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.

Rate constant for a reaction H_(2) + I_(2) to 2HI is 49 , then rate constant for reaction 2 HI to H_(2) + I_(2) is

Answer»

7
`1//49`
49
21

Solution :For REVERSIBLE reaction RATE constant is also REVERSE .
2.

Rateconstantdoesnotdepend uponunitof concentrationfor reactionwhoseorderis

Answer»

ZERO
first
fractional
infinite

Answer :B
3.

Rate constant.

Answer»

Solution :The rate constant of a chemical REACTION is DEFINED as the rate of the chemical reaction when the CONCENTRATION (or active masses) of each reactant has unit VALUE.
4.

Rate constant of a reaction depends upon:

Answer»

temparature
time
initial concentration
None of these.

Answer :A
5.

Rascing's process is employed for thecommerical preparation of:

Answer»

Ehthyl CHLORIDE
GRIGNARD's regent
Chlorobenzane
Ethanol

Solution :NA
6.

Rare earth metals belongs to the ....... column of moder periodic table

Answer»


SOLUTION :3RD COLUMN
7.

Rank the following subtances in order of increasing boiling points (lowestrarrhighest) : I. CH_(3)CH_(2)CH_(2)OH II. (CH_(3))_(2)CHOCH_(3) III. (CH_(3))_(3)COH"" IV. (CH_(3))_(4)C.

Answer»

`IltIIIltIIltIV`
`IIltIVltIIIltI`
`IVltIIltIIIltI`
`IIltIIIltIltIV`

SOLUTION :it is the CORRECT ANSWER.
8.

Rank the indicated C-C bonds in order of increasing bond length, and explain why you chose this order

Answer»


SOLUTION :N/A
9.

Rank the following dienes in order of increasin heat of hydrogenation.

Answer»


SOLUTION :N/A
10.

Rank the alkyl halides in orderof increasing reactivity in E_(2) reaction.

Answer»

SOLUTION :`(a) P GT R gt Q (B) gt R gt P (C)Q gt P gt R`
11.

Ranitidine is used as an ...

Answer»

antioxidant
antiseptic
ANTACID
ANTIBIOTIC

SOLUTION : antacid
12.

Range the following electrolytes in the increasing order of coagulating power for the coagulation of As_2S_3 sol (I) MgSO_4 (II) BaCl_2 (III) NaCl

Answer»

`III GT I GTII`
`IgtIIgtIII`
`IIIgtIIgtI`
`IgtIIIgtII`

ANSWER :C
13.

Ranitidine is a

Answer»

ANTIHISTAMINE
Antifertile
Analgesic drug
Tranquiliser

ANSWER :A
14.

Ranitidine belongs to which class of drugs.

Answer»

SOLUTION :RANITIDINE BELONGS of ANTACIDS.
15.

Radon was discovered by :

Answer»

Dorn
Ramsay
Rayleigh
None

Answer :A
16.

Random motion of colloidal particles is known as:

Answer»

DIALYSIS
Brownian MOVEMENT
Electro-osmosis
TYNDALL effect.

Answer :B
17.

Radon is a noble gas. It is radioactivity is used in the treatment of:

Answer»

Typhold
Cancer
Cough and cold
Thyroid

Answer :B
18.

Rancidity of oils and fats is due to:

Answer»

PARTIAL hydrolysis by the ACTION of atmospheric moisture and oxidation of fatty acid to foul SMELLING products
Aabsourbation of foul smelling ingredients from the air
Fermentaion caused by MICRO - organisms
Slow DECOMPOSITION of fatty acid

Answer :A
19.

Rancidity of oils is the effect of

Answer»

EXPOSURE to LIGHT
Exposure to OXYGEN of the ATMOSPHERE
Exposure to MOISTURE and oxygen ofatmosphere
Variation of atmosphere temperature

Answer :A
20.

Rancidity of butter is due to the formationof

Answer»

BUTYRIC ACID
Formaldehyde
Acetic acid
BENZOIC acid

Answer :C
21.

Ramsay - Rayleigh second method is based on

Answer»

CONVERSION of `O_2 " to " CuO`
Conversion of `N_2 " to " Mg_3N_2`
Conversion of `O_2 and N_2` into oxides of nitrogen
all

Answer :C
22.

Ramsay was awarded Noble prize for the discovery of rare gases in:

Answer»

1900
1902
1904
1910

Answer :C
23.

