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.

Aspirin and Analgin are ____ drugs.

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SOLUTION :non-narcotics
2.

Aspirin, an antipyretic drug is chemically :

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METHYL salicylate
Ethyl salicylate
Acetyl salicylic ACID
o-hydroxy benzoic acid

ANSWER :C
3.

Aspirin acts as an analgesic because it

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inhibits the synthesis of PROSTAGLANDINS which STIMULATE INFLAMMATION of the tissue
prevents the RELEASE of HCL in the stomach
prevents the interaction of histamine with its receptor
inhibits activities of enzymes

Answer :A
4.

Aspirin act as

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antioxidant
antibiotic
antisepti
ANALGESIC and antipyretic

Answer :D
5.

Aspiriin can be prepared by the reaction of:-

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Sallicylic acid with methanol in PRESENCE of `H_(2)SO_(4)`
Salicyldehydewith acetic anhydride in presence of `H_(2)SO_(4)`
SALICYLIC acid with acetic anhydride in presence of `H_(2)SO_(4)`
Cinnamic acid with acetic anhydride in presence of `H_(2)SO_(4)`

SOLUTION :N//A
6.

Asphalt used for construction of roads is .........

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an emulsion of ASPHALT in oil. 
an emulsion of asphalt in water. 
molten asphalt. 
a solution of asphalt in water. 

ANSWER :B
7.

Aspartame, the artificial sweetener is made of a dipeptide of the amino acids,

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aspartic ACID and phenylalanine
aspartic acid and GLYCINE
ALANINE and glycine
aspartic acid and glucatamic acid.

Answer :A
8.

Aspartame is used as ………..

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SOLUTION :An ARTIFICIAL SWEETENING AGENT.
9.

Aspartame is unstable at cooking temperature, where would you suggest aspartame to be used for sweetening ?

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SOLUTION :Since aspartame DECOMPOSES at cooking temperature, therefore, it is used as a SWEETENING agent in cold FOODS and SOFT drinks.
10.

Aspartame is less preferred than sucralose as sweetner. Why?

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Solution :Aspartame is not stable at cooking temperature but sucralose is stable. Aspartame is 160 times more sweeter than sucrose but sucralose is 550 times more sweeter than sucrose. For these reasons aspartame is less preferred than sucralose in FOODS as ARTIFICIAL sweetner.
11.

Aspartame is unstable at cooking temperature, where would you suggest aspartame to be used for sweetening?

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Solution :Aspartame is used in COLD FOODS and SOFT DRINKS because it is UNSTABLE at cooking temperature.
12.

Asparmatge, an artificial sweetener is a peptide and has the following structure. Which of the following is correct about the molecule ?

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It has FOUR FUNCTIONAL groups
It has three functional groups
on HYDROLYSIS it produces only one AMINO ACID
on hydrolysis it produces a mixture of amino acids

Answer :A::D
13.

[A]:Sometimes rate of reactio does not depend on concentration. [R]:The order of reaction can be negative.

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Assertion [A] and reason [R] both are correct and [R] gives correct explanation of [A]
Assertion [A] and reason [R] both are correct but [R] does not give correct explanation of [A]
Assertion [A] is wrong but Reason [R] is wrong
Assertion [A] is wrong but Reason [R]is correct

Solution :It can be possible that the rate of reaction does not DEPEND on concentration.Like some FAST reaction.
The ORDER of reaction can be NEGATIVE,so both statement are correct but [R] is not the correct explanation of[A.
14.

A(s)hArrH_(2)S(g)+B(g) At eq., pressure = 18 atm C(s)hArrH_(2)S(g)+D(g) At eq., pressure = 36 atm Calculation mole fraction of B in the mixture.

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<P>

Solution :`A(s)hArrunderset(x+y x)( H_(2)S(g))+B(g),K_(p_(1))=(9)^(2)=81`
`C(s)hArrunderset(x+y y)(H_(2)S(g))+D(g),K_(p_(2))=(18)^(2)=324`
mole fraction of B in mixture `=x/(2(x+y))`
`=(k_(p_(1)))/(2(K_(p_(1))+K_(p_(2))))`
15.

