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.

Depending upon concentration , phenol can be used both as an _______as well as a _______

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ANSWER :ANTISEPTIC , DISINFECTANT
2.

Deoxyribonucleic acid (DNA) consists of the following units:

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Peptides
Glucosides
Nucleotides
Deoxyribose

Answer :C
3.

Deoxyadenylic acid, deoxyguanylic acid, deoxythymidylic acid & deoxycytidylic acid are

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Necleotides of DNA
FATTY acids
Necleosides of DNA
Nucleotides of RNA

Answer :A
4.

Deoxyribonucleic acid (DNA) is a polymer of units called:

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Sugars
Ribose
Amino acids
Nucleosides

Answer :D
5.

Dental amalgam is composed of:

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Hg+Ag + Cd+Au+Fe
CU + Sn+Hg+Ag + Zn
Cd + Cu + Ni+Au + Fe
Cu + Sn+Au+Hg + CO

Answer :B
6.

Density order of alpha, beta, gamma forms in Sulphur

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`alpha gt BETA gt gamma`
`alpha LT beta lt gamma`
`alpha gt gamma gt beta`
`S_(gamma)gtS_(alpha)gtS_(beta)`

Answer :A
7.

Density order of alpha, beta , gamma forms in Sulphur

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`ALPHA GT BETA gt gamma`
`alpha lt beta lt gamma`
`alpha gt gamma gt beta `
`S_(gamma) gt S_(alpha) gt S_(beta)`

ANSWER :A
8.

Density of water at room tempearture is 1.0g cm^(-3). How many molecules are there in one drop of water if its volume is 0.1 cm^(3)?

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SOLUTION :Mass of one DROP `"vol".xxd=0.1xx1=0.1g`
No. of moles `=(0.1)/(18)`: No. of MOLECUELS `=6.02xx10^(23)xx(0.1)/(18)`.
9.

Density of solution prepared by dissolving 120 gm urea in 1000 gm water is 1.15 gm/mL then find out molarity of solution is …….

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0.50 M
1.78 M
1.02 M
2.05 M

ANSWER :D
10.

Density of solid is given by

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`d=(zM)/(a^(3)N_(A)`
`d=(a^(3)N_(A))/(zM)`
`d=(ZA^(3))/(MN_(A))`
`d=(zN_(A))/(a^(3)M)`

Answer :A
11.

Density of nitrogen gas prepared from air is slightly greater than that of nitrogen prepared by chemical reaction from a compound of nitrogen due to the presence of

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argon
carbon dioxide
some `N_(3)` molecules analogous to `O_(3)`
greater amount of `N_(2)` molecules derived from N-15 isotope

Solution :Air contains about 1% inert gases, MAINLY Ar (At. Wt=40). The atomic wt. of `N_(2)` is 28.
12.

Density of silver metal is 10.8 g c c^(-1), A 20 xx 10 cm thin iron foil is totally immersed in aqueous silver nitrate. Making it as cathode, electrolysis is done using a current strength of 1930 amp for a period of 100 sec. Calculate the thickness of silver metal electroplated on each face of iron foil.

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Solution :Quantity of ELECTRICITY (Q) = it = `1930 xx 100 = 193000` coul.
Amount of silver depsited using `193000 coul = (193000 xx 108)/(96500) = 216g`
Amount of silvre DEPOSITED on each face of the iron foil = 108 g
Volume of silver electroplated on each face of the iron foil = `w/d = (108)/(10.8) = 10 cc`
The volume is given as, `v = (l + b) xx t, ` if (t) is the thickness and `(l xx b)` is AREA.
thickness of silver electroplated = `("Volume")/("area") = 10/(20 xx 10) = 0.05 cm = 0.5 mm.`
13.

Density of nitrogen gas prepared from air is slightly greater than that of nitrogen prepared by chemical reaction from a compound ofnitrogen because aerial nitrogen contains

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`CO_2`
Argon
Some `N_2` MOLECULES analogous to `O_2`
GREATER amount of `N_2` molecules DERIVED from `N_15` isotope

Answer :B
14.

