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.

Flourine is highly reactive because

Answer»

F-F BOND ENERGY is high
F-F bond energy is law
It is gaseous at room temperature
F has a SMALLER size

Answer :B
Discussion
2.

Flocculation value is expressed in terms of -

Answer»

Millimole/litre
Mole/litre
Grams/litre
Mole/millilitre

Answer :A
Discussion
3.

Flocculation value of BaCl_(2) is much less than that of KCl for sol A and flocculation value of Na_(2)SO_(4) is much less than that of NaBr for sol B. The correct statement among the following is:

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Both the sols A and B are NEGATIVELY charged
SOL A is POSITIVELY charged and Sol B is negatively charged
Both the sols A and B are positively charged
Sol A is negatively charged and sol B is positively charged

Solution :In first case the GIVEN compounds have same anion but different cations having different charge hence they will precipitate negatively charged sol i.e., A.
In second case the given compounds have SIMILAR cation but differentcharge. Hence, they will precipitate positively charged sol i.e., B.
Discussion
4.

Flint glass is obtained from which of the following ?

Answer»

Zinc and barium borosilicate
Sand, red LEAD and POTASIUM carbonate
SODIUM ALUMINIUM borosilicate
Pure silica and zinc OXIDE

Solution :Flint glass is obtained from red lead and potassium carbonate.
Discussion
5.

Flash bulbs contain wire or foil of Mg packed in an atmosphere of:

Answer»

`SO_3`
`O_2`
AIR
`N_2`

ANSWER :B
Discussion
6.

Flase statement is

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rate of zero order RACTION is INDEPENDENT of INITIAL CONCENTRATION of ration is inversely concentration of the reactant
half life of a third order reaction is inversetly proportional to the square of intial concentration of the reactant
for a first order raction `t_(1//2) = (0.693)/(k) `
molecularity of a raction may be zero negative or evenfractional

SOLUTION :Molecularity of a reaction can never be zero , fractional or negative .
Discussion
7.

Flame test is not gives by

Answer»

`Mg^(2+)` IONS
`Ba^(2+)` ION
`Be^(2+)` ions
`Ca^(2+) `ions

Answer :a,C
Discussion
8.

Flame test is not given by :

Answer»

`Mg^(2+) `
`CA^(2+) `
`K^(+) `
`SR^(2+)`

ANSWER :A
Discussion
9.

Flagpole intersection is present in:

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BOAT FORM of cyclohexane
Chair form of cyclohexane
Anti form of n-butane
Fully ECLIPSED form of n-butane

Answer :A
Discussion
10.

Fixed mass of a gas is subjected to the changes as shown is diagram, calculate T_(3),T_(4),P_(1),P_(2) and V_(1) as shown is diagram .Considering gas obeys PV = nRT equation.

Answer»


Answer :(i)`P_(1) = 5 atm; ` (ii) `T_(3) =360 K`; (iii) `V_(1) = 16`lit; (iv) `P_(2) =1.5 atm`; (v) `T_(4) = 90 K`
Discussion
11.

Flag - pole interaction is present in

Answer»

BOAT form of CYCLOHEXANE
CHAIR form of cyclohexane
Anti form of n-butane
Fully ECLIPSED form on n-butane

Answer :A
Discussion
12.

Fixedamount of an ideal gas contained ina sealed rigidvessl(V=24.6 litre) at 1.0baris heatedreversiblyfrom 27^(@)to 127^(@) C. Determine charge in G ibb's energy (|DeltaG|in Joule)if entropys=10+10^(+2)T(J//K)

Answer»

<P>

SOLUTION :`dG= V.dP - S.DT"","at CONST. volume"`
`DeltaG = V.DeltaP- int(10+10^(-2)T).dT""(P_(1))/(T_(1))=(P_(2))/(T_(2)) rArrP_(2) = 1 xx 400//300`
`DeltaG=24.6xx((4)/(3)-1) xx 100 -[10xx100+10^(2)xx((T_(2)^(2))/(2)-(T_(1)^(2))/(2))]`
`DeltaG=24.6xx(1)/(3)xx 100 -[1000+10^(2)xx((160000)/(2)-(90000)/(2))]`
` = 24.6 xx(1)/(3)xx100-1000-350=-530 J`
Discussion
13.

Five valence electrons of .^(15)P re labelled asIf the spin quantum numbers of B and Z is+(1)/(2), the group of electron with three of the quantum number (n,l,m) same are .

