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.

Rain damages the monuments like Taj Mahal in Agra when industries are present nearby Why ?

Answer»

SOLUTION :Industries produce a lot of OXIDES of nitrogen and sulphur which through reactions produce compounds which dissolve in RAIN water to form `H_2SO_4` and `HNO_3` . The rain THUS becomes acid rain . The marble `(CaCO_3)` of the monuments is attacked bythese ACIDS.
Discussion
2.

Rahul Dravid wants to wear 6.023xx10^(21) Ag atms in the form of a ring. His Silver Gold Copper alloy ring consists of 20% of Silver. The mass of the is 0.9 x. What is x?

Answer»


Solution :`6.023xx10^(23)` atoms `Ag-108g`
`6.023xx10^(21)` atoms `Ag-108g`
`20gAg-100g Ag-Au` ALLOY
`1.08g Ag = (1.08xx100)/(20)=5.4g`
`0.9x = 5.4 implies x=6`
Discussion
3.

Radius of the earth is 6.40 xx 10^6 m. Find the diameter of the earth.

Answer»


ANSWER :`1.28 XX 10^(7)`m
Discussion
4.

Radius ratio of an ionic compound is 0.93. The structure of the above ionic compound is of

Answer»

NaCltype
CsCl type
ZnS type
None of these

Solution :RADIUS tatio range 0.732 - 0.99 signifies coordination NUMBER 8 .
In css, thecoordination number RATIO 8 : 8
Discussion
5.

Radius of 3^(rd) orbit of Li^(2+) ion is x cm then which is not the de Broglie wavelength of electrons in the first orbit is (in cm)

Answer»

`6pix `
`(2PI x)/(9)`
`(2pi x )/(3)`
`(8pi x)/(3)`

Solution :`r prop (n^2)/(Z) , (r_1)/(r_3) = (n_1^2)/(n_2^2) = 1/(3^2) = 1/9 , r_1 = x/9 [therefore r_3 = x]`
`therefore 2 pi r = n LAMBDA`
`therefore 2 pi xx x/9 = 1 xx lambda = (2pi x)/(9)`
Discussion
6.

Radius of Cs^(+) ion is less than that of A) Fluoride ion B) Chloride ion C) Bromide ion

Answer»

A is CORRECT
A and B are correct
B and C are correct
All are correct

ANSWER :D
Discussion
7.

Radius of 3rd Bohr orbit is

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`6.529Å`
`2.116Å`
`4.761Å`
`8.464Å`

ANSWER :C
Discussion
8.

Radium disintegrates at an average rate of 2.24 xx 10^13 alpha-particles per minute. Each a-particle takes up two electrons from the air and becomes a neutral helium atom. After 420 days, helium gas collected was 0.5 rnL, measured at 27°C and 750 mmHg. Calculate the Avogadro constant.

Answer»


ANSWER :`6.7 XX 10^(23)`
Discussion
9.

Radioactive (overset(82)(Br)-overset(82)(Br)) adds to 1-bromocyclohexene. The product is

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is 1,1,2-tribromocyclohexane
has RADIOACTIVE at vicinal positions
has radioactive BROMINE TRANS to each other
has radioactive bromine cis to each other.

Answer :A::B::C
Discussion
10.

Radioactive elements emit alpha, beta and gamma-rays and are characterised by their half-lives. The radioactive isotope of hydrogen is

Answer»

Protium
Deuterium
TRITIUM
Hydronium

Solution :NUCLEIDES with n/p neutron-proton) RATIO `GT` 1.5 are usuallyradioactive. For EXAMPLE, tritium (n=2, p=1)
Discussion
11.

Radioactive isotope of hydrogen is

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uranium
deuterium
TRITIUM
None of these

Solution :Tritium `(""_(1)^(3)H)` is the RADIOACTIVE isotope of hydrogen. It has `n/p` ratio = 2.0. So, it is unstable.
Discussion
12.

Radioactive elements emit alpha, beta and gamma rays and are characterised by their half-lives. The radioactive isotope of hydrogen is

Answer»

<P>protium
deuterium
tritium
hydronium

SOLUTION :NUCLIDES with n/p (neutron-proton) RATIO gt 1.5 are usually radioactive. For, tritium (n = 2, p =1), it is 2.
Discussion
13.