Density of methane at 250^@ C and 6 atm pressure is ( R = 0.821 atm) :

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2.236 g/L
8 g/L
12 g/L
16 g/L

Answer :A
15.

Density of ionic substance is unchanged in Frenkel defect, but the conductivity of ionic crystalline solid increases. Comment.

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Solution :In Frenkel defect the cations missing at lattice points APPEAR at interstitial SITES. The total number of ions is conserved and hence density is constant.
HOWEVER conductivity increases because of the lattice VACANCY and interstitial OCCUPANCY of the ions in the lattice of the crystal.
16.

Density of ideal gas of molecular mass 32 u at 1.2 atm pressure and 327 °C is

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`0.072 GL^- 1`
`0.78 gL^- 1`
`7.8 gL^- 1`
`7.2 gL^- 1`

ANSWER :4
17.

Density of ether is

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HIGHER than water
equal to water
lower than water
can't be predicted

Answer :C
18.

Density of ammonia is 0.77 g/litre at NTP. Its vapour density is :

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8.65
86.5
4.33
43.3

Answer :A
19.

Density of carbon monoxide is maximum at

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2 atm and 600K
0.5 atm and 273K
6 atm and 1092K
4 amt and 500K

Solution :DENSITY of a GAS, `d = ( p M )/( RT) ` or `d alpha( p )/( T)`
By using this RELATION, density is MAXIMUM at 4 atm and 500K.
20.

Density of a unit cell is given by

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`d=(N_(A))/(ZMA^(3))`
`d=(a^(3)N_(A))/(ZM)`
`d=(a^(3)M)/(ZN_(A))`
`d=(ZM)/(a^(3)N_(A))`

ANSWER :D
21.

Density of a crystal is given by :

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`(a^(3)xxM)/(zxxN_(0))`
`(N_(0)xxM)/(zxxa^(3))`
`(zxxM)/(a^(3)xxN_(0))`
`(a^(3)XXN^(0))/(zxxM)`

ANSWER :C
22.

Density of 2.05 M solution of acetic acid in water is 1.02g//mL. The molality of same solutionis:

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1.14 m
3.28 m
2.28 m
0.44 m

Solution :2.05 M solution of acetic acid means that `(2.05xx60=123 G)` of acid is present in 1000 mL of solutiion
Mass of solution=`Vxxd`
=`(1000mL) XX (1.02 g mol^(-1))`
=1020 g
`"Mass os SOLVENT"= 1020-123=897 g.`
`"Molality (m)"=((2.05 mol))/((0.897 KG))`
`=2.28 mol kg^(-1)=2.28 m`
23.

Density of a 2.05 M solution of acetic acid in water is 1.02 g/ml. The molarity of the solution is :

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3.28 mol `kg^(-1)`
2.28 mol `kg^(-1)`
0.44 mol `kg^(-1)`
2.14 mol `kg^(-1)`

SOLUTION :MASS of 1 L solution of 2.05 M ACETIC ACID
`=1.02xx1000=1020` G
Mass of solute in solution `=2.05xx60=123g`
Mass of solvent in solution
`=1020-123=897` g
Molality `=(2.05xx1000)/(897)=2.28 "mol kg"^(-1)`.
24.

Density of 2M NaCI solution is 1.25 g/cc.The molality of the solution is :

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2.79 molal
0.279 molal
1.279 molal
3,85 molal

Solution :`"Mass of 3 M NaCI solution" = "density" x "volume" =(1.25g/ccxx1000cc)=1250g `
MOLAR mass of NaCI =58.5 G/mol
Mass of 3 moles of NaCI=`(58.5 g mol)XX(3 mol)=1.074 kg`
`"Molality of solution"=("No.of moles of NaCI")/("Mass of solvent in kg")`
`=((3mol))/((1.074 kg))=2.79 mol kg^(-1)=2.79 molal`
25.

Density , malleability and ductility in coinage metals increase in the order :

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CU , AG , AU
Au, Ag, Cu
Ag, Au ,Cu
Ag ,Cu , Au

ANSWER :A
26.

Densities of transition metals are:

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Low
Very low
High
Very high

Answer :D
27.

Denatured spirit is mainly used as a

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good fuel
DRUG
solvent in PREPARING varnishes
material in the PREPARATION of oil

ANSWER :C
28.

