- A. the functional role played by the organism where it lives
- B. the range of temperature that the organism needs to live
- C. the physical space where an organism lives
- D. all the biological factors in the organism’s environment
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Log in to view solution →- A. O3
- B. SO2
- C. CO2
- D. CO
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Log in to view solution →- A. Oxygen
- B. Fe
- C. Cl
- D. Carbon
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Log in to view solution →- A. 22nd April
- B. 16th September
- C. 21st April
- D. 5th June
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Log in to view solution →- A. Upright pyramid of biomass
- B. Upright pyramid of numbers
- C. Pyramid of energy
- D. Inverted pyramid of biomass
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Log in to view solution →- A. Number of individuals entering a habitat
- B. Number of individuals leaving the habitat
- C. Birth rate
- D. Death rate
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Log in to view solution →- A. Parthenogenesis
- B. Parthenocarpy
- C. Mitotic divisions
- D. Meiotic divisions
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Log in to view solution →- A. Virus
- B. Plant
- C. Bacterium
- D. Fungus
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Log in to view solution →- A. Francois Jacob and Jacques Monod – Lac operon
- B. Matthew Meselson and F. Stahl – Pisum sativum
- C. Alfred Hershey and Martha Chase – TMV
- D. Alec Jeffreys – Streptococcus pneumoniae
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Log in to view solution →- A. Francois Jacob and Jacques Monod – Lac operon
- B. Matthew Meselson and F. Stahl – Pisum sativum
- C. Alfred Hershey and Martha Chase – TMV
- D. Alec Jeffreys – Streptococcus pneumoniae
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Log in to view solution →- A. Sporopollenin
- B. Oil content
- C. Cellulosic intine
- D. Pollenkitt
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Log in to view solution →- A. Sporopollenin
- B. Oil content
- C. Cellulosic intine
- D. Pollenkitt
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Log in to view solution →- A. T.H. Morgan : Linkage
- B. XO type sex determination : Grasshopper
- C. ABO blood grouping : Co-dominance
- D. Starch synthesis in pea : Multiple alleles
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Log in to view solution →- A. T.H. Morgan : Linkage
- B. XO type sex determination : Grasshopper
- C. ABO blood grouping : Co-dominance
- D. Starch synthesis in pea : Multiple alleles
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Log in to view solution →- A. Papaya
- B. Mango
- C. Jackfruit
- D. Bamboo species
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Log in to view solution →- A. Papaya
- B. Mango
- C. Jackfruit
- D. Bamboo species
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Log in to view solution →- A. Transduction was discovered by S. Altman.
- B. Spliceosomes take part in translation.
- C. Punnett square was developed by a British scientist.
- D. Franklin Stahl coined the term linkage.
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Log in to view solution →- A. Transduction was discovered by S. Altman.
- B. Spliceosomes take part in translation.
- C. Punnett square was developed by a British scientist.
- D. Franklin Stahl coined the term "linkage".
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Log in to view solution →- A. Activation of amino acid
- B. Respiration in bacteria
- C. Formation of secretory vesicles
- D. Fatty acid breakdown
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Log in to view solution →- A. Activation of amino acid
- B. Respiration in bacteria
- C. Formation of secretory vesicles
- D. Fatty acid breakdown
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Log in to view solution →- A. Zygotene
- B. Diakinesis
- C. Diplotene
- D. Pachytene
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Log in to view solution →- A. CO2 concentration
- B. O2 concentration
- C. Light
- D. Temperature
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Log in to view solution →- A. Barrel shaped
- B. Rectangular
- C. Kidney shaped
- D. Dumb-bell shaped
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Log in to view solution →- A. Oxygen
- B. NADPH
- C. NADH
- D. ATP
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Log in to view solution →- A. It is a site for active ribosomal RNA synthesis.
- B. It takes part in spindle formation.
- C. It is a membrane-bound structure.
- D. Larger nucleoli are present in dividing cells.
