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.

Electric field lines: A. are trajectories of a test charge B. are vectors in the direction of the electric field C. form closed loops D. cross each other in the region between two point charges E. are none of the above 

Answer»

E. are none of the above

2.

The diagram shows the electric field lines in a region of space containing two small charged spheres (Y and Z). Then: A. Y is negative and Z is positive B. the magnitude of the electric field is the same everywhere C. the electric field is strongest midway between Y and Z D. the electric field is not zero anywhere (except infinitely far from the spheres) E. Y and Z must have the same sign

Answer»

D. the electric field is not zero anywhere (except infinitely far from the spheres) 

3.

The diagram shows the electric field lines due to two charged parallel metal plates. We conclude that: A. the upper plate is positive and the lower plate is negative B. a proton at X would experience the same force if it were placed at Y C. a proton at X experiences a greater force than if it were placed at Z D. a proton at X experiences less force than if it were placed at Z E. an electron at X could have its weight balanced by the electrical force

Answer»

B. a proton at X would experience the same force if it were placed at Y

4.

The diagram shows the electric field lines in a region of space containing two small charged spheres (Y and Z). Then:A. Y is negative and Z is positive B. the magnitude of the electric field is the same everywhere C. the electric field is strongest midway between Y and Z D. Y is positive and Z is negative E. Y and Z must have the same sign

Answer»

D. Y is positive and Z is negative 

5.

KI in acetone, undergoes `S_(N)2` reaction with each of `P,Q ,R` and S The rates of the reaction very as `underset(P)(H_(3)C-Cl)` .A. P gt Q gt R gt SB. S gt P gt R gt QC. P gt R gt Q gt SD. R gt P gt S gt Q

Answer» Correct Answer - B
6.

The electric field due to a uniform distribution of charge on a spherical shell is zero: A. everywhere B. nowhere C. only at the center of the shell D. only inside the shell E. only outside the shell

Answer»

D. only inside the shell

7.

Two charged point particles are located at two vertices of an equilateral triangle and the electric field is zero at the third vertex. We conclude: A. the two particles have charges with opposite signs and the same magnitude B. the two particles have charges with opposite signs and different magnitudes C. the two particles have identical charges D. the two particles have charges with the same sign but different magnitudes E. at least one other charged particle is present

Answer»

E. at least one other charged particle is present

8.

A charged particle is placed in an electric field that varies with location. No force is exerted on this charge: A. at locations where the electric field is zero B. at locations where the electric field strength is 1/(1.6 × 10−19) N/C C. if the particle is moving along a field line D. if the particle is moving perpendicularly to a field line E. if the field is caused by an equal amount of positive and negative charge

Answer»

A. at locations where the electric field is zero

9.

An isolated charged point particle produces an electric field with magnitude E at a point 2 m away from the charge. A point at which the field magnitude is E/4 is: A. 1 m away from the particle B. 0.5 m away from the particle C. 2 m away from the particle D. 4 m away from the particle E. 8 m away from the particle

Answer»

D. 4 m away from the particle

10.

The hyperconjugative stabilities of tert-butyl cation and `2`-butene, respectively, are due toA. `sigmararrp` (empty) and `sigmararrpi^(**)` electron delocalizationsB. `sigmararrsigma^(**)` and `sigmararrpi` delectron delocalizationsC. `sigmararrp` (filled) and `sigmararrpi` electron delocalizationsD. p(fillled)`rarrsigma^(**)` and `sigma rarrpi^(**)` electron delocalizations

Answer» Correct Answer - A
11.

The stability order in the following carbocations, `CH_(3)CH_(2)^(+)"(I), "(CH_(3))_(2)overset(+)CH"(II), "(CH_(3))_(3)C^(+)"(III) "overset(+)CH_(3)"(IV)"` is :A. `IgtIVgtIIIgtII`B. `IgtIIgtIIIgtIV`C. `IIIgtIVgtIgtII`D. `IIIgtIIgtIgtIV`

Answer» Correct Answer - D
12.

The order of stability of the following carbocations: `underset("I")(CH_(2)C=CH-overset(o+)CH_(2))" , "underset("II")(CH_(3)-CH_(2)-overset(o+)CH_(2))" , "underset("III")(C_(6)H_(5)-overset(o+)CH_(2))`A. III gt I gt IIB. III gt II gt IC. II gt III gt ID. I gt II gt III

Answer» Correct Answer - A
13.