Densest transition element is :

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Fe
Sc
Os
Mn

Answer :C
29.

Denatured spirit in mainly used as a

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Goodfuel
Drug
Solvent in PREPARING varnishes
Material in the PREPARATION of oil

Answer :C
30.

Denatured spirit is mainly used as a :

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GOOD fuel
Drug
Solvent in PREPARING varnishes
Material in the PREPARATION of oil

Answer :C
31.

Denatured alcohol is

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ethanol + METHANOL
rectified spirit + methanol + naphthalene
undistilled ethanol
rectified spirit

Solution :DENATURED alcohol or methylated spirit `RIGHTARROW` mixture of `90%` ethanol, `9%` methanol and `1%` other poisonous substances like pyridine and acetone or petroleum oil.
Rectified spirit `rightarrow` 95% alcohol + 5% water.
32.

Denatured alcohol is :

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Recified spirit
Undistilled ETHANOL
Rectified spirit `+10-15%`methanol + NAPHTHA + pyridine
`50%` ethanol `+50%` methanol

Answer :C
33.

Denaturation of proteins leads to loss of its biological activity by

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FORMATION of amino ACIDS
LOSS of PRIMARY structure
loss of both primary and secondary structure
loss of both secondary and TERTIARY structure

Answer :D
34.

Denaturation of protein leads to loss of its biological activity by

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FORMATION of amino acids 
LOSS of primary STRUCTURE 
Loss of both primary & quaternary structures 
Loss of both SECONDARY and tertiary structures 

ANSWER :D
35.

Denaturation of protein

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DISRUPTS the `1 ^(@)` and `2^(@)` structure of proteins
disrupts the 2° and 3° structure of proteins
disrupts 1 °,2°,3° structure of proteins
is REVERSIBLE process

Answer :B
36.

Denaturation of alcohol is the

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mixing of `CuSO_(4)` (a FOUL SMELLING solid) and pyridine (to GIVE the colour) to make the commercial alcohol unfit for drinking
mixing of `CuSO_(4)` (to give the colour) and pyridine (a foul smelling solid) to make the commercial alcohol unfit for drinking
Mixing of `Cu(OAc)_(2)` and AMMONIA to make the commercial alcohol unfit for drinking.
Mixing of `Cu(OAc)_(2)` and pyridineto make the commercial alcohol unfit for drinking.

Solution :The commercial alcohol is made unfit for drinking by mixing in it some copper sulphate (to give it colour) and pyridine (a foul smelling liquid). lt is known as denaturation of alcohol.
37.

Denaturation of ethyl alcohol is made by adding

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methanol only
Phyride only
methanol and pyridine
ZINC SULPHATE

ANSWER :C
38.

Denaturation involves converstion of ...............proteins to ...........proteins.

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SOLUTION :globuular, FIBROUS
39.

Demineraisled H_(2)O is obtained when

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WATER is passed through `R-SO_(3)`H resin
Water is passed through `R-OVERSET(+)NH_(3)Ooverset(Theta)H` resin
Water is passed through both `RSO_(3)H` resin and `R" "overset(o+)NH_(3)OoversetTheta(H)` resin
We can't get demineralised `H_(2)O` artificially by resins

Answer :C
40.

Denaturation does not involve

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BREAKING up of H-bonding in proteins
the loss of BIOLOGICAL action of enzyme
he loss of secondary STRUCTURE.
loss of PRIMARY structure of proteins

Solution :loss of primary structure of proteins
41.

DeltaUis equal to

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ISOCHORIC work
isoboric work
adiabatic work
isothermal work

Answer :C
42.

DeltaU is equal to:

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ISOBARIC work
Adiabatic work
Isothermal work
Isochoric word

Answer :B
43.

DeltaU^(@) for combustion of methane is -X kJ mol^(-1). The value of DeltaH^(@) is

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`=DeltaU^(@)`
`gtDeltaU^(@)`
`LT DeltaU^(@)`
`=0`

Solution :`CH_(4)(g)+2O_(2)(g) to CO_(2)(g)+2H_(2)O(l)`
`Deltan_(g)=(n_(p)-n_(R))=1-3=-2`
`DeltaH^(@)=DeltaU^(@)+Deltan_(g)RT=-X-2RT`
Hence, `DeltaH^(@) lt Delta U^(@)`
44.

