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.

An acid (A) which is an important constituent of vinegar, on reaction with red P_4 and Br_2 gives a monobromoderivative (B) which on reaction with NH_3 gives a white solid (C). However, ethanal on reaction with a mixture of ammonium chloride and sodium cyanide undergoes strecker synthes is to give a product which on acidic hydrolysis gives another high melting solid (D).The structure of the compound (D) is

Answer»

`CH_3 (OH)CHCOOH`
`(+) CH_3 CH (NH_2)CHCOOH`
`(PM) CH_3 CH (NH_2)COOH`
`(pm)CH_3 CH(OH)CONH_2`

ANSWER :C
2.

Amylose is a polymer of:

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`ALPHA`-D glucopyranose
FRUCTOSE
`BETA`-fructose
`beta`-D fructose

Answer :D
3.

Orlon is a polymer of

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`ALPHA`-D glucopyranose
FRUCTOSE
`BETA`-fructose
`beta`-D fructose

Answer :D
4.

Amyloseis a polymerof

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`ALPHA` - D - GLUCOSE
`BETA` - D- glucose
`alpha` - D- FRUCTOSE
`beta` - D -fructose

ANSWER :A
5.

Amylose contain

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C-1`to`C-4 `BETA`-D-glycosidic bond
C-1`to`C-4 `ALPHA`-D-glycosidic bond
C-1`to`C-6 `beta`-D-glycosidic bond
C-1`to`C-4 `beta`-D-glycosidic bond

Answer :B
6.

Amylose and amylopectin are costituent of

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`ALPHA`-D-fructose
`alpha`-D-glucose
`BETA`-D-fructose
`alpha`-D-fructose

Answer :B
7.

Amylose consists of

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BRANCHED CHAIN of `alpha-D-` glucose UNITS 
Unbranched chain of `beta-D-` glucose units
Units of sucrose
Unbranched chain of `alpha-D-` glucose units 

Answer :D
8.

Amylopetcin in solublein waterand constitutes about 80-50% of starch.

Answer»


ANSWER :FALSE
9.

Amylopectin is a polymer of

Answer»

`beta-D` GLUCOSE
`ALPHA-D` glucose 
`beta-D` FRUCTOSE 
`alpha- D` fructose 

ANSWER :B
10.

Amylopectin is

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WATER SOLUBLE
water insoluble
FORMS COLLOIDAL solution with water
soluble in all SOLVENTS.

Answer :B
11.

AmtchList I withList I andselectthecorrectanswer fromthe givencodes{:("List I","List II"),("Reaction ", "Metals used "),("A. Wurtz reaction ","1.Ni"),("B.Sabatiersenderen's reaction ","2. Zn"),("c.Frankland reaction ","e. Li"),("D.Corey -Housesynthesis ","4. Na"):}

Answer»

`{:(A,B,C,D),(1,2,3,4):}`
`{:(A,B,C,D),(4,1,2,3):}`
`{:(A,B,C,D),(3,2,4,1):}`
`{:(A,B,C,D),(2,4,2,3):}`

ANSWER :B
12.

Ampicillin and amoxycillin are synthetic modifications of penicillin and have ________________.

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SOLUTION :BROAD SPECTRUM
13.

Ampicillin is _____.

Answer»

SOLUTION :ANTIBIOTICS
14.

Amphoterism Amphoteric oxides, such as aluminium oxide, are soluble both in strongly acidic and in strongly base solutions : In acid:Al_2O_3(s)+6H_3O^(+)(aq)hArr2Al^(3+)(aq)+9H_2O(l) In base:Al(OH)_3(s)+OH^(-)(aq)hArr Al(OH)_4^(-)(aq) Dissolution of Al(OH)_3 in excess base is just a special case of the effect of complex-ion formation on solubility.Al(OH)_3 dissolves because excess OH^(-) ions convert it to the soluble complex ion Al(OH)_4^(-)(aluminate ion)The effect of pH on the solubility of Al(OH)_3 is shown in figure. Other examples of amphoteric hydroxides include Zn(OH)_2,Cr(OH)_3,Sn(OH)_2 and Pb(OH)_2, which react with excess OH^(-) ions to form the soluble complex ion Zn(OH)_4^(2-)(zincate ion),Cr(OH)_4^(-)(chromite ion),Sn(OH)_3^(-) , Fe(OH)_2 and Fe(OH)_3 , dissolve in strong acid but not in strong base. Zn(OH)_2 is a amphoteric hydroxide and is involved in the following two equilbria in aqueous solutions Zn(OH)_2(s)hArrZn^(2+)(aq)+2OH^(-)(aq),K_(sp)=1.2xx10^(-17) Zn(OH)_2(s)+2OH^(-)(aq)hArr[Zn(OH)_4]^(2-)(aq),K_(@)=0.12 At what pH the solubility of Zn(OH)_2 be minimum ?

