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.

Due to presence of .......... Portland cement has ash color.

Answer»

FE
MG
CA
NA

ANSWER :A
Discussion
2.

Due to high electronegativity of alkali metals react with elements of high electronegative compound and formed ……….. Bond.

Answer»

METALLIC BOND
COVALENT bond
Coordinate covalent bond
Ionic bond

Answer :D
Discussion
3.

Due to high _______ of H_2O, H_2O has a very strong hydrating tendency.

Answer»


ANSWER :DIELECTRIC
Discussion
4.

Due to high bond energy of carbon , which of the following character does not possess by carbon.

Answer»

C can FORM normal chain.
C can form CYCLIC CHAINS.
Can carry more than 4 coordination number.
Can possess catenation PROPERTIES .

Answer :C
Discussion
5.

Due to Frenkel defect, the density of the crystal __________whereas due to Schottky defect, it..

Answer»


ANSWER :REMAINS UNCHANGED , DECREASES
Discussion
6.

Due to Frankel defect, the density of ionic solids

Answer»

decreases
increases
does not change
changes

Solution :SINCE no ions are MISSING from THECRYSTAL as a whole, there is no effect on density.
Discussion
7.

Due to crystal fractures , the slippage of the planes relative to each other is the easiest way of distorting the crystal . However,even along their most favoured direction distortion of the crystals required a large force it is due to

Answer»

displacement may bring ions of the same charge CLOSE to each other .
displacement may bring ions of the same charge away to each other .
coulombicforces
motion of these PLANES in a direction parallel to these ROWS of ions.

Answer :C
Discussion
8.

Dual nature of matter was proposed by de Broglie in 1923, it was experimentally verified by Davisson and Germer by diffraction experiment. Wave haracter of matter has significance only for microscopic particles. De Broglie wavelength (lambda) can be calculated using the relation, (lambda) = (h)/(m v) where 'm' and 'v' are the mass and velocity of the particle. Which among the following is not used to calculate the de Broglie wavelength ?

Answer»

`lambda = (C)/(v)`
`lambda = (h)/(MV)`
`lambda = (h)/(sqrt(2E m))`
`lambda = (h)/ (sqrt(2Q Vm))`

ANSWER :A
Discussion
9.

Dual nature of matter was proposed by de Broglie in 1923, it was experimentally verified by Davisson and Germer by diffraction experiment. Wave character of amtter has significance only for microscopic partiles. De Broglie wavelength or wavelength of matter wave can be calculated using the following relation: lamda=(h)/(mv) Where, 'm' and 'v' are te mass and velocity of the particle. de Broglie hypothesis suggested that electron waves were being diffracted by the target, much as X-rays are diffracted by planes of atoms in the crystals. Q. An electron microscope is used to probe the atomic arrangement to a resolution of 5Ã…. What should be the electric potential to which the electrons need to be accelerated?

Answer»

2.5 V
6V
2.5 kV
5 kV

Answer :B
Discussion
10.

Dual nature of matter was proposed by de Broglie in 1923, it was experimentally verified by Davisson and Germer by diffraction experiment. Wave character of amtter has significance only for microscopic partiles. De Broglie wavelength or wavelength of matter wave can be calculated using the following relation: lamda=(h)/(mv) Where, 'm' and 'v' are te mass and velocity of the particle. de Broglie hypothesis suggested that electron waves were being diffracted by the target, much as X-rays are diffracted by planes of atoms in the crystals. Q. Wave nature of electrons is shown by:

Answer»

PHOTOELECTRIC effect
compton effect
diffraction experiment
stark effect

Answer :C
Discussion
11.

Dual nature of matter was proposed by de Broglie in 1923, it was experimentally verified by Davisson and Germer by diffraction experiment. Wave character of amtter has significance only for microscopic partiles. De Broglie wavelength or wavelength of matter wave can be calculated using the following relation: lamda=(h)/(mv) Where, 'm' and 'v' are te mass and velocity of the particle. de Broglie hypothesis suggested that electron waves were being diffracted by the target, much as X-rays are diffracted by planes of atoms in the crystals. Q. de Broglie equation is obtained by combination of which of the following theories?