Radioactive elements emit alpha, beta" and "gamma rays and are characterised by their half - lives. The radioactive isotope of hydrogen is

Answer»

protium
deuterium
tritium
hydronium

Solution :`""_(1)^(3)T RARR ""_(-1)^(0)beta+""_(2)^(3)He`
Discussion
14.

Radioactive elements emit alpha, beta and gamma are characterised by their half-lives. The radioactive isotope of hydrogen is

Answer»

Protium
deuterium
tritium
hydronium

Solution :The radioactive isotope of hydrogen is tritium. Because in tritium (n = 3, p = 1), so, n/p (neutron/proton) RATION is 3 and this ratio `GT`1.5 are USUALLY radioactive.
Discussion
15.

Radioactive alkali metal is ………. And its outer electronic configuration is …………. .

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SOLUTION :FRANCIUM , `7S^(1)`
Discussion
16.

Radio (overset(82)(Br)-overset(82)(Br)) adds to 1-bromocyclohexene. The product is

Answer»

is 1,1,2-tribromocyclohexane
has radio active BROMINE at VICINAL position
has radioactive bromine at vicinal position
has radioactive bromine trans to each other

Answer :A::B::C
Discussion
17.

What is the correct order of radii of Fe^(2+), Fe^(3+) and neutral Fe atom. Why?

Answer»


Answer :neutral ATOM is LARGER in SIZE than cation due to increased nuclear ATTRACTION. Di positively charged ION is still smaller due lesser number of electrons
Discussion
18.

Radial wave function for an electron in hydrogen atom is Psi = (1)/(16 sqrtpi) ((1)/(a_(0))^(3//2)) [(x -1) (x^(2) - 8x + 12)] e^(-x//2) where x = 2r//a_(0), a_(0) = radius of first Bohr orbit. Calculate the minimum and maximum positions of radial nodes in terms of a_(0)

Answer»

Solution :`Psi = (1)/(16 sqrtpi) ((1)/(a_(0))^(3//2)) [(x -1) (x^(2) -8x - 12)] e^(-x//2)`
At radial node, `Psi = 0`. Hence, `(x -1) = 0 or (x^(2) -8x - 12) = 0`
If `x -1 = 0`, then `x = 1`, i.e., `(2r)/(a_(0)) = 1 or r = (a_(0))/(2)`
If `x^(2) -8x + 12 = 0`, i.e.,` (x -6) (x -2) = 0`, then `(x -6) = 0 or (x -2) = 0`
If `(x -2) = 0`, then `x = 2 or (2r)/(a_(0)) = 2 or r = a_(0)`
If `(x - 6) = 0`, then x = 6 or `(2r)/(a_(0)) = 6 or r = 3 a_(0)`
Thus, nodes exist at `(a_(0))/(2), a_(0) and 3a_(0)`. Minimum is at `(a_(0))/(2)` and maximum at `3a_(0)`
Discussion
19.

Radial probability distribution curve is shown for s-orbital. The curve is:

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1s
2s
3s
4s

Answer :A
Discussion
20.

Radial part of the wave function depends on quantum numbers

Answer»

n and s
1 and m
1 and s
n and 1

Answer :C
Discussion
21.

Racemnic mixture is formed by mixing two

Answer»

Isomeric compounds
Chiral compounds
Meso compounds
Optical ISOMERS

Solution :An equimolar mixture of two i.e., DEXTRO and laevorotatory optical isomers is termed as RACEMIC mixture of d&L form or `(+-)` mixture.
Discussion
22.

R, u, T can be

Answer»

HCOONa,NaOH,HCHO
`HCOOCu,CuCl_(2),HCOOEt`
`CH_(3)COONA,CuCl_(2),HCHO`
None

Answer :A
Discussion
23.

