Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Bond angle in alkenes is equal to

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`120^@ `
`109^@ 28'`
`180^@`
`60^@`

ANSWER :A
2.

Bond angle in alkene (C_2H_4) in equal to

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`120^(@) `
`109^(@) 28^(@)`
`180^(@) `
`60^@`

Solution :`12^(@) ` (`SP^(2)` hybridisation )
3.

Bond angle (H-O-H) in H_(2)O is

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`90^(@)`
`104^(@) 30`
`107^(@) 18`
`109^(@) 28`

Answer :B
4.

Bomb calorimeteris used to determine the value of "….........................".

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Solution :`DELTAU ` ( i.e.,internal energy CHANGE viz. heat of reaction at CONSTANT VOLUME)
5.

Boltzmann constant represents the gas constant per

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MOLE 
AVOGADRO number of MOLECULES 
any number of molecules 
molecule 

SOLUTION :`K = R/(N_0)`
6.

Boiling points of methanol, water and dimethyl ether are respectively 65^@ C, 100^@ C and 34.5^@C. Which of the following best explains the wide variations in boiling points?

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Density of water is 1g`ml^(-1)`, methanol is 0.79g`ml^(-1)` and dimethyl ether is 0.71g`ml^(-2)`
Molecular weight increases from water (18) to methanol (32) to dimethyl ether (46)
EXTENT of HYDROGEN bonding decreases from water to methanol, while it is absent in ether
The NUMBER of H ATOMS per molecule increases from water to methanol to ether.

Answer :C
7.

Boiling point of water is very high.

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Solution :Due to the presence of INTERMOLECULAR HYDROGEN bonding in water.
8.

Boiling point of water in Fahrenheit ?

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SOLUTION :`F = 9/5(""^(@)C) + 32` (b.p, of WATER `100 ""^(@)C`)
`= 9/5 xx 100+32=212F`
9.

Boiling point of heavy water is

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`100^@C`
`99^@C`
`101.42^@C`
`110^@C`

ANSWER :C
10.

Boiling point of covalent compound depends on intennolecular force. Intermolecular forces are the force of attraction and repulsion between interacting particles (atoms and molecules). This term does not include the electrostatic forces that exist between the two oppositely charged ions and the forces that hold atoms of a molecule together i.e., covalent bonds Which of the following hydrogen bonds is the strongest?

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O-H-----N
F- H-----F
O-H--0
0-H--F

Answer :B
11.

Boiling point of a liquid depends upon

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Temperature
VOLUME
PRESSURE
Moisture

ANSWER :C
12.

Boiling point is highest for

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HF
`H_2 O`
`NH_3`
`CH_4`

ANSWER :B
13.

Bohr.s theory is not applicable to

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H
`He^(+)`
`Be^(3+)`
`H^(+)`

ANSWER :D
14.

Bohr.s theory is applicable to

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`LI^(+2)`
`Li^(+)`
`He^(+)`
Both 1 and 3

Answer :D
15.

Bohr.s theory explains that when Hydrogen-like particles are excited, their electrons are shifted higher energy orbits. As such a state of atom is not stable, electrons return to lower energy orbits releasing quantum of energy. Thus different spectral lines are formed. The wave number of spectral line is given by bar(v) = (2pi^(2)me^(4)z^(2))/(c h^(3))[(1)/(n_(1)^(2)-(1)/(n_(2)^(2)))] For a hydrogen -like particle, ionization energy is 122.4 eV. Then what is the wave number of series limit spectral line of the particle of paschen seres (R = rydberg.s constant)

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`(R)/(9)`
`(R)/(3)`
R
4R

Answer :C
16.

