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.

If a mixtureof methane, ammonia andoxygenis passedoverPt - gauge at 973K thentheproducts will be :

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`HCOOH`
`HCN`
`CH_(3)NH_(2)`
all of these

Answer :B
2.

If a mixture containing 1-butene, cis-2-butene and trans-2-butene is treated with perbenzoic acid, how many different isomers of epoxides result?

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SOLUTION :N//A
3.

If a mixture of gas has a total pressure of 100 cm Hg and the partial pressure of nitrogen in the mixture is 25 mm Hg, then the per cent of nitrogen in the mixture is :

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`4%`
`40%`
`400%`
`2.5%`

ANSWER :D
4.

If a mixture 0.4 mole H_2 and 0.2 mole Br_2 is heated at 700 K at equilibrium, the value of equilibrium constant is 0.25xx10^10 then find out the ratio of concentrations of (Br_2) and (HBr) (Report your answer as (Br_2)/(HBr)xx10^11)

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Solution :`{:(,H_2+," "Br_2" "hArr,2HBr),(t=0,0.4," "0.2,-),(t=t_(EQ),0.2,UNDERSET("=negligible")(y),underset(=y)(0.4)):}`
`because 1/4xx10^10=(0.4xx0.4)/(0.2xxy)IMPLIES y=3.2xx10^(-10)implies (Br_2)/(HBr)xx10^11=3.2/0.2xx10^(-10)xx10^(11)=80`
5.

If a is the length of the side of the cube, the distance between the body-centred atom and one corner atom in the cube will be …….

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`2/(SQRT3) a`
`4/(sqrt3)a`
`(sqrt3)/4 a`
`(sqrt3)/2 a`

ANSWER :D
6.

If "A" is total number of meso compounds and "B" is total number of optically active isomers, then find (A+B) for

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SOLUTION :`A=1, B=2 RARR A+B=3.`
7.

If 'a' is the length of the side of the cube, the distance between the body centered atom and one corner atom in the cube will be ………….. .

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`((2)/(sqrt3))a`
`((4)/(sqrt3))a`
`((sqrt3)/(4))a`
`((sqrt3)/(2))a`

Solution :Hint : If a is the length of the SIDE, then the length of the leading diagonal passing through the body CENTERED ATOM is `SQRT3A`
`"Required distance "=((sqrt3)/(2))a`
8.

If ‘a’ is the length of the side of a cubic unit cell the distance between the body-centred atom and the corner atom in the cube will be

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`(2)/sqrt(3)a`
`sqrt(3)/(2)a`
`(4)/sqrt(3)a`
`sqrt(3)/(4)a`

Answer :B
9.

If a is the initial concentration of the reaction and t_(1//2) is the half change time. Which of the following in true of the n^(th) order reaction

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`t_(1//2) PROP a^(N-1)`
`t_(1//2) prop a^n`
`t_(1//2)prop a^(1-n)`
`t_(1//2)prop a^(n+1)`

Solution :For `n^(th)` ORDER REACTION , `t_(1//2) prop (1)/(a^(n-1)) or t_(1//2) prop a^(n-1)`
10.

If a is the initial concentration then time required to decompose half of the substance for nth order is inversely proportional to:

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`a^n`
`a^(n-1)`
`a^(1-n)`
`a^(n-2)`

ANSWER :B
11.

If a is the initial concentration of the reactant, the time taken for completion of the reaction if it is of zero order will be

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a/k
a/2k
2a/k
k/a

Answer :A
12.

If a is the initial concentration then time required to decompose half of the substance for nth order is inversely proportiional to :

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`a^(N)`
`a^(n-1)`
`a^(1-n)`
`a^(n-2)`

SOLUTION :`t_(1/2)ALPHA(1/(a^(n-1)))`
13.

If 'a' is the initial concentration of the reactant, the time taken for completion of the reaction, if it is of zero order, will be

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`a/k`
`a/(2K)`
`(2a)/k`
`k/a`

SOLUTION :For a zero order REACTION , `t = (1)/(k){[A]_(0) - [A] }`
When reaction is complete `[A] = 0`
Hence , `t = ([A_(0)])/(k) `
14.

If a is the edge length of a unit cell, then correct option (s) is/are

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for simple CUBIC lattice, RADIUS of METAL ATOM `=a/2`
for bcc lattice, radius of metal atom`=(sqrt(3)a)/2`
for fcc lattie, radius of metal atom `=a/(2sqrt(2))`
distance between nearest neighbours for fcc `d=(sqrt(3)Q)/2`

Solution :For bcc lattice, radius of metal atom `=(sqrt(3))/4`
For fcc lattice, distance between nearest neighbours d
`=a/(sqrt(2))`
15.

