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.

Assertioin: Platinum and gold occur innative state. Reason: Platinum and gold are expensive metals.

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Both A and R ar TRUE and R is the correct EXPLANATION of A
Both A and R are true but R is not a correct explanation ofA
A is true but R is false
A is false but R is true

Solution :Platinum and GOLD OCCUR in the native state because they are less reactive.
2.

Asseriton (A) : covalent bonds are stronger than ionic bonds . Reson (R) :covalant bonds are stronger than ionic bonds .

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if both (A) and ® are correct , and ® is correct explanartion of (A
if both (A) and® are correct , but ® is nmot the correct EXPLANATION of
if (A) is correct , but ® is incorrect.
if both (A) and ® are icorrect .

Solution :Assertion is correct since higher melting POINTS of convalent crystals are DUE to network strucute. Correct (R): convalent bonds are weaker than ionic bonds.
3.

Aspirin is widely used as an analgesic drug. It is optically inactive. The structure of aspirin is Which of the following is not correct name for Aspirin?

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2-O-Acetyl SALICYLIC ACID
2-Acetoxy benzoic acid
2-Acetoxy salicylic acid
2-Acetoxy benzene CARBOXYLICACID

ANSWER :C
4.

Aspirin is widely used as an analgesic drug. It is optically inactive. The structure of aspirin is Ratio of SP:SP^2 : SP^3carbon hybrid orbitals in the Aspirin is

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`0:6:1`
`5:2:3`
`4:1:6`
`1:6:2`

ANSWER :A
5.

Aspirin is known as

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PHENYL salicylate
ACETYL salicylate
Methyl SALICYLIC acid
Acetyl salicylic acid

SOLUTION :ASPIRIN is acetylsalicylic acid
6.

Aspirin is known as :

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ACETYL salicylic ACID
Phenyl SALICYLATE
Acetyl salicylate
Methyl salicylate acid

Solution :Aspirin is acetyl salicylic acid.
7.

Aspirin is

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ACETYL SALICYLIC ACID
Phenyl salicylic acid
Salicyclic acid
Benzoic acid

Solution :
Acetyl salicylic acid (ASPIRIN)
8.

Aspirin (acetyl salicyclic acid, molar mass 180 g "mol"^(-1)) used as analgesic has pK_(a) value of 2. Two tablets of aspirin eachweighing 90 mg are dissolved in 100 mL of water. The pH of the solution is

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0.5
`1.0`
`2.0`
`4.0`

SOLUTION :Aspirin is a weak ACID. Hence, its pH is given by
`pH=(1)/(2)[pK_(a)-log c]`
Here, `c=2xx0.09 g "in " 100 ML = 1.8 g L^(-1)`
`=(1.8)/(180)M=0.01 M`
`:. pH=(1)/(2) [2-log 0.01]=(1)/(2)[2-(-2)]=2`
9.

Aspartame an artificial sweetner contains 9.52% by weight of nitrogen. There are two nitrogen atoms per molecule, then the molecular weight of aspartame is:

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588
266
147
294

Answer :D
10.

Asnwer the following queestion based o given reaction. How many fractions are present in fractions distiallation?

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4
5
6
8

Solution :(III) and (VII)
11.

Asnwer the following queestion based o given reaction. Total number of theretically products (includeing stereo ) are :

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4
6
8
10

Solution :(i), and (VIII)
12.

Asnwer the following queestion based o given reaction. How many products are resolvable?

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2
4
5
6

Solution :(i),(IV),(VI) and (VIII)
13.

A(s)hArrB(g)+C(g) K_(P)=40atm^(2) X(s)hArrB(g)+E(g) Above equilibrium is allowed to attain in a closed container and pressure of B was found to be 10 atm. Calculate standard Gibb's free energy change for X(s)hArrB(g)+E(g) "at" 300K (takeR=2 cal//k//mol)

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`3.5 Kcal//mol`
`3 Kcal//mol`
`2.5 Kcal//mol`
`2 Kcal//mol`

SOLUTION :N//A
14.

