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.

Which of the following is monomer of PVC ?(A) CH3 – CH = CH2 (B) C6H5 – CH= CH2 (C) CH2=CH – Cl (D) CH2=CH2

Answer»

Correct answer is (C) CH2=CH – Cl 

2.

Metallic character ------ moving down to the group.

Answer» The answer is increases
3.

The common bases in DNA and RNA are – (A) Adenine, Cytosine, Uracil (B) Guanine, Adenine, Cytosine (C) Guanine, Uracil, Thymine (D) Adenine, Thymine, Guanine

Answer»

Correct answer is (B) Guanine, Adenine, Cytosine

4.

If solutbility of `CH_(3)` COOAg(s) in a buffer solution of pH equal to 5 is `1.414 xx10^(-y)` then calculate the value of y. Given : `K_(sp)(CH_(3)COOAg)=10^(-10),K_(a)(CH_(3)COOH)=10^(-5)`

Answer» Correct Answer - 5
`CH_(3)COOAg(s)iffCH_(3)COO^(-)(aq)+Ag^(+)(aq)`
`CH_(3)COO^(-)(aq)+H_(2)O iff CH_(3)COOH+OH^(-)`
`{:(x-z,,,,,,,,z,,,10^(-9)):}`
`K_(h)=(10^(-14))/(10^(-5))=(zxx10^(-9))/(x-z)" "," "x-z=zimplies z=(x)/(2)`
`K_(sp)=10^(-10)=(x-z)xx x" ",10^(-10)=(x)/(2)xx x`
Solubility `x=sqrt(2)xx10^(-5)`
Solubility `x=1.414xx10^(-y)`
y=5
5.

Which of the following is not normally produced by the body ? (A) Enzyme (B) Vitamin (C) Protein (D) Harmone

Answer»

Correct answer is (B) Vitamin

6.

What are ligands? Classify them with example.

Answer»

 Ligands: The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be simple ions such as Cl-, small molecules such as H2O or NH3, larger molecules such as H2N CH2CH2 NH2 or N(CH2CH2NH3)3 or even macromolecules such as unidentate proteins.

When a ligand is bound to a metal through a single donor atoms such as Cl-, H2O or NH3,  the ligand is said to be unidentate Eg.- Cl, H2O

Bidentate: When a ligand is bound through two donor atoms, the ligand is called bidentate ligands.

Eg.- H2N, CH2CH2NH2 or C2O42-

Polydentate: N(CH2CH2NH2) Hexadentate ligand - (EDTA4+)

7.

The vertical columns were called ------ ?Element with electronic configuration 2,8,8,2 is present in ---------- ?

Answer»

In the Mendeleev periodic table, the vertical columns are called groups and the horizontal rows are called periods.

Magnesium is the element with the electronic configuration of 2,8,2 as its atomic number is 12, since the atomic number is the sum of all the electrons present in the electronic configuration.

8.

Complete the following reactions:I. KBr + Cl2 → ......... + ............II. I2 + H2O + Cl2 → ............ + .............III. NaOH + Cl2 → .............. + ..............(cold and dilute)

Answer»

I. KBr + Cl2 → HCl(aq.) + Cl Br

II. I2 + H2O + Cl2 → 2ICl + H2O

III. 6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O

9.

Element with electronic configuration 2,8,8,2 is present in ---------- ?

Answer»

Magnesium is the element with the electronic configuration of 2,8,2 as its atomic number is 12, since the atomic number is the sum of all the electrons present in the electronic configuration.

10.

Element with electronic configuration 2,8,8,2 is present in ---------- ?

Answer»

The answer is Magnesium

11.

When `._(92)U^(238)` decauys it emits an a-particle. The new nuclide in turn emits a beta-particele to give another nuclide X. The mass number and atomic number of X are. Respectively .A. ` 234 ` and ` 91`B. ` 234 `and ` 96`C. ` 231` and ` 88`D. ` 234` and ` 88`

Answer» Correct Answer - A
`. _(92) U^(238) rarr _zX^A +_2 H_(e)^(4)+_(-1)e^0`
Equating mass number on both sides
` 238 = A+4 0`
` :. A= 238 - 4 = 234`
Equating atomic numer on both sides
` 92 = Z + 2 -1`
` :. Z= 92 -1 =91`.
12.

According to ------- every eight element had properties similar to that of the first.

Answer»

According to the Newlands law of octaves, every eighth element had properties similar to that of the first element.

