Saved Bookmarks
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 benefication processes is used for the mineral `Al_(2)O_(3).2H_(2)O` ?A. LiquationB. LeachingC. Froth flotationD. Magnetic separation |
|
Answer» Correct Answer - B Benefaction is concentration of ore. |
|
| 2. |
In Column I four questions are given. You have to match the correct answers from the options in column II.Column IColumn II1. BHC(a) Hypnotic2. Gelatin(b) Semiconductor3. Chloretone(c) Insecticide4. Silicon(d) Colloid |
|
Answer» 1. (d) 2. (d) 3. (a) 4. (b) |
|
| 3. |
Dolomite mineral contains(a) Al(b) Mg(c) K(d) Ca |
|
Answer» Answer (b,d) Mg, Ca |
|
| 4. |
Which one of the following ores is a chloride?A. Horn silverB. ZinciteC. BauxiteD. Feldspar |
|
Answer» Correct Answer - A Horn silver `(AgC1)` |
|
| 5. |
Dolomite is mineral whose formula isA. `CaMg(CO_(3))_(2)`B. `MgCO_(3)`C. `CaCO_(3).MgCO_(3)`D. Both `(a)` and `(c )` |
| Answer» Correct Answer - D | |
| 6. |
`10 mL` of a given solution of `H_(2)O_(2)` contains `0.91 g` of `H_(2)O_(2)`. Express its strength in volume. |
|
Answer» `underset(68 g)(2H_(2)O_(2))to2H_(2)O+underset("22400 mL at NTP")(O_(2))` 68 g of `H_(2)O_(2)` produces 22400 mL of `O_(2)` at NTP. `:.` 0.91 g of `H_(2)O_(2)` produces `(22400xx0.91)/68`=300 mL of `O_(2)` at NTP Volume strength =`300/10=30` |
|
| 7. |
In some toothpastes hydrogen peroxide is used. What is the role of `H_(2)O_(2)` in them? |
| Answer» `H_(2)O_(2)` acts as an antiseptic. | |
| 8. |
In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present?(a) 3 double bonds; 9 single bonds(b) 6 double bonds; 6 single bonds(c) 3 double bonds; 12 single bonds(d) Zero double bonds; 12 single bonds |
|
Answer» (a) 3 double bonds; 9 single bonds |
|
| 9. |
In how many of the following complex ions, the central metal ions use (n-1)d, ns and np orbitals for hybridisation ? `[Mn(CN)_6]^(4-), [Ni(NH_3)_6]^(3-), [Co( o x)_3]^(3-), [Cu(NO_2)_6]^(4-), [AgF_4]^(-), [Ni(CN)_4]^(2-), [PdCl_4]^(2-), [PdCl_4]^(2-), [Pd(CN)_4]^(2-),[Co(SCN)_4]^(2-)` |
|
Answer» Correct Answer - 6 `[Mn(CN)_6]^(4-)-3d^(5),d^2sp^(3),+2` oxidation state `[Ni(NH_3)_6]^(2+)-3d^(8),sp^(3)d^2,+2` oxidation state `[Co(o x)_3]^(3-)-3d^(6),d^2sp^(3),+3` oxidation state `[Cu(NO_2)_6]^(4-)-3d^(9),sp^(3)d^2,+2` oxidation state `[AgF_4]^(-)-4d^(8),dsp^(2),+3` oxidation state `[Ni(CN)_4]^(2-)-3d^(8),dsp^(2),+2` oxidation state `[PbCl_4]^(2-)-4d^(8),dsp^(2-),+2` oxidation state `[Pb(CN)_4]^(2-)-4d^(8),dsp^(2-),+2` oxidation state `[Co(SCN)_4]^(2-)-3d^(7),sp^(3-),+2` oxidation state |
|
| 10. |
The brown ring complex compound is formulated as `[Fe(H_(2)O)_(5)NO]SO_(4)`. The oxidation state of `Fe` is |
|
Answer» Correct Answer - 1 Here NO is `NO^(+)`, so in `[Fe(H_2O)_5NO^(+)]^(2+)` , the oxidation state of iron is +1. |
|
| 11. |
The brown ring complex compound of iron is formulated as `[Fe (H_(2)O)_(5) (NO)] SO_(4)`. The oxidation state of iron is |
|
Answer» Correct Answer - 1 `[Fe(H_2O)_6(NO^(+))]SO_4` Oxidation number of iron is one . |
|
| 12. |
In the chromyl chloride test, the reagent used is(a) \( K _{2} CrO _{4} \)(b) \( CrO _{3} \)(c) \( K _{2} Cr _{2} O _{7} \)(d) \( \left( NH _{4}\right)_{2} Cr _{2} O _{7} \) |
|
Answer» Answer: \((c)\,K_2Cr_2O_7\) |
|
| 13. |