As_(2)S_(2) sol

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POSITIVE colloid
negative colloid
neutral colloid
None of the above.

Answer :B
15.

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

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ANSWER :ACIDIC, AMPHOTERIC
16.

As we move down the group , the electro positive character of elements _______.

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Increases
DECREASES
Gradutly decreases at REGULAR intervals
Remains same

Answer :a
17.

As we move down the group covalent character decreaes, solubility increases and thermal stability increases for which of the following?

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Fluorides of alkali metal
carbonates of alkali metal
Sulphates of alkali EARTH metal
None of thes

Solution :All the specified PROPERTIES are presentin CAR bonates of alkali METALS
18.

As we move down in group 13 elements increase in atomic size is comparatively very less. Explain.

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SOLUTION :It is due to the poor SHIELDING effect of d and f - electrons as a RESULT of which effect nuclear charge increases.
19.

As we go up above the earth's surface , does temperature decrease continuously ? Explain

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Solution :No, temperature does not DECREASE CONTINUOUSLY , From 0-11 km (called troprosphere) , temperature decreases.From 11-50 km (called STRATOSPHERE), temperature increases . From 50-85 km (calledmesosphere ) , temperature again decreases and finally from 85-100 km (called THERMOSPHERE) , temperature again increases.
20.

As the temperature is raised from 20^(@)C to 40^(@)C the average kinetic energy of neon atoms changes by a factor of which of the following ?

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`1//2`
`sqrt(313//293)`
313/293
2

Solution :For monoatomic NEON ATOMS, `E(3)/(2)RT`
`:. (E_(40.))/(E_(20.))=(T_(40.))/(T_(20.))=(40+273)/(20+273)=(313)/(293)`.
21.

As the temperature is raised from 20^@C " to " 40^@C, the average kinetic energy of neon atoms changes by a factor of which of the following?

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`1/2`
`SQRT(313/293)`
`313/293`
2

Solution :`313/293`
22.

As the temperature is raised from 20^(@)C to 40^(@)C the average kinetic energy of neon atoms changes by a factor

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2
`sqrt((313)/(293))`
`(313)/(293)`
`(1)/(2)`

ANSWER :c
23.

As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant ?

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INCREASES
decreases
REMAINS same
becomes half

Solution :At CONSTANT volume, as the TEMPERATURE is increased, pressure increases.
24.

As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume isconstant ?

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Increases
DECREASES
REMAINS same
Becomes half

Solution :As the TEMPERATURE increases, average KINETIC energy of molecules increases. From Gay - Lussac.s law, at CONSTANT volume, as the temperature is increaes, pressure also increases.
25.

As the s-character of a hybrid orbital increases the bond angle

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INCREASES
Decreases
does not change
Becomes ZERO

ANSWER :A
26.

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

Solution :As the O.S for any metal increases , the TENDENCY to SHOW ionic NATURE decreases
27.

As the molecular weight of alkanes increases, how do the boiling point and melting point change?

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BOILING POINT INCREASES, melting point increases
Boiling point decreases, melting point decreases
Boiling point increases,melting point decreases
Boiling point increases, melting point increases sequentially for ALKANES over FOUR carbons

Answer :D
28.

As regards magnetic behaviour, TiO_2is _______

Answer»


ANSWER :DIAMAGNETIC
29.

As per the graph what is K_c of A harr nB

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1.25 M
1.48 M
2.16 M
0.8 M

Solution :
`:.` A reacted =0.5, B FORMED =1.0
`implies` n = 2
`K_(C)=([B]^(2))/([A])=(1^(2))/(0.8)=1.25`
30.

As per the first law of theromdyamics, which of the following statements would be appropriate?

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Energy of system remains CONSTANT
Energy of SURROUNDING remains constant
Energy of universe remains constant
Energy of universe ALWAYS increases

Answer :C
31.

