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.

SF_(2), SF_(4) and SF_(6)have hybridisations at sulphur atom respectively as

Answer»

`SP^(2), sp^(3), sp^(3)d^(2)`
`sp^(3)sp^(3), sp^(3)d^(2)`
`sp^(3), sp^(3)d, sp^(3)d^(2)`
`sp^(3), SPD^(2), d^(2)sp^(3)`

SOLUTION :
2.

Sewage water can be purified for recycling with the action of

Answer»

AQUATIC plants
Penicillin
Micro-organisms
Fishes

Solution :Because micro-organisms can DECOMPOSE COMPLEX organic compounds into SIMPLE organic compound
3.

Several short-lived radioactive species have been used to determine the age of wood or animal fossils. One of the most intresting substance is ""_(6)C""^(14)(half-life fossils, etc.). Carbon-14 is produced by the bombardment of nitrogen atoms present in the upper atmosphere with neutrons (from cosmic rays). ""_(7)N""^(14)+""_(0)n""^(1)to""_(6)C""^(14)+""_(1)H""_(1) Thus carbon-14 is oxidised, to CO""_(2) and eventually lingested by land animals. The death of plants or animals put an end to the intake of C""_(14) from the atmosphere. After this the amount of C""_(14) in the dead tissues starts decreasing due to its disintegration as per the following reaction : ""_(6)C""^(14)to""_(7)N""^(14)+""_(-1)beta""^(0) The C""^(14) isotope enters the biosphjere when carbon dioxide is taken up in plant photosynthesis. Plants are eaten by animals, which exhale C""^(14) as CO""_(2). Eventually, C""^(14) participates in many aspects of carbon cycle. The C""^(14) lost by radioactive decay-replenishment process, a dynamic rquillibrium is established whereby the ratio of C""^(14) to C""^(12) remains constant in living matter. But when an individual plant or an animal dies, the C""^(14) isotpe inn it is no longer replenished, so the ratio decreases as C""^(14) decays. So, ther number C""^(14) nuclei after time t (after the death of living matter) would be less than in a livingthe following formula, t""_(1//2)=0.693/? The intensity of the cosmic rays have remain the same for 30,000 years. But since some years changes in this are observed due to excessive burning of fossil fuel and nuclear test? A nuclear explosion has taken place leading to increase in concentration of C""^(14) in nearby areas. C""^(14) concentration is C""_(1) in nearly areas and C""_(2) in areas far away. If the age of the fossil is determined to be T""_(1) and T""_(2) at the respective places then

Answer»

The age of the FOSSIL will inrease at the where EXPLOSION has taken place and
`T""_(1)-T""_(2)=1/lamda"in"C""_(1)/(C""_(2))`
The age of the fossil will decrease at the place where explosion has taken place and
`T""_(1)-T""_(2)=1/lamda"in"C""_(1)/(C""_(2))`
The age of the fossil will DETERMINED to be same
`T""_(1)/T""_(2)=C""_(1)/C""_(2)`

Solution :`N//A`
4.

Sewagecontainingorganicwasteshouldbedisposedin waterbodiesbecauseit causesmajorwater pollution . Fishessucha pollutedwaterdie because of

Answer»

Larger NUMBEROF MOSQUITOES.
Increase in the amountofdissolvedoxygen
DECREASE in the amountof dissolvedoxygenin water.
Cloggingof gillsby mud.

ANSWER :C
5.