Bohr.s theory explains that when Hydrogen-like particles are excited, their electrons are shifted higher energy orbits. As such a state of atom is not stable, electrons return to lower energy orbits releasing quantum of energy. Thus different spectral lines are formed. The wave number of spectral line is given by bar(v) = (2pi^(2)me^(4)z^(2))/(c h^(3))[(1)/(n_(1)^(2)-(1)/(n_(2)^(2)))] The angular momentum of an electron in a Bohr.s orbit of H-atom is 4.2178 xx 10^(-34) kg m^(2) sec^(-1). What is the wave number of the spectral line emitted when electron falls from this level to the next lower level

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`(7R)/(144)`
`5.5 XX 10^(-4) CM^(-1)`
`(8)/(9)R`
`1.1 xx 10^(-5)cm^(-1)`

Answer :A
17.

Bohr.s theory explains that when Hydrogen-like particles are excited, their electrons are shifted higher energy orbits. As such a state of atom is not stable, electrons return to lower energy orbits releasing quantum of energy. Thus different spectral lines are formed. The wave number of spectral line is given by bar(v) = (2pi^(2)me^(4)z^(2))/(c h^(3))[(1)/(n_(1)^(2)-(1)/(n_(2)^(2)))] When electrons of excited H-atoms make transition from fifth excited state to second orbit, then maximum number of different types of photos observed is

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3
6
10
4

Answer :C
18.

Bohr's postulate that mv r = (nh)/(2pi) is proved mathematically by

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Pauli.s exclusion principle
de BROGLIE WAVE nature of the electron
Heisenberg uncertainity principle
Sommerfield theory

Answer :C
19.

Bohr's equation for energy of an electron in a hydrogen atom is given as _________.

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`E = (-1312)/(n^2) KJ MOL^(-1)`
`E = (-1312)/(n^2h^2) KJ mol^(-1)`
`E = hv`
`E = (4pi^2 m e^4)/(n^2h^2) KJ mol^(-1)`

ANSWER :A
20.

Bohr radius of 1^(st) orbital of hyddrogen atom is 0529A.Assuming that the position of an electron in this orbit is determined with the accuracy of 0.5% of the radius,calculate the uncertainty in the velocity of the electron in hydrogen atom.

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Solution :UNCERTAINTY in POSITION`=/_\x`
`=(0.5%)/(100%)xx05.29A`
`=(0.5)/(100)xx10^(-10)xx0.529m`
`/_\x=2.645xx10^(13)m`
From Heisenberg.s uncertainty principle,
`/_\x./_\.+h/(4pi)`
`/_\x.m./_\V.+h/(4pi)`
`/_\vgt=h/(/_\x.m.4pi)`
`/_\v=(6.626xx10^(-34)Khm^(2)s^(-1))/(2.645xx10^(13)mxx9.11xx10^(-31)kGxx4xx3.14)`
`/_\v=2.189xx10^(8)m`
21.

Bohr proposed his atomic model based on Planck.s quantum theory and derived following relations for one electron system in C.G.S. units : For H -atom : r_(n) = n^(2) xx r_(1), E_(n) = E_(1) xx Z^(2), u_(n) = (u_(1))/(n), u_(1) = 2.19 xx 10^(8) cm//sec., E_(1) = -13.6eV For 1 electron system, other than H r_(n) = (n^(2) xx r_(1H))/(Z), E_(N) = (E_(1H) xx Z^(2))/(n^(2)), u_(n) = (u_(1H) xx Z)/(n). Later on de Broglie proposed the dual nature of electron and put forward his wave concept. The wave length of electron in an orbit was given lambda = h/mu The potential energy of electron in3^(rd) Bohr orbit of H-atom is :

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`-242 xx 10^(-12)` erg
`-4.84 xx 10^(-12)` erg
`+4.84 xx 10^(-12)` erg
`+2.42 xx 10^(-12)` erg

Answer :B
22.

Bohr’s atomic model could explain the spectrum o f ............

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Atoms or IONS which have TWO electrons.
Only dihydrogen MOLECULE.
`He^(-)` atom only.
Atoms or ions which contain only ONE electron

Answer :D
23.

Bohr.satomicmodel.

