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.

Graphite in an example of- (A) Ionic solid (B) Covalent Solid (C) Vander waal’s Crystal (D) Metallic crystal

Answer»

Solution : (B)
Graphite is a COVALENT solid having `sp^(2)` hybridised CARBON atoms
Discussion
2.

Graphite has a two dimensional sheet-like structure in which each carbon atom is sp^3 hybridized .

Answer»

SOLUTION :GRAPHITE has a two dimentional SHEET LIKE structure in which each carton is `sp^2` HYBRIDIZED .
Discussion
3.

Graphite has each of the following properties except:

Answer»

ABILITY to CONDUCT electricity
grey BLACK colour
hardness
metallic lustre.

Solution :GRAPHITE is not HARD.
Discussion
4.

Graphite has

Answer»

2-dsheet structure
Vander WAALS force between successive LAYERS of CARBON SHEETS
`sp^(2)` hybridised carbon linked with other three carbon atoms in HEXAGONAL planar structure
all the above

Answer :D
Discussion
5.

Graphite conducts electricity because of

Answer»

weak van DER waal's forces between layers.
covalent bonding between carbon atoms of layers.
delocalized electrons in each layer
`sp^(2)` hybridisation of carbon in each atoms in layers.

Solution :Graphite is composed of flat two-dimensional sheets of carbon atoms. Each sheet is a hexagonal net of C atoms. The `3E^(-)` of C form `sigma` bond & 4th ELECTRON is a `pie^(-)` and is delocalized over the whole sheet Y is thus mobile. Connduction of electricity is due to these delocalized electrons within each layer. Conduction does not occur from ONE sheet to another.
Discussion
6.

Graphite conducts electricity because of :

Answer»

Weak van der Waals forces between layers
Covalent BONDING between CARBON atoms of layers
Delocalized ELECTRONS in each layer
`sp^2`-hybridisation of carbon atoms in each layer

Answer :C
Discussion
7.

Graphite cannot be classified as

Answer»

CONDUCTING SOLID
network solid
covalent solid
IONIC solid

Solution :ionic solid
Discussion
8.

Graphite and diamond will behave differently in which of the following reactions?

Answer»

Burning in sufficient air
Reaction with hot CONC. `HNO_(3)`
Reaction with `F_(2)`
Reaction with `NAOH (aq.)`

Solution :N//A
Discussion
9.

Graphite cannot be classified as .........

Answer»

CONDUCTING SOLID 
network solid 
covalent solid 
IONIC solid 

SOLUTION :Constituent particles in graphite are carbon atoms held together by covalent bonding.
Discussion
10.

Graphite and diamond are …………….

Answer»

COVALENT and MOLECULAR CRYSTALS
ionic and covalent crystals
both covalent crystals
both molecular crystals

Solution :both covalent crystals
Discussion
11.

Graphically the total number of fundamental spatial arrangements possible are

Answer»

3
7
10
14

Answer :D
Discussion
12.

Graph of ln S^(@) vs (1)/(T) is plotted for two gases A and B [S^(@) represents solubility in molarity and T is in Kelvin]. Compare Henry's constant K_(H) for the two gases at same temperature.

Answer»

`K_(H,A) gt K_(H,B)`
`K_(H,A) LT K_(H,B)`
`K_(H,A) = K_(H,B)`
`K_(H,A)` may be GREATER or LESS than `K_(H,B)`

ANSWER :A
Discussion
13.

Graphene has ………………….. Lattice.

Answer»


ANSWER :HONEY COMB
Discussion
14.

Graph between which of the following coordinate produce striaght line with reference to zero order reaction

Answer»

Rate vs time
`t_(1//2)` Vs initial concentration
Concentration of reactants Vs time
Rate Vs concentration of reactant

Solution :`(t_(1))/(t_(2))=(C_(0))/(2.K),t_(1/2)=1/(2K)xxC_(0)`
Discussion
15.

Graph between P & V below critical temperature is

Answer»




ANSWER :D
Discussion
16.

Graph between log x/ m and log P is a straight line inclined at an angle theta – 45^@. When pressure of 0.5 atm and log k = 0.699, the amount of solute adsorbed per g of adsorbent will be:

Answer»

1g/g adsorbent
1.5g/g adsorbent
2.5g/g adsorbent
0.25g/g adsorbent

Solution :`x/m=k.p^(1//n) since logk=0.699 HENCE k=5, Hence k=5`
Slope =`1/5= tan45^@=1 THUS, x/m=5 TIMES 0.5=2.5g//g` adsorbent
Discussion
17.

Graph between log k and 1//T[where K is rate constant in s^(-1) and T is the temperature (in K) is a straight line with Hence, E_(a) will be

Answer»

`2.303 xx 2` CAL
`2/(2.303) cal`
2 cal
None of these

Solution :log k = log A`-(E_(a))/(2.303 R) = TAN phi= 1/(2.303)` (Given)
`E_(a) = 2.303 R xx` Slope
`=2.303 R xx 1/(2.303) = R=2 Cal`
Discussion
18.