Several short-lived radioactive species have been used to determine the age of wood or animal fossils. One of the most intresting substance is ""_(6)C""^(14)(half-life fossils, etc.). Carbon-14 is produced by the bombardment of nitrogen atoms present in the upper atmosphere with neutrons (from cosmic rays). ""_(7)N""^(14)+""_(0)n""^(1)to""_(6)C""^(14)+""_(1)H""_(1) Thus carbon-14 is oxidised, to CO""_(2) and eventually lingested by land animals. The death of plants or animals put an end to the intake of C""_(14) from the atmosphere. After this the amount of C""_(14) in the dead tissues starts decreasing due to its disintegration as per the following reaction : ""_(6)C""^(14)to""_(7)N""^(14)+""_(-1)beta""^(0) The C""^(14) isotope enters the biosphjere when carbon dioxide is taken up in plant photosynthesis. Plants are eaten by animals, which exhale C""^(14) as CO""_(2). Eventually, C""^(14) participates in many aspects of carbon cycle. The C""^(14) lost by radioactive decay-replenishment process, a dynamic rquillibrium is established whereby the ratio of C""^(14) to C""^(12) remains constant in living matter. But when an individual plant or an animal dies, the C""^(14) isotpe inn it is no longer replenished, so the ratio decreases as C""^(14) decays. So, ther number C""^(14) nuclei after time t (after the death of living matter) would be less than in a livingthe following formula, t""_(1//2)=0.693/? The intensity of the cosmic rays have remain the same for 30,000 years. But since some years changes in this are observed due to excessive burning of fossil fuel and nuclear test? What should be the age of the fossil meaningful determination of its age?

Answer»

6 years
6000 years
60000 years
can be USED to DETERMINE any age

Solution :`N//A`
6.

Several isomeric amines are possible with the molecular formula C_(8)H_(11)N. On the basis of the reaction given by each of the isomers, identify the structure of the compound. a) A(C_(8)H_(11)N)overset(NaNO_(2)-HCl)underset(0-5^(@)C)to p-Nitroso compound b) B(C_(8)H_(11)N)overset(NaNO_(2)-HCl)underset(0-5^(@)C)to Diazonium salt formed c) C(C_(8)H_(11)N)overset(NaNO_(2)-HCl)underset(0-5^(@)C)to N-Nitroso compound formed The structure of 'C' is

Answer»




All the above

Solution :`1^(@)` AROMATIC AMINE form diazonium SALT , `2^(@)` aromatic amine FORMS N-nitroso compound `3^(@)` aromatic amine form , P-nitoso compound
7.

Several isomeric amines are possible with the molecular formula C_(8)H_(11)N. On the basis of the reaction given by each of the isomers, identify the structure of the compound. a) A(C_(8)H_(11)N)overset(NaNO_(2)-HCl)underset(0-5^(@)C)to p-Nitroso compound b) B(C_(8)H_(11)N)overset(NaNO_(2)-HCl)underset(0-5^(@)C)to Diazonium salt formed c) C(C_(8)H_(11)N)overset(NaNO_(2)-HCl)underset(0-5^(@)C)to N-Nitroso compound formed The structure of 'B' is

Answer»




All the above

SOLUTION :`1^(@)` AROMATIC amine form diazonium salt , `2^(@)` aromatic amine forms N-nitroso COMPOUND `3^(@)` aromatic amine form , P-nitoso compound
8.

Several isomeric amines are possible with the molecular formula C_(8)H_(11)N. On the basis of the reaction given by each of the isomers, identify the structure of the compound. a) A(C_(8)H_(11)N)overset(NaNO_(2)-HCl)underset(0-5^(@)C)to p-Nitroso compound b) B(C_(8)H_(11)N)overset(NaNO_(2)-HCl)underset(0-5^(@)C)to Diazonium salt formed c) C(C_(8)H_(11)N)overset(NaNO_(2)-HCl)underset(0-5^(@)C)to N-Nitroso compound formed The structure of 'A' is likely to be

Answer»




Solution :`1^(@)` AROMATIC amine FORM DIAZONIUM salt , `2^(@)` aromatic amine forms N-nitroso compound `3^(@)` aromatic amine form , P-nitoso compound
9.

Several alkaloids are extracted from the extracts of the plants called marijuana. Marijuana owes its activity to tetrahydro cannabinol, which contains 70% as many as carbon atoms as hydrogen atoms and 15 times as many hydrogen atoms as oxygen atoms. One gram of tetrahydro cannabinol is 0.00318. Percentage composition of carbon in the compound is:

Answer»

0.6046
0.7085
0.8025
0.5964

Answer :C
10.