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Log in to view solution →- A. Oscillatoria
- B. Nostoc
- C. Mycobacterium
- D. Saccharomyces
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Log in to view solution →- A. carbonyl and hydroxyl
- B. carbonyl and phosphate
- C. carbonyl and methyl
- D. hydroxyl and methyl
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Log in to view solution →- A. a-iii, b-iv, c-i, d-ii
- B. a-ii, b-iv, c-iii, d-i
- C. a-iii, b-ii, c-i, d-iv
- D. a-i, b-iv, c-iii, d-ii
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Log in to view solution →- A. Unicellular organism – Chlorella
- B. Gemma cups – Marchantia
- C. Biflagellate zoospores – Brown algae
- D. Uniflagellate gametes – Polysiphonia
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Log in to view solution →- A. Saccharomyces
- B. Agaricus
- C. Alternaria
- D. Neurospora
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Log in to view solution →- A. Pinus
- B. Mango
- C. Cycas
- D. Mustard
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Log in to view solution →- A. pBR 322
- B. \u03bb phage
- C. Ti plasmid
- D. Retrovirus
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Log in to view solution →- A. Basmati
- B. Lerma Rojo
- C. Sharbati Sonora
- D. Co-667
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Log in to view solution →- A. Bioexploitation
- B. Biodegradation
- C. Biopiracy
- D. Bio-infringement
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Log in to view solution →- A. G. Mendel – Transformation
- B. T.H. Morgan – Transduction
- C. F2 × recessive parent – Dihybrid cross
- D. Ribozyme – Nucleic acid
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Log in to view solution →- A. G. Mendel – Transformation
- B. T.H. Morgan – Transduction
- C. F2 × recessive parent – Dihybrid cross
- D. Ribozyme – Nucleic acid
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Log in to view solution →- A. Denaturation, Annealing, Extension
- B. Denaturation, Extension, Annealing
- C. Annealing, Extension, Denaturation
- D. Extension, Denaturation, Annealing
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Log in to view solution →- A. Denaturation, Annealing, Extension
- B. Denaturation, Extension, Annealing
- C. Annealing, Extension, Denaturation
- D. Extension, Denaturation, Annealing
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Log in to view solution →- A. Genetic Engineering Appraisal Committee (GEAC)
- B. Research Committee on Genetic Manipulation (RCGM)
- C. Council for Scientific and Industrial Research (CSIR)
- D. Indian Council of Medical Research (ICMR)
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Log in to view solution →- A. Genetic Engineering Appraisal Committee (GEAC)
- B. Research Committee on Genetic Manipulation (RCGM)
- C. Council for Scientific and Industrial Research (CSIR)
- D. Indian Council of Medical Research (ICMR)
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Log in to view solution →- A. It is the final electron acceptor for anaerobic respiration.
- B. It is a nucleotide source for ATP synthesis.
- C. It functions as an electron carrier.
- D. It functions as an enzyme.
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Log in to view solution →- A. It is the final electron acceptor for anaerobic respiration.
- B. It is a nucleotide source for ATP synthesis.
- C. It functions as an electron carrier.
- D. It functions as an enzyme.
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Log in to view solution →- A. Viola
- B. Banana
- C. Yucca
- D. Hydrilla
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Log in to view solution →- A. –160°C
- B. –196°C
- C. –80°C
- D. –120°C
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Log in to view solution →- A. Both ferric and ferrous
- B. Free element
- C. Ferrous
- D. Ferric
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Log in to view solution →- A. Syngamy and triple fusion
- B. Fusion of two male gametes with one egg
- C. Fusion of one male gamete with two polar nuclei
- D. Fusion of two male gametes of a pollen tube with two different eggs
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Log in to view solution →- A. Chara
- B. Cycas
- C. Nostoc
- D. Green sulphur bacteria
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Log in to view solution →- A. Calcium
- B. Potassium
- C. Sodium
- D. Magnesium
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Log in to view solution →- A. Submerged hydrophytes
- B. Carnivorous plants
- C. Free-floating hydrophytes
- D. Halophytes
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Log in to view solution →- A. Mitochondria are the powerhouse of the cell in all kingdoms except Monera.
- B. Pseudopodia are locomotory and feeding structures in Sporozoans.
- C. Mushrooms belong to Basidiomycetes.
- D. Cell wall is present in members of Fungi and Plantae.
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Log in to view solution →- A. Axillary meristems
- B. Phellogen
- C. Vascular cambium
- D. Apical meristems
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Log in to view solution →- A. Rhizome
- B. Tap root
- C. Adventitious root
- D. Stem
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Log in to view solution →- A. Stems are usually unbranched in both Cycas and Cedrus.
