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.

The kinetic energy of an electron is 5 xx 10^(5) eV (electron volts). Calculate the wavelength of the wave associated with the electron. The mass of the electron may be taken as 10^(-30) kg

Answer»


Solution :`1 EV = 1.6 xx 10^(-19)J, KE = (1)/(2) mv^(2) or (5 xx 10^(-5)) (1.6 xx 10^(-19)) = (1)/(2) xx 10^(-30) xx v^(2)`
or `v^(2) = 16 xx 10^(6) or v = 4 xx 10^(3) m s^(-1)`
Hence, `lamda = (h)/(mv) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/(10^(-30) kg xx 4 xx 10^(3) ms^(-1)) = 1.66 xx 10^(-7) m`
Discussion
2.

The kinetic energyof electronis 4.55 xx 10^(-25) Jandmas is 9.1 xx 10^(-31) kg thencalculatevelocity, momentum and wavelengthofelectron.

Answer»

Solution :VELOCITY `=1XX 10^(3)ms^(-1)` MOMENTUM = `9.1 xx 10^(28) kg ms^(-1)lambda = 7.28 xx 10^(7)m`
Discussion
3.

The massof anelectronis 9.1 xx 10^(31) kgif its K.E.Is 3.0 xx 10^(25) J calculateitswavelength .

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SOLUTION :`7.25 XX 10^(7) m`
Discussion
4.

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a_(0) is Bohr radius]

Answer»

`(h^(2))/(4pi^(2) m a_(0)^(2))`
`(h^(2))/(16pi^(2) m a_(0)^(2))`
`(h^(2))/(32pi^(2) m a_(0)^(2))`
`(h^(2))/(64 pi^(2) m a_(0)^(2))`

SOLUTION :By Bohr's postulate, `MV r= (nh)/(2pi)`
`:. v = (nh)/(2pi mr)`
`K.E. = (1)/(2) mv^(2) = (1)/(2) m (n^(2) h^(2))/(4pi^(2) m^(2) r^(2))`..(i)
But `r = (a_(0) n^(2))/(Z)`
for 2ND Bohor orbit of hydrogen, `n = 2, Z = 1`
`:. r = (a_(0) xx 4)/(1) = 4a_(0)`
Substituting in eqn (i)
`K.E. = (1)/(2) (m(2)^(2) h^(2))/(4pi^(2) m^(2) (16a_(0)^(2))) = (h^(2))/(32pi^(2) ma_(0)^(2))`
Discussion
5.

The kinetic energy of a sub-atomic particles is 5.85 xx 10^(-25)J. Calculate the frequency of the particle wave. (Planck's constant, h = 6.626 xx 10^(-34) kg m^(2) s^(-1))

Answer»

SOLUTION :K.E. `= (1)/(2) mv^(2) = 5.85 xx 10^(-25) J`
By de Broglie equation, `lamda = (h)/(mv)`
But `lamda = (u)/(V) :. (u)/(v) = (h)/(mv) or v (mv^(2))/(h) = (2 xx 5.85 xx 10^(-25)J)/(6.626 xx 10^(-34) Js) = 1.77 xx 10^(9) s^(-1)`
Discussion
6.

The Kinetic energy of a gas depends upon a) nature of the gas b) absolute temperature c) molecular weight of the gas d) number of moles of the gas

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B, d 
c, d 
a, d
a, b 

SOLUTION :`KE = 3/2 NRT`.
Discussion
7.

The kinetic energy (KR) of photoelectron emitted on irradiating a metal surface with frequency .vartheta. is related by KE = h vartheta - IE. The plots of K.E vs incident frequency is a straight line which shows

Answer»

SLOPE EQUAL to planck.s constant
intercept on x-axis equal to the product of threshold frequency & planck.s constant
with extrapoloated on y-axis equal to IE
with intercept on x-axis equal to threshold frequency

Answer :A::C::D
Discussion
8.

The kinetic energy of 1 mole of oxygen molecules in cal mol^(-1)at 27°C

Answer»

300
600
900
800

Answer :C
Discussion
9.