If 'a' is the initial concentration of a substance which reacts according to zero order kinetic and k k'is rate constant, the time for the reactant to go to completion is,

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`a//K`
`2//k`
`K//a`
`2K//a`

ANSWER :A
16.

If a is the initial concentration of a isubstance which reacts according to zero order kinetics and k is rate constant the time for the reaction to go to completion is :

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K/a
ak
a/2k
a/k

SOLUTION :(D) `[A]_(t) = [A]_(0)-KT `
kt `=[A]_(0) "or " t = (a)/(k)`
17.

If a is the degree of dissociation of Na_2 SO_4,the Van't Hof's factor 'i' used for calculating the molecular mass is:

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`1+ ALPHA`
`1-alpha`
`1+2ALPHA`
`1- 2alpha`

SOLUTION :`1+2alpha`
18.

If 'A' is C_(2)H_(5)NH_(2) 'B' " is" (C_(2) H_(5) )_(2) NH, 'C' " is" (C_(2) H_(5))_(3) N thenorder ofsolubilityin water is

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`A GT B gt C`
`B lt A lt C`
`C lt B lt A`
`C gt Blt A`

Solution :Primaryamineshas `- NH_(2)`DUE towhichH- bondingis easilypossiblehencethe solubilityorder ofaminesis `3^(@) LG 2^(@)lg 1^(@)`
19.

If [A] is the concentration of A at any time t and [A_(0)] is the concentration at t = 0, then for the first order reaction, the rate equation can be written as ___________.

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`K=(2.303)/(t)LOG[(A)/(A_(0))]`
`k_(t)=2.303 log[(A_(0))/(A)]`
`k=(2.303)/(t)log[(A_(0))/([A_(0)]-[A])]`
`k=(2.303)/(t)log[(A_(0))/(A)]`

ANSWER :B
20.

If A is the central atom of the molecule containing A and X atoms and E is the number of lone pairs around it, then VSEPR notation AX_(3)E will be for the molecules :

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`ICl_(4), PCl_(3)`
`NH_(3), H_(2)O`
`ClO_(3)^(-), NH_(3)`
`ClO_(3)^(-), ClO_(4)^(-)`

ANSWER :C
21.

If a homogeneous catalytic reaction can take place through three alternative paths as depicted below, the catalytic efficiency of P, Q, R representing the relative case would be

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<P> PgtQgtR
QgtPgtR
PgtRgtQ
RgtQgtP

Solution :The catalytic efficiency of a CATALYST is measured in terms of the decrease in ACTIVATION energy of the reaction. A catalyst is more efficient if it decreases the `E_a`, to a larger extent. Thus, CORRECT order for catalytic efficiency is R>Q>P.
22.

If a graph is plotted between log V and log T for 2 moles of a gas of constant pressure of 0.0821 atm, then which of the following statements are correct?

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<P>The curve is straight LINE with slope -1.
The curve is straight line with slope +1
The INTERCEPT on y-axis is equal to 2.
The intercept on y-axis is equal to 0.3010.

Solution :`PV=nRT`
`log V =logT+"log"(nR)/P`
slope =1
intercept `="log"(nR)/P=log[(2xx0.0821)/0.0821]=0.3010`
23.

When raisins ( Kismis ) are placed in water , they swell due to the process of

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SOLUTION :When a grape is placed in water, there is a flow of water into the grape DUE to osmosis, HENCE it swells.
24.

If a gas is heated at constant pressure, its density :

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Will INCREASE
Will decrease
Will REMAIN unchanged
May increase or decrease

Answer :B
25.

If a gas ia expanded at constant temperature

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the PRESURE decreases
the KINETIC energy of the molecules remains the same
the kinetic energy of the molecules decreases
the number of molecules of the gas increases.

Solution :At constant TEMPERATURE, the kinetic energy remains same (`:.K.E. PROP T`) i.e. b. is correct and c is incorrect.
At constant temperature
PV=Constant [Boyle.s LAW]
If a gas expands i.e., V increases then P decreases i.e. a is correct d is incorrect because there is no change in number of molecules of gas.
26.

If a gas gets half compressed, compared to an ideal gas, the compressibility factor Z is equal to

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1
2
`1/2`
NONE of these

Answer :B
27.