Answer»

PLANCK's QUANTUM THEORY
Einstein's theory of mass-energy equivalence.
Theory of interference
Theory of diffraction.

Answer :A::B
Discussion
12.

Dual character of an electron was explained by

Answer»

BOHR
Heiseberg
de-Broglie
Pauli

Answer :C
Discussion
13.

Dual nature of matter was proposed by de Broglie in 1923, it was experimentally verified by Davisson and Germer by diffraction experiment. Wave character of amtter has significance only for microscopic partiles. De Broglie wavelength or wavelength of matter wave can be calculated using the following relation: lamda=(h)/(mv) Where, 'm' and 'v' are te mass and velocity of the particle. de Broglie hypothesis suggested that electron waves were being diffracted by the target, much as X-rays are diffracted by planes of atoms in the crystals. Q. Planck's constant has same dimension as that of:

Answer»

work
energy
power
angular momentum

Answer :D
Discussion
14.

Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other types of material. If the velocity of the electron in this microscope is 1.6 xx 10^(6) ms^(-1), calculate de Broglie wavelength associated with this electron.

Answer»

SOLUTION :`lamda = (h)/(mv) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/((9.11 xx 10^(-31) kg) (1.6 xx 10^(6) MS^(-1))) m = 4.55 xx 10^(-10) m = 455` pm
Discussion
15.

Dual behavior of matter proposed by de Broglie led to the discovery of elcetron microscope often used for the highly magnified images of biological molecules and other type of material.If the velocity of the electron in this microscopeis 1.6xx10^(6)ms^(-1),calculate de Broglie wavelength associated with this electron.

Answer»

SOLUTION :`gamma=(H)/(mv),gamma=((6.626xx10^(-34)Js))/((9.1xx10^(-31)kg)XX(1.6xx10^(6)MS^(-1)))`
`=0.455xx10^(-34+25)m=0.455nm=455p m.`
Discussion
16.

Dualbehaviour of matterproposedby de Broglieled tothediscovery ofelectronmicroscope oftenusedfor thehighlymagnified imagesof biologicalmoleculas andothertypeof material.If thevelocityof theelectronin thismicrosocopeis 1.6 xx 10^(6) ms^(-1) Calculatede- Brogliewavelengthassociatedwith thiselectron (m_(e) = 9.1 xx 10^(31) kgh= 6.626 xx 10^(34) Js )

Answer»

Solution :Velocity( v) =1.6 `XX 10^(6)ms^(-1)h= 6.626 xx 10^(34)J`
`m= 9.1 xx 10^(31) kg` wavelength
`lambda = (h )/(MV) = (6.626 xx 10^(34)JS)/((9.1 xx 10^(31) kg 1.6 xx 10^(6)nm))`
`=0.4551 xx 10^(9) m`
`=0.455 nm = 455 pm`
Discussion
17.

Dsicuss amphoteric nature of water with an example.

Answer»

Solution :A substance or molecule which acts both as an ACID as WELL as base
`underset("base")(H_(2)O)+HCl to H_(3)O^(+)+Cl^(-)`
`underset("Acid")(H_(2)O)+NH_(3)toNH_(4)^(+)+OH^(-)`
Discussion
18.

Drying agent which react with CO_(2) and removes water vapours from ammonia is

Answer»

`CAO`
`CaCl_(2)`
`CaCO_(3)`
`Ca(NO_(3))_(2)`

SOLUTION :CaO. is called quick LIME which is used as DRYING AGENT to absorb `NH_(3)` gas.
Discussion
19.

Dry ice is effective in seeding clouds because

Answer»

`CO_(2)` and `H_(2)O` have similar crystal structure
It increases water content of the cloud
`CO_(2)` MOLECULES OFFER nucleus for condensation
Upon sublimation, it lowers the temperature of water

Solution :DUE to sublimation
Discussion
20.

Dry ice is....

Answer»

Solid `NH_3`
Solid `SO_2`
Solid `CO_2`
Solid `N_2`

Solution :Solid `CO_2` is CALLED dry ice because it is USED for making ice BATH for organic reaction in laboratory. It is prepared by cooling `CO_2` GAS at high PRESSURE.
Discussion
21.

