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.

During roasting of zinc, zinc blende , it convets to :

Answer»

`ZNO`
`ZnSO_4`
`ZnCO_3`
Zn

Answer :A
Discussion
2.

During roasting of copperr pyrites the major reaction observed is ___________.

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`2FES + 3O_(2) to 2FeO + 2SO_(2)`
`CuFeS_(2) + 3O_(2) to FeO + CUO + SO_(2) UARR`
`CuFeS_(2) + O_(2) to Cu_(2)S + 2FeS + SO_(2) uarr`
`2CuS + 3O_(2) to 2CuO + 2SO_(2) uarr`

SOLUTION :To remove the volatile impurity
Discussion
3.

During refining of liquid iron to steel , CaO plays various roles, out there which of the following is most important

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To MAKE a slag with `SiO_2` FORMED
To prevent oxidation of SULPHUR
To prevent higher concentration of phosphorus in steel
To make a very fluid slag

ANSWER :C
Discussion
4.

Draw the structure of BrO_(4)^(-)

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SOLUTION :
Discussion
5.

During respiration , food is oxidised to carbon dioxide in the presence of oxygen . This process is called :

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Aerobic
Anaerobic
Anabolism
Catabolism

Answer :A
Discussion
6.

During reduction of CuO, impurity of FeO can be removed by adding _________

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an acidic FLUX , `SiO_(2)`
a basic flux , LIMESTONE
a basic flux, `SiO_(2)`
an acidic flux , CaF

Solution :FeO is basic OXIDE.
Discussion
7.

Draw the structure of BrF_(3).

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SOLUTION :
Discussion
8.

Draw the structure of BrF_(5).

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Solution :The electronic configuration of Br (Z = 35) is `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(10) 4S^(2) 4p_(x)^(2) 4p_(y)^(2)4p_(z)^(1)`. It has only one half-filled orbital. But to form five Br-F bonds, we need five half-filled ORBITALS. To achieve this, one electron each of `4p_(x) and 4p_(y)` gets excited to 4d-orbitals. The resulting six orbitals of the fourth shell of Br in the 2ND

excited state undergo `SP^(3)d^(2)`-hydridization. As a result, `BeF_(5)` has octahedral (or square pyramidal) geometry with one position occupied by a lone pair.
Discussion
9.

During reduction of aldehydes with hydrazine and potassium hydroxide, the first is the formation of

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`R-CH=N-NH_2`
`R-C-=N`
`R-UNDERSET(O)underset(||)(C)-NH_(2)`
`R-CH=NH`

ANSWER :A
Discussion
10.

During reduction of aldehydes with hydrazine and potassium hydroxide, the first is formation of

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`R-C-=N`
`R-CO-NH_(2)`
`R-CH=NH`
`R-CH=N-NH_(2)`

SOLUTION :FIRST HYDRAZONES are FORMED.
Discussion
11.

Draw the structure of BrF_(3) molecule.

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SOLUTION :
Discussion
12.

During reduction of aldehydes with hydrazine and C_(2)H_(5)ONa the product formed is

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`R-CN=N-NH_(2)`
`R-C-=N`
`R-UNDERSET(O)underset(||)(C)-NH_(2)`
`R-CH_(3)`

ANSWER :D
Discussion
13.

During reduction of aldehyde with hydrazine andKOH , thefirst is the formation of

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`R - CH = NH`
`R - CH = N - NH_(2)`
`RCONH_(2)`
`R - C -= N`

Answer :B
Discussion
14.

Draw the structure of boric acid.

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Solution :Boric acid has a two dimensional LAYERED structure. It consists of `[BO_(3)]^(3-)` UNIT and these are linked to each other by HYDROGEN bonds.
Discussion
15.

During reaction of benzene with neopentyl chloride in the presence of AICI_(3) the major product is 2- methyl-2-phenylbutane and not neopentylbenzene. Explain?

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SOLUTION :The initially formed primary carbocation rearranges by methyl migration to form a more stable tertiary carbocation which ACTS as a electrophile and attack the benzene to GIVE the MAJOR product.
Discussion
16.

During recharging of car-battery, the battery acts as

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Electrolytic cell
Voltaic cell
dry cell
concentration cell

Solution :CAR battery is MAINLY lead ACCUMULATOR.
Discussion
17.