For the decomposition of `H_2O_2(aq)` it was found that `V_(O_2)` (t=15 min) was 100 mL (at `0^@C` and 1 atm) while `V_(O_2)` (maximum) was 200 mL (at `0^@C` and 2 atm). If the same reaction had been followed by the titration method of if `V_(KMnO_(4))^((cM))`(t=0) had been 40 mL, what would `V_(KMnO_(4))^((cM))`(t=15 min) have been ?A. 30 mLB. 25 mLC. 20 mLD. 15 mL |
|
Answer» Correct Answer - A `1/4th` reaction has completed upto 15 min.Hence `V_(KMnO_4)` will be `3/4xx40=30` mL |
|
| 14. |
Decomposition of `H_2O_2` (Ist order) `H_2O_2(aq)toH_2O(l)+1/2O_2(g)` Can be monitored by titration method by pressure measurement.If `{:("time(min.)",t=0,30,oo),("Pressure (mm of Hg)",200,375,400):}` While when progress of same reaction was monitored by titration method or the volume of titrant consumed after 10 min was found to be 20 ml.Then complete the following table. `{:("Time(min)",t=0,10,30),("Pressure(mm of Hg)",200,(b),(c )),("Volume of" KMnO_4,(a),20,(d)):}` consumed (ml) |
|
Answer» Correct Answer - a-40,b-300,c-375,d-5 `k=1/(30"min")ln{(400-200)/(400-375)}=1/(30"min")ln{200/25}=(ln2)/T_(1//2)` `T_(1//2)=(30 "min")(ln2)/(ln8)=10` min So, we will have `{:("Time(min)",t=0,10,30),("Pressure (in mm of Hg)",200,300,375),("Volume of" KMnO_4"consumed (ml)",40,20,5):}` |
|
| 15. |
Statement-1: ΔTf= kfmStatement-2: Molality is independent of temperature. (a) Both statements are correct(b) Statement 1 is correct, statement 2 is incorrect(c) Both statements are incorrect(d) Statement 2 is correct, statement 1 is incorrect |
|
Answer» (a) Both statements are correct |
|
| 16. |
The heat of combustion of `CH_(4) is -400 KJ mol^(-1)`. Calculate the heat released when 40g of `H_(2)O` is formed upon combustion :-A. `+444.4 KJ`B. `+888.8 KJ`C. `-444.4 KJ`D. `-888.8 KJ` |
|
Answer» Correct Answer - C `CH_(4)+2O_(2)toCO_(2)+2H_(2)O,=-400KJ//mol` `because` for 2mole or 36 gm water, heat released `=-400KJ//mol` `:.` for gm water, heat released will be `=(-400)/(36)xx40` `-444.44 KJ` |
|
| 17. |
2 mol of an ideal gas at `27^(@)`C expand isothermally and reversibly from a volume of 4L to 40L. The work done (in KJ) by the gas is :-A. `w=-28.72KJ`B. `w=-11.488KJ`C. `w=-5.736KJ`D. `w=-4.988KJ` |
|
Answer» Correct Answer - B `w=-2.303nRTlog(V_(2))/(V_(1))` `=-(2.303xx2xx8.314xx300log(40)/(4))/(1000)KJ` `w=-11.224KJ` |
|
| 18. |
Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x – 3y + 5z = 8. If the mirror image of the point (2, -1/2, 2) in the rotated plane is B(a, b, c), then :(A) a/8 = b/5 = c/-4(B) a/4 = b/5 = c/-2(C) a/8 = b/-5 = c/4(D) a/4 = b/5 = c/2 |
|
Answer» Correct option is (A) a/8 = b/5 = c/-4 Let equation of rotated plane be : (2x + 3y + z + 20) + λ (x – 3y + 5z – 8) = 0 (2 + λ )x + (3 – 3λ )y + (1 + 5l)z + 20 – 8λ = 0 Above plane is perpendicular to 2x + 3y + z + 20 = 0 So, (2 + λ ).2 + (3 – 3λ ).3 + (1 + 5λ).1 = 0 ⇒ λ = 7 ⇒ Equation of rotated plane : x – 2y + 4z – 4 = 0 Mirror image of A(2, -1/2, 2) in rotated plane is B(a, b, c) Equation of AB : \(\frac{x-2}1=\frac{(y+1)/2}{-2}=\frac{z-2}4=k\) Let coordinate of B be (2 + k/2, -1/2 - k, 2 + 2k) which will lie on the plane x - 2y + 4z - 4 = 0 Hence k = -2/3 Therefore B is (4/3, 5/6, -2/3) = (8/6, 5/6, -4/6) So, a/8, b/5 = c/-4 |
|
| 19. |
