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.

Process to get pure metals from ores by removing unwanted material or impurity is known as....

Answer»

METALLURGY 
ELECTROLYSIS 
REDUCTION 
PURIFICATION 

ANSWER :A
2.

.............process is used in the extraction of sliver and gold from their ores.

Answer»


ANSWER :Mac-Arthur-Forest CYANIDE
3.

Process followed before reduction of carbonate ore is

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CALCINATION
roasting
liquation
polling

Solution :Calcination is HEATING ore in absence of air to REMOVE moisture and volatile IMPURITIES. Carbonate ores decomposed to CORRESPONDING oxides as a result of calcination.
4.

Process, AtoB represent:

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isobaric
isochoric
isothermal
adiabatic

Answer :B
5.

Procaine, lidocaine, cemitidine, propofol, iso flurane.

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SOLUTION :CEMITIDINE. It is an antacid whereas others are ANAESTHETICS.
6.

Problems based on Faraday's laws of electrolysis.

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Solution :Determine the quantity of electricity (Q in coulomb) from the PRODUCT of electriccurrent (I in ampere) and time (t in second).
(2) Electrochemical EQUIVALENT of a SUBSTANCE can be calculated from the relation, `W=ZxxQ` if amount (W) of the substance deposited at the electrode is known.
(3) The amount of substance deposited at the electrode by 1 faraday of electricity = 1 gram - equivalent. If (a) W g of the substance is produced by Q coulomb of electricity, its gram - equivalent `=(W//Qxx96500)g`. (b) If 1 gram - equivalent of the substance is Eg, then Q coulomb electricity will produce `(E//96500xxQ)g` of the substance.
7.

Probability of finding electrons is nearly equalto 90- 60%for

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an ORBITAL
a SUB SHELL
an orbit
an ATOM

ANSWER :A
8.

Principles of thermodynamics can be applied to………………..

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SOLUTION :PYROMETALLURGY
9.

Principal, magnetic and azimuthal quantum numbers are respectively related to:

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SIZE, ORIENTATION, SHAPE
size, shape, orientation
shape, size and orientation
None of these

Answer :A
10.

Primary valency is satsified by

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only NEGATIVE ligands
only POSITIVE ligands
negative and NEUTRAL ligands
positive and neutral ligands

Answer :A
11.

Primary valency denotes

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coordination NUMBER
number of LIGAND
oxidation number
effective atomic number

Answer :C
12.

Primary, secondry, tertiary nitroalkanes can be identified by the action of :

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`NHO_2 + NAOH `AQ.
`CHCl_3 + NaOH` aq.
`CHCl_3 + KOH` alc.
None

Answer :A
13.

Primary, secondry, and tertiary amines may be scparated by using :

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ETHANOYL CHLORIDE
DIETHYL OXALATE
Thionyl chloride
None

Answer :B
14.

Primary structure of proteins refer to _______ .

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AMINO ACID present
the shape of protein MOLECULE
the peptide bond
the sequence of amino acid

SOLUTION :the sequence of amino acid
15.

Primary structure of protein is due to

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SEQUENCE in which `alpha`-amino acids are linked to one ANOTHER
sequence in which amino ACID of one polypeptide chain are joined to other chain
the folding patterns of polypeptide chains
the pattern in which the polypeptide chains are arranged.

Solution :The sequence in which the `alpha`-amino acids are linked to one another in a protein MOLECULE is called its PRIMARY structure.
16.

Primary , secondary , tertiary amines can be separated by the following except :

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Fractional distillation
Fractional METHOD ysubg duetgtk OXALATE
Hinsberg' s method using ` C_6 H-5 SOCl`
Selective crtystallisation

Solution :Statement is self EXPLANATORY
17.

Primary, secondary and tertiary amines can be differentiated by using _________

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thionyl CHLORIDE
NITROUS ACID
ethanoyl chloride
hydroxylamine

ANSWER :B
18.

Primary, secondary and tertiary alcohols may be distinguished by converting them into the corresponding nitroparaffins which are then treated with

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aqueous NaOH
CONC. `H_(2)SO_(4)`
conc. HCI
`NaNO_(2)+ "dilHCL"`

SOLUTION :The NITROALKANES are then treated with `HNO_(2)` i.e., `NaNO_(2) + "dilute" HCI.`
19.

Primary,secondary and tertiary alcohols may be distinguishedby employing

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HOFFMANN's test
Fehling SOLUTION. test
Lucas test
None of the above

Answer :C
20.

Primary, secondary and tertiary amines can be separated by using

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DIETHYL ether
Hinsberg's reagent
HOFMANN method
Filterpaper

Solution :Himsberg.s, Hofmann
21.

