis would fall to zero immediately, since any deficiency in one factor completely stops the process
- DThe rate of photosynthesis would increase because higher light intensity raises leaf temperature, which increases CO₂ solubility in the mesophyll cells
- BiologyPlant Growth and DevelopmentGibberellins and cytokininsEasy
- BiologyPlant Growth and DevelopmentEthylene and abscisic acidEasy
- BiologyPlant Growth and DevelopmentPlant growth regulators and their discoveryMediumQ268Match each plant growth regulator with the scientist or group whose work is linked to its discovery.Column IColumn IIP. Auxin1. F.W. WentQ. Gibberellin2. E. KurosawaR. Cytokinin (kinetin)3. Skoog and MillerS. Ethylene4. H.H. Cousins
- BiologyPlant Growth and DevelopmentArithmetic and geometric growthMediumQ269Growth can follow different mathematical patterns. Evaluate the statements.
- Statement I: In arithmetic growth, only one daughter cell continues to divide after mitosis, and the length increases as Lt = L0 + rt, which gives a linear curve.
- Statement II: Geometric growth has only a lag phase and a stationary phase, and its curve is a straight line.
- BiologyPlant Growth and DevelopmentAbsolute and relative growth rateMediumQ270Two leaves A and B have initial areas of 5 cm² and 10 cm². In one week A grows to 10 cm² and B grows to 15 cm². Which statement is correct?
- ABoth have the same absolute growth rate, but B has the higher relative growth rate
- BBoth have the same absolute and the same relative growth rate
- CBoth have the same absolute growth rate, but A has the higher relative growth rate
- DB has a higher absolute growth rate and also a higher relative growth rate
- BiologyPlant Growth and DevelopmentDifferentiation, dedifferentiation and redifferentiationMedium
- BiologyPlant Growth and DevelopmentGrowth and why it never stops in plantsMediumQ272Plant growth is termed open, in contrast to the determinate growth of most animals. Evaluate the Assertion and Reason.
- Assertion (A): Plants show an open form of growth.
- Reason (R): Meristems at certain locations in the plant body retain the capacity to divide throughout the life of the plant.
- BiologyPlant Growth and DevelopmentDevelopment and plasticityHardQ273Cotton, coriander and buttercup all show heterophylly (different leaf forms on the same plant), but not for the same reason. Which statement correctly gives the cause in each?
- AAll three: due to the environment
- BAll three: due to the different phases of life
- CCotton and coriander: different leaf shapes in juvenile and mature phases; buttercup: leaf shape differs in air and in water
- DCotton and coriander: leaf shapes differ in air and water; buttercup: differ with phase of life
- BiologyPlant Growth and DevelopmentEthylene and abscisic acidEasy
- BiologyPlant Growth and DevelopmentArithmetic and geometric growthHard
- BiologyPlant KingdomWhy algae matterEasy
- BiologyPlant KingdomBryophytes: amphibians of the plant kingdomEasy
- BiologyPlant KingdomAlgae: body and reproductionMediumQ278Sexual reproduction in algae is classified by the type of gametes that fuse. Match each type of sexual reproduction with the NCERT example.Column IColumn IIP. Isogamy with flagellated gametes1. Some species of Chlamydomonas, with gametes dissimilar in sizeQ. Isogamy with non-flagellated gametes2. Volvox and FucusR. Anisogamy3. Chlamydomonas, with gametes similar in sizeS. Oogamy4. Spirogyra
- BiologyPlant KingdomLiverworts and mossesMediumQ279Mosses show a clear alternation between two phases. Evaluate the following statements.
- Statement I: The moss gametophyte has two stages, a creeping protonema that arises from a spore and a leafy stage that develops from it as a lateral bud.
- Statement II: The sporophyte of a moss is free-living and independent of the gametophyte for water and food.
- BiologyPlant KingdomPteridophytesMedium
- BiologyPlant KingdomHomospory, heterospory and the seed habitMedium
- BiologyPlant KingdomHomospory, heterospory and the seed habitMediumQ282In some pteridophytes the female gametophyte stays on the plant that produced it. Evaluate the Assertion and Reason.
- Assertion (A): In heterosporous pteridophytes such as Selaginella and Salvinia, the retention of the female gametophyte on the parent sporophyte is regarded as a precursor to the seed habit.
- Reason (R): In these plants the zygote develops into a young embryo within the female gametophyte while the gametophyte is still retained on the parent sporophyte.
- BiologyPlant KingdomGymnosperms: naked seedsHard
- BiologyPlant KingdomLiverworts and mossesEasy
- BiologyPlant KingdomGreen, brown and red algaeHard
- ChemistryRedox ReactionsOxidation number rulesEasy
- ChemistryRedox ReactionsRedox as electron transferEasy
- ChemistryRedox ReactionsTypes of redox reactionsEasy
- ChemistryRedox ReactionsFractional oxidation numbersMedium
- ChemistryRedox ReactionsBalancing redox reactionsMedium
- ChemistryRedox ReactionsDisproportionation reactionsMediumQ291Chlorine, bromine and iodine disproportionate in alkali, but fluorine behaves differently. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): Fluorine does not undergo disproportionation reactions.
- Reason (R): Fluorine is the most electronegative element and cannot show a positive oxidation number.
- ChemistryRedox ReactionsCompetitive electron transfer reactionsMediumQ292Rods of zinc and copper are dipped in solutions of metal salts at room temperature. Which of the statements below are correct?
- Statement I: A zinc rod in copper sulphate solution becomes coated with copper while the blue colour of the solution fades.
- Statement II: A copper rod in silver nitrate solution becomes coated with silver while the solution turns blue.
- Statement III: A copper rod in zinc sulphate solution becomes coated with zinc.
- Statement IV: The tendency to release electrons decreases in the order Zn > Cu > Ag.
- ChemistryRedox ReactionsOxidation number rulesMediumQ293Several compounds break the usual oxidation-number rules. Match the element in each compound of List I with its oxidation number in List II.Column IColumn IIP. O in O₂F₂1. −1Q. C in C₃O₂2. +1R. O in KO₂3. +4/3S. H in CaH₂4. −½
- ChemistryRedox ReactionsRedox titrationsHard
- ChemistryRedox ReactionsRedox couples and electrode potentialsHard
- BiologyRespiration in PlantsGlycolysisEasy
- BiologyRespiration in PlantsTricarboxylic acid cycleEasy
- BiologyRespiration in PlantsAerobic respiration: pyruvate to acetyl CoAMediumQ298Match each reaction of respiration with the enzyme that catalyses it.Column IColumn IIP. Pyruvic acid to acetyl CoA1. HexokinaseQ. Acetyl CoA + OAA to citric acid2. InvertaseR. Glucose to glucose-6-phosphate3. Pyruvic dehydrogenaseS. Sucrose to glucose + fructose4. Citrate synthase
- BiologyRespiration in PlantsElectron transport system and oxidative phosphorylationMediumQ299Consider the following about the oxidation of acetyl CoA and the electron transport system.
- Statement I: In each turn of Krebs' cycle, three NADH, one FADH₂ and one GTP (which can be converted to ATP) are produced.
- Statement II: Oxygen is used in glycolysis as the final hydrogen acceptor, and the ETS operates on the outer mitochondrial membrane.
- BiologyRespiration in PlantsRespiratory quotientMedium
- BiologyRespiration in PlantsFermentationMedium
- BiologyRespiration in PlantsAmphibolic pathwayHardQ302Respiratory intermediates are drawn off for synthesis of other compounds. Evaluate the Assertion and Reason.
- Assertion (A): The respiratory pathway is considered amphibolic rather than purely catabolic.
- Reason (R): Fatty acids are broken down to acetyl CoA for respiration, and acetyl CoA is also withdrawn from the pathway to synthesise fatty acids.
- BiologyRespiration in PlantsElectron transport system and oxidative phosphorylationHard
- BiologyRespiration in PlantsElectron transport system and oxidative phosphorylationMedium
- BiologyRespiration in PlantsAmphibolic pathwayEasy
- ChemistrySome Basic Concepts of ChemistryMole concept and molar massesEasy
- ChemistrySome Basic Concepts of ChemistryUncertainty in measurement: significant figuresEasy
- ChemistrySome Basic Concepts of ChemistryLimiting reagentMedium
- ChemistrySome Basic Concepts of ChemistryReactions in solutions: molarityMedium
- ChemistrySome Basic Concepts of ChemistryEmpirical and molecular formulaMedium
- ChemistrySome Basic Concepts of ChemistryLaws of chemical combinationsEasyQ311Which observation is a direct illustration of the law of multiple proportions?
- ACarbon dioxide made by burning charcoal and carbon dioxide made by heating limestone both contain 27.3% carbon by mass.
- BIn N₂O, NO and NO₂, the masses of oxygen that combine with 28 g of nitrogen are 16 g, 32 g and 64 g respectively.
- CTwo volumes of hydrogen react with one volume of oxygen to give two volumes of water vapour at the same temperature and pressure.
- DEqual volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
- ChemistrySome Basic Concepts of ChemistryAtomic and molecular massesMediumQ312Consider the following statements: Which of the statements are correct?
- Statement I: One atomic mass unit (u) is defined as exactly one-twelfth of the mass of one carbon-12 atom.
- Statement II: The molar mass of a substance in g mol⁻¹ is numerically equal to its molecular (or formula) mass in u.
- Statement III: One mole of O₂ gas contains 6.022 × 10²³ oxygen atoms.
- Statement IV: The atomic mass of chlorine is 35.5 u because every chlorine atom has a mass of 35.5 u.
- ChemistrySome Basic Concepts of ChemistryReactions in solutions: molarity to molalityHard
- ChemistrySome Basic Concepts of ChemistryStoichiometry with purity and limiting reagentHard
- BiologyStructural Organisation in AnimalsExternal features of the frogEasy
- BiologyStructural Organisation in AnimalsCirculation, blood and lymphEasy
- BiologyStructural Organisation in AnimalsExternal features of the frogMediumQ317Match each structure of the frog with its role or description.Column IColumn IIP. Tympanum1. Receives sound signalsQ. Nictitating membrane2. Protects the eye while in waterR. Vocal sacs3. Sound-producing sacs present only in the maleS. Copulatory pad4. Present on the first digit of each fore limb of the male
- BiologyStructural Organisation in AnimalsRespirationMediumQ318The frog is amphibious and uses different respiratory surfaces. Evaluate the statements.
- Statement I: A frog does not drink water but absorbs it through its moist, mucus-covered skin.
- Statement II: On land a frog respires only through its skin, and the lungs and buccal cavity do not take part in gas exchange.
- BiologyStructural Organisation in AnimalsCirculation, blood and lymphMedium
- BiologyStructural Organisation in AnimalsExcretory systemMedium
- BiologyStructural Organisation in AnimalsDigestive systemMediumQ321The frog's alimentary canal is relatively short. Evaluate the Assertion and Reason.
- Assertion (A): The alimentary canal of a frog is short.
- Reason (R): The frog is a carnivore, and carnivores need only a short intestine.
- BiologyStructural Organisation in AnimalsNervous and endocrine controlHard
- BiologyStructural Organisation in AnimalsThe frog and its way of lifeEasy
- BiologyStructural Organisation in AnimalsCirculation, blood and lymphHard
- ChemistryStructure of AtomQuantum numbersEasy
- ChemistryStructure of AtomIsotopes and isobarsEasy
- ChemistryStructure of AtomPlanck's quantum theoryEasy
- ChemistryStructure of AtomBohr's model for hydrogen-like ionsMedium
- ChemistryStructure of AtomLine spectrum of hydrogenMedium
- ChemistryStructure of AtomDual behaviour of matterMedium
- ChemistryStructure of AtomStability of half-filled and fully filled orbitalsMediumQ331Chromium breaks the simple aufbau filling order in its ground state. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): The ground-state configuration of chromium (Z = 24) is [Ar]3d⁵4s¹ rather than [Ar]3d⁴4s².