R-OH+HX rarr RX+H_(2)O In the above reaction, the reactivity of alcohols is

Answer»

Tertiary `gt` secondary `gt` primary
Tertiary `lt` secondary `lt` primary
Teritary `gt` primary `gt` secondary
Secondary `gt` primary `gt` tertiary

Solution :The presence of three electron releasing GROUPS at `ALPHA`-carbon REPEL the BOND pair of `C-OH` bond and FACILITATE its replacement.
Discussion
24.

(R) is:

Answer»




ANSWER :C
Discussion
25.

R' in RMGX is

Answer»

`C_(3)H_(7)`
`CH_(3)`
`C_(2)H_(5)`
`C_(4)H_(9)`

ANSWER :C
Discussion
26.

R is :

Answer»

But-1-ene
2-methylbut-1-ene
2-muthylbut-2-ene
2-methyl propene

Solution :
Discussion
27.

R - CH_(2) - CH_(2)OH can be converted into RCH_(2)CH_(2)COOH. The correct sequence of reactants is

Answer»

`PBr_(3), KCN, H^(+)`
`PBr_(3), KCN, H^(+)`
`KCN, H^(+)`
`HCN, PBr_(3), H^(+)`

SOLUTION :`RCH_(2) CH_(2)OH OVERSET(PBr_(3))rarr RCH_(2)CH_(2)Br overset(KCN)rarr RCH_(2)CH_(2) CN overset(H^(+))rarr RCH_(2)CH_(2)COOH`.
Discussion
28.

R-CH_(2)-CH_(2)-OH can be converted into RCH_(2)CH_(2)COOH by the following sequence of steps

Answer»

`PBr_(3), KCN, H_(3)O^(+)`
`PBr_(3), KCN, H_(2)//PT`
`KCN, H_(3)O^(+)`
`HCN, PBr_(3), H_(3)O^(+)`

SOLUTION :The CONVERSION of `-CH_(2)OH` group to `-CH_(2)COOH` is done as under
Discussion
29.

R-CH = CH_2 overset(NOCl)to X, hence X is

Answer»

`R-UNDERSET(NO)underset(|)CH-CL`
`R-underset(NO)underset(|)CH-CH_2-Cl`
`R-CHO - CH_2Cl`
`R-underset(Cl)underset(|)CH-CH_2NO`

SOLUTION :`R - CH = CH_2 + NO - Cl to R - underset(Cl)underset(|)CH - CH_2NO`
Discussion
30.

R-C-=N underset(-40^(@)C)overset(DIBAL+H,H_(@)O)rarrProduct The product formed is :

Answer»

`R-CO-NH_(2)`
`R-CH_(2)-NH_(2)`
`R-CHO`
`R-CH_(2)-NO_(2)`

ANSWER :C
Discussion
31.

R-alanine is represented by

Answer»




SOLUTION :
Discussion
32.

Quicklime is _____________

Answer»

`CaCO_3`
CAO
`Ca(OH)_2`
`CaSiO_3`

ANSWER :B
Discussion
33.

Quicklime and slaked lime are the cheapest and the most widely used bases for neutralising unwanted acids. Lime is used to neutralise acidic soils. A important application of quicklime is in air pollution control for the removal of SO_(2) in electric power plants. Slaked lime is used in the manufacture of other alkalis and bleaching powder, in sugar refining, in tanning hides and in water softening The drying agent which absorbs CO_(2) and reacts violently with water is

Answer»

`CaCO_(3)`
`Ca(HCO_(3))_(2)`
`Ca(OH)_(2)`
`CaCI_(2)`

Solution :Temporary hardness is due to the PRESENCE of BICARBONATES of Ca and Mg. By using slaked lime `Ca(OH)_(2)` it was REMOVED
`Ca(OH)_(2) + Ca(HCO_(3))_(2) rarr 2CaCO_(3) darr + 2H_(2)O`
Discussion
34.

Quicklime is represented by the formula

Answer»

`Ca(OH)_(2)`
`CAO`
`CaCO_(3)`
`CaHCO_(3)`

Answer :B
Discussion
35.

Quick lime reacts with water to form

Answer»

`CaCO_(3)`
`Ca(OH)_(2)`
`CaH_(2)`
All

Solution :`CaO + H_(2)) RARR Ca(OH )_(2)`
Discussion
36.