Answer»

AB,XYZ,BY
AB
XYZ,AZ
AB,XYZ

Answer :2
Discussion
14.

Fixation of nitrogen means:

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Reaction of nitrogen with oxygen
Conversion of FREE atmospheric nitrogen into nitrogen COMPOUNDS
Decomposition of NITROGENOUS compounds to yield free nitrogen
The action of denitrifying bacteria on nitrogen compounds

Answer :B
Discussion
15.

Five thousand with three significant figures is written as:

Answer»

5000
`5.0 XX 10^3`
`5.00 xx 10^3`
0.50 X 10^4`

ANSWER :C
Discussion
16.

Five moles of an ideal gas undergoes reversible expansion from 1 L to 4 L at 27^oC. The work done and enthalpy change for the process respectively are

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`- 12.5 KJ and - 18.2 kJ`
`- 17.3 kJ and 0 kJ`
`+ 17.3 kJ and +22.5 kJ`
`+ 21.5 kJ and 0 kJ`

ANSWER :C
Discussion
17.

Five most abundant elements in the living cell are

Answer»

`C,H,O,N,Fe`
`C,H,O,N,P`
`C,H,N,Mg, Ca `
`C, H, Fe, Mg, Ca`

Solution :`C, H, O, N, P` are the five most abundant elements in the living CELL. C, H, O and N constitute about 95% of the weight of the cell contents. The biological importance of the principle elements is mainly due to their ability to FORM more stable covalent bonds than any other elements with the same valencies. They are found in the found in the form of carbohydrates, `H_2O`, proteins, nucleic acids (DNA/RNA), for life, both as a structural MATERIAL in ANIMALS and plants. About 60% bones and teeth are `Ca_(3)(PO_4)_(2)`.
Discussion
18.

Five isomeric para- disubtitutedaromaticcompounds (A) to (E) withmolecularformulaC_(6)H_(6)O_(2)were givefor identification . Basedon the followingobservations findstructureof thecompounds. (i)Both (A) and (B) form a silvermirrorwithTollen'sregent, also(B) givesa positivetest withFeCl_(3)solution . (ii) (C) givespositiveiodoform test .(iii)(D)isreadily extractedin aqueousNaHCO_(3) solution. (iv) (E) on acid hydrolysis gives 1,4- dihydroxybenzene.

Answer»

SOLUTION :
Discussion
19.

Five moles of an ideal gas at300 K, expandedisothermallyfroman intinalpressueof4 atmto a final pressureof 1 atm againsta cont. extpressureof 1 atm . Calculate q ,w, DeltaU & DeltaH. Calculatethe correspondingof allif theaboveprocessis carriedoutreversibly .

Answer»


Solution :`W_("IRR") = - P_("ext")[DeltaV]= -P_("ext")((NRT)/(P_(2))(nRT)/(P_(1)))`
` =- P_("ext") xx nRT [1-(1)/(4)]`
`= 9353.25 J`
`W_("rev") =- 2.303 nRT log .(P_(1))/(P_(2)) = -17288. 47J`
`DeltaU = 0,""DeltaH = 0`
Discussion
20.

Five hypothetical gases having name A,B,X,Y and Z are having boiling point in the range of 4.9 K to 6.9 K. Gases A and B are reacting to produce gases X,Y and Z as per the following stoichiometric equation in a container of volume 0.9 L pressure 1 atmosphere and at temperature 8.513K ( at equilibrium ) . A_((g)) + B_((g)) rarr X_((g)) + Y _((g)) + 2Z_((g)) At the condition of equilibrium, which of the following statement is correct ?

Answer»

`K_(P) = K_(C )`
`K_(P) gtK_(C )`
`K_(P) ltK_(C )`
`K_(P) GT gtK_(C )`

Answer :C
Discussion
21.

Fittig reaction.

Answer»

Solution :Fittig reaction : ARYL halides give analogus compounds
when treated with SODIUM in dry ether, in which TWO aryl groups
undergocouplng reaction to form diphenyl is called Fittig reaction.
Discussion
22.

Five amine synthesis are outlined below. In each reactions box enter a single letter designating the best reagent and conditions selected from the list at the bottom of the page. a) i) LiAlH_(4) in ether, ii) H_(2)O & base b) C_(2)H_(5)NH_(2)(cat. H^(+)) c) NaCN in alcohol d) H_(2) & Ni catalyst or H_(2) & Pd catalyst e) NaN_(3) in alcohol f) (CH_(3)CO)_(2) & pyridine g) C_(2)H_(5)Br h) (i) 2CH_(3)I & pyridine j) KOH in H_(2)O

Answer»


SOLUTION :
Discussion
23.