Several alkaloids are extracted from the extracts of the plants called marijuana. Marijuana owes its activity to tetrahydro cannabinol, which contains 70% as many as carbon atoms as hydrogen atoms and 15 times as many hydrogen atoms as oxygen atoms. One gram of tetrahydro cannabinol is 0.00318. Molecular formula of the compound is:

Answer»

`C_(21)H_(30)O_(2)`
`C_(21)H_(14)O_(3)`
`C_(21)H_(46)O`
NONE of these.

Answer :A
11.

Several alkaloids are extracted from the extracts of the plants called marijuana. Marijuana owes its activity to tetrahydro cannabinol, which contains 70% as many as carbon atoms as hydrogen atoms and 15 times as many hydrogen atoms as oxygen atoms. One gram of tetrahydro cannabinol is 0.00318. Number of oxygen atoms in 1 mol of the tetrahydro cannabinol is:

Answer»

`2N_(A)`
`N_(A)`
`3N_(A)`
`4N_(A)`

ANSWER :A
12.

Several alkaloids are extracted from the extracts of the plants called marijuana. Marijuana owes its activity to tetrahydro cannabinol, which contains 70% as many as carbon atoms as hydrogen atoms and 15 times as many hydrogen atoms as oxygen atoms. One gram of tetrahydro cannabinol is 0.00318. Molecular mass of the compound is:

Answer»

413amu
314amu
143amu
341amu

Answer :B
13.

Setting of plaster of paris involves:

Answer»

OXIDATION with ATMOSPHERIC oxygen
Combination with atmospheric `CO_2`
Dehydration
Hydration to YIELD ANOTHER hydrate

Answer :D
14.

Set up Nernst equation for the standard dry cell. Using this eqation show that the voltage of a dry cell has to decrease with use.

Answer»

Solution :The voltage of an electrochemical cell is related to the concentration of the solutions involved in the cell. A change in concentration of cell solutions results in a change in voltage that can be quantitatively described by the Nernst equation.
The voltage of an electrochemical cell involves an oxidation reaction and a reduction reaction. In the case of a copper-zinc cell the two half reactions are WRITTEN as REDUCTIONS , and their STANDARD cell potentials are :
`Cu^(2+)(aq) + 2e to Cu(s)` reduction =`+0.34V`
`Zn^(2+)(aq) + 2e to Zn`(s) reduction =-0.76 V
Copper will be reduction species in the copper-zinc cell due it's larger standard reduction potential. The voltage then for a Zn, `Zn^(2+)"||"Cu^(2+) + Zn^(2+)(aq)`
Equation 1 : ex.
`Cu^(2+)(aq) + 2e to Cu(s) E^(@)` reduction =+0.34 V
`Zn(s) + Zn^(2+)(aq) to2e^(-)E^(@)` oxidation =-0.76 V
`E_("cell") =E_("reduction")-E_("oxidation")`
`+1.10 V =+0.34 V-(-0.76 V)`
The voltage fo any electrochemical cell is a function of the molar concentrations of the compounds involved in the cell, as described by the Nernst equation :
`aA(s) + bB^(+) (aq) to cC(s)+ DD^(+)(aq)`
Equation 2:`E_("cell")=E^(@)-(0.06//n)log{D^(+]d)//[B^(+)]^(b)}`
Wquation 3: Using equation 1 as the balanced chemical equation for the `Zn, Zn^(2+)||Cu^(2+) , Cu` cell.
Equation 3 becomes :
`E_("cell") =E^(@)-(*0.03)log{Zn^(2+]//[Cu^(2+)]`
As the `Cu^(2+)` concentration decreases the coltage fo the cell changes . The `Cu^(2+)` concentration can change due to dilution, or due to the formation of a complex ion, `Cu(NH_(3))_(4)^(2+)` , upon addition of `NH_(3)`
15.

Setting of cement is an:

Answer»

EXOTHERMIC process
ENDOTHERMIC process
Neither endothermic nor exothermic
None of these

Answer :A
16.

Set of d-orbitals which is used by central metal during formation of MnO_(4)^(-)?