Primary , secondary and tertiary alcohols can be distinguished hy

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Baeyer's REAGENT
Fehling's SOLUTION
SULPHURIC acid
LUCAS reagent

ANSWER :D
22.

Primary, secondary and tertiary alcohols are distinguished by

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Cu/573 K
Victor Meyer's test
conc. HCl is presence of ANHYD. `ZnCl_(2)`
`Br_(2)-H_(2)O`

Answer :A::B::C
23.

Primary , secondary and tertiary alcohols can be distinguished by employing :

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OXIDATION
VICTOR MEYER's test
Lucas test
All of these

Answer :D
24.

Primary , Secondaryand Tertiary alcohols can be distinguished by

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Oxidation
Victormryer's test
Lucas reagent
All of these

Answer :D
25.

Primary, secondary and teritary amines can be distinguished using hinsberg's reagent. What is Hinsberg's reagent?

Answer»

Solution :Benzene sulphonyl chloride `(C_6H_5-SO_2-Cl)` is KNOWN as Hinsberg REAGENT.
26.

Primary (or) Secondary (or) Tertiary alcohols on oxidation finally give

Answer»

ESTERS
KETONES
ETHERS
Carboxylic acids

ANSWER :D
27.

Primary, secondaryamineforms insolublenitrosoamine with

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`HNO_(3)`
`HNO_(2)`
dil.HCl
dil.`H_(2)SO_(4)`

ANSWER :B
28.

Primary, seconary and tertiary alcohols can be distinguished by_____test.

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SOLUTION :LUCAS or Victor-Meyer's TEST.
29.

Primarynitrocompounds react withnitrousacidto fromnitrolicacidwhichdissolveinNaOH togive.

Answer»

YELLOW SOLUTION
red solution
bluesolution
colourlesssolution

Answer :B
30.

Primary nitroalkanes on hydrolysis gives:

Answer»

RCOOH + `NH_2OH`
RCOOH
`NH_2OH`
RCOR

Answer :A
31.

Primarynitroalkanes are obtainedin goodyieldby oxidisingaldoximeswith thehelpof

Answer»

trifluoroperoxyaceticacid
ACIDIFIEDPOTASSIUM PERMANGANATE
concentratednitricacid
potassiumdichromateand dilutesulphuricacid

SOLUTION :INFORMATION
32.

Primary and secondary valencies of platinum in the complex, [Pt(en)_(2)Cl_(2)] are

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4,6
2,6
4,4
6,4

Answer :B
33.

Primary nitroalkanes are obtained in good yield by oxidising aldoximes with the help of

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trifluoroperoxy acetic acid
acidified potassium permanganate
concentrated nitric acid
potassium dichromate and dilute SULPHURIC acid

Solution :Primary nitroalkanes are obtained in good yield by oxidising aldoximes with the help of trifluoroperoxy acetic acid.
`underset("ALDOXIME")(R CH = NOH) underset(CF_(3)COOOH)overset([O])RARR RCH = overset(O^(-))overset(|)(N^(+))-OH hArr underset((1^(@) "nitroalkane"))(R CH_(2)NO_(2))`
34.

Primary nitroalkanes are obtained in good by oxidising aldoximes with the help of

Answer»

trifluoroperoxyacetic ACID
acidified potassium permanganate
concentrated nitric acid
Potassium dichromate and DILUTE sulphuric acid

Solution :PRIMARY nitro ALKANES are obtained in good YIELD by oxidizing aldoximes with the help of Trifluoroperoxyacetic acid.
35.

Primary and secondary valences in coordination compounds are respectively

Answer»

both ionisable.
both non-ionisable.
both non-ionisable.
ionisable, non-ionisable.

Answer :D
36.

Primary and secondary nitroalkanescontaining alpha-Hatom show property of

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CHAIN isomerism
tautomerism
optical isomerism
geometrical isomerism

Answer :A::B
37.

Primary and secondary structures of nucleic acid reveals

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Nucleotide sequence & SINGLE or DOUBLE helix structure
Amino ACID sequence & 3D-folding
Amino acid sequence & SHAPE of protein
Single/double helix structure and Nucleotide sequence. 

Answer :D
38.

Primary and secondary nitroalkanes containing alpha-H atom show property of……………..

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CHAIN isomerism
TAUTOMERISM
OPTICAL isomerism
Geometrical isomerism

Answer :B
39.