The kinetic energy for 14 grams of nitrogen gas at 127^(@)C is nearly (mol. Mass of nitrogen=28 and gas constant=8.31J/mol/K.)

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1.0J
4.15J
2492.2J
3.3J

Answer :C
Discussion
10.

The kind of delocalization involving sigma bond and hybrid orbital's is called

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INDUCTIVE effect
Electrometric effect
Hyper conjugation
Mesomeric effect

Solution :DIRECT. BASED on DEFINITION
Discussion
11.

The kind of delocalization invloving sigma-bond orbitals is called............... .

Answer»


ANSWER :HYPERCONJUGATION
Discussion
12.

TheK_H for the solution of oxygen dissolved in water is4xx 10^(4) atm at a given temperature . If the partial pressure of oxygen in air is 0.4 atm the mole fraction of oxygen in solution is

Answer»

<P>`4.6 xx 10 ^(3)`
`1.6 xx 10 ^(4)`
`1xx 10^(-5)`
`1xx 10 ^(5)`

Solution :`K _(H) = 4 xx 10^(4) atm, (P _(O_(2)))_(air) = 0.4 atm, (x_(O_(2)))_("in solution")=` ?
`(P _(O_(2)))_(air) = K _(H) (x _(O_(2)))_("in solution ")`
`0.4 =4 xx 10 ^(4) (x _(O_(2)))_("in solution ")`
` therefore (x _(O_(2))) _(" in solution ")= (0.4)/(4 xx 10 ^(4)) =1 xx 10 ^(-5)`
Discussion
13.

The kind of delocalization involving sigma bond orbitals is called:

Answer»

INDUCTIVE EFFECT
HYPERCONJUGATION effect
Electromeric effect
Mesomeric effect

Answer :B
Discussion
14.

The K.E of N molecules of O_(2)is x Joules at -123°C. Another sample of O_(2) at 27°C has a KE of 2x Joules. The latter sample contains________N molecules of O_(2).

Answer»

SOLUTION :`(K.E_1)/(K.E_2) = (n_1T_1)/(n_2T_2)` .
Discussion
15.

The Kc for the reaction I_(2 (g)) hArr 2I_((g)) is 4 xx 10^(-3) . If the equilibrium concentration of atomic iodine is 4 xx 10^(-2) M . What is the concentration of molecular iodine ?

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0.8 M
0.3 M
0.4 M
0.2 M

ANSWER :C
Discussion
16.

The K_(a) values of formic acid and acetic acid are respectively 1.77xx10^(-4) and 1.75 xx 10^(-5). The ratio of acid strengthof 0.1 N acids is

Answer»

10
3.178
0.3
0.1

Answer :A::B
Discussion
17.

The K_(c) for given reaction will be A_(2(g))+2B_((g))hArrC_((g))+2D_((g))

Answer»

`K_(C)=([C][D]^(2))/([A_(2)][B]^(2))`
`K_(c)=([C])/([A_(2)][B]^(2))`
`K_(c)=([A_(2)][B]^(2))/([C][D]^(2))`
`K_(c)=([A_(2)][B]^(2))/([C])`

ANSWER :B
Discussion
18.

The K_(a) value of formic acid and acetic acid are respectively, 1.77 xx 10^(-4) " and 1.75 xx 10^(-5). The ratio of the acid strength of 0.1 N acids is

Answer»

10
3.178
0.3
0.1

Solution :`HA hArr H^(+) + A^(-), K_(a) = ([H^(+)][A^(-)])/([HA])`
LET `C MOL L^(-1)` be the conc. and `alpha` its DEGREE of dissociation
`:. K_(a) = (C alpha xx C alpha)/(C (1 - alpha)) = C alpha^(2)` (where `alpha` is very small)
`:. alpha = sqrt((K_(a))/(C))` where C is constant, `[H^(+)] prop alpha`
`:. (alpha_(1))/(alpha_(2)) = sqrt((K_(a_(1)))/(K_(a_(2)))) = sqrt((1.77 xx 10^(-4))/(1.75 xx 10^(-5)))`
`= sqrt(10) = 3.178`
Discussion
19.