Dry hydrogen was passed over 1.58 g of red hot copper oxide till all of it completely reduced to 1.26 g of copper (Cu). If in this process 0.36 g of H_2 O is formed, what will be the equivalent weight of Cu and O?

Answer»

Solution :Copper OXIDE `+H_2 to Cu + H_2 O`
`therefore` equivalent of copper oxide `=` equivalent of Cu `=` equivalent of `H_2 O`.
`therefore ("weight of copper oxide")/("eq. weight of copper oxide")=("weight of Cu")/("eq. wt. of Cu")=("weight of" H_2 O)/("eq. wt. of" H_2 O)`
Suppose eq. wt. of `Cu and O` are `X and y` RESPECTIVELY and SINCE,
eq. wt. of H is 1, we have,
equivalent weight of copper oxide `=x+y`
equivalent weight of copper `=x`
equivalent weight of `H_2 O = 1+y ( E_(H_2 O) = E_(H) + E_(O) )`
`therefore (1.58)/( x+y) = (1.26)/( x)= (0.36)/( 1+y)`
`therefore x=31.5`
and `y=8`.
Discussion
22.

Dry air was passed successively through a solution of 5 gm of solute dissolved in 80.0 gm of water and through pure water. The loss in weight of the solution was 2.5 gm and that of the pure solvent was 0.04 gm. What is the molecular weight of the solute?

Answer»

SOLUTION :`p ALPHA 2.5 GM,(p^(@)-p)alpha0.04 gm`
`{:("The relative lowring of"),("vapour pressure"):}}=(p^(@)-p)/(p^(@))`
`=(W_(2))/(M_(2)).(M_(1))/(W_(1))`
`THEREFORE (0.04)/(2.54)=(5xx18)/(M_(2)xx80)(thereforeM_(2)=71.43)`
`M_(2) =" mol.wt of solute "= 71.43`
Discussion
23.

Drinking is injurious of health. But for revenue purpose government has fixed some permissible value for alcohol . The permissible value for the alcohol content in the blood is 1 % mass . On analysis of blood sample of a driver of being drunk over than the pernissible value , it was obtained that 60 gm sample reacted with 30 mL of 8 M K_(2)Cr_(2)O_(7)(Acidic solution) . Reaction : 2K_(2)Cr_(2)O_(7) + 8 H_(2)SO_(4) + C_(2)H_(5)OH rarr 2Cr_(2) (SO_(4))_(3) + 11 H_(2)O + 2K_(2)SO_(4) + 2CO_(2) Assume K_(2)Cr_(2) O_(7) reacts only with the alcohol present in blood . What is the percentage of alcohol in the blood sample?

Answer»

0.092
0.088
0.008
0.072

Answer :a
Discussion
24.

Drinking is injurious of health. But for revenue purpose government has fixed some permissible value for alcohol . The permissible value for the alcohol content in the blood is 1 % mass . On analysis of blood sample of a driver of being drunk over than the pernissible value , it was obtained that 60 gm sample reacted with 30 mL of 8 M K_(2)Cr_(2)O_(7)(Acidic solution) . Reaction : 2K_(2)Cr_(2)O_(7) + 8 H_(2)SO_(4) + C_(2)H_(5)OH rarr 2Cr_(2) (SO_(4))_(3) + 11 H_(2)O + 2K_(2)SO_(4) + 2CO_(2) Assume K_(2)Cr_(2) O_(7) reacts only with the alcohol present in blood . Will the driver be prosecuted for drunken driving?

Answer»

yes
no
MAY or may not
DATE insufficient

Answer :a
Discussion
25.

Driving force of a reaction is the

Answer»

RESULTANT of enthalpy and ENTROPY change
resultant of enthalpy and internal ENERGY change
resultant of entropy and internal energy change
term concerned with change in free energy

Solution :`DELTA G` is the driving FORCE
Discussion
26.

Drew the M.O diagram of oxygen molecule. Calculate its bond order and magnetic character. (ii) Draw and explain the graph obtained by plotting solubility versus temperature for calcium chloride.