- B. Horsetails are gymnosperms.
- C. Selaginella is heterosporous, while Salvinia is homosporous.
- D. Ovules are not enclosed by ovary wall in gymnosperms.
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Log in to view solution →- A. Endodermis
- B. Cortex
- C. Pericycle
- D. Epidermis
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Log in to view solution →- A. Cycads
- B. Conifers
- C. Deciduous angiosperms
- D. Grasses
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Log in to view solution →- A. Free ribosomes and RER
- B. Nucleic acids and SER
- C. DNA and RNA
- D. Proteins and lipids
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Log in to view solution →- A. Oxidative phosphorylation takes place in outer mitochondrial membrane.
- B. Glycolysis operates as long as it is supplied with NAD that can pick up hydrogen atoms.
- C. Glycolysis occurs in cytosol.
- D. Enzymes of TCA cycle are present in mitochondrial matrix.
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Log in to view solution →- A. Nucleosome
- B. Plastidome
- C. Polyhedral bodies
- D. Polysome
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Log in to view solution →- A. Pleurodont, Diphyodont, Heterodont
- B. Pleurodont, Monophyodont, Homodont
- C. Thecodont, Diphyodont, Heterodont
- D. Thecodont, Diphyodont, Homodont
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Log in to view solution →- A. Pleurodont, Diphyodont, Heterodont
- B. Pleurodont, Monophyodont, Homodont
- C. Thecodont, Diphyodont, Heterodont
- D. Thecodont, Diphyodont, Homodont
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Log in to view solution →- A. Phospholipid synthesis
- B. Cleavage of signal peptide
- C. Protein glycosylation
- D. Protein folding
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Log in to view solution →- A. Phospholipid synthesis
- B. Cleavage of signal peptide
- C. Protein glycosylation
- D. Protein folding
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Log in to view solution →- A. Polytene – Oocytes of amphibians chromosomes
- B. Submetacentric – L-shaped chromosomes
- C. Allosomes – Sex chromosomes
- D. Lampbrush – Diplotene bivalents chromosomes
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Log in to view solution →- A. Polytene chromosomes – Oocytes of amphibians
- B. Submetacentric chromosomes – L-shaped chromosomes
- C. Allosomes – Sex chromosomes
- D. Lampbrush chromosomes – Diplotene bivalents
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Log in to view solution →- A. Estriol
- B. Estradiol
- C. Ecdysone
- D. Epinephrine
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Log in to view solution →- A. Estriol
- B. Estradiol
- C. Ecdysone
- D. Epinephrine
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Log in to view solution →- A. Corpus callosum: band of fibers connecting left and right cerebral hemispheres
- B. Hypothalamus: production of releasing hormones and regulation of temperature, hunger and thirst
- C. Limbic system: consists of fibre tracts that interconnect different regions of brain; controls movement
- D. Medulla oblongata: controls respiration and cardiovascular reflexes
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Log in to view solution →- A. Parathyroid hormone and Prolactin
- B. Estrogen and Parathyroid hormone
- C. Progesterone and Aldosterone
- D. Aldosterone and Prolactin
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Log in to view solution →- A. smooth muscles attached to the ciliary body
- B. smooth muscles attached to the iris
- C. ligaments attached to the iris
- D. ligaments attached to the ciliary body
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Log in to view solution →- A. pre-reproductive individuals are less than the reproductive individuals.
- B. reproductive and pre-reproductive individuals are equal in number.
- C. reproductive individuals are less than the post-reproductive individuals.
- D. pre-reproductive individuals are more than the reproductive individuals.
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Log in to view solution →- A. i ii iv iii
- B. iii iv i ii
- C. i iii iv ii
- D. ii i iii iv
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Log in to view solution →- A. Leaves
- B. Roots
- C. Latex
- D. Flowers
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Log in to view solution →- A. Amensalism
- B. Parasitism
- C. Mutualism
- D. Commensalism
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Log in to view solution →- A. Seed banks
- B. Botanical gardens
- C. Sacred groves
- D. Wildlife safari parks
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Log in to view solution →- A. Parietal cells
- B. Goblet cells
- C. Mucous cells
- D. Chief cells
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Log in to view solution →- A. ii iii i
- B. i iii ii
- C. i ii iii
- D. iii ii i
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Log in to view solution →- A. prevents the formation of bonds between the myosin cross bridges and the actin filament.