Ralate the electronic configuration of elements and their properties in the classification of elements.

Answer»

Solution :All the elements are divided into four types on the BASIS of their properties and electronic configurations. They are : (i) Noble gas elements (ii) Representative elements (iii) Transition elements and (iv) Inner transition elements.
(i)Noble gas elements : Elements in which the outermosts and p sub-shells are completely filled are called inert gas elements. "0" group elements belong to this type. Due to completely filled shells they show chemical inertness and possess more stability. The general outer configuration of these elements is `ns^(2) np^(6)` (except helium). The configuration of helium is `1s^(2)`. They are called noble GASES.
(ii) Representative elements : Elements in which the outermost s and p sub shells ae incompletely filled are called representative elements. These elements are so named because they represent most of the chemical reactions known.
Elements of s-block and p-block (except '0' group) belong to this type. The general outer electronic configuration of these elements is `ns^(1), ns^(2), ns^(2)np^(1)` to `ns^(2)np^(5)`. These elements enter into chemical reactions either by losing or gaining or sharing of electrons. Many of the non-metals, metalloids and some metals belong to this type.
(iii) Transition elements : Elements in which the outermost and penultimate shells are partially filled are called transition elements.
They beolong to d-block. Their general configuration is `(n-1)d^(1-9) ns^(1-2)`. They are so named because there is a gradation from electro +ve nature to electro -ve nature. They possess the following CHARACTERISTIC properties.
(1) They exhibt variable OXIDATION states.
(2) They form coloured compounds.
(3) They are all paramagnetic.
(4) They and their oxides acts as catalysts.
(5) They form alloys and interstitial compounds.
(6) They form complex compounds.
(iv) Inner tarnsition elements: Elements in which the outermost, penultimate and antipenultimate shells are partially filled are called Inner transition elements.
They belong to f-block and are placed separately at the bottom of the table. Their general outer configuration is `(n-2)f^(1-14) (n-1)d^(0,1)ns^(1-2)`. They are so named as they represent a transition of physical and chemical properties among them. There are two series of inner transition elements corresponding to 4f and 5f series. 4f series are called lanthanides and elements of 5f series are called ACTINIDES. Lanthanides are also called rare earths. Majority of actinides are synthetic.
24.

Raju and his father were going in a boat in the river. Raju's father was going to throw away the cell used in watches and hearing aids into the water. Raju prevented him doing so. (a) As a student of chemistry , why would you advise Raju's father not to throw the cell in the water body? (b) What is the value associated with the above decision ?

Answer»

Solution :(a) The watch cells are made up of MERCURY . This mercury will pollute the WATER. Water CONTAMINATED with mercury leads to accumulation of mercury in the body of fishes and other aquatic life.
(b) It helps in keeping the ENVIRONMENT safe from pollution DUE to mercury.
25.

Raising thetemperatue of an equlibrium system

Answer»

Favours the oxothermic REACTION only
Favours the endotermic reaction only
Favours both the exothermic and cindothermic reactions
Favours NEITHER the exothermic nor endotermic reactions

SOLUTION :According to Le chatelier's prnciple.
26.

Rain water sometimes contains NH_4NO_3 because lighting in the sky causes the air to react with produce oxides of nitrogen and

Answer»

`H_2`
`NH_3`
`CO_2`
NOBLE gas

Answer :B
27.

Rain containg dissolved air pollutants like oxides of S, N are referred to as :

Answer»

ARTIFICIAL RAIN
ACID rain
Hails
None of these

Answer :B
28.

Rahul visited the house of the friend Shayam and found that all the water laps were rusted. On equiry, he came to know that these were iron taps. Rahul advised his friend to use either chrome plated or nickel plated taps. Shyam accepted his advice. On the basis of above passage give the answer of the following questions: (i) Why did iron tap get rusted? (ii) What was the purpose of chrom plating or nickel plating? (iii) What is the value associated with this?

Answer»

Solution :(i) Iron is prone to rust. It GET rusted when kept in opne.
(ii) The purpose of depositng a layer of CHROMIUM or nickel or the surface of iron is to check RUSTING.
(iii) General AWARENESS
(b) Friendship
(c ) Knowledge of chemistry put to USE.
29.

Railway wagon axles are made by heating iron rods embedded in charcoal powder. This process is known as

Answer»

tempering
CASE HARDENING
sherardising
annealing

Solution :Case hardening.
30.