Answer»

SOLUTION : (i)
1. Electronic CONFIGURATION of O atom is `1s^2 2s^2 2p^4`
2. Electronic configuration of `O_2` molecule is
`sigma 1s^2 sigma^**1s^2 sigma2s^2 sigma^**2s^2 sigma2p_x^2 pi2p_y^2 pi2p_z^2 pi^**2p_y^1 pi^**2p_x^1`
3. Bond ORDER `=(N_b-N_a)/(2) = (10-6)/(2)=2`
4. Molecular has two unpaired electrons , hence it is paramagnetic
(ii) Even though the dissolution of calcium CHLORIDE is exothermic , the solubility increases moderately with increase in temperature.Here, the entropy FACTOR also plays a significant role in deciding the position of the equilibrium
Discussion
27.

Draw z vs PO graph for an ideal gas, and CO_(2) gas

Answer»

SOLUTION :
Discussion
28.

Drawback of DDT as pesticde is

Answer»

It BECOMES ineffective after some time
It is LESS effective than others
It is not easily/rapid degraded in natire
Its HIGH cost

Solution :DDT a pessticide is not easily and RAPIDLY degraded in nature
Discussion
29.

Draw the two chair conformers of each compounds and indicate which conformer is more stable. (a) cis-1-ethyl-3-methylcyclohexane (b) trans-1-ethyl-2-isopropylcyclohexane (c) trans-1-ethyl-2-methylcyclohexane (d) trans-1-ethyl-3-methylcyclohexane (e) cis-1-ethyl-3-isopropylcyclohexane (f) cis -1-ethyl-4-isopropylcyclohexane

Answer»


ANSWER :STABLE are (a) diequatorial, `(##BSL_CHM_ISO_E01_154_A01##)`
Discussion
30.

Draw the two chair conformers of each compounds and inicate which conformer is more stable . (a) cis-1-ethyl-3-methylcyclohexane (b) trans-1-ethyl-2-isopropylcyclohexane (c)trans-1-ethyl-2-methylcylcohexane (d) trans-1-ethyl-3-methylcyclohexane (e) cis-1-ethyl-3-isopropylcylcohexane (f) cis -1-ethyl-4-isopropylcyclohexane.

Answer»

54
55
56
57

Answer :STABLE are (a) diequatorial, `(##BSL_CHM_ISO_E01_148_A01##)`
Discussion
31.

What is the structure of BeCl_(2) ?

Answer»

SOLUTION :In the VAPOUR STATE, it EXISTS as a chlorobridged DIMER
Discussion
32.

Draw the structures of three isotopes of hydrogen.

Answer»

SOLUTION :
Discussion
33.

Draw the structures of the following compounds. (i) Hex-3-enoic acid, (ii) 2-chloro-2-methylbutan-1-ol (iii) 5, 5-Diethylnonan-3-ol, (iv) Propane-1, 3-diamine (v) 4-Nitropent-1-yne, (vi) But-2-ene-1, 4-dioic acid (vii) 4-Amino-2-ethylpent-2-enal, (viii) Ethyl-3-methoxy-4-nitrobutanoate (ix) 5-(1-Methylethyl)-3-methyloctane, (x) Hex-4-yn-2-one (xi) Hepta-1, 4 -dien-3-ol, (xii) 4-Chloropent-2-ene (xiii) 1-Phenylpropan-2-one, (xiv) 4-Ethyl-2, 6-dimethyldec-4-ene.

Answer»

SOLUTION :
Discussion
34.

Draw the structures of the following compounds. (i) m-Bromochlorobenzene (or 1-Bromo-3-Chlorobenzene) (ii) p-Chlorotoluene (iii) 4-Bromo-o-Xylene (iv) p-Iodo-o-cresol( or 4-Iodo-2-methylphenol) (v) 3-Bromo -p-hydroxybenzoic acid ( or 3-Bromo-4-hydroxybenzoic acid) (vi) 2-Nitro-p-toluidine or 4-amino--3-nitrotoluence or 1-amino-4-methyl -2-nitrobenzene (vii) 2-Chloro -1-metyl-4-nitrobenzene Straregy: First draw the strucuture of the parent compound. Then place the substituents at their positiions.

Answer»

SOLUTION :
Discussion
35.