- Reason (R): Exactly half-filled subshells are extra stable because of their symmetrical distribution and larger exchange energy.
- ChemistryStructure of AtomPhotoelectric effectMediumQ332Light of different frequencies and intensities falls on a clean metal surface. Which of the statements about the photoelectric effect are correct?
- Statement I: Increasing the intensity of light of a fixed frequency increases the kinetic energy of the ejected electrons.
- Statement II: Light below the threshold frequency ejects no electrons, however intense it is.
- Statement III: Above the threshold frequency, the number of electrons ejected per second rises with the intensity of light.
- Statement IV: Electrons are ejected as soon as suitable light strikes the metal, with no measurable time lag.
- ChemistryStructure of AtomShapes of atomic orbitalsHardQ333Match each hydrogen-like orbital in List I with the number of radial and angular nodes it has in List II.Column IColumn IIP. 3s1. 1 radial, 2 angularQ. 3p2. 0 radial, 3 angularR. 4d3. 2 radial, 0 angularS. 4f4. 1 radial, 1 angular
- ChemistryStructure of AtomHeisenberg's uncertainty principleHard
- PhysicsSystem of Particles and Rotational MotionCentre of massEasy
- PhysicsSystem of Particles and Rotational MotionTorque and angular momentumEasy
- PhysicsSystem of Particles and Rotational MotionMoment of inertiaEasyQ337A metre stick is easy to twirl about its own length but hard to swing about an axis through its middle at right angles. Judge the Assertion (A) and Reason (R) about the moment of inertia behind this.
- Assertion (A): The moment of inertia of a rigid body is not a fixed quantity; it depends on the axis of rotation.
- Reason (R): The moment of inertia depends on how the body's mass is distributed about the axis of rotation.
- PhysicsSystem of Particles and Rotational MotionKinematics of rotationMedium
- PhysicsSystem of Particles and Rotational MotionAngular momentum and its conservationMedium
- PhysicsSystem of Particles and Rotational MotionDynamics of rotationMedium
- PhysicsSystem of Particles and Rotational MotionMotion of the centre of massMediumQ341A shell flying along a parabolic path explodes in mid-air into several fragments. Air resistance is negligible. Consider statements I to IV and select the option that names every correct statement and no incorrect one.
- I. The centre of mass of the fragments continues along the original parabola until a fragment hits the ground.
- II. The total momentum of the fragments just after the explosion equals the shell's momentum just before.
- III. The total kinetic energy of the fragments just after the explosion equals the shell's kinetic energy just before.
- IV. The forces of the explosion can change the velocity of the centre of mass.
- PhysicsSystem of Particles and Rotational MotionPrinciple of moments and centre of gravityMedium
- PhysicsSystem of Particles and Rotational MotionVector product of two vectorsHard
- PhysicsSystem of Particles and Rotational MotionAngular momentum and its conservationHard
- BiologyThe Living WorldBinomial nomenclatureEasy
- BiologyThe Living WorldBinomial nomenclatureEasy
- BiologyThe Living WorldTaxonomic categories and hierarchyMedium
- BiologyThe Living WorldTaxonomic categories and hierarchyMediumQ348The hierarchy of taxonomic categories is built on shared characters. Read the two statements below and decide which are correct.
- Statement I: As one moves from species up to kingdom, the number of characters shared by the members of a category decreases.
- Statement II: A genus is a higher category than a family because it contains more kinds of organisms.
- BiologyThe Living WorldFamily, order and classHardQ349Match each organism with its family as given in the NCERT table of taxonomic categories.Column IColumn IIP. Housefly1. HominidaeQ. Man2. PoaceaeR. Mango3. MuscidaeS. Wheat4. Anacardiaceae
- BiologyThe Living WorldFamily, order and classMedium
- BiologyThe Living WorldPhylum, kingdom and the hierarchy tableMedium
- BiologyThe Living WorldClassification, taxonomy and systematicsMediumQ352Taxonomy and systematics are related but not identical. Evaluate the following Assertion and Reason.
- Assertion (A): Besides identification, nomenclature and classification, systematics also takes into account the evolutionary relationships between organisms.
- Reason (R): Linnaeus used 'Systema Naturae' as the title of his publication.
- BiologyThe Living WorldFamily, order and classHard
- BiologyThe Living WorldBinomial nomenclatureEasy
- PhysicsThermal Properties of MatterTemperature, heat and thermometer scalesEasy
- PhysicsThermal Properties of MatterRadiation and blackbody radiationEasy
- PhysicsThermal Properties of MatterAnomalous expansion of water and thermal stressEasyQ357A pond is cooled by winter air from 10 °C down to freezing. Which of the claims I to IV about how water's density behaves in this range are correct?
- I. Water has its maximum density at about 4 °C.
- II. When water is cooled from 4 °C to 0 °C, it contracts.
- III. A lake freezes from the top downward partly because water colder than 4 °C rises to the surface.
- IV. At 0 °C, ice is denser than liquid water.
- PhysicsThermal Properties of MatterThermal expansionMedium
- PhysicsThermal Properties of MatterCalorimetryMedium
- PhysicsThermal Properties of MatterConductionMedium
- PhysicsThermal Properties of MatterNewton's law of coolingHard
- PhysicsThermal Properties of MatterRadiation and blackbody radiationMedium
- PhysicsThermal Properties of MatterSpecific heat capacityMediumQ363Match each thermal quantity in List I with its SI unit in List II.Column IColumn IIP. Specific heat capacity1. J mol⁻¹ K⁻¹Q. Molar specific heat capacity2. J kg⁻¹R. Thermal conductivity3. J kg⁻¹ K⁻¹S. Latent heat4. W m⁻¹ K⁻¹
- PhysicsThermal Properties of MatterAnomalous expansion of water and thermal stressHard
- ChemistryThermodynamicsFirst law of thermodynamicsEasy
- ChemistryThermodynamicsEnthalpy: relation between ΔH and ΔUMedium
- ChemistryThermodynamicsHess's law of constant heat summationMedium
- ChemistryThermodynamicsGibbs energy and spontaneityEasy
- ChemistryThermodynamicsGibbs energy: threshold temperatureMedium
- ChemistryThermodynamicsWork: isothermal reversible expansion of an ideal gasMedium
- ChemistryThermodynamicsThermodynamic terms and enthalpyHardQ371Consider the following statements and choose the option that lists ALL the correct ones.
- Statement I: In an adiabatic process q = 0, so the change in internal energy equals the work done on the system (ΔU = wₐd).
- Statement II: The standard enthalpy of formation of O₃(g) is zero, because ozone is a form of the element oxygen.
- Statement III: Heat capacity of a sample is an extensive property, whereas its molar heat capacity is an intensive property.
- Statement IV: For N₂(g) + 3H₂(g) → 2NH₃(g) at constant temperature, with all gases behaving ideally, ΔH is less than ΔU.
- ChemistryThermodynamicsBond enthalpyHard
- ChemistryThermodynamicsInternal energy as a state functionEasy
- PhysicsThermodynamicsFirst law of thermodynamicsEasy
- PhysicsThermodynamicsCarnot engineEasy
- PhysicsThermodynamicsQuasi-static and isothermal processesEasyQ376An ideal gas expands slowly while kept in contact with a large reservoir at a fixed temperature. Assess the Assertion (A) about where the absorbed heat goes and the Reason (R) offered for it.
- Assertion (A): In an isothermal expansion of an ideal gas, all the heat absorbed by the gas is converted into work done by it.
- Reason (R): The internal energy of an ideal gas depends only on its temperature.
- PhysicsThermodynamicsSpecific heat capacityMedium
- PhysicsThermodynamicsQuasi-static and isothermal processesMedium
- PhysicsThermodynamicsAdiabatic processMedium
- PhysicsThermodynamicsIsochoric, isobaric and cyclic processesMedium
- PhysicsThermodynamicsSecond law of thermodynamicsMediumQ381Which of the claims I to IV about the second law of thermodynamics, Carnot engines and the nature of heat are correct?
- I. Heat cannot be drawn from a single reservoir and turned entirely into work as the only outcome of any process.
- II. Heat cannot pass from a colder body to a hotter body as the only outcome of any process.
- III. A Carnot engine working between two given temperatures can reach 100% efficiency if all friction is removed.
- IV. Heat is a thermodynamic state variable.
- PhysicsThermodynamicsAdiabatic processHard
- PhysicsThermodynamicsCarnot engineHard
- PhysicsUnits and MeasurementSignificant figuresEasy
- PhysicsUnits and MeasurementDimensions and dimensional formulaeEasy
- PhysicsUnits and MeasurementThe SI systemEasyQ386The SI system is built on seven base units. Which of the units named in statements I to IV are among those seven?
- I. The candela is an SI base unit.
- II. The radian is an SI base unit.
- III. The mole is an SI base unit.
- IV. The newton is an SI base unit.
- PhysicsUnits and MeasurementArithmetic with significant figuresMedium
- PhysicsUnits and MeasurementArithmetic with significant figuresMedium
- PhysicsUnits and MeasurementUncertainty in calculated resultsMedium
- PhysicsUnits and MeasurementDimensions and dimensional formulaeMediumQ390Match each quantity in List I with its dimensional formula in List II.Column IColumn IIP. Pressure1. [M L² T⁻³]Q. Power2. [M⁻¹ L³ T⁻²]R. Coefficient of viscosity3. [M L⁻¹ T⁻²]S. Gravitational constant G4. [M L⁻¹ T⁻¹]
- PhysicsUnits and MeasurementDeducing relations by dimensionsMediumQ391Can matching the dimensions of length, g and mass give the complete formula for a pendulum's period? Assess the Assertion (A) and Reason (R).
- Assertion (A): The formula T = 2π√(l/g) for a simple pendulum cannot be derived completely by dimensional analysis.
- Reason (R): Dimensional analysis gives no information about dimensionless constants.
- PhysicsUnits and MeasurementDeducing relations by dimensionsHard
- PhysicsUnits and MeasurementChecking dimensional consistencyHard
- PhysicsWavesDisplacement relation of a progressive waveEasy
- PhysicsWavesTransverse and longitudinal wavesEasyQ395Sound travels through air only as compressions and rarefactions. Judge whether the Assertion (A) about transverse mechanical waves in gases holds, and whether the Reason (R) about shearing stress explains it.
- Assertion (A): Transverse mechanical waves cannot travel through a gas.
- Reason (R): A gas cannot sustain a shearing stress.
- PhysicsWavesBeatsEasy
- PhysicsWavesWave speed on a stretched stringMedium
- PhysicsWavesHarmonics of strings and pipesMedium
- PhysicsWavesSpeed of sound: Newton and LaplaceMedium
- PhysicsWavesReflection and standing wavesMedium
- PhysicsWavesHarmonics of strings and pipesMediumQ401For each system in List I, the length is L and the wave speed is v. Match each with its frequency in List II.Column IColumn IIP. String fixed at both ends, fundamental1. 3v/4LQ. Pipe open at both ends, second harmonic2. v/4LR. Pipe closed at one end, fundamental3. v/2LS. Pipe closed at one end, first overtone4. v/L
- PhysicsWavesBeatsHard
- PhysicsWavesPrinciple of superpositionHard
- PhysicsWork, Energy and PowerWorkEasy
- PhysicsWork, Energy and PowerKinetic energyEasy
- PhysicsWork, Energy and PowerCollisionsEasyQ406Read the Assertion (A) and Reason (R) about a one-dimensional elastic collision between two bodies of different masses, and choose the correct option.