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4
10
6
8

Solution :Let solubility of `ZN(OH)_2`=s, some of will go in `Zn^(2+)` form and some in complex `[Zn(OH)_4]^(2-)`, of `[OH^(Theta)]=10^(-x)M`, then
`S_1(10^(-14)/10^(-x))^2=1.2xx10^(-17) `...(i)
`S_2/(10^(-14)/10^(-x))^2=0.12` [`S_1+S_2`=total solubility]
So , `S_1S_2=1.2xx1.2 10^(-18)`
Now we want `S=S_1+S_2` to be minimum we will have `S_1=S_2`
So, `S_1=S_2=1.2xx10^(-9)M`
Hencefrom `1^(st)`equation we GET
`(10^(-14)/10^(-x))^2=(1.2xx10^(-17)) /(1.2xx10^(-3))=10^(-10)`
Hence, x=10
15.

Amphoteric oxide (X) + 3C+ Cl_(2) to Poisonous gas + anhydrous chloride (Y) Hydrated chloride overset(Delta)to Z Element forming 'Y' other than 'CI' reacts with concentrated HCI but leads to passivation with conc. HNO_(3). Select the correct option.

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`X=Z` and `Y` on reacting with `LiH` FORMS strong OXIDISING AGENT
`X=Z` and `Y` on reacting with `LiH` forms strong reducing agent
`X!=Z `and `Y` is used as a CATALYST in Friedel crafts reaction
`X!=Z` and `Y` on reacting with `LiH` forms strong oxidising agent

Solution :N//A
16.

Amphoterism Amphoteric oxides, such as aluminium oxide, are soluble both in strongly acidic and in strongly base solutions : In acid:Al_2O_3(s)+6H_3O^(+)(aq)hArr2Al^(3+)(aq)+9H_2O(l) In base:Al(OH)_3(s)+OH^(-)(aq)hArr Al(OH)_4^(-)(aq) Dissolution of Al(OH)_3 in excess base is just a special case of the effect of complex-ion formation on solubility.Al(OH)_3 dissolves because excess OH^(-) ions convert it to the soluble complex ion Al(OH)_4^(-)(aluminate ion)The effect of pH on the solubility of Al(OH)_3 is shown in figure. Other examples of amphoteric hydroxides include Zn(OH)_2,Cr(OH)_3,Sn(OH)_2 and Pb(OH)_2, which react with excess OH^(-) ions to form the soluble complex ion Zn(OH)_4^(2-)(zincate ion),Cr(OH)_4^(-)(chromite ion),Sn(OH)_3^(-) , Fe(OH)_2 and Fe(OH)_3 , dissolve in strong acid but not in strong base. Which of the following curves best represents the variation of solubility of ferrous hydroxide Fe(OH)_2 with the concentration of [H^(+)] ions in the solution:

Answer»




Solution :On INCREASING CONCENTRATION of `[H^+]` IONS the SOLUBILITY of basic hydroxide `Fe(OH)_2` will increase.
17.

Ampicillin, amoxicillin, methiceillin, cetrizine, cephalosporin.

Answer»

Solution :CETRIZINE .It is an antihistamine WHEREAS OTHERS belongs to penicillin group.
18.

Amphoteric nature of aluminum is employed in which of the following process for extraction of Aluminum ?

Answer»

BAEYER's process
Hall's process
Serpek'sprocess
Dow's process.

Answer :A::B
19.