- B. detaches the myosin head from the actin filament.
- C. activates the myosin ATPase by binding to it.
- D. binds to troponin to remove the masking of active sites on actin for myosin.
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Log in to view solution →- A. Green – Orange – Violet – Gold
- B. Yellow – Green – Violet – Gold
- C. Yellow – Violet – Orange – Silver
- D. Violet – Yellow – Orange – Silver
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Log in to view solution →- A. 9
- B. 20
- C. 11
- D. 10
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Log in to view solution →- A. 256/81
- B. 81/256
- C. 3/4
- D. 4/3
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Log in to view solution →- A. F
- B. 4F
- C. 6F
- D. 9F
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Log in to view solution →- A. 84.5 J
- B. 42.2 J
- C. 208.7 J
- D. 104.3 J
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Log in to view solution →- A. r^4
- B. r^5
- C. r^2
- D. r^3
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Log in to view solution →- A. −7i − 4j − 8k
- B. −7i − 8j − 4k
- C. −4i − j − 8k
- D. −8i − 4j − 7k
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Log in to view solution →- A. 0.529 cm
- B. 0.053 cm
- C. 0.525 cm
- D. 0.521 cm
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Log in to view solution →- A. 1.5 m/s, 3 m/s
- B. 1 m/s, 3.5 m/s
- C. 1 m/s, 3 m/s
- D. 2 m/s, 4 m/s
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Log in to view solution →- A. a = g tan θ
- B. a = g cos θ
- C. a = g/sin θ
- D. a = g/ cos θ
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Log in to view solution →- A. −x direction
- B. −y direction
- C. +z direction
- D. −z direction
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Log in to view solution →- A. zero
- B. 30°
- C. 45°
- D. 60°
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Log in to view solution →- A. 13.89 H
- B. 1.389 H
- C. 138.88 H
- D. 0.138 H
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Log in to view solution →- A. 36 cm towards the mirror
- B. 30 cm towards the mirror
- C. 36 cm away from the mirror
- D. 30 cm away from the mirror
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Log in to view solution →- A. 36 cm towards the mirror
- B. 30 cm towards the mirror
- C. 36 cm away from the mirror
- D. 30 cm away from the mirror
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Log in to view solution →- A. 1 : −2
- B. 2 : −1
- C. 1 : −1
- D. 1 : 1
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Log in to view solution →- A. 1 : -2
- B. 2 : -1
- C. 1 : -1
- D. 1 : 1
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Log in to view solution →- A. λ0
- B. λ0 t
- C. λ0 (1 + eE0 t / (mV0))
- D. λ0 / (1 + eE0 t / (mV0))
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Log in to view solution →- A. 2 : 1
- B. 4 : 1
- C. 1 : 4
- D. 1 : 2
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Log in to view solution →- A. 15
- B. 30
- C. 10
- D. 20
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Log in to view solution →- A. 16 cm
- B. 12.5 cm
- C. 8 cm
- D. 13.2 cm
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Log in to view solution →- A. 12.5%
- B. 6.25%
- C. 20%
- D. 26.8%
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Log in to view solution →- A. 1.254 × 10^4 K
- B. 5.016 × 10^4 K
- C. 8.360 × 10^4 K
- D. 2.508 × 10^4 K
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Log in to view solution →- A. i = tan^-1(1/μ)
- B. i = sin^-1(1/μ)
- C. Reflected light is polarised with its electric vector perpendicular to the plane of incidence
- D. Reflected light is polarised with its electric vector parallel to the plane of incidence
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Log in to view solution →- A. 1.7 mm
- B. 2.1 mm
- C. 1.9 mm
- D. 1.8 mm
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Log in to view solution →- A. small focal length and small diameter
- B. large focal length and large diameter
- C. large focal length and small diameter
- D. small focal length and large diameter