- Assertion (A): In an elastic collision, the kinetic energy of each colliding body is the same before and after the collision.
- Reason (R): In an elastic collision, the total kinetic energy of the system is the same before and after the collision.
- PhysicsWork, Energy and PowerPotential energy of a springMedium
- PhysicsWork, Energy and PowerPowerMedium
- PhysicsWork, Energy and PowerWork-energy theoremMedium
- PhysicsWork, Energy and PowerConservative forces and potential energyMediumQ410The four claims I to IV below concern conservative forces, potential energy and mechanical energy. Pick the option listing exactly the true claims.
- I. The work done by a conservative force on a body moving around any closed path is zero.
- II. A potential energy function can be defined only for a conservative force.
- III. The total mechanical energy of a body remains constant even while kinetic friction acts on it.
- IV. The work done by a conservative force depends only on the initial and final positions of the body.
- PhysicsWork, Energy and PowerCollisions (elastic collision in one dimension)Hard
- PhysicsWork, Energy and PowerConservation of mechanical energy with frictionHard
- ChemistryAlcohols, Phenols and EthersReactions of alcoholsEasy
- ChemistryAlcohols, Phenols and EthersAcidity of alcohols and phenolsEasy
- ChemistryAlcohols, Phenols and EthersPreparation of ethersEasy
- ChemistryAlcohols, Phenols and EthersAcidity of alcohols and phenolsMedium
- ChemistryAlcohols, Phenols and EthersOxidation of alcoholsMedium
- ChemistryAlcohols, Phenols and EthersReactions of phenolsMedium
- ChemistryAlcohols, Phenols and EthersReactions of phenolsMediumQ419Phenol reacts with bromine under different conditions to give different products. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): Phenol with bromine water gives a white precipitate of 2,4,6-tribromophenol.
- Reason (R): The –OH group activates the benzene ring so strongly that, in a polar solvent such as water, bromine substitutes at every free ortho and para position without a Lewis acid.
- ChemistryAlcohols, Phenols and EthersMethanol and ethanolMediumQ420Methanol and ethanol are the two simplest alcohols of commercial importance. Which of the statements below are correct?
- Statement I: Methanol is made by catalytic hydrogenation of carbon monoxide.
- Statement II: Fermentation of sugars to ethanol needs a steady supply of air.
- Statement III: Yeast enzymes convert sugars into ethanol during fermentation.
- Statement IV: Ethanol for industrial use is denatured by mixing it with an equal volume of water.
- ChemistryAlcohols, Phenols and EthersReactions of ethersHardQ421Each ether in List I is heated with one equivalent of HI. Match it with the pair of products in List II.Column IColumn IIP. CH₃–O–C₂H₅1. C₆H₅OH + CH₃IQ. C₆H₅–O–CH₃2. C₆H₅OH + C₂H₅IR. (CH₃)₃C–O–CH₃3. CH₃I + C₂H₅OHS. C₆H₅–O–C₂H₅4. (CH₃)₃CI + CH₃OH
- ChemistryAlcohols, Phenols and EthersPreparation of alcoholsHard
- ChemistryAldehydes, Ketones and Carboxylic AcidsNucleophilic additionEasy
- ChemistryAldehydes, Ketones and Carboxylic AcidsOxidation and testsEasy
- ChemistryAldehydes, Ketones and Carboxylic AcidsReduction of carbonyl compoundsEasy
- ChemistryAldehydes, Ketones and Carboxylic AcidsReactions due to α-hydrogenMedium
- ChemistryAldehydes, Ketones and Carboxylic AcidsOxidation and testsMedium
- ChemistryAldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acidsMedium
- ChemistryAldehydes, Ketones and Carboxylic AcidsReactions of carboxylic acidsMediumQ429Carboxylic acids with α-hydrogens can be halogenated using red phosphorus. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): Ethanoic acid treated with Cl₂ and red phosphorus gives chloroethanoic acid.
- Reason (R): Carboxylic acids have no α-hydrogens, so halogen must replace the hydrogen of the –COOH group.
- ChemistryAldehydes, Ketones and Carboxylic AcidsNucleophilic additionMediumQ430Aldehydes and ketones react with many nucleophiles. Which of the statements below are correct?
- Statement I: Ethanal reacts with HCN to give a cyanohydrin, which on hydrolysis gives 2-hydroxypropanoic acid.
- Statement II: Aldehydes and ketones react with hydroxylamine to form oximes.
- Statement III: Sodium hydrogensulphite reduces aldehydes to primary alcohols.
- Statement IV: Propanone gives a red-brown precipitate with Fehling's solution.
- ChemistryAldehydes, Ketones and Carboxylic AcidsPreparation of aldehydes and ketonesHardQ431Each conversion in List I needs a specific reagent or set of conditions to make an aldehyde or ketone. Match each conversion with the reagents in List II.Column IColumn IIP. Benzoyl chloride → benzaldehyde1. CrO₂Cl₂ in CS₂, then H₃O⁺Q. Ethanenitrile → ethanal2. H₂, Pd on BaSO₄R. Toluene → benzaldehyde3. CH₃COCl, anhydrous AlCl₃S. Benzene → acetophenone4. SnCl₂ + HCl, then H₃O⁺
- ChemistryAldehydes, Ketones and Carboxylic AcidsCannizzaro reactionHard
- PhysicsAlternating CurrentRMS values and average powerEasy
- PhysicsAlternating CurrentAC through a capacitorEasy
- PhysicsAlternating CurrentSeries LCR circuit and impedanceMedium
- PhysicsAlternating CurrentResonanceMedium
- PhysicsAlternating CurrentAC through an inductorMediumQ437An ideal inductor (no resistance) is connected across an ac source. Read the Assertion (A) and Reason (R) and choose the correct option.
- Assertion (A): The average power delivered to the inductor over a complete cycle is zero.
- Reason (R): In an ideal inductor the current lags the voltage by π/2, so the power factor cos φ is zero.
- PhysicsAlternating CurrentPower factorMedium
- PhysicsAlternating CurrentResonanceHard
- PhysicsAlternating CurrentTransformersMedium
- PhysicsAlternating CurrentPhasorsEasyQ441Consider the following statements about ac circuits with a single ideal element. Which of them are correct?
- I. In a pure resistor, current and voltage are in phase.
- II. In a pure inductor, current leads the voltage by π/2.
- III. In a pure capacitor, current leads the voltage by π/2.
- IV. The length of a phasor represents the peak value of the quantity, and its projection on the vertical axis gives the instantaneous value.
- PhysicsAlternating CurrentSeries LCR circuit and impedanceHard
- ChemistryAminesReactions of aminesEasy
- ChemistryAminesPreparation of aminesEasy
- ChemistryAminesPreparation of aminesEasy
- ChemistryAminesBasic character of aminesMedium
- ChemistryAminesBasic character of aminesMedium
- ChemistryAminesElectrophilic substitution in anilineMedium
- ChemistryAminesHinsberg's testMediumQ449Benzenesulphonyl chloride (Hinsberg's reagent) is used to tell apart primary, secondary and tertiary amines. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): The sulphonamide formed from a secondary amine is insoluble in alkali.
- Reason (R): The sulphonamide from a secondary amine has no hydrogen on nitrogen that is acidic enough to be removed by alkali.
- ChemistryAminesDiazonium saltsMediumQ450Arenediazonium salts are prepared from aromatic amines and used widely in synthesis. Which of the statements below are correct?
- Statement I: Benzenediazonium chloride is prepared from aniline with NaNO₂ and HCl at 273–278 K.
- Statement II: The arenediazonium ion is stabilised by resonance with the benzene ring.
- Statement III: Primary aliphatic amines form diazonium salts that are stable at room temperature.
- Statement IV: Benzenediazonium chloride is usually dried and stored as a solid for later use.
- ChemistryAminesReactions of diazonium saltsHardQ451Benzenediazonium chloride is treated with each reagent in List I. Match each reagent with the main organic product in List II.Column IColumn IIP. Warm water1. BenzeneQ. Hypophosphorous acid (H₃PO₂)2. p-HydroxyazobenzeneR. Fluoroboric acid, then heat3. PhenolS. Phenol in alkaline solution4. Fluorobenzene
- ChemistryAminesDiazonium salts in synthesisHard
- PhysicsAtomsLine spectra of the hydrogen atomEasy
- PhysicsAtomsAlpha-particle scattering and Rutherford's nuclear modelEasy
- PhysicsAtomsEnergy levels of the hydrogen atomMedium
- PhysicsAtomsBohr model of the hydrogen atomMedium
- PhysicsAtomsAlpha-particle scattering and Rutherford's nuclear modelMedium
- PhysicsAtomsBohr model of the hydrogen atomMediumQ458A student tries to use Bohr's formula to predict the spectrum of a neutral helium atom, which has two electrons, and the predictions disagree with experiment. Read the Assertion (A) and Reason (R) and choose the correct option.
- Assertion (A): The Bohr model cannot give accurate energy levels for a neutral helium atom.
- Reason (R): The Bohr model considers only the attraction between one electron and the nucleus and ignores the repulsion between electrons.
- PhysicsAtomsLine spectra of the hydrogen atomMedium
- PhysicsAtomsDe Broglie's explanation of Bohr's second postulateHard
- PhysicsAtomsBohr model of the hydrogen atomHard
- PhysicsAtomsBohr model of the hydrogen atomEasyQ462Bohr proposed his postulates to rescue the nuclear atom from collapsing. Which of the following claims about an electron in a hydrogen atom agree with those postulates?
- I. The electron can occupy only certain orbits, called stationary states.
- II. In those orbits its angular momentum is a whole-number multiple of h/2π.
- III. While moving in a stationary orbit it radiates energy continuously, because it is accelerating.
- IV. It emits a photon of energy hν = Eᵢ − E_f when it jumps from a higher to a lower orbit.
- BiologyBiodiversity and ConservationWhat biodiversity isEasy
- BiologyBiodiversity and ConservationWhat biodiversity isEasy
- BiologyBiodiversity and ConservationThe evil quartetEasy
- BiologyBiodiversity and ConservationSpecies-area relationshipMediumQ466Two statements about the species-area relationship are given. Select the correct option.
- Statement I: When species richness is plotted against area on logarithmic axes, the relationship is a straight line.
- Statement II: For comparisons of areas within a single region, the slope Z usually lies between 0.6 and 1.2.
- BiologyBiodiversity and ConservationThe evil quartetMediumQ467Match the concept in Column I with the example or description in Column II, and choose the option with all matches correct.Column IColumn IIP. Alien species invasion1. Introduction of a predatory fish into a lake, followed by the loss of more than 200 cichlid speciesQ. Co-extinction2. Loss of an obligately linked species, such as a parasite, when its host disappearsR. Genetic diversity3. India's more than 50,000 distinct strains of a single cereal cropS. Ex situ conservation4. Storing gametes of threatened species at very low temperature
- BiologyBiodiversity and ConservationIn situ and ex situ conservationMedium
- BiologyBiodiversity and ConservationPatterns of biodiversityMediumQ469Read the Assertion and the Reason about the latitudinal gradient of species richness, then choose the correct option.
- Assertion (A): Species diversity generally decreases as we move from the equator towards the poles.
- Reason (R): Tropical regions receive more solar energy, which contributes to higher productivity and, in turn, indirectly to greater species diversity.
