- 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. zero
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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. 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. IB = 40 μA, IC = 10 mA, beta = 250
- B. IB = 25 μA, IC = 5 mA, beta = 200
- C. IB = 20 μA, IC = 5 mA, beta = 250
- D. IB = 40 μA, IC = 5 mA, beta = 125
Correct answer & explanation hidden
Log in to view solution →- A. IB = 40 μA, IC = 10 mA, β = 250
- B. IB = 25 μA, IC = 5 mA, β = 200
- C. IB = 20 μA, IC = 5 mA, β = 250
- D. IB = 40 μA, IC = 5 mA, β = 125
Correct answer & explanation hidden
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
Correct answer & explanation hidden
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. r^3
- B. r^2
- C. r^5
- D. r^4
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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. 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. 9 F
- B. 6 F
- C. 4 F
- D. F
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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. 10
- B. 11
- C. 20
- D. 9
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Log in to view solution →- 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. 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.
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Log in to view solution →- A. 0.5
- B. 0.25
- C. 0.8
- D. 0.4
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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. 330 m/s
- B. 339 m/s
- C. 350 m/s
- D. 300 m/s
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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
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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. 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.
Correct answer & explanation hidden
Log in to view solution →- A. (λ₀) / (1 + (eE₀t / mV₀))
- B. λ₀(1 + (eE₀t / mV₀))
- C. λ₀t
- D. λ₀
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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 : 2
- B. 1 : 4
- C. 4 : 1
- D. 2 : 1
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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. – 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. 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. 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. 13.2 cm
- B. 8 cm
- C. 12.5 cm
- D. 16 cm
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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. 2.508 × 10⁴ K
- B. 8.360 × 10⁴ K
- C. 5.016 × 10⁴ K
- D. 1.254 × 10⁴ K
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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. 40 Ω
- B. 25 Ω
- C. 250 Ω
- D. 500 Ω
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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. 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. 7 : 10
- B. 5 : 7
- C. 10 : 7
- D. 2 : 5
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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. 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.
Correct answer & explanation hidden
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
Correct answer & explanation hidden
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. 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. 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
Correct answer & explanation hidden
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
Correct answer & explanation hidden
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. 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. 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. 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. 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 2d z2 is zero.
Correct answer & explanation hidden
Log in to view solution →- A. NO
- B. CN–
- C. CN+
- D. CN
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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. Ga
- B. Al
- C. B
- D. In
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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. Fe
- B. Zn
- C. Mg
- D. Cu
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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
Correct answer & explanation hidden
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. 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. 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. 1.08 * 10^-10 mol2 L^-2
- B. 1.08 * 10^-12 mol2 L^-2
- C. 1.08 * 10^-14 mol2 L^-2
- D. 1.08 * 10^-8 mol2 L^-2
Correct answer & explanation hidden
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. NH3
- B. H2
- C. O2
- D. CO2
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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. N2O5
- B. NO2
- C. N2O
- D. NO
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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. 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. is halved
- B. is doubled
- C. is tripled
- D. remains unchanged
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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. Aniline
- B. Acetanilide
- C. Benzoic acid
- D. Glycine
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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. CrO4^-2
- B. Cr2O7^-2
- C. MnO4^-
- D. MnO4^-2
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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. tetranuclear
- B. mononuclear
- C. trinuclear
- D. dinuclear
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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. (1) iv v ii i
- B. (2) i ii iii iv
- C. (3) iv i ii iii
- D. (4) iii v i ii
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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. – 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. 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. 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. Fungus
- B. Bacterium
- C. Plant
- D. Virus
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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.
Correct answer & explanation hidden
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. Starch synthesis in pea : Multiple alleles
- B. ABO blood grouping : Co-dominance
- C. XO type sex determination : Grasshopper
- D. T.H. Morgan : Linkage
Correct answer & explanation hidden
Log in to view solution →- A. Bamboo species
- B. Jackfruit
- C. Mango
- D. Papaya
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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. Pollenkitt
- B. Cellulosic intine
- C. Oil content
- D. Sporopollenin
Correct answer & explanation hidden
Log in to view solution →- A. Temperature
- B. Light
- C. O2 concentration
- D. CO2 concentration
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Log in to view solution →- A. Pachytene
- B. Diplotene
- C. Diakinesis
- D. Zygotene
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Log in to view solution →- A. hydroxyl and methyl
- B. carbonyl and methyl
- C. carbonyl and phosphate
- D. carbonyl and hydroxyl
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Log in to view solution →- A. ATP
- B. NADH
- C. NADPH
- D. Oxygen
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Log in to view solution →- A. Dumb-bell shaped
- B. Kidney shaped
- C. Rectangular
- D. Barrel shaped
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Log in to view solution →- A. Saccharomyces
- B. Mycobacterium
- C. Nostoc
- D. Oscillatoria
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Log in to view solution →- A. Larger nucleoli are present in dividing cells.