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Solution :Nells Bohr (1913) was the firstto explainquqntitatively thegeneralfeaturesof hydrogenatomstructure andtisspectrum.
Pitulatesof model:
(i) Theelectron is the hydrogenatom can movearoundthenucleusin acircularpathcalledorbitsstationarystatesor allowedenergystates. Theseorbitsare arrangestconcentrically aroundthe nucleus.
(ii) The energyof anelectronin the orbit DOESNOT changewith timeHowever the electron willmovefroma lowerstationarystateto ahigherstationarystatewhenrequiredor energyis emitted whenelectronmovesdoesnottakeplace in acontinuouschange
(iii) The frequency of radiationabsorbedoremitted whentransitionoccursbetweentwostationarystatesthatdifferin energyby
Where`E_(1)`and `E_(2)`are the energies of lowerand higherallowedenergystatesrespectivelythisexpressionis commonlyknown asbohrfrequencyrule
Thusanelectroncan MOVE only in thoseorbitsfor whichitsangularmomentumisintegralmultiple of `((h)/( 2pi))` that iswhyonlycertain fixedorbitsare ALLOWED.
24.

Bohr proposed his atomic model based on Planck.s quantum theory and derived following relations for one electron system in C.G.S. units : For H -atom : r_(n) = n^(2) xx r_(1), E_(n) = E_(1) xx Z^(2), u_(n) = (u_(1))/(n), u_(1) = 2.19 xx 10^(8) cm//sec., E_(1) = -13.6eV For 1 electron system, other than H r_(n) = (n^(2) xx r_(1H))/(Z), E_(N) = (E_(1H) xx Z^(2))/(n^(2)), u_(n) = (u_(1H) xx Z)/(n). Later on de Broglie proposed the dual nature of electron and put forward his wave concept. The wave length of electron in an orbit was given lambda = h/mu The circumference (in m) of 3^(rd) Bohr orbit in H-atom is :

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`3.0 XX 10^(-7)`
`3.0 xx 10^(-8)`
`3.0 xx 10^(-6)`
`3.0 xx 10^(-9)`

ANSWER :D
25.

Bohr model of atom is contradicted by

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Pauli EXCLUSION PRINCIPLE
PLANCK QUANTUM theory
HEISENBERG uncertainty principle
All of these

Answer :C
26.

Bohr’s theory based on the Planck’s quantum theory.

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ANSWER :T
27.

Bohr’s theory cannot explain the spectra of multi-electron atoms.

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ANSWER :T
28.

Bohr model of hydrogen atom contradicts dual behaviour of matter and Heisenberg’s uncertainty principle. Justiy.

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SOLUTION :According to de Broglie concept, a moving electron has both particle and wave character. But Bohr considered electron only as particles. HEISENBERG’s uncertainty principles RULES out the existence of definite paths for electrons. But in Bohr’s model, electron in an atom is ASSUMED to be REVOLVING around the nucleus through definite paths called orbits.
29.

BOH_((aq)) + H_((aq))^(+) rarr B_((aq))^(+) + H_2O, DeltaH = -48 KJ.Heat of ionisation of the weak base is

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`9.3 KJ
`- 9.3 KJ`
`48 KJ`
`- 5 KJ`

ANSWER :A
30.

Bohr’s atomic theory is not able to explain the atomic spectra of atoms containing ________electron.

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SOLUTION :more than 1
31.

Bohr’s theory is based on ________ of radiation.

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SOLUTION :Planck.s THEORY
32.

B(OH)_(3)+NaOH rarr Na[B(OH)_(4)]+H_(2)O How can this reaction is made to proceeds in forward direction ?

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ADDITION of cis-1,2-diol
Addition of BORAX
Addition of trans-1,2-diol
Addition of `Na_(2)HPO_(4)`

Answer :A
33.

B(OH)_(3)+NaOH hArr NaBO_(2)+Na[B(OH)_(4)]+H_(2)OHow can this reaction is made to proceed in forward direction ?

