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.

If N_(0) is the initial number of nuclie , number of nuclei remaining undecayed at the end of nth half life is :

Answer»

`2^(-N) N_(0)`
`2^(n) N_(0)`
`n^(-2) N_(0)`
`n^(2) N_(0)`

SOLUTION :`N_(t) = (N_(0))/(2^(n))`
2.

If N_(0) and N are the number of radioactive particles at time t = 0 and t = t, then

Answer»

`LAMDA = (1)/(t) "LOG" (N_(0))/(N)`
`lamda = (2.303)/(t) "log"(N)/(N_(0))`
`lamda = (t)/(2.303)"log"(N_(0))/(N)`
`lamda = (2.303)/(t) "log" (N_(0))/(N)`

Answer :D
3.

If N^(0) is the initial number of nuclei, number of nuclei remaining undecayed at the end of n^(th) half-life is

Answer»

`2^(-N) N^(0)`
`2^(n) N^(0)`
`n^(-2) N^(0)`
`n^(2) N^(0)`

ANSWER :A
4.

If 'n' represents total number to asymmetric carbon atoms in a compound, the possible number of optical isomers of the compound is

Answer»

2n
`N^(2)`
`2^(n)`
`2n+2`

Solution :The POSSIBLE number of isomers of the compound is given by `2^(n)` where ',' represents TOTAL number of asymmetric carbon ATOMS in a compound.
5.

If 'n' is the number of ions given by 1 mole of electrolyte, the degree of dissociation 'alpha' of electrolyte is -

Answer»

`(i-1)/(n+1)`
`(i-1)/(n-1)`
`(n-1)/(i-1)`
`(n+1)/(i-1)`

ANSWER :B
6.

If N and S elements are present in organic compound, then during Lassaigne's test both change into

Answer»

`Na_(2)S " and NaCN`
`NaSCN`
`Na_(2)SO_(3) " and " Na_(2)CO_(3)`
`Na_(2)S" and " NaCNO`

Solution :SEE Lassaigne's test in UNIT 5.5.
7.

If n and l are respectively the principal and azimuthal quantum numbers, then the expression for the calculation of the total no. of electrons in any energy level is

Answer»

`sum_(l=0)^(l=n)2(2l+1)`
`sum_(l=n+1)^(l=n-1)2(2l+1)`
`sum_(l=0)^(l=n+1)2(2l+1)`
`sum_(l=0)^(l=n-1)2(2l+1)`

Solution :No. of sub-shells in `n^("th")` shell = 0 to n- 1. No of ORBITALS in a sub-shell = 2 l + l. No. of electrons in each orbital = 2. Hence no. of `E^(-)` in `n^("th")` shell
`=sum_(l=0)^(l=n-1)2(2l+1)`
8.

If n = 3, l= 0, m = 0, then atomic number is

Answer»

12 or 13
13 or 14
10 or 11
11 or 12

Solution :n = 3, l = 0 means last shell is 3s.
`therefore` ELECTRONIC configuration will be
`1s^(2)2s^(2)2P^(6)3s^(1-2)`
ATOMIC number is 11 or 12.
9.

If M,W and V represents molar mass of solute, mass of solute and volume of solution in litres respectively, which among following equations is true?

Answer»

`pi = (MWR)/(TV)`
`pi = (TMR)(WV)`
`pi = (TMR)/(VM)`
`pi = (TRV)/(WM)`

Solution :ACCORDING to van'f Hoff solution equation
`pi V = n RT` but `n = W/M`
Hence `pi V = W/M RT` or `pi = (WRT)/(MV)`
10.

If m_(p) is the mass of proton. m_(n) that of a neutron, M_(1) that of _(10)Ne^(20) nucleus and M_(2) that of _(20)Ca^(40) nucleus, then which of the following relations is//are not true?

Answer»

`M_(2) gt 2M_(1)`
`M_(2) lt 20 (m_(p)+m_(n))`
`M_(2)=2M_(1)`
`M_(2) lt 2M_(1)`

Solution :More the nucleons, more the more the BINDING energy & more REDUCED is the mas. So, `M_(2) lt 20(m_(p)+m_(n))` (Because some mass got reduced on release of binding energy).
`M_(2) lt 2M_(1)` (Because more mass is reduced for binding more nucleons)
11.

If moving wih equal speed,the longest wavelength of the following matter waves is that for (an)

Answer»

ELECRON
`ALPHA`- kparticle
PROTON
Neutron

Answer :A
12.

If molecules is pyramidal, X stereoisomers are possible for : ""C_(abed) Find the value of X.

Answer»


SOLUTION :x=6
13.

If molecular weight of compound is increased then sensitivity is decreased in which of the following methods

Answer»

Elevation in boiling point
Viscosity
OSMOSIS
DIALYSIS

Solution :According to the dialysis process molecular weight INCREASES but SENSITIVITY decreases.
14.