Primary and secondary nitroalkanes containing alpha-H atom show property of -

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Chain isomerism
Tautomerism
Optical isomerism
Geometrical isomerism

Solution :Tautomerism is the ABILITY of certain organic compounds to REACT in ISOMERIC structure that DIFFER from each other in he position of a HYDROGEN atom and double bond since, primary and secondary nitrolkanes differ from each other in the position of `alpha` - hydrogen atom, they show property of tautomerism.
40.

Primary and secondary amines can be distinguished by the action of:

Answer»

`CS_(2)//HgCl_(2)`
`NaNO_(2)//HCl`
`CHCl_(2)//KOH`
NaOH

Solution :NaOH cannot distinguish AMINES
41.

Name the test with which primary, secondary and tertiary amines can be distinguished.

Answer»

`Br_2//KOH`
`HCLO`
`HNO_2`
`NH_3`

ANSWER :C
42.

Primary and secondary amines can be distinguished by

Answer»

SCHIFF's reagent
Carbylamine reaction
Hofmann's bromamide reaction
Biuret reaction

Answer :2
43.

Primary and secondary amide exist as dimer in solid and pure liquid state.

Answer»


ANSWER :A::B::C::D
44.

Primary and secondary amines are distinguished by :

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`Br_2//KOH`
`HClO`
`NHO_2`
`NH_3`

Solution :` HNO_2` is USED to distinguish between ` 1^@, 2^@`, and ` 3^@` AMINES .
45.

Primary and secondary alcohols on action of reduced copper give

Answer»

ALDEHYDES and KETONES respectively
ketones and aldehydes respectively
only aldehydes
only ketones

Answer :A
46.

Primary and secondary alcohols on action of reduced copper gives

Answer»

ALDEHDES and KETONES resectively
Ketones and ALDEHYDES respectively
Only aldehydes
Only ketones

Answer :A
47.

Primary and secondary alcohols on heating with copper give :

Answer»

ALDEHDES and KETONES resectively
Ketones and ALDEHYDES respectively
Only aldehydes
Only ketones

Answer :A
48.

Primary and secondary alcohols on action is reduced copper give

Answer»

ALDEHYDES and Ketones respectively
Ketones and aldehydes respectively
Only aldehydes
Only ketones

Solution :`underset(1^(@)) "ALCOHOL")(R-CH_(2)OH) OVERSET(Cu)to CHO+H_(2)`
`underset(2^(@) "Alcohol")(R-underset(OH)underset(|)CH-Roverset(Cu) to R-underset(O)underset(||)C-R+H_(2)`
49.

Primary and secondary alcohols are dehydrogenated by copper at 573 K to aldehydes and ketones respectively. In contrast tertiary alcohols are dehydrated to alkenes by heating with copper at 573 K. Similarly, primary alcohols are easily oxidised to form first an aldehyde and then a carboxylic acid while secondary alcohols are oxidised to ketones which are further oxidised to form a mixture of acids. Tertiary alcohols are oxidised with difficulty and with strong oxidising agents in acidic medium. They form first ketones and then acids. In the case of alcohols containing carbon-carbon double bond, some oxidising agents oxidise both double bond and OH group while other reagents donot affect C-Chond. The product of the reaction : CH_(3)- underset(CH_(3)) underset(|)(CH)-CH_(2)OH overset(PC C) underset("Oxidation")to is

Answer»

2-Methylpropanal
2-Methylpropanoic
BUTANOIC acid
Butan-2-one

Solution :`CH_(3)underset(CH_(3))underset(|)(CH)CH_(2)OHunderset("Oxudation")overset(PC C)toCH_(3)underset(CH_(3))underset(|)(CH)CHO`
50.

Primary and secondary alcohols are dehydrogenated by copper at 573 K to aldehydes and ketones respectively. In contrast tertiary alcohols are dehydrated to alkenes by heating with copper at 573 K. Similarly, primary alcohols are easily oxidised to form first an aldehyde and then a carboxylic acid while secondary alcohols are oxidised to ketones which are further oxidised to form a mixture of acids. Tertiary alcohols are oxidised with difficulty and with strong oxidising agents in acidic medium. They form first ketones and then acids. In the case of alcohols containing carbon-carbon double bond, some oxidising agents oxidise both double bond and OH group while other reagents donot affect C-Chond. Butan-2-ol on heating with Cu at 573 K gives

Answer»

butanal
butan-2-one
PROPANONE
but-2-ene

Solution :`CH_(3)CH_(2)UNDERSET("Butan-2-ol")underset(OH)underset(|)(CHCH_(3))overset(Cu,573)toCH_(3)CH_(2)underset("Butan-2-one")underset(O)underset(||)(C CH_(3))`