Amphoteric oxide is:

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`Sb_4O_6`
`N_2O_5`
`Bi_2O_3`
`Na_2O`

ANSWER :A
20.

Amphiphilic molecules are normally associated with

Answer»

ISOPRENE based polymers
Soaps and detergents
Nitrogen based FERTILIZERS e.g. urea
Pain relieving medicines such as aspirin

Answer :B
21.

Amoxycillin is semi synthetic modification of

Answer»

Penicillin
Streptomycin
TETRACYCLINE
Chloraampheniol

ANSWER :A
22.

Amoxillin is semi-synthetic modification of

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PENICILLIN
Tetracycline
Streptomycin
Chloroampheniol

Solution :AMOXILLIN is SEMISYNTHETIC MODIFICATION of Penicillin.
23.

Amoxicillin is semi-synthetic modification of

Answer»

Penicillin
Streptomycin
Tetracycline
Chloramphenicol.

Answer :A
24.

Amount of oxygen required for complete combustion of 27 g Al IS :

Answer»

24G
12 g
20 g
6 g

Answer :A
25.

Amount of oxygen (in g) in 32.2g of Na_(2)SO_(4).10H_(2)O is

Answer»

20.8
22.4
2.24
2.08

Solution :1 mol `Na_(2)SO_(4).10H_(2)O = 14 g` atoms of OXYGEN
i.e. 32.2 g `Na_(2)SO_(4).10H_(2)O=22.4` g of oxygen.
26.

Amount of oxygen required for combustion of 1 kg of a mixture of butane and isobutane is :

Answer»

1.8 kg
2.7 kg
4.5 kg
3.58 kg

Answer :D
27.

Amount of oxalicacid present in a solution can bedetermined by its titration with KMnO_(4)solution in the presenceof H_(2)SO_(4) . The titration gives unsatisfactory rasult when carried out the presence of HCl because HCl

Answer»

reducs permanganate to `Mn^(2+)`
OXIDISES oxalic acid to carbon dioxide and water
gets oxidized by oxalic acid to chlorine
furnishes `H^(+)` ions in addition to those from oxalic acid.

Solution :In presence of `H_(2)SO_(4), KMnO_(4)` oxidises oxalic acid to `CO_(2)` .In presence of HCl, `KMnO_(4)` not only oxidises oxalic acid but ALSO oxidises HCl to `Cl_(2)` and itself it is reduced to `Mn^(2+)`
28.

Amount of oxalic acid present in a solution can be determined by its titration with KMnO_(4) solution in the presence of H_(2)SO_(4). The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl

Answer»

FURNISHES `H^(+)` ions in addition to those from oxalic acid
Reduces permanganate to `Mn^(2+)`
OXIDISES oxalic acid to carbon dioxide and water
Get oxidised by oxalic acid to chlorine

Answer :B
29.

Amount of oxalic acid present in a solution can be determined by its titration with KMnO_(4) solution in the presence of H_(2)SO_(4) . The titration gives unsatisfactory results when carried out in the presence of HCl because HCl

Answer»

OXIDISES oxalic ACID to CARBON dioxide and water
gets oxidised by oxalic acid to chlorine
furnishes `H^(+)` ions in addition to those from oxalic acid.
reduces permanganate to `Mn^(2+)`

Solution :`KMnO_(4)` can oxidise HCl also into `Cl_(2)` itself gets REDUCED to `Mn^(2+)`.
30.

Amount of oxalic acid present in a solution can be determined by its titration with KMnO_(4) solution in the presence of H_(2)SO_(4) . The titration gives unsatisfactory result when carried out in the presence of HCl because HCl:

Answer»

oxidises OXALIC acid to carbon dioxide and water
reduces PERMANGANATE to `Mn^(2+)`
GETS oxidised by oxalic acid to chlorine
furnishes `H^(+)` ions in ADDITION to those from oxalic acid.

Answer :B
31.

Amount of gas adsorbed per gram of adsorbent increases with pressure but after certain limit is reached, adsorption becomes constant. It is when

Answer»

MULTI layers are formed 
DESORPTION takes PLACE 
TEMPERATURE is increased 
absorption also starts

Answer :A
32.