Answer»

`d_(x^(2)-y^(2)),d_(x^(2)),d_(XY)`
`d_(xy),d_(YZ),d_(XZ)`
`d_(x^(2)-y^(2)),d_(xy),d_(xz)`
`d_(x^(2)-y^(2)),d_(z^(2)),d_(xz)`

Solution :Hybridisation of Mn in `MnO_(4)^(-):d^(3)s`
17.

Set of elements known as chalcogens is :

Answer»

<P> O, S, Se,<BR>CI, Br, I
N, P, S
C, Si, Ge

Answer :A
18.

Set of elements known as chalcogens is : 1) O, S, Se, 2) CI, Br, I 3) N, P, S 4) C, Si, Ge

Answer»

<P> O, S, Se,
CI, Br, I
N, P, S
C, Si, Ge

Answer :A
19.

Set -I (without catalyst) {:(,"Reaction",,"Temperature",,"E(activation)",,"k"),(,ArarrB,,T_(1)K,,Ea_(1),,k_(1)),(,A rarr B,,T_(2)K,,Ea_(2),,k_(2)):} Set-II (with catalyst) (consider +ve catalyst only) {:(,"Reaction",,"Temperature",,"E(activation)",,"k"),(,A rarr B,,T_(1)K,,Ea_(3),,k_(3)),(,A rarr B,,T_(2)K,,Ea_(4),,k_(4)):} For the Set-I:

Answer»

`Ea_(1)gt Ea_(2)if T_(1) gt T_(2)`
`Ea_(1) LT Ea_(2) if T_(1) gt T_(2)`
`Ea_(1) = Ea_(2)`
`Ea_(1)=0.5 Ea_(2)`

ANSWER :C
20.

Set -I (without catalyst) {:(,"Reaction",,"Temperature",,"E(activation)",,"k"),(,ArarrB,,T_(1)K,,Ea_(1),,k_(1)),(,A rarr B,,T_(2)K,,Ea_(2),,k_(2)):} Set-II (with catalyst) (consider +ve catalyst only) {:(,"Reaction",,"Temperature",,"E(activation)",,"k"),(,A rarr B,,T_(1)K,,Ea_(3),,k_(3)),(,A rarr B,,T_(2)K,,Ea_(4),,k_(4)):} Comparing Set-I and II:

Answer»

`k_(4) gt k_(3) and k_(2)k_(1), if T_(2) gt T_(1)` (endothermic)
`k_(4) LT k_(3) and k_(2) gt k_(1) if T_(2) lt T_(1)` (endothermic)
`k_(4) gt k_(3) and k_(2) gt k_(1) if T_(2) lt T_(1)` (EXOTHERMIC)
`k_(4) lt k_(3) and k_(2) lt k_(1) , if T_(2) gt T_(1)` (exothermic)

Answer :A
21.

Set -I (without catalyst) {:(,"Reaction",,"Temperature",,"E(activation)",,"k"),(,ArarrB,,T_(1)K,,Ea_(1),,k_(1)),(,A rarr B,,T_(2)K,,Ea_(2),,k_(2)):} Set-II (with catalyst) (consider +ve catalyst only) {:(,"Reaction",,"Temperature",,"E(activation)",,"k"),(,A rarr B,,T_(1)K,,Ea_(3),,k_(3)),(,A rarr B,,T_(2)K,,Ea_(4),,k_(4)):} For the Set-I:

Answer»

If `T_(1) gt T_(2),k_(1)gt k_(2)` always
If `T_(1) gt T_(2),k_(1) gt k_(2)` (for exothermic reaction)
If `T_(1) gt T_(2), k_(1) LT k_(2)` (for endothermic reaction)
`Ea_(1) NE Ea_(2)`

ANSWER :A
22.