Raghav's grandfather is not only obese but is also diabetic. Therefore, he suggested his grandfather not to take sugar but instead he should take artificial sweeteners. After reading the above passage, answer the following questions : (i) What values are expressed by Raghav? (ii) Name the artificial sweetener recommended for diabetic patients. What are its uses? (iii) Name an artificial sweetener which is a derivative of a dipeptide. Also write the names of the two alpha-amino acids from which this dipeptide is derived. What are its advantages and disadvantages?

Answer»

Solution :(i) Raghay expressed concern about the health and well being of his grandfather because sugar affects the functioning of the vital organs, i.e.,heart, liver and kidney of the BODY.
(ii)Saccharin is the artificial sweetener recommended for diabetic patients. It is not metabolized and do not have any calorific value of food. It is secreted as such in urine.
(III)Aspartame is the methyl ester of the dipeptide derived from the two AMINO acids: phenylalanine andaspartic acid.
Advantages. It is roughly 100 times as sweet as sucrose. Therefore, only a small amount is required
Disadvantages. People suffering from the DISEASE PHENYLKETONURIA cannot use aspartame since the metabolism causes the conversion of aspartame to phenylpyruvic acid. Accumulation of phenylpyruvic acid is harmful especially in infants due to brain damage and mental retardation
31.

Raffinose on hydrolysis gives

Answer»

GLUCOSE, FRUCTOSE, GALACTOSE
Fructose, Glucose, MALTOSE
Galactose, Fructose, MANNOSE
Maltose, Fructose, Galactose

Answer :A
32.

Raffinose is an example of

Answer»

trisaccharide
disaccharide
monosaccharide
polysaccharide

Answer :A
33.

Raffinose on hydrolysis forms :

Answer»

Glucose
Fructose
Galactose
All

Answer :D
34.

Raffinose is

Answer»

trisaccharide
disaccharide
monosaccharide
polysaccharide

Solution :Monosaccharides do not hydrolyse, disaccharides give two molecules of monosaccharides, trisaccharides give THREE molecules of monosaccharides and polysaccharides give many molecules of monosaccharides on HYDROLYSIS.
`UNDERSET"Raffinose"(C_18H_32O_16)+H_2O to ` glucose + fructose + galactose
`THEREFORE ` Raffinose is a trisaccharide
35.

Raffinose is :

Answer»

MONOSACCHARIDE
DISACCHARIDE
Trisaccharide
None

ANSWER :C
36.

Rae law for the reaction A+2BtoC is found to be Rate =K[A][B] concentration of reactant .B. is doubled ,keeping the concentration of .A. constant ,the value of rate constant will be……

Answer»

the same
doubled
quadrupled
halved

Solution :Initial RATE `r_(1)`=K[A][B]
If the concentration of B is double than new concentration [2B]
So `r_(2)`=k[A][B]=2K [A][B]
`therefore (r_(2))/(r_(1))=2` so the rate is double
37.

Radon undergoesdecay by alpha- emission ""_(86)^(222) Rn overset(t_(1//2) = 3.8 "days") (to) ""_(84)^(218) Po + ""_(2)^(4) He Which of the following statements will be true for this decay process ?

Answer»

If the initial AMOUNT of RADON was 1 mg , the amount of radon left after 11.4 days will 0.125 mg
Activity of radon after 7.6 days will be `N_(0) xx (5.3 xx 10^(-7)) s^-1` where `N_0` is the original number of atoms of the radon
The decay constant of radon is `2.1 xx 10^(-6) S^(-1)`
60% of the radon will decay in 5 days APPROXIMATELY