Draw the structure of beta-D-ribose and beta-D-2-deoxyribose.

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SOLUTION :
Discussion
18.

During reduction of aldehyde with(H_2N-NH_2)/(OH // "glycol") the first interinediate compound formed as

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RCN
`RCONH_(2)`
`R-CH=NH`
`R-CH=N NH_(2)`s

Answer :D
Discussion
19.

During purification of colloidal sol by ultracentrifugation which of the following is . Observed ?

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COLLODIAL particles are settled at THEBOTTOM of ultracentrifuge tube
Impurities are settled at the BOTTOM of theultracentrifuge tube
Impurities are removed through ULTRAFILTERS
Its rate can be increased by applying pressure

Answer :A
Discussion
20.

Draw the structure of borazole. B_(3)N_(3)H_(6).

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SOLUTION :
Discussion
21.

During preparation of phenol from cumene, side product obtained is

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acetone
alcohol
aldehyde
acid

Answer :A
Discussion
22.

Draw the structure of ampicillin.

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SOLUTION :
Discussion
23.

Draw the structure of an aromatic compound having molecular formula C_(12) O_(9) .

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SOLUTION :
Discussion
24.

Draw the structure of all isomeric alcohols of molecular formula C_(4)H_(10)O and give their IUPAC names.

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Solution :`{:("Imomers","IUPAC name"),((i)CH_(3)-CH_(2)-CH_(2)-CH_(2)-OH,"Butan-1-ol"),((ii)CH_(3)-UNDERSET(CH_(3))underset("|")(CH_(2))-CH_(2)-OH,"2-Methylpropan-1-ol"),((iii) CH_(3)-underset(OH)underset("|")"CH"-CH_(2)-CH_(3),"Butan-2-ol"),((iv) CH_(3)-OVERSET(CH_(3))overset("|")underset(OH)underset("|")"C "-CH_(3),"2-Methylpropan-2-ol"):}`
Discussion
25.

During prepartion of wrought iron, CO and SO_(2) formed escape whereas ________ to form slag.

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MnO and `SiO_(2)` COMBINE
MN and Si combine
MG and Si combine
MGO and `SiO_(2) ` combine

SOLUTION :Flux
Discussion
26.

Draw the structure of alpha -D fructose furanose and beta -D fructose furanose.

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SOLUTION :
Discussion
27.

During preparation of arenediazonium salts the excess of nitrous acid, if any, is removed by adding

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AQUEOUS `Na_2CO_3`
Aqueous NaOH
Aq.`NH_2CONH_2`
`NH_4OH`

ANSWER :3
Discussion
28.

Draw the structure of (a) N_(2)O_(4) (b) N_(2)O_(5).

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Solution :(a) `N_(2)O_(4)` (Nitrogen tetraoxide)
`OVERSET(":"overset(..)O":")overset(|)underset(underset(..)O)underset(||)N-underset(underset(..)(":"O":"))underset(|)overset(":"O":")overset(||)N`
(B) `N_(2)O_(5)`
`underset(underset(..)(":"O":"))underset(|)overset(":"O":")overset(||)N-overset(..)underset(..)O-underset(underset(..)(":"O":"))underset(|)overset(":"O":")overset(||)N`
Discussion
29.

During photochemical halogenation of alkanes the halogen molecule first gives:

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CHLORIDE ION
Chloronnium ion
Free radical
Carbonium ion

Answer :C
Discussion
30.

Draw the structure of (a) white phosphorous (b) red phosphorous

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Solution :(a) White PHOSPHOROUS (B)Red PHOSPHORUS
Discussion
31.

Draw the structure of (a) white phosphorous (b) red phsphorous

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SOLUTION :
Discussion
32.

During oxidation of oxalic acid by acidified KMnO_4 how many unpaired electrons are present in the autocatalyst ?

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Solution :`KMnO_4 underset(("Medium"))OVERSET(("Acidic"))(RARR) MN^(+2)`
`Mn^(+2) = 3d^5 "" n = 5`.
Discussion
33.

Draw the structure of (a) Pyrosulphuric acid (b) Peroromonosulphuric acid?

