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.

Conjugate acid of NH_(2)^(-)is :

Answer»

`NH_(4)OH`
`NH_(4)^(+)`
`NH^(2-)`
`NH_(3)`

Answer :D
Discussion
2.

Conjugate acid of HPO_4^2- is

Answer»

`H_2PO_4^-1`
`H_3PO_4`
`PO_4^(3-)`
`HPO_3^(2-)`

ANSWER :2
Discussion
3.

[Co(NH_3)_6SO_4]Br and [Co(NH_3)_6Br]SO_4 are a pair of _____ isomers .

Answer»

Ionization
Ligand
Co-ordination
Hydrate

Answer :A
Discussion
4.

[Co(NH_(3))_(6)][Cr(CN)_(6)] and [Cr(NH_(3))_(6)][Co(CN)_(6)] present an example of

Answer»

LINKAGE isomerism
ionisation isomerism
coordination isomerism
solvate isomerism

Answer :C
Discussion
5.

[Co(NH_3)_6]Cl_3 is called :

Answer»

Hexaammine COBALT (III) CHLORIDE
Amino cobalt chloride (III)
Cobalt chloride hexammine
Hexammine tricobalt chloride

Answer :A
Discussion
6.

[Co(NH_3)_6]^(3+)is diamagnetic where as [CoF_6]^(3-)is paramagnetic explain.

Answer»

SOLUTION :`[Co(NH_3) _6]^(3+) ` is DIAMAGNETIC because `NH_3` is a strong field ligand CAUSES PAIRING of electrons in d orbitals and there is no unpaired electrons
(ii)`[CoF_6]^(3+) ` is paramagnetic because `F^(-)` is a weak field ligand does not cause pairing of electrons and there are unpaired electrons.
Discussion
7.

[Co(NH_3)_5SO_4]Cl is an octahedral coordination compound Write the IUPAC name of the compound

Answer»

SOLUTION :Pentaamminesulphatocobalt(III0chloride
Discussion
8.

[Co(NH_3)_5SO_4]Cland [Co(NH_3)_5Cl]SO_4 arecoordination compounds.Write the IUPAC names of the above compounds.

Answer»

Solution :`[CO(NH_3)_5SO_4]CL`,pentaammine sulphato cobalt(III) chloride,`[Co(NH_3)_5Cl]SO_4`,Pentaamminechlorido cobalt(III) sulphate
Discussion
9.

[Co(NH_3)_5SO_4]Cl is an octahedral coordination compound write the formula of the ionisation isomer of the above compound

Answer»

SOLUTION :`[CO(NH_3)_5Cl]SO_4`
Discussion
10.

[Co(NH_3)_5SO_4]Cland [Co(NH_3)_5Cl]SO_4 arecoordination compounds.Identify the ligands in each of the above compounds.

Answer»

SOLUTION :In `[Co(NH_3)_5SO_4]CL`,the LIGANDS are `NH_3` and `SO-4` and in `[Co(NH_3)_5Cl)SO_4` The lignds rae NH_3 and CL:
Discussion
11.

Give evidence that [Co(NH_(3))_(5)Cl]SO_(4) and [Co(NH_(3))_(5)SO_(4)]Cl are ionization isomers.

Answer»

IONIZATION
Ligand
Co-ordination
Hydrate

ANSWER :A
Discussion
12.

Give evidence that [Co[NH_3)_5Cl]SO_4 and [Co(NH_3)_5SO_4]Cl are ionisation isomers.

Answer»

SOLUTION :IONISATION ISOMERISM
Discussion
13.

[Co(NH_(3))_(5)NO_(2)]Cl_(2)" and "[Co(NH_(3))_(5)ONO]Cl_(2) are related to each other as :

Answer»

GEOMETRICAL isomers
linkage isomers
coordination isomers
ionization isomers

Answer :B
Discussion
14.

[Co(NH_(3))_(5)NO_(2)]Cl_(2) and [Co(NH_(3))_(5)ONO]Cl_(2) are related to each other as :-

Answer»

GEOMETRICAL isomers
Linkage isomers
Coordination isomers
ionisation isomers

Answer :B
Discussion
15.

[Co(NH_3)_5Br]SO_4 and [Co(NH_3)_5SO_4]Br are the examples of :

Answer»

LINKAGE isomerism
Geometrical isomerism
Ionisation isomerism
Optical isomerism

Answer :C
Discussion
16.