Draw the structures of the following compounds : (i) Hex-3-en-1-oic acid (ii) 2-Chloro-2-methylbutane-1-ol (iii) 5, 5-Diethylnonan-3-ol (iv) 1-Bromo-3-chlorocyclohex-1-ene(v) 1, 3-Dimethylcyclohex-1-ene

Answer»

Solution :(i) `CH_(3)CH_(2)CH=CHCH_(2)COOH`(II) `{:(""CH_(3)),("|"),(CH_(3)CH_(2)-C-CH_(2)OH),("|"),(""CL):}`
(iii) `{:(""CH_(2)CH_(3)),("|"),(CH_(3)CH_(2)CH_(2)CH_(2)-C-CH_(2)-CH-CH_(2)-CH_(3)),("|""|"),(""CH_(2)CH_(3)""OH):}`
Discussion
36.

Draw the structures of the following compounds : (i) 4-Chloropent-2-ene (ii) Glyoxal (iii) Neohexane (iv) 2,3-Diethylbuta-1,3-diene (v) 4-Phenylbutanal (vi) 4,4-Dimethylpent-2-en-1-ol (vii) Tartaric acid (viii) Isoprene (xi) 4-Hydroxy-2-methylpent-2-enoic acid (x) 2-Ethyl-5-hydroxyhex-3-enal (xi) 5-(1,2-Dimethylpropyl) nonane (xii) 3-Oxopentanal (xiv) 1-Phenypentan-1-one (xv) 2-Methyl-3-ethoxypentane (xvi) 1-Phenylpropan-2-ol (xvii) 1-Phenylpropan-1-one (xviii) N,N-Diethylbutan-1-amine (xix) 3-Oxopentanal (xx) 4-Chloropentan-2-one (xxi) 2-Methylpropan-2-ol (xxii) Hex-1-en-3-ol (xxiii) p-Nitropropiophenone (xxiv) Hexane-1,6-dioic acid (xxv) 2-Aminotoluene (xxvi) Neopentyl alcohol (xxvii) 4,4-Dimenthylpentan-2-ol (xxviii) 1-Bromo-4-chlorobut-2-ene.

Answer»

SOLUTION :
Discussion
37.

Draw the structures of: (i) Bromobenzene (ii) 1,2-dichlorobenzene (iii) 1-chloro-3-methylbenzene

Answer»

SOLUTION :(i)
Discussion
38.

Draw the structures of (i) BeCl_(2) (vapour) (ii) BeCl_(2) (solid).

Answer»

Solution :In the VAPOUR state , it exists as a chlorine bridged dimer .

In the solid state , `BeCl_(2)` has polymeric STRUCTURE with chlorine BRIDGES.
Discussion
39.

Draw the structures of (i) o-xylene (ii) m-xylene (iii) p-xylene

Answer»

SOLUTION :(i)
Discussion
40.

Draw the structures of (i) Benzyl chloride (i) Benzal dichloride (iii) Benzotrichloride

Answer»

SOLUTION :(i)
Discussion
41.

Draw the structures of (i) 3-methylpentanal (ii) 5-hydroxy 2,2-dimethyl heptanoic acid (iii) 2-ethyl-4-propylpentane-dioic acid

Answer»

Solution :(i) `CH_(3)-CH_(2)-underset(CH_(3))underset(|)(CH)-CH_(2)-CHO` (ii) `HOOC-underset(CH_(3))underset(|)OVERSET(CH_(3))overset(|)C-CH_(2)-CH_(2)-underset(underset(CH_(3))underset(|)(CH_(2)))underset(|)(CH)-OH`
(III) `H_(3)C-CH_(2)-overset(COOH)overset(|)(CH)-CH_(2)-overset(COOH)overset(|)(CH)-CH_(2)-CH_(2)-CH_(3)`
Discussion
42.

Draw the structures of: (i) 3-methylhepta 1,3,5-triene, (ii) pent-1-yne, (iii) 2-methylpropan-2-ol

Answer»

Solution :(i) `CH_(3)-CH=CH-CH=UNDERSET(CH_(3))underset(|)C-CH=CH_(2)` (ii) `CH_(3)-CH_(2)-CH_(2)-C-=CH`
(III) `H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OH`
Discussion
43.

