- A. Violet – Yellow – Orange – Silver
- B. Yellow – Violet – Orange – Silver
- C. Yellow – Green – Violet – Gold
- D. Green – Orange – Violet – Gold
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Log in to view solution →- A. A graph of I versus n that is a straight line through the origin
- B. A graph of I versus n that is a rectangular hyperbola
- C. A graph of I versus n that is a parabola opening upward
- D. A graph of I versus n that is independent of n
Correct answer & explanation hidden
Log in to view solution →- A. 10
- B. 11
- C. 20
- D. 9
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Log in to view solution →- A. smaller
- B. 5 times greater
- C. 10 times greater
- D. equal
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Log in to view solution →- A. 330 m/s
- B. 339 m/s
- C. 350 m/s
- D. 300 m/s
Correct answer & explanation hidden
Log in to view solution →- A. 330 m/s
- B. 339 m/s
- C. 350 m/s
- D. 300 m/s
Correct answer & explanation hidden
Log in to view solution →- A. 2π s
- B. π s
- C. 2 s
- D. 1 s
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Log in to view solution →- A. 2π s
- B. π s
- C. 2 s
- D. 1 s
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Log in to view solution →- A. independent of the distance between the plates.
- B. linearly proportional to the distance between the plates.
- C. proportional to the square root of the distance between the plates.
- D. inversely proportional to the distance between the plates.
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Log in to view solution →- A. independent of the distance between the plates.
- B. linearly proportional to the distance between the plates.
- C. proportional to the square root of the distance between the plates.
- D. inversely proportional to the distance between the plates.
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Log in to view solution →- A. λ₀ / (1 + eE₀t / (mV₀))
- B. λ₀ (1 + eE₀t / (mV₀))
- C. λ₀ t
- D. λ₀
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Log in to view solution →- A. 1 : 2
- B. 1 : 4
- C. 4 : 1
- D. 2 : 1
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Log in to view solution →- A. 20
- B. 10
- C. 30
- D. 15
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Log in to view solution →- A. 1 : 1
- B. 1 : −1
- C. 2 : −1
- D. 1 : −2
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Log in to view solution →- A. affects only reverse resistance
- B. affects only forward resistance
- C. does not affect resistance of p-n junction
- D. affects the overall V-I characteristics of p-n junction
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Log in to view solution →- A. 3D/2
- B. D
- C. 5D/7
- D. 4D/5
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Log in to view solution →- A. WC > WB > WA
- B. WA > WB > WC
- C. WB > WA > WC
- D. WA > WC > WB
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Log in to view solution →- A. 0.5
- B. 0.25
- C. 0.8
- D. 0.4
Correct answer & explanation hidden
Log in to view solution →- A. Rolling friction is smaller than sliding friction.
- B. Limiting value of static friction is directly proportional to normal reaction.
- C. Frictional force opposes the relative motion.
- D. Coefficient of sliding friction has dimensions of length.
Correct answer & explanation hidden
Log in to view solution →- A. Rolling friction is smaller than sliding friction.
- B. Limiting value of static friction is directly proportional to normal reaction.
- C. Frictional force opposes the relative motion.
- D. Coefficient of sliding friction has dimensions of length.