Quicklimeisa/ an_______.

Answer»

Basicoxide
acidicoxide
amphoteric oxide
neutral oxide

Answer :A
Discussion
37.

Quicklime is ______

Answer»

`Ca (OH)_(2)`
`CaCO_(3)`
`CAO`
`CaSO_(4)`

Answer :C
Discussion
38.

Question Bank

Answer»

two
THREE
four
five

Solution :
Three CHAIN ISOMERS are POSSIBLE:
Discussion
39.

Quartz Is extensively used as a piezoelectricmaterial, it contains "………..".

Answer»

Pb
SI
Ti
Sn

Solution :QUARTZ is maed up of Si.
Discussion
40.

Quartz is extensively used as a piezoelectric material, it contains ……….

Answer»

PB
Si
Ti
Sn

Solution :Quartz is one of the CRYSTALLINE form of silica and at high temperature can be CONVERTED into other crystalline forms. It is EXTENSIVELY USED as a piezoelectric material.
Discussion
41.

Quartz is a crystalline variety of

Answer»

`Si`
`SiO_(2)`
`Na_(2)SiO_(3)`
`SIC`

ANSWER :B
Discussion
42.

Quantum numbers are assigned to get complete information of electrons regarding their energy, angular momentum, spectral lines etc. Four quantum numbers are principal quantum numbers which tell the distance of electron from nucleus , energy of electron in a particular shell and its angular momentum. Azimuthal quantum number tells about the subshells in a given shell and of course shape of orbital. Magnetic quantum number is the study of orientations or degeneracy ofa subshell. Spin quantum number which defines the spin of electron designated as +1/2 or -1/2 represented by respectively. Electrons are filled in orbitals following Aufbau rule, Pauli's exclusion principal and Hund's rule of maximum multiplicity. On the basis of this answer the following questions. Two unpaired electrons present in carbon atom are different with respect to their

Answer»

AZIMUTHAL QUANTUM number
Principal quantum number
MAGNETIC quantum number
Spin quantum number

Solution :TWO unpaired electrons PRESENT in carbon atom are in different orbitals. 'so' they have different magnetic quantum number
Discussion
43.

Quartz is an example of

Answer»

CHAIN silicate
sheet silicate
cyclic silicate
three DIMENSIONAL NETWORK silicate

Answer :D
Discussion
44.

Quantum numbers l=2 and m=0 represent which orbital?

Answer»

`d_(XY)`
`d_(X^(2)-y^(2))`
`d_(z^(2))`
`d_(zx)`

ANSWER :C
Discussion
45.

Quartz is

Answer»

PIEZOELECTRIC but not FERROELECTRIC
ferroelectric but not piezoelectric
both piezoelectric and ferroelectric
neither piezoelectric nor ferroelectric

Solution :QUARTZ is piezoelectric but not ferroelectric.
Discussion
46.

Quartz crystals are

Answer»

Prismatic
Hexagonal
CUBICAL
Needle SHAPED

Answer :B
Discussion
47.

Quantum number value for 2p sub shell are ?

Answer»

n=2, l=0
n=3, l=1
n=2, l=1
n=2, l=2

Answer :C
Discussion
48.

Quantity of electricity liberated during the formation of 1 mole water in a fuel cell is

Answer»

1F
2F
3F
4F

Answer :B
Discussion
49.

Quantitative estimation of C, H, and extra elements (e.g., N,S,P and halogens) is carried out by Liebig.s combustion, Carius, Dumas, and Kjeldahl.s method. In the quantitative estimation of oxygen by using I_2O_5,the formula used is:

Answer»

PERCENTAGE of `O = 44/32 XX (W xx 100)/(w)`
percentage of `O = 32/44 xx (W xx 100)/(w)`
percentage of `O = 44/32 xx (w xx 100)/(W)`
percentage of `O = 32/4 xx (w xx 100)/(W)`

Answer :B
Discussion
50.

Quantitive measurements of nitrogen in an organic compounds is done by the method.

Answer»

Barthelot method
Belstein method
Lassaigne TEST
Kjheldayhl's method

Answer :D
Discussion