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.

in a period from Li to F, ionisation potential

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INCREASES
DECREASES
Remain same
None of the above

Solution :Increases as the atomic size decreases and hence EFFECTIVE NUCLEAR CHARGE increases.
2.

In a period, elements are arranged strictly in sequence of

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DECREASING CHARGES in the nucleus
increasing charges in the nucleus
constant charges in the nucleus
equal charges in the nucleus

SOLUTION :Increasing charges in the nucleus (i.e, atomic number).
3.

In a particular reduction process, the concentration of a solution that is initially 0.24 M is reduced to 0.12 M in 10 hours and 0.06 M in 20 hours. What is the rate constant of this reaction ?

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Solution :As `t_(1//2)` is independent of INITIAL CONCENTRATION, it is a reaction of 1ST ORDER. `t_(1//2)=10` hours,
`k=0.693//t_(1//2)=0.693//10" hr"=6.93xx10^(-2)hr^(-1)" or "0.693//(10xx3600)s^(-1)=1.9xx10^(-5)s^(-1).`
4.

In a particular experiment , 272 g of phosphorus, P_(4) , reacted withexcess ofoxygen of form P_(4)O in89 . 5 %yield. In the secondstep of thereaction, a 97.8%yield of H_(3) PO_(4) was obtained. Whatmassof H_(3) PO_(4) was obtained ?

Answer»

<P>

Solution :ApplyingPOAC for P atomsfor the followingsteps. In the firststep, for `89.5%` yield,
moles of `P_(4) O_(10)` produced = moles of `P_(4) xx 0.895`
`= (272)/(124) xx 0.895 = 1.9632`
In the SECOND STEP,for `97.8% ` yield,
moles of `H_(3) PO_(4)` produced ` = 4 xx ` moles of `P_(4) O_(10) xx 0.978`
`= 4 xx 1.9632 xx 0.978 = 7.680`
`:.`wt. of `H_(3) PO_(4)` produced = `7 . 680 xx 98 g`
= 752 . 65 g
5.

In a particular case of bacterial growth following second order kinetics, concentraion of bacteria increases to 4 times initial concentration of 1 M in 24 minutes. What will be the generation time of the bacterial growth, if initial concentration is 2 M.

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24 min
16 min
8 min
12 min

Answer :C
6.

In a P -type semiconductor, germanium is doped with

Answer»

p-type semiconductor Ge Ge Ge Ge

No CHANGE in CONDUCTIVITY Ge Ge Ge Ge
It BECOMES SUPERCONDUCTOR Ge Ge Ge Ge

ANSWER :A
7.

In a particular isomer of [Co(NH_(3))_(4)Cl_(2)]^(0), the Cl-Co-Cl angle is 90^(@), the isomer is known as

Answer»

OPTICAL isomer
Cis-isomer
Position isomer
Linkage isomer

Solution :
8.

In a paper electrophoresis amino acids and peptides can be separated by their differential migration in an electric field. To the center of a strip of paper, wet with buffer at pH=6 is applied a mixture of the following three peptides in a single small spot: Gly-Ala, Gly-Asp and Gly-Arg. A positively charged electrode (anode) is attached to the left side of the paper and a negatively charged electrode (cathode) to the rightside. A voltage is applied across the ends of the paper for a time, after which the peptides have separated into three spots. One near the location of the original spot whtch peptide is in each spot? Explain.

Answer»

SOLUTION :Gly-Asp will MOVE towards ANODE while Gly-Arg will move towards calhode and Gly-Ald will remain HEAR SPOT.
9.

In a octahedral geometry the type of hybridisation involved is………….

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`SP^(3)d^(2)`
`d^(2)sp^(3)`
`DSP^(3)`
`a` or `b`

ANSWER :A::B::D
10.

In a nuleophilic substitution reaction : R-Br + Cl^(-) overset(DMF)(to) R-Cl+ Br^(-) ,which one of the followingundergoes complete inversion of configuration ?

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`C_6H_5CH_2Br`
`C_6H_5CHC_6H_5Br`
`C_6H_5CHCH_3Br`
`C_6H_5"CCH"_3 C_6H_5Br`

Solution : Due to secondary alkyl group with Aromatic rings GIVEN compound can UNDERGO nuclephilicsubstitution with complete INVERSION
11.

In a nucleotide the phosphate linkage is generally attached to

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C - 1 of sugar
C - 2 of sugar
C - 5 of sugar
N- of BASE 

Answer :C
12.

In a nucleoside, a base is attached at 1' position of sugar moiety. Nucleotide is formed by linking of phosphoric acid unit to the sugar unit of nucleoside. At which position of sugar unit is the phosphoric acid linked in a nucleoside to give a nucleotide ?

Answer»

Solution :The nucleotide is formed when the PHOSPHORIC ACID is linked at 5. POSITION of sugar moiety in a nucleoside.
13.