Correct answer & explanation hidden
Log in to view solution →- A. −8î − 4ĵ − 7k̂
- B. −4î − ĵ − 8k̂
- C. −7î − 8ĵ − 4k̂
- D. −7î − 4ĵ − 8k̂
Correct answer & explanation hidden
Log in to view solution →- A. -8i - 4j - 7k
- B. -4i - j - 8k
- C. -7i - 8j - 4k
- D. -7i - 4j - 8k
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Log in to view solution →- A. 2 m/s, 4 m/s
- B. 1 m/s, 3 m/s
- C. 1 m/s, 3.5 m/s
- D. 1.5 m/s, 3 m/s
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Log in to view solution →- A. 0.521 cm
- B. 0.525 cm
- C. 0.053 cm
- D. 0.529 cm
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Log in to view solution →- A. a = g cos θ
- B. a = g sin θ
- C. a = g cos θ
- D. a = g tan θ
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Log in to view solution →- A. 13.2 cm
- B. 8 cm
- C. 12.5 cm
- D. 16 cm
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Log in to view solution →- A. 2.508 × 10^4 K
- B. 8.360 × 10^4 K
- C. 5.016 × 10^4 K
- D. 1.254 × 10^4 K
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Log in to view solution →- A. 26.8%
- B. 20%
- C. 6.25%
- D. 12.5%
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Log in to view solution →- A. Reflected light is polarised with its electric vector parallel to the plane of incidence
- B. Reflected light is polarised with its electric vector perpendicular to the plane of incidence
- C. i = sin^-1(1/μ)
- D. i = tan^-1(1/μ)
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Log in to view solution →- A. 1.8 mm
- B. 1.9 mm
- C. 2.1 mm
- D. 1.7 mm
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Log in to view solution →- A. small focal length and large diameter
- B. large focal length and small diameter
- C. large focal length and large diameter
- D. small focal length and small diameter
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Log in to view solution →- A. small focal length and large diameter
- B. large focal length and small diameter
- C. large focal length and large diameter
- D. small focal length and small diameter
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Log in to view solution →- A. -z direction
- B. +z direction
- C. -y direction
- D. -x direction
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Log in to view solution →- A. −z direction
- B. +z direction
- C. −y direction
- D. −x direction
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Log in to view solution →- A. 60°
- B. 45°
- C. 30°
- D. 0°
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Log in to view solution →- A. 30 cm away from the mirror
- B. 36 cm away from the mirror
- C. 30 cm towards the mirror
- D. 36 cm towards the mirror
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Log in to view solution →- A. 0.138 H
- B. 138.88 H
- C. 1.389 H
- D. 13.89 H
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Log in to view solution →- A. 40 Ω
- B. 25 Ω
- C. 250 Ω
- D. 500 Ω
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Log in to view solution →- A. 7.14 A
- B. 5.98 A
- C. 14.76 A
- D. 11.32 A
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Log in to view solution →- A. 0.79 W
- B. 0.43 W
- C. 2.74 W
- D. 1.13 W
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Log in to view solution →- A. the current source
- B. the magnetic field
- C. the lattice structure of the material of the rod
- D. the induced electric field due to the changing magnetic field
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Log in to view solution →- A. r^3
- B. r^2
- C. r^5
- D. r^4
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Log in to view solution →- A. 4/3
- B. 3/4
- C. 81/256
- D. 256/81
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Log in to view solution →- A. 104.3 J
- B. 208.7 J
- C. 42.2 J
- D. 84.5 J
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Log in to view solution →- A. 9F
- B. 6F
- C. 4F
- D. F
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Log in to view solution →- A. 9 F
- B. 6 F
- C. 4 F
- D. F
Correct answer & explanation hidden
Log in to view solution →- A. Angular velocity
- B. Moment of inertia
- C. Rotational kinetic energy
- D. Angular momentum
Correct answer & explanation hidden
Log in to view solution →- A. Angular velocity
- B. Moment of inertia
- C. Rotational kinetic energy
- D. Angular momentum
Correct answer & explanation hidden
Log in to view solution →- A. K_A < K_B < K_C
- B. K_A > K_B > K_C
- C. K_B < K_A < K_C
- D. K_B > K_A > K_C
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Log in to view solution →- A. 7 : 10
- B. 5 : 7
- C. 10 : 7
- D. 2 : 5
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Log in to view solution →- A. Raindrops will fall faster.
- B. Walking on the ground would become more difficult.
- C. Time period of a simple pendulum on the Earth would decrease.
- D. g on the Earth will not change.
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Log in to view solution →- A. Epinephrine
- B. Ecdysone
- C. Estradiol
- D. Estriol
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Log in to view solution →- A. Medulla oblongata: controls respiration and cardiovascular reflexes.
- B. Limbic system: consists of fibre tracts that interconnect different regions of brain; controls movement.
- C. Hypothalamus: production of releasing hormones and regulation of temperature, hunger and thirst.
- D. Corpus callosum: band of fibers connecting left and right cerebral hemispheres.