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Log in to view solution →- A. I_B = 40 μA, I_C = 5 mA, β = 125
- B. I_B = 20 μA, I_C = 5 mA, β = 250
- C. I_B = 25 μA, I_C = 5 mA, β = 200
- D. I_B = 40 μA, I_C = 10 mA, β = 250
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Log in to view solution →- A. affects the overall V-I characteristics of p-n junction
- B. does not affect resistance of p-n junction
- C. affects only forward resistance
- D. affects only reverse resistance
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Log in to view solution →- A. affects the overall V–I characteristics of p-n junction
- B. does not affect resistance of p-n junction
- C. affects only forward resistance
- D. affects only reverse resistance
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Log in to view solution →- A. A + B
- B. A̅B + AB̅
- C. A̅B̅ + AB
- D. AB
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Log in to view solution →- A. 11.32 A
- B. 14.76 A
- C. 5.98 A
- D. 7.14 A
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Log in to view solution →- A. 1.13 W
- B. 2.74 W
- C. 0.43 W
- D. 0.79 W
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Log in to view solution →- A. the induced electric field due to the changing magnetic field
- B. the lattice structure of the material of the rod
- C. the magnetic field
- D. the current source
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Log in to view solution →- A. 500 Ω
- B. 250 Ω
- C. 25 Ω
- D. 40 Ω
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Log in to view solution →- A. 300 m/s
- B. 350 m/s
- C. 339 m/s
- D. 330 m/s
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Log in to view solution →- A. inversely proportional to the distance between the plates.
- B. proportional to the square root of the distance between the plates.
- C. linearly proportional to the distance between the plates.
- D. independent of the distance between the plates.
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Log in to view solution →- A. 1 s
- B. 2 s
- C. π s
- D. 2π s
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Log in to view solution →- A. equal
- B. 10 times greater
- C. 5 times greater
- D. smaller
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Log in to view solution →- A. K_B > K_A > K_C
- B. K_B < K_A < K_C
- C. K_A > K_B > K_C
- D. K_A < K_B < K_C
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Log in to view solution →- A. K_B > K_A > K_C
- B. K_B < K_A < K_C
- C. K_A > K_B > K_C
- D. K_A < K_B < K_C
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Log in to view solution →- A. 2 : 5
- B. 10 : 7
- C. 5 : 7
- D. 7 : 10
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Log in to view solution →- A. 2 : 5
- B. 10 : 7
- C. 5 : 7
- D. 7 : 10
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Log in to view solution →- A. ‘g’ on the Earth will not change.
- B. Time period of a simple pendulum on the Earth would decrease.
- C. Walking on the ground would become more difficult.
- D. Raindrops will fall faster.
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Log in to view solution →- A. 'g' on the Earth will not change.
- B. Time period of a simple pendulum on the Earth would decrease.
- C. Walking on the ground would become more difficult.
- D. Raindrops will fall faster.
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Log in to view solution →- A. Angular momentum
- B. Rotational kinetic energy
- C. Moment of inertia
- D. Angular velocity
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Log in to view solution →- A. 4/5 D
- B. 5/7 D
- C. D
- D. 2/3 D
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Log in to view solution →- A. W_A > W_C > W_B
- B. W_B > W_A > W_C
- C. W_A > W_B > W_C
- D. W_C > W_B > W_A
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Log in to view solution →- A. Coefficient of sliding friction has dimensions of length.
- B. Frictional force opposes the relative motion.
- C. Limiting value of static friction is directly proportional to normal reaction.
- D. Rolling friction is smaller than sliding friction.