If the plane 2x + y – 5z = 0 is rotated about its line of intersection with the plane 3x – y + 4z – 7 = 0 by an angle of π/2, then the plane after the rotation passes through the point : (A) (2, –2, 0) (B) (–2, 2, 0) (C) (1, 0, 2)(D) (–1, 0, –2) |
|
Answer» Correct option is (C) (1, 0, 2) (2x + y – 5z) + λ(3x – y + 4z – 7) = 0 Rotated by π/2 (2 + 3λ)x + (1 – λ)y + (–5 + 4λ)z – 7λ = 0 2x + y – 5z = 0 2(2 + 3 λ) + (1 – λ) – 5(–5 + 4 λ) = 0 ⇒ 4 + 6 λ + 1 – λ + 25 – 20λ = 0 30 = 15λ λ = 2 Required plane :- 8x – y + 3z – 14 = 0 |
|
| 20. |
If maximum possible range of a projectile is 100 m what will be the maximum possible height for same speed:(1) 100 m(2) 200 m(3) 50 m(4) 25 m |
|
Answer» Correct option is (1) 100 m Rmax = u2/g = 100 m Hmax = u2/2g = 50 m |
|
| 21. |
The energy of emitted photoelectrons from a metal is 0.9 eV and energy of incident photon is 3.1 eV then the work function of a metal is(a) 4.0 eV(b) 2.2 eV(c) 3.0 eV(d) 3.1 eV |
|
Answer» Correct option is (b) 2.2 eV E = ϕ + KEmax ϕ = 3.1 - 0.9 ϕ = 2.2 ev |
|
| 22. |
Arrange wavelengths of gamma, X Rays, Visible & Microwave in ascending order.(a) Gamma < X Rays < Visible < Microwave(b) Gamma > X Rays > Visible> Microwave(c) Gamma > Visible> X Rays > Microwave (d) Microwave < Gamma < X Rays < Visible |
|
Answer» (a) Gamma < X Rays < Visible < Microwave |
|
| 23. |
What type of plot is obtained for rate versus time for zero order reaction gt |
| Answer» A straight line parallel to time axis because rate is independent of concentration and hence time. | |
| 24. |
When is the rate of reaction equal to specific rate constants ? |
| Answer» When the concentration of each reactant is `1 mol L^(-1)` or order `w.r.t.` each reactant is zero. `i.e.,` reaction is of zero order. | |
| 25. |
When does the average rate become equal to instantaneous rate ? |
|
Answer» When `Deltat` approaches zero, `i.e.,` it becomes infinitesimally small. Thus, `r_("inst")`,i.e., `((dx)/(dt))=underset(Deltar rarr0)("lim")((Deltax)/(Deltat))` |
|
| 26. |
On the basis of enthalpy of formation, graphite is more stable than diamond, yet diamond does not change into graphite for years. Why ? |
|
Answer» `E_(a)` for the reaction `:` `C (` graphite `) rarr C (` diamond`)` is very high which is not available at room temperature. |
|
| 27. |
Can a reaction have negative or zero activation energy ? |
| Answer» Refer point (j) and (k), Overview, Chapter. | |
| 28. |
(i) An aqueous solution of a white coloured compound `(A)` on reaction with `HCl` gives a white precipitate of compound `(B)`. (ii) `(B)` becomes soluble in chlorine water with the formation of `(C)` (iii) `(C)` reacts with `KI` to give a precipitate which becomes solube in excess of it forming a compount `(D)`. the compound `(D)` is used for detecting ammonium salts. (iv) `(B)` and `(C)` both, on treatement with `SnCl_(2)` give a grey precipitate of `(E)`. (v) When conc. `H_(2)SO_(4)` is added slowly into a mixture of cold solutions of `(A)` and `FeSO_(4)` is added slowly into a mixture of cold solutions of `(A)` and `FeSO_(4)`, a brown ring of compound `(F)` is formed. The oxidation number of `FE` in compound `(F)` is:A. `+1`B. `+2`C. `+3`D. zero |
|
Answer» No in iron complex has oxidation no. `+1` `:. Fe` in `feSO_(4).NO` is in +1 |
|
| 29. |