The K_a value for HCN is 10^(-9). What is the pH of decimolar HCN solution ?

Answer»

SOLUTION :Dissociation constant, `K_a = 10^(-9)`, CONCENTRATION, `C=0.1 = 1 XX 10^(-3) mol L^(-1)`
`[H^+]= sqrt(K_aC) = sqrt(10^(-9) xx10^(-1) = 10^(-5)) mol L^(-1)`
pHof thesolution`pH=-log 10^(-5) = 5`
Discussion
20.

The K, of an indicator HIn is 9 x 10^(4), The percentage of the basic form of indicator is 10x in a solution of pH = 4. What is x ?

Answer»


SOLUTION :` PH =pK_(In)+LOG"" (In^(-))/( HI_n) , 4 =4- log 9 +log ""(In^(-))/(HI_n)`
` 9= ( In^(-))/( HI_n) RARR % " of " In^(-)=90 %`
Discussion
21.

The K for the reaction A + B hArr C + D is at 25^@C temperature is 2 xx 10^(-23) and At 50^@C temperature 2 xx10^(-12) this reaction will be endothermic or exothermic ?

Answer»

Solution :To INCREASE the temperature K will be increase i.e. FORWARD reaction occurs and increase PRODUCTS. Therefore reaction is ENDOTHERMIC.
Discussion
22.

The K, L and M shells of an atom are full. Its atomic number is ________.

Answer»

18
20
10
12

Answer :A
Discussion
23.

The Joule -Thomson coefficient for a gas is zero at . There are no intermolecular attractive forces in an ideal gas .

Answer»

SOLUTION :EXPLANATION is CORRECT REASON for STATEMENT .
Discussion
24.

The IVA element with highest and lowest first ionisation potential values

Answer»

C,Pb
C,SN
C,SI
Si,Pb

Solution :1st I.E. order is `C gt Si gt GE gt Pb gt Sn`
Discussion
25.

The IV A element with highest and lowest first ioniation potential values

Answer»

C, Pb
C, Sn
C, Si
Si, Pb

Solution :1ST I.E ORDER is `CgtSigtGegtPbgtSn`
Discussion
26.

The IUPAC of CH_(2)=CH-CH_(2)Cl is

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Ally] CHLORIDE
l-Chloro-3-propene
VINYL chloride
3-Chloro-1-propene

SOLUTION :IUPAC Nomenclature of RULES
Discussion
27.

Give IUPAC name of : CH_(3)-CO(CH_(2))_(2)-CH_(3)

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pentan-2-one
pentan -3-one
3-methybutan -2- one
3-methypentanal

Answer :A
Discussion
28.

The IUPAC name of the product formed when ethylene reacts with hypochlorous acid is

Answer»

ETHYLENE chlorohydrin
2-Chloroethanol
1-Chloroethanol
Hydroxyethyl CHLORIDE

ANSWER :B
Discussion
29.

The IUPAC name of the given structure (CH_3)_4C is

Answer»

DIMETHYL PROPANE
TERTIARY BUTANE
NEOPENTANE
Neobutane

Answer :A
Discussion
30.

The IUPAC name of the given compound CH_3-underset(Br)underset(|)(CH)-overset(CHO)overset(|)(CH)-overset(CONH_2)overset(|)(CH)-underset(COCl)underset(|)(CH)-COOH

Answer»

2-BROMO-4-carbamoyl-5-chloroformyl-3 - formyl hexanoic acid
5-Bromo-3-carbamoyl-2-chloroformyl- 4 - formyl hexanoic acid
4-Formyl-2-chloroformyl-3-carbamoyl - 5 -bromo hexanoic acid
2-Chloroformyl-3-carbamoyl-4-Formyl -5 bromo hexanoic acid

ANSWER :B
Discussion
31.