Amount of Br_2 required to react with 5g pentene to form monobromo derivative is:

Answer»

11.11 g
11.43 g
5.55 g
None of these

Answer :B
33.

Amount of80% pure NaOH sample which is required to completely react with 42.6 gm Chlorine in hot condition according to given reaction NaOH + Cl_(2) rarrNaCl + H_(2)O + O_(2) is -

Answer»

48
60 gm
24 gm
30 gm

Solution :`2NaOH + Cl_(2) RARR 2NaCl + H_(2)O + (1)/(2) O_(2)`
`1.2` mole `0.6` mole
`n_(NaOH "pure required") = 1.2`
`W_(NaOH "pure") = 1.2 XX 40 = 48 gm`
`W_("impure sample") xx (80)/(100) = W_(NaOH "pure")`
`W_("sample") = 60 gm`
34.

Amount (in g) of sample containing 80% NaOH required to prepare 60 litre of 0.5 M solution is

Answer»

1000
1200
1500
1600

Answer :C
35.

Amoung the following , the formula of saturated fatty acid is

Answer»

`C_(17)H_(29) COOH`
`C_(17)H_(35) COOH`
`C_(17)H_(31) COOH`
`C_(17)H_(33)COOH`

Solution :The general formula for saturated fatty acids is `C_(n) H_(2n + 1) COOH` ,
Among the given acids , only `C_(17)H_(35) COOH` SATISFY this formula , HENCE it is a saturated fatty acid.
36.

Amoungthe followingstatementson thenitrationofaromaticcompoundsthe falseone is

Answer»

the rateof NITRATIONOF benzeneis almostthe same asthat ofhexadeuterobenzne
the rateof nitration of tolueneis GREATERTHAN thatofbenzne
the rateof nitrationofbenzeneis greaterthan that ofhexadeuterobenzne
nitration is anelectrophilicsubstitutionreaction

Solution :Since `.^(+) NO_(2)` ISAN electrophile so nitratin iselectrophilicsubstitutionrection .
Nitration in the toluene inis greaterbecauseof electron- repelling(+ 1) N/Atureof `-CH_(3)`group .
Innitrationthe ratedetermining stepis the attackof `NO_(2) ^(+)`on benzeneand not thecleavageof `C- H ( orC- D^(2))` bond . the ratesof nitrationare almostsame.
37.

Amospheric air has 78% N_(2): 21% O_(2): 0.9% Ar and 0.1% CO_(2) by volume. What is the molecular mass air in the atmosphere?

Answer»

SOLUTION :MOLECULAR mass of mixture
`=(SUM %"of each")/(100)XX"Molar mass"`
`=(78)/(100)xx28+(21)/(100)xx32+(0.9)/(100)xx40+(0.1)/(100)=28.-964`
38.

Amorphous solids do not have sharp melting points. Explain.

Answer»

Solution :(1) Amorphous solids do not have perfectly ordered crystalline structure.
(2) They have short range order of REGULAR pattern hence periodically repeating regular pattern is over a short distance.
(3) Hence the temperature needed to melt the solid is not same, THEREFORE amorphous solids do not have SHARP melting points but melt over a range of temperature.
39.

Amorphous solids:

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Possess SHARP melting points
Undergo clean CLEAVAGE when CUT with knife
Do not undergo clean cleavage when cut with knife

Answer :C
40.

An amorphous solid is:

Answer»

POSSESS SHARP MELTING points
UNDERGO clean CLEAVAGE when cut with knife
Do not undergo clean cleavage when cut with knife

Answer :C
41.

Amorphous solids -

Answer»

Possess sharp MELTING points
UNDERGO clean CLEAVAGE when cut with knife
Do not undergo clean cleavage when cut with knife
Possess orderly arrangement over LONG distances

Answer :C
42.

Amorphous solid is :

Answer»

Rubber
Plastic
Glass
All

Answer :D
43.

Amorphous solid can also be called .........

Answer»

pseudo SOLIDS 
true solids 
SUPER cooled liquids 
super cooled solids 

Solution :AMORPHOUS solids have RANDOM arrangement of constituent particles and has TENDENCY to flow very slowly. Hence, they are called pseudo solids or super cooled liquids.
44.