As per second law of thermodynamics a process taken place spontaneously if and only if the entropy of the universe increases due to the process. Change in entropy is given by Delta S = (Q_(rev))/(T) What is the change in entropy of the universe due to the following reaction occuring at 27^(@)C ? A + 2B rarr C + 2D, Delta H = +1.8 kJ mol^(-1) Molar entropy values of A, B, C and D are 1,2,3 and 4 JK^(-1) mol^(-1) respectively.

Answer»

0
`6JK^(-1) mol^(-1)`
`8JK^(-1) mol^(-1)`
`1.2 JK^(-1) mol^(-1)`

Solution :`DELTA S_("UNIVERSE") = Detla S_("system") + Delta S_("surroundings")`
`= (Sigma S_(P)- Sigma S_(R )) + ((-Delta H_("system"))/(T))`
`= (2(4) + 3-2(2)-1) + ((-1.8 xx 10^(3))/(300))`
`= 6-6 = 0`
32.

As per second law of thermodynamics a process taken place spontaneously if and only if the entropy of the universe increases due to the process. Change in entropy is given by Delta S = (Q_(rev))/(T) A gas C_(V) = (0.2 T) Cal K^(-1). What is the change in its entropy when one mole of it is heated from 27^(@)C to 127^(@)C at constant volume ?

Answer»

`20 "cal "K^(-1) mol^(-1)`
`15 "cal "K^(-1) mol^(-1)`
`35 "cal "K^(-1) mol^(-1)`
`25 "cal "K^(-1) mol^(-1)`

Solution :`Delta S = UNDERSET(T_(1))OVERSET(T_(2))int (N C_(V) DT)/(T) + underset(V_(1))overset(V_(2))int (pdv)/(T)`
For the GIVEN condition
`Delta S= underset(T_(1))overset(T_(2))int (n C_(v)dT)/(T) = (0.2T)/(T) dT`
`= 02 (T_(2) - T_(1)) = 20` cal/mol- k
33.

As per MO theory last electron of N_(2) filled in which molecular orbitaJ ?

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`sigma_(MO)`
`pi_(MO)`
2p
sp

Solution :`sigma_(MO)`
34.

As per law of mass action, for NH_(4) HS(s) hArrNH_(3)(g) + H_(2) S(g) ratio of rate constants of forward (K_(f)) & backword (K_(b))reactions at equilibrium equals ot

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<P>`[NH_4HS]`
`P_(NH_3)+P_(H_2S)`
`[H_2S]+[NH_3]`
`[NH_3] [H_2S]`

SOLUTION :`(K_(f))/(K_(b))=K_(EQ)=[NH_(3)][H_(2)S]`
35.

As per Aufbau principle, arrange the orbitals in increasing order of energy.

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`4p GT 4D gt 5S gt 5p`
`4p lt 4d lt 5s lt 5p`
`4d lt 4p lt 5s lt 5p`
`4p lt 5s lt 4d lt 5p`

Answer :D
36.

As one moves along a given row in the periodic table, ionisation energy

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Remains same
Increasing from LEFT to righl
First increases and then DECREASES
deei'eascs from left LA right

Answer :B
37.

As observed from Andrew's isothermals of CO_(2), critical temperature of CO_(2) is "_____________"""^(@)C

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13.1
24.05
30.98
48.1

Answer :C
38.

As O_(2) (l)is colled at 1 atm pressure ,it freezes to form solid Iat54.5 K.At a lowertemperature, solid I rearrangesto solid II which has a different crystalstructure. Thermal measurementsshow thatform the phase transition of solidI to solid II, DeltaH =- 743.1J //mol and DeltaS= - 17.0 JK^(-1)mol^(-1).At what temperature aresolidsI andII in equilibrium ?

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2.06K
31.5K
43.7 K
53.4 K

Solution :For SOLID `I HARR `solid II EUILIBRIUM , `DeltaG=0` . Hence,from `DeltaG= DeltaH-T DeltaS ,0=DeltaH - T DeltaS`or `T= ( DeltaH )/( DeltaS)= ( - 743.15 mol^(-1))/( - 17.0 JK^(-1) mol^(-1))= 43.7 K`
39.