The IUPAC name of the given structure (CH_3)_2CH-CH(CH_3)_2 is

Answer»

2,2-Dimethyl BUTANE
ISOHEXANE
2,3-Dimethyl butane
DI isohexane

Answer :C
Discussion
32.

The IUPAC name of the given compound CH_3-CH=CH-COOC_2H_5 is :

Answer»

ETHYL 1-butenoate
Ethyl-2-butenoate
Ethyl propenoate
PROPENE ethyl methanoate

ANSWER :B
Discussion
33.

The IUPAC name of the given compound

Answer»

2-ethylhex-4-en-2-ol
4-ethylhex-2-en-2-ol
2-ethylhex-2-en-4-ol
4-ethylhex-4-en-2-ol

Solution :`overset(1)(C )H_(3)- underset(underset(OH)(|))overset(2)(C )H- overset(3)(C )H_(2)- underset(underset(CH_(2)CH_(3))(|))overset(4)(C )= overset(5)(CH)-overset(6)(C )H_(3)`
Discussion
34.

The IUPAC name of the following compound is : H_(3)C-overset(CH_(3))overset("|")"CH"-overset(OH)overset("|")"CH"-CH_(2)-COOH

Answer»

4-Methyl-3-hydroxypentanoic acid
2-Methyl-3-hydroxypentan-5-oic acid
3-Hydroxy-4-methylpentanoic acid
4,4-Dimethyl-3-hydroxybutanoic acid

Solution :`H_(3)underset(5)C-overset(CH_(3))overset("|")underset("4")"CH"-overset(OH)overset("|")underset("3")"CH"-underset(2)CH_(2)-underset(1)COOH`
Carboxylic group has HIGH priority than hydroxyl group so NUMBERING starts from carbon of carboxylic group.
3-hydroxy-4-methylpentanoic acid
Discussion
35.

The IUPAC name of the following compound is CH_(3)-underset(CH_(3))underset(|)CH-underset(CH_(2))underset(|)underset(CH_(2))underset(|)CH-CH_(2)-CH_(3)

Answer»

SOLUTION :3-Ethyl 2- METHYLPENTANE
Discussion
36.

The IUPAC name of the following compound is CH_(3)-underset(CH_(3))underset("|")("C")H-CH_(2)-CH=CH_(2)

Answer»

2-Methylpent-4-ene
4-Methylpent-1-ene
HEXENE
3-Methyl pent-1-ene

Solution :According to ALPHABETICAL order ene is COMING first when compared to mythyle hence numbering must be strated from left to RIGHT.
Discussion
37.

The IUPAC name of the following compound is CH_3 - undersetoverset(|)(CH_3)(C )H - CH_2 -CH = CH_2

Answer»

2- Methylpent -4-ene
4-Methylpent -1-ene
Hexene
3-Methyl pent-1-ene

ANSWER :B
Discussion
38.

The IUPAC name of the following compound is

Answer»

5-cyano-3-formylcyclohex-3-en-1-carboxylic ACID
3-cyano-5-formylcyclohex-4-ene-1-carboxylic acid
5-cyano-3-formylcyclohex-3-ene-1-carboxylic acid
5-carboxy-3-formylcyclohex-2-ene-1-carbonitrile

ANSWER :C
Discussion
39.

The IUPAC name of the following compound is {:(CH_(2)-CH--CH_(2)),("|""|""|"),(CN""CN""CN):}

Answer»

1, 2, 3-tricyanopropane
propane-1, 2, 3-trinitrile
3, cyanopentane-1, 5-dinitrile
1, 3, 5-pentanetrinitrile

Solution :3-Cyanopentane-1,5-dinitrile
Discussion
40.

The IUPAC name of the following compound is :

Answer»

3-ethyl-2, 6-dimethyloct-2, 6-dien-1-ol
6-ethyl-3-methyl-6-propyloct-2-en-1-ol
6-ethyl-3, 7-dimethyloct-2, 6-dien-1-ol
3-ehtyl-5, 8-dimethyloct-3, 7-dien-1-ol.