13.

` CH_3CH_2CHO overset (NaOH, Delta) underset((aldol)) (rarr)A,A` is.A. ` CH_3CH_2CH= CHCH_2CHO`

Answer» Correct Answer - D
`CH_(3)CH_(2)underset(O)underset(||)(C)H+alphaunderset(H)underset(|)overset(CH_(3))overset(|)(C)HCHOto CH_(3)CH_(2)underset(OH)underset(|)(C)Hoverset(CH_(3))overset(|)(C)HCHO`.
14.

You have given two species :- NOF and `NO_(2)F` and two dipole moments 1.81D and 0.46D. Then correct statement is:-A. 1.81 D for `NO_(2)F` and o.47D for NOFB. 0.47D for `NO_(2)F` and 1.81D for NOF because `NO_(2)F` is linear but NOF is nonlinear moleculeC. 0.47D for `NO_(2)F` because bond moments of NO bond and `NF` bonds are oriented in the opposite directionD. 0.46D for `NO_(2)F` and 1.81D for NOF

Answer» Correct Answer - C
15.

Which is correct :-A. `H_(2)O_(2)` oxidize `Mn^(2+)`B. `HOCl` reduce `H_(2)O_(2)` in acidic mediumC. Rxn of `KMnO_(4)` with `H_(2)O(2)` in both `(H^(+))` medium `& (OH^(-))` medium give `O_(3)`D. All

Answer» Correct Answer - C
16.

Choose the correct order for the energy barrier to rotation around the B-N bond: [`R=CH_(3)` is all above cases]A. `H_(2)B-NR_(2)ltBH(NR_(2))_(2)ltB(NR_(2))_(3)`B. `B(NR_(2))_(3)ltBH(NR_(2))_(2)ltBH_(2)-NR_(2)`C. `BH(NR_(2))_(2)gtB(NR_(2))_(3)gtBh_(2)-NR_(2)`D. `BH_(2)-NR_(2)gtB(NR_(2))_(3)gtBH-(NR_(2))_(2)`

Answer» Correct Answer - 1
17.

`SiO_(2)+Coverset(Delta)rarr` products is :-A. `SiC & CO_(2)`B. `SiO & CO`C. `SiC & CO`D. `Si & CO`

Answer» Correct Answer - C
18.

Double cahin structures are present in asbestos . Which of the anion are present in them ?A. `(Si_(2)O_(5)^(-2))_(n)`B. `(Si_(4)O_(11)^(-6))_(n)`C. `(SiO_(3)^(2-))_(n)`D. `SiO_(4)^(4-)`

Answer» Correct Answer - A
19.

Among the following compounds, the decreasing order of reactivity towards electrophilic substitution is : A. `IgtIIgtIIIgtIV`B. `IIgtIgtIIIgtIV`C. `IVgtIgtIIgtIII`D. `IIIgtIgtIIgtIV`

Answer» Correct Answer - D
20.

Arrange the following nucleophiles in the order of their nucleophilic strength.A. `OH^(-)gtCH_(3)COO^(-)gtCH_(3)O^(-)gtC_(6)H_(5)O^(-)`B. `CH_(3)COO^(-)ltC_(6)H_(5)O^(-)ltCH_(3)O^(-)ltOH^(-)`C. `C_(6)H_(5)O^(-)ltCH_(3)COO^(-)ltCH_(3)O^(-)ltOH^(-)`D. `CH_(3)COO^(-)ltC_(6)H_(5)O^(-)ltOH^(-)ltCH_(3)O^(-)`

Answer» Correct Answer - C
21.

State the Limitations of Ostwald dilution law.

Answer»

(1) It is not applicable for strong electrolyte. 

(2) It is not applicable for saturated solution.

22.

Difference between Common Ion Effect and Odd Ion Effect.

Answer»
Common Ion EffectOdd Ion Effect
NH4OH ⇌ NH4+ + OH 
On mixing NH4Cl
NH4Cl →NH4+ + Cl
NH4OH ⇌ NH4+ + OH 
On mixing HCl
HCl → H+ + Cl
Due to mixing of common ion concentration of ammonium ion will increase therefore equilibrium will shift in backward direction i.e. rate of backward reaction increases means α decreases.Due to mixing of odd ions concentration of OH will decrease 
∴ Equilibrium will shift in forward direction i.e. rate of forward reaction increases, means α increases

23.