- BiologyBiodiversity and ConservationDiversity and stabilityMedium
- BiologyBiodiversity and ConservationLoss of biodiversityHard
- BiologyBiodiversity and ConservationBiodiversity hotspotsHard
- ChemistryBiomoleculesDisaccharidesEasy
- ChemistryBiomoleculesVitaminsEasy
- ChemistryBiomoleculesNucleic acidsEasy
- ChemistryBiomoleculesStructure of glucoseMedium
- ChemistryBiomoleculesRing forms of glucoseMedium
- ChemistryBiomoleculesProteinsMedium
- ChemistryBiomoleculesRing forms of glucose and fructoseMediumQ479Fructose is a ketose, yet it is classed as a reducing sugar. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): Fructose reduces Tollens' reagent.
- Reason (R): Ketones are always oxidised more easily than aldehydes.
- ChemistryBiomoleculesAmino acidsMediumQ480Amino acids carry both an amino group and a carboxyl group. Which of the statements below are correct?
- Statement I: In aqueous solution an amino acid exists mainly as a dipolar ion called a zwitterion.
- Statement II: Amino acids are amphoteric, reacting with both acids and bases.
- Statement III: Amino acids that the body cannot synthesise are called essential amino acids.
- Statement IV: Glycine is optically active.
- ChemistryBiomoleculesVitaminsHardQ481Deficiency of each vitamin listed in List I causes a characteristic disease. Match each vitamin with the deficiency disease in List II.Column IColumn IIP. Vitamin B₁ (thiamine)1. ConvulsionsQ. Vitamin B₆ (pyridoxine)2. Increased blood clotting timeR. Vitamin E3. Beri-beriS. Vitamin K4. Increased fragility of red blood cells and muscular weakness
- ChemistryBiomoleculesNucleic acidsHard
- BiologyBiotechnology and its ApplicationsGenetically modified crops: Bt cottonEasy
- BiologyBiotechnology and its ApplicationsTissue culture — Meristem cultureEasy
- BiologyBiotechnology and its ApplicationsGene therapyMedium
- BiologyBiotechnology and its ApplicationsGenetically engineered insulinMediumQ486Two statements about production of human insulin by recombinant DNA technology are given. Choose the correct option.
- Statement I: In the first recombinant insulin, the A and B chains were produced separately and then joined by disulphide bonds.
- Statement II: Mature human insulin retains the C peptide of pro-insulin joining the A and B chains.
- BiologyBiotechnology and its ApplicationsRNA interferenceMediumQ487Study the Assertion and the Reason on engineered resistance to a root parasite in tobacco, then choose the correct option.
- Assertion (A): Transgenic tobacco plants making nematode-specific RNA are protected against Meloidogyne incognita.
- Reason (R): The RNAs produced in the plant form double-stranded RNA that silences a specific mRNA of the parasite.
- BiologyBiotechnology and its ApplicationsApplications of biotechnologyMediumQ488Match each application or product of biotechnology in Column I with its description in Column II.Column IColumn IIP. Pomato1. Somatic hybrid obtained by protoplast fusionQ. Golden rice2. Rice enriched with vitamin AR. Rosie3. Transgenic cow whose milk contained a human proteinS. ELISA4. Diagnosis based on antigen-antibody interaction
- BiologyBiotechnology and its ApplicationsTransgenic animalsEasy
- BiologyBiotechnology and its ApplicationsEthical issues and biopiracyMedium
- BiologyBiotechnology and its ApplicationsGene therapyHard
- BiologyBiotechnology and its ApplicationsRNA interferenceHard
- BiologyBiotechnology: Principles and ProcessesRestriction enzymesEasy
- BiologyBiotechnology: Principles and ProcessesGel electrophoresisEasy
- BiologyBiotechnology: Principles and ProcessesPolymerase chain reactionEasy
- BiologyBiotechnology: Principles and ProcessesRestriction enzymesMediumQ496Two statements about restriction enzymes are given. Select the correct option.
- Statement I: Cutting the two strands at positions slightly away from the centre of a palindromic site leaves single-stranded sticky ends.
- Statement II: Exonucleases cut DNA at specific internal recognition sites.
- BiologyBiotechnology: Principles and ProcessesCloning vectorsMediumQ497Read the Assertion and the Reason about selecting recombinants with a plasmid vector in which foreign DNA is inserted at the BamH I site, then choose the correct option.
- Assertion (A): Host cells carrying the recombinant plasmid fail to grow on tetracycline but grow on ampicillin.
- Reason (R): The insert at the BamH I site inactivates the tetracycline resistance gene.
- BiologyBiotechnology: Principles and ProcessesIsolation of the genetic materialMediumQ498Match each agent used while isolating DNA in Column I with its specific use in Column II.Column IColumn IIP. Lysozyme1. Breaks plant cellsQ. Cellulase2. Breaks bacterial cellsR. Chitinase3. Breaks fungal cellsS. Ribonuclease4. Removes RNA from the preparation
- BiologyBiotechnology: Principles and ProcessesCloning vectorsMedium
- BiologyBiotechnology: Principles and ProcessesIntroducing DNA into the hostMedium
- BiologyBiotechnology: Principles and ProcessesPolymerase chain reactionHard
- BiologyBiotechnology: Principles and ProcessesRestriction enzymesHard
- ChemistryChemical KineticsMolecularity of a reactionEasyQ503Which statement about the molecularity of a reaction is correct?
- AMolecularity is defined for an elementary reaction and is the number of species that collide simultaneously in that step.
- BReactions with molecularity greater than three are common.
- CMolecularity of a complex reaction equals the sum of stoichiometric coefficients of the overall equation.
- DMolecularity can be zero or fractional, like order.
- ChemistryChemical KineticsRate of a chemical reactionEasy
- ChemistryChemical KineticsOrder of a reactionEasy
- ChemistryChemical KineticsZero order reactionsMedium
- ChemistryChemical KineticsRate law and rate constantMedium
- ChemistryChemical KineticsOrder of a reaction and pseudo first orderMediumQ508Match each type of reaction in List I with an example of it in List II.Column IColumn IIP. Zero order reaction1. Acid hydrolysis of ethyl acetate in a large excess of waterQ. First order reaction2. Decomposition of NH₃ on a hot platinum surface at high pressureR. Pseudo first order reaction3. Decay of a radioactive nucleusS. Termolecular elementary reaction4. 2NO(g) + O₂(g) → 2NO₂(g)
- ChemistryChemical KineticsEffect of catalystMediumQ509A catalyst is added to a reversible reaction at constant temperature. Which of the statements below are correct?
- Statement I: The catalyst lowers the activation energy of the reaction.
- Statement II: The Gibbs energy change of the reaction is unaffected by the catalyst.
- Statement III: The catalyst increases the equilibrium constant of the forward reaction.
- Statement IV: The catalyst provides an alternative pathway, often through a temporary intermediate.
- ChemistryChemical KineticsTemperature dependence of rateMediumQ510Most reactions speed up noticeably when they are warmed. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): For many chemical reactions the rate constant nearly doubles for a 10 K rise in temperature.
- Reason (R): A rise in temperature increases the fraction of molecules whose kinetic energy equals or exceeds the activation energy.
- ChemistryChemical KineticsArrhenius equationHard
- ChemistryChemical KineticsIntegrated rate equation: first orderHard
- ChemistryCoordination CompoundsNomenclature of coordination compoundsEasy
- ChemistryCoordination CompoundsLigands and denticityEasy
- ChemistryCoordination CompoundsWerner's theoryEasy
- ChemistryCoordination CompoundsCrystal field theoryMedium
- ChemistryCoordination CompoundsValence bond theoryMedium
- ChemistryCoordination CompoundsOptical isomerismMedium
- ChemistryCoordination CompoundsColour in coordination compoundsMediumQ519Aqueous solutions of many titanium(III) salts have a characteristic colour. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): The complex [Ti(H₂O)₆]³⁺ is violet.
- Reason (R): [Ti(H₂O)₆]³⁺ absorbs light in the blue-green region, which promotes its single d electron from a t₂g to an e_g orbital.
- ChemistryCoordination CompoundsBonding in metal carbonylsMediumQ520Metal carbonyls contain only carbon monoxide as ligand. Which of the statements below are correct?
- Statement I: Ni(CO)₄ is tetrahedral.
- Statement II: Fe(CO)₅ is trigonal bipyramidal.
- Statement III: The metal–carbon bond in a metal carbonyl has both σ and π character.
- Statement IV: Cr(CO)₆ is square planar.
- ChemistryCoordination CompoundsStructural isomerismHardQ521Each pair in List I is a pair of structural isomers. Match each pair with its type of isomerism in List II.Column IColumn IIP. [Co(NH₃)₄(NO₂)Cl]Cl and [Co(NH₃)₄(ONO)Cl]Cl1. Ionisation isomerismQ. [Cu(NH₃)₄][PtCl₄] and [Pt(NH₃)₄][CuCl₄]2. Hydrate (solvate) isomerismR. [Pt(NH₃)₄Cl₂]Br₂ and [Pt(NH₃)₄Br₂]Cl₂3. Linkage isomerismS. [Cr(H₂O)₄Cl₂]Cl·2H₂O and [Cr(H₂O)₅Cl]Cl₂·H₂O4. Coordination isomerism
- ChemistryCoordination CompoundsCrystal field theoryHard
- PhysicsCurrent ElectricityDrift of electrons and the origin of resistivityEasy
- PhysicsCurrent ElectricityOhm's law and resistivityEasy
- PhysicsCurrent ElectricityTemperature dependence of resistivityEasy
- PhysicsCurrent ElectricityCells, emf, internal resistanceMedium
- PhysicsCurrent ElectricityWheatstone bridgeMedium
- PhysicsCurrent ElectricityCells in series and in parallelHard
- PhysicsCurrent ElectricityElectrical energy, powerMedium
- PhysicsCurrent ElectricityElectric currents in conductorsMediumQ530Read the Assertion (A) and Reason (R) about a copper wire whose cross-section tapers steadily from 2 mm² at one end to 1 mm² at the other end, carrying a steady current. Choose the correct option.
- Assertion (A): The drift speed of the free electrons at the 1 mm² end is twice their drift speed at the 2 mm² end.
- Reason (R): In steady state the same current I crosses every cross-section of the wire, and I = neAv_d with the free-electron density n the same throughout the copper.
- PhysicsCurrent ElectricityKirchhoff's rulesHard
- PhysicsDual Nature of Radiation and MatterParticle nature of light: the photonEasy
- PhysicsDual Nature of Radiation and MatterEffect of intensity on photocurrentEasy
- PhysicsDual Nature of Radiation and MatterEinstein's photoelectric equationMedium
- PhysicsDual Nature of Radiation and MatterParticle nature of light: the photonMedium
- PhysicsDual Nature of Radiation and MatterWave nature of matterMedium
- PhysicsDual Nature of Radiation and MatterEinstein's photoelectric equationHard
- PhysicsDual Nature of Radiation and MatterParticle nature of light: the photonMediumQ538A beam of monochromatic ultraviolet light is passed through a region with a strong electric field and then into a gas. Which of these statements about its photons are correct?
- I. All photons in the beam carry the same energy.
- II. The photons are deflected by the electric field because they carry charge.
- III. The total number of photons is conserved when the beam is absorbed by the gas.
- IV. Each photon carries momentum hν/c.
- PhysicsDual Nature of Radiation and MatterPhotoelectric effect and wave theory of lightMediumQ539A photocell is lit by extremely faint light above its threshold frequency, and a current starts to flow at once. Read the Assertion (A) and Reason (R) and choose the correct option.