If a gas is explained at constant temperature :

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Number of molecules of the GAS decrease
The KINETIC ENERGY of the molecules decrease
The kinetic energy of the molecules remains the same
The kinetic energy of the molecules increases

Answer :C
28.

If a gas expands at constant temperature , it indicates that

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KINETIC ENERGY of molecules remains the same
Number of the molecules of gas increases
Kinetic energy of molecules decreases
Pressure of the gas increase

Solution :Kinetic energy of GASEOUS molecules depend on TEMPERATURE only.
29.

If a fragment of DNA molecule has the base sequence : (a) TGATGCCGA (b) ATCTCGTAC (c) GTAATCCAG (d) AATTCCCGG.

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SOLUTION :(a) ACTACGGCT
(B) TAGAGCATG
(C) CATTAGGTC
(d) TTAAGGGCC.
30.

If a gas, at constant temperature and pressure expands, then its

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ENTROPY first increases and then decreases
internal energy increases
internal energy REMAINS the same
internal energy decreases

Solution :Internal energy of a gas depends upon its pressure ad temperature. THUS, if a gas expands at constant temperature and PRESURE, then its internal energy remains same.
31.

If a freshly formed ppt of SnO_2 is peptised by a small amount of NaOH, these colloidal particles may be represented as

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`[SnO_2]SnO_(3)^(2-) : Na^(+)`
`[SnO_2]SN^(4+) : O^(2-)`
`[SnO_2]Na^(+) : OH^(-)`
`[SnO_2]Sn^(4+) : OH^(-)`

SOLUTION :`SNO_(2) +NAOH rarr underset("SODIUM stannate ")(Na_2SnO_3) "" [SnO_2]SnO_(3)^(2-)`
32.

If a dilute solution of aqueous NH_(3) is saturated with H_(2)S then the product formed is :

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`(NH_(4))_(2)S`
`NH_(4)HS`
`(NH_(4))_(2)S_(x)`
`NH_(4)OH+S`

Answer :B
33.

If a cylinder at NTP contains a gas and it can withstand a temperature of 473K then pressure inside the cylinder at the time of explosion will be?

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Solution :Using Gay Lusaac’s Law `(P_(1))/(T_(1)) = (P_(2))/(T_(2))` . So, as per the data GIVEN `(1)/(273) = (P_(2))/(473) IMPLIES P_(2)= 1.732"ATM"` at the time of EXPLOSION.
34.

If a current of oneampere flows through a metal wire for one hour, how many electrons would flow through the wire?

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SOLUTION :Currentstrength (i) = 1 amp, TIME (t) = 1 hr = 3600s
Quantity of electricity `(Q)=1xx3600=3600` coulomb
Number of electrons FLOWN through the wire `=("Quantity of electricity")/(96500)xx6.023xx10^(23)`
`(3600xx6.023xx10^(23))/(96500)=2.24xx10^(22)`.
35.

If a current of 0.5 ampere flows through a metallic wire for two hours, then how many electrons flow through the wire?

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Solution :Time for which current is passed = `2 XX 60 xx 60 `seconds
Quantity of current =` Ixxt= 0.5xx2xx60xx60=3600C`
Number of ELECTRONS that flow when 3600 C of current is passed
= `6.02xx10^23 xx 3600/96500=2.24xx10^22`
36.

If a current of 0.5 ampere flows through a metallic wire for 2-hours, then how many electrons would flow through the wire?

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Solution :`Q("COULOMB") = 1("ampere") xx t("SEC")`
`= 0.5 "ampere" xx 2 xx 60 xx 60 = 3600 C`
A flow of 1F, i.e., 96500 C is equivalent to the flow of 1 mole of electrons i.e., `6.023 xx 10^23` electrons
3600 C is equivalent to flow electrons = `(6.02 xx 10^(23))/(96500) xx 3600 = 2.246 xx 10^(22)` electrons.
37.

If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire ?

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Solution :`{:(I,=,0.5 "AMPERE"),(t,=,2xx 60xx60),( ,=,7200sec ),( Q,=,It ),(,=,0.5 xx 7200),(,=,3600 C ),(96500 C ,=,1 "mole of " e ),(or 96500 C ,=,6.022 xx 10^(23)),(3600 C,=,(6.022 xx 10^(23))/((96500 )) xx 3600),( ,=,(6.022xx 360 )/(965 )xx 10^(22)),(,=,2.25xx 10^(22)"electrons"):}`
38.