Correct answer & explanation hidden
Log in to view solution →- A. ligaments attached to the ciliary body
- B. ligaments attached to the iris
- C. smooth muscles attached to the iris
- D. smooth muscles attached to the ciliary body
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Log in to view solution →- A. Aldosterone and Prolactin
- B. Progesterone and Aldosterone
- C. Estrogen and Parathyroid hormone
- D. Parathyroid hormone and Prolactin
Correct answer & explanation hidden
Log in to view solution →- A. using flagella for locomotion
- B. having a contractile vacuole for removing excess water
- C. using pseudopodia for capturing prey
- D. having two types of nuclei
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Log in to view solution →- A. Amphibia
- B. Reptilia
- C. Aves
- D. Osteichthyes
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Log in to view solution →- A. Dinoflagellates
- B. Diatoms
- C. Cyanobacteria
- D. Euglenoids
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Log in to view solution →- A. Macropus
- B. Chelone
- C. Camelus
- D. Psittacula
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Log in to view solution →- A. Earthworm
- B. Tunicate
- C. Moth
- D. Starfish
Correct answer & explanation hidden
Log in to view solution →- A. Presence of a boat shaped sternum on the 9th abdominal segment
- B. Presence of caudal styles
- C. Forewings with darker tegmina
- D. Presence of anal cerci
Correct answer & explanation hidden
Log in to view solution →- A. Presence of a boat-shaped sternum on the 9th abdominal segment
- B. Presence of caudal styles
- C. Forewings with darker tegmina
- D. Presence of anal cerci
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Log in to view solution →- A. Commensalism
- B. Mutualism
- C. Parasitism
- D. Amensalism
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Log in to view solution →- A. Commensalism
- B. Mutualism
- C. Parasitism
- D. Amensalism
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Log in to view solution →- A. Wildlife safari parks
- B. Sacred groves
- C. Botanical gardens
- D. Seed banks
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Log in to view solution →- A. Wildlife safari parks
- B. Sacred groves
- C. Botanical gardens
- D. Seed banks
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Log in to view solution →- A. a-ii, b-i, c-iii, d-iv
- B. a-i, b-iii, c-iv, d-ii
- C. a-iii, b-iv, c-i, d-ii
- D. a-i, b-ii, c-iv, d-iii
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Log in to view solution →- A. a-ii, b-i, c-iii, d-iv
- B. a-i, b-iii, c-iv, d-ii
- C. a-iii, b-iv, c-i, d-ii
- D. a-i, b-ii, c-iv, d-iii
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Log in to view solution →- A. pre-reproductive individuals are more than the reproductive individuals.
- B. reproductive individuals are less than the post-reproductive individuals.
- C. reproductive and pre-reproductive individuals are equal in number.
- D. pre-reproductive individuals are less than the reproductive individuals.
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Log in to view solution →- A. Flowers
- B. Latex
- C. Roots
- D. Leaves
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Log in to view solution →- A. hCG, hPL, progestogens, prolactin
- B. hCG, hPL, estrogens, relaxin, oxytocin
- C. hCG, hPL, progestogens, estrogens
- D. hCG, progestogens, estrogens, glucocorticoids
Correct answer & explanation hidden
Log in to view solution →- A. blocks estrogen receptors in the uterus, preventing eggs from getting implanted.
- B. increases the concentration of estrogen and prevents ovulation in females.
- C. is an IUD.
- D. is a post-coital contraceptive.
Correct answer & explanation hidden
Log in to view solution →- A. ectoderm and mesoderm
- B. endoderm and mesoderm
- C. mesoderm and trophoblast
- D. ectoderm and endoderm
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Log in to view solution →- A. In spermiogenesis spermatids are formed, while in spermiation spermatozoa are formed.
- B. In spermiogenesis spermatozoa are formed, while in spermiation spermatids are formed.
- C. In spermiogenesis spermatozoa from Sertoli cells are released into the cavity of seminiferous tubules, while in spermiation spermatozoa are formed.
- D. In spermiogenesis spermatozoa are formed, while in spermiation spermatozoa are released from Sertoli cells into the cavity of seminiferous tubules.