A solution of (-l)- chloro -1 phenyletane in toluene recemises slowly in the presence of a small amunt of `SbCI_(5)` due to the formation of .A. free radicalB. carbanionC. carbeneD. carbocation

Answer» Correct Answer - D
14.

Ksp = [A]3 [B]2 for the salt where A and B are the cation and anion as the case may be stand true for:(a) Ca3 (PO4)2 (b) As2S3(c) Bi2S3 (d) All are correct

Answer»

Correct option: (d)

Explanation: 

For each case Ksp = [A2+]3 [B3–]2

15.

A saturated solution of calcium fluoride contains 2 x 10–4 mole of the salt per litre of the solution. Its Ksp is:(a) 8 x 10–18 (b) 3.2 x 10–11(c) 4 x 10–6 (d) 1.43 x 10–9

Answer»

Correct option: (b) 3.2 x 10–11

Explanation: 

Ksp for CaF2 = 4s3 = 4 x (2 x 10–4)3 

= 3.2 x 10–11 

16.

If the solubility of Pb3(PO4)2 is s mol per litre, then the solubility product of Pb3(PO4)2 will be:(a) 6s2 (b) 6s5(c) s5 (d) 108s5

Answer»

Correct option: (d) 108 s5

Explanation: 

Ksp = [Pb2+] [PO42-]2 (3s)3 (2s)2

= 108 s5

17.

Compare the stability of following carbanion?

Answer»

The correct stability of carbanion is as follow: 

(3) > (1) > (2) > (4) 

18.

The solubility of AgCl (Ksp = 1.2 x 10–10) in a 0.10 M NaCl solution is:(a) 0.1 M (b) 1.2 x 10–6 M(c) 1.2 x 10–9 M (d) 1.2 x 10–10 M

Answer»

Correct option: (c) 1.2 x 10–9 M

Explanation: 

Ksp AgCl = 1.2 x 10–10 = [Ag+] [Cl] = [s][s + 0.1]

where s is solubility of AgCl

∴ Ksp = s x 0.1 = 1.2 x 10–10

∴ s = 1.2 x 10–9 M

19.

The 2 conformations shown below belongs to the compound cyclohexane. Infinite number of conformations are possible for cyclo hexane. But these 2 conformations given below has a peculiarity. Try to nd it. Also, dene the terms conformers and the phenomenon conformation?

Answer»

Infinite number of conformations are possible for cyclohexane. Out of it chair conformation is the most stable one and the boat conformation is the least stable form. 

The different arrangement of a compound which arises as a result of rotation about carbon single bond are called conformers and the phenomenon is called conformation

20.

Order of stability of given carbocation is(1) A > B > C(2) C > B > A(3) B > A > C(4) B > C > A

Answer»

Correct option is (3) B > A > C

B is more stable due to resonance.

21.

Arrange the following carbocation according to stability A. `R gt Q gt S gt P`B. `Q gt R gt P gt S`C. `Q gt R gt S gt P`D. `R gt Q gt P gt S`

Answer» Correct Answer - C
First electron gain enthalpy of all elements is negative except N & Be.
22.

in the above given compound how many functional group reduced by LAH (lithium aluminium hydride) and SBH (sodium Borohydride) respectively ?A. 4,4B. 4,3,C. 3,4D. 4,2

Answer» Correct Answer - D
SBH reduces only aldehyde & ketone while stronger LAH reducess aldehyde into appropriate alcohol.
23.

Which of the following alcohols gives a red colour in victor meyer test :-A. n- propyl alcoholB. isoproyl alcoholC. `(CH_(3))_(3)C-OH`D. Sec. Buty alcohol

Answer» Correct Answer - A
`I^(@)` alcohols give colour in victor meyer test.
24.

Which of the following alcohol gives the white turbidity imediately with `HCl+ZnCl_(2)` (anhy.)?A. `CH_(3)-underset(OH)underset(|)overset(Ph) overset(|)(C)-OC_(2)H_(5)`B. `Ph-underset(OH)underset(|)(CH)-CH_(3)`C. `Ph-underset(OH)underset(|)(CH)-CH=CH_(2)`D. `CH_(3)-CH_(2)CH_(2)OH`

Answer» Correct Answer - A::B::C
Alcohols that can from stable carbocation
25.