If molecular interaction betweentwo different liquid molecules are stronger than the molecular interactions between the pure liquid molecules, the mixture is expected to show:

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POSITIVE deviations
NEGATIVE deviations
No Deviations
Positive as WELL as negative deviations

Answer :B
15.

If molecular axis is X then which of the following overlapping will form pi bond p_(z)+p_(z),p_(x)+p_(x),p_(x)+p_(y),p_(s)+p_(z),p_(y)+p_(y)

Answer»
16.

If molarity = molality and density of the solution is 2gm/ml thenW_("solute"):W_("solvent") would be

Answer»

`1:2`
`2:1`
`1:1`
`1:3`

Solution :`:.` Given that
Molarity =molality
`("moles")/(V(sdn)(L))=("moles")/(W("SOLVENT")(kg))`
`IMPLIES W_(("solvent"))` is numerically EQUAL to volume of the solution.
`V_(("so In"))=(W("solution"))/(d("solution"))implies(w("solution"))/(d("solution"))=W` (solvent)
`((W_("solution"))/(W_("solvent"))-d_("solution"))`
As `d=2` gm/ml
`impliesW_("solution")=2W_("solvent")impliesW_("solute")=W_("solvent")`
17.

If molar heat of vaporization is 9698 cals mol^(-1) then entropy of vaporization of water at 100^(@)C will be

Answer»

20.0 CALS `MOL^(-1) k^(-1)`
24.0 cals `mol^(-1) k^(-1)`
26.0 cals `mol^(-1) k^(-1)`
28.0 cals `mol^(-1) k^(-1)`

ANSWER :C
18.

If mole fraction of the solvent in a solution decreases then:

Answer»

VAPOUR PRESSURE of SOLUTION increases
B.pt. decreases
Osmotic pressure increases
All are correct

Answer :C
19.

IF molalityof thedilutesolutionisdoubledthe valueof molaldepressionconstant(K_(f))will be______.

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Dobled
halved
tripled
unchanged.

Solution :VOLUE of `K_(f)` depends only UPON the natur of the SOLVENT. It is independent of the CONCENTRATION of the solution.
20.

If molality of the dilute solution is doubled, the value of the molal depression constant (K_(f)) will be

Answer»

doubled
halved
tripled
unchanged

Solution :MOLAL depression constant `(K_(F))` depends only on the solvent and is INDEPENDENT of the molality of the solution.
21.

If mirror image of the compouns is not superimposable on it, most appropriately it represent

Answer»

PAIR of OPTICAL ACTIVE isomers
pair of structuralisomers
pair of geometricalisomers
pair of keto-enol tautomers

Answer :A
22.

If molality of the dilute solution is doubled, the value of molal depression constant (K_f) will be

Answer»

unchanged
doubled
halved
tripled

Answer :A
23.

If M_("normal") is the normal molecular mass and alpha is the degree of ionsation of K_(3)[Fe(CN)_(6)], then the abnormal molecular mass of the complex in the solution will be

Answer»

`M_("NORMAL")(1+2alpha)^(-1)`
`M_("normal")(1+3alpha)^(-1)`
`M_("normal")(1+alpha)^(-1)`
EQUAL to `M_("normal")`

ANSWER :B
24.

If Mg^(2+) +2e rarrMg(s), E=-2.37 V,Cu^(2+) + 2e rarrCu(s),E=+0.34V then theemf of the cell Mg|Mg^(+2)||Cu^(2+)|Cu is :

Answer»

`2.71V`
`2.30V`
`2.80V`
`1.46V`

ANSWER :A
25.

If methyl group is on axial possition in product ( P-4),then what is the possition of Br-atoms on C_(1) "and"C_(2) respectively?

Answer»

Axial-equatrial
Equatorial-axial
Axial-axial
Equatorial-equatorial

SOLUTION :
26.

If methyl iodide and ethyl iodide are mixed in equal proportions, and the mixture is treated with metallic sodium in presence of dry ether, the number of possible products formed is:

Answer»

2
3
1
4

Answer :B
27.

If methyl bromide and ethyl bromide are mixed in equal proportion and the mixture is treated with sodium, the number of possible organic products is :

Answer»

1
2
3
4

Solution :`{:((i) 2CH_(3)Br+2Na,rarrCH_(3)CH_(3),"(Ethane)"),((II)2CH_(3)CH_(2)Br+2Na,rarrC_(4)H_(10),"(Butane)"),((iii)CH_(3)Br+C_(2)H_(5)Br+2Na,rarrC_(3)H_(8),"(PROPANE)"):}`
28.

If methyl bromide and ethyl bromide are mixed iequal proportion and themixture is treated with Na metal, the number of possible organic product are

Answer»

2
3
4
5

Solution :`3CH_(3)Br+6Na+3C_(2)H_(5)Br overset("dry ether") to C_(2)H_(6)C_(4)H_(10)+C_(3)H_(8)+6NaBr`
29.