In a nuclear reactor, the function of moderator is

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To stop the nuclear reaction
To PRODUCE excess of neutrons
To increase the SPEED of neutrons
To DECREASE the speed to neutrons

Answer :D
14.

In a nitrating mixture (conc. HNO_(3)+ conc. H_(2)SO_(4)) nitric acid acts as

Answer»

an ACID
base
a catalyst
an OXIDISING agent.

Solution :Nitric acid acts as a base as explained below
`H_(2)SO_(4) HARR H^(+) + HSO_(4)^(-)`
15.

In a Ni-Cd battery ( more than one correct )

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All the reactants and products in the overall REACTION are in the solid state.
The VOLTAGE of the cell changes rapidly.
The elctrolyte used is an ALKALI solution.
All of the above are true.

Solution :Refer Section
16.

In a nuclear reactor, chain reaction is controlled by introducing

Answer»

IRON rod
Cadmium rod
Graphite rod
Platinum rod

Solution :Cd and BORON rods are CONTROL rods used in reactors
17.

In a nature decay chain series starts with ""_(90)Th^(232) and finally terminates at ""_(82)Pb^(208). A thorium ore sample was found to contain 8 xx 10^(–5 ) ml of helium at 1 atm & 273 K and 5 xx 10^(–7) gm of Th^(232). Find the age of ore sample assuming that source of He to be only due to decay of Th^(232). Also assume complete retention of helium within the ore. (Half-life of Th^(232) = 1.39 xx 10^(10) Y)

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ANSWER :`t = 4.89 XX 10^(9)` YEARS
18.

In a neutron induced reaction of ""_(93)^(235)U, one of the products is ""_(37)^(95)Rb. In this process anther nucleide and three neutrons are also produced. The other nucleide is :

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`""_(56)^(137)Cs`
`""_(38)^(90)SR`
`""_(54)^(144)XE`
`""_(56)^(140)Ba`

Answer :A
19.

In a multi -electron atom, which of the following orbitals described by three quantum numbers will have the same energy in the absence of magnetic field ? (i) n = 1, I =0, m =0(ii) n = 2, I = 0 , m = 0 (iii) n = 1, I = 1, m =1(iv) n = 3, I = 2, m = 1 (v) n = 3, I=2, m = 0

Answer»

(I) and (II)
(II) and (III)
(III) and (IV)
(IV) and (V)

ANSWER :D
20.

In a multistep reaction, the overall rate of reaction is

Answer»

EQUAL to the rate of slowest step
equal to the rate of FASTEST step
equal to the average rate of VARIOUS steps
equal to the rate of the last step

SOLUTION :Rate determing step is slowest step.
21.

In a multistep reaction, the fastest step is the rate determining step.

Answer»
22.

In a monochlorination experiment, some of the monochloro products obtained have been mentioned (a to j) Find the total number of alkane reactants which can give these products

Answer»


SOLUTION :
23.

In a monoclinic unit cell :

Answer»

a = b = C, `ALPHA = BETA = gamma NE 90^(@)`
a = b = c, `alpha = beta = gamma = 120^(@)`
a = b = c, `alpha = beta = gamma = 90^(@)`
a ne b ne c, `alpha = gamma = 90^(@) beta ne 90^(@) `

Answer :D
24.

In a mono - keto compound , generally which from of tautomeric structure is more stable than other ?

Answer»

KETO form is more STABLE
Enol form is more stable
Equally stable
STABILITY cannot be predicted

ANSWER :A
25.

In a mixture of weak acid and its salt, the ratio of concentration of acid to salt is increased ten-fold. The pH of the solution

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Decreases by one
Increases by one-tenth
Increases by one
Increases ten-fold

SOLUTION :It is buffer solution so,
`pH = pK_(a) + log.("Salt")/("ACID")`
`pH = pK_(a) - log.("Salt")/("Acid")`(Given `(["Acid"])/(["Salt"]) = 10 :1`)
`pH = pK_(a) - log 10`
`pH = pK_(a) = - 1`.
26.

In a mixture, two enantiomers are found to be present in 85% and 15% respectively. The enatiomeric excess (e,e) is

Answer»

0.85
0.15
0.7
0.6

Solution :Since the TWO enantiomers are present in 85% and 15% RESPECTIVELY, therefore, 15% out of 85% will form a racemic mixture with the other ISOMER. Therefore, enantiomeric excess `(e,e)=85-15=70%`
27.

In a mixture of sample of H-atoms and He ions, electrons in all the H-atoms and He^+ ions are present in n = 4^"th" state. Then, find maximum number of different spectral lines obtained when all the electrons make transition from n=4 upto ground state

Answer»


ANSWER :11
28.

In a mixture of PbS, ZnS and FeS_2 each component is separated from other by using the reagents in the following sequence in froth floatation process.