- Assertion (A): Photoelectrons appear with no measurable delay, even when the light is very dim.
- Reason (R): Light energy arrives in discrete photons, and a single photon can give its whole energy hν to one electron at once.
- PhysicsDual Nature of Radiation and MatterWave nature of matterHard
- PhysicsDual Nature of Radiation and MatterElectron emissionEasyQ541Match each item in List I with what supplies or describes it in List II.Column IColumn IIP. Thermionic emission1. Light of frequency above the thresholdQ. Field emission2. h/eR. Photoelectric emission3. Heating the metalS. Slope of the stopping potential versus frequency graph4. A very strong electric field (about 10⁸ V m⁻¹)
- BiologyEcosystemProductivityEasy
- BiologyEcosystemEcological pyramidsEasy
- BiologyEcosystemFood chains and food websEasy
- BiologyEcosystemDecompositionMediumQ545Statements I and II below concern the rate and products of decomposition of detritus in an ecosystem. Which of the options is correct?
- Statement I: Humus is highly resistant to microbial action and undergoes decomposition at an extremely slow rate.
- Statement II: Detritus rich in lignin decomposes faster than detritus rich in nitrogen and water-soluble sugars.
- BiologyEcosystemEcological pyramidsMediumQ546Read the Assertion and the Reason about biomass pyramids in a sea and choose the correct option.
- Assertion (A): The pyramid of biomass in a sea is generally inverted.
- Reason (R): The standing crop of phytoplankton at any time is small compared with the biomass of the zooplankton and fishes it supports.
- BiologyEcosystemDecompositionMediumQ547Match the step of decomposition in Column I with the process in Column II.Column IColumn IIP. Fragmentation1. Enzymatic degradation of detritus into simpler substancesQ. Leaching2. Breaking of detritus into smaller particles by detritivoresR. Catabolism3. Movement of water-soluble inorganic nutrients into deeper soil layersS. Humification4. Formation of dark, resistant amorphous matter
- BiologyEcosystemTrophic levels and the 10 per cent lawMedium
- BiologyEcosystemProductivityMedium
- BiologyEcosystemProductivityHard
- BiologyEcosystemProductivityHard
- PhysicsElectric Charges and FieldsBasic properties of electric chargeEasy
- PhysicsElectric Charges and FieldsCoulomb's lawEasy
- PhysicsElectric Charges and FieldsElectric fieldMedium
- PhysicsElectric Charges and FieldsElectric dipoleMedium
- PhysicsElectric Charges and FieldsElectric field linesMediumQ556A student draws the field-line pattern for a pair of point charges and writes four rules beside it. Which of the rules are correct?
- I. Two field lines can never cross each other.
- II. Electrostatic field lines can form closed loops.
- III. Field lines start on positive charges and end on negative charges, or go to or come from infinity.
- IV. Where field lines are farther apart, the field is stronger.
- PhysicsElectric Charges and FieldsGauss's lawMedium
- PhysicsElectric Charges and FieldsApplications of Gauss's lawEasy
- PhysicsElectric Charges and FieldsElectric fieldHard
- PhysicsElectric Charges and FieldsDipole in a uniform external fieldMediumQ560An electric dipole sits in a uniform electric field with its moment tilted at an angle to the field. Read the Assertion (A) and Reason (R) about it and choose the correct option.
- Assertion (A): An electric dipole placed in a uniform electric field, with its moment at an angle to the field, experiences no net force but does experience a torque.
- Reason (R): The forces qE and −qE on the two charges are equal and opposite, but they act at different points, forming a couple.
- PhysicsElectric Charges and FieldsDipole in a non-uniform fieldHard
- ChemistryElectrochemistryGalvanic cellsEasy
- ChemistryElectrochemistryVariation of conductivity with concentrationEasy
- ChemistryElectrochemistryBatteriesEasyQ564A car battery and a watch battery work on different chemistry. Which of the statements below are correct?
- Statement I: When a lead storage battery discharges, PbSO₄ forms at both electrodes.
- Statement II: In a lead storage battery the anode is lead and the cathode is a lead grid packed with PbO₂.
- Statement III: The concentration of sulphuric acid in a lead storage battery rises during discharge.
- Statement IV: A mercury cell gives a nearly constant potential of about 1.35 V over its life.
- ChemistryElectrochemistryCell potential and Gibbs energyMedium
- ChemistryElectrochemistryNernst equationMedium
- ChemistryElectrochemistryQuantitative aspects of electrolysisMedium
- ChemistryElectrochemistryProducts of electrolysisMediumQ568The products of electrolysis depend on the electrolyte, its state and the electrodes. Match each electrolysis in List I with an observation in List II.Column IColumn IIP. Aqueous NaCl, inert electrodes1. Copper dissolves at the anode and deposits at the cathodeQ. Molten NaCl, inert electrodes2. Sodium metal forms at the cathodeR. Dilute H₂SO₄, platinum electrodes3. Oxygen is evolved at the anodeS. Aqueous CuSO₄, copper electrodes4. Hydrogen at the cathode and chlorine at the anode
- ChemistryElectrochemistryMolar conductivity of weak electrolytesMediumQ569Acetic acid and KCl behave very differently when their solutions are diluted. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): The molar conductivity of acetic acid increases steeply as its solution approaches infinite dilution.
- Reason (R): On dilution the degree of dissociation of a weak electrolyte increases, producing more ions per mole.
- ChemistryElectrochemistryMeasurement of conductivityHard
- ChemistryElectrochemistryKohlrausch's law and degree of dissociationHard
- PhysicsElectromagnetic InductionFaraday's law of inductionEasy
- PhysicsElectromagnetic InductionLenz's law and conservation of energyEasy
- PhysicsElectromagnetic InductionMotional emfMedium
- PhysicsElectromagnetic InductionRotating rods and wheelsMedium
- PhysicsElectromagnetic InductionSelf-inductance and magnetic energyMedium
- PhysicsElectromagnetic InductionMutual inductanceMediumQ577A coil-design worksheet lists four claims about self and mutual inductance. Pick the option that names every correct claim.
- I. The SI unit of inductance, the henry, equals one weber per ampere.
- II. For two long coaxial solenoids, the mutual inductance of the first with respect to the second equals that of the second with respect to the first.
- III. The self-induced emf in a coil opposes a rise in current but does not oppose a fall in current.
- IV. Inserting a soft-iron core into a solenoid increases its self-inductance.
- PhysicsElectromagnetic InductionAC generatorHard
- PhysicsElectromagnetic InductionLenz's law and conservation of energyMediumQ579A bar magnet is dropped, N pole first, along the axis of a fixed horizontal copper ring. Read the Assertion (A) and Reason (R) and choose the correct option.
- Assertion (A): While the magnet is above the ring and approaching it, its acceleration is less than g.
- Reason (R): The current induced in the ring opposes the change in magnetic flux that produces it.
- PhysicsElectromagnetic InductionMagnetic fluxHard
- PhysicsElectromagnetic InductionFaraday's law of inductionEasyQ581Match each expression in List I with what it gives in List II.Column IColumn IIP. −N dΦ/dt1. Energy stored in an inductor carrying a currentQ. Blv2. Emf across a straight rod moving at right angles to a fieldR. ½LI²3. Peak emf of a coil rotating in a uniform fieldS. NBAω4. Emf induced in a coil by a changing flux
- PhysicsElectromagnetic WavesNature of electromagnetic wavesEasy
- PhysicsElectromagnetic WavesThe electromagnetic spectrumEasy
- PhysicsElectromagnetic WavesDisplacement currentMedium
- PhysicsElectromagnetic WavesSpeed in vacuum and in mediaMedium
- PhysicsElectromagnetic WavesNature of electromagnetic wavesEasy
- PhysicsElectromagnetic WavesThe electromagnetic spectrumMediumQ587Match each band of the electromagnetic spectrum in List I with a matching fact about it in List II.Column IColumn IIP. Microwaves1. Produced when high-energy electrons strike a metal targetQ. Infrared2. Largely absorbed by the ozone layer in the upper atmosphereR. Ultraviolet3. Produced by klystrons, magnetrons and Gunn diodes; used in radarS. X-rays4. Used in TV remote controls and in physical therapy
- PhysicsElectromagnetic WavesSources of electromagnetic wavesMediumQ588Four claims are made about how electromagnetic waves are produced and how they travel. Which of them hold?
- I. Electromagnetic waves carry energy from one place to another.
- II. An accelerated charge radiates em waves, but a charge moving with constant velocity does not.
- III. In vacuum, higher-frequency em waves travel faster than lower-frequency ones.
- IV. Electromagnetic waves need a material medium to travel.
- PhysicsElectromagnetic WavesSpeed in vacuum and in mediaHard
- PhysicsElectromagnetic WavesThe charging capacitor puzzleMediumQ590A parallel plate capacitor is being charged by a steady current. Read the Assertion (A) and Reason (R) and choose the correct option.
- Assertion (A): A magnetic field exists in the region between the plates, even though no charge flows across the gap.
- Reason (R): The electric flux between the plates is changing, and a changing electric flux produces a magnetic field just as a conduction current does.
- PhysicsElectromagnetic WavesNature of electromagnetic wavesHard
- PhysicsElectrostatic Potential and CapacitancePotential due to a system of chargesEasy
- PhysicsElectrostatic Potential and CapacitanceElectrostatic potentialEasy
- PhysicsElectrostatic Potential and CapacitanceElectrostatics of conductorsMediumQ594A charged metal body is in electrostatic equilibrium. Which of the following statements about it are correct?
- I. The electric field everywhere inside the metal is zero.
- II. The electric potential everywhere inside the metal is zero.
- III. Just outside the surface, the field is normal to the surface and equals σ/ε₀, where σ is the local surface charge density.
- IV. Any excess charge resides on the surface.
- PhysicsElectrostatic Potential and CapacitancePotential energy of a system of chargesMedium
- PhysicsElectrostatic Potential and CapacitanceEffect of dielectric on capacitanceMedium
- PhysicsElectrostatic Potential and CapacitanceCombination of capacitorsMedium
- PhysicsElectrostatic Potential and CapacitanceEnergy stored in a capacitorHard
- PhysicsElectrostatic Potential and CapacitanceEquipotential surfacesMediumQ599A small test charge is carried around a circle centred on an isolated point charge. Read the Assertion (A) and Reason (R) about this and choose the correct option.
- Assertion (A): No work is done by the electric field on the test charge as it moves along the circle.
- Reason (R): The electric field at every point is perpendicular to the equipotential surface through that point.
- PhysicsElectrostatic Potential and CapacitancePotential energy in an external fieldHard
- PhysicsElectrostatic Potential and CapacitancePotential due to a dipoleEasyQ601Match each quantity in List I with how it depends on the distance r from the source in List II (for the dipole, r is much larger than its length).Column IColumn IIP. Potential of a point charge1. ∝ 1/rQ. Field of a short dipole2. ∝ 1/r²R. Potential inside a charged spherical conducting shell3. ∝ 1/r³S. Potential of a short dipole at a point on its axis4. Independent of r
- BiologyEvolutionChemical evolution and the first lifeEasy
- BiologyEvolutionHomology and analogyEasy
- BiologyEvolutionAdaptive radiationEasy
- BiologyEvolutionHardy-Weinberg principleMediumQ605Statements I and II below concern the Hardy-Weinberg principle for a gene with two alleles. Choose the correct option.
- Statement I: In a population in Hardy-Weinberg equilibrium, the frequency of heterozygotes is 2pq.
- Statement II: Strong gene flow into a population leaves its allele frequencies unchanged.