If metal ions of groups III are precipitated by NH_(4)Cl and NH_(4)OH without prior oxidation by conc. HNO_(3)....... is not completely precipitated.

Answer»


ANSWER :`FE^(2+)`
30.

If methanol vapour is passed over heated copper at 300^@C, it forms formaldehyde by:

Answer»

Hydrogenation
Dehydrogenation
Dehydration
Oxidation

Answer :B
31.

If mass % of oxygen in monovalent metal carbonate is 48% ,then find the number of atoms of metal present in 5mg of this metal carbonate sample is (N_(A) = 6.0 xx 10_(23) )

Answer»


ANSWER :`6XX10^(19)`
32.

If matel ionsof group II are precipitate by NH_(4)CI and NH_(4)OH without prior oxidationby conconirated HNO_(3) _____is not completelyprecipitate

Answer»


Solution :If METAL IONS of groupIII areprecipitate by `NH_(4)CI` and `Nh_(4)OH`withoutprioroxidationby CONCENTRATED `HNO_(3)` FERRIC ion notcompletelyprecipitate .Nitric oxideoxidisesferrous to ferric ions
33.

If mass of neutron is assumed to half of its original value. Whereas that of proton is assumed to be twice of its original value, then the atomic mass of ._(6)^(14C will be:

Answer»

same
14.28% more
14.28% less
28.56% less

Solution :In the ISOTOPE `._(6)^(14)C`: number of protons=6
Number of neutrons=8
New atomic MASS will be `=2xx6+(1)/(2)xx8=16`
% Increase in mass `=(16-14)/(14)xx100=14.28%`.
34.

If mercury is used as cathode in the electrolysis of aqueous NaCl solution, the ions dischaged at cathode are:

Answer»

`H^+`
`Na^+`
`OH^-`
`CI^-`

ANSWER :B
35.

If mass of KHC_(2)O_(4) (potassium acid oxalate) required to reduce 100 mL of 0.02 M KMnO_(4) in acidic medium is x g and to neutralise 100 mL of 0.05 M Ca(OH)_(2) is y g, then which of the following options may be correct :

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If x is 1 g then y is 2 g
If x is 5.5g then y is 11G
If x is 2g then y is 1g
If x is 11g then y is 5.5g

Answer :A::B
36.

If magnetic moment of [MnBr_(4)]^(2-) is 5.9 BM. Predict the number of electrons?

Answer»

2
3
6
5

Solution :MAGNETIC MOMENT can be calculated by using the relation `sqrt(n(n+2)B.M.)` where n=number of ELECTRONS and B.M. is Bohr MAGNETON.
`therefore 5.9=sqrt(n(n+2))`
`therefore n=5`.
37.

If m represents mass of substance (equivalent weight E) consumed or produced when quantity Q of electricity is passed

Answer»

`m PROP Q`
`m prop 1//Q`
`m prop 1//E`
`m prop Q .E`

SOLUTION :Combining Faradays 1ST and 2nd LAW.
38.

If m is the mass of a substance with equivalent weight E and Q is quantity of electicity passed , then

Answer»

`m PROP (1)/(Q)`
`m prop Q.E`
`m prop (1)/(E)`
`m prop Q`

ANSWER :B
39.

If M is the element of actinoid series, the degree of complex formation decreases in the order

Answer»

`M^(4+) GT M^(3+) gt MO_(2)^(2+) gt MO_(2)^(+)`
`MO_(2)^(+) gt MO_(2)^(2+) gt M^(3+) gt M^(4+)`
`M^(4+) gt MO_(2)^(+) gt M^(3+) gt MO_(2)^(+)`
`MO_(2)^(3+) gt MO_(2)^(+) gt M^(4+) gt M^(3+)`

ANSWER :B
40.

If 'M' is a monovalent metal , the order of ionic character of their halides respectively is

Answer»

`MFgtMClgtMBrgtMI`
`MIgtMBrgtMClgtMF`
`MFgtMClgtMIgtMBr`
`MClgtMFgtMBrgtMI`

ANSWER :A
41.

If l=length, R=resistance and A=cross sectional area, then. . . .

Answer»

`RPROP(l)/(A)`
`Rprop(A)/(l)`
`Rprop(l)/(Al)`
`Rprop lA`

SOLUTION :Resistance of any equal conductor is DIRECTLY PROPORTIONAL to its length and inversely proportional to its cross-sectional area (A).
So, `Rprop(l)/(A)`.
42.

If liquids A and B form an ideal solution,

Answer»

the ENTROPY of mixing is zero
the free energy of mixing is zero
the free energy as WELL as the entropy of mixing are each zero
the ENTHALPY of mixing is zero

Answer :D
43.