Answer»

Potassium ethyl xanthate, KCN. 
Potassium ethyl xanthate, KCN, NAOH, copper sulphate, acid. 
`KCN, CuSO_4` acid. 
Potassium ethyl xanthate, copper sulphate, NaOH, KCN, acid. 

Solution :Potassium ethyl xanthate is collector. KCN and NaOH depresses the floatation property of `FeS_2` and ZnS. Thus only PbS forms the froth. `CuSO_4` is then added to ACTIVATE the FROTHING character of ZnS and only ZnS forms the froth. The REMAINING slurry is acidified and Fes, floats along with the froth.
29.

In a mixture of closely related liquids (such as benzene and toluence), the Roult's law states that theratio P_(A) is proportional to the mole fraction of A in the liquid, P_(A)x_(A). Mixtures that obey the law throughout thecomposition ranges from pure A to pure B are called ideal solutions . In ideal solution thesolute also obeys Roult's law . Then the total pressure is given by P_("total")=P_(A)^(0)x_(A)+P_(B)^(0)x_(B)Benzeneand toluene froms nearly an ideal solution in which the mole fraction of the benzene is 0.4 . Ifv.p. of pure benzene is 40 mmHg and toluene is 30 mmHg at a given temperature , then what is the fraction of toluene in the vapour phase?

Answer»

0.6
0.47
0.53
0.74

Answer :C
30.

In a mixture of iso - octane and n-heptane, the percentage of n-heptane is 10. The octane number of the fuel is:

Answer»

110
90
10
zero

Answer :B
31.

In a mixture of closely related liquids (such as benzene and toluence), the Roult's law states that theratio P_(A) is proportional to the mole fraction of A in the liquid, P_(A)x_(A). Mixtures that obey the law throughout thecomposition ranges from pure A to pure B are called ideal solutions . In ideal solution thesolute also obeys Roult's law . Then the total pressure is given by P_("total")=P_(A)^(0)x_(A)+P_(B)^(0)x_(B) For the solution which obeys Roult's law , which of the following is incorrect ?

Answer»

<P>`DeltaV_("mix")=0`
`DeltaH_("mix")=0`
`DeltaS_("mix")=0`
`P_("total")=P_(A)^(0)x_(A)+P_(B)^(0)x_(B)`

ANSWER :C
32.

In a mixture of closely related liquids (such as benzene and toluence), the Roult's law states that theratio P_(A) is proportional to the mole fraction of A in the liquid, P_(A)x_(A). Mixtures that obey the law throughout thecomposition ranges from pure A to pure B are called ideal solutions . In ideal solution thesolute also obeys Roult's law . Then the total pressure is given by P_("total")=P_(A)^(0)x_(A)+P_(B)^(0)x_(B) Each of the followingsolution obeys Raoult's law except

Answer»

n-hexane +n-heptane
Methanol+Ethanol
carbon tetrachloride+carbon disulphide
chloroform+Acetone

Answer :D
33.

In a mixture of CH_3COOH and CH_3COONa the ratio of salt to acid concentration is increased by ten folds Thr ph of the solution will increase by :

Answer»

ZERO
1
2
3

Answer :B
34.

In a mixture of A and B, components show -ve deviations when

Answer»

`Delta V_(MIX) gt 0, Delta S_(mix) gt 0`
A-B interctions are weaker than A-A and B-B interactions
`Delta V_(mix) = 0,DeltaS_(mix) gt 0`
A-B interactions are stronger than A-A and B-B interactions

Solution :A-B interactions are stronger than A-A and B-B interactions
35.

In a mixture of 1 gm H_(2) and 8 gm O_(2), the mole fraction of hydrogen is

Answer»

`0.667`
`0.5`
`0.33`
None of these

Solution :MOLE FRACTION `=(N)/(n+N)=((W)/(m))/((w)/(m)+(W)/(M))=((1)/(2))/((1)/(2)+(8)/(32))=0.667`.
36.

In a mixture having nitrite and nitrate, nitrite can be destroyed by heating with

Answer»

`Na_(2)CO_(3)`
Urea
Oxalic acid
NaCl

Solution :`2NaNO_(2)+H_(2)SO_(4) to Na_(2)SO_(4)+2HNO_(2)`
`2HNO_(2) +underset("Urea")(NH_(2)CO NH_(2)) to 3H_(2)O +CO_(2) UARR+2N_(2)uarr`
37.

In a mixture having nitrite and nitrate , nitrate , nitrite can be destroyed by heating with H_2 SO_4 and

Answer»

`Na_2 CO_3`
UREA
oxalic acid
NaCl

Solution :`2 NaNO_2 + H_2 SO_4 to Na_2 SO_4 + 2 HNO_2NH_(2) UNDERSET("urea")(CON) H_(2) + 2 HNO_(2) overset(Delta)(to) 3 H_(2) O + CO_(2) UARR +2 N_(2) uarr`
38.