Which one is greater α1 or α2 for the following equations :(i) HCN + CCl4 → α1(ii) HCN + C6 H6 → α2 (A) α1 > α2 (B) α2 > α1(C) α1 = α2 (D)  none

Answer»

Correct option: (B) α2 > α1

Explanation: 

∴ µ(CCl4) = 0 and µ(C6H6) = 2.5

So, α2 > α1

24.

Oncotic pressure of plasma is due to

Answer»

Answer:Proteins

25.

In primary dehydration intracellular fluid volume is?

Answer»

In primary dehydration intracellular fluid volume is reduced

26.

The daily water allowance for normal adult (60kg) is about?

Answer»

Answer: 1800–2500 ml

27.

difference between semi permeable membrane and selectively permeable membrane

Answer»

The membrane which permits the passage of pure solvent molecules to pass through it and not the solute particles is called semipermeable membrane while the membrane which allows some substances to pass through it more readily than others is known as selectively permeable membrane .
 Moreover, a selectively permeable membrane requires constant energy for its maintenance, whereas a semipermeable membrane does not require constant energy for its maintenance.

28.

Is starch a dietary fiber

Answer»

Dietary fiber consists of non-starch polysaccharides and other plant components such as cellulose, resistant starch, resistant dextrins, inulin, lignins, chitins (in fungi), pectins, beta-glucans, and oligosaccharides.

29.

what do you understand by the term refraction of light?

Answer» the phenomenon, due to which a ray of light deviates from its original path, while travelling from one optical medium to another optical medium is called refraction of light.

Refraction is the bending of light as it passes from one substance to another. Here the light ray passes from air to glass and back to air. The bending is caused by the differences in density between the two substance. 

30.

D-galactose and D-glucose are

Answer»

D-Galactose is an epimer of D-glucose because the two sugars differ only in the configuration at C-4 . D-Mannose is an epimer of D-glucose because the two sugars differ only in the configuration at C-2 . When a molecule such as glucose converts to a cyclic form, it generates a new chiral centre at C-1

31.

The sugar residues of glycogen are

Answer»

Glycogen is as an important energy reservoir; when energy is required by the body, glycogen in broken down to glucose, which then enters the glycolytic or pentose phosphate pathway or is released into the bloodstream. Glycogen is also an important form of glucose storage in fungi and bacteria.

32.

Does methene exists? If yes give its structure

Answer»

Methene do not exist in nature. This is because the alkene family have the general formula C n H 2n where n is a natural number(which starts from 1). If we substitute 1 in place of n, then we get the formula CH2. CH2 has no independent existence, as the tetravalent nature of carbon is being violated.To have a double bond you need at least two atoms, so ethene is the smallest possible alkene.

33.

Does methyne exist? If yes give its structure and if no then explain why?

Answer»

Methyne do not exist in nature because the alkyne family have the general formula C n H 2n-2 where n is a natural number(which starts from 1). If we substitute 1 in place of n, then we get the formula CH0 which cannot exist. So, the first member of this series is ethyne. 

Hence, the structure of methyne is not possible to find out.

No, methyn does not exist because when we substitute 1 in place of n in the general formula we get CH0.
34.

All the following are composed exclusively of glucose except

Answer»

Answer: Lactose

Lactose is a disaccharide whose repeating unit is made up of glucose and galactose.

35.

All the following are composed exclusively of glucose except

Answer»

Maltose is a disaccharide whose repeating unit is made up of two units of glucose. Lactose is a disaccharide whose repeating unit is made up of glucose and galactose. Sucrose is a disaccharide whose repeating unit is made up of glucose and fructose. Galactose is a monosaccharide simple sugar. 

36.

The average of pH urine is

Answer»

The average urine sample tests at about 6.0.

37.

Why is pH 7 neutral

Answer»

pH is a measure of the amount of Hydrogen ions (H+) in a solution. Ions are just atoms that have an electric charge on them, so H+ is a hydrogen atom with charge of 1. Even in pure water ions tend to form due to random processes (producing some H+ and OH- ions). The amount of H+ that is made in pure water is about equal to a pH of 7. That's why 7 is neutral.

38.

A buffer solution contains 0.384 M KHCO3 and 0.239 M Na2CO3 . If 0.0464 moles of potassium hydroxide are added to 225.0 mL of this buffer, what is the PH of the resulting solution? (Assume the volume does not change upon adding potassium hydroxide.)