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addition of cis - 1, 2 DIOL
addition of borax
addition of trans-1, 2 diol
addition of `Na_(2)HPO_(4) Na_(2) HPO_(4)`

SOLUTION :
34.

B(OH)_3 is ……………… in nature.

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SOLUTION :ACIDIC
35.

Body centred cubic lattice has co-ordination number of

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8
12
6
4

Answer :A
36.

Body centred cubic lattic has a coordination number

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8
12
6
4

Solution :BCC has a COORDINATION NUMBER of 8.
37.

BOD_5 is

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waste DECOMPOSED in 5 days
oxygen USED in 5 days
microorganisms killed in 5 days
dissolved oxygen LEFT after 5 days

Answer :B
38.

BOD values less than 5 ppm indicates a water sample to be

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POOR in DISSOLVED oxygen
rich in dissolved oxygen
Highly polluted
Not SUITABLE for AQUATIC life

Answer :B
39.

BOD stands for____________.

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ANSWER :A::B::C::D
40.

B.O.D is connected with

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Organic MATTER
MICROBES
Microbes and organice matter
NONE of the above

Solution :BOD is CONNECTED with microbes and organic matter present in the WATER
41.

BOD is a measure of ......

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ORGANIC POLLUTANTS in WATER
Inorganic pollutants in water
Particulate MATTER in water
All of the above

Solution :Organic pollutants in water
42.

BOD is a measure of .................

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SOLUTION :ORGANIC POLLUTANT in WATER
43.

BOD is a measure of :

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Organic POLLUTANT in water
Oxides of SPN in air
INORGANIC pollutant in water
Particle matter in water.

Answer :A
44.

BOD is

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Biological oxygen DEFICIT
BIOSPHERE oxygen demand
Biological oxygen demand
NONE of the above

SOLUTION :B.O.D is biological oxygen demand
45.

Blue compound Q is:

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`Co(BO_(2))_(2)`
`NaCoPO_(4)`
`CoO.AI_(2)O_(3)`
`CoO.MgO`

Solution :
`Q RARR NaCoPO_(4)`,
`R rarr Mg(NH_(4))PO_(4)`
46.

Blue colour of C is due to the formation of :

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`Mn^(2+)`
`CrO_(5)`
`I_(2)`
`H_(2)TiO_(4)`

ANSWER :C
47.

Blue color of concentrated ammoniated solution of alkali metals get converted into ......... color.

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BLACK - BLUE
Black - green
Violet
Colorless

Answer :A
48.

Blood may be purified by

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Dialysis
Electroosmosis
Coagulation
Filtration

Answer :A
49.

Blood is a buffer of H_(2)CO_(3) and[HCO_(3)^(-)]with pH = 7.40. Given K_(1)of H_(2)CO_(3) = 4.5xx10^(-7). What will be the ratio of [HCO_(3)^(-)] to [H_(2)CO_(3)] in the blood ?

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SOLUTION :`H_(2)CO_(3)hArr H^(+)+HCO_(3)^(-), K_(1)=([H^(+)][HCO_(3)^(-)])/([H_(2)CO_(3)]) or ([HCO_(3)^(-)])/([H_(2)CO_(3)])=(K_(1))/([H^(+)])`
`PH = 7.40` means - log `[H^(+)] = 7.4` or log `[H^(+)] = -7.4= bar(8).6or [H^(+)]=3.981 xx 10^(-8)`
`([HCO_(3)^(-1)])/([H_(2)CO_(3)])=(4.5xx10^(-7))/(3.981xx10^(-8))=11.3`.
50.

Blister copper is

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Impure Cu
Cu alloy
PURE Cu
copper having 1% impurity

Solution :Theoretically AL can reduced MgO to Mg above 1623 K .SINCE Mg is costelier than Al and a higher TEMPERATURE (gt 1623 K) is difficult to maintain. THEREFORE, Mg is usually obtained by reduction of MgO by Al .