[Co(NH_3)_5Br]SO_4 and [Co(NH_3)_5SO_4]Br are related as

Answer»

LINKAGE isomers
ionisation isomers
Co-ordination isomers
none of these

Answer :B
Discussion
17.

[Co(NH_3)_5Br]SO_4 and [CO(NH_3)_5SO_4]Br are _____ isomers while [Co(NH_3)_5NO_2]Cl_2 and [Co(NH_3)_5ONO]Cl_2 are _____ isomers.

Answer»

SOLUTION :IONISATION, LINKAGE or FUNCTIONAL.
Discussion
18.

[Co(NH_(3))_(4)(SCN)_(2)]^(+)" and "[Co(NH_(3))_(4)(NCS)_(2)]^(+) are :

Answer»

IONISATION isomers
linkage isomers
coordination isomers
geometrical isomers

Answer :B
Discussion
19.

[Co(NH_(3))_(5)Br]SO_(4) and conc. [Co(NH_(3))_(5)SO_(4)]Br are example of which type of isomerism?

Answer»

LINKAGE
GEOMETRICAL
ionization
optical

ANSWER :C
Discussion
20.

[Co(NH_(3))_(4)(NO_(2))_(2)]Cl exhibits

Answer»

IONIZATION isomerism, geometrical isomerism and optical isomerism
LINKAGE isomerism, geometrical isomerism and optical isomerism
linkage isomerism, ionization isomerism and optical isomerism
linkage isomerism, ionization isomerism and geometrical isomerism

Solution :`{:([CO(NH_(3))_(4)(NO_(2))_(2)]Cl),([Co(NH_(3))_(4)(Cl)(NO_(2))]NO_(2)):}}rarr` ionization isomers
`{:([Co(NH_(3))_(4)(NO_(2))_(2)]Cl),([Co(NH_(3))_(4)(ONO)_(2)]Cl):}}rarr` linkage isomers.
Discussion
21.

[Co(NH_(3))_(4)(NO_(2))_(2)]Cl exhibits.

Answer»

linkage isomerism, ionization isomerism and geometrical isomerism
ionization isomerism, geometrical isomerism and optical isomerism
linkage isomerism, geometrical isomerism and optical isomerism
linkage isomerism, ionization isomerism and optical isomerism

Solution :The given compound may have linkage isomerism due to presence of `NO_(2)` grou pwhich may be in the FORM `-NO_(2)& -ONO`
it may have ionisation isomerism due to presence of two ionisable GROUP `-NO_(2) & -CL`. It may have geometrical isomerism in the form of ci-s trans form as follows:
`[CO(NH_(3))_(4)Cl(NO_(2))]NO_(2) and [Co(NH_(3))(NO_(2))_(2)]Cl`-ionisation isomers.
`[Co(NH_(3))_(5)(NO_(2))_(2)]Cl and [Co(NH_(3))_(5)(ONO)_(2)Cl` -Linkage isomers.

Geometrical isomers.
Discussion
22.

The complexes [Co(NH_(3))_(4)Cl_(2)]NO_(2) and : [Co(NH_(3))_(4)Cl.NO_(2)]Cl are isomers

Answer»

LINKAGE
coordination
ionisation
geometrical

ANSWER :C
Discussion
23.

[Co(NH_3)_4CI_2] possesses:

Answer»

SQUARE PLANAR geometry
Tetrahedral geometry
Tetrahedral nature
Octahedral geometry

Answer :D
Discussion
24.

[Co(NH_3)_4 (NO_2)_2]Cl exhibits

Answer»

IONIZATION ISOMERISM, GEOMETRICAL isomerism and OPTICAL isomerism
LINKAGE isomerism, geometrical isomerism and optical isomerism
linkage isomerism, ionization isomerism and optical isomerism
linkage isomerism, ionization isomerism and geometrical isomerism

Answer :D
Discussion
25.

{:("Configuration","Element"),("1) "5s^(1)4d^(5),"A) "Cu),("2) "6s^(1)5d^(10),"B) "Pd),("3) "4s^(1)3d^(1-),"C) "Mo),("4) "5S^(2)4d^(10),"D) "Cr),(,"E) "Au):} The correct match is

Answer»

`1-C, 2-A, 3-E, 4-B`
`1-C, 2-E, 3-B,4-A`
`1-C, 2-E, 3-A, 4-B`
`1-E, 2-C, 3-A, 4-B`

ANSWER :C
Discussion
26.