If Lamda_(NaOAc)^(@)=91 and Lamda_(HCl)^(@)=496.2" S "cm^(2)mol^(-1) then what is required to find out Lamda_(HOA c)^(@) ?

Answer»

`Lamda_(ClCH_(2)COOH)`
`Lamda_(CH_(3)COOH)`
`lamda_(H+)^(@)`
`Lamda_(NaCl)^(@)`

Solution :`Lamda_(NaAc)^(@)=(Lamda_(NaAc)^(@)+Lamda_(HCl)^(@))-(Lamda_(NaCl)^(@))`
i.e., by KNOWING the value of `^^_(NaCl)^(@)`, we can DETERMINE `^^_(HOA C)^(@)`.
44.

If lamdaClCH_(2)COONa=224Ohm^(-1)cm^(2)gm" "equ^(-1),lamdaNaCl=38.2Ohm^(-1)cm^(2)" gm "equ^(-1) and lamdaHCl=203Ohm^(-1)cm^(2)" gm "equ^(-1), then what is the value of lamdaClCH_(2)COOH ?

Answer»

`288.5Ohm^(-1)cm^(2)" gm "EQU^(-1)`
`289.5Ohm^(-1)cm^(2)" gm "equ^(-1)`
`388.5Ohm^(-1)cm^(2)" gm "equ^(-1)`
`59.5Ohm^(-1)cm^(2)" gm "equ^(-1)`

SOLUTION :`lamdaClCH_(2)COOH`
`=lamdaClCH_(2)COONa+lamdaHCl-lamdaNaCl`
`=224+203-38.2`
`=388.8Ohm^(-1)cm^(2)" gm "equ^(-1)`.
45.

If Lamda^(@)CH_(3)COONa=91" S "cm^(2)mol^(-1)," "Lamda^(@)HCl=462.2" S "cm^(2)mol^(-1), then choose correct option to determine Lamda^(@)CH_(3)COOH.

Answer»

`lamda^(@)CL^(-)`
`Lamda^(@)NaCl`
`Lamda^(@)H^(+)`
`Lamda^(@)ClCH_(2)COOH`

SOLUTION :`Lamda^(@)NaCl`
46.

If lamda=c_(2)[(n^(2))/(n^(2)-Z^(2))] for Balmer series what is the value of C_(2)?

Answer»

`r/(R_(H))`
`2/(R_(H))`
`2R_(H)`
`4R_(H)`

Solution :`LAMDA=C_(2)[(N^(2))/(n^(2)-Z^(2))]implies1/(lamda)=1/(C_(2))[(n^(2)-Z^(2))/(n^(2))]`
`implies1/(lamda)=1/(C_(2))[1-(Z^(2))/(n^(2))]`
`=(Z^(2))/(C_(2))[1/(Z^(2))-1/(n^(2))]`
`:.R_(Y)=(Z^(2))/(C_(2))impliesC_(2)=4/(R_(H))`
Hence a is the correct answer.
47.

If lambda_(0) is the threshold wavelnegth of a metal and lambda is the wavelength of the incident radiation, the maximum velocity of the ejected electron from the metal would be

Answer»

`[ ( 2hc)/( m )((lambda_(0) - LAMBDA)/( lambda_(0) lambda))]^(1//2)`
`[ ( 2hc)/( m )((lambda-lambda_(0) )/( lambdalambda_(0)))]^(1//2`
`[ ( 2hc)/( m ) ( lambda - lambda_(0) ) ]^(1//2)`
`[ ( 2hc)/( m ) ( lambda_(0) - lambda) ]^(1//2)`

ANSWER :A
48.

IfK_(sp)of M(OH)_(3) is 1 xx 10^(-12) then 0.001 M.M^(2+) isprecipitate in a pH gt 9

Answer»


Solution :`M(OH)_(3) `is PRECIPITATE
`[M^(3+)][OH^(Theta)]^(3) gt K_(sp) [OH^(Theta)]^(3)gt 1 XX 10^(-9)`
`[OH^(Theta)] gt 1 xx 10^(-3)`
Maximum `pOH = 3` Minimum `pH = 9`
Hence true.
49.

If lactobacillus bacteria (which converts milk to curd) multiply in milk as per the data given {:("Time(minute)",,0,,15,,45,,75),("No . of bacteria",,100,,200,,800,,3200):} 100g milk to curd requires 4xx10^(9) such bacteria. Atleast how many hours it will require to convert 1kg milk if initially 100 bacteria were mixed in it.

Answer»


ANSWER :`8HRS`
50.

If K_sp of HgSO_4, is 6.4 x 10^-5 M^2, then solubility of the HgSO_4, in water will be

Answer»

`5.4*10^-5 M`
`8*10^-3 M`
`8* 10^-4 M`
`6.4* 10^-3 M`

ANSWER :B