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SOLUTION :(a) Pyrosulphuric acid `H_(2)S_(2)O_(7)`
`HO-OVERSET(O)overset(||)underset(O)underset(||)S-O-overset(O)overset(||)underset(O)underset(||)S-OH`
(B) Peroromonosulphuric acid `H_(2)SO_(5)`
`HO-overset(O)overset(||)underset(O)underset(||)S-O-OH`
Discussion
34.

During osmosis, net flow of water through a semipermeable membrane is

Answer»

from both SIDES of semipermeable membrane with UNEQUAL flow rates
from SOLUTION having lower CONCENTRATION only
from solution having higher concentration only
from both sides of semipermeable menbrane with equal flow rates

Answer :B
Discussion
35.

Draw the structure of (a) Hyponitrous acid (b) Hydronitrous acid.

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SOLUTION :(a) Hyponitrous ACID : `H_(2)N_(2)O_(2)`
`HO-N=H-OH`
(b) Hydronitrous acid :
Discussion
36.

During osmosis, flow of water through a semipermable membrane is :

Answer»

FORM both sides of the SEMIPERMEABLE with equal flow rates.
from both sides of semipermeable membrane with unequal flow rate
form solution having lower concentraion only
form solution having HIGHER CONCERTRATION only.

Solution :is the correct ANSWER.
Discussion
37.

Draw the structure of (a) N_(2)O (b) N_(2)O_(3).

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SOLUTION :(a) `N_(2)O` (Nitrous oxide)
`:N-=overset(o+)N-overset(Theta)overset(..)UNDERSET(..)O:harroverset(Theta)overset(..)underset(*..)N=overset(o+)N=overset(..)underset(..)O`
(b) `N_(2)O_(3)` (Nitrogen sesquoxide)
`{:("O"O^(Theta)"OO"),("|||||||"),(N-N^(o+)harrN-N_(o+)),("|||"),("O"""O^(Theta)):}`
Discussion
38.

During nucleophilic addition reaction, the hybridisation of carbon changes from .....

Answer»

`SP^2` to `sp^3`
`sp^3`to `sp^2`
sp to `sp^3`
`dsp^2` to `sp^3`

SOLUTION :`sp^2` to `sp^3`
Discussion
39.

Draw the structure of (a) Marshall's acid (b) Polythionic acid (c ) Dithionic acid

Answer»

SOLUTION :
Discussion
40.

Duringnuclearexplosion , one of the productsis ""^(90)Sr withhalf-lifeof 28.1 years. If 1 mu gof ""^(90)Sr was absorebedin the bones of a newly born baby instead of calcium ,how much of it will remain after 10 years and ""^(14)C foundin a livingtree. Estimatethe ageof the sample.

Answer»


ANSWER :77.7 MIN
Discussion
41.

Draw the structure of a carbonyl group and indicate : the sigma and pi bonds.

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SOLUTION :.
Discussion
42.

Draw the structure of a carbonyl group and indicate : the electrophilic and nucleophilic centres

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SOLUTION :.
Discussion
43.

During osmosis, flow of water through a semi permeable membrane is

Answer»

from both sides of semi PERMEABLE MEMBRANE with EQUAL flow rates
from both sides of the semi permeable membrane with unequal flow rates
from solution having lower concenration only
from solution having HIGHER concentration only.

Answer :B
Discussion
44.

Draw the structure of a carbonyl group and indicate : hybridized state of carbon .

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SOLUTION :.
Discussion
45.

During nuclear explosion, one of the products is ""^(90)Sr with half-life of 28.1 years. If 1 mugof ""^(90)Srwas absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically?

Answer»

SOLUTION :Radioactive decay follow first order kinetics,
Decay constant of `""^(90)Sr(k)=(0.693)/(t_(1//2))=(0.693)/(28.1" years")=2.466xx10^(-2)"years"^(-1)`
Amount LEFT after 10 years.
`a=1 mug, t=10" years", k=2.446 xx 10^(-2)"years"^(-1)`
Substituting the values in the first order equation
`k=(2.303)/(t)"log"(a)/(a-x) or 2.466 xx 10^(-2)=(2.303)/(10)"log"(1)/((a-x))`
or `log(a-x)= -0.1071`
or `(a-x)="Antilog " bar1.8929=0.7814 mu G`.
Amount left after 60 years
`2.466 xx 10^(-2)=(2.303)/(60)"log"(1)/(a-x) or log(a-x)= -0.6425`
or `(a-x)="Antilog "bar1.3575=0.2278 mu g`.
Discussion
46.