Solution :It is the CORRECT ANSWER.
Discussion
41.

The IUPAC name of the element with atomic number 104 is..........

Answer»

SOLUTION :unnilquadrium.
Discussion
42.

The IUPAC name of the compound X in the following reaction (CH_(3))_(2)C = CHCH_(2) underset(H_(2)O_(2)//OH^(-))overset(B_(2)H_(6))rarrX is

Answer»

2-Methylbutane
2-Methylbutane-1-ol
2,2-Dimethylpropane
3-Methylbutane-2-ol

Solution :`(CH_(3))_(2)C=CHCH_(3)overset(B_(2)H_(6))RARR(CH_(3))_(2)CHCH-CH_(3)")"_(2)BH underset(OH^(-))overset(H_(2)O_(2))rarr underset("3-Methylbutan-2-ol")((CH_(3))_(2)CH-underset(OH)underset(|)(CH)-CH_(3))`
Hydroboration oxidation of ALKENES TAKES place according to anti-Markownikov's rule.
Discussion
43.

The IUPAC name of the compound, underset(OH)underset(|)(C)H_(2)-underset(NH_(2))underset(|)(C)H-COOH is

Answer»

2-amino-3-hydroxypropanoic acid
1-hydroxy-2-aminopropan-3-oic acid
1-amino-2-hydroxypropanoic acid
3-hydroxy-2-aminopropanoic acid.

Answer :A
Discussion
44.

The IUPAC name of the compound shown below is: H-underset(H)underset(|)overset(H)overset(|)C-underset(H)underset(|)overset(H)overset(|)C-underset(H)underset(|)overset(Cl)overset(|)C-Cl

Answer»

SOLUTION :1, 1-dicholoropropane
Discussion
45.

The IUPAC name of the compound shown below is ……

Answer»

2-bromo-6-chlorocyclohex-1-ene
6-bromo-2-chlorocyclohexene
3-bromo-1-chlorocyclohexene
1-bromo-3-chlorocyclohexene

Solution :
Discussion
46.

The IUPAC name of the compound shown below is :

Answer»

2-bromo-6-chlorocyclohex-1-ene
6-bromo-2-chlorocyclohexene
3-bromo-1-chlorocyclohexene
1-bromo-3-chlorocyclohexene

Solution : 3-Bromo-1-chlorocyclohexene
Discussion
47.

The IUPAC name of the compound is {:(HOOC - CH_(2) - CH - CH_(2) - CH_(2) - COOH),(""|),(""COOH):}

Answer»

2-(Carboxymethyl)pentane-1, 5-dioic acid
3-carboxyhexane-1, 6-dioic acid
Butane-1, 2, 4-tricarboxylic acid
4-Carboxyhexane-1, 6-dioic acid

Solution :`{:("COOHCOOH" " " COOH),("1|""2|"""3"4|"),(CH_(2) - CH - CH_(2) - CH_(2)),("Butane-1, 2, 4-tricarboxylic acid"):}` is
Discussion
48.

The IUPAC name of the compound OHC-CH_2-CH_2-underset(CHO)underset(|)(CH)-CH_3 is :

Answer»

2-Formyl pentanal
2-Methyl pentanedial
2,4-Diformyl BUTANE
1,3 - Diformyl butane

ANSWER :B
Discussion
49.

The IUPAC name of the compound is

Answer»

`1,1':3',1''-`Terycyclobutane
`1,3'`Dicyclobutylcyclobutane
`1,1':3',1''-`Tricyclobutane
`1,3-`Biscyclobutycyclobutane

Solution :
`1,1',3',1''-`Tercyclobutane
Discussion
50.

The IUPAC name of the compound having the formula CH-=C-CH=CH_(2) is

Answer»

1-butyn-3-ene
but-1-yne-3-ene
1-butene-3-yne
3-butene-1-yne

Solution :If both the double and triple BONDS are PRESENT the COMPOUND is regarded as derivative of alyne. Further if double and triple bonds are at equidistance from either side, the preference is given to double bond.
Discussion