Which of the following alcohols gives the best yield of dialkyl ether on being heated with a trace of sulphuric acid?  (1)  1-pentanol (2)  2-pentanol(3)  cyclopentanol (4)  2-propano

Answer»

1-pentanol alcohols gives the best yield of dialkyl ether on being heated with a trace of sulphuric acid.

26.

Statement-1: For the reaction: `Ni^(2+)+2e^(-)rarrNi` and `Fe^(2+)+2e^(-)rarrFe` `rArr E_(Fe^(2+)|Fe)^(@)ltE_(Ni^(2+)|Ni)^(@)` and `E_("Red")^(@)gt0` `rArr` So Fe electrode is cathode and Ni electrode is anode. Statement-2: If `DeltaG^(@) lt0` and `E_("Cell")^(@)lt0`, then , cell is feasible.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-2C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.

Answer» Correct Answer - D
`E_(Ni^(+2)|Ni)^@=-0.24 " " E_(Fe^(+2)|Fe)^@=-0.44`
`E_(Cell)^@=E_(Fe|Fe^(+2))^@+E_(Ni^(+2)|Ni)^@`
=0.44-0.24=0.20
Fe electrode Anode, Ni electrode cathode
`DeltaG lt 0 " " E_(Cell)gt0` For working of cell
27.

Diethyl ether combines with CO under specific conditions to form `:`A. acetic acidB. carbon dioxideC. ethyl propanoateD. acetyl chloride

Answer» Correct Answer - C
28.

Diethyl ether by regareded as anhydride of :A. `C_2H_5COOH`B. `C_2H_5OH`C. `C_2H_5CHO`D. `C_2H_5COOC_2H_5`

Answer» Correct Answer - B
29.

Write the Arrhenius equation and mention what each term stands for.

Answer»

\(\frac{-Ea}{RT}\)

K = A.e 

Where K = rate constant 

A = Arrhenices constant 

Ea = Energy of activation 

T = Temperature 

R = Gas constant

30.

Any two differences between lanthanides and Actinides.

Answer»

Lanthanides:

1. 4f – orbital is progressively filled 

2. Only Pm (promethion) is radio active 

3. They are less reactive

Actinides:

1. 5f – orbital os progressively filled 

2. All are radio active 

3. They are more reactive

31.

What are antacids? Give an example.

Answer»

The chemical substances which are used in the treatment of acidity are called antacids. 

Ex: Ranitidine (Zantac)

32.

Why is fluoroacetic acid a stronger acid than acetic acid?

Answer»

The larger the electron-withdrawing inductive effect the greater is the acidity.

−I effect of F > Cl. Therefore, Fluoroacetic acid is stronger acid than chloroacetic acid.

33.

What are the effects of the electron withdrawing and electron donating groups on acidity of carboxylic acids.

Answer»

Electron donating group decreases acidic strength and electron with drawing group increases acidic strength of carboxylic acids.

34.

Equivalent condictance of `BaCI_(2), H_(2)SO_(4)` and `HCI` are `x_(1) , x_(2)` and `x_(3) S cm^(2) "equiv"^(-1)` at infinite dilution , if specific condictance of structured `BaSO_(4)` solution is of `y S cm^(-1)` then `K_(p)` of `BaSO_(4)` isA. `(10^(3)y)/(2(x_(1) + x_(2)- 2x_(3)))`B. `(10^(6)y^(2))/((x_(1) + x_(2)- 2x_(3))^(2))`C. `(10^(6)y^(23))/(4(x_(1) x_(2)- 2x_(3))^(2))`D. `(x_(1) x_(2)- 2x_(3))/(10^(3)y^(2))`

