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.

Equal volumes of 0.1 M urea and 0.1 M glucose are mixed. The mixture will have :

Answer»

LOWER OSMOTIC pressure
sameosmotic pressure
higherosmotic pressure
none of these

Answer :B
2.

Equal quantities of electricity of passed through three voltameters containing FeSO_(4) , Fe_(2) (SO_(4))_(3) and Fe(NO_(3))_(3) . Consider the following statements in this regard : 1 . The amount of iron deposited in FeSO_(4) and Fe_(2)(SO_(4))_(3) are equal 2 . The amount of iron deposited in Fe(NO_(3))_(3) is two third of the amount of iron deposited in FeSO_(4) 3 . The amount of iron deposited is Fe_(2)(SO_(4))_(3) and Fe(NO_(3))_(3) is equal

Answer»

1 alone is correct
1 and 2 are correct
2 and 3 are correct
3 alone is correct .

SOLUTION :1. The O.S. of FE in `FeSO_(4)= + 2` and in `Fe_(2)(SO_(4))_(3) = + 3`
2 . The O.S. of Fe in `Fe(NO_(3))_(3) = + 3 ` and in `Fe(SO_(4)) = + 2`
3 . The O.S. of Fe in `Fe_(2)(SO_(4))_(3) = + 3` and in `Fe(NO_(3))_(3) = +3`
Hence STATEMENT 2 and 3 are correct .
3.

Equal quantities of electricity are passed through 3 votameters containing FeSO_(4),Fe_(2)(SO_(4))_(2) and Fe(NO_(3))_(3). Cosider the following statements (A) The amounts of iron deposited I FeSO_(4) and Fe_(2)(SO_(4))_(3) are equal (B) The amount of iron deposited in Fe(NO_(3))_(2) is 2//3^(rd) of the amount deposited in FeSO_(4) (C) The amount of iron deposited in Fe_(2)(SO_(4))_(3) and Fe(NO_(3))_(3) are equal

Answer»

(A) is CORRECT
(B) is correct
(C) is correct
Both (A) and (B) are correct

Solution :Suppose 2F of ELECTRICITY passed through `FeSO_(4),Fe_(2)(SO_(4))_(3) and Fe(NO_(3))_(3)` solution.
`FeSO_(4)toFe^(2+)+SO_(4)^(2-)`
`Fe^(2+)+underset(2F)(2e^(-))tounderset("1 mole")(Fe(S))`
`Fe_(2)(SO_(4))_(3)to2Fe^(3+)3SO_(4)^(2-)`
`Fe^(3+)+underset(3F)(3e^(-))to underset(1" mole")(Fe(S))`
`Fe(NO_(3))_(3)toFe_(3+)^(-)`
`Fe^(3+)+underset(3F)(3e^(-))tounderset("1 mole")(Fe(S))`
`therefore2F` will give`=(2)/(3)` mole
4.

Equal quantities of electricity are passed through three voltametrs containing FeSO_(4),Fe_(2)(SO_4)_3 and Fe(NO_3)_3 Consider the following statements in this regard : The amount of iron deposited in FeSO_(4) and Fe_(2)(SO_4)_3 are equal 2. The amount of iron deposited in Fe(NO_3)_3 is two third of the amount of iron deposited in FeSO_4. 3 The amount of iron deposited in Fe_(2)(SO_4)_3 and Fe(NO_3) is equal . Which of the following statements is /are correct ?

Answer»

only 1
1 and 2
2 and 3
Only 3

Solution :ASSUME 1 F CHARGE is PASSED through `FeSO_(4),Fe_(2)(SO_4)_3` and `Fe(NO_3)_3`
`{:(" ",FeSO_4,Fe_2(SO_4)_3,Fe(NO_3)_3),("Amount of Fe deposited ", 56/2,56/3,56/3):}`
5.

Equal quantities (50 ml) of the following four samples of water are placed in four beakers of 100 ml capacity . Their boiling points are determined accurately using the same thermometer . Which sample of water will having the lowermost boiling point as compared to other three samples ?

Answer»

DISTILLED water
Bottled mineral water
Well water
Seawater

SOLUTION :Distilled water
6.

Equal number of moles of A and B are allowed to react with each other till it reaches equilibrium. 2A+BhArrC+D The value of K_(c) for this equilibrium can never be

Answer»

`GT1`
`LT1`
`=1`
`OO`

Answer :C::D
7.

Equal molecules of N_2 and O_2 are kept in a closed container at pressure P. If N_2 is removed from the system, then the pressure of the container will be

Answer»

<P>P
P//2
`P/4`
2P

Answer :B
8.