Correct answer & explanation hidden
Log in to view solution →- A. Inflammation of bronchioles; Decreased respiratory surface
- B. Increased number of bronchioles; Increased respiratory surface
- C. Increased respiratory surface; Inflammation of bronchioles
- D. Decreased respiratory surface; Inflammation of bronchioles
Correct answer & explanation hidden
Log in to view solution →- A. a-iii, b-i, c-ii
- B. a-i, b-iii, c-ii
- C. a-i, b-ii, c-iii
- D. a-ii, b-i, c-iii
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Log in to view solution →- A. a-iii, b-ii, c-i, d-iv
- B. a-iii, b-i, c-iv, d-ii
- C. a-i, b-iv, c-ii, d-iii
- D. a-iv, b-iii, c-ii, d-i
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Log in to view solution →- A. a-iv, b-i, c-ii, d-iii
- B. a-i, b-ii, c-iii, d-iv
- C. a-ii, b-iii, c-i, d-iv
- D. a-iii, b-iv, c-ii, d-i
Correct answer & explanation hidden
Log in to view solution →- A. a b c d = iii ii iv i
- B. a b c d = i ii iii iv
- C. a b c d = ii iii i iv
- D. a b c d = iv i ii iii
Correct answer & explanation hidden
Log in to view solution →- A. a-iv, b-i, c-ii, d-iii
- B. a-iv, b-v, c-ii, d-iii
- C. a-v, b-iv, c-i, d-ii
- D. a-v, b-iv, c-i, d-iii
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Log in to view solution →- A. a b c d = iv v ii iii
- B. a b c d = iv i ii iii
- C. a b c d = v iv i ii
- D. a b c d = v iv i iii
Correct answer & explanation hidden
Log in to view solution →- A. Protein folding
- B. Protein glycosylation
- C. Cleavage of signal peptide
- D. Phospholipid synthesis
Correct answer & explanation hidden
Log in to view solution →- A. Enzymes of TCA cycle are present in mitochondrial matrix.
- B. Glycolysis occurs in cytosol.
- C. Glycolysis operates as long as it is supplied with NAD that can pick up hydrogen atoms.
- D. Oxidative phosphorylation takes place in outer mitochondrial membrane.
Correct answer & explanation hidden
Log in to view solution →- A. Proteins and lipids
- B. DNA and RNA
- C. Nucleic acids and SER
- D. Free ribosomes and RER
Correct answer & explanation hidden
Log in to view solution →- A. Thecodont, Diphyodont, Homodont
- B. Thecodont, Diphyodont, Heterodont
- C. Pleurodont, Monophyodont, Homodont
- D. Pleurodont, Diphyodont, Heterodont
Correct answer & explanation hidden
Log in to view solution →- A. Lampbrush – Diplotene bivalents chromosomes
- B. Allosomes – Sex chromosomes
- C. Submetacentric – L-shaped chromosomes
- D. Polytene chromosomes – Oocytes of amphibians
Correct answer & explanation hidden
Log in to view solution →- A. Polysome
- B. Polyhedral bodies
- C. Plastidome
- D. Nucleosome
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Log in to view solution →- A. an operator
- B. structural genes
- C. an enhancer
- D. a promoter
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Log in to view solution →- A. a b c = iii ii i
- B. a b c = i iii ii
- C. a b c = ii iii i
- D. a b c = iii i ii
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Log in to view solution →- A. Multiple step mutations
- B. Saltation
- C. Phenotypic variations
- D. Minor mutations
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Log in to view solution →- A. Only daughters
- B. Only sons
- C. Only grandchildren
- D. Both sons and daughters
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Log in to view solution →- A. AGGUAUCGCAU
- B. UGGTUTCGCAT
- C. ACCUAUGCGAU
- D. UCCAUAGCGUA
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Log in to view solution →- A. Chief cells
- B. Mucous cells
- C. Goblet cells
- D. Parietal cells
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Log in to view solution →- A. a b c = iii ii i
- B. a b c = i ii iii
- C. a b c = i iii ii
- D. a b c = ii iii i
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Log in to view solution →- A. Anthracis
- B. Silicosis
- C. Botulism
- D. Emphysema
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Log in to view solution →- A. Anthracis
- B. Silicosis
- C. Botulism
- D. Emphysema
Correct answer & explanation hidden
Log in to view solution →- A. binds to troponin to remove the masking of active sites on actin for myosin.
- B. activates the myosin ATPase by binding to it.
- C. detaches the myosin head from the actin filament.
- D. prevents the formation of bonds between the myosin cross bridges and the actin filament.
Correct answer & explanation hidden
Log in to view solution →- A. binds to troponin to remove the masking of active sites on actin for myosin.
- B. activates the myosin ATPase by binding to it.
- C. detaches the myosin head from the actin filament.
- D. prevents the formation of bonds between the myosin cross bridges and the actin filament.