Fitting reaction is used to prepare

Answer»

DIPHENYL
HIGHER ALKANE
ALKYL benzene
phenol

Answer :A
Discussion
24.

Fishes die by sewage because

Answer»

Of its bad smell
It REPLACES FOOD MATERIAL of fishes
It INCREASES oxygen competition among fishes
`CO_2` is mixed in LARGE amount in water

Answer :C
Discussion
25.

Fishes die in water bodies polluted by sewage due to :

Answer»

pathogens
clogging of GILLS by silt
reduction of `O_2`
FOUL smell

Answer :C
Discussion
26.

Fisher- Tropschprocessis usedin themanfactureof :

Answer»

ETHANE
BENZENE
SYNTHETIC PETROL
LPG

Answer :C
Discussion
27.

Fish do not grow in warm water as in cold water, why ?

Answer»

SOLUTION :The AMOUNT of dissolved oxygen in WARM water is less than in COLD water .
Discussion
28.

Fischer Tropsch process is used for the manufacture of

Answer»

SYNTHETIC petrol
Thermosetting plastics
Ethanol
Benzene

Answer :A
Discussion
29.

First three ionisation energies (in kJ/mol) of three representative elements are given below: {:("Element",IE_(1),IE_(2),IE_(3)),(P,495.8,4582,6910),(Q,737.7,1451,7733),(R,577.5,1817.2745):} then incorrect option is:

Answer»

Q: alkaline earth metal
P: alkali metals
R: s-block element
They BELONG to same period

Solution :R is p-block element, because DIFFERENCE between `IE_(2) and IE_(3)` is not very high as compared to between `IE_(1) and IE_(2)`, hence stable oxidation state of R will be higher than +2.
Discussion
30.

Fischer projection formula of this compound can be represented as :

Answer»




ANSWER :A::B::C::D
Discussion
31.

Fischer projection indicates:

Answer»

HORIZONTAL SUBSTITUENTS above the plane
VERTICAL substituents above the plane
Both horizontal and vertical substituents
Both horizontal and vertical substituents above the plane

Answer :A
Discussion
32.

First theory to explain the formation of complexes was proposed by

Answer»

WERNER
PAULING
SIDGWICK
Mullikan

Answer :A
Discussion
33.

First stable compound of inert gas was prepared by:

Answer»

RAYLEIGH and Ramsay
Bartlett
Gfrankland and Lockyer
Cavendish

Answer :B
Discussion
34.

First order reaction complete 50% in 16 minutes .How much time require to complete 87.5%?

Answer»

SOLUTION :165.58 MINUTES
Discussion
35.

First I.P. of Mg is …… than Al

Answer»

LESS
More
Equal
NONE of these

Solution :More because of STABLE CONFIGURATION of MG.
Discussion
36.

First group of basic radicals (cation) involve analysis of presence of Pb^(2+), Ag^(+)and Hg_(2)^(+2)ions in the salt. The chloride salts of these metal ions are sparingly soluble in water and their K_(sp) values are as follows. K_(sp) [PbCl_(2)] = 10^(-12), K_(sp)[AgCl] = 10^(-10), K_(sp)[Hg_(2)Cl_(2)] = 10^(-18) Then sequence of precipitation of these salts on adding dilute HCl to the equimolar mixture of these metal ions will be :

Answer»

`PbCl_(2), AgCl` and then `Hg_(2)Cl_(2)`
`Hg_(2)Cl_(2), PbCl_(2)` and then AgCl
`AgCl, PbCl_(2)` and then `Hg_(2)Cl_(2)`
`Hg_(2)Cl_(2), AgCl` and then `PbCl_(2)`

SOLUTION :On using equimolar mixture of these metal ions the required `CL^(-)` concentration is in order `Hg_(2)Cl_(2) lt AgCl lt PbCl_(2)`
HENCE SEQUENCE of ppt. will follow this order.
Discussion
37.

First, fourth and fifth periods of the long form the periodic table consist of elements respectively:

Answer»

1.2,8,8
2.8,8,18
3.2,18,18
4.2,8,18

Answer :C
Discussion
38.

First chlorinated insecticide is

Answer»

DDT
Gammaxene
BHC
Pyrene

ANSWER :A
Discussion
39.

First chemotherapeutic agent is

Answer»

analgin
organo aresnic COMPOUND
salol
dichlophenac sodium

Answer :B
Discussion
40.