{:("Set-I",, "Set-II") ,((A) "Noradrenaline" ,, (1) "Stable mental process" ),( (B) "Dopamine",, (2) "Regulation of control of movement "),( (C) "Serotonin",, (3) "Mood changes"),((D) "Histamine ",, (4) "Mild aches & pains "),((-),, (5) "Secretion of HCl "):}

Answer»

`{:(A,B,C,D) ,(1,2,3,5):}`
`{:(A,B,C,D) ,(3,2,1,5):}`
`{:(A,B,C,D) ,(3,2,4,5):}`
`{:(A,B,C,D) ,(2,3,4,1):}`

ANSWER :B
23.

Set containing isoelectronic species is/are

Answer»

`C_(2)^(2-), NO^(+), CN^(-), O_(2)^(2+)`
`CO, NO, O_2, CN`
`CO_2, NO_2, O_2, N_2O_5`
`CO,CO_2, NO, NO_2`

SOLUTION :`C_(2)^(2-) (12 + 2) , NO^(+) (7 + 8 - 1), CN^(-) (6 + 7 + 1)` and `O_(2)^(2+) (16 - 2)` are all isoelectronic SPEICES.
24.

{:("SET-1","SET-2"),((1)"Ostwald-walker",(A) "Osmotic pressure"),((2)"Cotrell's method",(B)"Depression of F.P"),((3)"Rast's camphor method","(C) Elevation of B.P"),((4)"Berkeley and Hartley's method",(D)"Lowering of vapour pressure" ):}

Answer»

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

ANSWER :A
25.

{:("List - I","List - II"),("(1) Steriod hormone","(a) Cytokinins"),("(2) None steroid hormone","(b) Estrogens"),("(3) Plant hormone","(c ) Auxins"),("(4) Peptide hormone","(d) Insulin"):}The correct match is

Answer»

`1 - B, 2- b, 3 - C, 4 - d`
`1-b, 2- a, 3 - d, 4-c`
`1 - b, 2- a, 3-c, 4-d`
`1 - c, 2-a, 3- d, 4 - b`

Answer :C
26.

{:("SET-1","SET-2"),((i)"RBC in 0.5% NaCl solution",(A)"Swells"),((ii)"RBC in 1% NaCl solution",(B)"Shrinks"),((iii)"egg (outer shell removed in water)",.),((iv)"egg(outer shell removed in NaCl solution)",.):}

Answer»

i-A
ii-B
iii-A
iv-A

Answer :D
27.

Sesquoxides of alkali metals may be represented by the formula :

Answer»

`M_(2)O_(5)`
`M_(2)O_(3)`
`MO_(2)`
`M_(2)O_(2)`

ANSWER :B
28.

Sesquioxide of lead is :

Answer»

`PBO`
`PbO_2`
`Pb_2O`
`Pb_2O_3`

ANSWER :D
29.

Serpeck's process is used for the extraction of :

Answer»

Mg
Ni
Al
Cu

Answer :C
30.

Serpeck's process is used for bauxite which has following main impurity :

Answer»

`SiO_(2)`
`Fe_(2)O_(3)`
`Fe_(3)O_(4)`
`CAO`

ANSWER :A
31.

Serpeck process may be used to prepare

Answer»

CYANIDE
ISOCYANIDE
ammonia
nitric ACID

ANSWER :3
32.

Sequence of acidic character is

Answer»

`SO_2 GT CO_2 gt CO gt N_2O_5`
`SO_2 gt N_2O_5 gt CO gt CO_2`
`N_2O_5 gt SO_2 gt CO gt CO_2`
`N_2O_5 gt SO_2 gt CO_2 gt CO`

ANSWER :D
33.

Separation of two layers are seen when Lucas reagent is treated with

Answer»

`CH_3OH`
`CH_3Cl`
`(CH_3)_3C-OH`
`(CH_3)_3C-Cl`

ANSWER :C
34.

Separation of organic compounds by column chromatography is due to:

Answer»

SELECTIVE ADSORPTION
Selective absorption
Solubilities
Selective adsorption and selective absorption

Answer :A
35.

Separationof basic radicals is based on (a) ____ and (b) ____.

Answer»


Answer :COMMON ION EFFECT,B. `K_(sp)` velues
36.