Answer» Correct Answer - c
`Delta_(BaSO_(4))^(@) = Delta_(BaCI_(2))^(@) + Delta_(H_(2)SO_(4))^(@) - 2 Delta_(HCI)^(@)`
`= (x_(1) + x_(2) - 2x_(3))`
`Delta_(BaSO_(4))^(@) = (1000 xx "sp. Conductance")/("solubility (in saturated solutions)")`
`(x_(1) + x_(2) xx 2x_(3)) = (1000 y)/("solubility")`
`:.` Solubilty of `BaSO_(4) = (1000y)/((x_(1)+x_(2) -2x_(3)))N`
`= (1000y)/(2(x_(1)+x_(2)-2x_(3)))M`
`BaSO_(4) rarr Ba^(2+) + SO_(4)^(2-)`
`K_(sp) (BaSO_(4)) = [Ba^(2+)] [SO_(4)^(2-)] M^(2) = (10^(6)y^(2))/(4(x_(1)+x_(2)-2x_(3))^(2))`
35.

Under standered condition `Delta G^(@)` for the reaction `2Cr(s)= 3Cd^(2+)(aq) rarr 2Cr_((a a))^(3+) +3Cd(s)` is `(E_(Cr^(3+)//Cr)^(@) = - 0.74 V, E_(Cd^(2+)//Cd)^(@) = - 0.4 V)`A. `-65.62 J` moleB. `-196.86 kJ` moleC. `- 98.43 kJ` moleD. `-96.86 J `mole

Answer» Correct Answer - b
`Delta G^(@) = - nFE_(cell)^(@) - 6 xx 96500 + 0.34 = - 196.86 KJ// mol`
36.

Calculate the standard cell potentials of galvanic cell in whiCHM the following reactions take place `:` `a. Cr(s) +3Cd^(2+)(aq) rarr 2Cr^(3+)(aq)+3Cd` `b. Fe^(2+)(aq)+Ag^(o+)(aq)rarr Fe^(3+)(aq)+Ag(s)` Calculate the `Delta_(r)G^(c-)` and equilibrium constant of the reactions .

Answer» Correct Answer - `a. Delta_(r)G^(c-)=-196.86 kJ mol^(-1),K_(c)=3.192xx10^(34)`
`b. Delta_(r)G^(c-)=-2.895 kJmol^(-1),K_(c)=3.22`
a. `E_("cell")^(c-)=E_("cathode")^(c-)-E_("anode")^(c-)`
`=-0.40V-(-0.74)=+0.34V`
`Delta_(r)G^(c-)=-nFE_("cell")^(c-)`
`=-6 molxx96400C mol^(-1)xx0.34V`
`=-196860J mol^(-1)=-196.86k J mol^(-1)`
`Delta_(r)G^(c-)=--2.303RTlogK`
`196860=2.303xx8.314xx298 log K`
`logK=34.5014`
`K="Antilog" (34.5014)=3.192xx10^(34)`
`b`. `E^(c-)``_(cell)=+0.80V-0.77V=+0.03V`
`Delta_(r)G^(c-)=-nFE_(cell)^(c-)`
`=-(1 mol)xx(96500 C mol^(-1))xx(0.03V)`
`=-2895J mol^(-1) Delta_(r)G^(c-)=-2.303RTlogK`
`-2895=-2.303xx8.314xx298xxlogK`
or log `K =0.5074` or `K ="Anitlog"(0.5074)=3.22`
37.

Compound (X)`C_9H_10O` is inert to `Br_2//C Cl_4`.Vigorous oxidation with hot alkaline `KMnO_4//OH` yield `C_6H_5COOH` (X) gives precipitate with 2,4-Dintrophenyl hydrazine. How can these isomers be distinguished by the usual chemical tests ? Following are possible isomers of X : (i)`C_6H_5-CH_2-CH_2-CHO` , (ii)`C_6H_5-undersetunderset(CH_3)(|)CH-CHO` (iii)`C_6H_5-CH_2-oversetoverset(O)(||)C-CH_3` , (iv)`C_6H_5-oversetoverset(O)(||)C-CH_2-CH_3`A. I gives red ppt. with Fehling solution and II & III can be distinguished by iodoform testB. I & II can be distinguished by simple chemical methodC. I & II give red ppt. with Fehling solution and III & IV can be distinguished by iodoform testD. II gives red ppt. with Fehling solution and I & IV can be distinguished by iodoform test

Answer» Correct Answer - A,C
I & IV both compounds do not give iodoform test and I & II both given similar test
38.