- B. It is a membrane-bound structure.
- C. It takes part in spindle formation.
- D. It is a site for active ribosomal RNA synthesis.
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Log in to view solution →- A. Fatty acid breakdown
- B. Formation of secretory vesicles
- C. Respiration in bacteria
- D. Activation of amino acid
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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. CO
- B. CO2
- C. SO2
- D. O3
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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. 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. 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. 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. 5th June
- B. 21st April
- C. 16th September
- D. 22nd April
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Log in to view solution →- A. 5th June
- B. 21st April
- C. 16th September
- D. 22nd April
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Log in to view solution →- A. Retrovirus
- B. Ti plasmid
- C. phage
- D. pBR 322
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Log in to view solution →- A. Retrovirus
- B. Ti plasmid
- C. λ phage
- D. pBR 322
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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)
Correct answer & explanation hidden
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. Bio-infringement
- B. Biopiracy
- C. Biodegradation
- D. Bioexploitation
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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. Apical meristems
- B. Vascular cambium
- C. Phellogen
- D. Axillary meristems
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Log in to view solution →- A. Halophytes
- B. Free-floating hydrophytes
- C. Carnivorous plants
- D. Submerged hydrophytes
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Log in to view solution →- A. Stem
- B. Adventitious root
- C. Tap root
- D. Rhizome
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Log in to view solution →- A. Ovules are not enclosed by ovary wall in gymnosperms.
- B. Selaginella is heterosporous, while Salvinia is homosporous.
- C. Horsetails are gymnosperms.
- D. Stems are usually unbranched in both Cycas and Cedrus.
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Log in to view solution →- A. Cell wall is present in members of Fungi and Plantae.
- B. Mushrooms belong to Basidiomycetes.
- C. Pseudopodia are locomotory and feeding structures in Sporozoans.
- D. Mitochondria are the powerhouse of the cell in all kingdoms except Monera.
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Log in to view solution →- A. Epidermis
- B. Pericycle
- C. Cortex
- D. Endodermis
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Log in to view solution →- A. Grasses
- B. Deciduous angiosperms
- C. Conifers
- D. Cycads
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Log in to view solution →- A. Uniflagellate gametes – Polysiphonia
- B. Biflagellate zoospores – Brown algae
- C. Gemma cups – Marchantia
- D. Unicellular organism – Chlorella
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Log in to view solution →- A. Herbarium - i. It is a place having a collection of preserved plants and animals. b. Key - ii. A list that enumerates methodically all the species found in an area with brief description aiding identification. c. Museum - iii. Is a place where dried and pressed plant specimens mounted on sheets are kept. d. Catalogue - iv. A booklet containing a list of characters and their alternates which are helpful in identification of various taxa.
- B. i iv iii ii
- C. iii ii i iv
- D. ii iv iii i
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Log in to view solution →- A. Mustard
- B. Cycas
- C. Mango
- D. Pinus
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Log in to view solution →- A. Mustard
- B. Cycas
- C. Mango
- D. Pinus
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Log in to view solution →- A. Neurospora
- B. Alternaria
- C. Agaricus
- D. Saccharomyces
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Log in to view solution →- A. Neurospora
- B. Alternaria
- C. Agaricus
- D. Saccharomyces
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Log in to view solution →- A. It functions as an enzyme.
- B. It functions as an electron carrier.
- C. It is a nucleotide source for ATP synthesis.
- D. It is the final electron acceptor for anaerobic respiration.
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Log in to view solution →- A. It functions as an enzyme.
- B. It functions as an electron carrier.
- C. It is a nucleotide source for ATP synthesis.
- D. It is the final electron acceptor for anaerobic respiration.
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Log in to view solution →- A. Green sulphur bacteria
- B. Nostoc
- C. Cycas
- D. Chara
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Log in to view solution →- A. Green sulphur bacteria
- B. Nostoc
- C. Cycas
- D. Chara
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Log in to view solution →- A. – 120°C
- B. – 80°C
- C. – 196°C
- D. – 160°C
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Log in to view solution →- A. Ferric
- B. Ferrous
- C. Free element
- D. Both ferric and ferrous
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Log in to view solution →- A. Fusion of two male gametes of a pollen tube with two different eggs
- B. Fusion of one male gamete with two polar nuclei
- C. Fusion of two male gametes with one egg
- D. Syngamy and triple fusion
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Log in to view solution →- A. Magnesium
- B. Sodium
- C. Potassium
- D. Calcium
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Log in to view solution →- A. Hydrilla
- B. Yucca
- C. Banana
- D. Viola
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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
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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.