Conformation in molecules is due to:

Answer»

ROTATION about a SINGLE bond
Change in DIRECTION of light
Structural changes
Restricted rotation about a DOUBLE bond

Answer :A
Discussion
27.

Configuration of a chiral molecule can be changed by

Answer»

ROTATION AROUND a sigma bond
cooling to `73^@`K
breaking a bond at chiral centre and reforming it
reacting it with an acid.

ANSWER :C
Discussion
28.

Cone HNO_(3) is addedbeforeproceedingto test forgroup II This is to

Answer»

CONVENT `Fe^(+2)` ion `Fe^(+3)` ion
Oxidise anyremaining `H_(2)S`
Fromnitratewhich give grandar precipitate
Increases IONISATION of `NH_(4)OH`

ANSWER :a
Discussion
29.

Cone H_(2)SO_(4)on aditionto dry KNO_(3) givesdrown fames of :

Answer»

`SO_(2)`
`SO_(3)`
`SO`
`NO_(2)`

Solution :`KNO_(3) + H_(2)SO_(4) rarr KHSO_(4) + HNO_(3)`
`4HNO_(3) rarr2H_(2)O + underset("BROWN")(4NO_(2))+O_(2)`
Discussion
30.

Cone H_(2)SO_(4) will not give any gas with

Answer»

`ZnSO_(4)`
barium PHOSPHATE
MEGNESIUM borate
sodium oxalate

Solution :`SO_(4)^(2-) ,PO_(4)^(2-) and BO_(3)^(3-)` do not react with cone `H_(2)SO_(4)`
Discussion
31.

Conductometric titration curveof a equimolar mixture of HCI and HCN with NaOH(aq) is

Answer»




SOLUTION :CONCEPTUAL
Discussion
32.

Conductors have electrical conductivity in the range of

Answer»

`10^(2)` to `10^(7)OHM^(-1)m^(-1)`
`10^(2)` to `10^(3)ohm^(-1)m^(-1)`
`10^(8)` to `10^(10)ohm^(-1)m^(-1)`
10 to 100 `ohm^(-1)m^(-1)`

ANSWER :A
Discussion
33.

Conductors allow the passage of electric current through them. Metallic and electrolytic are the two types of conductors. Current carriers in metallic and electrolytic conductors are free electrons and free ions respectively. Specific conductance or conductivity of the electrolyte solution is given by thefollowing relation: K= cx (l)/(A) where, c=1/R is the conductance and 1/A is the cell constant, Molar conductance (^^_m) andequivalence conductance (^^_e) of an electrolyte solution are calculated using the following similar relations: ^^_m = K xx (1000)/(M) ^^_(e) = K xx (1000)/(N) where, M and N are the molarity and normality of the solution respectively. Molar conductance of strong electrolyte depends on concentration : ^^_m = ^^_m^(0) - b sqrt(C) ^^_m^(0) = molar conductance at infinite dilution C = concentration of the solution b = constant The degrees of dissociation of weak electrolytes are calculated asalpha= (^^_m)/(^^_m^(0)) = (^^_e)/(^^_e^(0)) Which of the following equality holds good for the strong electrolytes?

Answer»

`LAMBDA= Lambda^(0) " as " C to 1 `
`Lambda= Lambda^(0) " as " C to 0`
`Lambda= Lambda^(0) " as " C to oo`
`Lambda= Lambda^(0) " as " C to SQRT(b)`

Solution :`^^_(m) = ^^_(M)^(0) - b sqrt(c) , " As " C to 0 , ^^ = ^^""^(0)`
Discussion
34.