Solution :1 mg `overset(3.8 "days")(to) (1)/(2) overset(3.8 "days") (to) (1)/(4) overset(3.8 "days") (to) (1)/(8) , N_(0) 0.5 ((N_(0))/(2)) 0.25 ((N_(0))/(4)) 0.125`
(b) `A_(0) = ((0.693)/(3.8)) xx N_(0) , N_(0) = (( 1 xx 10^(-3))/(222) xx 6 xx 10^(23)) . A_(t) = ((0.693)/(3.8)) xx N_(t) darr` 76 days
`A_(t) = ((0.693)/(3.8)) xx ((N_(0))/(4)) , N_(t) = ((N_0)/(4)) = ((1 xx 10^(-3))/(222) xx 6 xx 10^(-23) xx (1)/(4))`
`A_(t) = ((0.693)/(3.8 xx 24 xx 60 xx 60) xx (N_0)/(4)) = (5.277 xx 10^(-7) xx N_(0)) sec^(-1)`
(c) `lambda = ((0.693)/(t_(1//2))) = ((0.693)/(3.8 xx 24 xx 60 xx 60)) = (((0.693)/(3.046)) xx 10^(-6) ) sec^(-1) = 2.1 xx 10^(-6) sec^(-1)`
(d) `((0.693)/(3.8)) xx t = 2.303 xx log ((600)/(400)) , t = ((0.4 xx 3.8)/(0.3040)) = 5`
38.

Radon has low ionisation potential and empty d-orbitals in the valence shell. Still it does not form compounds with other elements. Why?

Answer»

Solution :Radon is radioactive element with EXTREMELY low half-life period. During this period it is difficult to form COMPOUNDS. Hence, it does not COMBINE with other ELEMENTS.
39.

radius of the bohr's orbit for hydrogen like spacies is given by which expression?

Answer»

`0.529n^2barA`
`0.529n^2/ZbarA`
`0.529Z/n^2barA`
`0.5291/n^2barA`

ANSWER :B
40.

Radium has atomic weight 226 and a half-life of 1600 years. The number of disintegrations produced per second from 1g are

Answer»

`4.8 XX 10^(10)`
`9.2 xx 10^(6)`
`3.7 xx 10^(10)`
Zero

Solution :`R = (0.693)/(t_(1//2)) xx N_(0) = (0.693)/(1600 xx 365 xx 24 xx 60 xx60) xx(6.023 xx 10^(23))/(226)`
`= 3.7 xx 10^(10)`dps
41.

Radium is obtained from

Answer»

Haematite
Pitchblende
Monazite
None of these

Answer :B
42.

Radium is obtained from:

Answer»

Pitchblende
Haematite
Monazite
None of these

Answer :A
43.

Radioisotopes have been used in treatment of certain diseases. For treatment of cancerous tumours, the radioisotope used was:

Answer»

Co-60
U-235
Pu-239
Th-231

Answer :A
44.

Radium disintegrates to give radon. Write chemical equation as well as rate equation.

Answer»

Solution :`.^(226)RA to .^(4)He+.^(222)RN`
Radioactive decay takes place by first ORDER KINETICS.
Rate =K[Ra]
45.

Radioactivity was discovered by

Answer»

Henry Becqueral
Rutherford
J.J. Thomson
MADAM Curie

Solution :Henry BECQUEREL noticed the emission of PENETRATING rays from potassium uranyl sulphate and Madam Curie NAMED it as RADIOACTIVITY
46.

Radioactivity of a radioactive element remains (1)/(10) of the original radioactivity after 2.303 seconds. The half-life period is

Answer»

2.303
0.2303
0.693
0.0693

Solution :`2.303 = (2.303)/(0.693) XX t_(1//2) LOG 10`
47.

Radioactivity of a sample (Z = 22) decreases 90% after 10 years. What will be the half-life of the sample ?

Answer»

5 years
2 years
3 years
10 years

SOLUTION :`k = (2.303)/(t) log (a)/(a - x) = (2.303)/(10) log 100/10`
`k = (2.303)/(10)`
`t_(1//2) = (0.693)/(2.303) XX 10 = 3` YRS.
48.

Radioactivity of neptunium stops when it is converted to

Answer»

Bi
Rn
Th
Pb

Solution :Bi is a STABLE END PRODUCT of Neptunium SERIES
49.

Radioactivity of a sample (Z =22) decrease 90% after 10 years . What will be the half -life of the sample

Answer»

5 YEARS
2 years
3 years
10 years

SOLUTION :`k= (2.303)/(t) "LOG" (a)/(a-x) = (2.303)/(10) "log" (100)/(10)`
`k = (2.303)/(10)`
`t_(1//2) = (0.693)/(2.303) xx 10 = 3 ` yrs.
50.

Radioactive substance has a constant activity of 2000 disintegrations per minute. The material is separated into two fractions, one of which has an initial activity of 1000 disintegration per second while the other fraction decays with t_(1//2) = 24 hours. The total activity in both sample after 48 hours of separation is :

Answer»

1500
1000
1250
2000

Answer :d