Draw the structures of (i) 3-ethyl-5-methylheptane, (ii) 3-ethyl-2-methylhexane (iii) 2,4-dimethylpent-2-ene

Answer»

Solution :(i) `CH_(3)-CH_(2)-UNDERSET(CH_(3))underset(|)(CH)-CH_(2)-underset(C_(2)H_(5))underset(|)(CH)-CH_(2)-CH_(3)` (ii) `CH_(3)-underset(CH_(3))underset(|)(CH)-underset(C_(2)H_(5))underset(|)(CH)-CH_(2)-CH_(3)`
(iii) `CH_(3)-underset(CH_(3))underset(|)C-CH-underset(CH_(3))underset(|)(CH)-CH_(3)`
Discussion
44.

Draw the structures of (i) 1-bromo-2 , 3-dichlorobutane (ii) 2-bromo-3-chloro-2 , 4-dimethyl pentane

Answer»

Solution :(i) 1-bromo-2, 3-dichlorobutane
`underset(Br)underset(|)(C)H_(2) - underset(Cl)underset(|)(C)H- underset(Cl)underset(|)(C)H- CH_(3)`
(ii) 2-bromo-3-chloro- 2 , 4 - dimethyl PENTANE
`CH_(3) - OVERSET(CH_(3))overset(|)underset(Br)underset(|)(C)- underset(Cl) underset(|)(C)H- underset(CH_(3))underset(|)(C)H- CH_(3)`
Discussion
45.

Draw the structures of: (i) 2-cyclobutyl propane (ii) 2-cyclopropyl butane (iii) chlorocyclo but-2-ene

Answer»

SOLUTION :(i)
Discussion
46.

Draw the structures of cis-and trans-isomers of the following compounds. Also write their IUPAC names. (i)CHCl=CHCl (ii)C_2H_5C(CH_3) = C(CH_3)C_2H_5

Answer»

SOLUTION :
Discussion
47.

Draw the structure of Becl_2(vapour)

Answer»

SOLUTION :
Discussion
48.

Alkali metals and alkaline earth metals belong to the s-block of the periodic table. Draw the chain structure of beryllium chloride in solid state.

Answer»

SOLUTION :
Discussion
49.

Draw the structures of BCl_3 . NH_3 and AlCl_3 (dimer).

Answer»

SOLUTION :In `BCl_3` , Boron is attached with chlorine by covalent bond but OCTET of boron is not complete, it requires two electrons or pair of electron to complete it octet.
Such electron deficient molecules have tendency to accept a pair of electrons to achieve stable electronic configuration and THUS, BEHAVE as Lewis acids.
The tendency to behave as Lewis acid decreases with the increase in the size down the group. `BCl_3`EASILY accepts a lone pair of electrons from ammonia to form `BCl_3NH_3` .
`H_3N : + BCl_3 to H_3N to BCl_3`
`AlCl_3`achieves stability by forming a dimer. In trivalent state, most of the compounds being covalent are hydrolyzed in water. For example, the trichlorides on hydrolysis in water form tetrahedral `M(OH)_4^(-)`. Element M is `sp^3` hybridized and has tetrahedral shape.
Discussion
50.

Draw the structures for the following alkenes (i) 6-brommo -2,3 -dimethyl -2-hexene (ii) 5-bromo -4-chloro -1-heptene (iv) 4-methyl -2 pentene

Answer»

SOLUTION :`(i) 6- `Bromo2,3dimethyl2-2 hexane
`CH_(2)underset( CH_(3)) underset(|) (C )= underset(CH_(3))underset(|) (C )- CH_(3)-underset( Br) underset(|) (CH_(2))`
(ii) 5- Bromo-4-chloro-1-heptene
`CH_(2)- CH_(2)- CH_(4)- underset(CH_(3) ) underset(|)(CH )- underset( Br)underset(|) (CH)- CH_(5)- CH_(4)`
(iii) 2.5dimethyl-4-Octaene
`CH_(4)- OVERSET(2) (CH) -overset(2) (CH ) - underset( CH_(2))underset( |) (CH)- CH_(2)`
Discussion