- BiologyEvolutionNatural selection in actionMediumQ606Read the Assertion and the Reason about the moth populations observed in industrial England, and choose the correct option.
- Assertion (A): After industrialisation, the proportion of dark-winged moths rose in polluted regions of England.
- Reason (R): Pollution killed the lichens and darkened tree trunks, so dark moths were better camouflaged against predators.
- BiologyEvolutionMutation and selectionMediumQ607Match the process in Column I with its effect in Column II.Column IColumn IIP. Stabilising selection1. Individuals with the mean trait value are favoured, narrowing the distributionQ. Directional selection2. Individuals at one extreme are favoured, shifting the mean to that sideR. Disruptive selection3. Individuals at both extremes are favoured, giving two peaksS. Genetic drift4. Random change in allele frequency, especially in small populations
- BiologyEvolutionHardy-Weinberg principleMedium
- BiologyEvolutionOrigin and evolution of manMedium
- BiologyEvolutionOrigin and evolution of manHard
- BiologyEvolutionHardy-Weinberg principleHard
- ChemistryHaloalkanes and HaloarenesSN2 mechanismEasy
- ChemistryHaloalkanes and HaloarenesPreparation of haloalkanesEasy
- ChemistryHaloalkanes and HaloarenesReactions of haloarenesEasy
- ChemistryHaloalkanes and HaloarenesSN1 mechanismMedium
- ChemistryHaloalkanes and HaloarenesElimination reactionsMedium
- ChemistryHaloalkanes and HaloarenesReactions with metalsMedium
- ChemistryHaloalkanes and HaloarenesStereochemistry of substitutionMediumQ618Optically active (R)-2-bromobutane is treated with hydroxide ion under conditions that favour the SN2 mechanism. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): The butan-2-ol formed has the inverted configuration.
- Reason (R): In SN2, the nucleophile attacks from the side opposite the leaving group, and as the new bond forms the other three groups on that carbon swing over to the other side.
- ChemistryHaloalkanes and HaloarenesPolyhalogen compoundsMediumQ619Several polyhalogen compounds were once widely used but are now restricted. Which of the statements below about them are correct?
- Statement I: Chloroform is slowly oxidised by air in the presence of light to a poisonous gas, phosgene.
- Statement II: Carbon tetrachloride released into the air helps rebuild the ozone layer.
- Statement III: DDT is not metabolised quickly by animals and builds up in their fatty tissue.
- Statement IV: Freons are quickly decomposed near the earth's surface and rarely reach the stratosphere.
- ChemistryHaloalkanes and HaloarenesNamed reactionsHardQ620Four named reactions start from a halogen compound or a diazonium salt. Match each reagent set in List I with the organic product it gives in List II.Column IColumn IIP. Bromobenzene + bromomethane + Na in dry ether1. ChlorobenzeneQ. Bromobenzene + Na in dry ether2. TolueneR. Benzenediazonium chloride + Cu₂Cl₂/HCl3. IodoethaneS. Bromoethane + NaI in dry acetone4. Biphenyl
- ChemistryHaloalkanes and HaloarenesReactions of haloarenesHard
- BiologyHuman Health and DiseaseCommon diseases in humansEasy
- BiologyHuman Health and DiseaseCommon diseases in humansEasy
- BiologyHuman Health and DiseaseAcquired immunityEasy
- BiologyHuman Health and DiseaseActive and passive immunityMediumQ625Statements I and II below compare ways of acquiring immunity. Which of the options is correct?
- Statement I: The antibodies a newborn receives through colostrum provide passive immunity.
- Statement II: Immunity produced by vaccination with weakened pathogens is passive immunity.
- BiologyHuman Health and DiseaseAIDSMediumQ626Read the Assertion and the Reason about the course of HIV infection and choose the correct option.
- Assertion (A): In an HIV-infected person, the number of helper T lymphocytes progressively decreases.
- Reason (R): The virus enters helper T lymphocytes, replicates in them and produces progeny viruses that attack more cells.
- BiologyHuman Health and DiseaseCommon diseases in humansMediumQ627Match the pathogen in Column I with the feature in Column II.Column IColumn IIP. Wuchereria bancrofti1. Dry, scaly lesions on skin, nails and scalpQ. Entamoeba histolytica2. Chronic inflammation of lymphatic vessels of the lower limbsR. Trichophyton3. Infection of the large intestine with mucus and blood in stoolsS. Plasmodium falciparum4. Most serious, malignant form of malaria
- BiologyHuman Health and DiseaseLymphoid organsMedium
- BiologyHuman Health and DiseaseCancerMedium
- BiologyHuman Health and DiseaseDrugs, alcohol and addictionHard
- BiologyHuman Health and DiseaseAIDSHard
- BiologyHuman ReproductionThe Male Reproductive SystemEasy
- BiologyHuman ReproductionThe Female Reproductive SystemEasy
- BiologyHuman ReproductionGametogenesis — SpermatogenesisEasy
- BiologyHuman ReproductionGametogenesis — OogenesisMedium
- BiologyHuman ReproductionMenstrual CycleMedium
- BiologyHuman ReproductionMenstrual CycleMediumQ637Consider the following statements about the phases of the human menstrual cycle. Which one/ones is/are correct?
- Statement I: During the follicular (proliferative) phase, the growing Graafian follicle secretes estrogen, which causes the uterine endometrium to regenerate and thicken.
- Statement II: During the luteal (secretory) phase, the corpus luteum secretes progesterone, which further prepares the endometrium to receive and support a possible embryo.
- BiologyHuman ReproductionFertilisation and ImplantationMediumQ638During fertilisation in humans, a sperm reaches the ovum in the ampullary region of the fallopian tube. Regarding the events that follow, which statement(s) is/are correct?
- Statement I: Entry of the sperm into the secondary oocyte induces it to complete its second meiotic division, which is unequal and forms a second polar body and a haploid ovum (ootid).
- Statement II: A sperm coming in contact with the zona pellucida layer of the ovum induces changes in the membrane that block the entry of additional sperms, ensuring that only one sperm fertilises an ovum.
- BiologyHuman ReproductionPregnancy and Embryonic DevelopmentHardQ639The placenta acts as an endocrine tissue during pregnancy. Read the Assertion and Reason below about human chorionic gonadotropin (hCG), judge whether each is true, and select the option that correctly relates them.
- Assertion (A): The presence of human chorionic gonadotropin (hCG) in a woman's body can be used as an indicator of pregnancy.
- Reason (R): hCG is secreted by the placenta and, like hPL and relaxin, is produced in women only during pregnancy.
- BiologyHuman ReproductionParturition and LactationMedium
- BiologyHuman ReproductionHormones of the Reproductive SystemHardQ641Match the hormone in Column I with its correct source and a key role or feature in Column II, then choose the option in which all four matches are correct.Column IColumn IIP. Follicle-stimulating hormone (FSH)1. Secreted by the placenta; produced in women only during pregnancyQ. Luteinising hormone (LH)2. Secreted by the anterior pituitary; acts on Leydig cells to stimulate androgen secretion and, at its mid-cycle surge, induces ovulationR. Human chorionic gonadotropin (hCG)3. Secreted by the anterior pituitary; acts on Sertoli cells and stimulates growth of ovarian folliclesS. Prolactin4. Secreted by the anterior pituitary; regulates growth of the mammary glands and formation of milk in them
- PhysicsMagnetism and MatterDipole in a uniform fieldEasy
- PhysicsMagnetism and MatterThe electrostatic analogMedium
- PhysicsMagnetism and MatterDiamagnetismEasy
- PhysicsMagnetism and MatterMagnetisation and magnetic intensityMedium
- PhysicsMagnetism and MatterParamagnetismMediumQ646Samples of a paramagnet, a diamagnet, a superconductor and a ferromagnet are studied in a lab. Which of the following observations about such materials are correct?
- I. The susceptibility of a paramagnetic material depends on its temperature.
- II. Diamagnetic materials have a small negative susceptibility.
- III. A superconductor expels a magnetic field completely and behaves as a perfect diamagnet, with μᵣ = 0.
- IV. A ferromagnetic material remains ferromagnetic however much it is heated.
- PhysicsMagnetism and MatterGauss's law for magnetismMediumQ647Read the Assertion (A) and Reason (R) about the magnetic field lines of a bar magnet and choose the correct option.
- Assertion (A): Unlike electric field lines, the magnetic field lines of a bar magnet have no starting point or end point; they do not stop at the poles but continue through the magnet from S to N.
- Reason (R): Isolated magnetic poles (monopoles) are not known to exist, so a magnetic field has no sources or sinks on which its lines could begin or end.
- PhysicsMagnetism and MatterMagnetisation and magnetic intensityHard
- PhysicsMagnetism and MatterThe electrostatic analogHard
- PhysicsMagnetism and MatterGauss's law for magnetismMedium
- PhysicsMagnetism and MatterFerromagnetismEasyQ651Match each material in List I with the description of its magnetic susceptibility χ in List II.Column IColumn IIP. Bismuth1. χ = −1Q. Gadolinium2. Small and negativeR. Aluminium3. Small and positiveS. A superconductor (below its critical temperature)4. Very large and positive (χ ≫ 1)
- BiologyMicrobes in Human WelfareMicrobes in household productsEasy
- BiologyMicrobes in Human WelfareMicrobes in industrial productsEasy
- BiologyMicrobes in Human WelfareBiogasEasy
- BiologyMicrobes in Human WelfareFermented beveragesMediumQ655Statements I and II below concern the production of alcoholic drinks by yeast. Which of the options is correct?
- Statement I: Wine and beer are produced without distillation of the fermented broth.
- Statement II: Whisky, brandy and rum are produced without distillation.
- BiologyMicrobes in Human WelfareSewage treatmentMediumQ656Read the Assertion and the Reason about the secondary stage of sewage treatment, and choose the correct option.
- Assertion (A): The biological treatment of effluent in aeration tanks greatly reduces its BOD.
- Reason (R): Vigorous aeration lets aerobic microbes grow as flocs that consume much of the organic matter.
- BiologyMicrobes in Human WelfareMicrobes in industrial productsMediumQ657Match the organic acid in Column I with the microbe that produces it in Column II.Column IColumn IIP. Citric acid1. Aspergillus nigerQ. Acetic acid2. Acetobacter acetiR. Butyric acid3. Clostridium butylicumS. Lactic acid4. Lactobacillus
- BiologyMicrobes in Human WelfareBioactive moleculesMedium
- BiologyMicrobes in Human WelfareSewage treatmentMedium
- BiologyMicrobes in Human WelfareBiofertilisersHardQ660Which of the following pairings of a microbe and its role is INCORRECT?
- AGlomus: forms mycorrhiza and helps the plant absorb phosphorus
- BRhizobium: forms symbiotic nitrogen-fixing nodules on roots of legumes
- CNostoc: a cyanobacterium that fixes atmospheric nitrogen and serves as a biofertiliser in paddy fields
- DAzospirillum: symbiotic nitrogen fixer in the root nodules of legumes
- BiologyMicrobes in Human WelfareBiocontrol agentsHard
- BiologyMolecular Basis of InheritanceStructure of DNAEasy
- BiologyMolecular Basis of InheritanceStructure of DNAEasy
- BiologyMolecular Basis of InheritanceThe Search for Genetic MaterialMediumQ664In Frederick Griffith's experiment, mice injected with a mixture of heat-killed S-strain (smooth, virulent) Streptococcus pneumoniae and live R-strain (rough, non-virulent) bacteria died, and living S-strain bacteria were recovered from the dead mice. What did Griffith conclude from this result?