How can we improve the success rate of fertilisation during artificial insemination in animal husbandry programmes?

Answer»

The technology is called MOET or Multiple Ovulation Embryo Transfer. During the procedure, a cow is given hormonal treatment so that more than one ovule (6-8 eggs) is produced per cycle. After mating or artificial insemination the embryos at 8-32 celled state are transferred to different surrogate mother cows. The method has been successfully used for cattle, sheep, buffalo etc.

39.

How are the colloidal solutions classified on the the basis of physical states of the dispersed phase and dispersion medium ?

Answer» Correct Answer - Refer to section
40.

Which of the following is a lanthanide ?A. CuriumB. CaliforniumC. UraniumD. Europium

Answer» Correct Answer - D
(A),(B) and (C ) are actinoids where as (D) europium is lanthanide
41.

Which of the following statement is not correct?A. `La(OH)_3` is less basic than `Li(OH)_3`B. In lanthaniode series, ionic radius of `Ln^(3+)` ion decreases.C. La is actually an element of transition series rather than lanthaniodes.D. Atomic radius of Zr and Hf are same because of lanthaniode contraction.

Answer» Correct Answer - A
`La(OH)_3` is more basic than `Li(OH)_3` In lanthaniodes the basic character of hydroxides decreases as the ionic radius decreases.
42.

The ionic radius of `._(57)La^(3+)` is 1.06Å. Which one of the following given values will be closest to the ionic radius of `._21Lu^(3+)`?A. 1.06ÅB. 1.56ÅC. 0.85ÅD. 1.35Å

Answer» Correct Answer - D
Ionic radius of `Lu^(3+)` is smaller than that of `La^(3+)`because of lanthanide contraction across the period.
43.

I am a number between 31 and 41. I am a multiple of 3. I am an even number. What number am I?

Answer» Number is 36
44.

An object starting from rest travels 20 m in the first 2s and 160 m in next 4s. What will be the velocity after 7s from the start?

Answer»

Using the second equation of motion: s = ut + \(\frac12\)at2 

where: s = distance covered = 20 m 

u = initial velocity = 0 m/s 

a = acceleration = ? m/s2

t = time = 2s 

s = ut + \(\frac12\)at2 

20 = 0 × 2 + \(\frac12\)× a × 22 

20 = 0 + 2 × a 

a = 10 m/s2 

Final velocity, v after 2s

The first equation of motion: v = u + at 

where: v = final velocity = ? 

u = initial velocity = 0 m/s 

a = acceleration = 10 m/s2 

t = time = 2s 

v = u + at 

v = 0 + 10 × 2 = 20 m/s 

Now it is given that in next 4s it covers 160 m. But now the vehicle has gained some velocity. So the final velocity of the previous case will become the initial velocity in this case.

Using the second equation of motion: s = ut + \(\frac12\)at2 

where: s = distance covered = 160 m 

u = initial velocity = 20 m/s 

a = acceleration = ? m/s2 

t = time = 4s 

s = ut + \(\frac12\)at2 

160 = 20 × 4 + \(\frac12\)× a × 4× 4 

160 = 80 + a × 8160 – 80 = a × 880 = a × 8a = 10 m/s2 

This shows that acceleration is uniform.

The first equation of motion: v = u + at 

where: v = final velocity = ? 

u = initial velocity = 0 m/s 

a = acceleration = 10 m/s2 t = time = 7s 

v = u + at 

v = 0 + 10 × 7 

v = 70 m/s

45.

What are the differences between Kinetic and Potential Energy?

Answer»

Difference Between Kinetic and Potential Energy:

Kinetic EnergyPotential Energy
Kinetic energy is the kind of energy present in a body due to the property of its motionPotential Energy is the type of energy present in a body due to the property of its state
It can be easily transferred from one body to anotherIt is not transferable
The determining factors for kinetic energy are Speed or velocity and massHere, the determining factors are Height/ distance and mass
Flowing water is one of the examples of kinetic energyWater present at the top of a hill is an example of potential energy
It is relative with respect to natureIt is non-relative with respect to nature
46.