In a mixing of acetic acid and sodium acetate the ratio of concentration of the salts to the acid is increased ten times. Then the pH of the solution

Answer»

INCREASE by ONE
Decreases by one
Decrease ten fold
Increase ten fold

Solution :If concentration of acid is increases ten times in a BUFFER then pH of the solution is increase by one.
39.

In a metallurgical process ,an acid flux is used for removing…..

Answer»

Slag
Basic flux
Acidic gangue
Basic gangue

Answer :D
40.

In a metal oxide, there is 20% oxygen by weight. Its equivalent weight is

Answer»

40
64
72
32

Solution :`O_(2)%=20%`
METAL%=80%=`(80)/(20)xx8=32`G of metal.
41.

In a metal oxide, oxide ions are arranged in hep array and the metal ion occupies octa­hedral holes. If one third of octahedral voids and all tetrahedral voids are empty the metal oxide formula is (M- metal)

Answer»

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

Answer :B
42.

In a metal M having bcc arrangement edge length of the unit cell is 400 pm . The atomic radius of the metal is

Answer»

`173` PM
`100` pm
`141` pm
`200` pm

Answer :A
43.

In a measurement of surface tension by the falling-drop method, 5 drops of a liquid of density 0.797 g/mL weighed 0.220 g. Calculate the surface tension of the liquid.

Answer»

Solution :Mass of the average drop = `0.22/5 = 0.044 g`
VOLUME of the drop `=("mass")/("DENSITY")`
`=(0.044)/(0.797) = 0.0552` mL
Assuming the drop to be spherical in SHAPE
Volume of the drop `=4/3 pir^(3) = 0.0552`
`r = root(3)((3 xx 0.0052)/(4 xx 3.14)) = 0.236 cm`
`lambda = (mg)/(2pir) = (0.044 xx 981)/(2 xx 3.14 xx 0.236)`
`=29.12 "dyne" cm^(-1)`
`=0.02912 N m^(-1)`
44.

In a lead strage battery, Pb(anode) and PbO_(2) (cathode) are used. Concentrated H_(2)SO_(4) is used as electrolyte. The battery holds 3.5 llitre acid with it. In the discharge process, the density of acid fell from 1.294 to 1.139 g/mL. The sulphuric acid of dencity 1.294 g mL^(-1) is 39% by mass and that of density 1.139 g/mL is 20% by mass. Number of ampere-hours for which the battery must been used is :

Answer»

26504 amp-hrs
2650.4 amp-hrs
265.04 amp-hrs
26.504 amp-hrs

Answer :C
45.

In a lead strage battery, Pb(anode) and PbO_(2) (cathode) are used. Concentrated H_(2)SO_(4) is used as electrolyte. The battery holds 3.5 llitre acid with it. In the discharge process, the density of acid fell from 1.294 to 1.139 g/mL. The sulphuric acid of dencity 1.294 g mL^(-1) is 39% by mass and that of density 1.139 g/mL is 20% by mass. Normalities of sulphuric acid before and after discharge are :

Answer»

`5.15, 2.32`
`2.32, 5.15`
`5.15,5.15`
`2.32,2.32`

ANSWER :A
46.

In a lime kiln, to get higher yield of CO_(2), the measure that can be taken is

Answer»

to remove CaO
to ADD more `CaCO_(3)`
to maintain HIGH temperature
to PUMP out `CO_(2)`

Solution :To remove `CaO`.
47.

In a kinetic study of the reduction of nitric oxide with hydrogen,the initial pressure of 340 mm, an equimolar mixture of gases was reduced to half the value in 102 seconds. In another experiment the initial pressure of 288 mm, under the same conditions was reduced to half the value in 140 sec. Calculate the order of the reaction.

Answer»


ANSWER :3
48.

In a lead storage battery

Answer»

`PbO_(2)` is reduced to `PbSO_(4)` at the cathode.
Pb is oxidised to `PbSO_4` at the anode.
Both ELECTRODES are IMMERSED in the same AQUEOUS SOLUTION of `H_2 SO_4`.
All the above are true.

Answer :D
49.

In a Lassaigne's test for sulphur in the organic compound with sodium nitroprusside solution the violet colour formed is due to :

Answer»

`Na_(4)[Fe(CN)_(5)NOS]`
`Na_(3)[Fe(CN)_(5)S]`
`Na_(2)[Fe(CN)_(5)NOS]`
`Na_(3)[Fe(CN)_(6)]`

Answer :C
50.

In a II order reaction when the concentration of both the reactants are equal. The reaction is completed 20% in 500 sec. How long (in seconds) it would take for the reactions to go to 60% completion?

Answer»


ANSWER :3000