Correct answer & explanation hidden
Log in to view solution →- A. Psoriasis
- B. Rheumatoid arthritis
- C. Alzheimer’s disease
- D. Vitiligo
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Log in to view solution →- A. Psoriasis
- B. Rheumatoid arthritis
- C. Alzheimer’s disease
- D. Vitiligo
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Log in to view solution →- A. Forelimbs of man, bat and cheetah
- B. Heart of bat, man and cheetah
- C. Brain of bat, man and cheetah
- D. Eye of octopus, bat and man
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Log in to view solution →- A. Forelimbs of man, bat and cheetah
- B. Heart of bat, man and cheetah
- C. Brain of bat, man and cheetah
- D. Eye of octopus, bat and man
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Log in to view solution →- A. Vitamin D
- B. Vitamin A
- C. Vitamin B12
- D. Vitamin E
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Log in to view solution →- A. Vitamin D
- B. Vitamin A
- C. Vitamin B12
- D. Vitamin E
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Log in to view solution →- A. Elephantiasis
- B. Ascariasis
- C. Ringworm disease
- D. Amoebiasis
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Log in to view solution →- A. Homology
- B. Analogy
- C. Convergent evolution
- D. Adaptive radiation
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Log in to view solution →- A. b, c and e
- B. a, b and c
- C. b, d and e
- D. a, c and e
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Log in to view solution →- A. Bamboo species
- B. Jackfruit
- C. Mango
- D. Papaya
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Log in to view solution →- A. Starch synthesis in pea : Multiple alleles
- B. ABO blood grouping : Co-dominance
- C. XO type sex determination : Grasshopper
- D. T.H. Morgan : Linkage
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Log in to view solution →- A. Franklin Stahl coined the term “linkage”.
- B. Punnett square was developed by a British scientist.
- C. Spliceosomes take part in translation.
- D. Transduction was discovered by S. Altman.
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Log in to view solution →- A. Fungus
- B. Bacterium
- C. Plant
- D. Virus
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Log in to view solution →- A. Meiotic divisions
- B. Mitotic divisions
- C. Parthenocarpy
- D. Parthenogenesis
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Log in to view solution →- A. Pollenkitt
- B. Cellulosic intine
- C. Oil content
- D. Sporopollenin
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Log in to view solution →- A. Alec Jeffreys – Streptococcus pneumoniae
- B. Alfred Hershey and Martha Chase – TMV
- C. Matthew Meselson and F. Stahl – Pisum sativum
- D. Francois Jacob and Jacques Monod – Lac operon
Correct answer & explanation hidden
Log in to view solution →- A. Indian Council of Medical Research (ICMR)
- B. Council for Scientific and Industrial Research (CSIR)
- C. Research Committee on Genetic Manipulation (RCGM)
- D. Genetic Engineering Appraisal Committee (GEAC)
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Log in to view solution →- A. Retrovirus
- B. Ti plasmid
- C. \u03bb phage
- D. pBR 322
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Log in to view solution →- A. Extension, Denaturation, Annealing
- B. Annealing, Extension, Denaturation
- C. Denaturation, Extension, Annealing
- D. Denaturation, Annealing, Extension
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Log in to view solution →- A. Co-667
- B. Sharbati Sonora
- C. Lerma Rojo
- D. Basmati
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Log in to view solution →- A. Ribozyme – Nucleic acid
- B. F2 × Recessive parent – Dihybrid cross
- C. T.H. Morgan – Transduction
- D. G. Mendel – Transformation
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Log in to view solution →- A. Ribozyme – Nucleic acid
- B. 2 × Recessive parent – Dihybrid cross
- C. T. H. Morgan – Transduction
- D. G. Mendel – Transformation
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Log in to view solution →- A. Bio-infringement
- B. Biopiracy
- C. Biodegradation
- D. Bioexploitation
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Log in to view solution →- A. Bio-infringement
- B. Biopiracy
- C. Biodegradation
- D. Bioexploitation
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Log in to view solution →- A. Death rate
- B. Birth rate
- C. Number of individuals leaving the habitat
- D. Number of individuals entering a habitat
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Log in to view solution →- A. Death rate
- B. Birth rate
- C. Number of individuals leaving the habitat
- D. Number of individuals entering a habitat
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Log in to view solution →- A. all the biological factors in the organism’s environment
- B. the physical space where an organism lives
- C. the range of temperature that the organism needs to live
- D. the functional role played by the organism where it lives
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Log in to view solution →- A. all the biological factors in the organism’s environment
- B. the physical space where an organism lives
- C. the range of temperature that the organism needs to live
- D. the functional role played by the organism where it lives
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Log in to view solution →- A. Inverted pyramid of biomass
- B. Pyramid of energy
- C. Upright pyramid of numbers
- D. Upright pyramid of biomass
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Log in to view solution →- A. Carbon
- B. Cl
- C. Fe
- D. Oxygen
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Log in to view solution →- A. All form monobasic oxyacids.