Statement-1: `CF_(3)-CHCl_(2)` when treated with `C_(2)H_(5)OD`, the major product formed is `CF_(3)-CDCl_(2)` rather than `CF_(2)=C Cl_(2)` Statement-2: `C_(2)H_(5)O^(-)` is a poor base.A. If both the statement are TRUE and Statement -2 is the correct explanation of Statement-1:B. If both the statement are TRUE but Statement-2 is not the correct explanation of Statement-1C. If statement-1 is TRUE and Statement-2 is FALSED. If statement -1 is FALSE and Statement-2 is TRUE |
| Answer» `C_(2)H_(5)O^(-)` is a very powerfull base. | |
| 30. |
For different aqueous solutions of `0.1N` urea, `0.1N NaCl, 0.1N Na_(2)SO_(4)` and `0.1N Na_(3)PO_(4)` solution at `27^(@)C`, select the correct statements:A. The order of osmotic pressure is, urea `gtNaClgtNa_(2)SO_(4)gtNa_(3)PO_(4)`B. The order of vapour pressure is , `Na_(3)PO_(4)gtNa_(2)SO_(4)gtNaClgt`ureaC. The order of boiling point `(T_(B))` is urea,`gt NaClgtNa_(2)SO_(4)gtNa_(3)PO_(4)`D. The order of freezing point `(T_(f))` is, urea `gtNaClgtNa_(2)SO_(4)gtNa_(3)PO_(4)` |
|
Answer» Correct Answer - 4 `pi =MRTi` Hence, `DeltaT_(b)=mK_(p)i," "Delta_(i)=mK_(t)xxi` |
|
| 31. |
The solutions of two compounds listed in column-I when taken in molecular proportions give coordination compounds on evaporation, Match these compounds with their properties listed in column-II. `{:("Column-I","Column-II"),((A)Fe(CN)_2. 4KCN,(p)"Coordination number of central metal ion is six"),((B)CuSO_4. 4NH_3,(q)"Coordination number of central metal ion is four"),((C)3NaF.AIF_3,(r)"Coordination number of central metal ion is two "),((C )3NaF. AIF_3, (s)"Oxidation state of central metal ion is +2"),(,(t)"Overall charge on complex ion is" +2):}` |
|
Answer» Correct Answer - A-p,s ; B-q,s,t ; C-p ; D-r `Fe(CN)_2+4KCNtoK_4[Fe(II)(CN)_6]^(4-)`, coordination number 6 `CuSO_4+4NH_3to[Cu(II)(NH_3)_4]^(2+)SO_4^(2-)`, coordination number 4 `3NaF+AlF_3toNa_3[Al(III)F_6]^(3-)`, coordination number 6 `AgCN+KCNtoK[Ag(I)(CN)_2]^(1-)`,coordination number 2 |
|
| 32. |
Which one of the following is a version of UDP with congestion control?(a) datagram congestion control protocol(b) stream control transmission protocol(c) structured stream transport(d) user congestion control protocol |
|
Answer» Right choice is (a) datagram congestion control protocol Easy explanation: The datagram congestion control is a transport layer protocol which deals with reliable connection setup, teardown, congestion control, explicit congestion notification, and feature negotiation. It is used in modern day systems where there are really high chances of congestion. The protocol was last updated in the year 2008. |
|
| 33. |
Match the pair of complexes given in Column-I and the characteristic given in column-II. `{:("Column-I","Column-II"),((A)(NH_4)_2[NiCl_4] and (NH_(4))_(2)[NI(CN)_4],(p)"Both show same electrical conductance"),((B)CoCl_3. 6NH_3 and PtCl_4. 5NH_3,(q)"Both show same effective atomic number "),((C)[Pt(NH_3)_2Cl_2] and (NH_4)_2[PtCl_4],(r)"Both show same primary valencies "),((D )K_2[Fe(H_2O)_6] and K_4[FeCl_6], (s)"Both gives white participate with" AgNO_3 "solution"):}` |
|