As its melting point, ice is lighter than water because:

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`H_(2)O` molecules are more closely packed in the SOLID state
ice crystals have follow hexagonal arrangement of `H_(2)O` molecules
on melting of ice `H_(2)O` molecules shrink in size
ice forms mostly heavy water on first melting.

Solution :(B) Ice crystals have `H_(2)O` molecules ARRANGED hexxagonally and linked by hydrogen bonds.
40.

As it passes into food chain the concentration of DDT

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Reamins same
Decreases
INCREASES
Unpredicatable

Solution :As DDT PASSES in to food CHAIN its concentrtion increases
41.

As ester is treated with excess of methyl magnesium bromide and after subsequent hydrolysis, the product formed is isopropyl alcoho. The ester can be

Answer»

METHYL formate
Ethyl fromate
Methyl acetate
Either (A) or (B)

SOLUTION :
42.

As electron moves away from the nucleus , its potential energy

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Increases
Decreases
Remains CONSTANT
NONE of these

ANSWER :B
43.

As electron moves away from the nucleus, its

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K.E. DECREASES by P.E. increases
Both K.E. and P.E. decrease
Both K.E. and P.E. increase
K.E. increases but P.E. decreases.

Solution :`K.E. = (ZE^(2))/(2r), P.E. = - (Ze^(2))/(r)`
44.

As DeltaG^(circ) for a reaction changes form a large negative value to a large positive value, K for the reaction will change form :

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a LARGE POSITIVE value to a large negative value.
a large positive value to a small positive value.
a large negative value to a large positive value.
a large negative value to a small negative value.

ANSWER :B
45.

Ascomparedtopotassium ,sodiumhas______.

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LOWER electronegativily
higherionizationpotential
GREATER ATOMIC radius
Lowermeltingpoint

ANSWER :B
46.

A_((s))+B_((g)) +" heat "Leftrightarrow 2C_((s))+2D_((g)). At equilibrium the concentration of B is doubled. By what factor the concentration of D should change to retain the equilibrium

Answer»

`SQRT2`
2
3
`SQRT3`

Solution :`K_(p)=(P_(D)^(2))/(P_(B)) ........(1), K_(P)=(x.P_(D))^(2)/(2P_(B)) .........(2)`
on solving `x=sqrt2`
47.

A_((s)) + B_((g)) + "heat" hArr 2C_((s)) + 2D_((g)) . At equilibrium the pressure of 'B is doubled. By what factor the concentration of 'D' should change, to reattain the equilibrium

Answer»

`sqrt(2)`
2
3
`sqrt(3)`

SOLUTION :`K_(C)=([D]_(1)^(2))/([B]_(1))=([D]_(2)^(2))/([B]_(2))`, `([D]_(1)^(2))/([B]_(1))=([D]_(2)^(2))/(2[B]_(1)) implies [D]_(2)=sqrt2 [D]_(1)`
48.

As an electron is brought from an infinite distance close to nucleus of atom, the energy of electron

Answer»

Increases to a GREATER +ve value
Decreases to a smaller +ve value
Increases to a smaller -ve value
Decreases to a greater -ve value

SOLUTION :As e APPROACHING towards nucleus its energy DECREASE and become more negative because of atractions.
49.

As an oxidising agent, hydrogen peroxide reacts rapidly in

Answer»

ACIDIC MEDIUM
BASIC medium
Neutral medium
Non AQUEOUS medium

Answer :B
50.

As a rubber band is stretched, it gets warmer, when released, it gets cooler. The correct sign of thermodynamic parameters for stretching of rubber band is

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`Delta H = -ve`
`Delta H = +ve`
`Delta S = -ve`
`Delta S = +ve`

Solution :As rubber band is STRETCHED, the polymeric MOLECULES come close together and HENCE heat is released. Close packing in streched rubber decreases the entropy against the LOOSELY held polymeric molecules in NORMAL relaxed rubber.