First and second ionisation energies of Mg (g) are 740 and 1450kJ mol ^(-1). Calculate percentage of Mg^(+)(g)and Mg ^(2+) (g),if 1 g of Mg(g) absorbs 50 kJ of energy.

Answer»

SOLUTION :Number of MOLES of 1G of `Mg = 1/24 = 0.0417`
Energy required to convert mg(g) to `Mg ^(+)(g) =0.0417 x 740 =30.83kJ`
Remaining energy `= 50 -30.83 =19.17 kJ`
Number of moles of `Mg ^(2+)` formed `= (19.17)/(1450) =0.0132`
Thus, remaining `Mg ^(+)` will be = `0.0417 - 0.0132=0.0285`
`%Mg ^(+) = (0.0285)/(0.0417) xx 100= 68.35%`
`%Mg ^(+2) =100 -68.35 =31. 65%`
Discussion
41.

First and second electron gain enthalpies of oxygen are -141 and + 702 kJ mol^(-1) How is large number of oxides accounted for ?

Answer»

SOLUTION :`O(g)+e^(-)rarrO^(-)(g),DeltaH=-141kJmol^(-1),O^(-)(g)+e^(-)rarrO^(2-)(g),DeltaH=+702kJ mol^(-1)`.
Formation of `O^(-)` is EXOTHERMIC and `O^(2-)` is endothermic. However, oxygen is divalent and forms OXIDE with stable `ns^(2)np^(6)` configuration. Further, the lattice ENERGIES of oxides are very HIGH on account of greater electrostatic forces of attraction.
Discussion
42.

Fire extinguishers contain H_2 SO_4 and

Answer»

`CaCO_(3)`
`Na_(2)CO_(3)`
`NaHCO_(3)`
`NaHCO_(3)` and `Na_(2)CO_(3)`

SOLUTION :`NaHCO_(3)` and `Na_(2)CO_(3)` decomposes to GIVE `CO_(2)` which extinguish fire.
Discussion
43.

Fire which results from the combustion of alkali metals can be extinguished by

Answer»

`C Cl_4`
Sand
Water
kerosene

Answer :A
Discussion
44.

Fire fighter uses clohts of

Answer»

LEXAN
NYLON
Terylene
Nomex

ANSWER :D
Discussion
45.

Fire extinguisher Pyrene is :

Answer»

`CO_(2)`
`C Cl_(4)`
`CHCl_(3)`
`H_(2)CO_(3)`

Solution :`C Cl_(4)`
Discussion
46.

Fire extinguiser contains:-

Answer»

`CaCO_(3)`
`H_(2)SO_(4) & NaHCO_(3)` solution
`NaHCO_(3)` solution
`CaCO_(3)&H_(2)SO_(4)` solution

Answer :B
Discussion
47.

Fire extinguisher is:

Answer»

`CO_(2)`
`CCl_(4)`
`CHCl_(3)`
`H_(2)CO_(3)`

SOLUTION :NA
Discussion
48.

Finkelstein reaction is used to prepare

Answer»

ALKYL chlorides
alkyl bromides
alkyl lodides
alkyl fluorides

Answer :C
Discussion
49.

Finkelstein reaction.

Answer»

Solution :Finkelstein reaction : When ALKYL chloride of bromide is
treated with SODIUM iodide in the presence of dry acetone, alkyl iodide
is obtained. This reaction is knownn as Finkelstein reaction.
`{:(R-Cl, + NaI underset("acetone")overset("dry")(rarr),RI,+NaCl),("alkyl chloride",,"alky iodide",):}`
`{:(R-Br, + NaI underset("acetone")overset("dry")(rarr),RI,+NaBr),("alkyl bromide",,"alky iodide",):}`
In the presence of dry acetone, sodium chloride of sodium bromide
gets precipitated.
`{:(C_(2)H_(5)Br, + NaI underset("acetone")overset("dry")(rarr),C_(2)H_(5)I,+NaBr),("ethyl bromide",,"ethyl iodide",):}`
`{:(C_(2)H_(5)Br, + NaI overset("dry acetone")(rarr),C_(2)H_(5)I,+NaBr),("ethyl chloride",,"ethyl chloride",):}`
Discussion
50.

Finelydivided platinum and palladium comonlyknown as platinumandpalladiumblack, maybereducingtheir solublesaltswith :

Answer»

`H_(2)O`
`C_(2)H_(5)IOH`
`HCHO`
`C_(6)H_(6)`

ANSWER :C
Discussion