Answer» Correct Answer - A-p,q,r ; B-q,s ; C-q,r ; D-q,r (A)`Ni^(2+)` ions in both complexes have same primary valencies, i.e, +2 Same number of ions i.e. 3 and so have same conductance. Both have same EAN i.e. 34. `Cl^(-)` ions in both complexes are in coordination sphere so no precipitate is obtained with `AgNO_3`. (B)`overset(III)([Co(NH_3)_6]Cl_3) and overset(IV)([Pt(NH_3)_5Cl])Cl_3` have different primary valencies but same number of ions. So same electrical conductance . Do not have same effective atomic number as in both complexes metals have different atomic number. As there are three `Cl^-` ions out side the coordination sphere both will gives 3 moles of the precipitate of AgCl(white). (C )`[Pt(NH_3)_2Cl_2] and (NH_4)_2 [PtCl_4]`. Pt is in +2 oxidation state so same primary valencies and same effective atomic number (84) and does not have same electrical conductance (former is neutral complex where as later one has 3 ions in aqueous solutons). No precipitate with `AgNO_3` as `Cl^-` ions in both complexes are in coordination sphere. (D)Both have Fe in +2 oxidation state so same primary valencies but have different number of ions.so different electrical conductance (former has 3 ions where as later one has 5 ions in aqueous solution). Both complexes have same effective atomic number (i.e. 36). |
|
| 34. |
Two broad categories of congestion control are(a) Open-loop and Closed-loop(b) Open-control and Closed-control(c) Active control and Passive control(d) Active loop and Passive loop |
|
Answer» Correct choice is (a) Open-loop and Closed-loop The explanation is: Open loop congestion control techniques are used to prevent congestion before it even happens by enforcing certain policies. Closed loop congestion control techniques are used to treat congestion after it has happened. |
|
| 35. |
`{:("Column-I","Column-II"),((A)[Ni(CO)_4],(p)"Octahedral paramagnetic"),((B)[NI(CN)_4]^(-2),(q)"Square planar diamagnetic "),((C)[NI(NH_3)_6]^(+2),(r)"Tetrahedral diamagnetic "),((D )[NICl_4]^(-2), (s)"Tetrahedral paramagnetic"):}` |
|
Answer» Correct Answer - A`to`r ; B`to`q ; C`to` p ; D`to`s (A)`[Ni(XO)_4]to` Tetrahedral diamagnetic (B)`[Ni(CN)_4]^(-2)to` Square planar diamagnetic (C )`[Ni(NH_3)_6]^(-2)to` Octahedral paramagnetic (D)`[NiCl_4]^(-2)to` Tetrahedral paramagnetic |
|
| 36. |
Name the two broad categories of organometallic compounds. |
|
Answer» (i) `sigma`-bonded `pi`-bonded organometallic compounds. |
|
| 37. |
Give names of two complexes which are used in medicines. |
|
Answer» (i). EDTA used in the treatment of lead poisoning. (ii). Cis-platin, `[Pt(NH_3)_2Cl_2]`, used in the treatment of cancer. |
|
| 38. |
Number of complexes that are paramagnetic in nature with number of unpaired are : 1.`[MnCl_4]^(2-)` 2.`[Mn_2(CO)_10]` 3.`[V(CO)_6]^(-)` 4. 5.`[Pt(NH_3)_4Cl_2]` 6.`[Co(NH_3)_2(H_2O)_4]Cl_2` 7.`[Ni(CN)_4]^(2-)` 8.`[Cu(H_2O)_4]^(2+)` 9. `K_3[Cr[CN)_5]` |
|
Answer» Correct Answer - 4 `[MnCl_4]^(2-),[Ni(H_2O)_6]Cl_2,[Co(NH_3)_2(H_2O)_4]Cl_2,K_3[Cr(CN)_6]` |
|
| 39. |
EAN rule is followed by the complexes. (i)`[Fe(pi-C_5H_5)_2]` (ii)`[Mn_2(CO)_10]` (iii)`[V(CO)_6]^(-)`A. only (i)B. only (ii)C. only (iii)D. all |
|
Answer» Correct Answer - B For all three EAN=36 |
|
| 40. |
The effective atomic number of iron in complex `[eta^5-C_5H_5)Fe(II)(CO)_2Cl]` is . |
|
Answer» Correct Answer - 36 `{:(Fe(II)=,"24 electrons"),((eta^(5)-C_5H_5^(-))=,"6 electrons"),(2(CO)=,"4 electrons"),(Cl^(-)=,"2 electrons"):}` 36 |
|
| 41. |
What is the distance between Na+ and Cl- in a NaCl crystal if its density is 2.165 g cm-3? NaCl crystallizes in fcc lattice. |