- B. All are oxidizing agents.
- C. All but fluorine show positive oxidation states.
- D. Chlorine has the highest electron-gain enthalpy.
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Log in to view solution →- A. Fe
- B. Zn
- C. Mg
- D. Cu
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Log in to view solution →- A. one
- B. two
- C. four
- D. three
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Log in to view solution →- A. B < Al < In < Ga < Tl
- B. B < Al < Ga < In < Tl
- C. B < Ga < Al < Tl < In
- D. B < Ga < Al < In < Tl
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Log in to view solution →- A. HNO3, NO, N2, NH4Cl
- B. HNO3, NO, NH4Cl, N2
- C. HNO3, NH4Cl, NO, N2
- D. NH4Cl, N2, NO, HNO3
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Log in to view solution →- A. Ga
- B. Al
- C. B
- D. In
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Log in to view solution →- A. C2H5OH, C2H6, C2H5Cl
- B. C2H5OH, C2H5Cl, C2H5ONa
- C. C2H5Cl, C2H6, C2H5OH
- D. C2H5OH, C2H5ONa, C2H5Cl
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Log in to view solution →- A. CH≡CH
- B. CH2=CH2
- C. CH3–CH3
- D. CH4
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Log in to view solution →- A. CH≡CH
- B. CH2=CH2
- C. CH3–CH3
- D. CH4
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Log in to view solution →- A. m-bromotoluene
- B. o-bromotoluene
- C. 3-bromo-2,4,6-trichlorotoluene
- D. p-bromotoluene
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Log in to view solution →- A. N2O5
- B. NO2
- C. N2O
- D. NO
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Log in to view solution →- A. N2O5
- B. NO2
- C. N2O
- D. NO
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Log in to view solution →- A. b
- B. a
- C. d
- D. c
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Log in to view solution →- A. b
- B. a
- C. d
- D. c
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Log in to view solution →- A. The magnitude of the charge on the ion alone
- B. Size of the ion alone
- C. Both magnitude and sign of the charge on the ion
- D. The sign of charge on the ion alone
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Log in to view solution →- A. 1.08 × 10^-10 mol^2 L^-2
- B. 1.08 × 10^-12 mol^2 L^-2
- C. 1.08 × 10^-14 mol^2 L^-2
- D. 1.08 × 10^-8 mol^2 L^-2
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Log in to view solution →- A. NH3
- B. H2
- C. O2
- D. CO2
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Log in to view solution →- A. iv v ii i
- B. i ii iii iv
- C. iv i ii iii
- D. iii v i ii
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Log in to view solution →- A. tetranuclear
- B. mononuclear
- C. trinuclear
- D. dinuclear
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Log in to view solution →- A. square planar geometry and diamagnetic
- B. tetrahedral geometry and diamagnetic
- C. square planar geometry and paramagnetic
- D. tetrahedral geometry and paramagnetic
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Log in to view solution →- A. [CrO4]2-
- B. [Cr2O7]2-
- C. [MnO4]-
- D. [MnO4]2-
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Log in to view solution →- A. Geometrical isomerism
- B. Coordination isomerism
- C. Ionization isomerism
- D. Linkage isomerism
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Log in to view solution →- A. Aniline
- B. Acetanilide
- C. Benzoic acid
- D. Glycine
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Log in to view solution →- A. HC≡C–C≡CH
- B. CH2=CH–C≡CH
- C. CH2=CH–CH=CH2
- D. CH3–CH=CH–CH3
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Log in to view solution →- A. –NH2 < –OR < –F
- B. –NR2 < –OR < –F
- C. –NH2 > –OR > –F
- D. –NR2 > –OR > –F
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Log in to view solution →- A. Mg2X3
- B. MgX2
- C. Mg2X
- D. Mg3X2
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Log in to view solution →- A. 2/3
- B. 23/34
- C. 24/33
- D. 2/1
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Log in to view solution →- A. 2/3
- B. 23/34
- C. 24/33
- D. 2/1
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Log in to view solution →- A. Total orbital angular momentum of electron in ‘s’ orbital is equal to zero.