- AHeat-killed S-strain bacteria had regained virulence on their own after injection
- BSome 'transforming principle' from the dead S-strain had been transferred to and had genetically transformed the living R-strain
- CThe R-strain bacteria mutated spontaneously into the S-strain due to the stress of infection
- DHeat-killed S-strain and live R-strain bacteria acted together as separate agents to kill the mouse without any exchange of material
- BiologyMolecular Basis of InheritanceThe Genetic Material is DNA — Hershey-Chase experimentMediumQ665Hershey and Chase used bacteriophages labelled separately with radioactive sulphur (³⁵S) and radioactive phosphorus (³²P) to infect E. coli, then agitated the mixture in a blender before centrifuging. Which observation confirmed that DNA, and not protein, is the genetic material?
- A³⁵S entered the bacterial cells and was passed on to the phage progeny
- BBacteria infected with ³²P-labelled phages became radioactive, while bacteria infected with ³⁵S-labelled phages did not
- CBoth ³⁵S and ³²P entered the bacterial cells in equal amounts, showing that protein and DNA act together as genetic material
- DNeither isotope entered the bacterial cells, showing that phage infection occurs without any genetic material entering the host
- BiologyMolecular Basis of InheritanceReplication — Meselson-Stahl experimentHard
- BiologyMolecular Basis of InheritanceReplication — enzymes and Okazaki fragmentsMediumQ667DNA polymerase can add nucleotides only in the 5'→3' direction. Because the two template strands at a replication fork are antiparallel, one new strand (the lagging strand) cannot be synthesised continuously. How is the lagging strand actually made?
- AAs one long continuous strand synthesised in the 3'→5' direction
- BAs short fragments called Okazaki fragments, synthesised discontinuously and later joined by DNA ligase
- CBy RNA polymerase, which replaces DNA polymerase on the lagging strand only
- DIt is not synthesised during replication; only the leading strand is copied, and the lagging strand is retained from the parent unchanged
- BiologyMolecular Basis of InheritanceTranscription UnitHardQ668RNA polymerase transcribes only one strand of a DNA segment, even though the DNA is double-stranded. Read the Assertion and Reason below on why transcription is restricted to a single strand, and choose the correct option.
- Assertion (A): Within a transcription unit, only one of the two DNA strands (the template strand) is transcribed into RNA.
- Reason (R): If both strands were transcribed at the same time, the two RNA molecules formed would be complementary to each other and would pair into a double-stranded RNA that could not be translated into protein.
- BiologyMolecular Basis of InheritanceGenetic CodeMediumQ669Two statements about the genetic code are given below. Which one/ones is/are correct?
- Statement I: The genetic code is degenerate, meaning a single amino acid can be specified by more than one codon.
- Statement II: The genetic code is ambiguous, meaning a single codon can specify more than one amino acid.
- BiologyMolecular Basis of InheritanceRegulation of Gene Expression — the lac operonMediumQ670Consider the following statements about regulation of the lac operon in E. coli. Which one/ones is/are correct?
- Statement I: In the absence of an inducer such as lactose, the repressor protein made by the i gene binds to the operator region and prevents RNA polymerase from transcribing the structural genes.
- Statement II: The inducer (lactose or allolactose) switches the operon on by binding directly to the operator DNA and physically pushing the repressor off it.
- BiologyMolecular Basis of InheritanceDNA FingerprintingEasy
- PhysicsMoving Charges and MagnetismMotion in a magnetic fieldEasy
- PhysicsMoving Charges and MagnetismMagnetic field and the Lorentz forceEasy
- PhysicsMoving Charges and MagnetismMotion in a magnetic fieldMedium
- PhysicsMoving Charges and MagnetismBiot–Savart lawMedium
- PhysicsMoving Charges and MagnetismForce between parallel currentsMedium
- PhysicsMoving Charges and MagnetismThe moving coil galvanometerMedium
- PhysicsMoving Charges and MagnetismTorque on a current loopHard
- PhysicsMoving Charges and MagnetismMotion in a magnetic fieldMediumQ679Electrons of different speeds are injected perpendicular to the same uniform magnetic field. Read the Assertion (A) and Reason (R) about their motion and choose the correct option.
- Assertion (A): All the electrons take the same time to complete one revolution.
- Reason (R): The radius of the circle is proportional to the speed, so the circumference and the speed increase in the same proportion.
- PhysicsMoving Charges and MagnetismAmpere's circuital lawEasyQ680Match each current arrangement in List I with the magnitude of the magnetic field it produces (List II).Column IColumn IIP. Long straight wire, at distance r1. μ₀nIQ. Centre of a circular coil of N turns and radius R2. μ₀IR²/2x³R. Inside a long solenoid with n turns per unit length3. μ₀I/2πrS. On the axis of a single loop of radius R, at distance x ≫ R4. μ₀NI/2R
- PhysicsMoving Charges and MagnetismAmpere's circuital lawHard
- PhysicsNucleiSize of the nucleusEasy
- PhysicsNucleiAtomic masses and composition of nucleusEasy
- PhysicsNucleiMass–energy and nuclear binding energyMedium
- PhysicsNucleiNuclear energy: fission and fusionMedium
- PhysicsNucleiNuclear forceMediumQ686A nucleus holds many protons together despite their mutual repulsion. Which of these statements about the force responsible are correct?
- I. It is much stronger than the Coulomb force at distances of a few femtometres.
- II. It acts almost equally between proton–proton, neutron–neutron and proton–neutron pairs.
- III. It falls off as the inverse square of distance, like gravity.
- IV. It becomes strongly repulsive when nucleons come closer than about 0.8 fm.
- PhysicsNucleiMass–energy and nuclear binding energyMediumQ687A uranium-235 nucleus captures a slow neutron and splits into two medium-mass fragments with a large release of energy. Read the Assertion (A) and Reason (R) and choose the correct option.
- Assertion (A): Splitting a nucleus of mass number about 240 into two nuclei of mass number about 120 releases energy.
- Reason (R): The binding energy per nucleon is higher for nuclei near A ≈ 120 than for nuclei near A ≈ 240.
- PhysicsNucleiNuclear energy: fission and fusionHard
- PhysicsNucleiNuclear energy: fission and fusionHard
- PhysicsNucleiMass–energy and nuclear binding energyEasyQ690Match each nuclear quantity in List I with its approximate value in List II.Column IColumn IIP. Energy equivalent of 1 atomic mass unit1. About 8.75 MeVQ. Constant R₀ in R = R₀A^(1/3)2. About 200 MeVR. Maximum binding energy per nucleon (near A ≈ 56)3. 931.5 MeVS. Energy released in one fission of ²³⁵U4. About 1.2 fm
- PhysicsNucleiSize of the nucleusMedium
- BiologyOrganisms and PopulationsPopulation growthEasy
- BiologyOrganisms and PopulationsPopulation interactionsEasy
- BiologyOrganisms and PopulationsPopulation attributesEasy
- BiologyOrganisms and PopulationsPopulation growthMediumQ695Statements I and II below compare logistic and exponential models of population growth. Which of the options is correct?
- Statement I: In logistic growth, the growth rate declines as the population size approaches the carrying capacity.
- Statement II: In exponential growth, the growth rate declines as the population size increases.
- BiologyOrganisms and PopulationsCompetitionMediumQ696Read the Assertion and the Reason about several closely related warbler species foraging on the same tree, and choose the correct option.
- Assertion (A): Several closely related warbler species can live together on the same tree.
- Reason (R): Each species forages in a different part of the tree, so they partition the resources.
- BiologyOrganisms and PopulationsPopulation interactionsMediumQ697Match the interaction in Column I with the example in Column II.Column IColumn IIP. Mutualism1. A fungus and an alga living together as a lichenQ. Commensalism2. Barnacles growing on the back of a whaleR. Parasitism3. A liver fluke needing a snail and a fish as intermediate hostsS. Competition4. One barnacle species excluding another from part of a rocky coast
- BiologyOrganisms and PopulationsPopulation growthMedium
- BiologyOrganisms and PopulationsPopulation growthMedium
- BiologyOrganisms and PopulationsPredationHard
- BiologyOrganisms and PopulationsCompetitionHard
- BiologyPrinciples of Inheritance and VariationMendel's Laws of InheritanceEasy
- BiologyPrinciples of Inheritance and VariationMendel's Laws of InheritanceEasy
- BiologyPrinciples of Inheritance and VariationInheritance of Two GenesMedium
- BiologyPrinciples of Inheritance and VariationDeviations from Mendelism — Incomplete DominanceMedium
- BiologyPrinciples of Inheritance and VariationDeviations from Mendelism — Multiple AllelesMedium
- BiologyPrinciples of Inheritance and VariationChromosomal Theory of Inheritance — Linkage and RecombinationMediumQ707With respect to linkage and recombination, which statement(s) is/are correct?
- Statement I: Genes located far apart on the same chromosome show a higher recombination frequency than genes located close together on that chromosome.
- Statement II: Linked genes located on the same chromosome always show 100% recombination in every generation, identical to genes located on different chromosomes.
- BiologyPrinciples of Inheritance and VariationSex DeterminationMediumQ708Regarding sex-determination mechanisms, which statement(s) is/are correct?
- Statement I: In humans, sex is determined by whether a sperm carrying an X or a Y chromosome fertilises the egg, making the male the heterogametic sex.
- Statement II: In birds, males are heterogametic (ZW) and females are homogametic (ZZ), the mirror image of the human XY system.
- BiologyPrinciples of Inheritance and VariationPedigree Analysis and Mendelian DisordersHardQ709Haemophilia is a sex-linked disorder in which a single protein of the blood-clotting cascade is affected, so even a small cut can bleed without stopping. Evaluate the Assertion about how often the disorder appears in each sex and the Reason offered for it.
- Assertion (A): Haemophilia is much more commonly observed in human males than in human females.
- Reason (R): Haemophilia is caused by a recessive allele located on the X chromosome, so a male needs only one copy of the allele to be affected, whereas a female needs two copies.
- BiologyPrinciples of Inheritance and VariationMendelian Disorders — Sickle-cell AnaemiaEasy
- BiologyPrinciples of Inheritance and VariationChromosomal DisordersHardQ711Match the condition in Column I with its correct karyotype/chromosomal basis in Column II, then choose the option in which all four matches are correct.Column IColumn IIP. Down syndrome1. 47, XXY (an extra X chromosome in a male)Q. Klinefelter syndrome2. 45, X0 (a single X chromosome, no second sex chromosome)R. Turner syndrome3. Trisomy 21 — 47 chromosomes, an extra copy of chromosome 21S. Normal human female karyotype4. 46, XX
- PhysicsRay Optics and Optical InstrumentsReflection of light by spherical mirrorsEasy
- PhysicsRay Optics and Optical InstrumentsTotal internal reflectionEasy
- PhysicsRay Optics and Optical InstrumentsPower of a lens and combination of thin lenses in contactEasy
- PhysicsRay Optics and Optical InstrumentsRefraction: apparent depthMedium
- PhysicsRay Optics and Optical InstrumentsRefraction by a lens: lensmaker's formulaMedium
- PhysicsRay Optics and Optical InstrumentsRefraction through a prismMedium
- PhysicsRay Optics and Optical InstrumentsOptical instruments: telescopeMediumQ718Consider the following statements about a refracting astronomical telescope used in normal adjustment (final image at infinity). Which of the statements are correct?
- I. The distance between the objective and the eyepiece equals f_o + f_e.
- II. Its magnifying power is f_o/f_e, so the objective should have a much larger focal length than the eyepiece.