Conductors allow the passage of electric current through them. Metallic and electrolytic are the two types of conductors. Current carriers in metallic and electrolytic conductors are free electrons and free ions respectively. Specific conductance or conductivity of the electrolyte solution is given by thefollowing relation: K= cx (l)/(A) where, c=1/R is the conductance and 1/A is the cell constant, Molar conductance (^^_m) andequivalence conductance (^^_e) of an electrolyte solution are calculated using the following similar relations: ^^_m = K xx (1000)/(M) ^^_(e) = K xx (1000)/(N) where, M and N are the molarity and normality of the solution respectively. Molar conductance of strong electrolyte depends on concentration : ^^_m = ^^_m^(0) - b sqrt(C) ^^_m^(0) = molar conductance at infinite dilution C = concentration of the solution b = constant The degrees of dissociation of weak electrolytes are calculated asalpha= (^^_m)/(^^_m^(0)) = (^^_e)/(^^_e^(0)) For which of the following electrolytic solution ^^_m and ^^_e are equal ?

Answer»

`BaCl_2`
`KCl`
`Al_2(SO_4)_3`
`CaCl_2`

SOLUTION :`K^(+) CL^(-) ,^^_(m) = ^^_(e)`
Discussion
35.

Conductors allow the passage of electric current through them. Metallic and electrolytic are the two types of conductors. Current carriers in metallic and electrolytic conductors are free electrons and free ions respectively. Specific conductance or conductivity of the electrolyte solution is given by thefollowing relation: K= cx (l)/(A) where, c=1/R is the conductance and 1/A is the cell constant, Molar conductance (^^_m) andequivalence conductance (^^_e) of an electrolyte solution are calculated using the following similar relations: ^^_m = K xx (1000)/(M) ^^_(e) = K xx (1000)/(N) where, M and N are the molarity and normality of the solution respectively. Molar conductance of strong electrolyte depends on concentration : ^^_m = ^^_m^(0) - b sqrt(C) ^^_m^(0) = molar conductance at infinite dilution C = concentration of the solution b = constant The degrees of dissociation of weak electrolytes are calculated asalpha= (^^_m)/(^^_m^(0)) = (^^_e)/(^^_e^(0)) Which of the following decreases on dilution of electrolytic solution?

Answer»

EQUIVALENT CONDUCTANCE
MOLAR conductance
SPECIFIC conductance
Conductance

Solution :specific conductance
Discussion
36.

Conductivity of saturated solution of BaSO_(4)" at 315 K is "3.648 xx 10^(-6)" ohm"^(-1)" cm"^(-1) and that of water is 1.25 xx10^(-6)" ohm"^(-1)" cm"^(-1). Ionic conductance of Ba^(2+) and SO_(4)^(2-)" are 110 and 136.6 ohm"^(-1)" cm"^(2)" mol"^(-1) respectively. Calculate the solubility of BaSO_(4) in g/L.

Answer»

Solution :`Lambda_(m)^(@)(BaSO_(4))=Lambda_(m)^(@)BA^(2+)+Lambda_(m)^(@)SO_(4)^(2-)=110+136.6="246.6 OHM"^(-1)"cm"^(-1)`
`K_(BaSO4)=K_(BaSO4)" (solution)"-K_("water")=3.648xx10^(-6)-1.25xx10^(-6)`
`=2.398xx10^(-6)"S cm"^(-1)`
`Lambda_(m)^(c )=(Kxx1000)/("Solubility")=(2.398xx10^(-6)xx1000)/(246.6)=9.72xx10^(-6)"mol/L"`
`"Solubility "=9.72xx10^(-6)xx233=2.26xx10^(-3)g//L`
Discussion
37.

Conductivity of an electrolytic solution depends on. . . .

Answer»

nature of electrolyte.
concentration of electrolyte
power of AC source.
distance between the electrodes.

SOLUTION :* Conductivity or SPECIFIC conductance `kappa` (kappa): It is the conductance of solution KEPT between TWO electrodes with 1 `m^(2)` area of cross section and distance of 1 m. it is the reciprocal of resistivity `(p)`.
`k=(1)/(p)`.
* SI unit of `k=S" "m^(-1)`. it depends on the nature of the electrolyte and concentration of the electrolyte.
* Conductivity does not depend on power of AC source and distance between two ions.
* The conductivity of the electrolytic solution depend on (i) nature of added electrolyte (ii) volume of produced ions and their solvation capacity (iii) nature of solvent and its viscosity and (iv) temperature.
* So, conductivity depend on (A) nature of electrolyte and (B) concentration of electrolyte.
Discussion
38.

Conductivity of a solution of [CoBr(NH_(3))_(5)]Cl_(2) corresponds to

Answer»

2 : 1 electrolyte.
1 : 2 electrolyte.
1 : 3 electrolyte.
2 : 3 electrolyte.