Separating of d and l enantiomorphs from a racemic mixture iscalled

Answer»

RESOLUTION
DEHYDRATION
Rotation
Dehydrohalogenation

ANSWER :A
37.

Separating of d and l enantiorphs from a racemic mixture is called

Answer»

Resolution
DEHYDRATION
ROTATION
Dehydrohalogenation

Solution :Recemic mixture can be separated into its two OPTICALLY acitve form by a process called resolution.
38.

Semiconductors are solids with conductivities in the range of _________________.

Answer»

Solution :`10^(-6)" to "10^(4)"OHM"^(-1)m^(-1)`
39.

Semiconductors are purified by ______ method.

Answer»

ZONE refining
Electrolytic refining
Mond's process
Beisemerisation

Answer :A
40.

Semi conservative method of DNA duplication means

Answer»

Newly synthesized DNA is CONSERVED only in ONE CELL cycle
Newly synthesized DNA molecules have one strand from the parant DNA molecule
Replication of DNA results in the formation of only one STRANDED DAUGHTER DNA
Only one strand of DNA molecule from RNA

Answer :B
41.

Semicarbzide is :

Answer»

`NH_2CONH_2`
`NH_2-NH_2`
`NH_2CONHNH_2`
None

Answer :C
42.

Self-ionisation of liquid ammonia occurs as, 2NH_(3) rarr NH_(4)^(+) + NH_(2)^(-), K = 10^(-10). In this solvent, an acid might be

Answer»

`NH_(4)^(+)`
`NH_(3)`
Any species that will form `NH_(4)^(+)`
All of these

Solution :A SUBSTANCE which an DONATE a proton is known as ACID so `NH_(4)^(+)` will be a acid.
43.

Selet correct statement.

Answer»

Saccharine is artificial SWEETENER.
SOAPS are Na or K salt HIGHER carboxylic acid.
Tetercycline is antibiotic.
Cetyltrimethyl AMMONIUM bromide is ANIONIC detergent.

Solution :N//A
44.

Self-condensation of two moles of ethyl acetate in presence of sodium ethoxide yields.

Answer»

ETHYL propionate
ethyl butyrate
acetoacetic ester
methyl acetoacetate

Answer :C
45.

Selectr true statement about this reaction :

Answer»

Degree of unsaturation is 10 for major product
Major product have 3 OXYGEN ATOM PER molecule
There is no S atom in major ORGANIC product
Major organic product is heterocyclic compound

Solution :Product is
46.

Selecting the correct statement if any from the following

Answer»

during zone rofining of silicon the metal is more soluble in the melt than the impurity
metals like Fe, Ag, Zn and HG are extracted/punfied by PYROMETALLURGY
white bauxite ore cannot be concentrated by serpeck's process
dolomite, magnesile and graphite are used as retractory materials fumaces

SOLUTION :In zone refining impurities are more soluble in melt than metal.
Metal like Fe, Zn, Hg are extracted by pyro metallurgy but not .Ag.
White bauxite is PURIFIED by Serpeck.s process .
Doilamite, Magnesite and graphite are used as REFRACTORY materials in furnace.
47.

Selected the correct option :

Answer»


can be differentiated by VICTOR MEYER's test.
can be differentiated by Bielstien test.

GIVE purple COLOUR with `FeCl_(3)`
withNaOI`Me-overset(O)overset(||)(C)-overset(O)overset(||)(C)-H` gives SODIUM oxalate.

Solution :N//A
48.

Selecthe correct statement(s):

Answer»

PHYSICAL chemistry
Physical Chemistry
Cation and ANION are called basic and acidic radical RESPECTIVELY
`[NiCl_4]^(2-)` is alow SPIN complex.

Answer :C
49.

Selected incorrect satment.

Answer»

p can turn blue LITMUS red
p can not give effervescence of `CO_(2)` with` NaHCO_(3)`
It is Dieckmann CONDENSATION
Product is a bicylo compound

SOLUTION :
50.

Selected method used to estimate methoxy group in organic compound :

Answer»

HERZIG Meryer METHOD
KJELDAHL method
Zeisel method
CARIUS method

Solution :N//A