If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

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SOLUTION :Given : 1 = 0.5 A, t = 2 h = 2 x 60 x 60 s = 7200 s
QUANTITY of electricity = I x t = (0.5 A) x (7200 s) = 3600 C A flow of 1 F, i.e., 96500 C is equivalent of flow of 1 MOLE of electrons, `6.02 xx 10^(23)` electrons
`therefore 3600 C` is equivalent to flow of `=(6.02 xx 10^(23))/96500 xx 3600` electrons `=2.246 xx 10^(22)` electrons.
39.

If a current of 0.5 ampere flows through a metallic wire for 2 hours then how many electrons would flow through the wire.

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Solution :Data: I=0.5 amp, `t=2 xx 60 xx 60 =7200 Q =It `
96500 Couloumbs `to 6.022 X 10^(23) e^(-)s""=0.5 xx 2 xx 60 xx 60`
`3600 Coulombus to ? ""=3600 C`
`=(6.022 xx 10^(23) xx 3600)/(96500) =0.224 xx 10^(23) e^(-)s`.
40.

If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons flow through the wire?

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Solution :Q (COULOMBS)=I(ampere)`XXT(sec)=(0.5" ampere")xx(2xx60xx60s)=3600C`
A flow of 1 F, i.e., 96500 C is equivaelent to flow of 1 mole of ELECTRONS, i.e., `6.02xx10^(23)` electrons
`therefore3600C` is EQUIVALENT to flow of electrons `=(6.02xx10^(23))/(96500)xx3600`
`=2.246xx10^(22)` electrons.
41.

If a compund absorbs orange colour from the white light, then the observed colour of the compound is

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YELLOW
orange
BLUE
VIOLET

Answer :C
42.

If a compounds has n chical centres and compound is unsymmetric then possible number of steroisomers (or optical isomers) is

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`2^(N)`
`2^(n+1)`
`2sqrt(n)`
`sqrt(2n)`

Solution :For compounds and show containing THREE chiral CENTRES, there could beeight optical isomers, for compounds containing four chira centres, there could be 16 optical isomers and so on. Hence the maximum number of optical isomers that can exist is equal `2^(n)`.In any CASE where meso compounds exist, there will be fewer optical isomers than this maximumnumber.
43.

If a current of 0.5A flows through a metallic wire for 2 hours, then howmany electrons would flow through the wire ?

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Solution :`q=ixxt=0.5xx2xx60xx60=3600C`
`"96500 Coulombs are EQUAL to 6.022"xx10^(23)E`.
So, 3600 Coulombs `=(6.022xx10^(23))/(96500)xx3600xx10^(22)" electrons"`
44.

If a compound on analysis was found to contain C = 18.5 H = 1.55% ,Cl= 55.04% and O = 24.81%. Then its empirical formula is

Answer»

`CHClO`
`CH_(3)ClO`
`C_(2)H_(2)Ocl`
`ClCH_(2)O`

Solution :`{:("ELEMENT"," NO. of moles"," Simple RATIO"),(C 18.5%,18.5//12=1.54,1),(H1.55%,1.55//1=1.55,1),(Cl55.04%,55.04//35.5=1.55,1),(O24.81%,24.81//16=1.55,1):}`
Hence,FORMULA `= CHClO`.
45.

If acompoundcontains two oxygen atoms four carbon atoms and number of hydrogen atoms is double of carbon atoms the vapour density of it is :

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88
44
132
72

Answer :B
46.

If a compound absorbs violet colour from light, it will be

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 yellow
orange
 blue
green

Answer :B
47.

If a complex, Ma_(4)b_(2) (where M is a central atom and a and b are mondentate ligands ) has trigonal prismatic geometry, the total number of possible geometrical isomers would be :

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4
2
3
5

Solution :
48.

If a compound absorbs violet colour from the sunlight, then the observed colour is:

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Yellow
Orange
Blue
Green

Answer :B
49.

If a closed container contain 5 moles of CO_(2) gas, then find total number of CO_(2) molecules in the container (N_(A)=6xx10^(23))

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ANSWER :`(##ALN_NC_CHM_MC_E01_002_A01##)`
`3XX10^(24)`
50.

If a charged particle (+q) is projected with certain velocity parallel to the magnetic field , then it will

Answer»

trace helical path
trace CIRCULAR path
continue its motion without any CHANGE
come to rest instantly

Solution : 'C' acts as a reducing agent in this reaction . A reducing agent is a SUBSTANCE that loses electron , making it possible for another substance to gain ELECTRONS and be REDUCED . The oxidized substance is always the reducing agent .