Amorphous boron is extracted from borax by following steps. "Borax" overset((A))(rarr)H_3BO_3overset((B))(rarr) B_2O_3 overset((C))(rarr) "Boron" (A) and (C) are

Answer»

`H_2SO_4, Al`
`HCl , C`
`H_2SO_4, Mg`
`HCl, Fe`

SOLUTION :`Na_2B_4O_7 + 2H_2SO_4 to Na_2SO_4 + H_2B_4O_7`
`H_2B_4O_7 + 5H_2O to 4H_3BO_3`
`3H_4BO_3 overset("HEAT (B)")(rarr)B_2O_3 + 3H_2O`
`B_2O_3 + overset((C))(3 Mg) rarr 2B + 3MgO`.
45.

Amont the following the most reactive towards alcoholic KOH is

Answer»

`CH_(2)=CHBr`
`CH_(3)COCH_(2)CH_(2)Br`
`CH_(3)CH_(2)Br`
`CH_(3)CH_(2)CH_(2)Br`

Solution :A ELIMINATION of `Br^(-)` become easier due to the PRESENCE of CARBONYL group.
46.

Amonst the followings, identify a copolymer?

Answer»

Orlon
PVC
PHBV
TEFLON

ANSWER :C
47.

Amonst the following the most stable complex is

Answer»

`[Fe(H_(2)O)_(6)]^(3+)`
`[Fe(NH_(3))]^(3+)`
`[Fe(C_(2)O_(4))_(3)]^(3-)`
`[FeCl_(6)]^(3-)`

Solution :`[Fe(C_(2)O_(4))_(3)]^(3-)` is the most stable complex due to chelate formation as `C_(2)O_(4)^(2-)` is a bidentate chelating ligand.
48.

Amongt the following, the total number of thermoplastics is___________. Polyester, bakelite, polyethene, PVC, teflon, PAN, PMMa nylon-6, melamine-formaldehyde.

Answer»
49.

Amongst TiF_(6)^(2-), CoF_(6)^(3-), Cu_(2)Cl_(2) and NiCl_(4)^(2-) the colourless species are:

Answer»

`CoF_(6)^(3-) and NiCl_(4)^(2-)`
`TiF_(6)^(2-) and CoF_(6)^(3-)`
`Cu_(2)Cl_(2) and NiCl_(4)^(2-)`
`TiF_(6)^(2-) and Cu_(2)Cl_(2)`

Answer :D
50.

Amongst the various ores of a metal (M) (sulphide, carbonates, oxides, hydrated or hydroxides) two ores [X] and [Y] show the following reactivity. (i)[X] on calcination gives a black solid (S), carbon dioxide and water. (ii)[X] Dissolved in dilute HCl, on reaction with Kl gives a white precipitate (P) and iodine. (iii)[Y] on roasting gives metal (M) and a gas (G_(1)) which turns acidified K_2Cr_2O_7 solution green. (iv)[Y] on reaction with dil.HCl givesa white precipitate (Q) and another gas (G_2) which turns lead acetate solution black, and also reacts with gas (G_1) to precipitate colloidal sulphur in presence of moisture. Reaction involved: (i)CuCO_3. Cu(OH)_2oversetDeltatoCuOdarr("black")+CO_2+H_2O CuCO_3.Cu(OH)_2+4HCl to2CuCl_2+3H_2O+CO_2 (ii)2Cu^(2+)+4I^(-)toCu_2l^2 darr (P)("white")+l_2uarr (iii)2Cu_2S+3O_2to2Cu_2O+2SO_2(G_1) Cu_2S+2Cu_2O to6Cu+SO_2 (iv)Cu_2S+2HCl to2CuCldarr(Q)("white")+H_2S(G_2) Pb^(2+)+H_2StoPbSdarr("black")+2H^(+) (v)SO_2(G_1)+2H_2S(G_2)overset("moisture")to3Sdarr+2H_2O The gas (G_1) acts as

Answer»

OXIDISING AGENT
REDUCING agent
Oxidising and reducing agent
Fluxing agent

Solution :`G_1=SO_2`, sulphur can increases & DECREASES its oxidation state.