Equal masses of oxygen hydrogen and methane are taken in a container under identical conditions. Find the ratio of their volumes.

Answer»


ANSWER :`1:16:2`
9.

Equal masses of nitrogen and ethyleneare mixed in an empty container at 27^@ C. The total pressure exerted by the gaseous mixture is 1 atm. The partial pressure exerted by ethylene gas is :

Answer»

0.67 atm
0.33 atm
0.50 atm
0.20 atm

Answer :C
10.

Equal masses of KClO_(3) undergoes different reaction in two different container: (i) 2KClO_(3) to 2KCl + 3O_(2) (ii) 4KClO_(3)to KCl + 3KClO_(4) Mass ratio of the KCl produced in respective reaction is x : 1. Value of 'x' will be

Answer»

4
2
0.25
3

Answer :A
11.

Equal masses of methane and oxygen are mixed in an empty container at 25^(@)C. The fraction of total pressure exerted by oxygen is

Answer»

`(1)/( 3) XX ( 273)/( 298) `
`(1)/( 3)`
`(1)/(2)`
`(2)/(3)`

Solution :Let the mass of METHANE and oxygen be x g each.
Moles of `CH_(4) = ( x )/( 16)`
Moles of `O_(2) = ( x )/( 32)`
TOTAL moles `= ( x )/( 16) + ( x)/( 32) = ( 3x )/( 32)`
`p_(O_(2)) = p_("total")xx x_(O_(2))`
`( p_(O_(2)))/( p_("total")) = x_(2)= ((x)/( 32))/( (3x)/(32)) = ( 1)/( 3)`
12.

Equal masses of a solute are dissolved in equal amount of two solvents A and B respective molecular masses being M_(A) and M_(B) . The relative lowering of vapour pressure of solution in solvent A is twice that of the solution in solvent B. If the solutions are dilute , M_(A) and M_(B) are related as

Answer»

`M_(A)=M_(B)`
`2M_(A)=M_(B)`
`M_(A)=2M`
`M_(A)=4M_(B)`

ANSWER :C
13.

Equal mass of oxygen and helium gases are mixed in a container at 27°C Fraction of total pressure exerted by helium gas is

Answer»

`1/8`
`8/9`
`1/4`
`3/4`

ANSWER :2
14.

Equal amounts of two gases of molecular weight 4 and 40 are mixed. The pressure of the mixture is 1.1 atm. The partial pressure of the light gas in this mixture is

Answer»

<P>0.55 atm
0.11 atm
1 atm
0.12 atm

Solution :No. of moles of lighter gas `= (m)/(4)`
No. of moles of heavier gas `=(m)/(40)`
Total no. of moles of `(m)/(4)+(m)/(40)=(11M)/(40)`
Mole fraction of lighter gas `((m)/(4))/((11m)/(40))=(10)/(11)`
Partial PRESSURE due to lighter gas `=P_(o)xx(10)/(11)=1.1 xx (10)/(11) =1 `atm.
15.

Equal amounts in grams of following sub-stances were dissolved in equal amount of water. Which of these will have the highest boiling point ?

Answer»

Urea `(NH_(2) CONH_(2))`
Glucose `(C_(6)H_(12)O_(6))`
Sodium chloride ( NaCl )
Calcium chloride `(CaCl_(2))`

Solution :The simple inspection of the question shows that (C) should be the correct choice because `CaCl_(2)` will give 3 mol of ions for 1 mol of `CaCl_(2)` DISSOLVED (assuming it to be completely dissociated). However, in the given question, the amounts are given in GRAMS. If we assume, `w` gram of all thesubstances are dissolved in 1000 g of solvent, then
Molal conc. of NaCl `=(w)/(58.5)m`
Molar conc. of `CaCl_(2)=(w)/(111)` m
Since NaCl gives two mole of ions for one mole of it so that
Molar concentration of ions in solution
`=(w)/(58.5)xx2=0.034w`
`CaCl_(2)` gives three moles of ions for 1 mol of it so that
Molar conc. of ions in solution
`=(w)/(111)xx3=0.027 w`
Therefore, molal conc. of solute in NaCl solution is more than in `CaCl_(2)` solution.
16.