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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.
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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. 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. Wildlife safari parks
- B. Sacred groves
- C. Botanical gardens
- D. Seed banks
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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. Eutrophication - Nutrient enrichment
- B. Sanitary landfill - Waste disposal
- C. Snow blindness - UV-B radiation
- D. Jhum cultivation - Deforestation
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Log in to view solution →- A. (1) ii i iii iv
- B. (2) i iii iv ii
- C. (3) iii iv i ii
- D. (4) i ii iv iii
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Log in to view solution →- A. (1) Commensalism
- B. (2) Mutualism
- C. (3) Parasitism
- D. (4) 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. (1) Protein folding
- B. (2) Protein glycosylation
- C. (3) Cleavage of signal peptide
- D. (4) Phospholipid synthesis
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Log in to view solution →- A. Protein folding
- B. Protein glycosylation
- C. Cleavage of signal peptide
- D. Phospholipid synthesis
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Log in to view solution →- A. (1) Enzymes of TCA cycle are present in mitochondrial matrix.
- B. (2) Glycolysis occurs in cytosol.
- C. (3) Glycolysis operates as long as it is supplied with NAD that can pick up hydrogen atoms.
- D. (4) Oxidative phosphorylation takes place in outer mitochondrial membrane.
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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.
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Log in to view solution →- A. (1) Polysome
- B. (2) Polyhedral bodies
- C. (3) Plastidome
- D. (4) Nucleosome
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Log in to view solution →- A. (1) Lampbrush – Diplotene bivalents
- B. (2) Allosomes – Sex chromosomes
- C. (3) Submetacentric – L-shaped chromosomes
- D. (4) Polytene – Oocytes of amphibians
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Log in to view solution →- A. (1) Proteins and lipids
- B. (2) DNA and RNA
- C. (3) Nucleic acids and SER
- D. (4) Free ribosomes and RER
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Log in to view solution →- A. (1) Thecodont, Diphyodont, Homodont
- B. (2) Thecodont, Diphyodont, Heterodont
- C. (3) Pleurodont, Monophyodont, Homodont
- D. (4) Pleurodont, Diphyodont, Heterodont
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Log in to view solution →- A. (1) iii ii iv i
- B. (2) i ii iii iv
- C. (3) ii iii i iv
- D. (4) iv i ii iii
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Log in to view solution →- A. (1) iv v ii iii
- B. (2) iv i ii iii
- C. (3) v iv i ii
- D. (4) v iv i iii
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Log in to view solution →- A. (1) Homology
- B. (2) Analogy
- C. (3) Convergent evolution
- D. (4) Adaptive radiation
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Log in to view solution →- A. (1) Psoriasis
- B. (2) Rheumatoid arthritis
- C. (3) Alzheimer’s disease
- D. (4) Vitiligo
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Log in to view solution →- A. (1) Forelimbs of man, bat and cheetah
- B. (2) Heart of bat, man and cheetah
- C. (3) Brain of bat, man and cheetah
- D. (4) Eye of octopus, bat and man
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Log in to view solution →- A. (1) b, c and e
- B. (2) a, b and c
- C. (3) b, d and e
- D. (4) a, c and e
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Log in to view solution →- A. (1) Elephantiasis
- B. (2) Ascariasis
- C. (3) Ringworm disease
- D. (4) Amoebiasis
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Log in to view solution →- A. (1) Vitamin D
- B. (2) Vitamin A
- C. (3) Vitamin B12
- D. (4) 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. (1) Epinephrine
- B. (2) Ecdysone
- C. (3) Estradiol
- D. (4) Estriol
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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. (1) Medulla oblongata : controls respiration and cardiovascular reflexes.
- B. (2) Limbic system : consists of fibre tracts that interconnect different regions of brain; controls movement.
- C. (3) Hypothalamus : production of releasing hormones and regulation of temperature, hunger and thirst.
- D. (4) Corpus callosum : band of fibers connecting left and right cerebral hemispheres.
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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.
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Log in to view solution →- A. (1) Aldosterone and Prolactin
- B. (2) Progesterone and Aldosterone
- C. (3) Estrogen and Parathyroid hormone
- D. (4) Parathyroid hormone and Prolactin
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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
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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. Earthworm
- B. Tunicate
- C. Moth
- D. Starfish
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Log in to view solution →- A. Amphibia
- B. Reptilia
- C. Aves
- D. Osteichthyes
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