DeltaU equal to

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Isobaric WORK
ADIABATIC work
Isothermal work
Isochoric work

Solution :For Adiabatic PROCESS, Q = 0
Now, `DeltaU=Q+Wimplies DeltaU=W`
45.

Delta_tG° for the following reactionI_2(s) + H_2S(g) rarr 2HI(g) + S(s) at 298 K is, given that Delta_IG°HI(g) = 1.8 kJ mol^-1. Delta_IG°H_2S(g) = 33.8 kJ mol^-1.

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30200 KJ
30.2 kJ
-30200J
-302 J

ANSWER :C
46.

Delta_(r)H^(Theta) of Cyclohexene (l) and benzene at 25^(@)C is -156 and + 46kJ mol^(-1), respectively. Delta_("hydrogenation")H^(Theta) of cyclohexene (l) at 25^(@)C is -119 kJ mol^(-1) Resonance energy of benzene is found to be -38x kJ mol^(-1). Find the value of x

Answer»


Solution :`+H_(2(g))rarrDeltaH^(THETA)=-119 kJ mol^(-1)`
This SHOWS that the generation of ONE (C=C) bond in cyclohexene requires 119 kJ `mol^(-1)` of enthalpy. To calculate RE resonance energy.

`DeltaH_(1)^(Theta)=DeltaH_(2)^(Theta)=DeltaH_(3)^(Theta)=119 kJ mol^(-1), DeltaH_(4)^(Theta)=RE`
`DeltaH_(5)^(Theta)=Delta_(f)H^(Theta)"(benzene)"-Delta_(f)H^(Theta)"(Cyclohexene)"`
`=46-(-156)=205kJ mol^(-1)""therefore" From Hess' LAW"`
`DeltaH_(4)^(Theta)=DeltaH_(5)^(Theta)-(DeltaH_(1)^(Theta)+DeltaH_(2)^(Theta)+DeltaH_(3)^(Theta))`
`=205-3xx119=-152 kJ mol^(-1)""therefore-38x=-152`
x=4.
47.

Delta_(r) H of which of the following reactions is zero ?

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`H_(2) (G) + 2 H^(+) (g) + 2E^(-)`
`2 H(g) + AQ to 2 H^(+) aq + 2e^(-)`
`2 H (g) to 2 H^(+) (g) + 2e^(-)`
`H_(2) (g) + aq to 2 H^(+) (aq) + 2e^(-)`

Answer :D
48.

DeltaH_("vap")=30kJ//"mol"andDeltaS_("vap")=75J"mol"^(-1)K^(-1). Find the temperature of the vapour at 1 atm.

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400 K
350 K
298 K
250 K

Solution :`T=(DELTAH)/(DELTAS)`
49.

Deltan, the change in the number of moles for the reactin, C_(12)H_(22)O_(11(s))+12O_(2(g))hArr12CO_(2(g))+11H_(2)O_((1))at 25^(@)C is

Answer»

0
2
4
`-1`

Solution :`C_(12)H_(22)O_(11(s))+12O_(2(g))to12CO_(2(g))+11H_(2)O`
`Deltan=12-12=0`
50.

DeltaH_(f)^(@) " for "H_(2)O_(2) and H_(2)O are -188 kJ/mole and -286 kJ/mole. What will be the enthalpy change of the reaction 2H_(2)O_(2)(l)rarr2H_(2)O(l)+O_(2)(g)

Answer»

`-196` kJ/mole
`146 kJ//"mole"`
`-494 kJ//"mole"`
`-98 kJ//"mole"`

Solution :`H_(2)+O_(2)rarrH_(2)O_(2) ""DeltaH_(f)^(@)=-188 kJ//"mole...(i)"`
`H_(2)+(1)/(2)O_(2)rarrH_(2)O_(2)" " DeltaH_(f)^(@)=-286 kJ//"mole...(II)"`
eq. (i) - eq. (ii) `xx2` gives the required result.