Equal amounts of two samples of carbon were burnt and the radioactivities ofCO_2 formed were measured. The activitied were 2100and 1400 counts per week respectively.What will be the age difference of the sample. [ t_(1//2) forC^(14) = 5600 years ]

Answer»

2002 years
2345 years
4250 years
3343 years

Solution :`t_(1//2) = 5600` years
Decay constant , `LAMBDA = (0.693)/(t_(1//2)) = (0.693)/(5600) = 1.24 xx 10^(-4) "year"^(-1)`
For first sample : `(N_0)/(N) = ("NUMBER of counts per week of CARBON INITIALLY")/("number of counts per week of carbon finally")`
`= (N_0)/(2100) "".......(i)`
For second sample : `(N_0)/(N) = (N_0)/(1400) "" .......(ii)`
The AGE difference of the sample :
`t = (2.303)/(lambda) ["log" (N_0)/(1400) - "log" (N_0)/(2100)]`
`= (2.303)/(1.24 xx 10^(-4))["log" N_0 - "log" 1400 - "log" N_0 + log 2100]`
`= (2.303)/(1.24 xx 10^(-4)) [log 2100 - log 1400]`
`= (2.303)/(1.24 xx 10^(-4)) [3.32 - 3.14]`
`= t = (2.303)/(1.24 xx 10^(-4)) xx 0.18 = 3343.06` years
17.

Equal amount of aqueous solution of CuSO_4 and alkaline sodium potassium tartarate is mixed and treated with aliphatic aldehyde, then

Answer»

RED BROWN PPT of `Cu_2O` is formed
Red brown ppt of CUO is formed
Blue ppt of CuO is formed
There will be no reaction

Answer :A
18.

Eq. wt. of an acid salt NaHSO_(4) is

Answer»

M/2
M/3
M/1
None

Answer :C
19.

Eq. wt of a metal , x g of which reacts with 1 eq. of an acid is

Answer»

1
x/2
2x
x

Answer :d
20.

Epsom salt or epsomite ore is:

Answer»

`MgSO_4.7H_2O`
`MgSO_4.2h_2O`
`MgSO_4.H_2O`
`CaSO_4.2H_2O`

ANSWER :A
21.

Epsom salt is MgSO_(4) H_(2)O What is the value of x' in epsom salt

Answer»


SOLUTION :Epsomsalt is `MgSO_(4).7H_(2)O`. The value of .X. is 7.
22.

Epsom salt has the following uses except :

Answer»

as a puragative in medicines
in dyeing and tanning industries
as a food preservative
a catalyst in the MANUFACTURE of SULPHURIC acid.

Answer :C
23.

Epoxy resins use a polymer made from bisphenol A and epichlorohydrin in basic medium. Which of the Statement are incorrect.

Answer»

REACTION proceeds VIA `SN_(2)`
The nucleophile attacks from hindered side.
Chirality of carbon in epichlorihidrin is retained
in each step two `SN_(2)` REACTIONS take place.

Answer :B
24.

Eplain why the stability of oxaacids of chlorine increasesin theorder given below ? HClO lt HClO_(2) lt HClO_(3) lt HClO_(4)

Answer»


Solution :(i) All these acids on losing a PROTON give their correspondingconjugate BASES (ie) `ClO^(-), ClO_(2)^(-), ClO_(3)^(-)` and `ClO_(4)^(-)`. Their structures are:

(ii) Since, oxygen is more electronegative than chlorine, therefore dispersalof negative CHARGE present on oxygen atom (singly bonded to Cl) increases as the number of oxygen atoms attached by a double bond to chlorine increases due to `P pi - d pi` back bonding.
(iii) Therefore stability of IONS will increasein the following order:
`ClO- LT ClO_(2)- lt ClO_(3)= lt ClO_(4)-`
(iv) Due to increase instability of conjugate base, acidic strength of correspondingacid increases in thesame order as :
`HClO- lt HClO_(2) lt HClO_(3) lt HClO_(4)`
25.

Epimers are pair of diastereoisomeric aldoses which differ only in configuration at position :

Answer»

`C_5`
`C_2`
`C_4`
`C_3`

ANSWER :B
26.

Epilepsy is caused due to deficiency of………………..

Answer»

VITAMIN `B_(6)`
Vitamin `B_(12)`
Vitamin E
Vitamin K

SOLUTION :Vitamin `B_(6)`
27.

E^(o) value of Mg^(2+)|Mg,Fe^(2+)|Fe and Zn^(2+) | Zn" are " - 2.37 V, - 0.44 and - 0.76 V respectively. The correct statement is :

Answer»

MG oxidizes Fe
Zn oxidizes. Fe
Zn REDUCES Mg
Zn reduces `Fe^(2+)`

ANSWER :D
28.