- III. Increasing the diameter of the objective (keeping its focal length unchanged) increases the magnifying power.
- IV. The final image seen through the telescope is erect.
- PhysicsRay Optics and Optical InstrumentsOptical instruments: compound microscopeHard
- PhysicsRay Optics and Optical InstrumentsRefraction by lenses: image by successive lensesHard
- BiologyReproductive HealthBirth control: oral pillsEasy
- BiologyReproductive HealthBirth control: IUDsEasy
- BiologyReproductive HealthMedical termination of pregnancyEasy
- BiologyReproductive HealthBirth control: sterilisationMediumQ724Statements I and II below describe surgical sterilisation. Which of the options is correct?
- Statement I: Vasectomy prevents pregnancy by cutting or tying the vas deferens, blocking the transport of sperms.
- Statement II: Tubectomy prevents pregnancy by stopping the formation of ova in the ovary.
- BiologyReproductive HealthBirth control: natural methodsMediumQ725Read the Assertion and the Reason about lactational amenorrhea as a contraceptive, and choose the correct option.
- Assertion (A): Lactational amenorrhea is an effective contraceptive method only for a limited period after parturition, up to about six months.
- Reason (R): During intense lactation, ovulation and the menstrual cycle do not occur.
- BiologyReproductive HealthAssisted reproductive technologiesMediumQ726Match each assisted reproductive technique in Column I with the way it is carried out in Column II.Column IColumn IIP. ZIFT1. Zygote or early embryo with up to 8 blastomeres transferred into the fallopian tubeQ. IUT2. Embryo with more than 8 blastomeres transferred into the uterusR. GIFT3. Ovum from a donor transferred into the fallopian tube of a female who cannot produce ovaS. ICSI4. A sperm injected directly into an ovum in the laboratory
- BiologyReproductive HealthBirth control: natural methodsMedium
- BiologyReproductive HealthSexually transmitted infectionsMedium
- BiologyReproductive HealthBirth control: pillsHard
- BiologyReproductive HealthBirth control: IUDsHard
- PhysicsSemiconductor Electronics: Materials, Devices and Simple CircuitsExtrinsic semiconductorEasy
- PhysicsSemiconductor Electronics: Materials, Devices and Simple CircuitsClassification of metals, conductors and semiconductorsEasy
- PhysicsSemiconductor Electronics: Materials, Devices and Simple CircuitsExtrinsic semiconductorMedium
- PhysicsSemiconductor Electronics: Materials, Devices and Simple CircuitsApplication of junction diode as a rectifierEasy
- PhysicsSemiconductor Electronics: Materials, Devices and Simple CircuitsSemiconductor diodeMediumQ735A silicon p-n junction with built-in barrier V₀ is studied first unbiased, then with an external voltage V in forward bias and in reverse bias. Which of these statements are correct?
- I. Under forward bias the effective barrier height falls to V₀ − V.
- II. Under reverse bias the depletion region becomes wider.
- III. The small reverse current is due to majority carriers diffusing across the junction.
- IV. With no bias, the diffusion current and drift current across the junction are equal and opposite.
- PhysicsSemiconductor Electronics: Materials, Devices and Simple CircuitsSemiconductor diodeMedium
- PhysicsSemiconductor Electronics: Materials, Devices and Simple CircuitsSemiconductor diodeHard
- PhysicsSemiconductor Electronics: Materials, Devices and Simple CircuitsExtrinsic semiconductorHard
- PhysicsSemiconductor Electronics: Materials, Devices and Simple CircuitsIntrinsic semiconductorMediumQ739Match each element in List I with its role or property in List II.Column IColumn IIP. Phosphorus added to silicon1. Acceptor impurityQ. Boron added to silicon2. Band gap about 0 eV, behaves as a metalR. Pure germanium3. Donor impurityS. Tin (group 14, like silicon)4. Intrinsic semiconductor, band gap about 0.7 eV
- PhysicsSemiconductor Electronics: Materials, Devices and Simple CircuitsIntrinsic semiconductorMediumQ740A strip of pure silicon is warmed from room temperature to 100 °C while its resistance is monitored, and the resistance drops. Read the Assertion (A) and Reason (R) and choose the correct option.
- Assertion (A): The electrical conductivity of an intrinsic semiconductor increases as its temperature rises.
- Reason (R): Thermal energy breaks more covalent bonds, lifting electrons into the conduction band and creating an equal number of holes.
- BiologySexual Reproduction in Flowering PlantsThe embryo sacEasy
- BiologySexual Reproduction in Flowering PlantsMicrosporangiumEasy
- BiologySexual Reproduction in Flowering PlantsDouble fertilisationEasy
- BiologySexual Reproduction in Flowering PlantsPollinationMediumQ744Statements I and II below concern pollination types and flower sexuality. Which of the options is correct?
- Statement I: Geitonogamy is functionally cross-pollination but genetically similar to autogamy.
- Statement II: In monoecious plants such as maize, geitonogamy is prevented.
- BiologySexual Reproduction in Flowering PlantsPollinationMediumQ745Read the Assertion and the Reason about cleistogamous flowers, and choose the correct option.
- Assertion (A): Cleistogamous flowers produce assured seed-set even in the absence of pollinators.
- Reason (R): These flowers never open, and pollen from the anthers pollinates the stigma within the closed flower.
- BiologySexual Reproduction in Flowering PlantsSeed and fruitMediumQ746Match each seed or fruit term in Column I with the correct description in Column II, then choose the right combination.Column IColumn IIP. Perisperm1. Persistent remnant of the nucellus in the seedQ. Coconut water2. Free-nuclear endospermR. Apple3. False fruit in which thalamus contributesS. Banana4. Fruit that develops without fertilisation
- BiologySexual Reproduction in Flowering PlantsPollinationMedium
- BiologySexual Reproduction in Flowering PlantsThe pollen grainMedium
- BiologySexual Reproduction in Flowering PlantsDouble fertilisationHard
- BiologySexual Reproduction in Flowering PlantsThe pollen grainHard
- ChemistrySolutionsHenry's lawEasy
- ChemistrySolutionsAbnormal molar masses and van't Hoff factorEasy
- ChemistrySolutionsIdeal and non-ideal solutionsEasy
- ChemistrySolutionsVapour pressure of liquid–liquid solutionsMedium
- ChemistrySolutionsElevation of boiling pointMedium
- ChemistrySolutionsOsmotic pressureMedium
- ChemistrySolutionsAbnormal molar masses and van't Hoff factorMediumQ757Two aqueous solutions, 0.1 M NaCl and 0.1 M glucose, are compared at the same temperature. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): The NaCl solution has a higher osmotic pressure than the glucose solution.
- Reason (R): Each formula unit of NaCl dissociates into two ions, so the NaCl solution holds nearly twice as many solute particles.
- ChemistrySolutionsIdeal and non-ideal solutionsMediumQ758Binary liquid mixtures can behave ideally or deviate from Raoult's law. Which of the statements below are correct?
- Statement I: A solution showing positive deviation from Raoult's law absorbs heat on mixing (ΔmixH > 0).
- Statement II: A large positive deviation leads to a minimum boiling azeotrope.
- Statement III: A solution showing negative deviation expands on mixing (ΔmixV > 0).
- Statement IV: An ideal solution obeys Raoult's law over the entire range of composition.
- ChemistrySolutionsAbnormal molar masses and van't Hoff factorHard
- ChemistrySolutionsVapour pressure of liquid–liquid solutionsHard
- ChemistryThe d- and f-Block ElementsFormation of coloured ionsEasy
- ChemistryThe d- and f-Block ElementsElectronic configuration of d-block elementsEasy
- ChemistryThe d- and f-Block ElementsLanthanoid contractionEasy
- ChemistryThe d- and f-Block ElementsMagnetic propertiesMedium
- ChemistryThe d- and f-Block ElementsTrends in M²⁺/M standard electrode potentialsMedium
- ChemistryThe d- and f-Block ElementsPotassium dichromateMedium
- ChemistryThe d- and f-Block ElementsTrends in M³⁺/M²⁺ electrode potentialsMediumQ767Cr²⁺ and Mn³⁺ both have a d⁴ configuration, but they behave in opposite ways. Judge the Assertion (A) and the Reason (R) below.
- Assertion (A): Cr²⁺ is a strong reducing agent.
- Reason (R): On losing an electron Cr²⁺ becomes Cr³⁺, which has the stable configuration with a half-filled t₂g level (d³).
- ChemistryThe d- and f-Block ElementsThe lanthanoidsMediumQ768Most lanthanoids prefer the +3 state, but a few show +2 or +4. Which of the statements below are correct?
- Statement I: The principal oxidation state of the lanthanoids is +3.
- Statement II: Ce⁴⁺ in aqueous solution is a strong oxidising agent.
- Statement III: The lanthanoid contraction arises because 4f electrons shield one another poorly from the nuclear charge.
- Statement IV: Eu²⁺ is a strong oxidising agent.
- ChemistryThe d- and f-Block ElementsPotassium permanganate and dichromateHardQ769The product of an oxidation by permanganate or dichromate depends on the substrate and the medium. Match each reaction in List I with the oxidation product of the substrate in List II.Column IColumn IIP. Acidified KMnO₄ with oxalate ion1. Sn⁴⁺Q. KMnO₄ in neutral or faintly alkaline solution with iodide ion2. SR. Acidified K₂Cr₂O₇ with H₂S3. IO₃⁻S. Acidified K₂Cr₂O₇ with Sn²⁺4. CO₂
- ChemistryThe d- and f-Block ElementsThe actinoidsHard
- PhysicsWave OpticsYoung's double-slit experimentEasy
- PhysicsWave OpticsRefraction of a plane waveEasy
- PhysicsWave OpticsMalus' law and crossed polaroidsMedium
- PhysicsWave OpticsCoherent and incoherent additionMedium
- PhysicsWave OpticsDiffraction at a single slitMedium
- PhysicsWave OpticsYoung's double-slit experimentHard
- PhysicsWave OpticsCoherent and incoherent additionEasyQ777Two separate sodium lamps of equal brightness illuminate the same region of a screen. Read the Assertion (A) and Reason (R) and choose the correct option.
- Assertion (A): No steady interference fringes are seen on the screen.
- Reason (R): The phase of light from each independent source changes randomly in times of about 10⁻¹⁰ s, so there is no constant phase difference between the two.
- PhysicsWave OpticsMalus' law and crossed polaroidsHard
- PhysicsWave OpticsWavefronts and Huygens principleMediumQ779A teacher writes four claims on the board about wavefronts and Huygens' construction. Which of the claims are correct?
- I. A point source in a uniform medium produces spherical wavefronts.
- II. Rays are perpendicular to the wavefronts, along the direction of energy flow.
- III. A thin convex lens turns a plane wavefront into a spherical wavefront converging to its focus.
- IV. The energy carried by a light wave depends on its speed, not on its amplitude.
- PhysicsWave OpticsInterference, diffraction and seeing themMediumQ780Match each situation in List I with the condition or result that describes it in List II (n is a whole number, a the slit width, I₀ the intensity from one source).Column IColumn IIP. Bright fringe in Young's experiment1. Path difference (n + ½)λQ. Dark fringe in Young's experiment2. Uniform intensity 2I₀, no fringesR. Minimum in single-slit diffraction3. a sin θ = nλ, with n ≠ 0S. Two incoherent sources of equal intensity4. Path difference nλ
© 2026 Lumi (Aikolumi Software Pvt Ltd). Original questions written by Lumi. Free to view and practise on this site. Copying, republishing or reproducing these questions, options or solutions, in print or online, is not permitted.