AsF_(5) molecule is trigonal bipyramidal. The orbitals of As atom involved in hybridisation are

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<P>`d_(x^(2)-y^(2)),d_(z^(2)),s,p_(x),p_(y)`
`d_(XY),s,p_(x),p_(y),p_(z)`
`s,p_(x),p_(y),p_(z),d_(z^(2))`
`d_(x^(2)-y^(2)),s,p_(x),p_(y),p_(z)`

Solution :For trigonal bipyramidal, the hybridisation is `sp^(3)`d which INVOLVES `s,p_(x),p_(y),p_(z)` and `d_(z)^(2)` ORBITALS.
16.

Asforption is accompained by

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DECREASE in enthalpy anc increase in ENTROPY
increase in enthalpy and increase in entropy
decrease in enthalpy and decrease in entropy
no CHANGE in enthalpy and entropy

Solution :Adsorption isaccompanied by decrease in enthalpy `(DeltaH=- ve , ` and decrease in entorpy `(DeltaS=-ve).`
17.

Ascorbic acid resembles the structure of

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VITAMIN
Glucose
Cellulose 
Vittamin D

Answer :B
18.

Ascorbic acid is a chemical name of:

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`"VITAMIN " B_(12)`
Vitamin D
Vitamin C
Vitamin K.

Answer :C
19.

Ascorbic acid is a……..

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Hormone
VITAMIN m
PROTEIN
NUCLEIC ACID

Solution :Vitamin
20.

Ascorbic acid is a/an :

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VITAMIN C
Enzyme
PROTEIN
NONE of these

Answer :A
21.

Ascorbic acid is a

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PROTEIN
Carbohydrate

Answer :A
22.

Ascorbic acid is a……….. .

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VITAMIN
ENZYME
PROTEIN
HORMONE

ANSWER :A
23.

Ascorbic acid belongs to which vitamin ?

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SOLUTION :VITAMIN C
24.

Ascorbic acid is:

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VITAMINS C
Enzyme
Proteins
Lipid

Answer :A
25.

Asbestos is chemically :

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SILICATE of CALCIUM and magnesium
Calcium alumino silicates
Magnesium alumino silicates
Calcium silicate + calcium aluminates

Answer :A
26.

As_2S_3 sol is negatively charged. Between sodium nitrate and aluminium nitrate which one is needed in large quantity to coagulate the above sol ?

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SOLUTION :SODIUM NITRATE
27.

As_(2)S_(3) sol is

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POSITIVE COLLOID
NEGATIVE colloid
neutral colloid
NONE of these

Solution :negative colloid
28.

As_2S_3 sol has a negative charge. Capacity to precipitate it is highest in:

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`AlCl_3`
`Na_3PO_4`
`CaCl_2`
`K_2SO_4`

ANSWER :A
29.

As_2S_3 is the example of

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POSITIVE colloid
negative colloid
neutral colloid
None of these.

Answer :B
30.

As_(2)S_(3) is solutionin(NH_(4))_(2)S_(2) (yellow ammonium sulphide) due to theformation of (a) _____.

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ANSWER :a.`(NH_(4))_(3)AnS_(4)`
31.

As_2S_3is solublein yellowammonium sulphide due to the formation of :

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`(NH_4)_2S`
`(NH_4)_3As S_4`
`H_3AsO_4`
`As_2S_3` sol.

SOLUTION :`As_2S_3 + 3(NH_4)_2 S + 2S to underset("AMM. thioarsenate")(2(NH_4)_2 As S_4)`
32.

As_(2)S_(3) is

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Black
Yellow
Orange
White

Answer :d
33.

As_(2)S_(3) and TiO_(2) sol are examples of

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NEGATIVITY CHARGES sols
POSITIVELY charged sols
Positively and NEGATIVELY charged sols respectively
Negatively and positively charged sold respectively

Answer :D
34.

As_(2)O_(5) is ............while Sb_(2)O_(5) is.................. .

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SOLUTION :ACIDIC, AMPHOTERIC
35.

As_2 S_3is ____shaped colloid.

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disc
plate
ROD
spherical

Answer :D
36.