Answer»

Calculate moles of bicarbonate and carbonate:
HCO3¯: (0.384 mol/L)(0.2250 L) = 0.0864 mol
CO32¯: (0.239 mol/L)(0.2250 L) = 0.053775 mol

The hydroxide reacts with the acid (the bicarbonate):
HCO3¯ decreases: 0.0864 - 0.0464 = 0.0400 mol
CO32¯ increases: 0.053775 + 0.0464 = 0.100175 mol

Need pKa of bicarbonate:
Ka of HCO3¯ is the same as the Ka2 of H2CO3.
Ka of HCO3¯ = 4.7 x 10¯11
pKa = 10.252

Use Henderson-Hasselbalch Equation:
pH = 10.328 + log (0.100 / 0.04)
pH = 10.328 + 0.398 = 10.726

39.

Why it is dangerous to carry liquid acetylene

Answer»

Acetylene poses unique hazards based on its high flammability, instability and unique storage and transportation requirements. Acetylene is highly unstable. High pressure or temperatures can result in decomposition that can result in fire or explosion.

40.

Write name of two ores of copper.

Answer»

The name of two ores of copper:–

(a) Copper Pyrites – CuFeS

(b) Azurite – 2CuCO3.Cu(OH)2

41.

Lucas test is used to distinguish– (a) Amine(b) Ethers(c) Alcohols(d) Alkyls halides

Answer»

Lucas test is used to  Alcohols

42.

Fehling test is positive for (a) Acetaldehyde (b) Acetone (c) Ether (d) Amine

Answer»

Fehling test is positive for Acetaldehyde 

43.

 A compound is formed by two elements X and Y. The element Y forms ccp arrangement and atoms of X occupy octahedral voids. What is the formula of the compound ?

Answer»

No. of Y (ccp) = 4

 No. of X (octahedral void) = 4

 X : Y = 4 : 4

 XY

44.

Which of the following will not undergo aldol condensation-(a) Acetaldehyde(b) Propanaldehyde(c) Benzaldehyde(d) Trideutero acetaldehyde 

Answer»

Benzaldehyde is not  aldol condensation.

45.

The genral formula of carbohydrate is–(a) Cx(H2O)y(b) Cx(H2O)y(c)  (CO)x (H2)y(d) (CO2)x (H2O)y

Answer»

The genral formula of carbohydrate is Cx(H2O)y

46.

The carbonyl compound formed when ethanol gets oxidised using this copper-based catalyst can also be obtained by ozonolysis of:

Answer»

The carbonyl compound formed when ethanol gets oxidised using this copper-based catalyst can also be obtained by ozonolysis of But-2-ene.

C2H5OH +Cu ->(at573k) = CH3CHO

But-2-ene + ozonolysis = CH3CHO

47.

What is he use of amorphous silica ?

Answer»

Used in photovoltanic cell

48.

The number of chiral C- atom in cyclic structure of glucose is-(a) 2 (b) 3(c) 4(d) 5

Answer»

The number of chiral C- atom in cyclic structure of glucose is  4.

49.

What are emulsion? What are their different types? Give examples of each type.

Answer»

An emulsion can be defined as a colloid consisting of two or more non-homogenous type of liquids wherein one of the liquid contains the dispersion of the different form of liquids.

Properties Of Emulsions:

  • Emulsions contain both a continuous and the dispersed with the boundary coming between the phases that are called “interface”.
  • Emulsions have a cloudy appearance due to many phase interfaces scattering light passing through the emulsions.
  • Emulsions appear in white colour when the light is dispersed in equal proportions.
  • If the emulsion is dilute, then higher-frequency and the low-wavelength type of light will be scattered in more fractions, and this kind of emulsion will appear in blue in colour. This is also referred to as the Tyndall effect.

Types of Emulsion:

Emulsions can be classified on the basis of the properties of the dispersed phase and the dispersion medium.

1) Oil in water (O/W):

In this type of emulsion, the oil will be the dispersed phase and water will be the dispersion medium. The best example for o/w emulsion is milk. In milk, the fat globules (which act as the dispersed phase) are suspended in water (which acts as the dispersion medium).

2) Water in oil (w/o):

In this type, water will be the dispersed phase and oil will be the dispersion medium. Margarine (a spread used for flavouring, baking and working) is an example of water in oil emulsion.

50.

Which block of elements are known as transition elements ?(a) p-block(b) s-block(c) d-block(d) f-block

Answer»

d-block block of elements are known as transition elements .