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Log in to view solution →- A. 0.4
- B. 0.8
- C. 0.25
- D. 0.5
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Log in to view solution →- A. dinuclear
- B. trinuclear
- C. mononuclear
- D. tetranuclear
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Log in to view solution →- A. iii v i ii
- B. iv i ii iii
- C. i ii iii iv
- D. iv v ii i
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Log in to view solution →- A. MnO4^2−
- B. MnO4−
- C. CrO7^2−
- D. CrO4^2−
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Log in to view solution →- A. tetrahedral geometry and paramagnetic
- B. square planar geometry and paramagnetic
- C. tetrahedral geometry and diamagnetic
- D. square planar geometry and diamagnetic
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Log in to view solution →- A. Linkage isomerism
- B. Ionization isomerism
- C. Coordination isomerism
- D. Geometrical isomerism
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Log in to view solution →- A. Glycine
- B. Benzoic acid
- C. Acetanilide
- D. Aniline
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Log in to view solution →- A. Glycine
- B. Benzoic acid
- C. Acetanilide
- D. Aniline
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Log in to view solution →- A. 5 16 2
- B. 2 16 5
- C. 2 5 16
- D. 16 5 2
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Log in to view solution →- A. 5 16 2
- B. 2 16 5
- C. 2 5 16
- D. 16 5 2
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Log in to view solution →- A. forces of attraction between the gas molecules
- B. electric field present between the gas molecules
- C. volume of the gas molecules
- D. density of the gas molecules
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Log in to view solution →- A. High temperature and low pressure
- B. High temperature and high pressure
- C. Low temperature and low pressure
- D. Low temperature and high pressure
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Log in to view solution →- A. High temperature and low pressure
- B. High temperature and high pressure
- C. Low temperature and low pressure
- D. Low temperature and high pressure
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Log in to view solution →- A. 400 kJ mol−1
- B. 800 kJ mol−1
- C. 100 kJ mol−1
- D. 200 kJ mol−1
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Log in to view solution →- A. 400 kJ mol^-1
- B. 800 kJ mol^-1
- C. 100 kJ mol^-1
- D. 200 kJ mol^-1
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Log in to view solution →- A. remains unchanged
- B. is tripled
- C. is doubled
- D. is halved
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Log in to view solution →- A. CH3 - CH = CH - CH3
- B. CH2 = CH - CH = CH2
- C. CH2 = CH - C ≡ CH
- D. HC ≡ C - C ≡ CH
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Log in to view solution →- A. -NR2 > -OR > -F
- B. -NH2 > -OR > -F
- C. -NR2 < -OR < -F
- D. -NH2 < -OR < -F
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Log in to view solution →- A. the rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations
- B. a first-order reaction can be catalyzed; a second-order reaction cannot be catalyzed
- C. the half-life of a first-order reaction does not depend on [A]0; the half-life of a second-order reaction does depend on [A]0
- D. the rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
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Log in to view solution →- A. BaH2 < BeH2 < CaH2
- B. BeH2 < BaH2 < CaH2
- C. CaH2 < BeH2 < BaH2
- D. BeH2 < CaH2 < BaH2
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Log in to view solution →- A. 10^-3 mol of water
- B. 0.00224 L of water vapours at 1 atm and 273 K
- C. 0.18 g of water
- D. 18 mL of water
Correct answer & explanation hidden
Log in to view solution →- A. C2H5OH, C2H5ONa, C2H5Cl
- B. C2H5Cl, C2H6, C2H5OH
- C. C2H5OH, C2H5Cl, C2H5ONa
- D. C2H5OH, C2H6, C2H5Cl
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Log in to view solution →- A. CH4
- B. CH3 - CH3
- C. CH2 = CH2
- D. CH ≡ CH
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Log in to view solution →- A. NO
- B. N2O
- C. NO2
- D. N2O5
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Log in to view solution →- A. NO
- B. N2O
- C. NO2
- D. N2O5
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Log in to view solution →- A. 4.4
- B. 2.8
- C. 3.0
- D. 1.4
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Log in to view solution →- A. 4.4
- B. 2.8
- C. 3.0
- D. 1.4
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Log in to view solution →- A. Amylose is made up of glucose and galactose
- B. Amylopectin have 1 → 4 α-linkage and 1 → 6 β-linkage
- C. Amylose have 1 → 4 α-linkage and 1 → 6 β-linkage
- D. Amylopectin have 1 → 4 α-linkage and 1 → 6 α-linkage
Correct answer & explanation hidden
Log in to view solution →- A. Amylose is made up of glucose and galactose
- B. Amylopectin have 1→4 α-linkage and 1→6 β-linkage
- C. Amylose have 1→4 α-linkage and 1→6 β-linkage
- D. Amylopectin have 1→4 α-linkage and 1→6 α-linkage
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Log in to view solution →- A. CaO
- B. BaO
- C. BeO
- D. MgO
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Log in to view solution →- A. In acidic (strong) medium aniline is present as anilinium ion.