- B. An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers.
- C. The electronic configuration of N atom is
- D. The value of m for 2z d is zero.
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Log in to view solution →- A. Total orbital angular momentum of electron in s orbital is equal to zero.
- B. An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers.
- C. The electronic configuration of N atom is
- D. The value of m for 2z d is zero.
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Log in to view solution →- A. NO
- B. CN−
- C. CN+
- D. CN
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Log in to view solution →- A. dichloromethyl cation (CHCl2+)
- B. formyl cation (CHO+)
- C. dichloromethyl anion (CHCl2−)
- D. dichlorocarbene (:CCl2)
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Log in to view solution →- A. formation of intramolecular H-bonding
- B. formation of carboxylate ion
- C. more extensive association of carboxylic acid via van der Waals force of attraction
- D. formation of intermolecular H-bonding
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Log in to view solution →- A. ethoxybenzene and iodine
- B. acetophenone and iodine
- C. ethyl phenyl ether and iodine
- D. 1-phenylethanol and iodine
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Log in to view solution →- A. 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
- B. 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
- C. a first-order reaction can be catalyzed; a second-order reaction cannot be catalyzed
- D. 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
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Log in to view solution →- A. BeH2 < CaH2 < BaH2
- B. CaH2 < BeH2 < BaH2
- C. BeH2 < BaH2 < CaH2
- D. BaH2 < BeH2 < CaH2
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Log in to view solution →- A. Br3O−
- B. Br4O−
- C. Br2
- D. HBrO
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Log in to view solution →- A. 18 mL of water
- B. 0.18 g of water
- C. 0.00224 L of water vapours at 1 atm and 273 K
- D. 10−3 mol of water
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Log in to view solution →- A. They contain covalent bonds between various linear polymer chains.
- B. They are formed from bi- and tri-functional monomers.
- C. Examples are bakelite and melamine.
- D. They contain strong covalent bonds in their polymer chains.
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Log in to view solution →- A. In spite of substituents nitro group always goes to only m-position.
- B. In electrophilic substitution reactions amino group is meta directive.
- C. In absence of substituents nitro group always goes to m-position.
- D. In acidic (strong) medium aniline is present as anilinium ion.
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Log in to view solution →- A. MgO
- B. BeO
- C. BaO
- D. CaO
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Log in to view solution →- A. MgO
- B. BeO
- C. BaO
- D. CaO
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Log in to view solution →- A. Amylopectin have 1→4 α-linkage and 1→6 α-linkage
- B. Amylose have 1→4 α-linkage and 1→6 β-linkage
- C. Amylopectin have 1→4 α-linkage and 1→6 β-linkage
- D. Amylose is made up of glucose and galactose
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Log in to view solution →- A. Amylopectin has 1→4 α-linkage and 1→6 α-linkage
- B. Amylose has 1→4 α-linkage and 1→6 β-linkage
- C. Amylopectin has 1→4 α-linkage and 1→6 β-linkage
- D. Amylose is made up of glucose and galactose
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Log in to view solution →- A. 1.4
- B. 3.0
- C. 2.8
- D. 4.4
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Log in to view solution →- A. 16 5 2
- B. 2 5 16
- C. 2 16 5
- D. 5 16 2
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Log in to view solution →- A. 16, 5, 2
- B. 2, 5, 16
- C. 2, 16, 5
- D. 5, 16, 2
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Log in to view solution →- A. density of the gas molecules
- B. volume of the gas molecules
- C. electric field present between the gas molecules
- D. forces of attraction between the gas molecules
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Log in to view solution →- A. density of the gas molecules
- B. volume of the gas molecules
- C. electric field present between the gas molecules
- D. forces of attraction between the gas molecules
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Log in to view solution →- A. Low temperature and high pressure
- B. Low temperature and low pressure
- C. High temperature and high pressure
- D. High temperature and low pressure
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Log in to view solution →- A. 200 kJ mol−1
- B. 100 kJ mol−1
- C. 800 kJ mol−1
- D. 400 kJ mol−1
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Log in to view solution →- A. is halved
- B. is doubled
- C. is tripled
- D. remains unchanged
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