Let `Delta^(2)` be the discriminant and `alpha,beta` be the roots of the equation `ax^(2)+bx+c=0` then `2aalpha+Delta` and `2abeta-Delta` can be roots of the equation.A. `x^(2)+2bx+b^(2)=0`B. `x^(2)-2bx+b^(2)=0`C. `x^(2)+2bx-3b^(2)+16ac=0`D. `x^(2)-2bx-3b^(2)+16ac=0`

Answer» Correct Answer - A::C::D
`alpha,beta=(-b+-Delta)/(2a)`
`alpha=(-b-Delta)/(2a),beta=(-b+Delta)/(2a)`
or `alpha=(-b+Delta)/(2a),beta=(-b-Delta)/(2a)`
`implies2aalpha+Delta=-b&2alphabeta-Delta=-b`
`implies2aalpha+Delta=-b&2abeta-Delta-b`
`implies2aalpha+Delta=-b+2Delta`
Sum of roots `=-2b`
sum of roots `=-2b`
product of root `=b^(2)`
product of root `=b^(2)-4`.`(b^(2)-4ac)`
Hence equation is `x^(2)+2bx+b^(2)=0`
`=-3b^(2)+16ac`
Hence equation is `x^(2)+2bx-3b^(2)+16ac=0`
47.

A particle is projected in such a way that it follows a curved path with constant acceleration `vec(a)`. For finite interval of motion. Which of the following option `(s)` may be correct `:` `vec(u)=` initial velocity `vec(a)=` acceleration of particle `vec(v)=` velocity at `tgt0`A. `|vec(a)xxvec(u)|cancel(=)0`B. `|vec(a)xxvec(v)|=0`C. `|vec(u)xxvec(v)|=0`D. `|vec(u).vec(v)|=0`

Answer» Correct Answer - A::D
If `|vec(a)xxvec(u)|=0` particle its will not forllow curved path. Above described motion is a projectile motion with parabolic path.
48.

A particle moves in xy plane with a velocity by `vec(v)=(8t-2)hat(i)+2hat(j)`. If it passes through the point (14, 4) at t = 2 sec, then give equation of the path.A. `x + y - y^(2) = 2`B. `y^(2)+x=4`C. `x-y-y^(2)=2`D. `y + x + y^(2)=2`

Answer» Correct Answer - A
`V = 8t -2`
`intdx = int(8t-2)dt`
`x = 4t^(2)-2t +c`
At `t =2,x = 14 implies " "14=4(4) -2(2)+c " implies "c = 2`
`x = 4t^(2)-2t+2,v_(y)=2`
`intdy int2dt`
`y = 2t + c`
At `t =2, y = 4 " implies "c = 0`
` y = 2t`
Equation `implies " " x = (2t)^(2)-2t +2`
`implies" " x=y^(2)-y+2`
`implies " "x + y-y^(2) =2`
49.

Trajectory of a particle in a projectile motion is given by `y=x-x^(2)/80` where x and y are in meters. Match the column-1 and column-2. `{:(,"Column-I",,"Column-II"),("(A)","x coordinate at height of 15 m",,"(P) 20 m"),("(B)",underset("point of projection at x= 100 m")"vertical distnce of particle from",,"(Q) 80 m"),("(C)","Horizontal range",,"(R) 60 m"),(,,,"(S) 25 m"):}`

Answer» Correct Answer - [(A)PR(B)S(C)Q]
(A)`15=x-x^(2)/80`
`1200=80x-x^(2)`
`x^(2)-80x+1200=0`
`x^(2)-60x-20x+1200=0`
`x=60,20`
(B) `y=100-(100)^(2)/80=100-(1000)/82=25`
(C) `Hz range:x-x^(2)/80=0Rightarrow x(x-x/80)=0,x=(0,80)`
50.

The velocity of a particle is given by `vec(v) = 2hat(i)-hat(j)+2hat(k)` in `m//s` for time interval `t = 0` to `t = 10` sec. If the distance travelled by the particle in given time interval `10n`, find value of `n` is.

Answer» Correct Answer - `3`
Speed of particle is
`|V| = sqrt((2)^(2) + (1)^(2) + (2)^(2))`
`|V| = 3m//sec`
`Soverset(10)underset(0)(int)|V|dt = overset(10)underset(0)(int)3dt = |3t|_(0)^(10)`
`S = 30 m`