Answer :B
Discussion
39.

Conductivity of a solution is directly proportional to

Answer»

Dilution
NUMBER of ions
Current density
Volume of the SOLUTION

Solution :CONDUCTIVITY of a solution is DIRECTLY proportional to the number of ions.
Discussion
40.

Conductivity of a saturated solution of a sparingly soluble salt AB at 298 K is 1.85 xx 10^(-5) S m^(-1).Solubility product of the salt AB at 298 K is Given ^^_m^(0)(AB) = 140 xx 10^(-4)S m^(2) " mol"^(-1)

Answer»

`5.7 xx 10^(-12)`
`1.32 xx 10^(-12)`
`7.5 xx 10^(-12)`
`1.74 xx 10^(-12)`

SOLUTION :`S_0 = (LAMBDA)/(lambda_0)`
`S = (k)/(1000 lambda_m) = (1.85 xx 10^(-5))/(1000 xx 140 xx 10^(-4))`
`S = 1.3 xx 10^(-6)`
`K_(sp) = S^(2) =(1.3 xx 10^(-6))^(2) = 1.69 xx 10^(-12)`
Discussion
41.

Conductivity of a saturated solution of a sparingly soluble salt AB(1:1 electrolyte) at 298 K is 1.85 times 10^(-5)Sm^(-1). Solubility product of the salt AB at 298 K(wedge_(m)^(@))_(AB)=14 times 10^(-3)S" "m^(2)mol^(-1).

Answer»

`5.7 TIMES 10^(-12)`
`1.32 times 10^(-12)`
`7.5 times 10^(-12)`
`1.74 times 10^(-12)`

ANSWER :D
Discussion
42.

Conductivity of a saturated of a sparingly soluble salt AB (1:1 electrolyte) at 298 K is 1.85 xx 10^(-5) S m^(-1). Solubility product of the salt AB at 298 (Lambda_m^@) = 14 xx 10^(-3) S m^(2) mol^(-1)

Answer»

`5.7xx 10^(-12)`
`1.32 XX 10^(-12)`
`7.5 xx 10^(-12)`
`1.74 xx 10^(-12)`

ANSWER :D
Discussion
43.

Conductivity of 2.5xx10^(-4) M methanoic acid is 5.25xx10^(-5)S com^(-1). Calculate its molar conductivity and degreeof dissociation. "Given ":lamda^(0)(H^(+))=349.5 S cm^(2) mol^(-1) and lamda^(0)(HCOO^(-))=50.5 S cm^(2) mol^(-1).

Answer»

Solution :`"Given, "K=5.25xx10^(-5) S CM^(-1)`
`C=2.5xx10^(-4)ML^(-1)`
Then molar conductivity, `^^_(m)=(K)/(C)`
`=(5.25xx10^(-5)S cm ^(-1))/(2.5xx10^(-4)ML^(-1))xx(1000 cm^(3))/(L)=210 cm^(2) M^(-1)`
`OVERSET(@)^^_(m)(HCOOH)=lamda^(@)(H^(+))+lamda^(@)(HCOO^(-))`
`=349.5 S cm^(2) mol^(-1)+50.5 S cm^(2) mol^(-1)=400 S cm^(2) mol^(-1)`
`"Now, "alpha=(^^_(m))/(overset(@)^^_(m))=(210)/(400)=0.525`
Discussion
44.

Conductivity of 2xx10^(-3)M methanoic acid is 8xx10^(-5)" S cm"^(-1). Calculate its molar conductivity and degree of dissociation if Lambda_(m)^(@) for methanoic acid is 404" S cm"^(2)" mol"^(-1).

Answer»

Solution :MOLAR conductivity `Lamda_(m)=(kxx1000)/(C)=(8xx10^(-5)" S CM"^(-1)xx1000)/(2XX10^(-3)" mol L"^(-1))`
`=(8xx10^(-2))/(2xx10^(-3))40" S cm"^(2)" mol"^(-1)`
Degree of dissociation `(Lamda_(m))/(Lamda_(m)^(@))=(40)/(404)=0.099`
Discussion
45.

Conductivity of 1 mol electrolytic solution present between two electrode having unity cross sectional area and unit length is known as. . . .