Graph between "log"(x/m) and log p is a st. line at angle 45^@ with intercept as shown in Fig. Hence (x/m) at a pressure of 0.2 atm is

Answer»

0.2 
0.4 
0.6 
0.8

Solution :`0.3010= "LOG"^2` : PRESSURE = `0.2, 2 XX 0.2 = 0.4`
Discussion
19.

Grape sugar is ……………….

Answer»

SUCROSE
GLUCOSE
FRUCTOSE
LACTOSE

SOLUTION :Glucose
Discussion
20.

Granulated Zn is obtained by:

Answer»

Suddenly colling MOLTEN ZN
Adding molted Zn to water
Haeating Zn to 100 to `150^@C`
Dropping molten Zn drop by drop

Answer :B
Discussion
21.

Grams of a sample of ferrussulphate wasdissolvedin dilutesulphuricacidand waterand itsvolumewas madeup to1 litre .25 mL of this solution required20 mLof N/10KmnO_(4)solutionfor completeoxidation. Calculate thepercentage of FeSO_(4). 7H_(2)Oin the sample .

Answer»

Solution :m.e of `KMnO_(4)"solution "= 1/10 xx 20 = 2 `
` :. ` m.eof 25 mLof `FeSO_(4) . 7H_(2)O " solution "= 2 "" …. (Eqn . 1)`
` :. ` m.eof 100m Lof `FeSO_(4). 7H_(2)O"solution " = 2/25 xx 1000 = 80`
Equivalent of `FeSO_(4) . 7H_(2)O = 80/1000 = 80 `
Equivalent of `FeSO_(4) . 7H_(2)O`solution = `2/25 xx 1000 = 80`
Equivalent of `FeSO_(4) . 7H_(2)O = 80/1000 ""...(Eqn . 3)`
` :. ` weightof `FeSO_(4) .7H_(2)O` = equivalent `xx` eq.WT
` = 80/100 xx278 = 22.4 G `
`{ "As " FE^(2+)to Fe^(3) ," eq . wt of " FeSO_(4) . 7H_(2)O= (mol . w.t )/(" changein ON ") = 278 / 1 } `
thusthe percentage of `FeSO_(4). 7H_(2)O ` in the sample` = (22.24)/25 xx 100`
` = 88.96 %`
Discussion
22.

Granulated zinc is obtained by:

Answer»

Suddenly COLLING MOLTEN ZINC
Adding molten zinc to water
Heating zinc to `100-150^@C`
Dropping molten zinc drop by drop

Answer :B
Discussion
23.

Gram molecular mass of CH_4 is

Answer»

16 g
16 u
32 g
32 u

Answer :A
Discussion
24.

Gram molecular volume of oxygen at STP is

Answer»

`3200 cm^(3)`
`5600 cm^(3)`
`22400 cm^(3)`
`11200 cm^(3)`

SOLUTION :Gram molecular VOLUME of OXYGEN at STP
GMV = 22.4 LITRE = `22.4= 22.4(10 cm)^(3)`
`therefore` GMV = 22400 `cm^(3)`.
Discussion
25.

Grain spirit is

Answer»

ISOPROPYL alcohol
ISOBUTYL alcohol
methyl alcohol
ethyl alcohol

ANSWER :D
Discussion
26.

Grain alcohol is the common name of :

Answer»

AMYL alcohol
Ethyl alcohol
Methanol
None

Answer :B
Discussion
27.

Graham's salt is one of the salt of Phosphorous acid having formula (NaPO_3)_nWhat is the value of n.

Answer»


SOLUTION :`(NaPO_3)` - is CALLED Graham.s salt.n = 6
Discussion
28.

Graham's salt is (NaPO_3)_n and is used in softening of hard water.

Answer»


ANSWER :T
Discussion
29.

Graham's salt is:

Answer»

SODIUM alumino silicate
Sodium hexameta phosphate
Ferrous AMMONIUM sulphate
Potassium CHROMIUM sulphate

Answer :B
Discussion
30.

Graham's law dealswih the relation between :

Answer»

Pressure and VOLUME
Density and RATE of DIFFUSION
Rate of diffusion and volume
Rate of diffusion and viscocity

Answer :B
Discussion
31.

Graham's law of diffusion gives better results at :

Answer»

HIGH pressure
High TEMPERATURE
LOW pressure
at all conditions

Answer :C
Discussion
32.

Global warming of the atmosphere due to trapping of long infrared radiations is called :

Answer»

AIR pollution
Air heating
Photosynthesis
green HOUSE effect

Answer :D
Discussion
33.

Gradual addition potassium iodide with nitric acid produces a dark brown precipitate 'A'. 'A' is soluble in excess KI and gives yellow solution 'B'. What are A and B?

Answer»

SOLUTION :`A=I_(2), B=KI_(3)`
Discussion
34.