E^(o) for the cell Zn|Zn_((aq))^(2+)||Cu_((aq))^(2+)|Cu is 1.10 V at 25^(@)C, the equilibrium constant for the reaction Zn+Cu_((aq))^(2+)hArrCu+Zn_((aq))^(2+) is of the order of

Answer»

`10^(-28)`
`10^(-37)`
`10^(+18)`
`10^(+17)`

Solution :`E_(cell)^(@)=(0.059)/(N)log" "K`
`log" "K=(1.10xx2)/(0.059)=37.2881` or `K=10^(-37)`
29.

Enzymes work best under ordinary conditions of temperature and pH

Answer»
30.

Enzymes trypsin converts

Answer»

Proteins into `alpha`-AMINO ACIDS
Starches into sugar
Glucose into glycogen
`alpha`-amino acids into proteins

Answer :A
31.

Enzymes, in the living systems:

Answer»

PROVIDE energy
Provide immunity
Transport oxygen
Catalysed BIOCHEMICAL process

Answer :D
32.

Enzymes in the living systems :

Answer»

provide energy
transport OXYGEN
CATALYSE biochemical REACTIONS
provide immunity.

SOLUTION :Enzymes in living systems catalyse biochemical reactions.
33.

Enzymes in the living systems…….. .

Answer»

PROVIDE energy
provideimmunity
CATALYSE BIOLOGICAL
TRANSPORT oxygen

Answer :C
34.

Enzymes in the living system

Answer»

Provide energy
Provide IMMUNITY
Transport OXYGEN
Catalyse biological PROCESS

ANSWER :D
35.

Enzymes greatly reduce the amount of ____________in their action.

Answer»

SOLUTION :ACTIVATION ENERGY
36.

Enzymes can be active in human body at a temperature of ............

Answer»

`98^(@)F`
`105^(@)F`
`37^(@)F`
`50^(@)F`

SOLUTION :`98^(@)F`
37.

Enzymes belong to which class of compounds ?

Answer»

Polysaccharides
Polypeptides
Polynitro HETEROCYCLIC compounds
Hydrocarbons 

ANSWER :B
38.

Enzymes belong to a category of :

Answer»

PROTEINS
CARBOHYDRATES
VITAMINS
harmones

Answer :A
39.

Enzymes are made up of ?

Answer»

PROTEIN
Lipid
Vitamin
Carbohydrate

Answer :A
40.

Enzymes are made up of .......

Answer»

Carbohydrate
Lipid
PROTEIN
Sucrose 

SOLUTION :Protein
41.

Enzymes are made of ………………..

Answer»

PROTEINS
VITAMIN
LIPID
Carbohydrate

Solution :Proteins
42.

Enzymes are made up of

Answer»

Edible proteins
Proteins with SPECIFIC structure
Nitrogen CONTAINING carbohydrates
Carbohydrates 

Answer :B
43.

Enzymes are known to increase the rate of reaction by :

Answer»

`10^2` TIMES
`10^-2` times
`10^5` times
`10^12` times

Answer :D
44.

Enzymes are known as biochemical catalysis. Write the two important characteristics of enzyme catalysis.

Answer»


Answer :Enzymes are highly specific in nature, i.e., one enzyme cannot CATALYSE more than one reaction.
Highly ACTIVE at OPTIMUM temperature (298-310K) and optimum pH (between 5 and 7).
45.

Enzymes are basically

Answer»

EDIBLE proteins
Carbohydrates
Carbodydrates CONTAINING nitrogen
Specially stuctured proteins

Answer :D
46.

Enzymes are :

Answer»

LIVING organisms
Dead organisms
Complex NITROGENOUS
SUBSTANCES PRODUCED
from living cells
None

Answer :C
47.

Enzymes :

Answer»

Have OPTIMUM ACTIVITY at BODY temperature
Consists of NUCLEIC acids
Carbohydrates
Have all these PROPERTIES

Answer :A
48.

Enzymes are

Answer»

COMPLEX nitrogenous substances produced in living CELLS 
Steroids
Living organisms
Dead organisms 

ANSWER :A
49.

Enzyme which breaks down starch to maltose is

Answer»

MALTASE
zymase
diastase
Glucose

SOLUTION :Maltase.
50.

Enzyme trypsin converts

Answer»

STARCH into sugar
proteins into `alpha`-AMINO ACIDS
glucose into glycogen
`alpha`- amino acids into proteins

Solution :Protein `OVERSET"Trypsin"to alpha`-amino ACID