Answer»

SPECIFIC CONDUCTIVITY
Specific resistance
Limiting MOLAR conductivity
Molar conductivity

Solution :Molar conductivity.
Discussion
46.

Conductivity of 0.01 M NaCl solution is 0.00147" ohm"^(-1) cm^(-1) . What happens to this conductivity if extra 100 ml of H_(2)O is added to the above solution?

Answer»

INCREASES
Decreases
Remains unchanged
First increases and then decreases.

Answer :D
Discussion
47.

Conductivity of 0.00241M acetic acid is 7.896 xx 10^(-5)" S cm"^(-1). Calculate its molar conductivity and if Lambda_(m)^(@)m for acetic acid is "390.5 S cm"^(2)" mol"^(-1), what is its dissociation constant ?

Answer»

SOLUTION :`Lambda_(m)^(@)=(kxx1000)/(M)`
`=(7.896xx10^(-5)"S CM"^(-1)xx1000cm^(3)L^(-1))/("0.0024 mol L"^(-1))`
`=32.76" S cm"^(2)" mol"^(-1)`
`alpha=(Lambda_(m))/(Lambda_(m)^(@))=(32.76)/(390.5)=8.39xx10^(-2)`
`K_(a)=(CALPHA^(2))/(1-alpha)=(0.00241xx(8.39xx10^(-2))^(3))/(1-8.39xx10^(-2))`
`=1.86xx10^(-5)`
Discussion
48.

Conductivity of 0.00241 M acetic acid solution is 7.896xx10^(-5)" S "cm^(-1). Calculate its molar conductivity in this solution. If wedge_(m)^(@) for acetic acid be 390.5 S cm^(2)mol^(-1), what would be its dissociation constant?

Answer»


Solution :`wedge_(m)^(C)=(kappaxx1000)/("Molarity")=(7.896xx10^(-5)" S "cm^(-1)xx1000cm^(3)L^(-1))/(0.00241" mol "L^(-1))=32.76" S "cm^(2)mol^(-1)`
`ALPHA=(32.76)/(390.5)=0.084,K=(calpha^(2))/(1-alpha)=(0.00241xx(0.084)^(2))/(1-0.084)=1.85xx10^(-5)`.
Discussion
49.

Conductivity of 0.00241 M acetic acid solution is 7.896 xx 10^(-5) S cm^(-1). Calculate its molar conductivity in this solution. If wedge_(M)^(@) for acetic acid be 390.5 S cm^(2) mol^(-1), what would be its dissociation constant?

Answer»

Solution :Conductivity of ACETIC acid,
`K = 7.896 XX 10^(-5) "S cm"^(-1), wedge_(m)^(@)` for acetic acid = `390.5 " S cm"^(2)mol^(-1)`
Molar conductivity,`wedge_(m)^(C) = (K xx 1000)/("Molarity")`
`= (7.896 xx 10^(-5)xx 1000)/(0.00241)= (789600xx1000xx10^(-5))/(241)`
`=32.76 " S cm"^(2) mol^(-1)`
Degree of dissociation,
`alpha = (wedge_(m)^(c))/(wedge_(m)^(@)) = (32.76)/(390.5) = 8.4xx10^(-2)`
Dissociation constant of acetic acid,
`Ka = (CALPHA^(2))/(1-alpha)=((0.00241)xx(8.4xx10^(-2))^(2))/(1-0.084) = 1.86 xx 10^(-5)`
Discussion
50.

Conductivity of 0.00241 M acetic acid is 7.896xx10^(-5)"S "cm^(-1). Calculate its molar conductivity and if wedge^(@) for acetic acid is 390.5 S cm^(2)mol^(-1), what is its dissociation constant?

Answer»

SOLUTION :`wedge_(m)^(@)=(kappaxx1000)/("MOLARITY")=((7.896xx10^(-5)" S "cm^(-1))xx1000cm^(3)L^(-1))/(0.00241" mol "L^(-1))=32.76" S "cm^(2)mol^(-1)`
`alpha=(wedge_(m)^(c))/(wedge_(m)^(@))=(32.76)/(390.5)=8.4xx10^(-2),K_(a)=(CALPHA^(2))/(1-alpha)=(0.00241xx(8.4xx10^(-2))^(2))/(1-0.084)=1.86xx10^(-5)`
Discussion