“Government that are seen to be unstable, quarrelsome and divisive have been severely punished”. Elucidate giving examples from 1975- 1977

Answer»

Solution :Emergency showed weakness and strength of the Indian democracy
1. Political crisis and a change in the party system
2. ELECTIONS of 77 took all by surprise
3. Opposition and coalition politics came into the foray.
4. Opposition fought on “save democracy”- felt the pulse of the nation. though it did not affect the southern STATES
5. Janata party referendum was Emergency and excesses during the emergency.
6. Also showed once in power how UNSTABLE parties are MORARJI Desai and Ch.Charan Singh. Stiff competition within the party and could not bring about the expected fundamental changes.
Discussion
35.

Gradual addition of KI solution to Bi(NO_(3))_(3) solution initially produces a dark brown precipitate which dissolves in excess of KI to give a clearyellow solution. In the above formed iodine, displaces chlorinefrom which one of the compound.

Answer»

KCl
`CaCl_(2)`
`C C l_(4)`
`KClO_(3)`

SOLUTION :`2KlO_(3) + I_(2) RARR 2KClO_(3) + Cl_(2)`
Discussion
36.

Gradual addition of KI solution to Bi(NO_(3))_(3) solution initially produces a dark brown precipitate which dissolves in excess of KI to give a clearyellow solution.The clear yellow solution is

Answer»

`NaI_(3)`
`BI(OH)(NO_(3))_(2)`
`KI_(3)`
NONE of these

Solution :Clear yellolw solution formed due to the formation of `KI_(3)`
Discussion
37.

Good yield of nitroalkane fromalkyl halideand KNO_(2) are obtained in presenceof solvent

Answer»

dimethlyether
DIMETHYLSULPHOXIDE
N,N-dimethylformanide
N,N-dimethylformation of dimethylsulphoxide

SOLUTION :SUITABLE solventmechanism
Discussion
38.

Gradual addition of KI solution to Bi(NO_(3))_(3) solution initially produces a dark brown precipitate which dissolves in excess of KI to give a clearyellow solution.In the above observation, the brown ppt is

Answer»

`Cl_(2)`<BR>`I_(2)`
`Br_(2)`
`F_(2)`

Solution :`I_(2)^(-)` - dissolve in `Bi(NO_(3))_(3)` solution produce dark brown ppt. Which dissolve in EXCESS ICI give yellow SOLVE.
Discussion
39.

Good reducing nature of H_(3)PO_(2) is attributed to the presence of :

Answer»

Two P - OH BONDS
ONE P - H bond
One P - OH bond
Two P - H bonds

SOLUTION :
Discussion
40.

Good conductor of heat and current is :

Answer»

Anthracite
Diamond
Charcoal
Graphite

Answer :D
Discussion
41.

Good quality bleaching powder contains available chlorine about:

Answer»

`10-20%`
`5-10%`
`35-38%`
`20-25%`

ANSWER :C
Discussion
42.

Good conductor of electricity is:

Answer»

YELLOW P
Red P
Violet P
Black P

Answer :D
Discussion
43.

Goldschmidt thermite process is used for .. .

Answer»

welding of BROKEN iron pieces. 
CONVERTING iron into steel. 
EXTRACTION of sulphur. 
reduction of metallic oxide by magnesium

Solution :`Fe_(2)O_(3) + 2Al to Al_2O_3 + 2Fe`
The reaction is exothermic. The molten iron PRODUCED, used to weld the broken iron pieces.
Discussion
44.

Gold sol is prepared by reduction of auric chloride using..................

Answer»

SOLUTION :FORMALDEHYDE
Discussion
45.

Gold when dissolved in aqua-regia gives :

Answer»


ANSWER :C
Discussion
46.

Gold sol is prepared by reduction of auric chloride using ...........

Answer»

water
HCHO
`CH_3CHO`
`CH_3COOH`

SOLUTION :HCHO
Discussion
47.

Gold of some lyophilic sols are : {:(I,:,"Casein",:,0.01),(II,:,"Haemoglobin",:,0.03),(III,:,"Gum arabic",:,0.15),(IV,:,"Sodium oleate",:,0.40):} Which has maximum protective power:

Answer»

I
II
III
IV

Solution :LOWER the GOLD number, higher the PRODUCING power of lyophillic colloid.
Discussion
48.

Gold sol is not a

Answer»

a macromolecular colloid
a LYOPHOBIC colloid
a multimolecularcolloid
negatively charged colloid.

Solution :STARCH, CELLULOSE, proteins and enzymes are the examples of macromolecular colloids.
Discussion
49.

Gold sol is an electronegative sol. The amount of electrolyte required to coagulate a certain amount of gold sol is minimum in case of

Answer»

`AlCl_(3)`
`Na_(2)SO_(4)`
`CaCl_(2)`
NaCl

Solution :This is because the oppositely charged `SO_(4)^(2-)` IONS have maximum charge. Coagulation is the conversion of a colloid into its INSOLUBLE PRECIPITATE. According to Hardy Schulze rules.
(i) A positively charged colloid is coagulated by a negative ION and vice-versa.
(ii) Increase in VALENCE of an ion decreases coagulation value of the colloid.
Discussion
50.

Gold ore is concentrated by…….

Answer»

CYANIDE leaching
Alkali leaching
Acid leaching
Hand picking

Answer :A
Discussion