- B. In absence of substituents nitro group always goes to m-position.
- C. In electrophilic substitution reactions amino group is meta directive.
- D. In spite of substituents nitro group always goes to only m-position.
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Log in to view solution →- A. They contain strong covalent bonds in their polymer chains.
- B. Examples are bakelite and melamine.
- C. They are formed from bi- and tri-functional monomers.
- D. They contain covalent bonds between various linear polymer chains.
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Log in to view solution →- A. c
- B. d
- C. a
- D. b
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Log in to view solution →- A. The sign of charge on the ion alone
- B. Both magnitude and sign of the charge on the ion
- C. Size of the ion alone
- D. The magnitude of the charge on the ion alone
Correct answer & explanation hidden
Log in to view solution →- A. 1.08 × 10−8 mol2 L−2
- B. 1.08 × 10−14 mol2 L−2
- C. 1.08 × 10−12 mol2 L−2
- D. 1.08 × 10−10 mol2 L−2
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Log in to view solution →- A. CO2
- B. O2
- C. H2
- D. NH3
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Log in to view solution →- A. Mg3X2
- B. Mg2X
- C. MgX2
- D. Mg2X3
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Log in to view solution →- A. 2/1
- B. 24/33
- C. 23/34
- D. 2/3
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Log in to view solution →- A. The value of m for 2dz2 is zero.
- B. The electronic configuration of N atom is
- C. An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers.
- D. Total orbital angular momentum of electron in ‘s’ orbital is equal to zero.
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Log in to view solution →- A. CN
- B. CN+
- C. CN−
- D. NO
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Log in to view solution →- A. Chlorine has the highest electron-gain enthalpy.
- B. All but fluorine show positive oxidation states.
- C. All are oxidizing agents.
- D. All form monobasic oxyacids.
Correct answer & explanation hidden
Log in to view solution →- A. Chlorine has the highest electron-gain enthalpy.
- B. All but fluorine show positive oxidation states.
- C. All are oxidizing agents.
- D. All form monobasic oxyacids.
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Log in to view solution →- A. In
- B. B
- C. Al
- D. Ga
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Log in to view solution →- A. In
- B. B
- C. Al
- D. Ga
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Log in to view solution →- A. three
- B. four
- C. two
- D. one
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Log in to view solution →- A. three
- B. four
- C. two
- D. one
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Log in to view solution →- A. Cu
- B. Mg
- C. Zn
- D. Fe
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Log in to view solution →- A. Cu
- B. Mg
- C. Zn
- D. Fe
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Log in to view solution →- A. B < Ga < Al < In < Tl
- B. B < Ga < Al < Tl < In
- C. B < Al < Ga < In < Tl
- D. B < Al < In < Ga < Tl
Correct answer & explanation hidden
Log in to view solution →- A. B < Ga < Al < In < Tl
- B. B < Ga < Al < Tl < In
- C. B < Al < Ga < In < Tl
- D. B < Al < In < Ga < Tl
Correct answer & explanation hidden
Log in to view solution →- A. NH4Cl, N2, NO, HNO3
- B. HNO3, NH4Cl, NO, N2
- C. HNO3, NO, NH4Cl, N2
- D. HNO3, NO, N2, NH4Cl
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Log in to view solution →- A. NH4Cl, N2, NO, HNO3
- B. HNO3, NH4Cl, NO, N2
- C. HNO3, NO, NH4Cl, N2
- D. HNO3, NO, N2, NH4Cl
Correct answer & explanation hidden
Log in to view solution →- A. dichlorocarbene (:CCl2)
- B. dichloromethyl anion (CHCl2−)
- C. formyl cation (CHO+)
- D. dichloromethyl cation (CHCl2+)
Correct answer & explanation hidden
Log in to view solution →- A. formation of intermolecular H-bonding
- B. more extensive association of carboxylic acid via van der Waals force of attraction
- C. formation of carboxylate ion
- D. formation of intramolecular H-bonding
Correct answer & explanation hidden
Log in to view solution →- A. C6H5CH(OH)CH3 and I2
- B. C6H5CH2CH2OH and I2
- C. C6H5COCH3 and Cl2
- D. C6H5CH(OH)CH3 and Br2
Correct answer & explanation hidden
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