Draw the structure of a carbonyl group and indicate clearly (i) the hybridised state of carbon, (ii) the sigma and pi it bonds present and (iii) the electrophilic and nucleophilic centres in it.

Answer»

SOLUTION :
(i) The hybridised state of .C. is `SP^(2)`, (II)
(iii)
Discussion
47.

During nuclear explosion, one of the products is ""^(90)"Sr" with half life of 28.1 years. If 1 mug of ""^(90)"Sr" was absorbed in bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically ?

Answer»

Solution :As radioactive disintegrations follow first order kinetics,
Decay CONSTANT of `""^(90)"Sr "(k)=(0.693)/(t_(1//2))=(0.693)/(28.1" y")=2.466xx10^(-2)y^(-1)`
To CALCULATE the amount left after 10 years.
`a=1 mug,t=10" years,"k=2.466xx10^(-2)y^(-1),(a-x)=?`
`k=(2.303)/(t)log""(a)/(a-x)`
`2.466xx10^(-2)=(2.303)/(10)log""(1)/((a-x))" or "log(a-x)=-0.1071`
or `(a-x)=" Antilog "=1.8929=0.7814 mug`
To Calculate the amount left after 60 years.
`2.466xx10^(-2)=(2.303)/(60)log""(1)/(a-x)" or "log(a-x)=-0.6425`
or `(a-x)=" Antilog T"*3575=0.2278" "mug.`
Discussion
48.

Draw the structure of 4- tertbutyl -3- iodoheptane .

Answer»

Solution :The structure of 4- tertbutyl -3- iodoheptaneis
`OVERSET(1)CH_(3) - overset(2)CH_(2) - overset(I)overset(3|)CH -underset(CH_(3) - underset(CH_(3)) underset(|) overset(|) C - CH_(3))overset(4)(CH)- overset(5)CH_(2) - overset(6)CH_(2)- overset(7)CH_(3)`
Discussion
49.

Draw the structure of 3 - methylbutanal.

Answer»

SOLUTION :`UNDERSET("3 - Methylbutanal")(CH_(3)-overset(CH_(3))overset("|")"CH"-CH_(2)-CHO)`
Discussion
50.

During nuclear explosion one of the products is ""^(90)Sr with half-life of 28.1 years .If 1 mug of ""^(90)Sr was absorbed in the boned of a newly born baby instead of calcium ,how much of it wil reman after 10 years and 60 years if it is not lost metabolically.

Answer»

Solution :Nuclear explosion is a first order reaction.
`1 mug ""^(90)Sr` is observed in NEW born baby.
`therefore[R]_(0)=1mug`
Time `t_(1)` =fter 10 years,so,`[R]_(t_(2))`=y `mug`
Constant of explosion reaction =k.If the reaction is of first order.
`k=(0.693)/(t_((1)/(2)))=(0.693)/(28.1 year)=2.4662xx10^(-2)years`
After 10 years ,the calculation of remianing `""^(90)Sr`:
`t=(2.303)/(k)` log `([R]_(0))/([R]_(t))` where ,t=10
`therefore 10=(2.303)/(2.4662xx10^(-2))(log 1 mug)/(x mug)`
`therefore (1)/(x)=(10xx0.024662)/(2.303)=0.1070 `
`therefore (1)/(x)`=ANTI log 0.1070
`therefore (1)/(2)=1.2794`
`therefore=0.7816 mu g ""^(90)Sr` will be remaining.
Suppose `""^(90)S=y mug` after t=60 year,
`t=(2.303)/(k)` log `([R]_(0))/([R]_(r))`
`therefore 60=(2.303)/(0.024662)` log`(1)/(y)`
`therefore (60xx0.024662)/(2.303)` =log `-log y
`therefore` 0.6425=0-log y=-log y
`therefore` log y=-0.6425
`therefore` y=Antilog (-0.6425)
Discussion