|
Answer» M is the molecular mass of NaCl crystal. And N is the Avogadro's number of atoms. Distance d = 281.4 pm |
|
| 42. |
The following table gives the TP schedule of labour. Find the corresponding average product and marginal product schedule:TPL01535504048L012345 |
||||||||||||
Answer»
|
|||||||||||||
| 43. |
Which one of the following is not a method of concentrain of ore?A. electromagnetic separationB. smeltingC. gravity separationD. froth floatation process |
|
Answer» Correct Answer - B In smelting the concentrated oxide ores like haematite, tin stone even after concentration, is heated with flux to remove the acidic or basic impurities froming the slag. All other processes are used fro removing the earth//siliceous impurities. |
|
| 44. |
Explain the indifference map with a diagram: |
|
Answer» If two or more indifference curves are shown in a single diagram then that diagram is termed as indifference map. It is also termed as family of indifference curves. It represents scale of preferences for the consumer also. Every indifference curve in that diagram represents different level of satisfaction. Curves which are nearer to the origin represents lower level of satisfaction and curves which are away from the origin represent higher level of satisfaction . |
|
| 45. |
Mention the three linkages of open economy. |
|
Answer» (a) Output market linkage (b) Financial market linkage (c) Labour market linkage |
|
| 46. |
Explain the differences between normal and inferior goods with examples. |
||||||||||
Answer»
|
|||||||||||
| 47. |
Briefly explain the central problems of an economy. |
|
Answer» The central problems of the economy (a) What to produce : The first problem which every economy has to face, is that which commodity they should produce and in what quantity it should be produced because available resources are limited and have alternative uses. An economy can produce a number of goods by the available limited resources. (b) How to produce : This problem is related with problem of choice of technique. After deciding the commodity and quantity of production. For eg. in re-construction sector we have to choose among labour intensive technique (technique where ratio of labour is more than capital) and capital intensive technique (technique where ratio of capital is more than labour). (c) For whom to produce: This problem is related with distribution of income generated or output produced. In this problem the economy has to decide that who should consume how much. So in this problem we studies how output produced or income generated is distributed among different means of production or different individuals. |
|
| 48. |
Write the differences between substitutes and complements. |
||||||||||||
Answer»
|
|||||||||||||
| 49. |
Why do people demand foreign exchange ? |
|
Answer» (a) Import of Goods and Services from Other Countries (b) Transfer Payments in Other Countries (c) To Invest in Other Countries (d) To Speculate on the Value of Foreign Currencies |
|
| 50. |
Why do people demand foreign exchange? |
|
Answer» People demand foreign exchange rate because of the following reasons: 1. To purchase goods and services from other countries. 2. To send gifts abroad. 3. To purchase financial assets abroad. |
|