As we shall soon see, sidium amide (NaNH_(2)) is useful, especially when a reaction requires a very strong base. Explain why a solvent such as methanol cannnot be used to carry out a reaction in which you might want to use sodium amide as a base.

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SOLUTION :An alcohol has `pK_(a)=16-17`, and AMMONIA has `pK_(a)=38`. This means that methanol is a significantly stronger acid than ammonia, and the conjugate base of ammonia (the `""^(-)NH_(2)` ion) is a significantly stronger base than an ALKOXIDE ion. Therefore , the following acid-base reaction would take place as soon as sodium amide is added to methanol.
`underset("Strongeracid")(CH_(3)OH)+underset("Strongerbase")(NaNH_(2)) underset(CH_(3)OH) rarrunderset("Weaker base")(CH_(3)ONa)+underset("Weakeracid")(NH_(3))`
With a `pK_(a)` different this LARGE, the sodium amide would convert all of the methanol to sodium methoxide, a much weaker base than sodium amide. (This is an example of what is called the leveling effect of a solvent.
37.

As we proceed from La(OH)_3 to Lu(OH)_3 basic strength increases / decreases.

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SOLUTION :DECREASES
38.

As the temperature increases, the pH of a KOH solution

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Will decreases
Will increases
Remains constant
Depends upon CONCENTRATION of KOH solution

Solution :PH will decrease because `[OH^(-)]` increased due to this pOH is DECREASED.
39.

As the speed of molecules increase, the number of collisions per second :

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Decrease
Increase
Does not change
None of these

Answer :B
40.

As the size of gold particle increases the colour of solution varies as

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PURPLE `to` BLUE `to` GOLDEN `to` RED
Golden `to` red `to` purple `to` blue
Red `to` purple `to` blue `to` golden
Blue `to` purple `to` golden `to` red

Answer :C
41.

As the s-character of hybridisation orbital increase, the bond angle

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INCREASE
Decreases
BECOMES ZERO
Does not change

Answer :A
42.

As the oxidation state for any metal increases,the tendency to show ionic nature:

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Decreases
Increases
Remains same
NONE of these

Answer :A
43.

As the number of carbon atoms in a chain increases the boiling point of alkanes

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INCREASES
Decreases
Remains same
May increase or decrease

Solution :BOILING point of alkanes increases with the number of CARBON atoms because surface area increases which increases the Vander WAAL forces.
44.

As the molecular weight increases the tensile strength of polymers

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Increases
Decreases
REMAINS UNCHANGED
Unecertain

ANSWER :A
45.

As the atomic number of the halogens increases, the halogens:

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LOSE the OUTERMOST ELECTRONS LESS readily
Become LIGHTER in colour
Become less dense
Gain electrons less readily

Answer :D
46.

As the atomic number of the halogens increases, the halogens :

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LOSE their OUTERMOST ELECTRONS LESS readily
become LIGHT dense
become light in colour
gain electrons less readily

Answer :D
47.

As Rhombic sulphur is heated in a test tube : {:(,"Process","Temperature"),((a),"Viscosity increases",T_(1)),((b),"Visconsity decrease",T_(2)),((c),"Paramagnetic molecules",T_(3)),((d),"Breakage of "S_(8) "rings",T_(4)),(,("Diradical formation in molten phase"),):} Correct order of temperature is

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`T_(1) LT T_(3) lt T_(4) lt T_(2)`
`T_(2) lt T_(4) lt T_(3) lt T_(1)`
`T_(4) lt T_(1) lt T_(2) lt T_(3)`
`T_(3) lt T_(4) lt T_(1) lt T_(2)`

Answer :C
48.

As temperature increases, vapour pressure of a liquid

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INCREASES linearly
decreases linearly
increases exponenyially
decreases exponentially

Answer :C
49.

As percentage of carbon increases in iron, its hardness,

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Decreases
Increases
Remains same
None

Answer :B
50.

As per sommerfeld 's extension which of these is correct ?

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orbits of all electrons in an atom are elliptical
NUCLEUS is present in the geometric center of the ellipse
`V_(transverse)=n_(phi).(H)/(2pi)` where `n_(phi) ne 0, n_(phi) in +I`
The fourth shellhas only circular orbits

Answer :C